[Fourier Transform] The Schwartz space / 슈바르츠 공간

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 1편이다.
 
https://whitemask.tistory.com/259

[Fourier Transform] Preliminaries

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1.3 The Schwartz space
 
The Schwartz space on $\mathbb{R}$, denoted by $\mathcal{S}(\mathbb{R})$ or simply $\mathcal{S}$, consists of the set of all indefinitely differentiable functions $f$ where $f$ and all its derivatives $f', f'', \cdots, f^{(l)}, \cdots$ are rapidly decreasing in the sense that

$$\sup_{x\in\mathbb{R}} |x|^k|f^{(l)}(x)|<\infty$$

for every integer $k, l\ge 0$. This means that $f^{(l)}$ decays faster than any polynomials, which makes $\mathcal{M}\subset\mathcal{S}$. It is obvious that $\mathcal{S}$ is a vector space over $\mathbb{C}$. Also, if $f\in\mathcal{S}$ then $f'$ and $x^n f(x)$ are both the element of $\mathcal{S}$ by definition. Thus, the Schwartz space is closed under differentiation and multiplication by polynomials.
 
Example 1. $f(x)=e^{-x^2}$ is a simple example of the Schwartz function. The bump function is a indefinitely differentiable and compactly supported function. A function $g$ is said to be compactly supported if for the compact set $K$, $g$ vanishes outside of $K$ and nonzero at $K$. A simple way to make a bump function is to use

$$g(x)=e^{-a/(x-a)}e^{-b/(x-b)}$$

for $a<b$. Now define

$$G(x)=\int_{-\infty}^x g(t)dt$$

Then $G(x)=0$ when $x\le a$, $G(x)=c$ when $x\ge b$ for some constant $c$ and $G$ strictly increases at the interval $(a, b)$. Then we observe that $G$ is indefinitely differentiable and

$$\sup_{x\le\frac{a+b}{2}}|x^k||G^{(l)}(x)|<\infty$$

Here, define

$$H(x)=\begin{cases}
    G(x) & \text{if}\; x\le\frac{a+b}{2}\\
    G\left(\frac{a+b}{2}-x\right) & \text{if}\; x>\frac{a+b}{2}
\end{cases}$$

then $H$ is a bump function in $\mathcal{S}$.

The choice of the Schwartz space $\mathcal{S}$ is motivated by the important principle which ties the decay of $\hat{f}$ to the continuity and differentiability properties of $f$ : the faster $\hat{f}(\xi)$ decreases as $|\xi|\to \infty$, the smoother $f$ must be. Note we had a similar relationships between the periodic function and its Fourier coefficients : if $f$ is a periodic function of period $2\pi$ which belongs to the class $C^k$ then

$$\hat{f}(n)=O(1/|n|^k)$$

This illustrates the connection between the Fourier coefficients and the Fourier transform, in the sense that the Fourier transform is a continuous version of the Fourier coefficients. Here is another example to keep in mind.

Example 2. Suppose that $f$ is a function of moderate decrease on $\mathbb{R}$ whose Fourier transform $\hat{f}$ is continuous and satisfies

$$\hat{f}(\xi)=O\left(\frac{1}{|\xi|^{1+\alpha}}\right)$$

as $|\xi|\to\infty$ for some $0<\alpha<1$. Then $f$ satisfies a Hölder condition of order $\alpha$ : that is, $f$ satisfies

$$|f(x+h)-f(x)|\le M|h|^\alpha$$

for all $x, h\in\mathbb{R}$. To prove the claim, we need a Fourier inversion formula.

1.4 The Fourier transform on $\mathcal{S}$

Now we are ready to define the Fourier transform of the function. For convenience, we first define the Fourier transform of the Schwartz function and then extend the range of definition.

Def. The Fourier transform of a function $f\in\mathcal{S}$ is defined by

    $$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi} dx$$
 
Here, we use the notation $f(x)\longrightarrow \hat{f}(\xi)$ to meant that $\hat{f}$ is the Fourier transform of $f$.

Here are some simple properties of the Fourier transform.

