marian mure¸san mathematical analysis and applications i draftchapter 1 sets the aim of this...

Marian Muresan

Mathematical Analysis and Applications I

Draft

Forword

These lecture notes have been written having in mind a computer scientist (not atypist) in our changing world. It is difficult to imagine the computer science withoutmathematics. For those interested mainly in software we recall as an argument a capi-tal (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 computerscience.

On the one side we have to notice the immutability of the mathematical world inthe following sense: a correct mathematical result remains true forever. Mathematicsenlarges continuously. The speed of this enlargement increases.

On the other side the needs of computer scientists change continuously. The timedesigned for study remains, more or less. unchanged. Hence, a question arises. Whatparts of mathematics and how to teach them in an optimal way to the computerscientist students? Not an easy question, indeed!

We are pressed to take into account some parts from mathematics and to neglectmany other.

The mathematical analysis offers a solid ground to many achievements in appliedmathematics and discrete mathematics. In spite of the facts that these lecture notesconcern with a part of the mathematical analysis on the real axis, we have tried toinclude useful and relevant examples, exercises, and results enlightening the reader onthe 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 11.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 The concept of a set . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Operations on sets . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Relations and functions . . . . . . . . . . . . . . . . . . . . . . . 41.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Sets of numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.1 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.2 The real number system . . . . . . . . . . . . . . . . . . . . . . 91.2.3 The extended real number system . . . . . . . . . . . . . . . . . 18

1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3.1 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Basic notions in topology 272.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.2 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 Numerical sequences and series 373.1 Numerical sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.1.1 Convergent sequences . . . . . . . . . . . . . . . . . . . . . . . . 373.1.2 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.1.3 Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 403.1.4 Monotonic sequences . . . . . . . . . . . . . . . . . . . . . . . . 413.1.5 Upper and lower limits . . . . . . . . . . . . . . . . . . . . . . . 453.1.6 Stoltz-Cesaro theorem and some of its consequences . . . . . . . 463.1.7 Some special sequences . . . . . . . . . . . . . . . . . . . . . . . 48

3.2 Numerical series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2.1 Series of nonnegative terms . . . . . . . . . . . . . . . . . . . . 573.2.2 The root and ration tests . . . . . . . . . . . . . . . . . . . . . . 603.2.3 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.2.4 Partial summation . . . . . . . . . . . . . . . . . . . . . . . . . 623.2.5 Absolutely and conditionally convergent series . . . . . . . . . . 65

v

3.2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4 Euclidean spaces 714.1 Euclidean spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5 Limits and Continuity 755.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.1.1 The limit of a function . . . . . . . . . . . . . . . . . . . . . . . 755.1.2 Right-hand side and left-hand side limits . . . . . . . . . . . . . 77

5.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2.1 Continuity and compactness . . . . . . . . . . . . . . . . . . . . 795.2.2 Uniform continuous mappings . . . . . . . . . . . . . . . . . . . 805.2.3 Continuity and connectedness . . . . . . . . . . . . . . . . . . . 825.2.4 Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.2.5 Monotonic functions . . . . . . . . . . . . . . . . . . . . . . . . 835.2.6 Darboux functions . . . . . . . . . . . . . . . . . . . . . . . . . 845.2.7 Lipschitz functions . . . . . . . . . . . . . . . . . . . . . . . . . 865.2.8 Convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . 885.2.9 Jensen convex functions . . . . . . . . . . . . . . . . . . . . . . 90

6 Differential calculus 916.1 The derivative of a real function . . . . . . . . . . . . . . . . . . . . . . 916.2 Mean value theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

6.2.1 Consequences of the mean value theorems . . . . . . . . . . . . 986.3 The continuity of derivatives . . . . . . . . . . . . . . . . . . . . . . . . 1026.4 L’Hospital theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.5 Higher order derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.6 Convex functions and differentiability . . . . . . . . . . . . . . . . . . . 104

6.6.1 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

7 Integral calculus 1097.1 The Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1097.2 The Gronwall inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 109References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

Author index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Subject index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

vi

Chapter 1

Sets

The aim of this chapter is to introduce several basic notions and results concerningsets.

1.1 Sets

1.1.1 The concept of a set

The basic notion of set theory which was first introduced by Cantor1 will occur con-stantly in our results. Hence it would be fruitful to discuss briefly some of the notionsconnected 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 setshas 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 appearthroughout to intuition and elementary logic. The so-called ”naive” theory of sets iscompletely satisfactory for us ([8]).

We will usually adhere to the following notational conventions. Elements of setswill be denoted by small letters: a, b, c, . . . , x, y, z, α, β, γ, . . . . Sets will bedenoted by capital Roman letters: A, B, C, . . . X, Y, . . . . Families of sets will bedenoted 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 meansthat 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,

1Georg 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 Ais called a subset of B and we write A ⊂ B or B ⊃ A. If A ⊂ B and B ⊂ A, wewrite A = B. A 6= B denies A = B. If A ⊂ B and A 6= B, we say that A is aproper subset of B and we write A ( B. We remark that under this idea of equalityof 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 callA ∪ B the union of A and B, see figure 1.1. Let A be a family of sets; then wedefine

∪A = {x | x ∈ A for some A ∈ A}.Similarly, if {Aα}α∈I is a family of sets indexed by α, we write

∪α∈IAα = {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 wecall A ∩ B the intersection of A and B, see figure 1.2. If A is any family of setswe define

∩A = {x | x ∈ A for all A ∈ A}.2An axiomatic introduction of natural numbers may be found in many textbooks, e.g., [22, Chap-

ter 1].3An 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∈IAα = {x | x ∈ Aα for all α ∈ I}.

Theorem 1.1. Let A, B, C be any sets. Then we have

(i) A ∪B = B ∪ A (i’) A ∩B = B ∩ A commutative law;

(ii) A ∪ A = A (ii’) A ∩ A = A idempotent law;

(iii) A ∪ ∅ = A (iii’) A ∩ ∅ = ∅;(iv) A ∪ (B ∪ C) = (A ∪B) ∪ C (iv’) A ∩ (B ∩ C) = (A ∩B) ∩ C associative law;

(v) A ⊂ A ∪B (v’) A ∩B ⊂ A;

(vi) A ⊂ B ⇐⇒ A ∪B = B (vi’) A ⊂ B ⇐⇒ A ∩B = A.

Theorem 1.2. Let A, B, C be any sets. Then we have

(i) A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C); distributive law;

(ii) A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C); distributive law.

If A ∩ B = ∅, the sets A and B are said to be disjoint, see figure 1.3. If Ais a family of sets such that each pair of distinct members of A are disjoint, A issaid to be pairwise disjoint. Thus an indexed family {Aα}α∈I is pairwise disjoint ifAα ∩ 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 {XA or {A. Other notation: X \ A.

Theorem 1.3. (de Morgan4 laws)

(a) {(A ∪B) = ({A) ∩ ({B);

(b) {(A ∩B) = ({A) ∪ ({B);

(c) {(∪α∈IAα) = ∩α∈I{Aα;

(d) {(∩α∈IAα) = ∪α∈I{Aα.

4de 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 consistingof those elements which are in exactly one of A and B, and that it may also bedefined 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 weconsider a pair (x1, x2) of elements in which we distinguish x1 as the first elementand x2 as the second element, then (x1, x2) is called an ordered pair. Thus, twoordered 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 ofall 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}. 4A (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 Qis 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 suchthat 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, ”ais congruent to b modulo n” if a − b = kn for a k ∈ Z. Here ”congruencemodulo n” is an equivalence relation on Z. 4

Let P be a set. A partial ordering on P is any relation ≤ ⊂ P × P such thatfor 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 eachx ∈ 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-orderedset.

Example 1.2. The set N of natural numbers with the usual ordering ≤ is awell-ordered set. 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 necessarilydistinct), we define max{x1, . . . , xn} := max{xn, max{x1, . . . , xn−1}}. Similarly, wedefine 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 tobe

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 isbounded below ;

(ii) an upper bound of A if y ≤ x for any y ∈ A; in this case we say that A isbounded 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 hasat most one infimum (denoted by inf A ) and at most one supremum (denoted bysup A ). 4Example. Let E consists of all numbers 1/n, where n = 1, 2, . . . . Then E isbounded, 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 setA 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 fis 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∈IAi) = ∪i∈If(Ai);

(b) f−1(∪j∈JBj) = ∪j∈Jf−1(Bj);

1.1. Sets of numbers 7

(c) f(∩i∈IAi) ⊂ ∪i∈If(Ai).

The following statements are true if f is a function, but may fail for arbitrary rela-tions

(c) f−1(∩j∈JBj) = ∪j∈Jf−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 tobe 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 existsan 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 ofX have distinct images in Y, that is, if no two two different elements in X have thesame image. A suitable way in deciding whether or not a given map is injective is iffor x, t ∈ X, f(x) = f(t) implies x = t. Figure 1.6 exhibits a surjective but not anone-to-one function, figure 1.7 exhibits an injective but not an onto function, whilefigure 1.8 presents a bijective function.

Theorem 1.5. Let X and Y be nonempty sets and f : X → Y be a function. Thenf is bijective if and only if it is injective and onto.

A sequence is a function having N∗ as its domain. Sometimes we will considersequences 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 thesequence. 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, conti-nuity, differentiation and integration) must be based on an accurately defined numberconcept.

We shall not, however, enter into any discussions of the axioms governing thearithmetic of the integers, but we take the rational number system as our startingpoint.

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 trianglehaving the length of a cathetus 1. Can we express the length of the hypotenuse as arational 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 canwrite p = m/n, where m and n are integers with n 6= 0, and we can further choosem 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 by4, 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 conclu-sion 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 righttriangle 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 nolargest 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 < 1and such that

h <2− p2

2p + 1.

Put q = p + h. Then q > p, and

q2 = p2 + (2p + h)h < p2 + (2p + 1)h < p2 + (2− p2) = 2,


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− p2 − 2

2p=

p

2+

1

p.

Then 0 < q < p and

q2 = p2 − (p2 − 2) +

(p2 − 2

2p

)2

> p2 − (p2 − 2) = 2,

so that q ∈ B. 4The 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 isanother (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 orderingsatisfying 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, suchthat 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 theinverse 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 propertythat x ≤ y for every x ∈ A and y ∈ B there exists an element z ∈ X suchthat

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 totallyordered field. (R 14) and (R 15) express the compatibility of the ordering relation withthe algebraic operations. (R 16) has a special role which will be clear a little bit later.

x−1 ∈ X \ {0} from (R 7) is denoted as1

x, too. Hence

y

x= yx−1. 4

For a while let us ignore assumption (R 16).

Proposition 2.1. From (R 1)-(R 15) it follows that

(1) (a) x1 ≤ x2 and y1 ≤ y2 imply x1 + y1 ≤ x2 + y2;

(b) x1 < x2 and y1 ≤ y2 imply x1 + y1 < x2 + y2;

(2) (a) x > 0 if and only if x−1 > 0;

(b) x ≥ 0 implies − x ≤ 0;

(c) x > 0 implies − x < 0;

(3) (a) x ≤ y and z > 0 imply xz ≤ yz;

(b) x < y and z > 0 imply xz < yz;

(c) x ≤ y and z < 0 imply xz ≥ yz;

(d) x < y and z < 0 imply xz > yz;

(4) if xy > 0, then x ≤ y if and only if 1/x ≥ 1/y;

(5) (a) 0 ≤ x1 ≤ x2 and 0 ≤ y1 ≤ y2 imply x1y1 ≤ x2y2;

(b) 0 < x1 < x2 and 0 < y1 ≤ y2 imply x1y1 < x2y2;

(c) x1 ≤ x2 ≤ 0 and y1 ≤ y2 ≤ 0 imply x1y1 ≥ x2y2;

(d) x1 < x2 ≤ 0 and y1 ≤ y2 < 0 imply x1y1 > x2y2.


The absolute value function is defined by

for x ∈ X, |x| =

{x, x ≥ 0,

−x, x < 0.

Proposition 2.2. From (R 1)-(R 15), it follows that

(1) (a) |x| ≥ 0;

(b) |x| = 0 ⇐⇒ x = 0;

(c) |x| = | − x|;(2) (a) |x + y| ≤ |x|+ |y|;

(b) |x− y| ≥ | |x| − |y| |;(3) (a) |x| ≤ a ⇐⇒ −a ≤ x ≤ a;

(b) |x| < a ⇐⇒ −a < x < a;

(4) (a) |xy| = |x| · |y|;

(b)

∣∣∣∣xy∣∣∣∣ =

|x||y|

;

(c) |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 arealgebraically and order isomorphic, [9, theorem 5.34].

We choose one of them and called it the set of real numbers and denoted it byR = (R, +, ·,≤) . Any element x ∈ R is called a real number. Any real number xsuch that 0 ≤ x is called non-negative, while 0 < x is called positive number. Anyreal number x such 0 ≥ x is called non-positive, while 0 > x is called negativenumber.

Proposition 2.4. Number 1 is positive.

Proof. Suppose that 1 is non-positive, i.e., 1 ≤ 0. Adding −1 to the both sideswe 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-negativeand 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 infimumis 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 everyx ∈ A0 and y ∈ A it holds x ≤ y. From (R 16) it follows there exists a real numberz 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 anonempty 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 anonempty subset of A, then

sup A ≥ sup B.

Remark. The proofs of the existence of an infimum and the existence of a supremumhave used axiom (R 16). At the same time it can be proved that (R 16) follows fromany 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 thesame 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 propertythat x ≤ y for any x ∈ A and y ∈ B. Then A is nonempty and bounded above (byany 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 suchthat 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 nonemptyand 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 realnumbers x and y such that y > 0 there exists a natural number n such that x < ny.

5Archimedes, 287-212


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 existsa natural n such that u < ny. Since v /∈ A and that the real number set is a totallyordered 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 numberz 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 nsuch 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 showsthat 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 eithera1 < 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 upperbound less then a. 2

Theorem 2.8. ([23, theorem 1.37, p. 11]) For every real x > 0 and every integern > 0, there is one and only one real y > 0 such that yn = 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 yn

1 < yn2 .

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 yn < x. Choose h such that 0 < h < 1 and such that

h <x− yn

(1 + y)n − yn.

We have

(y + h)n = yn +

(n

1

)yn−1h + · · ·+ hn ≤ yn + h

[(n

1

)yn−1 + · · ·+ 1

]= yn + h[(1 + y)n − yn] < yn + (x− yn) = x.

Thus y + h ∈ E, contradicting the fact that y is an upper bound of E.Suppose yn > x. Choose k such that 0 < k < 1 such that k < y and such that

k <yn − x

(1 + y)n − yn.

Then, for t ≥ y − k, we have

tn ≥ (y − k)n = yn −(

n

1

)yn−1k +

(n

2

)yn−2k2 − · · ·+ (−1)nkn

= yn − k

[(n

1

)yn−1 −

(n

2

)yn−2k + · · ·+ (−1)n−1kn−1

]≥ yn − k

[(n

1

)yn−1 +

(n

2

)yn−2 + · · ·+ 1

]= yn − k[(1 + y)n − yn] > yn + (x− yn) = x.

Thus y−k is an upper bound of E, contradicting the fact that y = sup E. It followsthat yn = x. 2

An interval A of the real number system is a subset of R so that for every x, y ∈ Aand z ∈ R satisfying x ≤ z ≤ y, we have z ∈ A. An interval bounded below andabove is said to be bounded. Otherwise it is called unbounded. For any nonempty andbounded interval A, the non-negative real number l(A) = sup A − inf A is said tobe the length of A.

We remark that for any real numbers a and b with a ≤ b the next sets areintervals

[a, b] = {x ∈ R | a ≤ x ≤ b} closed interval;

[a, b[ = {x ∈ R | a ≤ x < b} left closed right open interval;

]a, b] = {x ∈ R | a < x ≤ b} left open right closed interval;

]a, b[= {x ∈ R | a < x < b} open interval.


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 ∈ Aand bm ∈ B such that

bm < ak.

We have either m < k or k < m. Suppose m < k. Then

bm < ak ≤ bk,

thus contradicting (2.6).Axiom (R 16) supplies a real z such that

ak ≤ z ≤ bk, 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) integerk 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 existsa 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 thereexists a natural n such that 1 < n(y − x). Then

(2.7) 1/n < y − x.

From theorem 2.10 it follows that there exists an integer m such that

(2.8) m ≤ nx < m + 1.

