[Fourier Analysis] The Poisson summation formula / 푸아송 합 공식

참고문헌 : Stein, Elias M., Shakarchi, Rami 2003. Fourier Analysis. Princeton University Press.
 
Stein 해석학 시리즈 1권 [Fourier Analysis] 5단원 Fourier Transfrom on R을 공부한 내용을 TeX으로 적어 정리하는 김에 블로그에도 글을 옮겨 적는다. 책의 내용 중 보충설명이 필요한 부분 그리고 증명을 생략하고 연습문제로 넘긴 명제들 중 일부를 함께 설명하였다. 본문은 PC에 최적화되어있다.
 
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3. The Poisson summation formula

The Poisson summation formula illustrates a remarkable connection between the analysis of functions on the circle and on $\mathbb{R}$.

Definition. Given a function $f\in\mathcal{M}$, on the real line, define a new function by

    $$F_1(x)=\sum_{n=-\infty}^{\infty} f(x+n)$$

This function is called the periodization of $f$.

It is clear that the periodization $F_1$ is periodic of period $1$ on the real line.

Proposition 1. If $f\in\mathcal{M}$ then its periodization converges absolutely and uniformly on every compact subset of $\mathbb{R}$. Thus $F_1$ is continuous on $\mathbb{R}$.

So far, to show that a function defined by an infinite series is continuous, we have shown that its partial sum sequence forms an uniform Cauchy sequence and then used the fact that the uniform limit of continuous functions is continuous. However, the periodization of a rapidly decreasing function may not converge uniformly on $\mathbb{R}$. For example, the periodization of the Schwartz function $f(x)=e^{-x^2}$ does not converge uniformly on $\mathbb{R}$. Therefore, to show that the periodization is continuous we need a different method. Here we show that the periodization converges uniformly on every compact subset of $\mathbb{R}$, and then use the fact that uniform convergence on every compact subset implies continuity on $\mathbb{R}$.

proof. Since $f\in\mathcal{M}$, $f$ decreases moderately and we have a constant $C$ such that

    $$|f(x)|\le\frac{C}{1+|x|^2}$$

Now take any compact subset $K\subset\mathbb{R}$ then $K$ is bounded so we may take $M$ where $x\in K$ implies $|x|\le B$. Here we use the Weierstrass M-test to the series

    $$F_1(x)=\sum_{n=-\infty}^{\infty} f(x+n)$$

If $|n|\ge 2B$ then $|x|\le B\le |n|/2$ so

    $$|f(x+n)|\le\frac{C}{1+(x+n)^2}\le \frac{C}{1+(n/2)^2}<\frac{C'}{n^2}$$

since $|x+n|\ge \Big||x|-|n|\Big|\ge |n|/2$ holds. Now let

    $$M_n=\begin{cases}
        \sup_{x\in K} |f(x+n)| & \text{if}\;\; |n|<2B \\
        C'/n^2 & \text{if}\;\; |n|\ge 2B
    \end{cases}$$

then

    $$\sum_{n=-\infty}^\infty |f(x+n)|\le \sum_{n=-\infty}^\infty M_n<\infty$$

so $F_1$ converges absolutely hence uniformly on $K$. Then $F_1$ is continuous on $K$ and then it is continuous on the whole $\mathbb{R}$.

Definition. Let $f\in\mathcal{M}$ and $\hat{f}$ be the Fourier transform of $f$. Then

    $$F_2(x)=\sum_{n=-\infty}^{\infty} \hat{f}(n)e^{2\pi inx}$$

is an discrete analogue of the Fourier inversion formula and converges absolutely and uniformly on the any compact subspace of $\mathbb{R}$ hence continuous on $\mathbb{R}$.

