[Fourier Analysis] The Plancherel formula / 플랑쉐렐 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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1.7 The Plancherel formula

Proposition 1.  If $f, g\in\mathcal{S}$ then

(1) $f*g\in \mathcal{S}$
(2) $f * g = g * f$
(3) $\widehat{(f * g)}(\xi) = \widehat{f}(\xi)\widehat{g}(\xi)$

proof. For (1), we claim that $f*g$ decreases rapidly. First, we claim that

    $$\sup_{x\in\mathbb{R}}|x|^l|g(x-y)|\le A_l (1+|y|)^l$$

for all non-negative integer $l$. To prove this, we first verify the inequality

    $$|x-y+y|^l\le (|x-y|+|y|)^l\le C_l(|x-y|^l+|y|^l)$$

which comes immediately from the fact that $h(x)=x^k$ is convex if $k$ is non-negative integer, therefore satisfies

    $$h\left(\frac{a+b}{2}\right)\le\frac{h(a)+h(b)}{2} \Rightarrow \left(\frac{a+b}{2}\right)^k\le\frac{a^k+b^k}{2}$$

Thus we just let $C_l=2^{k-1}$ then the inequality holds. Then we have $|x|^l\le C_l(|x-y|)^l+|y|^l)$ so

    $$|x|^l|g(x-y)|\le C_l(|x-y|^l |g(x-y)+|y|^l|g(x-y)|)$$

Now from $g\in\mathcal{S}$, $|g(x-y)|$ and $|x-y|^l|g(x-y)|$ are bounded so we may write the inequality above as

    $$|x|^l|g(x-y)|\le C_l(C+C'|y|^l)\le A_l(1+|y|)^l$$

for $A_l=C_l(C+C')$. This proves the claim. Now consider $\sup_x |x|^l|(f*g)(x)|$ then

    $$\sup_{x\in\mathbb{R}}|x|^l|(f*g)(x)|\le\sup_x\int_{-\infty}^{\infty}|f(y)|\cdot|x|^l|g(x-y)| dy\le A_l\int_{-\infty}^{\infty}|f(y)|(1+|y|)^l dy$$

from our claim. But since $f\in\mathcal{S}$, $|f(y)|(1+|y|)^l$ decreases rapidly so the integral make sense. Therefore $\sup_x |x|^l|(f*g)(x)|<\infty$ for all $l\ge 0$ and since we have $(f*g)'=f'*g=f*g'$, we have, for example, $(f*g)^{(k)}=f^{(k)}*g$ so the argument is similar for the derivatives of $f*g$. Therefore, $f*g\in\mathcal{S}$. Now (2) comes directly from the simple change of variables, just as we did in the convolution of functions on the circle. (3) is also easy. Define $F(x, y)=f(y)g(x-y)e^{-2\pi ix\xi}$ then for

    $$F_1(x)=\int_{-\infty}^{\infty} F(x, y) dy\;\;\;\text{and}\;\;F_2(y)=\int_{-\infty}^{\infty} F(x, y) dx$$

we have that $F_1(x)$ and $F_2(y)$ are continuous and satisfies

    $$\int_{-\infty}^{\infty}F_1(x)dx=\int_{-\infty}^{\infty}F_2(y) dy$$

from $f*g\in\mathcal{S}$ adn the Proposition 1 of the last article. Since $F_1(x)=e^{-2\pi ix\xi}(f*g)(x)$ and $F_2(y)=f(y)e^{-2\pi iy\xi}\hat{g}(\xi)$ we get

    $$\int_{-\infty}^{\infty}e^{-2\pi ix\xi}(f*g)(x)=\widehat{(f*g)}(\xi)=\hat{g}(\xi)\cdot\int_{-\infty}^{\infty}f(y)e^{-2\pi iy\xi} dy=\hat{f}(\xi)\hat{g}(\xi)$$

and this completes the proof of the proposition.

The Plancherel's theorem is the analogue for functions in $\mathbb{R}$ of Parseval's identity for Fourier series. Note that the Schwartz space $\mathcal{S}$ is a Hermitian space equipped with the standard Hermitian inner product

$$(f, g)=\int_{-\infty}^{\infty} f(x)\overline{g(x)} dx$$

which has an associated norm $\|f\|=(f, f)^{1/2}$.

