Download Marian Muresan Mathematical Analysis and Applications I Draft

Marian Mureşan
Mathematical Analysis and Applications I
Draft
Forword
These lecture notes have been written having in mind a computer scientist (not a
typist) in our changing world. It is difficult to imagine the computer science without
mathematics. For those interested mainly in software we recall as an argument a capital (and classical) work of Knuth, [11]. Many results in discrete (and even continuous)
mathematics are designed to be used in several parts of the giant called the computer
science.
On the one side we have to notice the immutability of the mathematical world in
the following sense: a correct mathematical result remains true forever. Mathematics
enlarges continuously. The speed of this enlargement increases.
On the other side the needs of computer scientists change continuously. The time
designed for study remains, more or less. unchanged. Hence, a question arises. What
parts of mathematics and how to teach them in an optimal way to the computer
scientist students? Not an easy question, indeed!
We are pressed to take into account some parts from mathematics and to neglect
many other.
The mathematical analysis offers a solid ground to many achievements in applied
mathematics and discrete mathematics. In spite of the facts that these lecture notes
concern with a part of the mathematical analysis on the real axis, we have tried to
include useful and relevant examples, exercises, and results enlightening the reader on
the power of mathematical tools.
We recommend [1], [3], [4], [5], [6], [8], [9], [12], [14], [15], [16], [21], [23], [25].
iii
Contents
Forword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Sets
1.1 Sets . . . . . . . . . . . . . . . . . . . .
1.1.1 The concept of a set . . . . . . .
1.1.2 Operations on sets . . . . . . . .
1.1.3 Relations and functions . . . . . .
1.1.4 Exercises . . . . . . . . . . . . . .
1.2 Sets of numbers . . . . . . . . . . . . . .
1.2.1 An example . . . . . . . . . . . .
1.2.2 The real number system . . . . .
1.2.3 The extended real number system
1.3 Exercises . . . . . . . . . . . . . . . . . .
1.3.1 Inequalities . . . . . . . . . . . .
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1
1
1
2
4
7
8
8
9
18
19
19
2 Basic notions in topology
27
2.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.2 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3 Numerical sequences and series
3.1 Numerical sequences . . . . . . . . . . . . . . . . . . . . .
3.1.1 Convergent sequences . . . . . . . . . . . . . . . . .
3.1.2 Subsequences . . . . . . . . . . . . . . . . . . . . .
3.1.3 Cauchy sequences . . . . . . . . . . . . . . . . . . .
3.1.4 Monotonic sequences . . . . . . . . . . . . . . . . .
3.1.5 Upper and lower limits . . . . . . . . . . . . . . . .
3.1.6 Stoltz-Cesaro theorem and some of its consequences
3.1.7 Some special sequences . . . . . . . . . . . . . . . .
3.2 Numerical series . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Series of nonnegative terms . . . . . . . . . . . . .
3.2.2 The root and ration tests . . . . . . . . . . . . . . .
3.2.3 Power series . . . . . . . . . . . . . . . . . . . . . .
3.2.4 Partial summation . . . . . . . . . . . . . . . . . .
3.2.5 Absolutely and conditionally convergent series . . .
v
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37
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55
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62
65
3.2.6
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4 Euclidean spaces
71
4.1 Euclidean spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
5 Limits and Continuity
5.1 Limits . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 The limit of a function . . . . . . . . . .
5.1.2 Right-hand side and left-hand side limits
5.2 Continuity . . . . . . . . . . . . . . . . . . . . .
5.2.1 Continuity and compactness . . . . . . .
5.2.2 Uniform continuous mappings . . . . . .
5.2.3 Continuity and connectedness . . . . . .
5.2.4 Discontinuities . . . . . . . . . . . . . .
5.2.5 Monotonic functions . . . . . . . . . . .
5.2.6 Darboux functions . . . . . . . . . . . .
5.2.7 Lipschitz functions . . . . . . . . . . . .
5.2.8 Convex functions . . . . . . . . . . . . .
5.2.9 Jensen convex functions . . . . . . . . .
6 Differential calculus
6.1 The derivative of a real function . . . .
6.2 Mean value theorems . . . . . . . . . .
6.2.1 Consequences of the mean value
6.3 The continuity of derivatives . . . . . .
6.4 L’Hospital theorem . . . . . . . . . . .
6.5 Higher order derivatives . . . . . . . .
6.6 Convex functions and differentiability .
6.6.1 Inequalities . . . . . . . . . . .
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7 Integral calculus
109
7.1 The Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
7.2 The Gronwall inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 109
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Author index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
Subject index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
vi
Chapter 1
Sets
The aim of this chapter is to introduce several basic notions and results concerning
sets.
1.1
1.1.1
Sets
The concept of a set
The basic notion of set theory which was first introduced by Cantor1 will occur constantly in our results. Hence it would be fruitful to discuss briefly some of the notions
connected to it before studying the mathematical analysis.
We take the notion of a set as being already known. Roughly speaking, a set
(collection, class, family) is any identifiable collection of objects of any sort.
We identify a set by stating what its members (elements) are. The theory of sets
has been described axiomatically in terms of the notion ”member of ” ([10]).
We shall make no effort to built the complete theory of sets, but will appear
throughout to intuition and elementary logic. The so-called ”naive” theory of sets is
completely satisfactory for us ([8]).
We will usually adhere to the following notational conventions. Elements of sets
will be denoted by small letters: a, b, c, . . . , x, y, z, α, β, γ, . . . . Sets will be
denoted by capital Roman letters: A, B, C, . . . X, Y, . . . . Families of sets will be
denoted by capital script letters: A, B, C, . . . .
A set is often defined by some property of its elements. We will write {x | P (x)}
(where P (x) is some proposition about x ) to denote the set of all x such that P (x)
is true. Here | is read ”such that”.
If the object x is an element of the set A, we write x ∈ A; while x ∈
/ A means
that this x is not in A.
We write ∅ for the empty (void ) set; it has no member at all.
For any object x, {x} will denote the set whose only member is x. Then x ∈ {x},
but x 6= {x}. Similarly, {x1 , x2 , . . . , xn } is the set whose elements are precisely x1 ,
1
Georg Cantor, 1845-1918
1
2
1. Sets
x2 , . . . , xn . Let us emphasize that {x, x} = {x}.
Examples 1.1.
(a) The set of natural numbers, N = {0, 1, 2, 3, . . . }; 2
(b) The set of non-zero natural numbers, N∗ = {1, 2, 3, . . . };
(c) The set of integers Z = {0, ±1, ±2, ±3, . . . };
(d) The set of rational numbers Q = {p/q | p, q ∈ Z, q 6= 0}; 3
(e) The set of positive integers less than 7;
(f) The set of Romanian cities having more than five million of inhabitants;
(g) The set S of vowels in English alphabet. S may be written as S = {a, e, i, o, u}
or S = {x | x is a vowel in English alphabet}.
4
Let A and B be sets such that every element of A is an element of B. Then A
is called a subset of B and we write A ⊂ B or B ⊃ A. If A ⊂ B and B ⊂ A, we
write A = B. A 6= B denies A = B. If A ⊂ B and A 6= B, we say that A is a
proper subset of B and we write A ( B. We remark that under this idea of equality
of sets, the empty set is unique, i.e., if ∅1 and ∅2 are any two empty sets, we have
∅1 ⊂ ∅2 and ∅2 ⊂ ∅1 .
Let A be a set. By P(A) we denote the family of subsets of A. Thus P(∅) =
{∅, {∅}} . For A = {1, 2} , we have P(A) = {∅, {1} , {2} , {1, 2}} .
It is clear that if A is not a subset of B, the following statement has to be true:
”there is an element x such that x ∈ A and x ∈
/ B ”.
1.1.2
Operations on sets
If A and B are sets, we define A ∪ B as the set {x | x ∈ A or x ∈ B} and we call
A ∪ B the union of A and B, see figure 1.1. Let A be a family of sets; then we
define
∪A = {x | x ∈ A for some A ∈ A}.
Similarly, if {Aα }α∈I is a family of sets indexed by α, we write
∪α∈I Aα = {x | x ∈ Aα for some α ∈ I}.
For given sets A and B, we define A ∩ B as the set {x | x ∈ A and x ∈ B} and we
call A ∩ B the intersection of A and B, see figure 1.2. If A is any family of sets
we define
∩A = {x | x ∈ A for all A ∈ A}.
2
An axiomatic introduction of natural numbers may be found in many textbooks, e.g., [22, Chapter 1].
3
An axiomatic introduction of natural, integer, and rational numbers may be found, e.g., [17,
I.§2-§4].
1.1. Sets
3
Figure 1.1: Union
Figure 1.2: Intersection
Figure 1.3: Disjoint sets
Similarly, if {Aα }α∈I is a family of sets indexed by α, we write
∩α∈I Aα = {x | x ∈ Aα for all α ∈ I}.
Theorem 1.1. Let A, B, C be any sets. Then we have
(i)
(ii)
(iii)
(iv)
(v)
(vi)
A∪B =B∪A
A∪A=A
A∪∅=A
A ∪ (B ∪ C) = (A ∪ B) ∪ C
A⊂A∪B
A ⊂ B ⇐⇒ A ∪ B = B
(i’)
(ii’)
(iii’)
(iv’)
(v’)
(vi’)
A∩B =B∩A
commutative law;
A∩A=A
idempotent law;
A ∩ ∅ = ∅;
A ∩ (B ∩ C) = (A ∩ B) ∩ C associative law;
A ∩ B ⊂ A;
A ⊂ B ⇐⇒ A ∩ B = A.
Theorem 1.2. Let A, B, C be any sets. Then we have
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C);
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C);
(i)
(ii)
distributive law;
distributive law.
If A ∩ B = ∅, the sets A and B are said to be disjoint, see figure 1.3. If A
is a family of sets such that each pair of distinct members of A are disjoint, A is
said to be pairwise disjoint. Thus an indexed family {Aα }α∈I is pairwise disjoint if
Aα ∩ Aβ = ∅ whenever α, β ∈ I and α 6= β.
Let A and B be two sets. Then
A \ B := {x | x ∈ A and x ∈
/ B}
is said to be the difference of A and B , see figure 1.4.
If A ⊂ X, we define the complement of A (relative to X ) by the set {x | x ∈
X, x ∈
/ A}. This set is denoted by {X A or {A. Other notation: X \ A.
Theorem 1.3. (de Morgan4 laws)
(a) {(A ∪ B) = ({A) ∩ ({B);
(b) {(A ∩ B) = ({A) ∪ ({B);
(c) {(∪α∈I Aα ) = ∩α∈I {Aα ;
(d) {(∩α∈I Aα ) = ∪α∈I {Aα .
4
de Morgan
4
1. Sets
Figure 1.4: Difference
Figure 1.5: Symmetric difference
For sets A and B, the symmetric difference of A and B is the set (A\B)∪(B\A)
and we write A4B for this set, see figure 1.5. Note that A4B is the set consisting
of those elements which are in exactly one of A and B, and that it may also be
defined by
A4B = (A ∩ {B) ∪ ({A ∩ B).
1.1.3
Relations and functions
Sometimes it becomes significant to consider the order of the elements in a set. If we
consider a pair (x1 , x2 ) of elements in which we distinguish x1 as the first element
and x2 as the second element, then (x1 , x2 ) is called an ordered pair. Thus, two
ordered pairs (x, y) and (u, v) are equal if and only if x = u and y = v.
Let X and Y be sets. The Cartesian product of X and Y is the set X × Y of
all ordered pairs (x, y) such that x ∈ X and y ∈ Y. Hence, X × Y := {(x, y) | x ∈
X, y ∈ Y }.
Remark. (1, 2) 6= (2, 1) while {1, 2} = {2, 1}.
4
A (binary) relation R on two sets X and Y is any subset of the Cartesian
product of X and Y, i.e., R is a relation on X and Y ⇐⇒ R ⊂ X × Y.
Let R be a relation. The domain of R is the set DomR := {x | (x, y) ∈
R for some y}. The range of R is the set RangeR := {y | (x, y) ∈ R for some x}.
The symbol R−1 denotes the inverse of R, i.e., R−1 := {(y, x) | (x, y) ∈ R}.
Let R and Q be relations. The composition (product) of two relations R and Q
is the relation
Q ◦ R := {(x, z) | for some y, (x, y) ∈ R and (y, z) ∈ Q}.
The composition of R and Q may be empty. Q ◦ R 6= ∅ ⇐⇒ (RangeR) ∩ (DomQ) 6=
∅.
Let X be a set. An equivalence relation on X is any relation ∼ ⊂ X × X such
that for all x, y and z in X it hold
(i) x ∼ x (reflexive);
(ii) x ∼ y implies y ∼ x (symmetric);
(iii) x ∼ y and y ∼ z imply x ∼ z (transitive).
Examples 1.2.
(a) ”=” is an equivalence relation on the set of rational numbers Q.
1.1. Sets
5
(b) Let Z be the set of integers and fix a natural number n. For any a, b ∈ Z, ”a
is congruent to b modulo n” if a − b = kn for a k ∈ Z. Here ”congruence
modulo n” is an equivalence relation on Z.
4
Let P be a set. A partial ordering on P is any relation ≤ ⊂ P × P such that
for all x, y and z in X we have
(i) x ≤ x (reflexive);
(ii) x ≤ y and y ≤ x imply y = x (antisymmetric);
(iii) x ≤ y and y ≤ z imply x ≤ z (transitive).
Examples 1.3.
(a) Let X be a nonempty set and take A, B ⊂ X. Define A ≤ B provided A ⊂ B.
Then ” ≤ ” is a partial ordering on the class of subsets of X.
(b) For m, n ∈ N define m ≤ n provided there exists k ∈ N∗ such that m = kn.
Then ” ≤ ” is a partial ordering on N.
4
If ≤ also satisfies
(iv) x, y ∈ P implies x ≤ y or y ≤ x,
then ≤ is called a total ordering on P. If x ≤ y and x 6= y, then we write x < y.
The expression x ≥ y means y ≤ x and x > y means y < x.
Example 1.1. ≤ is a total ordering on Q.
4
If ≤ is a total ordering such that
(v) ∅ =
6 A ⊂ P implies there exists an element a ∈ A such that a ≤ x for each
x ∈ A ( a is the smallest element of A ),
then ≤ is called a well ordering on P.
A partially ordered set is an ordered pair (P, ≤), where P is a set and ≤ is a
partial ordering on P. If ≤ is a well ordering, the pair (P, ≤) is called a well-ordered
set.
Example 1.2.
well-ordered set.
The set N of natural numbers with the usual ordering ≤ is a
4
If ≤ is a total ordering on P, the pair (P, ≤) is called a totally ordered set.
Let P be a totally ordered set. For x, y ∈ P we define max{x, y} := y if x ≤ y,
and max{x, y} := x if y ≤ x. For a finite subset {x1 , . . . , xn } (not all xj ’s necessarily
distinct), we define max{x1 , . . . , xn } := max{xn , max{x1 , . . . , xn−1 }}. Similarly, we
define min{x, y}. That is, min{x, y} := x whenever x ≤ y and min{x1 , . . . , xn } :=
min{xn , min{x1 , . . . , xn−1 }}.
Let (P, ≤) be a partially ordered set and A ⊂ P. An element x ∈ P is said to
be
6
1. Sets
Figure 1.6:
Figure 1.7:
Figure 1.8:
(i) a lower bound of A if x ≤ y for any y ∈ A; in this case we say that A is
bounded below ;
(ii) an upper bound of A if y ≤ x for any y ∈ A; in this case we say that A is
bounded above;
(iii) the greatest lower bound of A or infimum of A if
(iii1) x is a lower bound of A,
(iii2) if x < y, then y is not a lower bound of A;
(iv) the least upper bound of A or supremum of A if
(iv1) x is an upper bound of A,
(iv2) if y < x, then y is not an upper bound of A.
A is bounded if it is bounded below and above.
Remark. A set A may have several lower and/or upper bounds. A set A has
at most one infimum (denoted by inf A ) and at most one supremum (denoted by
sup A ).
4
Example. Let E consists of all numbers 1/n, where n = 1, 2, . . . . Then E is
bounded, sup E = 1, inf E = 0, and 1 ∈ E while 0 ∈
/ E.
4
Let f be a relation and A be a set. The image of A under f is the set
f (A) := {y | (x, y) ∈ f for some x ∈ A}.
Observe that f (A) 6= ∅ ⇐⇒ A ∩ Domf 6= ∅. This is interpreted as ” f maps the set
A into the set B ”. The inverse image of A under f is the set
f −1 (A) := {x | (x, y) ∈ f for some y ∈ A},
[18, §3.2], [19, §1.1], and [20, §2.3].
A relation f is said to be single-valued if (x, y) ∈ f and (x, z) ∈ f imply y = z.
In this case we write f (x) = y. A single-valued relation is called a function (mapping,
application, transformation, operation). If f and f −1 are both single-valued, then f
is called a bijective function, figure 1.8.
Theorem 1.4. Let X and Y be sets and f ⊂ X × Y be a relation. Suppose that
{Ai }i∈I is a family of subsets of X and {Bj }j∈J is a family of subsets of Y. Then
(a) f (∪i∈I Ai ) = ∪i∈I f (Ai );
(b) f −1 (∪j∈J Bj ) = ∪j∈J f −1 (Bj );
1.1. Sets of numbers
7
(c) f (∩i∈I Ai ) ⊂ ∪i∈I f (Ai ).
The following statements are true if f is a function, but may fail for arbitrary relations
(c) f −1 (∩j∈J Bj ) = ∪j∈J f −1 (Bj );
(d) f −1 ({Y B) = {X (f −1 (B)), B ⊂ Y ;
(e) f (f −1 (B) ∩ A) = B ∩ f (A), A ⊂ X, B ⊂ Y.
Let f be a function such that Domf = X and Rangef ⊂ Y. Then f is said to
be a function from (on) X into (to) Y and we write f : X → Y. If Rangef = Y,
we say that f is onto, that is f (X) = Y. It means that to every y ∈ Y there exists
an x ∈ X such that y = f (x).
f is said to be injective or one-to-one if for f (x) = y and f (t) = y, then x = y.
In other words, a function f : X → Y is said to be one-to-one if distinct elements of
X have distinct images in Y, that is, if no two two different elements in X have the
same image. A suitable way in deciding whether or not a given map is injective is if
for x, t ∈ X, f (x) = f (t) implies x = t. Figure 1.6 exhibits a surjective but not an
one-to-one function, figure 1.7 exhibits an injective but not an onto function, while
figure 1.8 presents a bijective function.
Theorem 1.5. Let X and Y be nonempty sets and f : X → Y be a function. Then
f is bijective if and only if it is injective and onto.
A sequence is a function having N∗ as its domain. Sometimes we will consider
sequences having N as their domain. If x is a function, we frequently write xn
instead x(n) for the value of x at n. The value xn is called the nth term of the
sequence. The sequence whose nth term is xn will be denoted by (xn )∞
n=1 or (xn )n
or (xn ). A sequence (xn ) is said to be in X if xn ∈ X for each n ∈ N∗ .
1.1.4
Exercises
1. Let S = {0, ±1, ±2, 3} and Q (the set of rational numbers) be two sets. Then,
for a function f : S → Q, given by f (t) = t2 − 1, for all t ∈ S, find f (S).
2. Show whether or not each of the following functions is one-to-one and/or onto:
(i) the function f : N → N, defined by
f (n) = 2n,
n ∈ N;
(ii) the function f : Q × Q → Q, defined by
f (p, q) = p,
p, q ∈ Q;
(iii) the function f : Q × Q → Q × Q, defined by
f (p, q) = (p, −q) p, q ∈ Q.
8
1. Sets
1.2
Sets of numbers
A satisfactory discussion of the main concepts of analysis (e.g., convergence, continuity, differentiation and integration) must be based on an accurately defined number
concept.
We shall not, however, enter into any discussions of the axioms governing the
arithmetic of the integers, but we take the rational number system as our starting
point.
1.2.1
An example
It is well known that the rational number system is inadequate for many purposes.
Maybe the most frustrating case is the following. Let be an isosceles right triangle
having the length of a cathetus 1. Can we express the length of the hypotenuse as a
rational number?
Example. Let us begin by showing that the equation
(2.1)
p2 = 2
is not satisfied by any rational p. For, suppose that (2.1) is satisfied. Then we can
write p = m/n, where m and n are integers with n 6= 0, and we can further choose
m and n so that not both are even. Let us assume that this is done. Then (2.1)
implies
(2.2)
m2 = 2n2 .
This shows that m2 is even. Hence m is even (if m were odd, m2 would be odd),
and so m2 is divisible by 4. It follows that the right-hand side of (2.2) is divisible by
4, so that n2 is even, which implies that n is even.
Thus the assumption that (2.1) holds for a rational number leads us to the conclusion that both m and n are even, contrary to our choice of m and n. Hence (2.1)
is impossible for rational p. So, the length of the hypotenuse to an isosceles right
triangle with unitary cathetus is non-rational.
Let us examine the situation a little more closely.
Let A be the set of all positive rationals p such that p2 < 2, and let B be the
set of all positive rationals p such that p2 > 2. We shall show that A contains no
largest element, and B contains no smallest element.
More explicitly, for every p ∈ A we can find a rational q ∈ A such that p < q,
and for every p ∈ B we can find a rational q ∈ B such that q < p.
Suppose that p ∈ A. Then p2 < 2. Choose a rational h such that 0 < h < 1
and such that
2 − p2
.
h<
2p + 1
Put q = p + h. Then q > p, and
q 2 = p2 + (2p + h)h < p2 + (2p + 1)h < p2 + (2 − p2 ) = 2,
1.2. Sets of numbers
9
so that q is in A. This proves the first part of our assertion.
Next, suppose that p ∈ B. Then p2 > 2. Put
q =p−
p 1
p2 − 2
= + .
2p
2 p
Then 0 < q < p and
2
2
2
q = p − (p − 2) +
p2 − 2
2p
2
> p2 − (p2 − 2) = 2,
so that q ∈ B.
4
The purpose of the above discussion has been to show that the rational number
system has certain gaps, in spite of the fact that between any two rationals there is
another (since p < (p + q)/2 < q ).
1.2.2
