Is mollifier of compact support?

Question

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 $D$?

Answer 1

Convolution does not help in making the support of a function smaller; this is what multiplication is used for. Convolution is "fuzzying" the function, spreading its values around. Basically the opposite of making it compact.

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 $D$.