Seanie Lee

I love mathematics and statistics

Fourier Inversion Formula

2 minute read

Proposition (Multiplication formula)

Let $f,g\in\mathcal{S}(\mathbb{R})$. Then

\[\begin{align*} \int_{-\infty}^\infty f(x)\hat{g}(x) dx = \int_{-\infty}^\infty \hat{f}(y) d(y) dy \end{align*}\]

Remark

Recall Fubini’s theorem: if $F(x,y)$ is a continuous function on $\mathbb{R}^2$ satisfying the condition

\[\begin{align*} \lvert F(x,y) \rvert \leq \frac{A}{(1+x^2)(1+y^2)}, \end{align*}\]

then

\[\begin{align*} \int_{-\infty}^\infty \int_{-\infty}^\infty F(x,y) dxdy = \int_{-\infty}^\infty \int_{-\infty}^\infty F(x,y) dy dx. \end{align*}\]

<Proof>

Let $F(x,y)= f(x)g(y)e^{-2\pi i xy}$. Then $F$ satisfies the decay conditions for Fubini’s theorem. Applying the theorem to the function,

\[\begin{align*} \int_{-\infty}^\infty f(x)\hat{g}(x) dx = \int_{-\infty}^\infty\int_{-\infty}^\infty f(x) g(y) e^{-2\pi i xy}dy dx = \int_{-\infty}^\infty\int_{-\infty}^\infty g(y) f(x) e^{-2\pi i xy} dx dy = \int_{-\infty}^\infty g(y) \hat{f}(y) dy \end{align*}\] \[\tag*{$\square$}\]

Theorem (Fourier inversion)

If $f\in\mathcal{S}(\mathbb{R})$, then

\[\begin{align*} f(x) = \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x \xi} d\xi \end{align*}\]

<Proof>

First prove for $x=0$. Take $G_\delta(x)$ to be what should be the inverse Fourier transform of $K_\delta(x)=\delta^{-1/2} \exp(-\pi x^2/\delta)$, i.e.,

\[\begin{align*} G_\delta(x) = e^{-\pi\delta x^2}, \widehat{G_\delta}(\xi)=K_\delta(\xi). \end{align*}\]

By the multiplication formula,

\[\begin{align} \int_{-\infty}^\infty f(x) K_\delta(x) dx = \int_{-\infty}^\infty \hat{f}(\xi) G_\delta (\xi) d\xi. \label{eq:1} \end{align}\]

Since

\[\begin{align*} f*K_\delta(x) &= \int_{-\infty}^\infty f(x-t) K_\delta (t) dt \\ &=\int_{-\infty}^\infty f(y) K_\delta (y-x) dy \end{align*}\]

by the change of variable and uniform convergence of $f*K_\delta$, the left hand side of Equation $\ref{eq:1}$

\[\begin{align*} f*K_\delta (0)=\int_{-\infty}^\infty f(y) K_\delta(y)dy \to f(0) \end{align*}\]

as $\delta \to 0^+$. Clearly, right hand side of Equation $\ref{eq:1}$,

\[\begin{align*} \int_{-\infty}^\infty \hat{f}(\xi) e^{-\pi \delta \xi^2}d\xi \to \int_{-\infty}^\infty \hat{f}(\xi) d\xi \text{ as } \delta \to 0. \end{align*}\]

Thus,

\[\begin{align*} f(0)= \int_{-\infty}^\infty \hat{f}(\xi) d\xi. \end{align*}\]

For general $x\in\mathbb{R}$, let $F(y) = f(y+x)$. By the Proposition,

\[\begin{align*} f(x) = F(0) = \int_{-\infty}^\infty \hat{F}(\xi) d\xi = \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x\xi} d\xi. \end{align*}\] \[\tag*{$\square$}\]

Definition (Inverse Fourier transform)

Given $g\in\mathcal{S}(\mathbb{R})$, we define the inverse Fourier transform $\check{g}$ of $g$ by

\[\begin{align*} \mathcal{F}^*(g)(x) = \check{g}(x) = \int_{-\infty}^\infty g(\xi) e^{2\pi ix\xi}d\xi \end{align*}\]

Theorem

Let $\mathcal{F}: \mathcal{S}(\mathbb{R}) \to\mathcal{S}(\mathbb{R})$ be Fourier transform and let $\mathcal{F}^*: \mathcal{S}(\mathbb{R}) \to\mathcal{S}(\mathbb{R})$ be inverse Fourier transform. Then Fourier transform is a bijective mapping on the Schwartz spzce.

<Proof>

By Fourier inversion,

\[\begin{align*} (\mathcal{F}^* \circ \mathcal{F})(g)(y)&=\int_{-\infty}^\infty \left( \int^\infty_{-\infty}g(y)e^{-2\pi iy\xi}dy\right)e^{2\pi i y\xi}d\xi \\ &=\int_{-\infty}^\infty \hat{g}(\xi) e^{2\pi iy\xi}d\xi \\ &=g(y) \end{align*}\]

for all $y\in\mathbb{R}$. Thus $\mathcal{F}^* \circ \mathcal{F}=I$.

Since $\mathcal{F}(g)(y) = \mathcal{F}^*(g)(-y)$, by setting $\zeta=-\xi$,

\[\begin{align*} (\mathcal{F}\circ \mathcal{F}^*)(g)(y)&=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty g(y)e^{2\pi i y \xi}dy \right)e^{-2\pi i y \xi}d\xi \\ &=\int_{-\infty}^\infty \left(\int_{-\infty}^\infty g(y)e^{-2\pi i y \zeta}dx\right)e^{2\pi i y\zeta} d\xi \\ &=\int_{-\infty}^\infty \hat{g}(\zeta) e^{2\pi i y\zeta} d\zeta \\ &=(\mathcal{F}^* \circ \mathcal{F}) (g)(y). \end{align*}\]

Thusm $F^*$ ins the inverse of the Fourier transform on $\mathcal{S}(\mathbb{R})$.

\[\tag*{$\square$}\]

Plancherel’s theorem

For any $f,g\in\mathcal{S}(\mathbb{R})$, we have

\[\begin{align*} \lVert \hat{f}\rVert_{L^2(\mathbb{R})} = \lVert f\rVert_{L^2(\mathbb{R})} \end{align*}\]

<Proof>

By the multiplication theorem, we get

\[\begin{align*} \int_{-\infty}^\infty f(x)\hat{g}(x) dx = \int_{-\infty}^\infty \hat{f}(y)g(y). \end{align*}\]

Define $g$ such that $\hat{g}=\bar{f}$, i.e., $g=\check{\bar{f}}$.

\[\begin{align*} \int_{-\infty}^\infty f(x)\bar{f}(x) dx &= \int_{-\infty}^\infty \hat{f}(y) \check{\bar{f}}(y) dy \\ &=\int_{-\infty}^\infty \hat{f}(y) \bar{\hat{f}}(y) dy \end{align*}\]

\(\tag*{$\square$}\)

Reference

You may also enjoy