I am doing some self-study on Evans' PDE and I am stuck at some detail of Theorem 3, Chapter 6.5.2 (page 361). Let me provide some setup:
We will now consider a uniformly elliptic operator in the non-divergence form, i.e., $Lu := -\sum_{i,j=1}^n a^{ij} u_{x_i x_j} + \sum_{i=1}^n b^i u_{x_i} + cu$ for $a^{ij}, b^i, c \in C^{\infty}(\overline{U})$ for $U \subset \mathbb{R}^n$ bounded, open and connected. Suppose further $\partial U$ is smooth, $a^{ij} = a^{ji}$ and $c \geq 0$ on $U$.
The goal of the theorem is to prove for nonsymmetric elliptic operator, the principal eigenvalue $\lambda_1$ is real and simple; and $\lambda_1 \leq \text{Re}(\lambda)$ for any other eigenvalue $\lambda$. But I think the context of the theorem is not relevant yet, because I am stuck at the very beginning of the proof, which says
Choose $m = [\frac{n}{2}]+3$ and consider the Banach space $X = H^m(U) \cap H_0^1(U)$. According to the Sobolev inequality, we have $X \subset C^2(\overline{U})$. Define the linear, compact operator $A: X \to X$ such that $Af := u$, where $u$ is the unique solution to the following equation:
$$\begin{cases}
Lu = f \ \ \ \text{ in $U$} \\
u = 0 \ \ \ \ \ \ \text{ on $\partial U$}.
\end{cases}$$
My question: Why could we define $A$? My only thought is to use Lax-Milgram to conclude the existence of the solution, however, we may not be able to say $B[u,u] := (Lu, u) \geq \beta \|u\|_{H^1_0}$ for some $\beta > 0$. Could anyone give me some hint on this? If you need more context of the proof, please comment below.
Best Answer
Thanks for the hint in the comment. Instead of using Lax-Milgram, we should invoke Fredholm alternative(Theorem 4 of Chapter 6.2.3 in Evans). Consider the equation $$\begin{cases} Lu = 0 \ \ \ \text{ in $U$} \\ u = 0 \ \ \ \ \ \ \text{ on $\partial U$}. \end{cases}$$ Suppose $u$ is a weak solution to $Lu= 0$. Then note that by regularity theorem, $u \in C^\infty(\overline{U})$. By the strong maximum (and minimum) principle, $u \equiv 0$. Therefore, $Lu=0$ has only the trivial solution, which, by Fredholm alternative, says that for each $f \in X = H^m(U) \cap H_0^1(U)$, we have a unique solution $u \in H^1_0(U)$. Invoking again regularity theorem, $u \in H^{m+2}(U) \cap H^1_0(U) \subset X$.