show that $f_{\epsilon} \in D(\Omega)$; moreover, $f_{\epsilon} \to f$ uniformly as $\epsilon \to 0$.

Question

Let $K$ be a compact subset of $\Omega \subset \mathbb{R^m}$, $\Omega$ is open and nonempty and let $f \in C(\Omega)$ have support contained in $K$. For $\epsilon \gt 0$, let

$$f_{\epsilon}(x)=\frac{1}{C(\epsilon)}\int_{K} \phi_{y,\epsilon}(x) f(y) dy$$

where $$C(\epsilon)=\int_{\mathbb{R^m}}\phi_{y,\epsilon}(x) dy$$, For a fixed $y$

$$\phi_{y,\epsilon}(x)= \left\{ \begin{array}{ll} \exp(-\frac{\epsilon^2}{\epsilon^2-|x-y|^2}) & \mbox{if $|x-y| \lt \epsilon$};\\ 0 & \mbox{otherwise}.\end{array} \right. $$

If $\epsilon \lt \text{dist}(K,\partial \Omega)$, then show that $f_{\epsilon} \in D(\Omega)$; moreover, $f_{\epsilon} \to f$ uniformly as $\epsilon \to 0$.

I look at $$|\frac{1}{C(\epsilon)}\int_{K} \phi_{y,\epsilon}(x) f(y) dy-f(x)|=|\frac{\int_{K} \phi_{y,\epsilon}(x) f(y) dy-\int_{K}\phi_{y,\epsilon}(x)f(x) dy}{\int_{K}\phi_{y,\epsilon}(x) dy}| $$ $$ \le \frac{\int_{K}|f(y)-f(x)|\phi_{\epsilon,y}(x) dy}{\int_{K} \phi_{y,\epsilon}(x)dy}$$

For $y,x \in K$, as $\epsilon \to 0, |x-y| \lt \epsilon$ and continuity of $f$ forces $f(y)-f(x)| \lt \delta$ (for any given $\delta\gt 0$) and hence the difference can be arbitrarily small as well.

$f_{\epsilon}$ is compactly supported as $f$ is compactly supported. I am unable to show that $f_{\epsilon}$ is $C^{\infty}$ rigorously.

Thanks for the help!!

$f_\epsilon$ is $C^\infty$ because $\phi_{y,\epsilon}$ is. See here — Gregor de Cillia, Feb 07 '16 at 15:22
You can write $f_\epsilon$ as $f\star \mu_\epsilon$ with. $\mu_\epsilon(x,y)=\phi_{y,\epsilon}(x)$. From the theorem, you get $D^{\alpha}f = f\star (D^{\alpha}\mu_\epsilon)$ which exists for each multiindex $\alpha$ — Gregor de Cillia, Feb 07 '16 at 15:56
@GregordeCillia I should define $\mu_{\epsilon}=\exp^{-\frac{\epsilon^2}{\epsilon^2-|x|^2}}$ ?? — tattwamasi amrutam, Feb 07 '16 at 16:07
Yes, exactly. Then you have to show that $\frac{d}{dx}\int f(y)\mu_\epsilon(x-y)\ dy = \int f(y)\mu'_\epsilon(x-y)\ dy$. As far as i know, this can be done with the dominated convergence theorem. — Gregor de Cillia, Feb 07 '16 at 16:09
@is it not because I am differentiating on a compact set and can take the differential operator inside?? — tattwamasi amrutam, Feb 07 '16 at 16:15
I think you have to show more than that. Take a look at the accepted answer here — Gregor de Cillia, Feb 07 '16 at 16:20

show that $f_{\epsilon} \in D(\Omega)$; moreover, $f_{\epsilon} \to f$ uniformly as $\epsilon \to 0$.

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