The real number system
There are several ways for introducing the real number set.
We say that a set X is the real number set provided there are defined two operation
X × X 3 (x, y) 7→ x + y ∈ X,
X × X 3 (x, y) 7→ xy ∈ X
called addition and multiplication as well as a binary relation ≤ called ordering
satisfying the following axioms (conditions, assumptions)
(R 1) (x + y) + z = x + (y + z), ∀ x, y, z ∈ X;
(R 2) there exists an element 0 ∈ X, called zero or null such that x+0 = x, ∀ x ∈ X;
(R 3) for each x ∈ X there exists an element −x ∈ X, called the opposite to x, such
that x + (−x) = 0;
(R 4) x + y = y + x, ∀ x, y ∈ X;
(R 5) (xy)z = x(yz), ∀ x, y, z ∈ X;
(R 6) there exists an element 1 ∈ X \ {0}, called unity or identity such that x1 = x,
∀ x ∈ X;
(R 7) for each element x ∈ X \ {0} there exists an element x−1 ∈ X, called the
inverse of x, such that xx−1 = 1;
(R 8) xy = yx, ∀ x, y ∈ X;
(R 9) x(y + z) = xy + xz, ∀ x, y, z ∈ X;
10
1. Sets
(R 10) x ≤ x, for all x ∈ X;
(R 11) x ≤ y and y ≤ x imply x = y, for all x, y ∈ X;
(R 12) x ≤ y and y ≤ z imply x ≤ z, for all x, y, z ∈ X;
(R 13) for all x, y ∈ X we have x ≤ y or y ≤ x;
(R 14) x ≤ y implies x + z ≤ y + z, for all x, y, z ∈ X;
(R 15) if x ≥ 0 and y ≥ 0 imply xy ≥ 0;
(R 16) for every ordered pair (A, B) of nonempty subsets of X having the property
that x ≤ y for every x ∈ A and y ∈ B there exists an element z ∈ X such
that
x ≤ z ≤ y, for any x ∈ A and y ∈ B.
Remarks.
From (R 1)-(R 4) we have that (X, +) is an Abelian (commutative)
group; from (R 5)-(R 8) we have that (X \ {0}, ·) is an Abelian (commutative) group,
too; from (R 1)-(R 9) we have that (X, +, ·) is a field ; from (R 10)-(R 13) we have that
(X, ≤) is a totally ordered set; from (R 1)-(R 15) we have that (X, +, ·, ≤) is a totally
ordered field. (R 14) and (R 15) express the compatibility of the ordering relation with
the algebraic operations. (R 16) has a special rôle which will be clear a little bit later.
1
y
x−1 ∈ X \ {0} from (R 7) is denoted as , too. Hence
= yx−1 .
4
x
x
For a while let us ignore assumption (R 16).
Proposition 2.1. From (R 1)-(R 15) it follows that
(1) (a)
(b)
(2) (a)
(b)
(c)
(3) (a)
(b)
(c)
(d)
(4)
(5) (a)
(b)
(c)
(d)
x1 ≤ x2 and y1 ≤ y2 imply x1 + y1 ≤ x2 + y2 ;
x1 < x2 and y1 ≤ y2 imply x1 + y1 < x2 + y2 ;
x > 0 if and only if x−1 > 0;
x ≥ 0 implies − x ≤ 0;
x > 0 implies − x < 0;
x ≤ y and z > 0 imply xz ≤ yz;
x < y and z > 0 imply xz < yz;
x ≤ y and z < 0 imply xz ≥ yz;
x < y and z < 0 imply xz > yz;
if xy > 0, then x ≤ y if and only if 1/x ≥ 1/y;
0 ≤ x1 ≤ x2 and 0 ≤ y1 ≤ y2 imply x1 y1 ≤ x2 y2 ;
0 < x1 < x2 and 0 < y1 ≤ y2 imply x1 y1 < x2 y2 ;
x1 ≤ x2 ≤ 0 and y1 ≤ y2 ≤ 0 imply x1 y1 ≥ x2 y2 ;
x1 < x2 ≤ 0 and y1 ≤ y2 < 0 imply x1 y1 > x2 y2 .
1.2. Sets of numbers
11
The absolute value function is defined by
(
x,
for x ∈ X, |x| =
−x,
x ≥ 0,
x < 0.
Proposition 2.2. From (R 1)-(R 15), it follows that
(1) (a)
(b)
(c)
(2) (a)
(b)
(3) (a)
(b)
(4) (a)
(b)
(c)
|x| ≥ 0;
|x| = 0 ⇐⇒ x = 0;
|x| = | − x|;
|x + y| ≤ |x| + |y|;
|x − y| ≥ | |x| − |y| |;
|x| ≤ a ⇐⇒ −a ≤ x ≤ a;
|x| < a ⇐⇒ −a < x < a;
|xy| = |x| · |y|;
x |x|
=
y |y| ;
|xn | = |x|n ;
The distance function is defined by
for x, y ∈ X,
d(x, y) = |x − y|.
Proposition 2.3. From proposition 2.2 it follows that
d(x, y) = 0 ⇐⇒ x = y;
d(x, y) = d(y, x), ∀ x, y ∈ X;
d(x, y) ≤ d(x, z) + d(z, y), ∀ x, y, z ∈ X.
Warning. There exist several systems satisfying (R 1)-(R 16) axioms. But all are
algebraically and order isomorphic, [9, theorem 5.34].
We choose one of them and called it the set of real numbers and denoted it by
R = (R, +, ·, ≤) . Any element x ∈ R is called a real number. Any real number x
such that 0 ≤ x is called non-negative, while 0 < x is called positive number. Any
real number x such 0 ≥ x is called non-positive, while 0 > x is called negative
number.
Proposition 2.4. Number 1 is positive.
Proof. Suppose that 1 is non-positive, i.e., 1 ≤ 0. Adding −1 to the both sides
we have 0 ≤ −1. Multiplying both sides by the non-negative number −1 and using
(R 15), we get 0 ≤ (−1)(−1) ⇐⇒ 0 ≤ 1. Now, 1 is simultaneously non-negative
and non-positive, so 0 = 1. But this contradicts (R 6). Hence 1 > 0.
2
It is clear that any set of real number (i.e., any subset of R ) having an infimum
is nonempty and bounded below. The converse statement is true, too.
12
1. Sets
Theorem 2.1. Every nonempty and bounded below subset A of R has an infimum.
Proof. Denote by A0 the set of lower bounds of A. Since A is bounded below,
A0 6= ∅. Remark that the ordered system (A0 , A) has the property that for every
x ∈ A0 and y ∈ A it holds x ≤ y. From (R 16) it follows there exists a real number
z such that
x ≤ z ≤ y for any x ∈ A0 and y ∈ A.
It results that the number z is the greatest element in A0 , i.e., the infimum of A.
2
Corollary 2.1. If A is a nonempty and bounded below subset of R and B is a
nonempty subset of A, then
inf A ≤ inf B.
Theorem 2.2. Every nonempty and bounded above subset A of R has a supremum.
Corollary 2.2. If A is a nonempty and bounded above subset of R and B is a
nonempty subset of A, then
sup A ≥ sup B.
Remark. The proofs of the existence of an infimum and the existence of a supremum
have used axiom (R 16). At the same time it can be proved that (R 16) follows from
any one of these theorems. Thus (R 16) is equivalent to any one of these theorems.
Hence we may substitute (R 16) by any one of these statements in order to get the
same real number system.
4
Theorem 2.3. Suppose X is a totally ordered field (i.e., it satisfies axioms (R 1)(R 15)) and, moreover, every nonempty and bounded above subset of it has a supremum.
Then it is fulfilled axiom (R 16).
Proof. Consider an ordered pair (A, B) of nonempty subsets of X having the property
that x ≤ y for any x ∈ A and y ∈ B. Then A is nonempty and bounded above (by
any element of B ). It follows that there exists z ∈ X such that
(2.3)
z = sup A.
We have to show that z ≤ y, for every y ∈ B. For, suppose there exists y0 ∈ B such
that y0 < z. Then y0 is an upper bound of A strictly less then z, contradicting
(2.3).
2
A similar statement holds for infimum.
Theorem 2.4. Suppose X is a totally ordered field and, moreover, every nonempty
and bounded below subset of it has an infimum. Then it is fulfilled axiom (R 16).
Theorem 2.5. (Archimedes5 principle, [3, theorem 6.5.1, p. 72]) For any two real
numbers x and y such that y > 0 there exists a natural number n such that x < ny.
5
Archimedes, 287-212
1.2. Sets of numbers
13
Proof. Under the above-mentioned assumptions define
A = {u ∈ R | ∃ n ∈ N∗ , u < ny}.
and remark that A 6= ∅ (since, at least y ∈ A ). We wish to show that A = R. For,
suppose that A 6= R and denote B = R \ A. Obviously, B 6= ∅.
Note that for any u ∈ A and v ∈ B, u < v. Indeed, for any u ∈ A there exists
a natural n such that u < ny. Since v ∈
/ A and that the real number set is a totally
ordered set it follows that ny ≤ v. Then
u < ny ≤ v =⇒ u < v.
Axiom (R 16) implies that for the ordered pair (A, B) there exists a real number
z such that
(2.4)
u ≤ z ≤ v,
∀ u ∈ A, v ∈ B.
The real number z − y belongs to A, since otherwise z − y ∈ B, and then by (2.4)
z ≤ z − y =⇒ y ≤ 0,
contradicting the hypothesis. So z − y ∈ A. Then we can find a natural number n
such that z − y < ny. We also have
z + y = (z − y) + 2y < (n + 2)y,
and it follows that z + y ∈ A. Then z + y ≤ z, so y ≤ 0. The contradiction shows
that A = R and the theorem is proved.
2
Remark. It can be shown that axiom (R 16) is equivalent to the Archimedes principle.
4
Theorem 2.6. The supremum of a nonempty and bounded above set is unique.
Proof. For, suppose that sup A = a1 and sup A = a2 and a1 6= a2 . Then either
a1 < a2 or a2 < a1 . In both cases we get a contradiction.
2
Theorem 2.7. A real number a is the supremum of a set A if and only if
(i) for any x ∈ A, x ≤ a;
(ii) for any ε > 0 there is y ∈ A such that y > a − ε.
Proof. (i) says that a is an upper bound of A, while (ii) shows that there is no upper
bound less then a.
2
Theorem 2.8. ([23, theorem 1.37, p. 11]) For every real x > 0 and every integer
n > 0, there is one and only one real y > 0 such that y n = x.
14
1. Sets
√
Remark. This number y is written n x or x1/n .
Proof. That there is at most one such y is clear, since 0 < y1 < y2 implies y1n < y2n .
Let E be the set consisting of all positive reals t such that tn < x.
If t = x/(1 + x), then 0 < t < 1; hence tn < t < x, so E is not empty.
Put t0 = 1 + x. Then t > t0 implies tn ≥ t > x, so that t ∈
/ E, and t0 is an
upper bound of E. Let y = sup E (which exists, by theorem 2.2).
Suppose y n < x. Choose h such that 0 < h < 1 and such that
x − yn
h<
.
(1 + y)n − y n
We have
n n−1
n n−1
n
n
(y + h) = y +
y h + ··· + h ≤ y + h
y
+ ··· + 1
1
1
= y n + h[(1 + y)n − y n ] < y n + (x − y n ) = x.
n
n
Thus y + h ∈ E, contradicting the fact that y is an upper bound of E.
Suppose y n > x. Choose k such that 0 < k < 1 such that k < y and such that
yn − x
.
k<
(1 + y)n − y n
Then, for t ≥ y − k, we have
n n−1
n n−2 2
n
n
n
t ≥ (y − k) = y −
y k+
y k − · · · + (−1)n k n
1
2
n n−2
n n−1
n
n−1 n−1
=y −k
y
−
y k + · · · + (−1) k
1
2
n n−1
n n−2
n
≥y −k
y
+
y
+ ··· + 1
1
2
= y n − k[(1 + y)n − y n ] > y n + (x − y n ) = x.
Thus y − k is an upper bound of E, contradicting the fact that y = sup E. It follows
that y n = x.
2
An interval A of the real number system is a subset of R so that for every x, y ∈ A
and z ∈ R satisfying x ≤ z ≤ y, we have z ∈ A. An interval bounded below and
above is said to be bounded. Otherwise it is called unbounded. For any nonempty and
bounded interval A, the non-negative real number l(A) = sup A − inf A is said to
be the length of A.
We remark that for any real numbers a and b with a ≤ b the next sets are
intervals
[a, b] = {x ∈ R | a ≤ x ≤ b}
[a, b[ = {x ∈ R | a ≤ x < b}
]a, b] = {x ∈ R | a < x ≤ b}
]a, b[= {x ∈ R | a < x < b}
closed interval;
left closed right open interval;
left open right closed interval;
open interval.
1.2. Sets of numbers
15
Theorem 2.9. Let (Ik )k∈N∗ be a nested sequence of nonempty closed intervals in R,
i.e.,
(2.5)
Ik+1 ⊂ Ik ,
k ∈ N∗ .
Then
∩k∈N∗ Ik 6= ∅.
Proof. Denote Ik = [ak , bk ], k ∈ N∗ . From (2.5) it follows that
(2.6)
ak ≤ ak+1 ≤ bk+1 ≤ bk ,
k ∈ N∗ .
Denote A = {x | x = ak , for some k ∈ N∗ } and B = {y | y = bk , for some k ∈ N∗ }.
Then for any x ∈ A and any y ∈ B we have x ≤ y, since otherwise there are ak ∈ A
and bm ∈ B such that
bm < ak .
We have either m < k or k < m. Suppose m < k. Then
bm < a k ≤ bk ,
thus contradicting (2.6).
Axiom (R 16) supplies a real z such that
ak ≤ z ≤ b k ,
k ∈ N∗ .
Then z ∈ Ik , for every k ∈ N∗ , and therefore z ∈ ∩Ik∈N∗ .
2
Theorem 2.10. For every real x there exits a unique integer k such that k − 1 ≤
x < k.
Let x be a real number. Its integer part is the unique (by theorem 2.10) integer
k satisfying
k − 1 ≤ x < k,
and it is denoted by [x]. Hence, the integer part function is defined by
R 3 x 7→ [x] ∈ Z.
The fractional part of a real number x is defined as x − [x] and it is denoted by {x}.
So, the fractional part function is defined by
R 3 x 7→ {x} ∈ [0, 1[ .
Theorem 2.11. For every two real numbers x and y such that x < y there exists
a rational lying between them, i.e., x < u < y, for an element u ∈ Q.
16
1. Sets
Proof. Based on Archimedes principle (theorem 2.5) for the positive real y − x there
exists a natural n such that 1 < n(y − x). Then
1/n < y − x.
(2.7)
From theorem 2.10 it follows that there exists an integer m such that
m ≤ nx < m + 1.
(2.8)
Obviously, u = (m + 1)/n is a rational, and satisfies x < u. From the left-hand side
of (2.8) as well as from (2.7) we infer that u also satisfies
u=
m 1
1
+ ≤ x + < y. 2
n
n
n
Corollary 2.3. Given any two real numbers x and y such that x < y, there exists
an irrational number v (from R \ Q ) such that x < v < y.
√
Proof. Choose any irrational number v0 ( 2, for example). Then x − v0 < y − v0
and by theorem 2.11 there exists a rational u such that x − v0 < u < y − v0 , i.e.,
x < v0 + u < y. Finally, we remark that v = v0 + u is irrational, since otherwise it
follows that v0 itself is rational, and this is not the case.
2
Let A and B be two sets. If there exists a bijective mapping from A onto B, we
say that A and B have the same cardinal number or that A and B are equivalent,
and we write A ∼ B.
Theorem 2.12. The relation ∼ defined above is an equivalence relation.
For any positive integer n, let N∗n be the set whose elements are precisely the
integers 1, 2, . . . , n. For a set A we say that
(a) A is finite if A ∼ N∗n for some n (the empty set is, by definition, finite). The
number of elements of a nonempty finite set A is n provided A ∼ N∗n . In this
case we write |A| = n, and we read ”‘the number of elements of the nonempty
and finite set is equal to n”. By definition, |∅| = 0.;
(b) A is infinite if A is not finite;
(c) A is countable if A ∼ N∗ . We write |A| = ℵ0 ;
(d) A is uncountable if A is neither finite nor countable. We write |A| ≥ ℵ1 > ℵ0 ;
(e) A is at most countable (or denumerable) if A ∼ N∗ or A ∼ N∗n for some
n ∈ N∗ . We write |A| ≤ ℵ0 ;
Remarks 2.1. (a) For two finite sets A and B, we obviously have A ∼ B if and only
if A and B contain the same number of elements. For infinite sets, however, this is not
exactly so. Indeed, let M be the set of all even positive integers, M = {2, 4, 6, . . . }.
1.2. Sets of numbers
17
Figure 1.9: Infinite array
It is clear that M is a proper subset of N∗ . But N∗ ∼ M, since N∗ 3 n 7→ 2n ∈ M
is a bijection.
(b) The sets {1, −1, 2, −2, 3, −3, . . . } and {0, 1, −1, 2, −2, 3, −3, . . . } are equivalent. Indeed, the function
(
0,
k = 1,
bk =
ak−1 , k > 1
maps the k rank term ak of the first set to the k + 1 rank term in the second set in
a bijective way. We conclude that the sets N∗ and Z are equivalent. Therefore we
can write |Z| = ℵ0 .
4
Exercise. Let A be a finite set. Then |P(A)| = 2|A| .
Theorem 2.13. Every infinite subset of a countable set is countable.
Theorem 2.14. Let {An }, n = 1, 2, . . . , be a sequence of countable sets, and put
B = ∪∞
n=1 An .
(2.9)
Then B is countable.
Proof. Let every set An be arranged in a sequence (xn k )n , k = 1, 2, . . . , and
consider the infinite array in which the elements of An form the nth row, figure 1.9.
The array contains all elements of B. As indicated by the arrows, these elements can
be arranged in a sequence
(2.10)
x11 , x21 , x12 , x31 , x22 , x13 , . . . .
If any two of the elements An have elements in common, these will appear more than
once in (2.10). Hence there is a subset C of the set of all positive integers such that
C ∼ B, which shows that B is at most countable (theorem 2.13). Since A1 ⊂ B,
and A1 is infinite, B is infinite, and thus countable.
2
Corollary 2.4. Suppose A is at most countable and for every α ∈ A, Bα is at most
countable. Put
C = ∪α∈A Bα .
Then C is at most countable.
Proof. For C is equivalent to a subset of (2.9).
2
18
1. Sets
Theorem 2.15. Let A be a countable set, and let Bn be the set of all ordered
n -tuples (a1 , a2 , . . . , an ), where ak ∈ A (k = 1, . . . , n), i.e.,
Bn = |A × A ×
{z· · · × A} .
n times
Then Bn is countable.
Proof. That B1 is countable is obvious, since B1 = A. Suppose Bn−1 is countable
(n = 2, 3, . . . ). The elements of Bn are of the form
(2.11)
(b, a) (b ∈ Bn−1 , a ∈ A).
For every fixed b, the set of pairs (b, a) is equivalent to A, and hence, countable.
Thus Bn is a countable union of countable sets. By theorem 2.14, Bn is countable.
2
Corollary 2.5. The set of all rational numbers is countable.
Proof. We apply theorem 2.15 with n = 2, noting that every rational r is of the
form a/b, where a and b are integers and b 6= 0. The set of such pairs (a, b), and
therefore the set of fractions a/b, is countable.
2
Therefore we can write |Q| = ℵ0 .
Theorem 2.16. Let A be the set of all sequences whose elements are the digits 0
and 1. Then A is uncountable.
Proof. Let B be a countable subset of A, and let B consists of the sequences
s1 , s2 , . . . . We construct a sequence s as follows. If the nth digit in sn is 1 we let
the nth digit of s be 0, and vice versa. Then the sequence s differs from every
member of B in at least one place; hence s ∈
/ B. But clearly s ∈ A, so that B is a
proper subset of A.
We have shown that every countable subset of A is a proper subset of A. It
follows that A is uncountable (for otherwise A would be a proper subset of A,
which is absurd).
2
Corollary 2.6. The real number set is uncountable.
Proof. Use the binary representation of the real numbers and apply theorem 2.16.
2
Therefore we can write |R| = ℵ1 .
1.2.3
The extended real number system
The extended real number set consists of the real number set to which two symbols,
+∞ and −∞ have been adjoined, with the following properties
1.2. Exercices
19
(a) if x is real, −∞ < x < +∞, and
x
x
x + ∞ = +∞, x − ∞ = −∞,
=
= 0;
+∞
−∞
(b) if x > 0, x(+∞) = +∞, x(−∞) = −∞;
(c) if x < 0, x(+∞) = −∞, x(−∞) = +∞.
The extended real number system is denoted by R = R ∪ {+∞} ∪ {−∞} with the
above-mentioned conventions.
Any element of R is called finite while +∞ and −∞ are called infinities.
Let A be a nonempty subset of the extended real number set. If A is not bounded
above (i.e., for every real y there is an x ∈ A such that y < x ), we define sup A =
+∞. Similarly, if A is not bounded below (i.e., for every real y there is an x ∈ A
such that y > x ), we define inf A = −∞.
We define intervals involving infinities
[a, +∞[ = {x ∈ R | a ≤ x};
[a, +∞] = {x ∈ R | a ≤ x ≤ +∞};
] − ∞, a] = {x ∈ R | x ≤ a};
[−∞, a] = {x ∈ R | −∞ ≤ x ≤ a};
[−∞, ∞] = R;
] − ∞, ∞[ = R.
1.3
1.3.1
]a, +∞[ = {x ∈ R | a < x};
]a, +∞] = {x ∈ R | a < x ≤ +∞};
] − ∞, a[ = {x ∈ R | x < a};
[−∞, a[ = {x ∈ R | −∞ ≤ x < a};
Exercises
Inequalities
The proofs of Hölder6 inequalities are based on the W. H. Young7 inequality.
Theorem 3.1. (The integral form of Young inequality, [9, p.189]) Let f be a continuous and strictly increasing function defined on [0, ∞) such that limu→∞ f (u) = ∞
and f (0) = 0. Denote g = f −1 . For x ∈ [0, ∞) we define the following functions
Z x
Z x
(3.1)
F (x) =
f (u)du and G(x) =
g(v)dv.
0
6
7
Otto Hölder,
W. H. Young,
0
20
1. Sets
Then a, b ∈ [0, ∞) imply
ab ≤ F (a) + G(b),
(3.2)
and equality holds if and only if b = f (a).
Corollary 3.1. (Young inequality, [9, p. 90]) Suppose p > 1 and α and β are
non-negative reals. Then
αp β q
αβ ≤
+ ,
p
q
(3.3)
if
1 1
+ = 1.
p q
The equality holds if and only if αp = β q .
Proof. The first approach runs as follows. For u ∈ [0, ∞), define f (u) = up−1 and
apply theorem 3.1.
The second approach runs as follows. Consider the function f : ]0, ∞[ → R given
by
xp x−q
+
.
f (x) =
p
q
It has an absolute minimum at x = 1. The required inequality follows from f (1) ≤
1
1
f (α q β − p ).
2
Let a = (α1 , . . . , αn ) ∈ Rn or Cn . If r 6= 0, define the weighted mean with weight
r of the finite sequence a as
!1/r
n
X
1/r
X
r
r
(3.4)
Mr (a) =
|αk |
=
|αk |
.
k=1
Proposition 3.1. Suppose p > 1, 1/p + 1/q = 1, and there are given two finite
sequences a = (α1 , . . . , αn ) and b = (β1 , . . . , βn ) satisfying Mp (a) = Mq (b) = 1.
Then
M (ab) = M1 (ab) ≤ 1,
(3.5)
where ab = (α1 β1 , . . . , αn βn ).
Proof. Choose k ∈ {1, . . . , n}. We apply Young inequality (3.3) to |αk | and |βk |. It
follows
|αk βk | ≤
(3.6)
|αk |p |βk |q
+
.
p
q
Summing up (3.6) for k = 1, 2, . . . , n, we get
X
|αk βk | ≤
1X
1X
1 1
|αk |p +
|βk |q = + = 1. 2
p
q
p q
1.3. Exercises
21
Theorem 3.2. (Hölder inequality for p > 1 and q > 1 ) Suppose p > 1, 1/p+1/q =
1, a = (α1 , . . . , αn ), b = (β1 , . . . , βn ) are two finite sequences satisfying Mp (a) > 0
and Mq (b) > 0. Then
X
(3.7)
M (ab) =
|αk βk | ≤ Mp (a)Mq (b).
Proof. Define
αk =
αk
,
Mp (a)
βk =
a = (α1 , . . . , αn ),
βk
,
Mq (b)
b = (β 1 , . . . , β n ).
We remark that Mp (a) = Mq (b) = 1, and therefore we can apply proposition 3.1.
2
Thus we find that M (ab) ≤ 1, i.e., (3.7).
Theorem 3.3. Inequality (3.7) turns into an equality if and only if the fraction
does not depend upon k. (the fraction
|αk |p
|βk |q
0
is excluded)
0
Theorem 3.4. (Hölder inequality for positive weights, [21, Part 2, Chapter 2, problem 81.3]) Consider m ∈ N∗ , m ≥ 2, aj = (αj1 , . . . , αjn ), j = 1, . . . , m, and
P 1
p1 , p2 , . . . , pm > 0 so that
= 1. Suppose that Mpj (aj ) > 0, j = 1, . . . , m.
pj
Then
n Y
m
m
Y
X
(3.8)
Mpj (aj ).
αjk ≤
k=1 j=1
j=1
Proof. If m = 2, theorem 3.4 reduces to theorem 3.2. Suppose that m ≥ 3 and than
we prove (3.8) by induction. Consider that (3.8) is true for m − 1 and we prove it
for m. Using theorem 3.2 we have
 p1p−1

