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Page 1: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

Harmonic AnalysisCathal Ormond

1

Page 2: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

Contents

1 Fourier Analysis 41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Review of Lebesgue Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Summability Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.7 Cesaro Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Functional Analysis 152.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Holder’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 Minkowski’s Integral Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 The Poisson Kernel 183.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Definition: The Poisson Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 Definition: Abel Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 Hilbert Spaces 224.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.2 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.3 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.4 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.5 Definition: Gram-Schmidt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5 Pointwise Convergence 275.1 Divergence at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.2 Convergence of Partial Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

6 Weyl’s Equidistribution Theorem 296.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.2 Equidistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

7 The Fourier Transform on R 32

8 The Heat Equation 338.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2

Page 3: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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9 Poisson Summation and Heisenberg’s Uncertainty Principle 349.1 Poisson Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349.2 Heisenberg’s Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

10 The Fourier Transform on Rd 3710.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3710.2 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3710.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3810.4 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3810.5 The Wave Equaion in Rd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

11 The Radon Transform 4111.1 The Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4111.2 The Dual Radon Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

12 Finite Fourier Analysis 4312.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4312.2 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4412.3 Abelian Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

13 Dirichlet’s Theorem 4813.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

Page 4: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

Chapter 1

Fourier Analysis

1.1 Introduction

In the first section of this course, we will look at fourier analysis and its applications to Lp spaces.We’ll begin by examining the solutions at the boundary to the heat equation.

1.2 The Heat Equation

Given a straight metal bar, we denote by u(x, t) the heat of the bar at the point x ∈ [0, L] at timet ≥ 0, where L is the length of the bar. We know that the propogation of heat may be described bythe Heat Equation:

∂u

∂t= c2

∂2u

∂x2

for some constant c. If we know certain facts about the value of u or its derivatives at the minimumor maximum values for x (in this case when x = 0, x = L), we call these boundary conditions. If weknow the value of u or its derivatives when t = 0, we call these initial conditions. There are threedifferent boundary conditions we will use in our study of PDEs in Fourier Analysis:

• Dirichlet Conditions: we know the value of u(x, t) at the boundary.

• Neumann Conditions: we know the value of ux(x, t) at the boundary.

• Robin Conditions: we know that aux(x, t) + bu(x, t) = 0 at the boundary.

For the heat equation, we will assume that we know the following boundary and initial conditions:

u(0, t) = 0 ∀t ≥ 0ux(L, t) = 0 ∀t ≥ 0u(x, 0) = f(x)

In other words, we assume that no heat escapes at the initial point of the metal bar, the heat flowexiting the metal bar at the endpoint is constant and that we know the heat flow at all point at thebeginning.

We assume that seperable solutions exist, i.e. that there exist functions X(x) and T (t) such thatu(x, t) = X(x)T (t). Substituting this into our PDE gives us the following:

X(x)T ′(t) = c2X ′′(x)T (t) ⇒ T ′(t)T (t)

= c2X ′′(x)X(x)

= −λ λ ≤ 04

Page 5: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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where the equation on the left must be constant, as each side is independent of the other in terms ofvariables, and any change in t, for example, will not change the value of X(x) or X ′′(x).

Now, we consider the two ODEs. If λ = 0, this gives us the trivial solution u(x, t) = 0 when weconsider the initial and boundary conditions. Letting λ 6= 0, this gives us that T (t) = Ae−λt and

X(x) = α sin

(√λ

cx

)+ β cos

(√λ

cx

)where A,α, β ∈ R. This gives us:

u(x, t) = Ae−λt

[α sin

(√λ

cx

)+ β cos

(√λ

cx

)]

We’ll now consider our initial conditions:

• u(0, t) = 0: this gives us that Aβe−λt = 0. We assume that A 6= 0, so that we don’t get thetrivial solution again. Thus, β = 0.

• ux(L, t) = 0: this gives Aα

√λ

c

[cos

(√λ

cL

)]eλt = 0. Again, we assume that α 6= 0 to avoid

the trivial solution. This gives that cos

(√λ

cL

)= 0, which will be true when

√λ

cL = π

2 + πn

for n ∈ Z, which gives λ =c(π2 + πn)2

L2

Now, putting this back into our solution, we get:

u(x, t) = A

[α sin

( π2 + πn

L

)]exp

[−(c(π2 + πn)2

L2

)t

]for some n ∈ Z. We may rescale the length of the bar to be of length π, done by substituting y =

πx

L,

to simplify. Furthermore, because the PDE was linear and homogeneous, we may write the solutionas a sum. Thus:

u(x, t) =∞∑n=0

an

[sin( π

2 + πn

L

)]exp

[−(c(π2 + πn)2

L2

)t

]and, on rescaling, we see that:

f(y) = u(x, 0) =∞∑n=0

an sin(n+

12

)y

The main idea of Fourier Analysis is that (almost) any function f may be written in the above form,or in a similar form.

1.3 Review of Lebesgue Integration

For any [a, b] ⊂ R, and for 0 < p <∞ we define:

Lp([a, b]) ={f measurable on [a,b] :

ˆ|f(x)|p dx <∞

}/∼

where we have the equivalence relation f ∼ g iff f(x) = g(x) almost everywhere (a.e.) for x ∈ [a, b].

Page 6: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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We define a mapping (which we can show is a norm) on Lp, namely || · || : Lp → R, given by

||f ||p =[ˆ ∞−∞|f(x)|p dx

] 1p

On L2, we can define the following inner product, which makes the space a Hilbert space and a Banachspace with an inner product:

< f, g >=ˆ ∞−∞

f(x)g(x) dx ||f ||2 =√< f, f >

We may also define the above for p =∞ as follows:

L∞([a, b]) = {f | [a, b]→ C : ∃M ∈ R s.t. m({|f(x)| > M}) = 0}

where m(A) is the Lebesgue measure on the set A. The above M is define to be the norm of a functionf . The elements in L∞ are called the Essentially Bounded Functions. We have the followingtheorems about Lebesgue integrals and the Lp spaces:

1.3.1 Theorem

For all 1 ≤ p <∞, C1(C) is dense in Lp, where C1(C) is the set of all f : C→ C that are C1.

1.3.2 The Dominated Convergence Theorem

If (fn)∞n=1 are measurable on I with fn → f a.e. and there exists some g ∈ L1(I) such that |fn(x)| ≤|g(x)| a.e. for all n ∈ N, then:

limn→∞

ˆIfn(x) dx =

ˆIf(x) dx

1.3.3 Theorem

If (fn)∞n=1 are non-negative with fn ↑ f a.e., then:

limn→∞

ˆIfn(x) dx =

ˆIf(x) dx

1.3.4 The Fubini-Tonelli Theorem

If f(x, y) is measurable on I × I ′, with I, I ′ intervals, and if˜I×I′ |f(x, y)| dx dy <∞, then

¨I×I′|f(x, y)| dx dy =

¨I×I′|f(x, y)| dy dx

1.4 Fourier Coefficients

Let f : [−π, π] be periodic, i.e. f(x+ π) = f(x), then we define the Fourier Coefficients of f asfollows:

an = π

−πf(x) cos(nx) dx bn =

π

−πf(x) sin(nx) dx

Where we use the convention: f(x) dx =

12π

ˆf(x) dx

Page 7: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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We may also define the complex fourier coefficients of f by the mapping f : Z → C, given byf(n) = 1

´ π−π f(x)e−inx dx. We define the associated fourier series of f by

12a0 +

∞∑n=1

(an cosnx+ bn sinnx)

Three important questions arise when considering fourier series:

1. Does the fourier series converge to f?

2. Does every f have a fourier series?

3. Is the series (i.e. are the coefficients) unique?

To the first question, it can be shown that yes, the fourier series congerves to f . To the secondquestion, we restrict our study to f ∈ L1([−π, π]) with f periodic, and so each f will have a fourierseries, as Lebesgue integration must hold.

1.4.1 Proposition 1.1

If f ∈ L∞ and f(n) = 0 for all n ∈ Z, then f(x) = 0 wherever f is continuous at x.

Proof

Assume that f ∈ L∞(R) and WLOG, that f is continuous at 0, with f(0) > 0. (Note that we pickx = 0 because we can shift the periodic function to any value of x we wish by a change of variablesin the integral). By continuity, if |x| < δ, we have f(x) > 1

2f(0). Choose some ε < 13 such that

cos θ < 1− 3ε for all θ ∈ [−π,−δ] ∪ [δ, π].Let P (θ) = cos θ + 2ε. There exists some η > 0 such that |P (θ)| > 1 + ε on [−η, η]. Define Pk(θ) =[P (θ)]2k. This gives: ffl π

−π f(θ)Pk(θ) dθ = 0 as f(n) = 0

We’ll break the integral into three parts: |θ| ≥ δ, η ≤ |θ| ≤ δ and |θ| ≤ η.

• On δ ≤ |θ| ≤ π, we have |P (θ)| < 1− ε, giving |θ|≥δ

f(θ)Pk(θ) dθ ≤ ||f ||∞|1− ε|2k · 2π → 0 as k →∞

• On |θ| ≤ η, we have f(θ) > 12f(0) as η < δ, giving

|θ|≤η

f(θ)Pk(θ) dθ >f(0)

2|1 + ε|2ks · 2η →∞ as k →∞

Thus, we expected our integral to vanish, but one if the sections goes to +∞. If the final termdoes not go to −∞, we will have a contradiction.

