Some Applications of Fourier Series

傅里叶分析

Q1: The isoperimetric inequality（等周不等式）

Among all simple closed curves of length $l$ in the plane $\mathbb R^2$, which one encloses the largest area?

The geometry proof: wiki

1. Define the notion(概念) of curves, length and area.

A parametrized curve $\gamma$ is a mapping

$$\gamma:[a,b]\to \mathbb R^2$$

• The curve $\Gamma$ is simple and closed. $\iff$ $\gamma(a) \neq\gamma(b)$ unless $s_1=a ~~and~~ s_2=b$ .
• We assuming that $\gamma$ is of class $C^1$, and that its derivative $\gamma'$ satisfies $\gamma'(s) \neq 0.$
• The length of the curve $\Gamma: \gamma(s)=(x(s),y(s))$ is defined by $$l=\int_a^b|\gamma'(s)|ds=\int_a^b(x'(s)^2+y'(s)^2)^{\frac{1}{2}}ds$$
• The aera of the curve is defined by: $$A=\frac{1}{2}|\int_{\gamma}(xdy-ydx)|=\frac{1}{2}|\int_a^bx(s)y'(s)-y(s)x'(s)ds|$$
• Any $C^1$ bijective mapping(双射的) $s:[c,d]\to [a,b]$ gives raise to another parametrization of $\Gamma$ by the formula $\eta(t)=\gamma(s(t))$.(变换参数不改变曲线的性质,包括其长度)

2. Statement and preparation of the isoperimetric inequality

Theorem 1.1 Suppose that $\gamma$ is a simple closed curve in $\mathbb R^2$ of length $l$, and let A denote the area of the region enclosed by this curve. Then

$$A\leq \frac{l^2}{4\pi}$$
with equality if and only if $\gamma$ is a circle.

• Consider the mapping $\mathbb R^2 \to \mathbb R^2, (x,y)\to(\delta x,\delta y)$. By taking $\delta = 2\pi / l$, we see that it suffices to prove that if $l=2\pi$ then $A\leq \pi$, with equality only if $\gamma$ is a circle.
• We say that $\gamma$ is a parametrization by arc-
  length if $|\gamma'(s)|= 1$ for all s. This means that $\gamma(s)$ travels at a constant speed. Any curve admits a parametrization by arc-length (Exercise 1).
• $\gamma:[0,2\pi]\to \mathbb R^2$ with $\gamma(s)=(x(s),y(s))$ is a parametrization by arc-length of the curve $\Gamma$, that is, $x'(s)^2+y'(s)^2=1$ for all $s \in [0,2\pi].$
  This implies that $$\frac{1}{2\pi}\int_0^{2\pi}(x'(s)^2+y'(s)^2)ds=1 \tag{*}$$

• Consider $x(s)\sim \sum a_ne^{ins}$ and $y(s)\sim \sum b_ne^{ins}$
  we have $x'(s)\sim \sum a_nine^{ins}$ and $y'(s)\sim \sum b_nine^{ins}$
  Pareseval's identity applied to $(*)$ gives:
  

  $$\sum_{n=-\infty}^{\infty}|n|^2(|a_n|^2+|b_n|^2)=1$$

• reference: Parseval's identity
  在数学分析中，以Marc-Antoine Parseval命名的帕塞瓦尔恒等式是一个有关函数的傅里叶级数的可加性的基础结论。从几何观点来看，这就是内积空间上的毕达哥拉斯定理。

3. proof

Since $x(s)$ and $y(s)$ are real-valued, we have $a_n=\overline{a_{-n}}$ and $b_n=\overline{b_{-n}}$. So we find that:

$$A =\frac{1}{2}|\int_a^bx(s)y'(s)-y(s)x'(s)ds|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ =\frac{1}{2}|\int_0^{2\pi}(\Sigma a_ne^{ins})(\Sigma b_nine^{ins})-(\Sigma b_ne^{ins})(\Sigma a_nine^{ins})ds|\\ =\frac{1}{2}|\int_0^{2\pi}(\Sigma ina_nb_{-n}-\Sigma ina_{-n}b_n)ds|~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ =\pi|\sum_{n=-\infty}^\infty n(a_n\overline{b_n}-b_n\overline{a_n})|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \leq\pi\sum_{n=-\infty}^\infty|n|(|a_n|^2+|b_n|^2)\\ \leq\pi\sum_{n=-\infty}^\infty|n^2|(|a_n|^2+|b_n|^2)\\ <=\pi$$

