Trace operator on $W^{1,\infty}$ My question is rather short and simple really:
Is the trace operator well defined on $W^{1,\infty}(\Omega)$ for some bounded Lipschitz domain $\Omega$?
The reason I ask is because I have seen statements, such as on wikipedia, that it is well defined for $p\ge 1$, but I do not know if that includes $\infty$ or not.
Thanks!
 A: If you assume a $C^1$-boundary rather than just Lipschitz, we can use Morrey's inequality. (It might be possible to use this inequality for Lipschitz domains, I am not sure.) So there exists $M>0$ such that, for all $u\in W^{1,\infty}(\Omega)$, we have
$$\|u\|_{C^{0,1}(\Omega)}\le M\|u\|_{W^{1,\infty}(\Omega)}$$
and in particular $u$ is Lipschitz on $\Omega$. Since $u$ is bounded and Lipschitz, we can extend its domain to $\overline\Omega$ by continuity. Indeed, if $x_0\in\partial\Omega$ and $\delta_1:=\liminf_{x\to x_0}u(x)<\limsup_{x\to x_0}u(x)=:\delta_2$, then with $K$ as the Lipschitz constant of $u$, we have $x_1,x_2\in\Omega$ with $|x_1-x_0|,|x_2-x_0|<\frac{\delta_2-\delta_1}{4K}$ and $u(x_2)-u(x_1)>\frac{\delta_2-\delta_1}2$. The Lipschitz condition gives $|u(x_1)-u(x_2)|\le K|x_1-x_2|<\frac{\delta_2-\delta_1}2$, a contradiction.
Hence, we have an operator $W^{1,\infty}(\Omega)\rightarrow L^\infty(\partial\Omega):\,u\mapsto u|_{\partial\Omega}$ which is continuous since $\|u|_{\partial\Omega}\|_{L^\infty(\partial\Omega)}\le\|u\|_\infty\le\|u\|_{W^{1,\infty}(\Omega)}$. That is, we have a trace operator.
