Analysis

Course 221

David Simms

School of Mathematics

Trinity College

Dublin

Typesetting

Eoin Curran.

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Contents

1 Lebesgue Measure 71.1 Algebra of Subsets . . . . . . . . . . . . . . . . . . . . . . . . 71.2 The Interval Algebra . . . . . . . . . . . . . . . . . . . . . . . 91.3 The length measure . . . . . . . . . . . . . . . . . . . . . . . . 111.4 The σ-algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 The outer measure . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Extension of measure to σ-algebra, using outer measure . . . . 161.7 Increasing Unions, Decreasing Intersections . . . . . . . . . . . 201.8 Properties of Lebesgue Measure . . . . . . . . . . . . . . . . . 221.9 Borel Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 Integration 292.1 Measure Space, Measurable sets . . . . . . . . . . . . . . . . . 292.2 Characteristic Function . . . . . . . . . . . . . . . . . . . . . . 302.3 The Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.4 Monotone Convergence Theorem . . . . . . . . . . . . . . . . 342.5 Existence of Monotone Increasing Simple Functions converg-

ing to f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.6 ’Almost Everywhere’ . . . . . . . . . . . . . . . . . . . . . . . 392.7 Integral Notation . . . . . . . . . . . . . . . . . . . . . . . . . 452.8 Fundamental Theorem of Calculus . . . . . . . . . . . . . . . 462.9 Fatou’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 482.10 Dominated Convergence Theorem . . . . . . . . . . . . . . . . 492.11 Differentiation under the integral sign . . . . . . . . . . . . . . 50

3 Multiple Integration 533.1 Product Measure . . . . . . . . . . . . . . . . . . . . . . . . . 533.2 Monotone Class . . . . . . . . . . . . . . . . . . . . . . . . . . 563.3 Ring of Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . 573.4 Integration using Product Measure . . . . . . . . . . . . . . . 593.5 Tonelli’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 63

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3.6 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 64

4 Differentiation 714.1 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.2 Normed Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.3 Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.4 Topological space . . . . . . . . . . . . . . . . . . . . . . . . . 734.5 Continuous map of topological spaces . . . . . . . . . . . . . . 754.6 Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 774.7 Operator Norm . . . . . . . . . . . . . . . . . . . . . . . . . . 774.8 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.9 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.10 Cr Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 844.11 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

5 Calculus of Complex Numbers 895.1 Complex Differentiation . . . . . . . . . . . . . . . . . . . . . 895.2 Path Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.3 Cauchy’s Theorem for a triangle . . . . . . . . . . . . . . . . . 955.4 Winding Number . . . . . . . . . . . . . . . . . . . . . . . . . 985.5 Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . 1015.6 Term-by-term differentiation, analytic functions, Taylor series 104

6 Further Calculus 1076.1 Mean Value Theorem for Vector-valued functions . . . . . . . 1076.2 Contracting Map . . . . . . . . . . . . . . . . . . . . . . . . . 1086.3 Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . 109

7 Coordinate systems and Manifolds 1157.1 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . 1157.2 Cr-manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177.3 Tangent vectors and differentials . . . . . . . . . . . . . . . . . 1187.4 Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.5 Pull-back, Push-forward . . . . . . . . . . . . . . . . . . . . . 1257.6 Implicit funciton theorem . . . . . . . . . . . . . . . . . . . . 1277.7 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1307.8 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . 1327.9 Tangent space and normal space . . . . . . . . . . . . . . . . . 1337.10 ?? Missing Page . . . . . . . . . . . . . . . . . . . . . . . . . . 1347.11 Integral of Pull-back . . . . . . . . . . . . . . . . . . . . . . . 1357.12 integral of differential forms . . . . . . . . . . . . . . . . . . . 136

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7.13 orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

8 Complex Analysis 1458.1 Laurent Expansion . . . . . . . . . . . . . . . . . . . . . . . . 1458.2 Residue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 1478.3 Uniqueness of analytic continuation . . . . . . . . . . . . . . . 151

9 General Change of Variable in a multiple integral 1539.1 Preliminary result . . . . . . . . . . . . . . . . . . . . . . . . . 1539.2 General change of variable in a multiple integral . . . . . . . . 156

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Chapter 1

Lebesgue Measure

1.1 Algebra of Subsets

Definition Let X be a set. A collection A of subsets of X is called analgebra of subsets of X if

1. ∅ ∈ A

2. E ∈ A =⇒ E ′ ∈ A. E ′ = {x ∈ X : x 6∈ E}

3. E1, . . . , Ek ∈ A =⇒ E1 ∪ E2 ∪ · · · ∪ Ek ∈ A. i.e., A is closed undercomplements and under finite unions. Hence A is closed under finiteintersection. (because E1 ∩ · · · ∩ Ek = (E1

′ ∪ · · · ∪ Ek ′)′)

Example Let I be the collection of all finite unions of intervals in R of theform:

1. (a, b] = {x ∈ R : a < x ≤ b}

2. (−∞, b] = {x ∈ R : x ≤ b}

3. (a,∞) = {x ∈ R : a < x}

4. (−∞,∞) = R

I is an algebra of subsets of R, called the Interval Algebra.

�

�

�

�

�

�

7

We want to assign a ‘length’ to each element of the interval algebra I, sowe want to allow ‘∞’ as a length. We adjoin to the real no’s the 2 symbols∞ and −∞ to get the Extended Real Line:

Definition The Extended Real Line:

R ∪ {−∞,∞} = [−∞,∞]

We extend ordering, addition, multiplication to [−∞,∞] by:

1. −∞ < x <∞ ∀x ∈ R

2. Addition

(a) ∞ + ∞ = x+ ∞ = ∞ + x = ∞(b) (−∞) + (−∞) = x+ (−∞) = (−∞) + x = −∞

(Don’t define ∞ + (−∞) or (−∞) + ∞.)

3. Multiplication

(a) ∞∞ = ∞ = (−∞)(−∞)

(b) ∞(−∞) = (−∞)∞ = −∞

(c) x∞ = ∞x =

∞ x > 00 x = 0−∞ x < 0

(d) x(−∞) = (−∞)x =

−∞ x > 00 x = 0∞ x < 0

Definition If A ⊂ [−∞,∞] then

1. supA = ∞ if ∞ ∈ A or if A has no upper bound in R.

2. infA = −∞ if −∞ ∈ A or if A has no lower bound in R.

Definition Let A be an algebra of subsets of X. A function m : A −→ [0,∞]is called a measure on A if:

1. m(∅) = 0

2. if E =⋃∞j=1Ej is countably disjoint with Ej ∈ A ∀j then m(E) =

∑∞j=1m(Ej). (m is Countably Additive.)

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E

Figure 1.1: E is a disjoint union of sets

Note:

1. Countable means that E1, E2, . . . is either a finite sequence or can belabelled by (1, 2, 3, . . .).

2. Disjoint means that Ei ∩ Ej = ∅ ∀i 6= j

3. m(E) =∑∞

j=1m(Ej) means either:

(a)∑∞

j=1m(Ej) is a convergent series of finite no’s with the sum ofthe series as m(E)

(b)∑∞

j=1m(Ej) is a divergent series with m(E) = ∞(c) m(Ej) = ∞ for some j, and m(E) = ∞

1.2 The Interval Algebra

Definition For each E ∈ I, the interval algebra (See Section 1.1), write

E = E1 ∪ · · · ∪ Ek

(disjoint) where each Ei is of the form (a, b], (−∞, b], (a,∞), or(−∞,∞). Thelength of E is m(E) = m(E1) +m(E2) + · · ·+m(Ek) where m(a, b] = b− a.

m(−∞, b] = m(a,∞) = m(−∞,∞) = ∞

Definition a set V ⊂ R is called an Open Subset of R if:

for each a ∈ V ∃ε > 0 s.t.(a− ε, a+ ε) ⊂ V

�

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�

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�

�

a − ε a + εa

i.e., every point is an interior point : it has no end points.

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Theorem 1.2.1. (Heine-Borel-Lebesgue) The closed interval is compact.

Let {Vi}i∈I be a family of open sets in R which cover the closed interval[a, b]. Then ∃ a finite number of them: Vi1 , . . . , Vik (say) which cover [a, b](i.e. [a, b] ⊂ Vi1 ∪ · · · ∪ Vik .)

Proof. Put

K = {x ∈ [a, b] s.t. ∃( finite set {i1, . . . , ir} ⊂ I s.t. [a, x] ⊂ Vi1 ∪ · · · ∪ Vir)}

ca b

a ∈ K =⇒ K 6= ∅

Let

c = supK

Then

a ≤ c ≤ b c ∈ Vj (say)

Vj is open, therefore

∃ ε > 0 s.t. (c− ε, c+ ε) ⊂ Vj

and

∃ k ∈ K s.t. c− ε < k ≤ c,

[a, k] ⊂ Vi1 ∪ · · · ∪ Vir (say)

Then

[a,min(c +ε

2, b)] ⊂ (Vi1 ∪ · · · ∪ Vir ∪ Vj)

therefore

min(c +ε

2, b) ∈ K

But c+ ε26∈ K (as c = supK) =⇒ b ∈ K as required.

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1.3 The length measure

Theorem 1.3.1. The length function

m : I −→ [0,∞]

is a measure on the interval algebra IProof. We have to check that m is countably additive, which reduces toshowing that if

(a, b] =∞⋃

j=1

(aj, bj]

is a countable disjoint union then

∞∑

j=1

(bj − aj)

is a convergent series with sum b− a.Take the first n intervals and relabel them:

(a1, b1], . . . , (an, bn]

so that:

a a1 b1 a2 b2 a3 b3 an bn b

then∑n

j=1(bj − aj) = b1 − a1 +b2 − a2 +b3 − a3 + · · · +bn − an≤ a2 − a +a3 − a2 +a4 − a3 + · · · +b− an= b− a

therefore ∞∑

j=1

(bj − aj)

converges and has sum ≤ b− a

Let ε > 0 and put ε0 = ε2, ε1 = ε

4, . . . , εj = ε

2j+1 , . . .

so εj > 0 and∑∞

j=1 εj = ε

Then(a− ε0, a+ ε0), (a1, b1 + ε1), (a2, b2 + ε2), . . .

is a family of open sets which cover [a, b]

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By the compactness of [a, b] ∃ a finite number of these open sets whichcover [a, b], and which by renumbering and discarding some intervals if nec-essary we can take as:

(a− ε0, a+ ε0), (a1, b1 + ε1), . . . , (an, bn + εn)

so that:

a − ε0 a+ ε0

a a1

a2 b2 + ε2

b1 + ε1 a3 b3 + ε3 an−1 bn−1 + εn−1 b

an bn + εn

b− a = a1 − a +a2 − a1 + · · · +an − an−1 +b− an< ε0 +(b1 + ε1 − a1) + · · · +(bn−1 + εn−1 − an−1) +(bn + εn − an)< ε +

∑∞j=1(bj − aj)

is true ∀ ε > 0. Therefore

b− a ≤∞∑

j=1

(bj − aj)

Therefore ∞∑

j=1

(bj − aj) = b− a

as required.

Let m :−→ [0,∞] be a measure m on an algebra A of subsets of X.

E ⊂ F =⇒

Thenm(E) ≤ m(F )and

m(F ∩ E ′) = m(F ) −m(E) if m(F ) 6= ∞

ThenF = E ∪ (F ∩ E ′)

E

F

is a disjoint union, and

m(F ) = m(E) +m(F ∩ E ′)

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Theorem 1.3.2. (m is subadditive on countable unions)

m(

∞⋃

j=1

) ≤∞∑

i=1

m(Ei)

Proof. We proceed as follows:Let

Fn = E1 ∪ E2 ∪ · · · ∪ En

1E E2 E3 En

∞⋃

j=1

Ej = E1 ∪ (E2 ∩ F1′) ∪ (E3 ∩ F2

′) ∪ · · ·

m(⋃∞j=1Ej) = m(E1) +m(E2 ∩ F1

′) +m(E3 ∩ F2′) + · · ·

≤ m(E1) +m(E2) +m(E3) + · · ·hence the result:

m(∞⋃

j=1

Ej) ≤∞∑

j=1

m(Ej)

m is subadditive on countable unions.

1.4 The σ-algebra

Our aim is to extend the notion of length to a much wider class of subsets ofR. In particular to sets obtainable from I by a sequence of taking countableunions and taking complements.

Definition an algebra A of subsets of X is called a σ-algebra if for eachsequence

E1, E2, E3, . . .

of elements of A, their union

E1 ∪ E2 ∪ E3 ∪ · · ·is also an element of A.

So A is closed under countable unions, and hence also under countableintersections.

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1.5 The outer measure

Definition We define the outer measure m associated with m to be thefunction:

m : {all subsets of X} −→ [0,∞]

given by:

m(E) = inf

∞∑

j=1

m(Ej)

where the inf is taken over all sequences Ej of elements of A such that:

E ⊂∞⋃

j=1

Ej

Theorem 1.5.1. m(E) = m(E) ∀E ∈ A (i.e. m agrees with m on A)

Proof. 1. ifE ∈ A

and

E ⊂∞⋃

j=1

Ej with Ej ∈ A

then:

m(E) ≤∞∑

j=1

m(Ej)

thereforem(E) ≤ m(E)

2. ifE ∈ A

thenE, ∅, ∅, . . .

is a sequence of elements of A whose union contains E. Therefore:

m(E) ≤ m(E) + 0 + 0 + · · ·

thereforem(E) ≤ m(E)

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Theorem 1.5.2. E ⊂ F =⇒ m(E) ≤ m(F )

Proof. if E ⊂ F then each sequence in A which covers F will also cover E.Therefore:

m(E) ≤ m(F )

Theorem 1.5.3. m is subadditive on countable unions

m(∞⋃

j=1

Ej) ≤∞∑

j=1

m(Ej)

for all sequences E1, E2, . . . of subsets of X.

Proof. Let ε > 0. Choose:

ε0, ε1, ε2, . . . > 0

such that∞∑

j=0

εj = ε

choose Bij ∈ A s.t.

Ei ⊂ Bi1 ∪ Bi2 ∪ · · · ∪Bij ∪ · · ·

and s.t.∞∑

j=1

m(Bij) ≤ m(Ei) + εi

Then∞⋃

i=1

Ei ⊂∞⋃

i=1

∞⋃

j=1

Bij

and∞∑

i=1

∞∑

j=1

m(Bij) ≤∞∑

i=1

m(Ei) + ε

therefore

m(∞⋃

j=1

Ej) ≤∞∑

j=1

m(Ej)

as required.

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Definition We call a subset E ⊂ X measurable w.r.t m if:

m(A) = m(A ∩ E) + m(A ∩ E ′)

for all A ⊂ X

(E splits every set A into two pieces whose outer measures add up.)

E

A

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1.6 Extension of measure to σ-algebra, using

outer measure

We can now prove the central:

Theorem 1.6.1. Let m be a measure on an algebra A of subsets of X, mthe associated outer measure, and M the collection of all subsets of X whichare measurable with respect to m.

Then M is a σ-algebra containing A and m is a (countably additive)measure on M .

Corollary 1.6.2. the measure m on the algebra A can be extended to ameasure (also denoted by m) on the σ-algebra M by defining:

m(E) = m(E) ∀E ∈M

in particular:

Corollary 1.6.3. the length measure m on the interval algebra I can beextended to a measure (also denoted by m) on the σ-algebra M of measurablesets w.r.t. m. We call the elements of M the Lebesgue Measureable sets,and the extended measure the (one-dimensional) Lebesgue Measure.

I ⊂ {Lebesgue Measurable Sets} ⊂ all subsets of R

length measure m Lebesgue measure outer measure m(countably additive) (not countably additive

=⇒ not a measure)↘ ↓ ↙

[0,∞]

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Proof. 1. A ⊂M

Let E ∈ A, let A ⊂ X

Need to show:

m(A) = m(A ∩ E) + m(A ∩ E ′)

For ε > 0. Let

F1, F2, . . .

be a sequence in A s.t.

A ⊂∞⋃

j=1

Fj

and s.t.∞∑

j=1

m(Fj) ≤ m(A) + ε

E

AFi

Then

m(A) ≤ m(A ∩ E) +m(A ∩ E ′) since m is subadditive

≤ m(⋃∞j=1 Fj ∩ E) +m(

⋃∞j=1 Fj ∩ E ′)

≤∑∞j=1 m(Fj ∩ E) +

∑∞j=1 m(Fj ∩ E ′)

=∑∞

j=1m(Fj ∩ E) +∑∞

j=1m(Fj ∩ E ′) since Fj ∩ E, Fj ∩ E ′ ∈ Aand m = m on A

=∑∞

j=1m(Fj) since m is additive

≤ m(A) + ε

17

therefore,

m(A) ≤ m(A ∩ E) + m(A ∩ E ′) ≤ m(A) + ε

∀ε > 0, and therefore

m(A) = m(A ∩ E) + m(A ∩ E ′)

as required.

2. M is an algebra

let E, F ∈M ; Let A ⊂ X

4

A

F

E

1 2

3

Then

m(A) = m(1 + 2) +m(3 + 4) since E ∈M

= m(1 + 2) +m(3) +m(4) since F ∈M

= m(1 + 2 + 3) +m(4) since E ∈M

thereforeE ∪ F ∈M

and M closed under finite unions.

Also, M closed under complements by symmetry in the definition.

3. M is a σ-algebra

Let E =⋃∞i=1Ei be a countable disjoint union with Ei ∈ M . We need

to show that E ∈M . Put

Fn = E1 ∪ E2 ∪ · · · ∪ En

18

n

1

A

E En2

F

E

Fn ∈M since M is an algebra. Let A ⊂ X. Then

m(A) = m(A ∩ Fn) +m(A ∩ Fn′) since Fn ∈M

=∑n

k=1 m(A ∩ Ek) +m(A ∩ Fn′) since E1, . . . , En ∈M,

and are disjoint

≥∑nk=1 m(A ∩ Ek) +m(A ∩ E ′) since E ′ ⊂ Fn

′

is true ∀n. Therefore:

m(A) ≥∑∞k=1 m(A ∩ Ek) +m(A ∩ E ′)

≥ m(⋃∞k=1A ∩ Ek) +m(A ∩ E ′) since m is countably subadditive

(*)

= m(A ∩ E) +m(A ∩ E ′)

≥ m(A) since m is subadditive

All the above are equalities:

m(A) = m(A ∩ E) + m(A ∩ E ′)

thereforeE ∈M

and M is closed under countable disjoint unions. But any countableunion:

E1 ∪ E2 ∪ E3 ∪ · · ·

19

can be written as a countable disjoint union

E1 ∪ (E2 ∩ F1′) ∪ (E3 ∩ F2

′) ∪ · · ·

whereFn = E1 ∪ · · · ∪ En

therefore M is closed under countable unions as required.

4. m is countably additive on M

Put A = E in (*) to get

∞∑

k=1

m(Ek) = m

( ∞⋃

k=1

Ek

)

as required.

1.7 Increasing Unions, Decreasing Intersec-

tions

Definition We use the notation Ej ↑ E to denote that

21

E

EE

E1 ⊂ E2 ⊂ · · · ⊂ Ej ⊂ · · ·is an increasing sequence of sets such that

∞⋃

j=1

Ej = E

Definition We use the notation Ej ↓ E to denote that

1E

E2E

20

E1 ⊃ E2 ⊃ · · · ⊃ Ej ⊃ · · ·is a decreasing sequence of sets such that

∞⋂

j=1

Ej = E

Theorem 1.7.1. Ej ↑ E =⇒ limj→∞m(Ej) = m(E)

Proof. 1. if m(Ej) = ∞ for some j then the result holds

2. if m(Ej) is finite ∀j then

E = E1 ∪ (E2 ∩ E1′) ∪ (E3 ∩ E2

′) ∪ · · ·

is a countable disjoint union and

m(E) = m(E1) + [m(E2) −m(E1)] + [m(E3) −m(E2)] + · · ·

= limn→∞

{

m(E1) + [m(E2) −m(E1)] + · · · + [m(En) −m(En−1)]

}

= limn→∞m(En)

as required.

Theorem 1.7.2.

Ej ↓ Em(E1) 6= ∞

}

=⇒ limj→∞

m(Ej) = m(E)

Proof.(E1 ∩ Ej ′) ↑ (E1 ∩ E ′)

therefore:limj→∞

m(E1 ∩ Ej ′) = m(E1 ∩ E ′)

andlimj→∞

[m(E1) −m(Ej)] = m(E1) −m(E)

solimj→∞

m(Ej) = m(E)

as required.

21

Example

(a− 1

n, b] ↓ [a, b]

�

�a− 1

na b

for Lebesgue measure m:

m[a, b] = limn→∞m(a− 1n, b]

= limn→∞(b− a+ 1n)

= b− a

as expected.

