Evans' PDE: Theorem 3 of Chapter 6.5.2 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.

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, $u \in H^{m+2}(U) \cap H^1_0(U) \subset X$.