Since $f\in\mathcal{M}$ implies $\hat{f}$ is continuous, we can make a bound $M_n$ which bounds each $\hat{f}(n)e^{2\pi inx}$ on every compact subset and makes the sum $\sum M_n$ converges. Then by Weierstrass M-test, $F_2$ is continuous on every compact subset of $\mathbb{R}$ hence continuous on whole $\mathbb{R}$. The argument is exactly symmetric to that of $F_1$.

Theorem 1. Poisson summation formula - If $f\in\mathcal{M}$ then

    $$\sum_{n=-\infty}^{\infty} f(x+n)=\sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi inx}$$

In particular, setting $x=0$, we have

    $$\sum_{n=-\infty}^{\infty} f(n)=\sum_{n=-\infty}^{\infty}\hat{f}(n)$$

proof. Both $F_1$ and $F_2$ can be interpreted as the continuous functions on the circle. By the uniqueness of the Fourier series, it suffices to show that both sides have the same Fourier coefficients. Clearly, the $m^{\text{th}}$ Fourier coefficients of RHS is $\hat{f}(m)$. Now we compute the Fourier coefficients of LHS.

    $$\int_0^1\left(\sum_{n=-\infty}^{\infty} f(x+n)\right)e^{-2\pi imx} dx=\sum_{n=-\infty}^{\infty}\left(\int_0^1 f(x+n) e^{-2\pi imx} dx\right)$$

Since $f$ decrease moderately, we may apply Lesbegue's dominant convergence theorem to interchange the order of integral and the sum. Now changing variables $x+n=y$ we have

    $$\sum_{n=-\infty}^{\infty}\left(\int_0^1 f(x+n) e^{-2\pi imx} dx\right)=\sum_{n=-\infty}^{\infty}\left(\int_{n}^{n+1} f(y) e^{-2\pi imy}e^{2\pi imn} dy\right)$$

Here $e^{2\pi imn}=1$, so

    $$\sum_{n=-\infty}^{\infty}\left(\int_{n}^{n+1} f(y) e^{-2\pi imy} dy\right)=\int_{-\infty}^\infty f(y)e^{-2\pi imy} dy=\hat{f}(m)$$

Then from the uniqueness of the Fourier coefficients, the proof is completed.

Example 4. Let $g$ be the trigonometric wave function defined by

$$g(x)=\begin{cases}
    1-|x| & \text{if} \;\;|x|\le 1 \\
    0 & \text{otherwise}
\end{cases}$$

then we have shown that

$$\hat{g}(\xi)=\begin{cases}
    (\sin\pi\xi/\pi\xi)^2 & \text{if}\;\;\xi\ne 0 \\
    1 & \text{if}\;\; \xi=0
\end{cases}$$

Since $g$ is even, we have

$$\mathcal{F}(\hat{g})(\xi)=g(-\xi)=g(\xi)$$

from duality. Now apply the Poisson summation formula to $\hat{g}$ then we have

$$\sum_{n=-\infty}^{\infty}\frac{\sin^2\pi(n+x)}{\pi^2(n+x)^2}=\sum_{n=-\infty}^{\infty} g(n)=1$$

from the definition of $g$. Since $\sin \pi(n+x)=(-1)^n\sin\pi x$ we finally have

$$\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2}=\frac{\pi^2}{\sin^2\pi x}$$

if $x$ is a real number which is not an integer. Now integrate both sides then

$$\int\left(\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2} \right)dx=\int\frac{\pi^2}{\sin^2\pi x} dx$$

For LHS, we can interchange the order of integral and the sum from DCT. Then we have

$$\int\left(\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2} \right)dx=\sum_{n=-\infty}^{\infty}\int\frac{dx}{(n+x)^2}=-\sum_{n=-\infty}^{\infty}\frac{1}{n+x}+C$$

Also, for RHS we have

$$\int \pi^2\csc^2\pi x dx=-\pi\cot\pi x +C'$$

With the slight adjustment of constant, we get

$$\sum_{n=-\infty}^{\infty}\frac{1}{n+x}=\frac{\pi}{\tan \pi x}$$

also for real $x$ which is not an integer.