Analysis on the Real Line

Cam Quilici - Instructor Dr. Andrea Bonito

Fall 2023

Contents

1 The Real Numbers 3

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Axioms of the Real Numbers . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Bounded Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Supremum and Inﬁmum . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Sequences 8

2.1 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Algebraic Manipulations of Limits . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Comparison Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 Limit Inﬁmum and Limit Supremum . . . . . . . . . . . . . . . . . . . . 13

2.5 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.6 Bolzano-Weirstrass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.7 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.8 Divergent Sequences to ±∞ (Unbounded Case) . . . . . . . . . . . . . . 21

3 Continuous Functions 22

3.1 Sequential Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Algebraic Manipulation of Continuous Limits . . . . . . . . . . . . . . . 25

3.3 Comparison Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.4 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.5 Inﬁnite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.6 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.7 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.8 Inverses and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 Differentiation 42

4.1 Extremum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2.1 Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2.2 Advanced Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3 Mean Value Theorem and Applications . . . . . . . . . . . . . . . . . . . 49

4.4 Limited Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.4.1 Local Extremum (Re-Visited) . . . . . . . . . . . . . . . . . . . . . 56

5 Integration 57

5.1 Extension of Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.2 Properties of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.3 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5.4 Properties of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.5 Change of Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5.6 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.7 Taylor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.8 Weak Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

1 The Real Numbers

1.1 Introduction

Lemma 1.1.1. The equation x

−2 =0 has no solution in Q.

Proof. By contradiction, assume there exists some

/q ∈Q, p, q ∈N, q 6=0 such

that

−2 =0. (∗)

Without loss of generality, we can assume that the greatest common divisor

between p and q is 1. We rewrite (*) as p

=2q

which implies that p

is even.

This means that p is even as well.

• We say that N is well-ordered, but not Q since Q does not have a least element.

Proposition 1.1.2. There is no natural number such that 0 <n <1.

Proof. Left as an exercise.

1.1.1 Axioms of the Real Numbers

• Binary operations:

(+): R ×R 7→R x, y ∈R, x +y ∈R,

(·): R ×R 7→R x, y ∈R, x ·y ∈R.

• We have axioms for the real numbers as follows:

I Algebraic Axioms

(i) Associativity: For x, y, z ∈R, we have

x +(y +z) =(x +y) +z

x ·(y ·z) =(x ·y)·z

(ii) Commutativity: For x, y ∈R, we have

x +y = y +x

x ·y = y ·x

(iii) Distributivity: For x, y, z ∈R, we have

x ·(y +z) =x ·y +x ·z

(iv) Identity:

Addition: There exists some 0 ∈R such that for all x ∈R, x+0 = x.

Multiplication: There exists some 1 ∈ R such that for all x ∈ R,

x ·1 =x.

(v) Inverses:

Addition: For all x ∈R, there exists some y ∈ R such that

x +y = 0 ⇐⇒ y =−x.

Multiplication: For all x ∈R, there exists some y ∈R such that

x ·y = 1 ⇐⇒ y =

⇐⇒ y = x

−1

II Ordering

(i) For some x, y, z ∈R, we have

x ≤ y =⇒ x +z ≤ y +z.

(ii) For some x, y ∈R, we have

0 ≤x,0 ≤ y =⇒ 0 ≤ x ·y.

(iii) For some x, y, z ∈R, we have

x ≤ y, y ≤ z =⇒ x ≤ z.

(iv) For some x, y ∈R, we have

x ≤ y, y ≤ x =⇒ x = y.

(v) For some x, y ∈R, we have

x 6= y =⇒ x ≤ y or y ≤ x.

III No Hole (not satisﬁed by Q)

For any non-empty subset X of {x >0}, there exists some a ∈R such that

(1) a ≤ x ∀x ∈ X , and

(2) For all ε >0(ε 6=0,0 ≤ε), there exists some x

such that x

−a ≤ε.

1.1.2 Bounded Intervals

• Given a <b, the set X of real numbers is a bounded interval if it is in the form:

Open interval with endpoints a,b

{x ∈ X | a < x <b} =(a,b).

Closed intervals

{x ∈ X | a ≤ x ≤b} =[a,b].

Half-open intervals

{x ∈ X | a < x ≤b} =(a,b].

1.2 Supremum and Inﬁmum

Deﬁnition 1.2.1 (Boundedness). Let X ⊆ R be non-empty. A number m ∈ R

(not necessarily in X ) is a lower bound for X if for all x ∈ X , m ≤x.

If a set has a lower bound, it is called bounded below (for instance, Z).

• Remark. X either has no lower bounds or inﬁnitely many.

Deﬁnition 1.2.2 (Inﬁmum). Let X ⊆ R be non-empty. A number s ∈ R (not

necessarily in X ) is called an inﬁmum of the set X if it is

(1) A lower bound and

(2) For all other lower bounds r

of X , s ≥r

The inﬁmum is hence known as the “greatest lower bound." We denote “r is

the inﬁmum of X " as

r =inf(X ).

Deﬁnition 1.2.3 (Supremum). Let X ⊆ R be non-empty. A number s ∈R (not

necessarily in X ) is called a supremum of the set X if it is

(1) An upper bound and

(2) For all other upper bounds r

of X , s ≤r

The supremum is hence known as the “least upper bound." We denote “r is

the supremum of X " as

r =sup(X ).

Theorem 1.2.4. Let X ⊆ R be non-empty. X has an upper bound if and

only if X has a ﬁnite supremum and the ﬁnite supremum is unique.

Proof. If X has a supremum then that supremum is an upper bound for X .

Assume that b is an upper bound of X . Let X ={b+1−y | y ∈ X }, then X is non

empty and X ⊂{x > 0} (there is a 1 so it is surely greater than 0).

So, by Axiom 3, there is a real number a such that

(1) If z ∈ X , then a ≤ z (i.e., a is a lower bound for X ) and

(2) For every ε >0, there exists a z

∈ X such that

−a ≤ε.

Claim. b

=b +1 −a is the supremum for X .

There are two things to check:

(1) Is b

an upper bound for X ? Well, if y ∈ X , then a ≤b +1−y and we have

y ≤b +1 −a =b

(2) Is b

the least upper bound?

Let M be some upper bound for X . Let us show that b

≤ M. By way

of contradiction, assume that M <b

=b +1 −a. Let ε =b

−M >0.

By Axiom 3, there exists some z ∈ X such that z −a ≤ε. We have

z = b +1−y, y ∈ X , ε =b

−M =b +1 −a −M.

So, supposedly z − a = b +1 −y −a ≤ b +1 −a −M with M ≤ y (this is

a contradiction since M is assumed to be an upper bound, it cannot be

smaller than y) and therefore b

≤M and b

=sup(X ).

• Archimedean Axiom (satisﬁed in R). For every x >0, y ≥0, there exists n ∈N

∗

such that

n ·x > y.

Proof. By way of contradiction, assume that n ·x ≤ y ∀n ∈ N

∗

(for some x >

0, y ≥0). Then let S ={n ·x |n ∈N

∗

} ⊂R. Clearly S is bounded above by y and

S is non-empty. This implies that sup(S) exists and n ·x ≤sup(s) ∀n ∈N

∗

. This

all implies that (m +1) ·x ≤sup(S) ∀m ∈N

∗

Lemma 1.2.5. Let X ⊂R be non-empty and bounded above. Then for every

ε >0, there exists some x

such that

sup(X )−ε <x

≤sup(X ).

Proof. Since sup(X ) is an upper bound for X , then we know x ≤ sup(X ) ∀x ∈

X . This implies the second part of the inequality.

To prove the ﬁrst part of the inequality, assume by way of contradiction that

there exists some ε

> 0 such that for every x ∈ X , sup(x) −ε

≥ x. However,

this is not possible since sup(x) is an upper bound, i.e., subtract anything from

it and it will be less than x.

Lemma 1.2.6. Let x, y ∈ R with y > x, then there exists some r ∈ Q such

that x <r < y.

Proof. Take y −x > 0, the Archimidean principle guarantees the existence of

n ∈N such that

n ·(y −x) >1 ⇐⇒ y −x >

Recall,

bxc≤ x ≤cxb+1

bnxc≤

nx +1

Let r :=

(bnxc+1)

/n ∈ Q. Then we have

x =

bnxc+1

≤

nx +1

=x +

=x +y −x

< y.

• Corollary. Let p, q ∈ Q with p < q. There exists some z ∈ R \ Q such that

p <z < q.

Proof. By Lemma,

⇐⇒

< r

|{z}

∈Q

⇐⇒ p < r ·c

|{z}

∈R \ Q

<q.

1.2.1 Absolute Value

Deﬁnition 1.2.7 (Absolute value). For some x ∈R, the absolute value function

is deﬁned as follows:

|x|:=

(

−x , if x <0

x , if x ≥0.

Proposition 1.2.8. Let x,a ∈R. Then we have

(i) |x|=0 ⇐⇒ x =0.

(ii) Let a >0, then

|x|< a ⇐⇒ −a <x < a

2 Sequences

2.1 Convergence

Deﬁnition 2.1.1 (Sequential convergence). A sequence of real numbers (x

)

n∈N

is said to be convergent if there exists x ∈ R such that ∀ε > 0 ∃N (ε) ∈ N such

that

|x −x

|≤ε ∀n ≥ N(ε).

Furthermore, we write lim

n→∞

=x.

• Example. Show that lim

n→∞

1 +

/(n+1) = 1. We want

|x −x

1 −

1 +

n +1

≤ε ∀ε >0.

Let ε >0, set N (ε) =b

/εc=N(ε). This holds for

|x −x

n +1

≤

c+1

≤

≤ε.

Lemma 2.1.2. A sequence of real numbers has at most one limit.

Proof. Assume l

∈ R are two limits. We will show that l

= l

by showing

that for all ε >0, |l

−l

|≤ε. Let ε >0.

(1) (x

)

n∈N

converges to l

, i.e., for all ε

>0 ∃N

(ε

) such that

−l

|≤ε

∀n ≥ N

(ε

(2) (x

)

n∈N

converges to l

, i.e., for all ε

>0 ∃N

(ε

) such that

−l

|≤ε

∀n ≥ N

(ε

Let ε >0 and set ε

−

/2,ε

/2. Then we have

−l

|=|l

−x

−l

≤|l

−x

|+|l

−x

|≤ε

where |l

−x

| ≤ ε

∀n ≥ N

and |l

−x

| ≤ ε

n ≥ N

n ≥ max{N

, N

}. Alto-

gether, this implies that l

• Example of Divergent Sequence. Take x

= n ·sin(n ·

/2). By way of contra-

diction, assume that (x

)

n∈N

is converging to some x ∈R. Say

4n+1

=(4n +1) ·sin((4n +1) ·

)

| {z }

=4n +1

4n+5

=(4n +5) ·sin((4n +5) ·

)

| {z }

=4n +5

which implies that |x

−x

| = 4. From the contradiction assumption, for

ε =

/2 there exists N (

/2) such that

−x|≤

∀n ≥ N(

Take n such that 4n +1 ≥N (

/2). Then we have

−x|≤

4 =|x

−x

|=|x

−x +x −x

≤|x

−x|+|x

−x|

≤

≤1

but this is a contradiction, so (x

)

n∈N

must be divergent.

Lemma 2.1.3. Every convergent sequence is bounded.

Proof. Let (x

)

n∈N

be a convergent sequence, that is, there exists some x ∈ R

such that for all ε >0, there exists some N(ε) ∈N and

|x −x

|≤ε ∀n ≥ N(ε).

In particular, take for ε =1, there is N (1) such that

−x|≤1 ∀n ≥ N(1).

Recall the reverse triangle inequality,

|−|x|≤|x −x

| ≤1 n ≥ N(1)

=⇒ |x

|≤1 +|x| n ≥ N (1)

Let M =max{|x

|,|x

|,. .. ,|x

N(1)−1

|,1 +|x|}. Then certainly

|≤M n ≤ N(1)

|≤M n ≥ N(1)

=⇒ | x

|≤M ∀n ∈N.

• Recall. A sequence (x

)

n∈N

convergent =⇒ ∃M ∈N such that |x

|≤M ∀n ∈N.

Deﬁnition 2.1.4 (Sequential monotonicity). A sequence (x

)

n∈N

is said to be

Increasing if x

≥x

when m ≥n,

Decreasing if x

≤x

when m ≥n,

Monotone if (x

)

n∈N

is increasing or decreasing.

Lemma 2.1.5. Any increasing sequence that is bounded above is conver-

gent.

Proof. Let X = {x

| n ∈ N}. Then x

≤ M,n ∈ N =⇒ X bounded above =⇒

x = sup(X ).

We want to show that (x

)

n∈N

converges to x =sup(X ). This means (by deﬁni-

tion) that for every ε >0, there exists N (ε) ∈N such that

−x|≤ε ∀n ≥ N(ε).

Note that x =sup(X ) implies two important things:

(i) x

≤x ∀n ∈N,

(ii) ∀ε >0, ∃x

∈ X such that x−x

≤ε by the characterization of supremum.

Note, we can equivalently replace x

with x

∈N to obtain ∀ε >0, ∃x

∈

X such that x −x

≤ε.

So, let ε >0, compute

−x|= x −x

≤x −x

≤ε

with x

≤ x

and n ≥ n

((x

)

n∈N

increasing) and for all ε > 0, take N (ε) =

• Corollary. A decreasing sequence (x

)

n∈N

that is bounded below is conver-

gent.

• Remark. There are two cases:

(i) (x

)

n∈N

increasing, bounded above =⇒ lim

n→∞

=sup{x

|n ∈N},

(ii) (x

)

n∈N

decreasing, bounded below =⇒ lim

n→∞

=inf{x

|n ∈N}.

• Example. Take x

−n

decreasing, bounded below. Then

lim

n→∞

=inf{2

−n

|n ∈N} =0.

• Corollary. Let (x

)

n∈N

be a monotone sequence. Then (x

)

n∈N

is convergent

if and only if (x

)

n∈N

is bounded.

2.2 Algebraic Manipulations of Limits

Lemma 2.2.1. Let (x

)

n∈N

,(y

)

n∈N

be convergent sequences. Set lim

n→∞

x,lim

n→∞

= y. Three things are true:

(1) The sequence (x

)

n∈N

is convergent and li m

n→∞

) =x+y,

(2) The sequence (x

·y

)

n∈N

is convergent and lim

n→∞

·y

) =x ·y,

(3) Assume x

6=0, y

6=0, then the sequence (

)

n∈N

is convergent and

lim

n→∞

(

) =

/x.

Proof. of (2). We want to show that ∀ε >0 ∃N (ε) ∈N such that

−x y|≤ε ∀n ≥N (ε).

Note that

−x y|=|(x

−x)y

+x(y

−y)|

≤|x

−x||y

|+|x||y

−y|

by the triangle inequality.

Case 1: x 6= 0. Because (y

)

n∈N

is convergent, there exists N

∈ N such

that |y

−y|≤

/2·|x| ∀n ≥ N

. Also, (y

)

n∈N

is bounded, i.e., |y

|≤M ∀n ∈

N ∃M ≥1.

