I have some question :
Let $\Omega\subset{\mathbb{R^{2}}}$ bounded with lipschitz boundary, assme the function $f$ satisfy : $f\in{W^{2,\infty}(\Omega)}$.
I need to show that : $\nabla f \in{C^{0}(\partial{\Omega)}}$. That is how i proceed :
Since $f\in{W^{2,\infty}(\Omega)}$ implies $\nabla f\in{W^{1,\infty}(\Omega)}$ and by the trace theorem :
$\nabla f\in{W^{\frac{1}{2},\infty}(\partial\Omega)}$
then using the inclusion :$W^{\frac{1}{2},\infty}(\partial\Omega)\subset C^{m}(\partial\Omega)$ for $ m = k - \frac{N}{p}$ here $k=\frac{1}{2}, N=2, p = \infty$
I deduce that : $\nabla f\in { C ^{0}(\partial\Omega)}$
Question : Is that true ?
Since $f\in W^{2,\infty}$, $\nabla f\in W^{1,\infty}$. By Sobolev embedding, this implies $\nabla f \in C(\bar \Omega)$. Hence, the restriction of $\nabla f$ to the boundary is a continuous function as well.