Proving that $f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixy}\int_{-\infty}^{\infty}e^{-ixy}f(x)dxdy$.

Question

I need to show that if $f\in L^2(\mathbb{R})$ then $$\displaystyle f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixy}\left(\int_{-\infty}^{\infty}e^{-ixy}f(x)dx\right)dy$$

Answer 1

The standard proof uses the idea of gaussian regularization. First, write

$$ \mathcal{F}\{f\}(\xi) = \hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-i\xi x} \, \mathrm{d}x. $$

Since $\mathcal{F}$ is a continuous linear operator on $L^2(\mathbb{R})$, it suffices to establish the Fourier inversion formula on a dense subset $\mathcal{D}$ of $L^{2}(\mathbb{R})$. For the purpose of our proof, we will choose $\mathcal{D}$ such that $f \in C_b(\mathbb{R}) \cap L^{1}(\mathbb{R})$ and $\hat{f} \in L^1(\mathbb{R})$ for each $f \in \mathcal{D}$. (Note that such $\mathcal{D}$ indeed exists. For instance, we may choose $\mathcal{D}$ as the Schwartz space.)

Let $f \in \mathcal{D}$. Since $\hat{f} \in L^1(\mathbb{R})$, the dominated convergence theorem tells that

$$ \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \hat{f}(\xi) \, \mathrm{d}\xi = \int_{-\infty}^{\infty} e^{ix\xi} \hat{f}(\xi) \, \mathrm{d}\xi. $$

On the other hand, by the Fubini's Theorem and the gaussian integral,

\begin{align*} \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \hat{f}(\xi) \, \mathrm{d}\xi &= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \left( \int_{-\infty}^{\infty} e^{-i\xi y}f(y) \, \mathrm{d}y \right) \, \mathrm{d}\xi \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{i(x-y)\xi} e^{-\epsilon \xi^2} \, \mathrm{d}\xi \right) f(y) \, \mathrm{d}y \\ &= \frac{1}{\sqrt{4\epsilon\pi}} \int_{-\infty}^{\infty} e^{-\frac{(y-x)^2}{4\epsilon}} f(y) \, \mathrm{d}y \end{align*}

Substituting $y=x+\sqrt{4 \epsilon \pi} \, t$, it follows that

\begin{align*} \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} e^{-\epsilon \xi^2} \hat{f}(\xi) \, \mathrm{d}\xi &= \int_{-\infty}^{\infty} e^{-\pi t^2} f(x+\sqrt{4 \epsilon \pi} \, t) \, \mathrm{d}t \end{align*}

Since $f$ is bounded and continuous, this converges to

$$ \int_{-\infty}^{\infty} e^{-\pi t^2} f(x) \, \mathrm{d}t = f(x) $$

as $\epsilon \to 0^+$ by the dominated convergence theorem. This proves that

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

for all $f \in \mathcal{D}$ and $x \in \mathbb{R}$, hence establishing the Fourier inversion formula over $\mathcal{D}$. Now as mentioned in the beginning, this extends to all of $L^2(\mathbb{R})$ via the continuity of $\mathcal{F}$.