TITLE: What will be the support of the convolution of two test functions.
QUESTION [12 upvotes]: If $g\in C^{\infty}_c$ defined on $\Bbb R^n$ and K is the support of function $g$. I want to find the support of $g_\epsilon$. Where $g_\epsilon$ is regularization of $g$.

Regularization of $g$ is defined as:
$g_\epsilon$:=  $g*\omega_\epsilon$ (convolution of g and a test function) i.e. $g_\epsilon(x
)$=$\int_{\Bbb R^n}g(x-y).\omega_\epsilon(y)dy $
I think, this process is also known as mollification. 
Here 
$ \omega_\epsilon (x)= 
\begin{cases}
\frac{C^{-1}}{\epsilon^n} e^{{\frac{-\epsilon^2}{{\epsilon^2}-|x|^2}}},  &\text{for |x|< $\epsilon$ } \\
0, & \text{otherwise}  \\
\end{cases} $ 

I only know that support of convolution of two compactly supported functions is again a compact set. And, as support of $\omega_\epsilon $ is a ball $B(0,\epsilon)$ and support of g is a compact set K(given) then support of their convolution should be the intersection of the support of $g(x-y)$ and $\omega_\epsilon(y)$. 
Because for non zero function integration remains non zero...

REPLY [3 votes]: Reminder: The sum of two compact sets, e.g. K an the closed ball $M := B_\varepsilon(0)$ is compact again, as it is the continuous image of a
compact set $K \times M$ in the product domain. Therefore, if both supports are compact you can omit the closure on the right hand side.