A detail in Evans's prove of THEOREM 3 in section 6.5 (about principal eigenvalue)

Question

The following picture shows the statement of the theorem and the first part of Evans' proof. In his prove, Evans defines an operater A, which I know is properly defined (because of the maximum principle) and is linear on X, but what I don't understand is why A is compact. At first I thought it should use compact embedding like in section 6.2, but seems it's not the case, and I don't know what is the norm of space X as the intersection of two banach space. Any help is appreciated.

Answer 1

I realized that I have ignored the simple fact that $X=H^m(U)\cap H_0^1(U)$ is a closed subspace of $H^m(U)$: for a sequence $\{f_n\}\subset X$ converge to $f$ in $H^m(U)$, $\{f_n\}$ also converge in $H^1(U)$ and since $H_0^1(U)$ is a closed subspace of $H^1(U)$, $f\in X$. So the norm of X should be $H^m(U)$-norm.

Now we proof $A$ is compact. Let $\{f_n\}\subset X$ be a bounded sequance. Use the estimate in Thoerem 5 of section 6.3, we have $$ \Vert Af_n\Vert_{H^{m+2}(U)}\leq C_1(\Vert f_n\Vert_{H^{m}(U)}+\Vert Af_n\Vert_{L^2(U)}) $$ Moreover, by the theorem "boundness of inverse" in section 6.2, we have $\Vert Af_n\Vert_{L^2(U)}\leq C\Vert f_n\Vert_{L^2(U)}$ for some constant $C$. So we can deduce $\{Af_n\}$ is bounded in $H^{m+2}(U)$, what left is to show that a bounded sequence in $H^{m+2}(U)$ has subsequence converge in $H^m(U)$.

To prove this statement, we apply induction to $m$. Suppose the conclusion is true for natural number $k$ not larger than m. For $k=m+1$, Let $\{f_n\}\subset X$ be a bounded sequence in $H^{k+2}(U)=H^{m+3}(U)$, then by induction hypothesis, $\{f_n\}$ has subsequence $\{f_{n_k}\}$ converge in $H^m(U)$. For $\alpha$ satisfy $|\alpha|=m+1$, $\{D^{\alpha}f_{n_k}\}\subset H^2(U)\subset H^1(U)$ is bounded, since $H^1(U)$ can be compactly embedded in $L^2(U)$, we know $\{D^{\alpha}f_{n_k}\}$ converge in $L^2(U)$. Then $\{f_{n_k}\}$ is a Cauchy sequence in $H^{m+1}(U)$ and we have proved the conclusion for $k = m+1$. $\square$