Because (x

)

n∈N

is convergent, there exists some N

∈N such that

|x −x

|≤

∀n ≥ N

Therefore, for all ε >0, we set N

=max{N

, N

} and we have that

−x y|≤

≤ε ∀n ≥ N

which ultimately shows that lim

n→∞

·y

=x ·y.

Case 2: x = 0. Obviously it is true that

−x y|=|x

|≤|x

||y

Then, consider that

)

n∈N

convergent implies |y

|≤M , and

)

n∈N

convergent to 0 implies ∀ε > 0 ∃N

∈ N such that |x

| ≤

/2M ∀n ≥ N

. Again, we have

−x y|≤

·M ≤

≤ε ∀n ≥ N

which implies that lim

n→∞

=0.

2.3 Comparison Results

Lemma 2.3.1. Let (x

)

n∈N

and (y

)

n∈N

be two convergent sequences such

that there exists N

∈N such that

≤ y

∀n ≥ N

Then

lim

n→∞

≤ lim

n→∞

Proof. Let x := lim

n→∞

, y :=lim

n→∞

. For the sake of contradiction, assume

that y < x. Because lim

n→∞

= x and lim

n→∞

= y, we have that for ε =

(x−y)

/4, there exists some N ∈ N such that

−x|≤ε and |y

−y|≤ε ∀n ≥ N.

Note that

−x|≤ε =⇒ x ≤ x

+ε

−y|≤ε =⇒ y ≥ y

−ε.

This means that for n ≥max{N

, N}, we have

≤ y +ε = y +

x −y

< y +

x −y

| {z }

(x+y)/2

=x −

x −y

≤x −

x −y

=x −ε <x

which implies that y

which is a contradiction.

Theorem 2.3.2 (Squeeze Theorem). Let (x

)

n∈N

, (y

)

n∈N

, (z

)

n∈N

be se-

quences such that

Sequences (x

)

n∈N

and (z

)

n∈N

are both converging to some limit L,

and

There exists some N

∈N such that

≤ y

≤z

∀n ≥ N

Then (y

)

n∈N

is convergent and lim

n→∞

=L.

Proof. Since lim

n→∞

=L and lim

n→∞

=L, we have that ∀ε >0 ∃N (ε) ∈N

such that

−L|≤ε and |z

−L|≤ε ∀n ≥ N(ε).

Then, consider N ≥ max{N (ε), N

} and we have

L −ε ≤x

≤ y

≤z

≤L +ε

i.e., L−ε ≤ y

≤L+ε or |L−y

|≤ε ∀n ≥max{N(ε),N

} =⇒ lim

n→∞

=L.

2.4 Limit Inﬁmum and Limit Supremum

• Let (x

)

n∈N

be a bounded sequence. Construct a new sequence y

={x

: k ≥

n}. We want to show two things about this sequence:

(1) (y

)

n∈N

is decreasing:

Consider n <m, then we have y

=sup{x

: k ≥n}

{z }

and y

=sup{x

: k ≥m}

| {z }

It is clear that B ⊂ A =⇒ sup(B ) ≤ sup(A) (by an exercise) which alto-

gether implies that y

≤ y

(2) (y

)

n∈N

is bounded below:

Since (x

)

n∈N

is bounded below, so is (y

)

n∈N

Deﬁnition 2.4.1 (limsup and liminf). Let (x

)

n∈N

be a sequence that is bounded

above. Then we deﬁne the limit superior as

limsup

n→∞

= lim

n→∞







sup{x

: k ≥n}

| {z }







Similarly, when (x

)

n∈N

is bounded above, we deﬁne the limit inferior as

liminf

n→∞

= lim

n→∞





inf{x

: k ≥n}

| {z }





• Example. Consider the sequence x

=(−1)

. Then we have

limsup

n→∞

= lim

n→∞

sup{x

: k ≥n} =1

liminf

n→∞

= lim

n→∞

inf{x

: k ≥n} =−1.

Since limsup

n→∞

6=liminf

n→∞

, (x

)

n∈N

diverges.

Theorem 2.4.2. Let (x

)

n∈N

be a bounded sequence. The sequence (x

)

n∈N

is convergent if and only if

limsup

n→∞

=liminf

n→∞

Proof. Assume that (x

)

n∈N

is convergent and let x := lim

x→∞

. Further, de-

note

=inf{x

: k ≥n}

z = lim

n→∞

=liminf

n→∞

(we want to show that x = z.) Since x

converges to x, we know that for every

ε >0, there exists some N

(ε) ∈N such that

|x −x

|≤

∀n ≥ N

(ε)

and since z

converges to z we similarly have that there exists some N

(ε) ∈N

such that

|z −z

|≤

∀n ≥ N

(ε).

Recall that z

= inf{x

: k ≥ n}. Then, thanks to the characterization of in-

ﬁmums, we know that for every ε > 0 (pick our ε =

/4), there exists some

∈{x

: k ≥n} such that

−z

|=x

−z

≤

since m

≥n and z

≤x

Now, compute

|x −z|=|x −x

−z

−z|

≤|x −x

| {z }

≤

+|x

−z

| {z }

≤

+|z

−z|

| {z }

≤

≤ε ∀m ≥n ≥ N

(ε).

So we proved that ∀ε >0,

|x −z|≤ε

=⇒ x −z =0

=⇒ x = z.

From a similar argument, lim

n→∞

= lim sup

n→∞

= y, i.e., x = y =⇒ y =

The other direction is much simpler. Assume y =limsup

n→∞

=liminf

n→∞

z. Then, recall

y ← y

=sup{x

: k ≥n}

z ← z

=inf{x

: k ≥n}.

So, by our assumption we have z

≤ x

≤ y

which by the squeeze theorem

implies that z = y. Therefore, x

→ y = z i.e., lim

n→∞

= y = z.

2.5 Subsequences

Deﬁnition 2.5.1 (Subsequences). A subsequence (x

)

n∈N

is a sequence (y

)

n∈N

such that for all k ∈N, there exists n

,. .. ,n

∈N with n

<···<n

with

Lemma 2.5.2. Any subsequence (x

) of a convergent sequence (x

)

k∈N

convergent to the same limit.

Proof. Left as an exercise.

Lemma 2.5.3. Let (n

)

∞

k=1

be a sequence of increasing natural numbers.

Then

≥k.

Proof. By induction (left as an exercise).

Proof. (Of ﬁrst Lemma). For every ε >0, there exists some N(ε) ∈N such that

|x −x

|≤ε ∀k ≥ N(ε)

(because x

→x). From previous Lemma (above), n

≥k ≥N (ε) so we have

|x − x

|{z}

|≤ε ∀n

≥N (ε).

Lemma 2.5.4. Any subsequence of a bounded sequence is bounded.

Proof. Left as an exercise.

2.6 Bolzano-Weirstrass

Deﬁnition 2.6.1 (Peak points). A peak point of a sequence (x

)

n∈N

is a term

such that x

means q > p.

Lemma 2.6.2 (Monotone Subsequence). Every sequence has a monotone

subsequence.

Proof. We argue on whether the sequence has either (A) no peak point, (B)

ﬁnitely many peak points, or (C) inﬁnitely many peak points.

(A) We construct (x

)

k∈N

as follows:

=1, x

is not a peak point (no peak points)

=⇒ q >1 such that x

=q, x

≤x

≤···montonically increasing

(B) If x

is the ﬁnal peak point (p is the largest index where x

is a peak

point), then take n

=p +1, x

p+1

. Then proceed as before.

<···<p

<··· .

By deﬁnition, x

> x

for j > i . So we have a decreasing subsequence

taking x

Theorem 2.6.3 (Bolzano-Weirstrass). Every bounded sequence has a con-

vergent subsequence.

−M M

Proof. We have that (x

)

n∈N

implies that (x

)

k∈N

is monotone. Further, (x

)

n∈N

bounded implies that (x

)

k∈N

is bounded. So we have a montone and bounded

subsequence which is thus convergent.

Lemma 2.6.4. Let (x

)

n∈N

be a bounded sequence. There is a convergent

subsequence x

such that

lim

k→∞

=limsup

n→∞

Proof. We proceed by induction. Let y = lim

n→∞

, y

= sup{x

: k ≥ n}

such that y = lim sup

n→∞

P(k): There are n

<···<n

and x

, x

,. .. , x

such that

y −

|{z}

→y

< y +

|{z}

→y

∀1 ≤i ≤k

which implies that x

→ y as well by squeeze theorem.

P(1): There is n

∈N and x

such that y −1 <x

< y +1.

We have y

→ y =⇒ ∃N ∈ N such that |y

− y| ≤

/4 ∀n ≥ N. Further we

have y

= sup{x

: k ≥ n} =⇒ ∃x

,m ≥ n such that y

−x

≤

/4 by the

characterization of suprema. Then compute

|y −x

|=|y − y

−x

≤|y − y

|+|y

−x

≤

so y −

/2 ≤x

≤ y +

/2 =⇒ y −1 <x

< y +1.

Next, assume P (k) is true: There is n

<···<n

and x

, x

,. .. , x

such

that

y −

< y +

∀1 ≤i ≤k.

Since y

→ y, ∃N ∈ N such that |y − y

| ≤

/(4(k+1)), n ≥ max{N ,n

k+1

} > N

Further, we have y

= sup{x

: k ≥ n} which implies there exists some m ∈N

with m ≥n > n

such that

−x

4(k +1)

Therefore, we have

|y − y

|≤|y − y

|+|y

−x

≤

4(k +1)

2(k +1)

k +1

=⇒ y −

k +1

< y +

k +1

k+1

=m >n

Therefore, P(k +1) is true. Thus, we have shown that P (k) is true for all k ∈ N

and x

→ y by the squeeze theorem.

• Remark. Every bounded sequence (x

)

n∈N

has a convergent subsequence

)

k∈N

such that

lim

k→∞

=liminf

n→∞

(Proof is the same).

• Result (forgot to mention last class). If (x

)

n∈N

is bounded, then every conver-

gent subsequence (x

)

k∈N

satisﬁes

liminf

n→∞

≤ lim

k→∞

≤limsup

n→∞

• Example. Take x

=(−1)

. We want to ﬁnd the limsup.

Show that (x

)

n∈N

is bounded, i.e., x

≤ 1 =⇒ limsup

n→∞

≤ 1. Then, con-

sider the subsequence (x

) = (1)

k∈N

. Then show that (x

) →

k→∞

1 =⇒ 1 ≤

limsup

n→∞

Since we have shown that limsup

n→∞

≤ 1 ≤ lim sup

n→∞

, we know that

limsup

n→∞

=1.

2.7 Cauchy Sequences

Deﬁnition 2.7.1 (Cauchy sequences). A sequence (x

)

n∈N

is Cauchy if ∀ε >

0,∃N ∈N with

−x

|≤ε ∀m, n ≥ N.

• Example. Take x

/2(x

n−1

n−2

), x

=0, x

=1. Then we compute

−x

n−1

n−2

) −x

n−1

n−2

−x

n−1

)

−x

=−

=⇒ x

=−

+1 =

−x

(1 −

) =

Then, by induction we can show that

−x

n−1

−

n−1

Now take m ≥n and compute

−x

m−1

−x

m−2

±···±x

n+1

−x

|≤

m−1

m−2

+···+

1 +

+···+

m−n−1

1 −

m−2

1 −

≤2 (by geometric sum)

≤2

−n+1

Given ε >0, if we want 2

−n+1

≤ε, compute

−n+1

≤ε

−n +1 ≤ln(ε)

n ≥ln(ε) +1.

Therefore, take N = bln(ε)c+2, n ≥ ln(ε) +1. So, for m,n ≥ N =⇒ |x

−x

| ≤

ε =⇒ x

Cauchy. In fact, lim

n→∞

/3 (details left to reader).

Warning: Checking the difference between two subsequent terms of the se-

quence is not enough to ensure that the sequence is Cauchy.

For example, consider x

k=1

/k (which is not Cauchy).

Lemma 2.7.2. A sequence (x

)

n∈N

is convergent if and only if (x

)

n∈N

Cauchy.

Proof. First, assume that x

→x. Therefore, ∀ε >0, ∃N ∈N such that

−x|≤

∀n ≥ N.

Therefore, for ε >0 and m,n ≥N ,

|x −x

|≤

and |x −x

|≤

This implies that

−x

|=|x −x

+x −x

≤|x −x

|+|x −x

≤

=ε

=⇒ x

Cauchy.

For the other direction, assume (x

)

n∈N

is Cauchy. First, we must prove some

claims that will lead us to this proof.

(1) Claim. If (x

)

n∈N

is Cauchy, then (x

)

n∈N

bounded.

Proof. This proof is omitted because it is almost identical to the proof as

convergent implies bounded.

(2) Claim. If (x

)

n∈N

has a convergent subsequence (x

)

k∈N

, then (x

)

n∈N

is convergent (to the same limit).

Proof. To see this, assume there exists some (x

) convergent subsequence

and

(i) We know x

is Cauchy, i,e„ ∀ε >0, ∃N

∈N such that

−x

|≤

∀m, n ≥ N

(ii) There exists an x ∈N such that ∀ε >0, ∃N

∈N such that

|x −x

|≤

∀k ≥N

Now, we will show that the entire sequence (x

) converges to x.

For every ε >0, set N =max N

, N

. Then compute

|x −x

|≤|x −x

|+|x

−x

≤

∀k ≥N .

So, we have

) Cauchy

(1)

=⇒(x

) bounded

(B.W.)

===⇒(x

) has a convergent subsequence

(2)

=⇒(x

) convergent.

2.8 Divergent Sequences to ±∞ (Unbounded Case)

Deﬁnition 2.8.1 (Divergence to ±∞). We say that a sequence (x

)

n∈N

is di-

verging to +∞ if ∀M ∈R, ∃N ∈N such that

≥M ∀n ≥ M.

Similarly, we say that a sequence is diverging to −∞ if ∀M ∈ R, ∃N ∈ N such

that

≤M ∀n ≥ N.

Lemma 2.8.2. If (x

)

n∈N

is converging to ±∞, then every subsequence is

convering to ±∞ (respectively).

3 Continuous Functions

Deﬁnition 3.0.1 (Functional limits). Let a < b, x

∈ (a,b), f : (a,b) \ {x

} 7→ R.

We say that f has a limit l at x

if ∀ε >0 ∃δ >0 such that, for x ∈(a,b)

0 <|x −x

|≤δ =⇒ |f (x) −l|≤ε.

If f has a limit l, we write

lim

x→x

f (x) =l.

a −δ

a +δ

L −ε

L +ε

f (x)

Proposition 3.0.2. Let x

∈(a,b), f : (a,b)\ {x

} 7→R. If f has a limit at x

then the limit must be unique.

Proof. For the sake of contradiction, assume that l

are two limits of f at x

and further assume that l

. Take ε =(l

−l

)/4. Then we have

a limit means ∃δ

>0 such that

0 <|x −x

|≤δ

=⇒ |f (x) −l

|≤ε.

a limit means ∃δ

>0 such that

0 <|x −x

|≤δ

=⇒ |f (x) −l

|≤ε.

