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TRANSCRIPT
Harmonic AnalysisCathal Ormond
1
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
3
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
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
5
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].
6
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
7
We may also define the complex fourier coefficients of f by the mapping f : Z → C, given byf(n) = 1
2π
´ 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. �
8
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. �
9
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
10
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
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→∞
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 )
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.
14
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. �
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
16
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
17
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.
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
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
Cδ
which, in the limit, gives us our result. �
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
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. �
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
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. �
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
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. �
26
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
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
28
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. �
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
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
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. �
Chapter 7
The Fourier Transform on R
32
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
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
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.
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. �
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
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.
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
40
• 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.
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
*********41
42
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
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
2π
ˆ 2π
0f(eiθ)g(eiθ) dθ
Lemma 12.1
{ek}N−1k=1 is an orthonormal basis for V .
43
44
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 .
45
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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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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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
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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