Given a convex domain $\Omega$ with piecewise $C^1$ boundary data $g \in C(\partial \Omega)$ and $g \in C^1(\Gamma_i)$ with $\bigcup_{i} \Gamma_i = \partial \Omega$.

Now, I want to know if there exists a funcion $u \in C(\Omega)$ with furthermore $ \Vert u \Vert_{L^{\infty}(\Omega)} \leq C \Vert g \Vert_{L^{\infty}(\partial \Omega)} $ and(!) $ \Vert u \Vert_{W^{1,\infty}(\Omega)} \leq C \Vert g \Vert_{W^{1,\infty}(\partial \Omega)} $.

If it simplifies something, I am happy if things work on simplices in $\mathbb{R}^d$, $d=2,3$.

I thought about harmonic extensions or explicitly constructed extensions but none really worked out so far, although I expect the claim to hold.


You can extend a bounded Lipschitz function from any set whatsoever, while preserving both the Lipschitz constant and the supremum norm. This is called the McShane–Whitney extension (reference). Let $L$ be the Lipschitz constant of $g$. Define $$ h(x) = \inf_{y\in \partial \Omega} (g(y)+L|x-y|) $$ This is an $L$-Lipschitz function on all of $\mathbb{R}^d$, and it agrees with $g$ on $\partial\Omega$ (both things are easy to prove).

Finally, truncate $h$: $$ g(x) = \min(\sup_{\partial\Omega} g, \max(\inf_{\partial\Omega} g, h(x))$$ Truncation preserves the Lipschitz property (same $L$ again), and does not change the values of $h$ on $\partial\Omega$.


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