•fflη≤|θ|≤δ f(θ)Pk(θ) dθ ≥ 0 as f(θ) > 0 and Pk(θ) > 0 on η ≤ |θ| ≤ δ, which gives a contradiction.

Thus, our statement is true for all f ∈ L∞(R). If f ∈ L∞(C), then we have f(z) = u(z) + iv(z), withu, v ∈ L∞(R). We note that:

u(n) =ffl π−π

[f(eiθ)+f(eiθ)

2

]e−inθ dθ

= 12

ffl π−π f(eiθ)e−inθ dθ + 1

2

ffl π−π f(eiθ)e−inθ dθ

= 12 f(n) + 1

2 f(−n) = 0 ∀n ∈ Z

Similarly, v(n) = 0 for all n ∈ Z, giving us the required result. �

Page 8: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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1.4.2 Corollary 1.2

If f ∈ C1(T) where T = {z ∈ C : |z| < 1} and f(n) = 0 ∀n ∈ Z, then f = 0.

1.4.3 Corollary 1.3

If f ∈ C1(T) and∞∑

n=−∞|f(n)| <∞, then SN (f) =

N∑n=−N

|f(n)| converges uniformly to f .

Proof

Since we have∞∑

k=−∞|f(k)| < ∞, the Snf converges to some g ∈ C1(T). Given any ε we must have

some N ∈ N such that: ∑|k|≥N

|f(k)| < ε

Which gives: ∣∣∣∣∣∞∑

k=−∞|f(k)|eikx −

N∑k=−N

|f(k)|eikx∣∣∣∣∣ < ε

If we have g(n) = f(n) for all n ∈ Z, then g = f by corollary 1.2. We have:

g(k) =ffl π−π g(x)e−ikx dx

=ffl π−π

(limN→∞

SNf(x))e−ikx dx

= limN→∞

π

−πSNf(x)e−ikx dx

= limN→∞

f(k) if k ≤ N

= f(k)

as required. �

1.4.4 Corollary

If f ∈ C2(T), then f(n) is O(

1n2

), and hence SNf → f uniformly.

Proof

Integrating by parts, we have, for n 6= 0:

f(n) = π

−πf(x)e−inx dx =

f(x)e−inx

−in

∣∣∣∣π−π

+1in

π

−πf ′(x)e−inx dx =

1inf ′(n)

This gives n2f(n) = −f ′′(n). As f ′′ is continuous on T, it is bounded above, say by M . This givesthe following:

|f ′′(n)| =∣∣∣∣ π

−πf ′′(x)e−inx dx

∣∣∣∣ ≤ ∣∣∣∣ 12πM(π − (−π))

∣∣∣∣ = |M |

Thus, we have |f(n)| ≤∣∣∣∣M 1

n2

∣∣∣∣, as required. �

Page 9: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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1.5 Convolution

1.5.1 Definition

If f, g ∈ L1(T), the Convolution of f and g is give by:

(f ∗ g)(eix) = π

−πf(eit)g(ei(x−t)) dt

or, if f, g ∈ L1([−π, π]), then:

(f ∗ g)(x) = π

−πf(t)g(x− t) dt

1.5.2 Definition: The Dirichlet Kernel

We define the Dirichlet Kernel to be the sequence of functions in L1(T), (Dn)∞n=1 given by:

DN (x) =N∑

k=−Neikx

1.5.3 Proposition 1.5

Given any f ∈ L1([−π, π]), we have [SN (f)](x) = [f ∗DN ](x)

Proof

By direct calculation, we see that:

[f ∗DN ](x) =ffl π−π f(t)

[N∑

k=−Neik(x−t)

]dt

=N∑

k=−N

π

−πf(t)eik(x−t) dt

=N∑

k=−Neikx

π

−πf(t)e−ikt dt

=N∑

k=−Nf(k)eikx

= [SN (f)](x)

as required. �

Theorem 1.6

If f, g ∈ L1([−π, π]) then:

1. The convolution of f and g is linear.

2. ||f ∗ g||1 ≤ ||f ||1 · ||g||1

3. The convolution of f and g is commutative

4. The convolution of f and g is associative

Page 10: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

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5. f ∗ g is continuous if f ∈ L∞([−π, π])

6. f ∗ g(n) = f(n)g(n)

Proof

Part 1 follows from the linearity of Lebesgue integrals.Note that Part 2 implies that if fn → f for a sequence of functions (fn)∞n=1, then fn ∗ g → f ∗ g in L1.We have:

||f ∗ g||1 = π

−π

∣∣∣∣ π

−πf(y)g(x− y) dy

∣∣∣∣ dx≤ π

−π|f(y)| · |g(x− y)| dy dx

= π

−π|f(y)| · |g(x− y)| dx dy

= π

−π|f(y)| dy · ||g||1

= ||f ||1 · ||g||1Thus, if lim

n→∞fn = f , we have lim

n→∞||fn∗g−f ∗g||1 = lim

n→∞||(fn−f)∗g||1 ≤ lim

n→∞||fn−f ||1 ·||g||1 = 0

as required.For Part 3, we make a change of variables, z = x− y to see that:

(f ∗ g)(x) = π

−πf(y)g(x− y) dy

= − x−π

x+πf(x− z)g(z) dz

= x+π

x−πg(z)f(x− z) dz

= π

−πg(z)f(x− z) dz

= (g ∗ f)(x)

as required.Part 4 follows from the associativity of the integral, coupled with the Fubini-Tonelli Theorem.For Part 5, we note that:

|(f ∗ g)(x+ h)− (f ∗ g)(x)| =∣∣∣∣ π

−πf(y)g(x+ h− y) dy −

π

−πf(y)g(x− y) dy

∣∣∣∣=∣∣∣∣ π

−πf(y)[g(x+ h− y)− g(x− y)] dy

∣∣∣∣≤ π

−π|f(y)| · |g(x+ h− y)− g(x− y)| dy

≤ ||f ||∞ π

−π·|g(x+ h− y)− g(x− y)| dy

Page 11: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

11

and the right-hand side goes to 0, as required.For Part 6, we use a change of variable z = x− y for x:

f ∗ g(n) = π

−π(f ∗ g)(x)e−inx dx

= π

−π

[ π

−πf(y)g(x− y) dy

]e−inx dx

= π

−πf(y)g(z)e−in(z+y) dy dz

= π

−πf(y)e−iny

[ π

−πg(z)e−inz dz

]dy

= f(n)g(n)

as required. �

1.6 Summability Kernels

1.6.1 Definition

A sequence of functions (Kn)∞n=1 in L1([−π, π]) is called a Summability Kernel (or sometimes it iscalled an Approximation to the Identity) if:

•ˆ π

−πKn(x) dx = 1 for all n ∈ N

• There exists some M such thatˆ π

−π|Kn(x)| dx < M

• For all δ > 0, we have limn→∞

[ˆ|x|>δ

|Kn(x)| dx

]= 0

1.6.2 Theorem 1.7

Let (Kn)∞n=1 be a summability kernel and let f ∈ L1([−π, π]). Then:

1. If f is continuous at x ∈ [−π, π], then limn→∞

[f ∗Kn](x) = f(x).

2. If f ∈ C0(T), then (f ∗Kn)→ f uniformly.

3. (f ∗Kn)→ f in L1.

Proof

For Part 1, we note that, if f is continuous at x, then for all δ > 0:

|(f ∗Kn)(x)− f(x)| =∣∣∣∣ π

−πKn(y)[f(x− y)− f(x)] dy

∣∣∣∣≤∣∣∣∣ δ

−δKn(y)[f(x− y)− f(x)] dy

∣∣∣∣+∣∣∣ffl|y|>δKn(y)[f(x− y)− f(x)] dy

∣∣∣≤ δ

−δ|Kn(y)| · |f(x− y)− f(x)| dy +

|y|>δ

|Kn(y)| · |f(x− y)− f(x)| dy

< Mε+ 2||f ||∞ · |y|>δ

|Kn(y)| dy

< Mε as n→∞

Page 12: Harmonic Analysis - maths.tcd.ieormondca/notes/Harmonic Analysis.pdf · Harmonic Analysis Cathal Ormond 1. Contents 1 Fourier Analysis 4 1.1 Introduction . . . . . . . . . . .

12

For Part 2, we note that the above δ was independent of the value of x, and so the above convergenceis uniform, if f ∈ C0([−π, π]).For Part 3, given any ε > 0 we let g ∈ C0([−π, π]) be such that ||g − f ||1 <

ε

3M, where M is given in

the definition of the summability Kernel. Thus:

||f ∗Kn − f ||1 = ||(f ∗Kn − g ∗Kn + g ∗Kn − g + g − f ||1≤ ||((f − g) ∗Kn||1 + ||g ∗Kn − g||1 + ||g − f ||1≤ ||f − g||1||Kn||1 + ||g ∗Kn − g||∞ + ||g − f ||1< ε

3M ·M + ε3 + ε

3M< ε

as required. �

1.7 Cesaro Summation

1.7.1 Definition

We say that a complex series∞∑n=1

cn is Cesaro Summable to L ∈ C if the following limit exists and

is equal to L:

limN→∞

σN = limN→∞

1N

N−1∑n=1

cn

If f ∈ L1, then we have:

σN (f) =1N

[N−1∑n=1

SN (f)

]

=1N

[N−1∑n=1

f ∗DN

]

= f ∗

[1N

N−1∑n=1

DN

]= f ∗ FN

where FN =1N

N−1∑n=1

DN is called the Fejer kernel.

