Introduction to
One-parameter semigroups and evolution
equations
Lecture 1: Necessary background from Real Analysis
Seminar 1
Ivan Remizov
HSE Nizhny Novgorod, Russia
6 September 2022
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Language of teaching | ßçûê ïðåïîäàâàíèÿ
All the teaching will be in English, as prescibed on HSE website
https://nnov.hse.ru/en/ma/math/
Translation to Russian: Âñ¼ ïðåïîäàâàíèå áóäåò íà àíãëèéñêîì,
êàê îïèñàíî íà ñàéòå ÂØÝ https://nnov.hse.ru/ma/math/
However this year in your group we have students with dierent
language background: some of them know English and Russian,
some of them only Russian, some of them English and other
language (not Russian). This brings us some troubles but also some
possibilities.
Translation to Russian: Îäíàêî, â ýòîì ãîäó â âàøåé ãðóïïå
ïðèñóòñòâóþò ñòóäåíòû ñ ðàçëè÷íûì óðîâíåì ÿçûêîâîé
ïîäãîòîâêè: íåêîòîðûå çíàþò àíãëèéñêèé è ðóññêèé, íåêîòîðûå
òîëüêî ðóññêèé, íåêîòîðûå àíãëèéñêèé è äðóãîé ÿçûê (íå
ðóññêèé). Ýòî ïðåäñòàâëÿåò ñîáîé íåêîòîðóþ ïðîáëåìó, íî
âìåñòå ñ òåì è äà¼ò íåêîòîðûå âîçìîæíîñòè.
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Language of teaching | ßçûê ïðåïîäàâàíèÿ
Teaching in your language-mixed group I will do in the following
way: I will provide slides in English, but read them both in English
and Russian. You will have video-recording of all lections and
seminars so you can come back to them at any time. Also you are
welcome to interrupt me at any time and ask questions.
Translation to Russian: Ïðåïîäàâàíèå â âàøåé ñìåøàííîé â
ïëàíå ÿçûêà ãðóïïå ÿ áóäó âåñòè ñëåäóþùèì îáðàçîì: ñëàéäû
áóäó ïèñàòü íà àíãëèéñêîì, íî ÷èòàòü èõ áóäó ïî-àíãëèéñêè è
ïî-ðóññêè. Ó âàñ áóäóò âèäåîçàïèñè âñåõ ëåêöèé è ñåìèíàðîâ,
òàê ÷òî âû ìîæåòå âåðíóòüñÿ ê íèì â ëþáîå âðåìÿ. Òàêæå ÿ
ïðîøó âàñ îñòàíàâëèâàòü ìåíÿ è çàäàâàòü âîïðîñû â ëþáîå
âðåìÿ.
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Language of teaching | ßçûê ïðåïîäàâàíèÿ
We will proceed slower but we have chance to learn new foreign
words. De facto English is now the language of scientic
communication. Meanwhile Russian is a bright representitative of
the Slavic languages together with Ukranian, Belorussian, Czech,
Serbian, Polish etc. This all may be very interesting to touch, both
for Russians and foreigners.
Translation to Russian: Ìû áóäåì ïðîäâèãàòüñÿ ìåäëåííåå, íî
çàòî èìååì øàíñ âûó÷èòü íîâûå èíîñòðàííûå ñëîâà. Äå-ôàêòî
àíãëèéñêèé ñåé÷àñ ÿçûê íàó÷íîãî îáùåíèÿ. Ìåæäó òåì,
ðóññêèé ÿðêèé ïðåäñòàâèòåëü ñëàâÿíñêèõ ÿçûêîâ âìåñòå ñ
óêðàèíñêèì, áåëîðóññêèì, ÷åøñêèì, ñåðáñêèì, ïîëüñêèì è ò.ä.
Ïðèêîñíóòüñÿ ê ýòîìó ìîæåò áûòü î÷åíü èíòåðåñíî, êàê
ðóññêèì, òàê è èíîñòðàíöàì.
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Language of teaching | ßçûê ïðåïîäàâàíèÿ
I see my duty as follows: to teach everyone who is ready (and
wishes!) to accept new information and study actively both in
Mathematics and in English. I will do my best to teach those of you
who are ready to work hard. But I seriously need your help: I ask
you to interrupt me and ask questions, to ask for translation to
English and to Russian, to ask to repeat if needed etc. It is MY
duty to teach you, and it is YOUR duty to check that you
understand everything and ask questions if something is not clear.
Translation to Russian: Ñâîþ çàäà÷ó ÿ âèæó â òîì, ÷òîáû ó÷èòü
êàæäîãî, êòî ãîòîâ (è õî÷åò!) âîñïðèíèìàòü íîâóþ
èíôîðìàöèþ è àêòèâíî ó÷èòüñÿ êàê ìàòåìàòèêå, òàê è
àíãëèéñêîìó ÿçûêó. ß áóäó ñòàðàòüñÿ èçî âñåõ ìîèõ ñèë, ÷òîáû
íàó÷èòü òåõ, êòî ãîòîâ õîðîøåíüêî ïîòðóäèòüñÿ. Íî ìíå
òðåáóåòñÿ ñåðü¼çíàÿ ïîìîùü îò âàñ: ÿ ïðîøó âàñ ïðåðûâàòü
ìåíÿ è çàäàâàòü âîïðîñû, ïðîñèòü ïåðåâåñòè íà àíãëèéñêèé èëè
íà ðóññêèé, ïðîñèòü ïîâòîðèòü è ò.ä. ÌÎß îòâåòñòâåííîñòü ó÷èòü âàñ, ÂÀØÀ îòâåòñòâåííîñòü ñëåäèòü, ÷òî âû âñ¼
ïîíèìàåòå è çàäàâàòü âîïðîñû, åñëè ÷òî-òî íå ÿñíî.
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Language of teaching | ßçûê ïðåïîäàâàíèÿ
Now you see the last slide with full English-Russian translation.
Those who do not know English please start learning it. Start from
writing your vocabulary book, there will be a Google table for this,
the link will be in VK. If you have learned a new word plese insert
it to the table. If you need translation for some words please ask
me, I will give a translation.
Translation to Russian: Ýòî ïîñëåäíèé ñëàéä ñ ïîëíîé ïîäà÷åé
èíôîðìàöèè íà àíãëèéñêîì è ðóññêîì. Òå, êòî íå çíàåò
àíãëèéñêîãî ïîæàëóéñòà, íà÷íèòå åãî ó÷èòü. Íà÷àòü ñòîèò ñ
èçó÷åíèÿ íîâûõ ñëîâ, äëÿ ýòîãî áóäåò ãóãë-òàáëèöà, ññûëêà
áóäåò â ÂÊ. Åñëè Âû óçíàëè íîâîå ñëîâî - âïèñûâàéòå åãî â
òàáëèöó. Åñëè íóæåí ïåðåâîä êàêèõ-òî ñëîâ, îáðàòèòåñü êî ìíå,
ÿ ïåðåâåäó.
Good luck! Please don't hesitate to ask me.
Translation to Russian: Óäà÷è! Ïîæàëóéñòà, íå ñòåñíÿéòåñü
çàäàâàòü ìíå âîïðîñû.
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Curriculum
This course is all about exponential expressions e tL , where t is
time, and L is a linear (in most cases unbounded) operator in
Banach space. We will dene e tL , study for which operators L this
exponential exists, present some applications of e tL , and show some
methods of calculation of e tL approximately. We will see that e tL is
a ow in sense of dynamical systems and that it provides solution
to some so-called evolution equations that can be considered as
linear innite-dimensional dynamical systems.
Family (e tL )t≥0 satisfying some properties (we will study them
later) is called a one-parameter semigroup where t is the parameter,
and operator L is called the generator of this semigroup. Equation
d
dt U(t) = LU(t) is an example of linear evolution equation. The
initial-value problem U(0) = u0 for this equation has (unique in
some sense) solution U(t) = e tL u0 in agreement with trivial case
U : [0, +∞) → R and L, u0 being real constants. Now you have
rst brief understanding of all the words in the course's title...
... but before studying them we will have about 10 lessons to
refresh necessary background from real and functional analysis.
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Control and grades
Normally we have 1 lecture and 1 seminar (practice) in one day each
week, but rst two weeks we will have 2 lectures and 2 seminars
each week. After each seminar there is a homework of 5 problems
for solving at home. Each problem may be marked 0-2 points. No
solution or wrong solution with rude mistakes 0, solution with some
mistakes but correct answer and idea 1, completely correct solution
2. All together this is 10 points, and this is your cumulative mark
(íàêîï) for this seminar. Your total cumulative mark is an average
of cumulative marks for all seminars in the module. Final mark is an
average of total cumulative mark and exam mark after each
module. The course lasts 3 modules (3/4 of academic year).
Each homework has a deadline until which you should present
(defend) your solutions. If you defend at least one problem from the
current homework you get extra 1 point for this homework. So
maximally you can get 11 points for each homework. If you defend
the homework after the module ends you lose 2 points for this
homework. Is this clear for all of you? Please ask questions :)
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Numbers: rst natural numbers come
God made the integers; all else is the work of man.
Áîã ñîçäàë öåëûå ÷èñëà ÷èñëà. Âñ¼ îñòàëüíîå
ïðèäóìàëè ëþäè.
Leopold Kronecker.
Ëåïîëüä Êðîíåêåð.
Natural numbers N = {1, 2, 3, . . . } with all their properties are
dened by ve Peano axioms. Sometimes (not in our course)
natural numbers are strated from zero: {0, 1, 2, 3, . . . }.
The set {1, 2, 3, . . . } comes from ordering of indivisible objects:
rst line, second line, third line, etc. This is called the ordinal
approach to denition of natural numbers.
The set {0, 1, 2, 3, . . . } comes from measuring the quantity of
indivisible objects: I have no apples (I have zero apples), I have one
apple, I have two apples, etc. This is called the cardinal approach to
denition of natural numbers, because the cardinality of a nite set
is dened as the number of elements in this set.
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Numbers: rst natural numbers come
Peano axioms (denition of the set N):
Axiom 1: 1 ∈ N. This means that 1 is a natural number.
Axiom 2: for each natural number x there exists exactly one natural
number denoted by x 0 and called the successor of x .
Axiom 3: for all natural numbers x we have x 0 6= 1. This means
that 1 is the rst natural number, it is not a successor for any other
natural number.
Axiom 4: for each natural numbers x and y condition x 0 = y 0
implies condition x = y . This means that backward branching is
not allowed for natural numbers.
Axiom 5 (axiom of mathematical induction).
Suppose that we have a set A that posesses two properties:
Property 1 (this condition is called the base of induction): 1 ∈ A.
Property 2 (this condition is called the step of induction): if x ∈ A
and x is a natural number then x 0 ∈ A.
Then N ⊂ A, i.e. A contains all natural numbers.
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Numbers: rst natural numbers come
Theorem (denition of the addition). To every pair of natural
numbers x , y we can assign in exactly one way a natural number
denoted as x + y , this reads ¾x plus y¿, such that
1) x + 1 = x 0 for each x ∈ N
2) x + y 0 = (x + y )0 for each x ∈ N and each y ∈ N.
The number x + y is called the result of addition of y to x , or the
sum of x and y .
Proof is done by using Peano axioms, especially axiom 5. We have
no time to study it carefully in our course, but if you want to be a
real mathematician then it is higly recommended to do it yourself.
See the book ¾Edmund Landau. Foundations of Analysis.¿ The le
with the book is uploaded to VK.
Theorem (associativity and commutativity of addition). For
all natural numbers x, y , z we have x + (y + z) = (x + y ) + z and
x + y = y + x.
Proof. The same comment as for the above proof.
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Numbers: rst natural numbers come
Theorem. For each natural numbers x and y exactly one of the
statements holds:
1. x = y
2. There exists exactly one natural number u such that x = y + u .
3. There exists exactly one natural number v such that x + v = y .
Proof. The same comment as for the above proof.
Denition. The above theorem is used both for dening order on
N and substraction in N as follows:
* If there exists exactly one natural number u such that x = y + u ,
then by denition x > y and u = x − y .
* If there exists exactly one natural number v such that x + v = y
then by denition x < y and v = y − x .
Theorem. For all natural numbers x, y , z the following holds: if
x < y and y < z then x < z .
Exercise. Prove this theorem assuming that all other theorems
above are true.
Denition. We write x ≤ y if x = y or x < y .
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Numbers: more numbers come
Exercise. State (construct yourself of nd in the book ¾Edmund
Landau. Foundations of Analysis.¿) the theorem that denes the
multiplication of natural numbers x and y .
So, natural numbers N are made from Peano axioms, together with
addition, multiplication, order on N and their properties. The next
steps are the following:
I integer numbers Z = {. . . , −2, −1, 0, 1, 2, . . . } are made from
natural numbers
I rational numbers Q = r = m
n |m ∈ Z, n ∈ N are made from
integer and natural numbers
I real numbers R are made from rational numbers via the
procedure known as Dedekind completion, using Dedekind cuts
(Google what is this if you want to be a real mathematician!)
I complex numbers C are dened as pairs of real numbers (x, y )
with operations dened as (x, y ) + (u, v ) = (x + u, y + v ),
(x, y ) · (u, v ) = (x · u − y · v , x · v + y · u)
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Real numbers: axiomatic approach
Denition. Axiomatically R is any set with selected elements
dened as 0 and 1, which is equipped by order and by operations
+, -, ·, / which has the following properties:
1. The set R is a eld, meaning that addition and multiplication are
dened and have the usual properties.
2. The eld R is ordered, meaning that there is a linear order ≥
such that for all real numbers x , y and z :
* if x ≥ y , then x + z ≥ y + z ;
* if x ≥ 0 and y ≥ 0, then xy ≥ 0.
3. The order is Dedekind-complete, meaning that every non-empty
subset A of R with an upper bound in R has the least upper bound
(also called supremum) denoted as sup A.
Property 3 needs more comments, see the next page.
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Real numbers: upper bound and supremum
Denition. A number x ∈ R is called an upper bound for a set
A ⊂ R i for all a ∈ A we have a ≤ x . If a set has an upper bound
then this set is called bounded from above.
Denition. A number x ∈ R is called the least upper bound for a
set A ⊂ R, or supremum, which is denoted as x = sup A i two
conditions hold:
1. x is an upper bound for A
2. if y is an upper bound for A then x ≤ y
Equivalent denition. A number x ∈ R is called the least upper
bound for a set A ⊂ R, or supremum, which is denoted as
x = sup A i two conditions hold:
1. a ≤ x for all a ∈ A
2. for all ε > 0 number x − ε is not an upper bound for A, i.e. there
exists aε ∈ A such that x − ε < aε .
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Seminar 1
On the seminar 1 we recalled that real numbers can be dened
axiomatically, and noticed two models for these axioms: Dedekind
cuts (we do not have time to discuss it in details but you are
welcome to Google what it is) and innite decimal fractions (we
expect that everyone remembers it from school).
Also we introduced the following denition and theorem.
Denition. Real numbers that are not rational are called irrational.
The set of all irrational numbers is dened as I = R \ Q.
Theorem. Real number x is rational i decimal representation of x
is periodic.
On the seminar 1 we solved the following problems:
1. Prove that for all natural numbers x we have
1)
.
1 + 2 + · · · + x = x(x+
2
2. Represent decimal fraction 3.4444... as m
n where m is integer, n
is natural
√
3. Prove that 2 ∈ R is not a rational number, using Fundamental
theorem of arithmetic (èïîëüçóÿ îñíîâíóþ òåîðåìó àðèôìåòèêè)
4. Represent 23 as decimal fraction.
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Homework 1
Problems to solve at home (deadline September 12, 20:00):
1a. Exercise at p.12.
1b. Exercise at p.13.
2. Prove that for all natural numbers n we have:
2a. 12 + 23 + 32 + · · · + n2 = 61 n(n + 1)(2n + 1).
2b. 2n > n.
3. Represent decimal fraction x as m
n where m is integer, n is
natural for the following values of x :
3a. x = 0.99999... = 0.(9)
3b. x = 2.36363636... = 2.(36)
4. Represent x as decimal fraction for the following values of x :
4a. x = 72
4b. x = 11
6
√
5. Prove that 3 ∈ R is not a rational number, using Fundamental
theorem of arithmetic (èïîëüçóÿ îñíîâíóþ òåîðåìó àðèôìåòèêè)
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Literature to lesson 1
1. Edmund Landau. Foundations of Analysis.
(Ýäìóíä Ëàíäàó. Îñíîâû àíàëèçà.)
2. Ã.Ì.Ôèõòåíãîëüö. Îñíîâû ìàòåìàòè÷åñêîãî àíàëèçà, â
äâóõòîìàõ, òîì ïåðâûé, ãëàâà ïåðâàÿ: Âåùåñòâåííûå ÷èñëà.
(G. M.Fikhtengol'ts. The Fundamentals of Mathematical Analysis:
International Series in Pure and Applied Mathematics, volume 1,
Chapter 1: Real numbers)
3. Wiki on real numbers
https://en.wikipedia.org/wiki/Real_number
Ñì. òàêæå ðóññêóþ âåðñèþ ýòîé ñòàòüè â Âèêèïåäèè
4. Wiki on constructive methods of dening a real number
https://en.wikipedia.org/wiki/Construction_of_the_
real_numbers
Ñì. òàêæå ðóññêóþ âåðñèþ ýòîé ñòàòüè â Âèêèïåäèè
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Introduction to
One-parameter semigroups and evolution
equations
Lecture 2: Necessary background from Real Analysis
Seminar 2
Ivan Remizov
HSE Nizhny Novgorod, Russia
9 September 2022
1 / 39
Appendix to Lecture 1:
fundamental theorem of arithmetic
Ïðèëîæåíèå ê ëåêöèè 1:
îñíîâíàÿ òåîðåìà àðèôìåòèêè
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Denition. If a, b, c are natural numbers, and a · b = c , then a and
b are called divisors of c . Also in that case we say that c is divisible
by a and by b . (in Russain: ÷èñëà a è b íàçûâàþòñÿ äåëèòåëÿìè
÷èñëà c , ÷èñëî c äåëèòñÿ íà a è íà b ). Representation c = a · b is
called a factorization of c , in that context a and b are called factors
of c . (in Russain: ïðåäñòàâëåíèå ÷èñëà c â âèäå c = a · b
íàçûâàåòñÿ ôàêòîðèçàöèåé ÷èñëà c . Factor çíà÷èò ¾äåëèòåëü¿.)
Remark. For each natural number n we have 1 · n = n. Hence each
natural number is divisible by 1 and by itself.
Denition. If natural number c has exactly two divisors: 1 and c ,
then c is called prime number (ïðîñòîå ÷èñëî). If natural number
c has more than two divisors then c is called composite number
(ñîñòàâíîå ÷èñëî). Number 1 is not prime and is not composite
because it has only one divisor.
Exercise. Visit webpage
https://en.wikipedia.org/wiki/Prime_number
Remark. The rst 25 prime numbers (all the prime numbers less
than 100) are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
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Euclid's lemma. If a prime number p divides the product ab of
two natural numbers a and b , then p must divide at least one of
those numbers a or b .
Example. If p = 19, a = 133, b = 143, then
ab = 133 · 143 = 19019, and since this is divisible by 19 because
19019 = 19 · 1001, the Euclid's lemma implies that one or both of
133 or 143 must be as well. In fact, 133 = 19 · 7.
Exercise. Visit webpage
https://en.wikipedia.org/wiki/Euclid's_lemma
Fundamental theorem of arithmetic, also called the unique
factorization theorem and prime factorization theorem. Every
natural number greater than 1 can be represented uniquely as a
product of prime numbers, up to the order of the factors.
Example.
1200 = 24 · 31 · 52 = (2 · 2 · 2 · 2) · 3 · (5 · 5) = 5 · 2 · 5 · 2 · 3 · 2 · 2 = . . .
The theorem says two things about this example: rst, that 1200
can be represented as a product of primes, and second, that no
matter how this is done, there will always be exactly four 2s, one 3,
two 5s, and no other primes in the product.
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Remark. The requirement that the factors are prime is necessary:
factorizations containing composite numbers may not be unique
(for example, 12 = 2 · 6 = 3 · 4).
Exercise. Visit webpage https://en.wikipedia.org/wiki/
Fundamental_theorem_of_arithmetic
Euclid's theorem. There are innitely many prime numbers.
Exercise. Visit webpage
https://en.wikipedia.org/wiki/Euclid's_theorem
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Lecture 2
6 / 39
Îáùåå ïîíÿòèå òðåáóåò, ÷òîáû ôóíêöèåé îò x íàçûâàòü
÷èñëî, êîòîðîå äàåòñÿ äëÿ êàæäîãî x è âìåñòå ñ x
ïîñòåïåííî èçìåíÿåòñÿ.
General sense demands to call function of x such a
number, that is given for each x and changes gradually
when x changes.
Íèêîëàé Èâàíîâè÷ Ëîáà÷åâñêèé, 1834.
Nikolay Lobachevskii, 1834.
This denition is more or less contemporary. We should keep in
mind that when we deal with real or complex numbers denig
function is not an easy way. Even arithmetic operations are not easy
to dene. Indeed, x + y for natural numbers x and y is dened by
induction using Peano axioms. For rational numbers x = xx12 and
y = yy12 with integer x1 , y1 and natural x2 , y2 we dene
1 ·x2
x + y = x1 ·yx22+y
. But how we can dene sum of two irrational
·y2
real numbers x and y ? As we already know, irrational numbers are
represented via innite non-periodic decimal fractions. Does it mean
that to nd x + y we need to add innitely many numbers? How?
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Keeping in mind Lobachevskii's words, we can ask:
What is the general sense that for given real number y allows to
dene real number x + y , that is given for each real number x and
changes gradually when x changes?
√
What is the general sense that allows to dene real number x ,
that is given for each real number x and changes gradually when x
changes?
What is the general sense that allows to dene real number sin x ,
that is given for each real number x and changes gradually when x
changes?
These are deep questions that were precisely and accurately
answered only in the beginning of XX century in theory developed
by Peano, Borel, Dedekind and Lebesgue.
To deal with such questions we need to give contemporary
denition of a function. There are two denitions useful in the
beginning of XXI century: naive and axiomatical. We will give both
of them, but before we discuss the concept of a set.
8 / 39
We start from sets
From the middle of XX century there is a tradition to build
fundations of mathematics from sets. Now we know that this is not
the only possible base for building ¾the house of mathematics¿,
there are some other variants (categories, types and some other
things may serve as basic objects instead of sets) but we will follow
the traditional way.
The rst step is to answer the question what a set is. We can
answer it in two ways:
1. Take ¾set¿ and ¾be element of¿ as primitive notions that are
not dened, but the interplay between these notions is stated
explicitly in some statements that are called axioms of set theory.
Of course, dierent systems of axioms may generate dierent set
theories. Systems of axioms are called equivalent if they generate
the same theory. There are known Zermelo-Fraenkel (ZF) axioms,
von Neumann-Bernays-G
odel (NBG) axioms and some other
systems of axioms. We can also accept, reject or ignore (depending
on our purposes) some additional axioms independent on the initial
system of axioms e.g. Axiom of Choice, Continuum Hypothesis etc. 9 / 39
We start from sets
In axiomatic approach a set is a ¾black box¿ which you can ask
about any object and which always gives one of replies: a) this is
my element or b) this is not my element. Axioms of that particular
version of the set theory describe how this ¾black box¿ works.
2. Describe (not dene!) sets intuitively. Then a set is a collection,
crowd, family, unordered list, heap, unordered array of any objects
of material or intellectual world that are called elements of the set
and are distinct from each other and from other objects of material
or intellectual world. A set is only a collection of things, without
any order, relations and other interplay between these things.
In our introductory course on One-parameter semigroups it is not
relevant which approach to sets to choose. For example, it may be
naive, ZF or NBG. As usual in Functional Analysis, we accept
Axiom of Choice to have equivalence of Cauchy and Heine
continuity, Lebesgue unmeasurable sets, Hamel basis, HahnBanach
theorem, innite products of non-zero sets and other beautiful
things, but in our introductory course set theory will be ¾under the
10 / 39
hood¿ of the facts and notions that we discuss.
Some notation from logic
Denition. Logical statement is a meaningful declarative sentence
that is true or false.
Remark. Examples of sentences that are true statements:
I "Socrates is a man."
I "A triangle has three sides."
I "Madrid is the capital of Spain."
Examples of sentences that are also statements, even though they
aren't true:
I "All apples are made of solid gold."
I "Two plus two equals ve."
11 / 39
Some notation from logic
Examples of sentences that are not statements:
1. "Who are you?"
2. "Run!"
3. "Greenness is a heap."
4. "I have a grunch and a sad knife."
5. "The Queen of England is wise."
6. "Broccoli tastes good."
The rst two examples are not declarative sentences and therefore
are not statements. The third and fourth are declarative sentences
but, lacking meaning, are neither true nor false and therefore are
not statements. The fth and sixth examples are meaningful
declarative sentences, but are not statements but rather matters of
opinion or taste.
12 / 39
Some notation from logic
If a, b and c are logical statements then:
I ¬a, a is true i a is false. Statements ¬a, a both are called
¾negation of a¿.
I Statements a&b , a ∧ b are true i a and b both are true. Symbols
& and ∧ are called ¾logical AND¿, conjunction.
I Statement a ∨ b is true i at least one of a and b is true. This
means that a ∨ b is true only in all of three cases: 1) a and b both
are true, 2) a is true and b is false, 3) b is true and a is false.
Symbol ∨ is called ¾logical OR¿, disjunction.
I Statements a ⇒ b , a → b are true i a implies b . Symbols ⇒, →
are called ¾implication¿.
I Statements a ⇐⇒ b , a ↔ b are true i ( a is true i b is true).
Symbols ⇐⇒ , ↔ are called ¾logical equivalence¿, which means
the same as ¾logical biconditional¿.
Exercise. Visit the following webpages:
Notation.
https://en.wikipedia.org/wiki/Boolean_function
https://en.wikipedia.org/wiki/Truth_table
Exercise. Using truth tables prove that (a ⇐⇒ b) = (a ⇒ b)&(b ⇒ a).
13 / 39
Some notation from logic
Denition. Predicate is a statement that depends on parameter,
i.e. includes parameter as a letter e.g. x , y , z etc.
Example. Let us dene predicate P as follows: P(x) is true i x is
a man. Then, if Socretes is a man, we will have that P(x) is true
for x =Socrates. Meanwhile P(x) is false for x =Earth.
Example. Let us dene predicate Q as follows: Q(x, y ) is true i x
has property y . Then, if Socretes is a man, we will have that
Q(x, y ) is true for x =Socrates and y =to be a man". Meanwhile
Q(x, y ) is false for x =Earth and y =lifeless.
Remark. Very often people write just P(x) instead of ¾P(x) is
true¿. Example: we write just x = 2 instead of ¾x = 2 is true¿.
14 / 39
How to dene a set
Naive (intuitive) description of a set. A set is a collection,
crowd, family, unordered list, heap, unordered array of any objects
of material or intellectual world that are called elements of the set
and are distinct from each other and from other objects of material
or intellectual world. A set is only a collection of things, without
any order, relations and other interplay between these things.
Remark. The fact that object a is an element of the set A is
written as follows: a ∈ A and reads as ¾a is in A¿, ¾a belongs to
A¿. The same fact can be written as A 3 a which reads as ¾A
contains a.¿
Remark. The fact that object a is not an element of the set A is
written as follows: a ∈
/ A and reads as ¾a is not in A¿, ¾a do not
belong to A¿. The same fact can be written as A 63 a which reads
as ¾A do not contain a¿.
Denition. Empty set is an empty list of elements. Empty set is
denoted ∅. For all x it is false that x ∈ ∅ and it is true that x ∈
/ ∅.
15 / 39
How to dene a set
Remark. To dene a set, you can use direct listing of elements:
A = {a, b, c}, or introduce selection rule: A = {x|P(x)} (also
written as A = {x : P(x)}), or give inductive procedure:
a ∈ A ⇒ f (a) ∈ A, and some other ways.
Remark. Selection rule A = {x|P(x)} reads as follows: A is the set
(collection) of all abjects x such that P(x), i.e. the predicate P is
true on x , i.e. P(x) is true. Selection rule A = {x|P(x), Q(x)}
reads as follows: A is the set (collection) of all abjects x such that
P(x) ∧ Q(x), i.e. predicates P and Q are both true on x , i.e.
P(x)&Q(x) is true.
Denition. We say that A is a subset of B , and write A ⊂ B i
(x ∈ A) ⇒ (x ∈ B). This means that all elements of A are also
elements of B . In that case we also say that B is a supset of A, and
write B ⊃ A.
Remark. For all sets A it is true that ∅ ⊂ A.
Remark. Selection rule A = {x|x ∈ B, P(x)} is often written as
A = {x ∈ B|P(x)} and means the same: A is the set of all
elements x of the set B such that P(x) is true.
16 / 39
Making new sets from known sets
Remark. The following standard symbols for sets are widely used.
Suppose that A and B are sets. Then:
I The set A ∪ B is called the union of sets A and B . By
denition x ∈ A ∪ B ⇐⇒ (x ∈ A) ∨ (x ∈ B). I.e. x ∈ A ∪ B
i x belongs to at least one of the sets A and B .
I The set A ∩ B is called the intersection of sets A and B . By
denition x ∈ A ∩ B ⇐⇒ (x ∈ A) ∧ (x ∈ B). I.e. x ∈ A ∩ B
i x belongs to both sets A and B .
I The set A \ B is called the relative complement of B with
respect to a set A, also A \ B reads as ¾A minus B ¿. By
denition x ∈ A \ B ⇐⇒ (x ∈ A) ∧ (x ∈
/ B). I.e. x ∈ A \ B
i x belongs to A and do not belong to B .
I The set A4B is called the symmetric dierence of sets A and
B , also known as the disjunctive union of sets A and B . By
denition A4B = (A ∪ B) \ (A ∩ B). Element x belongs to
A4B i x belongs to exactly one of sets A and B .
17 / 39
After sets tuples come
Denition. Unordered pair of elements a, b is the set {a, b}. If
a = b then formally {a, a} = {a} because a set can not have
identical elements (all elements must be distinct).
Denition. Ordered pair (also called couple, 2-tuple) of elements
define
a, b is the set (a, b) = {{a}, {a, b}} where a is called the rst
element, and b is called the second element.
Remark. It is clear that for a 6= b we have {a, b} = {b, a} because
in a set there is no order (a set is just a list of elements), and it is
clear that (a, b) = {{a}, {a, b}} =
6 {{b}, {a, b}} = (b, a).
Denition. Inductive denition: for n ≥ 2 if n-tuple (a1 , . . . , an ) is
already dened then (n + 1)-tuple is dened by
define
(a1 , . . . , an , an+1 ) = ((a1 , . . . , an ), an+1 ).
Denition. 0-tuple () = ∅, 1-tuple (a) = a.
18 / 39
Cartesian product
Denition. Cartesian product A × B of sets A and B is dened by
equality A × B = {(a, b)|a ∈ A, b ∈ B}. For n sets A1 , . . . , An
A1 × · · · × An
denote
=
n
Y
define
Ak = {(a1 , . . . , an )|ak ∈ Ak , 1 ≤ k ≤ n}.
k=1
For (a1 , . . . , an ) ∈ A1 × · · · × An element ak is called the k -th
coordinate.
Denition. Particular case Cartesian square A × A = A2 ,
Cartesian cube A × A × A = A3 and Cartesian n-th power
Q
n
n
k=1 A = A .
Remark. For a nite set A the number of elements in A is denoted
as |A|. Then it is easy to show that for nite sets A1 , . . . , An we
have
n
n
Y
Y
Ak =
|Ak |.
k=1
k=1
19 / 39
Functions - naive approach
Naive general denition of a function. Let A, B, C be
non-empty sets and ∅ 6= C ⊂ A. Then we say that a function f on
C (and also partial function f on A) with values in B is given i for
each x ∈ C element denoted by f (x) is given, and f (x) ∈ B . In
that case sets A, B, C and R(f ) = f (C ) are called:
A the set of departure (Russian: ìíîæåñòâî îòïðàâëåíèÿ) of f
B the set of arrival, the codomain (Russian: ìíîæåñòâî
ïðèáûòèÿ, ìíîæåñòâî çíà÷åíèé) of f
D(f ) = C the domain (Russian: îáëàñòü îïðåäåëåíèÿ) of f
denote
R(f ) = f (D(f )) = {f (x)|x ∈ D(f )} = Im(f ) range, image
(Russian: ìíîæåñòâî ïðèíèìàåìûõ çíà÷åíèé, îáðàç) of f .
Comments:
1. It is important that f (x) exists for EACH x ∈ C , and it is
important that y = f (x) and z = f (x) imply y = z , i.e. f (x) is
UNIQUELY dened by x .
2. Writings ¾f : A → B ¿ without any comments about D(f ),
or ¾f is dened on A and takes values in B ¿, or ¾f is a B -valued
20 / 39
function on A¿ mean the same and all state that D(f ) = A.
Functions - naive approach
Comments:
3. Unfortunatelly, nowdays there is some mess in the terminology
concerning domains and codomains. The terminology presented
above is (in my opinion) not the best possible, but it is (in my
opinion) the best choice if you wish to understand and to be
understood in 2020's. Words as active (co)domain, (co)domain of
denition, active (co)domain of denition and some other terms for
A, B, D(f ), R(f ) are sometimes used.
4. I use words function, map, mapping, operator as synonyms.
f
5. Writings f : A → B , A → B reads as ¾function f maps the set A
into the set B ¿
f
6. Writings f : a 7−→ b , a 7−→ b , f (a) = b reads as ¾function f
maps the element a into the element b ¿, also ¾f takes value b on
argument a¿, or just ¾f of a is b ¿.
7. Dot and bracket notation: f = f (·) = [x 7−→ f (x)].
8. In functional analysis it is important NOT to confuse function f
the way of getting f (x) from x , and f (x) the value of function
21 / 39
f on argument x .
Functions - naive approach
Example. In functional analysis sin(x) is a NUMBER, depending
on number x . Meanwhile sin or sin(·) is the sine FUNCTION.
Example. Writing f = [y 7−→ y 2 ] reads ¾f is the function that
maps a value to its square¿ and means that f (x) = x 2 .
Remark. Why this is so important? Becuse in functional analysis
very often we consider functions that take values or have arguments
in some space of functions and this can produce misunderstanding
because x , f and f (x) may all be functions, but dierent functions!
Naive denition of a function in the restricted sense. The
same as naive general denition of a function, but A ⊂ R and
B ⊂ R. I.e. in the restricted sense function is always real-valued
and always have real argument.
Remark on functionals and operators. If all values of function f
are numbers, and domain of f is some set of functions then usually
we say that ¾f is a functional¿. If both domain and codomain of
function f are sets of functions then we usually say that ¾f is an
operator¿ and most likely will write capital F instead of small f .
This is informal explanation which helps to bulid intuition.
22 / 39
Functions - profound approach: denition via relations
Denition. Binary relation r between sets A and B (also called
binary relation on A × B ) is by denition any subset r ⊂ A × B . We
say that a and b are in relation r and write arb i (a, b) ∈ r .
Denition. For relation r ⊂ A × B two sets have special names:
D(r ) = {a ∈ A|∃b ∈ B : arb} the domain
R(r ) = {b ∈ B|∃a ∈ A : arb} the range
Denition. Relation r ⊂ A × B is called:
1. Injective, or left-unique i ((a1 rb)&(a2 rb)) ⇒ (a1 = a2 ).
2. Right-unique i ((arb1 )&(arb2 )) ⇒ (b1 = b2 ).
3. Left-total i D(r ) = A.
4. Right-total, or surjective, or ¾onto¿ i R(r ) = B .
Denition. Right-unique relation f ⊂ A × B is called a partial
function (Russian: ÷àñòè÷íàÿ ôóíêöèÿ) with the domain D(f ) and
range R(f ) i D(f ) 6= A.
Denition. Right-unique and left-total relation f ⊂ A × B is called
a function f : A → B . Traditionally we write b = f (a) which means
the same as (a, b) ∈ f and afb .
23 / 39
Functions - profound approach: further denitions
Remark. In the naive approach for each function f its graph G (f )
is dened as G (f ) = {(x, f (x))|x ∈ D(f )}. Meanwhile, in the
profound approach G (f ) = f i.e. we do not distinguish a function
and its graph. This holds true for partial functions even in a more
bright way.
Denition. For relation r ⊂ A × B the inverse relation
r −1 ⊂ B × A is dened by the rule arb ⇐⇒ br −1 a.
Denition. If relation f ⊂ A × B is a function, and f −1 is also a
function, then function f is called invertible, and inverse relation
f −1 is called the inverse function for f .
Denition. Function f is called bijective, bijection (also called
¾one-to-one¿ function) i f injective and surjective.
Exercise. Show that a function is invertible i it is bijective.
Denition. We say that sets A and B have the same cardinality
and write |A| = |B| i there is a bijection f : A → B .
Denition. If A, B are non-empty sets then AB denotes the set of
all A − valued functions dened on B , i.e. AB = {f |f : B → A}.
24 / 39
Denition of niteness and several useful cardinalities
Remark. Cardinality |{1, 2, 3, . . . , n}| is n. The same cardinality
has any nite set that consists of n elements. |∅| = 0.
Denition. The set A is called nite i there exists such n ∈ N
that there exists bijection f : {1, 2, 3, . . . , n} → A. In that case we
say that cardinality of A is n, and that A has n elements, and
denote this as |A| = n.
Denition. The set is called innite i it is not nite.
Theorem. The set A is innite i for all n ∈ N there exists a
subset An ⊂ A such that |An | = n.
Theorem. The set A is innite i there exists such a subset B ⊂ A
that B 6= A and there exists bijection f : B → A.
Denition. Cardinality |N| = ℵ0 is called innite countable. Finite
cardinalities and innite countable cardinality are together called
¾countable cardinality¿. Symbol ℵ is read as ¾aleph¿.
Denition. Cardinality |R| = c is called continuum. We say that
cardinality of A is continuum i there exists a bijection f : A → R.
