Intro Avram

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Lecture NotesBasic Concepts of AnalysisMATH 372

Instructor: Ivan Avramidi

Textbook: R. G. Bartle and D. R. Sherbert, (John Wiley, 2000)

New Mexico Institute of Mining and Technology

Socorro, NM 87801

June 3, 2004

Author: Ivan Avramidi; File: introanal.tex; Date: November 21, 2005; Time: 18:02


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Contents

1 Preliminaries 1

1.1 Sets and Functions . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 Set Operations . . . . . . . . . . . . . . . . . . . . . . . 21.1.2 Cartesian Products . . . . . . . . . . . . . . . . . . . . . 31.1.3 Mappings and Functions . . . . . . . . . . . . . . . . . . 31.1.4 Direct and Inverse Images . . . . . . . . . . . . . . . . . 41.1.5 Special Types of Functions . . . . . . . . . . . . . . . . 41.1.6 Inverse Functions . . . . . . . . . . . . . . . . . . . . . 41.1.7 Composition of Functions . . . . . . . . . . . . . . . . . 51.1.8 Restrictions of Functions . . . . . . . . . . . . . . . . . 5

1.2 Finite and Innite Sets . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 The Real Numbers 92.1 The Algebraic and Order Properties of R . . . . . . . . . . . . . . 9

2.1.1 Rational and Irrational Numbers . . . . . . . . . . . . . . 112.1.2 The order Properties of R . . . . . . . . . . . . . . . . . 122.1.3 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Absolute Value and Real Line . . . . . . . . . . . . . . . . . . . 152.2.1 The Real Line . . . . . . . . . . . . . . . . . . . . . . . 162.2.2 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3 The Completeness Property of R . . . . . . . . . . . . . . . . . . 172.3.1 Suprema and Inma . . . . . . . . . . . . . . . . . . . . 172.3.2 The Completeness Property of R . . . . . . . . . . . . . 192.3.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.4 Applications of Supremum Property . . . . . . . . . . . . . . . . 202.4.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 20

I


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II CONTENTS

2.4.2 The Archimedean Property . . . . . . . . . . . . . . . . 21

2.4.3 The Existence of 2 . . . . . . . . . . . . . . . . . . . . 222.4.4 Density of Rational Numbers in R . . . . . . . . . . . . . 232.4.5 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.5.1 Characterization of Intervals . . . . . . . . . . . . . . . . 242.5.2 Nested Intervals . . . . . . . . . . . . . . . . . . . . . . 252.5.3 The Uncountability of R . . . . . . . . . . . . . . . . . . 262.5.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Sequences and Series 293.1 Sequences and Their Limits . . . . . . . . . . . . . . . . . . . . . 29

3.1.1 The Limit of a Sequence . . . . . . . . . . . . . . . . . . 303.1.2 Tails of Sequences . . . . . . . . . . . . . . . . . . . . . 313.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 323.1.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2.2 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.3 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . . 383.3.1 Sequence dened recursively . . . . . . . . . . . . . . . . 393.3.2 Limit Superior and Limit Inferior . . . . . . . . . . . . . 403.3.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4 Subsequences and the Bolzano-Weierstrass Theorem . . . . . . . 423.4.1 The Existence of Monotone Subsequences . . . . . . . . 433.4.2 The Bolzano-Weierstrass Theorem . . . . . . . . . . . . 443.4.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.5 The Cauchy Criterion . . . . . . . . . . . . . . . . . . . . . . . . 463.5.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.6 Properly Divergent Sequences . . . . . . . . . . . . . . . . . . . 493.7 Introduction to Series . . . . . . . . . . . . . . . . . . . . . . . . 50

3.7.1 Comparison Tests . . . . . . . . . . . . . . . . . . . . . 50

4 Limits 514.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.1.1 The Denition of the Limit . . . . . . . . . . . . . . . . 524.1.2 Sequential Criterion for Limits . . . . . . . . . . . . . . 524.1.3 Divergence Criteria . . . . . . . . . . . . . . . . . . . . 53

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CONTENTS III

4.1.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2 Limits Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 56

5 Continuous Functions 575.1 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 57

5.1.1 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 595.2 Combinations of Continuous Functions . . . . . . . . . . . . . . 60

5.2.1 Composition of Continuous Functions . . . . . . . . . . 615.2.2 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.3 Continuous Functions on Intervals . . . . . . . . . . . . . . . . . 625.3.1 The Maximum-Minimum Theorem . . . . . . . . . . . . 635.3.2 Bisection Method . . . . . . . . . . . . . . . . . . . . . . 645.3.3 Bolzanos Theorem . . . . . . . . . . . . . . . . . . . . 645.3.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.4 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . 675.4.1 Lipshitz Functions . . . . . . . . . . . . . . . . . . . . . 685.4.2 The Continuous Extension Theorem . . . . . . . . . . . 695.4.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.5 Continuity and Gauges . . . . . . . . . . . . . . . . . . . . . . . 715.5.1 Existence of -ne Partitions . . . . . . . . . . . . . . . 71

5.6 Monotone and Inverse Functions . . . . . . . . . . . . . . . . . . 725.6.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . 755.6.2 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 76

6 Diff erentiation 776.1 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.1.1 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . 806.1.2 Inverse Functions . . . . . . . . . . . . . . . . . . . . . 816.1.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 82

6.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . 836.2.1 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 87

6.2.2 Intermediate Value Property of Derivatives . . . . . . . . 876.2.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 89

6.3 Taylors Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 906.3.1 Relative Extrema . . . . . . . . . . . . . . . . . . . . . . 916.3.2 Convex Functions . . . . . . . . . . . . . . . . . . . . . 926.3.3 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 93



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IV CONTENTS

7 The Riemann Integral 95

7.1 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . 957.1.1 Partitions and Tagged Partitions . . . . . . . . . . . . . . 957.1.2 Denition of Riemann Integral . . . . . . . . . . . . . . 967.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 977.1.4 Properties of the Integral . . . . . . . . . . . . . . . . . . 977.1.5 Boundedness Theorem . . . . . . . . . . . . . . . . . . . 987.1.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.2 Riemann Integrable Functions . . . . . . . . . . . . . . . . . . . 997.2.1 The Squeeze Theorem . . . . . . . . . . . . . . . . . . . 1007.2.2 Classes of Riemann Integrable Functions . . . . . . . . . 101

7.2.3 The Additivity Theorem . . . . . . . . . . . . . . . . . . 1027.2.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 1037.3 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . 104

7.3.1 The Fundamental Theorem (First Form) . . . . . . . . . 1047.3.2 The Fundamental Theorem (Second Form) . . . . . . . . 1057.3.3 Substitution Theorem . . . . . . . . . . . . . . . . . . . 1077.3.4 Lebesgues Integrability Criterion . . . . . . . . . . . . . 1077.3.5 Integration by Parts . . . . . . . . . . . . . . . . . . . . 1107.3.6 Homework . . . . . . . . . . . . . . . . . . . . . . . . . 111

8 Topology of Real Numbers 113

8.1 Open and Closed Sets in R . . . . . . . . . . . . . . . . . . . . . 1138.1.1 The Characterization of Closed Sets . . . . . . . . . . . . 1158.1.2 The Characterization of Open Sets . . . . . . . . . . . . 115

8.2 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1158.3 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 116

8.3.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . 1168.3.2 Preservation of Compactness . . . . . . . . . . . . . . . 116

8.4 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1168.4.1 Metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . 1168.4.2 Neighborhoods and Convergence . . . . . . . . . . . . . 117

8.4.3 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . 1178.4.4 Open Sets and Continuity . . . . . . . . . . . . . . . . . 118

Bibliography 118

Notation 121



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Chapter 1

Preliminaries

1.1 Sets and Functions

Reading : Appendix A (Logic and Proofs).

Set .

Elements (members).

Subset .

Proper subset .

Denition 1.1.1 Two sets A and B are equal if they contain the sameelements. Alternatively, A = B if A B and B A.

Description of sets.

Empty set .

Universal set .

Examples.1. Natural numbers N

2. Integer numbers Z

3. Rational numbers Q

1


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2 CHAPTER 1. PRELIMINARIES

4. Real numbers R

5. Complex numbers C

1.1.1 Set Operations

Denition 1.1.2 The union of the sets A and B is the set

A B = { x | x A or x B}The intersection of the sets A and B is the set

A

B =

{ x

| x

A and x

B

}The complement of the set B relative to the set A is the set A \ B = { x | x A and x B}

The sets A and B are disjoint if they have no common elements, that is

A B =

Theorem 1.1.1 De Morgans Laws

A \ ( B C ) = ( A \ B) ( A \ C ) A \ ( B C ) = ( A \ B) ( A \ C )

Proof: Diagrams.

Innite collection of sets .{ Ak }k = 1

Union of a collection of setsk = 1

Ak = { x | x Ak for some k N}

Intersection of a collection of setsk = 1

Ak = { x | x Ak for all k N}

Examples .



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1.1. SETS AND FUNCTIONS 3

1.1.2 Cartesian Products

Denition 1.1.3 The Cartesian product of two nonempty sets A and B is the set of all ordered pairs (a , b) with a A and b B, that is

A B = {(a , b) | a A, b B} Example .