Obviously, u = (m + 1)/n is a rational, and satisfies x < u. From the left-hand sideof (2.8) as well as from (2.7) we infer that u also satisfies

u =m

n+

1

n≤ x +

1

n< y. 2

Corollary 2.3. Given any two real numbers x and y such that x < y, there existsan 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 itfollows 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, wesay 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 nonemptyand 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 onlyif A and B contain the same number of elements. For infinite sets, however, this is notexactly so. Indeed, let M be the set of all even positive integers, M = {2, 4, 6, . . . }.


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 ∈ Mis a bijection.

(b) The sets {1,−1, 2,−2, 3,−3, . . . } and {0, 1,−1, 2,−2, 3,−3, . . . } are equiva-lent. Indeed, the function

bk =

{0, k = 1,

ak−1, k > 1

maps the k rank term ak of the first set to the k + 1 rank term in the second set ina bijective way. We conclude that the sets N∗ and Z are equivalent. Therefore wecan 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

(2.9) B = ∪∞n=1An.

Then B is countable.

Proof. Let every set An be arranged in a sequence (xn k)n, k = 1, 2, . . . , andconsider 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 canbe 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 thanonce in (2.10). Hence there is a subset C of the set of all positive integers such thatC ∼ 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 mostcountable. Put

C = ∪α∈ABα.

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 orderedn -tuples (a1, a2, . . . , an), where ak ∈ A (k = 1, . . . , n), i.e.,

Bn = A× A× · · · × 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 theform a/b, where a and b are integers and b 6= 0. The set of such pairs (a, b), andtherefore 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 0and 1. Then A is uncountable.

Proof. Let B be a countable subset of A, and let B consists of the sequencess1, s2, . . . . We construct a sequence s as follows. If the nth digit in sn is 1 we letthe nth digit of s be 0, and vice versa. Then the sequence s differs from everymember of B in at least one place; hence s /∈ B. But clearly s ∈ A, so that B is aproper subset of A.

We have shown that every countable subset of A is a proper subset of A. Itfollows 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 theabove-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 ∈ Asuch 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 | a ≤ x ≤ +∞}; ]a, +∞] = {x ∈ R | a < x ≤ +∞};

]−∞, a] = {x ∈ R | x ≤ a}; ]−∞, a[ = {x ∈ R | x < a};[−∞, a] = {x ∈ R | −∞ ≤ x ≤ a}; [−∞, a[ = {x ∈ R | −∞ ≤ x < a};

[−∞,∞] = R; ]−∞,∞[ = R.

1.3 Exercises

1.3.1 Inequalities

The proofs of Holder6 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 contin-uous 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

(3.1) F (x) =

∫ x

0

f(u)du and G(x) =

∫ x

0

g(v)dv.

6Otto Holder,7W. H. Young,

20 1. Sets

Then a, b ∈ [0,∞) imply

(3.2) ab ≤ F (a) + G(b),

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 arenon-negative reals. Then

(3.3) αβ ≤ αp

p+

βq

q, if

1

p+

1

q= 1.

The equality holds if and only if αp = βq.

Proof. The first approach runs as follows. For u ∈ [0,∞), define f(u) = up−1 andapply theorem 3.1.

The second approach runs as follows. Consider the function f : ]0,∞[→ R givenby

f(x) =xp

p+

x−q

q.

It has an absolute minimum at x = 1. The required inequality follows from f(1) ≤f(α

1q β−

1p ). 2

Let a = (α1, . . . , αn) ∈ Rn or Cn. If r 6= 0, define the weighted mean with weightr of the finite sequence a as

(3.4) Mr(a) =

(n∑

k=1

|αk|r)1/r

=(∑

|αk|r)1/r

.

Proposition 3.1. Suppose p > 1, 1/p + 1/q = 1, and there are given two finitesequences a = (α1, . . . , αn) and b = (β1, . . . , βn) satisfying Mp(a) = Mq(b) = 1.Then

(3.5) M(ab) = M1(ab) ≤ 1,

where ab = (α1β1, . . . , αnβn).

Proof. Choose k ∈ {1, . . . , n}. We apply Young inequality (3.3) to |αk| and |βk|. Itfollows

(3.6) |αkβk| ≤|αk|p

p+|βk|q

q.

Summing up (3.6) for k = 1, 2, . . . , n, we get∑|αkβk| ≤

1

p

∑|αk|p +

1

q

∑|βk|q =

1

p+

1

q= 1. 2

1.3. Exercises 21

Theorem 3.2. (Holder 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) > 0and Mq(b) > 0. Then

(3.7) M(ab) =∑

|αkβk| ≤ Mp(a)Mq(b).

Proof. Define

αk =αk

Mp(a), βk =

βk

Mq(b),

a = (α1, . . . , αn), b = (β1, . . . , βn).

We remark that Mp(a) = Mq(b) = 1, and therefore we can apply proposition 3.1.Thus we find that M(ab) ≤ 1, i.e., (3.7). 2

Theorem 3.3. Inequality (3.7) turns into an equality if and only if the fraction|αk|p

|βk|q

does not depend upon k. (the fraction0

0is excluded)

Theorem 3.4. (Holder inequality for positive weights, [21, Part 2, Chapter 2, prob-lem 81.3]) Consider m ∈ N∗, m ≥ 2, aj = (αj1, . . . , αjn), j = 1, . . . ,m, and

p1, p2, . . . , pm > 0 so that∑ 1

pj

= 1. Suppose that Mpj(aj) > 0, j = 1, . . . ,m.

Then

(3.8)n∑

k=1

∣∣∣∣∣m∏

j=1

αjk

∣∣∣∣∣ ≤m∏

j=1

Mpj(aj).

Proof. If m = 2, theorem 3.4 reduces to theorem 3.2. Suppose that m ≥ 3 and thanwe prove (3.8) by induction. Consider that (3.8) is true for m − 1 and we prove itfor m. Using theorem 3.2 we have

n∑k=1

∣∣∣∣∣m∏

j=1

αjk

∣∣∣∣∣ =n∑

k=1

|α1k|

∣∣∣∣∣m∏

j=2

αjk

∣∣∣∣∣ ≤ Mp1(a1)

n∑k=1

(m∏

j=2

|αjk|

) p1p1−1

p1−1

p1

= Mp1(a1)

[n∑

k=1

m∏j=2

|αjk|p1

p1−1

] p1−1p1

.(3.9)

We remark that

pj(p1 − 1)

p1

> 0, j = 2, . . . ,m, andm∑

j=2

p1

pj(p1 − 1)=

p1

p1 − 1

m∑j=2

1

pj

= 1.

22 1. Sets

Thus (3.9) is further evaluated as

≤Mp1(a1)

m∏

j=2

[n∑

k=1

|αjk|p1

p1−1

pj(p1−1)

p1

] p1pj(p1−1)

p1−1

p1

=Mp1(a1)m∏

j=2

(n∑

k=1

|αjk|pj

) 1pj

=m∏

j=1

Mpj(aj). 2(3.10)

Theorem 3.5. (Holder 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

(3.11) M(ab) ≥ Mp(a)Mq(b).

Proof. Take u = 1/p (> 1), 1/u + 1/v = 1. We define γk = β− 1

uk , δk = β

1uk α

1uk ,

k = 1, . . . , n. So γuk = βq

k. Using theorem 3.2, we get

n∑k=1

αpk =

n∑k=1

γkδk ≤(∑

δuk

) 1u(∑

γvk

) 1v

=(∑

αkβk

)p (∑β− v

uk

) 1v

=(∑

αkβk

)p (∑βq

k

) 1v.

HenceMp(a)Mq(b) ≤ M(ab) 2

Theorem 3.6. (Holder inequality for negative weights, [4]) Consider m ∈ N∗, m ≥2, finite and nonzero sequences aj = (αj1, . . . , αjn), j = 1, . . . ,m, and the weightsp1, p2, . . . , pm−1 < 0 and pm ∈ ]0, 1[ satisfying

∑mj=1 1/pj = 1. Then

(3.12)n∑

k=1

∣∣∣∣∣m∏

j=1

αjk

∣∣∣∣∣ ≥m∏

j=1

Mpj(aj).

Proof. If m = 2, this theorem reduces to theorem 3.5. Suppose that (3.12) holds foran m ≥ 2. We show that it also holds for m+1. Therefore consider p1, p2, . . . , pm < 0and pm+1 ∈ R such that

∑m+1j=1 1/pj = 1, and aj = (αj1, . . . , αjn) are nonzero for all

j = 1, . . . ,m + 1. Thus 0 < pm+1 < 1. Hence

n∑k=1

∣∣∣∣∣m+1∏j=1

αjk

∣∣∣∣∣ =n∑

k=1

|α1k|

∣∣∣∣∣m+1∏j=2

αjk

∣∣∣∣∣ ≥ Mp1(a1)

n∑k=1

(m+1∏j=2

|αjk|

) p1p1−1

p1−1

p1

= Mp1(a1)

[n∑

k=1

m+1∏j=2

|αjk|p1

p1−1

] p1−1p1

.(3.13)

1.3. Exercises 23

We remark that

pj(p1 − 1)

p1

< 0, j = 2, . . . ,m,pm+1(p1 − 1)

p1

> 0, and

m+1∑j=2

p1

pj(p1 − 1)=

p1

p1 − 1

m+1∑j=2

1

pj

= 1.

Then (3.13) is further evaluated as

≥Mp1(a1)

m+1∏j=2

[n∑

k=1

|αjk|p1

p1−1

pj(p1−1)

p1

] p1pj(p1−1)

p1−1

p1

=Mp1(a1)m∏

j=2

(n∑

k=1

|αjk|pj

) 1pj

=m∏

j=1

Mpj(aj). 2

Theorem 3.7. (Minkowski8 inequality for p ≥ 1. ) Suppose there are satisfied theassumptions of theorem 3.2. Then

(3.14) Mp(a + b) ≤ Mp(a) + Mp(b).

Proof. For p = 1 the above inequality follows from the inequality given in (2) (a) ofthe proposition 2.2.

Suppose that p > 1. We apply the Holder inequality (3.7) for the following twopairs of finite sequences

(ak)nk=1, (|ak + bk|p−1)n

k=1 and (bk)nk=1, (|ak + bk|p−1)n

k=1.

Then we have the following estimations

∑|ak + bk|p ≤

∑|ak| · |ak + bk|p−1 +

n∑k=1

|bk| · |ak + bk|p−1

≤ (Mp(a) + Mp(b))(∑

|ak + bk|p)1/q

.

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 itholds

(3.15) (1 + x)n ≥ 1 + nx.

8Hermann Minkowski, 1864-19099Bernoulli,

24 1. Sets

Proof. For x = −1 the left-hand side of (3.15) is null while the right-hand is non-positive. So we may suppose that x > −1.

For x > −1 we prove (3.15) by induction. If n = 1, we have an equality. Ifn = 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∗ andfor every x > −1, that is

(3.16) (1 + x)k ≥ 1 + kx.

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 thegeometric mean is less or equal to the arithmetic mean, i.e.

(3.18) m√

x1x2 . . . xm ≤ x1 + x2 + · · ·+ xm

m.

Corollary 3.2. Let x1, x2, . . . xm be positive reals. Then the harmonic mean isless or equal to the geometric mean, i.e.

(3.19)m

1x1

+ 1x2

+ · · ·+ 1xm

≤ m√

x1x2 . . . xm.

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

4√

x1x2x3x4 =√√

x1x2

√x3x4 ≤

√x1x2 +

√x3x4

2

≤x1+x2

2+ x3+x4

2

2=

x1 + x2 + x3 + x4

4.

10Augutin Louis Cauchy, 1789-1857

1.3. Exercises 25

Now, suppose inequality (3.18) holds for m = 2k, i.e.,

(3.20) 2k√x1x2 . . . x2k ≤ x1 + x2 + · · ·+ x2k

2k.

Then

2k+1√x1x2 . . . x2k+1 =

√2k√

x1 . . . x2k2k√

x2k+1 . . . x2k+1

≤2k√

x1 . . . x2k + 2k√x2k+1 . . . x2k+1

2.

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

m(> 0).

Take

xk+1 = · · · = x2k =: l

and consider inequality (3.20). Then we write

2k√x1x2 . . . xm · l

2k−m

2k ≤ ml + (2k −m)l

2k= l

2k√x1x2 . . . xm ≤ l

m

2k ⇐⇒ x1 . . . xm ≤ lm. 2

Second approach. This approach consists in considering a special case of (3.19) whichis, 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

x1x2 . . . xnxn+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 > 1and 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 + xnxn+1 ≥ n.

Then we continue the estimation

x1 + x2 + · · ·+ xn+1 ≥ n− xnxn+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 themean inequality, too. 2

Proposition 3.5. (Lagrange11 identity) Let ai and bi be real numbers, i = 1, . . . ,m.Then

(3.23)

(m∑

i=1

a2i

)(m∑

i=1

b2i

)=

(m∑

i=1

aibi

)2

+∑

1≤i<j≤m

(aibj − ajbi)2.

Proof. It follows at once by induction. 2

Corollary 3.3. Let ai and bi be real numbers, i = 1, . . . ,m. Then

(3.24)

(m∑

i=1

a2i

)(m∑

i=1

b2i

)≥

(m∑

i=1

aibi

)2

.

Corollary 3.4. Let ai and bi be real numbers, i = 1, . . . ,m. Then(m∑

i=1

a2i

)(m∑

i=1

b2i

)=

(m∑

i=1

aibi

)2

if and only if aibj − ajbi = 0 for all 1 ≤ i < j ≤ m.

Corollary 3.5. Let ai be real numbers, i = 1, . . . ,m. Then

a1 + a2 + · · ·+ am

m≤√

a21 + a2

2 + · · ·+ a2m

m.

Proof. Take bi = 1, i = 1, . . . ,m, in inequality (3.24). 2

11Joseph Louis Lagrange, 1736-1813

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 withany two points x and y of X there is associated a real number ρ(x, y), called thedistance 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 functionis symmetrical. (c) is called the triangle inequality.Examples. (a) Consider a nonempty set X and define on it the following metric

(1.1) ρ(x, y) =

{0, 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 calledthe Euclidean metric or the Euclidean distance on R.

(c) Consider the complex plane C and define on it the following distance function

ρ(z1, z2) = |z1 − z2| =√

(x1 − x2)2 + (y1 − y2)2,

zk = xk + iyk, xk, yk ∈ R, k = 1, 2, i2 = −1,(1.2)

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|;

( 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 uniformmetric.

(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 understoodto 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 pointy 6= x such that y ∈ A. If x ∈ A and x is not a limit point of A, x is said to bean 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 atopological space. Sometimes we say that X is a topological space when the topologyis irrelevant.Remarks. (a) All these topological notions (openess, closeness, etc.) are basedultimately 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 Rgiven by (1.1) coincides with the family of all subsets of R. Obviously, this is not thecase if we consider the Euclidean metric ρ2 on R. For, the set containing precisely apoint 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 basicnotion, [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 positivenumber 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 containsinfinitely many points of A.

Proof. Suppose there is a neighborhood V of x which contains only a finite numberof points of A. Let y1, . . . , yn be those points of V ∩ A, which are distinct from x,and put

r = min1≤k≤n

ρ(x, yk).

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 nopoint in A is a limit point of A; we stress the difference between a limit pointand containing one;

(f) the set of all complex numbers;

(g) the interval ]0, 1[ .


Let us note that (d), (e) and (g) can be regarded also as subsets of R.Some properties of these sets are tabulated below

closed open perfect bounded

(a) no yes no yes

(b) yes no yes yes

(c) yes no no yes

(d) yes no no no

(e) no no no no

(f) yes yes yes no

(g) no no yes 4

Theorem 1.3. A set A is open if and only if {XA is closed.

Proof. First, suppose that {XA is closed. Choose x ∈ A. Then x /∈ {XA and xis 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 neighborhoodof x contains a point of {A, so that x is not an interior point of A. Since A isopen, 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, ∪αGα is open.(b) For any family {Fα} of closed sets, ∩αFα is closed.(c) For any finite family G1, . . . , Gn of open sets, ∩n

i=1Gi is open.(d) For any finite family F1, . . . , Fn of closed sets, ∪n

i=1Fi is closed.

Proof. (a) Put G = ∪αGα. If x ∈ G, then x ∈ Gα for some α. Since x is an interiorpoint of Gα, x is also an interior point of G, and G is open.

