My question concerns one part of the proof of Theorem 3 in §6.5 in PDE Evans.
We consider in this section the boundary-value problem $$\begin{cases}Lw=\lambda w & \text{in }U \\ \, \, \, \, w=0 & \text{on }\partial U \end{cases} \tag{1}$$ where $U$ is open, bounded, and recall that $\lambda$ is an eigenvalue of $L$ provided there exists a nontrival solution of $(1)$.
The main theorem:
THEOREM 3 (Principal eigenvalue for nonsymmetric elliptic operators).
(i) There exists a real eigenvalue $\lambda_1$ for the operator $L$, taken with zero boundary conditions, such that if $\lambda \in \mathbb{C}$ is any other eigenvalue, we have $$\operatorname{Re}(\lambda) \ge \lambda_1.$$
(ii) There exists a corresponding eigenfunction $w_1$, which is positive within $U$.
(iii) The eigenvalue $\lambda_1$ is simple; that is, if $u$ is any solution of $(1)$, then $u$ is a multiple of $w_1$.
The full proof of Theorem 3 is too long, so I'm linking a PDF file of it. However, I will print part of the proof below, the part which I have a question about. This is towards the end of the paragraph of step 6 in the proof.
Consequently $$K(|v|^2) \le 2(\operatorname{Re} \lambda - (1-\epsilon)\lambda_1)|v|^2 \quad \text{in }U.$$ Thus if $\operatorname{Re} \lambda \le (1-\epsilon)\lambda_1$, then $K(|v|^2) \le 0$ in $U$. As $v=0$ on $\partial U$, according to $(33)$ and $(39)$, we deduce from the maximum principle that $v=\frac uw = 0$ in $U$. Thus, $u \equiv 0$ in $U$ and so $\lambda$ cannot be an eigenvalue. This conclusion obtains for each $\epsilon > 0$, and so $\operatorname{Re} \lambda \ge \lambda_1$ if $\lambda$ is any complex eigenvalue.
My question:
Given that $K(|v|) \le 0$ in $U$ and $v = 0$ on $\partial U$, how does the maximum principle apply here to show that $v = 0$ in $U$? This is needed to show that $u \equiv 0$ in $U$ (since $w$ is not identically zero and arbitrary).
I think Evans means the (weak) maximum principle for elliptic PDE's. Elliptic PDE generalizes Laplacian. Let's see how it is for the simplest case. If we have $-\Delta u\le 0$ in $U$ then the function $u$ is subharmonic, so it satisfies the Mean Value theorem with inequality $$ u(a)\le \frac{1}{|\partial B(a)|}\int_{\partial B(a)}u(\xi)\,dS_\xi, $$ with the straightforward conclusion that $\max_{\bar U} u=\max_{\partial U} u$.
For elliptic PDE there is a similar result: if $u\in C^2(U)\cap C(\bar U)$ and $$ Ku=-\sum_{i,j}a_{ij}u_{x_i,x_j}+\sum_{i}b_iu_{x_i}\le 0 $$ then $\max_{\bar U} u=\max_{\partial U} u$. /Check it, for example, here, page 67./
The argument in the paper goes: since $K(|v|^2)\le 0$ and $v|_{\partial U}=0$, we have that $|v|^2\le 0$ in $U$, which gives obviously $v\equiv 0$ in $U$. Perhaps, this kind of maximum principle is mentioned somewhere earlier in the Chapter.