[Fourier Transform] Preliminaries

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.

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1.1 Elementary theory for the Fourier transform

 

The most natural extension of integration of continuous functions on the real line $\mathbb{R}$ is

$$\int_{-\infty}^{\infty} f(x)dx=\lim_{N\to \infty} \int_{-N}^{N} f(x)dx$$

But the problem is, the limit may not exist. For the limit to exists, the integrand, $f$ must decay at suitable speed as $|x|$ goes to infinity. Note, since $f$ is continuous, we only need to check this decay condition to determine if the limit exist. Here is a useful criterion.

Def. A function $f$ defined on $\mathbb{R}$ is said to be of moderate decrease if $f$ is continuous and there exists $\epsilon>0$ and a constant $A>0$ such that

    $$|f(x)|\le\frac{A}{1+|x|^{1+\epsilon}}$$

 

for all $x\in\mathbb{R}$.

This inequality says that $f$ is bounded by $A$ and also that it decays at infinity as fast as $1/|x|^{1+\epsilon}$. If $f$ is of moderate decrease, the integral over the real line converges. To see this, define

$$I_N=\int_{-N}^{N} f(x)dx$$

 

Then since $f$ is continuous, $I_N$ is well defined for all $N\in\mathbb{N}$. Now we show that the sequence $\{I_N\}$ is a Cauchy sequence. For $N<M$,

\begin{align*}
    |I_M-I_N|&\le \int_{N\le |x|\le M} |f(x)|dx\le 2\int_N^M \frac{A}{1+x^{1+\epsilon}}dx\\
    &\le 2\int_N^M \frac{A}{x^{1+\epsilon}}dx<\frac{C}{N^\epsilon}
\end{align*}

for constant $C$. Since $\epsilon>0$ is fixed, $|I_M-I_N|$ goes $0$ as $N$ goes $\infty$. Then by completeness, the limit of $I_N$ exists and this proves the well-definedness. We shall denote $\mathcal{M}(\mathbb{R})$ or simply $\mathcal{M}$ as the collection of all functions of moderate decrease. It is easy to see that $\mathcal{M}$ is a vector space over the field $\mathbb{C}$. All we need to check is the closedness, but it is obvious from the triangular inequality. From now on, we shall take $\epsilon=1$ for convenience so when we say $f\in\mathcal{M}$ or $f$ is of moderate decrease, then it implies

$$|f(x)|\le\frac{A}{1+x^2}$$

We summarize some elementary properties of integration over the real line in the following proposition.

Proposition 1. The integral of a function of moderate decrease on the real line satisfies the following properties.
    1. Linearity : if $f, g\in\mathcal{M}$ and $a, b\in\mathbb{C}$, then

    $$\int_{-\infty}^{\infty} (af(x)+bg(x)) dx=a\int_{-\infty}^{\infty}f(x)dx+b\int_{-\infty}^{\infty}g(x)dx$$

    2. Translation invariance : for every $h\in\mathbb{R}$ we have

    $$\int_{-\infty}^{\infty} f(x-h) dx=\int_{-\infty}^{\infty} f(x)dx$$

    3. Scaling under dilations : if $\delta>0$, then

    $$\delta\int_{-\infty}^{\infty} f(\delta x) dx=\int_{-\infty}^{\infty} f(x)dx$$

    4. Continuity : if $f\in\mathcal{M}$, then

    $$\int_{-\infty}^{\infty}|f(x-h)-f(x)| dx \to 0\;\;\;\text{as}\;\; h\to 0$$

proof. 1 is immediate. 2 and 3 comes from the change of variables. For 4, suppose $|h|\le 1$. Since $f$ is of moderate decrease, we can take a large $N$ that makes

    $$\int_{|x|\ge N} |f(x)|dx\;\;\text{and}\; \int_{|x|\ge N} |f(x)|dx <\frac{\epsilon}{4}$$

 

The proof is obvious, so we omit here. For such $N$, $f$ is uniformly continuous on the compact set $[-N-1, N+1]$ so we have $\delta>0$ such that $|h|<\delta$ implies

    $$\sup_{|x|\le N} |f(x-h)-f(x)|<\frac{\epsilon}{4N}$$

 

from the definition of uniform continuity. Then if $|h|<\delta$,

 

    \begin{align*}
        \int_{-\infty}^{\infty}|f(x-h)-f(x)| dx&\le \int_{|x|\le N}|f(x-h)-f(x)| dx+\int_{|x|\ge N} |f(x-h)|+|f(x)|dx\\
        &<\frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{4}=\epsilon
    \end{align*}

 

and this ends the proof.

