Show that the following integral is equal to a double integral

Question

Given $\epsilon>0$, define $$g(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$ and $$g_\epsilon(x) = \frac{1}{\epsilon}g\left(\frac{x}{\epsilon}\right) = \frac{1}{\epsilon\sqrt{2\pi}}e^{-x^2/(2\epsilon^2)}.$$

Then for $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, show that the integral

$$\int_{-\infty}^{\infty}\hat{g_\epsilon}(\xi)\vert \hat{f}(\xi)\vert^2 d\xi = 2\pi\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x)\overline{f(y+x)}g_\epsilon(y) dydx$$

using Fubini's Theorem and the Fourier Inversion Theorem.

Here is my attempt:

\begin{align} \int_{-\infty}^{\infty}\hat{g_\epsilon}(\xi)\vert \hat{f}(\xi)\vert^2 d\xi &= \int_{-\infty}^{\infty}\hat{g_\epsilon}(\xi) \hat{f}(\xi)\overline{\hat{f}(\xi)} d\xi\\ &= \int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty}g_\epsilon(y)e^{-i\xi y} dy\right) \left(\int_{-\infty}^{\infty}f(x)e^{-i\xi x}dx\right) \left(\int_{-\infty}^{\infty}f(t)e^{-i\xi t}dt\right)d\xi \end{align}

Beyond this, I haven't made any progress. Any hints on how to apply both theorems?

Answer 1

Begin with $$ |\hat{f}(\xi)|^2 = \hat{f}(\xi) \overline{\hat{f}(\xi)} = \left( \int f(x) \, e^{-i\xi x} \, dx \right) \left( \int \overline{f(z)} \, e^{i\xi z} \, dz \right) . $$ Then change to the variable $y=z-x$ in the second integral: $$ |\hat{f}(\xi)|^2 = \left( \int f(x) \, e^{-i\xi x} \, dx \right) \left( \int \overline{f(y+x)} \, e^{i\xi (y+x)} \, dy \right) = \iint f(x) \overline{f(y+x)} \, e^{i\xi y} \, dy \, dx . $$

Now multiply with $\hat{g}_\epsilon(\xi),$ integrate over $\xi$ and swap the integrals: $$ \int |\hat{f}(\xi)|^2 \, \hat{g}_\epsilon(\xi) \, d\xi = \int \left( \iint f(x) \overline{f(y+x)} \, e^{i\xi y} \, dy \, dx \right) \, \hat{g}_\epsilon(\xi) \, d\xi \\ = \iint f(x) \overline{f(y+x)} \left( \int \hat{g}_\epsilon(\xi) \, e^{i\xi y} \, d\xi \right) \, dy \, dx \\ = \iint f(x) \overline{f(y+x)} \, 2\pi \, g_\epsilon(y) \, dy \, dx \\ = 2\pi \iint f(x) \overline{f(y+x)} \, g_\epsilon(y) \, dy \, dx . $$

Answer 2

We can show this while replacing $g_\varepsilon$ by any $L^1 \cap L^2$ function $h$. I'll also use interchangeably $\hat{u}$ and $\mathcal{F}[u]$ to denote the Fourier transform of a function $u$ for my own convenience, and $\mathcal{F}^{-1}[v]$ will be the inverse Fourier transform of a function $v$.

First, let's recall some facts about Fourier transforms (I won't prove them here, but you should be able to either find these proofs by yourself or find them in some books/online):

• If we have $u : x \mapsto \overline{v(x)}$ for two $L^1$ functions $u$ and $v$, then: $\forall \xi,\,\,\hat{u}(\xi) = \overline{\hat{v}(-\xi)}$.
  Therefore, here $\xi \mapsto \overline{\hat{f}(\xi)}$ is the Fourier transform of $x \mapsto \overline{f(-x)}$.
• With the normalisation convention used here, we have, if $u * v$ is the convolution of $u$ with $v$: $\mathcal{F}\left[u * v\right] = \mathcal{F}[u]\mathcal{F}[v]$.
• Finally, we have the following equality for any $L^2$ functions $u$ and $v$: $$\int_\mathbb{R} \mathcal{F}[u](t) v(t) \mathrm{d}t = \int_\mathbb{R} u(t) \mathcal{F}[v](t) \mathrm{d}t$$ If we replace $\mathcal{F}[v]$ by any $L^2$ function $w$ (which we can thanks to the Fourier Inversion Theorem [FIT]) this can be rewritten as: $$\int_\mathbb{R} \mathcal{F}[u](t)\mathcal{F}^{-1}[w](t) \mathrm{d}t = \int_\mathbb{R} u(t) w(t) \mathrm{d}t$$

Let's dive in. We have, for $\xi \in \mathbb{R}$: $$\begin{split} \left|\hat{f}(\xi)\right|^2 &= \hat{f}(\xi) \overline{\hat{f}(\xi)}\\ &= \mathcal{F}[f](\xi) \cdot \mathcal{F}\left[\overline{f(-\,\,\cdot)}\right](\xi)\\ &= \mathcal{F}\left[f * \overline{f(-\,\,\cdot)}\right](\xi)\\ &= \int_\mathbb{R} e^{-it\xi} \left(\int_\mathbb{R} f(x) \overline{f(-(t - x))}\right)\mathrm{d}t\\ &= \int_\mathbb{R} e^{iy\xi} \left(\int_\mathbb{R} f(x) \overline{f(y + x)}\right)\mathrm{d}y \quad \text{with the change } y := -t, \mathrm{d}y = -\mathrm{d}t\\ &= 2\pi \mathcal{F}^{-1}\left[\int_\mathbb{R} f(x) \overline{f({{} \cdot}+ x)} \mathrm{d}x\right]\end{split}$$ thanks to the FIT and the fact that $f$ is $L^1 \cap L^2$, providing that $\int_\mathbb{R} f(x) \overline{f(-({{} \cdot}- x))} \mathrm{d}x$ and thus $\int_\mathbb{R} f(x) \overline{f({{} \cdot}+ x)} \mathrm{d}x$ are $L^1 \cap L^2$.

Hence, we obtain: $$\begin{split} {}\\ \int_\mathbb{R} \hat{h}(\xi)\left|\hat{f}(\xi)\right|^2 \mathrm{d}\xi &= 2\pi \int_\mathbb{R} \mathcal{F}[h](\xi)\mathcal{F}^{-1}\left[\int_\mathbb{R} f(x) \overline{f({{} \cdot}+ x)} \mathrm{d}x\right](\xi) \mathrm{d}\xi\\ &= 2\pi \int_\mathbb{R} h(\xi)\left(\int_\mathbb{R} f(x) \overline{f(\xi + x)} \mathrm{d}x\right) \mathrm{d}\xi\\ &= 2\pi \int_\mathbb{R} \int_\mathbb{R} f(x) \overline{f(y+x)} h(y) \mathrm{d}y \mathrm{d}x\quad\text{using Fubini (left to be justified)}\end{split}$$ where I only renamed the $\xi$ into $y$ in the integral to have the desired formula.