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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 $D$?

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    $\begingroup$ 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. $\endgroup$ – RideTheWavelet Sep 24 '17 at 19:34
  • $\begingroup$ If we use this mollifier, can we say that the support of $f_{\epsilon}$ is the set $x\in D: \mathrm{dist}(x,\partial D)\geq\epsilon$? $\endgroup$ – M. Rahmat Sep 24 '17 at 19:51
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    $\begingroup$ 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. $\endgroup$ – RideTheWavelet Sep 24 '17 at 22:38
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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 $D$.

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