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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?

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    $\begingroup$ 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? $\endgroup$ – DisintegratingByParts 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. $\endgroup$ – DisintegratingByParts Nov 9 '17 at 6:03
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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.

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