Proposition 1. If $f\in\mathcal{S}$, then the following properties hold.
    1. $f(x+h)\longrightarrow \hat{f}(\xi)e^{2\pi ih\xi}$ whenever $h\in\mathbb{R}$.
    2. $f(x)e^{-2\pi ixh}\longrightarrow \hat{f}(\xi+h)$ whenever $h\in\mathbb{R}$.
    3. $f(\delta x)\longrightarrow \hat{f}(\xi/\delta)/\delta$ whenever $\delta>0$.
    4. $f'(x)\longrightarrow 2\pi i \xi\hat{f}(\xi)$
    5. $-2\pi ixf(x)\longrightarrow (\hat{f})'(\xi)$

proof. For 1,

    $$\int_{-\infty}^{\infty} f(x+h)e^{-2\pi ix\xi}dx=\int_{-\infty}^{\infty} f(t)e^{-2\pi i(t-h)\xi}dt=e^{2\pi ih\xi}\hat{f}(\xi)$$

Similarly, 2 and 3 are immediate from the simple change of variables. For 4, first consider the integral from $-N$ to $N$ then

    $$\int_{-N}^{N} f'(x) e^{-2\pi ix\xi} dx=f(x)e^{-2\pi ix\xi}\Bigg]_{-N}^N +2\pi i\xi\int_{-N}^N f(x)e^{-2\pi ix\xi}dx$$

from integration by parts. Now let  $N\to\infty$ then since $f$ decreases rapidly, the first term of RHS goes 0 so

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

For 5, use the definition of derivative for $\hat{f}(\xi)$ then

    \begin{align*}
        \frac{\hat{f}(\xi+h)-\hat{f}(\xi)}{h}-\left(\widehat{-2\pi ixf}\right)(\xi)&=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}\left(\frac{e^{-2\pi i xh}-1}{h}\right)+2\pi ixf(x)e^{-2\pi ix\xi} dx\\
        &=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}\left(\frac{e^{-2\pi i xh}-1}{h}+2\pi ix\right)dx
    \end{align*}

Since $f\in\mathcal{S}$, $xf(x)\in\mathcal{S}$ too. Then they both decrease rapidly, so there exists large $N$ such that

    $$\int_{|x|\ge N} |f(x)|dx<\epsilon\;\;\;\text{and}\;\int_{|x|\ge N} |xf(x)|dx<\epsilon$$

Note that the derivative of $e^{-2\pi ixh}$ in the variable $h$ is $-2\pi ix e^{-2\pi ixh}$ so at $h=0$, the value of the derivative is $-2\pi ix$. Since $e^{-2\pi ixh}$ is continuously differentiable,

    $$\Bigg|\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\Bigg|$$

goes 0 as $h$ goes 0. Now choose $x\in [-N, N]$ then since $[-N, N]$ is compact, the derivative is uniformly continuous so there exists uniform $h_0$ such that $|h|<h_0$ implies

    $$\Bigg|\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\Bigg|<\frac{\epsilon}{N}$$

for all $|x|\le N$. Now we have
    
    \begin{align*}
        \Bigg|\frac{\hat{f}(\xi+h)-\hat{f}(\xi)}{h}-\left(\widehat{-2\pi ixf}\right)(\xi)\Bigg|&=\Bigg|\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}\left(\frac{e^{-2\pi i xh}-1}{h}\right)+2\pi ixf(x)e^{-2\pi ix\xi} dx\Bigg|\\
        &\le \int_{-\infty}^{\infty}|f(x)|\cdot\Bigg|\frac{e^{-2\pi i xh}-1}{h}+2\pi ix\Bigg|dx\\
        &\le\int_{|x|\le N} +\int_{|x|\ge N}
    \end{align*}