Then, compute for 0 <|x −x

|≤min{δ

,δ

}

−l

≤|l

− f (x)|+|l

− f (x)|

≤ε +ε =2ε

−l

Of course, this is a contradiction so l

• Example. Take f (x) = x

. We wish to show that lim

x→2

f (x) =4. To show, we

have ∀ε >0 ∃δ >0 such that

0 <|x −2|≤δ =⇒ |x

−4|≤ε.

Then, compute

−4|=|x −2|

|{z}

≤δ

|x +2|

≤δ(|x|+2)

≤δ(4 +δ)

=⇒ | x|≤2 +δ

(since |x|= |x −2 +2| ≤|x −2|+2). Now we want δ (4 +δ) ≤ε. Further, restrict

δ ≤1 =⇒ δ

≤δ and δ(4 +δ) ≤5δ. We must now show that 5δ ≤ε.

So, ∀ε >0, set δ =min{1,

/5}. Then

0 <|x −2|≤δ =⇒ |x

−4|≤5δ ≤ε.

So indeed lim

x→2

f (x) =4.

• Example. Take f (x) =(x−1)/(x−1), x 6=1. This function is not deﬁned at x =1,

so take f : (−2,2) \ {1} 7→ R. However, we claim still that lim

x→∞

f (x) = 1. We

show ∀ε >0, set δ =1. Then we have

0 <|x −1|≤δ =⇒ |f (x) −1|=|1 −1|=0 ≤ε.

3.1 Sequential Limits

Proposition 3.1.1. Let x

∈(a,b), f : (a,b)\{x

} 7→R. Then f has the limit l

at x

if and only if for every sequence (z

)

n∈N

, z

∈(a,b)\{x

} with z

→x

then the sequence (f (z

))

n∈N

has

f (z

) →l.

Proof. First, suppose lim

x→x

f (x) = l and (z

)

n∈N

, z

∈ (a,b) \ {x

} with z

→

. Then, we know the following

(1) lim

x→x

f (x) =l =⇒ ∀ε > 0 ∃δ > 0 such that 0 < |x −x

| ≤ δ =⇒ |f (x) −

l|≤ε and

(2) z

→x

=⇒ ∀

ε ∃

N ∈N such that |z

−x

|≤

ε ∀n ≥

Then, let ε >0 be given. Set

ε = δ so that n ≥ N =⇒ 0 <|z

−x

|≤ δ and then

together with (1) this implies |f (z

)−l|≤ε. Of course, this means lim

n→∞

f (z

) =

l by deﬁnition.

For the other direction, assume by contradiction that lim

x→x

f (x) 6=l. By def-

inition, this implies that ∃ε

≥ 0 such that ∀δ > 0 and 0 < |x −x

| ≤ δ =⇒

|f (x) −l | > ε

. Now, let n ∈N and z

∈(x

−

/n, x

/n) \{x

}. By the Squeeze

Theorem (Theorem 2.3.2)

−

≤z

≤x

=⇒ z

→x

By assumption, lim

n→∞

f (z

) =l, i.e., ∀ε >0 ∃n ∈N such that ∀n ≥ N , |f (z

)−

l|≤ε. Taking δ =

/n, we obtain a contradiction. Therefore, our contradiction

assumption must be false and lim

x→x

f (x) =l.

• Example. Take f (x) =

/|x|, x 6= 0. How do we show that this does not have a

limit at x = 0? We could use the epsilon-delta proof technique, but this could

be long and tedious. Instead, consider

Let z

=−

→0

f (z

) =−1 →−1

Let

→0

f (

) =1 →1

=⇒ f (x) does not have a limit at x

• Corollary. Let x

∈ (a,b) \ {x

}, f : (a,b) 7→ R. Assume that every sequence

)

n∈N

, z

∈ (a,b) \ {x

}, z

→ x

, one has (f (z

))

n∈N

is convergent. Then f

has a limit at x

Proof. We will show that for (v

)

n∈N

, (w

)

n∈N

two sequences with v

, w

∈

(a,b) \ {z

} and v

→x

, w

→x

, one has lim

n→∞

f (v

) =lim

n→∞

f (w

For the sake of contradiction, assume l

=limn →∞f (v

) 6=lim

n→∞

f (w

) =

. Set

(

n even

n odd.

Then we have (1) y

∈ (a,b) \ {x

} and (2) y

→ x

together imply ( f (y

))

n∈N

convergent. However,

f (y

) =

(

f (v

) n even

f (w

) n odd

cannot be convergent since f (y

) → l

and f (y

2n+1

) → l

which contradicts

our assumption that l

6= l

. Therefore, our assumption must be false and

. Combining this with our previous result (Proposition 3.1.1) we get that

lim

x→x

f (x) =l

Lemma 3.1.2 (Cauchy Criteria for Functions). Let x

∈(a,b)\{x

}, f : (a,b)\

} 7→R. Then f has a limit at x

if and only if for every ε > 0, there exists

δ >0 such that for every x

, x

satisfying

0 <|x

−x

|≤δ 0 <|x

−x

|≤δ,

one has

|f (x

) − f (x

)|≤ε.

Proof. First, assume that lim

x→x

f (x) =l. This implies that ∀ε >0 ∃δ >0 such

that |0 < |x −x

| ≤ δ =⇒ |f (x) −l| ≤

/2. Let ε > 0 be given, then there exists

some δ >0 such that 0 <|x

−x

|≤ δ and 0 <|x

−x

|≤ δ =⇒ |f (x

) −l |≤

for i =1,2. Therefore,

|f (x

) − f (x

)|≤|f (x

) −l|+|f (x

) −l|

≤

=ε.

For the other direction, we will show that for every sequence (z

)

n∈N

, z

∈

(a,b) \ {x

}, (f (z

))

n∈N

is convergent (Corollary implies f has a limit at x

Recall the fact that (f (z

))

n∈N

convergent ⇐⇒ (f (z

))

n∈N

Cauchy. Thus, we

only need to show that ∀ε >0 ∃n ∈N such that ∀m,n ≥N , |f (z

) − f (z

)|≤ε.

Since z

→ x

by assumption, we also know that for every δ > 0, ∃N ∈ N such

that ∀n ≥ N, 0 <|z

−x

|≤δ.

Putting everything together, we get that ∀ε > 0, ∃N ∈ N and 0 < |x

−z

| ≤

δ ∀n ≥ N which in turn implies that |f (z

) − f (z

)| ≤ ε ∀m,n ≥ N. So, we

know that (f (z

))

n∈N

is Cauchy. Therefore, (f (z

))

n∈N

is convergent.

3.2 Algebraic Manipulation of Continuous Limits

• The theme of this subsection is: “everything we know about sequences holds

for limits of continuous functions."

Lemma 3.2.1. Let f , g : (a,b)\{x

} 7→R with lim

x→x

f (x) =l

and lim

x→x

g (x) =

. Then

(1) f +g has a limit at x

and lim

x→x

(f +g ) =l

(2) For λ ∈R, λ · f has a limit at x

and lim

x→x

(λ · f ) =λ·l

(3) f ·g has a limit at x

and lim

x→x

(f ·g ) =l

·l

(4) Assume f (x) 6=0,x ∈ (a,b) \ {x

} and l

6=0. Then

/f has a limit at x

and lim

x→x

(

/f ) =

Proof. (1) Let (z

)

n∈N

, z

∈(a,b) \ {x

}, z

→x

. We know that

lim

x→x

f (x) =l

=⇒ f (z

) →l

lim

x→x

g (x) =l

=⇒ g(z

) →l

Then, using the properties of sequences,

lim

n→∞

(f (z

) +g (z

)) =l

which implies that lim

x→x

(f +g )(x) =l

3.3 Comparison Results

Lemma 3.3.1. Let f , g : (a,b) \ {x

} 7→R. Assume further that

(1) lim

x→x

f (x) =l

and lim

x→x

g (x) =l

and

(2) ∃α >0 such that 0 <|x −x

|≤α =⇒ f (x) ≥g (x).

Then l

≥l

Proof. Let (z

)

n∈N

, z

∈(a,b)\{x

}, z

→x

. This tells us that ∃N ∈N such that

∀n ≥ N, 0 <|z

−x

|≤α

(2)

=⇒ f (z

) ≥g (z

) ∀n ≥ N.

Then, using the properties of sequences, we get that

= lim

n→∞

f (z

) ≥ lim

n→∞

g (z

) =l

Lemma 3.3.2. Let f : (a,b) \ {x

} 7→ R such that lim

x→x

f (x) = l . Then

lim

x→x

|f (x)|=|l|.

Proof. By the reverse triangle inequality, we have 0 ≤||f |−|l ||≤ |f −l|. Then,

let

f (x) =

(

1 x ∈ Q

−1 x ∈ R \ Q.

Then |f (x)|=1 =⇒ lim

x→0

|f |(x) =1. But δ >0 in (0,δ) there is always

∈Q ∩(0,δ)

∈R \ Q ∩(0,δ).

We know further that |f (x

) − f (x

)| = 2 which implies that f does not have

a limit at 0 since x

and x

do not satisfy the criteria of a Cauchy sequence

(Lemma 3.1.2).

Theorem 3.3.3 (Squeeze Theorem). Let f ,g,h : (a,b) \ {x

} 7→ R. Further,

assume

(1) lim

x→x

f (x) =l =lim

x→x

g (x) and

(2) ∃α >0 such that 0 <|x −x

|≤α.

Then lim

x→x

h(x) =l.

Proof. Let (z

)

n∈N

, z

∈ (a,b) \ {x

}, z

→ x

. This means ∃N ∈ N such that

∀n ≥ N, 0 <|z

−x

|≤α. By (2) of Lemma 3.3.1, we have that

f (z

) ≤h(z

) ≤g (z

) ∀n ≥ N.

Therefore, we have lim

n→∞

h(z

) =l and because (z

)

n∈N

is arbitrary, lim

x→x

h(x) =

• Example. Take f : (−

/2,

/2) \ {0} 7→ R deﬁned by f (x) =

sin x

/x, x > 0. How can

we show that lim

x→0

f (x) = 1? We could show this rigorously with the deﬁni-

tion of limits, however this would be tedious. Instead, consider that we know

sin x ≤ x. In particular,

|1 −cos x|=2

sin

≤

This implies that lim

x→0

cos x = 1. On the other hand,

sin x

≤π ·

2π

≤

tan x

and for x 6=0, we obtain

1 ≤

sin x

≤1.

Since cos x −→

x→0

1 and 1 −→

1→0

1, we have by the squeeze theorem for functions

that lim

x→0

sin x

/x =1.

3.4 Composition of Functions

Lemma 3.4.1 (Composition of Functions). Let f : (a,b)\{x

} 7→(c,d) such

that lim

x→x

f (x) = y

. Let g : (c,d)\{y

} 7→R and assume that lim

y→y

g (y) =

l. Then g ◦ f has a limit at x

and lim

x→x

(g ◦ f )(x) =l.

Note: we must assume ∃α > 0 such that 0 <|x −x

|≤α =⇒ f (x) = y

Proof. We know two things by assumption:

(1) lim

x→x

f (x) = y

means

∀ε

>0 ∃δ

>0 such that 0 <|x −x

|≤δ

=⇒ |f (x) −y

|≤ε

(2) lim

y→y

g (y) =l means

∀ε

>0 ∃δ

>0 such that 0 <|y −y

|≤δ

=⇒ | g (y) −l|≤ε

Here, we further assume that ∃α >0 such that 0 <|x −x

|≤α =⇒ f (x) 6=

. Altogether this implies that

0 ≤|x −x

|≤min{δ

,α} =⇒ 0 <|f (x) −y

|≤ε

So we want to show that ∀ε >0 ∃δ >0 such that

0 <|x −x

|≤δ =⇒ |g ( f (x)) −l| ≤ ε.

Let ε > 0 be given. Set ε

= ε =⇒ ∃δ

, set ε

= δ

=⇒ δ

and then set δ =

min{δ

,α}. Thus,

0 <|x −x

|≤δ =⇒ 0 <| f (x)

|{z}

−y

|≤ε

=δ

=⇒ | g ( f (x)) −l|≤ε.

3.5 Inﬁnite Limits

Deﬁnition 3.5.1 (Inﬁnite functional limits). We say that f : (a,b) \ {x

} 7→ R

tends to +∞ when x tends to x

if ∀M ∈R ∃δ >0 such that

0 <|x −x

|≤δ =⇒ f (x) ≥M.

We write lim

x→x

f (x) =+∞.

Similarly, we say that f tends to −∞ when x tends to x

if ∀M ∈ R ∃δ > 0

such that

0 <|x −x

|≤δ =⇒ f (x) ≤M.

We write lim

x→x

f (x) =−∞.

Deﬁnition 3.5.2 (Inﬁnite functional limits). We say that f : (a,+∞) 7→R has a

limit l when x tends to +∞ if ∀ε >0 ∃D ∈R such that

x ≥ D =⇒ |f (x) −l|≤ε.

We write lim

x→+∞

f (x) =l.

Similarly, we say that f : (−∞, b) 7→ R has a limit l when x tends to −∞ if

∀ε >0 ∃D ∈R such that

x ≤ D =⇒ |f (x) −l|≤ε.

We write lim

x→−∞

f (x) =l.

Deﬁnition 3.5.3 (Combined inﬁnite functional limits). We say that f : (a,+∞) 7→

R tends to +∞ when x →+∞ if ∀M ∈R ∃D ∈R such that

x ≥ D =⇒ f (x) ≥ M .

We write lim

x→+∞

f (x) =+∞.

Similarly, we say that f : (−∞,b) 7→ R tends to −∞ when x → −∞ if ∀M ∈

R ∃D ∈R such that

x ≤ D =⇒ f (x) ≤ M .

We write lim

x→−∞

f (x) =−∞.

3.6 Continuity

Deﬁnition 3.6.1 (Functional continuity). We say that f : (a,b) 7→ R is contin-

uous at x

∈(a,b) if

lim

x→x

f (x) = f (x

This is equivalent to:

(i) ∀ε >0 ∃δ(ε, x

) >0 such that |x −x

|≤δ =⇒ |f (x) − f (x

)|≤ε.

Proof. Left as an exercise.

(ii) ∀ε > 0 ∃δ(ε, x

) > 0 such that |x

−x

| ≤ δ and |x

−x

| ≤ δ =⇒ |f (x

) −

f (x

)|≤ε.

Proof. Left as an exercise.

