Approximate Sobolev function by smooth function - error estimate?

Question

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!

Answer 1

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*}

Answer 2

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.