1.8 Properties of Lebesgue Measure

We now show that the Lebesgue measure is the only way of extending the‘length’ measure on the interval algebra I to a measure on the Lebesguemeasurable sets.

We need:

Definition a measure m on an algebra A of subsets of X is called σ-finiteif ∃ a sequence Xi in A such that

Xi ↑ X

and m(Xi) is finite for all i.

Example the Lebesgue measure m is σ-finite because:

(−n, n] ↑ R

and m(−n, n] = 2n is finite

�

�−n O n

22

Theorem 1.8.1. (Uniqueness of Extension) Let m be a σ-finite measure onan algebra A of subsets of X.

Let M be the collection of measurable sets w.r.t. m.Let l be any measure on M which agrees with m on A.Then l(E) = m(E) ∀E ∈M

Proof. Let E ∈M . Then for each sequence {Ai}, Ai ∈ A covering E:

E ⊂∞⋃

j=1

Ai

we have

l(E) ≤∑∞i=1 l(Ai) =

∑∞i=1m(Ai) since l = m on A

Therefore,

(∗) l(E) ≤ m(E) by definition of the outer measure m

Now let Xi ↑ X and m(Xi) finite, Xi ∈ A. Consider

E

iX

l(Xi ∩ E) + l(Xi ∩ E ′) = l(Xi) = m(Xi) = m(Xi ∩ E) + m(Xi ∩ E)

By (*) it follows that

l(Xi ∩ E) = m(Xi ∩ E)

Take limi→∞ to get l(E) = m as required.

As a consequence we have:

Lemma 1.8.2. Let m be the Lebesgue measure on R. Then for each mea-surable set E and each c ∈ R we have:

1. m(E + c) = m(E). m is translation invariant

2. m(cE) = |c|m(E)

Proof. Let M be the collection of measurable sets

23

1. define a measure mc on M by

mc(E) = m(E + c)

then

mc(a, b] = m(a+ c, b + c] = (b + c) − (a + c) = b− a = m(a, b]

therefore mc agrees with m on the interval algebra I. Therefore mc

agrees with m on M .

2. define a measure mc on M by

mc(E) = m(cE)

Thenmc(a, b] = m(c(a, b])

=

m(ca, cb] c > 0m{0} c = 0m[cb, ca) c < 0

=

cb− ca

0ca− cb

= |c|(b− a)

= |c|m(a, b]

Therefore mc agrees with |c|m on the interval algebra. Therefore mc

agrees with |c|m on M .

1.9 Borel Sets

Definition if V is any collection of subsets of X, we denote by G(V) theintersection of all the σ-algebras of subsets of X which contain V. We have:

1. G(V) is a σ-algebra containing V

2. if W is any σ-algebra which contains V then

V ⊂ G(V) ⊂ W

24

Thus G(V) is the smallest σ-algebra of subsets of X which contains V. G(V)is called the σ-algebra generated by V.

Definition The σ-algebra generated by the open sets of R is called thealgebra of Borel Sets of R.

Theorem 1.9.1. The σ-algebra generated by the interval algebra I is thealgebra of Borel sets of R.

Proof. Let V be the collection of open sets of R

1. (a, b] =⋂∞n=1(a, b+ 1

n) is the intersection of a countable family of open

sets. Therefore (a, b] ∈ G(V).

Therefore

I ⊂ G(V)

and

G(I) ⊂ G(V)

2. let V be an open set in R. For each a ∈ V choose an interval Ia ∈ Iwith rational endpoints s.t. a ∈ Ia ⊂ V .

�

�

�

�a

�

�

Ia

V

V =⋃

a∈VIa

so V is the union of a countable family of elements of I. Therefore

V ∈ G(I)

implies

V ⊂ G(I) =⇒ G(V) ⊂ G(I)

So, combining these two results,

G(I) = G(V) = algebra of Borel sets

25

We have:

I ⊂ Borel Sets in R ⊂ Lebesgue measurable Sets in R

The following theorem shows that any Lebesgue measurable set in R canbe obtained from a Borel set by removing a set of measure zero.

Theorem 1.9.2. Let E be a Lebesgue measurable subset of R. Then thereis a Borel set B containing E such that

B − E = B ∩ E ′

has measure zero, and hence

m(B) = m(E)

B

EB-E

Proof. 1. Suppose m(E) is finite. Let k be an integer > 0. Choose asequence

I1, I2, I3, . . .

in I such that

E ⊂∞⋃

i=1

Ii

and s.t.

m(E) ≤∞∑

i=1

m(Ii) ≤ m(E) +1

k

Put Bk =⋃∞i=1 Ii then E ⊂ Bk, Bk is Borel, and

m(E) ≤ m(Bk) ≤ m(E) +1

k

Put B =⋂∞k=1Bk. Then E ⊂ B, B is Borel, and

m(E) ≤ m(B) ≤ m(E) +1

k

for all k. Som(E) = m(B) =⇒ m(B − E) = 0

since m(E) is finite.

26

2. Suppose m(E) = ∞.

−(k + 1) −k k k + 1

PutEk = {x ∈ E : k ≤ |x| < k + 1}

Then

E =

∞⋃

k=0

Ek

is a countable disjoint union and m(Ek) is finite.

For each integer k choose a Borel set Bk s.t.

Ek ⊂ Bk

andm(Bk − Ek) = 0

Put B =⋃∞k=1Bk. Then E ⊂ B, B is Borel, and

B − E ⊂∞⋃

k=1

(Bk − Ek)

therefore

m(B − E) ≤∞∑

k=1

m(Bk − Ek) =

∞∑

k=1

0 = 0

as required.

E1 E2

B1B2

27

28

Chapter 2

Integration

2.1 Measure Space, Measurable sets

Definition We fix a set X, a σ-algebra M of subsets of X, and a measurem on M . The triple (X,M,m) is then called a measure space, the elementsof M are called the measurable sets of the measure space.

Definition We call a function

f : X −→ [−∞,∞]

measurable iff−1(α,∞] = {x ∈ X : f(x) > α}

is a measurable ∀α ∈ R.

f

R

α

>α >α

Example The collection

{E ⊂ R : f−1(E) is measurable}

29

is a σ-algebra containing the sets

{(α, β] : α, β ∈ R}

therefore contains the Borel sets. Therefore f−1 is measurable for each Borelset B ⊂ R.

2.2 Characteristic Function

Definition If E ⊂ X, we denote by χE the function on X:

χE(x) =

{1 x ∈ E

0 x 6∈ E

χE is called the characteristic function of E.

E

R

1

E

Definition A real valued function

φ : X −→ R

is called simple if it takes only a finite number of distinct values

a1, a2, . . . , an

(say). Each simple function φ can be written in a unique way as:

φ = a1χE1 + · · · + anχEn

where a1, . . . , an are distinct and

X = E1 ∪ · · · ∪ Enis a disjoint union.

30

2.3 The Integral

Definition If φ is a non-negative measurable simple function with φ =a1χE1 + · · ·+ anχEn

; X = E1 ∪ · · · ∪En disjoint union we define the integralof φ w.r.t. the measure m to be

∫

φ dm = a1m(E1) + · · · + anm(En) ∈ [0,∞]

1

R

E E

a

1 2

2

a

Recall that 0.∞ = 0, a.∞ = ∞ if a > 0.

Definition if E is a measurable subset of X and φ is a non-negative mea-surable simple function on X, we define the integral of φ over E w.r.t. themeasure m to be: ∫

E

φ dm =

∫

φχE dm

We note that:

1.∫

(φ+ ψ) dm =∫φ dm+

∫ψ dm

2.∫cφ dm = c

∫φ dm ∀c ≥ 0

To see 1. we put:

φ =∑

aiχEi

ψ =∑

bjχFj

Let {ck} be the set of distinct values of {ai + bj}. Then

φ+ ψ =∑

ckχGk

where Gk =⋃Ei ∩ Fj, the union taken over {i, j : ai + bj = ck}.

31

∫(φ+ ψ) =

∑ckm(Gk)

=∑

k ck∑

{i,j:ai+bj=ck}m(Ei ∩ Fj)

=∑

i,j(ai + bj)m(Ei ∩ Fj)

=∑aim(Ei ∩ Fj) +

∑bjm(Ei ∩ Fj)

=∑

i aim(Ei) +∑

j bjm(Fj)

=∫φ dm+

∫ψ dm

which proves 1.

A very useful property of the integral is:

Theorem 2.3.1. Fix a simple non-negative measurable function φ on X.For each E ∈M put

λ(E) =

∫

E

φ dm

Then λ is a measure on M .

Proof. Let

φ = a1χE1 + · · ·anχEn

for ai ≥ 0. Then

λE =∫

Eφ dm

=∫φχE dm

=∫

(a1χE1∩E + · · · + anχEn∩E) dm

= a1m(E1 ∩ E) + · · · +anm(En ∩ E)

= a1mE1(E) + · · · +anmEn(E)

therefore

λ = a1mE1 + · · ·+ anmEn

is a linear combination of measures with non-negative coefficients. Thereforeλ is a measure.

32

Corollary 2.3.2. 1. If E =⋃∞n=1En is a countable disjoint union then

∫

E

φ dm =

∞∑

n=1

∫

En

φ dm

2. If E =⋃∞n=1En is a countable increasing union then

∫

E

φ dm = limn→∞

∫

En

φ dm

3. If E =⋂∞n=1En is a countable decreasing intersection and

∫

E1φ dm <

∞ then ∫

E

φ dm = limn→∞

∫

En

φ dm

We can now define the integral of any non-negative measurable function.

Definition Let f : X −→ [0,∞] be a measurable function. Then we definethe integral of f w.r.t. the measure m to be:

∫

f dm = supφ

∫

φ dm

where the sup is taken over all simple measurable functions φ such that:

0 ≤ φ ≤ f

fφ

Definition If E is a measurable subset of X then we define the integral off over E w.r.t. m to be:

∫

E

f dm =

∫

fχE dm

33

f

Ε

PSfrag replacements fχE

We then have:

1. f ≤ g =⇒∫f dm ≤

∫g dm

2. E ⊂ F =⇒∫

Ef dm ≤

∫

Ff dm

2.4 Monotone Convergence Theorem

We can now prove our first important theorem on integration.

Theorem 2.4.1. (Monotone Convergence Theorem, MCT) Let {fn} be amonotone increasing sequence of non-negative measurable functions. Then

∫

lim fn dm = lim

∫

fn dm

Proof. We write fn ↑ f to denote that {fn} is an increasing sequence offunctions with lim fn = f . We need to prove that:

lim

∫

fn dm =

∫

f dm

1.

fn ≤ fn+1 ≤ f

for all n. Therefore:∫

fn dm ≤∫

fn+1 dm ≤∫

f dm

for all n. Therefore:

lim

∫

fn dm ≤∫

f dm

34

2. Let φ be a simple measurable function s.t.:

0 ≤ φ ≤ f

f

f

φ

n

Let 0 < α < 1. For each integer n > 0 put

En = {x ∈ X : fn(x) ≥ αφ(x)}so En ↑ X. Now:

nf

En

fαφ

∫

En

αφ dm ≤∫

En

fn dm ≤∫

fn dm

Let n→ ∞:

α

∫

φ dm ≤ lim

∫

fn dm

is true ∀0 < α < 1. Therefore:∫

φ dm ≤ lim

∫

fn dm

so: ∫

f dm ≤ lim

∫

fn dm

hence ∫

f dm = lim

∫

fn dm

35

Definition In probability theory we have a measure space:

X , M , P

sample eventsspace

with P (X) = 1. X is the sure event.A measurable function:

f : X −→ R

is called a random variable.

P{x : f(x) ∈ B}is the probability that the random variable f takes value in the Borel setB ⊂ R. If

f = a1χE1 + · · ·+ anχEn

with a1, . . . , an;E1, . . . , En disjoint, and E1 ∪ · · · ∪ En = X, then the proba-bility that f takes value ai is

P{x ∈ X : f(x) = ai} = P (Ei)

and ∫

f dP = a1P (E1) + · · ·+ anP (En)

is the average value or the expectation of f .We define the expectation of any random variable f to be:

E(f) =

∫

f dP

2.5 Existence of Monotone Increasing Simple

Functions converging to f

In order to apply the MCT effectively we need:

Theorem 2.5.1. Let f be a non-negative measurable function f : X −→[0,∞]. Then there exists a monotone increasing sequence φn of simple mea-surable functions converging to f

36

Proof. Put each integer n > 0:

φn(x) =

{k2n

k2n ≤ f(x) < k+1

2n k = 0, 1, 2, . . . , n2n − 1n f(x) ≥ n

n

0

PSfrag replacements

k2n

k+12n

Then

1. 0 ≤ φn(x) ≤ φn+1(x)

2. each φn is simple and measurable

3. limn→∞ φn(x) = f(x)

Theorem 2.5.2. Let f, g be non-negative measurable functions mapping Xto [0,∞] and c ≥ 0. Then

1.∫cf dm = c

∫f dm

2.∫

(f + g) dm =∫f dm+

∫g dm

Proof. Let φn, ψn be monotonic increasing sequences of non-negative simplefunctions with f = limφn, g = limψn. Then

1.∫cf dm

MCT= lim

∫cφn dm = c lim

∫φn dm

MCT= = c

∫f dm

2.∫

(f + g) dmMCT= lim

∫(φn + ψn) = lim

∫φn dm + lim

∫ψn dm

MCT=

∫f dm+

∫g dm

This enables us to deal with series:

37

Theorem 2.5.3. Let fn be a sequence of non-negative measurable functionsX −→ [0,∞]. Then:

∫

(∞∑

n=1

fn) dm =∞∑

n=1

∫

fn dm

Proof. put sn = f1 + · · ·+ fn sum to n terms. sn is monotone increasing:∫

( limn→∞

sn) dmMCT= lim

n→∞

∫

sn dm = limn→∞

∫ n∑

r=1

fr dm = limn→∞

n∑

r=1

∫

fr dm

Therefore:∫

(∞∑

r=1

fr) dm =∞∑

r=1

∫

fr dm

Theorem 2.5.4. Let f : X −→ [0,∞] be a non-negative measurable and put

λ(E) =

∫

E

f dm

for each E ∈M . Then λ is a measure in M .

Proof. Let E =⋃∞k=1Ek be a countable disjoint union. Then

λ(E) =∫

Ef dm

=∫fχE dm

=∫

(∑∞

k=1 fχEk) dm

MCT=∑∞

k=1

∫fχEk

dm

=∑∞

k=1

∫

Ekf dm

=∑∞

k=1 λ(Ek)

therefore λ is countably additive, as required.

Corollary 2.5.5. 1. if E =⋃∞k=1Ek countable disjoint union then

∫

E

f dm =∞∑

k=1

∫

Ek

f dm

2. if Ek ↑ E then limk→∞∫

Ekf dm =

∫

Ef dm

3. If Ek ↓ E and∫

E1f dm <∞, then limk→∞

∫

Ekf dm =

∫

Ef dm

38

2.6 ’Almost Everywhere’

Definition Let f, g : X −→ [0,∞]. Then we say that f = g almost every-where (a.e.) or f(x) = g(x) almost all x ∈ X (a.a.x) if {x ∈ X : f(x) 6= g(x)}has measure zero.

Theorem 2.6.1. Let f : X −→ [0,∞] be non-negative measurable. Then∫f dm = 0 ⇔ f = 0 a.e.

Proof. Put E = {x ∈ X : f(x) 6= 0}

1. Put

En = {x ∈ X : f(x) >1

n}

for each integer n > 0, so

f >1

nχEn

E

PSfrag replacements

1n

En

Suppose that∫f dm = 0. Then

0 =

∫

f dm ≥ 1

nm(En)

som(En) = 0

for all n. But En ↑ E, so

m(E) = limm(En) = lim 0 = 0

thereforef = 0 a.e.

2. Suppose f = 0 a.e., so m(E) = 0. Now,

0 ≤ f ≤ limnχE

39

E

PSfrag replacements

nχE

therefore0 ≤

∫f dm

≤∫

limnχE dm

MCT= lim

∫nχE dm

= lim nm(E)

= lim 0 = 0

so ∫

f dm = 0

Corollary 2.6.2. Let f : X −→ [0,∞] be non-negative and measurable andlet E have measure zero. Then

∫

E

f dm = 0

Proof.fχE = 0 a.e.

therefore ∫

E

f dm =

∫

fχE dm = 0

Corollary 2.6.3. Let f, g : X −→ [0,∞] be non-negative and measurableand f = g a.e. then

∫f dm =

∫g dm

40

Proof. Let f = g on E and m(E ′) = 0 then

∫

f dm =

∫

E

f dm+

∫

E′

f dm =

∫

E

g dm+

∫

E′

g dm =

∫

g dm

Thus changing f on a set of measure zero makes no difference to∫f dm.

Also

Theorem 2.6.4. If fn are non-negative and fn ↑ f a.e. then

lim

∫

fn dm =

∫

f dm

Proof. suppose fn ≥ 0 and fn ↑ f on E with m(E ′) = 0. Then

∫f dm =

∫

Ef dm+

∫

E′ f dm

MCT= lim

∫

Efn dm+ 0

= lim[∫

Efn dm+

∫

E′ fn dm]

= lim∫fn dm

So far we have dealt with functions

X −→ [0,∞]

which are non-negative, but have allowed the value ∞.Now we look at functions

X −→ R

which may be negative and we do not allow ∞ as a value.

Definition Let f : X −→ R be measurable. Put

f+(x) =

{f(x) f(x) ≥ 00 f(x) ≤ 0

}

f−(x) =

{−f(x) f(x) ≤ 00 f(x) ≥ 0

}

41

fPSfrag replacements

f+

f

PSfrag replacements

f−

Thus f = f+ − f− and both f+, f− are non-negative. We say that f isintegrable w.r.t. m if

∫

f+ dm <∞ and

∫

f− dm <∞

and we write ∫

f dm =

∫

f+ dm−∫

f− dm

and call it the integral of f (w.r.t. measure m).If E is measurable we write

∫

E

f dm =

∫

fχE dm =

∫

E

f+ dm−∫

E

f− dm

Theorem 2.6.5. Let f = f1 − f2 where f1, f2 are non-negative and measur-able and ∫

f1 dm <∞ and

∫

f2 dm <∞

Then f is integrable and∫

f dm =

∫

f1 dm−∫

f2 dm

42

Proof. 1. f = f1 − f2, therefore f+ ≤ f1; f− ≤ f2. So

∫

f+ dm ≤∫

f1 dm <∞;

∫

f− dm ≤∫

f2 dm <∞

therefore f is integrable.

2. f = f1 − f2 = f+ − f−. So

f1 + f− = f+ + f2

therefore∫

f1 dm+

∫

f− dm =

∫

f+ dm+

∫

f2 dm

so ∫

f1 dm−∫

f2 dm =

∫

f+ dm−∫

f− dm =

∫

f dm

as required.

Theorem 2.6.6. Let f, g be integrable and f = g a.e. Then∫

f dm =

∫

g dm

Proof.f+ = g+ a.e. =⇒

∫f+ dm =

∫g+ dm

f− = g− a.e. =⇒∫f− dm =

∫g− dm

so∫

f dm =

∫

f+ dm−∫

f− dm =

∫

g+ dm−∫

g− dm =

∫

g dm

So when integrating we can ignore sets of measure zero.

Theorem 2.6.7. Let f : X −→ R be measurable. Then

1. f is integrable ⇔ |f | is integrable

2. if f is integrable then∣∣∣∣

∫

f dm

∣∣∣∣≤∫

|f | dm

43

Proof. 1.

f integrable ⇔∫f+ dm <∞ and

∫f− dm <∞

⇔∫

(f+ + f−) dm <∞

⇔∫|f | dm <∞

⇔ |f | integrable

2. ∣∣∫f dm

∣∣ =

∣∣∫f+ dm−

∫f− dm

∣∣

≤∫f+ dm+

∫f− dm

=∫|f | dm

Definition A complex valued function f : X −→ C with

f(x) = f1(x) + if2(x)

(say) (f1, f2 real) is called integrable if f1 and f2 are integrable and we define:∫

f dm =

∫

f1 dm+ i

∫

f2 dm

Thus

<∫

f dm =

∫

(<f) dm

=∫

f dm =

∫

(=f) dm

We have |f1| ≤ |f |, |f2| ≤ |f |, and therefore|f | integrable ⇔ |f1| and |f2| integrable ⇔ f1 and f2 integrable ⇔ f

integrable.We also have:f, g integrable and c ∈ C =⇒ f + g and cf integrable, and

∫

(f + g) dm =

∫

f dm+

∫

g dm

and ∫

(cf) dm = c

∫

f dm

44

Thus, the set L(X,R, m) of all integrable real valued functions on X isa real vector space, and the set L(X,C, m) of all integrable complex valuedfunctions on X is a complex vector space. And on each space:

f −→∫

f dm

is a linear form.