m
m
! p p−1
1
1
n Y
n
n
m
X
Y
1
X
X
Y


|α1k | αjk ≤ Mp1 (a1 )
|αjk |
αjk =
k=1 j=1
k=1
j=2
"
(3.9)
= Mp1 (a1 )
k=1
n Y
m
X
|αjk |
p1
p1 −1
j=2
# p1p−1
1
.
k=1 j=2
We remark that
pj (p1 − 1)
> 0, j = 2, . . . , m,
p1
and
m
X
j=2
m
p1
p1 X 1
=
= 1.
pj (p1 − 1)
p1 − 1 j=2 pj
22
1. Sets
Thus (3.9) is further evaluated as
≤Mp1 (a1 )
 "
m
n
Y
X

(3.10)
=Mp1 (a1 )
j=2
pj (p1 −1)
p1
p1
j (p1 −1)
 p1 −1
 p1

k=1
m
n
Y
X
j=2
p1
|αjk | p1 −1
#p
! p1
j
pj
|αjk |
=
m
Y
Mpj (aj ). 2
j=1
k=1
Theorem 3.5. (Hölder inequality for 0 < p < 1 ) Consider 0 < p < 1, 1/p+1/q = 1,
and two finite sequences of positive numbers a = (α1 , . . . , αn ) and b = (β1 , . . . , βn ).
Then
M (ab) ≥ Mp (a)Mq (b).
(3.11)
−1
1
1
Proof. Take u = 1/p (> 1), 1/u + 1/v = 1. We define γk = βk u , δk = βku αku ,
k = 1, . . . , n. So γku = βkq . Using theorem 3.2, we get
n
X
αkp
=
k=1
=
X
αk βk
n
X
γk δk ≤
X
δku
u1 X
γkv
v1
k=1
p X
v
−u
βk
v1
=
X
αk βk
p X
βkq
v1
.
Hence
Mp (a)Mq (b) ≤ M (ab) 2
Theorem 3.6. (Hölder inequality for negative weights, [4]) Consider m ∈ N∗ , m ≥
2, finite and nonzero sequences aj = (αj1 , . . . ,P
αjn ), j = 1, . . . , m, and the weights
m
p1 , p2 , . . . , pm−1 < 0 and pm ∈ ]0, 1[ satisfying
j=1 1/pj = 1. Then
n Y
m
m
Y
X
(3.12)
Mpj (aj ).
αjk ≥
k=1 j=1
j=1
Proof. If m = 2, this theorem reduces to theorem 3.5. Suppose that (3.12) holds for
an m ≥ 2. We show that it
holds for m+1. Therefore consider p1 , p2 , . . . , pm < 0
Palso
m+1
and pm+1 ∈ R such that
j=1 1/pj = 1, and aj = (αj1 , . . . , αjn ) are nonzero for all
j = 1, . . . , m + 1. Thus 0 < pm+1 < 1. Hence
n m+1
X
Y
k=1
j=1

n
n
X
m+1
X
Y
αjk =
|α1k | αjk ≥ Mp1 (a1 ) 
j=2
k=1
"
(3.13)
= Mp1 (a1 )
n m+1
X
Y
k=1 j=2
k=1
|αjk |
p1
p1 −1
#
p1 −1
p1
.
m+1
Y
j=2
 p1p−1
1
! p p−1
1
1
|αjk |

1.3. Exercises
23
We remark that
pm+1 (p1 − 1)
pj (p1 − 1)
< 0, j = 2, . . . , m,
> 0,
p1
p1
m+1
m+1
X
p1
p1 X 1
=
= 1.
p
(p
−
1)
p
−
1
p
j
1
1
j
j=2
j=2
and
Then (3.13) is further evaluated as
≥Mp1 (a1 )

