Refers to Fourier Analysis by Stein

# The isoperimetric inequality

Theorem 4.1.1 Suppose that Γ\Gamma is a simple closed curve in R2\mathbb R^2 of length ll, and let A\mathcal A denote the area of the region enclosed by this curve. Then we have the isoperimetric inequality

Al24π\mathcal A\leq \dfrac{l^2}{4\pi}

with equality if and only if Γ\Gamma 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\gamma is a mapping

γ:[a,b]R2\gamma:[a,b]\to \mathbb R^2

then Γ=γ([a,b])\Gamma=\gamma([a,b]) is a curve which is the image of γ\gamma. The curve Γ\Gamma 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)\gamma(s_1)\neq \gamma(s_2)

unless s1=a,s2=bs_1=a,s_2=b, where γ(a)=γ(b)\gamma(a)=\gamma(b). We also impose it is regular, which means

γC1([a,b])\gamma\in C^1([a,b])

γ0,s[a,b]\gamma'\neq 0,\ \forall s\in[a,b]

Clearly, the conditions that Γ\Gamma be closed and simple are independent of the chosen parametrization.

Definition 4.1.3 If Γ\Gamma is parametrized by γ(s)=(x(s),y(s))\gamma(s)=(x(s),y (s)), then the length ll of the curve Γ\Gamma is defined by

l=abγ(s)ds=ab(x(s)2+y(s)2)1/2dsl=\int^b_a|\gamma'(s)|\mathrm ds=\int^b_a(x'(s)^2+y'(s)^2)^{1/2}\mathrm ds

The equation is guaranteed by the smoothness of γ\gamma. And the length of Γ\Gamma is a notion intrinsic to the curve, not depending on its parametrization. So we can use a parametrization by arclength γ\gamma to parametrize Γ\Gamma if γ(s)=1|\gamma'(s)|=1 for all ss.

Definition 4.1.4 The area A\mathcal A of the simple closed curve Γ\Gamma is given by the formula of the calculus

A=12Γ(xdyydx)=12abx(s)y(s)y(s)x(s)ds\mathcal A=\dfrac 12\left|\int_{\Gamma}(x\mathrm dy-y\mathrm dx)\right|=\dfrac 12 \left|\int^b_ax(s)y'(s)-y(s)x'(s)\mathrm ds\right|

At the beginning of the proof, we give Wirtinger inequation

Lemma 4.1.5 Suppose fC1(S1)f\in C^1(S^1), then we have Wirtinger inequation

ππ(ffˉ)2dxππf2dx\int^\pi_{-\pi}(f-\bar f)^2\mathrm dx\leq \int^\pi_{-\pi}f'^2\mathrm dx

where

fˉ=12πππfdx\bar f=\dfrac1{2\pi}\int^\pi_{-\pi}f\mathrm dx

with equality if and only if its Fourier coefficient

an=bn,n2a_n=b_n,\ n\geq 2

that is to say

f=fˉ+acosx+bsinxf=\bar f+a\cos x+b\sin x

Proof of Lemma 4.1.5

Tips: Parseval's identity

Note that

ffˉ=f14a02||f-\bar f||=||f||-\dfrac 14a^2_0

Proof of Theorem 4.1.1

Tips: Wirtinger inequation, Arclength parametrization

Without loss of generality, let l=2πl=2\pi. Now choose tt to be arclength parametrization of γ=(x(t),y(t))\gamma=(x(t),y(t)), then

l=02πγ(t)dtl=\int^{2\pi}_0\mathrm |\gamma'(t)|\mathrm dt

Note that

02πxydt=1202π[(xxˉy)2+(xxˉ)2+y2]dt\int ^{2\pi}_0xy'\mathrm dt=\dfrac12\int^{2\pi}_0[-(x-\bar x-y')^2+(x-\bar x)^2+y'^2]\mathrm dt

the rest of the proof is trivial.

# Weyl's equidistribution theorem

Definition 4.2.1 If xRx\in \mathbb R, we let [x][x] denote the greatest integer less than or equal to xx and call the quantity [x][x] the integer part of xx. The fractional part of xx is then defined by

[0,1)x=x[x],xR[0,1)\ni \langle x\rangle=x-[x],\ \forall x\in\mathbb R

The rational number case is trivial, while for irrational number case Kronecker told that if γR\Q\gamma\in\mathbb R-\backslash \mathbb Q, then {nγ}\{\langle n\gamma \rangle \} is dense in the interval [0,1)[0,1). There is a more powerful result below.