When $A=\pi$, we see from the above argument that

$$x(s)=a_{-1}e^{-is}+a_0+a_1e^{is}~and~y(s)=b_{-1}e^{-is}+b_0+b_1e^{is}$$
because $|n|<|n^2|$ as soon as $|n|>2$.
Because of $a_{-1}=\overline{a_1}$, $b_{-1}=\overline{b_1}$, $2(|a_1|^2+|b_1|^2)=1$, $a_1=b_1$ finally we have $$x(s)=a_0+cos(\alpha+s)~and~y(s)=b_0\overline+sin(\alpha+s)$$

4. remarks

• $(p121.ex1)$ Prove that any curve $\Gamma$ admits a parameteization by arc-length.
• $(p122.ex4)$
• The isoperimetric inequality in 3D(difficult)
  zhihu:请教如何证明同等体积时球体表面积最小？

Q2: Weyl's equidistribution theorem

The fractional part of $x$ is defined by $\langle x\rangle =x-[x]$. Now Start with a real number $\gamma \neq 0$ and look at the sequence $\{\langle n\gamma\rangle\}$. Here are some simple observations.
(1) If $\gamma$ is rational, then only finitely many numbers appearing in $\langle n\gamma\rangle$are distinct.
(2) If $\gamma$ is irrational, then the numbers $\langle n\gamma\rangle$ are all distinct.

Defination of equidistribution

A sequence of numbers $\xi_1,\xi_2,...,\xi_n,...\in [0,1)$ is said to be equidistributed if for every interval $(a,b) \subset [0,1)$,

$$\lim_{N\to \infty} \frac{\#\{1\leq n\leq N:\xi_n \in (a,b)\}}{N}=b-a$$
where #A denotes the cardinality of the finite set A.(元素个数)

Theorem 2.1

If $\gamma$ is irrational, then the sequence of fractional parts $\langle n\gamma\rangle$ is equidistributed in [0,1).

In particular, $\langle n\gamma\rangle$ is dence in [0,1). Let $\chi_{(a,b)}(x)$ donate the characteristic function of the intercal (a,b), that is, the function equal to 1 in (a,b) and 0 in [0,1)-(a,b). As a consequence of the definitions we find that:

$$\#\{1\leq n \leq N:\langle n\gamma\rangle \in (a,b)\}=\sum_{n=1}^n \chi_{(a,b)}(n\gamma)$$
and the theorem can be reformulated as the statment that $$\frac{1}{N}\sum_{n=1}^N \chi_{(a,b)}(n\gamma) \to \int_0^1 \chi_{(a,b)}(x)dx,~~~~ as~ N \to \infty$$

Lemma 2.2

If f is continuous and periodic of period 1, and $\gamma$ is irrational, then

$$\frac{1}{N}\sum_{n=1}^N f(n\gamma) \to \int_0^1 f(x)dx,~~~~ as~ N \to \infty$$

proof of lemma 2.2

The proof of the lemma is divided into three steps.

• Step 1. If $f=1$, the limit surely holds. If $f=e^{2\pi ikx}$ with $k \neq 0$, then the integral is 0. Since $\gamma$ is irrational, we have $e^{2\pi ik\gamma} \neq 1$, therefore $$\frac{1}{N}\sum_{n=1}^N f(n\gamma) =\frac{e^{2\pi ik\gamma}}{N} \frac{1-e^{2\pi ikN\gamma}}{1-e^{2\pi ik\gamma}} \to 0 (N \to \infty)$$
• Step 2. It is clear that if $f$ and $g$ satisfy the lemma, then so does $Af + Bg$ for any $A,B \in \mathbb C$. Therefore, the first step implies that the lemma is true for all trigonometric polynomials.
• Step 3. Let $\epsilon>0$. If $f$ is any continous periodic function of period 1, choose a trigonometric polynomial $P$ so that $sup_{x \in \mathbb R}|f(x)-P(x)|<\epsilon/3$. Then, by step 1, for all large N we have $$|\frac{1}{N} \sum_{n=1}^{N}P(n\gamma)-\int_0^1 f(x)dx|<\epsilon / 3$$
  Therefore
  $$|\frac{1}{N} \sum_{n=1}^{N}f(n\gamma)-\int_0^1 f(x)dx|<\sum|f-P|+|\sum P-\int P|+\int|P-f|<\epsilon$$
  and the lemma is proved.

