Use Fourier Transform to Show that $f=0$ a.e.

Question

I was working through an old qualifier on my own when I ran across this following question that I was unable to crack. Here it is verbatim:

"Let $f\in L^2(\mathbb{R}, \mathcal{L}, m)$ and suppose that

$$\int_{\mathbb{R}} f(y) e^{-(x-y)^2/2} \, dy = 0$$

for all $x\in \mathbb{R}$. Prove that $f = 0$ a.e."

There is a hint to use the Fourier Transform as well.

Even with the hint, I am still confused after reading as much as I could from the various textbooks that I have. Am I supposed to use the transform to turn this into something that looks like the typical problem asking to show that $f=0$ a.e? In other words, I have dealt with questions asking me to show a function is zero given some conditions on the integral, but not like this. I can't see my way through this one.

If you have any hints, suggestions, or solutions, I would appreciate whatever you are willing to share.

Answer 1

Set $g(x):=e^{-x^{2}/2}$ and recall that

$$\widehat{g}(\xi)=\int_{\mathbb{R}}e^{-y^{2}/2}e^{-2\pi i y\xi}\mathrm{d}y=e^{-\xi^{2}/2},\qquad\forall \xi\in\mathbb{R}$$

Let $\left\{f_{n}\right\}$ be a sequence of functions in $L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$ converging to $f$ in $L^{2}$ and pointwise a.e. Applying the convolution theorem to $f_{n}$ and $g$, we have

$$\widehat{(f_{n}\ast g)}(\xi)=\widehat{f}_{n}(\xi)\widehat{g}(\xi)=\widehat{f}_{n}(\xi)e^{-\xi^{2}/2},\qquad\forall \xi\in\mathbb{R}$$

Since $f_{n}\rightarrow f$ in $L^{2}$, Young's convolution inequality gives $f_{n}\ast g\rightarrow f\ast g$ in $L^{2}$. Passing to a subsequence if necessary, we may assume that convergence is pointwise a.e. too. $\widehat{f_{n}}\rightarrow\mathcal{F}(f)$ in $L^{2}$ by definition of the $L^{2}$ Fourier transform $\mathcal{F}$. Passing to a subsequence if necessary, we may assume convergence is pointwise a.e. too. Since the $e^{-\xi^{2}/2}$ is bounded, we see that the RHS converges in $L^{2}$ to $\mathcal{F}(f)\widehat{g}(\xi)$.

Combining these observations with the continuity of $\mathcal{F}$, we have

\begin{align*} \mathcal{F}(f\ast g)=\lim_{n\rightarrow\infty}\mathcal{F}(f_{n}\ast g)=\lim_{n\rightarrow\infty}\widehat{(f_{n}\ast g)}=\lim_{n\rightarrow\infty}\widehat{f}_{n}\widehat{g}=\mathcal{F}(f)\widehat{g} \end{align*}

in $L^{2}$. Your hypothesis gives $$0=\mathcal{F}(f)(\xi)e^{-\xi^{2}/2} \qquad \text{ a.e. }\xi$$

whence $\mathcal{F}(f)(\xi)=0$ for a.e. $\xi$, since the second factor is nonzero. The Fourier transform is an isometry on $L^{2}(\mathbb{R})$, so $f=0$ a.e.

Answer 2

$$ \mathfrak{F}(\int_0^x f(y)g(x-y)dy)=\mathfrak{F}(f(t))\times\mathfrak{F}(g(t))=0$$

In your case $g(x)=e^{-x^2/2}$. $$\mathfrak{F}(g(t))=e^{-x^2/2}>0$$ $$\mathfrak{F}(f(t))=0$$ $$f(t)=0$$