25 / 39
Functions everywhere
Denition. If A is a set, then the set of all subsets of A is called
the power set of A and denoted as 2A = P(A) = {A1 |A1 ⊂ A}.
Denition (natural numbers as sets). In one of the models of
natural numbers it is dened that 0 = ∅, 1 = {0}, 2 = {0, 1},
n = {k|k < n} = {0, 1, . . . , n − 1}.
Denition (power set as set of indicator functions). With the
above denitions we have 2A = {0, 1}A = {1A1 |A1 ⊂ A}.
Denition (Cartesian product as a set of functions). Product
of
Q |B| copies Bof the set A is dened as
α∈B A = A = {f |f : B → A}.
Denition (tuples as functions). The n-tuple (a0 , . . . , an−1 ) can
be considered as function a : n = {k|k < n} → {a0 , . . . , an−1 }
dened by a(k) = ak .
Example (Rn as set of functions).QWith the above denitions:
· · × R} = Rn = R{0,1,...,n−1} = k∈{0,...,n−1} R.
|R × ·{z
n
Denition (sequence as function). A sequence (an )n∈N of
elements of the set A by denition is a function a ∈ AN , an = a(n). 26 / 39
Functions of two variables
Denition. Function of two variables is dened as function of one
variable dened on Cartesian product, f : A × B → C , f ∈ C A×B ,
and we often write ¾f (a, b) ∈ C for all a ∈ A, b ∈ B ¿ instead of
¾f ((a, b)) ∈ C for all (a, b) ∈ A × B ¿.
Remark. Function of two variables f ∈ C A×B can be considered as
a function-valued function in two ways:
f (a, ·) = [b 7−→ f (a, b)] ∈ C B for each a ∈ A,
f (·, b) = [a 7−→ f (a, b)] ∈ C A for each b ∈ B.
For every xed a the value f (a, b) is a function of b , f (a, ·) ∈ C B .
For every xed b the value f (a, b) is a function of a, f (·, b) ∈ C A .
Example. You have already met this construction if you had an
experience of changing order of integration in double integral:
Z
Z Z
Z Z
f (x, y )dxdy =
f (x, y )dy dx =
f (x, y )dx dy
X ×Y
X
f ∈ RX ×Y
Y
f (x, ·) ∈ RY
Y
X
f (·, y ) ∈ RX
27 / 39
Functions of two variables are also called operations
Remark. C A×B = (C B )A = (C A )B because J dened as
1
J1 : C A×B 3 f 7−→ J1 (f ) = [a 7→ [b 7→ f (a, b)]] ∈ (C B )A
is a natural bijection between C A×B and (C B )A . Also J2 dened as
J2 : C A×B 3 f 7−→ J2 (f ) = [b 7→ [a 7→ f (a, b)]] ∈ (C A )B
is a natural bijection between C A×B and (C A )B .
Denition. Functions f : A × A → A are often called operations in
A and admit three notations for f (a1 , a2 ):
1. Inx notation f (a1 , a2 ) = a1 fa2
2. Prex (Polish) notation f (a1 , a2 ) = fa1 a2
3. Postx (reverse Polish) notation f (a1 , a2 ) = a1 a2 f
Remark. Now it is clear what is meant when people say ¾+ is an
operation in R: they mean that + is a function + : R × R → R and
its values are written in the inx notation +(x, y ) = x + y .
Denition. Functions f : A × B → C are also often called
operations and the same notation is used.
28 / 39
Fields
Denition. Let F be a set having at least two elements (denoted 0
and 1) and suppose that there are two opeartions + and · in F . We
say that (F , 0, 1, +, ·) is a eld i the following conditions are true:
1. Associativity of addition and multiplication:
a + (b + c) = (a + b) + c , and a · (b · c) = (a · b) · c .
2. Commutativity of addition and multiplication: a + b = b + a,
and a · b = b · a.
3. Additive and multiplicative identity: there exist two dierent
elements 0 and 1 in F such that a + 0 = a and a · 1 = a.
4. Additive inverses: for every a in F , there exists an element in F ,
denoted −a, called the additive inverse of a, such that
a + (−a) = 0.
5. Multiplicative inverses: for every a 6= 0 in F , there exists an
element in F , denoted by a−1 or 1/a, called the multiplicative
inverse of a, such that a · a−1 = 1.
6. Distributivity of multiplication over addition:
a · (b + c) = (a · b) + (a · c).
29 / 39
Equivalence relations
Denition. Binary relation R ⊂ A × A is called:
1. Reexive i aRa for all a ∈ A
2. Symmetric i (aRb) ⇐⇒ (bRa) for all a, b ∈ A
3. Transitive i ((aRb)&(bRc)) ⇒ (aRc) for all a, b, c ∈ A
4. Equivalence relation i all conditions 1,2,3 hold. Equivalence
relations are usually denoted by symbols similar to ∼, ≃, ∼
=.
Denition. If A is a set with equivalence relation ∼ then for each
a ∈ A the class of equivalence of element a is dened as
Ka = {b ∈ A|a ∼ b}. Every b ∈ Ka is called a representative of the
class Ka . If B ⊂ A and there exists such a ∈ A that B = Ka then B
is called a class of equivalence of the relation ∼.
Exercise. Prove that a ∈ Ka and that Ka ∩ Kb 6= ∅ ⇒ Ka = Kb .
Remark. The above statement shows that A can be represented as
disjoint sum of classes of equivalence of the relation ∼. I.e. there
exists a partition of A into subsets such that each subset is a class
of equivalence of the relation ∼. Such representation is called
factorization of A with respect to ∼. The set of classes is called a
factor set and denoted A/ ∼.
30 / 39
Equivalence relations
Denition. Two sets A and B are called disjoint i A ∩ B = ∅.
Denition. Notation C = A t B means that C = A ∪ B and that
at the same time A and B are disjoint. Expression A t B is called
the disjoint sum (or disjoint union) of A and B .
Example. Consider A = Z, number p ∈ N and a ∼ b i there
exists such m ∈ Z that a − b = pm. Then there are exactly p
classes of equivalence Kj = {j + pm|m ∈ Z} for j = 0, 1, . . . , p − 1,
we have factorization Z = K0 t K1 t · · · t Kp−1 and factor set
Zp = {K0 , K1 , . . . , Kp−1 }.
31 / 39
Order relations
Denition. Binary relation R ⊂ A × A is called:
1. Anti-reexive i aRa is false for all a ∈ A
2. Anti-symmetric i ((aRb)&(bRa)) ⇒ a = b for all a, b ∈ A
3. Connected i for all a, b ∈ A such that a 6= b at least one of the
following is true: aRb or bRa.
4. Order relation i R is reexive, anti-symmetric, transitive.
(Notation: order relations are usually denoted by symbols similar to
≤, ≥, 4, <.)
5. Strict order relation i R is anti-reexive, anti-symmetric,
transitive. (Notation: strict order relations are usually denoted by
symbols similar to <, >, ≺, .)
Denition. Relation R is called linear order i R is an order
relation and R is connected.
Denition. Relation R is called strict linear order i R is a strict
order relation and R is connected.
32 / 39
So, what about adding real numbers?
We have now dened (I hope!) almost all logical symbols and
notation from the set theory that are used in other disciplines at
our Master's program.
Now we can come back to the question: how one can dene x + y ,
√
x , sin(x) for real x and y ? The answer is the following: this is
done via the limit procedure.
Denition. We say that sequence a : N → R, n 7−→ an , a(n) = an
has limit a ∈ R, and also we say in that case that an converges to
a, and write lim an = a i the following holds:
n→∞
I For each k ∈ N there exists such n0 ∈ N that condition n ≥ n0
implies |a − an | < k1 .
Remark. The above condition can be generalized to the following,
which produces the same notion of convergence:
I For each ε ∈ R such that ε > 0 there exists such n0 ∈ N that
condition n ≥ n0 implies |a − an | < ε.
33 / 39
So, what about adding real numbers?
Theorem. Suppose that x and y are real numbers given in their
decimal representation:
x = x1 x2 . . . xk , xk+1 . . . and y = y1 y2 . . . xm , ym+1 . . .
Denote sequences an and bn of rational numbers as follows:
an = x1 x2 . . . xk , xk+1 . . . xn 00000 . . .
bn = y1 y2 . . . ym , ym+1 . . . yn 00000 . . .
(Clearly an and bn are rational numbers for all n ≥ max(k, m).)
Then limit lim an + bn and limit lim an · bn both exist.
n→∞
This allows to dene
x + y = lim an + bn ,
n→∞
n→∞
x · y = lim an · bn .
n→∞
Theorem. Denote 0! = 1, 1! = 1, 2! = 1 · 2, 1 · 2 · 3 · · · · · m = m!
Pn (−1)k x 2k+1
for all natural m. Then for all x ∈ R limit lim
k=0 (2n+1)!
n→∞
Pn (−1)k x 2k+1
exists. This allows to dene sin(x) = lim
k=0 (2n+1)! .
n→∞
√
Question. What about x ? The answer will be in the next lesson. 34 / 39
Seminar 2
On seminar 2 we solved two problems.
Problem 1. Prove that sequence xn = n converges to 0.
1
Problem 2. Prove that sequence xn = (−1)n does not have a limit.
35 / 39
Question
During the lecture there was a question that I preferred not to
answer at the moment because of lack of lecture time. The question
was a request to provide example to the remark that I repeat below.
Remark on functionals and operators. If all values of function f
are numbers, and domain of f is some set of functions then usually
we say that ¾f is a functional¿. If both domain and codomain of
function f are sets of functions then we usually say that ¾f is an
operator¿ and most likely will write capital F instead of small f .
This is informal explanation which helps to bulid intuition.
Example. Let RR be the set of all functions x that map R to R,
i.e. x : R → R.
• For each x ∈ RR denote f (x) = 1, g (x) = x(0),
h(x) = sin(x(0) + |x(1)|3 ), w (x) = x(x(0)). Then f , g , h, w are
R-valued functionals dened on RR .
• For each t ∈ R and each x ∈ RR let us denote (F (x))(t) = 1,
(G (x))(t) = x(0)t + sin(t) + x(t)e t , (H(x))(t) = x(t + 1),
(W (x))(t) = x(t 2 + sin(x(0))). Then F , G , H, W are RR -valued
operators dened on RR .
36 / 39
Question
Sets of functions that are domain and codomain of operator may
be deerent.
Example. Let RR be as before, and R[ , ]×[ , ] be the set of all
0 1
0 1
R-valued functions u of two variables t1 , t2 ∈ [0, 1].
• For each t1 ∈ [0, 1], t2 ∈ [0, 1] and each x ∈ RR let us denote
(S(x))(t1 , t2 ) = 1 + t1 + x(t1 + t2 ).
Then S is an operator S : RR → R[0,1]×[0,1] .
• For each t ∈ R and each u ∈ R[0,1]×[0,1] dene
(K (u))(t) = 2 + t 2 u(| cos(t)|, 1/(1 + t 2 )).
Then K is an operator K : R[0,1]×[0,1] → RR .
37 / 39
Homework 2
Problems to solve at home (deadline September 15, 20:00):
1a. Exercises at p.3, p.4, p.5.
1b. Apply fundamental
theorem of arithmetic to numbers 72, 5040,
√
1, 0, -1, 13 and 101.
2a. Do the rst exercise at p.13. Let a, b and c be logical
statements. Write truth tables for the following statements a = a,
a ∧ b , a ∨ b , ¬a ∨ a, ¬a ∧ a, a → b , a ⇐⇒ b .
2b. Do the last exercise at p.13.
Denition. Suppose that U is a non-empty set, A ⊂ U and for
each x ∈ U dene χA (x) = 1 ⇐⇒ x ∈ A and
χA (x) = 0 ⇐⇒ x ∈
/ A. This denes function χA : U → {0, 1}
which is called the characteristic function of the set A.
3. Suppose that A ⊂ U and B ⊂ U . The task is to express
χA∪B , χA∩B , χA\B , χA4B in terms of χA and χB .
Example: for each x ∈ U we have χA∩B (x) = χA (x)χB (x).
38 / 39
Homework 2
3a. On a sheet of paper plot sets [0, 1] × [2, 3], (0, 1] × R,
[0, +∞) × (−∞, 1].
3b. Consider relation Q ⊂ R × R dened as follows:
xQy ⇐⇒ x · y ≥ 0. Plot the set Q . Is Q injective (left-unique),
right-unique, left-total, right-total (surjective)? Is Q a function? Is
Q a bijection? Is Q an equivalence relation? Is Q an order relation?
Plot Q −1 . Is Q −1 a function?
3c. Consider relation S ⊂ R × R dened as follows:
xSy ⇐⇒ y = x 2 . Plot the set S . Is S injective (left-unique),
right-unique, left-total, right-total (surjective)? Is S a function? Is
S a bijection? Is S an equivalence relation? Is S an order relation?
Plot S −1 . Is S −1 a function?
4a. Suppose that x ∈ R and x 6= 1. Prove that
n+1
for all n ∈ N.
1 + x + x 2 + · · · + x n = 1−x
1−x
4b. Suppose that x ∈ R and |x| < 1. Prove that the sequence
n+1
an = 1−x
converges to 1/(1 − x).
1−x
4c. Prove that the sequence bn = n does not have a limit.
5. Understand and explain answer to the question at pp. 36-37.
39 / 39
Introduction to
One-parameter semigroups and evolution
equations
Lecture 3: Necessary background from Real Analysis
Seminar 3
Ivan Remizov
HSE Nizhny Novgorod, Russia
13+16 September 2022
1 / 15
Real numbers: upper bound and supremum
Denition. A number x ∈ R is called an upper bound for a set
A ⊂ R i for all a ∈ A we have a ≤ x . If a set has an upper bound
then this set is called bounded from above.
Denition. A number x ∈ R is called the least upper bound for a
set A ⊂ R, or supremum, which is denoted as x = sup A i two
conditions hold:
1. x is an upper bound for A
2. if y is an upper bound for A then x ≤ y
Equivalent denition. A number x ∈ R is called the least upper
bound for a set A ⊂ R, or supremum, which is denoted as
x = sup A i two conditions hold:
1. a ≤ x for all a ∈ A
2. for all ε > 0 number x − ε is not an upper bound for A, i.e. there
exists aε ∈ A such that x − ε < aε .
One more equivalent denition. By denition, the exact upper
bound for the set A, or the supremum of the set A, is the smallest
(the minimal) upper bound for A.
2 / 15
Real numbers: lower bound and inmum
Denition. A number x ∈ R is called a lower bound for a set
A ⊂ R i for all a ∈ A we have a ≥ x . If a set has a lower bound
then this set is called bounded from below.
Denition. A number x ∈ R is called the greatest lower bound for
a set A ⊂ R, or inmum, which is denoted as x = inf A i two
conditions hold:
1. x is a lower bound for A
2. if y is a lower bound for A then x ≥ y
Equivalent denition. A number x ∈ R is called the greatest
lower bound for a set A ⊂ R, or inmum, which is denoted as
x = inf A i two conditions hold:
1. a ≥ x for all a ∈ A
2. for all ε > 0 number x + ε is not a lower bound for A, i.e. there
exists aε ∈ A such that x + ε > aε .
One more equivalent denition. By denition, the exact lower
bound for the set A, or the inmum of the set A, is the greatest
(the maximal) lower bound for A.
3 / 15
Monotone sequences
Denition. Suppose that (xn )n∈N is a sequence of real numbers. If
for all n ∈ N it is true that
I xn+ > xn , then the sequence (xn )n∈N is called strictly
increasing
I xn+ ≥ xn , then the sequence (xn )n∈N is called non-strictly
increasing (also called non-decreasing)
I xn+ < xn , then the sequence (xn )n∈N is called strictly
decreasing
I xn+ ≤ xn , then the sequence (xn )n∈N is called non-strictly
decreasing (also called non-increasing)
Denition. In all these cases the sequence is called monotone.
1
1
1
1
4 / 15
Bounded monotone sequence converges
Denition. A set is called bounded i it is bounded from above
and from below.
Denition. A sequence is called bounded (bounded from above,
bounded from below) i the set of all values of this sequence is
bounded (bounded from above, bounded from below).
Theorem. If a sequence of numbers is bounded and monotone,
then it has a limit. That means:
• if a sequence is bounded from above and (strictly or non-strictly)
increases, then it has a limit.
• if a sequence is bounded from below and (strictly or non-strictly)
decreases, then it has a limit.
Remark. This theorem is used to dene all arithmetic operations
+, -, ·, / for real numbers. In that approach real numbers are
assumed to be limits of monotone sequences of rational numbers,
e.g. the corresponding sequence for π = 3.14159265359 . . . is 3,
3.1, 3.14, 3.141, 3.1415, 3.14159, ...
5 / 15
Properties of the limit
Theorem. If we have real number c , two sequences of real
numbers xn , yn , and limn→∞ xn = x and limn→∞ yn = y then limits
limn→∞ xn + yn , limn→∞ xn − yn , limn→∞ c · xn all exist and the
following equalities hold:
lim xn + yn = lim xn + lim yn ,
n→∞
n→∞
n→∞
lim xn − yn = lim xn − lim yn ,
n→∞
n→∞
n→∞
lim xn · yn = lim xn · lim yn ,
n→∞
n→∞
n→∞
lim c · xn = c · lim xn .
n→∞
n→∞
If, additionally, yn 6= 0 for all n ∈ N, and limn→∞ yn 6= 0, then limit
limn→∞ xn /yn exists and
lim
xn
n→∞ yn
=
limn→∞ xn
.
limn→∞ yn
6 / 15
Limit of a function and continuity, existence of continuous inverse function
Denition. Suppose that we have a function f : (a, b) → R, and
x ∗ ∈ (a, b). If there exists such a number y ∈ R that for all
sequences (xn ) ⊂ (a, b) with xn 6= x ∗ condition limn→∞ xn = x ∗
implies condition limn→∞ f (xn ) = y , then we say that f (x) tends
to y as x tends to x ∗ , and write limx→x ∗ f (x) = y .
Denition. Function f : (a, b) → R is called continuous at point
x ∗ ∈ (a, b) i the limit limx→x ∗ f (x) exists and
limx→x ∗ f (x) = f (x ∗ ).
Denition. Function f : (a, b) → R is called continuous at the set
A ⊂ (a, b) i f is continuous at each point x ∗ ∈ A.
Theorem. If function f : (a, b) → R is continuous and strictly
increasing, then f is a bijection as a mapping f : (a, b) → f ((a, b)),
f −1 is a function, and f −1 is a continuous, strictly increasing
mapping f −1 : f ((a, b)) → (a, b).
Theorem. If function f : (a, b) → R is continuous and strictly
decreasing, then f is a bijection as a mapping f : (a, b) → f ((a, b)),
f −1 is a function, and f −1 is a continuous, strictly decreasing
mapping f −1 : f ((a, b)) → (a, b).
7 / 15
Continuity of composition
Denition. Suppose that A, B, C are non-empty sets, f : A → B
and g : B → C are functions, and D(f ) = A, D(g ) = B . Then
function h : A → C dened on the domain D(h) = A by equality
h(x) = g (f (x)) for all x ∈ A is called the composition of functions
f and g . Very often notation h = g ◦ f is used. In functional
analysis for operators notation h = gf is also widely used. The idea
is that for all x ∈ A the element f (x) ∈ B is well dened, so we can
calculate the value of g on this element, and this value is g (f (x)).
Remark. Recall that f (A) = {f (x)|x ∈ A}.
Theorem. Suppose functions f and g are given and continuous on
their domains f : (a, b) → R, g : (c, d) → R. Suppose that
f ((a, b)) ⊂ (c, d). Then the composition h : (a, b) → R,
h(x) = g (f (x)) is a continuous function on (a, b).
8 / 15
Elementary functions
Exercise. Visit web-page
https://en.wikipedia.org/wiki/Elementary_function
Using theorems on existence of inverse functions the basic
elementary functions are introduced.
Denition. Basic elementary functions of a single variable x :
• Constant functions: 2, π, e , etc.
√
2
m
• Rational powers of x : x, x 3 , x n for all R 3 x ≥ 0, all m ∈ Z
and all n ∈ N.
• Exponential functions: e x , ax for all a > 0 and all x ∈ R
• Logarithms: loga x , ln x = loge x for all x ∈ (0, +∞)
• Trigonometric functions sin x, cos x for all x ∈ R. Trigonometric
function tan x for all x ∈ R, x 6= π/2 + πk for all k ∈ Z. Other
triginometric functions are also sometimes called basic.
• Inverse trigonometric functions: arcsin x , arccos x for all
x ∈ [0, 1], and arctan x for all x ∈ R. Other inverse triginometric
functions are also sometimes called basic.
9 / 15
Elementary functions
Denition. Elementary functions are by denition are functions that can
be obtained from basic elementary functions using arithmetic operations
+, −, ·, / and composition, i.e. if f and g are elementary functions then
h(x) = f (g (x)) is also by denition an elementary function.
sin(x π +1)
√
x
+arctan(x)
√
Example. w (x) = log
, |x| = x 2 are elementary functions.
5 ( x+1000·e)−3
Theorem. All elementary functions are continuous in each point of their
domains.
Remark. Other properties of elementary functions e.g.
(sin(x))2 + (cos(x))2 = 1 are supposed to be known from school.
Depending on the approach which is used to dene basic elementary
functions, these properties may be theorems or may be part of the
denition. When an elementary function is dened then all its properties
are proven using only facts that we already mentioned in our course. This
relates also to the following theorem:
Theorem. All elementary functions are monotone on the intervals of
monotonicity that are known from school.
10 / 15
Innite limits
Denition. We say that limn→∞ xn = +∞ i for each E > 0 there
exists such n (E ) ∈ N that for all n ≥ n (E ) we have xn > E .
Exersice. Give denition of the following innite limits:
limn→∞ xn = −∞, limx→+∞ f (x) = +∞, limx→+∞ f (x) = −∞,
limx→−∞ f (x) = +∞, limx→−∞ f (x) = −∞.
0
0
11 / 15
Seminar 3
Seminar 3 was given in two parts. On the rst part (September 13)
we solved the following problems:
Problem 1. Second exercise from p.13 of lesson 2.
Problem 2. Suppose that A = {1, 2, ∗, ?, +},
B = {−1, −2, 0, 1, 7, 8}. Find A ∩ B , A ∪ B , A \ B , B \ A, A4B .
Problem 3. Let A be as in problem 2. Dene by explicitly giving
elements several sets, given by selection rule: {x ∈ A|x is a digit},
{n ∈ N|n < 10}, {y ∈ R|y < 100}, {y ∈ R|y < 9}.
2
Problem 3. Suppose that A = B = R and relation r is given as
r = {(x, y ) ∈ R |(x − 2) + y ≤ 1}. Test r for properties of binary
relation listed at p.23 of lesson 2 and p.30 of lesson 2, p.32 of
lesson 2. Find D(r ) and R(r ). Is r a function?
2
2
2
Problem 4. The same question as in problem 3, but for
r = {(x, y ) ∈ R |x + y + 1 = 0}.
2
12 / 15
Seminar 3
On the second part of seminar 3 (September 16) we solved the
following problems:
Problem 1. Suppose that binary relation ∼ on QN is given by the
rule x ∼ y i exists limn→∞ xn − yn = 0. Prove that ∼ is an
equivalence relation.
We also intruduced the following denition and made the following
remark.
Denition. Sequence x ∈ QN is called fundamental i for each
ε > 0 there exists n (ε) ∈ N such that for all m, n ∈ N condition
(m ≥ n (ε)) ∧ (m ≥ n (ε)) implies |xn − xm | < ε.
Remark. The set R can be dened as factor set
{x ∈ QN |x is fundamental}/ ∼.
0
0
0
See the next page
13 / 15
Seminar 3
Problem 2. For several sets A nd the set of all upper bounds, the set of
all lower bounds, sup A, inf A. We considered A = (0, 1],
A = { n2 : n ∈ N}, A = N.
Problem 3. Suppose that if x, y ∈ RN , and limn→∞ xn = a ∈ R,
limn→∞ yn = b ∈ R. Using the denition of a limit prove that
limn→∞ xn + yn = a + b . Comment: on the seminar we wrote x instead of
a and y instead of b , which is not the best choice of notation because it
is natural to accept that x is a sequence, i.e. a function of natural
number n dened as x(n) = xn , so it is an abuse of notation to write
limn→∞ xn = x dening a sequence and its limit by the same letter x .
Problem 4. Prove that function f : [0, +∞) → R, f (x) = x 2 is
continuous and strictly increasing. Comment: we again used bad notation
as in problem 3.
Remark. By the continuous inverse theorem (see p.7 of lesson 3)
function f from problem 4 has a continuous strictly increasing inverse
f√−1 . This is a standard denition of the square root function
x = f −1 (x) for all x ∈ [0, +∞).
14 / 15
Homework 3
Problems to solve at home (deadline September 22, 20:00):
1. Exercise at p. 30 of lesson 2.
2. For sets A given below nd the set of all upper bounds, the set of all
lower bounds, sup A, inf A:
2a. A = (0, 1] ∪ [2, 3)
2b. A = {− n12 : n ∈ N}
2c. A = Q.
3a. Prove that if a sequence of real numbers has a limit which is a real
number then this sequence is fundamental.
3b. Give an example of fundamental sequence of rational numbers which
has a limit equal to irrational number. Understand that this is a sequence
which is fundamental in Q but do not have limit in Q, i.e. this sequence
is fundamental even if real numbers are not introduced yet, and this
sequence has no limit if real numbers are not introduced yet.
4. Prove that if sequence of real numbers is fundamental then it is
bounded.
5a. Using the denition of a limit of a sequence (i.e. explicitly giving
n+1
n0 (ε) for each ε > 0) prove that limn→∞ 2n+
3 = 2.
x+2
5b. Find limx→1 x+3 by denition of a limit of a function.
2
5c. Find limx→1 x+
15 / 15
x+3 using properties of elementary funtions.
Introduction to
One-parameter semigroups and evolution
equations
Lesson 4: Necessary background from Real Analysis
Seminar 4
Ivan Remizov
HSE Nizhny Novgorod, Russia
20 September 2022
1 / 15
Neighbourhoods
Denition. For ε > 0 the open epsilon-neighbourhood (or just
ε-neighbourhood) of a point x ∈ R is the set
Uε (x) = (x − ε, x + ε) = {y ∈ R : |x − y | < ε}.
Denition. A deleted neighbourhood (sometimes called a
punctured neighbourhood) of a number x ∈ R is a neighbourhood
of x without x , i.e. the set
Ůε (x) = (x − ε, x) ∪ (x, x + ε) = {y ∈ R : 0 < |x − y | < ε}.
Denition. The E -neighbourhood of +∞ by denition is
(E , +∞) = {y ∈ R : y > E }, of −∞ by denition is
(−∞, E ) = {y ∈ R : y < E }, here E may be positive or negative.
For E > 0 the E -neighbourhood of unsigned innity ∞ by
denition is (−∞, E ) ∪ (E , +∞) = {y ∈ R : |y | > E }. For
innities usually it is assumed that there is no dierence between
punctured neighbourhood and usual neighbourhood.
Remark. Although +∞, −∞, ∞ are not real numbers (really!), it
is useful to call some sets of real numbers being their
neighbourhoods. Sometimes this allows to give some statements in
a unied way. The example will be shown on the next page.
2 / 15
Limits in terms of neighbourhoods
In many cases the following statement is useful:
Lemma. Let ∗ be real number or innity, and suppose that
real-valued function f is dened in some deleted neighbourhood of
∗. Then limx→∗ f (x) = 0 i limx→∗ |f (x)| = 0.
Proof. Indeed |f (x) − 0| = |f (x)| = ||f (x)| − 0|. (See the denition
of a limit!)
Exercise. Write the full proof and understand it.
Also with the notion of neighbourhood we can give an equivalent
denition of a limit:
Theorem. Let ∗ and a be real numbers or innities, and suppose
that real-valued function f is dened in some deleted
neighbourhood of ∗. Then limx→∗ f (x) = a which we dened
previously in terms of sequences IFF for each neighbourhood U(a)
of a there exists punctured neighbourhood V̊ (∗) of ∗ such that
f (x) ∈ U(a) for all x ∈ V̊ (∗).
Exercise. Give the denition of continuity of a function in the
point ∗ ∈ R in terms of ε-neighbourhoods and δ -neighbourhoods.
3 / 15
Several remarkable limits
Theorem. The following holds:
1. limx→0 sin(x)
= 1 the rst remarkable limit (in Russian
x
terminological tradition)
x
2. limx→+∞ 1 + x1 = e the second remarkable limit (in
Russian terminological tradition)
n
Here we must mention that the sequence 1 + n1 can be shown to
be strictly increasing and bounded, hence it has a limit, and that
n
fact is a standard denition of number e = limn→∞ 1 + n1 .
x
3. limx→0 e x−1 = 1
4. limx→0 ln(1x+x) = 1
1
5. limx→0 cos(x)−
= − 12
x2
µ
−1
6. limx→0 (1+x)
= µ for all µ ∈ R
x
√
n
1
7. limx→0 1+x−
= n1 for all n ∈ N (this follows from item 6
x
above for µ = 1/n)
α
8. limx→+∞ xe x = limx→+∞ ln(x)
x α = 0 for all α > 0
4 / 15
Limits of functions of two variables
Let us dene continuity and limits for functions of two variables.
Denition. For ε > 0 the ε-neighbourhood of a point (x0 , y0 ) ∈ R2
is by defnition the set Uε (x0 , y0 ) = Uε (x0 ) × Uε (y0 ). By denition
Ůε (x0 , y0 ) = Uε (x0 , y0 ) \ {(x0 , y0 )}.
Remark. See the dierence between Ůε (x0 ) × Ůε (y0 ) and
Uε (x0 , y0 ) \ {(x0 , y0 )}.
Denition. Suppose that we have number or innity ∗ and point
(x0 , y0 ) ∈ (a, b) × (c, d) ⊂ R2 . Then ∗ is called a limit of a
function f : (a, b) × (c, d) → R as (x, y ) → (x0 , y0 ), and notation
f (x, y ) = ∗
lim
(x,y )→(x0 ,y0 )
is used i for each neighbourhood U(∗) of ∗ there exists punctured
neighbourhood V̊ (x0 , y0 ) of (x0 , y0 ) such that f (x, y ) ∈ U(∗) for all
x ∈ V̊ (x0 , y0 ).
Denition. Function f : (a, b) × (c, d) → R is called continuous at
the point (x0 , y0 ) ∈ (a, b) × (c, d) i
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
5 / 15
Properties of the limit
Theorem. If exists
lim
(x,y )→(x0 ,y0 )
f (x, y ) = a ∈ R,
then two repeated limits exist and
lim lim f (x, y ) = lim lim f (x, y ) =
x→x0 y →y0
y →y0 x→x0
lim
f (x, y ) = a.
(x,y )→(x0 ,y0 )
Theorem (corollary of the previous theorem). If function f is
continuous at point (x0 , y0 ), then double limit and two repeated
limits all exist and
lim lim f (x, y ) = lim lim f (x, y ) =
x→x0 y →y0
y →y0 x→x0
lim
(x,y )→(x0 ,y0 )
f (x, y ) = f (x0 , y0 ).
Theorem. Functions f (x, y ) = x + y , g (x, y ) = x − y ,
h(x, y ) = x · y are continuous at each point of R2 . Function
w (x, y ) = yx is continuous at each point (x0 , y0 ) such that y0 6= 0.
Function z(x, y ) = x y is continuous at each point (x0 , y0 ) such
that x0 > 0.
6 / 15
Properties of the limit
Remark. Recall that function h that is dened on some
neighbourhood of a ∈ R is continuous at a i
limx→a h(x) = h (limx→a x), because h (limx→a x) = h(a). On the
language of sequences this means that f is continuous at a i
condition limn→∞ xn = a implies that
limn→∞ h(xn ) = h (limn→∞ xn ). The last phrase takes into account
all possible sequences xn such that limn→∞ xn = a. What if xn is
not arbitrary, but, in particular, xn = u(yn )? We will have
limn→∞ u(yn ) = a, and, if h is continuous, obtain
limn→∞ h(u(yn )) = h (limn→∞ u(yn )). This reasoning is higly
employed. Let us state a theorem about this.
Theorem. Let a, b be numbers or innities. Suppose that h is
dened in some neighbourhood of a, u is dened in some
neighbourhood of b , and there exist nite or innite
limy →b u(y ) = a, limx→a h(x). Then
limy →b h(u(y )) = limx→a h(x). If a ∈ R and h is continuous at a,
then limy →b h(u(y )) = limx→a h(x) = h(a) = h (limy →b u(y )) .
Remark. The same is true for functions of two variables.
7 / 15
Properties of the limit
Remark. The above theorems are highly used in calculation of
limits of functions of one variable. Suppose that f0 , g0 are real
numbers and f0 > 0. Then, because function z = u v is continuous
at point (f0 , g0 ) we have
lim
uv =
(u,v )→(f0 ,g0 )
lim
(u,v )→(f0 ,g0 )
z(u, v ) = z(f0 , g0 ) = f0 g0 .
Suppose that exists limx→x0 f (x) = f0 > 0 and exists
limx→x0 g (x) = g0 ∈ R then
f0 g0 =
lim
(u,v )→(f0 ,g0 )
uv
in particular
=
lim
(u,v )→(f0 ,g0 )
u v = lim f (x)g (x) .
x→x0
u=f (x),v =g (x)
So, continuity of z = u v at point (f0 , g0 ) implies
limx→x g (x)
0
g (x)
g0
lim f (x)
= f0 = lim f (x)
.
x→x0
x→x0
Exercise. Provide example of a function that is not continuous at
point (0, 0) but limits limx→0 limy →0 f (x, y ), limy →0 limx→0 f (x, y )
both exist and a) are equal b) are not equal.
8 / 15
Innitely small sequences
Denition. Innitely small functions, also called innitesimal
functions, that are some class of functions, should NOT be
confused with innitesimal numbers that are some class of objects
that are ¾arbitrary close to zero but not zero¿ and do not belong
to R. In our course we will use innitesimal functions but not
innitesimal numbers. Russian terminology is better to destingush
these two notions.
Denition. Suppose that real-valued function f is dened in some
neighbourhood of ∗ that may be a number or innity. We say that
f is an innitesimal function as x tends to ∗ i limx→∗ f (x) = 0.
Denition. We say about two innitesimal functions f and g that
• f and g are equivalent as x → ∗ i limx→∗ gf (x)
(x) = 1, in that case
we also write ¾f (x) ∼ g (x) as x → ∗¿
• f and g have the same order i limx→∗ gf (x)
(x) = c 6= 0, c ∈ R.
• for nite ∗ we say that f has order α > 0 at ∗ i
f (x) ∼ C |x − ∗|α for some C 6= 0
9 / 15
¾Small o¿ and ¾Big O¿ notation, innitely large functions
Denition. We say about two functions f and g , dened in some
neighbourhood of ∗, which can be a number or innity, that:
• f is a higher order innitesimal than g i f and g are
innitesimal functions as x → ∗ and for all ε > 0 there exists such
a neinghbourhood U(∗) that |f (x)| ≤ ε|g (x)| for all x ∈ U(∗), in
that case we also write ¾f (x) = o(g (x)) as x → ∗¿. If g (x) 6= 0
this means that limx→∗ gf (x)
(x) = 0.
• if property g (x) ≥ 0 is satised then we say that f (x) = O(g (x))
as x → ∗ i there exists such a number M > 0 and such a
neinghbourhood U(∗) that |f (x)| ≤ Mg (x) for all x ∈ U(∗).
Denition. Suppose that real-valued function f is dened in some
neighbourhood of ∗, which can be a number or innity. We say that
f is an innitely large function as x tends to ∗ i limx→∗ f (x) = ∞
or limx→∗ f (x) = +∞ or limx→∗ f (x) = −∞.
10 / 15
Seminar 4
Seminar 4 was in two parts. On the rst part we solved the
following two problems.
Problem 1. HW2, problem 4b. Suppose that x ∈ R and |x| < 1.
n+1
Prove that the sequence an = 1−x
converges to 1/(1 − x).
1−x
Solution. On the seminar we gave not the best solution and forgot
about 1-x in the denominator. Here we provide a better solution.
The idea was to apply the denition of a limit directly to see on
this example how this denition works.
Suppose that x ∈ (−1, 1) is xed, and arbitrary ε > 0 is given. We
need to prove that there exists such n0 (x, ε) ∈ N that for all n ≥ n0
the following inequality holds:
ε > |an − a| =
1
−x n+1
|x|n+1
1 − x n+1
−
=
=
.
1−x
1−x
1−x
1−x
It will be satised i |x|n+1 < ε(1 − x). Case x = 0 is trivial
because an = 1 = a for all n, and one can take n0 = 1 for all ε > 0.
Let us consider the case x 6= 0, then together with x ∈ (−1, 1) it
gives |x| ∈ (0, 1) and (1 − x) ∈ (0, 2).
11 / 15
Seminar 4
If we were aware of logarithms on lesson 2 (we are solving problem
from HW2) we could mention that |x|n+1 < ε(1 − x) is equivalent
to ln(|x|n+1 ) < ln(ε(1 − x)) because function x 7→ ln(x) is strictly
increasing. The last inequality is equivalent to
n + 1 > ln(ε(1 − x))/ ln(|x|) because ln(y z ) = z ln(y ) and
ln |x| < 0.
Denition. Let us denote by dy e the lowest integer that is not less
than real number y , examples are d1.1e = 2, d1.9e = 2,
d−1.1e = −1, dke = k for all k ∈ Z. Function y 7→ dy e is called
the ceiling function, and writing dy e reads as ¾ceiling of y ¿.
Keeping in mind that by denition
we could give the
l dy e ≥ y m
1−x))
∈ N because for this
answer in the form n0 = max 1, ln(ε(
ln(|x|)
number n0 for all n ≥ n0 the
holds:m
l following
mestimate
l
ln(ε(1−x))
ln(ε(1−x))
1−x))
n + 1 > n ≥ n0 = max 1,
≥
≥ ln(ε(
ln(|x|)
ln(|x|)
ln(|x|) ,
so n + 1 > ln(ε(1 − x))/ ln(|x|) and |an − a| < ε. This would be the
end of the solution. However, at lesson 2 we have not introduced
logarithms yet, so we need another solution. See the next page.