(b) Fα is closed ⇐⇒ {Fα open =⇒ ∪α{Fα open ⇐⇒ {(∪αFα) open⇐⇒ ∪αFα closed.

(c) Put H = ∩n1Gi. 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 xof 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 topologygenerated by the Euclidean metric.Example. In parts (c) and (d) of the preceding theorem, the finiteness of the familiesis essential. For, let Gn be the interval ]− 1/n, 1/n[ , n ∈ N∗. Then Gn is an opensubset 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. Lety = 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, sothat 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 opensubset of X means that to each point x ∈ A there is associated a positive numberr such that ρ(x, y) < r, y ∈ X imply that y ∈ A. First remark that (Y, ρ|Y×Y ) isa metric space, too. We say that A is open relative to Y if to each x ∈ A there isassociated 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 withoutbeing 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 onlyif 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 numberrx such that the conditions ρ(x, y) < rx and y ∈ Y imply y ∈ A. Let Vx be the setof all y ∈ X such that ρ(x, y) < rx, and define

G = ∪x∈AVx.

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 ofthe theorem is proved.

Conversely, if G is open in X and A = G∩Y, for every x ∈ A has a neighborhoodVx ⊂ 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 ′ be the set of limit points ofa set E. Then E ′ is closed.

Proof. By definition E ′ is closed if and only if (E ′)′ ⊂ E ′.Take y ∈ (E ′)′. Then for every ε > 0 there is an x ∈ E ′ such that 0 < ρ(x, y) <

ε/2. Since x ∈ E ′ there is a v ∈ E such that

ρ(x, v) < ρ(x, y).

Hence v 6= y and0 < 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 ′ and the proof iscomplete. 2

Let E be a set, let E ′ be the set of limit points of E, and define E = E ∪ E ′.


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 closedF ;

(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) subsetof R. 2

2.2 Compact spaces

A covering of a set A in a topological space X is a family of subsets {Gα} of Xsuch that A ⊂ ∪αGα.

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} isan open covering of K, then there are finitely many indices α1, . . . , αn such that

K ⊂ Gα1 ∪ · · · ∪Gαn .

Example. Every finite set is compact. 4Remark. We observed earlier that if A ⊂ Y ⊂ X, then A may be open relative toY without being open relative to X. The property of being open thus depends onthe space in which A is embedded. The same is true of the property of being closed.4Theorem 2.1. Suppose K ⊂ Y ⊂ X. Then K is compact relative to X if and onlyif 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 isa 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 {XK isopen.

If K = X, we are done. Suppose that {XK 6= ∅. Choose x ∈ X, x /∈ K. Ify ∈ K, let Vy and Wy be neighborhoods of x, respectively, y of radius less thanρ(x, y)/2 (> 0, since x 6= y ). Since K is compact, there are finitely many pointsy1, . . . , 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 ⊂ {XK, so that x is an interior point of {XK. So, {XK is open and Kis 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 ⊂ ∪αVα ∪ (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 coverF, it remains a finite number of Vα that covers F. We have thus shown that a finitemembers 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, bytheorem 2.3 F ∩K is compact. 2

Theorem 2.4. If {Kα} is a family of compact subsets of a metric space X such thatthe 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 iscompact, there are finitely many indices α1, . . . , αn such that

K1 ⊂ Gα1 ∪ · · · ∪Gαn .

This means thatK1 ∩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 limitpoint in K.

Proof. If no point of K were a limit point of A, then each x ∈ K would have aneighborhood Vx which contains at most one point of A (namely x, if x ∈ A ). It isclear that no finite subfamily of {Vx} can cover E; moreover K since A ⊂ K. Thiscontradicts 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 intervalsQ1 = [a, c] and Q2 = [c, b] then determine 2 subintervals whose union is I. Atleast 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


(i) I ⊃ I1 ⊃ I2 ⊃ . . . ;

(ii) In is not covered by any finite subfamily of {Gα};

(iii) if x, y ∈ In, then |x− y| ≤ 2−nδ.

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−nδ < 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 inR, 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 nota 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| wouldhave a constant positive value, for infinite many n ), M has x0 as a limit point, andM has no other limit point in R. For if y ∈ R, y 6= x0, then

|xn − y| ≥ |x0 − y| − |xn − x0| ≥ |x0 − y| − 1/n ≥ 1

2|x0 − y|

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 pointin R.

1Heine,2Emile, Borel, 1871-19563Karl Weierstrass, 1815-1897

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, anddenote the points of P by x1, x2, . . . . We shall construct a sequence Bn of balls, asfollows.

Let B1 be any ball with center x1. If B1 = B(x1, r1), the corresponding closedball 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 Pis 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 = Bn∩P. Since Bn is closed and bounded, Bn is compact. Since xn /∈ Kn+1,

no point of P lies in ∩∞1 Kn. Since Kn ⊂ P, this implies that ∩Kn is empty. Buteach 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 Rwhich contain no interval.

Let C0 be the interval [0, 1]. Remove the open interval ]1/3, 2/3[ and let C1 bethe 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.


The setC = ∩∞1 Cn

is called the Cantor set. C is clearly compact, and theorem 2.4 shows that C is notempty.

No interval of the form

(2.1)

]3k + 1

3m,3k + 2

3m

[,

where k and m are positive integers, has a point in common with C. Since everyinterval ]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 ofCn which contains x. Choose n large enough, so that In ⊂ I. Let xn be an endpoint of In, such that xn 6= x.

It follows from the construction of C that xn ∈ C. Hence x is a limit point ofC, and C is perfect. 4

Chapter 3

Numerical sequences and series

The present chapter is devoted introducing several results on numerical sequencesand series.

3.1 Numerical sequences

3.1.1 Convergent sequences

A sequence (xn)n in a metric space (X, ρ) is said to converge if there is a pointx ∈ X with the following property: for every ε > 0 there is an integer nε such thatn ≥ 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

limn→∞

xn = lim xn = x.

If (xn) does not converges, it is said to diverge.Remarks. (a) It might be well to point out that our definition of ”convergentsequence” 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,∞[ (withρ(x, y) = |x− y| ).

(b) We may say that

limn→∞

xn = a ⇐⇒ ∩ε>0 ∪m>0 ∩n≥m ]xn − a, xn + a[ = {a}. 4

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 allbut 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 Asuch that lim xn = x.

Proof. (a) Suppose lim xn = x and let V ∈ V(x). For some ε > 0, the conditionsρ(y, x) < ε, y ∈ X imply y ∈ V. Corresponding to this ε, there exists nε such thatn ≥ 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 xnyn = xy;

(d) lim1

xn

=1

x, provided xn 6= 0 (n = 1, 2, . . . ) and x 6= 0.

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

|xnyn − xy| ≤ |xn||yn − y|+ |y||xn − x|.Since (xn) converges, it is bounded and we can find an M > 1 such that |xn| < Mand |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

|xnyn − 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

|xn| >1

2|x|, n ≥ m.

Given ε > 0, there is an integer nε ≥ m such that n ≥ nε implies

|xn − x| < 1

2|xn − x|2ε.

Hence, for n ≥ nε ∣∣∣∣ 1

xn

− 1

x

∣∣∣∣ =

∣∣∣∣xn − x

xnx

∣∣∣∣ < 2

|x|2|xn − x| < ε. 2

3.1.2 Subsequences

Given a sequence (xn), consider a sequence (nk)k of positive integers, such thatn1 < 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 thatxn1 = 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. Choosen1 so that |xn1 − x| < 1. Having chosen n1, . . . , ni−1, we see by theorem 1.2, page29, that there is an integer ni > ni−1 such that |xni

− x| < 1/i. The sequence (xni)i

thus obtained converges to x. 2


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

1,1

2,2

2,3

2,1

4,2

4,3

4,4

4,5

4, . . . ,

1

2n, . . . ,

2n + 1

2n, . . .

is the closed interval [0, 1], while the set of subsequential limits of the followingsequence

1,1

2,2

2,3

2,1

4,2

4,3

4,4

4, . . . ,

9

4, . . . ,

1

2n,

2

2n, . . . ,

3n

2n, . . .

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 funda-mental sequence if for every ε > 0 there is an integer nε such that ρ(xn, xm) < εprovided n ≥ nε, m ≥ nε.

Let (X, ρ) be a metric space and E ⊂ X, E 6= ∅. The diameter of E is

diamE = supx,y∈E

ρ(x, y).

If (xn) is a sequence in X and En = {xn, xn+1, . . . }, (xn) is a Cauchy sequence ifand only if

limn→∞

diamEn = 0.

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 setsand if

limn→∞

diamKn = 0,

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 pointsx′, y′ ∈ E such that ρ(x, x′) < ε, ρ(y, y′) < ε. Hence

ρ(x, y) ≤ ρ(x, x′) + ρ(x′, y′) + ρ(y′, y) ≤ 2ε + ρ(x′, y′) ≤ 2ε + diamE.


It follows thatdiamE ≤ 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

En be the closure of En. By definition and theorem 1.5 we see that

(1.1) limn→∞

diamEn = 0.

In particular, the sets En are bounded. They are also closed. Hence each En iscompact. Also En ⊃ En+1. By theorem 1.5 there is a unique point x ∈ R which liesin every En.

Let ε > 0 be given. By (1.1) there is an integer n0 such that diamEn < ε, ifn ≥ n0. Since x ∈ En, this means that |y − x| < ε for all y ∈ En, hence for ally ∈ En. In other words, if n ≥ n0, then |xn − x| < ε. But this says precisely thatxn → x, and thus the proof is completed. 2

A metric space (X, ρ) in which every Cauchy sequence converges is said to becomplete.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 se-

quences.

Theorem 1.7. Suppose (xn) is monotonic. Then (xn) converges if and only if it isbounded.

Proof. Suppose xn ≤ xn+1. Let E be the range of (xn). If (xn) is bounded, let xbe the least upper bound of E. Then

xn ≤ x, n ∈ N∗.


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

2+

1

3+ · · ·+ 1

n, n ∈ N∗.

Then an →∞. Indeed, it is enough to remark that the sequence has positive terms, ismonotonically increasing, but divergent. To check this the last property it is enoughto see that it is not a Cauchy one. The sequence (an) is not fundamental since

a2n − an =1

n + 1+ · · ·+ 1

2n>

1

2n+ · · ·+ 1

2n=

1

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 thereexists a positive M such that |an| ≤ M, for every n. From the hypothesis we inferthat the open intervals ]an−1/(2n), an +1/(2n)[ are disjoint and their union satisfies

∪n ]an −1

2n, an +

1

2n[⊂ ]−M − 1

2, +M +

1

2[ .

Since the length of the interval ]−M − 12, +M + 1

2[ is 2M + 1, it follows that the

sum of the length to the disjoint intervals ]an − 1/(2n), an + 1/(2n)[ does not exceed2M +1. On the other side, the length of an interval ]an−1/(2n), an +1/(2n)[ is equalto 1/n. So the sum of the length of the first n intervals is

∑nk=1 1/k which tends

to +∞, accordingly to (a). The contradiction shows that our hypothesis regardingthe 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. Weshow that a ≥ b.

If a < b, denote c := b − a. Consider ε := c/2. Since a is the limit of thesequence (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


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. Thenlim xn ≤ lim yn.

Proof. The sequence (yn − xn)n is convergent and has non-negative terms. So, itslimit 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

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 isincreasing since

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

x1 <1 +

√1 + 4a

2,

xn <1 +

√1 + 4a

2=⇒ xn+1 <

1 +√

1 + 4a

2, ∀n ≥ 1

we infer that the sequence is bounded. Hence the given sequence is convergent. 4

Exercise 1.2. ([24, Probl. 7, p. 9]) Show that the sequence (xn) defined by

x1 =√

a1, x2 =√

a1 +√

a2, . . . , xn+1 =√

an+1 + xn, . . . , ai > 1,

converges if

limn→∞

1

nln(ln an) < ln 2.


We introduce other two sequences

bn =an

e2n , n ≥ 1,

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) isincreasing.

From the hypothesis of the exercise it follows that there exists an n0 ∈ N∗ suchthat for n ≥ n0

1

nln(ln an) < ln 2 ⇐⇒ an < e2n ⇐⇒ bn < 1.

Consider a = max{b1, b2, . . . , bn0 , 1} and define the following sequence

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) satis-fying xn ≤ an ≤ yn from a certain rank. If the sequences (xn) and (yn) converge tothe same limit, then the sequence (an) is convergent and its limit coincides with thelimit of (xn).

Proof. It is obvious since |an − xn| ≤ |yn − xn|. 2

Exercises 1.2. Find the limits

(a) limn→∞

(1

n2 + 1+

1

n2 + 2+ · · ·+ 1

n2 + n);

(b) limn→∞

1 + 22 + · · ·+ nn

nn.

The first limit is equal to 1 since

n

n2 + n≤ 1

n2 + 1+

1

n2 + 2+ · · ·+ 1

n2 + n≤ n

n2 + 1.

For the second limit we take into account the following inequalities

1 ≤1 + 22 + · · ·+ nn

nn≤ 1 + n + n2 + · · ·+ nn

nn=

nn+1 − 1

n− 1

1

nn→ 1.

Thus we get that the limit is 1. 4


3.1.5 Upper and lower limits

Let (xn) be a sequence of real numbers with the following property: for every real mthere 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 impliesxn ≤ m, we write

xn → −∞.

Let (xn) be a sequence of real numbers. Let E be the set of numbers x (in theextend real number system) such that xnk

→ x for some sequence (xnk)k. This set

E contains all subsequential limits, plus possibly, the elements +∞ and −∞. Weput

x∗ = sup E,

x∗ = inf E.

The elements x∗ and x∗ are called the upper and lower limits of (xn); we use thenotation

lim supn→∞

xn = x∗, lim infn→∞

xn = x∗.

Theorem 1.10. Let (xn) be a sequence of real numbers. Let E and x∗ as definedearlier. 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, andthere 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 subse-quential limit; hence for any real m, xn > m for at most a finite number of valuesof 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 infinitelymany 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 thatlim supn→∞ an = lim supn→∞ bn = +∞. Then there exist m, n such that |am−an| > 1and |bm − bn| > 1.


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 existsn3 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 numbersand (bn) be a strictly increasing and divergent sequence. Then

(1.2) limn→∞

an+1 − an

bn+1 − bn

= l (∈ [−∞, +∞])

implies

(1.3) limn→∞

an

bn

= l.

Proof. Suppose that l is finite. Choose a strictly positive ε. For ε/3 we find a ranknε such that for every n ≥ nε it holds∣∣∣∣an+1 − an

bn+1 − bn

− l

∣∣∣∣ < ε/3,

that is,

(1.4) (bn+1 − bn)(l − ε/3) < an+1 − an < (bn+1 − bn)(l + ε/3).

Taking in (1.4), successively, n = nε, n = nε + 1, . . . , n = nε + p − 1, and addingall these inequalities, we get

(bnε+p − bnε)(l − ε/3) < anε+p − anε < (bnε+p − bn)(l + ε/3)

or

(1.5) l − ε

3− (l − ε

3)

bnε

bnε+p

+anε

bnε+p

<anε+p

bnε+p

< l +ε

3− (l +

ε

3)

bnε

bnε+p

+anε

bnε+p

.

Notice that the sequences (bnε/bnε+p)p and (anε/bnε+p)p tend to 0. Then there existsa rank pε such that for every p ≥ pε there hold∣∣∣∣ bnε

bnε+p

(l ± ε

3)

∣∣∣∣ < ε

3and

∣∣∣∣ anε

bnε+p

∣∣∣∣ < ε

3.

1Otto Stoltz,2Ernesto Cesaro,


Hence, for every n > nε + pε, we finally have∣∣∣∣l − an

bn

∣∣∣∣ < ε.

Now suppose that l = +∞. Choose a positive ε as large as we like. There existsa rank nε such that for every n ≥ ε it holds

an+1 − an

bn+1 − bn

>ε

2or

an+p − an >ε

2(bn+1 − bn).

Adding the first p such inequalities, we may write

an+p

bn+p

>ε

2+

an − εbn

bn+p

.

Now, there exists a rank pε such that for every p > pε we have∣∣∣∣an − εbn

bn+p

∣∣∣∣ < ε

2.

Hence, for every n > nε + pε, we finally havean

bn

> ε. 2

Corollary 1.3. Suppose there is given a strictly positive sequence (an).

(a) If

limn→∞

an+1

an

= a (a > 0),

thenlim

n→∞n√

an = a.

