# Is mollifier of compact support?

Let $f$ be a locally integrable function on a bounded open set $D$ of $\mathbb{R}^{n}$, and let $f_{\epsilon}:=f\ast\phi_{\epsilon}$ be its mollifier. Can we say that $f_{\epsilon}$ has compact support in $D$?

• It depends on the mollifying kernel, but there are ones that preserve compact support, for example the function equal to $e^{-1/(1-\|x\|^{2})}$ when $\|x\|\leq 1$ and $0$ elsewhere. – RideTheWavelet Sep 24 '17 at 19:34
• If we use this mollifier, can we say that the support of $f_{\epsilon}$ is the set $x\in D: \mathrm{dist}(x,\partial D)\geq\epsilon$? – M. Rahmat Sep 24 '17 at 19:51
• If we use the mollifier $\phi_{\epsilon}(x):=\phi(x/\epsilon),$ where $\phi$ is the function I mentioned above, then the support of $f_{\epsilon}$ will be in the set $\{x\in\mathbb{R}^{n}:\mathrm{dist}(x,D)\leq\varepsilon\}$. If you know that $f$ has compact support strictly contained in $D$, then you may choose $\epsilon>0$ small enough that the support of $f$ is contained in $D$, however. – RideTheWavelet Sep 24 '17 at 22:38

In your example, one has to first ask how $f*\phi_\epsilon$ is even defined, given that the domain of definition of $f$ is $D$. I suppose it's done by letting $f=0$ outside of $D$, so that the integral defining the convolution makes sense. Still the support of $f$ will be typically larger than $D$. As a simple example, consider the constant function $f\equiv 1$ in $D$; when it's extended by $0$ outside of $D$, and then convolved, the support of $f_\epsilon$ also includes the $\epsilon$-neighborhood of $D$.