12 / 15
Seminar 4
Let us use only properties of real numbers to nd such n0 ∈ N that for all
n ≥ n0 we have |x|n+1 < ε(1 − x). First, let us mention that due to
|x| < 1 we have |x|n+1 < |x|n < ε(1 − x). We will establish the second
part of the two-sided inequality because it is easier to write n than to
1
1
> 1 hence if we dene y = |x|
− 1 we
write n + 1. Again, |x| < 1, so |x|
1
will have |x| = 1 + y and y > 0.
Let us consider (1 + y )n , open the brackets and estimate as follows:
(1 +y )n = (1 + y ) · . . . · (1 + y ) = 1 +ny +C2 y 2 +C3 y 3 +· · ·+Cn y n > ny
|
{z
}
n
because Ck and y are some strictly positive real numbers. Next,
1
1
1
|x| = 1+y
so |x|n = (1+y
)n < ny . If n is large enough to achieve
1
1
1
1
ny ≤ ε(1 − x), i.e. n ≤ y ε(1 − x), i.e. n ≥ y ε(1−x) = ( 1 −1)ε(1−x) then
|x|
|x|n+1 < ε(1 − x). Using decimal representation of positive real number
z = z1 . . . zm , zm+1 . . . , where zk ∈ {0, 1, . . . , 9}, it is clear how to dene
the ceilng of thus number: dze = z1 . . . zm , 000 . . . +1 if z is not integer,
and dze = z if z is integer. Finally we can set n0 = ( 1 −11)ε(1−x) .
|x|
13 / 15
Seminar 4
Problem 2. Find limx→0
tg(2x)
x
. Answer: 2.
On the second part of seminar 4 we solved the following ve problems.
x
Problem 1. Find limx→−∞ 1 + x1 . Answer: e .
y x
Problem 2. For all y ∈ R nd limx→+∞ 1 + x . Answer: e y . This
formula works both for y = 0 and y 26= 0.
√
Problem 3. Find limx→0 (cos(x))1/x . Answer 1/ e .
√
Problem 4. Find limn→∞ n n. Answer: 1.
Problem 5. Are functions f and g innitesimal as x → 0? If yes then are
√
they of the
√ same order? Find orders of f and g . Take f (x) = tg x ,
g (x) = x 2 + 1 − 1.
√
Answer for problem 5: this problem is not well posed because x is not
dened for x < 0, so function f is not dened in some punctured
neighbourhood of 0, hence limit limx→0 f (x) is not dened so we cannot
say is true that limx→0 f (x) = 0 or not, hence we cannot say if f (x) is
innitesimal as x → 0 or not.
p
However, we can consider dierent function f , namely f (x) = |x| and
solve the same problem for it, calling it 5'.
Answer for problem 5': both f and g are innitesimal fuctions as x → 0,
they are of dierent orders, g (x) = o(f (x)), f (x) ∼ |x|1/2 , g (x) ∼ 12 x 2 ,
14 / 15
order of f is 1/2, order of g is 2.
Homework 4
Problems to solve at home (deadline October 1, 20:00):
1a. First exercise at p.3. 1b. Second exercise at p.3.
2a. Exercise at p.8, part a). 2b. Exercise at p.8, part b).
cos 2x
3a. Find limx→0 xlnsin(
3x)
5x)
5x)
e sin(√
e sin(
√
3b. Find limx→0 cos
and limx→0 cos
x
|x|
4a. Find limx→∞ sinx x using the denition of a limit.
q
x
4b. Find limx→∞ 3 1 + x1 + cos(2/x) − cos(1/x 2 )
5. Are f (x) and g (x) innitesimal as x → ∗? If yes then are they of the
same order as x → ∗? If yes, nd some simple standard equivalent
innitesimal functions for f and g as x → ∗. Is one of then o-small or
O-big for another as x → ∗? Find orders of f and g if they are
well-dened.
√
5a. Take ∗ = 0, f (x) = x√3 ln 1 + x12 , g (x) = 3 x 4 + 1 − 1.
5b. Take ∗ = 1, f (x) = x − 1, g (x) = 2x − 2.
q
x
6. Find limx→∞ 3 1 + x1 + cos(2x) − cos(x 2 ) . This is an optional
competitive task, 5 points for the rst in class correct solution, 4 points
for the second in class correct solution, 0 points for other solutions.
15 / 15
Introduction to
One-parameter semigroups and evolution
equations
Lesson 5: Necessary background from Real Analysis
Seminar 5
Ivan Remizov
HSE Nizhny Novgorod, Russia
27+29 September 2022
1 / 23
One-sided neighborhoods
This we already know:
Denition. For ε > 0 the open epsilon-neighbourhood (or just
ε-neighbourhood) of a point x ∈ R is the set
Uε (x) = (x − ε, x + ε) = {y ∈ R : |x − y | < ε}.
This is a new denition:
Denition. For ε > 0 the right one-sided epsilon-neighbourhood
(or just right ε-neighbourhood) of a point x ∈ R is the set
Uε (x + 0) = [x, x + ε) = {y ∈ R : |x − y | < ε and y ≥ x}.
Remark. In the symbol Uε (x + 0) adding +0 is a symbolic
combination that can not be reduced arithmetically using equality
x + 0 = x between numbers. This happens because adding this
¾+0¿ in the denition of the right neighbourhood has not only
numerical sence but also is a way of notation, distinguishing normal
(two-sided) and one-sided neighbourhoods. So x + 0 = x but
Uε (x + 0) 6= Uε (x), which is rather funny, but this notation is still
widely used and I consider this notation being a good one.
2 / 23
One-sided neighborhoods and one-sided limits
Similarly are dened 3 other one-sided neighbourhoods:
Denition. For ε > 0 the left one-sided epsilon-neighbourhood (or
just left ε-neighbourhood) of a point x ∈ R is the set
Uε (x − 0) = (x − ε, x] = {y ∈ R : |x − y | < ε and y ≤ x}.
Denition. For ε > 0 the left deleted one-sided
epsilon-neighbourhood (or just left deleted ε-neighbourhood) of a
point x ∈ R is the set
Ůε (x − 0) = (x − ε, x) = {y ∈ R : |x − y | < ε and y < x}.
Denition. For ε > 0 the deleted one-sided right
epsilon-neighbourhood (or just right deleted ε-neighbourhood) of a
point x ∈ R is the set
Ůε (x + 0) = (x, x + ε) = {y ∈ R : |x − y | < ε and y > x}.
Denition. Let a, b be real numbers, and f be a real-valued
function, dened in some deleted right neighbourhood of a. We say
that limx→a+ f (x) = b + 0 i for each right neighbourhood
Uε (b + 0) of point b there exists right deleted neighbourhood
Ůδ (a + 0) of point a such that f (x) ∈ Uε (b + 0) for all
x ∈ Ůδ (a + 0).
3 / 23
0
One-sided neighborhoods and one-sided limits
Exercise. Give at least 5 other denitions of one-sided limit, e.g.
limx→a−0 f (x) = b − 0, limx→a−0 f (x) = b etc.
Exercise. Prove the following theorem. Theorem. Let a, b be real
numbers, and f be a real-valued function, dened in some deleted
neighbourhood of a. Then ∃ limx→a f (x) = b if and only if all three
conditions hold:
∃ lim f (x), ∃ lim f (x), lim f (x) = lim f (x) = b.
x→a+0
x→a−0
x→a+0
x→a−0
4 / 23
Derivative as a limit
Denition. Suppose that function f takes real values and is
dened in some neighbourhood of point a ∈ R. We say that f is
dierentiable at a i the limit
f (x) − f (x0 )
x→x0
x − x0
lim
exists and is equal to some real number (not innity). The value
(x0 )
f 0 (x ) = limx→x0 f (x)−f
is called at that case the derivative of f
x−x0
at point x .
0
0
5 / 23
Right and left derivatives
Denition. Suppose that function f takes real values and is
dened in some right neighbourhood of point a ∈ R. We say that f
(x0 )
is right-dierentiable at a i the limit limx→x0 + f (x)−f
exists
x−x0
and is equal to some real number (not innity). The value
(x0 )
f+0 (x ) = limx→x0 + f (x)−f
is called at that case the right
x−x0
derivative of f at point x .
Denition. Suppose that function f takes real values and is
dened in some left neighbourhood of point a ∈ R. We say that f
(x0 )
exists and
is left-dierentiable at a i the limit limx→x0 − f (x)−f
x−x0
is equal to some real number (not innity). The value
(x0 )
is called at that case the left
f−0 (x ) = limx→x0 − f (x)−f
x−x0
derivative of f at point x .
Remark. It follows from the second exercise at p.4 that f is
dierentiable at point x i all three conditions hold:
• f is left-dierentiable at point x
• f is right-dierentiable at point x
0
0
0
0
0
0
0
0
0
0
0
• f+0 (x0 ) = f−0 (x0 )
If all three conditions are true then f 0 (x ) = f+0 (x ) = f−0 (x ).
0
0
0
6 / 23
Straight lines going through the point (x0 , y0 )
Remark. Let us mention, that each (non-vertical) straight line on a
plane R , which goes through the point (x , y ) ∈ R , is described
by equation y − y = k · (x − x ), where (x , y ) is an arbitrary
point on the line. See the picture:
2
0
1
0
1
0
2
0
1
1
So, there is a one-to-one mapping Ψ that maps these lines
bijectively to real numbers:
Ψ:
(x1 , y1 ) ∈ R2 |y1 − y0 = k · (x1 − x0 ) 7→ k.
7 / 23
Limit of straight lines going through the point (x0 , y0 )
Remark. The existence of such bijection Ψ means that each such
line is dened by the number k uniquely. This allows to dene limit
procedure for straight lines using limit procedure for numbers.
Denition. Suppose that we have a sequence of straight lines Ln ,
and a straight line L∗ , all going through the point (x , y ):
0
0
Ln = (x1 , y1 ) ∈ R2 |y1 − y0 = kn · (x1 − x0 ) ,
L∗ = (x1 , y1 ) ∈ R2 |y1 − y0 = k∗ · (x1 − x0 ) .
We say that limn→∞ Ln = L∗ i limn→∞ kn = k∗ .
Denition. Similarly, suppose that we have a family of straight
lines Lx , parametrized by number x ∈ R, and a straight line L∗ , all
going through the point (x , y ):
0
0
Lx = (x1 , y1 ) ∈ R2 |y1 − y0 = k(x) · (x1 − x0 ) .
We say that limx→∗ Lx = L∗ i limx→∗ k(x) = k∗ .
8 / 23
Secant lines
Suppose that f : R → R is a continuous function, and we study
straight line that goes through two points on the graph of f , namely
(x , y ) with y = f (x ) and (x, f (x)). The equation of this line is
y − y = k(x) · (x − x ) where k(x) = (f (x) − f (x ))/(x − x )
and (x , y ) is an arbitrary point on the line, see the picture:
0
1
0
0
0
1
0
1
0
0
0
1
Denition. This line is called the secant line for the graph, going
through two points (x , f (x )) and (x, f (x)) that belong to the
graphof f . Denote
0
0
Lx = (x1 , y1 ) ∈ R2 |y1 − f (x0 ) = k(x) · (x1 − x0 ) .
9 / 23
Tangent line as a limit of secant lines
Denition. If there exists limx→x0 Lx = L then the graph of f is
called smooth at point (x , f (x )) and L is called the tangent line
to graph of f at point (x , f (x )).
Remark. The graph of f is called smooth at point (x , f (x )) i f
is dierentiable at x . In other words, the tangent line L exists
exactly when the derivative f 0 (x ) exists. This follows directly from
the deniton of limit procedure for secant lines going through point
(x , f (x )) and from the denition of the derivative.
Example. Consider f (x) = |x|. The right and left derivatives at
x = 0 exist, but their values are dierent so the derivative at
x = 0 does not exists. Indeed
|
= limx→+ xx = 1,
f+0 (0) = limx→ + |x|−|
x−
|
0
f−0 (0) = limx→ − |x|−|
= limx→− −x
x−
x = −1. The value of f (0) is
not dened for function f , and the tangent line at point (0, 0) is
not dened for the graph y = f (x). The reason is that one-sided
limits x → +0 and x → −0 are not equal, and secant lines for
x > 0 and x < 0 have constant but not equal slope.
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
10 / 23
Calculation of derivatives
Exercise 1. Using derivative's denition, nd f 0 (x ). 1a consider
0
f (x) = x 2 , x0 ∈ R. 1b consider f (x) = e x , x0 ∈ R.
Theorem (¾small table¿ of derivatives). This is true for all
x ∈ R:
(sin(x))0 = cos(x),
(cos(x))0 = − sin(x),
(e x )0 = e x .
This is true for all x > 0:
(ln x)0 =
1
x
,
(x a )0 = ax a−1 for all a ∈ R.
Remark. This theorem is proved using properties of functions that
participate in equations above and theorem on remarkable limits.
11 / 23
Calculation of derivatives
Theorem (basic ¾rules of dierentiation¿). Assuming that all
symbols used below are well dened, we have:
1. Dierentiation of a sum: (f (x) + g (x))0 = f 0 (x) + g 0 (x).
2. Product: (f (x) · g (x))0 = f 0 (x) · g (x) + f (x) · g 0 (x)
3. Multiplication by a constant: If C ∈ R is a constant that do not
depend on x , then (C · g (x))0 = C · g 0 (x). This is a particular case
of the previous rule: let f be a constant function f (x) ≡ C , then
f 0 (x) ≡ 0 and we obtain
(C · g (x))0 = 0 · g (x) + C · g 0 (x) = C · g 0 (x).
4. Composite function: (f (g (x)))0 = f 0 (g (x)) · g 0 (x)
5. Inverse function: suppose that y = f (x ), x = f − (y ), then
0
0
(f −1 )0 (y0 ) = (f −1 (y0 ))0 = (f 0 (x0 ))−1 =
1
0
1
f 0 (x )
=
0
0
1
f 0 (f −1 (y ))
.
0
Remark. There is a tricky moment in notation: g − (a) is the value
1
at point a of the function, that is inverse to g ; meanwhile
(g (a))− = g (a) is a number that is inverse to g (a).
1
1
12 / 23
Calculation of derivatives
Theorem (¾dierentiation of a fraction¿). Suppose that f 0 and
g 0 exist, and g 0 (x) 6= 0. Then
f (x)
g (x)
0
=
f 0 (x) · g (x) − f (x) · g 0 (x)
.
(g (x))2
− and apply the rule for
Proof. Let us rewrite gf (x)
(x) = f (x) · (g (x))
dierentiation of a product and of a composite function together
with the (1/x)0 = −1/x which is a particular case of
(x a )0 = ax a− .
1
2
1
(f (x) · (g (x))−1 )0 = f 0 (x) · (g (x))−1 + f (x) · ((g (x))−1 )0 =
= f 0 (x) · (g (x))−1 + f (x) · (−1) · ((g (x))−1 )−2 · g 0 (x) =
=
f 0 (x) f (x) · g 0 (x)
f 0 (x) · g (x) − f (x) · g 0 (x)
−
=
.
g (x)
(g (x))2
(g (x))2
13 / 23
Calculation of derivatives
Remark. Using the above properties and ¾small table¿ of
derivatives we can calculate e.g.
(sin(x 2 ))0 = cos(x 2 ) · 2x,
((sin(x))2 )0 = 2 sin(x) · cos(x).
Also, we can extend our ¾small table¿ of derivatives. For example,
if a > 0, a 6= 1 then
(ax )0 = ((e ln(a) )x )0 = (e x ln(a) )0 = e x ln a ln(a) = ax ln(a).
Similarly,
(tan(x))0 =
sin(x)
cos(x)
0
sin(x)))
1
= cos(x)·cos(x)−(sin(x)·(−
= (cos(x))
2.
(cos(x))2
Exercise. Calculate (cot(x))0 , (arctan(x))0 , (arcsin(x))0 , (loga (x))0
using ¾small table¿ of derivatives and rules of dierentiation.
14 / 23
Dierentials and tangent line
Denition. Suppose for simplicity that function f : R → R is continuous
everywhere and dierentiable at point x0 ∈ R. Let us denote y0 = f (x0 ).
Then by denition:
4x = x − x0 dierence in independent variable
dx = x − x0 dierential of the independent variable
4y = f (x) − f (x0 ) dierence in dependent variable via the graph of f
dy = f 0 (x0 )dx dierential of the dependent variable
dy = y − y0 dierential of the dependent variable, i.e. dierence in
dependent variable via the graph of tangent line to the graph of f at
point (x0 , f (x0 )).
df (x) = dy another notation for dierential of the dependent variable
With this notation we have
f 0 (x0 ) =
dy
4y
f (x) − f (x0 )
df (x)
=
= lim
= lim
.
x→x0
4x→0 4x
dx
dx
x − x0
Moreover the equation of the tangent line at point (x0 , y0 ) can be written
as dy = f 0 (x0 )dx , i.e. y − y0 = f 0 (x0 )(x − x0 ). This means that tangent
line is the set of all points (x, y ) that satisfy y − y0 = f 0 (x0 )(x − x0 ). See
the picture on the next page.
15 / 23
Dierentials and tangent line
16 / 23
Geometric interpretation of the derivative of inverse function
Remark. For the initial function y = f (x) the independent variable is x ,
so f 0 (x0 ) = tan α where α is the angle between the tangent line and
x -axis. Meanwhile for the inverse function x = f −1 (y ) the independent
variable is y , so (f −1 )0 (y0 ) = tan β where β is the angle between the
tangent line and y -axis. It is clear from the plot that α and β are two
acute angles of the same right triangle ABC, so tan α · tan β = 1, i.e.
f 0 (x0 ) · (f −1 )0 (y0 ) = 1.
17 / 23
Monotonicity and derivative
Denition. Suppose we have a function f : (a, b) → R. If for all
x, y ∈ (a, b) condition x < y implies
• f (x) > f (y ) then f is called strictly increasing
• f (x) ≥ f (y ) then f is called non-strictly increasing, also
non-decreasing
• f (x) < f (y ) then f is called strictly decreasing
• f (x) ≤ f (y ) then f is called non-strictly decreasing, also
non-increasing
In all these four cases f is called monotone.
Theorem. Suppose we have a dierentiable function
f : (a, b) → R. If for all x ∈ (a, b) we have
• f 0 (x) > 0 then it is true that f is strictly increasing
• f (x) ≥ 0 then it is true that f is non-strictly increasing
• f (x) < 0 then it is true that f is strictly decreasing
• f (x) ≤ 0 then it is true that f is non-strictly decreasing
Remark. Note that inequality sign in f (x)R 0 is the same as in
f (x)Rf (y ) where R ∈ {<, ≤, >, ≥}.
18 / 23
Types of maximums: local/global, non-strict/strict
Denition. Suppose that A ⊂ R, f : A → R, x0 ∈ A. Number x0 is called a
non-strict global maximum point of f on A, and number f (x0 ) is called the
non-strict global maximum value of f on A, i f (x0 ) ≥ f (x) for all x ∈ A. If,
moreover, f (x0 ) > f (x) for all x ∈ A such that x 6= x0 then we say that x0 is
the strict global maximim point of f on A, and f (x0 ) is the strict global
maximum value of f on A.
Denition. Suppose that A ⊂ R, f : A → R, x0 ∈ A. Number x0 is called a
non-strict local maximum point of f on A, and number f (x0 ) is called a
non-strict local maximum value of f on A, i there exists such number ε > 0
that f (x0 ) ≥ f (x) for all x ∈ A ∩ (x0 − ε, x0 + ε). If, moreover, f (x0 ) > f (x) for
all x ∈ A ∩ (x0 − ε, x0 + ε) such that x 6= x0 then we say that x0 is a strict local
maximim point of f on A, and f (x0 ) is a strict local maximum value of f on A.
Denition (equivalent to the previous). Suppose that A ⊂ R, f : A → R,
x0 ∈ A. Number x0 is called a non-strict local maximum point of f on A, and
number f (x0 ) is called a non-strict local maximum value of f on A, i there
exists such neighbourhood Uε (x0 ) that x0 is a point of non-strict global
maximum for the function f : A ∩ Uε (x0 ) → R on A ∩ Uε (x0 ). If, moreover x0 is
the strict global maximim point of f : A ∩ Uε (x0 ) → R on A ∩ Uε (x0 ), then we
say that x0 is a strict local maximim point of f on A, and f (x0 ) is a strict local
maximum value of f on A.
19 / 23
Extrema
Denition. If in the denition of non-strict/strict local/global
maximum we substitute > by <, and ≥ by ≤, then we will get the
denition of non-strict/strict local/global minimum.
Denition. Extremum is a word that means minimum or maximum.
Remark. This word can be combined with words non-strict/strict
and local/global. For example: points πk, k ∈ Z are points of strict
local extremum of function f (x) = cos(x) on R. These points are
also points of non-strict global extremum for the same function on
the same set. ¾Extrema¿ is multiple for ¾extremum¿. For
example: points πk, k ∈ Z are extrema of function f (x) = cos(x).
Exrecise. Give denition of strict local minimum in terms of
neighbourhoods.
Exrecise. Can a point of strict local minimum be also a point of
non-strict global maximum of a function f : A → R? Consider two
cases: A = (0, 1), A = (0, 1) ∪ N. Provide proofs and examples.
20 / 23
Extrema and derivative
Denition. Suppose that x ∈ A ⊂ R. Point x is called the
interior point of the set A i x belongs to A with some open
neighbourhood, i.e. if there exits such Uε (x ) that x ∈ Uε (x ) ⊂ A.
Example: (a, b) is the set of all interior points of [a, b].
Fermat's lemma (interior extremum theorem). If real-valued
function is dened on the segment of real line and has an extremim
at some interior point of this segment, and is dierentiable at that
point, then the derivative at that point is zero. In other words: if
function f : [a, b] → R has extremum at point x ∈ (a, b), and
f 0 (x ) exists, then f 0 (x ) = 0. Geometrical meaning: if function
f : [a, b] → R has extremum at point x ∈ (a, b), and tangent line
at point (x , f (x )) exists, then this tangent line is horisontal.
0
0
0
0
0
0
0
0
0
0
0
0
21 / 23
Seminar 5
On the seminar 5 we solved:
Exercise 1 at p.11
Problem 1. Find tangent line to the graph of function f at point
(x , f (x )).
1a. Consider f (x) = sin(x), x = π .
1b. Consider f (x) = e x + 1, x = 0.
Problem 2. Find f 0 (x).
2a. f (x) = x sin(ex ) 2b. f (x) = arctan esin(x)
2x +
0
0
0
0
2
1
22 / 23
Homework 5
Problems to solve at home (deadline will be given later):
1a. First exercise at p.4. 1b. Second exercise at p.4. x
x
2a. Find limx→+ 1 + x . 2b. Find limx→− 1 + x .
3. Exercise at p.14.
4. Exercises at p.20.
5. Find tangent line to the graph of function f at point (x , f (x )).
Consider f (x) = e x cos(x), x = 0. Sketch the graph.
0
1
0
1
0
0
0
23 / 23
Introduction to
One-parameter semigroups and evolution
equations
Lesson 6: Necessary background from Real Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
4+6 October 2022
1/1
22.10.04
30ÊсентябряÊ2022Êг.
15:51
22.10.06
6ÊоктябряÊ2022Êг.
12:28
Homework 6
Problems to solve at home (deadline 17 October 16:00):
1. For function f nd derivative, logarithmic derivative and elasticity Exf (x) ,
trying to make it in optimal way (the solution should not be too long). Äëÿ
ôóíêöèè f íàéòè ïðîèçâîäíóþ, ëîãàðèôìè÷åñêóþ ïðîèçâîäíóþ è
ýëàñòè÷íîñòü Exf (x) , æåëàòåëüíî ñäåëàòü ýòî îïòèìàëüíûì îáðàçîì
(ðåøåíèå íå äîëæíî áûòü ñëèøêîì äëèííûì).
√
√
1a. f (x) = (sin x)cos x , 1b. f (x) = x 3/2 3 x + 1 5 x + 2
2. Prove the statement. Äîêàæèòå óòâåðæäåíèå.
( lim an = a, lim bn = b, an ≤ bn ) ⇒ (a ≤ b).
n→∞
n→∞
3. Derive Lagrange's mean value theorem from Rolle's theorem. Âûâåñòè
òåîðåìó Ëàãðàíæà èç òåîðåìû Ðîëëÿ.
4. For function f nd derivative, extrema, extremal values. Find the correct
name for each extremum. Find intervals of monotonicity. Sketch the graph.
Äëÿ ôóíêöèè f íàéòè ïðîèçâîäíóþ, ýêñòðåìóìû, ýêñòðåìàëüíûå çíà÷åíèÿ.
Íàéäèòå ïðàâèëüíîå íàçâàíèå êàæäîãî ýêñòðåìóìà. Íàéäèòå èíòåðâàëû
ìîíîòîííîñòè. Íàðèñóéòå ãðàôèê.
4a. f (x) = x x , D(f ) = (0, +∞). 4b. f (x) = x 3 − 1, D(f ) = R.
5. Find the square of the gure F ⊂ R2 . Íàéòè ïëîùàäü ôèãóðû F ⊂ R2 .
5a. F = {(x, y )|0 ≤ y ≤ sin(2x), 0 ≤ x ≤ π/2}.
5b. F = {(x, y )|y ≥ 0, x ∈ R} ∩ {(x, y )|y ≤ 1 − x 2 , x ∈ R}.
1/1
Introduction to
One-parameter semigroups and evolution
equations
Lesson 7: Necessary background from Real Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
13+20 October 2022
1/1
22.10.13
13ÊоктябряÊ2022Êг.
17:57
Number series
Denition. Suppose that for each n P
∈ N number an ∈ R is given.
Then expressions a1 + a2 + . . . and ∞
n=1 an mean the same and
are called the number series with common term an .
Remark. The expression a1 + a2 + . . . is understood as the result
of adding terms one after another, i.e. at rst we have a1 , then we
add a2 and obtain a1 + a2 , then we add a3 and obtain a1 + a2 + a3
and continue in this manner. It is important to keep in mind that
the order of summation is important. This all brings us to the
following denition.
Denition. SupposePthat for each n ∈ N number an ∈ R is given.
k
If the limit limk→∞
P∞ n=1 an exists as a number, then we say that
number series n=1 an converges (isP
convergent). In that case we
assign a number value to expression ∞
a , denoting this value
P∞n=1 n
P
by the same symbolic combination: n=1 an = limk→∞ kn=1 an .
Denition. SupposePthat for each n ∈ N number an ∈ R is given.
If the limit limk→∞ kn=
P1 an does not exist as a number, then we
say that number series ∞
In that case
n=1 an diverges (is divergent).
P∞
no number value is assigned to the expression n=1 an .
1 / 17
Number series
Remark.
If limk→∞
P
Pk
+∞ then usually we write
n=1 an =P
∞
∞
a
=
+∞
and
say
that
¾
n
n=1
n=1 an diverges to +∞¿. Here
+∞ canP
be substituted by −∞ but not byP∞. And if
k
∞
limk→∞
P∞n=1 an = A ∈ R then we write n=1 an = A and say
that ¾ n=1 an converges to A¿. This all are just traditions of
usage of words ¾converge¿ and ¾diverge¿
in the context
Pkof
P
number series. In all cases we have ∞
a
=
lim
n
k→∞
n=1
n=1 an .
Theorem.
Suppose
that
for
each
n
∈
N
number
a
∈
R
is
given. If
n
P∞
P∞
n=1 |an | converges, then
n=1 an converges.
Denition.
Suppose
that
for
each n ∈ N number
P∞
P∞ an ∈ R is given.
• If n=1 |an | converges, then we say that n=1 an converges
absolutely.
P
P∞
• If ∞
n=1 |an | diverges, then we say that
n=1 an diverges
absolutely.
P
P∞
•PIf ∞
n=1 |an | diverges and
n=1 an converges, then we say that
∞
a
converges
conditionally.
n=1 n
Exercise. Visit webpage
https://en.wikipedia.org/wiki/Series_(mathematics)
2 / 17
First few terms do no change the convergence type but can
change the value of the sum
Theorem.
Suppose that m ∈ N is xed. Then series
P
P∞
∞
n=m an behave similarly:
P∞
P
• P∞
n=1 an converges i
n=m an converges
P
∞
• ∞
a
diverges
i
n
n=1
n=m an diverges
n=1 an and
Proof. This follows from the denition of the notion of
convergence and from the fact that for k > m we have
k
X
an =
n=1
where
m
X
an +
n=1
k
X
an
n=m+1
Pm
n=1 an is a constant that does not depend on k , so
∃ lim
k→∞
k
X
n=1
!
an
⇐⇒
∃ lim
k→∞
m
X
n=1
an +
k
X
!!
an
.
n=m+1
3 / 17
Number series
Remark. For simple cases it is possible to nd out if the series
converges or diverges just directly applying the denition. Even
more, sometimes it is possible to nd the sum following this way.
For example:
∞
X
n = lim
n=1
∞
X
k→∞
k
X
n=1
k(k + 1)
= +∞.
k→∞
2
n = lim
k
1
1
−
=
n n+1
n=1
n=1
1 1 1 1
1
1
1
= lim
− + − + ··· + −
= lim 1 −
=1
k→∞ 1
k→∞
2 2 3
k
k +1
k +1
X
1
= lim
n(n + 1) k→∞
For more complicated cases one needs to apply more profound
methods to test the series for convergence.
Exercise.
that for all C ∈ R, C 6= 0
P∞Using the denition, prove
P∞
series n=1 an converges i series n=1 C · an converges.
4 / 17
Harmonic series
1
Denition. The series ∞
n=1 n is called the harmonic series.
Theorem. Harmonic series P
diverges.
P
Proof. Let us denote Sk = kn=1 n1 and show that limk→∞ Sk does
not exist as a real number. The sequence Sk is strictly increasing
1
because Sk+1 − Sk = k+
1 . Let us show that sequence Sk is not
bounded from above, this will give us limk→∞ Sk = +∞.
For k = 2m let us estimate as follows:
1
1 1 1 1 1 1 1 1
+ + + + + +
+ + · · ·+ m
2 3 4 5 6 7 8 9
2
1 1 1 1 1 1 1 1
1
≥1+ + + + + + + +
+ · · ·+ m
2 4 4 8 8 8 8 16
2
1
1
1
1
1
1
1
1
1
= 1 + 1 + 2 + 2 + 3 + 3 + 3 + 3 +··· + m + ··· + m
2
2
2
2
2
2
2
2
2
| {z } |
{z
}
|
{z
}
S2m = 1 +
21
22
|
=1+
2m−1
{z
}
m
1
1
+ · · · + = 1 + m2 . So S2m ≥ 1 + m2 for all k ∈ N, hence
2
2}
|
{z
m
m
sup Sk ≥ sup S2m ≥ sup 1 +
= +∞. Theorem is proved.
k∈N
m∈N
m∈N
2
5 / 17
Geometric series
Denition. The series
P∞
n=0 x
n = 1 + x + x 2 + . . . is called the
geometric series.
Theorem. For |x| ≥ 1 geometric
diverges. For |x| < 1
Pseries
∞
1
n
.
geometric series converges and n=0 x = 1−x
Proof. For
x
=
1
we
have
P
limk→∞ kn=0 x n = limk→∞ (k + 1) = +∞. For x = −1 we have
Pk
Sk = n=0 x n = 1 − 1 + 1 − 1 + · · · + (−1)k = 12 (1 + (−1)k ), so
S0 = 1, S1 = 0, S2 = 1, S3 = 0, etc, so sequence Sk does not have
nite limit andPdoes not have innite limit as k → +∞.
k+1
If x 6= 1 then kn=0 x n = 1−x
1−x we proved this by induction in k
at lessons 1 and 2.
So for |x| < 1 we have
P
k+1
1−limk→∞ x k+1
−0
limk→∞ kn=0 x n = limk→∞ 1−x
= 11−x
.
1−x =
1−x
Pk
1−limk→∞ x k+1
1−x k+1
n
For x > 1 limk→∞ n=0 x = limk→∞ 1−x =
=
1−x
1−∞
1−x = +∞.
For x < −1 there is no nite or innite limit limk→∞ x k+1 so there
k+1
is no nite or innite limit limk→∞ 1−x
1−x . Theorem is proved.
6 / 17
Convergence tests for number series
Remark. There are many properties that guarantee that number
series converges and many properties that guarantee that number
series diverges. Let us mention only very few of them.
Exercise. Visit webpage
https://en.wikipedia.org/wiki/Convergence_tests
P
Theorem (limit of common term test). If ∞
n=1 an converges,
then limn→∞ an = 0. P
Proof. Suppose that ∞
n=1 an = A ∈ R. Then:
Pk
Pk−1 limk→∞ ak = limk→∞
n=1 an −
n=1 an =
Pk
Pk−1
limk→∞ n=1 an − limk→∞ n=1 an = A − A = 0. P
n
Example of application of this test. Let us study ∞
n=1 (−1)
n
for convergence. It is false that lim
Pn→∞ (−1)n = 0 so thanks to the
limit of common term test series ∞
n=1 (−1) diverges.
Remark
test is not universal). For both series
P∞ 1 (whyPthis
∞ 1
1
1
and
n=1 n we have limn→∞P
n=1 2n
2n = limn→∞ n = 0.
∞
1
However the rst series converges
and n=1 2n = 1, but the
P
1
second series diverges and ∞
n=1 n = +∞.
7 / 17
Convergence tests for number series
Theorem (direct comparison test). Assume that there exists n0 ∈ N
such that |an | ≤ |bn | for allPn ≥ n0 . If the series ∞
n=1 bn converges
∞
a
converges
absolutely.
If the series
absolutely,
then
the
series
n=1 n
P∞
P∞
b
diverges
absolutely.
a
diverges
absolutely,
then
the
series
n=1 n
n=1 n
Example of application of this test. Suppose that number x ∈ R is
P
sin(nx)
given. Let us study ∞
for convergence. For all n ∈ N and all
n=1
2n
P∞
sin(nx)
1
x ∈ R we have
≤ 2n , and series n=1 21n converges as we
2n
P
proved before. So thanks to the direct comparison test series
converges absolutely for all x ∈ R.
P
Another example of application of this test.
because √1n > n1 and
P∞
1
n=1 n = +∞.
P∞
n=1
sin(nx)
2n
∞
√1
n=1 n = +∞
Theorem (alternating series test, also known as the Leibniz
criterion). Suppose that limn→∞ an = 0, and for all n ∈ N we have
0 < an+1 ≤ an . Then
n
n→∞ (−1) an converges.
P
P∞ (−1)n
Example of application of this test. Let us study
for convergence.
n=1
n
1
1
We have limn→∞ n = 0 and for all n ∈ N we have 0 < n+1 ≤ 1n . So thanks to
P
P
(−1)n
(−1)n
the alternating series test series ∞
converges. Moreover: ∞
n=1
n=1
n
n
n
P
1
converges conditionally because (−n1) = 1n and ∞
n=1 n diverges.
8 / 17
Convergence tests for number series
Theorem (condensation convergence test). Suppose that
0P≤ f (n + 1) ≤ f (n) for P
all n ∈ N. Then
∞
∞
n
n
f
(n)
converges
i
n=1
P∞
P∞n=1 2k f (2k ) converges.
P
Moreover n=1 f (n) ≤ k=0 2 f (2 ) ≤ 2 ∞
n=1 f (n).
Remark. This is a very powerful test. It is proved by an argument
very similar to one that we used to prove that harmonic series
diverges.
P∞ 1
Example of application
of
this
test.
Let
us
study
n=1 n for
P∞ n 1
convergence. Series n=1 2 2n diverges because
common term
P the
n 1 diverges due
is constant 1 and do not converge to 0, so ∞
2
n=1
2n
to limitPof common term test, so thanks to the condensation test
1
series ∞
n=1 n diverges.
Another
this test. Let
P∞ 1 example of application
Pof
P∞us study
∞
1
1
n
n
n=1 n2 for convergence. Series
n=1 2 (2n )2 =
P∞ n=11 2
converges because it is the geometric series, so n=1 n2 converges
thanks to the condensation
test.
P∞
1
Exercise. Study n=2 n ln(n) for convergence.
9 / 17
Convergence tests for number series
Theorem (ratio limit test, also known as d'Alembert's
criterion).
1
= r . If
Suppose that there exists r ∈ R such that limn→∞ an+
an
P∞
r < 1, then theP
series n=1 an is absolutely convergent. If r > 1,
then the series ∞
If r = 1, the ratio test is
n=1 an diverges.
P
inconclusive, and the series ∞
a
n=1 n may converge or diverge.
Example of application
of
this
P∞ x n test. Suppose that number x ∈ R
is given. Let us study n=1 n! for convergence. We have
an+1
|x|n+1 |x|n
|x|
= lim
:
= lim
=0<1
n→∞
n→∞ (n + 1)!
n→∞ n + 1
an
n!
P
xn
so thanks to the ratio limit test series ∞
n=1 n! converges
absolutely for all x ∈ R.
Remark
test is not universal). For both series
P∞ 1 (whyPthis
∞ 1
1
1
and
n=1 n2
n=1 n we have limn→∞ (n+1)2 : n2 = 1 and
lim
1
limn→∞ n+
: 1 = 1. However the rst series converges and
P∞ 1 1 π 2 n
n=1 n2 = 6 , but the second series diverges.
10 / 17
Properties of converging number series
∞
Theorem. If P
n=1 an converges absolutely, and f : N → N is a
∞
P
bijection,
thenP n=1 af (n) converges absolutely and
P∞
∞
n=1 af (n) =
n=1 an .
Remark. The above theorem says that for absolutely convergent
series the order of summation is not important: the sum anyway
will be the same.