1.1.3 Mappings and Functions

Function as a mapping . f : A B, f maps A into B

a f

b, image of a under f

b = f (a ), value of f at a

Denition 1.1.4 A function from a set A to a set B is a rule that assigns to each element of the set A an element of the set B.

The set A is the domain of the function f , denoted by D ( f ) , and the set B is the codomain of the function f .

The range of the function f is the subset R ( f ) of the codomain B suchthat for each b R( f ) there exists a A such that b = f (a ).

Remark. The range of a function is a subset of the codomain: D( f ) = A, R( f ) B

Diagrams.

Function as a relation .

Denition 1.1.5 Afunction from a set A to a set B is a set f of ordered pairs (a , b) in A B such that for each a A there exists a unique b Bwith (a , b) f . That is, if (a , b) f and (a , b ) f , then b = b .



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1.1.4 Direct and Inverse Images

Denition 1.1.6 Let f : A B be a function. The direct image of asubset E A under f is the subset of B

f ( E ) = { f ( x) | x E }The inverse image of a subset H B under f is the subset of A

f 1( H ) = { x A | f ( x) H }.

Remark. The range of a function is the image of the domain: R( f ) = f ( A)

Examples . Diagrams.

1.1.5 Special Types of Functions

Denition 1.1.7 Let f : A B be a function. f is injective (one-to-one ) if whenever x 1 x2 then f ( x1) f ( x2).

f is surjective (maps A onto B) if f ( A) = B, or R( f ) = B.That is, if for any b B there exists a A such that f (a ) = b. f is bijective if it is both injective and surjective ( one-to-one and onto ).

Example.

1.1.6 Inverse Functions

Denition 1.1.8 Let f : A B be a bijection of A onto B. Theinverse function f 1 : B A is dened by f 1 = {(b, a ) B A | (a , b) f }

That is, for any b B, f 1(b) = a where a A is such that f (a ) = b.

Remarks . D( f ) = R( f 1), R( f ) = D( f 1) .



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1.1. SETS AND FUNCTIONS 5

1.1.7 Composition of Functions

Denition 1.1.9 Let f : A B and g : B C be two functions.Then the composite function g f : A C is the function from A intoC dened by

(g f )( x) = g( f ( x)), x A.

Remarks.g f f g

D(g f ) = D( f )

Examples

Theorem 1.1.2 Let f : A B and g : B C. If H C, then

(g f )1( H ) = f 1(g1( H )) . Proof: Exercise!

1.1.8 Restrictions of Functions

Let f : A

B and A1

A. The restriction of f to A1 is the function

f 1 : A1 B dened by f 1( x) = f ( x), x A1 .

Notation f 1 = f | A1



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1.2 Finite and Innite Sets

Reading: Proofs are in Appendix B.

Denition 1.2.1 Empty set has 0 elements.

A set S has n elements if there is a bijection f : Nn S from Nn ={1, 2, . . . , n} onto S . A set S is nite if it is either empty or has n elements for some n N . A set S is innite if it is not nite.

Theorem 1.2.1 Uniqueness Theorem . The number of elements of a nite set is unique.

Theorem 1.2.2 The set N is innite.

Theorem 1.2.3 1. Let A and B be disjoint sets. If A has m elementsand B has n elements, then the set A B has n + m elements.

2. Let C A. If A has m elements and C has 1 element, then A \C has m 1 elements.

3. If A is an innite set and B is a nite set, then the set A \ B isinnite.

Theorem 1.2.4 Let B A. Then:1. If A is nite, then B is nite.

2. If B is innite, then A is innite.

1.2.1 Countable Sets

Denition 1.2.2 A set S is denumerable (countably innite ) if thereis a bijection f : N S . A set S is countable if it is either nite or countably innite.

A set S is uncountable if it is not countable.



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1.2. FINITE AND INFINITE SETS 7

Examples . Remark. The union of countably innite sets is countably innite. Theorem 1.2.5 The set N N is countably innite.

Proof: Diagonal argument.

Theorem 1.2.6 Let B A. Then:1. If A is countable, then B is countable.

2. If B is uncountable, then A is uncountable.

Theorem 1.2.7 The following statements are equivalent:

1. S is a countable set.

2. There is a surjection g : N S .3. There is an injection f : S N .

Theorem 1.2.8 The set Q is countably innite. Proof:

1. Surjection g : N

N

Q + .

2. Bijection f : N N N .3. Surjection g f : N Q + .

Theorem 1.2.9 The union of a countable collection of countable sets is

countable. Proof: Diagonal argument.

Theorem 1.2.10 Cantors Theorem . Let A be a set and P( A) be the

set of all subsets of A. There is no surjection f : A

P( A).

1.2.2 Homework

Reading: Appendix A, B. Sec. 1.1-1.3. Exercises: 1.1[19], 1.3[1,5,7,9].



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Chapter 2

The Real Numbers

2.1 The Algebraic and Order Properties of R

R is a eld with respect to addition and multiplication.

Binary operation

B : R R R , (a , b) B(a , b)

Denition 2.1.1 Algebraic properties of R. There are two binaryoperations on R: addition, + , and multiplication, , satisfying the fol-lowing properties: a , b, c R

1. a + b = b + a

2. (a + b) + c = a + (b + c)

3. 0 R such that 0 + a = a

4. a R (a ) R such that a + (a ) = 0

5. ab = ba

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10 CHAPTER 2. THE REAL NUMBERS

Denition 2.1.2

1. (ab )c = a (bc )

2. 1 R , such that 1 0 and 1 a = a

3. a R , a 0 , a1 R such that aa 1 = 1

4. a (b + c) = ab + ac

Theorem 2.1.1 1. Let x, a R and x + a = a. Then x = 0.2. Let y, b R and yb = b. Then y = 1.3. For any a R , 0a = 0.

Proof:

1. x = x + 0 = x + (a + (a )) = ( x + a ) + (a ) = a + (a ) = 0.2. 0a + a = 0a + 1a = (0 + 1)a = 1a = a . So 0a = 0.

Theorem 2.1.2 1. Let a , b R , a 0 and ab = 1. Then b = a1 .

2. Let a , b R and ab = 0. Then either a = 0 or b = 0. Proof:

1. b = 1b = (aa 1)b = a1(ab ) = a1

2. If a 0 and b 0, then 1 = b

1a

1ab = b

1a

10 = 0. Contradiction.

Subtraction is dened by: a , b Ra b = a + (b)



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2.1. THE ALGEBRAIC AND ORDER PROPERTIES OF R 11

Division is dened by: a , b R , b 0,ab

= ab 1

2.1.1 Rational and Irrational Numbers

Identify N = Z + with the subset of R byn = 1 + + 1

n Dene Z as a subset of R as follows. Identify the zeros

0Z = 0R

Dene the negative integers as

n = (1) + + (1)

nor as the opposite to n. Then

Z = Z + {0} Z

Dene the rational numbers Q as the subset of R by

Q = r = ab

a , b Z , b 0

The set of irrational numbers is R \ Q

Theorem 2.1.3 There does not exist a rational number r Q such that

r 2 = 2. Proof: By contradiction.

1. Let r = p/ q and r 2 = 2.

2. Then p2

= 2q2.

3. So, p is even. So, p = 2m for some m N .4. Thus, q is odd.

5. q2 = 2m2 . So, q is even. Contradiction.



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2.1.2 The order Properties of R

Denition 2.1.3 The Order Properties of R. There is a nonemptysubset R + R , the set of positive real numbers , that satises the prop-erties: a , b R +

1. a + b R + ,2. ab R + ,3. a R + exactly one of the following holds:

a

R + , (

a )

R + , a = 0 .

The set of negative real numbersR

= {a | a R + } Trichotomy Property . R is a disjoint union of the following sets

R = R {0} R +that is

0 R + , 0 R, R R + = Positive real numbers a R + , a > 0

Non-negative real numbers

a R + {0}, a 0Negative real numbers

a R, a < 0Non-positive real numbers

a R {0}, a 0 .

Denition 2.1.4 Let a, b R .1. If a b > 0 , then a > b (or b < a).2. If a b 0 , then a b (or b a)



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2.1. THE ALGEBRAIC AND ORDER PROPERTIES OF R 13

Trichotomy Property . For any a, b R exactly one of the following is truea > b, a = b, a < b

Corollary .If a b and a b, then a = b.

Theorem 2.1.4 Let a , b, c R .1. If a > b and b > c, then a > c.

2. If a > b, then a + c > b + c.

3. If a > b and c > 0 , then ac > bc.

4. If a > b and c < 0 , then ac < bc.

Proof:

1. Since a b > 0 and bc > 0, then ( a b) + (bc) > 0. Thus, a c > 0and a > c.

Theorem 2.1.5 1. Let a

R , a 0. Then a 2 > 0.

2. 1 > 0.

3. For any n N , n > 0. Proof:

1. We have a > 0 or a < 0. If a > 0, then a2 > 0. If a < 0, then (a ) > 0and a2 = (a )(a ) > 0.

2. 1 = 12 . So 1 > 0.

3. n = 1 +

+ 1. So, n > 0.

Remark . There is no smallest positive real number. Theorem 2.1.6 Let a R . If for any > 0 , 0 a < , then a = 0.

Proof: By contradiction.



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1. Suppose a > 0. Let = a / 2. Get contradiction.

Theorem 2.1.7 Let a, b R and ab > 0. Then either

1. a > 0 and b > 0 , or

2. a < 0 and b < 0.