1.7.2 Lemma 1.8

The Fejer kernel is a summability kernel and satisfies:

FN (x) =1N

sin2(Nx2 )sin2(x2 )

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13

Proof

Let eix = w. This gives DN (x) =w−N − wN+1

1− w. Thus:

FN (x) =1N

11− w

N−1∑n = 1w−N − wN+1

=1N

11− w

[1− wN

1− w− w − wN+1

1− w

]=

1N

2− wN − wN

|1− w|2

Now, note that:

sin2(Nx2 ) =

(wN/2 − wN/2

2i

)2

sin2(x2 ) =

(w1/2 − w1/2

2i

)2

= −wN + wN − 2

4= −w + w − 2

4

which gives:1N

sin2(Nx2 )sin2(x2 )

=wN + wN − 2w + w − 2

= FN (x)

Consider the following:

• As FN (x) = 1N

N−1∑k=1

Dk(x) = 1 + · · ·

• FN (x) =1N

sin2(Nx2 )sin2(x2 )

, and so |FN (x)| = FN (x), as requried.

• For all δ ∈ (−π, π), consider:

ˆ δ

−π|FN (x)| dx+

ˆ π

δ|FN (x)| dx

On [δ, π], sin2(x/2) ≥ sin2(δ/2) = c > 0, and so FN (x) ≤ 1cN

, as required.

1.7.3 Definition: Trigonometric Polynomial

A Trigonometric Polynomial is an expression in the following form f(x) =N−1∑

k=1−Ncke

ikx.

1.7.4 Theorem 1.9

1. If f ∈ L1, f(n) is Cesaro summable to f in L1, i.e. limN→∞

||σN (f)− f ||1 = 0

2. If f ∈ L∞, then [σN (f)](x) = f(x) where f is continuous at x.

3. If f ∈ C0(T), then σN (f)→ f uniformly.

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Proof

1.7.5 Corollary 1.10

If g, h ∈ L1 and g(n) = h(n) for all n ∈ Z, then g = h a.e.

Proof

If f = g − h, then we have f(n) = 0 for all n ∈ Z. This gives that σN (f) = 1N

(N−1∑k=1

Sn(f)

)= 0.

Thus, by part 1 of Theorem 1.0, f = 0 a.e. �

1.7.6 Corollary 1.11

The trigonometric polynomials are dense in L1([−π, π]) and in the set of continuous, periodic functionsdefined on [−π, π].

1.7.7 Theorem 1.12: Fejer’s Theorem

If f ∈ L1, then limN→∞

[σN (f)](x) =12

limh→0

[f(x+ h) + f(x− h)], whenever the right-hand limit exists.

Proof

Define f(x) = 12 limh→0

[f(x + h) + f(x − h)], which we assume exists. For all ε > 0, we choose some δ

such that∣∣12(f(x+ y) + f(x− y))− f(x)

∣∣ < ε whenever |y| < δ. Also, choose some N ∈ N such thatmax|y|≥δ

|Fn(y)| ≤ ε for all n ≥ N . This gives the following:

|σN (f)− f(x)| =∣∣∣ffl π−π FN (y)[f(x− y)− f(x)] dy

∣∣∣=∣∣∣∣2 δ

0FN (y)

[f(x+ y) + f(x− y)

2− f(x)

]dy + 2

π

δFN (y)

[f(x+ y) + f(x− y)

2− f(x)

]dy

∣∣∣∣≤∣∣∣∣2 δ

0FN (y)

[f(x+ y) + f(x− y)

2− f(x)

]dy

∣∣∣∣+∣∣∣2 ffl πδ FN (y)

[f(x+y)+f(x−y)

2 − f(x)]dy∣∣∣

< 2 δ

0FN (y)ε dy + 2

π

δε

[f(x+ y) + f(x− y)

2− f(x)

]dy

≤ 2 π

0FN (y)ε dy + 2ε

[12

π

δf(x+ y) + f(x− y) dy +

12π|f(x)|π

]< ε[1 + ||f ||1 + |f |]

As ε was chosen arbitrarily, this gives the required result. �

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Chapter 2

Functional Analysis

2.1 Introduction

We will now look more closely at Lp spaces, and consider their applications to what we have alreadyproven.

2.1.1 Lemma 2.1

If a, b ≥ 0 and λ ∈ (0, 1), the we have:

aλb1−λ ≤ λa+ (1− λ)b

Proof

If b = 0, the result is trivial, so assume that b 6= 0, and set t = ab . Dividing across the inequality by b,

and replacing with t, we wish to prove that:

tλ ≤ λt+ (1− λ)

Now, we consider the function f(t) = tλ − λt. We can see that for t < 1, f ′(t) > 0 and for t > 1,f ′(t) < 0. This means that we have a maximum at t = 1, i.e. f(t) ≤ f(1) for all t. This gives us thedesired result. �

2.1.2 Definition: A Weight

A Weight w(x) on a measurable set E ⊆ Rn is a piecewise continuous function such that w(x) ≥ 0and w(x) > 0 a.e.

2.1.3 Definition: Lp given a Weight

Given a weight w(x), and for any 0 ≤ p <∞, we define the following:

Lp(w) ={f ∈ Lp(E) | f is measuable on E and

´E |f(x)|pw(x) dx <∞

}||f ||Lp(w) =

[ˆE|f(x)|pw(x) dx

] 1p

L∞(w) = L∞(E)

15

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2.2 Holder’s Inequality

2.2.1 Theorem: Holder’s Inequality

Let p ∈ [1,∞] and q be the dual exponent to p, i.e. let q be given by 1p + 1

q = 1. Let w be a weighton E, and let f, g be measurable on E. Then, we have:

||fg||L1(w) ≤ ||f ||Lp(w) · ||g||Lq(w)

Furthermore, we have equality:

• for 1 < p <∞, when there exists α, β ≥ 0 such that α|f |p = β|g|q a.e

• for p = 1, when´E |f(x)g(x)| dx =

(´E |f(x)| dx

)||g||∞ for f 6= 0 and

∣∣´E f(x)g(x)w(x) dx

∣∣ =||f ||Lp(w) · ||g||lq(w), if in addition sgn(fg) is constant a.e. for fg 6= 0

Proof

We’ll assume that ||f ||, ||g|| 6= 0. Without loss of generality, as the inequality is homogeneous, we maydivide across by constants, and so we’ll take:

||f ||Lp(w) = ||g||Lq(w) = 1

Letting a = |f(x)|p, b = |g(x)|q and λ = 1p , we apply Lemma 2.1 as above to see that |f(x)| · |g(x)| ≤

1p |f(x)|p + 1

q |g(x)|q, which gives:

||fg||L1(w) =´E |f(x)| · |g(x)|w(x) dx

≤´E

[1p |f(x)|p + 1

q |g(x)|q]w(x) dx

≤ 1p + 1

q = 1

and so, we have ||fg||L1(w) ≤ ||f ||Lp(w) · ||g||Lq(w).

For 1 < p <∞, equality is given by lemma 2.1 when a = b, or |f(x)|p = |g(x)|q a.e., subjected to

||f ||Lp(w) = ||g||Lq(w) = 1. If this is not the case, then we can set α =1

||f ||Lp(w)and β =

1||g||Lq(w)

,

which gives α|f(x)|p = β|g(x)|q a.e.

For p = 1, we firstly note that´|fg| =

∣∣´ fg∣∣ only if |fg| = fg a.e., i.e. if sgn(fg) is constanta.e. for fg 6= 0. We know that |g(x)| ≤ ||g||∞, and so we have ||fg||L1(w) ≤ ||g||∞||f ||L1(w). If f 6= 0,then we may cancel |f(x)| from the above equation giving equality only if |g(x)| = ||g||∞. �

2.2.2 Corollary 2.4: Minkowski’s Inequality

If w is a weight on E and f, g are measureable on E, then for 1 ≤ p ≤ ∞, we have:

||f + g||Lp(w) ≤ ||f ||Lp(w) + ||g||Lp(w)

Proof

The inequlity is clear when p = 1,∞ or when f + g = 0 a.e., so we’ll exclude those cases. By thetriangle inequality, we have:

|f(x) + g(x)|p = |f(x) + g(x)| · |f(x) + g(x)|p−1 ≤ (|f(x)|+ |g(x)|) |f(x) + g(x)|p−1

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This gives the following:

´E |f(x) + g(x)|pw(x) dx ≤

ˆE|f(x)| · |f(x) + g(x)|p−1 dx+

ˆE|g(x)| · |f(x) + g(x)|p−1 dx

≤ ||f ||Lp(w)||(|f + g|)p−1||Lq(w) + ||g||Lp(w)||(|f + g|)p−1||Lq(w)