proof of theorem 2.1

Now we can finish the proof of the theorem 2.1. Choose two contiuous periodic functions $f^-_{\epsilon},f^+_{\epsilon}$ of period 1 and agree with $\chi_{(a,b)}(x)$ except in intervals of total length $2\epsilon$.

Corollary 2.3

The conclusion of Lemma 2.2 holds for every function f which is Riemann integrable in [0,1], and periodic of period 1.(no need for continuous)

Proof

• consider a partition of the interval [0,1], say$0=x_0<x_1<...<x_N=1$
• define $f_U, f_L$(每段积分区间取最小值/最大值), and for a given $\epsilon > 0$,by making the partition suffiviently fine, we can have$\int_0^1f_U(x)-f_L(x)< \epsilon$ .
• $\frac{1}{N}\sum_{n=1}^N f_L(n\gamma) \to \int_0^1 f_L(x)dx,~~~~ as~ N \to \infty$, the same as $f_R$.

Weyl's criterion(外尔准则)

A sequence of real numbers $\xi_1,\xi_2,...in [0,1)$ is wquidistributed if and onlu if for all integers $k\neq0$ one has

$$\frac{1}{N} \sum_{n=1}^N e^{2\pi ik\xi_n} \to 0~ ,~~~as~ N \to \infty$$

Riemann integral criterion for equidistribution

Suppose $\{s_1, s_2, s_3, ...\}$ is a sequence contained in the interval $[a, b]$. Then the following conditions are equivalent:

• The sequence is equidistributed on $[a, b]$.
• For every Riemann-integrable (complex-valued) function $f : [a, b] → C$, the following limit holds:
  $$\lim_{n\to \infty}\frac{1}{N}\sum_{n=1}^Nf(s_n)=\frac{1}{b-a}\int_a^bf(x)dx$$
  proof

This criterion leads to the idea of Monte-Carlo integration, where integrals are computed by sampling the function over a sequence of random variables equidistributed in the interval. (随机算法的收敛性)

proof of Weyl's criterion

If the sequence is equidistributed modulo 1, then we can apply the Riemann integral criterion (described above) on the function $f(x)=e^{2\pi i\ell x}$, which has integral zero on the interval [0, 1]. This gives Weyl's criterion immediately.

reference

• p122. ex5
  prove that the sequence $\{\gamma_n\}^{\infty}_{n=1}$, where $\gamma_n$ is the fractional part of $(\frac{1+\sqrt{5}}{2})^n$ is not equidistributed in $[0,1]$.
• p123. ex7 Weyl's criterion
• p123. ex8/9

Q3: A continuous but nowhere differentiable function

Theorem 3.1 If $0 < \alpha < 1$, then the function

$$f_{\alpha}(x)=f(x)=\sum_{n=1}^{\infty}2^{-n\alpha}e^{i2^nx}$$ is continuous but nowhere differentiable.

• The continuity is clear because of the absolute convergence of the series.
• Three methods of summing a Fourier series.
  
  • Partial means
  • Cesaro means
  • Delayed means
• For series that havelacunary properties like those of $f$, the delayed means are essentially equal to the partial means. In particular, note that for our function $f=f_\alpha$, $$S_N(f)=\Delta_{N'}(f),$$
  Where $N'$ is the largest integer of the form $2^k$ with $N'<N$.
  Lemma 3.2 Let $g$ be any continuous function that is diferentiable at $x_0$. Then, the Cesaro means satisfy $$\ \sigma_N(g)'(x_0)=O(logN)$$
  Lemma 3.3 If $2N=2^n$, then $$\Delta_{2N}(f)-\Delta_{N}(f)=2^{-n\alpha}e^{i2^nx}$$

拓展阅读:

• Weierstrass function
• 处处不可导函数的稠密性分析:在配备了经典维纳测度的连续函数空间中, 至少有一点可导的函数所构成的集合的测度是0.