(b) Iflim

n→∞an = a (a > 0),

thenlim

n→∞n√

a1a2 . . . an = a.

(c) For p ∈ N∗, it holds

limn→∞

1p + 2p + · · ·+ np

np+1=

1

p + 1.

Proof. (a) Take xn = ln n√

an. Then xn = ln an/n. Now we apply theorem 1.11 ofStoltz-Cesaro and we get lim xn = lim ln an+1/an = ln lim an+1/an = ln a.

(b) Take xn = ln n√

a1a2 . . . an and reason as before.(c) Applying Stoltz-Cesaro theorem 1.11, we get

(n + 1)p

(n + 1)p+1 − (n)p+1=

np + pnp−1 + . . .

(p + 1)np + . . .→ 1

p + 1as n →∞.


3.1.7 Some special sequences

Proposition 1.1. The following limits hold

(a) If p > 0, limn→∞1

np= 0;

(b) if p > 0, lim n√

p = 1;

(c) lim n√

n = 1;

(d) if p > 0 and α is real, then limn→∞nα

(1 + p)n= 0;

(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

(1.6) an =

(1 +

1

n

)n

.

Then the sequence (an) is convergent.

Proof. First approach. Based on Newton3 formula we may write

an = 1 +

(1

n

)1

n+

(2

n

)1

n2+ · · ·+

(n

n

)1

nn

= 1 + 1 +n(n− 1)

2!

1

n2+ · · ·+ n(n− 1)(n− 2) . . . 2 · 1

n!

1

nn

= 1 + 1 +

(1− 1

n

)1

2!+ · · ·+

(1− 1

n

)(1− 2

n

)(1− n− 1

n

)1

n!

< 1 + 1 +1

2!+

1

3!+ · · ·+ 1

n!(1.7)

Since n! ≥ 2n−1, for every n ≥ 2, it follows that

2 < an < 1 +

(1 +

1

2+

1

22+ · · ·+ 1

2n−1

)< 3.

Thus the sequence (an) is bounded.We study now the monotonicity of the sequence (an).

an+1 = 1 + 1 + (1− 1

n + 1)1

2!+ · · ·+ (1− 1

n + 1)(1− 2

n + 1) . . . (1− n

n + 1)

1

(n + 1)!

> 1 + 1 + (1− 1

n + 1)1

2!+ · · ·+ (1− 1

n + 1)(1− 2

n + 1) . . . (1− n− 1

n + 1)

1

n!

> 1 + 1 + (1− 1

n)1

2!+ · · ·+ (1− 1

n)(1− 2

n) . . . (1− n− 1

n)

1

n!= an.

3Sir Issac Newton, 1642-1727


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 − yn+1 = (x− y)n∑

i=0

xn−iyi,

valid for any real x and y and n ∈ N∗. If x > y it results

(1.8) (n + 1)(x− y)yn < xn+1 − yn+1 < (n + 1)(x− y)xn.

Substitute x = 1 + 1/n and y = 1 + 1/(n + 1) and we get

x− y =1

n(n + 1), xn+1 = x · an, yn+1 = an+1

x =n + 1

n=

x− 1

y − 1> y =⇒ y2 < x + y − 1 = 1 +

1

n+

1

n + 1=⇒

yn+2 <

(1 +

1

n+

1

n + 1

)yn.(1.9)

The right-hand side of (1.8) supplies

x · an − an+1 <1

nan =⇒

(1 +

1

n

)an −

1

nan < an+1 =⇒ an < an+1.

Thus the sequence (an) is increasing.Denote un = (1 + 1

n)n+1. Then an < un, for every n ∈ N∗. From (1.9) it follows

un+1 = yn+2 <

(1 +

1

n+

1

n + 1

)yn =

(1

n+ y

)yn,

while from the left-hand side of (1.8) it results

(1.10)1

nyn < un − yn+1 =⇒ (

1

n+ y)yn < un =⇒ un+1 < (

1

n+ y)yn < un.

Thus the sequence (un) is decreasing and

an < un < · · · < u5 = (1 +1

5)6 = 2, 985984 . . . =⇒

an < 3, ∀n ∈ N∗.

Thus the sequence (an) is increasing and bounded above, hence it converges. 2

As usually we denote the limit of the sequence (an) by e, i.e.,

(1.11) limn→∞

(1 +

1

n

)n

=: e.


Corollary 1.4. There hold the inequalities(1 +

1

n

)n

< e <

(1 +

1

n

)n+1

, ∀n ∈ N∗.

Proposition 1.3. Below we introduce some properties of number e.

(i) It holds that

limn→∞

(1 +

1

1!+

1

2!+

1

3!+ · · ·+ 1

n!

)= e;

(ii) e can be written as

(1.12) e = 1 +1

1!+

1

2!+

1

3!+ · · ·+ 1

n!+

θn

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

2!+

1

3!+ · · ·+ 1

n!

the sequence (cn) defined as

cn = 1 + 1 +1

2!+

1

3!+ · · ·+ 1

n!, n ∈ N∗

is convergent and therefore the following inequality holds

(1.13) e = limn→∞

an ≥ limn→∞

cn.

On the other hand

an =

(1 +

1

n

)n

= 1 +

(1

n

)1

n+

(2

n

)1

n2+ · · ·+

(n

n

)1

nn

> 1 + 1 +

(1− 1

n

)1

2!+ · · ·+

(− 1

n

)(1− 2

n

). . .

(1− k − 1

n

)1

k!, ∀k ≤ n

=⇒ limn→∞

an = e ≥ 1 + 1 +1

2!+ · · ·+ 1

k!, ∀k ≥ 2

=⇒ e ≥ ck =⇒ e ≥ limk→∞

ck.(1.14)

Hence, from (1.13) and (1.14), we infer

limk→∞

ck = e.


(ii)

cn+m − cn =1

(n + 1)!+

1

(n + 2)!+ · · ·+ 1

(n + m)!

<1

(n + 1)!

∞∑k=0

1

(n + 2)k=

1

(n + 1)!

n + 2

n + 1<

1

n · n!.

For fixed n and passing m → +∞ it follows

0 < e− cn <1

n · n!.

Denote

0 < θn :=e− cn

1n·n!

< 1

and it follows (1.12).(iii) The inequality

0 < e− cn <1

n · n!< 10−5

is satisfied for any n ≥ 8; so

e ∼= 2 +1

2!+ · · ·+ 1

8!∼= 2.71828.

(iv) Suppose that e ∈ Q. Then we may write e = m/n, for some m, n ∈ N∗.Then

e =m

n= 1 +

1

1!+

1

2!+

1

3!+ · · ·+ 1

n!+

θn

n · n!, θn ∈ ]0, 1[ ,

=⇒ (n)! ·m− n!

(2 +

1

2!+

1

3!+ · · ·+ 1

n!

)=

θn

n.

The last equality is impossible since θn ∈ ]0, 1[ . Hence e ∈ R \Q. 2

Proposition 1.4. The sequence

an = 1 +1

2+ . . .

1

n− ln n, n ≥ 1,

is decreasing and bounded.

Proof. By corollary 1.4 successively follow(1 +

1

n

)n+1

> e >

(1 +

1

n

)n

=⇒ (n + 1)[ln(n + 1)− ln n] > 1 > n[ln(n + 1)− ln n]

=⇒ 1

n + 1< ln(n + 1)− ln n <

1

n.(1.15)


Inequality (1.15) follows from the Lagrange4 mean value theorem 2.3 at page 97, too.Taking n = 1, 2, . . . , k we get

(1.16)k∑

n=1

1

n + 1< ln(k + 1) <

k∑n=1

1

n.

Remark that from (1.15) it follows

an+1 − an =1

n + 1− ln(n + 1) + ln n < 0,

so that the sequence (an) is decreasing.From (1.16) we get

ln k < ln(n + 1) <k∑

n=1

1

n.

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

limn→∞

1 + 12

+ · · ·+ 1n

ln n= 1.

Proof. Pass to the limit in

1 + 12

+ · · ·+ 1n− ln n

ln n+ 1.

We may use equally well the Stoltz-Cesaro theorem 1.11. 2

Corollary 1.6. It holds

limn→∞

(1

n + 1+

1

n + 2+ · · ·+ 1

kn

)= ln k, ∀k ∈ {2, 3, . . . }.

Proof. Pass to the limit in

1

n + 1+

1

n + 2+ · · ·+ 1

kn

=

(1 +

1

2+ · · ·+ 1

kn− ln kn

)−(

1 +1

2+ · · ·+ 1

n− ln n

)+ ln k, n ≥ 1. 2

4J. L. Lagrange,5Leonhard Euler, 1707-17836Mascheroni


Corollary 1.7. Consider the limit

limn→∞

(1 +

1

2+ · · ·+ 1

n− α ln 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 andit is finite. Remark that M 6= ∅ since by proposition 1.4, 1 ∈ M. Now

1 +1

2+ · · ·+ 1

n− ln n + (1− α) ln n →

γ, α = 1,

+∞, α < 1,

−∞, α > 1.

Hence M = {1}. 2

Corollary 1.8. Denote

an = 1− 1

2+

1

3− 1

4+ · · ·+ (−1)n−1 1

n.

Then it holdslim

n→∞an = ln 2.

Proof. Recall Catalan7 identity

1− 1

2+

1

3− 1

4+ · · · − 1

2n=

1

n + 1+

1

n + 2+ · · ·+ 1

2n.

Then

a2n =1

n + 1+

1

n + 2+ · · ·+ 1

2n

= 1 +1

2+ · · ·+ 1

2n− ln 2n−

(1 +

1

2+ · · ·+ 1

n− ln n

)+ ln 2

→ ln 2. 2

Lemma 1.1. (Problem of Traian Lalescu,8 Gazeta Matematica, 1901, problem 579)Define

an =n+1√

(n + 1)!− n√n!.

Then limn→∞ an = e−1.

The result will follow from the next two proposition.

7E. Ch. Catalan, 1814-18948Traian Lalescu,


Proposition 1.5. The sequence

xn =n

n√n!

, n > 3,

satisfy the inequalities

(1.17) e1− 1√

n < xn < e1− 1n

Proof. We will prove them by induction. For n > 3 the next inequality holds

(1.18) xn > e1− 1√

n .

Indeed, since

e <

√32

3and x4 =

4

√32

3

it follows that (1.18) holds for n = 4.Suppose that (1.18) holds for some n ≥ 4. Then

xn+1 >

(1 +

1

n

) n

n + 1 · en√

n

n + 1 ,

and from corollary 1.4 it follows

xn+1 > e1−f(n), where f(n) =1 + (n + 1)

√n

(n + 1)2.

It is obvious that

n ≥ 4 =⇒ f(n) ≤ 1√n + 1

.

Soxn+1 > e

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

xn+1n+1 = xn

n ·(n + 1)n

nn<

(n + 1)n

nnen−1 =⇒ xn+1 <

(1 +

1

n

)n−1n+1

< e1− 1n+1 ,

ending in this way the proof. 2

Proposition 1.6. For every ε > 0 it is satisfied the next inequality∣∣∣∣an −1

e

∣∣∣∣ < ε

for every n > nε, where

nε = 1 +

[8 +

1

2ε2+

∣∣∣∣8− 1

2ε2

∣∣∣∣] .

3.1. Series 55

Proof. Using (1.17) and corollary 1.4, for n ≥ 16, we get

an =n( n√

xn+1 − 1)

xn

<n(e

1n+1 − 1)

e· e

1√n <

1

ee

1√n

<1

e

(1 +

1

[√

n]− 1

)<

1

e

(1 +

1√n− 2

)<

1

e+

1√n

.

At the same time

n( n√

xn − 1) > n[e

1n+1+

√n+1 − 1

]> n

(e

1[n+1+

√n+1] − 1

)>

n

[n + 2 +√

n + 1]≥ n

n + 2 +√

n + 1.

Thus

an >1

e· nen

n + 2 +√

n + 1>

1

e− 1√

n.

Further, for n ≥ 16, we have ∣∣∣∣an −1

e

∣∣∣∣ < 1√n

.

Since nε := 1 + [max{16, 1/ε2}] > max{16, 1/ε2}, for n ≥ nε we have 1/√

n. Hence∣∣∣∣an −1

e

∣∣∣∣ < ε ∀ n ≥ nε. 2

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, un-less the contrary is explicitly stated.

Given a sequence (xn) we use the notation

q∑n=p

xn (p ≤ q)

to denote the sum xp + xp+1 + · · · + xq. To (xn)n≥1 we associate a sequence (sn),where

sn =n∑

k=1

xk.


We also use the symbolic expression

x1 + x2 + . . . .

Or, more concisely,

(2.1)∞∑

n=1

xn.

The symbol (2.1) we call an infinite series or just a series. The numbers sn are calledthe partial sums of the series. If (sn) converges, say to s, then the series convergesand we write

∞∑n=1

xn = s.

The number s is called the sum of the series ; but it should be clearly understoodthat 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)∞∑

n=0

xn.

And frequently, when there is no possible ambiguity, or when the distinction is im-material, we shall simply write

∑xn in place of (2.1) or (2.2).

Theorem 2.1. (General criterion of Cauchy)∑

xn converges if and only if for everyε > 0 there is an integer nε such that

(2.3)

∣∣∣∣∣m∑

k=n

xk

∣∣∣∣∣ < ε

if m ≥ n ≥ nε.

Proof. Apply theorem 1.6 from page 41 to (sn). 2

In particular, for m = n (2.3) becomes

|xn| < ε, n ≥ nε,

that is

Theorem 2.2. (Necessary condition) If∑

xn converges, then lim xn = 0.

Remark. The condition xn → 0 is not, however, sufficient to ensure convergence of∑xn. For instance, the series

∞∑1

1

n

diverges as it results from (a) of exercises 1.1 from page 42 or it will become clear bytheorem 2.19. 4

3.2. Numerical series 57

Theorem 2.3. (Operations with series) Let∑

an and∑

bn be two convergent seriesand c a real number. Then the series

∑c an and

∑(an ± bn) are convergent and,

moreover,

(2.4)∑

c an = c∑

an

and

(2.5)∑

(an ± bn) =∑

an ±∑

bn.

Proof. Indeed,

c∑

an = c limn∑

ak = limn∑

c ak =∑

c an

and∑an ±

∑bn = lim

n∑ak ± lim

n∑bk = lim(

n∑ak ±

n∑bk) =

∑(an ± bn). 2

Remark. The assumption that both series are convergent is essential. Otherwise itcould happen that the left-hand side of (2.5) exists while the right-hand side of it hasno meaning. For,

0 =∑

(1− 1) versus∑

1−∑

1. 4

As for finite sums, we may group the terms of a convergent series in brackets (butno commutativity is allowed). For example

a1 + (a2 + a3) + (a4 + a5 + a6 + a7) + . . . .

Theorem 2.4. Let∑

an be a convergent series. Then by grouping its terms it resultsa convergent series having the same sum.

Proof. The sequence of partial sums of the transformed series is a subsequence of theconvergent 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 sumsform a bounded sequence.

Proof. The sequence of partial sums is increasing. It is convergent if and only if it isbounded, accordingly to the theorem 1.7 from page 41. 2

Theorem 2.6. Let∑

an be a convergent series with nonnegative terms. Then theseries

∑bn obtained from the former by rearranging (commuting) and renumbering

its terms is also convergent having the same sum.


Proof. Let s be the sum of the first series and sn its partial sum of rank n. Denoteby pn the partial sum of rank n of the series

∑bn. Fix a rank, let it be n and

consider pn. Thenpn = b1 + · · ·+ bn = ak1 + · · ·+ akn

and denote N = max{k1, . . . , kn}. Obviously, pn ≤ sN ≤ s. Thus we conclude thatthe series

∑bn converges and, if b is its sum, b ≤ s. Reasoning vice-verse, we get

that∑

an =∑

bn. 2

Theorem 2.7. (Comparison test) (a) If |xn| ≤ cn for n ≥ n0, where n0 is somefixed integer, and if

∑cn converges, then

∑xn converges.

(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∑k=n

ck ≤ ε,

by the Cauchy criterion. Hence∣∣∣∣∣m∑

k=n

xk

∣∣∣∣∣ ≤m∑

k=n

|xk| ≤m∑

k=n

ck ≤ ε,

and (a) follows.(b) follows from (a), for if

∑xn converges, so must

∑yn. 2

Theorem 2.8. If 0 ≤ x < 1,∞∑

n=0

xn =1

1− x.