However, ||(|f+g|)p−1||Lq(w) =[´

(|f(x) + g(x)|p−1)qw(x) dx] 1q =

[´|f(x) + g(x)|pw(x) dx

] 1q because

q(p− 1) = p. This gives that[ˆ|f(x) + g(x)|pw(x) dx

]≤[ˆ|f(x) + g(x)|pw(x) dx

] 1q

(||f ||Lp(w) + ||g||Lp(w))

and so ||f + g||Lp(w) ≤ ||f ||Lp(w) + ||g||Lp(w), as required. �

2.3 Minkowski’s Integral Inequality

2.3.1 Theorem 2.6: Minkowski’s Integral Inequality

Let w1 and w2 be weights on E1 and E2 respectively, and let F (x, y) be measurable on E1×E2. Then,for all 0 ≤ p <∞, we have[ˆ

E1

∣∣∣∣ˆE2

|F (x, y)|w2(y) dy∣∣∣∣pw1(x) dx

] 1p

≤ˆE2

[ˆE1

|F (x, y)|pw1(x) dx] 1q

w2(y) dy

2.3.2 Corollary 2.7

For all 1 ≤ p ≤ ∞, we have L1 ∗ Lp ⊆ Lp, or in other words, ||f ∗ g||p ≤ ||f ||1 · ||g||p

Proof

If p = 1,∞, the statement is equivalent to L1 ∗L∞ ⊆ C0(T), which is trivially true, so let 1 < p <∞.Thus, if f, g ∈ Lp, we have the following:

||f ∗ g||p =[ˆ ∣∣∣∣ˆ f(y)g(x− y) dy

∣∣∣∣p dx] 1p

≤ˆ [ˆ

|f(y)g(x− y)|p dx] 1p

dy

=ˆ|f(y)|

[ˆ|g(x− y)|p dx

] 1p

dy = ||f ||1 · ||g||p

2.3.3 Corollary 2.8

If (Kn)∞n=1 is a summability kernel and f ∈ Lp with 1 ≤ p <∞, then f ∗Kn → f in Lp.

Proof

This is true by considering the proof of theorem 1.7.1. �

2.3.4 Corollary 2.9

If (Kn)∞n=1 is the Fejer kernel, we have that for all f ∈ Lp with 1 ≤ p < ∞, σN (f) → f in Lp, andconsequently, the trigonometric polynomials are dense in Lp.

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Chapter 3

The Poisson Kernel

3.1 Introduction

In this short chapter, we will consider a particular infinite sequence of functions, and develop theirrelation to all functions that are harmonic on the unit disk in the complex plane.

3.2 Definition: The Poisson Kernel

Given r ≤ 1 and θ ∈ [0, 2π], we define the Poisson Kernel as follows:

Pr(θ) = P (r, θ) =∞∑

n=−∞r|n|einθ

We can see that by splitting the sum and letting z = reiθ that we have:

P (z) = P (r, θ) =1− |z|2

|1− z|2=

1− r2

1 + r2 − 2r cos θ

3.3 Definition: Abel Summation

Given an infinite compex series∞∑n=0

cn, we define for all real r ∈ [0, 1):

A(r) =∞∑n=0

cnrn

We say that the series is Abel Summable to some s if limr↗1

A(r) = s.

If f(x) =∞∑

n=−∞cne

inx, we define:

[Ar(f)](θ) =∞∑

n=−∞cnr|n|einθ

Note that [Ar(f)](θ) = f ∗ P (r, θ)

18

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19

3.3.1 Lemma 3.1

The Poisson Kernel has the following properties:

1.ffl π−π P (r, θ) dθ = 1

2. There exists some M such thatffl π−π |P (r, θ)| dθ < M

3. For all δ > 0, we have limr↗1

[ˆ|θ|>δ

|P (r, θ)| dx

]= 0

4. P (r, θ) ≥ 0

5. P (r,−θ) = P (r, θ)

6. For all δ > 0, we have limr↗1

[sup|θ|>δ

P (r, θ)

]= 0

Proof

For Part 1, we note that as the series for the Poisson Kernel is uniformly convergent, we may inter-change the integral with the summation, and integrate term by term. In doing so, each term (exceptthe n = 0 term) vanishes, leaving just 1.

For Part 2, again we interchange the integral with the summation, and let M =∞∑

n=−∞r|n|, we see

that: ffl π−π |P (r, θ)| dθ =

ffl π−π

∣∣∣∣∣∞∑

n=−∞r|n|einθ

∣∣∣∣∣ dθ≤ffl π−π

∞∑n=−∞

r|n||einθ| dθ

=∞∑

n=−∞

π

−πr|n||einθ| dθ

=∞∑

n=−∞r|n|

π

−π|einθ| dθ

≤∞∑

n=−∞|r|n|

π

−πdθ

= M

as required.For Part 3, apply Part 6.

For Part 4, this is clear from the closed form P (r, θ) =1− |z|2

|1− z|2.

For Part 5, this is also clear from the fact that in the closed form P (r, θ) =1− r2

1 + r2 − 2r cos θ, the only

term with θ is an even function itself.For Part 6, given any δ, if δ ≤ |θ| ≤ π there exists some Cδ such that cos θ ≤ 1− Cδ. Note also that1 + r2 − 2r cos θ = 1 + r2 − 2r + 2r(1− cos θ) ≥ (1− r)2 + 2rCδ ≥ 2rCδ. This gives that:

supδ≤|θ|≤π

P (r, θ) ≤ 1− r2

which, in the limit, gives us our result. �

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20

3.3.2 Theorem 3.2

For and 1 ≤ p <∞, we have the following:

1. If f ∈ Lp, then [Ar(f)](θ) = f ∗ Pr → f in Lp

2. If f is continuous, then Ar(f)→ f uniformly.

3. If f ∈ L1, then limr↗1

(f ∗ P (r, θ)) = limh→0

12

(f(θ + h) + f(θ − h))

Proof

3.3.3 Theorem 3.3

Let f(θ) be a continuous function on the open unit disk D in the complex plane. Then:

1. There exists some function u such that

(a) u ∈ C2(D) ∩ C(D).

(b) 4u = 0 within D.

(c) u |∂D= f

2. The above function u is unique.

3. u(r, θ) = Pr(θ) ∗ f

Proof

We will prove parts (a) and (c) in the same step. We define:

u(r, θ) = Pr(θ) ∗ f =∞∑

n=−∞r|n|f(n)einθ

We note that this series converges absolutely and uniformly for all r < 1, and so do the derivatives ofthe series, and so u ∈ C2(D) ∩ C(D).We will be differentiating in polar coordinates, so we note that the laplacian in polar coordinates isgiven by:

4 =∂2

∂r2+

1r

∂r+

1r2

∂2

∂θ2

For n > 0, we have:

4[r|n|f(n)einθ] = f(n)[n(n− 1)rn−2einθ + nrn−2einθ − n2rn−2einθ] = 0

The n = 0 term is constant, so it vanishes when differentiated. When n < 0, we have n = −m, whichgives us the above case, and so u is harmonic on the disk. We redefine u as follows:

u(r, θ) ={Pr(θ) ∗ f r < 1f(θ) r = 1

Thus, we have proved parts (a) and (c). Assume that there exists some vr(θ) which satisfies parts (a)and (c). This gives

vr(θ) =∞∑

k=−∞ak(r)einθ where ak(r) =

π

−πvr(θ)e−inθ

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21

We know thatffl π−π(4vr(θ))e−inθ dθ = 0, as v is harmonic. Expanding the laplacian in polar coordi-

nates, we get the following:

∂2

∂r2

π

−πv(r, θ)e−inθ dθ +

1r

∂r

π

−πv(r, θ)e−inθ dθ +

1r2

π

−π

∂2

∂θ2v(r, θ)e−inθ dθ = 0

⇒ ∂2

∂r2an(r) +

1r

∂ran(r)− n2

r2an(r) = 0

Where the final integral was evaluating using integration by parts twice. This gives us the ODE:

r2a′′n(r) + ra′n(r) + n2an(r) = 0

It may be shown that the solutions to the above are given by:

an(r) ={crn + dr−n n 6= 0c+ d log r n = 0

As an(r) is bounded, we simply get an(r) = cnr|n|. This gives us

limr↗1

an(r) = limr↗1

cnr|n| =

π

−πvr(θ)e−inθ dθ = f(n)

And so, u = v by the uniqueness of fourier coefficients. �

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Chapter 4

Hilbert Spaces

4.1 Introduction

We will now study the properties of certain kinds of normed complex vector spaces, called HilbertSpaces.

4.2 Inner Products

4.2.1 An inner Product

Let H be a complex vector space. An Inner Product on H is a mapping < ·, · >: H×H → C whichis a non-degenerate, conjugate-symmetric sesquilinear form, i.e. for all x, y, z ∈ H and α, β ∈ C:

• < αx+ βy, z >= α < x, z > +β < y, z >

• < x,αy + βz >= α < x, y > +β < x, z >

• < x, y >= < y, x >

• < x, x >≥ 0 with equality iff x = 0

With an inner product, we can define a mapping ||x|| = √< x, x >, which we can see is a norm bynoting the above and the following lemma. We say that H is an Inner Product Space if an innerproduct is defined on it.