" n
m+1
Y X

=Mp1 (a1 )
j=2
#p
p1
j (p1 −1)
 p1 −1
 p1

k=1
m
n
Y
X
j=2
|αjk |
p1 pj (p1 −1)
p1 −1
p1
! p1
j
pj
|αjk |
=
m
Y
Mpj (aj ). 2
j=1
k=1
Theorem 3.7. (Minkowski8 inequality for p ≥ 1. ) Suppose there are satisfied the
assumptions of theorem 3.2. Then
Mp (a + b) ≤ Mp (a) + Mp (b).
(3.14)
Proof. For p = 1 the above inequality follows from the inequality given in (2) (a) of
the proposition 2.2.
Suppose that p > 1. We apply the Hölder inequality (3.7) for the following two
pairs of finite sequences
(ak )nk=1 , (|ak + bk |p−1 )nk=1
and
(bk )nk=1 , (|ak + bk |p−1 )nk=1 .
Then we have the following estimations
X
|ak + bk |p ≤
X
|ak | · |ak + bk |p−1 +
n
X
|bk | · |ak + bk |p−1
k=1
≤ (Mp (a) + Mp (b))
X
|ak + bk |p
1/q
.
P
Now, dividing the both sides by ( |ak + bk |p )1/q , we get inequality (3.14).
2
Proposition 3.2. (Bernoulli9 inequality) For every n ∈ N∗ and every x ≥ −1 it
holds
(3.15)
8
9
Hermann Minkowski, 1864-1909
Bernoulli,
(1 + x)n ≥ 1 + nx.
24
1. Sets
Proof. For x = −1 the left-hand side of (3.15) is null while the right-hand is nonpositive. So we may suppose that x > −1.
For x > −1 we prove (3.15) by induction. If n = 1, we have an equality. If
n = 2, then
(1 + x)2 = 1 + 2x + x2 ≥ 1 + 2x,
so the inequality holds. Suppose now that inequality (3.15) holds for a k ∈ N∗ and
for every x > −1, that is
(1 + x)k ≥ 1 + kx.
(3.16)
We will prove that (3.15) holds for k + 1 and every x > −1, too. Since x > −1,
1 + x > 0. We multiply (3.16) by 1 + x > 0 getting
(1 + x)k+1 ≥ 1 + kx + x + kx2 = 1 + (k + 1)x + kx2 ≥ 1 + (k + 1)x.
So, (3.15) holds for k + 1 and every x > −1.
Thus we conclude inequality (3.15) holds for every n ∈ N∗ and every x ≥ −1.
2
Proposition 3.3. (Bernoulli inequality) Consider n real numbers xi > −1, i =
1, 2, . . . , n such that all of them have the same sign. Then
(3.17)
(1 + x1 )(1 + x2 ) . . . (1 + xn ) ≥ 1 + x1 + x2 + · · · + xn .
Proof. Similar to the one supplied for the previous proposition.
2
Proposition 3.4. (Mean inequality) Let x1 , x2 , . . . xm be positive reals. Then the
geometric mean is less or equal to the arithmetic mean, i.e.
(3.18)
√
m
x1 x2 . . . xm ≤
x1 + x2 + · · · + xm
.
m
Corollary 3.2. Let x1 , x2 , . . . xm be positive reals. Then the harmonic mean is
less or equal to the geometric mean, i.e.
√
m
m
(3.19)
x 1 x2 . . . xm .
1
1
1 ≤
+ x2 + · · · + xm
x1
Proof. Substitute xi → 1/xi in (3.18).
2
Proof of proposition 3.4.
Cauchy’s10 approach. First we prove the mean inequality introducing an extra
assumption, namely m = 2k . Later on we will remove this assumption.
If k = 1, the mean inequality is known. For k = 2 we follow the next way
√
√
q
x1 x2 + x3 x4
√
√
√
4
x1 x2 x3 x4 =
x1 x2 x3 x4 ≤
2
x1 +x2
x3 +x4
+
x
+
x
+
x3 + x4
1
2
2
≤ 2
=
.
2
4
10
Augutin Louis Cauchy, 1789-1857
1.3. Exercises
25
Now, suppose inequality (3.18) holds for m = 2k , i.e.,
√
2k
(3.20)
x 1 + x 2 + · · · + x 2k
.
2k
x 1 x 2 . . . x 2k ≤
Then
√
2k+1
√
√
k
x1 . . . x2k 2 x2k +1 . . . x2k+1
√
√
k
2k
x1 . . . x2k + 2 x2k +1 . . . x2k+1
.
≤
2
x1 x2 . . . x2k+1 =
q
2k
Now using (3.20), we get that (3.19) holds for any m equal to a power of 2.
It remained the case of unrestricted m. For, suppose that m is not a power of 2.
Take a k such that m < 2k and denote
l :=
x1 + x2 + · · · + xm
(> 0).
m
Take
xk+1 = · · · = x2k =: l
and consider inequality (3.20). Then we write
√
2k
√
2k
x1 x2 . . . xm · l
ml + (2k − m)l
=l
2k
⇐⇒ x1 . . . xm ≤ lm . 2
2k −m
2k
m
x 1 x 2 . . . x m ≤ l 2k
≤
Second approach. This approach consists in considering a special case of (3.19) which
is, actually, equivalent to it. This special case reads as
(3.21)
x1 , . . . , xm > 0, x1 . . . xm = 1 =⇒ x1 + x2 + . . . xm ≥ m.
If m = 1 or m = 2, (3.21) is trivial. Suppose that (3.21) holds for m = n, i. e.,
(3.22)
x1 , . . . , xn > 0, x1 . . . xn = 1 =⇒ x1 + x2 + . . . xn ≥ n.
Suppose also that there are given n + 1 positive real numbers satisfying
x1 x2 . . . xn xn+1 = 1.
We prove that their sum is at least n + 1.
If either all xi ’s are equal to 1, and then the conclusion follows at once, or there
are two numbers one greater than 1, the other less than 1. We suppose that xn > 1
and xn+1 < 1. Now we consider n positive numbers, namely
x1 , x2 , . . . , xn−1 , xn · xn+1 .
26
1. Sets
Based on the induction hypothesis, (3.22), we get the following lower estimation
x1 + x2 + · · · + xn−1 + xn xn+1 ≥ n.
Then we continue the estimation
x1 + x2 + · · · + xn+1 ≥ n − xn xn+1 + xn + xn+1 = n + 1 + (1 − xn+1 )(xn − 1)
≥ n + 1.
Thus we infer that implication (3.22) holds for any m ∈ N∗ . Hence it follows the
mean inequality, too.
2
Proposition 3.5. (Lagrange11 identity) Let ai and bi be real numbers, i = 1, . . . , m.
Then
! m !
!2
m
m
X
X
X
X
(3.23)
a2i
b2i =
ai b i +
(ai bj − aj bi )2 .
i=1
i=1
i=1
Proof. It follows at once by induction.
1≤i<j≤m
2
Corollary 3.3. Let ai and bi be real numbers, i = 1, . . . , m. Then
! m !
!2
m
m
X
X
X
(3.24)
a2i
b2i ≥
ai b i .
i=1
i=1
i=1
Corollary 3.4. Let ai and bi be real numbers, i = 1, . . . , m. Then
! m !
!2
m
m
X
X
X
a2i
b2i =
ai bi
i=1
i=1
i=1
if and only if ai bj − aj bi = 0 for all 1 ≤ i < j ≤ m.
Corollary 3.5. Let ai be real numbers, i = 1, . . . , m. Then
r
a1 + a2 + · · · + am
a21 + a22 + · · · + a2m
≤
.
m
m
Proof. Take bi = 1, i = 1, . . . , m, in inequality (3.24).
11
Joseph Louis Lagrange, 1736-1813
2
Chapter 2
Basic notions in topology
This chapter is dedicated introducing basic notions and results concerning, mainly,
metric spaces.
2.1
Metric spaces
A set X, whose elements are called points, is said to be a metric space if with
any two points x and y of X there is associated a real number ρ(x, y), called the
distance from x to y, such that
(a) ρ(x, y) > 0 if x 6= y; ρ(x, x) = 0;
(b) ρ(x, y) = ρ(y, x);
(c) ρ(x, y) ≤ ρ(x, z) + ρ(z, y), for any z ∈ X.
The distance function ρ is called a metric on X, too.
From (a) we have that the distance ρ is non-negative. From (b) we have that the
distance from x to y is same to the distance from y to x. So, the distance function
is symmetrical. (c) is called the triangle inequality.
Examples. (a) Consider a nonempty set X and define on it the following metric
(
0, x = y,
(1.1)
ρ(x, y) =
1, x 6= y.
Thus we notice on a nonempty set we can define at least a metric.
(b) We recall the definition of the distance function given on R at the page 11. It
follows that (R, ρ), where ρ(x, y) = |x − y|, is a metric space. This metric is called
the Euclidean metric or the Euclidean distance on R.
(c) Consider the complex plane C and define on it the following distance function
p
ρ(z1 , z2 ) = |z1 − z2 | = (x1 − x2 )2 + (y1 − y2 )2 ,
(1.2)
zk = xk + iyk , xk , yk ∈ R, k = 1, 2, i2 = −1,
27
28
2. Basic notions in topology
called the Euclidean distance on C.
(d) Consider the plane R2 and define on it the following metrics, for uk =
(xk , yk ) ∈ R2 , k = 1, 2,
(d1 ) ρ1 (u1 , u2 ) = |x1 − x2 | + |y1 − y2 |;
p
( d2 ) ρ2 (u1 , u2 ) = (x1 − x2 )2 + (y1 − y2 )2 ;
( d3 ) ρ3 (u1 , u2 ) = max{|x1 − x2 |, |y1 − y2 |};
Metric ρ2 is said to be the Euclidean metric on R2 while ρ3 is said to be the uniform
metric.
(e) From (d) it follows at once that on a given set we can define several metrics.
4
Let (X, ρ) be a metric space. All points and sets mentioned bellow are understood
to be elements and subsets of X.
The open ball with center at x and radius r > 0 is the set B(x, r) given by
B(x, r) = {y ∈ X | ρ(x, y) < r}.
A neighborhood of a point x is a set V such that there exists a ball B(x, r) with
B(x, r) ⊂ V.
We denote by V(x) the family of neighborhoods of x, i.e.,
V(x) = {V | V is a neighborhood of x}.
A point x is a limit point of the set A if every neighborhood x contains a point
y 6= x such that y ∈ A. If x ∈ A and x is not a limit point of A, x is said to be
an isolated point of A.
A is closed if every limit point of A is a point of A.
A point x is an interior of A if there is a neighborhood V of x such that V ⊂ A.
A is open if every point of A is an interior point of A. The empty set is open.
A is perfect if A is closed and if every point of A is a limit point of A.
A is bounded if there is a real number m such that
ρ(x, y) ≤ m,
for any x, y ∈ A.
A is dense if every point of X is a limit point of A or a point of A (or both).
The system of open sets corresponding to a metric space (X, ρ) is denoted by τ
and is called the topology generated by the metric ρ. The pair (X, τ ) is said to be a
topological space. Sometimes we say that X is a topological space when the topology
is irrelevant.
Remarks.
(a) All these topological notions (openess, closeness, etc.) are based
ultimately on the notion of metric. Thus changing the metric, the open (closed, etc.)
2.1. Metric spaces
29
sets change. For example, the family of open sets corresponding to the metric on R
given by (1.1) coincides with the family of all subsets of R. Obviously, this is not the
case if we consider the Euclidean metric ρ2 on R. For, the set containing precisely a
point in R is open in the first case whereas it is not in the second case.
(b) In our presentation we considered the metric as a primary notion. However,
it is possible and actually largely used, to consider the topology on a set as a basic
notion, [7].
4
Theorem 1.1. Every open ball is an open set.
Proof. Consider A = B(x, r) and let y be any point of A. Then there is a positive
number h such that
ρ(x, y) = r − h.
For all points z such that ρ(y, z) < h we have then
ρ(x, z) ≤ ρ(x, y) + ρ(y, z) < r − h + r + r,
so that z ∈ A. Thus y is an interior point of A.
2
Theorem 1.2. If x is a limit point of a set A, then every neighborhood of x contains
infinitely many points of A.
Proof. Suppose there is a neighborhood V of x which contains only a finite number
of points of A. Let y1 , . . . , yn be those points of V ∩ A, which are distinct from x,
and put
r = min ρ(x, yk ).
1≤k≤n
Note that r > 0.
The neighborhood V (x, r) contains no point y ∈ A such that y 6= x, so that x
is not a limit point of A. This contradiction establishes the theorem.
2
Examples 1.1. Let us consider the following subsets of R2 (bijective to C ).
(a) The set of all complex z such that |z| < 1;
(b) the set of all complex z such that |z| ≤ 1;
(c) a finite set;
(d) the set of all integers;
(e) the set A = {1/n | n = 1, 2, . . . }. A has a limit point (namely x = 0 ), but no
point in A is a limit point of A; we stress the difference between a limit point
and containing one;
(f) the set of all complex numbers;
(g) the interval ]0, 1[ .
30
2. Basic notions in topology
Let us note that (d), (e) and (g) can be regarded also as subsets of R.
Some properties of these sets are tabulated below
(a)
(b)
(c)
(d)
(e)
(f)
(g)
closed
no
yes
yes
yes
no
yes
no
open
yes
no
no
no
no
yes
perfect
no
yes
no
no
no
yes
no
bounded
yes
yes
yes
no
no
no
yes
4
Theorem 1.3. A set A is open if and only if {X A is closed.
Proof. First, suppose that {X A is closed. Choose x ∈ A. Then x ∈
/ {X A and x
is not a limit point of {A. Hence there exists a neighborhood V of x such that
{A ∩ V = ∅, that is V ⊂ A. Thus x is an interior point of A, and A is open.
Now, suppose A is open. Let x be a limit point of {A. Then every neighborhood
of x contains a point of {A, so that x is not an interior point of A. Since A is
open, this means that x ∈ {A. It follows that {A is closed.
2
Corollary 1.1. A set is closed if and only if its complement is open.
Theorem 1.4. (a) For any family {Gα } of open sets, ∪α Gα is open.
(b) For any family {Fα } of closed sets, ∩α Fα is closed.
(c) For any finite family G1 , . . . , Gn of open sets, ∩ni=1 Gi is open.
(d) For any finite family F1 , . . . , Fn of closed sets, ∪ni=1 Fi is closed.
Proof. (a) Put G = ∪α Gα . If x ∈ G, then x ∈ Gα for some α. Since x is an interior
point of Gα , x is also an interior point of G, and G is open.
(b) Fα is closed ⇐⇒ {Fα open =⇒ ∪α {Fα open ⇐⇒ {(∪α Fα ) open
⇐⇒ ∪α Fα closed.
(c) Put H = ∩n1 Gi . For every x ∈ H there exist some balls Bi of x with radii
ri such that Bi ⊂ Gi , i = 1, . . . , n. Put r = min1≤i≤n ri and let B be the ball of x
of radius r. Then B ⊂ Gi , i = 1, . . . , n, so that B ⊂ H and H is open.
(d) Similarly to (b).
2
Warning. Hereafter the real number system is considered endowed with the topology
generated by the Euclidean metric.
Example. In parts (c) and (d) of the preceding theorem, the finiteness of the families
is essential. For, let Gn be the interval ] − 1/n, 1/n[ , n ∈ N∗ . Then Gn is an open
subset of R. Put G = ∩∞
1 Gn . Thus G consists of a single point (that is, x = 0 )
which is not an open set in R.
4
2.1. Metric spaces
31
Theorem 1.5. Let A be a closed set of real numbers which is bounded above. Let
y = sup A. Then y ∈ A.
Proof. Suppose y ∈
/ A. For any ε > 0 there is a point x ∈ A such that y −ε ≤ x ≤ y,
theorem 2.7. Thus, every neighborhood of y contains a point x ∈ A, and x 6= y,
since y ∈
/ A. It follows that y is a limit point of A which is not a point of A, so
that A is not closed. This contradicts the hypothesis.
2
Suppose A ⊂ Y ⊂ X, where (X, ρ) is a metric space. To say that A is an open
subset of X means that to each point x ∈ A there is associated a positive number
r such that ρ(x, y) < r, y ∈ X imply that y ∈ A. First remark that (Y, ρ|Y ×Y ) is
a metric space, too. We say that A is open relative to Y if to each x ∈ A there is
associated an r > 0 such that y ∈ A whenever ρ(x, y) < r and y ∈ Y.
Example. Example 1.1 (g) showed that a set may be open relative to Y without
being an open subset of X. However, there is a simple relation between these concepts.
4
Theorem 1.6. Suppose Y ⊂ X. A subset A of Y is open relative to Y if and only
if A = Y ∩ G for some open subset G of X.
Proof. Suppose A is open relative to Y. To each x ∈ A there is a positive number
rx such that the conditions ρ(x, y) < rx and y ∈ Y imply y ∈ A. Let Vx be the set
of all y ∈ X such that ρ(x, y) < rx , and define
G = ∪x∈A Vx .
Then G is an open subset of X.
Since x ∈ Vx for all x ∈ A, it is clear that A ⊂ G ∩ Y. By our choice of Vx we
have Vx ∩ Y ⊂ A for x ∈ A, so that G ∩ Y ⊂ A. Thus A = G ∩ Y, and one half of
the theorem is proved.
Conversely, if G is open in X and A = G∩Y, for every x ∈ A has a neighborhood
Vx ⊂ G. Then Vx ∩ Y ⊂ A, so that A is open relative to Y.
2
Theorem 1.7. Let (X, ρ) be a metric space and let E 0 be the set of limit points of
a set E. Then E 0 is closed.
Proof. By definition E 0 is closed if and only if (E 0 )0 ⊂ E 0 .
Take y ∈ (E 0 )0 . Then for every ε > 0 there is an x ∈ E 0 such that 0 < ρ(x, y) <
ε/2. Since x ∈ E 0 there is a v ∈ E such that
ρ(x, v) < ρ(x, y).
Hence v 6= y and
0 < ρ(v, y) ≤ ρ(x, v) + ρ(x, y) < ε.
Since v ∈ E, this shows that y is a limit point of E, i.e., y ∈ E 0 and the proof is
complete.
2
Let E be a set, let E 0 be the set of limit points of E, and define E = E ∪ E 0 .
32
2. Basic notions in topology
Theorem 1.8. (a) E is a closed set, and that E ⊂ F if E ⊂ F and F is closed;
(b) Moreover
E = ∩E⊂F, F closed F ;
(c) E is closed if and only if E = E.
A metric space is said to be separable if it contains a countable dense set.
Theorem 1.9. R is separable.
Proof. Q is a countable (corollary 2.5, p. 18) and dense (theorem 2.11, p. 15) subset
of R.
2
2.2
Compact spaces
A covering of a set A in a topological space X is a family of subsets {Gα } of X
such that A ⊂ ∪α Gα .
An open covering is a covering consisting of open subsets.
A subset K of a topological space X is said to be compact if every open covering
contains a finite subcovering of it. More explicitly, the requirement is that if {Gα } is
an open covering of K, then there are finitely many indices α1 , . . . , αn such that
K ⊂ G α1 ∪ · · · ∪ Gαn .
Example. Every finite set is compact.
4
Remark. We observed earlier that if A ⊂ Y ⊂ X, then A may be open relative to
Y without being open relative to X. The property of being open thus depends on
the space in which A is embedded. The same is true of the property of being closed.
4
Theorem 2.1. Suppose K ⊂ Y ⊂ X. Then K is compact relative to X if and only
if K is compact relative to Y.
If K is a compact set of a metric space X and K = X, then we say that X is
a compact space.
Theorem 2.2. Every compact subset of a metric space is closed.
Proof. Let K be a compact subset of a metric space. We shall prove that {X K is
open.
If K = X, we are done. Suppose that {X K 6= ∅. Choose x ∈ X, x ∈
/ K. If
y ∈ K, let Vy and Wy be neighborhoods of x, respectively, y of radius less than
ρ(x, y)/2 (> 0, since x 6= y ). Since K is compact, there are finitely many points
y1 , . . . , yn in K such that
K ⊂ Wy1 ∪ · · · ∪ Wyn = W.
2.2. Compact spaces
33
If V = Vy1 ∪ · · · ∪ Vyn , then V is a neighborhood of x which does not intersect W.
Hence V ⊂ {X K, so that x is an interior point of {X K. So, {X K is open and K
is closed.
2
Theorem 2.3. Every closed subset of a compact set is compact.
Proof. Suppose F ⊂ K ⊂ X, F is closed (relative to X ), and K is compact. Let
{Vα } be an open covering of F. Then we have
F ⊂ K ⊂ ∪α Vα ∪ (X \ F ).
Since K is compact, there is a finite subcovering by open sets of the form {Vα } ∪
(X \ F ). This finite subcovering of K is a covering of F. Since X \ F does not cover
F, it remains a finite number of Vα that covers F. We have thus shown that a finite
members of {Vα } covers F.
Corollary 2.1. If F is closed and K is compact, then K ∩ F is compact.
Proof. By theorem 2.2 K is closed, by theorem 1.4 F ∩ K is closed, and, finally, by
theorem 2.3 F ∩ K is compact.
2
Theorem 2.4. If {Kα } is a family of compact subsets of a metric space X such that
the intersection of every finite family of {Kα } is nonempty, then ∩Kα is nonempty.
Proof. Fix a member K1 of {Kα } and put Gα = {Kα . Assume that no point of K1
belongs to every Kα . Then the sets Gα form an open covering of K1 . Since K1 is
compact, there are finitely many indices α1 , . . . , αn such that
K1 ⊂ G α 1 ∪ · · · ∪ G α n .
This means that
K1 ∩ Kα1 ∩ · · · ∩ Kαn = ∅,
contrary to our assumption.
2
Theorem 2.5. If A is an infinite subset of a compact set K, then it has a limit
point in K.
Proof. If no point of K were a limit point of A, then each x ∈ K would have a
neighborhood Vx which contains at most one point of A (namely x, if x ∈ A ). It is
clear that no finite subfamily of {Vx } can cover E; moreover K since A ⊂ K. This
contradicts the compactness of K.
2
Theorem 2.6. Every closed and bounded interval is compact.
Proof. Let I = [a, b] be a closed and bounded interval. Denote δ = |a − b| = b − a.
Then |x − y| ≤ δ, for any x, y ∈ I. Suppose there exists an open covering {Gα }
of I which contains no finite subcovering of I. Put c = (a + b)/2. The intervals
Q1 = [a, c] and Q2 = [c, b] then determine 2 subintervals whose union is I. At
least one of these sets, call it I1 , cannot be covered by any finite subfamily of {Gα }
(otherwise I could be so covered). We next subdivide I1 and continue the process.
We obtain a sequence {In } with the following properties
34
2. Basic notions in topology
(i) I ⊃ I1 ⊃ I2 ⊃ . . . ;
(ii) In is not covered by any finite subfamily of {Gα };
(iii) if x, y ∈ In , then |x − y| ≤ 2−n δ.
By (i) and theorem 2.9 there is a point z ∈ ∩In . For some α, z ∈ Gα . Since Gα
is open, there exists r > 0 such that B(z, r) ⊂ Gα . If n is so large that 2−n δ < r,
then (iii) implies that In ⊂ Gα , which contradicts (ii).
2
Theorem 2.7. (Heine1 -Borel2 ) If a set A ⊂ R has one of the following properties,
then it has the other two
(a) A is closed and bounded;
(b) A is compact;
(c) every infinite subset of A has a limit point in A.
Proof. (a) =⇒ (b). There exists a closed and bounded interval I such that A ⊂ I.
Then (b) follows from theorems 2.6 and 2.2.
(b) =⇒ (c). This is theorem 2.5.
(c) =⇒ (a). If is not bounded, Then A contains points xn with
|xn | > n,
(n = 1, 2, . . . ).
The set P consisting of these points xn is infinite and clearly has no limit point in
R, hence has none in A. Thus (c) implies that A is bounded.
If A is not closed, then there is a point x0 ∈ R which is a limit point of A but not
a point of A. For n = 1, 2, 3, . . . there are points xn ∈ A such that |xn − x0 | < 1/n.
Set M be the set of these points xn . Then M is infinite (otherwise, |xn − x0 | would
have a constant positive value, for infinite many n ), M has x0 as a limit point, and
M has no other limit point in R. For if y ∈ R, y 6= x0 , then
1
|xn − y| ≥ |x0 − y| − |xn − x0 | ≥ |x0 − y| − 1/n ≥ |x0 − y|
2
for all but finitely many n; this shows that y not a limit point of M (theorem 1.2).
Thus M has no limit point in A; hence A must be closed if (c) holds.
2
Remark. (b) ⇐⇒ (c) in any metric space. (a) does not, in general, imply (b) and
(c).
4
Theorem 2.8. (Weierstrass3 ) Every bounded infinite subset of R has a limit point
in R.
1
Heine,
Emile, Borel, 1871-1956
3
Karl Weierstrass, 1815-1897
2
2.2. Compact spaces
35
Proof. Being bounded, the set A in question is a subset of an interval [a, b] = I ⊂ R.
By theorem 2.6 I is compact, and so A has a limit point in I, by theorem 2.5.
2
Perfect sets have been defined at page 28.
Theorem 2.9. Let P be a nonempty perfect set in R. Then P is uncountable.
Proof. Since P has limit points, P must be infinite. Suppose P is countable, and
denote the points of P by x1 , x2 , . . . . We shall construct a sequence Bn of balls, as
follows.
Let B1 be any ball with center x1 . If B1 = B(x1 , r1 ), the corresponding closed
ball B1 is defined to be the set of all x ∈ R such that |x − x1 | ≤ r1 . ( {B1 is open,
hence closed balls are closed).
Suppose Bn has been constructed, so that Bn ∩ P 6= ∅. Since every point of P
is a limit point of P, there is a ball Bn+1 such that
(i) Bn+1 ⊂ Bn ;
(ii) xn ∈
/ Bn+1 ;
(iii) Bn+1 ∩ P 6= ∅.
By (iii) Bn+1 satisfies our induction hypothesis, and the construction can proceed.
/ Kn+1 ,
Kn = Bn ∩ P. Since Bn is closed and bounded, Bn is compact. Since xn ∈
∞
no point of P lies in ∩1 Kn . Since Kn ⊂ P, this implies that ∩Kn is empty. But
each Kn is nonempty, by (iii), and Kn ⊃ Kn+1 , by (i); this contradicts theorem 2.4.
2
Corollary 2.2. Every interval [a, b] (a < b) is uncountable.
Remark. We introduce the Cantor set showing that there exist perfect sets in R
which contain no interval.
Let C0 be the interval [0, 1]. Remove the open interval ]1/3, 2/3[ and let C1 be
the union of the intervals
[0, 1/3] ∪ [2/3, 1].
Remove the middle thirds of these intervals and let C2 be the union of the intervals
[0, 1/9] ∪ [2/9, 3/9] ∪ [6/9, 7/9] ∪ [8/9, 1].
Continuing in this way we obtain a sequence of compact sets Cn such that
(i) C1 ⊃ C2 ⊃ . . . ;
(ii) Cn is the union of 2n intervals of length 3−n .
36
2. Basic notions in topology
The set
C = ∩∞
1 Cn
is called the Cantor set. C is clearly compact, and theorem 2.4 shows that C is not
empty.
No interval of the form
3k + 1 3k + 2
(2.1)
,
,
3m
3m
where k and m are positive integers, has a point in common with C. Since every
interval ]α, β[ contains a segment of the form (2.1), if
3−m < (β − α)/6,
C contains no segment.
To show that C is perfect, it is enough to show that C contains no isolated point.
Let x ∈ C, and let I be any open interval containing x. Let In be that interval of
Cn which contains x. Choose n large enough, so that In ⊂ I. Let xn be an end
point of In , such that xn 6= x.
It follows from the construction of C that xn ∈ C. Hence x is a limit point of
C, and C is perfect.
4
Chapter 3
Numerical sequences and series
The present chapter is devoted introducing several results on numerical sequences
and series.
3.1
3.1.1
Numerical sequences
Convergent sequences
