Lecture 6

Today we study some corollaries of mean square convergence and applications.

Parsval identity¶

$\|f\|_{L^2}^2 = \|f - S_N f\|_{L^2}^2 + \|S_N f\|^2_{L^2} \geq \|S_N f\|_{L^2}^2 = \sum_{i = -N}^N |\hat{f}(n)|^2$. This is the special case of Bessel inequality.

Then the series $\sum_{n = -\infty}^{+\infty}|\hat{f}(n)|^2$ converges. Since $\|f - S_N f\|^2 \to 0$, we obtain

Parsval’s identity¶

Let $f\in L^2(\mathbb{T})$. Then $\|f\|^2 = \sum_{n = -\infty}^{+\infty}|\hat{f}(n)|^2$.

Riemann-Lebesgue lemma¶

The Riemann-Lebesgue lemma states that if $f\in L^1(\mathbb{T})$ then $|\hat{f}(n)|\to 0$ when $|n|\to \infty$. In particular, this implies that for an integrable function on the circle, $\int_{0}^{2\pi }f(x)\sin(nx) dx \to 0$ and $\int_{0}^{2\pi}f(x)\cos(nx)\to 0$ when $|n|\to \infty$.

Proof.

If $f$ is continuous, then the Parsval identity implies the positive infinite series $\sum_{N = -\infty}^{+\infty}|\hat{f}(n)| = \|f\|_{L^2}\leq +\infty$ so this implies $|\hat{f}(n)|\to 0$ when $|n|\to \infty$.

In general, let $g$ be a continuous function such that $\|f-g\|_{L^1}<\epsilon$. Then we have $|\hat{f}(n)-\hat{g}(n)| = |\int_{0}^{1}(f(x)-g(x))e^{-2\pi i n x}dx|\leq \int_{0}^1|f(x)-g(x)|dx = \|f-g\|_{L^1} < \epsilon$. So $|\hat{g}(n)|\to \infty$ implies $|\hat{f}(n)|\to \infty$.

Wirtinger inequality¶

Let $f$ be a $2\pi$-periodic function with mean value 0 (i.e. $\hat{f}(0) = \int_{0}^{2\pi}f(x)\frac{dx}{2\pi}$ = 0 ). Then $\|f'\|_{L^2}\geq \|f\|_{L^2}$, i.e. $\int_{0}^{2\pi}|f'(x)|^2\frac{dx}{2\pi}\geq \int_{0}^{2\pi}|f(x)|^2 \frac{dx}{2\pi}$.

Proof. By the derivative theorem $\hat{f'}(n) = i n \hat{f}(n)$. Then the Parsval identity implies $\|f'\|^2_{L^2} = \sum_{n\in \mathbb{Z}} n^2|\hat{f}(n)|^2$.

We apply the simple fact that multiplying a strictly positive number by $n^2$ will make it bigger for $|n|\geq 2$ and equal when $|n| = 1$, so when $\hat{f}(0) = 0$, $\|S_N(f)\|_{L^2}^2 = \sum_{|n|\leq N,n\neq 0}|\hat{f}(n)|^2\leq \sum_{n\in \mathbb{Z},n\neq 0}n^2|\hat{f}(n)|^2 = \|S_N(f')\|_{L^2}^2$. If $\hat{f}(n) = 0$ for all $|n|\geq 2$, the above inequality will be equality (converse also true). Then let $N\to \infty$, the left hand side goes to $\|f\|_{L^2}^2$, the right hand side goes to $\|f'\|_{L^2}^2$.

As remarked, the inequality holds if and only if $\hat{f}(n) = 0$ for all $n$ except $|n|=1$, so $f(x) = \hat{f}(1)e^{2\pi i x}+f(-1)e^{-2\pi i x}$. If $f$ is real-valued, then $f(x) = a\sin(x)+b\cos(x)$.

Isoperimetric inequality¶

The isoperimetric inequality says of all closed curves on the plane with the same length, the circle has the largest area.

Some differential geometry¶

A regular curve is a $\mathscr{C}^1$ mapping $\gamma:[a,b]\to \mathbb{R}^2$, $t\mapsto (x(t),y(t)):=\gamma(t)$, such that $\gamma'(t) =(x'(t),y'(t))\neq \mathbf{0}$ for every $t\in [a,b]$. Here the $\mathscr{C}^1$ means both $x(t)$ and $y(t)$ has continuous first derivatives.