If x ≥ 1, the series diverges.

Proof. If x 6= 1,

sn =n∑

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 2.9. (Cauchy condensation test) Suppose x1 ≥ x2 ≥ x3 ≥ · · · ≥ 0. Thenthe series

∑∞n=1 xn converges if and only if the series

(2.6)∞∑

k=0

2kx2k = x1 + 2x2 + 4x4 + 8x8 + . . .

converges.


Proof. By theorem 2.5 it suffices to consider the boundedness of the partial sums. Let

sn = x1 + x2 + · · ·+ xn,

tk = x1 + 2x2 + · · ·+ 2kx2k .

For n < 2k,

sn = x1 + · · ·+ xn (add terms)

≤ x1 + (x2 + x3) + · · ·+ (x2k + · · ·+ x2k+1−1)

≤ x1 + 2x2 + · · ·+ 2kx2k = tk,

so that

(2.7) sn ≤ tk.

On the other hand, if n > 2k,

sn = x1 + · · ·+ xn (neglect terms)

≥ x1 + x2 + (x3 + x4) + · · ·+ (x2k−1+1 + · · ·+ x2k)

≥ 1

2x1 + x2 + 2x4 + · · ·+ 2k−1x2k =

1

2tk,

so that

(2.8) 2sn ≥ tk.

By (2.7) and (2.8) the sequences (sn) and (tk) are either both bounded or bothunbounded. 2

Theorem 2.10. ∑ 1

npconverges if and only if p > 1.

Proof. If p ≤ 0, divergence follows from the necessary condition, i.e., theorem 2.2. Ifp > 0, the previous theorem is applicable, and we are led to the series

∞∑k=0

2k 1

2kp=

∞∑k=0

2(1−p)k.

Now, 21−p < 1 if and only if 1 − p < 0, and the result follows by comparison withthe geometric series (take x = 21−p ). 2

Remark. The series

1 +1

2+

1

3+

1

4+ . . .

is called the harmonic series and it is divergent. From corallary 1.5 we deduce thatthe speed of divergence of the harmonic series agrees to the speed of divergence of thelogarithmic function. 4


Theorem 2.11. If p > 1,

(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 applytheorem 2.9 to (2.9); this leads us to the series

(2.10)∞∑

n=1

2k 1

2k(ln 2k)p=

1

(ln 2)p

∞∑k=1

1

kp. 2

3.2.2 The root and ration tests

Theorem 2.12. (Root test or D’Alembert9 criterion) Given∑

xn, we put α =lim supn→∞

n√|xn|. Then

(a) if α < 1,∑

xn converges;

(b) if α > 1,∑

xn diverges;

(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

n√|xn| < β

for all n ≥ m. That is, n ≥ m implies

|xn| < βn.

Since 0 < β < 1,∑

βn converges. Convergence of∑

xn follows now from thecomparison test, theorem 2.7.

(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 forconvergence of

∑xn, does not hold.

(c) To prove (c), we consider the series∑ 1

n,∑ 1

n2.

For each of these series α = 1, but the first diverges, while the second converges.2

Theorem 2.13. (Ratio test or Cauchy criterion) The series∑

xn

9Jean LE Rond D’Alembert, 1717-1783


(a) converges if lim supn→∞

∣∣∣∣xn+1

xn

∣∣∣∣ < 1;

(b) diverges if

∣∣∣∣xn+1

xn

∣∣∣∣ ≥ 1, for n ≥ n0, where n0 is some fixed integer;

(c) if

lim infn→∞

∣∣∣∣xn+1

xn

∣∣∣∣ ≤ 1 ≤ lim supn→∞

∣∣∣∣xn+1

xn

∣∣∣∣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| < β2|xm|,

. . . . . .

|xm+p| < βp|xm|.

That is

|xn| < βn−m|xm|

for n ≥ m, and (a) follows from the comparison test, since∑

βn 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 ∑ 1

n,∑ 1

n2.

In any case we have

limn→∞

xn+1

xn

= 1,

but the first series diverges while the second one converges. 2

Theorem 2.14. For any sequence (xn) of positive numbers

lim infn→∞

xn+1

xn

≤ lim infn→∞

n√

xn

lim supn→∞

n√

xn ≤ lim supn→∞

xn+1

xn

.


3.2.3 Power series

Given a sequence (cn) of complex numbers, the series

(2.11)∞∑

n=0

cnzn = 1 +

∞∑n=1

cnzn,

is called a power series. The numbers cn are called the coefficients of the series; z isa complex number.

Theorem 2.15. (Cauchy-Hadamard10) Given the power series∑

cnzn, put

α = lim supn→∞

n√|cn|, R =

1

α.

(If α = 0, R = +∞; if α = ∞, R = 0. ) Then∑

cnzn converges if |z| < R; and

diverges if |z| > R.

Remark. R is called the radius of convergence of∑

cnzn.

Proof. Put xn = cnzn and apply the root test

lim supn→∞

n√|xn| = |z| lim sup

n→∞

n√|cn| =

|z|R

. 2

Examples. (a)∑

nnzn, R = 0;

(b)∑ zn

n!, R = +∞;

(c)∑

zn, R = 1. If |z| = 1, the series diverges, since zn does not tend to 0 asn →∞;

(d)∑ zn

n, R = 1. On the circle of convergence, the series diverges at z = 1; it

converges at all other points of |z| = 1. The last assertion follows from theorem2.19.

(e)∑ zn

n2, R = 1. The series converges at all points of the circle |z| = 1, by the

comparison test, since

∣∣∣∣zn

n2

∣∣∣∣ =1

n2if |z| = 1. 4

3.2.4 Partial summation

Lemma 2.1. Given two sequences (an), (bn). Put

An =n∑

k=0

ak

10Jacques Hadamard, 1865-1963


if n ≥ 0; put A−1 = 0. Then, if 0 ≤ p ≤ q, we have

(2.12)

q∑n=p

anbn =

p−1∑n=p

An(bn − bn−1) + Aqbq − Ap−1bp.

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,

(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, wehave |an + an+1 + · · ·+ am| < α.

Then∑∞

n=1 anbn 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

anbn + · · ·+ ambm = αn,nbn + (αn,n+1 − αn,n)bn+1 + · · ·+ (αn,m − αn,m−1)bm

= αn,n(bn − bn+1) + · · ·+ αn,m−1(bm−1 − bm) + αn,mbm.(2.13)

Now, we choose a positive ε. For ε/(2α) we find an nε ∈ N∗ that satisfies thefollowing two requirements

• it is greater or equal to the nε from the general criterion of Cauchy theorem2.1 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, anbn| ≤ α|bn| < α[ε/(2α)] < ε;

• if m > n, from (2.13) we get

|m∑

k=n

akbk| ≤ α(|bn − bn+1|+ · · ·+ |bm−1 − bm|) + α|bm| < αε

(2α)+ α

ε

(2α)= ε.

The conclusion follows from the the general criterion of Cauchy theorem 2.1. 2

Remark. If bn ↓ 0, the assumption that the series∑∞

n=1 |bn − bn+1| converges intheorem 2.16 is useless. It may be neglected since the n -rank partial sum of the series∑∞

n=1 |bn − bn+1| is equal to b1 − bn+1 which, in turn, tends to b1. 411Niels Henrik Abel, 1802-1829


Theorem 2.17. (Dirichlet12 theorem) Suppose

(i) the partial sums An of∑

an form a bounded sequence;

(ii) b0 ≥ b1 ≥ b2 ≥ . . . ;

(iii) limn→∞ bn = 0.

Then∑

anbn 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.

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

2p+

1

3p− 1

4p+ · · ·+ (−1)n+1 1

np+ . . .

converges. For p = 1, based on corollary 1.8, we can write the Leibnitz13 series

(2.14) 1− 1

2+

1

3− 1

4+ · · ·+ (−1)n+1 1

n+ · · · = ln 2. 4

Theorem 2.19. Suppose the radius of convergence of∑

cnzn is 1, and suppose

c0 ≥ c1 ≥ c2 . . . , limn→∞ cn = 0. Then cnzn converges at every point of the circle

|z| = 1, except possibly at z = 1.

Proof. Put an = zn, bn = cn. The hypotheses of theorem 2.17 are satisfied, since

|An| =

∣∣∣∣∣n∑

k=0

zk

∣∣∣∣∣ =

∣∣∣∣1− zn+1

1− z

∣∣∣∣ ≤ 2

|1− z|,

if |z| = 1, z 6= 1. 2

12Gustav Lejeune Dirichlet, 1805-185913Gottfried Wilhelm Leibnitz, 1646-1716


3.2.5 Absolutely and conditionally convergent series

Consider

(2.15)∑

an.

Series (2.15) with complex terms is said to be absolutely convergent if the series

(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 ≥ mit holds

|m∑

k=n

|ak| | =m∑

k=n

|ak| < ε.

But using a well known inequality, we get

|m∑

k=n

ak| ≤m∑

k=n

|ak| < ε.

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 absolutelyconvergent, too. But there are convergent series (with arbitrary terms) which arenot absolutely convergent. Indeed, as we already saw the series in example 2.1 isconvergent 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 whereasthe corresponding series of the absolute values (2.16) diverges.

A very important property to a sum of a finite number of real (complex) summandsis the commutative property, that is, a rearrangement of the terms does not affecttheir sum. Unfortunately, this is not the case for series. Consider the convergentseries (2.14), i.e.,

1− 1

2+

1

3− 1

4+ · · ·+ (−1)n+1 1

n. . . .

We write it, based on theorem 2.4, as

(2.17)

(1− 1

2

)+

(1

3− 1

4

)+ · · ·+

(1

2k − 1− 1

2k − 2

)+ . . . .


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

(2.18)

(1− 1

2− 1

4

)+

(1

3− 1

6− 1

8

)+ · · ·+

(1

2k − 1− 1

4k − 2− 1

4k

)+ . . . .

Denote by Sn the n -th order partial sum of series (2.18). Then we have

S3n =n∑

k=1

(1

2k − 1− 1

4k − 2− 1

4k

)=

n∑k=1

(1

4k − 2− 1

4k

)=

1

2

n∑k=1

(1

2k − 1− 1

2k

)=

1

2s2n.

Thus we can write

(2.19) S3n =1

2s2n.

Also

(2.20) S3n−1 =1

2s2n +

1

4nand S3n−2 =

1

2s2n +

1

4n− 2.

From (2.19) and (2.20) we conclude that

limn→∞

S3n = limn→∞

S3n−1 = limn→∞

S3n−2 =1

2s2n.

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) Therearrangement of the terms of an absolutely convergent series supplies another abso-lutely convergent series having the same sum as the original one.

Proof. Consider

(2.21)∞∑

an

an absolutely convergent series. Further, consider the positive, respectively, the neg-ative parts of its terms, more precisely,

(2.22) a+n =

{an, an ≥ 0

0, an < 0a−n =

{−an, an ≤ 0

0, an > 0.


We have

(2.23) an = a+n − a−n .

Consider, now, the following two series with nonnegative terms

(2.24)∑

a+n and

∑a−n .

Then both series in (2.24) converge, since a+n ≤ |an| and a−n ≤ |an|.

Consider∑

bn the rearranged series of (2.21). For it construct the series∑

b+n

of positive parts , respectively, the series∑

b−n of negative parts, as in (2.22). Then

∞∑an =

∞∑(a+

n − a−n ) =∞∑

a+n −

∞∑a−n =

∞∑b+n −

∞∑b−n =

∞∑(b+

n − b−n ) =∞∑

bn.

The first equality is obvious, the second equality follows from theorem 2.3, the thirdone follows from theorem 2.6. For the other equalities we argue in the same way. 2

Corollary 2.1. Suppose the series∑

an is absolutely convergent. Then the series∑a+

n and∑

a−n are absolutely convergent, too.

The converse statement is true since the difference of two convergent series bothof them having nonnegative terms is a convergent series, theorem 2.3.

Corollary 2.2. For series∑

an to be absolutely convergent it is necessary and suf-ficient that series (2.24) generated by it be convergent.

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 a−n → 0.

Proof. From corollary 2.2 it follows that at least one of the series (2.24) generated byit is divergent, that is

∑a+

n = ∞, since a+n ≥ 0.

(2.25)n∑

a−k =n∑

a+k −

n∑ak.

We examine the behaviour of the right-hand side of (2.25) as n →∞. The second sumtends 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, theother is divergent, too.

The sequences (a±n ) tend to zero since the series∑

an is convergent. 2

Theorem 2.22. (On rearrangement of conditionally convergent series of Riemann14)Let series

∑∞0 an and

∑∞0 bn be two divergent series with positive terms whose gen-

eral terms tend to zero, i. e., an, bn → 0 as n →∞.

14Bernhard Riemann, 1826-1866


Then for any s ∈ [−∞,∞] we can construct a series

a0 + a1 + · · ·+ am1 − b0 − b1 − · · · − bn1(2.26)

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∑∞

0 an and∑∞0 bn only once.

Proof. First we suppose that s ∈ R. The indices n1, n2, . . . and m1, m2, . . . canbe chosen in this case as the smallest natural numbers for which the correspondinginequalities written bellow are fulfilled

(i) α1 =∑n1

0 ak > s,

(ii) α2 = α1 −∑m1

0 bk < s,

(iii) α3 = α2 +∑n2

n1+1 bk > s,

(iv) α4 = α3 −∑m2

m1+1 bk < s, . . . .

At the pth step of this construction we indeed can choose the natural numbers np andmp satisfying the pth inequality since the series

∑∞0 an and

∑∞0 bn have positive

terms terms and diverge. The fact that the series thus constructed converges to sfollows from the above inequalities and from the assumption that an, bn → 0 asn →∞.

Now, suppose that s = +∞. We can replace s in the right-hand side of theinequalities (i), (ii), (iii), . . . by a divergent sequence of the form 2,1,4,3,5,. . . . 2

Corollary 2.3. Let∑

an be a convergent series but not absolutely. Choose ans ∈ [−∞,∞]. Then there is a rearrangement of the series such that the resultingseries converges to s.

Proof. We just split the series∑

an into two series∑

a+n and

∑a−n as it is indicated

by (2.22), then apply lemma 2.2 and theorem 2.22. 2


3.2.6 Exercises

Study the convergence of the following series and find their sum if any

∞∑n=1

√n

n;

∞∑n=1

n

(n + 1)2n;

∞∑n=1

1

2n− 1;

∞∑n=1

3√n2

n;

∞∑n=1

n + 1

n23n;

∞∑n=1

sin(nx)

5n;

∞∑n=1

2

3n;

∞∑n=1

cos(nx)− cos(n + 1)x

n;

∞∑n=1

1

n(n + 1);

∞∑n=1

(−1)n−1

n + 1

(2 +

1

n

);

∞∑n=1

arctg1

2n2;

∞∑n=1

(3√

n + 2− 3√n + 1 +

3√

n);

∞∑n=1

lnn2 + 5n + 6

n+5n + 4;

∞∑n=1

arcsin

√(n + 1)2 − 1−

√n2 − 1

n(n + 1);

∞∑n=1

1

npsin

π

n, p ∈ R;

∞∑n=1

(n + 1)n2

nn23n;

∞∑n=1

ln n

n(ln4 n + 1);

∞∑n=1

e−√

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 arecalled points or vectors, especially when k > 1. If y = (y1, . . . , yk) and α is a realnumber, let

x + y = (x1 + y1, x2 + y2, . . . , xk + yk), + : Rk × Rk → Rk;

αx = (αx1, . . . , αx2, . . . , αxk), · : R× Rk → Rk.

These define addition of vectors, and respectively, multiplication of a vector by a realnumber (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 thereals. The zero element 0 in Rk is called the origin or the null vector and all itscoordinates 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 = α1x+· · ·+αnx, 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

〈x, y〉 =k∑

i=1

xiyi

and the Euclidean norm of x by

‖x‖2 =√〈x, x〉 =

(k∑

i=1

x2i

)1/2

.

The Minkowski norm or the l1 -norm of x is defined by

‖x‖1 = |x1|+ · · ·+ |xk|,

while the uniform norm of x is defined by

‖x‖∞ = max{|x1|, . . . , |xk|}.

For 1 ≤ p < +∞ we define the lp -norm of x by

‖x‖p = (|x1|p + · · ·+ |xk|p)1/p .

In order to indicate which the norm we are refering to we denote (Rk, ‖ · ‖p) and(Rk, ‖ · ‖∞).