Lemma 3.6.2. Let x

∈(a,b) and f : (a, b) 7→R. Then f is continuous at x

if and only if for every sequence (z

)

n∈N

, z

∈(a,b), z

→x

f (z

) → f (x

Proof. First, assume that f is continuous at x

. In other words, lim

x→x

f (x) =

f (x

). Let the sequence (z

)

n∈N

, z

∈(a,b), z

→ x

. Let ε > 0 be given. Then

∃N ∈ N such that ∀n ≥N (z

6=x

)

∗

|f (z

) − f (x

)|≤ε

=⇒ f (z

) −→ f (x

For the other direction, assume that every sequence (z

)

n∈N

, z

∈(a,b), z

→

=⇒ f (z

) → f (x

). In particular, consider the sequences (z

)

n∈N

, z

∈

(a,b) \ {x

}, z

→ x

=⇒ f (z

) → f (x

) =⇒ lim

x→x

f (x) = lim

n→∞

f (z

) =

f (x

• Example. Take f : (0, +∞) 7→ R be deﬁned by f (x) =

x. We claim that f is

continuous at every x

∈(0,+∞). Set x ∈(0,+∞and compute

|f (x) − f (x

)|=|

x −

|x −x

|{z}

≤

|x −x

Now, ∀ε >0 set δ =ε·

, then we have

|x −x

|≤δ =⇒ |f (x) − f (x

)|≤

≤ε

=⇒

x is continuous at x

Proposition 3.6.3. Let f , g : (a,b) 7→R continuous at x

∈(a,b). Then

(i) f +g is continuous at x

(ii) For λ ∈R, λ · f is continuous at x

(iii) f ·g is continuous at x

(iv) f 6=0 on (a,b),

/f is continuous at x

(v) f : (a, b) 7→(c,d), g : (c, d) 7→R, f is continuous at x

∈(a,b) and g is

continuous at f (x

) ∈(c,d), then g ◦ f is continuous at x

Proof. Left as an exercise.

• Example. Let f : (0,+∞) 7→ R. We claim f (x) =

x is continuous at every

∈(0,+∞).

Consider f : R 7→R deﬁned by

f (x) =

(

1 x ∈ Q

0 x ∈ R \ Q

which is discontinuous everywhere. Then let x

∈R,n ∈N. We know that there

exists a

∈(x

−

/n, x

/n) such that a

∈Q,b

∈R\Q such that a

−→

n→∞

. Then we have

f (a

) =1 →1

f (b

) =0 →0

=⇒ f discontinuous at x

Deﬁnition 3.6.4 (Continuity on an interval). We say f : (a,b) 7→R is continu-

ous on (a,b) if it is continuous at every x

∈(a,b).

We write f ∈C

(a,b)

Deﬁnition 3.6.5 (Left/right limits). Let f : (a,b)\{x

} 7→R. We say that f has a

left-sided limit l ∈R at x

if ∀ε >0 ∃δ(x

,ε) >0 such that

0 <|x −x

|≤δ

{z }

x−δ≤x≤x

=⇒ |f (x) −l |≤ε.

We write lim

x→x

−

f (x) =x

Similarly, we say that f has a right-sided limit l ∈R at x

if ∀ε >0 ∃δ(x

,ε) >0

such that

0 <|x −x

|≤δ

| {z }

≤x≤x

+δ

=⇒ |f (x) −l |≤ε.

We write lim

x→x

f (x) =x

Deﬁnition 3.6.6 (Left/right continuous). We say that f : (a,b) 7→ R is left-

continuous at x

∈(a,b] if lim

x→x

−

f (x) = f (x

) ⇐⇒ ∀ε >0 ∃δ >0 such that

−δ ≤x ≤ x

=⇒ |f (x) −l |≤ε.

Similarly, we say that f : (a,b) 7→R is right-continuous at x

∈[a,b) if lim

x→x

f (x) =

f (x

) ⇐⇒ ∀ε >0 ∃δ >0 such that

≤x ≤ x

+δ =⇒ |f (x) −l|≤ε.

• Example. Let f (x) =

x. Is f (x) is right-continuous at 0? i.e., lim

x→x

f (x)

=0.

We note that f (x) is continuous on (0,1) but does not have a right limit at 0.

Lemma 3.6.7. Let f : (a,b) 7→ R, x

∈ (a,b). Then f is continuous at x

and only if

lim

x→x

−

f (x) = lim

x→x

f (x) = f (x

Proof. Left as an exercise.

Deﬁnition 3.6.8 (Continuity on a closed interval). We say f : [a, b] 7→R is con-

tinuous on a closed interval if it is continuous on (a,b) and lim

x→x

f (x) =

f (a),lim

x→x

−

f (x) = f (b).

We write f ∈C

[a,b]

Theorem 3.6.9 (Intermediate Value Theorem). Let f : (a,b) 7→R be a con-

tinuous function such that f (a) < f (b). Then for every y

∈ R such that

f (a) < y

< f (b), there exists x

∈ (a,b) such that f (x

) = y

(symmetrical

for f (b) < f (a)).

BCaution: y

may not be unique.

Proof. Let X ={x ∈ [a, b] | f (x) < y

} ⊂[a,b]. We know that X 6=; since a ∈ X

by deﬁnition. Further, since X is a subset of a bounded set, we know that X is

bounded. Therefore, we know that s =sup(X ) exists.

Now, we want to show

(i) a < s <b and

(ii) f (x) = y

If we can show these two things then we get that x

=s.

(i) We want to show that a < s. Recall that f is right-continuous at a means

that ∀ε >0 ∃δ >0 such that

a ≤ x ≤ a +δ =⇒ |f (x) − f (a)|≤ε.

In particular, for e =(y

− f (a))/2, we obtain

a ≤ x ≤ a +δ =⇒ |f (x) − f (a)|≤ε =

− f (a)

=⇒ f (x) ≤ f (a) +ε =

f (a)+y

< y

by construction. Now for the sake of contradiction, assume that sup(X ) =

s = a. Then take z =

(a+δ)

/2 and f (z) < y

by construction which fur-

ther implies that z ∈ X . However, z > a = s which contradicts the initial

assumption that s = sup(X ). Therefore, our contradiction assumption

must be false and a <s.

Next, we must show that s <b. Recall that since f is left-continuous at b,

we have ∀ε >0 ∃δ >0 such that

b −δ ≤x ≤b =⇒ |f (x) − f (b)|≤ε =⇒ f (x) ≥ f (b) −ε.

Therefore, take ε = f (b) − y

=⇒ f (x) ≥ y

by construction. For the

sake of contradiction, assume that sup(X ) = s = b. Then we have that

b−δ ≤ x ≤ b =⇒ x 6∈ X . So take z =b−

/2 which means that z ≥ x ∀x ∈ X .

However, this contradicts the initial assumption that s is the smallest up-

per bound of X . Therefore our contradiction assumption must be false

and s <b.

(ii) We want to show that f (s) = y

. First, we will show that f (s) ≥ y

. Let

)

n∈N

, z

∈ (s,s +

/n), z

→ s. Then we surely have that z

> s,z

6∈ X

which implies that (by the deﬁnition of X ) f (z

) ≥ y

. Note that by com-

parison theorems, we have that lim

n→∞

f (z

) ≥ y

(since f (z

) ≥ y

∀n ∈

N). Also, since f is continuous on the interval, we have

≤ lim

n→∞

f (z

) = f

lim

n→∞

z(n)

= f (s) =⇒ f (s) ≥ y

Next, we will show that f (s) ≤ y

. Recall the property of supremum:

∃z

∈ X such that z

→ s. Then we know that f (z

) < y

since z

∈ X .

Similarly to before, we have

≥ lim

n→∞

f (z

) = f

lim

n→∞

f (z

)

= f (s) =⇒ f (s) ≤ y

Together, (i) and (ii) imply that f (s) = y

• BWarning: If f : [a,b] 7→R such that f (a) < f (b) and ∀y

∈R with f (a) < y

f (b) =⇒ ∃x

∈(a,b) with f (x

) = y

6=⇒ f is continuous on [a,b]!

Deﬁnition 3.6.10 (Monotone functions). A function f : [a,b] 7→ R is mono-

tone increasing if

∀x, y ∈[a,b] with x ≤ y =⇒ f (x) ≤ f (y).

Similarly, such a function is monotone decreasing if

∀x, y ∈[a,b] with x ≤ y =⇒ f (x) ≥ f (y).

Lemma 3.6.11. Let f : [a,b] 7→ R be monotone and the conditions of the

Intermediate Value Theorem hold. Then f is continuous.

Proof. Details of proof omitted.

• Corollary. Let f : [a, b] 7→R be a continuous function such that f (a)· f (b) ≤0.

Then there exists x

∈[a,b] such that f (x

) =0.

Proof. There are two cases to consider:

(1) Either f (a) =0 or f (b) =0.

(2) Both f (a) 6=0 and f (b) 6=0 =⇒ f (a) · f (b) <0.

Without loss of generality, assume f (a) < 0 and f (b) > 0. Then by the Inter-

mediate Value Theorem, y

=0 and f (a) < y

< f (b) and ∃x

∈(a,b) such that

f (x

) = y

=0.

• Corollary (Fixed Point). Let f : [a,b] 7→ [a,b] be a continuous function. Then

there exists x

∈[a,b] such that x

= f (x

Proof. Let g (x) =x − f (x) be continuous on [a, b]. Then we have

g (a) =a − f (a) ≤0

g (b) =b − f (b) ≥0

since f : [a,b] 7→[a,b]. Then, by the previous corollary we have

g (a) ·g (b) ≤0 =⇒ ∃x

∈[a,b] ∧ g (x) =0 =⇒ x

= f (x

Deﬁnition 3.6.12 (Boundedness on a set). Let X ⊂ R be non-empty. A func-

tion f : X 7→R is said to be bounded on X iof there is M ≥0 such that

|f (x)|≤ M ∀x ∈ X .

Lemma 3.6.13. Let X ⊂ R be non-empty. If f : X 7→R is not bounded on X ,

then ∃(z

)

n∈N

∈ X such that |f (z

)|≥n.

Proof. Boundedness implies that ∃M ≥ 0 such that |f (x)| ≤ M ∀x ∈ X so by

negation unboundedness implies ∀M ≥ 0 ∃x ∈ X such that |f (x)| > M. Then

consider m =n,(z

)

n∈N

∈ X and we have |f (z

)|≥n.

Theorem 3.6.14 (Extreme Value Theorem). Let f : [a,b] 7→ R be a contin-

uous function. Then f is bounded on [a,b] and there are x

, x

∈[a,b] such

that

sup{f (x) : x ∈ [a,b]} = f (x

)

inf{f (x) : x ∈ [a,b]} = f (x

We denote these as

sup

x∈[a,b]

f (x) =sup{f (x) : x ∈[a,b]}

inf

x∈[a,b]

f (x) =inf{f (x) : x ∈[a,b]}.

Proof. We ﬁrst must show that f is bounded on [a,b].

For the sake of contradiction, assume that f is unbounded (not bounded).

Then there exists a sequence (z

)

n∈N

∈[a,b] with |f (z

)|≥n =⇒ f divergent.

However, (z

)

n∈N

is bounded and Bolzano-Weirstrass implies that ∃(z

)

k∈N

such that z

→l ∈[a,b]. Further, by comparison results we have

lim

k→∞

f (z

) = f ( lim

k→∞

) = f (l ).

But we know that ( f (z

))

k∈N

converges to f (l ) which implies that f is con-

vergent which is a contradiction. Therefore f must be unbounded.

Now we must show that f bounded =⇒ X bounded =⇒ M =sup(X ) and m =

inf(X ) exists. So we need to show that M = f (x

), x

∈[a,b] and m = f (x

), x

∈

[a,b]. From the characterization of inﬁmum, we know ∀ε > 0 ∃q

∈ X such

that q

−m ≤ε. Then we have that q

∈ X → f (x

), x

∈[a,b]. This means that

f (x

) −m ≤ ε. Taking ε =

/n we obtain m ≤ f (x

) ≤m +

/n =⇒ f (x

) →m by

the Squeeze Theorem.

But (x

)

n∈N

bounded implies there exists some subsequence (x

)

k∈N

such

that x

∈[a,b] =⇒ x

→l ∈[a,b]. So

m = lim

k→∞

f (x

) = f ( lim

k→∞

) = f (l ).

Take x

=l ∈[a,b].

Deﬁnition 3.6.15 (Supremum and max of a function). Let f : X → R, X non-

empty. If there is x ∈ X such that f (x) =sup{f (x) : x ∈ X }, we write sup

x∈X

f (x) =

max

x∈X

f (x). Similarly for inf.

For f : [a,b] 7→R, the EVT guarantees that sup

x∈[a,b]

f (x) =max

x∈[a,b]

f (x) and

inf

x∈[a,b]

f (x) =min

x∈[a,b]

f (x).

3.7 Uniform Continuity

• Remark. If (z

)

n∈N

is bounded and f is a continuous function, then (f (z

))

n∈N

is bounded.

• Remark. However, (z

)

n∈N

convergent 6=⇒ ( f (z

))

n∈N

convergent!

• Example. Consider f (x) =

/x which is continuous on (0,1). Indeed, x

∈

(0,1), ε >0 ∃δ =δ(ε, x

) such that

|x −x

|≤δ =⇒ |f (x) − f (x

)|≤ε

⇐⇒

−

x −x

x ·x

≤

4|x −x

≤ε ( by (*))

since

−δ ≤x ≤ x

+δ

δ ≤

=⇒ x ≥

(*)

However, since the deﬁnition of continuity is dependent on x

which is not

constant, we have a problem. Take z

/n →0. But f is continuous on (0,1).

So f (z

) =

/(1/n) = n is not convergent.

Deﬁnition 3.7.1 (Uniform continuity). Let X ⊂ R be non-empty. We say that

f : X 7→R is uniformly continuous on X if ∀ε >0 ∃δ =δ(ε) such that |X −y|≤

δ =⇒ |f (x) − f (y)|≤ε.

Lemma 3.7.2. Let f : X 7→ R, X ⊂ R non-empty where f is uniformly con-

tinuous. Then f is continuous.

Proof. To show f is continuous at x

∈ X , we need to show that ∀ε > 0 ∃δ =

δ(ε, x

) >0 such that

|x −x

|≤δ =⇒ |f (x) − f (x

)|≤ε.

Since f is assumed to be uniformly continuous, we know that ∀

ε >0 ∃

δ =

δ(

ε)

such that

|x −y|≤

ε =⇒ |f (x) − f (y)|≤

ε.

Then let ε >0 be given. Let

ε =ε and

δ =

δ(

ε) and write uniformly continuous

for y =x

. Then

|x −x

|≤

δ =⇒ |f (x) − f (x

)|≤

ε =ε.

It sufﬁces to take δ =

δ.

• Example (re-visited). Consider f (x) =

/x which is uniformly continuous on

(1,2). Because ∀ε >0, x, y ∈ (1, 2),

|f (x) − f (y)|=

−

|x −y|

x y

≤|x −y|.

Let δ =ε, then clearly

|x −y|≤δ =⇒ |f (x)− f (y)|≤ε.

Now, let us show that f is not uniformly continuous on (0,1).

Remark. A function f : X 7→ R, X ⊂ R non-empty is not uniformly continu-

ous on X if ∃ε

> 0 and (x

)

n∈N

,(y

)

n∈N

∈ X with |x

−y

| ≤

/n and |f (x

) −

f (y

)|≥ε

So, coming back to the example, let x

/n, y

/(n+1). Then compute

−y

n(n +1)

≤

|f (x

) − f (y

)|=|n −(n +1)|=1.

In other words, the difference between any terms of the same index of the

image sequences is 1 while the difference between any two terms of the same

index in the domain sequence tends to 0.