Proof:

1. We have a 0 and b 0.

2. If a > 0, then a1 > 0.3. Then b = a1(ab ) > 0.

Corollary 2.1.1 Let a , b R and ab < 0. Then either

1. a > 0 and b < 0 , or

2. a < 0 and b > 0.

2.1.3 Inequalities

Examples.



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2.2. ABSOLUTE VALUE AND REAL LINE 15

2.2 Absolute Value and Real Line

Denition 2.2.1 The absolute value of a real number a is dened by

|a| =a , if a > 00, if a = 0

a , if a < 0

Theorem 2.2.1 Properties the Absolute Value . Let a, b, c R and c 0. Then

1. |ab | = |a||b|2. |a|2 = a 23. |a| c if and only if c a c4. a |a| |a|

Proof: Exercise.

Theorem 2.2.2 Triangle Inequality. Let a, b R . Then

|a + b| |a|+ |b| Proof: Easy.

Corollary 2.2.1 Let a, b R . Then1. ||a| |b|| |a b| ,2. |a b| |a |+ |b|

Corollary 2.2.2 Let a1 , . . . , a n R . Then

|a 1 + + a n| |a 1|+ + |a n|

Examples.



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2.2.1 The Real Line

Denition 2.2.2 Let a, R and > 0. The -neighborhood of a isthe set

B (a ) = { x R | | x a | < } Theorem 2.2.3 Let a R . If for any > 0 , x B (a ) , then x = a.

Proof:

1. We have | x a | < . So | x a | = 0.

Examples.

2.2.2 Homework

Exercises: 2.1[13,20,23], 2.2[14,15]



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2.3. THE COMPLETENESS PROPERTY OF R 17

2.3 The Completeness Property of R

2.3.1 Suprema and Inma

R is a complete ordered eld .

Denition 2.3.1 Let S R be a nonempty set of real numbers.1. S is bounded above if there exists u R such that s S , s u.

Such u is an upper bound of S .

2. S is bounded below if there exists v R such that s S , s v.Such v is a lower bound of S .

3. S is bounded if it is bounded above and below.That is there is M R such that s S , |s| M.

4. S is unbounded if it is not bounded.That is for any M R there is s S such that |s| > M.

Example .

Denition 2.3.2 Let S R be a nonempty set of real numbers.

1. Let S be bounded above. Then u R is a supremum (leastupper bound ) of S if:(a) u is an upper bound of S , and

(b) if v is any upper bound of S , then u v.2. Let S be bounded below. Then t R is an inmum (greatestlower bound ) of S if:

(a) t is a lower bound of S , and

(b) if v is any lower bound of S , then t

v.

The supremum and inmum, when they exist, are unique.Proof: Exercise.

Notation.sup S , inf S



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Not every set S R has the supremum (or inmum). For a set to have the supremum it must be: 1) nonempty and 2) bounded

above.

For a set to have the inmum it must be: 1) nonempty and 2) boundedbelow.

For a set to have both the supremum and inmum it must be: 1) nonemptyand 2) bounded.

Equivalent statements about an upper bound:1. u is an upper bound of S 2. s S , s u.

Let u = sup S . Then the following are equivalent statements:1. if v is any upper bound of S , then u v,2. if w < u, then w is not an upper bound of S ,

3. if w < u, then s S such that s > w,4. for any > 0, s S such that u < s.

Alternative formulations for the supremum:

Lemma 2.3.1 Let S R be a nonempty set of real numbers. Thenu = sup S if and only if

1. s S , s u, and 2. if v < u, then there is s S such that v < s .

Proof: Exercise.

Lemma 2.3.2 Let S R be a nonempty set of real numbers and u Rbe an upper bound of S . Then u = sup S if and only if > 0 there iss S such that u < s.

Proof:

1.



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2.3. THE COMPLETENESS PROPERTY OF R 19

Remarks.

Supremum of a set does not have to belong to that set.

If u = sup S and u S , then u = max S is the largest element of the set S .

If S is nite, then sup S is its largest element, and inf S its least element.

Examples.

2.3.2 The Completeness Property of R

The nal axiom about R .

Denition 2.3.3 The Completeness Property of R . Every nonempty

set of real numbers bounded above has a supremum in R .

Corollary . Every nonempty set of real numbers bounded below has aninmum in R .

Proof: Exercise.

2.3.3 Homework

Exercises: 2.3[6,7,8,9,10]



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2.4 Applications of Supremum Property

Examples. Compatibility of sup and inf with algebraic properties of R . Let S R ,

S . Let a R and a + S be the seta + S = {a + s|s S }

If S is bounded above, then a + S is bounded above and

sup( a + S ) = a + sup S .

Proof: Exercise.

Proposition . Let A, B R , A, B . If a A, b B, a b, thensup A inf B .

Proof: Exercise.

2.4.1 Functions

Denition 2.4.1 Let f : D R .1. f is bounded above if f ( D) is bounded above.

2. f is bounded below if f ( D) is bounded below.

3. f is bounded if f ( D) is bounded.

Example. Let f , g : D R be bounded and x D, f ( x) g( x). Thensup f ( D) sup g( D)

Proof: Easy.

Note: there is no relation between sup f and inf g. Example. Let f , g : D R be bounded and x, y D, f ( x) g( y). Then

sup f ( D) inf g( D)Proof: Easy.



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2.4. APPLICATIONS OF SUPREMUM PROPERTY 21

2.4.2 The Archimedean Property

Theorem 2.4.1 For any x R , there exists n N such that x < n.That is N is not bounded above in R .

Proof:

1. By contradiction. Suppose x R such that n N , n x.2. Then x is an upper bound of N . So, a = sup N .3. Then a 1 is not an upper bound of N .4. m N such that m > a 1, or a < m + 1.5. Thus, a is not an upper bound.

Corollary 2.4.1 Let S = 1n n N . Then inf S = 0. Proof:

1. Let b = inf S . We have b 0.2. > 0, n N , such that 0 b 1n < .3. b = 0.

4. Or, by contradiction. Suppose b > 0. Get a contradiction.

Corollary 2.4.2 For any t R , t > 0 , there exists n N such that

0 < 1n

< t

Proof:

1. By Archimedean property.

Corollary 2.4.3 For any y R , y > 0 , there is n N such that

n 1 y < n . Proof:



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1. Let E = {m N| y < m}2. By Well Ordering Principle least element of E , say n.3. Then n 1 E .4. So, n 1 y < n.

2.4.3 The Existence of 2 Theorem 2.4.2 There exists x R + such that x 2 = 2.

Proof:

1. Let S = {s R|0 s < 2}.2. S and S is bounded above.3. x = sup S .4. We have x > 1.

5. Consider x2 . By Trichotomy Property x2 < 2 or x2 > 2 or x2 = 2.

6. Case x2 < 2. This contradicts the fact that x is an upper bound of S .

Since n N such that x + 1n2

< 2 and thus x + 1/ n S . Or x + 1/ n > x.

7. Case x2 > 2. This contradicts the fact that x is the least upper boundof S .Since m N such that x 1/ m is also an upper bound of S .Since ( x 1/ m)2 > 2 and s2 < 2 < ( x1/ m)2 . So s < x 1/ m.

8. Thus x2 = 2.

Corollary 2.4.4 For any a R+ there is a unique b > 0 such that

b2 = a.

b is the positive square root of a , b = a .

Corollary 2.4.5 For any a R + and any n N there is a unique b > 0

such that b n = a.

b is the positive n-th root of a , b = a 1/ n .



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2.4. APPLICATIONS OF SUPREMUM PROPERTY 23

2.4.4 Density of Rational Numbers in R

Theorem 2.4.3 The Density Theorem. Let x, y R and x < y. Then

there is r Q such that x < r < y. Proof:

1. Case 0 < x < y.

2. Case x < y < 0.

3. Case x < 0 < y.

Corollary 2.4.6 Let x, y R and x < y. Then there is an irrationalnumber z such that x < z < y. Proof:

1.

2.4.5 Homework

Exercises: 2.4[7,9,12], 2.5[3,6,10]



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2.5 Intervals

Open interval

Endpoints of an interval

Closed interval

Half-open (half-closed) intervals

Length of an interval

Innite open intervals

Innite closed intervals

Remark. R and are the only sets that are both open and closed .

2.5.1 Characterization of Intervals

Theorem 2.5.1 Characterization Theorem . If S R contains at least two points and

for any x , y S , x < y implies [ x, y] S ,then S is an interval.

Proof: Four cases.

1. Case I. S is bounded. Let a = inf S and b = sup S .

2. Then S [a , b].3. Claim: ( a , b) S .4. Thus, S is one of the bounded intervals.

5. Case II. S is bounded above and unbounded below.

6. Let b =

sup S .7. Then S (, b].8. Claim: ( , b) S .9. Thus, S is one of the two intervals unbounded below.

10. Case III. S is bounded below and unbounded above.



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2.5. INTERVALS 25

11. Similarly to II.

12. Case IV. S is unbounded above and below.

13. Claim: S = R .

2.5.2 Nested Intervals

Denition 2.5.1 A sequence of intervals { I n}n= 1 is nested if

I 1

I 2

I n

I n

+1

Examples.