P
Riemann rearrangement theorem. If ∞
n=1 an converges
conditionally, and A ∈ R ∪ {−∞,
P +∞}, then there exists a
bijection f : N → N such that ∞
n=1 af (n) = A. It is also possible to
P
nd such f that sequence Sk = kn=1 af (n) has no nite limit and
no innite limit.
Remark. The above theorem explains the meaning of the term
¾converges conditionally¿: the sum is equal to something under
condition that the summation is performed in a given order.
11 / 17
Taylor series
Remark. Recall that if function f : R → R for each n ∈ N has nth
derivative f (n) : R → R then for each x, y ∈ R and each k ∈ N one
can represent f (x) via Taylor's formula:
f (x) =
k
X
1
n=0
n!
f (n) (y ) · (x − y )n + Rk+1 (x, y ).
1 (n)
Expression n=0 n!
f (y ) · (x − y )n can be considered as a partial
P∞ 1 (n)
sum of the series n=0 n! f (y ) · (x − y )n , and one can ask if this
series converges or not. By denition of convergence of number
series, and taking into account Taylor's formula, we obtain
Pk
∞
X
1
n=0
n!
f
(n)
n
(y ) · (x − y ) = lim
k→∞
k
X
1
n=0
n!
f (n) (y ) · (x − y )n =
= lim (f (x) − Rk+1 (x, y )) = f (x) − lim Rk+1 (x, y ).
k→∞
k→∞
P∞ 1 (n)
If limk→∞ Rk+1 (x, y ) = 0 then n=0 n! f (y ) · (x − y )n = f (x).
12 / 17
Taylor series
Denition. Suppose that function f : (a, b) → R for each n ∈ N
has nth derivative f (n) : (a, b) → R, and x, y ∈ (a, b). Then
expression
∞
X
1 (n)
f (y ) · (x − y )n
n!
n=0
is called the Taylor series for function f with center at point y . If
for all x, y ∈ (a, b) we have
f (x) =
∞
X
1
n=0
n!
f (n) (y ) · (x − y )n
then function f is called analytic on (a, b), and notation
f ∈ A(a, b) is used.
13 / 17
Some notation
Denition. If function f : (a, b) → R
• is continuous at each point of (a, b) then f is called continuous
on (a, b) and notation f ∈ C (a, b) is used.
• is continuous at each point of (a, b), dierentiable at each point
of (a, b), and the derivative f 0 is continuous on (a, b) then f is
called continuously dierentiable on (a, b) and notation
f ∈ C 1 (a, b) is used.
• for each k = 0, 1, 2, . . . , n posesses property f (k) ∈ C (a, b) then
f is called n times continuously dierentiable on (a, b) and notation
f ∈ C n (a, b) is used.
• for each n ∈ N has nth derivative f (n) : (a, b) → R, then f is
called innitely smooth on (a, b), or innitely dierentiable on
(a, b), and notation f ∈ C ∞ (a, b) is used.
Remark. By denition of intersection of sets we have
n
C ∞ (a, b) = ∩∞
n=1 C (a, b).
Remark. With this notation we have:
A(a, b) ⊂ C ∞ (a, b) ⊂ . . . C n (a, b) ⊂ . . . C 2 (a, b) ⊂ C 1 (a, b) ⊂ C (a, b).
14 / 17
Smooth functions are not always analytic
Remark. A(a, b) ( C ∞ (a, b), i.e. if function f has derivatives of all
1 (n)
(y ) · (x − y )n , this
orders and one can write Taylor's series ∞
n=0 n! f
does not mean that the series converges to f (x). Example:
P
(
f : R → R,
f (x) =
0
e
− x12
x = 0,
if x 6= 0.
if
Then f (0) = 0 by denition of f . Next:
1
f (t) − f (0)
y
e − t2 − 0
f (0) = lim
= lim
= [1/t = y ] = lim y 2 = 0,
y →∞ e
t→0
t→0
t
t
0
−x −2
−2
For x 6= 0 we have f 0 (x) = e −x (−(−2))x −3 = 2 e x 3 . So
−t −2
−2
2 e t3 − 0
f 0 (t) − f 0 (0)
e −t
f (0) = lim
= lim
= 2 lim
=
t→0
t→0
t→0
t
t
t4
y4
= [1/t = y ] = 2 lim y 2 = 0,
y →∞ e
(n)
in the same way we prove that f (0) = 0 for each n ∈ N.
Then
all x 6= 0 and y = 0PTaylor's series converges,
but not to f (x):
P∞ for
P
∞ 1
1 (n)
(0) · (x − 0)n = n=0 n!
0 · xn = ∞
n=0 n! f
n=0 0 = 0 6= f (x).
00
15 / 17
Properties of analytic functions
Theorem. Sum, product, ratio (when dened) and composition of
analytic functions is an analytic function.
Theorem. If function is analytic, then Taylor's series can be
dierentiated and integrated termwise. This means that if
f ∈ A(a, b), and x, y , α, β ∈ (a, b) then
f (x) =
∞
X
1
n=0
0
f (x) =
∞
X
1
n=0
Z β
f (x)dx =
α
n!
n!
f (n) (y ) ·
∞
X
1
n=0
f (n) (y ) · (x − y )n
n!
f
(n)
d
(x − y )n
dx
Z β
(y ) ·
(x − y )n dx
α
16 / 17
Homework 7
Problems to solve at home (deadline 21 October 16:00):
1a.
R Check that
= ln |x| + C considering x > 0 and x < 0. Ïðîâåðüòå, ÷òî
x
= ln |x| + C ðàññìîòðåâ x > 0 è x < 0.
R
1b. Find indenite
Íàéòè íåîïðåäåë¼ííûé èíòåãðàë. x ln(x)dx
R integral.
1c. Find Íàéòè x 2 cos(3x + 1)dx
2. For function f nd intervals where f is convex, concave. Äëÿ ôóíêöèè f
íàéòè èíòåðâàëû, íà êîòîðûõ f âûïóêëà, âîãíóòà. f (x) = x 3 − 1, D(f ) = R.
3. Write Taylor's formula for f , with center at point y = 0 and Taylor
polynomial of order n = 5. Íàïèøèòå ôîðìóëó Òåéëîðà äëÿ f , ñ öåíòðîì â
òî÷êå y = 0 è ìíîãî÷ëåíîì Òåéëîðà ïîðÿäêà n = 5.
3a. f (x) = cos(x) 3b. f (x) = e x .
R dx
dx
x
4. Study number series for convergence, absolute convergence, conditional
convergence. Èññëåäóéòå ðÿä
ñõîäèìîñòü,
P∞àáñîëþòíóþ
P íà ñõîäèìîñòü,
(−1)n
sin(n)
óñëîâíóþ ñõîäèìîñòü. 4a. ∞
n=2 (ln(ln(n)))2
n=1 n3 . 4b.
5a. Prove that (äîêàæèòå, ÷òî) limn→∞ x n /n! = 0 for all x ∈ R.
5b.Using 5a, the denition of analytic function and Taylor's formula prove that
function f (x) = e 2x is analytic on R. Èñïîëüçóÿ 5a, îïðåäåëåíèå
àíàëèòè÷åñêîé ôóíêöèè è ôîðìóëó Òåéëîðà, äîêàæèòå, ÷òî ôóíêöèÿ
f (x) = e 2x ÿâëÿåòñÿ àíàëèòè÷åñêîé íà R.
6.
Elective competitive task (5 points
P∞
P∞for the3 rst correct solution). If series
a
converges,
is
it
true
that
n
n=1
n=1 (an ) converges?
P
1
7. Elective competitive task (10 points). Is series ∞
n=1 n3 sin(n) convergent?
17 / 17
Introduction to
One-parameter semigroups and evolution
equations
Lecture 8: Necessary background from Real Analysis
Seminar 8
Ivan Remizov
HSE Nizhny Novgorod, Russia
2 November 2022
1 / 18
Indexing
Denition. Let A and B be some non-empty sets. We say that A is
indexed by B , and call B the set of indexes, i there exists a
bijection f : B → A. In that case it is true that A = {f (b)|b ∈ B}.
In many cases the exact formula for f is not important, so even the
symbol of f is omitted: often in the context of indexing we write
a = ab instead of a = f (b). Keeping this in mind we also often
write A = {ab |b ∈ B} and A = {ab }b∈B assuming that B is a set
of indexes arbitrary elements that are used only for indexing.
Remark. Notation (ab )b∈B is also widely used, and many authors
(but not all) assume that (ab )b∈B = {ab }b∈B .
Remark. As for me, especially when natural linear order on B is
given, I understand (ab )b∈B as a generalization of the notion of the
tuple, i.e. indexing function itself, so for me (ab )b∈B is a set
{(b, ab )|b ∈ B} = (ab )b∈B = [B 3 b 7→ ab ∈ A].
So if g : B → B is a bijection which is not identical (at least for
one b ∈ B we have b 6= g (b)), then for me (ab )b∈B 6= (ag (b) )b∈B
and {ab }b∈B = {ag (b) }b∈B .
Summing up: I use {} when order is not important and () when
order is important.
2 / 18
Indexing
Example. For me (1, 2, 3) = (k)k∈{1,2,3} is a tuple, so
(1, 2, 3) = (k)k∈{1,2,3} 6= (4 − k)k∈{1,2,3} = (3, 2, 1). Meanwhile
{1, 2, 3} is a set, so
{1, 2, 3} = {k}k∈{1,2,3} = {4 − k}k∈{1,2,3} = {3, 2, 1}.
Example. Sequence (a function dened on N) is not the same as
values of this sequence. E.g. sequences 1,-1,1,-1,1,-1,. . . and
-1,1,-1,1,-1,1,. . . take the same values 1 and -1 but they are
dierent sequences. To stress this idea I use only () for sequences,
because sequence is an innite tuple, where order is important.
Let us start natural numbers from 1 for this example. For me
sequence (1, −1, 1, −1, 1, −1, . . . ) = ((−1)k−1 )k∈N is a tuple, so
(1, −1, 1, −1, 1, −1, . . . ) = ((−1)k−1 )k∈N 6= ((−1)k )k∈N =
(−1, 1, −1, 1, −1, 1, . . . ).
Meanwhile {(−1)k−1 }k∈N is a set, so
{1, −1} = {(−1)k−1 }k∈N = {(−1)k }k∈N = {−1, 1}.
Remark. Very often the following notation is used:
∞
(ak )k∈N = (ak )∞
k=1 , {ak }k∈N = {ak }k=1 .
Remark. Remember that many authors do not distinguish (ab )b∈B
and {ab }b∈B , so you should be careful and double check that you
understand the author well.
3 / 18
Innite unions and intersections of sets
Denition. If A is a set of sets, i.e. all elements of A are sets, then:
• ∩A is the intersection of A, i.e. x ∈ ∩A ⇐⇒ (∀A ∈ A|x ∈ A).
• ∪A is the union of A, i.e. x ∈ ∪A ⇐⇒ (∃A ∈ A|x ∈ A).
Denition (the same but with indexing). If A is a set of sets
indexed by a set of indexes B , A = {Aβ }β∈B where all Aβ are sets,
then:
T
• ∩A = β∈B Aβ is the intersection of A, i.e.
T
T
x ∈ β∈B Aβ ⇐⇒ (∀β ∈ B|x ∈ Aβ ). So x ∈ β∈B Aβ exactly
when x ∈SAβ for all β ∈ B .
• ∪A = β∈B Aβ is the union of A, i.e.
S
S
x ∈ β∈B Aβ ⇐⇒ (∃β ∈ B|x ∈ Aβ ). So x ∈ β∈B Aβ exactly
when x ∈ Aβ for at least one β ∈ B .
Remark. Very often the following notation is used:
∞
\
\
∞
∩n∈N An =
An = ∩n=1 An =
An ,
n=1
n∈N
∪n∈N An =
[
n∈N
An = ∪∞
n=1 An =
∞
[
n=1
An .
4 / 18
Sets of measure zero and a.e. true properties
Denition. A set A ⊂ R is said to be ¾Lebesgue measure zero
set¿ or just ¾measure zero set¿ i for each ε > 0 there exists
countable (innite or nite) family of closed intervals covering A
or N such that
and having total length less than ε, i.e. ∃([an , bn ])∞
n=1
∞S
or N
∞P
or N
A⊂
[an , bn ] and
(bn − an ) < ε. In that case the
n=1
n=1
notation µA = 0 is used. Of course µ∅ = 0.
Denition. A set A, where A ⊂ B ⊂ R, is said to be ¾full measure
set in B ¿ i µ(B \ A) = 0.
Denition. We say that some property P(x) holds ¾almost
everywhere (a.e.) in the set B ¿ i P(x) is true for all x from some
full measure subset of B , in other words: P(x) is false on the set of
measure zero, i.e. µ{x ∈ B|P(x) is false} = 0.
Remark. When it is clear from the context which set B we mean,
then we can use only letters a.e. and combine them with signs
a.e.
=, 6=, ≤, <, ≥, >. Example: f (x) = g (x) reads as ¾f (x) is almost
everywhere equal to g (x)¿ and means µ{x|f (x) 6= g (x)} = 0.
5 / 18
Sets of measure zero and a.e. true properties
Remark. Russian translation:
1. Full measure set = ìíîæåñòâî ïîëíîé ìåðû
2. Measure zero set = ìíîæåñòâî ìåðû íóëü
3. Almost everywhere, a.e. = ïî÷òè âñþäó, ï.â.
Theorem. If An are sets of measure zero, and Bn are full measure
sets, then:
∞
1. ∪∞
n=1 An and ∩n=1 An are measure zero sets.
∞
∞
2. ∪n=1 Bn and ∩n=1 Bn are full measure sets.
3. Bn ∪ An and Bn \ An are full measure sets.
Theorem. If A is a measure zero set and A1 ⊂ A then A1 is also a
measure zero set.
Theorem. If B is a full measure set and B1 ⊃ B then B1 is also a
full measure set.
Example. Q is a measure zero set, and R \ Q is a full measure
subset in R.
Example. Every countable set is a measure zero set.
Example. Classical Cantor set is not countable but measure zero.
We will prove properties given in these examples on the seminar.
6 / 18
Continuity of a real-valued function of real argument
Consider a function f : A → R, where A ⊂ R and A is an open,
closed or semiclosed interval, bounded or unbounded. Then we
dene the following continuity properties, starting from the weakest
and nishing with the strongest.
Denition. Function f : A → R is called:
1. Continuous in one point x0 ∈ A i f (x0 ) = limx→x0 f (x);
2. Continuous a.e. on A i µ{x0 |f (x0 ) 6= limx→x0 f (x)} = 0;
3. Continuous everywhere on A i ∀x0 ∈ A f is continuous in x0 ;
4. Uniformly continuous on A i
∀ε > 0∃δ > 0∀x 0 , x 00 ∈ A : (|x 0 − x 00 | < δ) ⇒ (|f (x 0 ) − f (x 00 )| < ε);
5. Absolutely continuous on A i
x100 < x20 < · · · < xn0 < xn00 ∈ A we have
∀ε
0∃δ > 0∀n ∈ N∀x10 <P
P>
n
0
00
( k=1 |xk − xk | < δ) ⇒ ( nk=1 |f (xk0 ) − f (xk00 )| < ε).
Remark. It follows directly from these denitions that
5 ⇒ 4 ⇒ 3 ⇒ 2.
Question. What is the relation between 1 and 2?
7 / 18
Continuity in terms of the graph of function
Denition. We say that the graph of function f : (a, b) → R enters
the 2ε − 2δ box at point x0 ∈ (a, b) properly i |x − x0 | < δ ⇒
|f (x) − f (x0 )| < ε.
Remark. On the next slide we will dene continuity in terms of
¾entering the box properly¿ property.
Question. How one can dene Lipschitz and Holder continuity in
these terms? (I will tell you on the next lesson, but try to develop
the idea yourself.)
8 / 18
Continuity in terms of the graph of function
Denition in terms of entering the box properly. Function f : A → R
is called:
1. Continuous in one point x0 ∈ A i for each ε > 0 there exists such
δ > 0 that the graph of f enters the 2ε − 2δ box at point x0 ∈ A properly.
Remember short version: Function f is continuous in one point x0 ∈ A
means that the graph enters the box properly at one point x0 ∈ A.
2. Remember short version: Continuous a.e. on A i the graph enters the
box properly at almost all points x0 ∈ A.
3. Remember short version: Continuous everywhere on A i the graph
enters the box properly at all points x0 ∈ A.
4. Uniformly continuous on A i for all ε > 0 there exists δ > 0 NOT
depending on x0 such that the graph of f enters the 2ε − 2δ box at each
point x0 ∈ A properly. This means that the box can ¾slide¿ through the
graph properly.
5. Absolutely continuous on A i for all ε > 0 there exists δ > 0 NOT
depending on N such that for
all N ∈ N and all
families of 2εk − 2δk
P
P
boxes satisfying conditions Nk=1 εk < ε and Nk=1 δk < δ we have that
each box slides throug the graph properly. Remember short version: nite
family of boxes slide through the graph properly.
9 / 18
Types of discontinuity points
Denition. Point x0 ∈ R for a function f is called a point of
discontinuity i f is not continuous at x0 . Points of discontinuity
are classied as follows:
a) point of removable discontinuity (òî÷êà óñòðàíèìîãî ðàçðûâà)
i ∃ nite limx→x0 f (x) but f (x0 ) 6= limx→x0 f (x) or f (x0 ) is not
dened
b) point of jump discontinuity (òî÷êà ñêà÷êÀ) i ∃ nite
lim f (x), ∃ nite lim f (x) but they are not equal
x→x0 +0
x→x0 −0
c) point of essential discontinuity (òî÷êà ñóùåñòâåííîãî ðàçðûâà)
i at least one of the limits lim f (x), lim f (x) does not exist
or is innite
x→x0 +0
x→x0 −0
Russian terminology. Òî÷êè ðàçðûâà òèïà a) è b) â
ðóññêîÿçû÷íîé ëèòåðàòóðå íàçûâàþò òî÷êàìè ðàçðûâà ïåðâîãî
ðîäà, à òî÷êè òèïà ñ) òî÷êàìè ðàçðûâà âòîðîãî ðîäà
10 / 18
Riemann integral
Proper (ñîáñòâåííûé) Riemann integral (R) (a,b) f (x)dx , or,
R
Rb
which is the same, (R) a f (x)dx
Interval (a, b) and function f are both bounded:
|a| < ∞, |b| < ∞, supx∈(a,b) |f (x)| < ∞. Example:
R π/2
(R) 0 sin(x)dx = 1.
Theorem (Lebesgue's criterion for Riemann integrability).
Function f : [a, b] → R is Riemann integrable i both conditions
are fullled:
1. f is bounded
2. f is a.e. continuous
Remark. This theorem is about Riemann integral, not about
Lebesgue integral. This theorem was proved by Lebesgue, this is
why it is named after him.
Theorem. (Main theorem of calculus for Riemann integral). If
function f : [a, b] → R is bounded and continuous, and we have
continuous function F : [a, b] → R satisfying F 0 (x) = f (x) for all
Rb
x ∈ (a, b) then (R) a f (x)dx = F (b) − F (a).
11 / 18
Riemann integral
Improper Riemann integral of the 1st kind: function is
bounded on each segment, but interval is unbounded in one of the
R +∞
Rb
endpoints. Denition: (R) a f (x)dx = lim (R) a f (x)dx . If
b→+∞
the limit exists then we say that integral converges (synonym:
exists). If the limit does not exist then we say that integral diverges.
Rb
If ∃ lim (R) a |f (x)|dx < +∞ then we say that integral
b→+∞
Rb
converges absolutely, if lim (R) a |f (x)|dx = +∞ then we say
b→+∞
that integral diverges absolutely. (The word ¾absolutely¿ appears
because |f (x)| is called the absolute value of f (x).)
Theorem. If an integral converges absolutely then it converges.
Denition. If an integral converges, but diverges absolutely then
we say that this integral converges conditionally.
Denition.
in the left-hand
R +∞ Improper integral
R0
R +∞side
(R) −∞ f (x)dx = (R) −∞ f (x)dx + (R) 0 f (x)dx by
denition exists i both integrals in the right-hand side exist.
12 / 18
Riemann integral
Improper Riemann integral of the 2nd kind: interval is
bounded, but function is unbounded in any neighbourhood of one
of the endpoints (let it be point b below) and bounded outside any
neighbourhood of this point. Denition:
Rb
R b−ε
(R) a f (x)dx = lim (R) a f (x)dx . If the limit exists then we
ε→+0
say that integral converges (exists), if the limit does not exist then
R b−ε
we say that integral diverges. If ∃ lim (R) a |f (x)|dx then we
ε→+0
say that integral converges absolutely, if
R b−ε
∃ lim (R) a |f (x)|dx = +∞ then we say that integral diverges
ε→+0
absolutely.
Theorem. If an integral converges absolutely then it converges.
Denition. If an integral converges, but diverges absolutely then
we say that this integral converges conditionally.
Denition. Improper integral for function f that is not bounded in
neighbourhood of x0 ∈ [a, b] is dened as follows:
Rb
Rx
Rb
(R) a f (x)dx = (R) a 0 f (x)dx + (R) x0 f (x)dx by denition
exists i both integrals in the right-hand side exist.
13 / 18
Riemann integral
Improper Riemann integral of mixed type (i.e. integral having
several points of unboundness of function or of the interval) must
be separated into several integrals of 1st and 2nd kinds. Integral of
mixed type exists i all the integrals in this decomposition exist.
R
Example. (R) 0+∞ √dxx is an improper integral of mixed type. By
denition we must represent it onto two simple improper integrals:
Z +∞
Z 1
Z +∞
dx
dx
dx
√ = (R)
√ + (R)
√ .
(R)
x
x
x
0
0
1
In the right-hand
R +∞ dx side rst integral exists, but the second does not,
√ does not exist.
so (R) 0
x
Denition. For absolutely divergent improper Riemann integrals of mixed type
we can dene ¾principal value in sense of Cauchy¿ denoted as v.p. as follows:
1. For bounded
R +∞ function f dene
RM
v .p.(R) −∞ f (x)dx = limM→+∞ (R) −M f (x)dx .
2. For function f unbounded only
inR any neighbourhoodR of x0 ∈ (a, b)
dene
Rb
x −ε
b
v .p.(R) a f (x)dx = limε→+0 (R) a 0 f (x)dx + (R) x0 +ε f (x)dx .
14 / 18
Measurable functions
Denition. Function f : [a, b] → R is called Lebesgue measurable
(or just measurable) i any of these properties hold:
1. f is continuous;
a.e.
2. f (x) = lim fn (x) where functions fn are continuous.
n→∞
Theorem. If f (x) a.e.
= lim fn (x) where functions fn are measurable
n→∞
then f is measurable.
Corollary. If g is measurable and f (x) a.e.
= g (x) then f is
measurable.
Question. If function f is a.e. dened and a.e. continuous, then is
it true that it is measurable? Answer: yes. Why?
Theorem. If f and g are measurable then f + g , f · g are
a.e.
measurable. If additionally g (x) 6= 0 then x 7−→ f (x)/g (x) is also
measurable.
Theorem. If f is measurable then x 7−→ |f (x)| is also measurable.
The reverse is not true, please nd an example.
15 / 18
Seminar 8
Problems to be solved on the Seminar 8:
1. Prove that [a) empty set b) one-point; c) nite; d) innite
countable] subset of R is a measure zero set.
2. Prove that classical Cantor set is a measure zero set.
3. Cantor function (¾Cantor ladder¿, ¾Cantor stairs¿
https://en.wikipedia.org/wiki/Cantor_function) to be introduced.
16 / 18
Homework 8
Problems to solve at home, deadline Wednesday November
9, 16:00
1. Prove that union of [a) two; b) n; c) countable many] measure
zero sets is a measure zero set.
2. Prove that Cantor function C : [0, 1] → [0, 1] is continuous,
uniformly continuous, but not absolutely continuous.
3. Find function f : R → R which: a) is nowhere continuous but
x 7−→ |f (x)| is continuous; b) is continuous only in one point
x0 ∈ R; c) continuous, but not uniformly continuous.
4. Study function f : [0, +∞) → R, for continuity,
√ uniform
continuity, absolute continuity where a) f (x) = x ; b) f (x) = 2x ;
c) f (x) = x 2 d) f (x) = x sin(1/x) for x 6= 0 and f (0) = 0.
5. Classify (proper, improper of 1st kind, improper of 2nd kind,
improper of mixed type), study for convergence, absolute
convergence, conditional convergence Riemann integrals, in case of
convergence in any sense calculate the value of this integral a)
R1 α
R +∞
R +∞ sin(x)
x dx for all α ∈ R; b) 1 x α dx for all α ∈ R; c) 0
x dx ;
0 R
+∞
2
d) 0 sin(x )dx .
17 / 18
Literature for homework 8
1. Vladimir I. Bogachev, Oleg G. Smolyanov. Real and Functional
Analysis. Springer, Scholtech, HSE (2020)
2. À.Í.Êîëìîãîðîâ, Ñ.Â.Ôîìèí. Ýëåìåíòû òåîðèè ôóíêöèé è
ôóíêöèîíàëüíîãî àíàëèçà. Ëþáîå èçäàíèå.
3. Bernard R. Gelbaum, John M. H. Olmsted. Counterexamples in
Analysis. Any edition
4. Íèêèòèí À.À., Ôîìè÷åâ Â.Â. Ìàòåìàòè÷åñêèé àíàëèç.
Óãëóáëåííûé êóðñ : ó÷åáíèê è ïðàêòèêóì äëÿ àêàäåìè÷åñêîãî
áàêàëàâðèàòà. 2-å èçä., èñïð. è äîï. Ìîñêâà : Èçäàòåëüñòâî
Þðàéò, 2018. 460 ñ.
5. Íàòàíñîí È. Ï. Òåîðèÿ ôóíêöèé âåùåñòâåííîé ïåðåìåííîé.
Ëþáîå èçäàíèå
6. Any other books on Real Analysis, on Calculus, on Mathematical
Analysis.
7. Any other study materials, Internet, mathoverow,
mathexchange, ...
18 / 18
Introduction to
One-parameter semigroups and evolution
equations
Lesson 9: Necessary background from Real Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
3 November 2022
1 / 16
Some useful theorems on continuous functions
Theorem. If function f : [a, b] → R is continuous, then:
1. ∃xmin ∈ [a, b] such that f (xmin ) = inf x∈[a,b] f (x) and
∃xmax ∈ [a, b] such that f (xmax ) = supx∈[a,b] f (x); this is the
Extreme value theorem.
2. f takes all values between inf x∈[a,b] f (x) and supx∈[a,b] f (x); this
is the Intermediate value theorem.
3. f is uniformly continuous; this is the Heine-Cantor theorem.
Remark. Items 1-2 can be replaced with the following fact:
¾Continuous image of a segment is a segment¿ where segment
means closed bounded interval [α, β].
Exercise. Substitute [a, b] above with [a, b) and construct
examples to show that statements 1 and 3 become wrong.
Theorem. If function f : R → R is monotone, then f is a.e.
dierentiable, may have only countable many discontinuities and
only of a jump type.
Theorem. If function f : R → R is absolutely continuous then it is
a.e. dierentiable, i.e. f 0 (x) exists for almost all x .
Remark. The reverse is not true, example: Cantor function.
2 / 16
Continuity revisited
Denition. Let a ∈ R or a = −∞, let b ∈ R or b = +∞. Then
f : (a, b) → R is called:
1. Lipschitz-continuous i ∃C > 0 such that
|f (x1 ) − f (x2 )| ≤ C |x1 − x2 | for all x1 , x2 ∈ (a, b); here C is called
the Lipschitz constant of f .
2. H
older-continuous i ∃C > 0, ∃α ∈ (0, 1] such that
|f (x1 ) − f (x2 )| ≤ C |x1 − x2 |α for all x1 , x2 ∈ (a, b); here α is called
the Holder exponent of f .
Exercise. Prove that if f is Holder-continuous with α > 1 then
f (x) ≡ const.
Exercise. Find examples showing that inclusions above are not
equalities. Hint: use Wikipedia, but understand the proofs.
3 / 16
Geometrical meaning of Lipschitz- and Holder-continuity
Function f : (a, b) → R is Lipschitz-continuous i there exists a
Lipschits box, that can slide through the graph of f properly:
∃C > 0 such that |f (x0 ) − f (x)| ≤ C |x0 − x| for all x0 , x ∈ (a, b)
Function f : (a, b) → R is H
older-continuous i there exists a
Holder box, that can slide through the graph of f properly:
∃C > 0, ∃α ∈ (0, 1] such that |f (x0 ) − f (x)| ≤ C |x0 − x|α for all
x0 , x ∈ (a, b)
Question. Is Cantor function Lipschitz-continuous,
Holder-continuous?
4 / 16
Convergence of function sequences
Denition. We say that sequence of functions fn : (a, b) → R
converges to f
a.e.
1. a.e. on (a, b) i limn→∞ fn (x) = f (x);
[for almost all x ∈ (a, b) ∀ε > 0
∃N(ε, x) ∈ N : n > N ⇒ |fn (x) − f (x)| < ε]
2. pointwise on (a, b) i limn→∞ fn (x) = f (x);
[∀x ∈ (a, b)∀ε > 0∃N(ε, x) ∈ N : n > N ⇒ |fn (x) − f (x)| < ε]
3. uniformly on (a, b) i limn→∞ supx∈(a,b) |fn (x) − f (x)| = 0.
[∀ε > 0∃N(ε) ∈ N∀x ∈ (a, b) : n > N ⇒ |fn (x) − f (x)| < ε.]
Exercise. Prove that if a sequence of uniformly continuous
functions fn : (a, b) → R converges to f uniformly then f is
uniformly continuous. Use the ¾ε/3¿-method, i.e. represent the
expression, which must be less than ε, as the sum of three terms,
each of which is less than ε/3.
Corollary. If a sequence of continuous functions fn : [a, b] → R
converges to f uniformly then f is continuous.
5 / 16
Measurable functions (recall from the previous lesson)
Denition. Function f : [a, b] → R is called Lebesgue measurable
(or just measurable) i any of these properties hold:
1. f is continuous;
a.e.
2. f (x) = lim fn (x) where functions fn are continuous.
n→∞
Theorem. If f (x) a.e.
= lim fn (x) where functions fn are measurable
n→∞
then f is measurable.
Corollary. If g is measurable and f (x) a.e.
= g (x) then f is
measurable.
Question. If function f is a.e. dened and a.e. continuous, then is
it true that it is measurable? Answer: yes. Why?
Theorem. If f and g are measurable then f + g , f · g are
a.e.
measurable. If additionally g (x) 6= 0 then x 7−→ f (x)/g (x) is also
measurable.
Theorem. If f is measurable then x 7−→ |f (x)| is also measurable.
The reverse is not true, please nd an example.
6 / 16
Lebesgue integral for nonnegative functions
a.e.
Denition. Real-valued function f dened a.e. on [a, b], f (x) ≥ 0,
is called Lebesgue integrable i all three conditions hold:
1. f is measurable;
a.e.
2. f (x) = lim fn (x) where functions fn are bounded,
n→∞
non-negative, a.e. continuous and a.e. monotonically
non-decreasing in n (the last property means that for almost all
x ∈ [a, b] for all n ∈ N we have fn+1 (x) ≥ fn (x));
Rb
3. Riemann integrals (R) a fn (x)dx are bounded, i.e.
Rb
supn∈N (R) a fn (x)dx < +∞.
If f is Lebesgue integrable then Lebesgue integral of f is dened as
follows:
Z b
Z b
Z b
definition
(L)
f (x)dx = sup (R)
fn (x)dx = lim (R)
fn (x)dx.
a
n∈N
a
n→∞
a
7 / 16
Lebesgue integral for arbitrary real-valued function
Lemma. Each function f : R → R can be represented in the form
f (x) = f+ (x) − f− (x) where f+ (x) ≥ 0 and f− (x) ≥ 0.
Remark. For example, one can set f+ (x) = 12 (f (x) + |f (x)|),
f− (x) = 12 (−f (x) + |f (x)|) because |f (x)| ≥ f (x) for all x .
Denition. Function f : [a, b] → R is called Lebesgue integrable i both
conditions hold:
1. f is measurable;
2. f is represented in the way f (x) = f+ (x) − f− (x) where f+ (x) ≥ 0 and
f− (x) ≥ 0, f+ and f− are measurable and Lebesgue integrable;
In that case by denition
Z b
(L)
Z b
f+ (x)dx − (L)
f (x)dx = (L)
a
Z b
a
f− (x)dx.
a
Improper Lebesgue integral. If lim (L) ab |f (x)|dx < +∞ then
R
b→+∞
Z +∞
(L)
f (x)dx
a
definition
=
Z b
lim (L)
b→+∞
f (x)dx.
a
Similarly
R +∞
R0
Rb
definition
(L) −∞ f (x)dx = lima→−∞ (L) a f (x)dx + limb→+∞ (L) 0 f (x)dx
if BOTH limits exist.
8 / 16
Relation between Riemann and Lebesgue integrals
Theorem.
Rb
1. If proper Riemann integral (R) a f (x)dx exists and
a.e.
f (x) = g (x) then
Z b
Z b
(L)
g (x)dx = (R)
f (x)dx.
a
a
Rb
2. If improper Riemann integral (R) a f (x)dx converges absolutely
a.e.
and f (x) = g (x) then
Z b
Z b
(L)
g (x)dx = (R)
f (x)dx.
a
a
Rb
3. If improper Riemann integral (R) a f (x)dx diverges absolutely
Rb
a.e.
and f (x) = g (x) then (L) a g (x)dx does not exist.
Remark. Item 3 is true even if improper Riemann integral
converges conditionally or in sense of principle value (v.p.), or in
some other sense!
9 / 16
Passing to limits under the sign of Lebesgue integral
Lebesgue's dominated convergence theorem. Let functions
fn : (a, b) → R be Lebesgue integrable on (a, b) and
a.e.
limn→∞ fn (x) = f (x). Suppose there exists Lebesgue integrable on
a.e.
(a, b) function Φ such that |fn (x)| ≤ Φ(x) for all n ∈ N.
Then f is Lebesgue integrable on (a, b) and the following equalities
hold:
Z b
Z b
lim (L)
fn (x)dx = (L)
f (x)dx,
n→∞
a
a
Z b
lim (L)
n→∞
|fn (x) − f (x)|dx = 0.
a
10 / 16
Passing to limits under the sign of Lebesgue integral
Beppo Levi theorem for series. Let functions
fn : (a, b) → [0, +∞), n ∈ N be Lebesgue integrable on (a, b), and
number series converges:
∞
X
Z b
(L)
Then series of functions
integrable function and
∞
X
n=1
fn (x)dx < +∞.
a
n=1
P∞
n=1 fn (x) a.e. converges to Lebesgue
Z b
(L)
fn (x)dx = (L)
a
Z bX
∞
fn (x)dx.
a n=1
11 / 16
Passing to limits under the sign of Lebesgue integral
Corollary of Lebesgue dominated convergence theorem.
Consider function f : (a, b) × (α1 , α2 ) → R, suppose function
x 7−→ f (x, α) is Lebesgue integrable on (a, b) for all α ∈ (α1 , α2 ).
Dene
Z
b
J(α) = (L)
f (x, α)dx.
a
1. If for almost all x function α 7−→ f (x, α) is continuous and there exists
Lebesgue integrable function Φ : (a, b) → R such that for all α and almost all x
we have |f (x, α)| ≤ Φ(x), then function J is continuous on (α1 , α2 ), i.e. for all
α ∈ (α1 , α2 ) we have
Z b
lim J(β) = lim (L)
β→α
β→α
Z b
f (x, β)dx = (L)
a
lim f (x, β)dx = J(α).
a
β→α
2. If for almost all x function α 7−→ f (x, α) is dierentiable and there exists
Lebesgue integrable function Ψ : (a, b) → R such that for all α and almost all x
we have |∂f (x, α)/∂α| ≤ Ψ(x), then function J is dierentiable on (α1 , α2 )
and for all α ∈ (α1 , α2 ) we have
J 0 (α) =
d
(L)
dα
Z b
Z b
f (x, α)dx = (L)
a
a
∂
f (x, α)dx.
∂α
12 / 16
Main theorem of calculus for the Lebesgue integral
Theorem. If f : R → R is absolutely continuous then f is a.e.
dierentiable, i.e. ∃f 0 (x) for a.e. x ∈ R.
Remark. The reverse is not true. Cantor's function is an example.
Theorem. If f : [a, b] → R then three conditions are equivalent:
1. f is absolutely continuous;
2. a.e. ∃f 0 (x), f 0 is Lebesgue integrable, and
Z x
f (x) = f (a) + (L)
f 0 (t)dt for all x ∈ [a, b].
a
3. ∃ Lebesgue integrable function g : [a, b] → R such that
Z x
f (x) = f (a) + (L)
g (t)dt for all x ∈ [a, b].
a
Remark. If these equivalent conditions are satised then necessarily
a.e.
g (x) = f 0 (x).
Remark. This theorem shows that absolutely continuous functions
is exactly the classRof functions f for which Newton-Leibniz formula
x
f (x) = f (a) + (L) a f 0 (t)dt holds for Lebesgue integral.
13 / 16
Seminar 9
Problems to be solved:
1. Consider function f : R → R dened as follows:
0 for x ∈ R \ Q,
f (x) = n1 for x ∈ Q, x 6= 0, x = m
n irreducible fraction,
1 for x = 0.
We proved that f is continuous in each point of the set R \ Q and
discontinuous in each point of the set Q. As f is bounded then via
Lebesgue's criterion of Riemann integrability we obtain that f is
Riemann integrable on each segment [a, b]. Also we proved that f
is Lebesgue integrable (because it is almost everywhere equal to a
Lebesgue integrable function - constant 0).
2. First part of Exercise on p.2.
3. Upper Exercise on p.3
14 / 16
Homework 9
Problems to solve at home (deadline 16:00 November 11):
1. Consider function f : R → R dened as follows:
0 for x ∈ R \ Q,
f (x) = n1 for x ∈ Q, x 6= 0, x = m
n irreducible fraction,
1 for x = 0.