4.2.2 Lemma 4.1

If H is a complex vector space with an inner product, then for all x, y ∈ H:

| < x, y > | ≤ ||x|| · ||y||

where ||.|| is defined above.

Proof

This is solves trivially when x = y = 0, so exclude this case. Set α = sgn(< x, y >) =< x, y >

| < x, y > |and

z = αy.For any t ∈ R, we have 0 ≤< x − tz, x − tz >= ||x||2 − 2t < x, z > +t2||z||2. As this quadratic

22

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23

polynomial is always greater than or equal to 0, its roots must be complex, and so the discriminant(b2 − 4ac) must be non-positive. This gives:

| < x, z > | ≤ ||x|| · ||z||

Now, noting that < x, z >=< x,αy >= α < x, y > and that ||z|| = √< αy, αy > =√αα < y, y > =

||y||, we have:| < x, y > | ≤ ||x|| · ||y||

as required. �.

4.2.3 Corollary 4.2

Given an inner product space H, the mapping ||x|| = √< x, x > is a norm.

Proof

For all x, y ∈ H and α ∈ C, we have:

• ||αx|| = √< αx, αx > =√αα√< x, x > = |α| · ||x||

• ||x|| = √< x, x > ≥ 0 for all x

• If x 6= 0, then ||x|| > 0, as√y > 0 for all y 6= 0.

• The triangle inequality is given by noting:

||x+ y|| =√< x+ y, x+ y > =

√< x, x > + < x, y > + < y, x > + < y, y >

and so ||x+ y||2 ≤ ||x||2 + ||y||2 + 2||x|| · ||y|| = (||x||+ ||y||)2

4.3 Hilbert Spaces

4.3.1 Definition

A Hilbert Space is an inner product space that is complete with respect to the norm ||x|| =√< x, x >.

4.3.2 Lemma 4.3

Let H be a Hilbert space. If xn → x and yn → y, then < xn, yn >→< x, y >.

Proof

Given ε > 0 let N ∈ N be such that | < xn, y > − < x, y > | < ε2 and | < x, yn > − < x, y > | < ε

2 forall n ≥ N . This gives us the following:

| < xn, yn > − < x, y > | = | < xn, yn > − < x, yn > + < x, yn > + < x, y > |≤ | < xn, yn > − < x, yn > |+ | < x, yn > + < x, y > |< ε

as required. �

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24

4.3.3 Proposition 4.4: The Parallelogram Law

For any x, y ∈ H, ||x+ y||2 + ||x− y||2 = 2||x||2 + 2||y||2.

Proof

||x + y||2 + ||x − y||2 = ||x||2 + 2||x|| · ||y|| + ||y||2 + ||x||2 − 2||x|| · ||y|| + 2||y||2 = 2||x||2 + ||y||2, asrequired. �

4.3.4 Proposition 4.5: The Polarisation Identity

For all x, y ∈ H, < x, y >= 14 [||x+ y||2 + i||x+ iy||2 − ||x− y||2 − ||x− iy||2]

Proof

The proof follows immediately by expansion of the powers of the above.

4.4 Orthogonality

4.4.1 Definition

We say that two vectors are Orthogonal, denoted x ⊥ y if < x, y >= 0. Also, we say that x ⊥ E ifx ⊥ y for all y ∈ E.

Given any E ⊆ H, we have E⊥ = {x ∈ H | x ⊥ E}, called the Orthogonal Complement of E

4.4.2 Proposition 4.6

E⊥ is a closed, linear subsepace of H.

Proof

If x1, x2 ∈ E⊥ and α1α2 ∈ C, then, for any y ∈ E, we have:

< α1x1 + α2x2, y >= α1 < x1, y > +α2 < x2, y >= 0

and so the space is linear. If (xn)→ x, we have:

< x, y >=< limn→∞

xn, y >= limn→∞

< xn, y >= 0

and so x ∈ E⊥, so the space is closed.

4.4.3 Theorem 4.7: The Pythagorean Theorem

If x1, x2, . . . , xn ∈ H are mutually orthogonal, then∣∣∣∣∣∣∣∣∣∣n∑k=1

xk

∣∣∣∣∣∣∣∣∣∣2

=n∑k=1

||xk||2

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25

Proof

By direct calculation, we have the following:∣∣∣∣∣∣∣∣∣∣n∑k=1

xk

∣∣∣∣∣∣∣∣∣∣2

=<n∑k=1

xk,

n∑j=1

xj >

=n∑k=1

n∑j=1

< xk, xj >

=n∑k=1

< xk, xk >

=n∑k=1

||xk||2

where the third step follows due to the orthogonality of the vectors.

4.4.4 Definition: The Direct Sum

If M1 and M2 are two orthogonal subspaces of H, then we define the Direct Sum of M1 and M2 tobe:

M1 ⊕M2 = {z ∈ H | z = x1 + x2, x1 ∈M1, x2 ∈M2}

4.4.5 Theorem 4.8

Let M be a complete, closed subspace of H. Then

H = M ⊕M⊥

Furthermore, if z = x+ y ∈M ⊕M⊥, then x and y are the closest vectors in M1 and M2 to z, and

||z|| = ||x||+ ||y||

Proof

Let d = inf {||z − x|| : x ∈M}. We know that there exists some (xn)∞n=1 such that ||xn − z|| → d, asd is the infimum.For all ε > 0, we can choose some N ∈ N such tha ||xn− z|| < ε

4 + d2. Thus, noting that 2||z−xn||2 +2||z−xm||2 = ||xn−xm||2 + ||xn+xm−2z||2, we have ||xn−xm||2 < 2||xn−z||2 +2||xm−z||−4d2 < εThus, the sequence is cauchy and so there exists some x such ||x−z|| = d, as M is closed and complete.

4.4.6 Corollary 4.9

Let E ⊆ H. Then (E⊥)⊥ is the smallest closed subspace containing E.

Proof

Assume that there exists some N ⊆ H such that E ⊆ N ( (E⊥)⊥ and let β ∈ (E⊥)⊥ \N . Then, wehave β = x+ α with x ∈ N and α ∈ N⊥. We’ll consider the following cases:

• If α ⊥ N then α ⊥ E and so α ∈ E⊥, which can only be true if α = 0.

• If α ∈ (E⊥)⊥, then we again have α = 0.

This gives that x = β ∈ N , which is a contradiction, as required. �

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4.4.7 Definition

We say that a collection {uα}α∈I is orthonormal if:

< uα, uβ >= δαβ ={

1 α = β0 α 6= β

4.5 Definition: Gram-Schmidt

Given a sequence of linearly independent vectors (wn)∞n=1 we may construct an orthonormal collectionas follows:

• Let v1 = w1

• Let un =vn||vn||

for all integers n > 0.

• Define vn = wn −n−1∑k=1

< wn, wk > uk for integers n > 1

4.5.1 Proposition 4.10: Bessel’s Inequality

If {uα}α∈I is an orthonormal set in a Hilbert space H, and x ∈ H then:∑α∈I| < x, uα > |2 ≤ ||x||

Proof

Let F ⊆ I be finite, and consider:

0 ≤

∣∣∣∣∣∣∣∣∣∣x−∑

α∈I< x, uα > uα|

∣∣∣∣∣∣∣∣∣∣

=< x−∑α∈I

< x, uα > uα, x−∑β∈I

< x, uβ > uβ >

=< x, x > −∑α∈I

< x, uα >< uα, x > −∑β∈I

< x, uβ >< uβ, x >

+∑α∈I

∑β∈I

< x, uα >< uβ, x >< uα, uβ >

= ||x||2 −∑α∈I| < x, uα > |2 −

∑β∈I| < x, uβ > |2 +

∑β∈I| < x, uβ > |2

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Chapter 5

Pointwise Convergence

5.1 Theorem 5.1

There exists some f ∈ C0(T) whose fourier series diverges at 0.