If f : X −→ C is integrable then∫

f dm =

∣∣∣∣

∫

f dm

∣∣∣∣eiθ (say) θ real

therefore(real)∣

∣∫f dm

∣∣ = e−iθ

∫f dm

=(real)∫e−iθf dm

=∫<[e−iθf ] dm

≤∫ ∣∣e−iθf

∣∣ dm

=∫|f | dm

therefore ∣∣∣∣

∫

f dm

∣∣∣∣≤∫

|f | dm

2.7 Integral Notation

Definition When dealing with 1-dimensional Lebesgue measure we write

∫

[a,b]

f dm =

∫ b

a

f(x) dx = −∫ a

b

f(x) dx

if a ≤ b. It follows that

∫ b

a

f(x) dx+

∫ c

b

f(x) dx =

∫ c

a

f(x) dx

and ∫ b

a

dx = b− a

45

∀a, b, c.Notice that x is a dummy symbol and that

∫ b

a

f(x) dx =

∫ b

a

f(y) dy =

∫ b

a

f(t) dt = · · ·

just asn∑

i=1

aibi =n∑

j=1

ajbj = · · ·

2.8 Fundamental Theorem of Calculus

Theorem 2.8.1. Fundamental Theorem of Calculus Let f : R −→ R becontinuous and put

F (t) =

∫ t

a

f(x) dx

thenF ′(t) = f(t)

Proof. Let t ∈ R, let ε > 0. Then ∃δ > 0 s.t.

|f(t+ h) − f(t)| ≤ ε ∀|h| ≤ δ

by continuity of f . Therefore

∣∣∣F (t+h)−F (t)

h− f(t)

∣∣∣

=∣∣∣1h

∫ t+h

tf(x) dx− 1

h

∫ t+h

tf(t) dx

∣∣∣

= 1|h|

∣∣∣

∫ t+h

t[f(x) − f(t)] dx

∣∣∣

≤ 1|h|ε|h| = ε

for all |h| ≤ δ; h 6= 0, and hence the result.

Corollary 2.8.2. If G : R −→ R is C1 (i.e. has a continuous derivative)

then∫ b

aG′(x) dx = G(b) −G(a).

Proof. put F (t) =∫ t

aG′(x) dx. Then

F ′(t) = G′(t)

46

for all t. ThereforeF (t) = G(t) + c

for all t, where c is constant. Therefore

G(b) −G(a) = F (b) − F (a) =

∫ b

a

G′(x) dx

as required.

Theorem 2.8.3. (Change of Variable)Let

[t3, t4]g−→ [t1, t2]

f−→ R

with f continuous and g ∈ C1; g(t3) = t1, g(t4) = t2. Then

∫ t2

t1

f(x) dx =

∫ t4

t3

f(g(y))g′(y) dy

Proof. Put

G(t) =

∫ t

t1

f(x) dx

then G′(t) = f(t), and therefore:

∫ t4t3f(g(y))g′(y) dy =

∫ t4t3G′(g(y)g′(y) dy

=∫ t4t3

[ ddyG(g(y))] dy

= G(g(t4)) −G(g(t3))

= G(t2) −G(t1)

=∫ t2t1f(x) dx

as required.

To deal with sequences which are not monotone we need the concepts oflim inf and lim sup.

Definition Let{an} = a1, a2, a3, . . .

47

be a sequence in [−∞,∞]. Put

bn = inf{an, an+1, an+2, . . .}

cn = sup{an, an+1, an+2, . . .}Then

b1 ≤ b2 · · · ≤ bn ≤ bn+1 ≤ · · · ≤ cn+1 ≤ cn ≤ · · · ≤ c2 ≤ c1

Define:lim inf an = lim bn

lim sup an = lim cn

Thenlim inf an ≤ lim sup an

and an converges iff lim inf an = lim sup an(= lim an).

2.9 Fatou’s Lemma

Theorem 2.9.1. (Fatou’s Lemma) Let fn be a sequence of non-negativemeasurable functions:

fn : X −→ [0,∞]

Then ∫

lim inf fn dm ≤ lim inf

∫

fn dm

Proof. Putgr = inf{fr, fr+1, . . .}

so that· · · ≤ gr ≤ gr+1 ≤ · · ·

Now:

fn ≥ gr ∀n ≥ r

=⇒∫fn ≥

∫gr ∀n ≥ r

=⇒ lim inf∫fn ≥

∫gr ∀r

=⇒ lim inf∫fn ≥ lim

∫gr

MCT=∫

lim gr =∫

lim inf fn

as required.

48

2.10 Dominated Convergence Theorem

Theorem 2.10.1. Lebesgue’s Dominated Convergence Theorem, DCT Letfn be integrable and fn → f . Let

|fn| ≤ g

for all n where g is integrable. Then f is integrable and∫f = lim

∫fn

Proof. |f | = lim |fn| ≤ g. Therefore |f | is integrable, and therefore f isintegrable. Now

g ± fn ≥ 0

therefore ∫

lim inf[g ± fn] ≤FATOU lim inf

∫

[g ± fn]

so ∫

[g ± f ] ≤∫

g + lim inf ±∫

fn

so

±∫

f ≤ lim inf ±∫

fn

Which gives:� :

∫f ≤ lim inf

∫fn

� : −∫f ≤ − lim sup

∫fn

therefore ∫

f ≤ lim inf

∫

fn ≤ lim sup

∫

fn ≤∫

f

which is equivalent to:

lim

∫

fn =

∫

f

as required.

For series this leads to:

Theorem 2.10.2. Dominated convergence theorem for series Let fn be asequence of integrable functions such that

∞∑

n=1

∫

|fn| dm <∞

Then∑∞

n=1 fn is (equal a.e. to) an integrable function and∫

(

∞∑

n=1

fn) dm =

∞∑

n=1

∫

fn dm

49

Proof. Put

sn = f1 + f2 + · · ·+ fn

sum to n terms. Then

|sn| ≤ |f1| + · · · + |fn| ≤∞∑

r=1

|fr| = g

(say). Then∫g dm =

∫ ∑∞r=1 |fr| dm

MCT=∑∞

r=1

∫|fr| dm <∞

Therefore g is integrable (a.e. equal to an integrable fn). Therefore lim sn isintegrable and

∫

lim sn dmDCT= lim

∫

sn dm

therefore∫ ∞∑

r=1

fr dm =

∞∑

r=1

∫

fr dm

as required.

2.11 Differentiation under the integral sign

Another useful application.

Theorem 2.11.1. Differentiation under the integral sign Let f(x, t) be anintegrable function of x ∈ X for each a ≤ t ≤ b and differentiable w.r.t t.

Suppose∣∣∣∣

∂f

∂t(x, t)

∣∣∣∣≤ g(x)

for all a ≤ t ≤ b, where g is integrable. Then

d

dt

∫

f(x, t) dx =

∫∂f

∂t(x, t) dx

Proof. Put

F (t) =

∫

f(x, t) dx

50

Let a ≤ t ≤ b. Choose a sequence tn in [a, b] s.t. lim tn = t and tn 6= t. Then,by the Mean Value Theorem:

|f(x, tn) − f(x, t)| =|tn − t|∣∣∂f∂t

(x, c(n, x, t))∣∣

≤ |tn − t|g(x)

Therefore ∣∣∣∣

f(x, tn) − f(x, t)

tn − t

∣∣∣∣≤ g(x)

solim F (tn)−F (t)

tn−t = lim∫

f(x,tn)−f(x,t)tn−t dx

DCT=∫

lim f(x,tn)−f(x,t)tn−t dx

=∫

∂f

∂t(x, t) dx

and thereforedF

dt=

∫∂f

∂t(x, t) dx

as required.

51

52

Chapter 3

Multiple Integration

3.1 Product Measure

We have established the Lebesgue measure and Lebesgue integral on R. Toconsider integration on

R2 = R × R

we use the concept of a product measure.We proceed as follows:

Definition Let l be a measure on a σ-algebra L of subsets of X. Let m bea measure on a σ-algebra M of subsets of Y .

Call{A× B : A ∈ L, B ∈ M}

the set of rectangles in X × Y

B

A

PSfrag replacements

A×B

53

Lemma 3.1.1. Let A×B =⋃∞i=1Ai×Bi be a rectangle written as a countable

disjoint union of rectangles. Then

l(A)m(B) =

∞∑

i=1

l(Ai)m(Bi)

Proof. We have

χA×B =

∞∑

i=1

χAi×Bi

=⇒ χA(x)χB(y) =∞∑

i=1

χAi(x)χBi

(y)

Fix x and integrate w.r.t. m term by term using the Monotone ConvergenceTheorem:

χA(x)m(B) =∞∑

i=1

χAi(x)m(Bi)

Now integrate w.r.t. x using MCT to get:

l(A)m(B) =

∞∑

i=1

l(Ai)m(Bi)

as required.

Definition Let A be the collection of all finite unions of rectangles in X×Y .Each element of A is a finite disjoint union of rectangles. For each E ∈ Asuch that

E =∞⋃

i=1

Ai × Bi

is a countable disjoint union of rectangles we define:

π(E) =∞∑

i=1

l(Ai)m(Bi)

Theorem 3.1.2. π is well-defined and is a measure on A.

Proof. 1. well-defined

Suppose

E =

∞⋃

i=1

Ai × Bi =

∞⋃

j=1

Cj ×Dj

54

then

Ai ×Bi =∞⋃

j=1

(Ai ∩ Cj) × (Bi ∩Dj)

Therefore, by Lemma 3.1.1

l(Ai)m(Bi) =∞∑

j=1

l(Ai ∩ Cj)m(Bi ∩Dj)

and hence,∑∞

i=1 l(Ai)m(Bi) =∑∞

i=1

∑∞j=1 l(Ai ∩ Cj)m(Bi ∩Dj)

=∑∞

j=1 l(Cj)m(Dj)

Dj

BiPSfrag replacements

Ai Cj

Di

Bj

2. countably additive Let E =⋃∞i=1Ei be a countable disjoint union with

E,Ei ∈ A.

Ei =

ni⋃

j=1

Aij × Bij

is a finite disjoint union (say). Then

E =∞⋃

i=1

ni⋃

j=1

Aij × Bij

Therefore

π(E) =∞∑

i=1

ni∑

j=1

l(Aij)m(Bij) =∞∑

i=1

π(Ei)

Therefore π is countably additive, as required.

55

Definition The measure π on A extends to a measure (also denoted by π

and called the product of the measures l and m) on the σ-algebra (denotedL ×M) of all subsets of X × Y which are measurable w.r.t. π.

Similarly, if(X1,M1, m1), . . . , (Xn,Mn, mn)

is a sequence of measure spaces then we have a product measure on a σ-algebra of subsets of

X1 × · · · ×Xn

s.t.π(E1 × · · · × En) = m1(E1)m2(E2) . . .mn(En)

In particular, starting with 1-dimensional Lebesgue measure on R we getn-dimensional Lebesgue Measure on

R × · · · × R = Rn

Example If we have two successive, independent events with independentprobability measures, P1, P2, then the probability of the first event E andthe second event F is:

P (E × F ) = P [(E × Y ) ∩ (X × F )]

= P (E × Y )P (X × F ) by independence

= P1(E)P2(F )

(i.e. product measure)

3.2 Monotone Class

Definition A non-empty collection M of subsets of X is called a monotoneclass if

1. En ↑ E,En ∈M =⇒ E ∈M

2. En ↓ E,En ∈M =⇒ E ∈M

i.e. M is closed under countable increasing unions and under countabledecreasing intersections.

We have: M is a σ-algebra =⇒ M is a monotone class. Therefore, thereare more monotone classes than σ-algebras.

56

Definition If V is a non-empty collection of subsets of X and if M is theintersection of all the monotone classes of subsets of X which contain V, thenM is a monotone class, called the monotone class generated by V.

M is the smallest monotone class containing V and

V ⊂ monotone class ⊂ σ-algebragenerated by V generated by V

3.3 Ring of Subsets

Definition A non-empty collection R of subsets of X is called a ring ofsubsets of X if:

E, F ∈ R =⇒

E ∪ F ∈ R

E ∩ F ′ ∈ R

i.e. R is closed under finite unions and under relative complements.

Example The collection of all finite unions of rectangles contained in theinterior X of a fixed circle in R2 is a ring of subsets of X.

Note:

1. if R is a ring then R is closed under finite intersections since E ∩ F =E ∩ (E ∩ F ′)′.

2. if R is a ring then ∅ ∈ R since E ∩ E ′ = ∅3. if R is a ring of subsets of X and if X ∈ R then R is an algebra of

subsets of X since E ′ = X ∩ E ′, so R is closed under complements.

Theorem 3.3.1. (Monotone class lemma)Let R be a ring of subsets of X and let M be the monotone class generated

by R.Then M is a ring.

Proof. We have to show M is closed under finite unions and relative comple-ments. i.e. that:

E ∩ F ′, E ∪ F, E ′ ∩ F ∈ M

for all E, f ∈M . So for each E ∈M put

ME = {F ∈M : E ∩ F ′, E ∪ F,E ′ ∩ F ∈M}and we must show

ME = M

for all E ∈M . Now

57

1. ME ⊂M by definition

2. ME is a monotone class, because:

Fn ↑ F, Fn ∈ ME

=⇒ (E ∩ Fn′) ↓ (E ∩ F ′) (E ∪ Fn) ↑ (E ∪ F ) (E ′ ∩ Fn) ↑ (E ′ ∩ F )

with E ∩ Fn′ E ∪ Fn E ′ ∩ Fn all ∈M

=⇒ E ∩ F ′ E ∪ F E ′ ∩ F all ∈M

(M i monotone)

=⇒ F ∈ME

therefore ME is closed under countable increasing unions, and similarlyME is closed under countable decreasing intersections.

3. E ∈ R

=⇒ R ⊂ ME since R is a ring. Therefore ME = M by (i), (ii) sinceM is smallest monotone class containing R.

4. F ∈ ME ∀E ∈ R,F ∈ M by (iii). Therefore E ∈ MF ∀E ∈ R,F ∈M .

Therefore

R ⊂MF ∀F ∈M

and therefore

MF = M ∀F ∈M

as required.

Corollary 3.3.2. Let M be the monotone class generated by a ring R, andlet X ∈M .

Then M = G(R) the σ-algebra generated by R.

Proof. M is a ring and X ∈ M

E ∈M =⇒ E ′ = E ′ ∩X ∈M

therefore M closed under compliments, and therefore M is an algebra.

58

Also, if En is a sequence in M and E =⋃∞

1 En put

Fn = E1 ∪ · · · ∪ En ∈M

then Fn ↑ E, and therefore E ∈M .

Therefore, M is a σ-algebra, hence the result.

Corollary 3.3.3. Let M be the monotone class generated by an algebra A.Then M = G(A) the σ-algebra generated by A.

Proof. X ∈ A =⇒ X ∈M , and hence the result by previous corollary.

3.4 Integration using Product Measure

Let l be a measure on a σ-algebra L of subsets of X.

Let m be a measure on a σ-algebra M of subsets of Y .

Let A be the algebra of finite unions of rectangles A×B; A ∈ L, B ∈ M.

Let π be the product measure on the σ-algebra G(A).

If

f : X −→ [0,∞]

g : Y −→ [0,∞]

F : X × Y −→ [0,∞]

are non-negative measurable, write:

∫f dl =

∫

Xf(x) dx

∫f dm =

∫

Yg(y) dy

∫F dπ =

∫

X×Y F (x, y) dx dy

If E ⊂ X × Y write

Ex = {y ∈ Y : (x, y) ∈ E}

Ey = {x ∈ X : (x, y) ∈ E}

59

Ex

y

x Ey

Theorem 3.4.1. if E ∈ G(A) and l, m are σ-finite then:

π(E) =

∫

X

m(Ex) dx =

∫

Y

l(Ey) dy

Proof. To show

π(E) =

∫

X

m(Ex) dx (3.1)

1. Suppose l, m are finite measures: l(X) < ∞, m(Y ) < ∞. Let N bethe collection of all E ∈ G(A) s.t. Equation (3.1) holds. We will showthat N is a monotone class containing A:

A ⊂ N ⊂ G(A)

and hence N = G(A) since by the corollary to the monotone classlemma, G(A) is the monotone class generated by A.

(a) A ∈ N .

If E = A× B is a rectangle then

m(Ex) = χA(x)m(B)

60

B

E

A

x

then∫

X

m(Ex) dx = l(A)m(B) = π(E)

therefore E ∈ N . Now each element of A can be written as afinite disjoint union of rectangles, therefore

A ⊂ N

(b) Let En ↑ E,En ∈ N . So,

(En)x ↑ Ex

for each x ∈ X.

x

E

En

61

Then∫

Xm(Ex) dx =

∫

Xlimm((En)x) dx

MCT= lim

∫

Xm((En)x) dx

= lim π(En)

= π(E)

therefore E ∈ N , and N is closed under converging unions.

(c) Let E ∈ N . Then

∫

Xm((E ′)x) dx =

∫

X[m(Y ) −m(Ex)] dx

= l(X)m(Y ) − π(E)x

= π(X × Y ) − π(E)

= π(E ′).

Therefore E ′ ∈ N , and N is closed under complements. ThereforeN is a monotone class, and therefore N = G(A).

2. Suppose l, m are σ-finite,

An ↑ X, Bn ↑ Y

(say) with

l(An) <∞, m(Bn) <∞

Then

Zn ↑ (X × Y )

where Zn = An ×Bn and π(Zn) <∞.

Let E ∈ G(A). Then (E ∩ Zn) ↑ E, and

(E ∩ Zn)x ↑ Ex

for all x ∈ X.

62

x

EZn

So: ∫

Xm(Ex) dx =

∫

Xlimm((E ∩ Zn)x) dx

MCT= lim

∫

Xm((E ∩ Zn)x) dx

(since Zn has finite measure)

= lim π(E ∩ Zn)

= π(E)

as required.

3.5 Tonelli’s Theorem

Theorem 3.5.1. (Repeated integral of a non-negative funtcion)

Let F : X × Y −→ [0,∞] be non-negative measurable. Then

∫

X

[∫

Y

F (x, y) dy

]

dx =

∫

X×YF (x, y) dx dy =

∫

Y

[∫

X

F (x, y) dx

]

dy

(notation as before.)

63

Proof. To show

∫

X

[∫

Y

F (x, y) dy

]

dx =

∫

X×YF (x, y) dx dy (3.2)

1. Equation 3.2 holds for F = χE, since

∫

X

[∫

Y

χE(x, y) dy

]

dx =

∫

X

m(Ex) dx = π(E) =

∫

X×YχE(x, y) dx dy

therefore Equation 3.2 also holds for any simple function.

2. (General Case)

∃ monotone increasing sequence of non-negative measurable functionswith F = limFn

∫

X

[∫

YF (x, y) dy

]dx =

∫

X

[∫

YlimFn(x, y) dy

]dx

MCT=

∫

X

[lim∫

YFn(x, y) dy

]dx

MCT= lim

∫

X

[∫

YFn(x, y) dy

]dx

(1.)= lim

∫

X×Y Fn(x, y) dx dy

MCT=

∫

X×Y limFn(x, y) dx dy

=∫

X×Y F (x, y) dx dy

3.6 Fubini’s Theorem

Theorem 3.6.1. (Repeated Integral of an integrable function)Let F be an integrable function on X × Y . Then:

∫

X

[∫

Y

F (x, y) dy

]

dx =

∫

X×YF (x, y) dx dy =

∫

Y

[∫

X

F (x, y) dx

]

dy

Where∫

YF (x, y) dy is equal to an integrable function of x a.e., and

∫

XF (x, y) dx

is equal to an integrable function of y a.e.

64

Proof.

∫

X×Y F (x, y) dx dy =∫

X×Y F+(x, y) dx dy−

∫

X×Y F−(x, y) dx dy

Tonelli=

∫

X

[∫

YF+(x, y) dy

]dx−

∫

X

[∫

YF−(x, y) dy

]dx

Therefore∫

YF+(x, y) dy, and

∫

YF−(x, y) dy are each finite a.a.x., and each

has a finite integral w.r.t. x. (*)So:

F (x, y) = F+(x, y) − F−(x, y)

is integrable w.r.t y a.e., and:

∫

Y

F (x, y) dy =

∫

Y

F+(x, y) dy −∫

Y

F−(x, y) dy

for a.a.x, and is an integrable function of x (by (*)), with:

∫

X

[∫

YF (x, y) dy

]dx =

∫

X

[∫

YF+(x, y) dy

]dx−

∫

X

[∫

YF−(x, y) dy

]dx

=∫

X×Y F (x, y) dx dy

Theorem 3.6.2. Let f be an integrable function R −→ R. Then

1.∫f(x+ c) dx =

∫f(x) dx

2.∫f(cx) dx = 1

|c|∫f(x) dx (c 6= 0)

Proof. These are true for f = χE, because:

1.∫

χE(x + c) dx =

∫

χE−c(x) dx = m(E − c) = m(E) =

∫

χE(x) dx

2.∫

χE(cx) dx =

∫

χ 1cE(x) dx = m(

1

cE) =

1

|c|m(E) =1

|c|

∫

χE(x) dx

Therefore, these are true for f simple, and therefore true for f non-negativemeasurable (by MCT), since ∃fn simple, s.t. fn ↑ f . Therefore, these aretrue for f = f+ − f− integrable.