Using denition of Riemann integral in terms of integral sums for
Rb
all a < b prove that ∃(R) a f (x)dx and nd its value.
2. a) Second part of Exercise on p.2 b) Give an example of a
function f that is dened on [a, b] and discontinuous only in one
point of [a, b], but f ([a, b]) is not equal to one interval.
3. Second exercise on p.3.
4. Exercise on p.5.
5. Example to Lebesgue's dominated convergence theorem: nd
such fn : [0, +∞) → [0, +∞) that functions fRn are Lebesgue
+∞
integrable, fn (x) → 0 uniformly but lim (L) 0 fn (x)dx = +∞.
n→∞
15 / 16
Literature for homework
1. Vladimir I. Bogachev, Oleg G. Smolyanov. Real and Functional Analysis.
Springer, Scholtech, HSE (2020)
2. À.Í.Êîëìîãîðîâ, Ñ.Â.Ôîìèí. Ýëåìåíòû òåîðèè ôóíêöèé è
ôóíêöèîíàëüíîãî àíàëèçà. Ëþáîå èçäàíèå.
3. Bernard R. Gelbaum, John M. H. Olmsted. Counterexamples in Analysis.
Any edition
4. Íèêèòèí À.À., Ôîìè÷åâ Â.Â. Ìàòåìàòè÷åñêèé àíàëèç. Óãëóáëåííûé
êóðñ : ó÷åáíèê è ïðàêòèêóì äëÿ àêàäåìè÷åñêîãî áàêàëàâðèàòà. 2-å èçä.,
èñïð. è äîï. Ìîñêâà : Èçäàòåëüñòâî Þðàéò, 2018. 460 ñ.
5. Íàòàíñîí È. Ï. Òåîðèÿ ôóíêöèé âåùåñòâåííîé ïåðåìåííîé. Ëþáîå
èçäàíèå
6. Óëüÿíîâ Ï.Ë., Áàõâàëîâ À.Í. Äåéñòâèòåëüíûé àíàëèç â çàäà÷àõ. Ëþáîå
èçäàíèå
7. Athreya, Krishna B.; Lahiri, Soumendra N. (2006), Measure theory and
probability theory, Springer, ISBN 0-387-32903-X
8. Leoni, Giovanni (2009), A First Course in Sobolev Spaces, Graduate Studies
in Mathematics, AMS. ISBN 978-0-8218-4768-8
9. Nielsen, Ole A. (1997), An introduction to integration and measure theory,
Wiley-Interscience, ISBN 0-471-59518-7
10. Royden, H.L. (1988), Real Analysis (third ed.), Collier Macmillan, ISBN
0-02-404151-3
11. Any other books on Real Analysis, on Calculus, on Mathematical Analysis
16 / 16
Introduction to
One-parameter semigroups and evolution
equations
Lesson 10: Necessary background from Real Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
10 November 2022
1 / 13
Convergence of function sequences
Denition. We say that a sequence of Lebesgue integrable
functions fn : (a, b) → R
1. is fundamental in mean i for all ε > 0 exists N ∈ N such that
Z b
(L)
|fn (x) − fk (x)|dx < ε for all n ≥ N, k ≥ N.
a
2. converges in mean to Lebesgue integrable function f i
Z b
lim (L)
n→∞
|fn (x) − f (x)|dx = 0.
a
In that case also some people write f (x) = l.i.m.n→∞ fn (x) where
l.i.m. is an abbreviation for ¾limit in mean¿.
Theorem. If sequence of Lebesgue integrable functions fn is
fundamental in mean then it converges in mean to a Lebesgue
integrable function.
2 / 13
Convergence of function sequences
Remark. Convergece in mean is dierent from pointwise
convergence: there exists a sequence of Lebesgue integrable
functions fn : [0, 1] → R such that fn → 0 in mean but
6 ∃ limn→∞ fn (x) for all x ∈ [0, 1].
Example can be constructed as follows: f1 = 1[0,1/2] , f2 = 1[1/2,1] ,
f3 = 1[0,1/3] , f4 = 1[1/3,2/3] , f5 = 1[2/3,1] , f6 = 1[0,1/4] ,
f7 = 1[1/4,2/4] ,f8 = 1[2/4,3/4] ,f9 = 1[3/4,1] , f10 = 1[0,1/5] ,
f11 = 1[1/5,2/5] ,...
R1
We have limn→∞ 0 |fn (x) − 0|dx = 0 but for each x sequence
fn (x) has innitely many values 0 and innitely many values 1, so
limit limn→∞ fn (x) cannot exist.
Theorem. If sequence of Lebesgue integrable functions fn is
fundamental in mean and a.e. converges to f , then f is Lebesgue
integrable and fn → f in mean.
Exercise. Find a sequense of Lebesgue integrable functions
fn : [0, 1] → R that a.e. converges to f : [0, 1] → R but
a) f is not Lebesgue integrable
b) f is Lebesgue integrable but fn 6→ f in mean
3 / 13
Lebesgue integrability in terms of simple functions
Denition. Function is called simple if it takes only nite number
of values.
Theorem. Function f : (a, b) → R is Lebesgue integrable i there
is a sequence fn of simple, Lebesgue integrable functions that a.e.
converges to f and fundamental in mean. In that case fn → f in
mean.
4 / 13
Lebesgue measure
Denition. A set A ⊂ [a, b] is called Lebesgue measurable i
function 1A dened as 1A (x) = 1 for x ∈ A and 1A (x) = 0 for
x∈
/ A is Lebesgue integrable. By denition
Z b
µA =
1A (x)dx.
a
Denition. A set A ⊂ R is called Lebesgue measurable i for all
k ∈ Z all sets A ∩ [k, k + 1] are measurable. If this holds then by
denition
+∞
0
X
X
µA =
µ(A ∩ [k, k + 1]) +
µ(A ∩ [k, k + 1])
k=1
k=−∞
where we accept value +∞ for some sets if one or two of the series
diverge. Some measurable sets have innite Lebesgue measure.
Denition. We say that a sequence of Lebesgue measurable
functions fn : (a, b) → R converges in measure to f i
∀ε > 0 lim µ{x : |fn (x) − f (x)| ≥ ε} = 0.
n→∞
5 / 13
Lebesgue spaces
Denition. We say that two functions f , g : R → R are versions of each
other i f a.e.
= g.
Denition. We say that function f : (a, b) → R belongs to the space
p
Lver
p (a, b) for some number p ∈ [1, +∞) i function x 7−→ |f (x)| is
Rb
Lebesgue integrable on (a, b) with (L) a |f (x)|p dx < ∞.
Exercise. Prove that a.e.
= is an equivalence relation (i.e. reexive,
symmetric, transitive) on Lver
p (a, b).
a.e.
Denition. Lp (a, b) is a factor set of Lver
p (a, b) over relation = .
Remark. However, often people write for function f things like
f ∈ Lp (a, b) meaning that f ∈ Lver
p (a, b) and assuming that symbol f
represents not function f but any version of f .
Denition. We say that a sequence functions fn ∈ Lver
p (a, b) for some
p ∈ [1, +∞) converges in Lp to f ∈ Lver
(a,
b)
i
p
Z b
lim (L)
n→∞
|fn (x) − f (x)|p dx = 0.
a
Remark. For p = 1 convergence in L1 is convergence in mean. For p = 2
convergence in L2 is called ¾convergence in mean square¿.
6 / 13
Lebesgue spaces
Theorem (H
older inequality). Let p > 1, q > 1, 1/p + 1/q = 1,
f ∈ Lp (a, b), g ∈ Lq (a, b). Then fg ∈ L1 (a, b) and
Z b
Z b
|f (x)g (x)|dx ≤
a
p
p1 Z b
|f (x)| dx
a
q
|g (x)| dx
q1
.
a
Theorem (Minkowski inequality). Let p ∈ [1, +∞),
f , g ∈ Lp (a, b). Then (f + g ) ∈ Lp (a, b) and
Z b
a
|f (x) + g (x)|p dx
p1
Z b
p1 Z b
p1
|f (x)|p dx +
|g (x)|p dx
.
a
a
≤
7 / 13
Lebesgue spaces
Denition. We say that function f : (a, b) → R is essentially
bounded from above i it has a bounded from above version. The
essential supremum of an essentiallly bounded from above function
is denoted as ¾ess sup¿ or ¾vrai sup¿ and dened as
ess sup f (x) = inf
x∈(a,b)
a.e.
sup g (x), or, equivalently, as
g = f x∈(a,b)
n
o
ess sup f (x) = inf y ∈ R : µ{x ∈ (a, b) : f (x) > y } = 0 .
x∈(a,b)
Remark. sup and ess sup can be dened and compared as follows:
M = sup f (x) means that M is the smallest number such that
x∈(a,b)
the set f −1 ((M, +∞)) is empty.
M = ess sup f (x) means that M is the smallest number such that
x∈(a,b)
the set f −1 ((M, +∞)) has measure zero.
Remark. f is bounded from above on (a, b) i sup f (x) < +∞,
x∈(a,b)
f is essentially bounded from above on (a, b) i ess sup f (x) < +∞.
x∈(a,b)
8 / 13
Lebesgue spaces
Denition. We say that function f : (a, b) → R is essentially
bounded from below i it has a bounded from below version. The
essential inmum of an essentiallly bounded from below function is
denoted as ¾ess inf¿ or ¾vrai inf¿ and dened as
ess inf f (x) = sup
x∈(a,b)
a.e.
g =f
inf g (x), or, equivalently, as
x∈(a,b)
n
o
ess inf f (x) = sup y ∈ R : µ{x ∈ (a, b) : f (x) < y } = 0 .
x∈(a,b)
Remark. inf and ess inf can be dened and compared as follows:
m = inf f (x) means that m is the biggest number such that the
x∈(a,b)
set f −1 ((−∞, m)) is empty.
m = ess inf f (x) means that m is the biggest number such that the
x∈(a,b)
set f −1 ((−∞, m)) has measure zero.
Remark. f is bounded from below on (a, b) i
inf f (x) > −∞,
x∈(a,b)
f is essentially bounded from below on (a, b) i ess inf f (x) > −∞.
x∈(a,b)
9 / 13
Lebesgue spaces
Denition. We say that function f : (a, b) → R is essentially
bounded i it is essentially bounded from above and from below,
which is equivalent to ess sup |f (x)| < +∞.
x∈(a,b)
Denition. We say that function f : (a, b) → R belongs to the
space Lver
∞ (a, b) i f is essentially bounded.
a.e.
Denition. L∞ (a, b) is a factor set of Lver
∞ (a, b) over relation = .
Remark. Now we have dened Lp spaces for all p ∈ [1, +∞].
Exercise. Suppose that 1 ≤ p1 < p2 ≤ +∞ and consider sets
a) A = Lp1 [a, b] and B = Lp2 [a, b]
b) A = Lp1 [a, +∞) and B = Lp2 [a, +∞).
Question: nd how A and B are related as sets:
1. is it true that A ⊂ B or A ⊃ B ?
2. is it true that A \ B = ∅ or B \ A = ∅?
Please give examples proving your opinion. Hint: use Holder
inequality to prove inclusions.
10 / 13
Convergence revisited
Let us study two statements that are examples of ¾selection
principle¿-type theorems:
Egorov (Egoro) selection theorem. If a, b ∈ R, and sequence
of measurable functions fn : (a, b) → R a.e. converges to f , then for
each ε > 0 there exists A ⊂ (a, b) such that µA < ε and
fn (x) → f (x) uniformly with respect to x ∈ (a, b) \ A.
F. Riesz selection theorem. If fn : (a, b) → R converges in
measure to f , then there is a subsequence (fnk ) converging to f
a.e.
almost everywhere: limk→∞ fnk (x) = f (x).
11 / 13
Seminar 10
Problems to be solved on seminar 10:
1. Exercise on p. 6. - done
2. Explain example on p.3 in details. - partially done
3. Prove in details that Cantor function C : [0; 1] → [0; 1] is
continuous, uniformly continuous, but not absolutely continuous.
(HW8, problem 2) - hope you have done it in HW8
4. Cantor set of positive Lebesgue measure. - to be discussed next
time
12 / 13
Homework 10
Problems to solve (deadline 16:00 Friday November 18 ):
1. Find such a sequence of Lebesgue integrable functions
fn : [0, 1] → R that fn → 0 pointwise (i.e. limn→∞ f (x) = 0 for all
x ∈ [0, 1]) and:
R1
a) 0 fn (x)dx = 1 for all n ∈ N
R1
b) for any given an ≥ 0 we have 0 fn (x)dx = an
R1
c) limn→∞ 0 fn (x)dx = +∞
R1
d) 6 ∃ nite or innite limn→∞ 0 fn (x)dx =
R1
e) limn→∞ 0 fn (x)dx = 0 but function x 7−→ sup fn (x) is not
Lebesgue integrable on [0, 1].
2. Exercise on p.3
3. Exercise on p.10
4. Find such f : [0, 1] → R that ∃(L)
P∞
R 1n
n∈N
R 1n
1
n+1
f (x)dx and
1 f (x)dx < ∞ but f is not Lebesgue integrable on [0, 1].
Rn+1 1
5. Find 0 C (x)dx where C is the Cantor function.
13 / 13
n=1
Introduction to
One-parameter semigroups and evolution
equations
Lecture 11: Necessary background from Real Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
22 November 2022
1 / 12
Properties of Lebesgue integral
Theorem (Chebyshev inequality). If f : (a, b) → R is Lebesgue
integrable then for all R > 0 we have
Z b
n
o
1
µ x : |f (x)| ≥ R ≤ (L)
|f (x)|dx.
R
a
Lebesgue dierentiation theorem. If f : [a, b] → R is Lebesgue
integrable then for almost all x ∈ (a, b) we have
Z x+h
1
f (x) = lim
(L)
f (y )dy ,
h→0 2h
x−h
Z x+h
1
lim
(L)
|f (y ) − f (x)|dy = 0.
(∗)
h→0 2h
x−h
Denition. Points x posessing property (∗) are called Lebesgue
points for function f . (Comment: informally this means that f does
not oscillate too much near x , in an average sense.)
Remark. With this denition, the above theorem says: for
Lebesgue integrable function almost all points in the domain are
Lebesgue points.
2 / 12
Properties of Lebesgue integral
Theorem (integration by parts in Lebesgue integral). If
real-valued functions f and g are absolutely continuous on [a, b],
then:
Z b
Z b
b
0
f (x)g 0 (x)dx
f (x)g (x)dx = f (x)g (x) − (L)
(L)
a
a
where we denoted f (x)g (x)
b
a
a
= f (b)g (b) − f (a)g (a).
Example. Let a = −1, b = 1, f (x) = |x|, g (x) = x . Then
f 0 (x) = sign(x) a.e., g 0 (x) = 1 and it is easy to check (please do
it) that
Z 1
Z 1
1
(L)
sign(x)xdx = |x|x
− (L)
|x| · 1 · dx.
−1
−1
−1
Exercise 1. Make a plot illustrating this situation.
Remark. Here all functions are Riemann integrable, but f 0 is
discontinuous and dened not everywhere, so usual formula for
integration by parts in Riemann integral is not applicable.
3 / 12
Properties of Lebesgue integral
Theorem (change of variable in Lebesgue integral). Let
ϕ : [a, b] → R be monotone absolutely continuous function and
ϕ([a, b]) ⊂ [c, d]. Then for each f ∈ Lver
1 [c, d] function
y 7−→ f (ϕ(y ))ϕ0 (y ) is Lebesgue integrable on [a, b] and
Z ϕ(b)
(L)
Z b
f (x)dx = (L)
ϕ(a)
f (ϕ(y ))ϕ0 (y )dy .
a
Remark. Here we make change of variable x = ϕ(y ).
Remark. The statement holds true if one substitutes [a, b] by
(−∞, b] or by [a, +∞) or by R.
Example. If we take Cantor function as ϕ then the equality breakes
because Cantor function in not absolutely continuous.
4 / 12
Properties of Lebesgue integral
Denition. If A ⊂ R, then the closure of A is denoted as A or
Cl(A) and dened as follows: (x ∈ Cl(A)) ⇐⇒ (there exists a
sequence (xn ) ⊂ A such that limn→∞ xn = x ).
Denition. Support of a function f : (a, b) → R is dened as
suppf = Cl({x|f (x) 6= 0}).
Remark. Function has compact support i it is zero outside some
interval of nite length.
Denition. We use symbol Cc∞ (a, b) to denote the space of all
functions (a, b) → R that are continuous, bounded, have bounded
derivatives of all orders and support suppf ⊂ (a, b) is a bounded
set. Those functions are often called innitely smooth compactly
supported functions.
Theorem (fundamental lemma of calculus of variations). If
f ∈ Lver
Z
1 (a, b) and
b
(L)
f (x)h(x)dx = 0
a
a.e.
for all h ∈ Cc∞ (a, b) then f (x) = 0.
5 / 12
Weak derivative
Remark.
As you know, the conventional derivative of a function
(x)
f : (a, b) → R is dened by equality f 0 (x) = limt→0 f (x+t)−f
. This is a
t
¾pointwise denition¿ because it uses values of f in all points. However,
another notion of derivative is possible to give using Lebesgue integral of
a function which uses only values of function in almost all points.
Denition. The weak derivative of a function f ∈ Lver
1 (a, b) is such
a function g : (a, b) → R that for all h ∈ Cc∞ (a, b) we have
Z b
Z b
f (x)h0 (x)dx = −(L)
g (x)h(x)dx.
(1)
(L)
a
a
If f is dened on [a, b] instead of (a, b) then h needs to be
continuous, innitely smooth and h(a) = h(b) = 0.
Remark. Recall the ¾dierentiation by parts¿ formula:
Z b
Z b
0
(L)
f (x)h (x)dx = f (b)h(b) − f (a)h(a) − (L)
g (x)h(x)dx.
a
a
Remark. Changing f or g on a measure zero set keeps (1), so
weak derivative is dened up to set of measure zero.
6 / 12
Weak derivative
Exercise 2. If function is absolutely continuous then its weak and
conventional derivatives coincide a.e. Hint: use integration by parts
formula.
Example. Cantor function is a.e. dierentiable but it does not have
weak derivative.
Exercise 3. Prove that constant zero is not a weak derivative of
Cantor function.
Remark. There is another notion of dierentiability the
distributional derivative. It is ¾even weaker¿ than the weak
derivative, but it involves generalized functions (distributions) and
we will not study this. But in that weaker sense every locally
Lebesgue integrable function (including Cantor function) has a
derivative.
7 / 12
Lebesgue spaces
Denition. Function f : (a, b) → R is called a version of function
a.e.
g : (a, b) → R i f (x) = g (x).
Denition (Lp norm). The Lp norm of a function f : (a, b) → R
for p ∈ [1, +∞] is dened as follows:
Z b
1/p
p
kf kp = (L)
|f (x)| dx
, p ∈ [1, +∞),
a
kf k∞ = ess sup |f (x)|.
x∈(a,b)
Remark. With this denition we can redene (equivalently to what
we have done in the previous lesson) spaces Lver
p and Lp as follows:
Denition. For xed p ∈ [1, +∞] by deition Lver
p (a, b) is the
space of all functions f : (a, b) → R such that kf kp < ∞.
Denition. For xed p ∈ [1, +∞] by deition Lp (a, b) is the
a.e.
factor-space of Lver
p (a, b) with respect to equivalence relation = . If
F ∈ Lp (a, b) then by denition kF kp = kf kp where f ∈ Lver
p (a, b)
is any function in the equivalence class F , i.e.
a.e.
f ∈ F = {g ∈ Lver
8 / 12
p (a, b)|f = g }.
Lebesgue spaces
Important remark. Usually people are too lasy to distinguish
between Lver
p and Lp , and write just Lp in both cases and hope that
this will not lead to misunderstanding. This hope is based on the
following:
Convention. Writing f ∈ Lp (a, b) by convention means that f is
any version of the class of equivalence of f ∈ Lver
p (a, b).
So if some property of f ∈ Lp (a, b) is stated then by this
convention it is assumed that it is additionally stated that this
property holds also for all g such that f and g are versions of each
a.e.
other, i.e. for all g such that f (x) = g (x).
Remark. This is important for proof writing/reading/checking!
9 / 12
Sobolev spaces
Denition (Sobolev spaces). Given numpers p ∈ [1, +∞] and
k = 1, 2, 3, . . . we say that function f : (a, b) → R belongs to
Sobolev space Wpk (a, b) i f ∈ Lp (a, b) has weak derivatives of
orders 1, . . . , k and they all belong to Lp (a, b).
Theorem. f ∈ Wpk (a, b) i f has conventional derivatives up to
order k − 1, f (k−1) is absolutely continuous and conventional
derivatives of orders 1, . . . , k belong to Lp (a, b).
Denition (Sobolev norm). For f ∈ Wpk (a, b) the number
kf kkp =
k
X
j=0
1/p
kf
(j)
kp
,
kf kk∞ =
k
X
kf (j) k∞
j=0
is called the Sobolev norm of f .
Remark. For k = 0 we have Wp0 (a, b) = Lp (a, b) so Lebesgue
space is a particular case of Sobolev space. Sobolev spaces can also
be dened as spaces of functions for which weak derivatives exist,
and Sobolev norm is nite. The notation is used: W2k = Hk .
10 / 12
Seminar 11
On seminar 11 we will solve the following problems:
1. Discuss Cantor-type set with positive Lebesgue measure.
2. Find all Lebesgue points of the function x 7−→ 1Q (x).
R1
3. In the integral (L) −1 xdx make variable change x = ϕ(y ) where
(
2y for y > 0,
ϕ(y ) =
y for y ≤ 0.
4. Find weak derivative of f (x) = |x|.
5. Prove that f : [0, 2] → R does not have a weak derivative where
(
1 for x ∈ [0, 1),
f (x) =
2 for x ∈ [1, 2].
11 / 12
Homework 11
Problems to solve at home (deadline 16:00 November 29):
1. Find all Lebesgue points for function f : [0, 2] → R where
(
1 for x ∈ [0, 1),
f (x) =
2 for x ∈ [1, 2].
2. Exercise 2 on p.7
3. Exercise 3 on p.7
4. Prove that f (x) = sign(x) does not have a weak derivative.
5. Find weak derivative of the function f : [0, 1] → R where
(
0 for x ∈ Q,
f (x) =
x 2 for x ∈ R \ Q.
12 / 12
Introduction to
One-parameter semigroups and evolution
equations
Lesson 12: Necessary background from Functional Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
1 December 2022
1 / 18
Linear theory: prelude
I have a theorem with a marvellous proof as an epigraph,
but the epigraph box is too narrow to contain it.
Your lecturer
Theorem. If there is at least one mathematician in Hell, then the
Hell is nonlinear.
Proof. Suppose there is a mathematician in Hell, but the Hell is
linear. Then for this mathematician the Hell is the Heaven!
Contradiction.
Remark. In the proof we implicitly used the ¾HH-conjecture¿
which says that Hell and Heaven are disjoint.
Comment. This theorem, as well as the proof, was an oral
epigraph that my rst teacher of functional analysis, Professor
Alexander Abrosimov, gave to his lecture dedicated to the linear
theory. Indeed, linearity provides a lot of nice properties which leads
to wide applicability of linear theory. However, linear
innite-dimansional objects can be as complicated, as nonlinear
nite-dimensional objects, and in some sense they are equivalent.
2 / 18
Field - as set of objects that are called numbers or scalars
Denition. Let F be a set having at least two elements (denoted 0
and 1) and suppose that there are two opeartions + and · in F . We
say that (F , 0, 1, +, ·) is a eld i the following conditions are true:
1. Associativity of addition and multiplication:
a + (b + c) = (a + b) + c , and a · (b · c) = (a · b) · c .
2. Commutativity of addition and multiplication: a + b = b + a,
and a · b = b · a.
3. Additive and multiplicative identity: there exist two dierent
elements 0 and 1 in F such that a + 0 = a and a · 1 = a.
4. Additive inverses: for every a in F , there exists an element in F ,
denoted −a, called the additive inverse of a, such that
a + (−a) = 0.
5. Multiplicative inverses: for every a 6= 0 in F , there exists an
element in F , denoted by a−1 or 1/a, called the multiplicative
inverse of a, such that a · a−1 = 1.
6. Distributivity of multiplication over addition:
a · (b + c) = (a · b) + (a · c).
3 / 18
Linear space - a set of objects that are called vectors
Denition. Let (F , 0F , 1F , +F , ·F ) be a eld and let V be a set having at
least one element (denoted 0) and suppose that there are two opeartions
+F and ·F in F , and operations + : V × V → V , · : F × V → V . We say
that (V , (F , 0F , 1F , +F , ·F ), 0, +, ·) or just V to make it shorter, is a
linear space (synonym: vector space) over the eld F , and say that
elements of F are scalars, and elements of V are vectors, i the following
conditions are true for all scalars a, b and all vectors u, v , w :
1. Associativity of addition: u + (v + w ) = (u + v ) + w
2. Commutativity of addition: u + v = v + u
3. Identity element of addition: there exists an element 0 ∈ V , called the
zero vector, such that v + 0 = v for all v ∈ V .
4. Inverse elements of addition: for every v ∈ V , there exists an element
−v ∈ V , called the additive inverse of v , such that v + (−v ) = 0.
5. Compatibility of scalar multiplication with eld multiplication:
a · (b · v ) = (a ·F b) · v .
6. Distributivity of scalar multiplication with respect to vector addition:
a · (u + v ) = a · u + a · v .
7. Distributivity of scalar multiplication with respect to eld addition:
(a +F b) · v = a · v + b · v .
8. Identity element of scalar multiplication: 1F · v = v .
4 / 18
Linear spaces (vector spaces)
Remark. Axiom 8 makes the whole denition of vector space
non-trivial. Indeed, if we dene a · v = 0 for all scalars a and
vectors v , then all axioms will hold true but 8 will fail if V 6= {0}.
Remark. Very often signs of multiplication ·F and · are omitted
but presumed: av = a · v .
Remark. In functional analysis most commonly used are linear
spaces over elds R and C. Sometimes linear spaces over Q are
used, mostly for constructing examples. Also linear spaces over the
eld Qp of p -adic numbers (google what it is it is interesting!) for
prime p are considered in the branch of mathematics called
ultrametric analysis. Linear spaces over Z2 are used in the coding
theory and practice.
Remark. In our course we use only linear spaces over eld R (they
are called real vector spaces) and over eld C (they are called
complex vector spaces).
Notation. From now, we will use one letter F dening R or C
in the cases where it is possible to use both R or C.
5 / 18
Vector subspaces and isomorphisms
Denition. Subset V1 ⊂ V of a vector space V is called a vector
subspace of V i V1 is a vector space with operations from V .
Exercise. Prove that if V1 ⊂ V , V is a vector space then: (V1 is a
vector subspace of V ) ⇐⇒ (for all scalars a and all vectors
u, v ∈ V1 we have av ∈ V1 , v + u ∈ V1 ).
Denition. Two vector spaces V1 and V2 are called isomorphic i
two conditions are satised:
1. V1 and V2 are vector spaces over the same eld of scalars
2. There exists a bijective function f : V1 → V2 such that:
2.1. a ·V1 v = u ⇐⇒ a ·V2 f (v ) = f (u) for all v ∈ V1 and all
scalars a
2.2. v +V1 u = w ⇐⇒ f (v ) +V2 f (u) = f (w ) for all u, v , w ∈ V1
Function f is called a linear isomorphism.
6 / 18
Methods of construction of vector spaces
Exercise (the simplest vector space). Prove that F is a vector
space over F with + and · operations in F.
Lemma (method 1: scalar functions). Let X be a set, then FX
is a linear space with linear operations introduced as follows:
1. For function f ∈ FX and number a ∈ F function af is dened as
follows: (af )(x) = af (x) for all x ∈ X .
2. For functions f , g ∈ FX function f + g is dened as follows:
(f + g )(x) = f (x) + g (x) for all x ∈ X .
Exercise. Prove that this lemma is a particular case of the
following lemma:
Lemma (method 2: vector functions). Let X be a set, and V be
a vector space then V X is a linear space with linear operations
introduced as follows:
1. For function f ∈ V X and number a ∈ F function af is dened as
follows: (af )(x) = af (x) for all x ∈ X .
2. For functions f , g ∈ V X function f + g is dened as follows:
(f + g )(x) = f (x) + g (x) for all x ∈ X .
Exercise. Prove the lemma.
7 / 18
Methods of construction of vector spaces
Exercise. Prove that for any nite set A = {a1 , . . . , an } vector
spaces Fn and FA are isomorphic and write down the isomorphism.
Remark (method 3: selecting a subspace by property). For
example, if we have a space F[a,b] of all functions f : [a, b] → F
then we can select a subset that appears to be a linear subspace
C [a, b] = {f : [a, b] → F|f is continuous} ⊂ F[a,b] .
Lemma (method 4: factorization over subspace). Let V be
vector space and H ⊂ V be a linear subspace of V . Then:
1. ∼ is an equivalence relation in V if by denition
u ∼ v ⇐⇒ (u − v ) ∈ H .
2. The factor set V / ∼ is a vector space with operations dened as
aKv = Kav , Ku + Kv = Ku+v for all numbers a and all vectors u, v .
Exercise. Prove the lemma.
Remark. The majority of denitions of vector spaces in functional
analysis are combination of those 4 methods.
Remark. It is possible to prove that any vector space is isomorphic
to some space of functions.
8 / 18
One more method of construction of vector spaces
Denition (method 5: direct sum). We say that vector space V
is a direct sum of its subspaces V1 ⊂ V and V2 ⊂ V , and write
V = V1 ⊕ V2 i for each v ∈ V there exists unique couple
(v1 , v2 ) ∈ V1 × V2 such that v = v1 + v2 .
Remark. If vector spaces V1 , V2 over the same eld F are given
then V1 ⊕ V2 is dened as follows. Consider the set V = V1 × V2
with operations
a · (v1 , v2 ) = (a · v1 , a · v2 ),
(v1 , v2 ) + (u1 , u2 ) = (v1 + u1 , v2 + u2 ).
Then U1 = {(v1 , 0) : v1 ∈ V1 } is a linear subspace in V which is
isomorphic to V1 , and isomorphism is given by
U1 3 (v1 , 0) 7−→ v1 ∈ V1 . In the same way U2 = {(0, v2 ) : v2 ∈ V2 }
is a linear subspace in V which is isomorphic to V2 , and
isomorphism is given by U2 3 (0, v2 ) 7−→ v2 ∈ V2 . By construction
V = U1 ⊕ U2 but up to isomorphism V = V1 ⊕ V2 .
Remark. This construction may be used for V1 = V2 and may be
define
inductively repeated: V1 ⊕ V2 ⊕ V3 = V1 ⊕ (V2 ⊕ V3 ).
9 / 18
Examples of vector spaces
Denition. There are some useful subspaces of FN :
c = {x ∈ FN : ∃ limn→∞ xn } converging sequences,
c0 = {x ∈ FN : ∃ limn→∞ xn = 0} converging to zero sequences,
c00 = {x ∈ FN : ∃n0 ∈ N : xn = 0 for all n ≥ n0 } eventually zero
sequences.
Denition. Let us consider subspaces in FA for uncountable set A:
C [a, b] = {f ∈ F[a,b] : f is continuous}
C n [a, b] = {f ∈ F[a,b] : f , f 0 , f 00 , . . . , f (n) ∈ C [a, b]}
C (R) = {f ∈ FR : f is continuous}
Cb (R) = {f ∈ FR : f is continuous and bounded}
UCb (R) = {f ∈ FR : f is uniformly continuous and bounded}
C n (R) = {f ∈ FR : f , f 0 , f 00 , . . . , f (n) ∈ C (R)}
Cbn (R) = {f ∈ FR : f , f 0 , f 00 , . . . , f (n) ∈ Cb (R)}
M(a, b) = {f ∈ F(a,b) : f is Lebesgue measurable}
(a,b) : f ∈ M(a, b) and kf k < ∞}
Lver
p
p (a, b) = {f ∈ F
a.e.
(a,b)
AEZ (a, b) = {f ∈ F
: f (x) = 0} a.e. zero functions
Lp (a, b) = Lver
(a,
b)/AEZ
(a, b) Lp space as a factor space
p
10 / 18
Some more examples
Denition. Let V be a vector space and let us give two simple
examples of subspaces in V A for uncountable set A:
C (R, Rn ) = {f ∈ (Rn )R : f is continuous} continuous Rn -valued
functions on A = R,
m
C (Rm , Rn ) = {f ∈ (Rn )R : f is continuous} continuous
Rn -valued functions on A = Rm .
Remark. There are many other spaces of vector-valued functions,
but we stop here for now.
Denition. For x ∈ FN dene
!1/p
n
X
kxkp =
|xn |p
for p ∈ [1, +∞), and kxk∞ = sup |xn |.
n=1
n∈N
Then for p ∈ [1, +∞] by denition `p = {x ∈ FN : kxkp < ∞}.
Remark. Here we create a linear space `p by selecting a subset in
FN that appears to be a linear subspace thanks to Minkowski
inequality.
11 / 18
Linear notions which do not involve additional structures
Denition. Linear combination of vectors v1 , . . . , vn with scalar
coecients a1 , . . . , an by denition is the expression
a1 v1 + · · · + an vn . Linear combination is called trivial i
a1 = · · · = an = 0.
Remark. Trivial linear combination is always equal to zero vector.
Denition. Finite set of vectors v1 , . . . , vn is called linearly
independent i only trivial linear combination of these vectors is
equal to zero vector. The set is called linearly dependent i it is not
linearly independent.
Denition. Innite set of vectors is called linearly independent i
each its nite subset is linearly independent.
Denition. Linearly independent subset X ⊂ V of vector space V
is called maximal linearly independent subset (and also called
Hamel basis of V , algebraic basis of V ) i for each v ∈ V the set
{v } ∪ X is linearly dependent.
Remark. In the context of linear dependence sets of vectors are
often called systems of vectors. We also sometimes will do this.
Remark. The plural of ¾basis¿ is ¾bases¿, not ¾basises¿!
12 / 18
Linear notions which do not involve additional structures
Exercise. Prove that if X is a Hamel basis of vector space V then
for each non-zero vector v ∈ V there exists unique number n ∈ N,
unique nite subset {v1 , . . . , vn } ⊂ X and unique nite set of
scalars {a1 , . . . , an } such that v = a1 v1 + · · · + an vn .
Theorem. In every non-trivial (not consisting only of zero vector)
linear space there is a Hamel basis.
Remark. Hamel basis is not unique and there are innitely many
Hamel bases in each non-trivial vector space. Moreover, usually it is
impossible to obtain a Hamel basis explicitly because its existence
relies heavily on the Axiom of Choice.
Theorem. If B1 and B2 are Hamel bases in vector space V , then
B1 and B2 have the same cardinality, i.e. there is a bijective
mapping B1 → B2 .
Denition. Algebraic dimension (or just dimension) of vector space
V is denoted as dimV and by denition is equal to the cardinality
of any Hamel basis of V .
Example. Consider Cn as a vector space over C, then dimCn = n.
Consider Cn as a vector space over R, then dimCn = 2n.
13 / 18
Linear notions which do not involve additional structures
Denition. Vector space is called innite-dimensional i it does
not have a nite Hamel basis. In other words, i for every n ∈ N in
that vector space there exists a linearly independent set consisting
of n vectors.
Remark. Georg Hamel proposed his notion of a basis in 1905,
considering R as a vector space over Q. This space, of course, is
innite-dimensional. Hamel suggested this construction while
solving the following problem stated in 1821 by Cauchy: nd all
functions f : R → R posessing the property f (x + y ) = f (x) + f (y ).
Hamel described all the solutions in terms of Hamel bases and
showed that along with f (x) = Cx there are innitely many
discontinuous (and not Lebesgue measurable!) solutions.
Remark. Hamel bases may have arbitrary high cardinality. For
example, if X is an innite set, then the Hamel basis of the space
{f ∈ FX |f (x) 6= 0 only for nite number of elements x ∈ X } is
the set {1{x} |x ∈ X } of characteristic functions of one-point
subsets of X . This set has the same cardinality as X . This example
shows that there exist vector spaces of arbitrary high dimension.
14 / 18
Linear notions which do not involve additional structures
Denition. For a subset X ⊂ V of vector space V the linear span
(also called linear hull) of X is denoted as spanX and by denition
is equal to the set of vectors such that each of them is
representable as linear combination of vectors from the set X .
Remark. If X is a Hamel basis in the vector space V then
spanX = V .
Example. Consider two vector spaces:
c00 = {x ∈ FN : ∃n0 ∈ N : xn = 0 for all n ≥ n0 } eventually zero
sequences,
c0 = {x ∈ FN : ∃ limn→∞ xn = 0} converging to zero sequences.
It follows from denitions that the set B = {x n ∈ FN |xkn = δkn } =
{(0, . . . , 0, 0, 1, 0, 0, . . . )| 1 at n-th place and 0 at other places} is
a Hamel basis for c00 . However, B is not a Hamel basis for c0 .
Ineeded, directly from denitions we have spanB = c00 6= c0 . For
example, vector (1, 1/2, 1/3, . . . , 1/n, . . . ) belongs to c0 but not to
c00 and can not be represented as a linear combination of vectors
from B because every such linear combination is a vector which is a
sequence that has only nite number of non-zero values.
15 / 18
What we plan to do on the Seminar 12:
Exercise p.8. Prove that for any nite set A = {a1 , . . . , an } vector
spaces Fn and FA are isomorphic and write down the isomorphism.
Solution. Let us start for simplicity from the case A = {1, . . . , n}.
Elements from Fn are n-tuples (x1 , . . . , xn ). Elements from FA are
functions f : A → F which are dened by their values f (1), . . . , f (n)
on the elements of the set A. Those values can be organized as
n-tuple (f (1), . . . , f (n)) and we obtain the element of the set Fn .
So we introduce bijection J : F{1,...,n} → Fn by the rule
J : F{1,...,n} 3 f 7−→ J(f ) = (f (1), . . . , f (n)) ∈ Fn ,
J −1 : Fn 3 (f1 , . . . , fn ) 7−→ f ∈ F{1,...,n} , f (k) = fk for k = 1, . . . , n.
