# Gilbarg & Trudinger: Why does this Theorem imply equicontinuity of first and second derivative?

I recently started to read Chapter 4 from " Elliptic Partial Differential Equation of Second Order" by Gilbarg and Trudinger. Below is what you need to know:

QUESTION:

Why does Theorem 4.8 imply the equicontinuity of solutions and of their first and second derivatives on compact subsets?

I understand why a function which satisfy a Holder condition is equicontinuous. But I don't see why this Theorem implies $\;u\in C^{0,\alpha}\;,\;u\in C^{1,\alpha}\;,\;u\in C^{2,\alpha}\;$. I suppose the fact that these norms are very confusing to me, is important here...

I've been stuck to this so any help would be valuable.

• I'm not sure exactly where your confusion is: the theorem tells you very directly that $u$ is bounded in the Hölder norm. Is your issue with the distinction between $|u|_{2,\alpha}$ and $|u|^*_{2,\alpha}$? – Anthony Carapetis Sep 20 '17 at 11:52
• @AnthonyCarapetis I understand neither of these norms completely, to be honest... As I see it, only the second derivative of $\;u\;$ satisfies a Holder condition by the Theorem... – kaithkolesidou Sep 23 '17 at 10:56
• The $C^2$ part of the norm already implies Holder control (Lipschitz even!) for $u$ and its first derivative. – Anthony Carapetis Sep 24 '17 at 3:53
$d_x=\text{dist}(x,\partial \Omega)$ and $d_{x,y}=\min\{d_x,d_y\}$. If you consider a set $U$ compactly contained in $\Omega$, then $\text{dist}(U,\partial \Omega)=c_0>0$. Hence, if $x,y\in U$, from the formula just above (4.20) and the one just below you get $$c_0|Du(x)|+c_0^2 |D^2 u(x)|\le RHS$$ and $$c_0^{2+\alpha}\frac{|D^2 u(x)-D^2 u(y)|}{|x-y|^\alpha}\le RHS,$$ so you get $C^{2,\alpha}$ bounds in $U$.