Proof

We know that there exists (ψn)n∈N ⊂ C0(T) such that

||psin||∞ ≤ 1 |[Sm(ψn)](0)| > 12||Dn||1 >

110

log n

as we proved this in an assignment. Let φn = σn2(ψn), which is a trigonometric polynomial of degreen2. We note that

||φn||∞ = ||ψn ∗ Fn2 ||∞ ≤ ||ψn||∞ · ||Fn2 ||∞ ≤ 1

where Fn is the Fejer Kernel. We also note that:

|[Sm(φn)](t)− [Sm(ψn)](t)| =

∣∣∣∣∣m∑

k=−mψ(k)

m2 − |k|m2

eikt −m∑

k=−mψ(k)eikt

∣∣∣∣∣=

∣∣∣∣∣m∑

k=−mψ(k)

|k|m2

eikt

∣∣∣∣∣≤ 2

whence|[Sm(φn)](0)| ≥ 1

10log n− 2

Now, we consider a rapidly increasing sequence defined by λn = 24n . We see that λ3n < λn+1. Also,

this gives the following

φλn(t) =λ2n∑

k=−λ2n

akeikt

φλn(λnt) =λ2n∑

k=−λ2n

akeikλnt

φλn+1(λnt) =λ2n+1∑

k=−λ2n+1

akeikλn+1t

27

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5.2 Theorem 5.2

Let f ∈ L1[−π, π] and suppose that f is differentiable at θ0. Then, we have

limN→∞

[SN (f)](θ0) = f(θ0)

Proof

Let g(t) be defined as follows:

g(t) =

f(θ0 + t)− f(θ0)t

|t| ∈ (0, π)

f ′(θ0) t = 0

Then, there exists some δ > 0 such that for all t ∈ [−δ, δ], g is bounded. On |t| > δ, we have|g(t)| ≤ 1

δ [f(θ0 + t)− f(θ0)] ∈ L1. Hence g ∈ L1[−π, π]. We note the following:

[Sn(f)](θ0)− f(θ0) = (f ∗Dn)(θ0)− f(θ0)=ffl π−π[f(θ0 − t)− f(θ0)]Dn(t) dt

=ffl π−π[−tg(−t)]Dn(t) dt

=ffl π−π[−tg(−t)]

(t

sin t/2

)[sinnt cos t/2 + cosnt sin t/2] dt

We see that as g ∈ L1,t

sin t/2∈ L∞ and that sinnt, cosnt are fourier coeficcients, we may appply the

Riemann-Lebesgue Lemma to see that the integral vanishes as t→∞, we required. �

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Chapter 6

Weyl’s Equidistribution Theorem

6.1 Notation

For all real number x, we make the following notational defintions:

• [x] = bxc = max{y ∈ Z | y ≤ x}

• < x >= x− [x]

Observation

If γ ∈ R, we consider the sequence (< nγ >)n∈N ⊆ [0, 1). This will be periodic if γ ∈ Q, andnon-periodic otherwise.

6.1.1 Proposition 6.1

If γ /∈ Q, then the sequence (< nγ >)n∈N is dense in [0, 1).

Proof

Any sequence in a compact metric space has a limit point iff it has a convergence subsequence. Weknow that there exists some x0 such that for all sub-intervals (a, b) ⊆ [0, 1) with x0 ∈ (a, b), then|{nγ | n ∈ Z, nγ ∈ (a, b)}| =∞.

For all ε > 0, there exists some m > 0 such that | < mγ > −x0| < ε2 . Also, there exists some

n > m such that | < nγ > −x0| < ε2 . Hence, we have | < mγ > − < nγ > | < ε.

Letting p = m − n and A = {< npγ > | n ∈ N}, we see that for all x ∈ [0, 1), |x − y| < ε forsome y ∈ A. Hence, the sequence is dense in [0, 1), as required.

6.2 Definition: Equidistribution

Let (ζn)n∈N ⊆ [0, 1). The sequence is called Equidistributed if for all intervals [a, b) ⊆ [0, 1), wehave

limN→∞

|{ζn | ζn ∈ [a, b), n ∈ {1, 2, . . . , N}|N

= b− a

6.2.1 Theorem 6.2: Weyl’s Theorem

If γ /∈ Q, then < nγ > is equidistributed in [0, 1).29

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30

Proof

The theorem is true iff limN→∞

1N

N∑k=1

χ[a,b)(< kγ >) =ˆ 1

0χ[a,b)(x) dx

6.2.2 Lemma 6.3

If f ∈ C[0, 1] is periodic and γ /∈ Q, then we have

limN→∞

1N

N∑k=1

f(< kγ >) =ˆ 1

0f(x) dx

Proof

1. If f(x) = e2πinx, then we have the following

1N

N∑k=1

e2πinkγ =1N

[e2πikγ − e2πik(N+1)γ

1− e2πikγ

]

The denominator of the fraction is non-zero, and its numerator is bounded, so as N → ∞, thesum on the left goes to 0, as required.

2. If f is a trigonometric polynomial, we can split up the sums, and reduce the problem to part 1),as required.

3. If f is periodic, then there exists some trigonometric polynomial p such that ||f−p|| < ε3 . Hence,∣∣∣∣ˆ f − 1

N

∑f(nγ)

∣∣∣∣ ≤ ∣∣∣∣ˆ (f − p)∣∣∣∣+∣∣∣∣ˆ p− 1

N

∑p(nγ)

∣∣∣∣+∣∣∣∣ 1N

∑p(nγ)− 1

N

∑f(nγ)

∣∣∣∣ < ε

for large enough N , where the integrals are on [0, 1) with respect to x, and the sums are summedover k = 1 to N . Hence, result. �

Return to Proof of Theorem 6.2

We can find some f+, f− such that f− ≤ χ ≤ f+ ≤ 1, and equal on (a − ε, a + ε) and (b − ε, b + ε.Hence,

(b− a)− 2ε <ˆf− ≤

ˆχ(a,b) ≤

ˆf+ ≤ (b− a) + 2ε

Applying the previous lemma, we see that

(b− a)− 2ε <1N

∑f−(nγ) ≤

ˆχ(a,b) ≤

1N

∑f+(nγ) ≤ (b− a) + 2ε

as required. �.

6.2.3 Theorem 6.3: Weyl’s Equidistribution Theorem on [0, 1]

Let (ζn)n∈N ⊆ [0, 1]. This sequence will be equidistributed iff

limN→∞

N∑n=1

e2πinζk = 0 ∀k ∈ N

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31

Proof

We proved the first section above. Hence to prove the converse, we assume that (ζn) is equidistributed.Thus, as we know that for characteristic functions f

limN→∞

N∑k=1

f(ζk) =ˆf(x) dx

we know the same must be true for simple functions. Let f ∈ C0([0, 1]). For all ε > 0 , define

hε(x) = f(x) +1ε

We know that ||hε − f || < ε. If hε oscillates infinitely, then we can choose another function, but onedefined along the y-axis instead of the x-axis. Hence this makes hε simple. Thus∣∣∣∣∣

ˆf −

N∑k=1

f(ζk)

∣∣∣∣∣ <∣∣∣∣ˆ (f − hε)

∣∣∣∣+

∣∣∣∣∣ˆhε −

N∑k=1

hε(ζk)

∣∣∣∣∣+

∣∣∣∣∣N∑k=1

[hε(ζk)− f(ζk)]

∣∣∣∣∣ < 3ε

as required. �

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Chapter 7

The Fourier Transform on R

32

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Chapter 8

The Heat Equation

8.1 Introduction

We began this course by looking at the general Dirichlet problem for the heat equation, and showingthat solutions were in the form of trigonometric polynomials.

33

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Chapter 9

Poisson Summation and Heisenberg’sUncertainty Principle

9.1 Poisson Summation

Theorem 9.1

Let ε > 0. If we have the following

|f(x)| ≤ 1(1 + |x|)1+ε

|f(ζ)| ≤ 1(1 + |ζ|)1+ε

e.g. if f ∈ S(R), then∞∑

n=−∞f(x+ n) =

∞∑n=−∞

f(n)e2πixn

Proof

If F1 and F2 are continuous and periodic with the same period, we know that it is sufficient to provethat F1 = F2 to show that F1 = F2. Define the following

F1(x) =∞∑

n=−∞f(x+ n) F2(x) =

∞∑n=−∞

f(n)e2πinx

We can readily see thatF2(m) = f(m) ∀m ∈ Z

We note that ˆ 1

0

∞∑n=−∞

|f(x+ n)| dx ≤ˆ 1

0

∞∑n=−∞

c

(1 + |x+ n|)1+εdx

=ˆ ∞−∞

∞∑n=−∞

c

(1 + |y|)1+εdy

<∞

34

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35

We’ll compute the Fourier transform of F1:

F1(m) =ˆ 1

0F1(x)e−2πimx dx

=ˆ 1

0

∞∑n=−∞

f(x+ n)e2πimx dx

=∞∑

n=−∞

ˆ 1

0f(x+ n)e2πimx dx

=∞∑

n=−∞

ˆ 1

0f(x+ n)e2πim(x+n) dx

=ˆ ∞−∞

f(y)e2πimy dy

= f(m) = F2(m)

as required. �

Before, we had that the Poisson kernel was given by

Pr(θ) =∞∑i=0

zi +∞∑i=1

zi

where z = reiθ. We can apply the above result to the Poisson Summation Formula to the PoissonKernel:

Py(x) =1π

y

x2 + y2Py(ζ) = e−2πy|ζ|

to see that∞∑

n=−∞Py(x+ n) =

∞∑n=−∞

e−2πy|n|e2πinx

=∞∑n=0

e−2πn(y−ix) +∞∑n=1

e−2πn(y+ix)

= Pr(2πx)

for r = e−2πy.

9.2 Heisenberg’s Uncertainty

Suppose that ψ ∈ L2(R) with ||ψ||2 = 1 describles the path of a particle x along the real line. Wedefine the following statistical functions:

P [Particle is in [a, b]] =ˆ b

a|ψ(x)|2 dx

E[x] =ˆ ∞−∞

x|ψ(x)|2 dx = x

Var[x] =ˆ ∞−∞

(x− x)2|ψ(x)|2 dx

Results from physics tell us that the Fourier Transform of ψ decribes the momentum p of the particlex. We can define similar statistical functions for the momentum as above by replacing ψ(x) with ψ(p).We know that by Plancherel’s Theorem, ||ψ||2 = ||ψ||2 = 1. This gives us the following theorem.

