We consider in this section the boundary-value problem
$$\left\{\begin{aligned} L w &=\lambda w \quad \text { in } U \ \ \ \ \ \ (1)\\ w &=0 \quad \text { on } \partial U \end{aligned}\right.$$ where $U$ is open and bounded, $$L u=-\sum_{i, j=1}^{n} a^{i j} u_{x_{i} x_{j}}+\sum_{i=1}^{n} b^{i} u_{x_{i}}+c u .$$
Let us for simplicity assume that $a^{i j}, b^{i}, c \in C^{\infty}(\bar{U})$ , that $U$ is open, bounded and connected, and that $\partial U$ is smooth. We suppose also $a^{i j}=a^{j i}(i, j= 1, \ldots, n)$ and
$$c \geq 0 \quad \text { in } U$$
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
$$\operatorname{Re}(\lambda) \geq \lambda_{1}$$
(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 $w_{1}$ .
Proof . 1. Choose $m=\left[\frac{n}{2}\right]+3$ and consider the Banach space $X= H^{m}(U) \cap H_{0}^{1}(U)$ . According to Sobolev Embedding Theorem, $X \subset C^{2}(\bar{U})$ . We define the linear, compact operator $A: X \rightarrow X$ by setting $A f=u$ , where $u$ is the unique solution of
$$\left\{\begin{aligned} L u=f & \text { in } U \\ u=0 & \text { on } \partial U \end{aligned}\right.$$
Next define the cone
$$C=\{u \in X \mid u \geq 0 \text { in } U\} .$$
According to the maximum principle, $A: C \rightarrow C$
- Hereafter fix any function $w \in C$, $w \not \equiv 0$ . Employing the strong maximum principle and Hopf's Lemma, we deduce $$v>0 \ \ in \ \ U, \frac{\partial v}{\partial \nu}<0 \ \ on\ \ \partial U \ \ \ (2)$$ for $v=A(w)$ . Remember that $w=0$ on $\partial U$ . So in view of (2) there exists a constant $\mu>0$ so that
$$\mu v \geq w \quad \text { in } U$$
My problem is: why such constant $\mu>0$ exists? I know we only need to consider the point near the boundary of $U$ and notice that $w=0$ on $\partial U$,we may need to apply Tailor's formula, and I believe the conclusion is true, but I just can't prove it clearly. Can someone help me? I would be appreciated if you could help me.
Since $v > 0$ (strictly positive) you can just define the function $$ g = \frac{w}{v}, $$ and then take $\mu = \sup_U g$ so that $\mu \geq g$ a.e. in $U$ and therefore $$ \mu \geq \frac{w}{v} \implies \mu v \geq w. $$ Remark that since $w \in C$ and $v > 0$, it holds $g \geq 0$ and therefore $\mu \geq 0$. We can see that $g = 0$ a.e. only if $w=0$ a.e., but in this case since $v > 0$ we can take any $\mu > 0$ and $\mu v > w$ would hold.