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.

  • $\begingroup$ Did you try to see if the proof of the first equivalence works for the second as well? I'm pretty sure it does. $\endgroup$ – 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).

Reference: Theorem 4.1 in Lectures on Lipschitz Analysis by Heinonen.

See also Sobolev embedding for $W^{1,\infty}$?

  • $\begingroup$ +1 for finding 5pm's answer. I remember GT used the exterior cone condition to prove the embedding, it seems quasiconvexity is even weaker. Any reference? $\endgroup$ – Shuhao Cao Jul 27 '13 at 2:09
  • $\begingroup$ @ShuhaoCao Theorem 4.1 in Lectures on Lipschitz Analysis. Quasiconvexity appears several times throughout the text. $\endgroup$ – 40 votes Jul 27 '13 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.