Theorem 2.5.2 Nested Intervals Property. The intersection of anested sequence of closed bounded intervals is nonempty.

or

If { I n}n= 1 is a nested sequence of closed bounded intervals, then thereexists

R , such that

I n for each n

N .

Proof:

1. Completeness of R is essential for this!

2. Let I n = [a n , bn].

3. Then

a n a n+ 1 b1 and bn bn+ 1 a 14. Let S =

{a n

|n

N

}.

5. Let = sup S .

6. Then

a n bn , n N



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Theorem 2.5.3 If { I n}n= 1 is a nested sequence of closed bounded inter-vals, I n = [a n , bn] , such that

inf {(bn a n) | n N} = 0,then there exists a unique real number R , such that I n for eachn N .

Proof:

1. Let

= inf {bn|n N} and = sup{a n|n N}.2. Then

a n bn for any n N .3. Next, > 0, m N such that n m,

bn a n < .4. We have

0 bn a n < .5. Thus = .

2.5.3 The Uncountability of R

Theorem 2.5.4 The set R is uncountable. Proof:

1. Consider I = [0, 1] R .2. Claim: I is uncountable.

3. Assume I =

{ x1 , x2 , . . . } is countable.4. Let I 1 be a closed bounded interval such that I 1 I and x1 I 1 .5. Let I n be a closed bounded interval such that I n I n1 and xn I n .6. Then I n for each n N .7. Thus, xn for each n N .



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2.5. INTERVALS 27

8. Contradiction.

Remark. The set of irrational numbers R \ Q is uncountable.

2.5.4 Homework

Exercises: 2.5[3,6,10]



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Chapter 3

Sequences and Series

3.1 Sequences and Their Limits

Sequence ( a n), an S , or f : N S .

Denition 3.1.1 A sequence of real numbers is a a function X : N

R .

Elements (terms) of a sequence xn = X (n).

Notation X = ( xn) = ( xn|n N ) = ( xn)n N = ( xn)n= 1

Remark . The set { xn|n N} (non-ordered collection of elements) is di ff er-ent from the sequence ( xn|n N ) (an ordered list of elements).

Description of sequences:1. list,

2. formula,

3. recursion.

Examples.

Denition 3.1.2 A sequence X = ( xn) is constant if xn = x for each

n N .

29


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30 CHAPTER 3. SEQUENCES AND SERIES

3.1.1 The Limit of a Sequence

Denition 3.1.3 Convergence, Limit, Divergence. A sequence X =( xn) converges to x R (or has a limit x) if > 0 , K N such that

n K,

| xn x| < .The number x is the limit of the sequence X.

If X has a limit, then X converges , or is convergent .

If X does not have a limit, then X diverges , or is divergent .

Notation. lim X = x, lim xn = x, xn x

Theorem 3.1.1 Uniqueness of Limit . A sequence can have at most

one limit.

Proof:

1. Suppose X has two limits x and y.

2. Show that > 0, | x y| < .3. So, x = y.

Theorem 3.1.2 Let X = ( xn) be a sequence in R and x R. The fol-lowing statements are equivalent:

1. lim X = x,

2. > 0 , K N such that n K ,

| xn x| < ,

3. > 0 , K N such that n K , x < xn < x + ,

4. for any -neighborhood B ( x) of x, K N such that n K, xn B ( x).



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3.1. SEQUENCES AND THEIR LIMITS 31

Proof: Follows from denitions.

Corollary 3.1.1 Let X = ( xn) be a sequence in R and x R. Thenlim X = x if for any -neighborhood B ( x) of x all but a nite number of terms of X belong to B ( x).

Examples. Remark. To show that X = ( xn) does not converge to x R it is enough to

nd 0 > 0 such that K N , n K such that| xn x| .

Theorem 3.1.3 Let X = ( xn) be a sequence of real numbers, x Rbe a real number, and A = (a n) be a sequence of positive real numbersconverging to zero, that is lim A = 0. Suppose that there exists a positivereal number C R + and some integer M N such that n M,

| xn x| Ca n .Then lim X = x.

Proof:

1. Easy.

3.1.2 Tails of Sequences

Denition 3.1.4 Let X = ( xn)n= 1 be a sequence in R and M N . The

M-tail of the sequence X is the sequence X M = ( xn)n= M + 1 = ( x M + k )k = 1 .

Theorem 3.1.4 Let X = ( xn) be a sequence in R and m N. Then the

m-tail X m converges if and only if X converges and lim X m = lim X.

Proof:1. Let X M = Y = ( yn) = ( xn+ M ).

2. I. Assume X x. Show Y x.3. II. Assume Y x. Show X x.



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3.1.3 Examples

3.1.4 Homework

Exercises: 3.1[3(b),8,10,13,14]



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3.2. LIMIT THEOREMS 33

3.2 Limit Theorems

Denition 3.2.1 Let X = ( xn) , Y = ( yn) and Z = ( zn) be sequences inR , c R and f : R R be a function.The multiple of X by c is the sequence

cX = (cxn)

We say that

X c if xn c for all n N .We say that

X 0 if xn 0 for all n N .The absolute value of the sequence X is the sequence

| X | = (| xn|)We say that

X Y if xn yn for all n N .The function of a sequence X is the sequence

f ( X ) = ( f ( xn))

The sum and the diff erence of X and Y are the sequences

X + Y = ( xn + yn) and X Y = ( xn yn)The product of X and Y is the sequence

XY = ( xn yn)

If Z 0 , then the quotient of X and Z is the sequence

X / Z = xn zn



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3.2. LIMIT THEOREMS 35

2. For XY use

| xn yn xy| | xn|| yn y|+ | y|| xn x|3. For X / Z use

xn zn

x z |

xn x|| zn|

+ | x|| z|

| z zn|| zn|

4. We have1

| zn| 0.

3. We estimate

| xn x| 1 x| xn x|.

Theorem 3.2.9 Let X = ( xn) be a convergent sequence in R and X > 0. Let Y = ( yn) be a sequence dened by y n = xn+ 1/ xn . If Y converges and lim Y < 1 , then X converges and lim X = 0.

Proof:

1. We have Y > 0.



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3.3 Monotone Sequences

Denition 3.3.1 Let X = ( xn) be a sequence in R .

X is increasing if

xn xn+ 1 for any n N . X is decreasing if

xn xn+ 1 for any n N . X is monotone if it is either increasing or decreasing.

Examples. Notation . Let X = ( xn) be a sequence in R dened by a function X : N R .

The set of all elements of the sequence is the range of this function

X (N ) = { xn|n N}.

Theorem 3.3.1 Monotone Convergence Theorem. Let X be a mono-tone sequence in R .

X is convergent if and only if it is bounded.(i) If X is bounded and increasing, then lim X = sup X (N ).

(ii) If X is bounded and decreasing, then lim X = inf X (N ).

Proof:

1. Claim: Convergence implies boundedness. Proved before.

2. Claim: Boundedness and monotonicity implies convergence.

3. Let X be increasing and bounded.

4. Then the set X (N ) is bounded.5. Let b = sup X (N ).

6. Claim: b = lim X .

7. Let > 0.

8. Then b is not an upper bound of X (N ).introanal.tex; November 21, 2005; 18:02; p. 38


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3.3. MONOTONE SEQUENCES 39

9. So, there exists k N such that xk > b .10. Hence, n k , xn > b .11. Thus, lim X = b.

12. Similarly, if X is decreasing.

3.3.1 Sequence dened recursively

Proposition 3.3.1 (Square Roots. ) Let a > 0 and X = ( xn) be a se-quence dened by

x1 = 1, xn+ 1 = 12

xn + a xn

Then X converges and lim X = a .

Proof:

1. We have

x2n 2 xn+ 1 xn + a = ( xn xn+ 1)2 x2n+ 1 + a = 02. Therefore, x2n a .3. Next, we have

xn xn+ 1 = xn 12

xn + a xn

= x2n a

2 xn 0

4. Therefore, a

xn+ 1

xn

5. Compute the limit from

x = 12

x + a x



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Example. Let x1 = 1, xn+ 1 = 2 + 1 xn

Show, X is bounded above and increasing, therefore, converging.

Compute the limit from

x = 2 + 1 x

3.3.2 Limit Superior and Limit Inferior

Denition 3.3.2 Let X be a bounded sequence in R and X n be the n-tailof the sequence X. Let X

n(N ) be the set of elements of the tail X

n

X n = { xn+ k |k N} Let A and B be the sequences dened by

a n = inf X n(N ), bn = sup X n(N ).

The limit superior of the sequence X is the limit of the sequence B, that is

lim sup X = limn

sup X n(N )

The limit inferior of the sequence X is the limit of the sequence A, that is

lim sup X = limn

inf X n(N )

Then the sequences A and B are bounded and monotone, A is increasing and B is decreasing.

Therefore, the sequences A and B are convergent. Notation.

limsup X = lim X , liminf X = lim X .

Denition 3.3.3 Let X be a sequence in R. A real number a R isa limit point of the sequence X if every neighborhood of a containsinnitely many elements of the sequence X.

A limit point of a sequence does not have to be an element of the sequence.introanal.tex; November 21, 2005; 18:02; p. 40


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3.3. MONOTONE SEQUENCES 41

Every point that appears innitely many times in a sequence is a limit pointof the sequence.