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36

Theorem 9.2: Heisenberg’s Uncertainty Principle

Let ψ ∈ S(R),´∞−∞ |ψ(x)|2 dx = 1. Then for all x0, ζ0 ∈ R, we have[ˆ ∞

−∞(x− x0)2|ψ(x)|2 dx

] [ˆ ∞−∞

(ζ − ζ0)2|ψ(ζ)|2 dζ]≥ 1

16π2

with equality whenever ψ(x) = Ae−Bx2, where A =

(2Bπ

)1/4

and B > 0.

Proof

It is sufficient to prove the above result for particular values of x0 and ζ0, because we can shift ourfunctions accordingly for different values. We’ll take x0 = ζ0 = 0 . We note that the Cauchy-Schwarzinequality gives us ∣∣∣∣ˆ fg

∣∣∣∣ ≤√ˆ

|f |2ˆ|g|2

with equality iff f = Ag for some constant A. This gives us

1 =ˆ ∞−∞|ψ(x)|2 dx

=ˆ ∞−∞

xd

dx|ψ(x)|2 dx

= −ˆ ∞−∞

x[ψ′(x)ψ(x) + ψ(x)ψ′(x)] dx

≤ 2ˆ ∞−∞|xψ(x)ψ′(x)| dx

≤ 2

√ˆ ∞−∞x2|ψ(x)| dx

√ˆ ∞−∞|ψ′(x)| dx

The Plancherel theorem also tell us that∣∣F(ψ′)(ζ)∣∣2 = 4π2ζ2|ψ′(ζ)|2

which gives us

1 ≤ 2

√ˆ ∞−∞x2|ψ(x)| dx

√ˆ ∞−∞|ψ′(x)| dx

= 2

√ˆ ∞−∞x2|ψ(x)| dx

√4π2

ˆ ∞−∞|ζ2ψ(ζ)| dζ

= 4π

√ˆ ∞−∞x2|ψ(x)| dx

√4π2

ˆ ∞−∞|ζ2ψ(ζ)| dζ

On squaring the final line, we have the result. We have equality when we have equality by theCauchy-Schwarz, i.e. when

ψ′(x) = Bψ(x)

which happens when ψ(x) = Ae−Bx2. We have eqality in the other ineqality above whenever ||ψ||2 = 1,

which gives

||ψ||2 = A2

ˆ ∞−∞

e−2Bx2dx = 1

i.e. A2

√π

2B= 1, as required. �

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Chapter 10

The Fourier Transform on Rd

10.1 Introduction

We now turn our attention to the Foureier Transform in more than one dimension. Before we begin,however, we’ll need some new notation. We define a multi-index α = (α1, . . . , αd) to be an element ofNd with the following conventions for al x ∈ Rd:

α! = α1!α2! · · ·αd!

xα = xα11 xα2

2 · · ·xαdd(

∂x

)α=

∂α1

∂xα1

∂α2

∂xα2· · · ∂

αd

∂xαd

|α| = α1 + α2 + · · ·+ αd

Taylor’s Theorem

We may re-state Taylor’s Theorem as

f(x+ h) =∑α∈Nd

1α!∂αf

∂xαhα

10.2 Orthogonality

Definition

A d× d matrix is said to be Orthogonal if any of the following properties hold for all x, y ∈ Rd

• RTR = I

• < Rx,Ry >=< x, y >

• < Rx,Rx >=< x, x >

We note that detR = ±1 for an orthogonal matrix. We say that R is

• a Proper Rotation if detR = 1.

• an Improper Rotation if detR = −1.37

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38

In general, if T is a d × d matrix with real coefficients and non-zero determinant (the space of suchmatrices will be denoted GL(d,R)), we have

ˆf(Tx) dx =

∣∣∣∣ 1detT

∣∣∣∣ˆ f(y) dy

which is easily provable by considering the Jacobian of the change of coordinates. Hence, orthogonalmatrices preserve the integral, up to a change of sign. The above also means that if δ ∈ R, we have

ˆf(δx) dx =

∣∣∣∣ 1δd

∣∣∣∣ˆ f(y) dy

10.3 Polar Coordinates

We define the Polar Coordinates in d dimensions byˆ

Rdf(x) dx =

ˆSd−1

ˆ ∞0

f(rγ)rd−1 dr dσ(γ)

where Sd−1 is the d−1-dimensional unit ball, γ is a point on Sd−1 and σ(γ) is the surface area measure.

Case: d = 2

When d = 2, we note that a point on S1 is given by (cos θ, sin θ). Also, the surface measure amountsto the change in the angle, so dσ(γ) = dθ, giving

ˆR2

f(x, y) dA =ˆ 2π

0

ˆ ∞0

rf(r cos θ, r sin θ) dr dθ

which is indeed consistant with our knowledge or polar coordinates.

Definition: Schwarz Space

As for R, we define the Schwarz Space on Rd as

S(Rd) ={f : Rd → R | ∀α, β ∈ Nd∃M s.t.

∣∣∣∣xβ ∂α∂xα f(x)∣∣∣∣ < M

}

10.4 The Fourier Transform

Definition

Given f ∈ L1(Rd), we define

[F(f)](ζ) = f(ζ) =ˆ

Rde−2πix·ζf(x) dx

Note the dot product in the exponential.

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39

Proposition 10.1

If f ∈ S(Rd), then talking the fourier transform, we have for all h ∈ Rd and T ∈ GL(d,R)

1. f(x+ h)→ f(ζ)e2πiζ·h

2. f(x)e−2πx·h → f(ζ + h)

3.(∂

∂x

)αf → (2πiζ)αf(ζ)

4. (−2πiζ)αf →(∂

∂ζ

)αf(ζ)

5. f(Tx)→ 1detT f((T t)−1ζ)

Proof

We’ll only prove part (5), as the others follow directly from the definitions, and techniques analogousto the Fourier Transform on R. Define g(x) = f(Tx). Hence g(ζ) =

´f(Tx)e−2πix·ζ dx. Noting that

x = T−1y x · ζ =< x, ζ >=< T−1y, ζ >=< y, (T t)−1ζ >

we haveg(ζ) =

1| detT |

ˆf(y)e2πi(T

−1y)·ζ =1

| detT |f((T t)−1ζ)

as required.

Corollary 10.2

F : S(Rd)→ S(Rd)

Corollary 10.3

If f is radial, i.e. if f(Rx) = f(x) for all R ∈ GL(d,R), then so is f .

Theorem 10.4

If f, f ∈ L1(Rd), then f(x) =´f(ζ)e2πix·ζ agrees a.e. with a continuous function.

Theorem 10.5

If f ∈ L1 ∩ L2, then f ∈ L2. Furthermore, F extends to a unitary isomorphism from L2 onto L2.

Proof

The following is a sketch of the proof.

• Step 1: We have to show that F(e−π||x||2) = e−π||ζ||

2, which can be showed by diect calculation.

• Step 2: By proposition 10.1, we have F(e−πδ||x||2) = δ−d/2e−(π||ζ||2)/δ

• Step 3: Letting Kδ(x) = δ−d/2e−(π||x||2)/δ, we show that Kδ is an “extra nice” summabilitykernel, in so far as it is even, positive and decays rapidly.

• Step 4: We need to show that´f(x)g(x) =

´f(x)g(x) over Rd

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• Repeat the proof of Theorem 7.10

• Repeat the proof of Theorem 7.15

10.5 The Wave Equaion in Rd

The Wave Equation in Rd is given by

∂2u

∂x1+ · · ·+ ∂2u

∂xd=

1c2∂2u

∂t

where u = u(~x, t), and we impose some initial conditions

u(x, t) = f(x) ut(x, 0) = g(x)

for f, g ∈ S(Rd).

If we consider the Fourier Transform of, say∂2u

∂xiwith respect to x, we see that

∂2u

∂xi= (2πiζi)2u

However, if we sum these terms over all i, we can see that

(2πi|ζ|)2u =∂2

∂t2u(ζ, t)

Viewing this as an ODE with initial conditions u(ζ, 0) = f(ζ) and ut(ζ, 0) = g(ζ), we can solve thisto get

u(ζ, t) = f(ζ) cos(2π|ζ|t) +g(ζ)2π|ζ|

sin(2π|ζ|t)

10.5.1 Theorem 10.6

If f, g ∈ S(Rd), then the function

u(x, t) =ˆ

Rd

[f(ζ) cos(2π|ζ|t) +

g(ζ)2π|ζ|

sin(2π|ζ|t)]e2πix·ζ dζ

satisfies the above wave equation.

Proof

This is easily proved by direct calculation.

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Chapter 11

The Radon Transform

11.1 The Transform

Definition

Given a plane P ⊆ R3 and f : P → R, we define the Radon Transform of f by

R(f)(P ) =ˆPf

This definition gives rise to some questions:

• If R(f1)(P ) = R(f2)(P ) for all planes P , does this imply that f1 = f2?

• If we know R(f)(P ) for all planes P , can we find f?

• If we know R(f)(P ) for pome finite collections of planes P , can we approximate f?