In general case A = {a1 , . . . , an } we have the same situation:
J(f ) = (f (a1 ), . . . , f (an )), (J −1 ((x1 , . . . , xn )))(ak ) = xk for k = 1, . . . , n.
The bijectivity of J and conditions a · f = g ⇐⇒ a · J(f ) = J(g ),
f + g = h ⇐⇒ J(f ) + J(g ) = J(h) follow directly from the
formulas above and denitions of operations +, · in Fn and FA . So
J is an isomorphism.
16 / 18
What we plan to do on the Seminar 12:
• Finish the solution of a problem from the previous seminar: prove
that f : [0, 2] → R does not have a weak derivative where
(
1 for x ∈ [0, 1),
f (x) =
2 for x ∈ [1, 2].
• Prove that spaces of functions of real argument that include all
polynomials or all polynomilals with positive weight multiplier are
innite-dimensional.
• Prove that spaces of sequences that include all indicator of a
number sequences are innite-dimensional.
• Solve some exercises from the lecture or from the HW at the
choice of students.
17 / 18
Homework 12
Problems to solve at home (deadline 16:00 December 12):
1. a) exercise at page 6. b) third exercise at p.7.
2. a) rst and second exercises at p.7 b) prove the lemma at p.8
3. Suppose that f : A → B is a bijective function. Write down a formula
for isomorphism of linear spaces CA and CB . Prove that it is indeed an
isomorphism.
4. Read about Minkowski inequality in Wikipedia:
https://en.wikipedia.org/wiki/Minkowski_inequality Both
English and Russian versions are ok. Using Minkowski inequality, prove
that real `p is a linear subspace in RN for all p ∈ [1, ∞].
5. Prove that bijection between Hamel bases naturally extends to a linear
isomorphism. In other words, suppose L and E are linear spaces over the
same eld F, B ⊂ L is a Hamel base in L, H ⊂ E is a Hamel base in E ,
and f : B → H P
is a bijection. Let us dene f (x) for arbitrary x ∈ L as
follows: if x = nk=1 αk bk for some n ∈ N, some numbers αk ∈ F and
some vectors bP
k ∈ B , keeping in mind that n, αk , bk depend on x , we
dene f (x) = nk=1 αk f (bk ). Problem: prove that f is an isomorphism of
linear spaces L and E .
6 (bonus task, 4 points). Prove that R and C are isomorphic as additive
groups, i.e. there exists a bijection f : R → C such that for all x, y , z ∈ R
it is true that x + y = z ⇐⇒ f (x) + f (y ) = f (z). Use problem 5.
18 / 18
Introduction to
One-parameter semigroups and evolution
equations
Lecture 13: Necessary background from Functional Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
10 January 2023
1 / 12
Linear operator in vector space without any other structures
(only addition of vectors and multiplication of a number by
a vector are used)
Denition. Let V1 , V2 be non-empty sets, and let D(f ) be a
subset of V1 . Function f : V1 ⊃ D(f ) → V2 is called linear i four
conditions are satised:
1. V1 , V2 be vector spaces over the same eld F.
2. D(f ) is a linear subspace in V1 .
3. For all a ∈ F and all v ∈ D(f ) we have f (av ) = af (v ) homogenity.
4. For all v1 , v2 ∈ D(f ) we have f (v1 + v2 ) = f (v1 ) + f (v2 ) additivity.
Denition. Contemporary denition of linear operator: linear
operator is a linear function in the above-mentioned sence.
Contemporary denition of linear functional: linear functional is a
linear function in the above-mentioned sence, satisfying V2 = F.
Exercise. If f : V1 ⊃ D(f ) → V2 is a linear function then the range
of function f , i.e. the set R(f ) = f (D(f )) = {f (x)|x ∈ D(f )} is a
linear subspace in V2 .
2 / 12
Linear operator in vector space without any other structures
Denition. If f : V1 ⊃ D(f ) → V2 is a linear function then the set
kerf = {x ∈ D(f )|f (x) = 0} is called a kernel of f .
Exercise. Prove that if f : V1 ⊃ D(f ) → V2 is a linear function
then kerf is a linear subspace in V1 and in D(f ).
Exercise. Prove that if f : V1 ⊃ D(f ) → V2 is a linear function
then kerf = {0} ⇐⇒ f is injective.
Exercise. Prove that if f : V1 ⊃ D(f ) → V2 is an injective linear
function then the inverse function f −1 : V2 ⊃ R(f ) → D(f ) is
linear, kerf −1 = {0}, and f −1 maps D(f −1 ) = R(f ) onto
R(f −1 ) = D(f ) bijectively.
Denition. A subset X ⊂ V is called invariant subset for linear
function f : V ⊃ D(f ) → V i f (X ) = {f (x)|x ∈ X } ⊂ X .
Denition. Suppose that A : V ⊃ D(A) → V is a linear operator
in vector space V over the eld F. Vector x ∈ V is called an
eigenvector for A corresponding to eigenvalue λ ∈ F i two
conditions hold: x 6= 0 and Ax = λx .
Remark. Obviously if x is an eigenvector for A corresponding to
eigenvalue λ, and C 6= 0, then Cx is also an eigenvector for A
corresponding to eigenvalue λ.
3 / 12
Linear operator in vector space without any other structures
Denition. Suppose that there are two linear operators
A : V1 ⊃ D(A) → V2 and B : V1 ⊃ D(B) → V2 . Then the following four
conditions are equivalent by denition:
1. D(A) ⊂ D(B) and Ax = Bx for all x ∈ D(A).
2. A ⊂ B in the sense that A ⊂ V1 × V2 , B ⊂ V1 × V2 .
3. (B, D(B)) is called a continuation of (A, D(A)) from D(A) to D(B).
4. (A, D(A)) is called a restriction of (B, D(B)) on D(A).
Remark. If any of the above conditions hold then B is also called an
extension of A, and if the word ¾extension¿ is used then D(B)
sometimes is not mentioned explicitly.
Remark. Important! Properties of linear operator A : V1 ⊃ D(A) → V2
suciently depend not only on the way of obtaining Ax from x , but also
on D(A) which may be bigger or smaller than the set of vectors x on
which formula for Ax is well-dened in the traditional way. It may happen
that for some x you calculate Ax via a formula, and for some x you obtain
Ax via continuation of any kind using values Ay for some y 6= x . For that
reason symbol A is not enough to dene a linear operator. Instead,
operator is completely dened only by the couple (A, D(A)). This is the
same as thinking of operator as of a subset of V1 × V2 because the set
A ⊂ V1 × V2 has information about both D(A) and Ax for all x ∈ D(A).
4 / 12
Linear operator in vector space without any other structures
Denition. If V is a vector space and V1 ⊂ V is a linear subspace
then the codimension of V1 in V is dened as
codimV1 = dimV /V1 .
Theorem. Let V be a vector space over the eld F and
f : V = D(f ) → F be a linear functional. Then:
1. (f (x) = 0 for all x ∈ V ) ⇐⇒ (R(f ) = {0}) ⇐⇒
(codim kerf = 0).
2. (∃x ∈ V : f (x) 6= 0) ⇐⇒ (R(f ) = F) ⇐⇒ (codim kerf = 1).
Remark. In the above theorem kerf plays the role of V1 in the
denition of codimension.
Example. Consider V = FN and
V1 = {x = (x1 , x2 , x3 , . . . ) ∈ FN | x2n = 0 for all n ∈ N}. Then
V /V1 is isomorphic to
{x = (x1 , x2 , x3 , . . . ) ∈ FN | x2n−1 = 0 for all n ∈ N}. In that
example we have dimV = dimV1 = codimV1 = ∞. As
codimV1 = ∞ then by the above theorem there is no such linear
functional f : FN → F such that V1 = kerf .
5 / 12
Linear operator in vector space without any other structures
Denition. Let V be a vector space over the led C and
A : V ⊃ D(A) → V be a linear operator. Then the point spectrum
(Russian: òî÷å÷íûé ñïåêòð) of (A, D(A)) by denition is the set of
all eigenvalues of (A, D(A)) and denoted as σp (A). That is, for all
λ ∈ C we have by denition:
(λ ∈ σp (A)) ⇐⇒ (∃x ∈ D(A), x 6= 0 : Ax = λx).
Remark. The point spectrum may be dened in the same way for
linear operators in vector spaces over the eld R or over some other
eld. But traditionally the spectrum (and point spectrum in
particular) is studied for linear operators in vector spaces over C.
The reason for that is that operator theory is widely used in
quantum physics where all vector spaces are over the eld C which
is stated explicitly in the axioms of quantum mechanics.
Remark. Quantum theory in 1930-1932 gave a birth to the
C0 -semigroup theory. Famous Schrodinger equation is an example
of evolution equation, and solutions of the Schr
odinger equation are
given by a one-parameter group of linear operators.
6 / 12
Linear operator in vector space without any other structures
Remark. Recall that for linear operator A : V ⊃ D(A) → V the
range of A and the kernel of A are dened as follows:
R(A) = A(D(A)) = {Ax|x ∈ D(A)},
ker(A) = A−1 (0) = {x|x ∈ D(A), Ax = 0}.
Recall that ker(A) and R(A) are linear subspaces in V .
Denition. The kernel ker(A) is called trivial i ker(A) = {0}.
Remark. It follows directly from the denitions that the set of all
non-zero vectors in the set ker(A) coincides with the set of all
eigenvectors corresponding to eigenvalue λ = 0. This gives the left
⇐⇒ below:
(0 6= σp (A)) ⇐⇒ (ker(A) = {0}) ⇐⇒ (A is injective),
and the right ⇐⇒ is a general fact about linear operators.
Denition. For an eigenvalue λ the geometrical multiplicity
(Russian: ãåîìåòðè÷åñêàÿ êðàòíîñòü) gm(λ) is dened as
dimension of the span of all eigenvectors corresponding to λ.
Remark. Above we have actually shown that gm(0) = dim ker(A).
7 / 12
Seminar 13
Problem 1. Consider non-empty set X , linear space CX , a function
g ∈ CX and operator of multiplication Mg dened as follows:
Mg : CX = D(Mg ) → CX ,
(Mg f )(x) = g (x)f (x) for all x ∈ X .
Find range, kernel, eigenvalues and eigenvectors of the operator
(Mg , CX ). For each eigenvalue nd its geometrical multiplicity.
Solution. Suppose that number λ ∈ C is given, and recall that the ¾full
preimage of λ with respect to g ¿ is the set that is dened as
g −1 (λ) = {x ∈ X |g (x) = λ} ⊂ X . By denitions Mg f = λf i
g (x)f (x) = λf (x) for all x ∈ X . This is equivalent to the following: if for
some x ∈ X we have g (x) = λ then f (x) may be arbitrary number,
meanwhile if for another x ∈ X we have g (x) 6= λ then f (x) = 0. So f is
dened by arbitrary function ϕ : g −1 (λ) → C as follows:
(
ϕ(x) for x ∈ g −1 (λ),
f (x) =
0 for x ∈ X \ g −1 (λ).
We have λ ∈ σp (Mg ) ⇐⇒ g −1 (λ) 6= ∅, so σp (Mg ) = g (X ). If
|g −1 (λ)| = n ∈ N then gm(λ) = |g −1 (λ)| = n, if |g −1 (λ)| = ∞ then
gm(λ) = |g −1 (λ)| = ∞. The kernel of Mg is trivial i g (x) is never zero;
otherwise ker(Mg ) is the set of functions f that are dened above
assuming λ = 0. Clearly R(Mg ) = {h ∈ CX |g −1 (0) ⊂ h−1 (0)}.
8 / 12
Seminar 13
Problem 2. Consider V1 = V2 = C [a, b] as a linear space over R,
A : V ⊃ D(A) = C 1 [a, b] → V , (Af )(x) = f 0 (x). Study A for
linearity and nd kerA, R(A), eigenvectors and eigenvalues.
Solution. Let us solve this problem in a very detailed manner so
you can see what is the full solution of it.
V = C [a, b] is a subset of linear space R[a,b] , and from the course
of calculus we know that sum of continuous functions is again a
continuous function, continuous function multiplied by a scalar is
again contiuous function, so V = C [a, b] is a linear subspace in
R[a,b] , so V = C [a, b] is itself a linear space.
Similarly D(A) = C 1 [a, b] is a linear subspace in V = C [a, b], we
only need to substitute ¾continuous function¿ by ¾function with a
continuous derivative¿.
(See the next page)
9 / 12
Seminar 13
For all f , g ∈ C 1 [a, b] and all C ∈ R we have
A(f + g ) = [x 7−→ (f + g )0 (x)] = [x 7−→ f 0 (x) + g 0 (x)] = [x 7−→
(Af )(x) + (Ag )(x)] = Af + Ag and A(Cf ) = [x 7−→ (Cf )0 (x)] =
[x 7−→ Cf 0 (x)] = [x 7−→ C (Af )(x)] = CAf so A is a homogenious
and additive function.
Summing up, we conclude that (A, D(A)) is a linear operator.
Let us now nd the kernel of A. By denitions f ∈ kerA ⇐⇒
(f ∈ D(A) and Af = 0) ⇒ f 0 (x) = 0 for all x ∈ [a, b]. As
f ∈ D(A) = C 1R[a, b] by the main theorem
of calculus we have
Rx
x
f (x) = f (a) + a f 0 (x)dx = f (a) + a 0dx = f (a) for all x ∈ [a, b],
so f is a constant function. Reversely, every constant function has
zero derivative hence lies in the kernel of A. We have proved that
kerA = {f : [a, b] → R|∃C ∈ R∀x ∈ [a, b] we have f (x) = C }.
Denitions of A, C [a, b], C 1 [a, b] imply that the range
R(A) = C [a, b].
(See the next page)
10 / 12
Seminar 13
Now let us nd eigenvectors and eigenvalues. Equation Af = λf is
equivalent to dierential equation f 0 (x) = λf (x). As we know, all
its solutions (assuming f ∈ C 1 [a, b]) are given by the formula
f (x) = Ce λx where λ, C are arbitrary numbers. But we are
interested not in all the solutions. As eigenvector is a nonzero vector
then C must not be equal to zero. But the case λ = 0 is possible,
and constants are eigenvectors corresponding to the eigenvalue 0.
Answer: all λ ∈ R are eigenvalues for A, and all eigenvectors fλ
corresponding to eigenvalue λ is given by fλ (x) = Ce λx where
0 6= C ∈ R is arbitrary constant that do not depend on λ. For all
λ ∈ R we have gm(λ) = 1.
Exercises at p. 2 and 3 - done, see seminar notes.
Problem 3. Consider V1 = V2 = D(A) = C (R, R) as a linear
space over R, (Af )(x) = sin(x) · f (x). Study A for linearity and
nd kerA, R(A) this was done on the seminar, see seminar notes.
Find eigenvectors and eigenvalues, for each λ ∈ σp (A) nd gm(λ) 11 / 12
this is given as a homework.
Homework 13
Problems to solve at home, deadline Monday January 30,
20:00
1a. Exercise at p.2. 1b. First (upper) exercise at p.3.
2a. Second (middle) exercise at p.3. 2b. Third (bottom) exercise at
p.3.
3a. Problem 1 at p.8. 3b. Problem 2 at p.9.
4a. Problem 3 at p.11 - linearity, kernel, range. 4b. Problem 3 at
p.11 - eigenvalues, eigenvectors, geometrical multiplicity.
Remark. You are welcome to use solutions of the above problems
from the seminar 13. There is no need to copy-paste or rewrite all
the text, it is enough to uderstand and explain.
5. Consider V1 = V2 = D(A) = RR as a linear space over R,
(Af )(x) = f (x) + f (−x) for all x ∈ R. Study A for linearity
explaining what you use at each step of solution (each equality
sign), and nd spaces kerA, R(A) and call their elements properly.
12 / 12
Introduction to
One-parameter semigroups and evolution
equations
Lecture 14: Necessary background from Functional Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
24+31 January + 7 February 2023
1 / 29
Additional structures on vector space
Remark. Lonely linear structure (i.e. vector operations: operation
of addition of vectors and operation of multiplying a vector by a
scalar) generates not so rich theory. This is why vector spaces (also
known as linear spaces) are very often used with additional
structures, most commonly used structures are:
I scalar product (allows to dene length of a vector and angle
between two vectors)
I norm (allows to dene length of a vector)
I metric (allows to dene distance between vectors)
I topology (allows to dene neighbourhood of a vector)
I convergence (allows to dene limit of a sequence of vectors)
We will briey touch all of them and discuss how they are related.
2 / 29
Additional structures on vector space: scalar product
Denition. We say that (V , h·, ·i) is an Euclidean space over the
eld F ∈ {R, C} i two conditions are satised:
1. V is a vector space over F (8 axioms).
2. Function h·, ·i = b : V × V → F is a scalar product in V , i.e. :
2.1. For all v ∈ V : b(v , v ) ≥ 0 non-negativity,
b(v , v ) = 0 ⇐⇒ v = 0 non-degeneracy
2.2. For all u, v , w ∈ V and all a ∈ F: b(au, v ) = ab(u, v ) homogenity in the rst argument, b(u + v , w ) = b(u, w ) + b(v , w )
additivity in the rst argument
2.3. For all u, v ∈ V : b(u, v ) = b(v , u) conjugate symmetry (x is
a complex conjugate of x )
Remark. These exioms imply for all u, v , w ∈ V and all a ∈ F:
b(u, av ) = ab(u, v ) conjugate homogenity in the second
argument, b(u, v + w ) = b(u, v ) + b(u, w ) additivity in the
second argument.
Exercise. Prove the CauchyBunyakovskySchwarz inequality:
p
p
|b(u, v )| ≤ b(u, u) b(v , v ).
3 / 29
Additional structures on vector space: scalar product
Exercise. Prove that function b is a scalar product on linear space V
over the eld F if
R
1. V = C ([a, b], R), F = R, b(x, y ) = ab x(t)y (t)dt ,
R
2. V = C ([a, b], C), F = C, b(x, y ) = ab x(t)y (t)dt
R
3. V = C ([a, b], C), F = R, b(x, y ) = ab x(t)y (t)dt
Denition. If b is a scalar product on the vector space V , then angle
angle(x, y ) between vectors x and y by denition is
b(x, y )
p
,
b(x, x) b(y , y )
p
and length of the vector x by denition is length(x) = b(x, x).
Remark. This is motivated by the case V = R2 where length and angle
angle(x, y ) = arccos p
are given by classical Euclid's plan geometry axioms and standard scalar
product is dened as
b(x, y ) = length(x) · length(y ) · cos(angle(x, y )).
Exercise. Consider V = C ([a, b], R) as a vector eld over R with scalar
R
product b(x, y ) = ab x(t)y (t)dt . Consider a triangle with vertexes 0,
x(t) = 1, y (t) = t . Find the length of the median of this triangle which
starts from the vertex y .
4 / 29
Additional structures on vector space: norm
Denition. We say that (V , k · k) is a normed vector space over
the eld F ∈ {R, C} i two conditions are satised:
1. V is a vector space over F (8 axioms).
2. Function k · k = n : V → R is a norm on V , i.e. :
2.1. For all v ∈ V : n(v ) ≥ 0 non-negativity,
n(v ) = 0 ⇐⇒ v = 0 non-degeneracy
2.2. For all u ∈ V and all a ∈ F: n(au) = |a|n(u) absolute
homogenity
2.3. For all u, v ∈ V : n(u + v ) ≤ n(u) + n(v ) the triangle
inequality
Exercise. Using the triangle inequality prove the so-called ¾second
triangle inequality¿:
|n(u) − n(v )| ≤ n(u + v ).
Denition. If (V , k · k) is a normed vector space, then the length
of the vector x is dened as length(x) = kxk.
5 / 29
Compatipility of linear and additional structures
Convergence of sequences (or, more genearal pseudotopology,
convergence of lters), topology and metric can de dened on
arbitrary set without any linear structure.
This is why to get convergence/topological/metric vector space it
is not enough to have just convergence/topology/metric on vector
space: we need to state additionally that topology/metric is
compatible with the linear structure. This compatibility usually is
expressed in the fact that vector operations (addition of vectors and
multiplication of a scalar and vector) are continuous in some sense.
Due to play of words ¾operations in V are continuous with respect
to metric d ¿ people often as a joke say ¾metric d respects
opeartions in V ¿, even in Russian: ìåòðèêà óâàæàåò âåêòîðíûå
îïåðàöèè.
Meanwhile, norm and scalar product use vector operations in their
denition and are compatible with vector operations by
construction.
6 / 29
Additional structures on vector space: metric
Denition. We say that (V , d) is a metric vector space over the
eld F ∈ {R, C} i three conditions are satised:
1.V is a vector space over F (8 axioms).
2. (V , d) is a metric space, i.e. d : V × V → R is a metric
(synonim: distance function) on V , i.e for all u, v , w ∈ V :
2.1. Non-negativity d(u, v ) ≥ 0, and non-degeneracy
d(u, v ) = 0 ⇐⇒ u = v
2.2. Symmetry d(u, v ) = d(v , u)
2.3. Triangle inequality d(u, v ) ≤ d(u, w ) + d(w , v )
3. Vector operations · and + are continuous with respect to
metric d in the following sence:
3.1. For each a0 ∈ F, each v0 ∈ V and each ε > 0 there exists
δ1 > 0 and δ2 > 0 such that conditions |a − a0 | < δ1 and
d(v , v0 ) < δ2 imply d(av , a0 v0 ) < ε.
3.2. For each v1 , v2 ∈ V and each ε > 0 there exists δ1 > 0 and
δ2 > 0 such that conditions d(u1 , v1 ) < δ1 and d(u2 , v2 ) < δ2
imply d(u1 + u2 , v1 + v2 ) < ε.
7 / 29
Additional structures on vector space: topology
Denition. We say that (V , τ ) is a topological vector space over
the eld F ∈ {R, C} i three conditions are satised:
1. V is a vector space over F (8 axioms).
2. (V , τ ) is a topological space, i.e. τ ⊂ 2V is a topology on V :
2.1. ∅ ∈ τ, V ∈ τ .
2.2. If A ∈ τ and B ∈ τ then A ∩ B ∈ τ .
2.3. If Aα ∈ τ for all α ∈ J then (∪α∈J Aα ) ∈ τ .
(Topological terminology: the set U ⊂ V is called open in topology
τ i U ∈ τ , the set U is called a neighbourhood of point x ∈ V i
x ∈ U ∈ τ .)
3. Vector operations · and + are continuous with respect to
topology τ in the following sence:
3.1. For each a0 ∈ F, each v0 ∈ V and each neighbourhood
U0 3 a0 v0 there exists ε > 0 and a neighbourhood U 3 v0 such
that conditions |a − a0 | < ε and v ∈ U imply av ∈ U0 .
3.2. For each v1 , v2 ∈ V and each neighbourhood U0 3 (v1 + v2 )
there exist neighbourhoods U1 3 v1 , U2 3 v2 such that conditions
u1 ∈ U1 and u2 ∈ U2 imply (u1 + u2 ) ∈ U0 .
8 / 29
Additional structures on vector space: topology
Denition. If (V , τ ) is a topological space then β ⊂ τ is called a
base of τ i for each U ∈ τ there exists β1 ⊂ β such that
U = ∪H∈β1 H . In other words: β is such a collection of open sets,
that every open set U can be represented as a union of some
subcollection β1 from this collection β . In that context often we say
that topology τ is generated by the base β .
Remark. Often it is dicult to give a simple description of all
elements of τ , but possible to give simple description of some base
of τ . Of course one toplogy can have many dierent bases. Also,
there is only one topology generated by a given base. Each
topology is a base for itself.
Theorem. Let V be a non-empty set and β ⊂ 2V . Then β
generates a topology (i.e. there exists a topology τ ⊂ 2V such that
β is a base of τ ) i two conditions hold:
1. β covers V , i.e. V = ∪H∈β H .
2. For each U1 , U2 ∈ β and each x ∈ U1 ∩ U2 there exists U3 ∈ β
such that x ∈ U3 ⊂ U1 ∩ U2 .
Exercise. Prove the above theorem (or nd the proof in a book). 9 / 29
Additional structures on vector space: convergence
Remark. We will consider only the notion of a convergent
sequence, i.e. sequence that has a limit. Actually not only
sequences can be said to have a limit. One can also consider
generalized sequences that are called nets and consider convergence
of lters. This leads to the notion of a pseudotopological space and
to an even more general notion of a generalized convergence space.
Interested students are referred to Wikipedia:
https://en.wikipedia.org/wiki/Convergence_space
As usual, we build mathematics from sets, and every new object is
introduced as a specic set with sertain properties. Let us recall
how this is done with functions: they are binary relations, and
binary relations are subsets of Cartesian product of two sets.
10 / 29
Recall some properties of binary relations
Denition. Binary relation r between sets A and B (also called
binary relation on A × B ) is by denition any subset r ⊂ A × B . We
say that a and b are in relation r and write arb i (a, b) ∈ r .
Denition. For relation r ⊂ A × B two sets have special names:
D(r ) = {a ∈ A|∃b ∈ B : arb} the domain
R(r ) = {b ∈ B|∃a ∈ A : arb} the range
Denition. Relation r ⊂ A × B is called:
1. Injective, or left-unique i ((a1 rb)&(a2 rb)) ⇒ (a1 = a2 ).
2. Right-unique i ((arb1 )&(arb2 )) ⇒ (b1 = b2 ).
3. Left-total i D(r ) = A.
4. Right-total, or surjective, or ¾onto¿ i R(r ) = B .
Denition. Right-unique relation f ⊂ A × B is called a partial
function (Russian: ÷àñòè÷íàÿ ôóíêöèÿ) with the domain D(f ) and
range R(f ) i D(f ) 6= A.
Denition. Right-unique and left-total relation f ⊂ A × B is called
a function f : A → B . Traditionally we write b = f (a) which means
the same as (a, b) ∈ f and afb .
11 / 29
Additional structures on vector space: convergence
Denition. A sequence (xn )n∈N of elements of the set X by
denition is a function x ∈ X N , dened as xn = x(n) for all n ∈ N.
Denition. A sequence x ∈ X N is called a subsequence of
sequence y ∈ X N i there exists such a function f : N → N that it
is strictly increasing (i.e. (a > b) ⇒ (f (a) > f (b)) for all for all
a, b ∈ N) and x = y ◦ f (i.e. xn = yf (n) for all n ∈ N).
Denition. Let X be a non-empty set. We say that a convergence
of sequences (X , C , L) is given on X i all conditions hold:
1. C is a collection of sequences of elements of X , i.e. C ⊂ X N
2. L is a binary relation L ⊂ X N × X , it relates sequence to a point.
3. C = D(L) and the following properties hold:
3a. If y ∈ C and x is a subsequence of y , then x ∈ C and
{a|xLa} ⊂ {a|y La}.
3b. If x is a constant sequence, i.e. xn = x1 for all n ∈ N, then
x ∈ C and {x1 } ⊂ {a|xLa}.
Remark. This is a formal denition. Let us explain in more detail
its informal meaning, see the next page.
12 / 29
Additional structures on vector space: convergence
Denition. Let X be a non-empty set. We say that a convergence
of sequences (X , C , L) is given on X i all conditions hold:
1. C is a collection of sequences of elements of X , i.e. C ⊂ X N .
C is the set of sequences that are called convergent in convergence
of sequences (X , C , L).
2. L is a binary relation L ⊂ X N × X , it relates sequence to a point
that is called one of the limit points of this sequence.
3. C = D(L) and the following properties hold:
3a. If y ∈ C and x is a subsequence of y , then x ∈ C and
{a|xLa} ⊂ {a|y La}. This means that limit point of a subsequence
is also a limit point of a sequence.
3b. If x is a constant sequence, i.e. xn = x1 for all n ∈ N, then
x ∈ C and {x1 } ⊂ {a|xLa}. This means that constant sequence
converges at least to this constant.
Denition. If x ∈ C , i.e. there exists such a ∈ X that xLa, then
one says that sequence x = (xn )n∈N is convergent to a, that a is
one of the limit points of x , and one writes lim xn 3 a.
n→∞
13 / 29
Additional structures on vector space: convergence
Remark. Note that limit of a sequence in general is not unique (relation
L is not right-unique), but if lim xn = 1 then it is common to write
n→∞
lim xn = a instead of lim xn 3 a, considering lim xn not as a singleton
n→∞
n→∞
n→∞
{a} but as the only element a of this singleton.
Denition. If X = V is a vector space over the eld F ∈ {R, C}, then
convergence of sequences (X , C , L) is called compatible with vector
operations · and + i for all sequences x, y and all scalars s ∈ F we have
• xLa and y Lb imply (x + y )L(a + b)
In other words: if lim xn 3 a and lim yn 3 b then
n→∞
n→∞
( lim xn + yn ) 3 (a + b).
n→∞
• xLa imply (s · x)L(s · a)
In other words: if lim xn 3 a then lim s · xn 3 s · a.
n→∞
n→∞
Remark. The term ¾convergence of sequences vector space¿ is not
widely used, meanwhile term ¾pseudotopological vector space¿ is widely
used. Pseudotopology describes not only convergence of sequences, but
also convergence of lters (which we will not consider in our introductory
level course). However, convergence with respect to base of lter is an
intersting concept which includes for functions R → R limits at a nite
14 / 29
point, at innity, one-sided limits and also Riemann integral.
Inheriting (standard generation) of structures
Remark. If you have specic structure, then naturally you have
also more general structure. On vector spaces there is the following
chain of generation (inheriting) of structures:
scalar product → norm → metric → topology → convergence
With this notion we can speak of e.g. Eucledian metric thinking
that it is a metric generated by norm generated by scalar product.
Also we can say e.g. ¾convergence in norm¿ meaning that
sequences converge in topology, generated by metric which is
generated by norm.
Remark. Two most simple steps of inheriting are from scalar
product to norm, and from norm to metric.
Exercise. Let V be a vector space. Prove that:
p
1. If h·, ·i is a scalar product in V then kxk =
hx, xi is a norm on V .
2. If k · k is a norm on V then d(x, y ) = kx − y k is a metric on V .
Remark. Two other steps (from metric to toplogy and from
topology to convergence) are not so straightforward and need some
supplementary results. Let us discuss them.
15 / 29
Inheriting (standard generation) of structures
Denition. In a metric space (M, d), for x ∈ M and r > 0 the set
Br (x) = {y ∈ M : d(x, y ) < r } is called an open ball with radius r
and center x , and the set Br (x) = {y ∈ M : d(x, y ) ≤ r } is called
a closed ball with radius r and center x .
Theorem. If (M, d) is a metric space then the set of all open balls
is a base of some topology on M . This topology is called the metric
topology generated by metric d .
Remark. This theorem says that the set is open in the metric
topology i this set can be represented as a union of open balls.
Exercise. Prove the above theorem. Hint: use theorem on p.9.
Denition. If (X , τ ) is a topological space, then the set A ⊂ X is
called closed in (X , τ ) i the complement of A is open, i.e. i
(X \ A) ∈ τ .
Exercise. Prove that in metric topology each open ball is an open
set, and each closed ball is a closed set.
Exercise. Suppose that (V , d) is a metric vector space and τ is the
topology generated by metric. Prove that (V , τ ) is a topological
vector space.
16 / 29
Inheriting (standard generation) of structures
Remark. Topology generates not only one, but two convergences of
sequences. One of them we will call standard convergence (which is
widely used) and the other is the subsequential convergence (which
should be taken to account sometimes).
Theorem. Let (X , τ ) be a topological space, and L ⊂ X N × X be a
binary relation dened as follows: xLa (in other words: limn→∞ xn 3 a) i
for all U ⊂ X such that a ∈ U ∈ τ there exists n0 ∈ N such that for all
n ≥ n0 we have xn ∈ U (in other words: outside any open neighbourhood
of a there are only nitely many items of the sequence (xn )). Dene
C = D(L). Then (X , C , L) is a convergence of sequences on X .
Denition. Above (X , C , L) is called the standard convergence of
sequences in topological space (X , τ ). Elements of the set {a|xLa} are
called limits (and limit points) of sequence x .
Exercise. Prove the above theorem.
Exercise. Prove that it is enough to check the property not for all
neighbourhoods but only for neighbourhoods from any base of topology.
I.e. let (X , τ ) be as above and β ⊂ τ is a base of τ . Dene xL1 a i for
all U ⊂ X such that a ∈ U ∈ β there exists n0 ∈ N such that for all
n ≥ n0 we have xn ∈ U . Then L1 = L.
17 / 29
Inheriting (standard generation) of structures
Theorem. Let (X , τ ) be a topological space, and S ⊂ X N × X be
a binary relation dened as follows: xSa i x has a subsequence y
such that y La (in other words: limn→∞ yn 3 a). Dene C = D(S).
Then (X , C , S) is a convergence of sequences on X .
Denition. Above (X , C , S) is called the subsequential
convergence of sequences in topological space (X , τ ). Elements of
the set {a|xSa} are called subsequential limits (and subsequential
limit points, partial limits) of sequence x .
Exercise. Prove that if (X , τ ) is a topological vector space then the
standard convergence of sequences and subsequential convergence
of sequences both are compatible with operations · and +.
Exercise. Let (X , d) be a metric space, τ be the metric topology
generated by d , and (X , C , L) be the standard convergence of
sequences in topological space (X , τ ). Prove that relation L is
right-unique (i.e. sequence can have at most one limit).
Remark. Unique limit of sequence is also a property of T2
topological spaces, that are also called Hausdor spaces.
18 / 29
Additional structures on vector space: reverse inheriting
Remark. The reverse way is not possible in the general setting:
scalar product 6← norm 6← metric 6← topology 6← convergence
However, there are some assumptions under which this reverse way
is possible on the each arrow. For example:
Theorem. The norm in real or complex vector space V is
generated by a scalar product i the following parallelogram law
holds for all x, y ∈ V :
kx − y k2 + kx + y k2 = kxk2 + kxk2 + ky k2 + ky k2 ,
which means that the sum of squares of lengths of all diagonals of
any parallelogram is equal to sum of squares of lengths of its sides.
This scalar product is given by the following polarization identity:
kx + y k2 − kxk2 − ky k2
for real vector spaces, and
2
kx + y k2 − kx − y k2
kx − iy k2 − kx + iy k2
hx, y i =
+i
4
4
hx, y i =
for complex vector spaces.
19 / 29
Additional structures on vector space: reverse inheriting
scalar product 6← norm 6← metric 6← topology 6← convergence
Remark.All arrows suer from the possible non-generability, for
example:
• Norm kxk = supt∈[0,1] |x(t)| on C [0, 1] is not generated by any
scalar product. This norm generates metric that generates topology
that generates convergece which is known as uniform convergence
of functions.
R 1 |x(t)−y (t)|
• Metric d(x, y ) = 0 1+|x(t)−y
(t)| dt on the space M[0, 1] of all
Lebesgue measurable functions is not generated by any norm. This
metric generates topology that generates convergece which is
known as convergence of functions in measure.
• Topology of pointwise convergence (see https://en.
wikipedia.org/wiki/Topology_of_pointwise_convergence)
on the set R[0,1] is not generated by any metric. This topology
generates convergece which is known as pointwise convergence of
functions.
• Convergence almost everywhere on the space M[0, 1] of all
Lebesgue measurable functions is not generated by any topology. 20 / 29
Additional structures on vector space: reverse inheriting
scalar product 6← norm 6← metric 6← topology 6← convergence
Remark.All arrows suer from the possible non-generability,
however in the two left arrows there is unique generability:
• If the norm is generated by a scalar product then this scalar
product is unique
• If the metric is generated by a norm then this norm is unique
Remark. The right two arrows suer also from multiple
generability:
• If (X , d) is a metric space then metric r (x, y ) = 2d(x, y )
generates the same topology as metric d . If d was compatible with
linear structure then r is compatible as well.
• On the space D = Cc∞ (R, R) of so-called test functions there is a
standard convergence of sequences but there are several dierent
topologies that generate the same convergence. All of them are
compatible with linear structure. The space D is remarkable
because it is used to construct so-called generalized functions (also
known as distributions) https://en.m.wikipedia.org/wiki/
Spaces_of_test_functions_and_distributions.
21 / 29
Additional structures on vector space: reverse inheriting
topology 6← convergence
Remark. Let us consider one more particular example of multiple
generability. In that case convergence and topology are not
compatible with linear structure, but they are simple so it is easy to
analyse the situation.
Theorem. On the set X = R consider two topologies. The rst
one τ1 = 2R is the so-called discrete topology. The second one τ2 is
the so-called co-countable topology, by denition A ∈ τ2 i X \ A is
nite, countable or equal to X . Then τ1 and τ2 generate the same
convergence that is given by the following rule: (xn ) is convergent
i (xn ) is eventually constant, i.e. there exists such n0 ∈ N that for
all n ≥ n0 we have xn = xn0 .
Exercise. Prove the theorem. Do not forget to prove that τ1 and τ2
are indeed topologies.
Exercise. Prove that τ1 and τ2 are not compatible with operations
· and + in R.
22 / 29
Norm of a linear operator
Denition. Let (V1 , k · k1 ) and (V2 , k · k2 ) be normed vector
spaces over the same eld R or C. Let A : V1 ⊃ D(A) → V2 be a
linear operator. Then by denition the norm of A is given by the
following equality:
kAk =
kAxk2
.
06=x∈D(A) kxk1
sup
Exercise. Under the above assumptions, prove that
kAxk2
=
sup
kAxk2 =
06=x∈D(A) kxk1
kxk1 ≤1,x∈D(A)
sup
=
sup
kxk1 =1,x∈D(A)
kAxk2 = inf{C ∈ R : ∀x ∈ D(A) : kAxk2 ≤ C kxk1 }.
This equality gives four equivalent denitions of the norm of a
linear operator.
23 / 29
Norm of a linear operator
Remark. Let (V1 , k · k1 ) and (V2 , k · k2 ) be normed vector spaces over
the same eld R or C. Let A : V1 ⊃ D(A) → V2 be a linear operator.
