1
$\begingroup$

I'm working through the book Elliptic Parial Differential Equations of Second Order by D. Gilbarg and N. S. Trudinger. Unfortunately I get stuck at some point. On page 23 they prove the following Theorem:

Let $u$ be harmonic in $\Omega$ and let $\Omega'$ be any compact subset of $\Omega$. Then for any multi-index $\alpha$ we have $$\sup_{\Omega'}|D^\alpha u|\le \left(\frac{n|\alpha|}{d}\right)^{|\alpha|} \sup_{\Omega}|u|$$ where $d=\operatorname{dist}(\Omega',\partial\Omega)$.

Now they conclude:

An immediate consequence of the bound above is the equicontinuity on compact subdomains of the derivatives of any bounded set of harmonic functions.

How could they conclude that?

Let $\{u_i\}$ a family of of bounded harmonic functions: why are the $u_i$ equicontinuous on compact subdomains?

Thanks for your help,

hulik

$\endgroup$
  • $\begingroup$ If you take a compact ball then the derivatives of first order are uniformly bounded. You can apply an inequality which involves $|u(x)-u(y)|$ and the norm of $Du$ on this ball. $\endgroup$ – Davide Giraudo Mar 8 '12 at 18:08
2
$\begingroup$

If $\{u_i\}_{i\in \mathcal{I}}$ is a bounded family of harmonic functions defined in $\Omega$ (i.e., there exists $M\geq 0$ s.t. $|u_i(x)|\leq M$ for $x\in \Omega$) then inequality: $$\sup_{\Omega'}|D^\alpha u|\le \left(\frac{n|\alpha|}{d}\right)^{|\alpha|} \sup_{\Omega}|u|$$ with $|\alpha|=1$ implies: $$\sup_{\Omega'}|\nabla u_i|\le C(\Omega^\prime)\ \sup_{\Omega}|u_i| \leq C(\Omega^\prime)\ M$$ for each $i\in \mathcal{I}$ (here $C(\Omega^\prime)\geq 0$ is a suitable constant depending on $\Omega^\prime$). Therefore the family $\{u_i\}_{i\in \mathcal{I}}$ is equi-Lipschitz on each compact subdomain $\Omega^\prime \subseteq \Omega$, for: $$\forall i \in \mathcal{I},\quad |u_i(x)-u_i(y)|\leq C(\Omega^\prime)\ M\ |x-y|$$ for all $x,y\in \Omega^\prime$, and equi-continuity follows.

$\endgroup$
  • 1
    $\begingroup$ I guess the inequality $|u_i(x) - u_i(y)| \leq C(\Omega') M |x - y|$ follows from the mean value inequality, but aren't we supposed to assume the segment $[x,y]$ lies in $\Omega$? $\endgroup$ – Eduardo Longa Jan 31 '16 at 17:53

Your Answer

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