# relation between $W^{1,\infty}$ and $C^{0,1}$

I know that $f \in C^{0,1}_{loc}(U)\Leftrightarrow f \in W^{1,\infty}_{loc}(U)$ and I have a reference for this. I would like a reference or a explanation for $C^{0,1} = W^{1,\infty}$ on domain convex.

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Did you try to see if the proof of the first equivalence works for the second as well? I'm pretty sure it does. –  user31373 Jul 19 '12 at 1:36
Suppose $f\in C^{0,1}(U)$. Then $f$ is Lipschitz on every segment parallel to coordinates axis (and on other segments, too). Hence, it is absolutely continuous on every segment, with bounded derivative. This qualifies it as a member of $W^{1,\infty}(U)$.
Conversely, suppose $W^{1,\infty}(U)$. This means that $f$ is absolutely continuous on almost every coordinate-aligned segment, with bounded derivative. That is to say, $f$ is Lipschitz on such segments, with a uniform bound on Lipschitz constant. Let $E$ be the union of these "good" segments. Because $U$ is convex, any two points $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)$ in $E$ can be connected by a polygonal line contained within $E$ with total length at most $$2\sum |x_i-y_i|\le 2\sqrt{n} \|x-y\|$$ Therefore, $f$ is Lipschitz on $E$. Since $U\setminus E$ is a null set, we can redefine $f$ there to make it Lipschitz on $U$ (extending $f$ to $U$ by continuity).