Support of a regularized function

Question

Let $f$ be a function such that $supp(f)=K$. Compute $supp(f_\varepsilon)$ where $f_\varepsilon$ is the regularization of $f$. Im not sure how to do this, since we have no information of the support of the convolution of two functions. We do know that $supp(\omega_\varepsilon)=B(0,\varepsilon)$ and $\omega_\varepsilon=e^{\frac{-1}{1-||x/\varepsilon||^2}} $ for all $\|x\|<\varepsilon$ and 0 otherwise.

Answer 1

The convolution is $$ (f\ast \omega_{\varepsilon})(x)=\int\ f(y)\omega_{\varepsilon}(x-y)\ dy $$ Then note that $f(y)\omega_{\varepsilon}(x-y)\neq 0$ implies $y\in{\rm supp}(f)$ and $x-y\in \bar{B}(0,\varepsilon)$ the closed ball around he origin of radius $\varepsilon$. So the support of the convolution is contained in ${\rm supp}(f)+\bar{B}(0,\varepsilon)$, i.e., the epsilon thickening of the support of $f$. I don't think one can in general compute the support of the convolution exactly because of possible sign cancellations. If $f$ and $\omega_{\varepsilon}$ are nonnegative then the above sum of sets is the support of the convolution.