Theorem 1.2. Let ‖ · ‖ be any of the norms defined above on Rk. Suppose x, y, z ∈Rk, and α ∈ R. Then(a) ‖x‖ ≥ 0;(b) ‖x‖ = 0 ⇐⇒ x = 0;(c) ‖αx‖ = |α|‖x‖;(d) |〈x, y〉| ≤ ‖x‖2‖y‖2;(e) ‖x + y‖ ≤ ‖x‖+ ‖y‖;(f) ‖x− z‖ ≤ ‖x− y‖+ ‖y − z‖.

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 ≤ 〈x + λy, x + λy〉 = 〈x, x〉+ 2λ〈x, y〉+ λ2〈y, y〉.

Taking λ = −〈x, y〉〈y, y〉

we get

0 ≤ 〈x, x〉 − 2〈x, y〉2

〈y, y〉+〈x, y〉2〈y, y〉〈y, y〉2

=〈x, x〉〈y, y〉 − 〈x, y〉2

〈y, y〉

and (d) follows.

4.1. Euclidean spaces 73

(e) If the norm is an lp -norm, p ≥ 1, then our inequality is actually the Minkowskiinequality, (see page 23). If the norm under consideration is the uniform one, then astraightforward evaluation proves this inequality.

(f) follows from (e). 2

Any function ‖ · ‖ : Rk → R satisfying (a), (b), (c), and (e) of theorem 1.2 is saidto be a norm on Rk.

Given a norm on Rk, we assign to it a metric by

(1.1) ρ(x, y) = ‖x− y‖.

Theorem 1.3. If ‖ · ‖ is a norm on Rk, then the metric ρ defined by (1.1) is ametric on Rk; thus every norm induces a metric.

Theorem 1.4. Any two metrics on Rk induced by any of the norms ‖ · ‖p, p ≥ 1,or ‖ · ‖∞ generate the same open sets.

Corollary 1.1. A sequence is convergent in one of the metrics mentioned in theabove 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) limn→∞

αjn = αj, 1 ≤ j ≤ k.

(b) Suppose (xn)n, (yn)n are sequences in Rk, (βn)n is a sequence of real num-bers, and xn → x, yn → y, βn → β. Then

lim(xn + yn) = x + y,lim 〈xn, yn〉 = 〈x, y〉,lim βnxn = βx.

Proof. We use the Euclidean metric.(a) If xn → x, the inequalities

|αjn − αj| ≤ ‖xn − x‖2, 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 thatn ≥ nε implies

|αjn − αj| <ε√k, 1 ≤ j ≤ k.

74 4. Euclidean spaces

Hence n ≥ nε implies

‖xn − x‖2 =

(k∑

j=1

|αjn − αj|2)1/2

< ε,

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, ‖ · ‖p), 1 ≤ p < ∞, and (Rk, ‖ · ‖∞) are complete.

We consider the case of (Rk, ‖ · ‖2). The other cases can be proved similarly. Weprove the following claim: a sequence (xn)n = (α1n, . . . , αkn)n in Rk is ‖ · ‖2 -Cauchyif and only if every sequence (αj,n)n, 1 ≤ j ≤ k, is a Cauchy sequence.

Proof of the claim. Suppose (xn) is ‖ · ‖2 -Cauchy. Then there is an integer nε

such that n, m ≥ nε imply‖xn − xm‖2 < ε.

It follows that|αjn − αjm| ≤ ‖xn − xm‖2 < ε, j = 1, . . . , k.

Conversely, suppose that for every ε > 0 there is an integer nε such that n,m ≥ nε

imply

|αjn − αjm| <ε√k, j = 1, . . . , k.

Hence, n, m ≥ nε imply‖xn − xm‖2 < ε.

Proof. Suppose (xn) is ‖ · ‖2 -Cauchy. Then every sequence (αjn)n, 1 ≤ j ≤ k, isCauchy, 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 pointin Rk.

Chapter 5

Limits and Continuity

The aim of this chapter is to introduce notions and results on limits and continuity.

5.1 Limits

5.1.1 The limit of a function

Let X and Y be metric spaces; suppose ∅ 6= E ⊂ X, f maps E into Y, and p isa limit point of E. We write

f(x) → q as x → p

or equivalently,

(1.1) limx→p

f(x) = q

if there is a point q ∈ Y with the following property: for every ε > 0 there exists aδ > 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) limx→p

f(x) = q

75

76 5. Limits and Continuity

if and only if

(1.5) limn→∞

f(pn) = q

for every sequence (pn)n in E such that

(1.6) pn 6= p, limn→∞

pn = p.

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 but0 < 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. 2

Suppose Y = Rk and f, g : E → Rk. Define f + g and 〈f, g〉 by

(f + g)(x) = f(x) + g(x), 〈f, g〉(x) = 〈f(x), g(x)〉, 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 ′, f, g are functions definedon E with values in Rk and

limx→p

f(x) = A, limx→p

g(x) = B.

Then

(a) limx→p(f + g)(x) = A + B;

(b) limx→p 〈f, g〉(x) = 〈A, B〉;

(c) If Y = R, and B 6= 0, then limx→p

(f

g

)(x) =

A

B.

Proof. 2

5.1. Limits 77

5.1.2 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

limx→p

f(x) = q ⇐⇒ f(x−) = f(x+) = q.

Proof. The necessity part is immediate.Suppose now that f(x−) = f(x+) = q. We use the characterization in the theo-

rem 1.1 and consider a sequence (xn) in I , xn → p and having infinitely many termsgreater than p, (denoted by (yn) ), and infinitely many terms less than p (denotedby (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 suchthat

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 tobe continuous at p. If p is an isolated point of E, then our definition implies thatevery function f which has E as its domain of definition is continuous at p. 4Theorem 2.1. In the above setting, assume also that p is a limit point of E. Thenf 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


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), thenh is continuous at p.

Proof. Let ε > 0 be given. Since g is continuous at f(p), there exists η > 0 suchthat 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 Xinto a metric space Y is continuous on X if and only if f−1(V ) is open in X forevery open set V in Y.

Proof. Suppose f is continuous on X and V is an open set in V. We have to showthat 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 sincef 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 ∈ Xand 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 thatρ(f(x), f(p)) < ε. 2

Theorem 2.4. Let f and g be complex-valued continuous functions on a metricspace X. Then f +g, fg, 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 state-ment 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 fbe 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 contin-uous.

(b) If f and g are continuous mappings of X into Rk, then f + g and 〈f, g〉are continuous on X.

Functions f1, . . . , fk are called components of f.Proof. Part (a) follows from the inequalities

|fj(x)− fj(y)| ≤ ‖f(x)− f(y)‖2 =

(k∑

i=1

|fi(x)− fi(y)|2)1/2

,

for j = 1, . . . , k. Part (b) follows from (a) and theorem 1.2. 2

5.2.1 Continuity and compactness

A mapping f of a set E into Rk is said to be bounded if there is a real number Msuch that |f(x)| < M, for all x ∈ E.

Theorem 2.6. Suppose f is a continuous mapping from a compact metric space Xinto 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 thesets f−1(Vα) is open (see theorem 2.3). Since X is compact, there are finitely manyindices, 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 Xinto 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 spaceX, and

M = supp∈X

f(p), m = infp∈X

f(p).

Then there exist points p, q ∈ X such that f(p) = M, f(q) = m.


Proof. By theorem 2.7, f(X) is a closed and bounded set of real numbers; hencef(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 metricspace onto a metric space Y. Then the inverse function f−1 defined on Y is acontinuous mapping of Y onto X.

Proof. Applying theorem 2.3 to f−1 in place of f, we see that it suffices to provethat f(V ) is an open set in Y for every open set V in X. Fix such a set V. Thecomplement X \ V is closed in X, hence compact (theorem 2.3, page 33). It followsthat f(X \ V ) is a compact subset of Y which implies that it is closed in Y. Sincef 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 satisfy-ing

(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(m

m) = mϕ(

1

m)

it follows that

ϕ(1

m) =

1

mϕ(1) m ∈ N∗.

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 thesystem of rational numbers.

Consider an arbitrary but fixed x ∈ R \Q. Then there exists a sequence (qn) ofrational numbers such that qn → x. Then using the continuity hypothesis one caninfer that

ϕ(x) = lim ϕ(qn) = lim qnϕ(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 saythat 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, whereascontinuity can be defined at a single point. Asking whether a given function is uni-formly continuous at a certain point is meaningless. Secondly, if f is continuous onX, it is possible to find, for each ε > 0 and each point p ∈ X, a number δ > 0having 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. Thatthe 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 spaceX 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 pointp ∈ 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

(2.4) I(p) = B(p,1

2φ(p)).

Since p ∈ I(p), the family of all sets I(p) is an open cover of X and since X iscompact, there is a finite set of points p1, . . . , pn in X, such that

(2.5) X = I(p1) ∪ · · · ∪ I(pn).

Let

(2.6) δ =1

2min{φ(p1), . . . , φ(pn)}(> 0).

Now, let x and p be points of X such that ρ(x, p) < δ. By (2.5), there is an integerm , 1 ≤ m ≤ n, such that p ∈ I(pm). It follows that

(2.7) ρ(p, pm) <1

2φ(pm),

and also that

ρ(x, pm) ≤ ρ(x, p) + ρ(p, pm) < δ +1

2φ(pm) ≤ φ(pm).

This, together with (2.3) imply that

ρ(f(p), f(x)) ≤ ρ(f(p), f(pm)) + ρ(f(pm), f(x)) < ε. 2


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 opensets 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 intoa metric space, then f(X) is connected.

Proof. If f(X) is not connected, there are disjoint open sets V and W in Y, bothof which intersect f(X), such that f(X) ⊂ V ∪W. Since f is continuous, the setsf−1(V ) and f−1(W ) are open in X; they are clearly disjoint and nonempty, and theirunion 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 continuousreal function on the interval [a, b]. If f(a) < f(b) and if c is a number such thatf(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 thetheorem 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 simplediscontinuity. Otherwise the discontinuity is said to be of the second kind. There aretwo 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

f(x) =

{1, x ∈ Q,

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

f(x) =

{x, x ∈ Q,

0, x ∈ R \Q.

1Gaston Garboux, 1842-1917

5.2. Continuity 83

Then f is continuous at x = 0 and has a discontinuity of the second kind at everyother point.

(c) Define

f(x) =

{sin 1

x, x 6= 0,

0, x = 0.

Since neither f(0+) nor f(0−) exist, f has a discontinuity of the second kind atx = 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[ ifa < x < y < b implies f(x) ≤ f(y). If the last inequality is reversed, we obtainthe definition of a monotonically decreasing function. The set of monotonic functionsconsists of both the increasing and the decreasing functions.

Theorem 2.14. Let f be monotonically increasing on ]a, b[ . Then f(x+) andf(x−) exist at every point x ∈]a, b[ . More precisely,

(2.8) supa<t<x

f(t) = f(x−) ≤ f(x) ≤ f(x+) = infx<t<b

f(t).

Furthermore, if a < x < y < b, then

(2.9) f(x+) ≤ f(y−).

Proof. The set of numbers f(t) with a < t < x is bounded above by f(x). Thereforeit 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 thereexists δ > 0 such that a < x− δ < x and

(2.10) A− ε < f(x− δ) ≤ A.

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+) = infx<t<b

f(t) = infx<t<y

f(t).


Similarly

(2.13) f(y−) = supx<t<b

f(t) = supx<t<y

f(t).

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 theset 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), ifx1 6= x2.

We have thus established an one-to-one correspondence between the set E and asubset of the rational numbers. The latter is countable. 2

Exercise 2.2. ([24, Probl. 11, p. 9]) Let f be a continuous and increasing functiondefined on an interval [a, b] and such that f(a) ≥ a and f(b) ≤ b. Choose anarbitrary 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 includedin a compact set and thus it is bounded.

Function f being increasing, the sequence (xn) is monotone. So we have estab-lished 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∗. 4

5.2.6 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 Darbouxfunction on an interval I by DI .Remark. Geometrical, the Darboux property says that for any a, b ∈ I, a < b andany λ between f(a) and f(b), the parallel to the Ox axis through y = λ meetsthe 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 intervalis 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 thaty1 < y2 ) and any λ ∈ ]y1, y2[ by (a), there is c ∈ ]t1, t2[ such that λ = f(c). Since Jis 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 thatc /∈ {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. Thentheir 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 mostcountable set. The function f is constant.

Proof. f(I) is an interval and is at most countable. Thus | f(I) |= 1, that is, f isconstant. 2

Corollary 2.4. Let f : I → R be a Darboux mapping. Then f is injective if andonly 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) orf(t1) > f(t2) < f(t3). Let us suppose that the first case holds. The other case runssimilarly. The first case has two subcases.• f(t1) < f(t3) < f(t2). Then there is c ∈ ]t1, t2[ with f(c) = f(t3). Thus functionf 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 Darbouxfunction 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


(a) |f |, f 2, and√|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,g3(t) =

√|t|, respectively, g(t) = 1/t. 2

Remark. We introduce two Darboux function whose sum is not a Darboux function.

f(t) =

{sin 1

t, t 6= 0,

0, t = 0,g(t) =

{− sin 1

t, t 6= 0,

1, t = 0,

are two Darboux functions while their sum

(f + g)(t) =

{0, t 6= 0,

1, t = 0,

is not a Darboux mapping. 4

Corollary 2.6. Let f : I → R be a Darboux mapping. Then f has no discontinuitypoint of the first kind.

Proof. [15, p. 52]. 2

Theorem 2.15. (Sierpinski2) Any mapping f : R → R is a sum of two discontinuousDarboux mappings, i.e., exist f1, and f2 Darboux and discontinuous functions suchthat

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. 4Let 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

2Wac law Sierpinski, 1882-19693Lipschitz

5.2. Continuity 87

Theorem 2.16. (Banach4 fixed point theorem) Let (X, ρ) be a complete metric spaceand 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 byxn+1 = f(xn), n = 0, 1, . . . . For every n ∈ N∗ and every integer p, p ≥ 1, we havethe estimations

ρ(xn+1, xn) = ρ(f(xn), f(xn−1) ≤ αρ(xn, xn−1) ≤ · · · ≤ αnρ(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)

≤ αn 1

1− αρ(x1, x0).

Thus we may conclude that (xn) is a Cauchy sequence. The metric space X iscomplete, thus the sequence (xn) is convergent. Let x ∈ X be its limit. Passing pto +∞ in the previous estimations we get

(2.15) ρ(x, xn) ≤ αn 1

1− αρ(x1, x0).

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 samelimit 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) toits limit x.Example. Suppose we have to find a real roots of the equation

(2.16) x3 + 2x− 1 = 0.

4stefan Banach,


First we note that if we denote the left-hand side of (2.16) by g(x), we get a polyno-mial function; this function is strictly increasing. Hence it has at most one real root.At the same time from

limx→−∞

g(x) = −∞ and limx→+∞

g(x) = +∞,

and, moreover, taking into account theorem 2.13, we conclude that (2.16) has preciselyone real root.

In order to find this real root we rewrite the given equation as

(2.17) x =1

x2 + 2.

Denote the right-hand side of (2.17) by f(x) and thus we get a real rational functiondefined on R.

We are checking the assumption of theorem 2.16. For x, y ∈ R we have

|f(x)− f(y)| = |x + y|(2 + x2)(2 + y2)

|x− y|.

However

|t| = 1√2

√2t2 ≤ 2 + t2

2√

2.

Thus

|x + y| ≤ |x|+ |y| ≤ 2 + x2 + 2 + y2

2√

2≤ (2 + x2)(2 + y2)

2√

2.

Now we conclude that the function f is a contraction. From (2.17) easily follows thatthe solution has to be strictly positive. Since the function f is strictly decreasingon the positive semi-axis and f([0, 1]) = [1/3, 1/2] ⊂ [0, 1], it follows that it is acontraction on the compact interval [0, 1]. Thus we may apply the Banach fixedpoint 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 convexprovided

(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

(2.20) f(α1x1 + · · ·+ αnxn) ≤ α1f(x1) + · · ·+ αnf(xn)

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 forn − 1. Consider x1, . . . , xn ∈ I, α1, . . . , αn ≥ 0, and α1 + · · · + αn = 1. Then∑n−1

i=1 αi/(1− αn) = 1, so by hypothesis

f

(n−1∑i=1

αi

1− αn

)≤

n−1∑i=1

αi

1− αn

f(xn).