Theorem 1. Plancherel - If $f\in\mathcal{S}$, then $\|\hat{f}\|=\|f\|$.

proof. For $f\in\mathcal{S}$, define $f^\flat (x)=\overline{f(-x)}$. Then

    \begin{align*}
        \hat{f^\flat}(\xi)&=\int_{-\infty}^{\infty} \overline{f(-x)}e^{-2\pi ix\xi} dx=\int_{-\infty}^{\infty} \overline{f(x)}e^{2\pi ix\xi} dx \\
        &=\overline{\left(\int_{-\infty}^{\infty} f(x)e^{-2\pi ix\xi} dx\right)}=\overline{\hat{f}(\xi)}
    \end{align*}

Now let $h=f*f^\flat$ then $\hat{h}(\xi)=\hat{f}(\xi)\hat{f^\flat}(\xi)=|\hat{f}(\xi)|^2$. Also from the definition, we have

    $$h(0)=\int_{-\infty}^{\infty} f(x)f^\flat(-x) dx=\int_{-\infty}^{\infty} |f(x)|^2 dx$$

But from the Fourier inversion formula, we have

    $$h(0)=\int_{-\infty}^{\infty} \hat{h}(\xi) d\xi=\int_{-\infty}^{\infty} |\hat{f}(\xi)|^2 d\xi$$

so

    $$h(0)=\int_{-\infty}^{\infty} |f(x)|^2 dx=\int_{-\infty}^{\infty} |\hat{f}(\xi)|^2 d\xi$$

and this implies $\|f\|=\|\hat{f}\|$.

We have noted that the Fourier transform $\mathcal{F} : \mathcal{S}\to\mathcal{S}$ has its inverse $\mathcal{F}^* : \mathcal{S}\to\mathcal{S}$. The Plancherel's theorem illustrates that $\|f\|=\|\mathcal{F}(f)\|$ and this implies that the Fourier transform is an unitary transform between Hermitian space $\mathcal{S}$.

We can also prove that the Plancherel's theorem holds for the functions of moderate decrease. To complete the proof, we need $f*g\in\mathcal{M}$ and then $\widehat{(f*g)}=\hat{f}\hat{g}$ for $f, g\in\mathcal{M}$.

Proposition 2. If $f, g\in \mathcal{M}$ then $f*g\in\mathcal{M}$.

proof. To prove $f*g\in\mathcal{S}$, we show that $|(f*g)(x)|$ decrease moderately in $x$. Now consider

    $$|(f*g)(x)|\le\int_{-\infty}^{\infty} |f(x-y) g(y)| dy=\int_{|y|\le|x|/2}+\int_{|y|>|x|/2}$$

then for the first integral, if $|y|\le |x|/2$ then

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

holds. Thus we have

    $$\int_{|y|\le|x|/2} |f(x-y)g(y)|dy\le \frac{4A}{4+x^2}\int_{|y|\le|x|/2} |g(y)|dy<\frac{C}{1+x^2}$$

for some constant $C$. Now for the second integral, if $|y|>|x|/2$ then

    $$|g(y)|\le \frac{A'}{1+y^2}\le\frac{A'}{1+(x/2)^2}$$

holds. Thus we have

    $$\int_{|y|>|x|/2} |f(x-y)g(y)| dy\le \frac{4A'}{4+x^2}\int_{|y|>|x|/2} |f(x-y)| dy<\frac{C'}{1+x^2}$$

for some constant $C'$. Then $(f*g)(x)=O(1/1+x^2)$ so $f*g\in\mathcal{M}$.

Then for $f, g\in\mathcal{M}$ we have $\widehat{(f*g)}=\hat{f}\hat{g}$ from the argument similar to Proposition 1 of the last article, as mentioned earlier. Thus the Plancherel's theorem can be extended to functions of moderate decrease.

Theorem 2. Planhcherel - If $f\in\mathcal{M}$, then $\|\hat{f}\|=\|f\|$.