[Fourier Transform] The Gaussians / 가우시안 그리고 커널

참고문헌 : Stein, Elias M., Shakarchi, Rami 2003. Fourier Analysis. Princeton University Press.

 

Stein 해석학 시리즈 1권 [Fourier Analysis] 5단원 Fourier Transfrom on $\mathbb{R}$을 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. 아래는 2편이다.

https://whitemask.tistory.com/260

[Fourier Transform] The Schwartz space

 

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1.5 The Gaussians as good kernels

In the study of Fourier transform, the Gaussian $e^{-ax^2}$ plays an important role. So we study the Gaussian carefully and show that these are good kernels when properly determined.

The integral of the Gaussian $e^{-ax^2}$ is normalized when $a=\pi$ by

$$\int_{-\infty}^{\infty} e^{-\pi x^2} dx=1$$

Here we omit the proof of this integral. There is well-known proof using the polar coordinate and Fubini's theorem for changing order of integrals.

Theorem 1. If $f(x)=e^{-\pi x^2}$ then $\hat{f}(\xi)=f(\xi)$. That is, $f$ is a Fourier transform of itself.

proof. Define $F(\xi)=\hat{f}(\xi)$ then

    $$F(0)=\int_{-\infty}^{\infty}e^{-\pi x^2} dx=1$$

We will prove the theorem by showing that $G(\xi)=F(\xi)e^{\pi\xi^2}$ is constant. Then we must calculate $G'(\xi)=F'(\xi)e^{\pi\xi^2}+2\pi\xi F(\xi)e^{\pi\xi^2}$. From Proposition 1 of 'The Schwartz space' on the link, we have

    $$ -2\pi ix f(x)=-2\pi ix e^{-\pi x^2}\longrightarrow F'(\xi)=\frac{d}{d\xi}\hat{f}(\xi)$$

But since $f'(x)=-2\pi x e^{-\pi x^2}$,

    $$F'(\xi)=\int_{-\infty}^{\infty} -2\pi ix e^{-\pi x^2} e^{-2\pi ix\xi}dx=i\int_{-\infty}^{\infty} f'(x)e^{-2\pi ix\xi}dx$$

Also since $f'(x)\longrightarrow 2\pi i\xi \hat{f}(\xi)$,

    $$\int_{-\infty}^{\infty} f'(x)e^{-2\pi ix\xi}dx=2\pi i\xi\int_{-\infty}^{\infty} f(x)e^{-2\pi ix\xi}dx$$

so

    $$F'(\xi)=\frac{d}{d\xi}\hat{f}(\xi)=-2\pi\xi\hat{f}(\xi)$$

Now define $G(\xi)=F(\xi)e^{\pi \xi^2}$ as desired then

    $$G'(\xi)=e^{\pi\xi^2}(F'(\xi)+2\pi\xi F(\xi))=0$$

so $G'(\xi)=0$. But since $G(0)=F(0)=1$, $G(\xi)=1$ and this proves the theorem.

Now for $\delta>0$, define $K_\delta(x)$ by

$$K_\delta(x)=\delta^{-1/2} e^{-\pi x^2}$$

then we have $\widehat{K_\delta}(\xi)=e^{-\pi\delta\xi^2}$. This is simple corollary from the previous theorem and 3 of the Proposition 1, mentioned earlier. Here, observe that as $\delta$ tends to $0$, the function $K_\delta$ peaks at the origin while its Fourier transform $\widehat{K_\delta}$ gets `flatter'. In this particular example, we see that $K_\delta$ and $\widehat{K_\delta}$ cannot both be concentrated at the origin. This is an example of a general phenomenon called the Heisenberg uncertainty principle. We will discuss this at the end of the chapter.

Theorem 2. The collection $\{K_\delta(x)\}_{\delta>0}$ is a family of good kernels as $\delta\to 0$.

proof. First, observe that

$$\int_{-\infty}^{\infty} K_\delta(x)dx=\int_{-\infty}^{\infty}\delta^{-1/2}e^{-\pi x^2/\delta}dx=\int_{-\infty}^{\infty}e^{-\pi t^2} dt=1$$

from the change of variables $x/\delta^{1/2}=t$ for $\delta>0$. Also, $K_\delta(x)> 0$ for all $x\in\mathbb{R}$. Now for $\eta>0$, consider

    \begin{align*}
        \int_{|x|\ge\eta} K_\delta(x) dx&=\int_{|x|\ge\eta} \delta^{-1/2} e^{-\pi x^2/\delta} dx=\int_{|t|\ge\eta/\delta^{1/2}} e^{-\pi t^2} dt
    \end{align*}

then the integral tends to $0$ as $\delta\to 0$ since $e^{-\pi t^2}$ is a Schwartz function. Therefore the collection $\{K_\delta\}_{\delta>0}$ is a family of good kernels.

After we prove that some family is a good kernels, we have always showed that the convolution of the good kernels and $f$ uniformly converges to $f$. Here we have analogous results. Note, the convolution between the functions $f, g$ on the real line is defined by

$$(f*g)(x)=\int_{-\infty}^{\infty} f(x-t)g(t)dt$$

given that the integral converges. Here the natural questions arise. If $f, g\in\mathcal{M}$ then $f*g\in\mathcal{M}$? What if $f, g\in \mathcal{S}$? Later, we will discuss these issues.

 

Proposition 1. If $f\in\mathcal{S}$, then $(f*K_\delta)(x)$ uniformly converges to $f(x)$ as $\delta\to 0$.

proof. From the first property of good kernels, we have

    $$|(f*K_\delta)(x)-f(x)|=\Bigg|\int_{-\infty}^{\infty}K_\delta(t)(f(x-t)-f(x))dt\Bigg|\le \int_{-\infty}^{\infty} K_\delta(t)\cdot|f(x-t)-f(x)| dt$$

Here we divide the integral into two,

    $$\int_{-\infty}^{\infty} K_\delta(t)\cdot|f(x-t)-f(x)| dt=\int_{|t|\le \eta} + \int_{|t|\ge \eta}$$

for $\eta>0$ to control each term. The second one goes $0$ as $\delta\to 0$ from the third property of the good kernels and the boundedness of $f\in\mathcal{S}$. However, the first integral is in question. To answer this, we must show that $f$ is uniformly continuous on $\mathbb{R}$. Since $f\in\mathcal{S}$, we have large $R$ such that $|f(x)|<\epsilon$ whenever $|x|\ge R$. Note that $f$ is uniformly continuous on the interval $[-R, R]$. Then there exists $\delta>0$ where $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$ for $x, y\in [-R, R]$. If $|x|, |y| > R$ then $|f(x)-f(y)|<2\epsilon$. If $|x-y|<\delta$ and $x<R<y$ then

    $$|f(x)-f(y)|\le |f(x)-f(R)|+|f(R)-f(y)|<\epsilon+2\epsilon$$

 

so by proper adjustment of the scalar, we can make $|x-y|<\delta$ imply $|f(x)-f(y)|<\epsilon$ for all $x, y\in \mathbb{R}$. Thus $f\in\mathcal{S}$ is uniformly continuous. Actually, this can be extended to the functions of moderate decrease. Now let our $\eta=\delta$ then

    $$\int_{|t|\le \eta} K_\delta(t)\cdot|f(x-t)-f(x)| dt<\epsilon\int_{|t|\le \eta} K_\delta(t)dt<\epsilon$$

so

    $$\sup_{x\in\mathbb{R}} |(f*K_\delta)(x)-f(x)|\to 0\;\;\;\;\text{as}\;\;\delta\to 0$$

This ends the proof.