Proposition 3.3.2 Let X be a bounded sequence in R and P be the set of all limit points of the sequence X. Then

limsup X = sup P, and lim inf X = inf P

A sequence X converges if and only if lim sup X = liminf X

In this caselim X = lim sup X = liminf X .

3.3.3 Homework

Exercises: 3.3[1,4,5,7,9,10]



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3.4 Subsequences and the Bolzano-Weierstrass The-orem

Lemma 3.4.1 Let A be a strictly increasing function A : N N suchthat

A(k + 1) > A(k ), k N .Then

A(k ) k , k N . Proof: Obvious.

Example.

Denition 3.4.1 Let X = ( xn) be a sequence in R and A = (nk ) be astrictly increasing sequence in N . A subsequence of X is the sequence

X = ( x A(k )) = ( xnk ) = ( xn1 , xn2 , . . . ) = ( xnk )k = 1 = ( xnk )k N .

A subsequence X : N R is the composition of two sequences A : N Nand X : N R , that is

X = X A, or X (k ) = X ( A(k )), k N

A tail of a sequence is a subsequence.

Example.

Theorem 3.4.1 Let X be a sequence in R and X be a subsequence of X. If X converges, then X also converges and

lim X = lim X

Proof: Easy.

Example.introanal.tex; November 21, 2005; 18:02; p. 42


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3.4. SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 43

Theorem 3.4.2 Let X be a sequence in R and a R . The following areequivalent statements:

1. X does not converge to a.

2. there exists 0 such that k N , nk N such that n k k and

| xnk a| 0 .3. there exists 0 and a subsequence X = ( xnk ) of X such that k

N ,

| xnk a| 0 . Proof:

1. (1) implies (2). Use negation of the denition of lim X = a .

2. (2) implies (3). Construct a subsequence X from (2).

3. (3) implies (1). By contradiction.

Proposition 3.4.1 (Divergence Criteria. ) Let X be a sequence in R . If X is: either (i) unbounded or (ii) X has two convergent subsequences X and X such that lim X lim X , then X diverges.

Proof: Exercise.

Examples.

3.4.1 The Existence of Monotone Subsequences

Theorem 3.4.3 Monotone Subsequence Theorem. Any sequence in

R has a monotone subsequence.

Proof:

1. We say that xm is a peak if n m, xm xn

That is xm is an upper bound for the m-tail X m, or

X m xm .introanal.tex; November 21, 2005; 18:02; p. 43


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2. Case I. X has innitely many peaks.

3. Then the sequence of peaks is a decreasing subsequence of X .

4. Case II. X has nitely many peaks.

5. Then the tail beyond the last peak does not have any peaks.

6. Since all the elements in the tail are not peaks, construct an increasingsubsequence of X .

3.4.2 The Bolzano-Weierstrass Theorem

Theorem 3.4.4 The Bolzano-Weierstrass Theorem. A bounded se-

quence in R has a convergent subsequence.

Proof:

1. (I) Indirect. X has a monotone subsequence X .

2. X is bounded and monotone, hence, convergent.

3. (II) Constructive.

4. Construct a sequence of closed bounded intervals I k = [a k , bk ] such

that I k I k +

1 with the lengths bk a k =

2k

(b1 a 1) with the propertythat X (N ) I k

5. Construct a subsequence X such that xnk I k .6. Then there exists a unique element I k , k N .7. So,

| xnk | < 2k (b1 a 1)8. Thus, lim X = .

Theorem 3.4.5 Let X be a bounded sequence in R and a R. If ev-ery convergent subsequence of X converges to a, then the sequence X converges to a.

Proof:



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3.4. SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS THEOREM 45

1. By contradiction. Suppose that X does not converge to a .

2. Then there exists 0 > 0 and a subsequence X such that k N,| xnk a| 0 .

3. Since X is bounded, X has a convergent subsequence X .

4. Then X converges to a .

5. This contradicts the above.

3.4.3 Homework

Exercises: 3.4[2,3,5,6,9]



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3.5 The Cauchy Criterion

Denition 3.5.1 A sequence X = ( xn) in R is a Cauchy sequence if

> 0 , K N such that n, m K,

| xn xm| < . Examples. Lemma 3.5.1 Every convergent sequence is Cauchy.

Proof: Easy.

Lemma 3.5.2 Every Cauchy sequence is bounded. Proof: Easy.

Theorem 3.5.1 Cauchy Convergence Criterion. A sequence of real

numbers is convergent if and only if it is Cauchy.

Proof:

1. (I). Proved above.

2. (II). Let X be Cauchy.

3. Then X is bounded.4. So, there exists a convergent subsequence X .

5. Claim: lim X = lim X .6. Use the fact that X is Cauchy and X converges to a = lim X .

7. Get an estimate for some K N| xn a| | xn xK |+ | xK a | <

Examples.

Denition 3.5.2 A sequence X is contractive if C R such that 0 < C < 1 and n N

| xn+ 2 xn+ 1| C | xn+ 1 xn|.The number C is the constant of the contractive sequence.



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3.5. THE CAUCHY CRITERION 47

Theorem 3.5.2 Every contractive sequence is Cauchy and, therefore,

convergent. Proof:

1. We have

| xn+ 2 xn+ 1| C n| x2 x1|2. Let m > n. Then

| xm xn| | xm xm1|+ + | xn+ 1 xn| (C m2 + + C n1)| x2 x1| C

n

1 1

C mn

1 C | x2 x1|

C n1 1

1 C | x2 x1|

3. So, X is Cauchy.

Corollary 3.5.1 Let X = ( xn) be a contractive sequence with constant C, 0 < C < 1 , and a = lim X. Then n N

1.

|a xn| C n11 C |

x2 x1|,2.

|a xn+ 1| C 1 C |

xn+ 1 xn|. Proof:

1. (1). We have m > n

| xm

xn

| C n1 1

1 C | x2

x1

|2. As m , we get (1).3. (2). We have for m > n

| xm xn| | xm xm1|+ + | xn+ 1 xn|introanal.tex; November 21, 2005; 18:02; p. 47


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3.6. PROPERLY DIVERGENT SEQUENCES 49

3.6 Properly Divergent Sequences

Denition 3.6.1 Examples.

Theorem 3.6.1 Proof:

1.

Theorem 3.6.2 Proof:

1.

Theorem 3.6.3

Proof:

1.

Examples.



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3.7 Introduction to Series

Denition 3.7.1 Examples.

Theorem 3.7.1 The n-th Term Test. Proof:

1.

Theorem 3.7.2 Cauchy Criterion for Series. Theorem 3.7.3 The n-th Term Test.

Proof:

1.

Examples.

3.7.1 Comparison Tests

Theorem 3.7.4 Comparison Test. Proof:

1.

Theorem 3.7.5 Limit Comparison Test. Proof:

1.

Examples.



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Chapter 4

Limits

4.1 Limits of Functions

Denition 4.1.1 Let A R. A point c R is a cluster point (or accumulation point , or limit point ) of A if for every > 0 there existsa point x A such that x c and | x c| < . A point c R is a cluster point of A if every -neighborhood B (c) of ccontains an element of A distinct from c.

The point c does not have to be an element of A.

Examples .

Let X = ( xn) be a sequence and c R . We say

X c if xn c for any n N .

Theorem 4.1.1 Let A

R. A number c

R is a cluster point of A if

and only if there exists a sequence X in A such that X c and lim X = c.

Proof: Easy.

Examples.

51


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52 CHAPTER 4. LIMITS

4.1.1 The Denition of the Limit

Denition 4.1.2 Let A R , c be a cluster point of A, f : A R be a function, and L R be a real number. We say that L is a limit of f at cif for any > 0 there exists > 0 such that

if | x c| < and x c then | f ( x) L| < .Then we say that f converges to L at c.

If the limit of f does not exists then we say that f diverges at c.

Notation.

lim xc f ( x) = L, lim xc

f = L, f ( x) x

c

L

Theorem 4.1.2 Let A R , f : A R and c be a cluster point of A.

Then f can have only one limit at c.

Proof: Easy.

Theorem 4.1.3 Let A R , c be a cluster point of A, f : A R be a function, and L R be a real number. Then lim xc f ( x) = L if and onlyif for any -neighborhood B ( L) of L, there exists a -neighborhood B(c) of c such that for any x A, x c, if x B(c) then f ( x) B ( L).

Proof: Exercise.

Examples.

4.1.2 Sequential Criterion for Limits

Theorem 4.1.4 Sequential Criterion. Let A R , c be a cluster point of A and f : A R. Then lim xc f ( x)

= L if and only if for everysequence X in A such that X c and lim X = c, the sequence f ( X )

converges to L, i.e. lim f ( X ) = L.

Proof:

1. (I). Suppose lim xc f ( x) = L.



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4.1. LIMITS OF FUNCTIONS 53

2. Then, > 0, > 0 such that x A, x c, if | x c| < then| f ( x) L| < .3. Let X = ( xn) be a sequence in A such that X c and lim X = c.

4. Then, for any > 0, K N such that n K , | xn c| < .5. Therefore, | f ( xn) L| < .6. Thus lim f ( X ) = L.

7. (II). Suppose that for every sequence X in A such that X c andlim X = c the sequence f ( X ) converges to L.

8. Suppose that f does not converge to L at c.

9. Then there exists 0 such that

> 0

x

A such that x c and

| x c| < and | f ( x) L| 0 .10. Then n N , xn A, xn c such that | xn c| < 1n but | f ( xn) L| 0 .11. Therefore f ( X ) does not converge to L (contradiction).