65

We now see how to deal with integration on change of variable on Rn.Recall

∫f(cx) dx = 1

|c|∫f(x) dx.

Theorem 3.6.3. (Linear change of variable)

Let Rn A−→ Rn be an invertable matrix, then

∫

f(Ax) dx =1

| detA|

∫

f(x) dx

Proof. we can reduce A to the unit matrix I by a sequence of elementaryrow operations.

1. To replace row i by row i + c row j, multiply A by:

N =

1...

. . ....

· · · · · · 1 · · · c · · · · · ·...1...

with c in the i,j position.

2. To interchange row i and row j, multiply A by:

P =

1...

...

1...

...· · · · · · 0 · · · · · · 1 · · ·

... 1...

. . .... 1

...· · · · · · 1 · · · · · · 0 · · ·

1. . .

1

66

3. To replace row i by c row j, multiply A by:

D =

1. . .

1c

1. . .

1

Therefore, there exists matrices B1, . . . , Bk, each of type N,P or D s.t.

B1B2 · · ·BkA = I

and therefore,

A = B−1k · · ·B−1

2 B−11

is a product of matrices of type N,P or D.Now, if the theorem holds for matrices A,B then it also holds for AB

since:∫

f(ABx) dx =1

| detB|

∫

f(Ax) dx =1

| detB|| detA|

∫

f(x) dx =1

| detAB|

∫

f(x) dx

therefore, it is sufficient to prove it for matrices of type N , P , D.

1. Let

N =

1 c 0 · · · 00 1 0 · · · 00 0 1 0 · · ·

. . .. . .

1

(say). detN = 1. Then

∫f(Nx) dx =

∫f(x1 + cx2, x2, . . . , xn) dx1 dx2 · · · dxn

=∫f(x1, x2, . . . , xn) dx1 dx2 · · · dxn

(by Fubini, and translation invariance)

= 1|detN |

∫f(x) dx

67

2. Let

P =

0 1 0 · · · 01 0 0 · · · 00 0 1 0 · · ·

. . .. . .

1

(say). detP = −1. Then∫f(Px) dx =

∫f(x2, x1, . . . , xn) dx1 dx2 · · · dxn

=∫f(x1, x2, . . . , xn) dx1 dx2 · · · dxn

(by Fubini)

= 1|detP |

∫f(x) dx

3. Let

D =

c 0 0 · · · 00 1 0 · · · 00 0 1 0 · · ·

. . .. . .

1

(say). detN = c. Then∫f(Dx) dx =

∫f(cx1, x2, . . . , xn) dx1 dx2 · · · dxn

= 1|c|∫f(x1, x2, . . . , xn) dx1 dx2 · · · dxn

(by Fubini)

= 1| detD|

∫f(x) dx

Corollary 3.6.4. if E ⊂ Rn is measurable and R

n A−→ Rn is a linear homo-

morphism thenm(A(E)) = | detA|m(E)

Proof.m(E) =

∫χE(x) dx

=∫χA(E)(AX) dx

= 1| detA|

∫χA(E)(x) dx

= 1| detA|m(A(E))

68

as required.

69

70

Chapter 4

Differentiation

4.1 Differentiation

If Rf−→ R is a real valued function od a real variable then the derivative of

f at a is defined to be:

f ′(a) = limh−→0

f(a+ h) − f(a)

h(4.1)

We want to define the derivative f ′(a) when f is a vector-valued function ofa vector variable:

f : M −→ N

where m,N are real (or complex) vector spaces.We cannot use Equation (4.1) directly since we don’t know how to divide

f(a+ h) − f(a), which is a vector in N , by h, which is a vector in M

So, we rewrite Equation (4.1) as:

f(a+ h) = f(a)︸︷︷︸

(constant)

+ f ′(a)h︸ ︷︷ ︸

(linear in h)

+ φ(h)︸︷︷︸

(remainder)

where

limh−→0

φ(h)

h=f(a+ h) − f(a)

h− f ′(a)

This suggests that we take M,N to be normed spaces and define f ′(a) tobe a linear operator such that

f(a+ h) = f(a) + f ′(a)h+ φ(h)

where

lim‖h‖−→0

‖φ(h)‖‖h‖ = 0

71

(h)

a

h

a+h

f

f(a)

f’(a) h

a+h

f(a+h)

φ

This f ′(a)h is the linear approx to the change in f when variable changes byh from a to a+ h.

4.2 Normed Space

Definition let M be a real or complex vector space. Then M is called anormed space if a function ‖.‖ exists,:

M −→ R

x −→ ‖x‖is given on M (called the norm on M), such that:

1. ‖x ≥ 0

2. ‖x‖ = 0 ⇔ x = 0

3. ‖αx‖ = |α| ‖x‖ ∀ scalar α

4. ‖x+ y‖ ≤ ‖x‖ + ‖y‖ (triangle inequality)

Example 1. Rn with ‖(α1, . . . , αn)‖ =√

α21 + · · · + α2

n is called the Eu-clidean Norm on Rn.

2. Cn with ‖(α1, . . . , αn)‖ =√

|α1|2 + · · ·+ |αn|2 is called the HilbertNorm on Cn.

3. Cn with ‖(α1, . . . , αn)‖ = max{|α1|, . . . , |αn|} is called the sup normon C

n.

4. if (X,M,m) is a measure space the the set of integrable functionsL′(X,R, m) with

‖f‖ =

∫

|f | dm

is called the L′-norm (functions are to be regarded as equal if they areequal a.e.)

72

4.3 Metric Space

Definition a set X is called a metric space if a function

D : X ×X −→ R

is given (called a metric on X) such that

1. d(x, y) ≥ 0

2. d(x, y) = 0 ⇔ x = y

3. d(x, z) ≤ d(x, y) + d(y, z). This is known as the triangle inequality

Definition If a ∈ X and r > 0 the we write

BX(a, r) = {x ∈ X : d(a, x) < r}

and call it the ball in X centre a, radius r.

Example if M is a normed space then M is also a metric space with

d(x, y) = ‖x− y‖

4.4 Topological space

Definition a set X is called a topological space id a collection V of subsetsof X is given (called the topology on X) such that:

1. ∅ and X belong to V

2. if {Vi}i∈I is any family of elements of V then⋃

i∈I vi belongs to V. i.e.V is closed under unions

3. if U and V belong to V then U ∩V belongs to V. i.e. V is closed underfinite intersections.

We call the elements of V the open sets of the topological space X, or openin X.

Example Let X be a metric space. Then X is a topological space where wedefine a set V to be open in V if V ⊂ X and each a ∈ V ∃ r > 0 s.t.

BX(a, r) ⊂ V

73

ar

V

Theorem 4.4.1. if X is a metric space then each c ∈ X and s > 0 the ballBX(c, s) is open in X.

Proof. Let a ∈ BX(c, s). Put r = s− d(a, c) > 0.

Then

x ∈ BX(a, r)

=⇒ d(x, a) < r = s− d(a, c)

=⇒ d(x, a) + d(a, c) < s

=⇒ d(x, c) < s

=⇒ x ∈ BX(c, s)

74

ar

c

xs

PSfrag replacements

BX(a, r)

BX(c, s)

and, therefore BX(a, r) ⊂ BX(c, s), as required.

4.5 Continuous map of topological spaces

Definition a map f : X −→ Y of topological spaces is called continuous if:

V open in Y =⇒ f−1V open in X

Theorem 4.5.1. let X, Y be metric spaces, then:

f : X −→ Y

is continuous if and only if,

for each a ∈ X, ε > 0∃ δ > 0 s.t.

d(x, a) < δ =⇒ d(f(x), f(a)) < ε

(4.2)

i.e.

fBX(a, δ) ⊂ BY (f(a), ε)

Proof. 1. Let f be continuous. Let a ∈ X, ε > 0. Then BY (f(a), ε) isopen in Y . Therefore, f−1BY (f(a), ε) is open in X.

75

aδ

εf(a)

PSfrag replacements

f−1BY (f(a), ε) BY (f(a), ε)

∃δ > 0 s.t.

BX(a, δ) ⊂ f−1BY (f(a), ε)

and

fBX(a, δ) ⊂ BY (f(a), ε)

therefore Equation (4.2) holds.

2. Let Equation (4.2) hold. Let V be open in Y , a ∈ f−1V , and sof(a) ∈ V .

Now, ∃ε > 0 s.t.

BY (f(a), ε) ⊂ V

f(a)

V

a

PSfrag replacements

f−1V

and ∃δ > 0 s.t.

fBX(a, δ) ⊂ BY (f(a), ε) ⊂ V

therefore,

BX(a, δ) ⊂ f−1V

so, f−1V is open in X, and f is continuous.

76

4.6 Homeomorphisms

Definition a map f : X −→ Y of topolgical spaces is called homeomorphismif:

1. f is bijective

2. f and f−1 are continuous

Definition X is homeomorphic to Y (topologically equivalent) if ∃ a home-omorphism X −→ Y . i.e. ∃ a beijective map under which the open sets ofX correspond to the open sets of Y .

Definition a property P of a topological space is called a topological prop-erty if X has property P and X homeomorphic to Y =⇒ Y has propertyP .

Example ’compactness’ is a topolgical property

Note: we have a category with topological spaces as objects, and con-tinuous maps as morphisms.

4.7 Operator Norm

Definition Let M,N be finite dimensional normed spaces. Then we canmake the vector space

L(M,N)

of all linear operators T : M −→ N , into a normed space by defining:

‖T‖ = supx∈M

x6=0

‖Tx‖‖x‖

‖T‖ is called the operator norm of T .

We have:

1. If x 6= 0 and y = x‖x‖ then

‖y‖ =1

‖x‖‖x‖ = 1

77

x

PSfrag replacements

x‖x‖

i.e. y has unit norm. Therefore

‖Tx‖‖x‖ =

∥∥∥∥T

x

‖x‖

∥∥∥∥

= ‖Ty‖

and

‖T‖ = supy∈M

‖y‖=1

‖Ty‖

T yy

PSfrag replacements

‖T‖

2. if α is a scalar then

‖αT‖ = sup‖y‖=1

‖αTy‖ = |α| sup‖y‖=1

‖Ty‖ = |α|‖T‖

3. if S, T : M −→ N then

‖S+T‖ = sup‖y‖=1

‖(S+T )y‖ ≤ sup‖y‖=1

(‖Sy‖+‖Ty‖) ≤ sup ‖Sy‖+sup ‖Ty‖ = ‖S‖+‖T‖

78

4.‖T‖ = 0 ⇐⇒ supx6=0

‖Tx‖‖x‖ = 0

⇐⇒ ‖Tx‖ = 0 ∀x

⇐⇒ Tx = 0 ∀x

⇐⇒ T = 0

(by 2,3,4 the operator norm is a norm).

Theorem 4.7.1. 1. if T : M −→ N and x ∈M then

‖Tx‖ ≤ ‖T‖ ‖x‖

2. If LT−→M

S−→ N then

‖ST‖ ≤ ‖S‖ ‖T‖Proof. 1.

‖Tx‖‖x‖ ≤ ‖T‖

∀x 6= 0, be definition. therefore,

‖Tx‖ ≤ ‖T‖ ‖x‖for all x

2.‖ST‖ = sup‖y‖=1 ‖STy‖

≤ sup ‖S‖ ‖Ty‖ (by 1.)

≤ sup‖y‖=1 ‖S‖ ‖T‖ ‖y‖ (by 1.)

= ‖S‖ ‖T‖

Note:

1. if M is finite dimensional then every choice of norm on M defines thesame topology on M

2. a sequence xr of points in a topological space X is said to converge toa ∈M if, for each open set V containing a ∃N s.t. xr ∈ V ∀ r ≥ N .

For a normed space M this is the same as: for each ε > 0 ∃N s.t.‖xr − a‖ < ε ∀ r ≥ N .

79

4.8 Differentiation

Definition let

M ⊃ Vf−→ N

where M,N are normed spaces and V is open in M . Let a ∈ V . THen f isdifferentiable at a if ∃ a linear operator

Mf ′(a)−→ N

such that

f(a+ h) = f(a0) + f ′(a)h+ φ(h)

where‖φ(h)‖‖h‖ −→ 0 as ‖h‖ −→ 0

Theorem 4.8.1. if f is differentiable at a the the operator f ′(a) is uniquelydetermined by:

f ′(a)h = limt−→0f(a+th)−f(a)

t

= directional derivative of f at a along h

= ddtf(a+ th)

∣∣t=0

a

a+th

a+h

f ′(a) is called the derivative of f at a

Proof.

f(a+ th) = f(a) + f ′(a)th + φ(th)

therefore,

∥∥∥∥

f(a+ th) − f(a)

t− f ′(a)h

∥∥∥∥

=

∥∥∥∥

φ(th)

t

∥∥∥∥

=‖φ(th)‖‖th‖ ‖h‖

tends to 0 as t −→ 0.

80

f’(a) h

a

a+th

a+hh

f

f(a)

f(a+th)

f(a+h)

Example 1. f : R −→ R, f(x) = x3. Then

f(a+ h) = (a+ h)3 = a3︸︷︷︸

f(a)

+ 3a2h︸︷︷︸

f ′(a)h

+ 3ah2 + h3

︸ ︷︷ ︸

φ(h)

so f ′(a) = 3a2.

2. f : Rn×n −→ Rn×n, f(X) = X3.

f(A+H) = (A +H)3

= A3︸︷︷︸

f(A)

+A2H + AHA+HA2

︸ ︷︷ ︸

f ′(A)H

+AH2 +HAH +H2A +H3

︸ ︷︷ ︸

φ(H)

taking (say) Euclidean norm on Rn and operator norm on R

n×n =L(Rn,Rn).

‖φ(H)‖‖H‖ = ‖AH2+HAH+H2A+H3‖

‖H‖

≤ 3‖A‖ ‖H‖2

‖H‖

= 3‖A‖ ‖H‖ −→ 0as ‖H‖ −→ 0

therefore, f is differentiable at A, and

f ′(A) : Rn×n −→ R

n×n

is given by

f ′(A)H = A2H + AHA+HA2

81

Theorem 4.8.2. Let Rn ⊃ Vf−→ Rm be a differentiable function with

f = (f 1, . . . , fm) = (f i)

Then

f ′ =

(∂f i

∂xj

)

Proof. For each a ∈ V we have

f ′(a) : Rn −→ R

m

is a linear operator. Therefore, f ′(a) is an m×n matrix whose jth column is

f ′(a)ej = limt−→0f(a+tej )−f(a)

t

= ∂f

∂xj (a)

=(∂f1

∂xj (a), . . . ,∂fm

∂xj (a))

therefore,

f ′(a) =

(∂f i

∂xj(a)

)

for all a ∈ V . Therefore,

f ′ =

(∂f i

∂xj

)

which is the Jacobian matrix.

Theorem 4.8.3. Let M ⊃ Vf−→ N1 × · · · ×Nk where

f(x) =(f 1(x), . . . , f k(x)

)

Then f 1, . . . , fk differentiable at a ∈ V =⇒ f is differentiable at a, and

f ′(a)h =(

f 1′(a)h, . . . , f k′(a)h

)

.

Proof. (k = 2)

M ⊃ Vf−→ N1 ×N2

f(x) =(f 1(x), f 2(x)

)

Take any norms on M,N1, N2 and define a norm on N! ×N2 by

‖(y1, y2)‖ = ‖y1‖ + ‖y2‖

82

Then

f(a+ h) = (f 1(a+ h), f 2(a+ h))

=(f 1(a) + f 1′(a)h+ φ1(h), f 2(a) + f 2′(a)h + φ2(h)

)

= (f 1(a), f 2(a)) +(

f 1′(a)h, f 2′(a)h)

︸ ︷︷ ︸

linear in h

+(φ1(h), φ2(h)

)

︸ ︷︷ ︸

remainder

Now:‖ (φ1(h), φ2(h)) ‖

‖h‖ =‖φ1(h)‖‖h‖ +

‖φ2(h)‖‖h‖

tends to 0 as h −→ 0. Therefore, f is differentiable at am and

f ′(a)h =(

f 1′(a)h, f 2′(a)h)

as required.

4.9 Notation

1. Given a function of n variables

Rn ⊃ V

f−→ R

V open, we shall denote by

x1, x2, . . . , xn

the usual co-ordinate functions on Rn and shall often denote the partial

derivative of f at a w.r.t. jth variable by:

∂f

∂xj (a) = limt−→0f(a1 ,a2,...,aj+t,...,an)−f(a1 ,a2,...,aj ,...,an)

t

= ddtf(a+ tej)

∣∣t=0

Notice that the symbol xj does not appear in the definition: it is a’dummy symbol’ indicating deriv w.r.t. jth ’slot’, sometimes written asf,j (a).

2. given a function

R2 ⊃ V

f−→ R

83

V open, we often denote by x, y the usual co-ordinate functions insteadof x1, x2 and write

∂f

∂xfor

∂f

∂x1

∂f

∂yfor

∂f

∂x2

etc.

3. given

M ⊃ Vf−→ N

V open, if f is differentiable at a, ∀a ∈ V , we say that f is differentiableon V and call the function on V :

f ′ : a −→ f ′(a)

the derivative of f. We write:

f (2) = (f ′)′

f (3) = ((f ′)′)′

and call f Cr if f (r) exists and is continuous, f C∞ if f (r) exists ∀r.

4.10 Cr Functions

Theorem 4.10.1. Let

Rn ⊃ V

f−→ R

V open. Then f is C1 ⇐⇒ ∂f

∂xi exists and is continuous for i = 1, . . . , n.

Proof. (case n = 2)

R2 ⊃ V

f−→ R, f(x, y)

1. if f is C1, then ∂f

∂x, ∂f∂y

exist and

f ′ =

(∂f

∂x,∂f

∂y

)

therefore, f ′ is continuous, and therefore, ∂f

∂xand ∂f

∂yare continuous

84

2. Let ∂f

∂x, ∂f∂y

exists and be continuous on V . Then

f(a+ h) = f(a) +∂f

∂x(a)h1 +

∂f

∂y(a)h2 + φ(h)

(say) where h = (h1, h2). Now

φ(h) = f(a+ h) − f(a+ h2e2) − ∂f

∂x(a)h1

+f(a+ h2e2) − f(a) − ∂f

∂y(a)h2

= ∂f

∂x(m1)h

1 − ∂f

∂x(a)h1

+∂f

∂y(m2)h

2 − ∂f

∂x(a)h2

(say) by mean value theorem.

PSfrag replacements

m1

m2

a

a+ ha+ h2e2

Therefore,

‖φ(h)‖‖h‖ ≤

∣∣∣∣

∂f

∂x(m1) −

∂f

∂x(a)

∣∣∣∣

|h1|‖h‖ +

∣∣∣∣

∂f

∂y(m2) −

∂f

∂y(a)

∣∣∣∣

|h2|‖h‖ −→ 0

as h −→ 0 since ∂f

∂x, ∂f∂y

are continuous at a.

Therefore, f is differentiable at a and f ′(a) =(∂f

∂x(a), ∂f

∂y(a))

. Also,

f ′ =(∂f

∂x, ∂f∂y

)

is continuous.

85

Corollary 4.10.2. f is Cr ⇐⇒ ∂rf

∂xi1 ···∂xirexists and is continuous for each

i1, . . . , ir.

Theorem 4.10.3. if

Rn ⊃ V

f−→ R V open

is C2 then ∂2f

∂xi∂xj is a symmetric matrix.

