Let's say we know that $u \in C^1(U)$ where $U$ is open, bounded and convex, so that we can apply the mean value theorem on $U$. Moreover $u$ has a continuous extension onto $\overline{U}.$ Now, assume that also $Du$ has such an extension onto $\overline{U}$. Does this mean that $u$ is globally Lipschitz on $\overline{U}$ if $\text{sup}_{\overline{U}}|Du|< \infty$?

I think globally Lipschitz on $U$ can directly be concluded from the mean-value theorem, the question is whether an extension onto $\overline{U}$ is possible?

If anything is unclear, please let me know.

  • $\begingroup$ Any uniformly continuous function on $U$ has an extension to $\bar{U}$ (to see this note that if $x_n\to x\in \partial U$ then $u(x_n)$ is Cauchy. It's easy to see that the limit does not depend on the sequence and so we can define $u(x)$ so that this extension to $\bar{U}$ is continuous.) $\endgroup$ – Jose27 Sep 12 '15 at 4:54
  • $\begingroup$ @Jose27 the question was whether the extension to $\overline{U}$ is then Lipschitz continuous $\endgroup$ – user167575 Sep 12 '15 at 12:21
  • $\begingroup$ It is. You can see this by how we defined the extension. $\endgroup$ – Jose27 Sep 12 '15 at 18:13

Your Answer

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

Browse other questions tagged or ask your own question.