For the first integral,

    \begin{align*}
        \int_{|x|\le N} |f(x)|\cdot\Bigg|\frac{e^{-2\pi i xh}-1}{h}+2\pi ix\Bigg|dx \le B\int_{|x|\le N} \frac{\epsilon}{N}dx=2B\epsilon
    \end{align*}

where $B$ is the bound of $f\in\mathcal{S}$. For the second integral, first note the useful inequality $|e^{it}-1|\le |t|$ for all $t\in\mathbb{R}$. Then we have

    $$\Bigg|\frac{e^{-2\pi ixh}-1}{h}\Bigg|\le 2\pi|x|$$

so we have

    \begin{align*}
        \int_{|x|\ge N} |f(x)|\cdot\Bigg|\frac{e^{-2\pi i xh}-1}{h}+2\pi ix\Bigg|dx&\le 2\pi \int_{|x|\ge N} |xf(x)dx|+2\pi \int_{|x|\ge N} |xf(x)| dx\\
        &< C_0 \epsilon
    \end{align*}

Thus if $|h|<h_0$ then

    $$\Bigg|\frac{\hat{f}(\xi+h)-\hat{f}(\xi)}{h}-\left(\widehat{-2\pi ixf}\right)(\xi)\Bigg|<C\epsilon$$

and this proves the proposition.

The crucial parts of this proposition are 4 and 5. They implies that the Fourier transform interchanges differentiation and multiplication. From these properties, we use the Fourier transform when solving the complex differential equation. Fourier transform changes the differential equation to simple algebraic equation, which is much easier to solve. The need for Fourier inversion formula then comes naturally to restore the original function from its Fourier transform version.

The inequality $|e^{it}-1|<|t|$ can be interpreted geometrically. Consider a point $(1, 0)$ on the unit circle on $\mathbb{C}$ and $e^{it}=(\cos t, \sin t)$ then $|e^{it}-1|$ is a length of segment between $e^{it}=(\cos t, \sin t)$ and $(1, 0)$. Here, $|t|$ is the length of arc between $e^{it}=(\cos t, \sin t)$ and $(1, 0)$. Then the length of arc is always larger than that of segment so the inequality holds. From this we realize that the equality holds if and only if $t=2n\pi$.

Theorem 1. If $f\in\mathcal{S}$ then $\hat{f}\in\mathcal{S}$.

proof. To show that $\hat{f}\in\mathcal{S}$, we first verify that $\hat{f}$ is indefinitely differentiable and next its derivatives are all decrease faster than any polynomials. But from the previous proposition, we have

    $$-2\pi ixf(x) \longrightarrow \frac{d}{d\xi}\hat{f}(\xi)$$

and this implies

    $$(-2\pi ix)^l f(x)\longrightarrow \left(\frac{d}{d\xi}\right)^l \hat{f}(\xi)$$

so $\hat{f}$ is indefinitely differentiable. Note if $f\in\mathcal{M}$ then $\hat{f}$ is continuous, so $l^{\text{th}}$ derivative of $\hat{f}$ is well-defined. Now from (4),

    $$\frac{1}{{(2\pi i)^k}}\left(\frac{d}{dx}\right)^k f(x)\longrightarrow \xi^k\hat{f}(\xi)$$

so $\xi^k\hat{f}(\xi)$ is bounded for all $k$. Then $\hat{f}\in\mathcal{S}$ by definition.

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본문에서도 언급했지만, Proposition 1의 4, 5는 무척 중요하다. 어떤 함수의 미분은 푸리에 변환에서 변수와 원래 함수의 푸리에 변환의 곱이 되고, 반대로 푸리에 변환의 미분은 변수와 원래 함수의 곱이 변환된 결과이다. 이 때문에 미분 방정식을 해결할 때 푸리에 변환이 무척 유용한데, 미분 방정식을 푸리에 변환으로 바꾸면 미분 방정식의 차수를 낮출 수 있고 따라서 우리가 쉽게 해결할 수 있는 미분 방정식의 꼴로 바꿀 수 있을 가능성이 높기 때문이다. 그런데 문제는, 푸리에 변환을 통해 미분 방정식을 해결하여 얻은 해는 어떤 함수의 푸리에 변환 꼴로 주어지는데, 이 푸리에 변환을 만드는 원래 함수가 무엇이냐 하는 것이다. 이때 바로 그 원래 함수를 찾을 수 있도록 하는 것이 푸리에 역변환이다. 특히 Schwartz space $\mathcal{S}$에서는 푸리에 변환과 푸리에 역변환이 일대일 대응이 된다. 다음 글에서는 푸리에 역변환에 대한 이론을 공부한다.