[Fourier Analysis] Fourier inversion formula / 푸리에 역변환 공식

참고문헌 : Stein, Elias M., Shakarchi, Rami 2003. Fourier Analysis. Princeton University Press.
 
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1.6 The Fourier inversion formula

We are now in the heart of the theory of Fourier transform. To prove Fourier inversion formula, we first need the following proposition.

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

    $$\int_{-\infty}^{\infty} f(x)\hat{g}(x)dx=\int_{-\infty}^{\infty}\hat{f}(y) g(y)dy$$

Before we prove the proposition, we need to discuss the interchange of the order of integration for double integrals. Suppose $F(x, y)$ is a continuous function in the plane. If $F(x, y)$ is absolutely integrable, that is, if

$$\iint|F(x, y)| dxdy <\infty$$
 
then we have

$$\iint F(x, y) dxdy=\iint F(x, y) dydx$$

from Fubini's theorem. Finally, we define

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

then they are both continuous if $F$ is continuous on the plane and decreases moderately on both variables : that is, if $F$ satisfies

$$|F(x, y)|\le \frac{A}{(1+x^2)(1+y^2)}$$.

In this case, first note that

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

from the above argument. Next, it is obvious that $F_1(x)$ and $F_2(y)$ decrease moderately on the variable $x$ and $y$ respectively. Now we show that they are both continuous functions.

From the decay condition, we may take large $N$ such that

$$\int_{|y|\ge N} |F(x, y)|dy <\epsilon$$

for all $x\in\mathbb{R}$. Then we have

\begin{align*}
    |F_1(x+h)-F_1(x)|&\le\int_{-\infty}^{\infty}|F(x+h, y)-F(x, y)| dy \\
    & = \int_{|y|\le N} |F(x+h, y)-F(x, y)| dy+\int_{|y|\ge N} |F(x+h, y)-F(x, y)| dy\\
\end{align*}

The second integral is bounded by $2\epsilon$. For the first integral, since $F$ is continuous, it is uniformly continuous on the compact interval $[-N-1, N+1]$. Then there exists uniform $0<h_0<1$ such that $|h|<h_0$ implies $|F(x+h, y)-F(x, y)|<\epsilon/N$ for all $x\in[-N, N]$. This bounds the first integral by $2\epsilon$. Thus $F_1$ is uniformly continuous on $\mathbb{R}$. The same argument applies to $F_2$, concluding it is also uniformly continuous on $\mathbb{R}$.

Now we are ready to prove the proposition.

proof. Define $F(x, y)=f(x)g(y)e^{-2\pi ixy}$ then $F_1(x)=f(x)\hat{g}(x)$ and $F_2(y)=\hat{f}(y)g(y)$. If $f, g\in\mathcal{M}$ then $F$ decreases moderately on both variables in the sense that

    $$|F(x, y)|\le \frac{A}{(1+x^2)(1+y)^2}$$

then $F_1$ and $F_2$ are both continuous and satisfies

    $$\int_{-\infty}^{\infty} F_1(x) dx=\int_{-\infty}^{\infty} F_2(y) dy \;\Rightarrow\; \int_{-\infty}^{\infty} f(x)\hat{g}(x)dx=\int_{-\infty}^{\infty}\hat{f}(y) g(y)dy $$

which is precisely the assertion of the proposition.

Theorem 1. Fourier Inversion : If $f\in\mathcal{S}$ then

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

proof. First, claim that

    $$f(0)=\int_{-\infty}^{\infty}\hat{f}(\xi) d\xi$$
 
Let $G_\delta(x)=e^{-\pi\delta x^2}$ then for $F(x)=e^{-\pi x^2}$, we have $G_\delta(x)=F(\delta^{1/2} x)$. Therefore,

    $$G_\delta(x)=F(\delta^{1/2} x)\longrightarrow \delta^{-1/2}\widehat{F}(\xi/\delta^{1/2})=\delta^{-1/2}e^{-\pi\xi^2/\delta}=K_\delta(\xi)$$

so $\widehat{G_\delta}(\xi)=K_\delta(\xi)$. By the multiplication formula, we have

    $$\int_{-\infty}^{\infty} f(x)K_\delta(x)dx=\int_{-\infty}^{\infty}\hat{f}(\xi) G_\delta(\xi) d\xi$$

Since $K_\delta(-x)=K_\delta(x)$, we have

    $$(f*K_\delta)(0)=(K_\delta*f)(0)=\int_{-\infty}^{\infty} f(x)K_\delta(x) dx$$
  
then as $\delta\to 0$, the integral tends to $f(0)$ since $K_\delta(x)$ is a good kernel. Note that $\hat{f}(\xi)G_\delta(\xi)$ decreases rapidly since $|\hat{f}(\xi)G_\delta(\xi)|\le |\hat{f}(\xi)|$ and the integrand converges for every $\xi$ to $\hat{f}(\xi)$ as $\delta\to 0$. So we may apply Lesbegue's dominant convergence theorem to RHS concluding that

    $$\lim_{\delta\to 0}\left(\int_{-\infty}^{\infty} \hat{f}(\xi)G_\delta(\xi) d\xi\right)=\int_{-\infty}^{\infty}\left(\lim_{\delta\to 0} \hat{f}(\xi)G_\delta(\xi) d\xi \right)=\int_{-\infty}^{\infty} \hat{f}(\xi)d\xi $$

and this proves the claim. In general, let $F(y)=f(y+x)$ then

    $$f(x)=F(0)=\int_{-\infty}^{\infty} \widehat{F}(\xi) d\xi=\int_{-\infty}^{\infty}\hat{f}(\xi)e^{2\pi ix\xi} d\xi$$

so the theorem is proved.

