A question about a series of solutions to an elliptic PDE in $B_R$ which is compactly convergent as $R \rightarrow +\infty$

Question

My question arises from Here.

I have a series of eigenvalue equations in $B_R$. $$ -\Delta \phi_R+H(x) \phi_R=\lambda_R \phi_R, $$ where $\lambda_R \geq 0$ is the first nonzero eigenvalue, with $\phi_R>0$. And $\lambda_R$ is non-increasing, so for bigger ball ($R\rightarrow+\infty$), we have a limit $\lambda_{\infty}$. So I want to discuss that let the $R$ tend to $+\infty$, can we also get a limit function $\Phi$, which satisfies that $$ -\Delta \Phi+H(x) \Phi=\lambda_\infty \Phi, $$ the answer is that we can find a $\phi_{\infty}$, $\phi_R$ is compactly convergent to $\phi_{\infty}$.

My question looks very naive, we did the bootstrap process on the selected compact set $K$ (since $K$ can be covered by a ball, so we get the $C^0$ estimate on $K$ by Harnack inequality, meanwhile take $R \rightarrow +\infty$ then use some regularity results and Arzela-Ascoli to get a convergent subsequence), we note as $$ -\Delta \Phi_K+H(x) \Phi_K=\lambda_\infty \Phi_K, $$

But the answer is that we can find a $\phi_{\infty}$, $\phi_R$ is compactly convergent to $\phi_{\infty}$, which means on every compact set the limit functions are the same, how to prove every $\Phi_K$ is the same, it seems that I misunderstood some essential conceptions, but I can't figure out, could you help me find where I misunderstood?

My attempt is:

If on two compact sets $K_1$ and $K_2$, $\phi_R$ converges to different functions $\phi_{K_1}$ and $\phi_{K_2}$, then we find a bigger compact sets to cover $K_1$ and $K_2$, so we get $\phi_{K_1}=\phi_{K_2}$.

Answer 1

The limit functions are the same basically by construction, there is no claim on uniqueness. Take an exhaustion $K_i \nearrow \mathbb{R}^n $ of $\mathbb{R}^n$ by compact sets. Then on $K_1$ by elliptic estimates and some compactness (Ascoli-Arzelà) some subsequence (call it $\Lambda_1 \subset \mathbb{N}$) of $(\phi_{R_k})_{k\ge 1}$ converges to $\phi_{\infty, K_1}$ on $K_1$ and $\phi_{\infty, K_1}$ is a function only defined on $K_1$. Then take $K_2 \supset K_1$ and by the same argument a subsequence of $\Lambda_1$ (call it $\Lambda_2$) converges to $\phi_{\infty, K_2}$ which is defined on $K_2$. Since $\Lambda_2 \subset \Lambda_1$ clearly $\phi_{\infty, K_2}|_{K_1} = \phi_{\infty, K_1}$. Go on with the exhaustion to infinity and this defines a limit function $\phi_\infty$ on $\bigcup_{I} K_i$ which is the entire $\mathbb{R}^n$.