Problem: Let $D \subset \mathbf{R}^n$ be open, bounded and let $\lambda \in \mathbf{R}$. a) Suppose there exists a solution $u \in C^2(D) \cap C(\bar{D}), u$ not identically zero, such that $$ \left\{\begin{array}{ccc} -\frac{1}{2} \Delta u=\lambda u & \text { in } & D \\ u=0 & \text { on } & \partial D \end{array}\right. $$ Then $\lambda>0$. b) It can be shown that if $D$ is smooth then there exist $0<\lambda_0<\lambda_1<$ $\cdots<\lambda_n<\cdots$ where $\lambda_n \rightarrow \infty$ such that condition holds for $\lambda=\lambda_n$, $n=0,1,2, \ldots$, and for no other values of $\lambda$. The numbers $\left\{\lambda_n\right\}$ are called the eigenvalues of the operator $-\frac{1}{2} \Delta$ in the domain $D$ and the corresponding (nontrivial) solutions $u_n$ of condition are called the eigenfunctions.
1). Put $\tau=\tau_D=\inf \left\{t>0 ; B_t \notin D\right\}$, choose $\rho>0$ and define $$ w_\rho(x)=E^x[\exp (\rho \tau)] ; \quad x \in D . $$ Prove that if $w_\rho(x)<\infty$ for all $x \in D$ then $\rho$ is not an eigenvalue for $-\frac{1}{2} \Delta$. 2). Conclude that $$ \lambda_0 \geq \sup \left\{\rho ; E^x[\exp (\rho \tau)]<\infty \text { for all } x \in D\right\} . $$
For 1).Let u be a solution with $\lambda=\rho$. Applying Dynkin's formula to the process $d(Y_t)=(dt,dB_t)$ and the function $f(t,x)=e^{\rho t}u(x)$, we get $$E^{(t,x)}[f(Y_{\tau \wedge n})]=f(t,x)+E^{(t,x)}[\int_0^{\tau \wedge n} Lf(Y_s) ds]$$. Since $$Lf(t,x)=\rho e^{\rho t} u(x)+\frac{1}{2}e^{\rho t}\triangle u(x)=0$$, then $$E^{(t,x)}[e^{\rho \tau \wedge n} u(B_{\tau \wedge n})]=e^{\rho t}u(x).$$ Let t=0 and $n \rightarrow \infty$, is it true we have the boundary solution and so $u(x)=0$?
For 2).Since eigenvalues are monotonically increasing and $\lambda_i>0$, we have the claim. Is it right?