Now define the maps $\mathcal{F}:\mathcal{S}\to\mathcal{S}$ and $\mathcal{F}^*:\mathcal{S}\to\mathcal{S}$ by

$$\mathcal{F}\{f(x)\}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi} dx\;\;\;\text{and}\;\;\; \mathcal{F}^*\{g(\xi)\}(x)=\int_{-\infty}^{\infty} g(\xi)e^{2\pi ix\xi} d\xi$$

then $\mathcal{F}$ is a Fourier transform and $\mathcal{F}^*$ is a Fourier inverse transform. Note again that $f\in\mathcal{S}$ guarantees $\hat{f}\in\mathcal{S}$ so $\mathcal{F}$ is well defined and the Fourier inversion formula guarantees that $\mathcal{F}^*$ is well defined. The inversion formula also implies that $\mathcal{F}^*\circ\mathcal{F}=I$ where $I:\mathcal{S}\to\mathcal{S}$ is an identity. Moreover, we have $\mathcal{F}\circ\mathcal{F}^*=I$. To prove this, we need the duality of the Fourier transform. If we write $\mathcal{F}\{g(x)\}(\xi)=\hat{g}(\xi)=h(\xi)$ then

\begin{align*}
    \mathcal{F}\{h(x)\}(-\xi)&=\int_{-\infty}^{\infty}h(x)e^{2\pi ix\xi}d\xi=\int_{-\infty}^{\infty}\hat{g}(x)e^{2\pi ix\xi}dx=g(\xi)
\end{align*}

so $\hat{h}(-\xi)=g(\xi) \Rightarrow \hat{h}(\xi)=g(-\xi)$. We can summary this as

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

This property of Fourier transform is called the duality. From this,

$$f(x)\underset{\mathcal{F}}{\longrightarrow} \hat{f}(\xi) \underset{\mathcal{F}}{\longrightarrow} f(-x) \underset{\mathcal{F}}{\longrightarrow} \hat{f}(-\xi) \underset{\mathcal{F}}{\longrightarrow} f(x)$$

so $\mathcal{F}^4=\mathcal{F}\circ\mathcal{F}\circ\mathcal{F}\circ\mathcal{F}=I$ and $\mathcal{F}(\mathcal{F}^*(f))=\mathcal{F}(\hat{f}(-\xi))=f$ therefore $\mathcal{F}\circ\mathcal{F}^*=I$. This implies that $\mathcal{F}^*$ is the inverse of $\mathcal{F}$ in $\mathcal{S}$. This leads to the following theorem.

Theorem 2. The Fourier transform is a bijective mapping on the Schwartz space.

Example 1. Define

$$\Pi(x)=\chi_{[-1/2, 1/2]}\;\;\;\text{and}\;\;\;\Lambda(x)=\begin{cases}
    1-|x| & \text{if}\;\; |x|\le 1 \\
    0 & \text{if}\;\;\text{otherwise}
\end{cases}$$

then $\Lambda(x)=(\Pi*\Pi)(x)$ since

$$(\Pi*\Pi)(x)=\int_{-\infty}^{\infty} \Pi(x-y)\Pi(y) dy=\int_{-1/2}^{1/2} \Pi(x-y) dy=\int_{x-1/2}^{x+1/2} \Pi(y) dy=\Lambda(x)$$

We will prove later that $\widehat{(f*g)}(\xi)=\hat{f}(\xi)\hat{g}(\xi)$ thus we have $\widehat{\Lambda}(\xi)=\widehat{\Pi}(\xi)\cdot\widehat{\Pi}(\xi)$. Then $\widehat{\Pi}(\xi)$ is given by

$$\widehat{\Pi}(\xi)=\int_{-\infty}^{\infty}\chi_{[-1/2, 1/2]} e^{-2\pi ix\xi} dx=\int_{-1/2}^{1/2} e^{-2\pi ix\xi} dx=\begin{cases}
    \sin \pi\xi/\pi\xi & \text{if}\;\;\; \xi\ne 0 \\
    1 & \text{if}\;\;\; \xi=0
\end{cases}$$

so

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

Then since $\Lambda (x) \to \widehat{\Lambda}(\xi)$, we have

$$\left(\frac{\sin \pi x}{\pi x}\right)^2 \to \Lambda(-\xi)=\Lambda(\xi)$$

from duality. Later, we will define the Fejér kernel on the real line as

$$\mathcal{F}_N(t)=\begin{cases}
    N(\sin N\pi t/N\pi t)^2 & \text{if}\;\;\; t\ne 0 \\
    N & \text{if}\;\;\; t=0
\end{cases}$$

then $\mathcal{F}_N(t)=N\widehat{\Lambda}(Nt)$ so by duality,

$$\widehat{\mathcal{F}}_N(\xi)=\Lambda(-\xi/N)=\Lambda(\xi/N)=\begin{cases}
    1-|\xi|/N & \text{if}\;\;\; |\xi|\le N \\
    0 & \text{if}\;\;\; \text{otherwise}
\end{cases}$$

Note that $f(\delta x)\rightarrow \delta^{-1}\hat{f}(\xi/\delta)$ from simple change of variables.