Proposition 3.7.3. Let f : (a,b) 7→ R be uniformly continuous. If (z

)

n∈N

is Cauchy with z

∈(a,b), then (f (z

))

n∈N

is Cauchy.

Proof. Since f is uniformly continuous, we have that ∀

ε >0 ∃

δ =

δ(

ε) >0 such

that

|x −y|≤

δ =⇒ |f (x) − f (y)|≤

ε x, y ∈ (a,b).

Also, since (z

)

n∈N

Cauchy, we have that ∀ε >0 ∃N ∈N such that ∀m,n ≥ N ,

−z

|≤ε.

Next, we need to show that ∀ε > 0 ∃N ∈ N such that ∀m,n ≥ N , |f (z

) −

f (z

)| ≤ ε. Let ε > 0 be given. Set

ε = ε which implies that there exists

δ > 0

such that

|x −y|≤

δ =⇒ |f (x) − f (y)|≤ε.

Then, set ε =

δ =⇒ ∃N ∈ N such that ∀m,n ≥ N,|z

−z

|≤ ε =

δ. Finally, set

N = N and we get that, for all m,n ≥ N,

|f (z

) − f (z

)|≤ε.

Theorem 3.7.4 (Uniform continuity). Let f : [a,b] 7→ R be a continuous

function. Then f : [a,b] 7→R is uniformly continuous.

Proof. For the sake of contradiction, assume f is not uniformly continuous.

Then ∃ε

>0 and sequences (x

)

n∈N

,(y

)

n∈N

such that x

, y

∈[a,b] with

−y

|≤

and |f (x

) − f (y

)|≥ε

Further, since x

∈ [a,b], we know that x

is bounded and by the Bolzano-

Weirstrass Theorem, there exists a subsequence (x

)

k∈N

such that x

→l

∈

[a,b].

Next, consider another subsequence (y

)

k∈N

which not necessarily conver-

gent (this is because, we are taking the indices the same as x

so we know the

subsequence exists but not that it converges). However, since y

is bounded,

the Bolzano-Weirstrass Theorem implies that it also has a convergent sub-

subsequence, say y

of the subsequence y

such that y

→l

∈[a,b]. Also,

we can similarly say that (x

)

k∈N

is a subsequence of x

so x

→l

We now must show that l

= l

. Consider |x

− y

| ≤

. As k → ∞, the

difference (x

− y

) → 0 and

→ 0. Therefore, lim

k→∞

= l

lim

k→∞

Now, we know f is continuous, so lim

k→∞

f (x

) = f (l

) and lim

k→∞

f (y

) =

f (l

) and l

= l

. Together, these two things imply that f (l

) = f (l

). This

means that lim

k→∞

|f (x

−y

)|= 0 which contradicts the assumption that

|f (x

) − f (y

)| ≥ ε

. Therefore, our initial assumption is false and f must

be uniformly continuous.

Theorem 3.7.5 (Continuity Extension). Let f : (a,b) 7→ R be a function.

Then f is uniformly continuous on (a,b) if and only if there is some con-

tinuous function g : [a,b] 7→R such that g (x) = f (x) on (a,b).

Proof. First, assume that g : [a,b] 7→ R is continuous. From Theorem 3.7.4,

g is uniformly continuous which implies that g is uniformly continuous on

(a,b) ⊂[a,b] which implies that f is continuous on (a,b) since f = g .

Next, assume that f is uniformly continuous on (a,b). Take g(x) = f (x) when-

ever x ∈ (a,b), i.e., g is also continuous on (a,b). Now, for g to be continuous

on [a,b], we need to decide the values of g (a) and g (b). In particular, it must

be that

g (a) = lim

x→a

g (x) = f (x) and g (b) = lim

x→b

−

g (x) = f (x)

by the deﬁnition of right and left continuity. By symmetry, we will only show

that g (a) =lim

x→a

g (x) = f (x), and showing the g (b) case is nearly identical.

By the characterization of sequential limits, we have that lim

x→a

g (x) =l ex-

ists and is ﬁnite if for every sequence (z

)

n∈N

, z

∈(a,b), we have

→a and f (z

) →l.

Take (z

)

n∈N

, z

∈ (a,b),z

→ a. Since (z

)

n∈N

is converging to a limit, it is

Cauchy. Now since f is uniformly continuous and z

is Cauchy, Proposition

3.7.3 tells us that (f (z

))

n∈N

is also Cauchy and is thus convergent to some

limit, say l

. We now show that the limit l

is independent of the sequence.

For the sake of contradiction, suppose there is some sequence (y

)

n∈N

, y

∈

(a,b), y

→ a, then we know that f (y

) →l

6=l

. Further, our initial assump-

tions yield some properties which we will list below:

(1) f (z

) →l

means that ∀ε

>0 ∃N

∈N such that ∀n ≥ N

|f (z

) −l

|≤ε

(2) f (y

) →l

means that ∀ε

>0 ∃N

∈N such that ∀n ≥ N

|f (y

) −l

|≤ε

(3) f being uniformly continuous means that ∀ε

> 0 ∃δ

> 0 such that, for

x, y ∈(a,b)

|x −y|≤δ

=⇒ |f (x) − f (y)|≤ε

(4) z

→a means that ∀ε

>0 ∃N

∈N such that ∀n ≥ N

−a|≤ε

(5) y

→a means that ∀ε

>0 ∃N

∈N such that ∀n ≥ N

−a|≤ε

Let ε > 0 be given. Further, set ε

/3,ε

/3 =⇒ ∃δ

by (3). Then,

set ε

/2,ε

/2. Why make these particular choices? Because then we

have

−a|≤

∀n ≥ N

, (by (4))

−a|≤

∀n ≥ N

, (by (5))

=⇒|z

−a −y

+a|≤|z

−a|+|y

−a|

=δ

=⇒ |z

−y

|≤δ

∀n ≥max{N

, N

Then, by (3), this implies that |f (z

) − f (y

)| ≤

/3. Therefore, by the triangle

inequality we get

−l

|≤|l

− f (z

| {z }

≤

/3 ∀n≥N

+ |f (z

) − f (y

| {z }

≤

/3 ∀n≥max{N

}

+|f (y

) −l

| {z }

≤

/3 ∀n≥N

≤

=ε.

So |l

−l

| ≤ ε,n ≥ max{N

, N

} ∀ε > 0 =⇒ l

= l

. However, this con-

tradicts our assumption that l

6= l

, therefore l

= l

and we can set g (a) =

3.8 Inverses and Continuity

• Example. Take the function f (x) =sin(x), which is continuous on the interval

(

−π

/2,

/2).

−π/2 π/2

−1

sin(x)

f (x)

−1 1

−π/2

π/2

sin

−1

(x)

f (x)

Although the picture makes it look obvious, we will rigorously show that the

inverse of a continuous function is also continuous.

Proposition 3.8.1. Let I ⊂ R be an interval (could be closed, open, etc.). If

f : I 7→R is injective and continuous, then f is strictly increasing or strictly

decreasing.

Proof. We will show for the case where I = (a,b) and the other cases are the

same. For the sake of contradiction, assume that there are

a < x

such that

f (x

) ≤ f (x

) and f (x

) ≥ f (x

i.e., not strictly increasing. Since f is injective, f (x

) ≤ f (x

) =⇒ f (x

) < f (x

)

and f (x

) ≤ f (x

) =⇒ f (x

) < f (x

). Therefore, ∃α ∈R such that

f (x

) <α < f (x

) and f (x

) <α < f (x

Now, since f is continuous on [x

, x

] and [x

, x

], the Intermediate Value The-

orem (3.6.9) implies that ∃

x ∈(x

, x

) such that f (

x) =α and ∃

x ∈(x

, x

) such

that f (

x) =α. By the conditions of the IVT, it is clear that

x 6=

x. But this gives

us that f (

x) = f (

x) which contradicts the fact that f is injective. Therefore,

our contradiction assumption must be false and f is either strictly increasing

or strictly decreasing.

Theorem 3.8.2 (Inverse continuity). Let f : [a,b] 7→ R be continuous and

injective. Then f ([a,b]) is a closed interval I and f

−1

: I 7→[a, b].

Proof. Never got around to this in class.

4 Differentiation

Deﬁnition 4.0.1 (Locally differentiable). Let f : (a,b) 7→ R and x

∈ (a,b). If

lim

x→x

(f (x) − f (x

))/(x −x

) exists and is ﬁnite, then we say that f is locally

differentiable at x

and write

) = lim

x→x

f (x) − f (x

)

x −x

• Remark.

(1) By function composition and taking g (x) = x

+h, we note that

lim

x→x

f (x) − f (x

)

x −x

= lim

h→0

f (x

+h) − f (x

)

(2) Since f is differentiable at x

, it is also continuous at x

. This is because

you can always write

f (x) = f (x

) +

f (x) − f (x

)

x −x

(x −x

) (x 6= x

)

Deﬁnition 4.0.2 (Globally differentiable). Let f : (a,b) 7→ R. We say f is glob-

ally differentiable on the interval (a,b) if f is differentiable on every x

∈

(a,b). In that case, we have

x 7→ lim

x→x

f (x) − f (x)

x −x

= f

(x)

is a well-deﬁned function on (a,b) called the derivative of f on (a,b).

• Examples.

(1) Let f (x) =α. We claim this is differentiable on R.

Let x

∈R, x

6=x. Compute

f (x) − f (x

)

x −x

α −α

x −x

=0.

So lim

x→x

f (x)−f (x

)

x−x

exists and is ﬁnite and equals 0.

(2) Let f (x) = x

. We claim this is differentiable on R. Let x

∈ R, x

6= x.

Compute

f (x) − f (x

)

x −x

−x

x −x

x +x

So lim

x→x

f (x)−f (x

)

x−x

exists and is ﬁnite and equals 3x

(3) Let f (x) = |x|. We claim this is differentiable on (−∞,0) ∪(0,∞) but not

at 0. First, consider x

>0. Subsequently, f (x) =x

>0. Then compute

f (x) − f (x

)

x −x

=1 =⇒ f

) =1.

Similarly, for x

<0, f

) =−1. When x

=0, we have

lim

x→x

f (x) − f (x

)

x −x

= lim

x→x

|x|

As previously established (similar to Example 3.1), this limit does not ex-

ist, so f is not differentiable at 0.

(4) Let f (x) =sin(x). We claim this is differentiable on R. Take x

∈R. Com-

pute

f (x) − f (x

)

x −x

sin(x) −sin(x

)

x −x

sin(

x−x

) ·cos(

x+x

)

x −x

sin(

x−x

)

x−x

·cos(

x +x

Then, we have

lim

x→x

sin(

x−x

)

x−x

·cos

x +x

lim

x→x

sin(

x−x

)

lim

x→x

x−x

· lim

x→x

cos(

x +x

)

=1 ·cos(x

Since this limit exists and is ﬁnite, we have f

) = cos(x

). Similarly,

f (x) =cos(x) =⇒ f

) =sin(x

Theorem 4.0.3 (Caratheodory’s Theorem). Let f : (a,b) 7→ R, x

∈ (a,b).

The function f is differentiable at x

if and only if there is ϕ : (a,b) 7→R that

is continuous at x

and f (x) = f (x

)+ϕ(x)(x−x

) ∀x ∈(a,b). Furthermore,

) =ϕ(x

Proof. First, suppose such a ϕ exists. Further, suppose x 6= x

. Then we have

f (x) − f (x

)

x −x

=ϕ(x).

Furthermore,

lim

x→x

f (x) − f (x

)

x −x

= lim

x→x

ϕ(x) =ϕ(x

) ←−ϕ continuous at x

→ f

) =ϕ(x

Next, suppose f

) is well-deﬁned. In particular, set

ϕ =

(

f (x)−f (x

)

x−x

x 6= x

) x = x

Clearly, ϕ(x) is continuous at x

, so

lim

x→x

ϕ(x) = lim

x→x

f (x) − f (x

)

x −x

= f

) =ϕ(x

On the other hand, when x 6= x

, we have

ϕ(x) =

f (x) − f (x

)

x −x

⇐⇒ f (x) = f (x

) +ϕ(x

)(x −x

)

⇐⇒ f (x) = f (x

) +ϕ(x)(x −x

Lemma 4.0.4. Let f : (a,b) 7→ R, If f is differentiable at x

∈ (a,b), then f

is continuous at x

Proof. If f is differentiable at x

, then there exists ϕ which is continuous at

such that f (x) = f (x

) +ϕ(x)(x −x

) and lim

x→x

f (x) = f (x

) +ϕ(x

) ·0 =

f (x

• Remark. There exists functions whose derivatives are not continuous. For

example, consider

f (x) =

(

sin

x 6= 0

0 x = 0.

4.1 Extremum

Deﬁnition 4.1.1 (Local extremum). Let f : I 7→ R. We say that c ∈ I is a local

minimum/maximum if there is δ >0 such that f (x) ≥/ ≤c ∀x ∈(c −δ,c +δ).

An extremum is a minimum or a maximum.

Theorem 4.1.2 (Fermat). Let f : (a,b) 7→ R. If f has an extremum at c ∈

(a,b) and is differentiable at c, then f

Proof (Minimum). Let c be such an element. We have c is a local minimum

which implies that there exists δ >0 such that f (x) ≥c ∀x ∈(c −δ,c +δ). Take

x ∈ (c, c +δ) and compute the ration

f (x) − f (c)

x −c

≥0 (x bigger than c by construction)

comp.

⇐⇒ lim

x→c

f (x) − f (x)

x −c

≥0.

Similarly, for x ∈(c −δ,c), we have

lim

x→c

−

f (x) − f (c)

x −c

≤0.

Next, we have that f is differentiable at c which implies

0 ≤ lim

x→c

f (x) − f (c)

x −c

= lim

x→c

f (x) − f (c)

x −c

= lim

x→c

−

f (x) − f (c)

x −c

≤0

⇐⇒ lim

x→c

f (x) − f (c)

x −c

=0 = f

(c).

4.2 Rules of Differentiation

4.2.1 Basic Rules

Proposition 4.2.1 (Basic rules of differentiation). Let f ,g : (a,b) 7→ R dif-

ferentiable at x

∈(a,b), λ ∈R.

