$\displaystyle g(x):=\int_{\mathbb R^n} e^{2\pi ix\cdot \xi} e^{-\frac{|\xi|^2}{2}} \widehat{f}(\xi)\ d\xi$ is not compact supported? Which function $f\in C^\infty_0(\mathbb R^n)$ should I choose to show that $$\displaystyle g(x):=\int_{\mathbb R^n} e^{2\pi ix\cdot \xi} e^{-\frac{|\xi|^2}{2}} \widehat{f}(\xi)\ d\xi$$ is not compactly supported?
Above, $f$ might be real- or complex-valued, $\widehat{f}$ denotes the Fourier transform of $f$, $|\xi|$ is the Euclidean norm of $\xi$.
Obs: For those who know about the theory of pseudo-differential operators $g$ is nothing but the pseudo-differential operator with symbol $$\sigma(\xi):=e^{-\frac{|\xi|^2}{2}}$$ and what I'm asking for is an example showing that those operators do not take $C^\infty_0(\mathbb R^n)$ into $C^\infty_0(\mathbb R^n)$. 
 A: Consider the convolution
$$
G(x) \equiv \int_{\mathbb R^n}e^{-|x-y|^2/2}f(y) dx = \left[e^{-|y|^2/2}\ast f(y)\right](x);
$$
for $f\in C^\infty_0(\mathbb R^n)$, then $G\in C^{\infty}(\mathbb R^n)$.
Now, the Fourier transform of the convolution of $L^1(\mathbb R^n)$ functions is the product of the Fourier transforms of the functions themselves
$$
\widehat G(\xi) = \widehat{e^{-|y|^2/2}}(\xi) \widehat{f}(\xi)=e^{-|\xi|^2/2}\widehat f(\xi);
$$
since $f$ was $L^1(\mathbb R^n)$ and hence $\widehat f$ is limited, $\widehat G$ is itself $L^1(\mathbb R^n)$: we may then apply the reverse Fourier transform, getting
$$
G(x)=\int_{\mathbb R^n}e^{+i2\pi \xi\cdot x}e^{-|\xi|^2/2}\widehat f(\xi)d\xi=g(x).
$$
We are now left with the task to prove that $G(x)$ was not compactly supported to begin with. By the Paley-Wiener theorem, if $\tau$ is a distribution on $\mathbb R^n$ which is compactly supported in the ball of radius $\rho$, then
$
\widehat \tau
$ is an analytic (in fact, entire) function and,
for some $N\in\mathbb N$ and an appropriate constant $C_N$,
$$
\left|\widehat \tau(\zeta)\right|<C_N(1+|\zeta|)^Ne^{\Im(\zeta)\rho}.
$$
Letting $\zeta=\xi+i\eta$, however,
$$ 
\left|e^{-\zeta^2}\right|
=\left|e^{-(\xi^2-\eta^2+i2\eta\cdot \xi)}\right|
=e^{-\xi^2+\eta^2},$$ 
and above estimates fail to hold for $|\widehat G(\zeta)|$, since the exponent is quadratic in $\Im(\zeta)=\eta.$ Hence, $G(x)$ cannot be of compact support.
Another way: let $\rho_\varepsilon(y)$ be a sequence of mollifiers, i.e. $\rho\in C^{\infty}_0(\mathbb R^n)$, $\rho(x)\ge 0$, 
$$
\int_{\mathbb{R}^n}\rho(x)dx=1,\qquad \rho_\varepsilon(y)\equiv \frac{1}{\varepsilon^n}\rho(x/\varepsilon);
$$
then
$$
G_\varepsilon(x) = \left[ e^{-|y|^2/2}\ast \rho_\varepsilon(y)\right](x)
$$
satisfies, for each $x\in\mathbb R^n$,
$$
\lim_{\varepsilon\to0}G_\varepsilon(x) = e^{-|x|^2/2}
$$
and the limit is nowhere vanishing. This means that, for each $x$, there exists $\varepsilon$ sufficiently small as to ensure $G_\varepsilon(x)>0$.
A: $g$ is essentially, up to constants, the convolution of $e^{-ax^2}$ (for some $a>0$), and $f$. If $f(x)>0$ for all $x$ in the interior of the support of $f$, then $g(x)>0$ for all $x\in\mathbb{R}^n$.
