reference request: trace/lifting operator for $L^{\infty}$ data in bounded $\Omega\subset R^d$

Question

I know the answer to my question must be somewhere in the literature. Probably in [Adams-Fournier], [Dautray-Lions] or something alike, but I don't have access to a library right now so I'll ask anyway.

Assume I have a nice bounded open set $\Omega\subset R^d$ with $\mathcal{C}^{2,\alpha}$ boundary $\Gamma=\partial\Omega$, and a function $v\in H^{1/2}\cap L^{\infty}(\Gamma)$. Can I lift it to a function $u\in H^1\cap L^{\infty}(\Omega)$ function, i-e if $T$ is the standard trace operator does there exists $$ u\in H^1\cap L^{\infty}(\Omega):\qquad v=T u? $$ If so, can we expect to select $u$ and define a lifting map $v\mapsto u$ that is continuous for the $L^{\infty}(\Gamma)\to L^{\infty}(\Omega)$ topology (I'm of course aware of the continuity $H^{1/2}(\Gamma)\to H^1(\Omega)$). I suspect that Riesz-Thorin interpolation should be helpful.

Thank you in advance, I apologize again if the question is too easy. Please feel free to migrate to SE if you deem it necessary.

Answer 1

The answer is yes, of course. Indeed one standard way to lift $u\in H^{1/2}(\Gamma)$ to $\Omega$ is to solve the Dirichlet problem $$ \left\{ \begin{align*} -\Delta v=0\qquad(\Omega),\\ v=u\qquad (\Gamma). \end{align*} \right. $$ For given $u\in H^{1/2}(\Gamma)$ we know that this problem has a unique solution $v\in H^1(\Omega)$. If we further assume $u\in L^{\infty}(\Gamma)$ we know by the weak maximum principle that $$ |v|_{L^{\infty}(\Omega)}\leq |u|_{L^{\infty}(\partial \Omega)}=|v|_{L^{\infty}(\Gamma)}. $$ As a consequence this particular lift $v\mapsto u$ is continuous form $H^{1/2}\cap L^{\infty}(\Gamma)$ to $L^{\infty}\cap H^1(\Omega)$.