(i) f +g is differentiable at x

with

(f +g )(x

) = f

) +g

(ii) λ · f is differentiable at x

with

(λ · f )

=λ · f

Proof. (i) Take x

∈(a,b) and compute

lim

x→x

(f +g )(x) −(f +g)(x

)

x −x

= lim

x→x

f (x) +g (x) − f (x

) −g (x

)

x −x

= lim

x→x

f (x) − f (x

)

x −x

+ lim

x→x

g (x) −g(x

)

x −x

= f

) +g

(ii) Compute

lim

x→x

(λf )(x)−(λf )(x

)

x −x

= lim

x→x

f (x) − f (x

)

x −x

=λ · lim

x→x

f (x) − f (x

)

x −x

=λ · f

4.2.2 Advanced Rules

Proposition 4.2.2 (Product Rule). Let f ,g : (a,b) 7→R be differentiable at

∈(a,b). Then f ·g is differentiable at x

with

(f ·g )

) = f

)g (x

) + f (x

Proof. Let x ∈(a,b)\{x

} and compute

lim

x→x

(f ·g )(x) −(f ·g)(x

)

x −x

= lim

x→x

f (x)g (x) − f (x

)g (x

)

x −x

= lim

x→x

f (x) − f (x

)

g (x) + f (x

)

g (x) −g(x

)

x −x

= lim

x→x

f (x) − f (x

)

x −x

·g (x) +

g (x) −g(x

)

x −x

· f (x

)

= lim

x→x

f (x) − f (x

)

x −x

·g (x) + lim

x→x

g (x) −g(x

)

x −x

· f (x

)

= f

) ·g (x

) + f (x

) ·g

). (*)

(*) Note: Since g , f are differentiable at x

, they are also continuous at x

and

thus lim

x→x

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

), f (x

Proposition 4.2.3 (Quotient Rule). Let f : (a,b) 7→ R be differentiable at

∈(a,b) such that f (x

) 6=0. Then there exists δ >0 such that f (x) 6=0 for

all x ∈(x

−δ, x

+δ) (f well-deﬁned on (x

−δ, x

+δ)) and

) =

−f

)

f (x

)

Then, assume g (x

) 6=0. Then

/g is differentiable at x

and

) =

)g (x

) − f (x

)

(g (x

))

Proof. We will ﬁrst show the ﬁrst part of the proposition and the second part

follows immediately. Take x ∈(x

−δ, x

+δ) and compute

lim

x→x

(x) −

)

x −x

= lim

x→x

f (x)

−

f (x

)

x −x

= lim

x→x

f (x)−f (x

)

f (x)·f (x

)

x −x

= lim

x→x

−

f (x) − f (x

)

x −x

f (x) · f (x

)

=− lim

x→x

f (x) − f (x

)

x −x

· lim

x→x

f (x) · f (x

)

=−f

) ·

(f (x

))

−f

)

(f (x

)

For the second part, we note that

/g = f ·

/g and compute

) = f

) ·

g (x

)

+ f (x

) ·

−g

)

(g (x

))

| {z }

product rule + quotient rule

)g (x

) + f (x

)

(g (x

))

Proposition 4.2.4 (Chain Rule). Let f : (a,b) 7→ (c,d) be differentiable at

∈(a,b) and let g : (c, d) 7→R be differentiable at f (x

) ∈(c,d). Then g ◦f

is differentiable at x

with

(g ◦ f )

) =g

(f (x

))f

Proof. Recall Caratheodory’s Theorem (4.0.3) which states that a function f is

differentiable at x

if and only if there exists a continuous function ϕ : (a,b) 7→

R such that f (x) = f (x

) +ϕ(x)(x −x

) and ϕ(x

) = f

). We can tweak this

equation for g as follows:

The function g is differentiable at f (x

) if and only if there exists a

continuous function Φ : (c,d) 7→R such that g (y) =g (f (x

)) +Φ(y)(y − f (x

))

and Φ(f (x

)) =g

(f (x

)).

Now compute for x ∈(a,b)\{x

(g ◦ f )(x) −(g ◦ f )(x

)

x −x

g (f (x)) −g( f (x

))

x −x

Φ(f (x))(f (x) − f (x

))

x −x

Now,

lim

x→x

Φ(f (x)) · lim

x→x

f (x) − f (x

)

x −x

=Φ(f (x

)) · f

)

(f (x

)) · f

)

since Φ is continuous and f is differentiable at x

. All of this together implies

that (g ◦ f )(x

) =g

(f (x

)) · f

4.3 Mean Value Theorem and Applications

Theorem 4.3.1 (Rolle’s Theorem). Let f : [a, b] 7→R be continuous on [a, b]

and differentiable on (a,b). If f (a) = f (b), then there exists c ∈ (a,b) such

that f

Proof. First note that f continuous on [a,b] implies (by the Extreme Value

Theorem) that there exists r, s ∈ [a, b] such that f (r )

|{z}

≤ f (x) ≤ f (s)

|{z}

∀x ∈ [a,b].

Then consider three cases:

(1) If m = M, then f (x) = m = M ∀x ∈ [a,b] which further implies that

(x) =0 ∀x ∈ (a,b) taking any c ∈ (a,b).

(2) If m < M, then either r ∈ (a,b) or s ∈ (a,b) because f (a) = f (b) by as-

sumption.

(i) If r ∈ (a,b), then f achieves its minimum at r and is thus differen-

tiable at r , then Fermat’s Theorem (4.1.2) implies that f

(r ) =0, tak-

ing c =r we get our desired result.

(ii) Take s ∈ (a,b) and the same result is obtained.

(3) The case where m >M is symmetric and left as an exercise.

Theorem 4.3.2 (Mean Value Theorem). Let f : [a,b] 7→ R be continuous

on [a,b] and differentiable on (a,b). Then there exists a c ∈(a,b) such that

f (b) − f (a)

b −a

Proof. Deﬁne h(x) = f (x)(b−a)+(a −x)(f (b)−f (a)). Since h is a composition

of two differentiable and continuous functions, h is continuous on [a,b] and

differentiable on (a, b). Then compute

h(a) = f (a) ·b − f (a) ·a

h(b) = f (b)(b −a) −(b −a)(f (b) − f (a))

=(b −a) · f (a) =h(a)

Then Rolle’s Theorem (4.3.1) implies that there exists a c ∈ (a,b) such that

(x) = f

(x)(b −a) −(f (b) − f (a))

0 =h

(c)(b −a) −(f (b) − f (a))

⇐⇒ f

f (b) − f (a)

b −a

Theorem 4.3.3 (Mean Value Theorem (Cauchy)). Let f ,g : [a, b] 7→ R be

continuous on [a,b] and differentiable on (a,b). Further assume g

(x) 6=

0 ∃x ∈(a,b). Then there exists c ∈(a,b) such that

(c)

f (b) − f (a)

g (b) −g(a)

Proof. First, note that g(a) 6=g (b) because otherwise, Rolle’s Theorem implies

that there exists some d ∈ (a, b) such that g

(d) = 0, which is a contradiction.

Thus, our ratio is well-deﬁned.

Let h(x) = f (x)(g (b)−g (a))−(g (x)−g(a))(g (b)− f (a)). As before, h us contin-

uous and differentiable on [a,b]/(a,b), respectively. Then

h(a) = f (a)g (b)− f (a)g (a)

h(b) = f (b)(g (b) −g(a))−(g (b) −g(a))(f (b) − f (a))

=g (b)f (a) −g (a)f (a) =h(a).

Then Rolle’s Theorem implies that there exists some c ∈(a,b) such that h

0 and we have

(x) = f

(x)(g (b) −g(a))−(g

(x)(f (b)− f (a)))

0 =h

(c)(g (b)−g (a)) −g

(c)(f (b)− f (a))

⇐⇒

(c)

f (b) − f (a)

g (b) −g(a)

Proposition 4.3.4 (L’Hospital’s Rule). Let f , g : (a, b) 7→R be two continu-

ous functions differentiable on (a,b) \ {x

} with

(x) 6=0 ∀x ∈ (a,b)\{x

lim

x→x

f (x) =lim

x→x

g (x) =0,

and there is l ∈R such that lim

x→x

(x)/g

(x) =l.

Then

lim

x→x

f (x)

g (x)

=l.

Proof. We show that lim

x→x

f (x)/g (x) =l =lim

x→x

−

f (x)/g (x). We ﬁrst show

that g cannot vanish (become zero) more than once on the interval (a,x

). Let

a < x

a x

What can be said about g : [x

, x

] 7→ R? Certainly, g is continuous since it is

differentiable on the entire interval, so g is continuous on [x

, x

] and differ-

entiable on (x

, x

). Then, the MVT (Theorem 4.3.2) implies that there exists a

c ∈(x

, x

) ⊂(a,b) such that

0 6=g

g (x

) −g (x

)

−x

which implies that g cannot vanish at x

or x

Say that g does vanish on γ such that a <γ < x, if it doesn’t vanish, then γ =a.

By assumption, we now that ∀ε >0 ∃δ > 0 such that if x ∈(a,x

) and (WLOG)

γ <x

−δ ≤x ≤ x

, then

(x)

−l

≤ε.

Let γ < x

−δ ≤ x < y < x

. Then, since f and g are continuous on the entire

interval, the MVT Cauchy Version (Theorem 4.3.3) implies that there exists a

c ∈(x, y) such that

(c)

f (y) − f (x)

g (y)−g (x)

−δ

x y x

Then consider

(x)

−l

f (y) − f (x)

g (y)−g (x)

−l

≤ε

comp.

=⇒ lim

y→x

(x)

−l

≤ε

f (x)

g (x)

−l

≤ε

=⇒ lim

x→x

−

f (x)

g (x)

=l.

Similarly, lim

x→x

f (x)/g (x) =l.

4.4 Limited Expansion

Deﬁnition 4.4.1. For n ∈N, we say that f : (a, b) 7→R is n-times differentiable

on (a,b) if

f is differentiable, f

is differentiable, .. ., f

(n)

is differentiable.

Theorem 4.4.2 (Taylor). Let f : (a,b) 7→R be (n +1) times differentiable (n

derivatives are continuous, but (n +1)

may not be). Then

f (x) =P

(x) +ε(x)(x −a)

where

(x) = f (a) +

(a)

(x −a) +···+

(n)

(a)

(x −a)

and ε(x) =

(n+1)

(c)

/(n+1)!(x −a) ∃c ∈ (a,b).

Proof. Since P

is a polynomial of degree n in x, rewrite as follows:

(x) =

i=0

(i)

(a)

(x −a)

where f

(0)

= f .

Then

(x) =

i=1

(i)

(a)

i ·(x −a)

i−1

i=1

(i)

(a)

(i −1)!

(x −a)

i−1

and similarly (for 1 ≤ j ≤n)

(j )

(x) =

i=j

(i)

(a)

(i − j )!

(x −a)

i−j

where P

(j )

(a) = f

(j )

(a) for j =0,...,n.

Let

g (x) = f (x) −P

(x) +

(y)− f (y)

(y −a)

n+1

(x −a)

n+1

∃y ∈(a,b).

Then g is (n +1) times differentiable and

0 =g (a) =g

(a) =···=g

(n)

(a).

In addition, g (y) =0, so by Rolle’s Theorem (4.3.1), since

g (a) =g (y) =0, ∃c

∈(a,b) such that g

) =0

(a) =g

) =0, ∃c

∈(a,b) such that g

) =0

(n)

(a) =g

(n)

) =0, ∃c ∈(a,b) such that g

(n+1)

Then we have

(n+1)

(x) = f

(n+1)

(x) +

(y)− f (y)

(y −a)

n+1

(n +1)!

⇐⇒ 0 = f

(n+1)

(y)− f (y)

(y −a)

n+1

(n +1)!

⇐⇒ f (y) =P

(y)+

(n+1)

(c)

(n +1)!

(y −a)

n+1

| {z }

ε(y)(y−a)

This is the result for x = y.

• Example. Compute the Taylor Expansion for f (x) =sin(x).

sin(x) =0 +x −

| {z }

(x)

sin(c)

| {z }

ε(x)x

∃c ∈(0, x).

Lemma 4.4.3 (Taylor Expansion uniqueness). Let f : (a,b) 7→R be (n +1)

times differentiable such that

f (x) =b

(x −a) +···+b

(x −a)

ε(x)(x −a)

∃b

∈R,

ε(x) −→

x→0

Then

(i)(a)

,i =0,. .. ,n and

ε(x) =ε(x) =

(n+1)

(c)

(n +1)!

(y −a) ∃c ∈(a,x).

Proof. We have two expansions of f :

(i) f (x) =

i=0

(x −a)

ε(x)(x −a) and

(ii) f (x) =

i=0

(i)

(a)/i!·(x −a)

+ε(x)(x −a)

Consider f (x) − f (x) =0

0 =

i=0

(i)

(a)

(x −a)

(

ε(x) −

ε(x)

)

(x −a)

Assume by contradiction that

(k)

(a)

/k! 6= b

0 ≤k ≤ n. Without loss of general-

ity, we take the smallest of such k’s and get

0 =

i=k

(i)

(a)

−b

(x −a)

+(ε(x) −

ε(x))(x −a)

0 =

i=k

(i)

(a)

−b

(x −a)

i−k

+(ε(x) −

ε(x))(x −a)

n−k

(*)

=⇒ lim

x→a

i=k

(i)

(a)

−b

(x −a)

i−k

+(ε(x) −

ε(x))(x −a)

n−k

=0 =

(k)

(a)

−b

Therefore,

(i)

(a)

/i! =b

∀i =0,...,n and by (*) we have

0 =(ε(x) −

ε(x))(x −a)

⇐⇒ε(x) =

ε(x).

• Remark. Let f , g : (a, b) 7→R be (n +1) times differentiable functions such that

f (x) =

i=0

(x −a)

+ε(x)(x −a)

∈R, ε(x) −→

x→a

g (x) =

i=0

(x −a)

+ε(x)(x −a)

∈R,

ε(x) −→

x→a

Then

(f +g ) =

i=0

)(x −a)

+(ε(x) +

ε(x))(x −a)

(f ·g ) =

i=0

·d

)(x −a)

+(ε(x) ·

ε(x))(x −a)

(f /g ) =

i=0

)(x −a)

+(ε(x)/

ε(x))(x −a)

g 6=0.

• Example. Consider the Taylor expansion of f (x) =

/(1+x). First realize that

(1 +x)(1 −x +x

+···) =1.

The point is:

f (x) =

1 +x

=1 −x +x

−x

+···+) −1)

+ε(x)x

where ε(x) =(−1)

n+1

x +(−1)

n+2

+···. Then f

(n)

(0) =(−1)

·n!.

• Example. Consider

f (x) =

(1 +x

)

=1 −x

−x

+···+(−1)

+ε(x)x

• Example. Consider f (x) =(sin

(x)−x

cos(x))/(x

(1−cos(x)) deﬁned on (−π/2,0)∪

(π/2,0). How can we compute lim

x→0

f (x)? One could employ L’Hopital’s rule,

but this would be extremely messy. A slightly more convenient way to com-

pute this limit is the following. Let us consider the Taylor expansions of the

terms that comprise f :

sin(x) = x −

+ε(x)x

(sin(x))

−

+ε(x)x

cos(x) =1 −

+ε(x)x

cos(x) = x

+ε(x)x

(1 −cos(x)) =

+ε(x)x

Now compute

lim

x→0

sin

(x) −x

cos(x)

(1 −cos(x))

= lim

x→0

+ε(x)x

+ε(x)

+ε(x) ·x

since ε(x) −→

x→0

Lemma 4.4.4. Let f : (a,b) 7→R be n +1 times differentiable. Suppose

(x) =b

(x −a) +b

(x −a)

+···+b

n−1

(x −a)

n−1

+ε(x)(x −a)

n−1

Let

g (x) = f (a)+b

(x −a) +

(x −a)

+···+

n−1

(x −a)

ε(x)(x −a)

Then f (x) =g (x).