4.1.3 Divergence Criteria

Theorem 4.1.5 Divergence Criteria. Let A R , c be a cluster point of A, L

R , and f : A

R .

1. The function f does not have limit L at c if and only if there existsa sequence X in A such that X c and lim X = c but the sequence f ( X ) does not converge to L.

2. The function f diverges at c if and only if there exists a sequence X in A such that X c and lim X = c but the sequence f ( X )diverges.

Proof: Exercise.

Examples.

4.1.4 Homework

Exercises: 4.1[3,6,11(a),12,14]



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4.2 Limits Theorems

Denition 4.2.1 Let A R , c be a cluster point of A, f : A R . The function f is said to be bounded on a neighborhood of c if there existsa -neighborhood B (c) of c and a constant M > 0 such that

if x A B(c) , then | f ( x)| M.That is if there exist > 0 and M > 0 such that

if x A and | xc| < , then | f ( x)| M.

Theorem 4.2.1 Let A

R , c be a cluster point of A and f : A

R. If

f has a limit at c, then f is bounded on a neighborhood of c.

Proof:

1. Let L = lim xc f . Let = 1.

2. We have

| f ( x)| L | f ( x) L| < 13. If c A, let M = max{| f (c)|, L + 1}.4. Then x A,

| f ( x)| M .

Examples.

Denition 4.2.2 Let A R , a R , and f , g, h, : A R be functionssuch that h ( x) 0 and ( x) 0 for all x A. The sum f + g, thediff erence f g, the product fg, the multiple a f , the absolute value| f | , the square root f and the quotient f / h are the functions dened by: x A

( f g)( x) = f ( x) g( x), ( f g)( x) = f ( x)g( x),

(a f )( x) = a f ( x), f h

( x) = f ( x)h( x)

,

| f |( x) = | f ( x)|, ( x) = ( x).introanal.tex; November 21, 2005; 18:02; p. 53


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4.2. LIMITS THEOREMS 55

Theorem 4.2.2 Let A R , c be a cluster point of A, a R , and f , g, h : A R be functions such that h ( x) 0 for all x A and lim xc f ( x) = L, lim xc g( x) = M, lim xc = K, lim xc h( x) = H 0. Then

lim xc

( f g) = L M lim xc( f g) = LM ,

lim xc

(a f ) = aL, lim xc

f h

= L H

.

lim xc | f |

= | L|, lim xc = K

Proof: Easy. Use sequential criterion.

Remarks.

If lim xc h = 0, the theorem is not valid.

Similarly, we havelim xc

( f ( x))n = Ln

c does not have to be in the domain.

If f is unbounded at c, then it does not have a limit at c. Examples.

Theorem 4.2.3 Let A R , a, b R , c be a cluster point of A, and f : A R be a function such that x A, x c

a f ( x) b.Then if f has a limit at c, then

a lim xc f b. Proof:

1. Use a sequence X in A such that X c and lim X = c



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Theorem 4.2.4 Squeeze Theorem. Let A R , c be a cluster point of A, and f , g, h : A R be functions such that x A, x c

f ( x) g( x) h( x).Then if lim xc f

= lim xc h, then

lim xc

g = lim xc

f = lim xc

h.

Proof: Exercise. By sequences.

Examples.

Theorem 4.2.5 Let A R , c be a cluster point of A, and f : A R bea function such that

lim xc

f > 0.

Then there exists > 0 such that for any x A, x c, if | xc| < , then f ( x) > 0.

That is, there is a -neighborhood B (c) of c such that f ( x) > 0 for all x A B(c) , x c.

Proof: Easy.

Note . f (c) may be negative.

Extension of the Limit Concept.

Right-hand limit.

Left-hand limit.

Limit at innity.

4.2.1 Homework

Exercises: 4.2[4,5,11(b,c),12]



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Chapter 5

Continuous Functions

5.1 Continuous Functions

Denition 5.1.1 Let A R , c A, and f : A R. Then f iscontinuous at c if for any > 0 there exists > 0 such that x A,if | x c| < then | f ( x) f (c)| < .

If f is not continuous at c, then f is said to be discontinuous at c.

Theorem 5.1.1 Let A R , c A, and f : A R . Then f is continuousat c if and only if for any -neighborhood B ( f (c)) of f (c) there exists a-neighborhood B (c) such that

if x A B(c) then f ( x) B ( f (c)) ,that is

f ( A B(c)) B ( f (c))

Remarks .

If c A is not a cluster point of A, then f is automatically continuous at c.

If c is a cluster point of A then f is continuous at c if and only if f (c) = lim

xc f ( x)

57


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58 CHAPTER 5. CONTINUOUS FUNCTIONS

Theorem 5.1.2 Sequential Criterion for Continuity. Let A R , c A, and f : A R. Then f is continuous at c if and only if for anysequence X in A converging to c, the sequence f ( X ) converges to f (c).

Theorem 5.1.3 Discontinuity Criterion. Let A R , c A, and f : A R. Then f is discontinuous at c if and only if there existsa sequence X in A such that X converges to c but the sequence f ( X )does not converge to f (c).

Denition 5.1.2 Let A R , B A, and f : A R. Then f is

continuous on the set B if f is continuous at every point of B.

Examples. The Dirichlet Function f : R R dened by

f ( x) = 0, if x is irrational1, if x is rational

is discontinuous at every point.

Proof: Use sequences.

The Thomae Function f : R

R

dened by

f ( x) =

0, if x is irrational

1q

, if x = pq

is rational , q > 0

and p and q are relatively prime

1, if x = 0

is discontinuous at every rational number and continuous at every irrationalnumber.

Proof:

1. Use sequences for rational numbers.

2. For any irrational number c, given > 0 there exists a -neighborhood B(c) of c such that it does not contain any rational numbers x = p/ qwith q < 1/ .



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5.1. CONTINUOUS FUNCTIONS 59

3. Therefore, x B(c), | f ( x)| < .4. So, lim xc f = f (c) = 0.5. Thus f is continuous at every irrational number.

Continuous extension of a function. Let f : A R and c be a clusterpoint of A such that c A. If f has a limit at c, then one can dene acontinuous extension F : A {c} R R of f by

F ( x) = f ( x), if x Alim xc f , if x

= c

5.1.1 Homework

Exercises: 5.1[3,7,11,12,13]



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5.2 Combinations of Continuous Functions

Let A R and f : A R . We say that f 0 if x A, f ( x) 0.

We say that

f 0 if x A, f ( x) 0. Let A R and f : A R . The absolute value of f is a function | f | : A R

dened by

| f

|( x) =

| f ( x)

|,

x

A.

Let A R, f : A R and f 0. The square root of f is a function f : A R dened by f ( x) = f ( x), x A.

Theorem 5.2.1 Let A R , a R , c A, and f , g, h, : A R becontinuous functions at c such that h 0 and 0. Then the functions f g, fg, af, f / h, | f | and are continuous at c.

Proof:

1. If c is an isolated point of A, then it is automatic.2. If c is a cluster point of A, then the assertions follow from the limit

theorems.

Theorem 5.2.2 Let A R , a R , and f , g, h, : A R be continuous functions on A such that h 0 and 0. Then the functions f g, fg,a f , f / h, | f | and are continuous on A.

Denition of Quotients. Let A R and f , h : A R . Let A

=

{ x A | h( x) 0}.The quotient f / h : A R is dened for any x A by f h

( x) = f ( x)h( x)

.

Then if f and h are continuous on A , then f / h is continuous on A .



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5.2. COMBINATIONS OF CONTINUOUS FUNCTIONS 61

Examples.

Polynomials.

Rational Functions.

Trigonometric Functions.

5.2.1 Composition of Continuous Functions

Theorem 5.2.3 Let A, B R , f : A B and g : B R . Let c A and b = f (c) B. Then if f is continuous at c and g is continuous at b, thenthe composition h

= g f : A R is continuous at c. Proof:

1. Let c A, b = f (c) and a = g(b).2. Let > 0.

3. There exists > 0 such that if y B B(b), then g( y) B (a ).4. There exists > 0 such that if x A B (c), then f ( x) B(a ).5. Therefore, if x A B (c), then h( x) = (g f )( x) = g( f ( x)) B (a ).6. Thus, h = g

f is continuous at c

A.

Theorem 5.2.4 Let A, B R , f : A B and g : B R. Then if f is continuous on A and g is continuous on B, then the compositionh = g f : A R is continuous on A.

Proof: Follows from the previous theorem.

Examples.

5.2.2 Homework

Exercises: 5.2[5,6,10,12,13]



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5.3 Continuous Functions on Intervals

Denition 5.3.1 Let A R and f : A R . We say that f is boundedon A if there exists M R such that | f | M, that is x A, | f ( x)| M.We say that f is unbounded on A if M > 0 there exists x A suchthat | f ( x)| > M.

A function is bounded on a set if its range is bounded and unbounded if itsrange is unbounded.

Theorem 5.3.1 Boundedness Theorem . Let I = [a , b] be a closed bounded interval and f : I R be continuous on I. Then f is bounded on I.

