# Proving an operator is compact

This is from Gilbarg-Trudinger Theorem 6.15. It is using the Fredholm alternative argument to prove existence of solutions.

We are concerned with the elliptical operator $L \equiv a^{ij}D_{ij} + b^iD_i + c$, which is strictly elliptic with coefficients in $C^{\alpha}(\Omega)$ in a $C^{2,\alpha}$ domain $\Omega$. Define $L_\sigma \equiv L - \sigma$ for $\sigma \geq \sup_{\Omega}c$.

It can be shown be previous results in the book that $L_{\sigma} : \{u\in C^{2,\alpha}(\bar{\Omega}) | u = 0 \quad \text{on} \quad \partial\Omega\} \rightarrow C^{\alpha}(\bar{\Omega})$ is invertible.

Here's the part of the theorem I don't get: "Claim : $L_{\sigma}^{-1}$ is a compact mapping from $C^{\alpha}(\bar{\Omega}) \rightarrow C^{2}(\bar{\Omega})$ and hence is also compact as a mapping from $C^{\alpha}(\bar{\Omega}) \rightarrow C^{\alpha}(\bar{\Omega})$"

According to the author, he writes one sentence saying the proof of the claim depends on two previous estimates. Here are the respective estimates :

Estimate 1 : Let $Lu \geq f$ in a bounded domain $\Omega$, where $L$ is elliptic, $c \leq 0$, and $u \in C^0(\bar{\Omega})\cap C^2(\Omega)$. Then $$\sup_{\Omega}{u(|u|)} \leq \sup_{\partial \Omega}{u^+(|u|)} + C\sup_{\Omega}{\frac{|f^-|}{\lambda}}\left(\frac{|f|}{\lambda}\right)$$

Estimate 2: Let $\Omega$ be a $C^{2,\alpha}$ domain in $\mathbb{R^n}$ and let $u$ be a $C^{2,\alpha}(\bar{\Omega})$ solution of $Lu=f$ in $\Omega$, (with a standard boundedness condition on the coefficients of $L$, that we have satisfied by our assumption). Let $\phi(x) \in C^{2,\alpha}(\bar{\Omega})$ and suppose that $u=\phi$ on $\partial \Omega$. Then $$|u|_{2,\alpha , \Omega} \leq C(|u|_{0,\Omega} + |\phi|_{2,\alpha,\Omega} + |f|_{0,\alpha,\Omega})$$

To show compactness of an operator $T$, we have to show that $T$ maps bounded sequences into sequence which contain convergent subsequences. I really don't see how this result follows from the two estimates given. Also, how does the compactness result to $C^2$ imply the compactness result to $C^{\alpha}$?

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en.wikipedia.org/wiki/… – Will Jagy Sep 12 '12 at 1:32
Very funny. That's what they want you to think. – Euler....IS_ALIVE Sep 12 '12 at 3:03
People usually get grumpy as they get older. If Euler is still alive, he is in a really foul mood. – Will Jagy Sep 12 '12 at 3:27
Don't you have $L_\sigma^{-1}: C^\alpha \to C^{2,\alpha}$? (If this is the case, Arzela-Ascoli gives the compactness) – Jose27 Sep 12 '12 at 4:34
Ahhh, I think that might be it. It is true that $L_{\sigma}^{-1} : C^{\alpha} \rightarrow C^{2,\alpha}$. From the first estimate we get boundedness, and from the second estimate we get equicontinuity. Am I correct in this thinking? – Euler....IS_ALIVE Sep 12 '12 at 4:58