Showing $f=0$ almost everywhere. 
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.

I can say 
$$f*g(x)=\int_\mathbb{R}f(y)e^{-(x-y)^2/2}dy=0,$$
where $g=e^{-x^2/2}$. Hence, $\widehat{f*g}=0$.
Now I'm not sure if it's good enough to say $\widehat{f*g}=\hat{f} \cdot \hat{g}$. As a result, if $\hat{g}\neq 0$, then $\hat{f}=0$ and then $f=0$.     
Is it a valid argument? 
 A: With your notation, we have $\hat{g}=\sqrt{2\pi} g$ and hence it is posiitve everywhere. Actually, we have a general formula as follows: 
Let $A$ be a SPD matrix, then
$$
\mathcal{F}(e^{-\frac 1 2\langle Ax,x\rangle})(\xi):=\frac{(2\pi)^{n/2}}{\sqrt{Det A}}e^{-\frac{1}{2}\langle A^{-1}\xi,\xi\rangle},
$$
where $\mathcal{F}$ is a Fourier transformation in $\mathbb{R}^n$.
Thus 
$$
0=\mathcal{F}({f\star g})(\xi)=\hat{f}(\xi) \hat{g}(\xi)=\sqrt{2\pi}\hat{f}(\xi) g(\xi),
$$
whence $\hat{f}(\xi)=0$. Thus, $f(x)=\mathcal{F}_{x\to \xi}^{-1}({\hat{f}})(x)=0$.
A: \begin{align}
& \int_{\mathbb R} f(y)e^{-(x-y)^2/2} \, dy \\[10pt]
= {} & \int_{\mathbb R} f(y) e^{-x^2/2} e^{xy -y^2/2} \, dy \\[10pt]
= {} & e^{-x^2/2} \int_{\mathbb R} \Big( f(y) e^{-y^2/2} \Big) e^{xy} \, dy \quad (\text{since } e^{-x^2/2} \text{ does not depend on } y) \\[10pt]
= {} & e^{-x^2/2} \int_{\mathbb R} h(y) e^{xy} \, dy.
\end{align}
Since $e^{-x^2/2}$ cannot be $0,$ this product cannot be $0$ unless this last integral is $0.$ This integral is a two-sided Laplace transform of $h$ evaluated at $x$ (or at $-x$ if you want to do it that way). A theorem says that transform cannot be identically $0$ unless $h$ is almost everywhere $0.$ It seems I've lost track of a reference to the theorem; I'll see if I can find it . . .  And $e^{-y^2/2}$ can never be $0,$ so $f$ is almost everywhere $0.$