Proof:

1. By contradiction. Suppose f is not bounded.

2. Then there exists a sequence X = ( xn) in I such that

f ( xn) > n

3. Since X is bounded, there exists a convergent subsequence X = ( xnk ).

4. Since I is closed c = lim X I .5. Since f is continuous lim f ( X ) = f (c).

6. Hence f ( X ) is bounded, which is a contradiction since

| f ( xnk )| > nk k .

Examples .



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5.3. CONTINUOUS FUNCTIONS ON INTERVALS 63

5.3.1 The Maximum-Minimum Theorem

Denition 5.3.2 Let A R and f : A R. We say that f hasan absolute maximum on A if there exists x 1 A, called an absolutemaximum point of f on A, such that f ( x1) f ( x) for any x A.

We say that f has an absolute minimum on A if there exists x 2 A,called an absolute minimum point of f on A, such that

f ( x2) f ( x) for any x A.

Remarks. x1 and x2 must be in A. x1 and x2 are not unique, in general. Example.

Theorem 5.3.2 Maximum-Minimum Theorem. Let I = [a , b] be aclosed bounded interval and f : I R be continuous on I. Then f hasan absolute maximum and an absolute minimum on I.

Proof:

1. Let u = sup f ( I ).

2. Claim: c I such that u = f (c).3. There exists a sequence X = ( xn) in I such that n N

u 1n

< f ( xn) u4. There exists a convergent subsequence X = ( xnk ).

5. Since I is closed c = lim X I .6. Since f is continuous, lim f ( X ) = f (c).7. On the other hand lim f ( X ) = u.

8. Therefore, f (c) = sup f ( I ) and c is a maximum point of f on I .

9. Similarly, for the minimum point.



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5.3.2 Bisection Method

Theorem 5.3.3 Location of Roots Theorem. Let I = [a , b] and f : I R be continuous on I. If f (a ) < 0 < f (b) or f (b) < 0 < f (a ) , thenthere exists c (a , b) such that f (c) = 0.

Proof:

1. Let f (a ) < 0 < f (b).

2. Dene a nested sequence ( I k = [a k , bk ]) of intervals dened by bisec-tion so that

I k + 1 I k , bk a k =

b

a

2k 1 , f (a k ) < 0 < f (bk )

3. If at some step we nd a midpoint pk = (a k + bk )/ 2 such that f ( pn) = 0,then just let c = pn . Otherwise, we get an innite sequence of nestedintervals.

4. Then there exists c I such thatlim a n = lim bn = c

5. Since f is continuous,

f (c) = lim f (a n) 0 and f (c) = lim f (bn) 06. Thus f (c) = 0.

Example.

5.3.3 Bolzanos Theorem

Theorem 5.3.4 Bolzanos Intermediate Value Theorem. Let I be an

interval and f : I R be continuous on I. Let a , b I and there exist k R such that f (a ) < k < f (b). Then there exists c I between a and b such that f (c) = k.

Proof:

1. Let a < b.



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5.3. CONTINUOUS FUNCTIONS ON INTERVALS 65

2. Let g( x) = f ( x) k .3. Use Location of Roots Theorem.

Corollary 5.3.1 Let I = [a , b] be a closed bounded interval and f : I R be continuous on I. If there exists a k R such that

inf f ( I ) k sup f ( I ),then there exists c I such that f (c) = k.

Proof:

1. Follows from the Maximum-Minimum Theorem and Bolzanos Theo-rem.

Theorem 5.3.5 Let I be a closed bounded interval and f : I R becontinuous on I. Then the range f ( I ) of f is a closed bounded interval.

A continuous image of a closed and bounded interval is a closed and bounded interval.

Proof:

1. Let m = inf f ( I ) and M = sup f ( I ).

2. Then m, M f ( I ) and f ( I ) [m, M ].3. Let k [m, M ].4. Then c I such that f (c) = k .5. So k f ( I ) and [m, M ] f ( I ).6. Thus f ( I ) = [m, M ].

Remarks. Closed and bounded intervals are compact sets . Continuous functions preserve compactness. Not true for not closed or unbounded intervals.



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5.4. UNIFORM CONTINUITY 67

5.4 Uniform Continuity

Denition 5.4.1 Let A R and f : A R. Then f is uniformlycontinuous on A if > 0 > 0 such that x, y A

if | x y| < , then | f ( x) f ( y)| < .

Remarks. = ( ) depends on but not on x, y! If f is just continuous, but not uniformly continuous, then = (, x) de-

pends on both and x

A.

Let ( ) = inf x A (, x). If ( ) > 0, then f is uniformly continuous. If ( ) = 0, then f is continuous but not uniformly continuous.

Not every continuous function is uniformly continuous. Examples.

Theorem 5.4.1 Nonuniform Continuity Criteria. Let A R and f : A R . The following are equivalent statements:

1. f is not uniformly continuous on A.

2. there exists 0 > 0 such that > 0 x, y A such that

| x y| < and | f ( x) f ( y)| 0 .3. there exists 0 > 0 and two sequences X and Y in A such that

lim( X Y ) = 0 and | f ( X ) f (Y )| 0 . Proof: Exercise.

Examples.

Theorem 5.4.2 Uniform Continuity Theorem. Let I be a closed bounded interval and f : I R be continuous on I. Then f is uniformly continuous on I.

Proof:



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5.4. UNIFORM CONTINUITY 69

1. Easy.

Not all uniformly continuous functions are Lipshitz. Examples.

5.4.2 The Continuous Extension Theorem

Theorem 5.4.4 Let A R and f : A R be uniformly continuous.

Then if X is a Cauchy sequence in A, then f ( X ) is a Cauchy sequence.

Proof:

1. Let > 0.

2. Let X = ( xn) be a Cauchy sequence in A.

3. There is > 0 such that x, y A,if x y| < , then | f ( x) f ( y)| < .

4. There is K N such that n, m K ,

| xn xm| < .5. Thus, n, m K ,

| f ( xn) f ( xm)| < and f ( X ) is Cauchy.

Theorem 5.4.5 Continuous Extension Theorem. A function f :(a , b) R on an open interval (a , b) is uniformly continuous if and only if it can be extended to a continuous function F : [a , b] R on theclosed interval [a , b].

Proof:1. (I). Obviously, if F is continuous then its restriction f is uniformly

continuous.

2. (II). Suppose f is uniformly continuous.

3. Claim: The limit lim xa f ( x) exists.



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4. Let X = ( xn) be a sequence in ( a , b) such that lim X = a .

5. Then X is Cauchy and f ( X ) is Cauchy.

6. So, f ( X ) converges.

7. Let Y = ( yn) be any other sequence in ( a , b) such that lim Y = a .

8. Then lim( X Y ) = 0.9. Since f is uniformly continuous,

lim f (Y ) = lim[ f (Y ) f ( X )] + lim f ( X ) = lim f ( X ).10. Therefore, lim x

a f ( x) exists.

11. Similarly, lim xb f ( x) exists.12. Dene F : [a , b] R by

F ( x) =lim xa f ( x), x

= a f ( x), x (a , b)lim xb f ( x), x

= b

13. Then F is continuous on [ a , b].

Examples.

5.4.3 Homework

Exercises: 5.4[2,3,6,7,8,12]



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5.5. CONTINUITY AND GAUGES 71

5.5 Continuity and Gauges

Denition 5.5.1 Let I = [a , b] be an interval. Let x k I, (k =0, . . . , n) , be a set of points in I, called the partition points , such that

a = x0 < < xk 1 < xk < . . . xn = band I k = [ xk 1 , xk ] , (k

= 1, . . . , n) , be a set of non-overlapping closed subintervals of I whose union is [a , b].

The set P = { I k | (k = 1, . . . , n)} of the intervals I k is a partition of theinterval I .

Let t k I k , (k = 1, . . . , n) , be xed points in each interval I k , called thetags . Then the set of ordered pairs

P = {( I k , t k ) | k = (1, . . . , n)}is a tagged partition of I.

Denition 5.5.2 Let I be an interval. A gauge on I is a strictly posi-tive function : I R + on I. Let P = {( I k , t k )|k = (1, . . . , n)} be a tagged partition of I . Then P is-ne if

I k [t k (t k ), t k + (t k )], (k = 1, . . . , n) .

Lemma 5.5.1 Let I = [a , b] be an interval, x I, and : I R+ bea gauge on I. The every tagged -ne partition P of I has a tag t k suchthat | x t k | (t k ).

5.5.1 Existence of -ne Partitions

Theorem 5.5.1 Let I = [a , b] and : I R+ be a gauge on I. Thenthere exists a -ne partition of I .



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5.6 Monotone and Inverse Functions

Denition 5.6.1 Let A R and f : A R . We say that: f is increasing on A if x, y A,

if x < y, then f ( x) f ( y). f is strictly increasing on A if x, y A,

if x < y, then f ( x) < f ( y).

f is decreasing on A if x, y A,if x < y, then f ( x) f ( y).

f is strictly decreasing on A if x, y A,if x < y, then f ( x) > f ( y).

f is monotone on A if it is either increasing or decreasing on A.

f is strictly monotone on A if it is either strictly increasing or strictlydecreasing on A.

If f is increasing, then ( f ) is decreasing and vice versa. Recall the left- and the right-handed limits.

Theorem 5.6.1 Let I be an interval and f : I R be increasing on I. Let c I be an interior point of I. Then:

1. lim xc f = sup f ( I (, c)) ,

2. lim xc+ f = inf f ( I (c, )).