Proof. Compute g

(x) and realize that g

(x) = f

(x) and then use the result

that says if two derivatives are equal on an interval, then the functions are

equal up to a constant (Homework 9, Exercise 1) and we obtain

g (x) = f (x) +c

g (a) = f (a) +α

=⇒ α =0.

4.4.1 Local Extremum (Re-Visited)

Lemma 4.4.5. Let x

∈ (a,b), f : (a, b) 7→ R be n times differentiable. As-

sume that f

(n)

is continuous at x

. Furthermore, assume that

) = f

) =···= f

(n−1)

) =0 and f

(n)

) 6=0.

Then x

is a local minimum if f

(n)

) >0 and a local maximum if f

(n)

Proof (f

(n)

) >0). Since f

(n)

) > 0 and f

(n)

is continuous at x

, there is

δ >0 such that f

(n)

(x) >0 ∃x ∈ (x

−δ, x

+δ). By Taylor’s Theorem,

f (x) = f (x

) + f

)(x −x

)

| {z }

+···+

(n+1)

)

(n −1)!

(x −x

)

n−1

| {z }

(n)

(c)

(x −x

)

| {z }

=⇒ f (x) ≥ f (x

) ∃x ∈(x

−δ, x

+δ)

which implies that x

is a local minimum by deﬁnition.

• Example. Let f : (a,b) 7→ R be twice differentiable and x

∈ (a,b) with f

)

continuous and f

) = 0. If f

) > 0 then x

is a local minimum and if

) <0 then x

is a local maximum.

For instance, consider f (x) = −x

, f

(x) = −2x, f

(x) = −2, f

(0) = 0, f

(0) <

0 =⇒ 0 is a local maximum.

5 Integration

Deﬁnition 5.0.1. Let a < b. A partition σ of a closed interval (a,b) is a list of

ordered pairs such that

a = x

<···<x

=b.

We write

={x

,. .. , x

} =(x

)

i=1

Deﬁnition 5.0.2. The step of a partition σ

is deﬁned as

h(σ

) = max

i=1,...,n

−x

i−1

The particular case when

+i ·

b −a

i =0,...,n

is called the regular/uniform partition with

h(σ

) =

b −a

Deﬁnition 5.0.3. Let f : [a,b] 7→R be continuous on [a,b] and σ

=(x

)

i=1

a partition of [a, b]. The real number

(f ) =

i=1

−x

i−1

)

where m

=min

x∈[x

i−1

]

f (x) is called the lower Darboux sum.

Similarly,

(f ) =

i=1

−x

i−1

)

where M

=max

x∈[x

i−1

]

f (x) is called the upper Darboux sum.

• Remark. We have that

m(b −a) ≤S

(f ) ≤S

(f ) ≤ M (b −a)

where m and M are deﬁned as above.

This implies that

S(f ) =inf{S

(f ) : σ a partition of [a,b]}

S(f ) =sup{S

(f ) : σ a partition of [a,b]}

are well-deﬁned.

Lemma 5.0.4. Let σ,σ

∗

be two partitions of [a,b] such that σ ⊂σ

∗

. Then

∗

(f ) ≤S

(f ) and S

≤S

∗

(f ).

Proof. Consider σ = {x

,. .. , x

} and σ

∗

= {x

∗

,. .. , x

∗

}. Let x

i−1

be the point in

∗

such that

i−1

≤x

i−1

≤x

j =0, ...,k

Then

(f ) =

i=1

min

x∈[x

i−1

]

f (x)

−x

i−1

)

≤

i=1

j =1

min

x∈[x

i−1

]

i−1

−x

j −1

i−1

)

∗

i=1

min

x∈[x

∗

i−1

∗

]

f (x)(x

∗

−x

∗

i−1

) =S

∗

(f ).

Lemma 5.0.5. Suppose a < b and let f : [a,b] 7→ R be a continuous func-

tion. Then S(f ) =S(f ).

Proof. Since f is continuous on [a,b], f is uniformly continuous by Theorem

3.7.4. Recall that this means that ∀ε >0 ∃δ >0 such that

|x −y|≤δ =⇒ |f (x)− f (y)|≤

3(b −a)

Let σ be a partition with h(σ) ≤δ and σ ={x

,. .. , x

}. Note that max

x∈[x

i−1

]

f (x) =

f (α

) ∃α

∈ [x

i−1

, x

] by the Extreme Value Theorem (Theorem 3.6.14). Simi-

larly, lim

x∈[x

i−1

]

f (x) = f (β

) ∃β

∈[x

i−1

, x

]. This implies that

max

x∈[x

i−1

]

f (x) − min

x∈[x

i−1

]

f (x) = f (α

) − f (β

) ≤

3(b −a)

since

|α

−β

|≤δ

since α

,β

∈[x

i−1

, x

] and h(σ) =max{x

−x

i−1

} ≤δ.

Therefore,

(f ) −S

(f ) =

i=1

−m

)(x

i−1

−x

)

≤

Additionally, by deﬁnition

(f ) =sup{S

(f ) : σ a partition of [a,b]}.

Therefore, by the characterization of suprema, there exists σ

, a partition of

[a,b], such that

(f ) −S

≤

Similarly, there exists σ

, a partition of [a, b], such that

(f ) −S(f ) ≤

Set σ =σ

∪σ

such that σ

⊂σ and σ

⊂σ (which is even smaller than σ

and

, i.e., a reﬁnement of the two). Using the result from the previous lemma,

we have

(f ) ≥S

(f ) and S

(f ) ≤S

(f )

which implies that

(f ) −S(f ) ≤

and S(f ) −S

(f ) ≤

Without loss of generality, we can assume that h(σ) ≤δ (if it were not, we can

simply consider the uniform partition such that

b− a

/n ≤ δ and perform the

same "trick"). Now, for every ε >0, we have

S(f ) −S(f ) =S(f ) −S

(f )

| {z }

≤

(f ) −S

(f ) +S

(f ) −S(f )

| {z }

≤

=⇒ S(f ) =S( f ).

Deﬁnition 5.0.6 (Integral). Let a < b and f : [a,b] 7→ R be a continuous func-

tion. We call S( f ) =S(f ) the integral of f over [a,b]. We use the notation

f (x) dx =S(f ) =S(f ).

• Extensions:

If b < a, then

f (x) dx =−

f (x)d x.

If b = a, then

f (x)d x =0.

• Remark. The integral

f (x)d x is really just notation for the sum

i=1

−x

i−1

Lemma 5.0.7. Let a < b, f : [a,b] 7→ R be a continuous function. Further,

suppose (σ

)

∞

n=1

be a sequence of partitions of [a,b] such that h(σ

) −→

n→∞

Then

lim

n→∞

(f ) =S(f ) and lim

n→∞

(f ) =S(f )

where of course

S(f ) =S( f ) =

f (x)d x.

Proof. Since f : [a,b] 7→R is continuous, f is uniformly continuous and ∀ε >

0 ∃δ >0,h(σ

) ≤δ such that

−S

≤ε (*)

≡

i=1

−m

| {z }

≤ε

)(x

−x

i−1

) ≤ε.

Note that because h(σ

) −→

n→∞

0, there exists N >0 such that h(σ

) ≤δ ∀n ≥ N.

In addition,

≤S(f )

f (x)d x

=S(f ) ≤S

(f ).

This means that ∀n ≥ N,

0 ≤

(f ) −S(f ) ≤ε

and

0 ≤S

(f ) −S(f ) ≤ε

i.e.,

lim

n→∞

(f ) =S(f ) and lim

n→∞

(f ) =S(f ).

• Remark. The proof of the previous lemma can also be used to show that if

(f ) =

i=1

f (

)(x

−x

i−1

)

∈[x

i−1

, x

]

and h(σ

) −→

n→∞

0, then

lim

n→∞

(f ) =

f (x)d x.

• Example. Let f (x) =e

, x ∈ [a,b]. Show that

d x =e

−e

First we want to compute S( f ). Consider











, a +

b −a

, a +2 ·

b −a

,. .. , a +i ·

b −a

,. .. , a +n ·

b −a











Clearly, h(σ

) =

b− a

/n and in particular h(σ

) −→

n→∞

0. Then compute

(f ) =

i=1

−x

i−1

)

b−a

b −a

i=1

a+(i −1)·

b−a

b −a

·e

i=1

b−a

i−1

b −a

·e

n−1

i=0

b−a

b −a

·e

1 −e

b−a

·n

1 −e

b−a

(1 −e

b− a

) ·

b− a

1 −e

b−a

where m

=min

x∈[x

i−1

]

i−1

. Now consider

lim

n→∞

(1 −e

b− a

) ·

b− a

1 −e

b−a

= lim

x→0

1 −e

= lim

x→0

−e

=−1.

Therefore, lim

n→∞

(f ) =(e

b− a

−1)e

−e

f (x)d x.

5.1 Extension of Integrability

• If σ is a partition of f bounded on [a,b] (not necessarily continuous), we can

deﬁne

(f ) =

i=1

−x

i−1

)

where

=sup

x∈[x

i−1

]

f (x) and similarly

(f ) =

i=1

−x

i−1

)

where

=inf

x∈[x

i−1

]

f (x).

Because f is bounded, we have m ≤ f (x) ≤ M ∀x ∈ [a,b] ∃m,M ∈ R which

subsequently means

m(b −a) ≤S

(f ) ≤S

(f ) ≤ M (b −a).

Then we have that the following are well-deﬁned:

S(f ) =inf{S

(f ) : σ a partition of [a,b]}

S(f ) =sup{S

(f ) : σ a partition of [a,b]}.

However, note that S(f ) 6=S(f ) in general.

Deﬁnition 5.1.1 (Riemann integrability). Let f : [a, b] 7→R be a bounded func-

tion. We say f is Riemann integrable when S(f ) =S( f ).

5.2 Properties of Integrals

Lemma 5.2.1. Let a < b and f : [a,b] 7→R be a continuous function. Fur-

ther, let c ∈[a,b]. Then

f (x)d x =

f (x)d x +

f (x)d x.

Proof. Assume that c 6= a 6=b. Then consider

a, a +

c −a

, a +2 ·

c −a

,. .. ,c

={x

}

i=0

β =

c,c +

b −c

,c +2 ·

b −c

,. .. ,b

={x

}

2n+1

i=n

where both α

and β

are regular partitions of [a,c]/[c,b], respectively. Then

consider

=α

∪β

which is a regular partition of [a,b] with h(σ

) = max{h(α

),h(β

)} −→

n→∞

Then compute

f (x)d x = lim

n→∞

(f ) = lim

n→∞

i=1

−x

i−1

2n+1

i=n+1

−x

i−1

)

= lim

n→∞

(f ) +S

(f )

f (x)d x +

f (x)d x.

Lemma 5.2.2. Let f : [a,b] 7→ R be a continuous function and let λ ∈ R.

Then

f (x)d x =λ

f (x)d x.

Proof. Let {σ

}

n∈N

be a sequence of partitions with h(σ

) −→

n→∞

0. Then we

have

(λf )(x)d x = lim

n→∞

(λf )

= lim

n→∞

i=1

max

x∈[x

i−1

]

(λf )(x)(x

−x

i−1

)

=λ · lim

n→∞

i=1

max

x∈[x

i−1

]

f (x)

| {z }

(f )

−x

i−1

)

=λ

f (x)d x.

Lemma 5.2.3 (Integral Mean Value Theorem). Let f : [a,b] 7→R be a con-

tinuous function. Then there exists c ∈ [a,b] such that

f (c) =

b −a

f (x)d x.

Proof. Since f is continuous on [a,b], the Extreme Value Theorem (Theorem

3.6.14) implies that

m = min

x∈[a,b]

f (x) ≤ f (x) ≤ max

x∈[a,b]

=M .

Let f (α) =min

x∈[a,b]

f (x) and f (β) =max

x∈[a,b]

. Then we have

m(b −a) ≤

f (x)d x ≤ M (b −a)

⇐⇒ m ≤

b −a

f (x)d x ≤ M .

Then, by the Intermediate Value Theorem (Theorem 3.6.9), there exists some

c ∈[a,b] such that

f (c) =

b −a

f (x)d x.

5.3 Antiderivatives

Deﬁnition 5.3.1 (Antiderivative). Let f : [a,b] 7→R be a continuous function.

We say F : [a,b] 7→R is an antiderivative of f on [a,b] if:

1. F is continuous on [a,b] and

2. F

(x) = f (x) ∀x ∈(a,b).

• Remark. If F and G are two antiderivatives of f on [a,b], then F

= f and

= f implies that F

on (a,b).

Lemma 5.3.2. Let f : [a,b] 7→R be a continuous function. Then

F (x) =

f (t)d t = lim

n→∞

(x)

(f )

is an antiderivative of f on [a,b].

Proof. We ﬁrst show that F

) = f (x

), x

∈ (a,b). Let x

∈ (a,b),x ∈ (a,b) \

}. The IMVT (Theorem 5.2.3) on the interval (x

, x) or (x, x

) implies that

there exists some c(x) ∈[x

, x] such that

f (c(x))(x −x

) =

f (t)d t.

We have

f (t)d t =

f (t)d t −

f (t)d t

=F (x) −F (x

)

⇐⇒ f (c(x)) =

F (x) −F (x

)

x −x

⇐⇒ lim

x→x

f (c(x)) = f

lim

x→x

c(x)

= f (x

)

⇐⇒ f (x

) = lim

x→x

F (x) −F (x

)

x −x

= f

)

and we have f (x) =F

(x), x ∈(a,b) and F is continuous on (a,b).

It remains to show that

lim

x→a

F (x) =F (a) and lim

x→b

−

F (x) =F (b).

Take x ∈ [a,b]. By the IMVT, there exists some c(x) ∈[a,x] such that f (c(x))(x−

a) =

f (t)d t =F (x) and

lim

x→a

f (x) = lim

x→a

f (c(x))(c −a) =0 =F(a).

The argument is identical for F (b) = f (b).

Lemma 5.3.3. Let f : [a,b] 7→R be a continuous function. Further, let h,g :

(c,d) 7→[a,b] be differentiable on (c,d). Then

K (x) =

g (x)

h(x)

f (t)d t

is differentiable on (c, d) and K

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

(x) − f (h(x))h

(x) for some

x ∈ (c, d).

Proof. Let

f (x) =











f (x) x ∈[a,b]

f (a) x ∈[a −1, a]

f (b) x ∈[b, b +1]

f is continuous on the interval [a −1, b +1]. Then we have

F (x) =

a−1

f (t)d t and K (x) =F (g (x))−F(h(x))

with

(x) =F

(g (x))g

(x) −F

(h(x))h

(x)

f (g (x))g

(x) −

f (h(x))h

(x)

= f (g (x))g

(x) − f (h(x))h

(x).

Theorem 5.3.4 (Fundamental Theorem of Calculus). Let f : [a, b] 7→R be

continuous and G be an antiderivative of f on [a,b]. Then

f (t)d t =G(b)−G(a) =G

and

(t )d t =G(b) −G(a).