The algorithm of nding kAk is the following:
1. Find such C that ∀x ∈ D(A) : kAxk2 ≤ C kxk1 . This gives us
kAxk2
2
sup kAxk
kxk1 ≤ C hence kAk =
kxk1 ≤ C . So kAk ≤ C .
06=x∈D(A)
2a. If it is possible, nd such vector x0 ∈ D(A) that kAx0 k2 = C kx0 k1 .
kAx0 k2
2
Then kAk = sup kAxk
kxk1 ≥ kx0 k1 = C . So kAk ≥ C . In case 2a we
06=x∈D(A)
say that the norm of A is reached on the vector x0 .
2b. If you failed to success in the case 2a, then nd such a sequence
n k2
xn ∈ D(A) that kAxn k2 < C kxn k1 but lim kAx
kxn k1 = C . Then
kAk =
sup
06=x∈D(A)
n→∞
kAxk2
kAxn k2
≥
lim
n→∞
kxk1
kxn k1 = C
. So kAk ≥ C . In case 2b we
say that the norm of A is reached on the sequence (xn ).
Remark. If indeed kAk = C then the case 2b is always possible, this
follows directly from the denition of supremum. Meanwhile case 2a
sometimes is not possible at all, or it is not easy to nd such a vector x0 .
Remark. It follows directly from the denition of supremum that
n k2
kAk = ∞ i ∃ a sequence of vectors xn ∈ D(A) that limn→∞ kAx
kxn k1 = ∞.
24 / 29
Seminar 14
On the seminar 14, we solved the following problems:
1. Consider V = C [0, 1] endowed with the uniform norm (also
called L∞ -norm) kxk = supt∈[0,1] |x(t)|. Consider operator
A : V = D(A) → V dened by the equality (Ax)(t) = tx(0) for all
t ∈ [0, 1]. Check that A is linear, nd kerA, R(A) and kAk.
Answer: kerA = {x ∈ C [0, 1] : x(0) = 0},
R(A) = {[t 7→ kt] ∈ C [0, 1] : k ∈ C}, kAk = 1.
2. Consider V = C [0, 1] endowed with the L1 -norm
R1
kxk = 0 |x(t)|dt . Consider operator A : V = D(A) → V dened
by the equality (Ax)(t) = tx(0) for all t ∈ [0, 1]. Check that A is
linear, nd kerA, R(A) and kAk.
Answer: kerA = {x ∈ C [0, 1] : x(0) = 0},
R(A) = {[t 7→ kt] ∈ C [0, 1] : k ∈ C}, kAk = ∞.
25 / 29
Seminar 14
1 -norm
3. Consider V1 = C 1 [0, 1] endowed with the W∞
0
kxkW∞
1 = sup |x(t)| + sup |x (t)|.
t∈[0,1]
t∈[0,1]
Consider V2 = C [0, 1] endowed with the L∞ -norm
kxkL∞ = sup |x(t)|.
t∈[0,1]
Consider A : V1 = D(A) → V2 , (Ax)(t) = x 0 (t) for all t ∈ [0, 1].
Check that A is linear, nd kerA, R(A) and kAk.
Answer: kerA = {[t 7→ k] ∈ C [0, 1] : k ∈ C}, R(A) = C [0, 1],
kAk = 1.
4. Consider A, V1 , V2 as in problem 3, but V1 is endowed with
L∞ -norm. The same question as in problem 3.
Answer: the kernel and range as in problem 3, but kAk = ∞.
26 / 29
Seminar 14
5 (problem for self-study). Consider V = C [0, 1] over R endowed
with the uniform norm (also called L∞ -norm)
kxk = supt∈[0,1] |x(t)|. Consider operator A : V = D(A) → V
R1
dened by the equality (Ax)(t) = 0 x(u) sin(t + u)du for all
t ∈ [0, 1]. Check that A is linear, nd kerA, R(A) and kAk.
Answer:
R1
R1
kerA = {x ∈ C [0, 1]| 0 x(t) sin(t)dt = 0, 0 x(t) cos(t)dt = 0},
R(A) = {[t 7→ α cos(t) + β sin(t)] ∈ C [0, 1]|α, β ∈ R}.
kAk = 2 sin(0.5).
Explanation why kAk = 2 sin(0.5):
Z 1
kAxk = sup
x(u) sin(t + u)du ≤
t∈[0,1]
0
Z 1
≤ sup
t∈[0,1] 0
Z 1
|x(u)|·| sin(t+u)|du ≤ sup |x(u)| sup
u∈[0,1]
t∈[0,1] 0
·| sin(t+u)|du =
= kxk sup | cos(t) − cos(t + 1)| = kxk2 sin(0.5).
t∈[0,1]
27 / 29
Seminar 14
So we have proved that for all x 6= 0 we have kAxk
kxk ≤ 2 sin(0.5)
hence kAk = sup kAxk
kxk ≤ 2 sin(0.5).
x6=0
Now let us try to nd such x0 that kAx0 k = 2 sin(0.5)kx0 k. Let us
try x0 (t) ≡ 1, then kx0 k = 1.
R1
(Ax0 )(t) = 0 sin(t + u)du = cos(t) − cos(t + 1). Analysis of this
function shows that maxt∈[0,1] | cos(t) − cos(t + 1)| = 1 so
kAx0 k = 1.
kAx0 k
Then kAk = sup kAxk
kxk ≥ kx0 k = 2 sin(0.5).
x6=0
Finally, we have proved that kAk = 2 sin(0.5).
28 / 29
Homework 14
Problems to solve at home (deadline 20:00 February 16):
1. 1a) Exercise on p. 3, considering both spaces over R and C, 1b)
rst exercise at p.4, 1c) last exercise at p.4, 1d) exercise at p.5.
2. 2a) Exercise on p. 9, 2b) rst exercise at p.16, 2c) second
exercise at p.16, 2d) rst exercise at p.17
3. 3a) exercise at p.15, 3b) last exercise at p.18
4. Exercise at p.23
5. Consider V = C [0, 1]. Consider operator A : V = D(A) → V
dened by the equality (Ax)(t) = (2t − 1)x(t) for all t ∈ [0, 1].
Check that A is linear, nd kerA, R(A) and kAk assuming that V is
R1
endowed with the 5a) L1 -norm kxk = 0 |x(t)|dt , 5b) L2 -norm
R
1/2
1
kxk = 0 |x(t)|2 dt
, 5c) L∞ -norm kxk = supt∈[0,1] |x(t)|, 5d
weighted L∞ -norm kxk = supt∈[0,1] |x(t)w (t)|, where w is called
the weight function; consider w (t) = e t .
29 / 29
Introduction to
One-parameter semigroups and evolution
equations
Lecture 15: Necessary background from Functional Analysis
Ivan Remizov
HSE Nizhny Novgorod, Russia
14 February 2023
1 / 20
Classication of points in a metric space
Denition. In a metric space (V , d) open ball with center x ∈ V
and radius r > 0 is the set Br (x) = {y ∈ V |d(x, y ) < r }.
Denition. Let (V , d) be metric space and A ⊂ V be a subset.
The point x ∈ V is called:
1. interior point (Russian: âíóòðåííÿÿ òî÷êà) for A i there exists
an open ball B such that x ∈ B ⊂ A.
2. exterior point (Russian: âíåøíÿÿ òî÷êà) for A i there exists an
open ball B such that x ∈ B ⊂ (V \ A).
3. isolated point (Russian: èçîëèðîâàííàÿ òî÷êà) for A i x ∈ A
and there exists an open ball B such that B ∩ A = {x}.
4. boundary point (Russian: ãðàíè÷íàÿ òî÷êà) for A i for each
open ball B 3 x there exist two points x1 ∈ A, x2 ∈ (V \ A) such
that x1 ∈ B 3 x2 .
5. adherent point (Russian: òî÷êà ïðèêîñíîâåíèÿ) for A i for
each open ball B 3 x there exists x1 ∈ A such that x1 ∈ B .
6. limit point (Russian: ïðåäåëüíàÿ òî÷êà) for A i for each open
ball B 3 x there exists x1 ∈ A such that x1 ∈ B and x 6= x1 .
2 / 20
Interior, exterior, closure and frontier in a metric space
Denition. If V is a metric space and A ⊂ V then by denition:
1. Int(A), the interior of A, is the set of all interior points of A.
2. Ext(A), the exterior of A, is the set of all exterior points of A.
3. Fr (A) = ∂A, the frontier of A, also called the boundary of A, is
the set of all boundary points of A.
4. Cl(A) = A, the closure of A, is the set of all adherent points of A
Example. Suppose V = [0, 2], d(x, y ) = |x − y |, A = (0, 1]. Then
Int(A) = (0, 1), Ext(A) = (1, 2], Fr (A) = {0} ∪ {1}, A = [0, 1].
Denition. The set A ⊂ V in a metric space V is called open if A
can be represented as a union of open balls.
Denition. The set A ⊂ V in a metric space V is called closed if
its complement V \ A is open, i.e. V \ A can be represented as a
union of open balls.
Lemma. If V is a metric space and A ⊂ V then:
1. Int(A) is the biggest open subset of A.
2. A and Fr (A) are closed sets.
3. A = Fr (A) t Int(A).
4. (A is closed) ⇐⇒ (A = A).
3 / 20
Limit and closure in a metric space
Denition. Let (V , d) be metric space and (xn ) ⊂ V 3 x . We say
that lim xn = x i lim d(xn , x) = 0.
n→∞
n→∞
Denition. Let V be metric space and A ⊂ V be a subset. The
closure of A is denoted as A and can be dened equivalently in the
following variants:
1. A is the intersection of all closed supsets of A.
2. A is the smallest closed supset of A, i.e. if F is closed and A ⊂ F
then A ⊂ F .
3. A is the set of all adherent points of A.
4. A is the set of all limits of sequences in A, i.e. x ∈ A i there
exists (xn ) ⊂ A such that limn→∞ xn = x .
Denition. Let (V , d) be metric space and A ⊂ V , B ⊂ V . The
fact that A is dense in B can be equivalently dened in three ways:
1. B ⊂ A.
2. ∀b ∈ B ∀ε > 0 exists a point a ∈ A such that d(a, b) < ε.
3. ∀b ∈ B exists a sequence (an ) ⊂ A such that limn→∞ an = b.
Denition. Let (V , d) be a metric space and A ⊂ V . We say that
4 / 20
A is everywhere dense i A is dense in B = V .
Continuous and linear functions in metric spaces
Denition. Let (M1 , d1 ) and (M2 , d2 ) be metric spaces and
∅ 6= D(f ) ⊂ M1 . Function f : D(f ) → M2 is called:
1. Continuous in point x ∈ D(f ) i
((xn ) ⊂ D(f ), limn→∞ xn = x )⇒ (limn→∞ f (xn ) = f (x)).
2. Uniformly continuous on D(f ) i for all ε > 0∃δ > 0 such that
(x, y ∈ D(f ), d1 (x, y ) < δ ) imply that d2 (f (x), f (y )) < ε.
Denition. The subset of metric space is called bounded if it is a
subset of a ball. (Recall that a ball by denition has nite radius.)
Theorem. Let V1 , V2 be normed spaces and A : V1 ⊃ D(A) → V2
be a linear operator. Then the following conditions are equivalent:
1. A is continuous at 0.
2. A is continuous in each point of D(A).
3. A is uniformly continuous on D(A).
4. kAk < ∞.
5. The set {Ax|x ∈ D(A), kxk ≤ 1} is bounded in V2 .
6. A maps each bounded subset of D(A) into a bounded subset of
V2 . (In that case A is called a bounded operator.)
Remark. This easy-to-prove theorem is very useful in linear theory. 5 / 20
Separable and complete metric spaces
Denition. Metric space (V , d) is called separable i it has a
countable, everywhere dense subset.
Denition. Let (V , d) be a metric space. Sequence (xn ) ⊂ V is
called fundamental i ∀ε > 0∃n ∈ N : (k, m ≥ n) ⇒ d(xm , xk ) < ε.
Exercise. Let (V , d) be a metric space. Prove that sequence
(xn ) ⊂ V is fundamental i ∀ε > 0 there exists an open ball B with
radius ε such that outside B there are only nite number of
members of sequence (xn ), i.e. ∃n ∈ N : (k ≥ n) ⇒ (xk ∈ B).
Denition. Metric space (V , d) is called complete i every
fundamental sequence (xn ) ⊂ V has a limit in V . In other words, i
((xn ) ⊂ V is fundamental) ⇒ (∃x ∈ V : limn→∞ xn = x ).
Completion theorem. Let (V , d) be a metric space (complete or
not). Then there exists complete metric space (M, ρ) (which is
called a completion of (V , d)) and a mapping f : V → M such that:
1. f is injective.
2. for all x, y ∈ V we have d(x, y ) = ρ(f (x), f (y )).
(1 and 2 mean that f is an isomorphism of (V , d) and (f (V ), ρ))
3. The set f (V ) is dense in M .
6 / 20
Banach spaces and Hilbert spacee
Theorem. Let (V , d) be a metric space and let (M1 , ρ1 ) and
(M2 , ρ2 ) be two dierent completions of (V , d). Then (M1 , ρ1 ) and
(M2 , ρ2 ) are isomorphic as metric spaces.
Remark. It follows from the above theorem that if (V , d) is a
complete metric space then it is isomorphic to all its completions.
Denition. Banach space is a normed space which is complete as
metric space. Hilbert space is an Euclidean space which is complete
as metric space.
Theorem. If metric space is a normed space then its completion
can be chosen also to be a normed space, and metric isomorphism
on a dense subset can be chosen to be linear isomorphism.
Corollary. Every normed space can be considered as dense linear
subspace in some (unique up to an isomorphism) Banach space.
Every Euclidean space can be considered as dense linear subspace
in some (unique up to an isomorphism) Hilbert space.
Remark. It follows directly from the denitions that each Hilbert
space is also a Banach space. The reverse is not true because not
all norms can be generated by a scalar product.
7 / 20
Schauder basis and orthonormal basis in Hilbert space
Denition. Schauder basis in a normed vector space V over
F ∈ {R, C} by denition is a sequence (en ) ⊂ V such that for all
x ∈ V there exists a unique sequence (xn ) ⊂ F such that
x=
∞
X
n=1
xn en i.e. lim
k→∞
x−
k
X
xn e n = 0 .
(1)
n=1
Denition. Schauder basis (en ) is called unconditional i the above
series converges unconditionally, i.e. i every permutation of (en ) is
a Schauder basis with coecients that are permutation of (xn ).
Theorem. If in a normed space a Schauder basis exists then this
normed space is separable.
Theorem (Eno 1972). There is an example of separable Banach
space which do not have Schauder basis.
Theorem. Each separable Hilbert space H has an orthonormal
basis (also called Hilbert basis), i.e. unconditional Schauder basis
(en ) ⊂ H such that ken k = 1 and hei , ej i = 0 for i P
6= j . For all
x ∈ H formula (1) reads as the Fourier series
x= ∞
n=1 hx, en i en ,
P
2
and the Parseval's identity holds: kxk2 = ∞
|
hx,
e
ni | .
n=1
8 / 20
Product of normed and Euclidean spaces
Denition. If V1 and V2 are linear spaces over the same eld R or
C, then standard scalar product (or norm) on the linear space
V1 × V2 can be introduced as follows, assuming that scalar product
(or norm) are already dened in V1 , V2 :
h(x1 , x2 ), (y1 , y2 )iV1 ×V2 = hx1 , y1 iV1 + hx2 , y2 iV2 ,
1/2
.
k(x1 , x2 )kV1 ×V2 = kx1 k2V1 + kx2 k2V2
Exercise. Prove that:
1. The above-dened scalar product on V1 × V2 is indeed a scalar
product.
2. The above-dened norm on V1 × V2 is indeed a norm.
3. The above-dened norm is generated by the above-dened scalar
product.
4. If V1 , V2 are Banach spaces then V1 × V2 with the above-dened
norm is a Banach space.
5. If V1 , V2 are Hilbert spaces then V1 × V2 with the above-dened
scalar product is a Hilbert space.
9 / 20
Closure of a linear operator
Remark. As we know, for a linear operator A : V1 ⊃ D(A) → V2
the graph G (A) = {(x, Ax) : x ∈ D(A)} ⊂ V1 × V2 is a linear
subspace in V1 × V2 . Also we know that A = G (A) in a profound
approach to denition of a function A.
Denition. Linear operator A : V1 ⊃ D(A) → V2 between normed
spaces V1 and V2 is called:
1. closed (Russian: çàìêíóòûé) i the graph G (A) of A (or A, in
other words) is a closed subset in V1 × V2 .
2. closable (Russian: çàìûêàåìûé) i the closure A of the set A in
V1 × V2 is a graph of a linear operator (which is called the closure
of A and denoted as A).
3. densely dened (Russian: ïëîòíî îïðåäåë¼ííûé) i D(A) is
dense in V1 , i.e. i D(A) = V1 .
Remark. In the item 2 it is important that relation A ⊂ V1 × V2 is
right-unique, i.e. if (x, y1 ) ∈ A and (x, y2 ) ∈ A then y1 = y2 .
Banach closed graph theorem. Consider linear operator
A : V1 = D(A) → V2 between Banach spaces V1 and V2 . Then A is
continuous i A is closed. (It is important here that D(A) = V1 .) 10 / 20
Linear functionals
Let us discuss some properties
of linear functionals
11 / 20
Linear functionals on a vector space
Remark. Recall that linear functional is a linear function that is dened
on a vector space and takes only number values.
Denition. Let V be a vector space over the eld F ∈ {R, C}. Then the
0
algerbraic dual space of V is denoted as V and dened as follows:
0
V
V = f ∈ F |f is linear . It is imporatant here that D(f ) = V .
Exercise. Prove that V00 is a linear subspace in FV . Hence V 00 = (V 0 )0 is
a linear subspace in FV because we can consider linear space U = V 0 ,
then due to V 00 = U 0 the set V 00 will be a linear subspace in FU .
Exercise. Let V be a0 linear space over the eld F. Let us consider
function J : V → FV dened for each x ∈ D(J) = V as
0
J(x) = [V 0 3 w 7→ w (x) ∈ F] ∈ FV , i.e. (J(x))(w ) = w (x).
Using denitions and lemmas proven before, please prove that:
a) for each x ∈ V we have J(x) ∈ V 00 , hence J(V ) ⊂ V 00 .
b) J is linear and injective. (Hint: use Hamel basis for injectivity)
c) J is a linear isomorphism of the space V and space J(V ) ⊂ V 00 .
Remark. If dimV < ∞ then dimV = dimV 0 = dimV 00 ; in linear algebra
course J is called the canonical isomorphism of V onto V 00 = J(V ).
Remark. Remember, I told you that each linear space can be represented
as a space of functions? This is how it works.
12 / 20
Properties of linear functionals on a normed space
Remark. The denition of the norm of a linear functional is a
particular case of denition of the norm of a linear operator. Linear
functionals are quite often everywhere dened, so if D(f ) is not
explicitly given then we assume that D(f ) is the whole space.
Denition. Let V be a normed vector space over the eld
F ∈ {R, C}. Then the norm kf k of a linear functional
f : V ⊃ D(f ) → F is dened as
kf k
definition
=
|f (x)| please prove
=
sup
|f (x)|.
x∈D(f ),kxk=1
06=x∈D(f ) kxk
sup
Remark. Linear functional f is continous i kf k < ∞. (see p.5)
Denition. Let V be a normed vector space over the eld
F ∈ {R, C}. Then the continuous dual space V ∗ of the normed
vector space V is dened as follows: V ∗ = {f ∈ V 0 : kf k < ∞}.
Exercise. Prove that V ∗ is a linear subspace in V 0 .
Exercise. Prove that V ∗ is a normed vector space with the norm
kf k dened above. This norm is called the standard norm on V ∗ .
13 / 20
Properties of linear functionals and hyperplanes
Denition. Vector hyperplane in a vector space is a subspace of
codimension 1.
Lemma. Vector hyperplanes are exactly the kernels of
non-everywhere-zero linear functionals.
Denition. Ane hyperplane in a vector space V is a set dened
as {x + y |y ∈ V1 } where x ∈ V is a vector, and V1 ⊂ V is a
subspace of codimension 1, where V1 is dened uniquely and x is
dened up to arbitrary additive perturbation by vectors from V1 .
Lemma. Ane hyperplanes are exactly the sets represendable in
the form {z|f (z) = a} where f is a non-everywhere-zero linear
functional, and a is a number. Here in the previous denition we
have V1 = kerf and a = f (x).
Lemma. There are only two possible variants:
a) Ane hyperplane is a closed proper nowhere dense subset in V ,
this case corresponds to 0 6= kf k < ∞.
b) Ane hyperplane is an everywhere dense subset in V , this case
corresponds to kf k = ∞.
14 / 20
Properties of linear functionals
Denition. In a normed space V the distance between the set S
and the point x ∈/ S is dened as d(x, S) = inf y ∈S kx − y k.
Exercise. Prove that d(x, S) = d(x, S) where S is the closure of S .
Exercise. In a normed space V for each x ∈ V and each closed
subset S ⊂ V the following holds: x ∈ S ⇐⇒ d(x, S) = 0.
Exercise (geometrical meaning of the norm of a linear
functional). Let f be a continuous linear functional on a normed
space V . Then the distance between point zero and hyperplane
{x ∈ V |f (x) = 1} is given by the formula:
d(0, {x ∈ V |f (x) = 1}) =
1
.
kf k
Hahn-Banach extension theorem. Let V be a normed space over
the eld F and V1 be a linear subspace. Suppose that there is a
continuous linear functional f dened on D(f ) = V1 . Then there
exists such an everywhere dened linear functional g : V → F such
that f (x) = g (x) for all x ∈ V1 and kf k = kg k.
Exercise. The extension in the Hahn-Banach may be not unique.
Please provide an example.
15 / 20
Properties of linear functionals
RieszFrechet representation theorem. For each continuous
everywhere dened linear functional f on a Hilbert space H with
scalar product h·, ·i there exists unique vector yf ∈ H such that
f (x) = hx, yf i for all x ∈ H . Moreover, kf k = kyf k.
Remark. The above theorem means that J : f 7→ yf is a linear
isomorphism of H ∗ and H . Moreover, J preserves the norm, so H ∗
and H are isomorphic as normed spaces. But the norm in Euclidean
spaces denes the scalar product uniuely, so H ∗ and H are
isomorphic as Euclidean spaces.
Theorem. If V is a normed space (complete or not complete),
then V ∗ is always a Banach space.
Denition. Banach space is called reexive i for canonical linear
isomorphism J : V → V 00 we have that J(V ) = V ∗∗ ⊂ V 00 and
kxkV = kJxkV ∗∗ for each x ∈ V . In other words: V is reexive i
V is canonicaly isomorphic to V ∗∗ .
Remark. Words ¾dual space¿ and ¾adjoint space¿ (Russian:
¾ñîïðÿæ¼ííîå ïðîñòðàíñòâî¿) have the same meaning.
16 / 20
Properties of linear functionals
Theorem. Dual spaces for some normed spaces:
I (`p )∗ = `q for each 1 < p < ∞ where 1/p + 1/q = 1. If X ∈ (`p )∗
P∞
then there exists such x = (xn )∞
n=1 ∈ `q that X (y ) =
n=1 xn yn for
each y = (yn )∞
∈
`
,
and
kX
k
=
kxk
.
p
n=1
I (`1 )∗ = `∞ . If X ∈ (`1 )∗ then there exists such x = (xn )∞
n=1 ∈ `∞
P
∞
x
y
for
each
y
=
(y
)
∈
`
,
and
kX
k = kxk.
that X (y ) = ∞
n n=1
1
n=1 n n
∗
∗
∞
I (c0 ) = `1 . If X ∈ (c0 ) then there exists such x = (xn )n=1 ∈ `1
P
∞
that X (y ) = ∞
n=1 xn yn for each y = (yn )n=1 ∈ c0 , and kX k = kxk.
I (Lp (a, b))∗ = Lq (a, b) for each 1 < p < ∞ where 1/p + 1/q = 1. If
X ∈ (Lp (a, b))∗ then there exists such x ∈ Lq (a, b) that
Rb
X (y ) = a x(t)y (t)dt for each y ∈ Lp (a, b), and kX k = kxk.
I (L1 (a, b))∗ = L∞ (a, b).
I (C [a, b])∗ is the set of all bounded Borel measures on [a, b]. If
Rb
X ∈ (C [a, b])∗ then there exists such µ that X (y ) = a y (t)µ(dt)
for each y ∈ C [a, b], and kX k = kµk.
See textbook V.I.Bogachev, O.G.Smolyanov. Real and functional
analysis, for details.
17 / 20
Seminar 15
On the Seminar 15, we will solve the following problems:
1. Prove that C [a, b] with the uniform norm, also called L∞ -norm
kxk = supt∈[a,b] |x(t)| is a Banach space.
R
2. Prove that C [a, b] with the L1 -norm 01 |x(t)|dt is not a Banach
space (nd fundamental sequence which do not have a limit).
3. Consider bounded linear operator A : V1 = D(A) → V2 between
normed spaces V1 and V2 . Prove that A is closed.
18 / 20
Homework 15
Problems to solve at home (deadline 20:00 February 24):
1. Prove theorem on p.5; 2. Exercise on p.6; 3. Exercise on p.9.
4. Consider linear operator A : V1 ⊃ D(A) → V2 between normed
spaces V1 and V2 . Prove that: a) (A, D(A)) is closed i
((xn ) ⊂ D(A), xn → x, Axn → y ) ⇒ (x ∈ D(A), y = Ax).
b) (A, D(A)) is closable i conditions
(xn ) ⊂ D(A), xn → x, Axn → y , (xn0 ) ⊂ D(A), xn0 → x, Axn0 → y 0
imply y = y 0 .
c) if (A, D(A)) is closable then
D(A) = {x ∈ V1 |∃(xn ) ⊂ D(A) : xn → x, ∃ limn→∞ Axn } and
Ax = limn→∞ Axn for all x ∈ D(A) where x and xn are from the
above formula for D(A).
5. Consider Hilbert space H = L2 [0, 1] with the L2 -norm
R
1/2
1
kxk = 0 |x(t)|2 dt
. Consider operator A : H ⊃ D(A) → H
dened on D(A) = C [0, 1] by equality (Ax)(t) = x(1)t for all
t ∈ [0, 1]. Is (A, D(A)) linear, bounded, closable, closed?
19 / 20
Additional problems Homework 15
Problems to solve at home (each problem costs 2 points for
full solution):
6. Exercises at p. 12
7. Exercises at p. 13
8. Exercises at p. 15
20 / 20
Introduction to
One-parameter semigroups and evolution
equations
Lesson 16: Basic denitions and facts of C0-semigroups theory
Ivan Remizov
HSE Nizhny Novgorod, Russia
28 February 2023
1 / 21
Useful propositions
Proposition (short wording: if A is invertible then A and A−1
are closed or not closed simultaneously). Consider injective
linear operator A : V1 ⊃ D(A) → V2 between normed spaces V1
and V2. Then (A, D(A)) is closed i (A−1, R(A)) is closed.
Proof. i) As A is injective then it is a bijection between D(A) and
R(A), so operator (A−1 , D(A−1 )) with D(A−1 ) = R(A) is
well-dened. Let us also mention that for all x ∈ V1 and all y ∈ V2
we have (x, y ) ∈ G (A) ⇐⇒ (y , x) ∈ G (A−1) and k(x, y )kV1×V2 =
(kxk2V1 + ky k2V2 )1/2 = (ky k2V2 + kxk2V1 )1/2 = k(y , x)kV2 ×V1 .
ii) Condition that (A, D(A)) is closed by denition means that the
set G (A) ⊂ V1 × V2 is a closed subset of normed space V1 × V2
with respect to norm on V1 × V2. This means: if (xn , yn ) ∈ G (A)
and n→∞
lim k(xn − x, yn − y )kV1 ×V2 = 0 then (x, y ) ∈ G (A).
iii) Independently of closedness of A or A−1 we have
(xn , yn ) ∈ G (A) ⇐⇒ (yn , xn ) ∈ G (A−1 ) and
k(xn − x, yn − y )kV1 ×V2 = k(yn − y , xn − x)kV2 ×V1 . This allows to
take the denition of closedness of (A, D(A)) and rewrite it as the
2 / 21
denition of closedness of (A−1, R(A)). Useful propositions
Proposition (short wording: sum of a closed and everywhere
dened bounded operator is closed). Consider closed linear
operator A : V1 ⊃ D(A) → V2 between normed spaces V1 and V2.
Consider bounded linear operator B : V1 = D(B) → V2. Then the
sum (A + B, D(A + B)) dened as (A + B)x = Ax + Bx for all
x ∈ D(A + B) = D(A) is a closed operator.
Proof. Having (xn , yn ) ∈ G (A + B) and
lim k(xn − x, yn − y )kV1 ×V2 = 0 where yn = Axn + Bxn , we wish
n→∞
to prove (x, y ) ∈ G (A + B).
In other words: having xn ∈ D(A), xn → x ∈ V1,
Axn + Bxn → y ∈ V2 we wish to prove x ∈ D(A) and Ax + Bx = y .
Let us now use what we have to prove what we wish to prove.
Operator B is continuous on V1 because B is bounded, so
Bxn → Bx . From kAxn + Bxn − y k → 0 and Bxn → Bx we infer
k limn→∞ Axn + Bx − y k = 0, i.e. limn→∞ Axn = y − Bx . So
sequence (xn , Axn ) ∈ G (A) converges, and (A, D(A)) is closed,
which gives us x ∈ D(A) and Ax = limn→∞ Axn = y − Bx . From
3 / 21
Ax = y − Bx we derive Ax + Bx = y . Spectrum of linear operator in normed space
Denition. Consider densely dened linear operator
in complex normed space V . Denote Ix = x .
is called regular point for (A, D(A)) i two
A : V ⊃ D(A) → V
λ∈C
The point
conditions hold:
1. Operator (λI − A) : D(A) → V is bijective.
2. Operator (λI − A)−1 : V → D(A) is bounded.
Denition. For densely dened linear operator A the set of all
regular points is called the resolvent set and denoted ρ(A). The set
σ(A) = C \ ρ(A) is called the spectrum of A.
Exercise. Consider densely dened linear operator
A : V ⊃ D(A) → V in complex normed space V . Prove that point
λ ∈ C is a regular point for (A, D(A)) i three conditions hold:
1. λ ∈/ σp (A), i.e. λ is not an eigenvalue of A, i.e. for each y ∈ V
equation λx − Ax = y has at most one solution x ∈ D(A).
2. R(λI − A) = V , i.e. (λI − A)(D(A)) = V , i.e. for each y ∈ V
equation λx − Ax = y has at least one solution x ∈ D(A).
3. There exists a > 0 that kλx − Axk ≥ akxk for all x ∈ D(A).
4 / 21
Resolvent operator
Denition. Consider densely dened linear operator
in complex normed space V . If λ ∈ C is a
regular point for
, i.e. λ belongs to the resolvent set ρ(A),
then we can dene the following operator:
A : V ⊃ D(A) → V
(A, D(A))
Rλ : V = D(Rλ ) → V , Rλ y = (λI − A)−1 y for all y ∈ V .
Operator (Rλ, V ) is called the resolvent operator for A. We will
write Rλ,A instead of Rλ if it is not clear what operator (A, D(A))
is considered.
Remark. By denition of a regular point, Rλ is bounded and
everywhere dened linear operator in V , the range of Rλ is dense in
V and is equal to R(Rλ ) = D(A).
Remark. Other notations that are used for the resolvent operator:
R(λ), R(λ, A), RA (λ), Rλ (A), RA or sometimes just R . They all
mean the same as Rλ dened above. We will not use them in order
not to confuse with R used to dene the range (the set of all taken
values) of operator: R(B) = B(D(B)) = {Bx|x ∈ D(B)} where
D(B) is the domain of some operator B .
5 / 21
The word "resolvent"
Remark. Consider densely dened linear operator
in complex normed space V . If we, given
, wish to solve (or resolve, in other words) equation
A : V ⊃ D(A) → V
y ∈V
λx − Ax = y , i.e. (λI − A)x = y ,
then it is solvable (resolvable, in other words) for all y ∈ V i λ
belongs to the resolvent set ρ(A), and in that case the solution is
unique and is given by the resolvent operator:
x = Rλ y = (λI − A)−1 y .
This explains why the word "resolvent" is used for the set ρ(A) and
the operator Rλ.
6 / 21
Banach theorems: closed graph, bounded inverse, open
mapping
Banach closed graph theorem. Consider linear operator
between Banach spaces V1 and V2. Then A is
bounded i is closed. (It is important here that D(A) = V1.)
Banach bounded inverse theorem. Consider bijective linear
operator A : V1 = D(A) → V2 between Banach spaces V1 and V2.
Then A is bounded then i A−1 is bounded. (It is important here
that D(A) = V1.)
Denition. A mapping f between topological spaces is called open
i it maps open set to an open set, i.e. if U is open then f (U) is
open.
Banach-Schauder open mapping theorem. Consider linear
operator A : V1 = D(A) → V2 between Banach spaces V1 and V2.
If A is bounded and surjective then A is an open mapping. (It is
important here that D(A) = V1.)
A : V1 = D(A) → V2
A
7 / 21
Spectrum of linear operator in Banach space
Remark. Always σp (A) ⊂ σ(A). In nite-dimensional spaces V all
linear mappings are bounded, closed and have only point spectrum:
σp (A) = σ(A). If dimV = ∞ then often σp (A) 6= σ(A).
Remark. If the domain D(A) ⊂ V is not dense in Banach space V
then the closure D(A) is a closed linear subset in V hence D(A) is
a Banach space with the norm from V , and D(A) is dense in D(A).
So if A(D(A)) ⊂ D(A) then (A, D(A)) is a densely dened operator
in Banach space D(A). This explains why only densely dened
operators in Banach spaces are sudied: this restriction do not reduce
the generality too much but gives some nice instruments to use.
Remark. The following theorem explains that spectrum says
something about linear operator only if the operator is closed.
Theorem. Consider linear operator A : V ⊃ D(A) → V in complex
Banach space V . If (A, D(A)) is not closed, then σ(A) = C.
Proof. Suppose λ ∈ ρ(A) then operator (λI − A)−1 : V → D(A) is
bounded, hence its graph is closed, hence the graph of
(λI − A) : D(A) → V is closed, hence the graph of A : D(A) → V
is closed. Contradiction. So condition λ ∈ ρ(A) is impossible. 8 / 21
Properties of closed densely dened linear operator in
Banach space
Theorem. Consider closed, densely dened linear operator
in complex Banach space . Then:
1.
is a closed subset of .
2. If for some
the operator
is bijective
then the operator
is bounded.
3.
i is bounded.
4. If is bounded then
.
is closed hence
is closed, hence
is closed and thanks to
Banach closed graph theorem is bounded.
Exercise. Prove item 3. Hint: use denitions and theorems from
this and the previous lesson.
A : V ⊃ D(A) → V
V
σ(A)
C
λ∈C
(λI − A) : D(A) → V
(λI − A)−1 : V → D(A)
D(A) = V
A
A
σ(A) 6= ∅
Proof of item 2. (A, D(A))
(λI − A) : D(A) → V
−
1
(λI − A) : V → D(A)
9 / 21
How to nd the spectrum
General algorithm of ndind the spectrum of an operator.
1. Check that we deal with a linear operator (A, D(A)) in complex Banach
space F . What we can do if not? If the operator is not linear - we have nothing
to do. If F is not a Banach space, then we can: a) select another norm on F ;
b) consider F as a dense subset in the completion of F which is always a
Banach space because it is complete. If F is a space over R (instead of being
over C), then we can consider complexication F ⊕ iF which is a space over C.
2. If it is easy, check if (A, D(A)) is closed or not. If (A, D(A)) is not closed
then σ(A) = C and the task is nished. If we proved (or suspect) that
(A, D(A)) is closed then proceed next.
3. Find all λ ∈ C such that operator λI − A : D(A) → F is not injective. In
other words, such λ ∈ C that ker(λI − A) 6= {0}. Those λ are called
eigenvalues and the point spectrum σp (A).
4. Find all λ ∈ C such that λ ∈
/ σp (A) and operator λI − A : D(A) → F is not
surjective.
5. For at least one λ which is not mentioned in steps 3 or 4 prove that there
exists a > 0 that kλx − Axk ≥ akxk for all x ∈ D(A). This λ will be a regular
point of A, hence A is closed. We can skip this step if we proved that A is
closed on step 2.
6. Write the answer: σ(A) is the set of all λ found in steps 3 and 4.
10 / 21
Parts of the spectrum
Denition. It is possible to divide σ(A) \ σp (A) into parts
depending on the behavior on the step 4:
4.1. Continuous spectrum σc (A) is the set of all λ ∈ C such that
λ∈
/ σp (A) and the set (λI − A)(D(A)) is not equal to F
HOWEVER it is dense in F .
4.2. Residual spectrum σr (A) is the set of all λ ∈ C such that
λ∈
/ σp (A) and the set (λI − A)(D(A)) is not equal to F AND is
not dense in F .
Denition. Representation
σ(A) = σp (A) t σc (A) t σr (A)
is called the decomposition of the spectrum of operator (A, D(A))
into parts of the spectrum.
Remark. Sometimes the decomposition into essential spectrum and
discrete spectrum is considered. Also the following parts of the
spectrum for self-adjoint operators in Hilbert space are considered:
absolutely continuous spectrum, singular continuous spectrum, and
pure point spectrum. We will not use them in our short course. 11 / 21
Spectrum of a bounded linear operator
Denition. Let A be bounded, everywhere dened linear operator
in Banach space. Then the number
r (A) = max |λ|
λ∈σ(A)
is called the spectral radius of A.
Theorem. Let A be bounded, everywhere dened linear operator in
Banach space. Then:
1. There exists limn→∞ kAn k1/n = r (A).
2. The spectrum σ(A) is a closed, not empty bounded subset of C,
located inside the closed disc of radius kAk centered at zero.
