# Support of a regularized function

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.

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.