Proof. We know that

F (x) =

f (t)d t

is an antiderivative of f which implies that F(x) =G(x)+c. Then we have that

F (b) =G(b)+c

F (a) =G(a)+c

and subsequently

f (t)d t =G(b)−G(a)

since F (a) is zero by deﬁnition.

5.4 Properties of Integrals

Lemma 5.4.1. Let f , g : [a,b] 7→ R be continuous functions and let λ ∈ R.

Then

(λf +g )(x)dx =λ

f (x)d x +

g (x)dx.

Proof. Take

K (x) =

(λf +g )(t)d t −λ

f (t)d t −

g (t )d t.

Then consider that K

(x) = (λf +g )(x) −λ f (x) −g(x) = 0 and K (x) = c ∃c ∈R

by the IMVT. Then we have K (b) =K (a) =0.

Lemma 5.4.2. Let f : [a,b] 7→R be a continuous function such that f (x) ≥

0 ∀x ∈[a,b]. Then, for every c ∈[a,b],

0 ≤

f (x)d x ≤

f (x)d x

and furthermore

f (x)d x =0 ⇐⇒ f (x) =0 ∀x ∈[a,b].

Proof. Left as an exercise.

5.5 Change of Variable

Lemma 5.5.1 (Change of Variable). Let f : [a, b] 7→R be a continuous func-

tion and I be an open interval. Further, let ϕ : I 7→ R be a continuous and

differentiable function on I (ϕ

is also continuous on I) such that such that

[α,β] ⊂I with ϕ([a,b]) ⊂[a,b]. Then

ϕ(β)

ϕ(α)

f (x)d x =

f (ϕ(t))ϕ

(t )d t.

We write x =ϕ(t ) is a change of variable.

Proof. Let G(x) =

ϕ(x)

f (t)d t and g (x) = f (ϕ(x))ϕ

(x). Both of these func-

tions are continuous on [α,β]. Furthermore, we have G

(x) =g (x) and G is an

antiderivative of g . So by the Fundamental Theorem of Calculus, we have

G(β) −G(α) =

g (t )d t

ϕ(β)

f (x)d x −

ϕ(α)

f (x)d x =

f (ϕ(t))ϕ

(t )d t

ϕ(β)

ϕ(α)

f (x)d x =

f (ϕ(t))ϕ

(t )d t.

• Example. Compute

J =

+1)

In this context, we have f : [0,1] 7→ R, f (x) = 1/(x

+1)

3/2

. Take ϕ(t) = tan(t )

such that ϕ : (−π/2,π/2) 7→R and ϕ

(t ) =1+tan

(t ). Apply change of variables:

J =

+1)

ϕ(0)=0

ϕ(

/4)=1

1 +tan

(t )

(tan

(t ) +1)

d t

(1 +tan

(t )

)

d t

cos(t )

d t

cos(t )d t =sin(t )

• Remark. The function ϕ does not need to be invertible.

5.6 Integration by Parts

Lemma 5.6.1 (Integration by Parts). Let f , g : [a,b] 7→R be two continuous

and differentiable functions with both f

and g

continuous on [a,b]. Then

(x)g (x)dx =−

f (x)g

(x)dx + f (x)g (x)

Proof. Note that

(f ·g )

= f

·g + f ·g

Thus, f ·g is an antiderivative of f

·g + f ·g

. Therefore,

f (b)g (b)− f (a)g (a) =

(x)g (x) + f (x)g

(x)

d x.

So, integration by parts is nothing more than the Fundamental Theorem of

Calculus applied to a product of functions.

• Examples.

1. Compute

J(x) =

(t )

=⇒

g (t)=e

cos(t )

f (t)

d t =−

(t )

(sin(t ))

f (t)

d t +cos(x) ·e

−cos(0) ·e

| {z }

=−

·cos(t )

| {z }

J(x)

+sin(x) ·e

−sin(0) ·e

+cos(x) ·e

−1

=⇒ 2J(x) =sin(x) ·e

+cos(x) ·e

−1

⇐⇒ J(x) =

(sin(x) +cos(x))·e

−

2. Take n ≥1. Compute

(x) =

+1)

d t.

Take g (t ) =1/(t

+1)

and f (t) =t such that g

(t ) =−n(t

+1)

−(n+1)

and

(t ) =1. Then we have

(x) =−

t(−2t n)

+1)

n+1

d t +

+1)

−0

=2n

+1)

n+1

d t −2n

+1)

n+1

d t +

+1)

=2n ·J

(x) −2n ·J

n+1

(x) +

+1)

⇐⇒ J

n+1

(x) =

2n −1

·J

(x) +

+1)

5.7 Taylor Expansion

Lemma 5.7.1. Let f : [a,b] 7→ R be (n +1) times differentiable and f

(n+1)

be continuous on [a,b]. Then

f (x) = f (a)+f

(a)(x −a)+···+

(n)

(a)

(x −a)

(x −t)

(n+1)

(t )d t.

Proof. Using integration by parts, compute for k ∈N,n ≥1

(x −t )

k−1

| {z }

(t )

(k)

(t )

| {z }

v(t)

d t =

(x −t )

(k+1)

(t )d t +

−

(x −t )

(k)

(t )

(x −t )

(k+1)

(t )d t +

(x −a)

(k)

(a).

We now proceed by induction on n. For n = 0, we can simply use the Funda-

mental Theorem of Calculus to obtain

f (x) − f (a) =

(t )d t

and P(0) is true. Now, assume that P(k −1) is true, i.e.,

f (x) = f (a)+···+

(k−1)

(a)

(k −1)!

(x −a)

k−1

(k −1)!

(x −t )

k−1

(k)

(t )d t.

Using the formula we just proved, we have that

f (x) = f (a)+···+

(k−1)

(a)

(x −a)

k−1

(k −1)!

(x −a)

(k)

(a)

(x −t )

(k+1)

(t )d t

which implies that P(k) is true.

• Example. Apply this for n =1 and obtain

f (x) = f (a)+ f

(a)(x −a) +

(x −a)f

(t )d t

⇐⇒ f

(a) =

f (x) − f (a)

x −a

−

x −a

(x −t )f

(t )d x

| {z }

“error"

⇐⇒

(a) −

f (x) − f (a)

x −a

≤

x −a

|(x −t )|·|f

(t )|d t ≤sup

[a,b]

(t )

2(x −a)

(x −a)

⇐⇒ ≤sup

[a,b]

(t ) ·

x −a

Take the ODE:

(t ) = f (t , y(t)), y(0) =α,t ∈[0,1].

Let σ

be a regular partition of [0,1] such that h(σ

) =

/n. Then we have that

) ≈

y(x

i+1

) −y(x

)

i+1

−x

and

) −

y(x

i+1

) −y(x

)

≤sup

[a,b]

(x)

This is a valid solution to the ODE since

y(x

i+1

) −y(x

)

≈ f (x

, y(x

)).

Let Y

be an approximation of y(x

) deﬁned by:

(

i+1

−Y

= f (x

) ∀i =0,1,...,n −1

=α.

This is equivalent to

(

i+1

+h · f (x

) ∀i =0,1,...,n −1

=α.

5.8 Weak Derivatives

Deﬁnition 5.8.1 (Weak derivative). Let f , g : [a,b] 7→ R be two continuous

functions with

(x)g (x)dx =−

f (x)g

(x)dx + f g

Now assume that g is smooth and g (a) = g (b) = 0. With this condition, we

now have

(x)g (x)dx =−

f (x)g

(x)dx.

We say that p : [a,b] 7→ R is the weak derivative of f provided that p is inte-

grable and

pg =−

f g

for every g smooth with g (c) =g (b) =0.

• Remark. If p exists, it is unique and if f is differentiable, then p = f

• Example. Take f (x) =|x|, x ∈ [−1, 1]. We have

−1

=−

−1

|x|·g

(x)dx ∀g smooth ,g(−1) =g (1) =0.

But

−

−1

|x|·g

(x)dx =

−1

x ·g

(x) −

x ·g

(x)

−1

1 ·g (x) +



x ·g (x)

−1

g (x)dx +



x ·g (x)

(IBP)

−1

p(x)g (x)d x

where

p(x) =

(

−1 x < 0

1 x ≥ 0.

Deﬁnition 5.8.2 (Diraq measure). Consider the function

(x) =











0 x >

0 x <−

−

≤x ≤

Compute

−1

f (x)δ

(x)dx =

−1

f (x)·

d x =

−1

n·f (x) −→

n→∞

f (0) =

−1

f (x)δ

(x)dx.

This is called the Dirac measure.

• Example. Now take

f (x) =

(

1 x ≥ 0

0 x < 0.

Now try to make sense of a weak derivative. Consider g smooth such that

g (−1) =g (1) =0. Then we have

−

−1

f (x)g

(x)dx =−

(x)dx

=−g (1) +g(0) (FTC)

=g (0) =

−1

(x)g (x)dx.

List of Theorems

1.1.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Deﬁnition (Boundedness) . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.2 Deﬁnition (Inﬁmum) . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.3 Deﬁnition (Supremum) . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.4 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.5 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.6 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.7 Deﬁnition (Absolute value) . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.8 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 Deﬁnition (Sequential convergence) . . . . . . . . . . . . . . . . . . 8

2.1.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.3 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.1.4 Deﬁnition (Sequential monotonicity) . . . . . . . . . . . . . . . . . 10

2.1.5 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.2 Theorem (Squeeze Theorem) . . . . . . . . . . . . . . . . . . . . . . 13

2.4.1 Deﬁnition (limsup and liminf) . . . . . . . . . . . . . . . . . . . . . 14

2.4.2 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5.1 Deﬁnition (Subsequences) . . . . . . . . . . . . . . . . . . . . . . . . 16

2.5.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.5.3 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.5.4 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.6.1 Deﬁnition (Peak points) . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.6.2 Lemma (Monotone Subsequence) . . . . . . . . . . . . . . . . . . . 16

2.6.3 Theorem (Bolzano-Weirstrass) . . . . . . . . . . . . . . . . . . . . . . 17

2.6.4 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.7.1 Deﬁnition (Cauchy sequences) . . . . . . . . . . . . . . . . . . . . . 19

2.7.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.8.1 Deﬁnition (Divergence to ±∞) . . . . . . . . . . . . . . . . . . . . . 21

2.8.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.0.1 Deﬁnition (Functional limits) . . . . . . . . . . . . . . . . . . . . . . 22

3.0.2 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.1.1 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.1.2 Lemma (Cauchy Criteria for Functions) . . . . . . . . . . . . . . . . 25

3.2.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3.3 Theorem (Squeeze Theorem) . . . . . . . . . . . . . . . . . . . . . . 27

3.4.1 Lemma (Composition of Functions) . . . . . . . . . . . . . . . . . . 28

3.5.1 Deﬁnition (Inﬁnite functional limits) . . . . . . . . . . . . . . . . . . 28

3.5.2 Deﬁnition (Inﬁnite functional limits) . . . . . . . . . . . . . . . . . . 29

3.5.3 Deﬁnition (Combined inﬁnite functional limits) . . . . . . . . . . . 29

3.6.1 Deﬁnition (Functional continuity) . . . . . . . . . . . . . . . . . . . 29

3.6.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.6.3 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.6.4 Deﬁnition (Continuity on an interval) . . . . . . . . . . . . . . . . . 31

3.6.5 Deﬁnition (Left/right limits) . . . . . . . . . . . . . . . . . . . . . . . 31

3.6.6 Deﬁnition (Left/right continuous) . . . . . . . . . . . . . . . . . . . 31

3.6.7 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.6.8 Deﬁnition (Continuity on a closed interval) . . . . . . . . . . . . . . 32

3.6.9 Theorem (Intermediate Value Theorem) . . . . . . . . . . . . . . . . 32

3.6.10 Deﬁnition (Monotone functions) . . . . . . . . . . . . . . . . . . . . 34

3.6.11 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.6.12 Deﬁnition (Boundedness on a set) . . . . . . . . . . . . . . . . . . . 34

3.6.13 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.6.14 Theorem (Extreme Value Theorem) . . . . . . . . . . . . . . . . . . . 35

3.6.15 Deﬁnition (Supremum and max of a function) . . . . . . . . . . . . 36

3.7.1 Deﬁnition (Uniform continuity) . . . . . . . . . . . . . . . . . . . . . 36

3.7.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.7.3 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.7.4 Theorem (Uniform continuity) . . . . . . . . . . . . . . . . . . . . . 38

3.7.5 Theorem (Continuity Extension) . . . . . . . . . . . . . . . . . . . . 39

3.8.1 Proposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.8.2 Theorem (Inverse continuity) . . . . . . . . . . . . . . . . . . . . . . 42

4.0.1 Deﬁnition (Locally differentiable) . . . . . . . . . . . . . . . . . . . . 42

4.0.2 Deﬁnition (Globally differentiable) . . . . . . . . . . . . . . . . . . . 43

4.0.3 Theorem (Caratheodory’s Theorem) . . . . . . . . . . . . . . . . . . 44

4.0.4 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.1.1 Deﬁnition (Local extremum) . . . . . . . . . . . . . . . . . . . . . . . 45

4.1.2 Theorem (Fermat) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2.1 Proposition (Basic rules of differentiation) . . . . . . . . . . . . . . . 46

4.2.2 Proposition (Product Rule) . . . . . . . . . . . . . . . . . . . . . . . . 46

4.2.3 Proposition (Quotient Rule) . . . . . . . . . . . . . . . . . . . . . . . 47

4.2.4 Proposition (Chain Rule) . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.3.1 Theorem (Rolle’s Theorem) . . . . . . . . . . . . . . . . . . . . . . . . 49

4.3.2 Theorem (Mean Value Theorem) . . . . . . . . . . . . . . . . . . . . 49

4.3.3 Theorem (Mean Value Theorem (Cauchy)) . . . . . . . . . . . . . . 50

4.3.4 Proposition (L’Hospital’s Rule) . . . . . . . . . . . . . . . . . . . . . . 51

4.4.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.4.2 Theorem (Taylor) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.4.3 Lemma (Taylor Expansion uniqueness) . . . . . . . . . . . . . . . . 53

4.4.4 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.4.5 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.0.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.0.2 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.0.3 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.0.4 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.0.5 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.0.6 Deﬁnition (Integral) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.0.7 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.1.1 Deﬁnition (Riemann integrability) . . . . . . . . . . . . . . . . . . . 62

5.2.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.2.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.2.3 Lemma (Integral Mean Value Theorem) . . . . . . . . . . . . . . . . 64

5.3.1 Deﬁnition (Antiderivative) . . . . . . . . . . . . . . . . . . . . . . . . 64

5.3.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5.3.3 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

5.3.4 Theorem (Fundamental Theorem of Calculus) . . . . . . . . . . . . 66

5.4.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.4.2 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5.5.1 Lemma (Change of Variable) . . . . . . . . . . . . . . . . . . . . . . . 67

5.6.1 Lemma (Integration by Parts) . . . . . . . . . . . . . . . . . . . . . . 68

5.7.1 Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.8.1 Deﬁnition (Weak derivative) . . . . . . . . . . . . . . . . . . . . . . . 71

5.8.2 Deﬁnition (Diraq measure) . . . . . . . . . . . . . . . . . . . . . . . . 71