Approximate Sobolev function by smooth function - error estimate? I wondered if there is, in general, a way to estimate the error (in $L^2$) between a Sobolev function and its mollified version.
Let's say $\Omega\subset\mathbb{R}^n$ is bounded with smooth boundary and $u\in H^1_0(\Omega)$. Ideally, I'd like to have an estimate of the form
$$\|u-u_\epsilon\|_{L^2(\Omega)} \leq a_\epsilon \|u\|_{H^1(\Omega)},$$
where $u_\epsilon$ is a mollified version of $u$ and $a_\epsilon\to 0$ as $\epsilon\to 0$.
Is there any chance of such a statement being true?
Thanks in advance!
 A: It is true almost in the way you stated it with $a_\epsilon = \epsilon$. You only have to restrict the norm on the left hand side to $\Omega_\epsilon =\{x \in \Omega \colon dist(x, \Omega^c)>\epsilon \}$, since this is where $u_\epsilon$ is defined. On the plus side, boundary conditions and boundary regularity are irrelevant.
The proof is rather straightforward: Write $$u_\epsilon(x) - u(x) = \int_{B_\epsilon (0)} \int_0^1\eta_\epsilon(y) \nabla u(x+ty)\cdot y\,dt\,dy. $$
Then take the squares, use Jensen' inequality and integrate over $\Omega_\epsilon$:
\begin{align*}
\int_{\Omega_\epsilon} \lvert u_\epsilon(x) - u(x)  \rvert^2 \,dx &\leq \epsilon^2 \int_{\Omega_\epsilon} \int_{B_\epsilon (0)} \int_0^1 \eta_\epsilon(y) \lvert \nabla u(x+ty) \rvert^2 \,dt \,dy \,dx \\
&\leq \epsilon^2 \int_\Omega \lvert \nabla u(x) \rvert^2 \,dx
\end{align*}
A: One approach is to use the Fourier transform. Since $u$ is in $H^1_0$, its zero extension to $\mathbb{R}^n$ is in $H^1(\mathbb{R}^n)$. Consider the Fourier transform $\hat u$. The squared $H^1$ norm of $u$ is 
$$\int (1+|\xi|^2) |\hat u|^2\tag1$$
The squared $L^2$ norm of $u-u_\epsilon$ is 
$$\int |1-\hat \eta_\epsilon|^2 |\hat u|^2\tag2$$
where $\eta_\epsilon$ is a mollifier used. If $\eta_\epsilon$ is a Gaussian localized at scale $\epsilon$, then its transform is a very flat Gaussian that is close to $ 1$ in a ball of radius $\sim 1/\epsilon$. 
Split (2) into integrals over $|\xi|<R$ and $|\xi|>R$. The former is much smaller than (1) because  $|1-\hat \eta_\epsilon|$ is small. The latter is much smaller than (1) because  $|\xi|^2 $ is large.
