# Mollifiers restricted on the boundary

Consider the function $\varphi$ defined on the domain $\mathcal{D}\subset \mathbb{R}^d$ and its trace $\gamma(\varphi)$ on $\partial \mathcal{D}$ (I assume a smooth boundary). We have that $\varphi \in L^2(\mathcal{D})$ and $\gamma(\varphi)\in L^2(\partial \mathcal{D})$. Now we introduce the mollifier $\rho_\varepsilon \in \overline{\mathcal{D}}$ and we note the regularisation $\varphi_\varepsilon = \varphi * \rho_\varepsilon$ defined on $\mathcal{D}$.

Question: Is it possible to show that the trace of $\varphi_\varepsilon$ on $\partial \mathcal{D}$ converges towards $\gamma(\varphi)$ on $L^2(\partial \mathcal{D})$ with just the given hypothesis or do I need more regularity (such as continuity of $\varphi$ on $\mathcal{D}$ but also up to $\partial \mathcal{D}$)?

• the problem is that the trace operator $\gamma$ isn't bounded $L^2(D) \to L^2(\partial D)$, and your question is if it is bounded on a subspace containing $\varphi$ and $\varphi_\epsilon$ – reuns Sep 12 '16 at 16:15
• @user1952009 thanks for your reply. Instead of demanding more regularity, could I then ask for a weaker result such as the $L^2-$norms of the trace of $\varphi_\varepsilon$ (or a subsequence of norms) converge towards the norm of $\gamma(\varphi)$, on $\partial \mathcal{D}$? – user3371583 Sep 12 '16 at 16:31
• a simple question : why don't you define the unbounded operator $\gamma : L^2(D) \to L^2(\partial D)$ as the limit in $L^2(\partial D)$ of $\gamma(\varphi \ast \rho_\epsilon)$ (where $\gamma : C^\infty_c(\mathbb{R}^d) \to L^2(\partial D)$ is well-defined) ? and hence how do you define $\gamma$ ? – reuns Sep 12 '16 at 16:34
• @user1952009 The problem comes from a hypoelliptic PDE (note $L$ the operator) so for any $f \in C^\infty_c$, I define $\gamma(\varphi)$ as: $\int_{\partial D} \gamma(\varphi) f := \int_{D} \varphi L^* f$. So I can't really redefine it. – user3371583 Sep 12 '16 at 16:42
• what I mean is that I'm thinking you are assuming $\gamma(\varphi \ast \rho_\epsilon) \to \gamma(\varphi)$ in $L^2(\partial D)$, in the very definition of $\gamma$ – reuns Sep 12 '16 at 16:45