12 / 21
Banach space of linear bounded operators in Banach space
Theorem (short wording: extension "by continuity" of a
bounded linear operator is unique). Consider linear bounded
operator A : V ⊃ D(A) → V in real or complex Banach space V .
Consider linear bounded operator B : V ⊃ D(B) → V in complex
Banach space V such that D(A) ⊂ D(A) = D(B) and Ax = Bx for
all x ∈ D(A). Then A is closable and B = A.
In particular, if A is densely dened (i.e. D(A) = V ) then
continuation of A on the whole space V without loss of continuity
is unique and coincides with the closure of A.
Remark. This theorem is the reason why bounded linear operators
are ususally considered being everywhere dened.
Exercise. Prove the theorem.
Denition. Let V be a Banach space. The symbol L (V ) denotes
the set of all linear bounded everywhere dened operators in V , i.e.
A ∈ L (V ) i A : V = D(A) → V is a linear bounded operator.
Exercise. If V is a Banach space, then L (V ) is a Banach space
over the same led R or C as V , with operations (αA)(x) = α(Ax),
(A + B)x = Ax + Bx and norm kAk = supkxk=1 kAxk.
13 / 21
Denition of a C0 -semigroup
Denition. Let F be a Banach space over the eld R or C. Let
be a set of all bounded linear operators in F . Suppose we
have a mapping Q : [0, +∞) → L (F), i.e. Q(t) is a bounded
linear operator Q(t) : F → F for each t ≥ 0. The mapping Q is
called a C0-semigroup, or a strongly continuous one-parameter
semigroup of operators if it satises the following conditions:
1) Q(0) is the identity operator I , i.e. ∀f ∈ F : Q(0)f = f ;
2) Q maps the addition of numbers in [0, +∞) into a composition
of operators in L (F), i.e. ∀t ≥ 0, ∀s ≥ 0 : Q(t + s) = Q(t) ◦ Q(s),
where (A ◦ B)(f ) = A(B(f )) = ABf ;
3) Q is continuous at zero with respect to the strong operator
topology in L (F), i.e. ∀f ∈ F function t 7−→ Q(t)f is continuous
at 0 as a mapping [0, +∞) → F , i.e. limt→0 kQ(t)f − f k = 0 for
all f ∈ F .
The denition of a C0-group is obtained by substitution of [0, +∞)
with R in the paragraph above.
L (F)
14 / 21
Denition of the generator of a C0 -semigroup
Denition. If (Q(t))t≥0 is a C0 -semigroup in Banach space F ,
then dene the set
Q(t)ϕ − ϕ
ϕ ∈ F : ∃ lim
t→+0
t
denote
= D(L)
.
The operator L dened on the domain D(L) by the equality
Lϕ = lim
t→+0
Q(t)ϕ − ϕ
t
is called an innitesimal generator (or just generator for short) of
the C0-semigroup (Q(t))t≥0.
Theorem. The generator is a densely dened closed linear operator.
15 / 21
C0 -semigroup is a generalization of exponent
Denition. If L is the generator of a C0 -semigroup Q then the
notation Q(t) = e tL is used.
Remark. If L is a bounded operator and D(L) = F , then e tL is
indeed the exponent dened by the power series e tL = P∞k=0 t k!L
converging with respect to the norm topology in L (F). In most
interesting cases the generator is an unbounded dierential
operator such as Laplacian ∆.
Theorem. Let (L1 , D(L1 )) be the generator of C0 -semigroup Q1 ,
and let (L2, D(L2)) be the generator of C0-semigroup Q2, in the
same Banach space. Then:
1. (L1, D(L1)) = (L2, D(L2)) ⇒ Q1(t) = Q2(t) for all t ≥ 0.
2. Q1(t) = Q2(t) for all t ≥ 0 ⇒ (L1, D(L1)) = (L2, D(L2)).
Remark. The previous theorem shows that the generator denes
the C0-semigroup uniquely and is dened by the C0-semigroup
uniquely, i.e. the mapping L 7→ e tL is a bijection between the set of
all generators of C0-semigroups in Banach space F and the set of
all C0-semigroups in F . Let us now describe partially these two sets.
k k
16 / 21
Hille-Yosida theorem
Denition. Linear bounded operator A is called a contraction i
. -semigroup Q is called a contraction semigroup i
for all t ≥ 0.
Hille-Yosida theorem (1948). Let A be a linear operator dened
on a linear subspace D(A) of the Banach space F . Then (A, D(A))
generates a contraction C0-semigroup i four conditions hold:
1. (A, D(A)) is densely dened, i.e. D(A) = F .
2. (A, D(A)) is closed.
3. (0, +∞) ⊂ ρ(A), i.e. every real λ > 0 belongs to the resolvent
set of A, i.e. the resolvent operator Rλ = (λI − A)−1 is a
well-dened linear bounded operator on F = D(Rλ).
4. kRλk ≤ λ1 for all λ > 0.
kAk ≤ 1 C0
kQ(t)k ≤ 1
17 / 21
Integral representation of the resolvent
Theorem. Suppose that (Q(t))t≥0 is a C0 -semigroup in Banach
space F having the generator (A, D(A)), and the resolvent for the
generator is Rλ = (λI − A)−1. Suppose that kQ(t)k ≤ Me wt for all
t ≥ 0. Suppose that Reλ > w . Then λ ∈ ρ(A) and for all f ∈ F
Z +∞
Rλ f =
0
e −λs Q(s)fds.
(1)
The above integral is an improper Riemann integral for
vector-valued function [0, +∞) 3 s 7−→ e −λs Q(s)f ∈ F . This function is
continuous and bounded on each segment [0, t], hence it is uniformly
continuousR on this segment, hence the limit of Riemann integral sums
t
exists for 0 e −λs Q(s)fds . After that we dene
Remark.
Z +∞
0
Remark.
e −λs Q(s)fds = lim
Z t
t→+∞
0
e −λs Q(s)fds.
If F = R and A ∈ R then after division by f ∈ R (1) reads as
1
=
λ−A
Z +∞
0
e
−λs sA
which is a simple fact of calculus.
e ds =
Z +∞
0
e (A−λ)s ds,
18 / 21
Seminar 16
On the Seminar 16, we will prove the following statement:
Theorem. Let Q be a C0 -semigroup in Banach space F . Then:
1. Q(t1)Q(t2) = Q(t2)Q(t1) for all t ≥ 0.
2. limt→t0 kQ(t)f − Q(t0)f k = 0 for each t0 ≥ 0 and each f ∈ F .
3. ∃ dtd Q(t)f t=0 = Lf for each f ∈ D(L).
4. ∃ dtd Q(t)f t=t0 = Q(t0)Lf for each f ∈ D(L) each t0 ≥ 0.
5. If t ≥ 0 and f ∈ D(L) then Q(t)f ∈ D(L).
6. If t ≥ 0 and f ∈ D(L) then Q(t)Lf = LQ(t)f .
19 / 21
Homework 16
Problems to solve at home (deadline 20:00 March 6):
1. Exercise on p.4.
2. Exercise on p.9.
3. First exercise on p.13.
4. Second exercise on p.13.
Denition. For linear operator A : F ⊃ D(A) → F with the
domain D(A)
⊂ F and n ∈ N we dene the domain D(An ) of
n
operator A inductively as follows:
(f ∈ D(An )) ⇐⇒ (f ∈ D(A), Af ∈ D(A), A2 f ∈ D(A), . . . , An−1 f ∈ D(A)),
which implies D(A) ⊃ D(A2) ⊃ · · · ⊃ D(An ).
5. Let Q be a C0-semigroup with generator (L, D(L)) in Banach
space F . Prove that for all t0 ≥ 0, all n = 1, 2, 3, . . . and all
f ∈ D(Ln ):
∃
dn
Q(t)f t=t = Q(t0 )Ln f = Ln Q(t0 )f .
0
dt n
20 / 21
Literature on C0 -semigroups
The short list of books is presented below. Any of these books may
bring you enough understanding of the subject to pass the nal
exam. If you ask me to select only one then it will be the rst one
by Engel and Nagel, 2000.
I K.-J. Engel, R. Nagel. One-Parameter Semigroups for Linear
Evolution Equations. Springer, 2000.
I K.-J. Engel, R. Nagel. A Short Course on Operator
Semigroups. Springer, 2006.
I A. Pazy. Semigroups of Linear Operators and Applications to
Partial Dierential Equations. Springer, 1983.
I E. Hille, R. S. Phillips: Functional Analysis and Semi-Groups.
American Mathematical Society, 1975.
I Ô.Êëåìåíò, Õ. Õåéìàíñ, Ñ. Àíãåíåíò, Ê. âàí Äóéí, Á. äå
Ïàõòåð, Îäíîïàðàìåòðè÷åñêèå ïîëóãðóïïû. Ì.: Ìèð, 1992.
I Ð.Â. Øàìèí. Ïîëóãðóïïû îïåðàòîðîâ. Ì.: ÐÓÄÍ, 2008.
21 / 21
Introduction to
One-parameter semigroups and evolution
equations
Lesson 17: interplay between C0 -semigroups and linear evolution
equations
Ivan Remizov
HSE Nizhny Novgorod, Russia
7 March 2023
1 / 19
Properties of the norm of a linear bounded operator
Lemma. Let A, B be linear bounded operators dened everywhere
in normed space F . Then:
1. kAxk ≤ kAk · kxk for all x ∈ F .
2. kABk ≤ kAk · kBk
3. kAn k ≤ kAkn for all n ∈ N.
Proof. 1. If kxk = 0 then x = 0, Ax = 0, kAxk = 0 and inequality
is trivially true. By denition of the norm of linear operator
kAk = supkxk=1 kAxk, so for all x with kxk = 1 we have
kAxk ≤ kAk. If 0 6= kx1 k =
6 1 then consider x = kx11 k x1 , we have
kxk = 1 hence kAk ≥ kAxk = A kx11 k x1 = kx11 k kAx1 k hence
kAk · kx1 k ≥ kAx1 k.
2. kABk = supkxk=1 kABxk ≤ supkxk=1 kAk · kBxk =
kAk supkxk=1 kBxk = kAk · kBk.
3. For n = 1 then the assertion is trivially true. For n = 2 set
A = B and get kA2 k = kAAk ≤ kAk · kAk = kAk2 . Then induction
in n: if kAn−1 k ≤ kAkn−1 is true then set B = An−1 and get
kAn k = kABk ≤ kAk · kBk = kAk · kAn−1 k ≤ kAk · kAkn−1 . 2 / 19
C0 -semigroups with bounded generator
Theorem. Let F be Banach space and A : F = D(A) → F be
linear bounded operator. Assume t 0 = 1 for all t ∈ R. Then:
1. For all t ∈ R with respect to norm in L (F)
k
∞
X
t n An denote X t n An denote tA
∃ lim
=
= e .
k→∞
n!
n!
n=0
n=0
2. ke tA k ≤ e |t|·kAk for all t ∈ R.
3. (e tA )t∈R is a C0 -group in F , (e tA )t≥0 is a C0 -semigroup in F .
4. The generator of this group and semigroup
.
P is (A,n F)
n
Proof. 1. Let us prove that sequence ak = kn=0 t n!A is
fundamental in L (F). Suppose ε > 0 is given, then
nk
n kAkn
Pk
P
P
t n An
≤ kn=m k(tA)
≤ kn=m |t| n!
< ε for some m
n=m n!
n!
Pk |t|n kAkn
is fundamental
large enough because sequence bk = n=0 n!
P∞ |t|n kAkn
|t|·kAk
in R because power series n=0 n! = e
converges in R.
Then ak has a limit because L (F) is a Banach space.
n kAkn
Pk t n An
P
2.
≤ kn=0 |t| n!
, then take limk→∞ .
n=0 n!
3 / 19
C0 -semigroups with bounded generator
0
3. By denitions
I , t 0 = 1, 0! = 1 so
P∞ 0n AnA =
2 2
00 A0
0
A
e = n=0 n! = 0! + 01A! + 0 2A! + · · · = 11·I + 0 + 0 + · · · = I .
tA e sA = e (t+s)A . Indeed, when multiplying for
Now let us prove eP
P∞ s n an
t n an
some a ∈ R series ∞
n=0 n! by series
n=0 n! we get a series
P∞ (t+s)n an
ta
sa
, because for numbers e e = e (t+s)a . Moreover,
n=0
n!
series for e ta e sa and for e (t+s)a must have the same Taylor's
expansion for a → 0, so an is multiplied by (t + s)n /n! in both
series. But An commutes with Ak and with I so if we perform the
standard construction of multiplication of series with a substituted
by A we get that e tA e sA = e (t+s)A .
P
|t|n An
Also ke tA x − xk ≤ ke tA − I k · kxk = k ∞
n=1 n! k · kxk ≤
P
P
|t|n−1 kAkn−1
|t|n−1 kAkn−1
· kxk ≤ |t|kAk ∞
· kxk =
|t|kAk ∞
n=1
n=1
n!
(n−1)!
|t|kAke |t|kAk kxk → 0 as t → 0Pfor all x ∈ F .
1
1
t n An
4. |t|
ke tA x − x − tAxk = |t|
k ∞
n=2 n! xk ≤
P
t n−2 kAkn−2
|t|kAk2 ∞
k · kxk ≤ |t|kAk2 e |t|kAk kxk → 0 as t → 0
n=2
n!
for all x ∈ F . So the domain of generator of (e tA )t∈R is F and the
4 / 19
generator is equal to A. C0 -semigroups with bounded generator
Denition. One-parameter semigroup of operators
(Q(t))t≥0 ⊂ L (F) in Banach space F is called uniformly
continuous or norm-continuous i limt→0 kQ(t) − I kL (F ) = 0.
Remark. The word ¾uniformly¿ here means ¾uniformly in x from
the unit ball {x ∈ F : kxk ≤ 1}¿. Indeed, for all x from this ball we
have kQ(t)x − xkF ≤ kQ(t) − I kL (F ) kxkF ≤ kQ(t) − I kL (F ) .
This shows that function t 7−→ Q(t)x is continuous in t uniformly
in x from the unit ball. The words ¾norm-continuous¿ above
means that function t 7−→ Q(t) ∈ L (F) is continuous in topology
generated by norm in L (F), i.e. by the operator norm.
Theorem. Each uniformly continuous semigroup of operators is a
C0 -semigroup.
Proof. See the estimate in the above remark.
Theorem. C0 -semigroup of operators (Q(t))t≥0 ⊂ L (F) in
Banach space F is uniformly continuous i its generator is bounded.
Remark. This theorem shows that uniformly continuous
semigroups are not very interesting. However, they are a useful tool
for proof making.
5 / 19
The norm bound for C0 -semigroups
Banach-Steinhaus theorem (uniform boundedness principle),
1927. Let X be a Banach space and Y be a normed space over the
same eld R or C. Let L (X , Y ) be the set of all linear bounded
everywhere dened operators X → Y . Suppose that
∅ 6= A ⊂ L (X , Y ), i.e. A is a collection of linear bounded
operators.
Then condition
∀x ∈ X ∃ sup kAxkY = C (x) < ∞
A∈A
imply that
∃ sup kAk =
A∈A
sup
A∈A,kxk=1
kAxkY = sup C (x) < ∞.
kxk=1
In other words: if a family of linear bounded operators is bounded
pointwise (i.e. in each point x ∈ X ) then it is bounded uniformly
(i.e. it is a bounded subset of normed space L (X , Y )).
Remark. It is important here that X is a Banach space.
6 / 19
The norm bound for C0 -semigroups
Theorem. Let (Q(t))t≥0 ⊂ L (F) be a C0 -semigroup of operators
in Banach space F . Then there exist constants M ≥ 1 and ω ≥ 0
such that kQ(t)k ≤ Me ωt for all t ≥ 0.
Proof. The set [0, 1] ⊂ R is compact, and function t 7→ kQ(t)xk is
continuous, so for each x ∈ F the set {Q(t)x|t ∈ [0, 1]} ⊂ F is
also compact because continuous image of a compact set is a
compact set. In metric space each compact set is bounded, so for
each x ∈ F we have supt∈[0,1] kQ(t)xk = C (x) < ∞. Then the set
of operators A = {Q(t)|t ∈ [0, 1]} satises the Banach-Steinhaus
theorem and ∃ supt∈[0,1] kQ(t)k = M < ∞.
For each t ≥ 0 there exist unique number [t] ∈ {0, 1, 2, . . . } integer part of t , {t} ∈ [0, 1) fractional part of t such that
t = [t] + {t}. Then Q(t) = Q([t] + {t}) = Q([t])Q({t}) =
Q(1)[t] Q({t}), so kQ(t)k = kQ(1)[t] Q({t})k ≤
≤ kQ(1)k[t] kQ({t})k ≤ kQ(1)k[t] M . If kQ(1)k ≤ 1 then theorem
is proved for ω = 0. If kQ(1)k > 1 then kQ(t)k ≤ MkQ(1)k[t] ≤
MkQ(1)k[t]+{t} = MkQ(1)kt = Me t log kQ(1)k and theorem is
proved for ω = log kQ(1)k. 7 / 19
Generalized Lumer-Phillips theorem
Denition. C0 -semigroup (Q(t))t≥0 is called:
1. Contraction semigroup i kQ(t)k ≤ 1 for all t ≥ 0.
2. Quasi-contraction semigroup i exists ω ≥ 0 such that
kQ(t)k ≤ e ωt for all t ≥ 0.
Denition. Linear operator A : F ⊃ D(A) → F in Banach space F
is called:
1. Dissipative i for all λ > 0 and all x ∈ D(A) we have
k(λI − A)xk ≥ λkxk.
2. Quasi-dissipative i there exists ω ≥ 0 such that operator A − ωI
is dissipative, i.e. i for all λ > 0 and all x ∈ D(A) we have
k((λ + ω)I − A)xk ≥ λkxk.
Generalized Lumer-Phillips theorem. Linear operator
A : F ⊃ D(A) → F in Banach space F generates:
1) a contraction semigroup i A is densely dened, closed,
dissipative, and (λ0 I − A, D(A)) is surjective for some λ0 > 0.
2) a quasi-contraction semigroup i A is densely dened, closed,
quasi-dissipative, and (λ0 I − A, D(A)) is surjective for some λ0 > ω
where ω is taken from the denition of quasi-dissipativity.
8 / 19
The most general theorem on generators of C0 -semigroups
FellerMiyaderaPhillips theorem (also known as generalized
Hille-Yosida theorem). Let A be a linear operator dened on a
linear subspace D(A) of the Banach space F . Suppose that M ≥ 1
and ω ∈ R. Then (A, D(A)) generates a C0 -semigroup (Q(t))t≥0
that satises kQ(t)k ≤ Me ωt for all t ≥ 0 i four conditions hold:
1. (A, D(A)) is densely dened, i.e. D(A) = F .
2. (A, D(A)) is closed.
3. (ω, +∞) ⊂ ρ(A), i.e. every real λ > ω belongs to the resolvent
set of A, i.e. the resolvent operator Rλ = (λI − A)−1 is a
well-dened linear bounded operator on F = D(Rλ ).
4. For all n = 1, 2, 3, . . . and all λ > ω we have
M
.
kRλn k = k(λI − A)−n k ≤
(λ − ω)n
Remark. Condition 4 usually is dicult to obtain because it is
usually dicult to nd the resolvent operator Rλ explicilty. So
usually the most easy way to get condition 4 is to prove estimate
kQ(t)k ≤ Me ωt and then get 4 via the above theorem.
9 / 19
Well-posed problems
Denition. Suppose we have a binary relation f ⊂ X × Y (in
particular, f may be a function f : X → Y ), where X and Y are
topological spaces (sometimes some other spaces). Suppose that
y ∈ Y is given and consider the problem of nding such x ∈ X that
(x, y ) ∈ f (which means f (x) = y if f is a function).
The problem is called:
1. Well-posed (Russian: êîððåêòíî ïîñòàâëåííàÿ) in sense of
Hadamard i for all y ∈ Y solution x = f −1 (y ) to the problem
exists, is unique and depends continuously on y ∈ Y .
2. Well-posed in sense of A.N.Tikhonov i there exists X1 ⊂ X and
Y1 ⊂ Y such that the problem associated with binary relation
f ⊂ X1 × Y1 (in particular, with a function f : X1 → Y1 ) is
well-posed in sense of Hadamard. In other words: for all y ∈ Y1
solution x = f −1 (y ) exists, is unique in X1 and depends
continuously on y ∈ Y1 .
Denition. Sets X1 and Y1 above are called sometimes classes of
correctness for the above-mentioned problem.
Denition. A problem is called ill-posed if it is not well-posed.
10 / 19
Cauchy problem for linear evolution equation
Denition. Suppose we have a linear operator A : F ⊃ D(A) → F
in Banach space F and u0 ∈ F . Function u : [0, +∞) → F is
called a classical solution of the Cauchy problem
(
d
dt u(t) = Au(t) for all t ≥ 0
u(0) = u0
(1)
i function u is continuous on [0, +∞), dierentiable on [0, +∞),
the derivative u 0 is continuous on [0, +∞), u(t) ∈ D(A) for all
t ≥ 0 and (1) holds.
Theorem. With notions from the above denition, if (A, D(A))
generates C0 -semigroup and u0 ∈ D(A) then u(t) = e tA u0 is the
classical solution of (1), which is unique in C 1 ([0, +∞), F).
Proof. For existence and formula u(t) = e tA u0 see the statement
that we proved on seminar 16. Continuity in t and dierentiability
in t follows from the denition of C0 -semigroup and statement
from seminar 16. Continuiuty in u0 follows from the fact that linear
bounded operator x 7→ e tA x is continuous. Uniqueness is more
dicult to prove, we will not do it in our introduction-level course. 11 / 19
Cauchy problem for linear evolution equation
Remark. Theorem above means, that for xed t ≥ 0 and given
y = u0 ∈ F = Y = X problem of nding x = u(t) ∈ X being
classical solution of (1) is well-posed in sense of Tikhonov with
correctness classes X1 = Y1 = D(A) ⊂ F , and x = f −1 (y ) = e tA y .
Remark. However, in this setting the problem of nding u0 if you
know u(t) usually is not well-posed in sense of Tikhonov for
arbitrary chosen C0 -semigroup (e tA )t≥0 . This is typical: inverse
problems are often ill-posed.
Remark. Meanwhile, if we deal with the C0 -group (e tA )t∈R then it
is easy to prove that all operators e tA are invertible and
(e tA )−1 = e −tA . In that case the problem of nding u0 if you know
u(t) is well-posed in sense of Tikhonov and u0 = e −tA u(t) with the
same correctness classes.
12 / 19
General denition of a dynamical system
Denition. Let (M, d) be metric space and T be the time interval
T ∈ {R, [0, +∞), Z, N0 = {0, 1, 2, 3, . . . }}. One says that
dynamical system in M is a continuous (in some sense) function
Φ : T × M → M , which satises two properties:
1. Φ(0, x) = x for all x ∈ M
2. Φ(t1 , Φ(t2 , x)) = Φ(t1 + t2 , x) for all x ∈ M and all t1 , t2 ∈ T .
Depending on T , dynamical systems are divided into two classes,
each with two subclasses:
I continuous time dynamical system: ow in M for T = R,
semiow in M for T = [0, +∞).
I discrete time dynamical system: cascade in M for T = Z,
semicascade in M for T = N0 .
Remark. In the discrete case
Φ(2, x) = Φ(1 + 1, x) = Φ(1, Φ(1, x)) = (Φ(1, ·) ◦ Φ(1, ·))(x),
so if we denote Φ(1, x) = ϕ(x) then Φ(n, ·) = ϕn = ϕ ◦ · · · ◦ ϕ so
| {z }
n
only function ϕ : M → M is really needed to dene Φ(n, x).
13 / 19
C0 -semigroups and semiows
Theorem (C0 -semigroup is a specic type of semiows.).
Suppose (e tA )t≥0 is a C0 -semigroup in Banach space F , denote
Φ(t, f ) = e tA f for all t ≥ 0, f ∈ F . Then Φ is a semiow in F .
Exercise. Prove the theorem.
Theorem (semiow in M naturally denes a specic type of
C0 -semigroups in space of functions on M ). Let M be a metric
space, and F = UCb (M) be the a Banach space of all uniformly
continuous and bounded functions f : M → R with the uniform
norm kf k = supx∈M |f (x)|. Suppose that φ : [0, +∞) × M → M is
a semiow in M with the following notion of continuity: φ is
continuous as a mapping [0, +∞) × M → M and φ(t, x) is
continuous in t iniformly with respect to x ∈ M , i.e.
limτ →0 supx∈M |φ(t + τ, x) − φ(t, x)| = 0 for all t ≥ 0, x ∈ M .
Let us dene Φ : [0, +∞) × F → F by equality
(Φ(t, f ))(x) = f (φ(t, x)) for all t ≥ 0, x ∈ M,
and Q(t)f = Φ(t, f ). Then Q(t) ∈ L (F) for each t ≥ 0, and
(Q(t))t≥0 is a C0 -semigroup in F . Exercise. Prove the theorem.
14 / 19
Regular and chaotic dynamics
Remark. Well-posednsess of Cauchy problem means that the value
u(t) = φ(t, u0 ) is determined by t and u0 uniquely (this is called
the determinism), and for each xed t ∈ T depends on u0
continuously (this is called the continuous dependence on inital
value). However, it is also important how continuity in u0 is related
with long-term dependence on t . This leads to the fact that
well-posed problems may produce two dierent behaviors of
solutions: the so-called regular and the so-called chaotic.
Denition. A short and informal denition of chaos was given by
one of the creators of the chaos theory Edward Lorenz:
Chaos is ¾When the present determines the future, but the
approximate present
does not approximately determine the future¿.
Denition. Dynamical system is called regular (or its trajectories
are called having regular behavior) if it is not chaotic.
Further reading on chaos theory:
https://en.wikipedia.org/wiki/Chaos_theory
https://en.wikipedia.org/wiki/Butterfly_effect
15 / 19
Seminar 17
On the Seminar 17, we briey proved theorem on p.11 and
theorems on p.14.
16 / 19
Homework 17
Deadline: 13 March 2023, 20:00.
1a. Consider R as a normed space over itself, with the norm
kxk = |x|. Is this space Banach? Is this norm generated by a scalar
product, why? If yes nd this scalar product. Is this space a
Hilbert space? Find the algebraic dimension of this space.
1b. Consider C as a normed space over R, with the norm
kxk = |x|. Is this space Banach? Is this norm generated by a scalar
product, why? If yes nd this scalar product. Is this space a
Hilbert space? Find the algebraic dimension of this space.
1c. Consider
Rn as a normed space over R, with the norm
q
kxk = x12 + · · · + xn2 . Is this space Banach? Is this norm
generated by a scalar product, why? If yes nd this scalar
product. Is this space a Hilbert space? Find the algebraic dimension
of this space.
1d. Consider Cn as a vector space over C. What natural norm can
you propose to this space? Will this space be Banach, Hilbert? Find
the algebraic dimension of this space.
17 / 19
Homework 17
Denition. Operator A is called accretive i operator −A is
dissipative.
2 a. Consider V = R as a normed space over R, with the norm
kxk = |x|. Suppose that a number a ∈ R is given. Consider
operator A dened on D(A) = V as follows: Ax = a · x . Please nd
kAk, σ(A), e tA . Is A accretive, disipative, quasi-dissipative? Is
(e tA )t≥0 contractive, quasi-contractive? Please study all possible
cases a ∈ R.
2b. Consider V = C as a normed space over R, with the norm
kxk = |x|. Suppose that a number a ∈ C is given. Consider
operator A dened on D(A) = V as follows: Ax = a · x . Please nd
kAk, σ(A), e tA . Is A accretive, disipative, quasi-dissipative? Is
(e tA )t≥0 contractive, quasi-contractive? Please study all possible
cases a ∈ C.
Please see the next page.
18 / 19
Homework 17
3a. Consider V = C2 as a normed space over C, and operator A given by
the matrix
A=
1
0
0
.
2
Please nd kAk, σ(A), e tA . Is A accretive, disipative, quasi-dissipative? Is
(e tA )t≥0 contractive, quasi-contractive?
3b. Consider V = C2 as a normed space over C, and operator A given by
the matrix
2 1
A=
.
0 2
Please nd kAk, σ(A), e tA . Is A accretive, disipative, quasi-dissipative? Is
(e tA )t≥0 contractive, quasi-contractive?
4. Prove in detail theorem on p.11.
5a. Prove in detail upper theorem on p.14.
5b. Prove in detail bottom theorem on p.14.
19 / 19
Introduction to
One-parameter semigroups and evolution
equations
Lecture 18: applications and approximations of semigroups
Ivan Remizov
HSE Nizhny Novgorod, Russia
14 March 2023
1 / 10
Applications of semigroups
This is the contents of the famous book K.J.Engel, R.Nagel.
One-parameter semigroups for linear evolution equations
(Springer, 2000):
We will discuss only few of applications of semigroups, and will
select only some of those that your lecturer uses or created.
2 / 10
Suppose that (A, D(A)) generates a C0 -semigroup (e tA )t≥0 in
Banach space F . Then:
1. For each u0 ∈ D(A) Cauchy problem
Theorem.
(
U 0 (t) = AU(t), t ≥ 0
U(0) = u0
(1)
has a solution U ∈ C 1 ([0, +∞), F) which is unique in C 1 ([0, +∞), F)
and is given by the formula U(t) = e tA u0 .
2. For each u0 ∈ D(A), f ∈ C ([0, +∞), F) Cauchy problem
(
U 0 (t) = AU(t) + f (t), t ≥ 0
U(0) = u0
has a solution U ∈ C 1 ([0, +∞),RF) which is unique in C 1 ([0, +∞), F)
and is given as U(t) = e tA u0 + 0t e (t−s)A f (s)ds .
3. For each u0 ∈ F Cauchy problem (1) in the integral form
Rt
U(t) = u0 + A 0 U(s)ds has a solution U ∈ C ([0, +∞), F) which is
unique in C ([0, +∞), F) and is given as U(t) = e tA u0 . This solution is
called the mild solution of (1) and exists for all u0 ∈ F .
tA
wt
4. If ke k ≤ Me
, then for each λ ∈ C satisfying Reλ > w and for each
g ∈ F equation λf − Af = g has a solution f ∈ D(A), which is unique in
R +∞
D(A) and is given by the formula f = R(λ, A)g = 0 e −λt e tA gdt.
3 / 10
Example 1.
Consider
F = UCb (R) = {f ∈ RR |f is uniformly continuous and bounded}
which is a Banach space with the so-called ¾uniform norm¿
kf k = supx∈R |f (x)|. Suppose that functions a, b, c ∈ UCb (R) are some
known parameters, and dene operator A by equality
(Af )(x) = a(x)f 00 (x) + b(x)f 0 (x) + c(x)f (x) for all x ∈ R, f ∈ D(A).
Let (A, D(A)) be closed linear operator with D(A) satisfying
UCb2 (R) = {f ∈ UCb (R)|f 0 , f 00 ∈ UCb (R)} ⊂ D(A) ⊂ UCb (R).
Suppose that (A, D(A)) generates a C0 -semigroup (e tA )t≥0 . Then Cauchy
problem for second order linear parabolic PDE for u : [0, +∞) × R × R
(
ut (t, x) = a(x)uxx (t, x) + b(x)ux (t, x) + c(x)u(t, x), t ≥ 0, x ∈ R
u(0, x) = u0 (x), x ∈ R
has solution u(t, x) = (e tA u0 )(x), U(t) = u(t, ·) = [x 7→ u(t, x)].
Moreover second order linear ODE for f : R → R
a(x)f 00 (x) + b(x)f 0 (x) + (c(x) − λ)f (x) = −g (x), x ∈ R
R
has solution f (x) = 0+∞ e −λt (e tA g )(x)dt.
4 / 10
Example 2.
Consider dimension d ∈ N, x = (x1 , x2 , . . . , xd ) ∈ Rd and
d
F = UCb (Rd ) = {f ∈ RR |f is uniformly continuous and bounded}
which is a Banach space with the so-called ¾uniform norm¿
kf k = supx∈Rd |f (x)|. Suppose that functions aij , bi , c ∈ UCb (Rd ) are
some known parameters, and dene operator A by equality
(Af )(x) =
d
X
aij (x)fxi xj (x) +
i,j=1
d
X
bi (x)fxi (x) + c(x)f (x) for all x ∈ Rd .
i=1
Suppose that (A, D(A)) generates a C0 -semigroup (e tA )t≥0 . Then Cauchy
problem for second order linear parabolic PDE, u : [0, +∞) × Rd × R
d
u (t, x) = P a (x)u
t
i,j=1
xi xj (t, x) +
ij
d
P
i=1
bi (x)uxi (t, x) + c(x)u(t, x), t ≥ 0, x ∈ Rd
u(0, x) = u (x), x ∈ Rd
0
has solution u(t, x) = (e tA u0 )(x), U(t) = u(t, ·) = [x 7→ u(t, x)].
Moreover second order linear elliptic PDE for f : Rd → R
d
X
i,j=1
aij (x)fxi xj (x) +
d
X
bi (x)fxi (x) + (c(x) − λ)f (x) = −g (x), x ∈ Rd
i=1
has solution f (x) = 0
R +∞
e −λt (e tA g )(x)dt.
5 / 10
Approximations of semigroups
Denition. The function G is called Cherno-tangent to the
operator L i:
(CT0)Let L (F) be the set of all linear bounded operators in a
Banach space F . Let the operator L : F ⊃ Dom(L) → F be linear
and closed.
(CT1) G is dened on [0, +∞), takes values in L (F), and the
function t 7−→ G (t)f is continuous for each f ∈ F .
(CT2) G (0) = I , i.e. G (0)f = f for each f ∈ F .
(CT3) There exists such a dense subspace D ⊂ F that for each
f ∈ D there exists a limit
G 0 (0)f = lim
t→0
G (t)f − f
.
t
(CT4) The closure of the operator (G 0 (0), D) is equal to
(L, Dom(L)).
Remark. Informal meaning: G (t) = I + tL + o(t) as t → 0.
6 / 10
Approximations of semigroups
Remark. In the denition of the Cherno tangency the family
(G (t))t≥0 usually does not have a semigroup composition property,
which in fact is a reason why we
often can nd a simple formula for
G (t). Each C0 -semigroup e tL t≥0 is Cherno-tangent to its
generator L, but if L is a dierential operator with variable
coecients then usually we do not have a simple formula for e tL .
We should not expect to have such a formula beacause the Cauchy
problem for parabolic equation [ut0 (t) = Lu(t), u(0) = u0 ] has the
solution u(t) = e tL u0 , so nding a formula for e tL is equivalent to
nding a formula that solves this Cauchy problem for each u0 ∈ F ,
which is usually not an easy task. However, we can obtain
approximations to e tL u0 via the following Cherno theorem.
7 / 10
Approximations of semigroups
Theorem (P. R. Chernoff, 1968). Let F and L (F) be as
before. Suppose that the operator L : F ⊃ Dom(L) → F is linear
and closed, and function G takes values in L (F). Suppose that
these assumptions are fullled:
(E) There exists a C0 -semigroup (e tL )t≥0 with the generator
(L, Dom(L)).
(CT) G is Cherno-tangent to (L, Dom(L)).
(N) There exists such ω ∈ R, that kG (t)k ≤ e ωt for all t ≥ 0.
Then for each f ∈ F we have (G (t/n))n f → e tL f as n → ∞ with
respect to norm in F locally uniformly in t , i.e. for each T > 0
lim
sup
n→∞ t∈[0,T ]
e tL f − (G (t/n))n f
= 0.
(C )
Remark. Expressions (G (t/n))n are called Cherno approximations
to the semigroup e tL . If condition (C) holds then G is called: a
Cherno function for operator L and (sometimes) a Cherno
function for the semigroup (e tL )t≥0 , also in that case family
(G (t))t≥0 is called Cherno-equivalent to the semigroup (e tL )t≥0 .
8 / 10
Seminar 18
We discussed results of two papers:
1. I.D. Remizov. Approximations to the solution of Cauchy problem
for a linear evolution equation via the space shift operator
(second-order equation example). // Applied Mathematics and
Computation. 328 (2018), 243246.
2. O. E. Galkin and I. D. Remizov. Rate of Convergence of Cherno
Approximations of Operator C0 -Semigroups// Mathematical Notes,
2022, Vol. 111, No. 2, pp. 305307.
Also, we solved problem 3 from the HW18 below.
9 / 10
Homework 18
Deadline March 22, 20:00.
1. Explain why items 1 and 4 of theorem on p.3 holds.
2. Explain examples on p.4 and p.5.
3. Prove one-dimensional version of the Cherno theorem: suppose that
s : [0, +∞) → R, s(0) = 1, ∃s 0 (0) = a, then ∃ lim (s(t/n))n = e ta .
n→∞
4. Explain proofs of theorem 2 and theorem 3 in the following paper:
I.D. Remizov. Approximations to the solution of Cauchy problem for a
linear evolution equation via the space shift operator (second-order
equation example). // Applied Mathematics and Computation. 328
(2018), 243246.
5. Consider F = UCb (R), kf k = supx∈R |f (x)|
√ . For each f ∈√F , x ∈ R,
t ≥ 0 dene (S(t)f )(x) = 23 f (x) + 16 f (x + 6t) + 16 f (x − 6t).
5a. For each t ≥ 0 prove that S(t) ∈ L (F) and kS(t)k = 1.
5b. Suppose that ϕ ∈ Cb∞ (R) ⊂ F , and dene Lf = f 00 . Which of the
statements below are true?
A. (I − S(t)) ϕ = o(1), t → 0
B. (I + tL − S(t)) ϕ = o(t),
t→0
C. I + tL + 12 t 2 L2 − S(t) ϕ = o(t 2), t → 0
D. I + tL + 12 t 2 L2 + 31! t 3 L3 − S(t) ϕ = o(t3 ), t → 0
E. I + tL + 12 t 2 L2 + 31! t 3 L3 + 41! t 4 L4 − S(t) ϕ = o(t 4 ), t → 0
10 / 10