Proof. Let

Rn ⊃ V

f−→ R

be C2. Need to show:

∂2f

∂x∂y=

∂2f

∂y∂x

V

a+h

b

b+h

a x

+

-+

-

Let (a, b) ∈ V . Let h 6= 0, k 6= 0 be such that the closed rectangle

(a, b), (a+ h, b), (a+ h, b+ k), (a, b+ k)

is contained in V . Put

g(x) = f(x, b+ k) − f(x, b)

86

Then

f(a+ h, b+ k) − f(a+ h, b)−f(a, b + k) + f(a, b) = g(a+ h) − g(a)

= hg′(c) some a ≤ c ≤ a+ h by MVT

= h[∂f

∂x(c, b+ k) − ∂f

∂x(c, b)

]

= hk ∂2f

∂y∂x(c, d) some b ≤ d ≤ b + k by MVT

and, similarly:

f(a+ h, b+ k) − f(a+ h, b)

−f(a, b+ k) + f(a, b) = kh ∂2f

∂x∂y(c′, di′)

for some a ≤ c′ ≤ a+ h and b ≤ d′ ≤ b + k. Therefore

∂2f

∂y∂x(c, d) =

∂2f

∂x∂y(c′, d′)

Now, let (h, k) −→ (0, 0). Then (c, d) −→ (a, b), and (c′, d′) −→ (a, b).Therefore,

∂2f

∂y∂x(a, b) =

∂2f

∂x∂y(a, b)

by continuity of ∂2f

∂y∂xand ∂2f

∂x∂y

4.11 Chain Rule

Theorem 4.11.1. (Chain Rule for functions on finite dimensional real orcomplex vector spaces)

Let L,M,N be finite dimensional real or complex vector spaces. Let U beopen in L, V open in M and let

Ug−→ V

f−→ N

Let g be differentiable at a ∈ U , f be differentiable at g(a). Then the compo-sition f · g is differentiable at a and

(f · g)′(a) = f ′(g(a)) g′(a) operator product

87

Proof. we have

f [g(ah)] = f [g(a) + Th+ φ(h)] where T = g′(a)

and ‖φ(h)‖‖h‖ −→ 0

as ‖h‖ −→ 0

= f [g(a) + y] where y = Th+ φ(h)so ‖y‖ ≤ ‖T‖ ‖h‖ + ‖φ(h)‖ −→ 0as ‖h‖ −→ 0

= f(g(a)) + Sy + ‖y‖ψ(y) where S = f ′(g(a))and ‖ψ(h)‖ −→ 0as ‖y‖ −→ 0

= f(g(a) + S(Th+ φ(h) + ‖y‖ψ(y)

= f(g(a)) + STh+ Sφ(h) + ‖y‖ψ(y)

Now

‖Sφ(h) + ‖y‖ψ(y)‖‖h‖ ≤ ‖S‖ ‖φ(h)‖

‖h‖ +

(

‖T‖ +‖φ(h)‖‖h‖

)

‖ψ(y)‖

tends to 0 as ‖h‖ −→ 0. Therefore, f · g is differentiable at a, and

(f · g)′(a) = ST = f ′(g(a)) g′(a)

Example

d

dtf(g1(t), . . . , gn(t)) = (f · g)′(t) = f ′(g(t))

︸ ︷︷ ︸

1×n

g′(t)︸︷︷︸

n×1

whereR ⊃ U

g−→ Vf−→ R

V open in Rn, and f, g differentiable. Then

f ′(g(t))g′(t) =(∂f

∂x1 (g(t)), . . .∂f

∂xn (g(t)))

ddtg1(t)...

ddtgn(t)

= ∂f

∂x1 (g(t))ddtg1(t) + · · ·+ ∂f

∂xn (g(t)) ddtgn(t)

is the usual chain rule.

88

Chapter 5

Calculus of Complex Numbers

5.1 Complex Differentiation

Definition let

C ⊃ Vf−→ C

be a complex valued function of a complex variable defined on open V . C isa 1-dimensional complex normed space.

Let a ∈ V . Then f is differentiable at a as a function of a complexvariable if ∃ a linear map of complex spaces:

Cf ′(a)−→ C

s.t.

f(a+ h) = f(a) + f ′(a)h+ φ(h)

where |φ(h)||h| −→ 0 as h −→ 0. f ′(a) is a 1× 1 complex matrix, i.e. a complex

number, and ∣∣∣∣

f(a+ h) − f(a)

h− f ′(a)

∣∣∣∣=

|φ(h)||h|

which tends to 0 as h −→ 0. Therefore

f ′(a) = limh−→0

f(a+ h) − f(a)

h

We call f ′(a) the derivative of f at a. f ′ is called holomorphic on V iff ′(a) exists ∀ a ∈ V . We write f = df

dy.

Example

f : C −→ C, f(z) = zn

89

then,

limh−→0f(z+h)−f(z)

h= limh−→0

(z+h)n−zn

h

= limh−→0[nzn−1 + n(n−1)

2zn−2h + higher powers h]

= nzn−1

therefore, f is holomorphic in C and f ′(z) = nzn−1.

Examplef(z) = z

then,f(z + h) − f(z)

h=h

h=

{1 h real−1 h pure imaginary

which does not converge as h −→ 0.

The usual rules for differentiation apply:

1. ddy

(f + g) = df

dy+ dg

dy

2. ddyfg = f dg

dy+ g df

dy

3. ddy

f

g=

gdfdy

−f dgdy

g2if g 6= 0

4. ddzf(g(z)) = f ′(g(z)) g′(z) chain rule

By definition C = R2

z = x + iy = (x, y)

and the operation of mult by i =√−1 C

i−→ C is the operator R2 i−→ R2

with matrix

(0 −11 0

)

, because

ie1 = i = (0, 1) = 0e1 + e2

ie2 = ii = −1 = −e1 + 0e2 = (−1, 0)

Let f is (real) differentiable on V then each a ∈ v the operator

Cf ′(a)−→ C i.e R

2 f ′(a)−→ R2

90

preserves addition and commutes with mult by real scalars, for each a ∈ V ,and f is holomorphic in V iff f ′(a) also commutes with multiplication bycomplex scalars. ⇐⇒ f ′(a) also commutes with multiplication by i.

⇐⇒( ∂u

∂x∂u∂y

∂v∂x

∂v∂y

)(0 −11 0

)

=

(0 −11 0

)( ∂u∂x

∂u∂y

∂v∂x

∂v∂y

)

⇐⇒( ∂u

∂y−∂u∂x

∂v∂y

−∂v∂x

)

=

( −∂v∂x

−∂v∂y

∂u∂x

∂u∂y

)

⇐⇒∂u∂x

= ∂v∂y

∂v∂x

= −∂u∂y

Cauchy-Riemann Equations

so, therefore: f = u + iv is holomorphic in V ⇐⇒ f is real-differentiableon V and satisfies the Cauchy-Riemann Equations.

Example f(z) = z2 = (x+ iy)2 = x2 − y2 + 2ixy. then

u = x2 − y2 ∂u∂x

= 2x = ∂v∂y

v = 2xy ∂v∂x

= 2y = −∂u∂x

Note, if f = u+ iv is holomorphic in V then

∂f

∂y(a) = limh−→0

f(a+h)−f(a)h

= limt−→0f(a+t)−f(a)

t= limt−→0

f(a+it)−f(a)it

= ∂∂x

[u+ iv] = 1i∂∂y

[u+ iv]

therefore,∂f

∂z=∂u

∂x+ i

∂v

∂x=∂v

∂y− i

∂u

∂y

this also gives the Cauchy-Riemann Equations.

5.2 Path Integrals

Definition C1 map

[t1, t2]α−→ V ⊂ R

2 V open

t −→ α(t)

91

is called a (parametrical) path in R fom α(t1) to α(t2) and if f, g are complex-valued continuous on V we write

∫

α

(f dx + g dy) =

∫ t2

t1

[f(α(t))d

dtx(α(t)) + g(α(t))

d

dty(α(t))] dt

and call it the integral of the differential form f dx+ g dy over the path α. If

[s1, s2]σ−→ [t1, t2]

is C1 with σ(s1) = t1, σ(s2) = t2 and β(s) = α(σ(s)) then β is called areparameterisation of α.

We have∫

β(f dx+ g dy) =

∫ s2s1

[f(β(s)) ddsx(β(s)) + g(β(s)) d

dsy(β(s))] ds

=∫ s2s1

[f(α(σ(s))) ddsx(α(σ(s))) + g(α(σ(s))) d

dsy(α(σ(s)))] ds

=∫ s2s1

[f(α(σ(s)))(x · α)′(σ(s)) + g(α(σ(s)))(y · α)′(σ(s))]σ′(s) ds

=∫ t2t1

[f(α(t)) ddtx(α(t)) + g(α(t)) d

dty(α(t))] dt

=∫

α(f dx + g dy)

therefore, the integral over α is independent of parmaterisation.If we take

[t1, t2]σ−→ [t1, t2]

with σ(s) = t1 + t2 − s, and β(s) = α(σ(s)) we have σ(t2) = t1, σ(t1) = t2then β is same path but traversed in the opposite direction and

∫

β(f dx+ g dy) =

∫ t1t2

[f(α(t)) ddtx(α(t)) + g(α(t)) d

dty(α(t))] ds

= −∫

α(f dx+ g dy)

Definition If f is C1 on V we write

df =∂f

∂xdx+

∂f

∂ydy

A differential form of this type is called exact, and is called the differentialof f .

92

If α is a path from a to b, α(t1) = a, α(t2) = b, then

∫

αdf =

∫ t2t1

[f(α(t)) ddtx(α(t)) + g(α(t)) d

dty(α(t))] dt

=∫ t2t1

[ ddtf(α(t))] dt

= f(α(t2)) − f(α(t1))

= f(b) − f(a) so we have path independence

= change of value of f along α

If α is a closed path, (i.e. a = b) then∫

α

df = 0

If f(z) = u(x, y) + iv(x, y), for z = x + iy with u, v real, put dz = dx + idy

andf(z) dz = [u+ iv][dx + idy]

= (u dx− v dy) + i(v dx + u dy)

then∫

αf(z) dz =

∫

α(u dx− v dy) + i

∫

α(v dx+ u dy)

=∫ t2t1

{[u(α(t)) + iv(α(t))] d

dt[x(α(t)) + iy(α(t))]

}dt

=∫ t2t1f(α(t))α′(t) dt

If f is holomorphic then

df = du+ i dv = ∂u∂xdx + ∂u

∂ydy + i ∂v

∂xdx+ i ∂v

∂ydy

= ∂u∂xdx− ∂v

∂xdx+ i ∂v

∂xdx+ i ∂u

∂xdy

=(∂u∂x

+ i ∂v∂x

)( dx+ i dy)

= f ′(z) dz

therefore, if f is holomorphic on V open then

1. for any path α in V from a to b:∫

α

f ′(z) dz = f(b) − f(a)

path independence

93

2. for any closed path α in V :

∫

α

f ′(z) dz = 0

Example 1. if α path from a to b

∫

α

zn dz =

∫

α

[d

dz

zn+1

n + 1] dz =

bn+1 − an+1

n+ 1n 6= 1

in particular,

2. if α is a closed path,

∫

α

zn dz = 0 n 6= 1

But:

3. for the closed path α(t) = eit, 0 ≤ t ≤ 2π

∫

α

dz

z=

∫ 2π

0

1

eitieit dt =

∫ 2π

0

i dt = 2πi 6= 0

therefore, 1z

cannot be the derivative of a holomorphic function on C.

Definition if [t1, t2]α−→ C is a path in C then we write

L(α) =

∫ t2

t1

|α′(t)| dt

and call it the length of α.

If β = α · σ is a reparameterisation of α with α′(s) > 0

L(β) =∫ s2s1

|β ′(s)| ds

=∫ s2s1

|α′(σ(s))σ′(s)| ds

=∫ t2t1

|α′(t)| dt

= L(α)

94

Theorem 5.2.1. (Estimating a complex integral)Let f be bounded on α:

|f(α(t))| ≤M

(say), t1 ≤ t ≤ t2. then

∣∣∣∣

∫

α

f(z) dz

∣∣∣∣≤ML(α)

Proof.∣∣∫

αf(z) dz

∣∣ =

∣∣∣

∫ t2t1f(α(t))α′(t) dt

∣∣∣

≤∫ t2t1

|f(α(t))| |α′(t)| dt

≤ M∫ t2t1

|α′(t)| dt

= ML(α)

5.3 Cauchy’s Theorem for a triangle

If f(z) = u(x, y) + i v(x, y) is holomorphic then

∫

α

f(z) dz =

∫

α

(u dx− v dy) + i

∫

(v dx + u dy)

and ∂∂yu = ∂

∂x(−v), ∂

∂yv = ∂

∂xu (by Cauchy-Riemann).

Therefore, the neccessary conditions for path-independence are satisfied,but not always sufficient, e.g.f(z) = 1

z. However, we have

Theorem 5.3.1. (Cauchy’s Theorem for a triangle)

Let C ⊃ Vf−→ C be holomorphic on open V , and let T be a triangle

(interior plus boundary δT ) ⊂ V . Then

∫

δT

f(z) dz = 0

Proof. write

i(T ) =

∫

δt

f(z) dz

95

and join the mid-points of the sides to get 4 triangles

S1, S2, S3, S4

then

i(T ) =

4∑

j=1

i(Sj)

and therefore

|i(T )| ≤4∑

j=1

|i(Sj)| ≤ 4|i(T1)|

(say) where T1 is one of S1, . . . , S4.

1

T

T

Repeat the process to get a sequence of triangles

T1, T2, . . . , Tn, . . .

with

|i(T )| ≤ 4n|i(Tn)|Let

⋂∞j=1 Tj = {c}. Then

i(Tn) =∫

δTnf(z) dz

=∫

δTn[f(c) + f ′(c)(z − c) + |z − c|φ(z − c)] dz

=∫

δTn|z − c|φ(z − c) dz where |φ(z − c)| −→ 0 as |z − c| −→ 0

96

nc

z

T

Let ε > 0. Choose δ > 0 s.t. |φ(z− c)| < ε ∀ |z− c| < δ. Let L =length of T .Choose n s.t. length of Tn = l

”n < δ.Then

|i(Tn)| ≤l

2nεl

2n=

l2

4nε

therefore |i(T )| ≤ l2ε, and therefore i(T ) = 0.

Definition A set V ⊂ C is called star-shaped if ∃a ∈ V s.t.

[a, z] ⊂ V ∀ z ∈ V

a

z

V

Theorem 5.3.2. Let C ⊃ Vf−→ C be a holomorphic function on an open

star-shaped set. Then f = F ′ for some complex-differentable function F onV and hence: ∫

α

f(z) dz = 0

for each closed curve α in V .

Proof. Choose a ∈ V s.t. [a, z] ⊂ V ∀ z ∈ V . Put F (w) =∫ w

af(z) dz. Then

97

w+h

a

w

F (w + h) =∫ w+h

af(z) dz

=∫ w

af(z) dz +

∫ w+h

wf(z) dz by Cauchy

= F (w) + f(w)h+

∫ w+h

w

[f(z) − f(w)] dy

︸ ︷︷ ︸

φ(h)

Let ε > 0. Choose δ > 0 w.t. |f(z) − f(w)| < ε ∀|z − w| < δ. Therefore

|φ(h)| ≤ |h| ε

for all |h| < δ, and therefore

limh−→0

|φ(h)||h| = 0

Therefore, F is differentiable at w and F ′(w) = f(w) as required.

5.4 Winding Number

We have seen that ∫

circle about o

dz

z= 2πi

More generally:

98

Theorem 5.4.1. Let a ∈ C ad α : [t1, t2] −→ C− {a} be a closed path. The

1

2πi

∫

α

dz

z − a

is an integer, called the winding number of α about a

Proof. put

β(t) =

∫ t

t1

α′(s)

α(s) − ads

so β ′(t) = α′(t)α(t)−a . Then

ddt

[[α(t) − a]e−β(t)

]= [α′(t) − β ′(t){α(t) − a}] e−β(t)

= 0

therefore, [α(t)−a]e−β(t) is a constant function of t and is equal to [α(t1)−a],since β(t1) = 0. Therefore

eβ(t) =α(t) − a

α(t1) − a

therefore,

eβ(t2) =α(t2) − a

α(t1) − a= 1

and therefore, β(t2) = 2nπi, with n an integer.

a

PSfrag replacements

eβ(t)

β(t)

2nπi

99

Therefore,

∫

α

dz

z − a=

∫ t2

t1

α′(s)

α(s) − ads = β(t2) = 2nπi

as required.

Theorem 5.4.2. Let C be a circle and a ∈ C. Then

1. if a is inside C, then C has winding number 1 about a:∫

C

dz

z − a= 2πi

a

C

2. if a is outside C then winding number about a is 0.∫

C

dz

z − a= 0

a

C

Proof. 1. Let a be inside C, let C1 be a circle, centre a inside C. Letα, β, γ be the closed paths shown, each is contained in an open starshaped set on which 1

z−a is holomorphic.

100

1

aα

β

γ

C

C

∫

αdzz−a +

∫

βdzz−a +

∫

γdzz−a =

∫

Cdzz−a −

∫

C1

dzz−a

0 + 0 = 0 =∫

C−∫

C1

therefore,

∫

C

dz

z − a=

∫

C1

dz

z − a=

∫ 2π

0

ireiθ

reiθdθ = 2πi

(put z = a+ reiθ).

2. Let a be outside C, then C is contained in an open star-shaped set onwhich 1

z−a is holomorphic. Therefore,

∫

C

dz

z − a= 0

5.5 Cauchy’s Integral Formula

Theorem 5.5.1. (Cauchy’s Integral Formula)Let f be holomorphic on open V in C. Let w ∈ V . Then, for any circle

C around w, such that C and it’s interior is contained in V , we have:

f(w) =1

2πi

∫

C

f(z)

z − wdz

(Thus the values of f on any circle uniquely determine the values of f insidethe circle)

101

wC

V

Proof. Let ε > 0. Choose a circle C1 centre w, radius r (say) inside C suchthat |f(z) − f(w)| ≤ ε ∀z ∈ C1 by continuity of f .

wα

β

γ

C

C1

Let α, β, γ be as indicated. Then, by integrating f(z)z−w :

∫

C

−∫

C1

=

∫

α

+

∫

β

+

∫

γ

= 0 + 0 + 0 = 0

since each of α, β, γ are contained in a star-shaped set on which f(z)z−w is a

holomorphic function of z. Therefore,

∫

C

f(z)z−w dz =

∫

C1

f(z)z−w dz

=∫

C1

f(w)z−w dz +

∫

C1

f(z)−f(w)z−w dz

= 2πi f(w) + 0

because: ∣∣∣∣

∫

C1

f(z) − f(w)

z − wdz

∣∣∣∣≤ ε

r2πr = 2πε

for all ε > 0. Hence result.

102

Let C ⊃ Vf−→ C be holomorphic on V open. Let w ∈ V . Pick a circle

C around w such that C and it’s interior ⊃ V . We have:

f(w) =1

2πi

∫

C

f(z)

z − wdz

differentiating w.r.t. w under the integral sign:

f ′(w) =1

2πi

∫

C

f(z)

(z − w)2dz

w

r z

CPSfrag replacements

w1

Justified since ∃ r > 0 s.t.∣∣∣∣

f(z)

(z − w1)2

∣∣∣∣≤ |f(z)|

r2

for all w1 on an open set containing w1, which is integrable w.r.t. z.Repeating n times gives:

f (n)(w) =n!

2πi

∫

C

f(z)

(z − w)n+1dz (5.1)

Thus, f holomorphic on open V =⇒ f has derivatives of all orders andf (n)(w) is given by Equation(5.1) for any suitable circle C (or any suitableclosed curve) around w.

103

5.6 Term-by-term differentiation, analytic func-

tions, Taylor series

Theorem 5.6.1. Let {fn(z)} be a sequence of functions which is uniformlyconvergent to the function f(z) on a path α. Then

limn−→∞

∫

α

fn(z) dz =

∫

α

f(z) dz

Proof. Let ε > 0. Choose N s.t.

|fn(z) − f(z)| ≤ ε ∀n ≥ N ∀ z = α(t) t1 ≤ t ≤ t2

(definition of uniform convergence). Then∣∣∫

αfn(z) dz −

∫

αf(z) dz

∣∣ ≤

∫

α|fn(z) − f(z) dz|

≤ ε l(α) ∀ n ≥ N

hence the result.

Theorem 5.6.2. (term by term differentiation)Let

∑∞i=1 fn(z) be a series of holomorphic functions on an open set V ,

which converges uniformly on a circle C which, together with it’s interior, iscontained in V .

Then the series converges inside C to a holomorphic function F (say)and

f (k) =∞∑

n=1

f (k)n

inside C

Proof. Let w be inside C, then λ = infz∈C |z − w| > 0.

=⇒ 1

|z − w|k+1≤ 1

λk+1∀ z ∈ C

Therefore,∑∞

n=1fn(z)

(z−w)k+1 converges uniformly to f(z)(z−w)k+1 z ∈ C. Therefore,

∞∑

n=1

∫

C

fn(z)

(z − w)k+1dz =

∫

C

f(z)

(z − w)k+1dz

and therefore,∞∑

n=1

f (k)n (w) = f (n)(w)

as required.

104

Definition Let C ⊃ Vf−→ C with V opn. Then f is called analytic on V

if, for each a ∈ V ∃ r > 0 and constants c0, c1, c2, . . . ∈ C, such that:

f(z) = c0 + c1(z − a) = c2(z − a)2 + · · · ∀ z s.t. |z − a| < r

The R.H.S. is called the Taylor series of f about a.