Proof:

1. (1). Let x I and x < c.2. The set f ( I (, c)) is bounded by f (c).3. Let L = sup f ( I (, c)).4. Let > 0.



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5.6. MONOTONE AND INVERSE FUNCTIONS 73

5. Then y0 I , y0 < c, such that L < f ( y0) L.

6. Let = c y0 .7. Then for any y between y0 and c

L < f ( y0) f ( y) L.

8. So, 0 < c y < implies | f ( y) L| < .9. Thus, lim xc f = L.

10. (2). Similarly.

Corollary 5.6.1 Let I be an interval and f : I R be increasing on I. Let c I be an interior point of I. Then the following are equivalent statements:

1. f is continuous at c.

2. f (c) = lim xc f = lim xc+

f .3. f (c) = sup f ( I (, c)) = inf f ( I (c, )).

Proof: Exercise.

Remarks.

Let I = [a , b] and f : I R be monotone on I . Then

f is continuous at a if and only if f (a ) = lim xa + f .

Similarly,

f is continuous at b if and only if f (b) = lim xb f .



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5.6. MONOTONE AND INVERSE FUNCTIONS 75

4. Then

D = n= 1

Dn

5. Claim: the set Dn is nite.

6. Suppose Dn is innite.

7. Let P = { x1 , . . . xm} Dn be a set of points in Dn such thata x1 < < xm b

8. We havemn

m

k = 1 j f ( xk ) f (b) f (a ).

9. By choosing m > n( f (b) f (a )) we get a contradiction.10. Thus Dn is nite, and the set D is countable.


Theorem 5.6.4 Continuous Inverse Theorem. Let I , J R be in-tervals and f : I J be strictly monotone and continuous function from I onto J = f ( I ). Then f has an inverse and the inverse function f 1 : J I is strictly monotone and continuous function from J onto I.

Proof:

1. Let f be strictly increasing.

2. Then f is an bijection and f 1 exists.3. Claim: f 1 is strictly increasing.4. Let y1 , y2 J and y1 < y2 .5. Then x1 , x2 I such that f ( x1)

= y1 and f ( x2)

= y2 .

6. Since f is strictly increasing, x1 < x2 .

7. Therefore, f 1( y1) = x1 < x2 = f 1( y2).8. Claim: f 1 is continuous on J .9. Suppose that f 1 is discontinuous at some c J .



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10. Then lim yc f 1 < lim yc

+ f 1 .

11. Then for any x I such that x f 1(c) and lim yc f 1 < x


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Chapter 6

Diff erentiation

6.1 The Derivative

Denition 6.1.1 Let I be an interval, f : I R and c I. Let : I R be a function dened by

( x) = f ( x) f (c)

x c, x c

(c), x = c,

where (c) is an arbitrary real number. The function f is diff erentiableat c if the function has a limit at c. In this case the limit lim xc isdenoted by f (c) and called the derivative of f at c.

That is, the function f is di ff erentiable at c if there exists a real number f (c) such that > 0 > 0 such that x I, x c,

if | xc| < , then f ( x) f (c)

x c f (c) < .

Notation. The derivative is also denoted byd f dx

, or just D f .

Remarks.

It is allowed that c is the endpoint of the interval.77


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78 CHAPTER 6. DIFFERENTIATION

One can dene the derivative of functions f : A R on any domain A witha cluster point c.

Theorem 6.1.1 Caratheodorys Theorem. Let I be an interval, c I, f : I R . Let : I R be a function dened by

( x) = f ( x) f (c)

x c, x c

(c), x = c

Then f is di ff erentiable at c if and only if the value (c) can be chosensuch that is continuous at c, that is (c) = lim

xc ( x). In this case

f (c) = (c).

Proof:

1. (I). If f is diff erentiable, choose (c) = f (c). Then is continuous.

2. (II). If is continuous at c, then (c) = lim xc ( x) = f (c).

Let I 1

I be the set of points where the function f is diff erentiable. The

derivative f : I 1 R is a function on I 1 .

Theorem 6.1.2 Let I be an interval, f : I R , and c I. If f is

di ff erentiable at c, then it is continuous at c.

Proof: Easy.

Not all continuous functions are di ff erentiable.

Example. There are functions that are continuous everywhere but are not di ff erentiable

at any point.

Example. (Weierstrass function).introanal.tex; November 21, 2005; 18:02; p. 75


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6.1.1 The Chain Rule

Theorem 6.1.4 Chain Rule. Let I and J be intervals and f : I Rand g : J R be functions such that f ( I ) J. Let h = g f : I R bethe composite function. Let c I and d = f (c) J, f be di ff erentiableat c and g be di ff erentiable at d. Then h is di ff erentiable at c and

h (c) = g (d ) f (c).

Proof:

1. Let : I R be the function dened by

( x) = f ( x) f (c) xc

, x c

f (c), x = c

2. Similarly, let : J R be the function dened by

( y) =g( y) g(d )

xc, y f (c)

g (d ), y = d

3. Then the function h = ( f ) : I R dened byh( x) = ( f ( x)) ( x)

is continuous at c and

h(c) = g (d ) f (c).

Examples.

f is not diff erentiable but continuous.

f is diff erentiable everywhere but f is not continuous.

f is diff erentiable, f is continuous but f is not diff erentiable.



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6.1. THE DERIVATIVE 81


Theorem 6.1.5 Let I be an interval, f : I J be strictly monotoneand continuous function from I onto J. Let f 1 : J I be the strictlymonotone and continuous function inverse to f . Let c I and d = f (c) J. If f is di ff erentiable at c, then f 1 is di ff erentiable at d and

( f 1) (d ) = 1 f (c)

.

Proof:

1. Let : I R be the function dened by

( x) = f ( x) f (c)

xc, x c

f (c), x = c

2. Then (c) = f (c) 0.

3. Thus, there exists a -neighborhood B(c) = (c , c + ) of c such that x B(c) I ,

( x) 0.

4. Let U = f ( B(c)

I ).

5. Let g = f 1 . Then y U , f (g( y)) = y. In particular g(d ) = c.6. Let h = 1/ ( g) : U R .7. Then y U

h( y) = 1(g( y))

=

g( y) g(d ) yd

, y d

1(c)

, y = d

8. It is continuous at d and

g (d ) = 1

(c) =

1 f (c)

.

Examples.



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6.1.3 Homework

Exercises: 6.1[4,9,10,11(c),13]



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6.2. THE MEAN VALUE THEOREM 83

6.2 The Mean Value Theorem

Denition 6.2.1 Let I be an interval, c I, and f : I R . Then f hasa relative maximum (minimum) at c if there exists a -neghborhood B(c) of c such that x B(c) I ,

f ( x) f (c), ( f ( x) f (c)).We say that f has a relative extremum at c if it has either a relativemaximum or relative minimum at c.

Theorem 6.2.1 Interior Extremum Theorem. Let I be an interval,

c I be an interior point of I, and f : I R . Let f be di ff erentiable at c and have a relative extremum at c. Then f (c) = 0. Proof:

1. We have f (c) = lim

xc f ( x) f (c)

x c2. Show that it is impossible to have f (c) > 0 and f (c) < 0.

Corollary 6.2.1 Let I be an interval, c I be an interior point of I, and f : I R . Let f be continuous on I and have a relative extremum at c.Then either f is not di ff erentiable at c or f (c) = 0.

Theorem 6.2.2 Rolles Theorem. Let I = [a , b] be a closed and bounded interval, and f : I R. Let f be continuous on theclosed interval [a , b] and di ff erentiable on the open interval (a , b) , and f (a ) = f (b) = 0. Then there exists a point c (a , b) such that

f (c) = 0.

Proof:1. Let

M = sup f ( I ), m = inf f ( I ) .

2. Thenm 0 M .



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3. If both m = M = 0 then f = 0.

4. If M > 0, then there is c (a , b) such that f (c) = M (maximum).5. If m < 0, then there is d (a , b) such that f (d ) = m (minimum).6. Thus, there is a point in ( a , b) where f vanishes.

Theorem 6.2.3 Mean Value Theorem. Let I = [a , b] be a closed and bounded interval, and f : I R. Let f be continuous on the closed interval [a , b] and di ff erentiable on the open interval (a , b). Then thereexists a point c (a , b) such that

f (c) = f (b) f (a )(b a ) .

Proof:

1. Consider the function

g( x) = f ( x) f (a ) f (b) f (a )

(b a ) ( x a )

and apply Rolles Theorem.

2. Then g (c) =

0 and the theorem follows.

Theorem 6.2.4 Let I = [a , b] be a closed and bounded interval, and f : I R. Let f be continuous on the closed interval [a , b] , di ff erentiableon the open interval (a , b) , and x (a , b)

f ( x) = 0.

Then f is constant on I.

Proof:

1. Let x (a , b).2. Then there is c (a , x) such that

f (c) = f ( x) f (a )

x a= 0



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6.2. THE MEAN VALUE THEOREM 85

3. Thus, f is constant.

Corollary 6.2.2 Let I = [a , b] be a closed and bounded interval, and f , g : I R. Let f , g be continuous on the closed interval [a , b] ,di ff erentiable on the open interval (a , b) , and x (a , b)

f ( x) = g ( x).

Then there exists a constant C such that f = g + C on I.

Proof: Easy.

Theorem 6.2.5 Let I be an int

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