1.2 Definition of the Fourier transform

If $f\in\mathcal{M}$, we define its Fourier transform for $\xi\in\mathbb{R}$ by

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

The integral makes sense since $|f(x)e^{-2\pi ix\xi}|=|f(x)|$ and $f$ is of moderate decrease. Roughly speaking, the Fourier transform is a continuous version of the Fourier coefficients. Recall that the Fourier coefficients are given as follow

$$\hat{f}(n)=\int_0^1 f(x)e^{-2\pi i nx} dx$$

If we replace the discrete $n\in\mathbb{Z}$ to continuous counterparts $\xi\in\mathbb{R}$ and change the domain of integration, we get a definition for the Fourier transform. Note that this is just a rough heuristics, but we will discover similar analogy as we proceed.

Proposition 2. Suppose $f\in\mathcal{M}$. Then $\hat{f}$ is continuous and $\hat{f}(\xi)$ goes $0$ as $|\xi|$ tends to $\infty$.

proof. We need the Lesbegue dominant convergence theorem to prove the proposition. For the continuity, consider

    $$|\hat{f}(\xi+h)-\hat{f}(\xi)|=\Bigg|\int_{-\infty}^{\infty}f(x)\big(e^{-2\pi ix(\xi+h)}-e^{-2\pi ix\xi}\big)dx\Bigg|\le \int_{-\infty}^{\infty}|f(x)||e^{-2\pi ixh}-1|dx$$

 

Since $|f(x)||e^{-2\pi ixh}-1|$ converges to $0$ as $h$ tends to $0$ and is dominated by $2|f(x)|$ since $|e^{-2\pi ixh}-1|\le 2$, we may apple DCT, concluding

    $$\lim_{h\to 0} \left(\int_{-\infty}^{\infty}|f(x)||e^{-2\pi ixh}-1|dx\right)=\int_{-\infty}^{\infty}\left(\lim_{h\to 0}|f(x)||e^{-2\pi ixh}-1|dx\right)=0$$

 

so $\hat{f}$ is continuous for all $\xi\in\mathbb{R}$. For the second assertion, we have

    $$\int_{-\infty}^{\infty} f(x-1/2\xi)e^{-2\pi ix\xi} dx=-\hat{f}(\xi)$$

 

by the change of variables $x-1/2\xi \to t$. Then we may write

    $$\hat{f}(\xi)=\frac{1}{2}\int_{-\infty}^{\infty}(f(x)-f(x-1/2\xi))e^{-2\pi ix\xi} dx$$

 

But from 4 of Proposition 1.,

    $$\int_{-\infty}^{\infty}|f(x)-f(x-1/2\xi)| dx \to 0\;\;\;\text{as}\;\;\Bigg|\frac{1}{2\xi}\Bigg|\to 0$$

 

But since

    $$|\hat{f}(\xi)|\le\frac{1}{2}\int_{-\infty}^{\infty}|f(x)-f(x-1/2\xi)| dx$$

the proof is done.

However, $f\in\mathcal{M}$ does not guarantee if $\hat{f}\in\mathcal{M}$. Even worse, it does not guarantee anything about the decay of $\hat{f}$. Our main theorem of the chapter is Fourier inversion formula, which says that

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

 

for 'suitable' $f$ on the real line. But this integral does not make sense unless we have information about the decay of $\hat{f}$ at infinity. To solve this problem, we first introduce a more refined space of functions considered by Schwartz, which is very useful in establishing the initial properties of the Fourier transform.