Lemma 5.6.3. If 0 < ρ < r then the series converges uniformly on

{z : |z − a| ≤ ρ}

r

R

a

PSfrag replacements

ρ

Proof. Let 0 < ρ < R < r.∑cn(z − a)n converges for z − a = R, and

therefore,∑cnR

n converges. Therefore, ∃K s.t. |cnRn| ≤ K ∀n. Therefore

|cn(z − a)n| ≤ |cn| |z − a|n ≤ K

Rnρn = K

( ρ

R

)n

for all |z − a| ≤ ρ.

Therefore, we can differentiate term by term n times:

f (n)(z) = n!cn + (n + 1)!cn+1|z − a| + · · · higher powers of z − a

Therefore, f (n)(a) = n!cn, and therefore, cn = 1n!f (n)(a).

The Taylor coefficents are uniquely determined, f is C∞, and

f(z) =∞∑

n=1

1

n!f (n)(z − a)n on |z − a| < r

Thus, f analytic on V =⇒ f holomorphic on V .Conversely,

105

Theorem 5.6.4. Let C ⊃ Vf−→ C be holomorphic on open V . Let C be

any circle, centre a s.t. C and it’s interior are contained in V . Then f hasa Taylor series about a convergent inside C. Hence f is analytic on V .

Proof. Let w be inside C. Then

f(w) = 12πi

∫

C

f(z)z−w dz

= 12πi

∫

C

f(z)(z−a)−(w−a) dz

= 12πi

∫

C

f(z)

(z−a)[1−w−az−a ]

dz

= 12πi

∫

C

f(z)z−a∑∞

n=1

(w−az−a)n

dz

=∑∞

n=1(w − a)n1

2πi

∫

C

f(z)

(z − a)n=1dz

︸ ︷︷ ︸

cn

=∑∞

n=01n!f (n)(c)(w − a)n

as required.

106

Chapter 6

Further Calculus

6.1 Mean Value Theorem for Vector-valued

functions

Theorem 6.1.1. (Mean value theorem for vector valued functions)Let M,N be finite dimensional normed spaces and let

M ⊃ Vf−→ N

be a C1 function, where V is open in M . Let x, y ∈ V and [x, y] ⊂ V . Then

‖f(x− f(y)‖ ≤ k ‖x− y‖

where k = supz∈[x,y] ‖f ′(z)‖Proof.

f(y) − f(x) =∫ 1

0ddtf [ty + (1 − t)x] dt

=∫ 1

0f ′[ty + (1 − t)x]︸ ︷︷ ︸

operator

(y − x)︸ ︷︷ ︸

vector

dt

Therefore,

‖f(y) − f(x)‖ ≤∫ 1

0‖f ′[ty + (1 − t)x]‖ ‖y − x‖ dt

≤∫ 1

0k ‖y − x‖ dt

= k ‖y − x‖

as required.

107

6.2 Contracting Map

Theorem 6.2.1. Let M be a finite dimensional real vector space with norm.Let

M ⊃ Vf−→M

be a C1 function, V open in M , f(0) = 0, f ′(0) = 1.Let 0 < ε < 1, and let B be a closed ball centre O s.t.

‖1 − f ′(x)‖ ≤ ε ∀x ∈ B

Then

1.‖f(x) − f(y)‖ ≥ (1 − ε) ‖x− y‖ ∀ x, y ∈ B (6.1)

Thus fB is injective.

2. (1 − ε)B ⊂ f(B) ⊂ (1 + ε)B

r

B

f

PSfrag replacements

(1 − ε)r

(1 + ε)r

f(B)

Proof. Let r =radius of B

1.‖f ′(x)‖ = ‖1 + (f ′(x) − 1)‖ ≤ 1 + ε ∀x ∈ B

therefore,

‖f(x)‖ = ‖f(x) − f(0)‖MV T

≤ (1 + ε)‖x‖ ≤ (1 + ε)r

for all x ∈ B. Therefore, f(B) ⊂ (1 + ε)B

108

2.

‖(1 − f)(x) − (1 − f)(y)‖MV T

≤ ε‖x− y‖ ∀ x, y ∈ B

therefore, ‖x− y| − ‖f(x) − f(y)‖ ≤ ε ‖x− y‖, and so

‖f(x) − f(y)‖ ≥ (1 − ε)‖x− y‖

and hence Eqn(6.1).

3. To show (1− ε)B ⊂ f(B). Let a ∈ (1− ε)B, define g(x) = x−f(x)+a.Then

‖g′(x)‖ = ‖1 − f ′(x)‖ ≤ ε ∀x ∈ B

therefore,

‖g(x) − g(y)‖MV T

≤ ε‖x− y‖therefore, g is a contracting map (shortens distances).

Also,‖g(x)‖ = ‖g(x) − g(0) + a‖ since a = g(0)

≤ ‖g(x) − g(0)‖ + ‖a‖

≤ ε‖x‖ + ‖a‖

≤ εr + (1 − ε)r ∀ x ∈ B

= r

therefore, g(x) ∈ B ∀ x ∈ B. So g maps B into B and is contracting.Therefore, by the contraction mapping theorem ∃x ∈ B s.t. g(x) = x.

i.e. x− f(x) + a = x

i.e. f(x) = a. Therefore, a ∈ f(B), and (1 − ε)B ⊂ f(B) as required.

6.3 Inverse Function Theorem

Definition Let M ⊃ Vf−→ W ⊂ N where M,N are finite dimensional

normed spaces. Then f is called a Cr diffeomorphism if

1. V open in M , W open in N

2. Vf−→W is bijective

109

3. f and f−1 are Cr

V is Cr diffeomorphic to W if ∃ a Cr-diffeomorphism V −→W .

Example

Rf−→ (0,∞)

f(x) = ex is C∞, and f−1(x) = lnx is C∞. Therefore, f is a C∞ diffeomor-phism.

PSfrag replacements

x = ln y

y = ex

Theorem 6.3.1. (Inverse Function Theorem)

Let M ⊃ Vf−→ N be Cr where M,N are finite dimensional normed

spaces and V is open in M . Let a ∈ V be a point at which

f ′(a) : M −→ N

is invertible. Then ∃ open neighbourhood W of a such that

fW : W −→ f(W )

is a Cr diffeomorphism.

110

M

N

f(W)

aW

Proof. Let T be the inverse of f ′(a) and let F be defined by

F (x) = Tf(x+ a) − Tf(a)

We have:

F (0) = Tf(a) − Tf(a) = 0

0

U

+a

a f

F

f(a)

0

T f(a)

- T f(a)

T

F(U)

U+a f(U+a)

We prove that F maps an open neighbourhood U of 0 onto an openneighbourhood F (U) of 0. It the follows that f maps open U + a onto open

111

f(U + a) by a Cr diffeomorphism. Now

F ′(x) = T · f ′(x+ a)

=⇒ F ′(0) = T · f ′(a) = 1M

Choose a closed ball B centre 0 of positive radius s.s.

‖F ′(x) − 1M‖ ≤ 1

2∀ x ∈ B

and also s.t. detF ′(x) 6= 0. Then by the previous theorem (with ε = 12) we

have:

FB is injective

and

‖F (x) − F (y)‖ ≥ 1

2‖x− y‖ ∀ x, y,∈ B

and1

2B ⊂ F (B)

B

U

F(B)F

PSfrag replacements

12B

F−1

therefore, F−1 : 12B −→ B is well-defined and continuous. Let B0 be the

interior of B, an open set. Put U = F−1(12B0) ∩ B0, an open set. F (U) is

open since F−1 is continuous. So FU : U −→ F (U) is a homeomorphism ofopen U onto open F (U).

Let G be it’s inverse. To show G is Cr.

112

F

G

U

y

y+l

x

x+h

F(U)

Let x, x + h ∈ F (U), G(x) = y, G(x+ h) = y + l (say), and ‖h‖ ≥ 12‖l‖.

Let F ′(y) = S. Then F (y + h) = F (y) + Sl + φ(l), where ‖φ(l)‖‖l‖ −→ 0 as

‖l‖ −→ 0.=⇒ x + h = x + Sl + φ(l),=⇒ l = S−1h− S−1φ(l)=⇒ G(x + h) = y + l = G(x) + S−1h− S−1φ(l)Now,

‖S−1φ(l)‖‖h‖ ≤ ‖S−1‖ ‖φ(l)‖

‖l‖‖l‖‖h‖ −→ 0

as ‖h‖ −→ 0 since ‖l‖ ≤ 2‖h‖ =⇒ ‖l‖‖h‖ ≤ 2.

So, G is differentiable at x ∀x ∈ F (U) and

G′(x) = S−1 = [F ′(y)]−1 = [F ′(G(x))]−1

It follows that if G is Cs for some 0 ≤ s < r then G′ is Cs since G′ is acomposition of Cs functions

F ′, G, [·]−1

and therefore G is Cs+1. So

G Cs =⇒ G Cs+1 ∀0 ≤ s < r

therefore G is Cr as required.

113

Example Let f, g be Cr on an open set containing (a, b)

W

x a

y

b

W’

g(a,b)

t

f(a,b) z

Let

∂(f, g)

∂(x, y)=

∣∣∣∣∣∣

∂f

∂x

∂f

∂y

∂g

∂x

∂g

∂y

∣∣∣∣∣∣

6= 0

at (a, b). Then (f, g) maps an open neighbourhood W of (a, b) onto an openneighbourhood W ′ of (f(a, b), g(a, b) by a Cr diffeomorphism. Therefore, foreach z, t ∈ W ′ ∃ unique (x, y) ∈ W such that

z = f(x, y)

t = g(x, y)

and x = h(z, t), y = k(z, t) where h and k are Cr.

114

Chapter 7

Coordinate systems andManifolds

7.1 Coordinate systems

Definition Let X be a topological space. Let V be open in X. A sequence

y = (y1, . . . , yn)

of real-valued functions on V is called an n-dimensional coordinate systemon X with domain V if

X ⊃ Vy−→ y(V ) ⊂ R

n

x :−→ y(x) = (y1(x), . . . , yn(x))

is a homeomorphism of V onto an open set y(V ) in Rn.

x V

X

y(V)

y(x)

PSfrag replacements

Rn

yn(x)

y1(x)

115

Example On the set V = {y 6= 0 or x > 0} in R2 the functions r, θ givenby

x = r cos θ

y = r sin θ

−π < θ < π

map V homomorphically onto an open set in R2.

r

(x,y)

vPSfrag replacements

−πθ

π

(r, θ)θ

θ

Therefore, (r, θ) is a 2-dimensional coordinate system on R2 with domainV .

Definition If y = (y1, . . . , yn) is a coordinate system with domain V and iff is a real-valued function on V then

f = F (y1, . . . , yn)

for a unique function F on y(V ) s.t. F = f · y−1

We call f a Cr function of y1, . . . , yn if F is Cr and we write

∂f

∂yi=∂F

∂xi(y1, . . . , yn)

and call it the partial derivative w.r.t Y i in the coord system y1, . . . , yn.

116

Note:

∂f

∂yi (a) = ∂F∂xi (y

1(a), . . . , yn(a))

= ddtF (y(a) + tei)

∣∣t=0

= ddtf(αi(t))

∣∣t=0

= rate of change of f along curve αi in V

a

PSfrag replacements

αi(t)

y(a)

y(a) + tei

where αi is the curve given by:

y(αi(t)) = y(a) + tei = (y1(a), . . . , yi(a) + t, . . . , yn(a))

therefore, along curve αi all coords y1, . . . , yn are constant except ith coordyi and αi(t) is parameterised by change t in ith coord yi.

αi is called the ith coordinate curve at a.

7.2 Cr-manifold

Definition Let y1, . . . , yn with domain V , and z1, . . . , zn with domain W

be two coordinate systems on X. Then these two systems are called Cr-compatible if each zi is a Cr function of y1, . . . , yn, and each yi is a Cr

function of z1, . . . , zn on V ∩W .We call X an n-dimensional Cr-manifold if a collection of n-dimensional

coordinate systems is given whose domains cover X and which are Cr-compatible.

117

Example the 2-sphere S2 given by:

x1 + y1 + z2 = 1 in R3

The function x, y with domain {z > 0} is a 2-dimensional coordinate systemon S2.

y

z

x

Similarly the functions x, z with domain {y > 0} is a 2-dimensional co-ordinate system

On the overlap {y > 0, z > 0} we have:

x = x x = x

y =√

1 − x2 − y2 z =√

1 − x2 − y2

these systems are C∞-compatible. In this way, we make S2 into a C∞-manifold.

7.3 Tangent vectors and differentials

Definition Let a ∈ X, X a manifold. Let y1, . . . , yn be co-ords on X at a(i.e. domain an open neighbourhood of a). Then, the linear operators

∂

∂y1

∣∣∣∣a

, · · · , ∂

∂yn

∣∣∣∣a

(7.1)

118

act on the differentiable functions f at a (i.e. functions f which are real-valued differentiable functions of y1, . . . , yn on an open neighbourhood of a)and are defined by:

∂

∂yj

∣∣∣∣a

f =∂f

∂yj(a)

The operators (7.1) are linearly independent since[

αj∂

∂yj

∣∣∣∣a

]

yi = αj∂yi

∂yj(a) = αjδij = αi (7.2)

=⇒ αj∂

∂yj

∣∣∣∣a

= 0 =⇒ αi = 0 ∀i (7.3)

The real vector space with basis (7.1) is denoted TaX and is called the tangentspace of X at a. If v ∈ TaX then (7.2) shows that v has components vyi

w.r.t. basis (7.1)

Definition If α(t) is a curve in X then the velocity vector α(t) ∈ Tα(t)X isthe tangent vector at α(t) given by taking rate of change along α(t) w.r.t. t:

PSfrag replacements

α(t)

α(t)

i.e.

α(t)f = ddtf(α(t))

= ddtF (y1(α(t)), . . . , yn(α(t)) if f = F (y1, . . . , yn)

= ∂F∂xj (y

1(α(t)), . . . , yn(α(t))) ddtyj(α(t))

= ∂f

∂yj (α(t)) ddtyj(α(t))

=

[

ddtyj(α(t)) ∂

∂yj

∣∣∣α(t)

]

f

119

therefore, ˙α(t) is the tangent vector with components

d

dtyj(α(t)) = yj(t)

i.e.

α(t) = yj(t)∂

∂yj

∣∣∣∣α(t)

The tangent space TaX does not depend on the choice of (compatible) coor-dinates at a since if z1, . . . , zn is another coordinate system at a, then

∂∂zj

∣∣a

= velocity vector of jth coordinate curve of z1, . . . , zn at a

= ∂yi

∂zj∂∂yi

∣∣∣a

∈ TaX

therefore,∂

∂z1

∣∣∣∣a

, · · · , ∂

∂zn

∣∣∣∣a

is also a basis for TaX, and ∂yi

∂zj (a) is the transition matrix from z1, . . . , zn toy1, . . . , yn.

Note also that the curve α(t) with coordinates:

y(α(t)) = (y1(a) + α1t, . . . , yn(a) + αnt)

has velocity vector

α(0) = α1 ∂

∂y1

∣∣∣∣a

+ · · ·αn ∂

∂yn

∣∣∣∣a

Therefore, every tangent vector (at a) is the velocity vector of some curve,and vice versa.

Definition Let a ∈ X and f be a differentiable function at a. Then for eachv ∈ TaX with v = α(t) (say), define

〈dfa, v〉 = vf = α(t)f =d

dtf(α(t)) = rate of change of f along v

Thus dfa is a linear form on TaX, called the differential of f at a dfa measuresthe rate of change of f at a. If y1, . . . , yn are coordinates on X at a then

⟨

dyia,∂

∂yj a

⟩

=∂

∂yj

∣∣∣∣a

yi =∂yi

∂yj(a) = δij

120

therefore, dy1a, . . . , dy

na is the basis of TaX

∗ which is dual to the basis ∂∂y1

∣∣∣a, · · · , ∂

∂yn

∣∣∣a

of TaX.Thus the linear form dyia gives the ith component of tangent vectors at a:

〈dyia, v〉 = ith component of v = vyi

Also, dfa has components:

⟨

dfa,∂

∂yj a

⟩

=∂

∂j af =

∂f

∂yj(a)

therefore,

dfa =∂f

∂yj(a) dyja

(which is called the chain rule for differentials.)

7.4 Tensor Fields

From now on assume manifolds, functions are C∞.

Definition Let W be an open set in a manifold X. A tensor field S on X

with domain W is a function on W :

x 7−→ Sx

which assigns to each x ∈ W a tensor of fixed type over the tangent spaceTaX at x. e.g.

Sx : TxX × (TxX)∗ × TxX −→ R

We can add tensor fields, contract them, form tensor products and wedgeproducts by carrying out these operations at each point x ∈ W . e.g.

(R + S)x = Rx + Sx

(R⊗ S)x = Rx ⊗ Sx

Definition A tensor field with no indices is called a scalar field (i.e. a real-valued function); a tensor field with one upper index is called a vector field ;a tensor field with r skew-symmetric lower indices is called a differential r-form; a tensor field with two lower indices is called a metric tensor if it issymmetric and non-singular at each point.

121

If y1, . . . , yn is a coordinate system with domain W and S is a tensor fieldon W with (say) indices of type down-up-down, then

S = αijkdy

i ⊗ ∂

∂yk⊗ dyk

where the scalar fields αijk are the components of S w.r.t. the coordinates

yi.If f is a scalar field, then df is a differential 1-form. If v is a vector field,

then vf is the scalar field defined by:

(vf)x = vxf = 〈dfx, vx〉 = 〈df, v〉x

i.e. vf = 〈df, v〉 = rate of change of f along v.If (·|·) is a metric tensor on X then for any two vector fields u, v with

common domain W we define the scalar field (u|v) by

(u|v)x = (ux|vx)x

if v is a vector field then we can lower it’s index to get a differential 1-formω such that

〈ω, u〉 = (v|u)Conversely, raising the index of a 1-form gives a vector field. Raising theindex of df gives a vector field gradf called the gradient of f :

(gradf |u) = 〈df, u〉 = uf = rate of change of f along u

If (·|·) is positive definite then for ‖u‖ = 1 we have:

|(gradf |u)| ≤ ‖gradf‖

Thus the maximum rate of change of f is ‖gradf‖ and is attained in thedirection of gradf .

A metric tensor (·|·) defines a field ds2 of quadratic forms called theassociated line element by:

ds2(v) = (v|v) = ‖v‖2 if metric is positive definite

If yi are coordinates with domain W then, on W :each vector field u = αi ∂

∂yi with components αi

each differential r-form ω = ωi1...irdyi1 ∧ · · · ∧ dyir

each differential 1-form ω = ωidyi

〈ω, u〉 = αiωi

122

if f is a scalar field df = ∂f

∂yidyi

〈df, u〉 = uf = αi∂f

∂yi

if (∂

∂yi

∣∣∣∣

∂

∂yj

)

= gij

then(·|·) = gijdy

i ⊗ dyj

andds2 = gijdy

idyj

gradf has components

gij∂f

∂yj

therefore,

gradf = gij∂f

∂yj∂

∂yi

so

‖gradf‖ = (gradf |gradf) = 〈df, gradf〉 = gij∂f

∂yi∂f

∂yj

Example On R3 with the usual coordinate functions x, y, z the usual metrictensor is

(·|·) = dx⊗ dx+ dy ⊗ dy + dz ⊗ dz

with components

gij =

1 0 0

0 1 0

0 0 1

= gij

∂∂x, ∂∂y, ∂∂z

are orthonormal vector fields

df =∂f

∂xdx +

∂f

∂ydy +

∂f

∂ydy

and

gradf =∂f

∂x

∂

∂x+∂f

∂y

∂

∂y+∂f

∂z

∂

∂z

The line element isds2 = (dx)2 + (dy)2 + (dz)2

123

If r, θ, φ are spherical polar cordinates on R3 then

x = r sin θ cosφ

y = r sin θ sinφ

z = r cos θ

therefore,

ds2 = (dx)2 + (dy)2 + (dz)2

= (sin θ cosφdr + r cos θ cos φdθ − r sin θ sin φdφ)2

+(sin θ sinφdr + r cos θ sin φdθ + r sin θ cosφdφ)2

+(cos θdr − r sin θdθ)2

= (dr)2 + r2(dθ)2 + r2 sin2 θ(dφ)2

therefore,

gij =

1 0 00 r2 00 0 r2 sin2 θ

and

gij =

1 0 00 1

r20

0 0 1r2 sin2 θ

so

df =∂f

∂rdr +

∂f

∂θdθ +

∂f

∂φdφ

and df has components: (∂f

∂r,∂f

∂θ,∂f

∂φ

)

therefore, gradf has components

(∂f

∂r,

1

r2

∂f

∂θ,

1

r2 sin2 θ

∂f

∂φ

)

therefore,

gradf =∂f

∂r

∂

∂r+

1

r2

∂f

∂θ

∂

∂θ+

1

r2 sin2 θ

∂f

∂φ

∂

∂φ

124

7.5 Pull-back, Push-forward

Definition Let X and Y be manifolds and

Xφ−→ Y

a continuous map. Then for each scalar field f on Y we have a scalr field

φ∗f = f · φon X (we assume that φ∗f is C∞ for each C∞ f).