A sequence (xn )n in a metric space (X, ρ) is said to converge if there is a point
x ∈ X with the following property: for every ε > 0 there is an integer nε such that
n ≥ nε implies that ρ(xn , x) < ε.
In this case we also say that (xn ) converges to x or that x is the limit of (xn ) ,
and we write xn → x, or
lim xn = lim xn = x.
n→∞
If (xn ) does not converges, it is said to diverge.
Remarks.
(a) It might be well to point out that our definition of ”convergent
sequence” depends not only on (xn )n but also on X; for instance, the sequence
(xn ), xn = 1/n converges in R to 0, but fails to converge in the set ]0, ∞[ (with
ρ(x, y) = |x − y| ).
(b) We may say that
lim xn = a ⇐⇒ ∩ε>0 ∪m>0 ∩n≥m ]xn − a, xn + a[ = {a}. 4
n→∞
A sequence (xn )n in a metric space (X, ρ) is said to be bounded if the set {xn | n}
is bounded (for bounded set see page 28).
Theorem 1.1. Let (xn ) be a sequence in a metric space (X, ρ).
(a) (xn ) converges to x ∈ X if and only if every neighborhood of x contains all
but finitely many of the terms of (xn );
(b) if x, y ∈ X and (xn ) converges to x and y, x = y;
37
38
3. Numerical sequences and series
(c) if (xn ) converges, {xn } is bounded;
(d) if A ⊂ X and if x is a limit point of A, then there is a sequence (xn ) in A
such that lim xn = x.
Proof. (a) Suppose lim xn = x and let V ∈ V(x). For some ε > 0, the conditions
ρ(y, x) < ε, y ∈ X imply y ∈ V. Corresponding to this ε, there exists nε such that
n ≥ nε implies ρ(xn , x) < ε. Thus n ≥ nε implies xn ∈ V.
(b) Let ε > 0 be given. There exists integers nε and mε such that
n ≥ nε =⇒ ρ(xn , x) < ε/2,
n ≥ mε =⇒ ρ(xn , y) < ε/2.
Hence if n ≥ max{nε , mε }, we have
0 ≤ ρ(x, y) ≤ ρ(xn , x) + ρ(xn , y) < ε.
Since ε has been chosen arbitrary, we conclude that ρ(x, y) = 0.
(c) Suppose lim xn = x. There is an integer m such that n > m implies
ρ(xn , x) < 1. Put
r = max{1, ρ(x1 , x), ρ(x2 , x), . . . , ρ(xn , x)}.
Then ρ(xn , x) ≤ r, for n = 1, 2, 3 . . . .
(d) For each positive integer n, there is a point xn ∈ A such that ρ(xn , x) < 1/n.
Given ε > 0, choose nε so that εnε > 1. If n > nε , it follows that ρ(xn , x) < ε.
Hence xn → x.
2
Theorem 1.2. Suppose (xn ), (yn ) are real sequences, and lim xn = x, lim yn = y.
Then
(a) lim(xn + yn ) = x + y;
(b) lim c · xn = cx, lim(c + xn ) = c + x, for any real c;
(c) lim xn yn = xy;
(d) lim
1
1
= , provided xn 6= 0 (n = 1, 2, . . . ) and x 6= 0.
xn
x
Proof. (a) Given ε > 0 there exist nε and mε such that
n ≥ nε =⇒ |xn − x| < ε/2,
n ≥ mε =⇒ |yn − y| < ε/2.
Put nε = max{nε , mε }. If n ≥ nε , then
|(xn + yn ) − (x + y)| ≤ |xn − x| + |yn − y| < ε/2 + ε/2 = ε.
3.1. Numerical sequences
39
(b) The first claim follows from (c), while the second claim follows from (a).
(c) We write
|xn yn − xy| ≤ |xn ||yn − y| + |y||xn − x|.
Since (xn ) converges, it is bounded and we can find an M > 1 such that |xn | < M
and |y| < M. Given ε > 0, there exist nε and mε such that
n ≥ nε =⇒ |xn − x| < ε/(2M ),
n ≥ mε =⇒ |yn − y| < ε/(2M ).
Put nε = max{nε , mε }. If n ≥ nε , then
|xn yn − xy| ≤ M |xn − x| + M |yn − y| < ε/2 + ε/2 = ε.
(d) Choosing m such that |xn − x| < (1/2)|x| if n > m, we see that
1
|xn | > |x|,
2
n ≥ m.
Given ε > 0, there is an integer nε ≥ m such that n ≥ nε implies
1
|xn − x| < |xn − x|2 ε.
2
Hence, for n ≥ nε
1
xn − x 1
2
− =
xn x xn x < |x|2 |xn − x| < ε. 2
3.1.2
Subsequences
Given a sequence (xn ), consider a sequence (nk )k of positive integers, such that
n1 < n2 < . . . . Then the sequence (xnk )k is called a subsequence of (xn )n . If (xnk )k
converges, its limit is called a subsequential limit of (xn )n .
Remark. It is clear that (xn ) converges to x if and only if every subsequence of
(xn ) converges to x.
Theorem 1.3. Every bounded sequence in R contains a convergent subsequence.
Proof. Let E be the range of the bounded sequence (xn ).
If E is finite, there is at least one point in E, say x, and a sequence (nk )k with
n1 < n2 < . . . such that
xn1 = xn2 = · · · = x.
The subsequence (xnk )k so obtained evidently converges.
If E is infinite, then E has a limit point x ∈ R, theorem 2.8 page 34. Choose
n1 so that |xn1 − x| < 1. Having chosen n1 , . . . , ni−1 , we see by theorem 1.2, page
29, that there is an integer ni > ni−1 such that |xni − x| < 1/i. The sequence (xni )i
thus obtained converges to x.
2
40
3. Numerical sequences and series
Theorem 1.4. The subsequential limits of a sequence (xn ) in a metric space (X, ρ)
form a closed set in X.
Proof. Apply theorem 1.7, page 31.
2
Example. The set of subsequential limits of the following sequence
1 2 3 1 2 3 4 5
1
2n + 1
1, , , , , , , , , . . . , n , . . . ,
,...
2 2 2 4 4 4 4 4
2
2n
is the closed interval [0, 1], while the set of subsequential limits of the following
sequence
9
1 2
3n
1 2 3 1 2 3 4
1, , , , , , , , . . . , , . . . , n , n , . . . , n , . . .
2 2 2 4 4 4 4
4
2 2
2
is the closed interval [0, ∞).
4
3.1.3
Cauchy sequences
A sequence (xn ) in a metric space (X, ρ) is said to be a Cauchy sequence or fundamental sequence if for every ε > 0 there is an integer nε such that ρ(xn , xm ) < ε
provided n ≥ nε , m ≥ nε .
Let (X, ρ) be a metric space and E ⊂ X, E 6= ∅. The diameter of E is
diamE = sup ρ(x, y).
x,y∈E
If (xn ) is a sequence in X and En = {xn , xn+1 , . . . }, (xn ) is a Cauchy sequence if
and only if
lim diamEn = 0.
n→∞
Theorem 1.5. (a) If E is the closure of a set E in a metric space X, then
diamE = diamE.
(b) (Cantor lemma) If (Kn ) is a sequence of nonempty, nested, and compact sets
and if
lim diamKn = 0,
n→∞
then
∩∞
1 Kn
consists in exactly one point.
Proof. (a) Since E ⊂ E, it is clear that
diamE ≤ diamE.
Fix ε > 0 and choose x ∈ E and y ∈ E. By the definition of E there are points
x0 , y 0 ∈ E such that ρ(x, x0 ) < ε, ρ(y, y 0 ) < ε. Hence
ρ(x, y) ≤ ρ(x, x0 ) + ρ(x0 , y 0 ) + ρ(y 0 , y) ≤ 2ε + ρ(x0 , y 0 ) ≤ 2ε + diamE.
3.1. Numerical sequences
41
It follows that
diamE ≤ 2ε + diamE,
and since ε was arbitrary, (a) is proved.
(b) Put K = ∩∞
1 Kn . By theorem 2.4, p. 33, K 6= ∅. If K contains more than
one element, then diamK > 0. But for each n, Kn ⊃ K, so that diamKn ≥ diamK.
This contradicts the assumption that diamKn → 0.
2
Theorem 1.6. (a) Every convergent sequence in a metric space is a Cauchy sequence.
(b) Every Cauchy sequence in R converges.
Proof. (a) If limn→∞ xn = x and ε > 0, there is an integer nε such that ρ(xn , x) <
ε/2 whenever n ≥ nε . Hence, if n, m ≥ nε , we have
ρ(xn , xm ) ≤ ρ(xn , x) + ρ(xm , x) < ε,
so that (xn ) is a Cauchy sequence.
(b) Suppose (xn ) is a Cauchy sequence in R. Let En = {xn , xn+1 , . . . } and let
E n be the closure of En . By definition and theorem 1.5 we see that
(1.1)
lim diamE n = 0.
n→∞
In particular, the sets E n are bounded. They are also closed. Hence each E n is
compact. Also E n ⊃ E n+1 . By theorem 1.5 there is a unique point x ∈ R which lies
in every E n .
Let ε > 0 be given. By (1.1) there is an integer n0 such that diamE n < ε, if
n ≥ n0 . Since x ∈ E n , this means that |y − x| < ε for all y ∈ E n , hence for all
y ∈ En . In other words, if n ≥ n0 , then |xn − x| < ε. But this says precisely that
xn → x, and thus the proof is completed.
2
A metric space (X, ρ) in which every Cauchy sequence converges is said to be
complete.
Remark. R is a complete metric space, while Q is not.
4
3.1.4
Monotonic sequences
A sequence (xn ) of real numbers is said to be
(a) monotonically increasing if xn ≤ xn+1 , n ∈ N∗ ;
(b) monotonically decreasing if xn ≥ xn+1 , n ∈ N∗ .
The class of monotonic sequences consists of the increasing and decreasing sequences.
Theorem 1.7. Suppose (xn ) is monotonic. Then (xn ) converges if and only if it is
bounded.
Proof. Suppose xn ≤ xn+1 . Let E be the range of (xn ). If (xn ) is bounded, let x
be the least upper bound of E. Then
xn ≤ x,
n ∈ N∗ .
42
3. Numerical sequences and series
For every ε > 0 there is an integer nε such that
x − ε < xnε ≤ x,
for otherwise x − ε would be an upper bound of E. Since (xn ) increases, n ≥ nε
therefore implies
x − ε < xn ≤ x < x + ε ⇐⇒ |xn − x| < ε,
which shows that (xn ) converges to x.
2
Exercises 1.1. (a) Consider the sequence (an )n≥1 defined by
an = 1 +
1
1 1
+ + ··· + ,
2 3
n
n ∈ N∗ .
Then an → ∞. Indeed, it is enough to remark that the sequence has positive terms, is
monotonically increasing, but divergent. To check this the last property it is enough
to see that it is not a Cauchy one. The sequence (an ) is not fundamental since
a2n − an =
1
1
1
1
1
+ ··· +
>
+ ··· +
= .
n+1
2n
2n
2n
2
(b) Consider a numerical sequence (an ) satisfying |an −am | > 1/n for any n < m.
Then the sequence (an ) is unbounded. Suppose that (an ) is bounded. Then there
exists a positive M such that |an | ≤ M, for every n. From the hypothesis we infer
that the open intervals ]an − 1/(2n), an + 1/(2n)[ are disjoint and their union satisfies
∪n ]an −
1
1
1
1
, an + [ ⊂ ] − M − , +M + [ .
2n
2n
2
2
Since the length of the interval ] − M − 12 , +M + 12 [ is 2M + 1, it follows that the
sum of the length to the disjoint intervals ]an − 1/(2n), an + 1/(2n)[ does not exceed
2M + 1. On the other side, the length of an interval ]an −1/(2n),
Pnan +1/(2n)[ is equal
to 1/n. So the sum of the length of the first n intervals is
k=1 1/k which tends
to +∞, accordingly to (a). The contradiction shows that our hypothesis regarding
the boundedness of the sequence (an ) is not true. Hence the sequence is unbounded.
4
Theorem 1.8. Suppose that beginning with a certain rank, the terms of a convergent
(xn ) satisfy the inequality xn ≥ b (xn ≤ b). Then the limit a of the sequence (xn )
satisfies the inequality a ≥ b (a ≤ b).
Proof. Suppose that there exists N ∈ N∗ such that for every n ≥ N, xn ≥ b. We
show that a ≥ b.
If a < b, denote c := b − a. Consider ε := c/2. Since a is the limit of the
sequence (xn ), there is a rank nε ∈ N∗ such that |xn − a| < ε, for all n ≥ nε , i.e.,
xn < a + ε < b for all n ≥ nε , contrary to our assumption. Hence a ≥ b.
2
3.1. Numerical sequences
43
Corollary 1.1. Suppose that beginning with a certain rank, the terms xn and yn
of the convergent sequences (xn ) and (yn ) satisfy the inequality xn ≤ yn . Then
lim xn ≤ lim yn .
Proof. The sequence (yn − xn )n is convergent and has non-negative terms. So, its
limit is non-negative. It means that lim yn − lim xn = lim(yn − xn ) ≥ 0.
2
Theorem 1.9. Suppose that beginning with a certain rank N, the terms xn and yn
of the convergent sequences (xn ) and (yn ) satisfy
(i) xn ≤ yn , for all n ≥ N ;
(ii) yn − xn → 0, as n → ∞.
Then lim xn = lim yn .
Proof. Consider the nonempty and compact intervals In defined by In := [xn , yn ],
n ≥ N. By (b) of theorem 1.5 we get the conclusion.
2
Exercise 1.1. ([24, Probl. 7, p. 9]) Let (xn ) a sequence defined by
q
√
√
√
x1 = a, x2 = a + a, . . . , xn+1 = a + xn , . . . , a > 0.
Show that the sequence (xn ) converges.
For the beginning we remark that the sequence (xn ) has positive terms and it is
increasing since
q
√
√
x2 = a + a > a = x1 ,
(xn+2 − xn+1 )(xn+2 + xn+1 ) = xn+1 − xn ,
∀ n ≥ 1.
We are interested now to see if the sequence is bounded above or not. Since
√
1 + 1 + 4a
,
x1 <
2
√
√
1 + 1 + 4a
1 + 1 + 4a
xn <
=⇒ xn+1 <
, ∀n ≥ 1
2
2
we infer that the sequence is bounded. Hence the given sequence is convergent.
Exercise 1.2. ([24, Probl. 7, p. 9]) Show that the sequence (xn ) defined by
q
√
√
√
x1 = a1 , x2 = a1 + a2 , . . . , xn+1 = an+1 + xn , . . . , ai > 1,
converges if
lim
n→∞
1
ln(ln an ) < ln 2.
n
4
44
3. Numerical sequences and series
We introduce other two sequences
an
bn = 2n , n ≥ 1,
e
q
p
p
p
y1 = b1 , y2 = b1 + b2 , . . . , yn+1 = bn+1 + yn , . . .
and we note that for every n ∈ N∗ yn = xn /e, i. e., the sequences (xn ) and
(yn ) converge or diverge simultaneously. We also remark that the sequence (yn ) is
increasing.
From the hypothesis of the exercise it follows that there exists an n0 ∈ N∗ such
that for n ≥ n0
1
n
ln(ln an ) < ln 2 ⇐⇒ an < e2 ⇐⇒ bn < 1.
n
Consider a = max{b1 , b2 , . . . , bn0 , 1} and define the following sequence
q
√
√
√
z1 = a, z2 = a + a, . . . , zn+1 = a + zn , . . . .
Then yn ≤ zn . Based on the exercise 1.1, we deduce that the sequence (zn ) converges.
Therefore the sequence (yn ) converges and, at the end, the sequence (xn ) converges,
too.
4
Corollary 1.2. Suppose there are given three sequences (xn ), (an ), and (yn ) satisfying xn ≤ an ≤ yn from a certain rank. If the sequences (xn ) and (yn ) converge to
the same limit, then the sequence (an ) is convergent and its limit coincides with the
limit of (xn ).
Proof. It is obvious since |an − xn | ≤ |yn − xn |.
2
Exercises 1.2. Find the limits
(a)
(b)
1
1
1
+ 2
+ ··· + 2
);
+1 n +2
n +n
1 + 22 + · · · + nn
lim
.
n→∞
nn
lim (
n→∞ n2
The first limit is equal to 1 since
n2
n
1
1
1
n
≤ 2
+ 2
+ ··· + 2
≤ 2
.
+n
n +1 n +2
n +n
n +1
For the second limit we take into account the following inequalities
1≤
1 + 22 + · · · + nn
1 + n + n2 + · · · + nn
nn+1 − 1 1
≤
=
→ 1.
nn
nn
n − 1 nn
Thus we get that the limit is 1.
4
3.1. Numerical sequences
3.1.5
45
Upper and lower limits
Let (xn ) be a sequence of real numbers with the following property: for every real m
there is an integer nm such that n ≥ nm implies xn ≥ m. We then write
xn → ∞.
Similarly, if for every real m there is an integer nm such that n ≥ nm implies
xn ≤ m, we write
xn → −∞.
Let (xn ) be a sequence of real numbers. Let E be the set of numbers x (in the
extend real number system) such that xnk → x for some sequence (xnk )k . This set
E contains all subsequential limits, plus possibly, the elements +∞ and −∞. We
put
x∗ = sup E,
x∗ = inf E.
The elements x∗ and x∗ are called the upper and lower limits of (xn ); we use the
notation
lim sup xn = x∗ , lim inf xn = x∗ .
n→∞
n→∞
Theorem 1.10. Let (xn ) be a sequence of real numbers. Let E and x∗ as defined
earlier. The x∗ has the following two properties
(a) x∗ ∈ E;
(b) if y > x∗ , there is an integer m such that n ≥ m implies xn < y.
Moreover, x∗ is the only number with the properties (a) and (b).
Proof. If x∗ = +∞, E is not bounded above; hence (xn ) is not bounded above, and
there is a subsequence (xnk ) such that xnk → ∞.
If x∗ is real, then E is bounded above, and at least one subsequential limit exists,
so that (a) follows from the theorem 1.4 and theorem 1.5, p. 30.
If x∗ = −∞, E contains only one element, namely −∞, and there is no subsequential limit; hence for any real m, xn > m for at most a finite number of values
of n, so that xn → −∞. This establishes (a) in all cases.
To prove (b) suppose there is a number y > x∗ such that xn ≥ y for infinitely
many values of n. In that case, there is a number z ∈ E such that z ≥ y > x∗ ,
contradicting the definition of x∗ .
Thus x∗ satisfies (a) and (b).
To show the uniqueness, suppose there are two numbers, p and q, which satisfy
(a) and (b) and suppose p < q. Choose x such that p < x < q. Since p satisfies
(b), we have that xn < x for n > m. But then q cannot satisfy (a).
2
Exercise 1.3. Let (an ) and (bn ) be two sequences of real numbers such that
lim supn→∞ an = lim supn→∞ bn = +∞. Then there exist m, n such that |am −an | > 1
and |bm − bn | > 1.
46
3. Numerical sequences and series
We note that the two sequences are unbounded. Then there exist n1 , n2 satisfying
|an1 − an2 | > 2 (since otherwise the sequence (an ) is bounded). Further, there exists
n3 such that |bn1 − bn3 | > 1 and |bn2 − bn3 | > 1 (since otherwise the sequence (bn )
is bounded). Now, if |an1 − an3 | > 1, then n := n1 and m := n3 . If |an1 − an3 | ≤ 1,
then |an2 − an3 | > |an1 − an2 | − |an1 − an3 | > 1, and hence n := n2 and m := n3 .
4
3.1.6
Stoltz-Cesaro theorem and some of its consequences
Theorem 1.11. (Stoltz1 -Cesaro2 theorem) Let (an ) be a sequence of real numbers
and (bn ) be a strictly increasing and divergent sequence. Then
an+1 − an
= l (∈ [−∞, +∞])
n→∞ bn+1 − bn
(1.2)
lim
implies
an
= l.
n→∞ bn
(1.3)
lim
Proof. Suppose that l is finite. Choose a strictly positive ε. For ε/3 we find a rank
nε such that for every n ≥ nε it holds
an+1 − an
< ε/3,
−
l
bn+1 − bn
that is,
(bn+1 − bn )(l − ε/3) < an+1 − an < (bn+1 − bn )(l + ε/3).
(1.4)
Taking in (1.4), successively, n = nε , n = nε + 1, . . . , n = nε + p − 1, and adding
all these inequalities, we get
(bnε +p − bnε )(l − ε/3) < anε +p − anε < (bnε +p − bn )(l + ε/3)
or
(1.5)
l−
ε
ε bn
anε
an +p
ε
ε bn
anε
− (l − ) ε +
< ε < l + − (l + ) ε +
.
3
3 bnε +p bnε +p
bnε +p
3
3 bnε +p bnε +p
Notice that the sequences (bnε /bnε +p )p and (anε /bnε +p )p tend to 0. Then there exists
a rank pε such that for every p ≥ pε there hold
bnε
anε ε
ε ε
bn +p (l ± 3 ) < 3 and bn +p < 3 .
ε
ε
1
2
Otto Stoltz,
Ernesto Cesaro,
3.1. Numerical sequences
47
Hence, for every n > nε + pε , we finally have
a
n
l − < ε.
bn Now suppose that l = +∞. Choose a positive ε as large as we like. There exists
a rank nε such that for every n ≥ ε it holds
an+1 − an
ε
>
bn+1 − bn
2
or
ε
an+p − an > (bn+1 − bn ).
2
Adding the first p such inequalities, we may write
an+p
ε an − εbn
> +
.
bn+p
2
bn+p
Now, there exists a rank pε such that for every p > pε we have
an − εbn ε
bn+p < 2 .
Hence, for every n > nε + pε , we finally have
an
> ε. 2
bn
Corollary 1.3. Suppose there is given a strictly positive sequence (an ).
(a) If
lim
n→∞
an+1
= a (a > 0),
an
then
lim
n→∞
√
n
an = a.
(b) If
lim an = a (a > 0),
n→∞
then
lim
n→∞
√
n
a1 a2 . . . an = a.
(c) For p ∈ N∗ , it holds
1p + 2p + · · · + np
1
=
.
p+1
n→∞
n
p+1
lim
√
Proof. (a) Take xn = ln n an . Then xn = ln an /n. Now we apply theorem 1.11 of
Stoltz-Cesaro and we get lim xn = lim ln an+1 /an = ln lim an+1 /an = ln a.
√
(b) Take xn = ln n a1 a2 . . . an and reason as before.
(c) Applying Stoltz-Cesaro theorem 1.11, we get
(n + 1)p
np + pnp−1 + . . .
1
=
→
p+1
p+1
p
(n + 1)
− (n)
(p + 1)n + . . .
p+1
as n → ∞.
48
3. Numerical sequences and series
3.1.7
Some special sequences
Proposition 1.1. The following limits hold
1
(a) If p > 0, limn→∞ p = 0;
n
√
n
(b) if p > 0, lim p = 1;
√
(c) lim n n = 1;
(d) if p > 0 and α is real, then limn→∞
nα
= 0;
(1 + p)n
(e) if |x| < 1, limn→∞ xn = 0.
Proof. (a), (b), and (c) follow from theorem 1.11 and corollary 1.3.
2
Proposition 1.2. Let the sequence (an ) be defined as
n
1
(1.6)
an = 1 +
.
n
Then the sequence (an ) is convergent.
Proof. First approach. Based on Newton3 formula we may write
1 1
2 1
n 1
an = 1 +
+
+ ··· +
2
n n
n n
n nn
n(n − 1) 1
n(n − 1)(n − 2) . . . 2 · 1 1
=1+1+
+ ··· +
2
n
2! n
n!
n
1 1
1
2
n−1 1
=1+1+ 1−
+ ··· + 1 −
1−
1−
n 2!
n
n
n
n!
1
1
1
(1.7)
< 1 + 1 + + + ··· +
2! 3!
n!
Since n! ≥ 2n−1 , for every n ≥ 2, it follows that
1
1
1
2 < an < 1 + 1 + + 2 + · · · + n−1 < 3.
2 2
2
Thus the sequence (an ) is bounded.
We study now the monotonicity of the sequence (an ).
1
1
1
2
n
1
) + · · · + (1 −
)(1 −
) . . . (1 −
)
n + 1 2!
n+1
n+1
n + 1 (n + 1)!
1
1
1
2
n−1 1
> 1 + 1 + (1 −
) + · · · + (1 −
)(1 −
) . . . (1 −
)
n + 1 2!
n+1
n+1
n + 1 n!
1 1
1
2
n−1 1
> 1 + 1 + (1 − ) + · · · + (1 − )(1 − ) . . . (1 −
) = an .
n 2!
n
n
n n!
an+1 = 1 + 1 + (1 −
3
Sir Issac Newton, 1642-1727
3.1. Numerical sequences
49
Thus the sequence (an ) is increasing. Taking into account that it is also bounded,
by theorem 1.7, we infer that the sequence (an ) is convergent.
Second approach. We start with the identity
xn+1 − y n+1 = (x − y)
n
X
xn−i y i ,
i=0
valid for any real x and y and n ∈ N∗ . If x > y it results
(1.8)
(n + 1)(x − y)y n < xn+1 − y n+1 < (n + 1)(x − y)xn .
Substitute x = 1 + 1/n and y = 1 + 1/(n + 1) and we get
1
, xn+1 = x · an , y n+1 = an+1
n(n + 1)
x−1
1
1
n+1
=
> y =⇒ y 2 < x + y − 1 = 1 + +
=⇒
x=
n
y−1
n n+1
1
1
n+2
y
< 1+ +
yn.
n n+1
x−y =
(1.9)
The right-hand side of (1.8) supplies
1
1
1
x · an − an+1 < an =⇒ 1 +
an − an < an+1 =⇒ an < an+1 .
n
n
n
Thus the sequence (an ) is increasing.
Denote un = (1 + n1 )n+1 . Then an < un , for every n ∈ N∗ . From (1.9) it follows
1
1
1
n+2
n
y =
+ y yn,
un+1 = y
< 1+ +
n n+1
n
while from the left-hand side of (1.8) it results
(1.10)
1
1 n
1
y < un − y n+1 =⇒ ( + y)y n < un =⇒ un+1 < ( + y)y n < un .
n
n
n
Thus the sequence (un ) is decreasing and
1
an < un < · · · < u5 = (1 + )6 = 2, 985984 . . . =⇒
5
an < 3,
∀n ∈ N∗ .
Thus the sequence (an ) is increasing and bounded above, hence it converges.
As usually we denote the limit of the sequence (an ) by e, i.e.,
n
1
(1.11)
lim 1 +
=: e.
n→∞
n
2
50
3. Numerical sequences and series
Corollary 1.4. There hold the inequalities
n
n+1
1
1
1+
<e< 1+
,
n
n
∀n ∈ N∗ .
Proposition 1.3. Below we introduce some properties of number e.
(i) It holds that
lim
n→∞
1
1
1
1
1 + + + + ··· +
1! 2! 3!
n!
= e;
(ii) e can be written as
(1.12)
e=1+
1
1
1
1
θn
+ + + ··· +
+
,
1! 2! 3!
n! n · n!
where θn ∈ ]0, 1[ ;
(iii) e can be approximated with an accuracy up to 10−6 , if n ≥ 8;
(iv) e is a non-rational number, e ∈ R \ Q.
Proof. (i) We saw by (1.7) that if
an < 1 + 1 +
1
1
1
+ + ··· +
2! 3!
n!
the sequence (cn ) defined as
cn = 1 + 1 +
1
1
1
+ + ··· + ,
2! 3!
n!
n ∈ N∗
is convergent and therefore the following inequality holds
e = lim an ≥ lim cn .
(1.13)
n→∞
n→∞
On the other hand
n
1
1 1
2 1
n 1
an = 1 +
=1+
+
+ ··· +
2
n
n n
n n
n nn
1 1
1
2
k−1 1
>1+1+ 1−
+ ··· + −
1−
... 1 −
,
n 2!
n
n
n
k!
1
1
=⇒ lim an = e ≥ 1 + 1 + + · · · + , ∀k ≥ 2
n→∞
2!
k!
(1.14)
=⇒ e ≥ ck =⇒ e ≥ lim ck .
k→∞
Hence, from (1.13) and (1.14), we infer
lim ck = e.
k→∞
∀k ≤ n
3.1. Numerical sequences
51
(ii)
1
1
1
+
+ ··· +
(n + 1)! (n + 2)!
(n + m)!
∞
X
1
1
1
n+2
1
<
=
<
.
k
(n + 1)! k=0 (n + 2)
(n + 1)! n + 1
n · n!
cn+m − cn =
For fixed n and passing m → +∞ it follows
0 < e − cn <
Denote
0 < θn :=
1
.
n · n!
e − cn
1
n·n!
<1
and it follows (1.12).
(iii) The inequality
0 < e − cn <
1
< 10−5
n · n!
is satisfied for any n ≥ 8; so
e∼
=2+
1
1
+ ··· + ∼
= 2.71828.
2!
8!
(iv) Suppose that e ∈ Q. Then we may write e = m/n, for some m, n ∈ N∗ .
Then
e=
m
1
1
1
1
θn
= 1 + + + + ··· +
+
, θn ∈ ]0, 1[ ,
n
1! 2! 3!
n! n · n! 1
1
1
θn
=⇒ (n)! · m − n! 2 + + + · · · +
= .
2! 3!
n!
n
The last equality is impossible since θn ∈ ]0, 1[ . Hence e ∈ R \ Q.
Proposition 1.4. The sequence
an = 1 +
1
1
+ . . . − ln n,
2
n
n ≥ 1,
is decreasing and bounded.
Proof. By corollary 1.4 successively follow
n+1
n
1
1
1+
>e> 1+
n
n
=⇒ (n + 1)[ln(n + 1) − ln n] > 1 > n[ln(n + 1) − ln n]
1
1
(1.15)
=⇒
< ln(n + 1) − ln n < .
n+1
n
2
52
3. Numerical sequences and series
Inequality (1.15) follows from the Lagrange4 mean value theorem 2.3 at page 97, too.
Taking n = 1, 2, . . . , k we get
k
X
(1.16)
n=1
k
X1
1
< ln(k + 1) <
.
n+1
n
n=1
Remark that from (1.15) it follows
an+1 − an =
1
− ln(n + 1) + ln n < 0,
n+1
so that the sequence (an ) is decreasing.
From (1.16) we get
k
X
1
ln k < ln(n + 1) <
.
n
n=1
So, the sequence (an ) is bounded below by 0. Hence our sequence is convergent.
Its limit is denoted by γ = 0.5772156649 . . . and it is called the Euler5 -Mascheroni6
constant.
Corollary 1.5. It holds
1 + 12 + · · · +
lim
n→∞
ln n
1
n
= 1.
Proof. Pass to the limit in
1 + 12 + · · · +
ln n
1
n
− ln n
+ 1.
We may use equally well the Stoltz-Cesaro theorem 1.11.
Corollary 1.6. It holds
1
1
1
lim
+
+ ··· +
= ln k,
n→∞
n+1 n+2
kn
2
∀k ∈ {2, 3, . . . }.
Proof. Pass to the limit in
1
1
1
+
+ ··· +
n +1 n + 2
kn
1
1
1
1
= 1 + + ··· +
− ln kn − 1 + + · · · + − ln n + ln k, n ≥ 1. 2
2
kn
2
n
4
J. L. Lagrange,
Leonhard Euler, 1707-1783
6
Mascheroni
5
3.1. Numerical sequences
53
Corollary 1.7. Consider the limit
1
1
lim 1 + + · · · + − α ln n
n→∞
2
n
and determine the set of constants α for which the above limit exists and it is finite.
Proof. We denote by M the set of constants α for which the above limit exists and
it is finite. Remark that M 6= ∅ since by proposition 1.4, 1 ∈ M. Now


