Refers to Fourier Analysis by Stein
# The isoperimetric inequality
Theorem 4.1.1 Suppose that Γ is a simple closed curve in R2 of length l, and let A denote the area of the region enclosed by this curve. Then we have the isoperimetric inequality
A≤4πl2
with equality if and only if Γ is a circle.
The heart of the Theorem 4.1.1 is parametrization, and we now declare some concepts mentioned above.
Definition 4.1.2 A parametrized curve γ is a mapping
γ:[a,b]→R2
then Γ=γ([a,b]) is a curve which is the image of γ. The curve Γ is said to be simple if it does not intersect ifself, and closed if its two end-points coincide. That is to say
γ(s1)=γ(s2)
unless s1=a,s2=b, where γ(a)=γ(b). We also impose it is regular, which means
γ∈C1([a,b])
γ′=0, ∀s∈[a,b]
Clearly, the conditions that Γ be closed and simple are independent of the chosen parametrization.
Definition 4.1.3 If Γ is parametrized by γ(s)=(x(s),y(s)), then the length l of the curve Γ is defined by
l=∫ab∣γ′(s)∣ds=∫ab(x′(s)2+y′(s)2)1/2ds
The equation is guaranteed by the smoothness of γ. And the length of Γ is a notion intrinsic to the curve, not depending on its parametrization. So we can use a parametrization by arclength γ to parametrize Γ if ∣γ′(s)∣=1 for all s.
Definition 4.1.4 The area A of the simple closed curve Γ is given by the formula of the calculus
A=21∣∣∣∣∣∫Γ(xdy−ydx)∣∣∣∣∣=21∣∣∣∣∣∣∫abx(s)y′(s)−y(s)x′(s)ds∣∣∣∣∣∣
At the beginning of the proof, we give Wirtinger inequation
Lemma 4.1.5 Suppose f∈C1(S1), then we have Wirtinger inequation
∫−ππ(f−fˉ)2dx≤∫−ππf′2dx
where
fˉ=2π1∫−ππfdx
with equality if and only if its Fourier coefficient
an=bn, n≥2
that is to say
f=fˉ+acosx+bsinx
Proof of Lemma 4.1.5
Tips: Parseval's identity
Note that
∣∣f−fˉ∣∣=∣∣f∣∣−41a02
Proof of Theorem 4.1.1
Tips: Wirtinger inequation, Arclength parametrization
Without loss of generality, let l=2π. Now choose t to be arclength parametrization of γ=(x(t),y(t)), then
l=∫02π∣γ′(t)∣dt
Note that
∫02πxy′dt=21∫02π[−(x−xˉ−y′)2+(x−xˉ)2+y′2]dt
the rest of the proof is trivial.
# Weyl's equidistribution theorem
Definition 4.2.1 If x∈R, we let [x] denote the greatest integer less than or equal to x and call the quantity [x] the integer part of x. The fractional part of x is then defined by
[0,1)∋⟨x⟩=x−[x], ∀x∈R
The rational number case is trivial, while for irrational number case Kronecker told that if γ∈R−\Q, then {⟨nγ⟩} is dense in the interval [0,1). There is a more powerful result below.
Definition 4.2.2 A sequence of numbers {ξn∈[0,1)}n=1∞ is said to be equidistributed if for every interval (a,b)⊂[0,1)
N→∞limN#{1≤n≤N:ξn∈(a,b)}=b−a
where #A denotes the cardinality of the finite set A.
Theorem 4.2.3 If γ∈R\Q, then the sequence of fractional parts {⟨nγ⟩} is equidistributed in [0,1), which is Weyl's equidistribution theorem.
Proof of Theorem 4.2.3
Tips: Characteristic function
Note that if we fix (a,b)⊂[0,1) and let χ(a,b)(x) denote the characteristic function of the interval (a,b), and extend this function to R by periodicity. We find
#{1≤n≤N:⟨nγ⟩∈(a,b)}=n=1∑Nχ(a,b)(nγ)
Then to apply Lemma 4.2.4 to prove the theorem, we choose two continuous periodic functions which fit both
fϵ−(x)≤χ(a,b)(x)≤fϵ+(x)
and
b−a−2ϵ≤∫01fϵ−(x)dx,∫01fϵ+(x)dx≤b−a+2ϵ
then
N1n=1∑Nfϵ−(nγ)≤N1n=1∑Nχ(a,b)(nγ)≤N1n=1∑Nfϵ+(nγ)
Let N→∞, and then we proof the theorem.
Lemma 4.2.4 If f∈C(S1(1)), and γ∈R\Q, then
N1n=1∑Nf(nγ)→∫01f(x)dx, as N→∞
Proof of Lemma 4.2.4
Tips: Interpolation, Uniform approximation
Note that for all trigonometric polynomial P whose period is 1 and γ∈R\Q
∫01P(x)dx=0
and since it's continuous
N1n=1∑NP(nγ)=N1n=1∑Nm=−∞∑∞P^(m)e2πimnγ=N1m=−∞∑∞P^(m)1−e2πimγe2πimγ(1−e2πiNmγ)
With Parseval identity, it is obvious that as N→∞, LHS→0.
Then apply Corollary 2.5.7.
Corollary 4.2.5 If f∈R(S1(1)), and γ∈R\Q, then
N1n=1∑Nf(nγ)→∫01f(x)dx, as N→∞
Proof of Corollary 4.2.5