Assuming that f ∈ S(R3), we can represent a plane in the form

t~γ + u1 ~e2 + u2 ~e2

where ~γ, ~e2, ~e2 are orthonormal. This gives us the Radon Transform of f as

R(f)(t, γ) =ˆ ∞−∞

ˆ ∞−∞

f(t~γ + u1 ~e2 + u2 ~e2) du1 du2

Lemma 11.1

If f ∈ S(R3), then as a function of t, R(f)(t, γ) ∈ S(R3) for all γ. Moreover,

R(f)(s, γ) = f(sγ)

where the Fourier Transform on the left is one dimensional and the Fourier Transform on the right isthree dimensional. Note the following notation for the above:

F1 (R(f)(t, γ)) |t=s = F3(f(t)) |t=sγ

Proof

For all N ∈ N, there exists some AN ∈ R such that

(1 + |t|)N (1 + |~u|)N |f(t~γ + ~u| ≤ AN

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Corollary

If f1, f2 ∈ S(R3) and R(f1)(P ) = R(f2)(P ) for all planes P , then f1 = f2.

Proof

We know that (f1 − f2)(sγ) = [R(f1)−R(f2)](s, γ) = 0. Hence, by the Fourier inversion theorem, wehave to have f1 − f2 = 0 as required.

11.2 The Dual Radon Transform

Definition

Given a plane R× S2, we define the Dual Radon Transform of F denoted R∗F : R3 → R by

R∗F (x) =ˆS2

F (x · γ, γ) dσ(γ)

Theorem 11.3

If f ∈ S(R3), then

f = − 18π24(R∗Rf)

Proof

Since we have R(f)(s, γ) = f(sγ), we apply the Fourier inversion theorem to see that

Rf(t, γ) =ˆ

Rf(sγ)e2πisx ds

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Chapter 12

Finite Fourier Analysis

12.1 Introduction

We will consider Fourier Analysis on finite groups, instead of infinite ones. From here on in, we willmake use of the following notation:

Z(N) = {1, ω, . . . , ωN−1} ω = e2πi/N

Z/NZ = {0, 1, . . . , N − 1}

Here, ω is a complex root of unity. To perform fourier analysis, we want analogues to the functionsen(x) = e2πinx, defined on [0, 1). We note the following about the above funcions:

• {en | n ∈ Z} forms an orthonormal basis for L2[0, 1].

• Finite linear combinations of these functions are dense in Cper[0, 1].

• Since en(x + y) = en(x)en(y), we have that en : ([0, 1],+) → (T,×) is a group homomorphism,where T is the unit circle in the complex plane.

• en(x)em(x) = en+m(x)

On Z(N), we define similar functions as el(ωk) = ωlk, where 0 ≤ n, k ≤ N . Let V be the space of allmappings f : Z(N) → C. This space is an N dimensional vector space over R. We define an innerproduct on this space to make it an inner product space, namely

< F,G >=1N

N−1∑k=0

F (ωk)G(ωk)

This is analogous to the standard inner product on Lp(T) given by

< f, g >=1

ˆ 2π

0f(eiθ)g(eiθ) dθ

Lemma 12.1

{ek}N−1k=1 is an orthonormal basis for V .

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Proof

By direct computation, we have

< ei, ej > =1N

N−1∑k=0

ei(ωk)ej(ωk)

=1N

N−1∑k=0

ωikωjk

=1N

N−1∑k=0

(ωi−j

)k=

1N

1− (ωi−j)N

1− ωi−j

={

1 i = j0 i 6= j

as required �.

Corollary 12.1

If f ∈ V , then we have f =N−1∑k=0

< f, ek > ek.

Also, we have the Plancherel formula: ||f ||2 =1N

N−1∑k=0

|F (ωk)|2 =N−1∑k=0

| < f, ek > |2

12.2 The Fourier Transform

Given some f ∈ V , we define its Fourier Transform as

f(k) =1N

N−1∑j=0

f(ωj)ωjk

Theorem 12.3

If N = 2n for some n, then we can calculate the fourier transform of a function in 4n2n steps, i.e. inO(N logN) time.

Before we prove this theorem, we note the following definition and lemma:

Definition: Notation

We’ll denote by #(M) the minimum number of operations needed to calculate the fourier transformon Z(M).

Lemma 12.3.1

#(2M) ≤ 2#(M) + 8M , given ωM = e2πi/M .

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Proof

To calculate ω22M , ω

32M , . . . , ω

2M−12M , we require 2M − 2 multiplications. Suppose F is a function on

Z(2M). Define F0(k) and F1(k) on Z(M) by

F0(k) = F (2k) F1(2k + 1)

We note the following

a2Mk (F ) =

12M

2M−1∑l=0

F (l)ωlk2M

=1

2M

M−1∑j=0

F0(j)ωjkM +M−1∑j=0

F1(j)ωjkMωk2M

=

12[aMk (F0) + aMk (F1)ωk2M

]Thus, to calculate #(2M), this amounts at most to calculating #(M) twice. However, inside out lastexpression above, we have 3× (2M) + (2M − 2) operations. Hence, result, as required �.

Proof of Theorem 12.3

We’ll prove this by induction on n. For n = 1, we have ω2 = −1. On Z/2Z, we have

F (0) =12

[F (0) + F (1)] F (1) =12

[F (0)(−1)2 + F (1)(−1)1]

The above can be calculated in 5 operations, and as 5 ≤ 8, we have the theorem is true for n = 1.

Assume that the theorem holds for n. We have

#(2n+1) ≤ 2#(2n) + 8 · 2n ≤ 2[4n2n] + 8 · 2n ≤ 4(n+ 1)2n+1

as required �.

12.3 Abelian Groups

Henceforth, let G be an Abelian group.

Definition: Character

A character on a topological Abelian group G is a (continuous) homomorphism from G to T. The setof all characters of G, called the dual group of G, will be denoted G. We’ll denote by χ0 the trivialcharacter χ0(x) = 1 for all x ∈ G.

Examples

Some examples of characters and their dual groups are:

• On R, eζ(x) = eiζx, so R ∼= R

• On T, en(λ) = λn, so (T ) ∼= Z

• On Z/NZ, ek(n) = e2πikn/N , so Z/NZ ∼= Z/NZ

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46

Lemma 12.4

G is a group.

Proof

First, we must define an operation on the group. We define the product of two characters ei, ej ∈ Gas

[ei · ej ](x) = ei(x)ej(x) ∀x ∈ GAssociativity follows from the associativity of G. Clearly, the trivial character is an identity element,and we define the inverse of a character by

e−1i (x) = ei(x) ∀x ∈ G

which gives us inverses �.

It is not too difficult to prove that G = G for all locally compact Abelian group. Note that wetopologize G with the topology of uniform convergence.

Theorem 12.5

The characters of G form an orthonormal basis for l2(G) with respect to the inner product

< f, g >=1|G|

∑x∈G

f(x)g(x)

Proof

Since |e(x)| = e(x)e(x) = 1 for any non-trivial character e, we have

< e, e >=1|G|

∑x∈G

e(x)e(x) = 1

If ei 6= ej , then we have

< ei, ej >=1|G|

∑x∈G

ei(x)ej(x) =1|G|

∑x∈G

[eie−1j ](x)

We finish the proof by a lemma.

Lemma 12.5.1

If e is a non-trivial character of an Abelian group G, then∑x∈G

e(x) = 0

Proof

If x, y ∈ G, as G is a group, there exists a bijection x 7→ xy. Thus,∑x∈G

e(x) =∑x∈G

e(xy) =∑x∈G

e(y)e(x) = e(y)∑x∈G

e(x)

However, this can only hold when∑x∈G

e(x) = 0, as e is non-trivial.

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47

Definition: Fourier Coefficients

The Fourier Coefficients of f ∈ V are

f(e) =1|G|

∑x∈G

f(x)e(x)

Theorem 12.6

The following relations are analogues of the Fourier Inversion Theorem and Plancherel’s Theorem:

f(x) =∑e∈G

f(e)e(x)

and||f ||2 =

∑e∈G

|f(e)|2

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Chapter 13

Dirichlet’s Theorem

13.1 Introduction

We’ll adopt the notation that the set of all primes is P.

Euclid’s Theorem 13.1

|P| =∞

Proof

Assume that P = {p1, p2, . . . , pn}, written in ascending order. We consider the number M =p1p2 · · · pn + 1. By the fundamental theorem of arithmetic, we have that either M is expressableas a product pf prime powers, or is prime itself.

As M > pn we have M /∈ P. Thus, M must have a prime factor, say pi. However, if pi|M , thisimplies that pi|1 as pi|p1p2 · · · pn, which is a contradiction.

Proposition 13.2

The set S3 = {x ∈ P ‖ x ≡ 3 mod 4} is infinite.

Proof

As above, assume the contrary, that S3 = {p1, . . . , pn} and let M = p2p3 · · · pn+ 3. Note that we startfrom 2 because p1 = 3, and we exclude this case. We note that M ≡ 3 mod 4.

We have 2 - M , as 2 - 3. Also, as 3 - p2p3 · · · pn, 3 - M also. We note that M /∈ S3, as it isbigger than all of the elements of S3. Thus, all of the prime factors of M are equivalent to 1 mod 4.However, if a = 4n + 1 and b = 4m + 1, then ab = 4(4nm + n + m) + 1, and so ab ≡ 1 mod 4. Thisimplies that M ≡ 1 mod 4, which is a contradiction.

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