Download - Measure Theory Notes - David Simms
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.
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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.)
8
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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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
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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
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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.
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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.
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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
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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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