α = 1,
γ,
1
1
1 + + · · · + − ln n + (1 − α) ln n → +∞, α < 1,

2
n

−∞, α > 1.
Hence M = {1}.
2
Corollary 1.8. Denote
an = 1 −
1 1 1
1
+ − + · · · + (−1)n−1 .
2 3 4
n
Then it holds
lim an = ln 2.
n→∞
Proof. Recall Catalan7 identity
1−
1 1 1
1
1
1
1
+ − + ··· −
=
+
+ ··· + .
2 3 4
2n
n+1 n+2
2n
Then
1
1
1
+
+ ··· +
n+1 n+2
2n 1
1
1
1
= 1 + + ··· +
− ln 2n − 1 + + · · · + − ln n + ln 2
2
2n
2
n
→ ln 2. 2
a2n =
Lemma 1.1. (Problem of Traian Lalescu,8 Gazeta Matematică, 1901, problem 579)
Define
p
√
n
n+1
an =
(n + 1)! − n!.
Then limn→∞ an = e−1 .
The result will follow from the next two proposition.
7
8
E. Ch. Catalan, 1814-1894
Traian Lalescu,
54
3. Numerical sequences and series
Proposition 1.5. The sequence
n
,
xn = √
n
n!
n > 3,
satisfy the inequalities
1− √1n
(1.17)
e
1
< xn < e1− n
Proof. We will prove them by induction. For n > 3 the next inequality holds
(1.18)
xn > e
1− √1n
.
Indeed, since
r
r
4 32
32
e<
and x4 =
3
3
it follows that (1.18) holds for n = 4.
Suppose that (1.18) holds for some n ≥ 4. Then
√
n n
n
1 n+1 n+1
·e
,
xn+1 > 1 +
n
and from corollary 1.4 it follows
xn+1 > e
1−f (n)
√
1 + (n + 1) n
, where f (n) =
.
(n + 1)2
It is obvious that
n ≥ 4 =⇒ f (n) ≤ √
So
xn+1 > e
1
1− √n+1
1
.
n+1
,
and (1.18) is completely proved.
Now we suppose that for some n ≥ 4, xn < e1−1/n . Then applying corollary 1.4
we get
n−1
n
1
(n + 1)n n−1
1 n+1
1− n+1
n+1
n (n + 1)
xn+1 = xn ·
<
e
=⇒
x
<
1
+
<
e
,
n+1
nn
nn
n
ending in this way the proof.
2
Proposition 1.6. For every ε > 0 it is satisfied the next inequality
1
an − < ε
e
for every n > nε , where
1
1 n ε = 1 + 8 + 2 + 8 − 2 .
2ε
2ε
3.1. Series
55
Proof. Using (1.17) and corollary 1.4, for n ≥ 16, we get
1
√
n( n xn+1 − 1)
n(e n+1 − 1) √1n
1 √1
an =
<
·e < e n
e
xn
e
1
1
1
1
1
1
1+ √
<
< +√ .
<
1+ √
e
e
e
n
n−2
[ n] − 1
At the same time
h
i
1√
1√
√
n( n xn − 1) > n e n+1+ n+1 − 1 > n e [n+1+ n+1] − 1
n
n
√
√
>
≥
.
[n + 2 + n + 1]
n+2+ n+1
Thus
an >
1
nen
1
1
√
·
> −√ .
e n+2+ n+1
e
n
Further, for n ≥ 16, we have
an −
1 1
<√ .
e
n
√
Since nε := 1 + [max{16, 1/ε2 }] > max{16, 1/ε2 }, for n ≥ nε we have 1/ n. Hence
1
an − < ε ∀ n ≥ n ε . 2
e
Remark. Different approaches of the above problem may be found in [5, page 140],
[2, Problem 3.20, page 437].
3.2
Numerical series
All the results under consideration in the present section are complex-valued, unless the contrary is explicitly stated.
Given a sequence (xn ) we use the notation
q
X
xn
(p ≤ q)
n=p
to denote the sum xp + xp+1 + · · · + xq . To (xn )n≥1 we associate a sequence (sn ),
where
n
X
sn =
xk .
k=1
56
3. Numerical sequences and series
We also use the symbolic expression
x1 + x2 + . . . .
Or, more concisely,
(2.1)
∞
X
xn .
n=1
The symbol (2.1) we call an infinite series or just a series. The numbers sn are called
the partial sums of the series. If (sn ) converges, say to s, then the series converges
and we write
∞
X
xn = s.
n=1
The number s is called the sum of the series; but it should be clearly understood
that s is the limit of a sequence of sums, and not obtained simply by addition.
If (sn ) diverges, the series is said to diverge.
Sometimes, for convenience of notation, we shall consider series of the form
(2.2)
∞
X
xn .
n=0
And frequently, when there is no
Ppossible ambiguity, or when the distinction is immaterial, we shall simply write
xn in place P
of (2.1) or (2.2).
Theorem 2.1. (General criterion of Cauchy)
xn converges if and only if for every
ε > 0 there is an integer nε such that
m
X (2.3)
xk < ε
k=n
if m ≥ n ≥ nε .
Proof. Apply theorem 1.6 from page 41 to (sn ).
In particular, for m = n (2.3) becomes
|xn | < ε,
2
n ≥ nε ,
that is
Theorem 2.2. (Necessary condition) If
P
xn converges, then lim xn = 0.
Remark.
The condition xn → 0 is not, however, sufficient to ensure convergence of
P
xn . For instance, the series
∞
X
1
1
n
diverges as it results from (a) of exercises 1.1 from page 42 or it will become clear by
theorem 2.19.
4
3.2. Numerical series
57
P
P
Theorem 2.3. (Operations with series)P
Let
an and
P bn be two convergent series
and c a real number. Then the series
c an and
(an ± bn ) are convergent and,
moreover,
X
X
(2.4)
c an = c
an
and
X
(2.5)
(an ± bn ) =
X
an ±
X
bn .
Proof. Indeed,
c
X
an = c lim
n
X
ak = lim
n
X
c ak =
X
c an
and
X
an ±
X
bn = lim
n
X
ak ± lim
n
X
bk = lim(
n
X
ak ±
n
X
bk ) =
X
(an ± bn ). 2
Remark. The assumption that both series are convergent is essential. Otherwise it
could happen that the left-hand side of (2.5) exists while the right-hand side of it has
no meaning. For,
X
X
X
0=
(1 − 1) versus
1−
1. 4
As for finite sums, we may group the terms of a convergent series in brackets (but
no commutativity is allowed). For example
a1 + (a2 + a3 ) + (a4 + a5 + a6 + a7 ) + . . . .
P
Theorem 2.4. Let
an be a convergent series. Then by grouping its terms it results
a convergent series having the same sum.
Proof. The sequence of partial sums of the transformed series is a subsequence of the
convergent sequence of partial sums to the original sequence.
2
3.2.1
Series of nonnegative terms
Theorem 2.5. A series of nonnegative terms converges if and only if its partial sums
form a bounded sequence.
Proof. The sequence of partial sums is increasing. It is convergent if and only if it is
bounded, accordingly to the theorem 1.7 from page 41.
2
P
Theorem
an be a convergent series with nonnegative terms. Then the
P 2.6. Let
series
bn obtained from the former by rearranging (commuting) and renumbering
its terms is also convergent having the same sum.
58
3. Numerical sequences and series
Proof. Let s be the sum of the first series and snPits partial sum of rank n. Denote
by pn the partial sum of rank n of the series
bn . Fix a rank, let it be n and
consider pn . Then
p n = b 1 + · · · + b n = ak 1 + · · · + ak n
and denotePN = max{k1 , . . . , kn }. Obviously, pn ≤ sN ≤ s. Thus we conclude that
the series
bP
n converges and, if b is its sum, b ≤ s. Reasoning vice-verse, we get
P
that
an = b n .
2
Theorem 2.7. (Comparison
test) (a) If |x
P
Pn | ≤ cn for n ≥ n0 , where n0 is some
fixed integer, and if
cn converges, then P xn converges.
P
(b) If xn ≥ yn ≥ 0 for n ≥ n0 and if
yn diverges, then
xn diverges.
Proof. (a) Given ε > 0 there exists nε ≥ n0 such that m ≥ n ≥ nε implies
m
X
ck ≤ ε,
k=n
by the Cauchy criterion. Hence
m
m
m
X X
X
xk ≤
|xk | ≤
ck ≤ ε,
k=n
k=n
k=n
and (a) follows.
P
P
(b) follows from (a), for if
xn converges, so must
yn .
Theorem 2.8. If 0 ≤ x < 1,
∞
X
xn =
n=0
2
1
.
1−x
If x ≥ 1, the series diverges.
Proof. If x 6= 1,
sn =
n
X
k=0
xk =
1 − xn+1
.
1−x
The result follows if we let n → ∞. For x = 1, we get
1 + 1 + 1 + ...,
which evidently diverges.
2
Theorem P
2.9. (Cauchy condensation test) Suppose x1 ≥ x2 ≥ x3 ≥ · · · ≥ 0. Then
∞
the series
n=1 xn converges if and only if the series
(2.6)
∞
X
k=0
converges.
2k x2k = x1 + 2x2 + 4x4 + 8x8 + . . .
3.2. Numerical series
59
Proof. By theorem 2.5 it suffices to consider the boundedness of the partial sums. Let
sn = x 1 + x 2 + · · · + x n ,
tk = x1 + 2x2 + · · · + 2k x2k .
For n < 2k ,
sn = x1 + · · · + xn (add terms)
≤ x1 + (x2 + x3 ) + · · · + (x2k + · · · + x2k+1 −1 )
≤ x1 + 2x2 + · · · + 2k x2k = tk ,
so that
sn ≤ tk .
(2.7)
On the other hand, if n > 2k ,
sn = x1 + · · · + xn (neglect terms)
≥ x1 + x2 + (x3 + x4 ) + · · · + (x2k−1 +1 + · · · + x2k )
1
1
≥ x1 + x2 + 2x4 + · · · + 2k−1 x2k = tk ,
2
2
so that
2sn ≥ tk .
(2.8)
By (2.7) and (2.8) the sequences (sn ) and (tk ) are either both bounded or both
unbounded.
2
Theorem 2.10.
X 1
converges if and only if p > 1.
np
Proof. If p ≤ 0, divergence follows from the necessary condition, i.e., theorem 2.2. If
p > 0, the previous theorem is applicable, and we are led to the series
∞
X
∞
X
1
2 kp =
2(1−p)k .
2
k=0
k=0
k
Now, 21−p < 1 if and only if 1 − p < 0, and the result follows by comparison with
the geometric series (take x = 21−p ).
2
Remark. The series
1 1 1
1 + + + + ...
2 3 4
is called the harmonic series and it is divergent. From corallary 1.5 we deduce that
the speed of divergence of the harmonic series agrees to the speed of divergence of the
logarithmic function.
4
60
3. Numerical sequences and series
Theorem 2.11. If p > 1,
∞
X
(2.9)
n=2
1
n(ln n)p
converges; if p ≤ 1, the series diverges.
Proof. The logarithmic function increases, hence 1/[n(ln n)p ] decreases. We apply
theorem 2.9 to (2.9); this leads us to the series
(2.10)
3.2.2
∞
X
∞
1
1 X 1
2 k
=
. 2
k )p
p
p
2
(ln
2
(ln
2)
k
n=1
k=1
k
The root and ration tests
Theorem 2.12.
(Root test or D’Alembert9 criterion) Given
p
lim supn→∞ n |xn |. Then
P
(a) if α < 1,
xn converges;
P
(b) if α > 1,
xn diverges;
P
xn , we put α =
(c) if α = 1, the test gives no information.
Proof. (a) If α < 1, we can choose β so that α < β < 1 and an integer m such that
p
n
|xn | < β
for all n ≥ m. That is, n ≥ m implies
|xn | < β n .
P n
P
Since 0 < β < 1,
β converges. Convergence of
xn follows now from the
comparison test, theorem 2.7.
p
(b) If α > 1, then there is a sequence (nk ) such that nk |xnk | → α. Hence
|xn | > 1 for infinitely
many values of n, so the condition xn → 0, necessary for
P
convergence of
xn , does not hold.
(c) To prove (c), we consider the series
X1
,
n
X 1
.
n2
For each of these series α = 1, but the first diverges, while the second converges.
2
P
Theorem 2.13. (Ratio test or Cauchy criterion) The series
xn
9
Jean LE Rond D’Alembert, 1717-1783
3.2. Numerical series
61
xn+1 < 1;
(a) converges if lim supn→∞ xn xn+1 ≥ 1, for n ≥ n0 , where n0 is some fixed integer;
(b) diverges if xn (c) if
xn+1 xn+1 lim inf ≤ 1 ≤ lim sup n→∞
xn x
n→∞
n
the test gives no information.
Proof. (a) If condition (a) holds, we can find β < 1 and an integer m such that
xn+1 xn < β
for n ≥ m. In particular
|xm+1 | < β|xm |,
|xm+2 | < β|xm+1 | < β 2 |xm |,
......
|xm+p | < β p |xm |.
That is
|xn | < β n−m |xm |
P n
for n ≥ m, and (a) follows from the comparison test, since
β converges.
(b) If |xn+1 | > |xn | for n ≥ m, it is easily seen that the necessary condition
xn → 0 does not hold, and (b) follows.
(c) We again consider series
X1
,
n
In any case we have
lim
n→∞
X 1
.
n2
xn+1
= 1,
xn
but the first series diverges while the second one converges.
Theorem 2.14. For any sequence (xn ) of positive numbers
√
xn+1
≤ lim inf n xn
n→∞
n→∞
xn
√
xn+1
lim sup n xn ≤ lim sup
.
xn
n→∞
n→∞
lim inf
2
62
3. Numerical sequences and series
3.2.3
Power series
Given a sequence (cn ) of complex numbers, the series
∞
X
(2.11)
n
cn z = 1 +
n=0
∞
X
cn z n ,
n=1
is called a power series. The numbers cn are called the coefficients of the series; z is
a complex number.
P
Theorem 2.15. (Cauchy-Hadamard10 ) Given the power series
cn z n , put
α = lim sup
p
n
|cn |,
R=
n→∞
(If α = 0, R = +∞; if α = ∞, R = 0. ) Then
diverges if |z| > R.
P
Remark. R is called the radius of convergence of
Proof. Put xn = cn z n and apply the root test
lim sup
p
n
|xn | = |z| lim sup
n→∞
Examples. (a)
P
p
n
1
.
α
cn z n converges if |z| < R; and
P
cn z n .
|cn | =
n→∞
|z|
. 2
R
nn z n , R = 0;
P zn
, R = +∞;
n!
P n
(c)
z , R = 1. If |z| = 1, the series diverges, since z n does not tend to 0 as
n → ∞;
(b)
P zn
, R = 1. On the circle of convergence, the series diverges at z = 1; it
n
converges at all other points of |z| = 1. The last assertion follows from theorem
2.19.
P zn
(e)
, R = 1. The series converges at all points of the circle |z| = 1, by the
n2
n
z 1
comparison test, since 2 = 2 if |z| = 1.
4
n
n
(d)
3.2.4
Partial summation
Lemma 2.1. Given two sequences (an ), (bn ). Put
An =
n
X
k=0
10
Jacques Hadamard, 1865-1963
ak
3.2. Numerical series
63
if n ≥ 0; put A−1 = 0. Then, if 0 ≤ p ≤ q, we have
(2.12)
q
X
an b n =
n=p
p−1
X
An (bn − bn−1 ) + Aq bq − Ap−1 bp .
n=p
The identity (2.12) is said to be the partial summation formula.
Theorem 2.16. (Abel11 theorem) Suppose there are given two sequences (an )n and
(bn )n . Moreover,
P∞
(i) the series
n=1 |bn − bn+1 | converges;
(ii) limn→∞ bn = 0;
(iii) there exists a positive α such that for every n ∈ N∗ and m ∈ N∗ , m ≥ n, we
have |an + an+1 + · · · + am | < α.
P∞
Then
n=1 an bn converges.
Proof. For every n ∈ N∗ and m ∈ N∗ , m ≥ n, we denote αn,m = an +an+1 +· · ·+am .
Then based on a variant of the partial summation formula (2.12) we have
an bn + · · · + am bm = αn,n bn + (αn,n+1 − αn,n )bn+1 + · · · + (αn,m − αn,m−1 )bm
(2.13)
= αn,n (bn − bn+1 ) + · · · + αn,m−1 (bm−1 − bm ) + αn,m bm .
Now, we choose a positive ε. For ε/(2α) we find an nε ∈ N∗ that satisfies the
following two requirements
• it is greater or equal to the nε fromPthe general criterion of Cauchy theorem
∞
2.1 applied to the convergent series
n=1 |bn − bn+1 |;
• for every n ≥ nε , |bn | < ε/(2α).
For every m ≥ n > nε we have
• if m = n, an bn | ≤ α|bn | < α[ε/(2α)] < ε;
• if m > n, from (2.13) we get
|
m
X
ak bk | ≤ α(|bn − bn+1 | + · · · + |bm−1 − bm |) + α|bm | < α
k=n
ε
ε
+α
= ε.
(2α)
(2α)
The conclusion follows from the the general criterion ofPCauchy theorem 2.1.
2
∞
Remark. If bn ↓ 0, the assumption that the series
n=1 |bn − bn+1 | converges in
theorem
2.16 is useless. It may be neglected since the n -rank partial sum of the series
P∞
|b
4
n=1 n − bn+1 | is equal to b1 − bn+1 which, in turn, tends to b1 .
11
Niels Henrik Abel, 1802-1829
64
3. Numerical sequences and series
Theorem 2.17. (Dirichlet12 theorem) Suppose
P
(i) the partial sums An of
an form a bounded sequence;
(ii) b0 ≥ b1 ≥ b2 ≥ . . . ;
(iii) limn→∞ bn = 0.
P
Then
an bn converges.
Proof. It follows from the previous remark.
2
Theorem 2.18. Suppose
(i) |c1 | ≥ |c2 | ≥ . . . ;
(ii) c2m−1 ≥ 0, c2m ≤ 0, m = 1, 2, . . . ;
(iii) limn→∞ cn = 0.
P
Then
cn converges.
Proof. Apply theorem 2.17 with an = (−1)n+1 , bn = |cn |.
2
Example 2.1. For p > 0 the series
1−
1
1
1
1
+ p − p + · · · + (−1)n+1 p + . . .
p
2
3
4
n
converges. For p = 1, based on corollary 1.8, we can write the Leibnitz13 series
1 1 1
1
+ − + · · · + (−1)n+1 + · · · = ln 2. 4
2 3 4
n
P
Theorem 2.19. Suppose the radius of convergence of
cn z n is 1, and suppose
c0 ≥ c1 ≥ c2 . . . , limn→∞ cn = 0. Then cn z n converges at every point of the circle
|z| = 1, except possibly at z = 1.
(2.14)
1−
Proof. Put an = z n , bn = cn . The hypotheses of theorem 2.17 are satisfied, since
n
X
1 − z n+1 2
k
≤
|An | = z = ,
1−z
|1 − z|
k=0
if |z| = 1, z 6= 1.
12
13
2
Gustav Lejeune Dirichlet, 1805-1859
Gottfried Wilhelm Leibnitz, 1646-1716
3.2. Numerical series
3.2.5
65
Absolutely and conditionally convergent series
Consider
X
(2.15)
an .
Series (2.15) with complex terms is said to be absolutely convergent if the series
X
(2.16)
|an |
converges.
Theorem 2.20. The convergence of series (2.16) implies the convergence of series
(2.15).
Proof. We apply the Cauchy general criterion, theorem 2.1. For arbitrary, but fixed,
ε > 0 there exists a rank nε such that for every ranks n and m, n ≥ nε and n ≥ m
it holds
m
m
X
X
|
|ak | | =
|ak | < ε.
k=n
k=n
But using a well known inequality, we get
|
m
X
k=n
ak | ≤
m
X
|ak | < ε.
k=n
Now, we apply once again theorem 2.1.
2
The converse statement is not exactly true. It is enough to consider the series
(2.14) and the series in theorem 2.10 for p = 1.
Remarks. It is obvious that a convergent series with nonnegative terms is absolutely
convergent, too. But there are convergent series (with arbitrary terms) which are
not absolutely convergent. Indeed, as we already saw the series in example 2.1 is
convergent for p > 0, but it is absolutely convergent only for p > 1, theorem 2.10.
4
Series (2.15) is said to be conditionally convergent if the series converges whereas
the corresponding series of the absolute values (2.16) diverges.
A very important property to a sum of a finite number of real (complex) summands
is the commutative property, that is, a rearrangement of the terms does not affect
their sum. Unfortunately, this is not the case for series. Consider the convergent
series (2.14), i.e.,
1
1 1 1
1 − + − + · · · + (−1)n+1 . . . .
2 3 4
n
We write it, based on theorem 2.4, as
1
1 1
1
1
(2.17)
1−
+
−
+ ··· +
−
+ ....
2
3 4
2k − 1 2k − 2
66
3. Numerical sequences and series
and we notice, as already we saw, that its sum is s := ln 2.
Now we rearrange series (2.14) according to the rule: two negative terms after a
positive one. We find
1 1
1 1 1
1
1
1
+
− −
+ ··· +
−
−
+ ....
(2.18)
1− −
2 4
3 6 8
2k − 1 4k − 2 4k
Denote by Sn the n -th order partial sum of series (2.18). Then we have
X
n n X
1
1
1
1
1
S3n =
−
−
=
−
2k − 1 4k − 2 4k
4k − 2 4k
k=1
k=1
n 1X
1
1
1
=
−
= s2n .
2 k=1 2k − 1 2k
2
Thus we can write
1
S3n = s2n .
2
(2.19)
Also
(2.20)
1
1
S3n−1 = s2n +
2
4n
1
1
and S3n−2 = s2n +
.
2
4n − 2
From (2.19) and (2.20) we conclude that
1
lim S3n = lim S3n−1 = lim S3n−2 = s2n .
n→∞
n→∞
n→∞
2
Thus we have proved that series (2.18) converges and its sum is equal to s/2.
Hence, by rearrangement of a conditionally convergent series we got a convergent
series whose sum does not agree with the sum of the initial series.
The previous example illustrates that the commutativity is not longer valid for
arbitrary series. But let us see the positive case.
Theorem 2.21. (On rearrangement of absolutely convergent series of Cauchy) The
rearrangement of the terms of an absolutely convergent series supplies another absolutely convergent series having the same sum as the original one.
Proof. Consider
(2.21)
∞
X
an
an absolutely convergent series. Further, consider the positive, respectively, the negative parts of its terms, more precisely,
(
(
a
,
a
≥
0
−an , an ≤ 0
n
n
a−
(2.22)
a+
n =
n =
0,
an > 0.
0, an < 0
3.2. Numerical series
67
We have
−
an = a+
n − an .
(2.23)
Consider, now, the following two series with nonnegative terms
X
X
and
a−
(2.24)
a+
n
n.
+
−
Then both series
P in (2.24) converge, since an ≤ |an | and an ≤ |an |.
P +
Consider
bn the rearranged series of
(2.21).
For
it
construct
the
series
bn
P −
of positive parts , respectively, the series
bn of negative parts, as in (2.22). Then
∞
X
an =
∞
X
(a+
n
−
a−
n)
=
∞
X
a+
n
−
∞
X
a−
n
=
∞
X
b+
n
−
∞
X
b−
n
=
∞
X
(b+
n
−
b−
n)
=
∞
X
bn .
The first equality is obvious, the second equality follows from theorem 2.3, the third
one follows from theorem 2.6. For the other equalities we argue in the same way.
2
P
Corollary
2.1. Suppose the series
an is absolutely convergent. Then the series
P +
P
−
an are absolutely convergent, too.
an and
The converse statement is true since the difference of two convergent series both
of them having nonnegative terms is a convergent series, theorem 2.3.
P
Corollary 2.2. For series
an to be absolutely convergent it is necessary and sufficient that series (2.24) generated by it be convergent.
P
Lemma 2.2. If series
an is convergent but not absolutely, both series (2.24)
−
generated by it are divergent but a+
n → 0 and an → 0.
Proof. From corollary 2.2
P it+follows that at+least one of the series (2.24) generated by
it is divergent, that is
an = ∞, since an ≥ 0.
(2.25)
n
X
a−
k =
n
X
a+
k −
n
X
ak .
We examine the behaviour of the right-hand
P side of (2.25) as n → ∞. The second sum
tends to a finite number, since the series
an is convergent. The first sum increases
to ∞. Therefore the sum in the left-hand side of (2.25) increases to ∞ as n → ∞.
As a conclusion, if one of the series (2.24) is divergent, under our assumptions, the
other is divergent, too.
P
The sequences (a±
an is convergent.
2
n ) tend to zero since the series
Theorem P
2.22. (On rearrangement
of conditionally convergent series of Riemann14 )
P∞
∞
Let series
0 an and
0 bn be two divergent series with positive terms whose general terms tend to zero, i. e., an , bn → 0 as n → ∞.
14
Bernhard Riemann, 1826-1866
68
3. Numerical sequences and series
Then for any s ∈ [−∞, ∞] we can construct a series
(2.26)
a0 + a1 + · · · + am1 − b0 − b1 − · · · − bn1
am1 +1 + am1 +2 + · · · + am2 − bn1 +1 − bn1 +2 − · · · − bn2 + . . .
whose sum is equal to s. The series (2.26) contains all the terms of
P
∞
0 bn only once.
P∞
0
an and
Proof. First we suppose that s ∈ R. The indices n1 , n2 , . . . and m1 , m2 , . . . can
be chosen in this case as the smallest natural numbers for which the corresponding
inequalities written bellow are fulfilled
(i) α1 =
Pn1
0
ak > s,
(ii) α2 = α1 −
Pm1
(iii) α3 = α2 +
Pn2
> s,
(iv) α4 = α3 −
Pm2
< s, . . . .
0
bk < s,
n1 +1 bk
m1 +1 bk
At the pth step of this construction we indeed can choose
P∞ the natural
P∞ numbers np and
mp satisfying the pth inequality since the series
a
and
n
0
0 bn have positive
terms terms and diverge. The fact that the series thus constructed converges to s
follows from the above inequalities and from the assumption that an , bn → 0 as
n → ∞.
Now, suppose that s = +∞. We can replace s in the right-hand side of the
inequalities (i), (ii), (iii), . . . by a divergent sequence of the form 2,1,4,3,5,. . . .
2
P
Corollary 2.3. Let
an be a convergent series but not absolutely. Choose an
s ∈ [−∞, ∞]. Then there is a rearrangement of the series such that the resulting
series converges to s.
P
P +
P −
Proof. We just split the series
an into two series
an and
an as it is indicated
by (2.22), then apply lemma 2.2 and theorem 2.22.
2
3.2. Numerical series
3.2.6
69
Exercises
Study the convergence of the following series and find their sum if any
∞ √
X
n
∞
X
1
;
2n − 1
n=1
n
;
n
(n
+
1)2
n=1
3
∞ √
X
n2
;
n
n=1
∞
X
n+1
∞
X
sin(nx)
n=1
∞
X
n=1
n
;
n2 3n
;
∞
X
2
;
3n
n=1
1
;
n(n + 1)
n=1
∞
X
1
arctg 2 ;
2n
n=1
2
n + 5n + 6
;
ln +
n 5n + 4
n=1
∞
X
1
π
sin
, p ∈ R;
p
n
n
n=1
∞
X
;
∞
X
cos(nx) − cos(n + 1)x
∞
X
∞
X
n=1
5n
ln n
;
4
n(ln
n
+
1)
n=1
n
(−1)n−1
1
2+
;
n+1
n
n=1
;
n=1
∞
X
∞
X
√
√
√
3
3
3
( n + 2 − n + 1 + n);
n=1
∞
X
p
arcsin
n=1
∞
X
2
√
(n + 1)2 − 1 − n2 − 1
;
n(n + 1)
(n + 1)n
;
n2 3n
n
n=1
√
∞ −
X
e
√
n=1
n
n
;
Chapter 4
Euclidean spaces
Some basic facts regarding the Euclidean spaces are presented in this chapter.
4.1
Euclidean spaces
For each positive integer k, let Rk be the set of all ordered k -tuples
x = (x1 , x2 , . . . , xk ),
where x1 , . . . , xk are real numbers, called coordinates of x. The elements of Rk are
called points or vectors, especially when k > 1. If y = (y1 , . . . , yk ) and α is a real
number, let
x + y = (x1 + y1 , x2 + y2 , . . . , xk + yk ),
αx = (αx1 , . . . , αx2 , . . . , αxk ),
+ : Rk × Rk → R k ;
· : R × Rk → Rk .
These define addition of vectors, and respectively, multiplication of a vector by a real
number (a scalar).
Theorem 1.1. (a) (Rk , +) is a commutative group;
(b) α(x + y) = αx + αy, for every α ∈ R, x, y ∈ Rk ;
(c) (α + β)x = αx + βx, for every α, β ∈ R, x ∈ Rk ;
(d) α(βx) = (αβ)x, for every α, β ∈ R, x ∈ Rk .
These two operations make Rk into a vector (linear ) space over the field of the
reals. The zero element 0 in Rk is called the origin or the null vector and all its
coordinates are 0.
Proposition 1.1. (Calculus rules in a linear space)
(a) 0 · x = 0, ∀x ∈ Rk (the first 0 is a scalar, while the second is a vector);
(b) α · 0 = 0, ∀α ∈ R (the two 0 coincide, the null vector);
(c) (−1)x = (−x1 , . . . , −xn ), ∀x ∈ Rk ;
(d) α(x1 +· · ·+xn ) = αx1 +· · ·+αxn , for every α ∈ R, and xi ∈ Rk , i = 1, 2, . . . , n;
(e) (α1 +· · ·+αn )x = α1 x+· · ·+αn x, for every αi ∈ R, i = 1, 2, . . . , n, and x ∈ Rk .
71
72
4. Euclidean spaces
We define the inner product of x, y ∈ Rk by
hx, yi =
k
X
xi yi
i=1
and the Euclidean norm of x by
kxk2 =
p
hx, xi =
k
X
!1/2
x2i
.
i=1
The Minkowski norm or the l1 -norm of x is defined by
kxk1 = |x1 | + · · · + |xk |,
while the uniform norm of x is defined by
kxk∞ = max{|x1 |, . . . , |xk |}.
For 1 ≤ p < +∞ we define the lp -norm of x by
kxkp = (|x1 |p + · · · + |xk |p )1/p .
In order to indicate which the norm we are refering to we denote (Rk , k · kp ) and
(Rk , k · k∞ ).
Theorem 1.2. Let k · k be any of the norms defined above on Rk . Suppose x, y, z ∈
Rk , and α ∈ R. Then
(a) kxk ≥ 0;
(b) kxk = 0 ⇐⇒ x = 0;
(c) kαxk = |α|kxk;
(d) |hx, yi| ≤ kxk2 kyk2 ;
(e) kx + yk ≤ kxk + kyk;
(f) kx − zk ≤ kx − yk + ky − zk.
Proof. (a), (b), and (c) are trivial.
(d) If y = 0, we have equality. Suppose y 6= 0 and consider a real λ. Then
0 ≤ hx + λy, x + λyi = hx, xi + 2λhx, yi + λ2 hy, yi.
Taking λ = −
hx, yi
we get
hy, yi
0 ≤ hx, xi − 2
and (d) follows.
hx, xihy, yi − hx, yi2
hx, yi2 hx, yi2 hy, yi
=
+
hy, yi
hy, yi
hy, yi2
4.1. Euclidean spaces
73
(e) If the norm is an lp -norm, p ≥ 1, then our inequality is actually the Minkowski
inequality, (see page 23). If the norm under consideration is the uniform one, then a
straightforward evaluation proves this inequality.
(f) follows from (e).
2
Any function k · k : Rk → R satisfying (a), (b), (c), and (e) of theorem 1.2 is said
to be a norm on Rk .
Given a norm on Rk , we assign to it a metric by
ρ(x, y) = kx − yk.
(1.1)
Theorem 1.3. If k · k is a norm on Rk , then the metric ρ defined by (1.1) is a
metric on Rk ; thus every norm induces a metric.
Theorem 1.4. Any two metrics on Rk induced by any of the norms k · kp , p ≥ 1,
or k · k∞ generate the same open sets.
Corollary 1.1. A sequence is convergent in one of the metrics mentioned in the
above theorem if and only if it converges in any other of them.
Theorem 1.5. (a) Suppose xn ∈ Rk , n = 1, 2, · · · , and
xn = (α1n , α2n , · · · , αkn ).
Then (xn ) converges to x = (α1 , · · · , αk ) if and only if
(1.2)
lim αjn = αj ,
n→∞
1 ≤ j ≤ k.
(b) Suppose (xn )n , (yn )n are sequences in Rk , (βn )n is a sequence of real numbers, and xn → x, yn → y, βn → β. Then
lim(xn + yn ) = x + y,
lim hxn , yn i = hx, yi,
lim βn xn = βx.
Proof. We use the Euclidean metric.
(a) If xn → x, the inequalities
|αjn − αj | ≤ kxn − xk2 ,
j = 1, . . . , k
follow immediately from the definition of the norm in Rk . These imply that (1.2)
holds.
Conversely, if (1.2) holds, then to each ε > 0 corresponds an integer nε such that
n ≥ nε implies
ε
|αjn − αj | < √ , 1 ≤ j ≤ k.
k
74
4. Euclidean spaces
Hence n ≥ nε implies
kxn − xk2 =
k
X
!1/2
|αjn − αj |2
< ε,
j=1
so that xn → x. This proves (a).
(b) follows from (a) and from theorem 1.2, page 38.
2
Theorem 1.6. Every space (Rk , k · kp ), 1 ≤ p < ∞, and (Rk , k · k∞ ) are complete.
We consider the case of (Rk , k · k2 ). The other cases can be proved similarly. We
prove the following claim: a sequence (xn )n = (α1n , . . . , αkn )n in Rk is k · k2 -Cauchy
if and only if every sequence (αj,n )n , 1 ≤ j ≤ k, is a Cauchy sequence.
Proof of the claim. Suppose (xn ) is k · k2 -Cauchy. Then there is an integer nε
such that n, m ≥ nε imply
kxn − xm k2 < ε.
It follows that
|αjn − αjm | ≤ kxn − xm k2 < ε,
j = 1, . . . , k.
Conversely, suppose that for every ε > 0 there is an integer nε such that n, m ≥ nε
imply
ε
|αjn − αjm | < √ , j = 1, . . . , k.
k
Hence, n, m ≥ nε imply
kxn − xm k2 < ε.
Proof. Suppose (xn ) is k · k2 -Cauchy. Then every sequence (αjn )n , 1 ≤ j ≤ k, is
Cauchy, hence convergent to, say, αj ∈ R. Based on (a) of theorem 1.5, we infer that
(xn ) converges to x = (α1 , · · · , αn ).
2
A closed and bounded interval I in Rk is defined as
I = [a1 , b1 ] × · · · × [ak , bk ],
where each [ai , bi ] is a closed and bounded interval.
Theorem 1.7. Every closed and bounded interval I in Rk is compact.
Theorem 1.8. Let E be a set in Rk . Then the following statements are equivalent
(a) E is closed and bounded;
(b) E is compact;
(c) every infinite subset of E has a limit point in E.
Theorem 1.9. (Weierstrass) Every bounded infinite subset of Rk has a limit point
in Rk .
Chapter 5
Limits and Continuity
The aim of this chapter is to introduce notions and results on limits and continuity.
5.1
5.1.1
Limits