Hence

f

(n∑1

αixi

)=f((1− αn)

n−1∑1

αi

1− αn

xi + αnxn)

≤(1− αn)f(n−1∑

1

αi

1− αn

xi) + αnf(xn)

≤(1− αn)n−1∑

1

αi

1− αn

f(xi) + αnf(xn) =n∑1

αif(xi).

Corollary 2.7. Suppose f : I → R is a convex mapping. Then

(2.21) f

(α1x1 + · · ·+ αnxn

α1 + · · ·+ αn

)≤ α1f(x1) + · · ·+ αnf(xn)

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 =⇒ sf,x0 is increasing on I \ {x0};5Jensen, 1859-1925


(b) f is strictly convex on I =⇒ sf,x0 is strictly increasing on I \ {x0};Proof. 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 Jensenconvex or J-convex provided

(2.22) x, y ∈ I =⇒ f(x + y

2) ≤ f(x) + f(y)

2.

Proposition 2.3. A continuous and J-convex function f : I → R is convex.

Proof. From (2.22), by induction, we get

(2.23) f(1

2k(x1 + · · ·+ x2k)) ≤ 1

2k[f(x1) + · · ·+ f(x2k)],

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

α =α1

2+

α2

22+ · · ·+ αk

2k+ . . . ,

where αi ∈ {0, 1}, for all i ∈ N∗. Denoting

α(k) =α1

2+

α2

22+ · · ·+ αk

2k=

α12k−1 + · · ·+ αk

2k=

βk

2k,

it result

limk→∞

α(k) = α, 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

f(α(k)x + (1− α(k)y) =f(βkx + (2k − βk)

2ky)

≤βkf(x) + (2k − βk)f(y)

2k= α(k)f(x) + (1− α(k)f(y).

Now passing k → ∞ and taking into account the continuity of f, the conclusionfollows. 2

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

(1.2) f ′(x) = limt→x

φ(t)

provided this limit exists.We thus associate to the function f a function f ′ whose domain of definition is

the set of points x at which limit (1.2) exists; f ′ is called the derivative of f.If f ′ is defined at a point x, we say that f is differentiable on a point at x. If

f ′ 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(t− x) → f ′(x) · 0 = 0. 2

Remark. The converse is not true. Consider f(t) =| t |, t ∈ R and x = 0. 4It holds a stronger result, namely

Theorem 1.2. (Weierstrass) There exists a continuous function on R having nopoint of differentiability.

91

92 6. Differential calculus

Proof. Consider the function h(t) = |t|, t ∈ [−1, 1], and extend it by periodicity onR such that

g(t) =

h(t), t ∈ [−1, 1],

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

f(t) =∞∑

n=0

(3

4

)n

g(4nt).

Since 0 ≤ g ≤ 1, the series is uniformly convergent. We have uniform convergenceof continuous functions, so f is continuous on R.

Now we show that function f is nowhere differentiable. Choose an arbitrary, butfixed x ∈ R.

Define δm = ±124−m, where the sign is chosen such that between 4mx and 4m(x+

δm) there is no integer.Define

γn =g(4n(x + δm))− g(4nx)

δm

.

If n > m, 4nδm is an even number and thus γn = 0. If 0 ≤ n ≤ m, from (1.3) itfollows that |γn| ≤ 4n. Since |γm| = 4m, it follows the estimation∣∣∣∣f(x + δm)− f(x)

δm

∣∣∣∣ =

∣∣∣∣∣∞∑

n=0

(3

4

)n

γn

∣∣∣∣∣ ≥ 3m −m−1∑n=0

3n =1

2(3m + 1).

If m → ∞, δm → 0. Hence function f is not differentiable on x, and the theoremis 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 apoint x ∈ [a, b]. Then f + g, f · g and f/g are differentiable at x, and

(a) (f + g)′(x) = f ′(x) + g′(x);

(b) (fg)′(x) = f ′(x)g(x) + f(x)g′(x);

(c)

(f

g

)(x) =

f ′(x)g(x)− f(x)g′(x)

g2(x)(we assume that g(x) 6= 0).

6.1. The derivative of a real function 93

Proof. (a) is trivial.(b) Let h = f · g. Then

h(t)− h(x)

t− x=

f(t)g(t)− f(x)g(x)

t− x=

f(t)g(t)− f(x)g(t) + f(x)g(t)− f(x)g(x)

t− x

=g(t)f(t)− f(x)

t− x+ f(x)

g(t)− g(x)

t− x.

(c)(f

g

)(t)−

(f

g

)(x)

t− x=

f(t)g(x)− g(t)f(x)

(t− x)g(x)g(t)

=1

g(t)g(x)

[f(t)− f(x)

t− xg(x)− f(x)

g(t)− g(x)

t− x

]. 2

Theorem 1.4. Suppose f is continuous on [a, b], f ′(x) exists at some point x ∈[a, b], g is defined on a closed interval I that contains the range of f, and g isdifferentiable at the point f(x). If

h(t) = g(f(t)) a ≤ t ≤ b,

h is differentiable at x, and

(1.4) h′(x) = g′(f(x)) · f ′(x).

Proof. Let y = f(x). By the definition of the derivative, we have

f(t)− f(x) = (t− x)[f ′(x) + u(t)](1.5)

g(s)− g(y) = (s− y)[g′(y) + v(s)](1.6)

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′(y) + v(s)]

=(t− x)[f ′(x) + u(t)][g′(y) + v(s)]

or, if t 6= x,

(1.7)h(t)− h(x)

t− x= [g′(y) + v(s)][f ′(x) + u(t)].

Let t → x and see that s → y, by the continuity of f. Thus the right-hand side of(1.7) tends to g′(y)f ′(x), which is actually (1.4). 2


Examples 1.1. (a) Consider

f(x) =

{x sin 1

x, x 6= 0

0, x = 0.

Then f ′(x) = sin 1x− 1

xcos 1

x, x 6= 0 and does not exist f ′(0).

(b) Consider

f(x) =

{x2 sin 1

x, x 6= 0

0, x = 0.

Then f ′(x) = 2x sin 1x− cos 1

x, x 6= 0. Now f ′(0) = 0.

(c) The above cases are particular cases of the next one. Let a ≥ 0, c > 0 beconstants and f : [−1, 1] → R be a function defined by

f(x) =

{xa sin(x−c), x 6= 0

0, x = 0.

Then

(c 1 ) f is continuous ⇐⇒ a > 0;

(c 2 ) ∃ f ′(0) ⇐⇒ a > 1;

(c 3 ) f ′ is bounded ⇐⇒ a ≥ 1 + c;

(c 4 ) f ′ is continuous ⇐⇒ a > 1 + c;

(c 5 ) ∃ f ′′(0) ⇐⇒ a > 2 + c;

(c 6 ) f ′′ is bounded ⇐⇒ a ≥ 2 + 2c;

(c 7 ) f ′′ 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). Choosetwo sequences (αn)n and (βn)n satisfying −1 < αn < βn < 1 such that αn → 0 andβn → 0 as n →∞. Define

dn :=f(βn)− f(αn)

βn − αn

.

Then

(a) if αn < 0 < βn, limn→∞ dn = f ′(0);

(b) if 0 < αn and the sequence(

βn

βn−αn

)n

is bounded, limn→∞ dn = f ′(0);

6.1. The derivative of a real function 95

(c) if f ′ is continuous on ]− 1, 1[ , limn→∞ dn = f ′(0);

(d) there exists a differentiable function f on ]− 1, 1[ (with discontinuous deriva-tive) such that αn → 0, βn → 0, and ∃ limn→∞ dn, but limn→∞ dn 6= f ′(0).

Proof. (a) We write dn as a convex combination of the following form

f(βn)− f(αn)

βn − αn

=f(βn)− f(0)

βn

· βn

βn − αn

+f(0)− f(αn)

−αn

· −αn

βn − αn

,

λ1 =βn

βn − αn

, λ2 =−αn

βn − αn

, λ1, λ2 > 0, λ1 + λ2 = 1.

It follows

min

{f(βn)− f(0)

βn

,f(0)− f(αn)

−αn

}≤f(βn)− f(αn)

βn − αn

≤ max

{f(βn)− f(0)

βn

,f(0)− f(αn)

−αn

}.

The assumption guarantees us that

limn→∞

f(βn)− f(0)

βn

= limn→∞

f(0)− f(αn)

−αn

= f ′(0).

Now invoking theorem 1.9 from page 43 we get the conclusion.(b) Consider the following identity

f(βn)− f(αn)

βn − αn

− f(0)− f(αn)

−αn

=βn

βn − αn

(f(βn)− f(0)

βn

− f(0)− f(αn)

−αn

).

The sequence(

βn

βn−αn

)n

is bounded and since the difference from the parenthesis

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

nπ. 2

Theorem 1.5. Let I, J ⊂ R be compact intervals and f : I → J be a continuousand bijective mapping. Suppose function f is differentiable on a point x0 ∈ I andf ′(x0) 6= 0. Then the inverse function f−1 : J → I is differentiable at y0 = f(x0)and it holds

[f−1(y0)]′ =

1

f ′(x0).


Proof. Since function f is continuous and bijective and the interval I is compact itfollows 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)

y − y0

=1

f(x)− f(x0)

x− x0

, ∀x ∈ I \ {x0}.

From (1.5) it follows that y → y0 ⇐⇒ x → x0, hence

limy→y0

f−1(y)− f−1(y0)

y − y0

= limx→x0

1

f(x)− f(x0)

x− x0

=1

f ′(x0). 2

6.2 Mean value theorems

Let f be a real function defined on a metric space X. We say that f has a localmaximum at a point p ∈ X if there exists δ > 0 such that f(q) ≤ f(p) for allq ∈ B(p, δ). Analogously, we say that f has a local minimum at a point p ∈ X ifthere 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 pointx ∈ ]a, b[ and if there exists f ′(x), then f ′(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 abovedefinition, so that a < x− δ < x < x + δ < b. If x− δ < t < x,

f(t)− f(x)

t− x≥ 0.

Letting t → x, we see that f ′(x) ≥ 0.If x < t < x + δ,

f(t)− f(x)

t− x≤ 0.

Letting t → x, we see that f ′(x) ≤ 0. Hence f ′(x) = 0. 2

1Pierre de Fermat, 1601-1665

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] whichare differentiable in ]a, b[ , then there exists a point x ∈]a, b[ at which

[f(b)− f(a)]g′(x) = [g(b)− g(a)]f ′(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 h′(x) = 0 for some x ∈ ]a, b[ .If h is constant, h′(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[ , whileby 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 apoint 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 dif-ferentiable in ]a, b[ , then there exists a point x ∈ ]a, b[ at which

f(b)− f(a) = (b− a)f ′(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 ′(t) ≥ 0 for any t ∈ ]a, b[ , f is monotonically increasing on ]a, b[ ;

(b) if f ′(t) = 0 for any t ∈ ]a, b[ , f is constant on ]a, b[ ;

(c) If f ′(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′(x),

which is valid for each pair of numbers x1, x2 ∈ ]a, b[ , for some x between x1 andx2. 2


6.2.1 Consequences of the mean value theorems

Lemma 2.1. Consider I = [a, b[ , a < b ≤ +∞. Suppose f, g : I → R arecontinuous functions. Moreover,

(i) f and g are differentiable on ]a, b[ and f ′(t) ≤ g′(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 h′(t) ≥ 0 for anyt ∈ ]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 continuousfunctions. Moreover,

(i) f and g are differentiable on ]a, b[ ;

(ii) |f ′(t)| ≤ g′(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, h2 : I → R,

{h1(t) = g(t)− f(t)− (g(a)− f(a)),

h2(t) = f(t) + g(t)− (f(a) + g(a)).

We note that h1(a) = h2(a) = 0 and{h′1(t) = g′(t)− f ′(t) ≥ 0,

h′2(t) = f ′(t) + g′(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 acontinuous function on I, differentiable on I \ {t0}.


(a) If f ′ has a left hand side limit at t0, then φ in (1.1) has a left hand side limitat t0 and

limt→t0t<t0

f ′(t) = limt→t0t<t0

φ(t).

(b) If f ′ has a right hand side limit at t0, then φ in (1.1) has a right hand sidelimit at t0 and

limt→t0t>t0

f ′(t) = limt→t0t>t0

φ(t).

(c) If f ′ has a finite limit at t0, then f is differentiable at t0 and f ′ is continuousat t0.

Proof. (a) Let t < t0, t ∈ I. By the mean value theorem (2.3) there exists ct ∈ ]t, t0[such that

φ(t) =f(t)− f(t0)

t− t0= f ′(ct).

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 differentiablefunction on I. If f ′ is bounded on I, then f is Lipschitz on I.

Proof. Since f ′ is bounded on I, there is a positive L such that |f ′(t)| ≤ L, for anyt ∈ I. Then for any t1, t2 ∈ I, t1 < t2 there is c ∈ ]t1, t2[ such that

|f(t2)− f(t1)| = |f ′(c)(t2 − t1)| ≤ L(t2 − t1),

hence function f is Lipschitz on I. 2

Theorem 2.5. (Denjoy2-Bourbaki3theorem, [13, vol.2, p. 77]) Let I ⊂ R be anonempty 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 differentiableon ]a, b[ \A .

Then

inft∈ ]a,b[ \A

φ(t+) ≤ f(b)− f(a)

b− a≤ sup

t∈ ]a,b[ \Aφ(t+).

2Arnaud Denjoy, 1884-19743Nikolas Bourbaki, collective pseudonym of several French mathematicians, 1939 ↑


Proof. We prove that

f(b)− f(a)

b− a≤ sup

t∈ ]a,b[ \Aφ(t+) =: M.

The other side follows in a similar way.If M = +∞, the inequality is obvious. Suppose M < ∞ and consider the

functiong : I → R g(t) := Mt− f(t).

We remark that function g fulfills the assumptions of lemma 2.6. Thus g is increasingon 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 suchthat 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 ⊂ Isuch 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 theoremmay 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 δthere 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;


•• 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

(2.3)sn ∈ Sλ

f continuous at M

}=⇒ [λ ≤ f(sn) =⇒ λf(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, dueto the continuity of f,

(2.4) f(M) ≤ λ.

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 definedby (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)

h− φ(t+)

∣∣∣∣ < ε,

which means that

−ε < φ(t+)− ε <f(t + h)− f(t)

h< φ(t+) + ε

and sof(t + h)− f(t) + hε > 0, for every h ∈ ]0, δ[ , t ∈ I \ A,


that is

f(t + h) + (t + h)ε− [f(t) + tε] > 0, for every h ∈ ]0, δ[ , t ∈ I \ A.

Denotegε(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. 2

6.3 The continuity of derivatives

We have already seen

f(x) =

{x2 sin 1

x, x 6= 0

0, x = 0,

that a function f may have a derivative f ′ which exists at every point, but is dis-continuous 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 incommon with functions which are continuous on an interval: intermediate values areassumed.Theorem 3.1. (Darboux property of derivatives) Suppose f is a real differentiablefunction an [a, b] and suppose f ′(a) < λ < f ′(b). Then there is a point x ∈ ]a, b[such that f ′(x) = λ.

A similar result holds of course if f ′(a) > f ′(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 ′(a), as t → a, g(t) → f ′(b), as t → b,and so theorem 2.12 page 82 implies that g(t0) = λ for some t0 ∈ ]a, b[ . Fix t0. Bytheorem 2.3 page 97 there is a point x such that α(t0) < x < β(t0) and such thatf ′(x) = g(t0). Hence f ′(x) = λ. 2

Corollary 3.1. If function f is differentiable on [a, b], then f ′ cannot have anydiscontinuities 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′(x) 6= 0 for all x ∈ ]a, b[ , where −∞ ≤ a < b ≤ +∞. Suppose

(4.1)f ′(x)

g′(x)→ A 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)

g(x)→ A as x → a.

Proof. We first consider the case when −∞ ≤ A < +∞. Choose a real number qsuch that A < q, and the choose r such that A < r < q. By (4.1) there is a pointc ∈ ]a, b[ such that a < x < c implies

(4.5)f ′(x)

g′(x)< r.

If a < x < y < c, then Cauchy theorem 2.2 shows that there is a point t ∈ ]x, y[ suchthat

(4.6)f(x)− f(y)

g(x)− g(y)=

f ′(t)

g′(t)< r.

