The following is straight out of Gilbarg and Trudinger, theorem 4.8:

Let $u \in C^2(\Omega)$, $f \in C^\alpha(\Omega)$ satisfy $\Delta u = f$ in an open subset $\Omega$ of $\mathbb{R}^n$. Then $\lvert u\rvert^*_{2, \alpha; \Omega} \leq C(\lvert u\rvert_{0;\Omega} + \lvert f\rvert_{0,\alpha; \Omega})$, where $C= C(n, \alpha)$.

The text goes on to claim that this interior estimate implies that, on compact subsets of $\Omega$, any bounded set of solutions $u$ (as well as their first and second derivatives) are equicontinuous. How can I see this?

  • 1
    $\begingroup$ Are you able to show that a set of functions with a uniformly bounded $C^{\alpha}$ norm on a suitable set forms an equicontinuous family? $\endgroup$ Nov 9 '17 at 0:22
  • $\begingroup$ I can, but wouldn't that only give equicontinuity of the second derivatives in my example? $\endgroup$
    – Mathmank
    Nov 9 '17 at 0:52
  • $\begingroup$ Looks like you have your answer below. $\endgroup$ Nov 9 '17 at 6:03

On any compact subset $\Omega' \subset \Omega$, the distance to the boundary is bounded below by a positive constant and thus a bound on the weighted Hölder norm $|u|^*_{2,\alpha;\Omega}$ implies a bound on the standard Hölder norm $$|u|_{2,\alpha;\Omega'} = \sup |u| + \sup |Du| + \sup |D^2 u| + [D^2 u]_{\alpha;\Omega'},$$ where all the supremums are taken over $\Omega'$. The part providing the equicontinuity of $u$ here is the uniform bound on $|Du|$: for any points $x,y \in \Omega'$ we have $|u(x) - u(y)| \le (\sup |Du|) |x-y|$ by e.g. the mean value theorem; so we have uniform Lipschitz control of $u$ and thus equicontinuity.

Similarly the bound on $|D^2 u|$ provides equicontinuity of the first derivatives, and the bound on $[D^2 u]_\alpha$ provides equicontinuity of the second derivatives.

This is why we typically only include the Hölder seminorm of $D^2 u$ in the $C^{2,\alpha}$ norm - the bounds on $|Du|$ and $|D^2 u|$ already imply Lipschitz (and thus Hölder for any $\alpha \in (0,1]$) control of $u$ and $Du$ respectively.


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.