Definition 4.2.2 A sequence of numbers {ξn[0,1)}n=1\{\xi_n\in [0,1)\}^\infty_{n=1} is said to be equidistributed if for every interval (a,b)[0,1)(a,b)\subset [0,1)

limN#{1nN:ξn(a,b)}N=ba\lim_{N\to\infty}\dfrac{\#\{1\leq n\leq N:\xi_n\in(a,b)\}}{N}=b-a

where #A\# A denotes the cardinality of the finite set AA.

Theorem 4.2.3 If γR\Q\gamma\in\mathbb R\backslash\mathbb Q, then the sequence of fractional parts {nγ}\{\langle n\gamma\rangle\} is equidistributed in [0,1)[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)(a,b)\subset [0,1) and let χ(a,b)(x)\chi _{(a,b)}(x) denote the characteristic function of the interval (a,b)(a,b), and extend this function to R\mathbb R by periodicity. We find

#{1nN:nγ(a,b)}=n=1Nχ(a,b)(nγ)\#\{1\leq n\leq N:\langle n\gamma\rangle \in (a,b)\}=\sum^N_{n=1}\chi_{(a,b)}(n\gamma)

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)f^-_\epsilon(x)\leq \chi_{(a,b)}(x)\leq f_\epsilon^+(x)

and

ba2ϵ01fϵ(x)dx,01fϵ+(x)dxba+2ϵb-a-2\epsilon\leq \int^1_0f^-_\epsilon(x)\mathrm dx,\quad \int^1_0f^+_\epsilon(x)\mathrm dx\leq b-a+2\epsilon

then

1Nn=1Nfϵ(nγ)1Nn=1Nχ(a,b)(nγ)1Nn=1Nfϵ+(nγ)\dfrac1N\sum^N_{n=1}f^-_\epsilon(n\gamma)\leq \dfrac1N\sum^N_{n=1}\chi_{(a,b)}(n\gamma)\leq \dfrac1N\sum^N_{n=1}f^+_\epsilon(n\gamma)

Let NN\to \infty, and then we proof the theorem.

Lemma 4.2.4 If fC(S1(1))f\in C(S^1(1)), and γR\Q\gamma\in \mathbb R\backslash \mathbb Q, then

1Nn=1Nf(nγ)01f(x)dx,asN\dfrac 1N\sum^N_{n=1}f(n\gamma)\to \int^1_0f(x)\mathrm dx,\text{ as } N\to\infty

Proof of Lemma 4.2.4

Tips: Interpolation, Uniform approximation

Note that for all trigonometric polynomial PP whose period is 1 and γR\Q\gamma\in\mathbb R\backslash \mathbb Q

01P(x)dx=0\int^1_0P(x)\mathrm dx=0

and since it's continuous

1Nn=1NP(nγ)=1Nn=1Nm=P^(m)e2πimnγ=1Nm=P^(m)e2πimγ(1e2πiNmγ)1e2πimγ\dfrac1N\sum^N_{n=1}P(n\gamma)=\dfrac1N\sum^N_{n=1}\sum^\infty_{m=-\infty}\hat P(m)e^{2\pi imn\gamma}=\dfrac1N\sum^\infty_{m=-\infty}\hat P(m)\dfrac{e^{2\pi im\gamma}(1-e^{2\pi iNm\gamma})}{1-e^{2\pi im\gamma}}

With Parseval identity, it is obvious that as NN\to \infty, LHS0\mathrm{LHS}\to 0.

Then apply Corollary 2.5.7.

Corollary 4.2.5 If fR(S1(1))f\in\mathcal R(S^1(1)), and γR\Q\gamma\in\mathbb R\backslash\mathbb Q, then

1Nn=1Nf(nγ)01f(x)dx,asN\dfrac1N\sum^N_{n=1}f(n\gamma)\to \int^1_0f(x)\mathrm dx,\text{ as }N\to\infty

Proof of Corollary 4.2.5

Tips: Riemann sum