The limit of a function
Let X and Y be metric spaces; suppose ∅ 6= E ⊂ X, f maps E into Y, and p is
a limit point of E. We write
f (x) → q
as
x→p
or equivalently,
(1.1)
lim f (x) = q
x→p
if there is a point q ∈ Y with the following property: for every ε > 0 there exists a
δ > 0 such that
(1.2)
ρ(f (x), q) < ε
for all points x ∈ E for which
(1.3)
0 < ρ(x, p) < δ.
Remark. It should be noted that p ∈ X, but that p need not be a point of E.
Moreover, even if p ∈ E, we may very well have f (p) 6= limx→p f (x).
4
We can recast this definition in terms of sequences.
Theorem 1.1. Let X, Y, E, f, and p be as in the definition given above. Then
(1.4)
lim f (x) = q
x→p
75
76
5. Limits and Continuity
if and only if
(1.5)
lim f (pn ) = q
n→∞
for every sequence (pn )n in E such that
pn 6= p,
(1.6)
lim pn = p.
n→∞
Proof. Suppose (1.4) holds. Choose (pn ) in E satisfying (1.6). Let ε > 0 be given.
Then there exists δ > 0 such that ρ(f (x), q) < ε if x ∈ E and 0 < ρ(x, p) < δ.
Also, there exists nε such that n ≥ nε implies 0 < ρ(pn , p) < δ. Thus for n ≥ nε
we have ρ(f (pn ), q) < ε, which shows that (1.5) holds.
Conversely, suppose (1.4) is false. Then there exists some ε > 0 such that for
every δ > 0 there is a point x ∈ E (depending on δ ), for which ρ(f (x), q) > ε but
0 < ρ(x, p) < δ. Taking δn = 1/n, n ∈ N∗ , we thus find a sequence in E satisfying
(1.6) for which (1.5) is false.
2
Corollary 1.1. If f has a limit at p, this limit is unique.
Proof. Follows from theorem 1.1, (b), page 37 and from theorem 1.1.
Suppose Y = Rk and f, g : E → Rk . Define f + g and hf, gi by
(f + g)(x) = f (x) + g(x),
hf, gi(x) = hf (x), g(x)i,
2
x∈E
and if λ is a real number, (λf )(x) = λf (x).
Theorem 1.2. Suppose E ⊂ X is a metric space, p ∈ E 0 , f, g are functions defined
on E with values in Rk and
lim f (x) = A,
x→p
lim g(x) = B.
x→p
Then
(a) limx→p (f + g)(x) = A + B;
(b) limx→p hf, gi(x) = hA, Bi;
f
A
(c) If Y = R, and B =
6 0, then limx→p
(x) = .
g
B
Proof.
2
5.1. Limits
5.1.2
77
Right-hand side and left-hand side limits
Let f be defined on ]a, b[ . Consider any point x such that a ≤ x < b. We write
f (x+) = q
if f (tn ) → q as n → ∞ for all sequences (tn ) in ]x, b[ such that tn → x.
Consider any point x such that a < x ≤ b. We write
f (x−) = q
if f (tn ) → q as n → ∞ for all sequences (tn ) in ]a, x[ such that tn → x.
Theorem 1.3. Let I be a nonempty interval and p ∈ I. If f : I → R, then
lim f (x) = q ⇐⇒ f (x−) = f (x+) = q.
x→p
Proof. The necessity part is immediate.
Suppose now that f (x−) = f (x+) = q. We use the characterization in the theorem 1.1 and consider a sequence (xn ) in I , xn → p and having infinitely many terms
greater than p, (denoted by (yn ) ), and infinitely many terms less than p (denoted
by (zn ) ). By hypothesis, f (yn ) → q and f (zn ) → q. Then f (xn ) → q.
2
5.2
Continuity
Suppose X and Y are metric spaces, E ⊂ X, p ∈ E, and f maps E into Y.
Then f is said to be continuous at p if for every ε > 0, there exists a δ > 0 such
that
ρ(f (x), f (p)) < ε,
for all points x ∈ E for which ρ(x, p) < δ.
If f is continuous at every point of E, then f is said to be continuous on E.
Remark. It should be noted that f has to be defined at the point p in order to
be continuous at p. If p is an isolated point of E, then our definition implies that
every function f which has E as its domain of definition is continuous at p.
4
Theorem 2.1. In the above setting, assume also that p is a limit point of E. Then
f is continuous at p if and only if limx→p f (x) = f (p).
Proof.
2
Now we prove that a continuous function of a continuous function is continuous,
more precisely the following theorem holds
78
5. Limits and Continuity
Theorem 2.2. Suppose X, Y, and Z are metric spaces, E ⊂ X, f maps E in Y,
g maps the range of E, f (E), into Z, and h maps E into Z defined by
h(x) = g(f (x)),
x ∈ E.
If f is continuous at a point p ∈ E and if g is continuous at the point f (p), then
h is continuous at p.
Proof. Let ε > 0 be given. Since g is continuous at f (p), there exists η > 0 such
that if ρ(y, f (p)) < η and y ∈ f (E), we have
ρ(g(y), g(f (p))) < ε .
Since f is continuous at p, there exists δ > 0 such that
ρ(f (x), f (p)) < η,
if ρ(x, p) < δ and x ∈ E. It follows that
ρ(h(x), h(p)) = ρ(g(f (x)), g(f (p))) < ε,
if ρ(x, p) < δ and x ∈ E. Thus h is continuous at p.
2
Theorem 2.3. (Characterization of continuity) A mapping f of a metric space X
into a metric space Y is continuous on X if and only if f −1 (V ) is open in X for
every open set V in Y.
Proof. Suppose f is continuous on X and V is an open set in V. We have to show
that every point of f −1 (V ) is an interior point of f −1 (V ). Suppose p ∈ X, f (p) ∈ V.
Since V is open, there exists ε > 0 such that y ∈ V if ρ(f (p), y) < ε, and since
f is continuous at p there exists δ > 0 such that ρ(f (x), f (p)) < ε if ρ(x, p) < δ.
Thus x ∈ f −1 (V ) as soon as ρ(x, p) < δ.
Conversely, suppose f −1 (V ) is open in X for every open V in Y. Let p ∈ X
and ε > 0. Let
V = {y ∈ Y | ρ(y, f (p)) < ε}.
Then V is open; hence f −1 (V ) is open; hence, there exists δ > 0 such that x ∈
f −1 (V ) as soon as ρ(p, x) < δ. However, if x ∈ f −1 (V ), then f (x) ∈ V, so that
ρ(f (x), f (p)) < ε.
2
Theorem 2.4. Let f and g be complex-valued continuous functions on a metric
space X. Then f + g, f g, and f /g are continuous on X (g(x) 6= 0 for all x ∈ X).
Proof. At isolated points of X there is nothing to prove. At limit points, the statement follows from theorems 1.2 and 2.1.
5.2. Continuity
79
Theorem 2.5. (a) Let f1 , . . . , fk be real functions on a metric space X, and let f
be the mapping of X into Rk defined by
f (x) = (f1 (x), . . . , fk (x)),
x ∈ X.
Then f is continuous if and only if each of the functions f1 (x), . . . , fk (x) is continuous.
(b) If f and g are continuous mappings of X into Rk , then f + g and hf, gi
are continuous on X.
Functions f1 , . . . , fk are called components of f.
Proof. Part (a) follows from the inequalities
|fj (x) − fj (y)| ≤ kf (x) − f (y)k2 =
k
X
!1/2
2
|fi (x) − fi (y)|
,
i=1
for j = 1, . . . , k. Part (b) follows from (a) and theorem 1.2.
5.2.1
2
Continuity and compactness
A mapping f of a set E into Rk is said to be bounded if there is a real number M
such that |f (x)| < M, for all x ∈ E.
Theorem 2.6. Suppose f is a continuous mapping from a compact metric space X
into a metric space Y. Then f (X) is compact.
Proof. Let {Vα } be an open covering of f (X). Since f is continuous, each of the
sets f −1 (Vα ) is open (see theorem 2.3). Since X is compact, there are finitely many
indices, say α1 , . . . , αn , such that
(2.1)
X ⊂ f −1 (Vα1 ) ∪ f −1 (Vα2 ) ∪ · · · ∪ f −1 (Vαn ).
Since f (f −1 (E)) = E, for every E ⊂ Y, (2.1) implies that
f (X) ⊂ Vα1 ∪ Vα2 ∪ · · · ∪ Vαn .
This completes the proof.
2
Theorem 2.7. Suppose f is a continuous mapping from a compact metric space X
into Rk . Then f (X) is closed and bounded. Thus f is bounded.
Theorem 2.8. Suppose f is a continuous real function on a compact metric space
X, and
M = sup f (p), m = inf f (p).
p∈X
p∈X
Then there exist points p, q ∈ X such that f (p) = M, f (q) = m.
80
5. Limits and Continuity
Proof. By theorem 2.7, f (X) is a closed and bounded set of real numbers; hence
f (X) contains its sup and inf, (see theorem 1.5, page 30).
Theorem 2.9. Suppose f is a continuous bijective mapping of a compact metric
space onto a metric space Y. Then the inverse function f −1 defined on Y is a
continuous mapping of Y onto X.
Proof. Applying theorem 2.3 to f −1 in place of f, we see that it suffices to prove
that f (V ) is an open set in Y for every open set V in X. Fix such a set V. The
complement X \ V is closed in X, hence compact (theorem 2.3, page 33). It follows
that f (X \ V ) is a compact subset of Y which implies that it is closed in Y. Since
f is bijective, f (X \ V ) = Y \ f (V ). Hence f (V ) is open.
2
Exercise 2.1. (Cauchy functional equation) Let ϕ be a continuous function satisfying
(2.2)
ϕ(x + y) = ϕ(x) + ϕ(y),
∀x, y ∈ R.
Note that the solution is an odd function. Taking x = y = 0, we find ϕ(0) = 2ϕ(0),
i.e., ϕ(0) = 0. By induction one can show that ϕ(n) = nϕ(1), n ∈ N∗ . Then from
ϕ(1) = ϕ(
1
m
) = mϕ( )
m
m
it follows that
ϕ(
1
1
) = ϕ(1) m ∈ N∗ .
m
m
By induction one can further show that ϕ(p/q) = (p/q)ϕ(1), p ∈ Z, q ∈ Z \ {0}.
Now we know the representation of the solution if we restrict the problem to the
system of rational numbers.
Consider an arbitrary but fixed x ∈ R \ Q. Then there exists a sequence (qn ) of
rational numbers such that qn → x. Then using the continuity hypothesis one can
infer that
ϕ(x) = lim ϕ(qn ) = lim qn ϕ(1) = [lim qn ]ϕ(1)] = xϕ(1).
Denote ϕ(1) =: a. Then the solution of the Cauchy functional equation is ϕ(x) = ax,
for every x ∈ R.
4
5.2.2
Uniform continuous mappings
Let f be a mapping on a metric space X with values in a metric space Y. We say
that f is uniformly continuous on X if for every ε > 0 there exists δ > 0 such that
ρ(f (p), f (q)) < ε
for all p, q ∈ X for which ρ(p, q) < δ.
5.2. Continuity
81
Remark. Uniform continuity is the property of a function on the whole set, whereas
continuity can be defined at a single point. Asking whether a given function is uniformly continuous at a certain point is meaningless. Secondly, if f is continuous on
X, it is possible to find, for each ε > 0 and each point p ∈ X, a number δ > 0
having the property specified in definition of continuity; thus δ = δ(ε, p).
However, if f is uniformly continuous on X, then it is possible, for each ε > 0,
to find one number δ > 0 which will do for all points p ∈ X.
It is trivial to see that every uniformly continuous function is continuous. That
the two concepts are equivalent on compact sets follows from the next theorem.
Theorem 2.10. (Cantor) Let f be a continuous mapping of a compact metric space
X into a metric space Y. Then f is uniformly continuous on X.
Proof. Let ε > 0 be given. Since f is continuous, we can associate to each point
p ∈ X, a positive number φ(p) such that
(2.3)
x ∈ X, ρ(p, x) < φ(p) =⇒ ρ(f (p), f (x)) < ε/2.
Let I(p) be the ball defined as
1
I(p) = B(p, φ(p)).
2
(2.4)
Since p ∈ I(p), the family of all sets I(p) is an open cover of X and since X is
compact, there is a finite set of points p1 , . . . , pn in X, such that
X = I(p1 ) ∪ · · · ∪ I(pn ).
(2.5)
Let
(2.6)
δ=
1
min{φ(p1 ), . . . , φ(pn )}(> 0).
2
Now, let x and p be points of X such that ρ(x, p) < δ. By (2.5), there is an integer
m , 1 ≤ m ≤ n, such that p ∈ I(pm ). It follows that
(2.7)
1
ρ(p, pm ) < φ(pm ),
2
and also that
1
ρ(x, pm ) ≤ ρ(x, p) + ρ(p, pm ) < δ + φ(pm ) ≤ φ(pm ).
2
This, together with (2.3) imply that
ρ(f (p), f (x)) ≤ ρ(f (p), f (pm )) + ρ(f (pm ), f (x)) < ε. 2
82
5. Limits and Continuity
5.2.3
Continuity and connectedness
A set E in a metric space X is said to be connected if there are no two disjoint open
sets A and B, A ⊂ E, B ⊂ E such that E ⊂ A ∪ B. 4
Theorem 2.11. On the real axis, a set is connected if and only if it is an interval.
Similar to theorem 2.6, we have
Theorem 2.12. If f is a continuous mapping of a connected metric space X into
a metric space, then f (X) is connected.
Proof. If f (X) is not connected, there are disjoint open sets V and W in Y, both
of which intersect f (X), such that f (X) ⊂ V ∪ W. Since f is continuous, the sets
f −1 (V ) and f −1 (W ) are open in X; they are clearly disjoint and nonempty, and their
union is X. This implies that X is not connected, in contradiction to the hypothesis.
2
Theorem 2.13. (Darboux1 property to continuous functions) Let f be a continuous
real function on the interval [a, b]. If f (a) < f (b) and if c is a number such that
f (a) < c < f (b), there exists a point x ∈ ]a, b[ such that f (x) = c.
A similar result holds if f (a) > f (b).
Proof. By theorem 2.11, [a, b] is connected; hence, theorem 2.12 shows that f ([a, b])
is a connected subset of R1 , and the assertion follows if we appeal once more to the
theorem 2.11.
2
5.2.4
Discontinuities
If x is a point in the domain of definition of a function f at which f is not continuous,
we say that f is discontinuous at x, or that f has a discontinuity at x.
Let f be defined on ]a, b[ . If f is discontinuous at a point x, and if f (x+)
and f (x−) exist, then f is said to have a discontinuity of the first kind or a simple
discontinuity. Otherwise the discontinuity is said to be of the second kind. There are
two ways in which a function can have a simple discontinuity: either f (x+) 6= f (x−)
(in which case the value f (x) is immaterial), or f (x+) = f (x−) 6= f (x).
Examples. (a) Define
(
1, x ∈ Q,
f (x) =
0, x ∈ R \ Q.
Then f has a discontinuity of the second kind at the point x, since neither f (x+)
nor f (x−) exists.
(b) Define
(
x, x ∈ Q,
f (x) =
0, x ∈ R \ Q.
1
Gaston Garboux, 1842-1917
5.2. Continuity
83
Then f is continuous at x = 0 and has a discontinuity of the second kind at every
other point.
(c) Define
(
sin x1 , x 6= 0,
f (x) =
0,
x = 0.
Since neither f (0+) nor f (0−) exist, f has a discontinuity of the second kind at
x = 0. However, f is continuous at every point x 6= 0.
4
5.2.5
Monotonic functions
Let f be real on ]a, b[ . Then f is said to be monotonically increasing on ]a, b[ if
a < x < y < b implies f (x) ≤ f (y). If the last inequality is reversed, we obtain
the definition of a monotonically decreasing function. The set of monotonic functions
consists of both the increasing and the decreasing functions.
Theorem 2.14. Let f be monotonically increasing on ]a, b[ . Then f (x+) and
f (x−) exist at every point x ∈]a, b[ . More precisely,
(2.8)
sup f (t) = f (x−) ≤ f (x) ≤ f (x+) = inf f (t).
x<t<b
a<t<x
Furthermore, if a < x < y < b, then
f (x+) ≤ f (y−).
(2.9)
Proof. The set of numbers f (t) with a < t < x is bounded above by f (x). Therefore
it has a least upper bound which we shall denote by A. It is obvious that A ≤ f (x).
We have to show that A = f (x−).
Let ε > 0 be given. It follows from the definition of A as supremum that there
exists δ > 0 such that a < x − δ < x and
A − ε < f (x − δ) ≤ A.
(2.10)
Since f is monotonic, we have
(2.11)
f (x − δ) ≤ f (t) ≤ A,
x − δ < t < x.
Combining (2.10) and (2.11), we obtain
|f (t) − a| < ε,
x − δ < t < x.
Hence f (x−) = A.
The second half of (2.8) is proved in precisely the same way. Next, if a < x < y < b
we infer from (2.8) that
(2.12)
f (x+) = inf f (t) = inf f (t).
x<t<b
x<t<y
84
5. Limits and Continuity
Similarly
(2.13)
f (y−) = sup f (t) = sup f (t).
x<t<y
x<t<b
Comparing (2.12) and (2.13), we get (2.9).
2
Corollary 2.1. Monotonic functions have no discontinuity of the second kind.
Proof. Suppose, for the sake of definiteness, that f is increasing, and let E be the
set of points at which f is discontinuous.
With every point x ∈ E, we associate a rational number r(x) such that
f (x−) < r(x) < f (x+).
Since x1 < x2 , it follows that f (x1 +) ≤ f (x2 −). We note that r(x1 ) 6= r(x2 ), if
x1 6= x2 .
We have thus established an one-to-one correspondence between the set E and a
subset of the rational numbers. The latter is countable.
2
Exercise 2.2. ([24, Probl. 11, p. 9]) Let f be a continuous and increasing function
defined on an interval [a, b] and such that f (a) ≥ a and f (b) ≤ b. Choose an
arbitrary x1 ∈ [a, b] and consider the sequence (xn ) obtained as xn+1 = f (xn ),
n ≥ 1. Show that the limit lim xn =: x∗ exists and that f (x∗ ) = x∗ .
Since f is continuous, f ([a, b]) is a compact set. Thus the sequence (xn ) is included
in a compact set and thus it is bounded.
Function f being increasing, the sequence (xn ) is monotone. So we have established that the sequence (xn ) converges, let x∗ ∈ [a, b] be its limit.
From the estimations
0 ≤ |f (xn ) − x∗ | = |xn+1 − x∗ | → 0 as n → ∞
we conclude that f (x∗ ) = x∗ .
5.2.6
4
Darboux functions
Let I ⊂ R be a nonempty interval and f : I → R be a mapping.
We say that f is a Darboux function if for any a, b ∈ I, a < b and any λ between
f (a) and f (b), there is c ∈ ]a, b[ such that f (c) = λ. We denote the set of Darboux
function on an interval I by DI .
Remark. Geometrical, the Darboux property says that for any a, b ∈ I, a < b and
any λ between f (a) and f (b), the parallel to the Ox axis through y = λ meets
the restriction of f to the open interval ]a, b[ at least in one point.
4
Proposition 2.1. For a mapping f : I → R the next statements are equivalent
5.2. Continuity
85
(a) f is a Darboux function;
(b) if J ⊂ I is an interval, then f (J) is an interval, too (the image of an interval
is an interval);
(c) if a, b ∈ I, a < b, f ([a, b]) is an interval.
Proof. (a) =⇒ (b). For any y1 = f (t1 ), y2 = f (t2 ) ∈ f (J) (we may suppose that
y1 < y2 ) and any λ ∈ ]y1 , y2 [ by (a), there is c ∈ ]t1 , t2 [ such that λ = f (c). Since J
is an interval and t1 , t2 ∈ J, it follows that c ∈ J and therefore λ = f (c) ∈ inf(J).
Thus f (J) is an interval.
(b) =⇒ (c). It is trivial.
(c) =⇒ (a). Let a, b ∈ I, a < b and any λ between f (a) and f (b). Then
λ ∈ f ([a, b]) and by (c), there is c ∈ [a, b] such that λ = f (c). We note that
c∈
/ {a, b} and λ = f (c). Hence f is a Darboux mapping on I. 2
Corollary 2.2. Let f : I → R and g : f (I) → R be two Darboux mappings. Then
their composition g ◦ f is a Darboux mapping.
Proof. If f, g are Darboux mappings and J is a subinterval of I, (g◦f )(J) = g(f (J))
is an interval. Thus, by proposition 2.1, the mapping g ◦ f is a Darboux mapping.
2
Corollary 2.3. Let f : I → R be a Darboux mapping whose range is an at most
countable set. The function f is constant.
Proof. f (I) is an interval and is at most countable. Thus | f (I) |= 1, that is, f is
constant.
2
Corollary 2.4. Let f : I → R be a Darboux mapping. Then f is injective if and
only if it is strictly monotone.
Proof. Sufficiency. If f is strictly monotone, it is injective.
Necessity. Suppose f : I → R is a Darboux and injective function. If f is not
strictly monotone, there exist t1 < t2 < t3 such that either f (t1 ) < f (t2 ) > f (t3 ) or
f (t1 ) > f (t2 ) < f (t3 ). Let us suppose that the first case holds. The other case runs
similarly. The first case has two subcases.
• f (t1 ) < f (t3 ) < f (t2 ). Then there is c ∈ ]t1 , t2 [ with f (c) = f (t3 ). Thus function
f is not injective which is a contradiction.
•• f (t3 ) < f (t1 ) < f (t2 ). Reasoning as above, we find c ∈ ]t2 , t3 [ with f (c) = f (t1 ).
Thus it is contradicted the injectivity of f.
2
Corollary 2.5. Let f : I → R be a Darboux mapping. Then f is also a Darboux
function on any subinterval J ⊂ I.
Proof. It follows from (ii) of proposition 2.1.
2
Proposition 2.2. Suppose f : I → R is a Darboux mapping. Then
86
5. Limits and Continuity
(a) |f |, f 2 , and
p
|f | are Darboux mappings on I;
(b) if f (t) 6= 0 for every t ∈ I, then 1/f is a Darboux mapping on I.
Proof. Apply
corollary 2.2 to mapping f and to mappings g1 (t) = |t|, g2 (t) = t2 ,
p
g3 (t) = |t|, respectively, g(t) = 1/t.
2
Remark. We introduce two Darboux function whose sum is not a Darboux function.
(
(
sin 1t , t 6= 0,
− sin 1t , t 6= 0,
f (t) =
g(t) =
0,
t = 0,
1,
t = 0,
are two Darboux functions while their sum
(
0, t 6= 0,
(f + g)(t) =
1, t = 0,
is not a Darboux mapping.
4
Corollary 2.6. Let f : I → R be a Darboux mapping. Then f has no discontinuity
point of the first kind.
Proof. [15, p. 52].
2
Theorem 2.15. (Sierpinski2 ) Any mapping f : R → R is a sum of two discontinuous
Darboux mappings, i.e., exist f1 , and f2 Darboux and discontinuous functions such
that
f (t) = f1 (t) + f2 (t), ∀t ∈ R.
A proof of the previous theorem may be found, e.g., in [15, p. 46-48].
5.2.7
Lipschitz functions
Let I be a real interval and f : I → R be a function. Then f is said to be Lipschitz3
on I if there exists a nonnegative L such that
(2.14)
|f (x) − f (y)| ≤ L|x − y|,
∀ x, y ∈ I.
Remark. Every Lipschitz function on an interval is uniformly continuous on it.
4
Let A be a nonempty set and f : A → A be a mapping. A point x ∈ A is said
to be a fixed point of f if f (x) = x.
Let (X, ρ) be a metric space and f : X → X be a mapping. f is said to be a
contraction if there exists a constant α ∈ ]0, 1[ so that
ρ(f (x), f (y)) ≤ αρ(x, y),
for every x, y ∈ X.
In this very case the mapping f is said to be an α -contraction.
Remark. Every α -contraction is a Lipschitz function.
4
2
3
Waclaw Sierpinski, 1882-1969
Lipschitz
5.2. Continuity
87
Theorem 2.16. (Banach4 fixed point theorem) Let (X, ρ) be a complete metric space
and f : X → X be an α -contraction. Then f has a unique fixed point.
Proof. Choose an arbitrary, but fixed, point x0 ∈ X. Define the sequence (xn )n≥1 by
xn+1 = f (xn ), n = 0, 1, . . . . For every n ∈ N∗ and every integer p, p ≥ 1, we have
the estimations
ρ(xn+1 , xn ) = ρ(f (xn ), f (xn−1 ) ≤ αρ(xn , xn−1 ) ≤ · · · ≤ αn ρ(x1 , x0 ),
and by the triangle inequality
ρ(xn+p , xn ) ≤ ρ(xn+p , xn+p−1 ) + ρ(xn+p−1 , xn+p−2 ) + · · · + ρ(xn+1 , xn )
≤ (αn+p−1 + αn+p−2 + · · · + αn )ρ(x1 , x0 )
= αn (αp−1 + αp−2 + · · · + 1)ρ(x1 , x0 )
1
≤ αn
ρ(x1 , x0 ).
1−α
Thus we may conclude that (xn ) is a Cauchy sequence. The metric space X is
complete, thus the sequence (xn ) is convergent. Let x ∈ X be its limit. Passing p
to +∞ in the previous estimations we get
(2.15)
ρ(x, xn ) ≤ αn
1
ρ(x1 , x0 ).
1−α
Now, substituting xn by f (xn−1 ) and passing n to +∞, we get
ρ(x, f (x)) = 0.
Thus x = f (x), i.e., x is a fixed point of f.
Suppose that there are points x and y (not necessarily distinct) such that x =
f (x) and y = f (y). Then
ρ(x, y) = ρ(f (x), f (y)) ≤ αρ(x, y)
implies that
(1 − α)ρ(x, y) ≤ 0.
Hence x = y, i.e., the fixed point of the function f is unique.
2
Remark. We notice that for any starting point x0 ∈ X we get precisely the same
limit point and this limit point is the unique fixed point of f.
4
The sequence (xn ) is said to be the sequence of successive approximations of x.
Based on (2.15), we can establish the speed of convergence of the sequence (xn ) to
its limit x.
Example. Suppose we have to find a real roots of the equation
(2.16)
4
stefan Banach,
x3 + 2x − 1 = 0.
88
5. Limits and Continuity
First we note that if we denote the left-hand side of (2.16) by g(x), we get a polynomial function; this function is strictly increasing. Hence it has at most one real root.
At the same time from
lim g(x) = −∞
and
x→−∞
lim g(x) = +∞,
x→+∞
and, moreover, taking into account theorem 2.13, we conclude that (2.16) has precisely
one real root.
In order to find this real root we rewrite the given equation as
(2.17)
x=
x2
1
.
+2
Denote the right-hand side of (2.17) by f (x) and thus we get a real rational function
defined on R.
We are checking the assumption of theorem 2.16. For x, y ∈ R we have
|f (x) − f (y)| =
However
|x + y|
|x − y|.
(2 + x2 )(2 + y 2 )
1 √
2 + t2
|t| = √ 2t2 ≤ √ .
2
2 2
Thus
|x + y| ≤ |x| + |y| ≤
2 + x2 + 2 + y 2
(2 + x2 )(2 + y 2 )
√
√
≤
.
2 2
2 2
Now we conclude that the function f is a contraction. From (2.17) easily follows that
the solution has to be strictly positive. Since the function f is strictly decreasing
on the positive semi-axis and f ([0, 1]) = [1/3, 1/2] ⊂ [0, 1], it follows that it is a
contraction on the compact interval [0, 1]. Thus we may apply the Banach fixed
point theorem and if we start from x0 = 0, then
x1 =0.5, x2 = 0.444444 . . . , x3 = 0.455056 . . . , x4 = 0.453088 . . . ,
x5 =0.453455 . . . , x6 = 0.453386 . . . , x7 = 0.453399 . . . . 4
5.2.8
Convex functions
Let I be a real interval and f : I → R be a function. Then f is said to be convex
provided
(2.18)
x, y ∈ I, α ∈ [0, 1] =⇒ f ((1 − α)x + αy) ≤ (1 − α)f (x) + αf (y).
f is said to be strictly convex provided
(2.19)
x, y ∈ I, α ∈ [0, 1] =⇒ f ((1 − α)x + αy) < (1 − α)f (x) + αf (y).
Figure 5.1 represents a convex function.
5.2. Continuity
89
Figure 5.1:
Lemma 2.1. (Jensen5 inequality) Suppose f is a convex function. Then
f (α1 x1 + · · · + αn xn ) ≤ α1 f (x1 ) + · · · + αn f (xn )
(2.20)
for any n ∈ N∗ , any x1 , . . . , xn ∈ I, and any α1 , . . . , αn ≥ 0 with α1 + · · · + αn = 1.
Proof. If n = 2, the claim is true. Suppose n ≥ 2 and the claim is true for
n
1. Consider x1 , . . . , xn ∈ I, α1 , . . . , αn ≥ 0, and α1 + · · · + αn = 1. Then
P−
n−1
i=1 αi /(1 − αn ) = 1, so by hypothesis
! n−1
n−1
X
X αi
αi
f
≤
f (xn ).
1
−
α
1
−
α
n
n
i=1
i=1
Hence
f
n
X
!
αi xi
=f ((1 − αn )
1
n−1
X
1
n−1
X
≤(1 − αn )f (
≤(1 − αn )
1
n−1
X
1
αi
xi + αn xn )
1 − αn
αi
xi ) + αn f (xn )
1 − αn
n
X
αi
f (xi ) + αn f (xn ) =
αi f (xi ).
1 − αn
1
Corollary 2.7. Suppose f : I → R is a convex mapping. Then
α1 x1 + · · · + αn xn
α1 f (x1 ) + · · · + αn f (xn )
(2.21)
f
≤
,
α1 + · · · + αn
α1 + · · · + αn
for any x1 , . . . , xn ∈ I, α1 , . . . , αn ≥ 0 with α1 + · · · + αn > 0.
Let f be a real function defined on an interval I. Then
sf,x0 (x) =
f (x) − f (x0 )
,
x − x0
x ∈ I \ {x0 }
is said to be the slope of f.
Lemma 2.2. Suppose f : I → R is a mapping. Then
(a) f is convex on I
5
Jensen, 1859-1925
=⇒ sf,x0 is increasing on I \ {x0 };
90
5. Limits and Continuity
(b) f is strictly convex on I
Proof.
=⇒ sf,x0 is strictly increasing on I \ {x0 };
2
Theorem 2.17. Suppose f : I → R is a convex mapping on an interval I. Then
(a) f is continuous on the interior of I;
(b) f is Lipschitz on every compact interval contained in I.
Proof.
2
5.2.9
Jensen convex functions
Let I be a real interval and f : I → R be a function. Then f is said to be Jensen
convex or J-convex provided
x+y
f (x) + f (y)
)≤
.
2
2
Proposition 2.3. A continuous and J-convex function f : I → R is convex.
x, y ∈ I =⇒ f (
(2.22)
Proof. From (2.22), by induction, we get
1
1
(x
+
·
·
·
+
x
[f (x1 ) + · · · + f (x2k )],
k )) ≤
1
2
2k
2k
for every k ∈ N∗ , and x1 , . . . , x2k ∈ I.
Consider x, y ∈ I and α ∈ ]0, 1[ . We show that (2.18) is satisfied.
Write α as
α1 α2
αk
α=
+ 2 + ··· + k + ...,
2
2
2
∗
where αi ∈ {0, 1}, for all i ∈ N . Denoting
(2.23)
f(
α(k) =
α1 α2
αk
α1 2k−1 + · · · + αk
βk
+ 2 + ··· + k =
=
,
2
2
2
2k
2k
it result
lim α(k) = α,
k→∞
1 − α(k) =
2k − βk
.
2k
We consider in (2.23) that
x1 = x2 = · · · = xβk = x and xβk +1 = xβk +1 = · · · = x2k = y.
Then we get
βk x + (2k − βk )
y)
2k
βk f (x) + (2k − βk )f (y)
≤
= α(k) f (x) + (1 − α(k) f (y).
2k
Now passing k → ∞ and taking into account the continuity of f, the conclusion
follows.
2
f (α(k) x + (1 − α(k) y) =f (
Chapter 6
Differential calculus
This chapter is devoted to introduce some basic results on differential calculus.
6.1
The derivative of a real function
Let f be defined on [a, b] and real-valued. For any x ∈ [a, b] form the quotient
(1.1)
φ(t) =
f (t) − f (x)
,
t−x
(a < t < b, t 6= x)
and define
f 0 (x) = lim φ(t)
(1.2)
t→x
provided this limit exists.
We thus associate to the function f a function f 0 whose domain of definition is
the set of points x at which limit (1.2) exists; f 0 is called the derivative of f.
If f 0 is defined at a point x, we say that f is differentiable on a point at x. If
f 0 is defined at every point of a set E ⊂ [a, b], we say that f is differentiable on E.
Theorem 1.1. Let f be defined on [a, b]. If f is differentiable at a point x ∈ [a, b],
then f is continuous at x.
Proof. As t → x, we have
f (t) − f (x) =
f (t) − f (x)
(t − x) → f 0 (x) · 0 = 0. 2
t−x
Remark. The converse is not true. Consider f (t) =| t |, t ∈ R and x = 0.
It holds a stronger result, namely
4
Theorem 1.2. (Weierstrass) There exists a continuous function on R having no
point of differentiability.
91
92
6. Differential calculus
Proof. Consider the function h(t) = |t|, t ∈ [−1, 1], and extend it by periodicity on
R such that