The length of the curve is given by the line integral $\int_{a}^b |\gamma'(t)|dt = \int_{a}^b \sqrt{(x'(t)^2+y'(t)^2)}dt$.

A reparametrization of the curve is determined by composition $\gamma(t(s)):[c,d]\mapsto \mathbb{R}^2$, where $t(s)$ is a $\mathscr{C}^1$ function $s\in [c,d]\mapsto t(s)\in [a,b]$, such that $t(c) = a, t(b) = d$ and $t'(s)\neq 0$ for every $s$. In particular, $t(s)$ is invertible, and either strictly increasing or strictly decreasing. If $t'(s)> 0$, the reparametrization keeps orientation, and reverses the orientation if $t'(s)< 0$.

The length $l$ of a curve is invariant under reparametrization.

An arc-length parametrization of the curve $C$ is $\gamma(s):[0,l]\to C$ such that $|\gamma'(s)|\equiv 1$. Any regular curve can be arc-length parametrized. To see this, let $\gamma(t):[a,b]\to C$ be any parametrization of the curve. Let $s(t) = \int_{a}^t |\gamma'(x)|dx$ be the length up to $\gamma(t)$. For every $t_0\in (a,b)$, $s'(t_0) = |\gamma'(t_0)|\neq 0$ because the curve is regular, then by the inverse function theorem, the function $s(t)$ has inverse, denoted by $t(s)$ such that $s(t(s)) = s$, on a neighbourhood of $s(t_0)$. Repeating this step we get a reparemetrization $\gamma(t(s))$, which is piecewisely defined on $[0,l]$, such that $|\frac{d}{ds}\gamma(t(s))| = |\gamma'(t(s))||t'(s)| = |\gamma'(t(s))|\frac{1}{|\gamma'(t(s))|} = 1$. This is the arc-length parametrization of $C$.

A curve $\gamma(t)$ is closed provided that $\gamma(a) = \gamma(b)$. A simple closed curve is a closed curve without self intersection. Let $C$ be a regular simple closed curve, then $C$ surronds a domain $D$ with $\partial D = C$. By the Green formula, the area of $D$ can be calculated by the line integral $\frac{1}{2}|\int_Cxdy - ydx|$. One can also verify this by Stokes theorem.

Statement and proof of isoperimetric inequality¶

Theorem¶

Let $C$ be a simple closed curve on $\mathbb{R}^2$ with length $l$. Let $A$ be the area of the region enclosed by $C$. Then $l^2\geq 4\pi A$, and the equality holds if and only if $C$ is a circle.

Proof of Isoperimetric inequality¶

Proof of inequality¶

We prove the basic case when $l=2\pi$, so the inequality reduces to $A\leq \pi$. The general case follows directly from the basic case by scaling.

If $C$ is the unit circle, then $l=2\pi$, $A = \pi$ so $4\pi A = 4\pi^2 = l^2$.

Let $C$ be a $\mathscr{C}^1$ curve.

Let $\gamma(s):[0,2\pi]$ be the arc-length parametrization of $C$. Write $\gamma(s) = (x(s),y(s))$, then $x(s), y(s)\in \mathscr{C}^1(S^1)$. Since $\gamma(s)$ is arc-length parametrization, we have $|\gamma'(s)|\equiv 1$ i.e. $x'(s)^2 + y'(s)^2 \equiv 1$, so $\frac{1}{2\pi}\int_{0}^{2\pi}x'(s)^2 + y'(s)^2 dx = 1$, this implies

where $\|\cdot\|$ denote the $L^2$-norm on the circle.

By (1),

where $(\cdot)$ is the inner product of functions on the circle. We want to show that $A\leq 2\pi$, so it reduces to show that $|(x,y') - (y,x')|\leq 1$ and find condition of equality.