φ∗f is called the pull-back of f to X under φ.

X Y

a

f

PSfrag replacements

φ

φ∗f

φ(a)

(φ∗f)(x) = f(φ(x))

For each a ∈ X we have a linear operator

TaXφ∗−→ TaY

If v = α(t) ∈ TaX then we define φ∗v by:

[φ∗v]f =d

dtf(φ(α(t))) = α(t)[f · φ] = v[f · φ] = v[φ∗f ]

for each scalar field f on Y at φ(a).

PSfrag replacements

a

α(t)α(t)

φ

φ(α(t))

φ∗α(t)

φ(a)

125

Thus the velocity vector of α(t) pushes forward to the velocity vector ofφ(α(t))

Theorem 7.5.1. Let Xφ−→ Y and let y1, . . . , yn be coordinates on X at a

and letφ1(x), . . . , φn(x)

be the coordinates of φ(x) w.r.t. a coordinate system z1, . . . , zn on Y at φ(a).Then the push-forward

TaXφ∗−→ Tφ(a)Y

has matrix∂φi

∂yj(x)

Proof.

ith component of φ∗∂∂yj

a=

[

φ∗∂∂yj

a

]

zi

= ∂∂yj

a[φ∗zi]

= ∂φi

∂yj (a)

since φi(x) = zi(φ(x)), so φi = φ∗zi.

Thus the matrix of φ∗ is the same as the matrix of the derivative in thecase Rn −→ Rn. The push-forward φ∗ is often called the derivative of φ ata.

The chain rule for vector spaces

(ψ · φ)′(x) = ψ′(φ(x))φ′(x)

corresponds, for manifolds, to:

Theorem 7.5.2. (Chain rule for maps of manifolds; functorial property ofthe push-forward)

Let Xφ−→ Y

ψ−→ Z be maps of manifolds. Then

(ψ · φ)∗ = ψ∗ · φ∗

Proof. Let α(t) be a tangent vector on X, then:

(ψ · φ)∗α(t) = velocity vector of ψ(φ(α(t)))

= ψ∗[velocity vector of φ(α(t))]

= ψ∗[φ∗α(t)]

as required.

126

Definition Let Xφ−→ Y be a map of manifolds and ω a tensor field on Y

with r lower indices. Then we define the pull-back of ω to X under φ to bethe tensor field φ∗ω on X having r lower indices and defined by:

(φ∗ω)x[v1, . . . , vr] = ωφ(x)[φ∗v1, . . . , φ∗vr]

all v1, . . . , vr ∈ TxX, ∀x ∈ X. i.e.

(φ∗ω)x = φ∗[ωφ(x)]

We note the φ∗ on tensor fields preserves all the algebraic operations suchas additions, tensor products, wedge products.

We have:

Theorem 7.5.3. if Xφ−→ Y and f is a scalar field on Y , then

φ∗df = dφ∗f

i.e., the pull-back commutes with differentials

Proof. Let v ∈ TxX. Then

〈(φ∗df)x, v〉 = 〈(df)φ(x), φ∗v〉

= [φ∗v]f

= v[φ∗f ]

= 〈(dφ∗f)x, v〉

therefore, φ∗df)x = (dφ∗f)x ∀x ∈ X, and therefore φ∗df = dφ∗f .

7.6 Implicit funciton theorem

Theorem 7.6.1 (Implicit Function Theorem). (on the solution spacesof l equations in n variables)

Let f = (f 1, . . . , f l) be a sequence of Cr real-valued functions on an openset V in Rn. Let c = (c1, . . . , cl) ∈ Rl and let

X = {x ∈ V : f(x) = c, rankf ′(x) = l}

Then for each a ∈ X we can select n − l of the usual coordinate functionsx1, . . . , xn:

xl+1, . . . , xn

127

(say) so that on an open neighbourhood of a in X they form a coordinatesystem on X. Any two such coordinate systems are Cr-compatible. Thus Xis an (n− l)-dimensional Cr manifold.

Proof.

f ′ =

∂f1

∂x1 · · · ∂f1

∂xl · · · ∂f1

∂xn

......

...∂f l

∂x1 · · · ∂f l

∂xl · · · ∂f l

∂xn

l×n

Let a ∈ X. THen f ′(a) has rank l, so the matrix f ′(a) has l linearly inde-pendent columns: the first l columns (say). Put

F = (f 1, . . . , f l, xl + 1, . . . , xn)

then

F ′ =

∂f1

∂x1 · · · ∂f1

∂xl 0 · · ·...

...∂f l

∂x1 · · · ∂f l

∂xl 0 · · ·

0 · · · 0 1 0...

.... . .

0 · · · 0 0 1

l×n

therefore,

detF ′(a) =

∣∣∣∣∣∣∣

∂f1

∂x1 (a) · · · ∂f1

∂xl (a)...

...∂f l

∂x1 (a) · · · ∂f l

∂xl (a)

∣∣∣∣∣∣∣

6= 0

PSfrag replacements

Rl

Rn−l

RlRl

Rn−l Rn−l

a

W

F (W )

G

F

X

U

c

128

By the inverse function theorem, F maps an open neighbourhoodW of a ontoan open neighbourhood F (W ) by a Cr-diffeomorphism with inverse G (say).Note that F , and hence also G leave the coordinates xl+1, . . . , xn unchanged.

F maps W ∩X homeomorphically onto {c}×U where U is open in Rn−l.Thus (xl+1, . . . , xn) maps W ∩X homeomorphically onto U and is there-

fore an (n − l)-dimensional coordinate system on X with domain W ∩ X.Also, if

G = (G1, . . . , Gl, xl+1, . . . , xn)

on F (W ) thenx1 = G1(c1, . . . , cl, xl+1, . . . , xn)

......

xl = Gl(c1, . . . , cl, xl+1, . . . , xn)

on W ∩X. Therefore x1, . . . , xl and Cr functions of xl+1, . . . , xn on W ∩X.Hence any two such coordinate systems on X are Cr compatible.

Note: the proof shows that if (say) the first l columns of the matrix∂f i

∂xj (a) are linearly independent, then the l equations in n unknowns

f 1(x1, . . . , xn) = c1

......

f l(x1, . . . , xn) = cl

determine x1, . . . , xn as Cr functions of xl+1, . . . , xn on an open neighbour-hood of a in the solution space.

Example 1.

Rn ⊃ V

f−→ R

a Cr function.

f ′ =

(∂f

∂xi

)

1×n=

(∂f

∂x1, . . . ,

∂f

∂xn

)

= ∇f

are the components of the gradient vector field.

If ∇f is non-zero at each point of the space X of solutions of theequations:

f(x1, . . . , xn) = c

(one equation in n unknowns) then X is an (n − 1)-dimensional Cr-manifold.

129

If (say) ∂f

∂x1 (a) 6= 0 at a ∈ X then

x2, x3, . . . , xn

are coordinates on an open neighbourhood W of a in X and x1 is a Cr

function of x2, . . . , xn on W .

2. Rn ⊃ Vf1,f2

−→ R two Cr functions.

(∂f i

∂xj

)

=

(∂f1

∂x1 · · · ∂f1

∂xn

∂f2

∂x1 · · · ∂f2

∂xn

)

If the two rows ∇f 1,∇f 2 are linearly independent at each point of thespace X of solutions of the two equations:

f 1(x1, . . . , xn) = c1

f 2(x1, . . . , xn) = c2

(two equations in n unknowns) then X is an (n − 2)-dimensional Cr

manifold. If (say)

∂(f 1, f 2)

∂(x1, x2)6= 0 at a ∈ X

then x3, x4, . . . , xn are coordinates on an open neighbourhood W of ain X and x1 and x2 are Cr functions of x3, . . . , xn on W .

7.7 Constraints

If f 1, . . . , f l are Cr functions on an open set V in Rn and c = (c1, . . . , cn) weconsider the equations

f 1 = c1, . . . , f l = cl

as a system of constraints which are satisfied by points on the constraintmanifold :

X =

{

x ∈ V : f 1(x) = c1, . . . , f l(x) = cl, rank∂f i

∂xj(x) = l

}

Let Xi−→ Rn be the inclusion map i(x) = x ∀x ∈ X. Then for each scalar

field f on V :(i∗f)(x) = f(i(x)) = f(x) ∀x ∈ X

130

the pull-back i∗f is the restriction of f to X. We call i∗f the constrainedfunction F . If ω is a differential r-form on V the i∗ω is called the constrainedr-form ω. Similarly if (·|·) is a metric tensor on V and ds2 the line elementthe i∗(·|·) is the constrained matrix and i∗ds2 is the constrained line element.

For each a ∈ X the push-forward

TaXi∗−→ TaR

n

is injective and enables us to identify TaX with a vector subspace of TaRn.

The constrained metric and constrained line element are just the restrictionsto TaX of the matrix and line element, for each a ∈ X.

Example Let X be the constraint surface in R3:

f(x, y, z) = c f Cr

We have:

df =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz unconstrained

Pulling back to X, where f is constant, we get:

0 =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz constrained

If (say) ∂f

∂z6= 0 at a ∈ X then, by the implicit function theorem, the con-

strained functions x, y are coordinates on X in a neighbourhood W of a andconstrained z is a Cr function of x, y on W .

z = F (x, y)

say. Now

dz = −(∂f

∂x

/∂f

∂z

)

dx−(∂f

∂y

/∂f

∂z

)

dy

Therefore, (∂z

∂x

)

y

=∂F

∂x= −∂f

∂x

/∂f

∂z(∂z

∂y

)

x

=∂F

∂y= −∂f

∂y

/∂f

∂z

The usual line element on R3 is

ds2 = (dx)2 + (dy)2 + (dz)2 unconstrained

131

pulling back to X gives:

ds2 = (dx)2 + (dy)2 +[

−(∂f

∂x

/∂f

∂z

)

dx−(∂f

∂y

/∂f

∂z

)

dy]2

=

[

1 +(∂f

∂x

/∂f

∂z

)2]

(dx)2 + 2∂f∂x

∂f∂y

( ∂f∂y )

2dx dy +

[

1 +(∂f

∂y

/∂f

∂z

)2]

(dy)2

The coefficients give the components of the constrained metric on X withrespect to the coordinates x and y

Example using spherical polar coordinates on R3 the usual line element is

ds2 = (dr)2 + r2(dθ)2 + r2 sin2 θ(dφ)2

Pulling back to the sphere S2 by the constraint:

r = const

we have:ds2 = r2(dθ)2 + r2 sin2(dφ)2

Thus the constrained line element has components:(r2 00 r2 sin2 θ

)

7.8 Lagrange Multipliers

A scalar field f on a manifold X has a critical point a ∈ X if dfa = 0.i.e. ∂f

∂yi (a) = 0 for a coordinate system yi at a. (e.g. a local maximum or

minimum or saddle point)

Problem given a scalar field F on V open in Rn, to find the critical pointof constrained F , where the constraints are:

f 1 = c1, . . . , f l = cl

and f i are Cr functions on V

Method Take f 1, . . . , f l, xl+1, . . . , xn (say) as coordinates on R3 as in theproof of the implicit function theorem, so

dF =

l∑

i=1

∂f

∂f idf i +

n∑

i=l+1

∂f

∂xidxi unconstrained

132

Thus

dF = 0 +

n∑

i=l+1

∂F

∂xidxi constrained

This is zero at a critical point a, so ∂F∂xi (a) = 0, i = l + 1, . . . , n, and

hence

dFa =

l∑

i=1

∂F

∂f i(a)df i

at a critical point a of constrained F . If we put

λi =∂F

∂f i(a) = rate of change of F at a with respect to the ith constraint

we have:dF = λ1df

1 + · · ·+ λldfl

at a where the scalars λ1, . . . , λl are called Lagrange multipliers.

Since dF is a linear combination of df 1, . . . , df l at a we can take thewedge product to get the equations:

dF ∧ df 1 ∧ · · · ∧ df l = 0

f 1 = c1, . . . , f l = cl

which must hold at any critical point of constrained F .

7.9 Tangent space and normal space

If f is constant on the constraint manifold X then

df = 0 constrained

therefore,〈df, α(t)〉 = 0

for all curves in X. Hence the system of l linear equations

df 1 = 0, . . . , df l = 0

give the tangent space to X at each point. Also if (·|·) is a metric tensorwith domain V then

(gradf |α(t)) = 〈df, α(t)〉 = 0

133

for all curves in X. Therefore gradf is orthogonal to the tangent space to Xat each point. Hence

gradf 1, . . . , gradf l

is a basis for the normed space to X at each point.At a critical point of constrained F we have:

dF = λ1df1 + · · · + λldf

l

and raising the index gives:

gradF = λ1gradf 1 + · · ·+ λlgradf l

thus gradF is normal to the constraint manifold X at each critical point ofconstrained F .

7.10 ?? Missing Page

1. =⇒ ω is closed.

However ω is not exact because it’s integral around circle α(t) =(cos t, sin t) 0 ≤ t ≤ 2π is

∫

α

ω = 2π0

[cos t cos t − (sin t)(− sin t)]

cos2 t + sin2 tdt =

∫ 2π

0

dt = 2π 6= 0

Note: on R2 − {negative x-axis} we have:

x = r cos θ − π < θ < π

y = r sin θ

therefore,

ω =r cos θ[sin θ dr + r cos θ dθ] − r sin θ[cos θ dr − r sin θ dθ]

r2 cos2 θ + r2 sin2 θ= dθ

so.∫

αω is path-independent provided α does not cross the -ve x-axis.

∫

αω = θ(b) − θ(a), and ω is called the angle-form

2. Let ω be a differential n-form with domain V open in Rn

ω = f(x1, . . . , xn)dx1 ∧ · · · ∧ dxn (say) xi usual coord

then we define:∫

V

ω =

∫

V

f(x1, . . . , xn)dx1 dx2 . . . dxn Lebesgue integral xi dummy

134

7.11 Integral of Pull-back

Theorem 7.11.1. Let Vφ−→ φ(V ) be a C1 diffeomorphism of open V in Rn

onto open φ(V ) in Rn, with detφ′ > 0. Let ω be an n-form on φ(V ). Then

∫

V

φ∗ω =

∫

φ(V )

ω

[so writing 〈ω, V 〉 to denote integral of ω over V we have: 〈φ∗ω, V 〉 = 〈ω, φV 〉adjoint]

Proof.

PSfrag replacementsφ∗ω

V

ω

φV

φ

Letω = f(x1, . . . , xn)dx1 ∧ · · · ∧ dxn

φ = (φ1, . . . , φn)

i.e. φi(x) = xi(φ(x)) i.e. φi = xi · φ = φ∗xi.Then

φ∗ω = f(φ1, . . . , φn)dφ1 ∧ · · · ∧ dφn

= f(φ(x)) ∂φ1

∂xi1dxi1 ∧ · · · ∧ ∂φn

∂xindxin

= f(φ(x)) det(∂φi

∂xj (x))

dx1 ∧ · · · ∧ dxn

therefore,∫

V

φ∗ω =

∫

V

f(φ(x)) detφ′(x)dx1 dx2 . . . dxn =

∫

φ(V )

f(x) dx

by the general change of variable theorem for multiple integrals (still to beproved).

135

Corollary 7.11.2. Let y1, . . . , yn be positively oriented coordinates with do-main V open in Rn. Then

∫

V

f(y1, . . . , yn)dy1 ∧ · · · ∧ dyn︸ ︷︷ ︸

non-dummy yi with wedge

=

∫

y(V )

f(y1, . . . , yn)dy1 dy2 . . . dyn

︸ ︷︷ ︸

Lebesgue integral over y(V ). y1, . . . , yn dummy. No wedge

Proof.

PSfrag replacementsy∗xi

V y(V )

y = (y1, . . . , yn)

xi

y∗xi = xi · y = yi

therefore,

LHS =∫

Vy∗[f(x1, . . . , xn) dx1 ∧ cdots ∧ dxn ]

=∫

y(V )f(x1, . . . , xn) dx1 ∧ · · · ∧ dxn

=∫

y(V )f(x1, . . . , xn) dx1 dx2 . . . dxn x1, . . . , xn dummy

= RHS

7.12 integral of differential forms

To define∫

Xω where ω is an n-form and X is an n-dimensional manifold

(e.g. ω is a 2-form, X a surface) we need some topological notions:

1. A topological space X is called Hausdorff if, for each a, b ∈ X, a 6= b, ∃open disjoint V,W s.t. a ∈ V, b ∈ W .

136

PSfrag replacements

a b

VW

X

2. A set A ⊂ X is called closed in X if it’s complement in X

A′ = {x ∈ X : x 6∈ Z}

is open in X.

PSfrag replacements A

X

A′

3. If A ⊂ X then the intersection of all the closed subsets of X whichcontain A is denoted A and is called the closure of A in X; A is thesmallest closed subset of X which contains A.

4. Let S be a tensor field on a manifold X. THe closure in X of the set

{x ∈ X : Sx 6= 0}

is called the support of S, denoted supp S.

PSfrag replacements S 6= 0

X

suppS

Theorem 7.12.1. Let X be a Hausdorff manifold and let A ⊂ X be compact.Then ∃ scalar fields

F1, . . . , Fk

on X s.t.

137

1. Fi ≥ 0

2. F1 + · · · + Fk = 1 on A (partition of unity)

3. each suppFi is contained in the domain Vi of a coordinate system onX.

Proof. (sketch) Let

g(t) =

{

e−1t t > 0

0 t ≤ 0

}

PSfrag replacements

1

g(t)

let

b(t) =g[1 − t2]

g(1)

PSfrag replacements

11

1

−1

b(t)

b is C∞, supp b = [−1, 1], 0 ≤ b ≤ 1. (b is for bump).Let a ∈ A. Pick a coordinate system y on X at a with domain V . Pick

a ball centre y(a) radius 2r > 0 contained in y(V ). Put

h(x) =

{

b(

‖y(x)−y(a)‖r

)

if x ∈ V

0 if x 6∈ V

}

138

then h is C∞, supp h ⊂ V , 0 ≤ h ≤ 1, h(a) = 1; ’bump’ at a.

Let W = {x ∈ X : h(x) > 0}. then W is an open neighbourhood of aand h > 0 on W .

Thus, for each a ∈ A we have an open neighbourhood Wa of a and ascalar field fa s.t. ha > 0 on Wa, ha is C∞, supp ha ⊂ a coord domain Va,0 ≤ ha ≤ 1.

PSfrag replacementsa

Wa

Va

y(a)

r

2r

Rn

ha is a bump at a, for each a ∈ A. Since A is compact we can select afinite number of points a1, . . . , ak such that Wa1 , . . . ,Wak

cover A. THen put

Fi =

{hai

ha1+···+hak

ifhai6= 0

0 ifhai= 0

}

to get the required scalar fields F1, . . . , Fk.

139

PSfrag replacements

ha1ha2

X

F1 F2

F1 + F2 = 1

7.13 orientation

Definition Let y1, . . . , yn with domain V , z1, . . . , zn with domain W be twocoordinate systems on a manifold X. Then they have the same orientationif

∂(y1, . . . , yn)

∂(z1, . . . , zn)= det

∂yi

∂zj> 0

on V ∩W .

Since∂

∂zj=∂yi

∂zj∂

∂yi

this means that ∂∂z1 a

, · · · , ∂∂zn a

has the same orientation in TaX as ∂∂y1 a

, · · · , ∂∂yn

a

each a ∈ V ∩W .

We call X oriented if a family of mutually compatible coordinate systemsis given on X, whose domains cover X and any two of which have the sameorientation.

Example x = r cos θ, y = r sin θ. Then ∂(x,y)∂(r,θ)

= r > 0. Therefore x, y andr, θ have the same orientation.

140

Definition Let ω be a differential n-form with compact support on an ori-ented n-dimensional Hausdorff manifold X. We define

∫

X

ω

the integral of ω over X as follows:

1. Suppose suppω ⊂ V where V is the domain of positively orientedcoordinates y1, . . . , yn and ω = f(y1, . . . , yn) dy1∧· · ·∧dyn on V . Thendefine:

∫

x

ω =

∫

y(V )

f(y1, . . . , yn) dy1 . . . dyn dummy yi

PSfrag replacements

VW

X

z(W )

ω2ω1

φ

y z

y(V )

This is independent of the choice of coordinates since if suppω ⊂ W

where W is the domain of positively oriented coordinates z1, . . . , zn

then

ω = f(y1, . . . , yn) dy1 ∧ · · · ∧ dyn = g(z1, . . . , zn) dz1 ∧ · · · ∧ dzn (say) on V ∩W

= y∗[f(x1, . . . , xn) dx1 ∧ · · · ∧ dxn = z∗[g(x1, . . . , xn) dx1 ∧ · · · ∧ dxn

= y∗ω1 = z∗ω2

141

(say). So ω1 = φ∗ω2 where φ = z · y−1 : y(V ∩ W ) −→ z(V ∩ W ).Therefore,

∫

y(V )f(x1, . . . , xn) dx1 . . . dxn =

∫

y(V )ω1 =

∫

y(V ∩W )ω1 =

∫

z(V ∩W )ω2

=∫

z(W )ω2 =

∫

z(W )g(x1, . . . , xn) dx1 . . . dxn

as required.