Suppose (4.2) holds. Letting x → a in (4.6) we see that

(4.7)f(y)

g(y)≤ r < q (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)f(x)

g(x)< r − r

g(y)

g(x)+

f(y)

g(x), (a < x < c1).

4Guillaume de L’Hospital, marquis de Sainte-Mesme, 1661-1704


If we let x → a in (4.8), (4.3) shows that there is a point c2 ∈ ]a, c1[ such that

(4.9)f(x)

g(x)< q (a < x < c2).

Summing up, (4.7) and (4.9) show that for any q , subject only to the condition

A < q, there is a point c2 such that f(x)g(x)

< r if a < x < c2.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. 2

6.5 Higher order derivatives

Theorem 5.1. (Leibniz formula) Let u and v be two functions having derivativesup to the n -order on an interval. Then

(uv)(n) = unv +

(1

n

)u(n−1)v′ + · · ·+

(n− 1

n

)u′v(n−1) + uv(n).

Proof. By induction. 2

Proposition 5.1. The following identities hold on R

(ex)(n) = ex, sin(n)(x) = sin(x + n

π

2

), cos(n)(x) = cos

(x + n

π

2

).

6.6 Convex functions and differentiability

Theorem 6.1. Suppose f : I → R is convex, a = inf I, b = sup I, I being aninterval. Then

(a) f has side derivatives on ]a, b[ and for any t1, t2 ∈ ]a, b[ , t1 < t2, we have

f ′l (t1) ≤ f ′r(t1) ≤ f ′l (t2) ≤ f ′r(t2);

(b) if a ∈ I (b ∈ I), then f is right-hand differentiable in a ( respectively it isleft-hand differentiable in b) and

f ′r(a) ≤ f ′l (t) (respectively f ′r(t) ≤ f ′l (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

f ′l (t1) = sf,t1(t1−) ≤ sf,t1(t1+) = f ′r(t1) ≤ sf,t2(t2−) = f ′l (t2) ≤ sf,t2(t2+) = f ′r(t2).

(b) If a ∈ I , we repeat the previous proof for a < t1 < u.(c) From (a) it follows that f ′l (t) is increasing on ]a, b[. Hence there is an at mostcountable set A such that f ′l (t) is continuous on ]a, [ \A.

Choose t0 ∈ ]a, b[ \A. Then f ′l (·) is continuous on t0, and for t > t0

f ′l (t0) ≤ f ′r(t0) ≤ f ′l (t).

Letting t → t0, we get f ′l (t0) = f ′r(t0), that is f is differentiable on t0. Hence f isdifferentiable 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

f : [0, 1] → R, f(t) =

{0, t ∈ ]0, 1[

1, t ∈ {0, 1}.4

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 ∈ [f ′l (t0), f′r(t0)].

Proof. If t > t0,

sf,t0(t) =f(t)− f(t0)

t− t0≥ inf

s∈Is>t0

sf,s(s) = f ′r(t0) ≥ m.

Similarly, if t < t0,

sf,t0(t) =f(t)− f(t0)

t− t0≤ sup

s∈Is<t0

sf,s(s) = f ′l (t0) ≤ m.

Now the conclusion follows. 2


Corollary 6.3. Suppose f : I → R is differentiable on I. Then f is convex if andonly if

(6.1) f(t) ≥ f(t0) + f ′(t0)(t− t0), 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 ′((1− α)a + αb)α(a− b),

f(b) ≥f((1− α)a + αb)− f ′((1− α)a + αb)(1− α)(a− b).

Multiply the first inequality by (1 − α), the second by α, and then sum them. Itresults 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 convexand differentiable on I. Then for any point t0 ∈ int (I) the following statements areequivalent

(a) (t0, f(t0)) is the global minimum of f ;

(b) (t0, f(t0)) is a local minimum of f ;

(c) f ′(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 ofI. 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 anycompact [a, b] in I.

Proof. Consider an arbitrary subinterval [a, b] ⊂ I and u, v ∈ [a, b], u < v. DenoteM := max{|f ′r(a)|, |f ′l (b)|}. Then

−M ≤ f ′r(a) ≤ f ′r(u) ≤ f(u)− f(v)

u− v≤ f ′l (v) ≤ f ′l (b) ≤ M.

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 andonly 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)

a− b>

f(c)− f(b)

c− b.

By the Lagrange mean value theorem it follows that there are t1 ∈ ]a, b[ and t2 ∈ ]b, c[with f ′(t1) > f ′(t2). But this contradicts that the first derivative is increasing. 2

Corollary 6.6. A differentiable function f : I → R is concave (strictly concave) ifand 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 derivativeon I. Then f is convex (concave) if and only if f ′′ ≥ 0 (respectively, f ′′ ≤ 0 ).

Proof. The claim follows from the remark that f ′ is increasing (decreasing) if andonly if f ′′ is positive (negative). 2

Corollary 6.8. Let f : I → R be a function with second order derivative on I. Thenf is strictly convex (strictly concave) if and only if f ′′ > 0 (respectively, f ′′ < 0 ).

6.6.1 Inequalities

Proposition 6.1. (Young generalized inequality) Take n ∈ N∗, yk > 0, pk > 0, fork = 1, . . . , n, and

∑nk=1 1/pk = 1. Then

(6.2)n∏

k=1

yk ≤n∑

k=1

1

pk

ypk

k .

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 ′′(x) > 0, for everyx ∈ R, we infer that function f is convex. Based on Jensen inequality, (2.18) at thepage 89, taking αk = pk and xk = ln ypk

k we may write

exp

(n∑

k=1

1

pk

ln ypk

k

)≤

n∑k=1

1

pk

exp (ln ypk

k ) .

But

exp

(n∑

k=1

1

pk

ln ypk

k

)= exp

(n∑

k=1

ln yk

),

n∑k=1

1

pk

exp (ln ypk

k ) =n∑

k=1

1

pk

ypk

k ,

and the generalized Young inequality follows. 2


Proposition 6.2. (Generalized mean inequality) Take n ∈ N∗, xi > 0, and αi ≥ 0,i = 1, . . . , n satisfying α1 + · · ·+ αn = 1. Then

(6.3) xα11 xα2

2 . . . xαnn ≤ α1x1 + . . . αnxn.

Proof. Consider the function f(x) = ln x, x > 0. Since f ′′(x) < 0, for every x > 0,we infer that function f is (strictly) concave. So

ln

(n∑

i=1

αixi

)≥

n∑i=1

αi ln xi,

and (6.3) follows. 2

Corollary 6.9. Taking α1 = · · · = αn = 1/n in (6.3) it follows

(6.4) n√

x1 . . . xn ≤x1 + x2 + . . . xn

n.

The genuine Cauchy’s proof of the above inequality is given at page 24 and it isavailable 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

x1

+ · · ·+ αn

xn

)−1

≤ xα11 xα2

2 . . . xαnn .

Corollary 6.11. Taking α1 = · · · = αn = 1/n in (6.5) it follows

(6.6)n

1x1

+ · · ·+ 1xn

≤ n√

x1 . . . xn.

Chapter 7

Integral calculus

The aim of the present chapter is to introduce some basic results on integralcalculus.

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 isthe following one.Lemma 2.1. Suppose a continuous function x : [a, b] → R satisfies

(2.1) 0 ≤ x(t) ≤ c +

∫ t

a

h(s)x(s)ds, t ∈ [a, b]

for some constant c and some nonnegative integrable function h : [a, b] → R. Then

(2.2) 0 ≤ x(t) ≤ c + c

∫ t

a

h(r) exp(

∫ t

r

h(s)ds)dr, t ∈ [a, b].

Proof. Denote

w(t) := c +

∫ t

a

h(s)x(s)ds.

Then

w′(t) = h(t)x(t), w(t) > 0, w(a) = c.

109

110 7. Integral calculus

From (2.1) it follows that x(t) ≤ w(t), for any t ∈ [a, b]. Thus it holds the followingsequence of implications

w′(t) = h(t)x(t) ≤ h(t)w(t) =⇒ w′(t)

w(t)≤ h(t) =⇒∫ t

a

w′(s)

w(s)ds ≤

∫ t

a

h(s)ds =⇒ ln w(t) ≤∫ t

a

h(s)ds + ln c =⇒

w(t) ≤ c · exp(

∫ t

a

h(s)ds =⇒ x(t) ≤ c · exp(

∫ t

a

h(s)ds.

Substituting in (2.1), we get (2.2). 2

Bibliography

[1] M. BALAZS and I. KOLUMBAN, Matematikai analizis, Dacia, Cluj-Napoca,1978 (Hungarian).

[2] D. M. BATINETU, Siruri (Romanian).

[3] W. W. BRECKNER, Analiza matematica. Topologia spatiului Rn , Universitateadin Cluj-Napoca, Cluj-Napoca, 1985 (Romanian).

[4] W.-S. CHEUNG, Generalizations of Holder’s inequality, Internat. J. Math. &Math. Sci. 26 (2001), no. 1, 7–10.

[5] S. COBZAS, Analiza matematica, Presa Universitara Clujeana, Cluj-Napoca,1997 (Romanian).

[6] D. I. DUCA and E. DUCA, Culegere de probleme de analiza matematica, 1,2,GIL, Zalau, 1997 (Romanian).

[7] R. ENGELKING, General Topology, Monografie Matematyczne, PWN,Warszawa, 1977.

[8] P. R. HALMOS, Naive Set Theory, Van Nostrand, Princeton, New Jersey, 1967.

[9] E. HEWITT and K. STROMBERG, Real and Abstract Analysis, Springer-Verlag,New York, 1975, A modern treatment of the theory of functions of a real variable,Third printing, Graduate Texts in Mathematics, No. 25.

[10] T. J. JECH, Lectures on Set Theory with Particular Emphasis on the Method ofForcing, Lecture Notes on Mathematics, vol. 217, Springer, Berlin, 1971.

[11] D. E. KNUTH, Seminumerical algorithms, second ed., The Art of ComputerProgramming, vol. 1-7, Addison-Wesley, Reading, Massachusetts, 1973.

[12] I. MARUSCIAC, Analiza matematica I, Universitatea Babes-Bolyai, Cluj-Napoca, 1980 (Romanian).

[13] M. MEGAN, Bazele analizei matematice, Matuniv, vol. 4, Eurobit, Timisoara,1997 (Romanian).

[14] , Analiza matematica, vol. 1, Mirton, Timisoara, 1999 (Romanian).

111

[15] , Analiza matematica, vol. 2, Mirton, Timisoara, 1999 (Romanian).

[16] M. MEGAN, A. L. SASU, and B. SASU, Calcul diferential ın R, prin exercitiisi probleme, Editura Universitatii de Vest, Timisoara, 2001 (Romanian).

[17] C. MEGHEA, Foundation of Mathematical Analysis. Tretise of Analysis, Ed.Stiintifica si Enciclopedica, Bucuresti, 1977 (Romanian).

[18] M. MURESAN, Introducere in Control Optimal, Risoprint, Cluj-Napoca, 1999(Romanian).

[19] , Introduction to Set-Valued Analysis, Cluj University Press, Cluj-Napoca,1999.

[20] , Analiza Neneteda si Aplicatii, Risoprint, Cluj-Napoca, 2001 (Romanian).

[21] G. POLYA and G. SZEGO, Problems and Theorems in Analysis I, Springer-Verlag, Berlin, 1972.

[22] T. POPOVICIU, Numerical Analysis. Basic Notions of Approximative Calculus,Calculus Theory, Numerical Analysis, and Computer Science, vol. 1, Acad. R. S.R., Bucharest, 1975 (Romanian).

[23] W. RUDIN, Principles of Mathematical Analysis, McGraw-Hill, New York, 1976.

[24] V. A. SADOVNICHII, A. A. GRIGORIAN, and S. B. KONIAGIN, Problems fromStudents’s Mathematics Olympiads, Ed. Moscow University, 1987 (Russian).

[25] G. SIRETCHI, Calcul diferential si integral, I, II, EDP, Bucuresti, 1985 (Roma-nian).

112

Author index

Abel, N. H., 64Archimedes, 13, 14, 17

Banach, S., 87Bernoulli, , 24, 25Borel, E., 36Bourbaki, N., 101

Cantor, G., 1, 37, 38, 42, 81Catalan, E. Ch., 1814-1894, 55Cauchy, A., 80, 98Cauchy, A. L., 25, 42, 43, 58, 60, 62,

64, 65, 68Cesaro, ., 48, 54

D’Alembert, J. le R., 62Darboux, , 82, 84Denjoy, , 101Dirichlet, P. G. L., 65

Euler, L., 54

Fermat, P., 98, 108

Holder, O., 20, 22, 23Hadamard, J., 64Heine, , 36

Jensen, , 89, 90, 109

Lagrange, J. J., 99Lagrange, J. L., 27, 54Lalescu, T., 55Leibnitz, G., 106Lipschitz, , 86

Minkowski, H., 24Morgan, de, , 4

Newton, I., 50

Riemann, B., 69

Sierpinski, W., 86

Stoltz, ., 48, 54

Weierstrass, K., 36, 93

Young, , 109Young, W. H., 20, 21

113

Subject index

addition, 10application, 7

ball, 30closed, 37open, 30

cardinal number, 17components, 79constant

of Euler, 54contraction, 86coordinate, 71covering, 34

open, 34

diameter, 42discontinuity

first kind, 82second kind, 82simple, 82

distance, 29Euclidean, 29

on C , 30

element, 1bound

greatest lower, 6least upper, 6lower, 6upper, 6

identity, 10infimum, 6inverse, 10null, 10smallest, 6supremum, 6unity, 10zero, 10

equationfunctional

Cauchy, 80

familypairwise disjoint, 3

fixed point, 86formula

of Leibnitz, 106function, 7

absolute value, 12antiderivative, 110bijective, 7continuous at a point, 77continuous on a set, 77convex, 88

Jensen, 90stricly, 88

Darboux, 84derivative, 93differentiable, 93

on a set, 93discontinuous, 82distance, 12fractional part, 16from, 8injective, 8integer part, 16Lipschitz, 86monotonic, 83monotonically decreasing, 83monotonically increasing, 83one-to-one, 8onto, 8primitive, 110slope, 90

identityLagrange, 27

inequalityBernoulli, 24, 25generalized

mean, 110

115

Jensen, 89mean, 25of Young, 20triangle, 29

infimum, 6inner product, 72integral

Darbouxlower, 113

interval, 15bounded, 15length, 15unbounded, 15

limit point, 30linear space, 71

mapping, 7maximum

local, 98mean

weighted, 21member, 1metric, 29

Euclidean, 29Euclidean on R2 , 30uniform on R2 , 30

minimumlocal, 98

multiplication, 10

neighborhood, 30norm, 73

Euclidean, 72l1 − norm , 72lp -norm, 72Minkowski, 72uniform, 72

null vector, 71number

fractional part, 16integer part, 16

operation, 7

ordered pair, 4origin, 71

partial summation formula, 64partition, 112point, 29, 71

interior, 30isolated, 30

radius of convergence, 64relation, 4

composition, 5domain, 5equivalence, 5image, 7inverse, 5inverse image, 7ordering

partial, 5total, 6well, 6

product, 5range, 5single-valued, 7

sequence, 8bounded, 39Cauchy, 42convergent, 39divergent, 39fundamental, 42in, 8limit, 39

lower, 47upper, 47

monotonic, 43monotonically

decreasing, 43increasing, 43

of successive approximations, 87speed of convergence, 87term, 8

seriesabsolutely convergent, 66

116

conditionally convergent, 67convergent, 58divergent, 58power, 63

coefficient, 63radius of convergence, 64sum, 58

set, 1at most countable, 17bounded, 6, 30

above, 6below, 6

Cantor, 38closed, 30compact, 34connected, 82countable, 17dense, 30denumerable, 17empty, 1finite, 17infinite, 17open, 30operation

Cartesian product, 4intersection, 2symmetric difference, 4union, 2

orderedpartially, 6totally, 6

perfect, 30uncountable, 17void, 1well-ordered, 6

setsdifference, 3disjoint, 3

spacecompact, 34complete, 43metric, 29

distance, 29

separable, 34triangle inequality, 29

topological, 30subset, 2

proper, 2sum

Darbouxlower, 113

supremum, 6

theoremStolz-Cesaro, 48

topology, 30transformation, 7

uniform continuous mapping, 80

vector, 71vector space, 71

weightof a mean, 21

117

marian mure¸san mathematical analysis and applications i draftchapter 1 sets the aim of this...

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