I want to ask about a question about the convergence of eigenvalue $\lambda_R$ and the eigenfunction $\phi_R$ of a schrodinger operator in $B_R$.
Given an eigenvalue equation (2nd order elliptic PDE) in a ball $B_R$, how to prove the convergent of the eigenfunction $\phi_R$?
Actually I'm reading the Proposition4.2 in On a Long-Standing Conjecture of E. De Giorgi: Symmetry in 3D for General Nonlinearities and a Local Minimality Property, it said that:
Let $H: \mathbf{R}^n \rightarrow \mathbf{R}$ be a bounded continuous function. Then $$\tag{1} \int_{\mathbf{R}^n}\left\{|\nabla \xi|^2+H(x) \xi^2\right\} \mathrm{d} x \geqslant 0, \quad \forall \xi \in C_c^{\infty}\left(\mathbf{R}^n\right) $$ if and only if there exists a continuous function $\varphi: \mathbf{R}^n \rightarrow(0, \infty)$ such that $\Delta \varphi=H(x) \varphi$ in the sense of distributions.
Proof. Condition (1) implies that the first eigenvalue of the Schrödinger operator $-\Delta+H(x)$ in each ball $B_R$ is nonnegative
this can be simply verified by the real eigenvalue of self-adjoint operator.
Since the first eigenvalue in $B_R$ is a decreasing function of $R$, it follows that all these first eigenvalues are positive
How to get the fact that the first eigenvalue in $B_R$ is a decreasing function of $R$? This is simple for Laplacian (just take $\Phi=\phi(x/a)$ in $B_{aR}$), but we have another $H$, how to deal with this difference?
This implies that, for every constant $c_R>0$, there exists a unique solution $\phi_R$ of $$ \begin{cases}\Delta \phi_R=H(x) \phi_R, & \text { in } B_R, \\ \phi_R=c_R, & \text { on } \partial B_R,\end{cases} $$ and, moreover, $\phi_R>0$ in $B_R$.
I think this step simplifies many steps, first, the the positive $\phi_R$, this might be related to the property that the eigenfunction related to the first nonzero eigenvalue of $-\Delta$ is either strictly positive or strictly negative or you can directl read Theorem8.38 in Gilbarg&Trudinger's book. Second, we should have $$-\Delta \phi_R + H(x)\phi_R=\lambda_R\phi_R,$$ but he directly dropped the $\lambda_R\phi_R$ term, maybe this is because they proved $lim_{R\rightarrow+\infty}\lambda_R=0$. But the I can't figure out the detail.
We choose the constant $c_R$ such that $\varphi_R(0)=1$. Then, by the Harnack inequality a subsequence of $\left(\phi_R\right)$ converges locally to a solution $\phi>0$ of $\Delta \phi=H(x) \phi$.
$sup_{B_R} \phi < inf_{B_{R}} \phi \le 1$, this convergence part is a little vague for me, can you help me figure out some details? Since we already have the $C^0$ estimate, the $C^1$, $C^2$...should be nature but what exactly theorem we need here? Is $R$ going to big enough? Besides, why we need "converges locally".
Conversely, multiplying the equation $\Delta \phi=H(x) \phi$ by $\xi^2 / \phi$, integrating by parts, and using the Cauchy-Schwarz inequality, we obtain (1).