t ∈ [−1, 1],
h(t),
g(t) = g(t − 2),
t > 1,


g(t + 2),
t < −1.
Then for any s, t ∈ R
(1.3)
|g(t) − g(s)| ≤ |t − s|.
Thus function g is Lipschitz on R, so it is continuous on R.
Define
∞ n
X
3
f (t) =
g(4n t).
4
n=0
Since 0 ≤ g ≤ 1, the series is uniformly convergent. We have uniform convergence
of continuous functions, so f is continuous on R.
Now we show that function f is nowhere differentiable. Choose an arbitrary, but
fixed x ∈ R.
Define δm = ± 12 4−m , where the sign is chosen such that between 4m x and 4m (x+
δm ) there is no integer.
Define
g(4n (x + δm )) − g(4n x)
.
γn =
δm
If n > m, 4n δm is an even number and thus γn = 0. If 0 ≤ n ≤ m, from (1.3) it
follows that |γn | ≤ 4n . Since |γm | = 4m , it follows the estimation
∞ n m−1
X
f (x + δm ) − f (x) X
3
1
m
=
γ
≥
3
−
3n = (3m + 1).
n
δm
4
2
n=0
n=0
If m → ∞, δm → 0. Hence function f is not differentiable on x, and the theorem
is proved.
2
We introduce some arithmetic properties of differentiable functions.
Theorem 1.3. Suppose f and g are defined on [a, b] and are differentiable at a
point x ∈ [a, b]. Then f + g, f · g and f /g are differentiable at x, and
(a) (f + g)0 (x) = f 0 (x) + g 0 (x);
(b) (f g)0 (x) = f 0 (x)g(x) + f (x)g 0 (x);
f
f 0 (x)g(x) − f (x)g 0 (x)
(c)
(x) =
(we assume that g(x) 6= 0).
g
g 2 (x)
6.1. The derivative of a real function
93
Proof. (a) is trivial.
(b) Let h = f · g. Then
f (t)g(t) − f (x)g(t) + f (x)g(t) − f (x)g(x)
h(t) − h(x) f (t)g(t) − f (x)g(x)
=
=
t−x
t−x
t−x
g(t) − g(x)
f (t) − f (x)
=g(t)
+ f (x)
.
t−x
t−x
(c)
f
f
(t) −
(x)
f (t)g(x) − g(t)f (x)
g
g
=
t−x
(t − x)g(x)g(t)
1
f (t) − f (x)
g(t) − g(x)
g(x) − f (x)
. 2
=
g(t)g(x)
t−x
t−x
Theorem 1.4. Suppose f is continuous on [a, b], f 0 (x) exists at some point x ∈
[a, b], g is defined on a closed interval I that contains the range of f, and g is
differentiable at the point f (x). If
h(t) = g(f (t)) a ≤ t ≤ b,
h is differentiable at x, and
(1.4)
h0 (x) = g 0 (f (x)) · f 0 (x).
Proof. Let y = f (x). By the definition of the derivative, we have
(1.5)
(1.6)
f (t) − f (x) = (t − x)[f 0 (x) + u(t)]
g(s) − g(y) = (s − y)[g 0 (y) + v(s)]
where t ∈ [a, b], s ∈ I, u(t) → 0 as t → x, v(s) → 0 as s → y. First we use (1.6)
and then (1.5). We obtain
h(t) − h(x) =g(f (t)) − g(f (x)) = [f (t) − f (x)][g 0 (y) + v(s)]
=(t − x)[f 0 (x) + u(t)][g 0 (y) + v(s)]
or, if t 6= x,
(1.7)
h(t) − h(x)
= [g 0 (y) + v(s)][f 0 (x) + u(t)].
t−x
Let t → x and see that s → y, by the continuity of f. Thus the right-hand side of
(1.7) tends to g 0 (y)f 0 (x), which is actually (1.4).
2
94
6. Differential calculus
Examples 1.1. (a) Consider
(
x sin x1 ,
f (x) =
0,
x 6= 0
x = 0.
Then f 0 (x) = sin x1 − x1 cos x1 , x 6= 0 and does not exist f 0 (0).
(b) Consider
(
x2 sin x1 , x =
6 0
f (x) =
0,
x = 0.
Then f 0 (x) = 2x sin x1 − cos x1 , x 6= 0. Now f 0 (0) = 0.
(c) The above cases are particular cases of the next one. Let a ≥ 0, c > 0 be
constants and f : [−1, 1] → R be a function defined by
(
xa sin(x−c ), x 6= 0
f (x) =
0,
x = 0.
Then
(c 1 ) f is continuous ⇐⇒ a > 0;
(c 2 ) ∃ f 0 (0) ⇐⇒ a > 1;
(c 3 ) f 0 is bounded ⇐⇒ a ≥ 1 + c;
(c 4 ) f 0 is continuous ⇐⇒ a > 1 + c;
(c 5 ) ∃ f 00 (0) ⇐⇒ a > 2 + c;
(c 6 ) f 00 is bounded ⇐⇒ a ≥ 2 + 2c;
(c 7 ) f 00 is continuous ⇐⇒ a > 2 + 2c.
4
Now we study the behaviour of the ratio from (1.1).
Proposition 1.1. Consider a function f : ] − 1, 1[ → R such that ∃ f 0 (0). Choose
two sequences (αn )n and (βn )n satisfying −1 < αn < βn < 1 such that αn → 0 and
βn → 0 as n → ∞. Define
dn :=
f (βn ) − f (αn )
.
βn − αn
Then
(a) if αn < 0 < βn , limn→∞ dn = f 0 (0);
n
is bounded, limn→∞ dn = f 0 (0);
(b) if 0 < αn and the sequence βnβ−α
n
n
6.1. The derivative of a real function
95
(c) if f 0 is continuous on ] − 1, 1[ , limn→∞ dn = f 0 (0);
(d) there exists a differentiable function f on ] − 1, 1[ (with discontinuous derivative) such that αn → 0, βn → 0, and ∃ limn→∞ dn , but limn→∞ dn 6= f 0 (0).
Proof. (a) We write dn as a convex combination of the following form
f (βn ) − f (0)
βn
f (0) − f (αn )
−αn
f (βn ) − f (αn )
=
·
+
·
,
βn − αn
βn
βn − αn
−αn
βn − αn
βn
−αn
λ1 =
, λ2 =
, λ1 , λ2 > 0, λ1 + λ2 = 1.
βn − αn
βn − αn
It follows
f (βn ) − f (0) f (0) − f (αn )
f (βn ) − f (αn )
min
,
≤
βn
−αn
βn − αn
f (βn ) − f (0) f (0) − f (αn )
≤ max
,
.
βn
−αn
The assumption guarantees us that
f (βn ) − f (0)
f (0) − f (αn )
= lim
= f 0 (0).
n→∞
n→∞
βn
−αn
lim
Now invoking theorem 1.9 from page 43 we get the conclusion.
(b) Consider the following identity
f (βn ) − f (αn ) f (0) − f (αn )
βn
−
=
βn − αn
−αn
βn − αn
f (βn ) − f (0) f (0) − f (αn )
−
βn
−αn
.
n
The sequence βnβ−α
is bounded and since the difference from the parenthesis
n
n
tends to 0, the conclusion follows.
(c)
(d) Consider the function given in (b) of examples 1.1 and the sequences
βn =
2
,
(2n + 1)π
αn =
1
. 2
nπ
Theorem 1.5. Let I, J ⊂ R be compact intervals and f : I → J be a continuous
and bijective mapping. Suppose function f is differentiable on a point x0 ∈ I and
f 0 (x0 ) 6= 0. Then the inverse function f −1 : J → I is differentiable at y0 = f (x0 )
and it holds
1
[f −1 (y0 )]0 = 0
.
f (x0 )
96
6. Differential calculus
Proof. Since function f is continuous and bijective and the interval I is compact it
follows that f −1 is continuous (theorem 2.9, page 80). Therefore
(1.8)
xn ∈ I \ {x0 }, xn → x0 ⇐⇒ yn = f (xn ) ∈ J \ {y0 }, yn → y0 .
Take y = f (x). Then
f −1 (y) − f −1 (y0 )
1
,
=
f (x) − f (x0 )
y − y0
x − x0
∀x ∈ I \ {x0 }.
From (1.5) it follows that y → y0 ⇐⇒ x → x0 , hence
lim
y→y0
f −1 (y) − f −1 (y0 )
1
1
= lim
= 0
. 2
x→x0 f (x) − f (x0 )
y − y0
f (x0 )
x − x0
6.2
Mean value theorems
Let f be a real function defined on a metric space X. We say that f has a local
maximum at a point p ∈ X if there exists δ > 0 such that f (q) ≤ f (p) for all
q ∈ B(p, δ). Analogously, we say that f has a local minimum at a point p ∈ X if
there exists δ > 0 such that f (q) ≥ f (p) for all q ∈ B(p, δ).
Theorem 2.1. (Fermat1 ) Let f be defined on [a, b]. If f has a local maximum point
x ∈ ]a, b[ and if there exists f 0 (x), then f 0 (x) = 0.
The analogous statement for local minimum is true, too.
Proof. The idea is suggested in figure 6.1. Choose δ in accordance with the above
definition, so that a < x − δ < x < x + δ < b. If x − δ < t < x,
f (t) − f (x)
≥ 0.
t−x
Letting t → x, we see that f 0 (x) ≥ 0.
If x < t < x + δ,
f (t) − f (x)
≤ 0.
t−x
Letting t → x, we see that f 0 (x) ≤ 0. Hence f 0 (x) = 0.
1
Pierre de Fermat, 1601-1665
2
6.2. Mean value theorems
97
Figure 6.1:
Figure 6.2:
Theorem 2.2. (Cauchy) If f and g are continuous real functions on [a, b] which
are differentiable in ]a, b[ , then there exists a point x ∈]a, b[ at which
[f (b) − f (a)]g 0 (x) = [g(b) − g(a)]f 0 (x).
Proof. Put
h(t) := [f (b) − f (a)]g(t) − [g(b) − g(a)]f (t),
a ≤ t ≤ b.
Then h is continuous on [a, b] and differentiable on ]a, b[ , and
(2.1)
h(a) = f (b)g(a) − f (a)g(b) = h(b).
We have to show that h0 (x) = 0 for some x ∈ ]a, b[ .
If h is constant, h0 (x) = 0 for every x ∈ ]a, b[ .
If h(t) > h(a) for some t ∈ ]a, b[ , let x be a point in [a, b] at which h attains
its maximum value, theorem 2.6, page 79. By (2.1) it follows that x ∈ ]a, b[ , while
by theorem 2.1 we get the conclusion.
If h(t) < h(a) for some t ∈ ]a, b[ , the same reasoning applies if we choose as x a
point in [a, b] at which h attains its minimum value.
2
Theorem 2.3. (Lagrange) If f are continuous real function on [a, b] which is differentiable in ]a, b[ , then there exists a point x ∈ ]a, b[ at which
f (b) − f (a) = (b − a)f 0 (x).
Proof. Take g(t) = t in theorem 2.3. The idea is suggested in figure 6.2.
2
Theorem 2.4. Suppose f is differentiable on ]a, b[ .
(a) If f 0 (t) ≥ 0 for any t ∈ ]a, b[ , f is monotonically increasing on ]a, b[ ;
(b) if f 0 (t) = 0 for any t ∈ ]a, b[ , f is constant on ]a, b[ ;
(c) If f 0 (t) ≤ 0 for any t ∈]a, b[ , f is monotonically decreasing on ]a, b[ ;
Proof. All conclusions can be read off from the relation
f (x2 ) − f (x1 ) = (x2 − x1 )f 0 (x),
which is valid for each pair of numbers x1 , x2 ∈ ]a, b[ , for some x between x1 and
x2 .
2
98
6.2.1
6. Differential calculus
Consequences of the mean value theorems
Lemma 2.1. Consider I = [a, b[ , a < b ≤ +∞. Suppose f, g : I → R are
continuous functions. Moreover,
(i) f and g are differentiable on ]a, b[ and f 0 (t) ≤ g 0 (t) for every t ∈]a, b[ ;
(ii) f (a) ≤ g(a).
Then f (t) ≤ g(t) for any t ∈ I.
Proof. Define h = g − f. Function h is differentiable on ]a, b[ and h0 (t) ≥ 0 for any
t ∈ ]a, b[ . From theorem 2.4 it follows that h is increasing and hence
h(t) ≥ h(a) ≥ 0,
for all t ∈ [a, b[ . 2
Lemma 2.2. Consider I = [a, b], a < b. Suppose f, g : I → R are continuous
functions. Moreover,
(i) f and g are differentiable on ]a, b[ ;
(ii) |f 0 (t)| ≤ g 0 (t) for every t ∈ ]a, b[ .
Then
|f (b) − f (a)| ≤ g(b) − g(a).
Proof. The conclusion is equivalent to
(2.2)
g(a) − g(b) ≤ f (b) − f (a) ≤ g(b) − g(a),
that is
g(b) − f (b) − (g(a) − f (a)) ≥ 0 and f (b) + g(b) − (f (a) + g(a)) ≥ 0.
These two inequalities suggest us to consider the next auxiliary functions
(
h1 (t) = g(t) − f (t) − (g(a) − f (a)),
h1 , h2 : I → R,
h2 (t) = f (t) + g(t) − (f (a) + g(a)).
We note that h1 (a) = h2 (a) = 0 and
(
h01 (t) = g 0 (t) − f 0 (t) ≥ 0,
h02 (t) = f 0 (t) + g 0 (t) ≥ 0.
Then h1 (t) ≥ h1 (a) and h2 (t) ≥ h2 (a), on [a, b], and hence (2.2) follows.
2
Lemma 2.3. Let I ⊂ R be a nonempty interval, t0 ∈ I. Consider f : I → R a
continuous function on I, differentiable on I \ {t0 }.
6.2. Mean value theorems
99
(a) If f 0 has a left hand side limit at t0 , then φ in (1.1) has a left hand side limit
at t0 and
lim f 0 (t) = lim φ(t).
t→t0
t<t0
t→t0
t<t0
(b) If f 0 has a right hand side limit at t0 , then φ in (1.1) has a right hand side
limit at t0 and
lim f 0 (t) = lim φ(t).
t→t0
t>t0
t→t0
t>t0
(c) If f 0 has a finite limit at t0 , then f is differentiable at t0 and f 0 is continuous
at t0 .
Proof. (a) Let t < t0 , t ∈ I. By the mean value theorem (2.3) there exists ct ∈ ]t, t0 [
such that
f (t) − f (t0 )
φ(t) =
= f 0 (ct ).
t − t0
Letting t → t0 we get the conclusion.
(b) Similar to (a).
(c) It follows from (a) and (b).
2
Lemma 2.4. Let I ⊂ R be a nonempty interval and f : I → R be a differentiable
function on I. If f 0 is bounded on I, then f is Lipschitz on I.
Proof. Since f 0 is bounded on I, there is a positive L such that |f 0 (t)| ≤ L, for any
t ∈ I. Then for any t1 , t2 ∈ I, t1 < t2 there is c ∈ ]t1 , t2 [ such that
|f (t2 ) − f (t1 )| = |f 0 (c)(t2 − t1 )| ≤ L(t2 − t1 ),
hence function f is Lipschitz on I.
2
Theorem 2.5. (Denjoy2 -Bourbaki3 theorem, [13, vol.2, p. 77]) Let I ⊂ R be a
nonempty interval, a, b ∈ I, a < b, and f : I → R be a function on I. Suppose
(i) function f is continuous on [a, b];
(ii) there is an at most countable set A ⊂ ]a, b[ such that f is right differentiable
on ]a, b[ \A .
Then
inf
t∈ ]a,b[ \A
2
3
φ(t+) ≤
f (b) − f (a)
≤ sup φ(t+).
b−a
t∈ ]a,b[ \A
Arnaud Denjoy, 1884-1974
Nikolas Bourbaki, collective pseudonym of several French mathematicians, 1939 ↑
100
6. Differential calculus
Proof. We prove that
f (b) − f (a)
≤ sup φ(t+) =: M.
b−a
t∈ ]a,b[ \A
The other side follows in a similar way.
If M = +∞, the inequality is obvious. Suppose M < ∞ and consider the
function
g : I → R g(t) := M t − f (t).
We remark that function g fulfills the assumptions of lemma 2.6. Thus g is increasing
on I. Then g(b) ≥ g(a), that is M (b − a) ≥ f (b) − f (a).
2
Corollary 2.1. Suppose f, g : I → R are continuous on a nonempty interval I.
Then
(a) f is constant on I if and only if there is an at most countable set A ⊂ I such
that f is right-hand side differentiable on I \ A and
φ(t+) = 0,
for any t ∈ I \ A;
(b) f − g is constant on I if and only if there is an at most countable set A ⊂ I
such that f and g are right-hand side differentiable side on I \ A and
φf (t+) = φg (t+) = 0,
for any t ∈ I \ A,
where φf and φg are the φ functions corresponding to f, respectively to g.
Remark. The right-hand side differentiability from the Denjoy-Bourbaki theorem
may be substituted by the left-hand side differentiability.
4
Lemma 2.5. Let I ⊂ R be a nonempty interval. Suppose
(i) function f is continuous on I;
(ii) there is an at most countable set A ⊂ I such that for every s ∈ I \ A and δ
there is t ∈ ]s, s + δ[ satisfying f (t) ≥ f (s).
Then function f is increasing on I.
Proof. Consider a, b ∈ I, a < b. We show that f (a) ≤ f (b).
Let λ ∈
/ f (A), λ < f (a). Consider the set
Sλ := {s ∈ [a, b] | λ ≤ f (s)}.
Note that
• Sλ 6= ∅, since a ∈ Sλ ;
6.2. Mean value theorems
101
•• set Sλ is bounded above, since Sλ ⊂ [a, b].
Hence
∃ M := sup Sλ (∈ [a, b]).
Since M is a limit point of Sλ , there is sn ∈ Sλ such that sn → M. From
)
sn ∈ Sλ
(2.3)
=⇒ [λ ≤ f (sn ) =⇒ λf (M ).]
f continuous at M
So, M ∈ Sλ .
Now we show that M = b. Suppose M < b. Then M is a limit point for the
complement of Sλ , so there is tn ∈
/ Sλ , tn → M. It follows that f (tn ) < λ, and, due
to the continuity of f,
f (M ) ≤ λ.
(2.4)
From (2.3)and (2.4) it follows that f (M ) = λ and thus M ∈
/ A.
On the other side, since M < b and (ii), there is t ∈ ]M, b[ such that λ = f (M ) ≤
f (t) and so t ∈ Sλ and t > M, that is a contradiction. Thus M = b.
Hence for any λ ∈
/ f (A) such that λ < f (a) it follows that λ ≤ f (b). So we have
that f (a) ≤ f (b).
2
Lemma 2.6. Let I ⊂ R be a nonempty interval. Suppose
(i) function f is continuous on I;
(ii) there is an at most countable set A ⊂ I such that there exists φ(t+) (φ defined
by (1.1) ) for any t ∈ I \ A;
(iii) φ(t+) ≥ 0 for any t ∈ I \ A.
Then function f is increasing on I.
Proof. From (iii) it follows that for every ε > 0 and t ∈ I \ A there is δ = δ(ε, t)
such that for every h ∈ ]0, δ[ it holds
f (t + h) − f (t)
< ε,
−
φ(t+)
h
which means that
−ε < φ(t+) − ε <
f (t + h) − f (t)
< φ(t+) + ε
h
and so
f (t + h) − f (t) + hε > 0,
for every h ∈ ]0, δ[ , t ∈ I \ A,
102
6. Differential calculus
that is
f (t + h) + (t + h)ε − [f (t) + tε] > 0,
for every h ∈ ]0, δ[ , t ∈ I \ A.
Denote
gε (t) := f (t) + tε.
Then the last inequality is equivalent with
gε (t + h) − gε (t) > 0,
for every h ∈ ]0, δ[ , t ∈ I \ A.
From lemma (2.5) it follows that gε is increasing on I, that is for any t1 , t2 ∈ I,
t1 < t2 we have
gε (t1 ) ≤ gε (t2 ) ⇐⇒ f (t1 ) ≤ f (t2 ) + ε(t2 − t1 ).
Letting ε → 0 we get that f (t1 ) ≤ f (t2 ), meaning that f is increasing.
6.3
2
The continuity of derivatives
We have already seen
(
x2 sin x1 ,
f (x) =
0,
x 6= 0
x = 0,
that a function f may have a derivative f 0 which exists at every point, but is discontinuous at some point. However, not every function is a derivative. In particular,
derivatives which exist at every point of an interval have one important property in
common with functions which are continuous on an interval: intermediate values are
assumed.
Theorem 3.1. (Darboux property of derivatives) Suppose f is a real differentiable
function an [a, b] and suppose f 0 (a) < λ < f 0 (b). Then there is a point x ∈ ]a, b[
such that f 0 (x) = λ.
A similar result holds of course if f 0 (a) > f 0 (b).
Proof. Put c = (a + b)/2. If a ≤ t ≤ c, define α(t) = a, β(t) = 2t − a. If c ≤ t ≤ b,
define α(t) = 2t − b, β(t) = b. Then a ≤ α(t) < β(t) ≤ b in ]a, b[ . Define
g(t) :=
f (β(t)) − f (α(t))
,
β(t) − α(t)
a < t < b.
then g is continuous on ]a, b[ , g(t) → f 0 (a), as t → a, g(t) → f 0 (b), as t → b,
and so theorem 2.12 page 82 implies that g(t0 ) = λ for some t0 ∈ ]a, b[ . Fix t0 . By
theorem 2.3 page 97 there is a point x such that α(t0 ) < x < β(t0 ) and such that
f 0 (x) = g(t0 ). Hence f 0 (x) = λ.
2
Corollary 3.1. If function f is differentiable on [a, b], then f 0 cannot have any
discontinuities of the first kind on [a, b].
6.4. L’Hospital theorem
103
6.4
L’Hospital theorem
Theorem 4.1. (L’Hospital4 ) Suppose f and g are real and differentiable in ]a, b[
and g 0 (x) 6= 0 for all x ∈ ]a, b[ , where −∞ ≤ a < b ≤ +∞. Suppose
(4.1)
f 0 (x)
→A
g 0 (x)
as x → a.
If
(4.2)
f (x) → 0 and g(x) → 0 as x → a,
or if
(4.3)
g(x) → +∞
as x → a,
then
(4.4)
f (x)
→A
g(x)
as x → a.
Proof. We first consider the case when −∞ ≤ A < +∞. Choose a real number q
such that A < q, and the choose r such that A < r < q. By (4.1) there is a point
c ∈ ]a, b[ such that a < x < c implies
(4.5)
f 0 (x)
< r.
g 0 (x)
If a < x < y < c, then Cauchy theorem 2.2 shows that there is a point t ∈ ]x, y[ such
that
(4.6)
f (x) − f (y)
f 0 (t)
= 0
< r.
g(x) − g(y)
g (t)
Suppose (4.2) holds. Letting x → a in (4.6) we see that
(4.7)
f (y)
≤r<q
g(y)
(a < y < c).
Next, suppose (4.3) holds. Keeping y fixed in (4.6) we can choose a point c1 ∈
]a, y[ such that g(x) > g(y) and g(x) > 0 if a < x < c1 . Multiplying (4.6) by
[g(x) − g(y)]/g(x), we get
(4.8)
4
f (x)
g(y) f (y)
<r−r
+
,
g(x)
g(x) g(x)
(a < x < c1 ).
Guillaume de L’Hospital, marquis de Sainte-Mesme, 1661-1704
104
6. Differential calculus
If we let x → a in (4.8), (4.3) shows that there is a point c2 ∈ ]a, c1 [ such that
f (x)
<q
g(x)
(4.9)
(a < x < c2 ).
Summing up, (4.7) and (4.9) show that for any q , subject only to the condition
(x)
< r if a < x < c2 .
A < q, there is a point c2 such that fg(x)
In the same manner, if −∞ < A ≤ +∞, and p is chosen so that p < A, we can
find a point c2 such that
(4.10)
p<
f (x)
g(x)
a < x < c2 ,
and (4.4) follows from these two statements.
6.5
2
Higher order derivatives
Theorem 5.1. (Leibniz formula) Let u and v be two functions having derivatives
up to the n -order on an interval. Then
1 (n−1) 0
n − 1 0 (n−1)
(n)
n
(uv) = u v +
u
v + ··· +
uv
+ uv (n) .
n
n
Proof. By induction.
2
Proposition 5.1. The following identities hold on R
π
π
(ex )(n) = ex , sin(n) (x) = sin x + n
, cos(n) (x) = cos x + n
.
2
2
6.6
Convex functions and differentiability
Theorem 6.1. Suppose f : I → R is convex, a = inf I, b = sup I, I being an
interval. Then
(a) f has side derivatives on ]a, b[ and for any t1 , t2 ∈ ]a, b[ , t1 < t2 , we have
fl0 (t1 ) ≤ fr0 (t1 ) ≤ fl0 (t2 ) ≤ fr0 (t2 );
(b) if a ∈ I (b ∈ I), then f is right-hand differentiable in a ( respectively it is
left-hand differentiable in b) and
fr0 (a) ≤ fl0 (t)
(respectively fr0 (t) ≤ fl0 (b)),
t ∈ int (I);
6.6. Convex functions and differentiability
105
(c) there is an at most countable set A ⊂ I such that f is differentiable on I \ A.
Proof. (a) Suppose t ∈ ]a, b[ . Since sf,t is increasing on I \ {t}, it follows that sf,t
has finite side limits at t, and hence f has side derivatives on t.
Suppose t1 , t2 ∈ ]a, b[ , t1 < t2 , and choose u, v, w satisfying a < u < t1 < v <
t2 < w < b. Then from
sf,t1 (u) ≤ sf,t1 (v) = sf,v (t1 ) ≤ sf,v (t2 ) = sf,t2 (v) ≤ sf,t2 (w)
it follows
fl0 (t1 ) = sf,t1 (t1 −) ≤ sf,t1 (t1 +) = fr0 (t1 ) ≤ sf,t2 (t2 −) = fl0 (t2 ) ≤ sf,t2 (t2 +) = fr0 (t2 ).
(b) If a ∈ I , we repeat the previous proof for a < t1 < u.
(c) From (a) it follows that fl0 (t) is increasing on ]a, b[. Hence there is an at most
countable set A such that fl0 (t) is continuous on ]a, [ \A.
Choose t0 ∈ ]a, b[ \A. Then fl0 (·) is continuous on t0 , and for t > t0
fl0 (t0 ) ≤ fr0 (t0 ) ≤ fl0 (t).
Letting t → t0 , we get fl0 (t0 ) = fr0 (t0 ), that is f is differentiable on t0 . Hence f is
differentiable on I \ A.
2
Corollary 6.1. If f : I → R is convex, f is continuous on the interior of I.
Remark. A convex function need not be convex on the extreme points of I. Indeed
(
0, t ∈ ]0, 1[
f : [0, 1] → R, f (t) =
4
1, t ∈ {0, 1}.
Corollary 6.2. Suppose f : I → R is convex, a = inf I, , b = sup I, and t0 ∈ ]a, b[ .
Then
f (t) ≥ f (t0 ) + m(t − t0 )
for every t ∈ I and m ∈ [fl0 (t0 ), fr0 (t0 )].
Proof. If t > t0 ,
sf,t0 (t) =
f (t) − f (t0 )
≥ inf sf,s (s) = fr0 (t0 ) ≥ m.
t − t0
s∈I
s>t0
Similarly, if t < t0 ,
sf,t0 (t) =
f (t) − f (t0 )
≤ sup sf,s (s) = fl0 (t0 ) ≤ m.
t − t0
s∈I
s<t0
Now the conclusion follows.
2
106
6. Differential calculus
Corollary 6.3. Suppose f : I → R is differentiable on I. Then f is convex if and
only if
f (t) ≥ f (t0 ) + f 0 (t0 )(t − t0 ),
(6.1)
for any t, t0 ∈ I.
Proof. The necessity part follows from the previous corollary.
Sufficiency. From (6.1) for any a, b ∈ I and α ∈ [0, 1] follow
f (a) ≥f ((1 − α)a + αb) + f 0 ((1 − α)a + αb)α(a − b),
f (b) ≥f ((1 − α)a + αb) − f 0 ((1 − α)a + αb)(1 − α)(a − b).
Multiply the first inequality by (1 − α), the second by α, and then sum them. It
results that
(1 − α)f (a) + αf (b) ≥ f ((1 − α)a + αb),
hence f is convex.
2
Corollary 6.4. (Fermat theorem for convex functions) Suppose f : I → R is convex
and differentiable on I. Then for any point t0 ∈ int (I) the following statements are
equivalent
(a) (t0 , f (t0 )) is the global minimum of f ;
(b) (t0 , f (t0 )) is a local minimum of f ;
(c) f 0 (t0 ) = 0.
Proof. (a) =⇒ (b) is obvious.
(b) =⇒ (c) follows without any convexity assumption, by theorem 2.1.
(c) =⇒ (a). This implication follows from the previous corollary.
2
We have seen that a convex function f : I → R is continuous on the interior of
I. It holds even a stronger statement on any compact subinterval of I.
Corollary 6.5. Let f : I → R be a convex function. Then f is Lipschitzean on any
compact [a, b] in I.
Proof. Consider an arbitrary subinterval [a, b] ⊂ I and u, v ∈ [a, b], u < v. Denote
M := max{|fr0 (a)|, |fl0 (b)|}. Then
−M ≤ fr0 (a) ≤ fr0 (u) ≤
f (u) − f (v)
≤ fl0 (v) ≤ fl0 (b) ≤ M.
u−v
Thus |f (u) − f (v)| ≤ M |u − v| for any u, v ∈ [a, b].
2
Theorem 6.2. A differentiable function f : I → R is convex (strictly convex) if and
only if its first derivative is increasing (respectively, strictly increasing).
6.6. Convex functions and differentiability
107
Proof. The necessity part follows from theorem 6.1.
Sufficiency. Suppose that there exists a differentiable function whose first order
derivative is increasing and it is not convex. Then there exist a, b, c ∈ I, a < b < c,
and sf,b (a) > sf,b (c), that is
f (a) − f (b)
f (c) − f (b)
>
.
a−b
c−b
By the Lagrange mean value theorem it follows that there are t1 ∈ ]a, b[ and t2 ∈ ]b, c[
with f 0 (t1 ) > f 0 (t2 ). But this contradicts that the first derivative is increasing.
2
Corollary 6.6. A differentiable function f : I → R is concave (strictly concave) if
and only if its derivative is decreasing (respectively, strictly decreasing).
Proof. Apply the previous theorem to −f.
2
Corollary 6.7. (Jensen) Let f : I → R be a function with second order derivative
on I. Then f is convex (concave) if and only if f 00 ≥ 0 (respectively, f 00 ≤ 0 ).
Proof. The claim follows from the remark that f 0 is increasing (decreasing) if and
only if f 00 is positive (negative).
2
Corollary 6.8. Let f : I → R be a function with second order derivative on I. Then
f is strictly convex (strictly concave) if and only if f 00 > 0 (respectively, f 00 < 0 ).
6.6.1
Inequalities
Proposition 6.1. P
(Young generalized inequality) Take n ∈ N∗ , yk > 0, pk > 0, for
n
k = 1, . . . , n, and
k=1 1/pk = 1. Then
(6.2)
n
Y
n
X
1 pk
yk ≤
yk .
p
k
k=1
k=1
For n = 2 the above inequality reduces to (3.3) at page 20.
Proof. Consider the function f (x) = exp(x), x ∈ R. Since f 00 (x) > 0, for every
x ∈ R, we infer that function f is convex. Based on Jensen inequality, (2.18) at the
page 89, taking αk = pk and xk = ln ykpk we may write
!
n
n
X
X
1
1
pk
exp
ln yk
≤
exp (ln ykpk ) .
p
p
k
k
k=1
k=1
But
n
X
1
exp
ln ykpk
p
k
k=1
!
n
X
= exp
!
ln yk
k=1
n
n
X
X
1
1 pk
pk
exp (ln yk ) =
y ,
pk
pk k
k=1
k=1
and the generalized Young inequality follows.
2
,
108
6. Differential calculus
Proposition 6.2. (Generalized mean inequality) Take n ∈ N∗ , xi > 0, and αi ≥ 0,
i = 1, . . . , n satisfying α1 + · · · + αn = 1. Then
xα1 1 xα2 2 . . . xαnn ≤ α1 x1 + . . . αn xn .
(6.3)
Proof. Consider the function f (x) = ln x, x > 0. Since f 00 (x) < 0, for every x > 0,
we infer that function f is (strictly) concave. So
!
n
n
X
X
ln
αi xi ≥
αi ln xi ,
i=1
and (6.3) follows.
i=1
2
Corollary 6.9. Taking α1 = · · · = αn = 1/n in (6.3) it follows
√
n
(6.4)
x1 . . . xn ≤
x1 + x2 + . . . xn
.
n
The genuine Cauchy’s proof of the above inequality is given at page 24 and it is
available in many books, let as mention only one, namely [21, Part 2, Chapter 2].
Corollary 6.10. Substituting xi by 1/xi in (6.3) we get
(6.5)
α1
αn
+ ··· +
x1
xn
−1
≤ xα1 1 xα2 2 . . . xαnn .
Corollary 6.11. Taking α1 = · · · = αn = 1/n in (6.5) it follows
(6.6)
1
x1
n
+ ··· +
1
xn
≤
√
n
x1 . . . xn .
Chapter 7
Integral calculus
The aim of the present chapter is to introduce some basic results on integral
calculus.
7.1
The Riemann integral
Let f be defined on [a, b] and real-valued.
7.2
The Gronwall inequality
A basic tool in many results connected to differential equations and inclusions is
the following one.
Lemma 2.1. Suppose a continuous function x : [a, b] → R satisfies
Z
(2.1)
t
0 ≤ x(t) ≤ c +
h(s)x(s)ds,
t ∈ [a, b]
a
for some constant c and some nonnegative integrable function h : [a, b] → R. Then
Z
(2.2)
0 ≤ x(t) ≤ c + c
t
Z
h(r) exp(
a
t
h(s)ds)dr,
t ∈ [a, b].
r
Proof. Denote
Z
t
w(t) := c +
h(s)x(s)ds.
a
Then
w0 (t) = h(t)x(t),
w(t) > 0,
109
w(a) = c.
110
7. Integral calculus
From (2.1) it follows that x(t) ≤ w(t), for any t ∈ [a, b]. Thus it holds the following
sequence of implications
w0 (t)
w (t) = h(t)x(t) ≤ h(t)w(t) =⇒
≤ h(t) =⇒
w(t)
Z t
Z t 0
Z t
w (s)
ds ≤
h(s)ds =⇒ ln w(t) ≤
h(s)ds + ln c =⇒
a w(s)
a
a
Z t
Z t
w(t) ≤ c · exp( h(s)ds =⇒ x(t) ≤ c · exp( h(s)ds.
0
a
Substituting in (2.1), we get (2.2).
a
2
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112
Author index
Stoltz, ., 48, 54
Abel, N. H., 64
Archimedes, 13, 14, 17
Weierstrass, K., 36, 93
Banach, S., 87
Bernoulli, , 24, 25
Borel, E., 36
Bourbaki, N., 101
Young, , 109
Young, W. H., 20, 21
Cantor, G., 1, 37, 38, 42, 81
Catalan, E. Ch., 1814-1894, 55
Cauchy, A., 80, 98
Cauchy, A. L., 25, 42, 43, 58, 60, 62,
64, 65, 68
Cesaro, ., 48, 54
D’Alembert, J. le R., 62
Darboux, , 82, 84
Denjoy, , 101
Dirichlet, P. G. L., 65
Euler, L., 54
Fermat, P., 98, 108
Hölder, O., 20, 22, 23
Hadamard, J., 64
Heine, , 36
Jensen, , 89, 90, 109
Lagrange, J. J., 99
Lagrange, J. L., 27, 54
Lalescu, T., 55
Leibnitz, G., 106
Lipschitz, , 86
Minkowski, H., 24
Morgan, de, , 4
Newton, I., 50
Riemann, B., 69
Sierpinski, W., 86
113
Subject index
addition, 10
application, 7
Cauchy, 80
family
pairwise disjoint, 3
fixed point, 86
formula
of Leibnitz, 106
function, 7
absolute value, 12
antiderivative, 110
bijective, 7
continuous at a point, 77
continuous on a set, 77
convex, 88
Jensen, 90
stricly, 88
Darboux, 84
derivative, 93
differentiable, 93
on a set, 93
discontinuous, 82
distance, 12
fractional part, 16
from, 8
injective, 8
integer part, 16
Lipschitz, 86
monotonic, 83
monotonically decreasing, 83
monotonically increasing, 83
one-to-one, 8
onto, 8
primitive, 110
slope, 90
ball, 30
closed, 37
open, 30
cardinal number, 17
components, 79
constant
of Euler, 54
contraction, 86
coordinate, 71
covering, 34
open, 34
diameter, 42
discontinuity
first kind, 82
second kind, 82
simple, 82
distance, 29
Euclidean, 29
on C , 30
element, 1
bound
greatest lower, 6
least upper, 6
lower, 6
upper, 6
identity, 10
infimum, 6
inverse, 10
null, 10
smallest, 6
supremum, 6
unity, 10
zero, 10
equation
functional
identity
Lagrange, 27
inequality
Bernoulli, 24, 25
generalized
mean, 110
115
ordered pair, 4
origin, 71
Jensen, 89
mean, 25
of Young, 20
triangle, 29
infimum, 6
inner product, 72
integral
Darboux
lower, 113
interval, 15
bounded, 15
length, 15
unbounded, 15
partial summation formula, 64
partition, 112
point, 29, 71
interior, 30
isolated, 30
radius of convergence, 64
relation, 4
composition, 5
domain, 5
equivalence, 5
image, 7
inverse, 5
inverse image, 7
ordering
partial, 5
total, 6
well, 6
product, 5
range, 5
single-valued, 7
limit point, 30
linear space, 71
mapping, 7
maximum
local, 98
mean
weighted, 21
member, 1
metric, 29
Euclidean, 29
Euclidean on R2 , 30
uniform on R2 , 30
minimum
local, 98
multiplication, 10
sequence, 8
bounded, 39
Cauchy, 42
convergent, 39
divergent, 39
fundamental, 42
in, 8
limit, 39
lower, 47
upper, 47
monotonic, 43
monotonically
decreasing, 43
increasing, 43
of successive approximations, 87
speed of convergence, 87
term, 8
series
absolutely convergent, 66
neighborhood, 30
norm, 73
Euclidean, 72
l1 − norm , 72
lp -norm, 72
Minkowski, 72
uniform, 72
null vector, 71
number
fractional part, 16
integer part, 16
operation, 7
116
separable, 34
triangle inequality, 29
topological, 30
subset, 2
proper, 2
sum
Darboux
lower, 113
supremum, 6
conditionally convergent, 67
convergent, 58
divergent, 58
power, 63
coefficient, 63
radius of convergence, 64
sum, 58
set, 1
at most countable, 17
bounded, 6, 30
above, 6
below, 6
Cantor, 38
closed, 30
compact, 34
connected, 82
countable, 17
dense, 30
denumerable, 17
empty, 1
finite, 17
infinite, 17
open, 30
operation
Cartesian product, 4
intersection, 2
symmetric difference, 4
union, 2
ordered
partially, 6
totally, 6
perfect, 30
uncountable, 17
void, 1
well-ordered, 6
sets
difference, 3
disjoint, 3
space
compact, 34
complete, 43
metric, 29
distance, 29
theorem
Stolz-Cesaro, 48
topology, 30
transformation, 7
uniform continuous mapping, 80
vector, 71
vector space, 71
weight
of a mean, 21
117