2. Choose a partition of unity

F1 + · · ·+ Fk = 1

on suppω. Put ωi = Fiω. Then

ω1 + · · ·+ ωk = ω

and suppωi ⊂ suppFi ⊂ Vi where Vi is the domain of a coordinatessystem. Define ∫

X

ω =

∫

X

ω1 + · · · +∫

X

ωk

using (1.). This is independent of the choice of partition of unity sinceif

G1 + · · ·+Gl = 1 on suppω

then ∑lj=1

∫

XGjω =

∑lj=1

∫

X

∑ki=1 FiGjω

=∑l

j=1

∑ki=1

∫

XFiGjω

similarly=

∑k

i=1

∫

XFiω

Definition If A ⊂ X is a Borel set we define∫

A

ω =

∫

X

χAω

Example to find the area of the surface X:

x2 + y2 + z = 2, z > 0

We have:2x dx+ 2y dy + dz = 0

142

constrained to X. Therefore,

ds2 = (dx)2 + (dy)2 + (2x dx+ 2y dy)2 constrained to X

= (1 + 4x2)(dx)2 + 8xy dx dy + (1 + 4y2)(dy)2

gij =

(1 + 4x2 4xy

4xy 1 + 4y2

)

therefore,√g =

√det gij =

√

1 + 4x2 + 4y2. So, area element is√

1 + 4x2 + 4y2 dx∧dy. Therefore:

area =∫ √

1 + 4x2 + 4y2 dx ∧ dy

=∫ √

1 + 4r2r dr ∧ dθ

=∫ 2π

0

[∫ √2

0r√

1 + 4r2dr]

dθ

= 2π[

23

18(1 + 4r2)

32

]√2

0

= π6[27 − 1]

= 133π

143

144

Chapter 8

Complex Analysis

8.1 Laurent Expansion

Theorem 8.1.1. Let f be holomorphic on V − {a} where V is open anda ∈ V . Let C be a circle, centre a, radius r s.t. C and it’s interior iscontained in V . Then ∃ {cn}n = 0,±1,±2, . . . ∈ C s.t.

f(z) =∞∑

n=−∞cn(z − a)n in 0 < |z − a| < r

The RHS is called the Laurent series of f about a. The coefficient cn areuniquely determined by:

cn =1

2πi

∫

C

f(z)

(z − a)n+1dz

Proof. Let w be inside C. Choose circles C1, C2 as shown with centres a, w:

PSfrag replacements

Ca

w

C1

C2

145

Then

f(w) = 12πi

∫

C2

f(z)z−w dz

= 12πi

∫

C

f(z)z−a d− 1

2πi

∫

C1

f(z)z−w dz

= 12πi

∫

C

f(z)(z−a)−(w−a) dz + 1

2πi

∫

C1

f(z)(w−a)−(z−a) dz

= 12πi

∫

C

f(z)

(z−a)[1−w−az−a ]

dz + 12πi

∫

C1

f(z)

(w−a)[1− z−aw−a ]

dz

= 12πi

∫

C

f(z)(z−a)

∑∞n=0

(w−az−a)n

dz + 12πi

∫

C1

f(z)(w−a)

∑∞n=0

(z−aw−a)n

dz

=∑∞

n=0(w − a)n1

2πi

∫

C

f(z)

(z − a)n+1dz

︸ ︷︷ ︸

+ve powers (w − a)

+∑∞

n=0(w − a)−n−1 1

2πi

∫

C!

f(z)

(z − a)−ndz

︸ ︷︷ ︸

-ve powers (w − a)

as required.

Definition If f is holomorphic in V − {a} anf

f(z) =

∞∑

n=−∞cn(z − a)n = · · ·+ c−2

(z − a)2+

c−1

z − a︸ ︷︷ ︸

p(z)

+c0 + c1(z − a) + · · ·

is the Laurent series of f at a. p(z) =∑−1

n=−∞ cn(z−a)n is called the principalpart of f at a. p(z) is holomorphic on C − {a}.

f(z) − p(z) is holomorphic in V (defining it’s value at 0 to be c0).

For any closed curve α in C − {a}:

∫

αp(z) dz =

∫

α

[

· · ·+ c−2

(z−a)2 + c−1

z−a

]

dz

= · · ·+ 0 + c1∫

αdzz−a

= 2πiRes(f, a)W (α, a)

where W (α, a) = 12πi

∫

αdzz−a is the winding number of α about a. and

Res(f, a) = c−1 is the residue of f at a

146

8.2 Residue Theorem

Theorem 8.2.1 (Residue Theorem). Let f be holomorphic on V−{a1, . . . , an}where a1, . . . , an are distinct points in a star-shaped (or contractible) open setV . Then for any closed curve α in V − {a1, . . . , an} we have:

∫

α

f(z) dz = 2πi

n∑

j=1

Res(f, aj)W (α, aj)

Proof. Let p1, . . . , pn be the principal parts of f at a1, . . . , an respectively.Then f−(p1+· · ·+pn) is holomorphic on V . Therefore,

∫

α[f − (p1 + · · ·+ pn)] dz =

0. So,

∫

f(z) dz =n∑

j=1

∫

α

pj(z) dz = 2πin∑

j=1

Res(f, aj)W (α, aj)

as required.

Definition If f is holomorphic on a neighbourhood of a, excluding possiblya itself, and if the Laurent expansion has at most a finite number of negativepowers:

f(z) =∑∞

r=n cr(z − a)r

= cn(z − a)n + cn+1(z − a)n+1 + · · · cn 6= 0

= (z − a)n [cn + cn+1(z − a) + · · · ]

= = (z − a)nf1(z) f1 holomorphic on a nbd of a and f1(a) 6= 0

then we say that f has order n at a.e.g. order 2:

f(z) = c2(z − a)2 + c3(z − a)3 + · · · c2 6= 0

order -3:

f(z) =c−2

(z − a)3+

c−2

(z − a)2+

c1

(z − a)+ c0 + c1(z − a) + · · · c−3 6= 0

If n > 0 we call a a zero of f .If n < 0 we call a a pole of f .n = 1: simple zero; n = 2: double zero.

147

n = −1: simple pole; n = −2: double pole.If f has order n and g has order m:

f(z) = (z − a)nf1(z) f1(a) 6= 0

g(z) = (z − a)ng1(z) g1(a) 6= 0

then:

f(z) g(z) = (z − a)n+mf1(z) g1(z) fg has order n+m

f(z)g(z)

= (z − a)n−mf1(z)g1(z)

f

ghas order n−m

The residue theorem is very useful for evaluating integrals:

Example to evaluate∫∞0

x sinxx2+a2

, a > 0 we put

f(z) = z eiz

z2+a2[eiz is easier to handle than sin z]

= z eiz

(z−ia)(z+ia)

Simple poles at ia,−ia.

PSfrag replacements

ia

−ia

−R R

α

Choose a closed contour that goes along x-axis −R to R (say) then loopsaround a pole, upper semi-circle α (say).

∫ R

−R

x eix

x2 + a2dx+

∫

α

z eiz

z2 + a2dz = 2πiRes(f, ia)

148

1. if f(z) = cz−ia + c0 + c1(z − ia) + · · · on neighbourhood of ia then

(z − ia)f(z) = c + c0(z − ia) + c1(z − ia)2 + · · · on neighbourhood ofia. Then limz−→ia(z − ia)f(z) = c = Res(f, ia). Therefore

Res(f, ia) = c = limz−→iaf(z)z−ia

= limz−→iaz eiz

z+ia= ia

2iae−a

= 12e−a

2. α(t) = Reit 0 ≤ t ≤ π = R(cos t + i sin t). Therefore

∣∣∣

∫

αz eiz

z2+a2dz∣∣∣ =

∣∣∣

∫ π

0ReiteiR(cos t+i sin t)iReit

R2e2it+a2dt∣∣∣

≤ R2

R2−a2∫ π

0e−R sin t dt

which −→ 0 as R −→ ∞ since

limR→∞

∫ π

0

e−R sin t dtDCT=

∫ π

0

limR→∞

e−R sin t dt = 0

Therefore ∫ ∞

−∞

x(cos x+ i sin x)

x2 + a2dx + 0 = 2πi

e−a

2

so ∫ ∞

−∞

x sin x

x2 + a2dx = πe−a

Example to calculate∫∞−∞

sinxxdx put f(z) = eiz

zholomorphic except for

simple pole at z = 0.

PSfrag replacements

−R R

α

−r r

β

149

Choose a closed contour along x-axis from −R to −r, loops around 0 byβ(t) = reit then r to R then back along α(t) = Reit.

∫ −r

−R

eix

xdx+

∫

β

eiz

zdz +

∫ R

r

eix

xdx+

∫

α

eiz

zdz = 0

1. Res(f, 0) = limz→0(z − 0)f(z) = limz→0 eiz = 1

2. ∣∣∣

∫

αeiz

zdz∣∣∣ =

∣∣∣

∫ π

0eiR(cos t+i sin t)iReit

Reit dt∣∣∣

≤∫ π

0e−R sin t dt

which −→ 0 as R −→ ∞.

3. eiz

z= 1

z+ g(z), g holomorphic in neighbourhood of 0. Therefore

∫

β

eiz

zdz =

∫

β

dz

z+

∫

β

g(z) dz

Let sup |g(z)| = M (say on a closed ball centre 0 redius δ > 0 (say).

∣∣∣∣

∫

β

g(z) dz

∣∣∣∣≤Mπr −→ 0 as r −→ 0

if r ≤ δ∫

β

dz

z= −

∫ π

0

ireit

reitdt = −

∫ π

0

i dt = −iπ ∀r

therefore

limr−→0

∫

β

eiz

zdz = −iπ

therefore∫ 0

−∞

eix

xdx− iπ +

∫ ∞

0

eix

xdx+ 0 = 0

so ∫ ∞

−∞

cos x + i sin x

xdx = iπ

and ∫ ∞

−∞

sin x

xdx = π

150

8.3 Uniqueness of analytic continuation

Theorem 8.3.1 (Uniqueness of analytic continuation). Let f, g be holo-morphic on a connected open set V . Let a ∈ V and let {zk} be a sequence inV (6= a) converging to a s.t.

f(zk) = g(zk) ∀k

Then f = g on V .

Proof. Put F = f − g, so F (zk) = 0. To show F = 0 on V .

1. F is holomorphic at a. Therefore F (z) = b0 +b1(z−a)+b2(z−a)2 + · · ·on |z − a| < R (say). F (a) = limF (zk) = 0. Therefore b0 = 0.

Suppose we know that b0, b1, . . . , bm−1 are all zero.

F (z) = (z − a)m[bm + bm+1(z − a) + bm+2(z − a)2 + · · · ]

= (z − a)mFm(z)

(say). F (zk) = 0, so Fm(zk) = 0 and Fm(a) = 0. Therefore bm = 0.br = 0∀r and F = 0 on an open ball centre a.

2. Put V = W∪W ′ where W = {z ∈ V : F = 0 on an open ball centre z}and W ′ = V −W . Then W is open, and a ∈ W by 1. Therefore Wis non-empty. Suppose c ∈ W ′ and c is a non-interior point of W ′.Then each integer r > 0∃wr ∈ W s.t. |wr − c| < 1

r, F (wr) = 0 and

limwr = c. Therefore F = 0 on an open ball centre c by 1., and c ∈ W .Therefore W ′ is empty.

151

152

Chapter 9

General Change of Variable ina multiple integral

9.1 Preliminary result

Theorem 9.1.1. Let Rn ⊃ Vφ−→ Rn be C1, V open, a ∈ V , detφ′(a) 6= 0.

Then

limm(φB)

m(B)= | detφ′(a)|

where the limit is taken over cubes B containing a with radius B −→ 0.

Proof. Let ‖ · ‖ be the sup norm on Rn.

‖(α1, . . . , αn)‖ = max(|α1|, . . . , |αn|)

so a ball, radius r is a cube, with side 2r.Let 0 < ε < 1. Put T (x) = [φ′(x)]−1. Fix a closed cube J containing

a and put k = supx∈J ‖T (x)‖. Choose δ > 0 s.t. ‖φ′(x) − φ′(y)‖ ≤ εk

all‖x− y‖ ≤ 2δ; x, y ∈ J .

If B is a cube ⊂ J containing a of radius ≤ δ. and centre c (say).Consider:

T (c)φ has derivative T (c)φ′(x) which equals 1 at x = c and

‖T (c)φ′(x) − 1‖ = ‖T (c)φ′(x) − T (c)φ′(c)‖

≤ ‖T (c‖ ‖φ′(x) − φ′(c)‖

≤ k εk

= ε

153

therefore (1 − ε)B1 ⊂ T (c)φB ⊂ (1 + ε)B1, where B1 is a translate of B tonew centre T (c)φ(x). Therefore

(1 − ε)nm(B) ≤ | detT (c)|m(φB) ≤ (1 + ε)nm(B)

=⇒ (1 − ε)n ≤ | detT (c)|m(φB)

m(B)≤ (1 + ε)n

=⇒ lim | detT (c)|m(φB)

m(B)= 1

=⇒ | detT (a)| lim m(φB)

m(B)= 1

Therefore:

limm(φB)

m(B)=

1

| detT (a)| = | detφ′(a)|

as required

Theorem 9.1.2. Let f be a continuous real valued function on an openneighbourhood of a in Rn. THen

lim

∫

Bf(x) dx

m(B)= f(a)

where the limit is taken over cubes B containing a with radius B −→ 0.

Proof. Let ε > 0. Choose δ > 0 s.t. |f(x) − f(a)| < ε ∀ ‖x − a‖ < δ. Theneach cube B containing a of radius ≤ δ

2we have:

∣∣∣∣

∫

Bf(x) dx

m(B)− f(a)

∣∣∣∣=

∣∣∫

B[f(x) − f(a)] dx

∣∣

m(B)≤ εm(B)

m(B)= ε

hence result.

Recall that the σ-algebra generated by the topology of Rn is called the

collection of Borel Sets in Rn.

Theorem 9.1.3. Let A be a Borel set in Rr, B be a Borel set in Rs. ThenA× B is a Borel set in R

r+s.

Proof. For fixed V open in Rr, the sets:

{V ×W : W open in Rs} (9.1)

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are all open in Rr+s. Therefore the σ-algebra generated by Eqn(9.1):

{V × B : B Borel in Rs}

consists of Borel sets in Rr+s. Hence for fixed B Borel in Rs, the set:

{V × B : V open in Rr (9.2)

are all Borel sets in Rr+s. Therefore the σ-algebra generated by Eqn(9.2):

{A× B : A Borel in Rr}

consists of Borel sets in Rr+s. Therefore A× B is Borel for each A Borel inRr, B Borel in Rs.

Theorem 9.1.4. Let E be a lebesgue measurable subset of Rn. THen there

is a Borel set B containing E such that

B − E = B ∩ E ′

has measure zero, and hence m(B) = m(E).

Proof. We already know this is true for n = 1. So we use induction on n.Assume true for n− 1. Let E ⊂ Rn, E measurable.

1. Let m(E) <∞. Let k be an integer > 0.

E ⊂ Rn−1 × R

choose a sequence of rectangles

A1 ×B − 1, A2 × B2, . . .

covering E with Ai ⊂ Rn−1 measurable and Bi ⊂ R measurable, andsuch that

m(E) ≤∞∑

i=1

m(Ai)m(Bi) ≤ m(E) +1

k

By the induction hypothesis choose Borel sets Ci, Di s.t.

Ai ⊂ Ci, Bi ⊂ Di, m(Ai) = m(Ci), m(Bi) = m(Di)

Put Bk =⋃∞i=1 Ci × Di. Then E ⊂ Bk, Bk is Borel and m(E) ≤

m(Bk) ≤∑∞

i=1m(Ci)m(di) ≤ m(E) + 1k.

Put B =⋂∞k=1Bk. Then E ⊂ B, B is Borel and m(E) = m(B).

Therefore m(B ∩ E ′) = m(B) −m(E) = 0 as required.

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2. Let m(E) = ∞. Put Ek = {x ∈ E : k ≤ |x| < k + 1}. E =⋃∞k=0Ek is

a countable disjoint union, and m(Ek) <∞. For each integer k chooseby 1. Borel Bk s.t. Ek ⊂ Bk and m(Bk ∩ E ′

k) = 0

Put B =⋃Bk. Then E ⊂ B and m(B ∩ E ′) = m(B) − m(E) as

required. ???????????

9.2 General change of variable in a multiple

integral

Theorem 9.2.1 (General change of variable in a multiple integral).

Let Rn ⊃ V

φ−→ W ⊂ Rn be a C1 diffeomorphism of open V onto open W .

Let f be integrable on E. Then∫

W

f(x) dx =

∫

V

f(φ(x))| detφ′(x)| dx (Lebesgue Integrals) (9.3)

Proof. 1. we have f = f+−f− with f+, f− ≥ 0. Therefore, it is sufficientto prove Eqn(9.3) for f ≥ 0.

2. if f ≥ 0 then ∃ a monotone increasing sequence of non-negative simplefunctions {fn} such that f = lim fn so, using the monotone convergencetheorem, it is sufficient to prove Eqn(9.3) for f simple.

3. if f is simple then f =∑k

i=1 aiχEiwith {Ei} Lebesgue measurable,

so it is sufficient to prove Eqn(9.3) with f = χE with E Lebesguemeasurable.

4. if E Lebesgue measurable ∃ Borel B s.t. E ⊂ B and Z = B − E hasmeasure zero. Therefore, χE = χB − χZ , and it is sufficient to proveEqn(9.3) for f = χB, B Borel, and for f = χZ , Z measure zero.

5. If f = χE with E Borel then E = φF with F Borel and Eqn(9.3)reduces to ∫

W

χφF (x) dx =

∫

V

χF (x)| detφ′(x)| dx

i.e.

m(φF ) =

∫

F

| detφ′(x)| dx (9.4)

If Eqn(9.4) holds for each rectangle F ⊂ V then Eqn(9.4) holds for F ∈ring R of finite disjoint unions of rectangles ⊂ V .

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Therefore Eqn(9.4) holds for F ∈ monotone class generated by R.

Therefore Eqn(9.4) holds for F ∈ σ-algebra generated by R (monotoneclass lemma). So Eqn(9.4) holds for any Borel set F ⊂ V

6. to show Eqn(9.4) holds for any rectangle F ⊂ V : for each rectangleF ⊂ V put

λ(F ) = m(φF ) −∫

F

| detφ′(x)| dx

Then, λ is additive:

λ(⋃

Bj) =∑

λ(Bj)

for a disjoint union.

Must prove λ(F ) = 0 for each rectangle F . Suppose B is a rectanglefor which λ(B) 6= 0.

Suppose B is a cube. |λ(B)| > 0. Therefore, ∃ ε > 0 s.t. |λ(B)| ≥εm(B).

Divide B into disjoint subcubes, each of 12

the edge of B. For one such,B1 say,

|almbda(B1)| ≥ εm(B1)

Sivide again|λ(B2)| ≥ εm(B2)

continuing we get a decreasing sequence of cubes {Bk} converging to asay, with

limh−→∞

|λ(Bk)|m(Bk)

≥ ε > 0

But

limh→∞λ(Bk)m(Bk)

= limh→∞m(φBk)−

∫

Bk|det φ′(x)| dx

m(Bk)

= | detφ′(a)| − | detφ′(a)|

= 0

a contradiction. Therefore λ(B) = 0 for all cubes B. Therefore,m(φB) =

∫

B| detφ′(x)| dx for all cubes B as required.

7. Now suppose Z has measure zero, and choose a Borel set B s.t.

Z ⊂ B ⊂ V

and s.t. B has measure zero.

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Then

m(φZ) ≤ m(φB) =

∫

B

| detφ′(x)| dx = 0

since B has measure zero. Therefore φZ has measure zero.

Therefore under a C1 diffeomorphism we have:

Z of measure zero =⇒ φZ of measure zero

therefore f = χZ with Z of measure zero =⇒ f · · ·φ = χφ−1Z withφ−1Z of measure zero.

Therefore, Eqn(9.3) holds for f = χZ since then both sides are zero.This completes the proof.

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