Since $x(s),y(s)$ are $\mathscr{C}^1$ functions, $x(s) = \sum_{n = -\infty}^{+\infty}a_ne^{ins}$ and $y(s) = \sum_{n = -\infty}^{+\infty}b_n e^{ins}$. By the derivative theorem, $x'(s) = \sum_{n = -\infty}^{+\infty}a_nin e^{ins}$ and $y'(s) = \sum_{n = -\infty}^{+\infty}b_n in e^{ins}$. Note that here writing “=” is an abuse of notation because the Fourier series of $x'$ and $y'$ may not converge pointwise but here we just use it to calculate integrals, which will not cause problem.

Then the Parsval identity and (2) implies $\sum_{n = -\infty}^{+\infty}n^2(|a_n|^2+|b_n|^2) = 1$. Moreover, $(x,y') = \sum_{n = -\infty}^{+\infty}a_n \overline{inb_n}$, $(y,x') = \sum_{n = -\infty}^{+\infty} b_n \overline{in a_n}$ (this is a generalization of the finite dimensional identity, but can be proved using Paraval and the polarization identity), see Page 81, lemma 1.5 of Stein-Shakarchi).

Now we can write

The inequality has been proved.

Condition of equality¶

Now we analyze the condition when the equality holds. First, if $C$ is a circle, we know the equality holds. Conversely, suppose the equality hold, necessarily the (a)(b)(c) part should take equal.

When $(c)$ takes “=”, the summation should not have any term with $|n|\geq 2$. So $x(s) = a_0 + a_1e^{is} + a_{-1}e^{-is}$, $y(s) = b_0 + b_1 e^{is} + b_{-1}e^{-is}$. Note that $x,y$ are real functions, so $a_{-1} = \overline{a_1}$, $b_{-1} = \overline{b_1}$, $a_0$ and $b_0$ are real. Let $a_1 = r_1 e^{i\alpha}$, $b_1 = r_2 e^{-i\beta}$. Then $x(s) = a_0 + r_1 \cos(\alpha + s)$, $y = b_0 + r_2 \cos(\beta + s)$. Now to show that $\gamma(s) = (x(s),y(s))$ is a circle, it reduces to show that $r_1 = r_2$ and $\alpha - \beta = \frac{k\pi}{2}$. These shall follow from equality (b) and (a).

When $(b)$ takes “=”, we conclude that $|a_1| = |b_1|$, this implies $r_1 = r_2 = \frac{1}{2}$.

When $(a)$ takes “=”, by writing the equation explicitly we get $|a_1\overline{b_1} - b_1 \overline{a_1}| = 2|a_1||b_1| = 2\cdot \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{2}$. Then plug in $a_1 = \frac{1}{2}e^{i\alpha}$ and $b_1 = \frac{1}{2}e^{i\beta}$ to the left hand side we get $|\frac{1}{4}\cdot 2i \sin(\alpha- \beta)| = \frac{1}{2}$, so $|\sin(\alpha - \beta)| = 1$. This implies $\alpha - \beta = \frac{k\pi}{2}$. The proof is complete.

Proof of isoperimetric inequality using Wirtinger’s inequality¶

Still, we consider the basic case $l = 2\pi$ and take arc-length parametrization $(x(t),y(t))$ of $C$. The idea is to apply Wirtinger’s inequality to $x(t)$, so we need its mean value to be zero, this can be done by replacing $x(t)$ by $x(t)-\int_{0}^{2\pi}x(t)dt$.

Also, the area $A = \frac{1}{2}\int_{C}xdy - ydx = \int_{0}^{2\pi}xdy$. To see this, integration by parts implies $\int_C xdy = \int_{0}^{2\pi}x(t)dy(t) = x(t)y(t)|_{0}^{2\pi} - \int_{0}^{2\pi}y(t)dx(t) = 0 - \int_C ydx$.

Hence

The first term is $\geq 0$ by Wirtinger inequality, the second is automatically $\geq 0$. If the curve $C$ acheives equality $l^2 = 4\pi A$, then both terms should be zero. The first term is zero when $x(t) = a \sin t + b \cos t$; The second term is zero when $y' = x$, i.e. $y = -a \cos t + b \sin t$. Then $x'^2+y'^2 = 1$ implies $a^2+b^2 = 1$, so $a = \sin \alpha, b = \cos\alpha$ for some α. So $x(t) = \cos(t-\alpha), y(t) = \sin(t - \alpha)$.