Proving finite essential supremum of $L^\infty \cap H^1$ traces

Question

I want to solve Exercise 4.1 in Optimal Control of Partial Differential Equations by Fredi Troltzsch. It states:

Prove $$||y||_{L^\infty(\Gamma)} \leq ||y||_{L^\infty(\Omega)}, \quad \forall y \in H^1(\Omega)\cap L^\infty(\Omega).$$

Here $\Gamma$ is the boundary of the Lipschitz domain $\Omega$. Following the hint from the book, I've managed to do the following:

Since $H^1(\Omega)\cap C(\overline{\Omega})$ is dense in $H^1(\Omega)$, we start by extracting an approximating sequence $\{y_k\}_{k \in \mathbb{N}}$ for which $y_k \to y$ in $H^1(\Omega)$ as $k \to \infty$.

Next, we consider the continuous projection operator $P:H^1(\Omega)\cap C(\overline{\Omega}) \to H^1(\Omega)\cap C(\overline{\Omega})$ which projects functions onto $[-c,c]$ where $c=||y||_{L^\infty(\Omega)}$. Now any $v \in H^1(\Omega)\cap C(\overline{\Omega})$ has a trace that equals its restriction to the boundary, hence $||P v||_{L^\infty(\Gamma)} \leq ||P v||_{C(\overline\Omega)} \leq c$. Applying this inequality to the above sequence yields $$ ||P y_k||_{L^\infty(\Gamma)} \leq ||P y_k||_{C(\overline\Omega)} \leq c.$$

I would like to argue using sequential continuity that $||P y_k||_{L^\infty(\Gamma)} \to ||y||_{L^\infty(\Gamma)}$ as $k \to \infty$, but I'm stuck doing so.

Answer 1

I'll answer this question myself using a different approach.

Let $T:H^1(\Omega) \to L^2(\Gamma)$ denote the trace operator, which is continuous since $\Omega$ is Lipschitz. Recall that $Py_k \in H^1(\Omega) \cap C(\overline \Omega)$ and thus $TPy_k = Py_k \in [-c,c]$ on $\Gamma$.

Since $y_k \to y$ in $H^1(\Omega)$, we also have that $Py_k \to Py=y$ in $H^1(\Omega)$ by continuity of $P$. By continuity of $T$ we thus have $TPy_k \to Ty$ in $L^2(\Gamma)$, and hence we may extract a subsequence $\{y_{k'}\}_{k' \in \mathbb{N}}$ such that $TPy_{k'}(x) \to Ty(x)$ pointwise for almost every $x \in \Gamma$.

Now since $TPy_{k'}(x) \in [-c,c]$ for all $k'$, it must also be contained in the limit. Thus $Ty(x) \in [-c,c]$ a.e. on $\Gamma$.

In other words $Ty \in L^\infty(\Gamma)$ with $||Ty||_{L^\infty(\Gamma)} \leq c = ||y||_{L^\infty(\Omega)}$.