Given a fixed value $\lambda>0$ let $u\in \dot{H}^1(\mathbb{R}^n)$ be a weak solution (or eigenfunction with eigenvalue $\lambda$) in dimension $n\geq 3$ to the following PDE, $$-\Delta u = \lambda \rho u$$ where $\rho\in L^{n/2}(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$ and $\lambda >0.$ Can we conclude that $u\in L^{\infty}(\mathbb{R}^n)?$
I know by standard elliptic regularity theory we can deduce that $u\in C^{\alpha}_{\operatorname{loc}}(\mathbb{R}^n)$, but does this imply that the solution is in $L^{\infty}(\mathbb{R}^n)$ as well?
Using the embedding $\dot{H}^1\hookrightarrow L^{p}_{\rho}$ (weighted Lp space) for $p\in [1, 2^*]$ we know that the solution $u$ is integrable. I am trying to combine the fact that $\phi \in L^{1}(\mathbb{R}^n)\cap C^{\alpha}_{\operatorname{loc}}(\mathbb{R}^n)$ to arrive at a contradiction, but this seems not trivial at the moment.
Any references/hints will be much appreciated.
PS. Note that $\dot{H}^1(\mathbb{R}^n)$ is the closure of $C^{\infty}_{c}(\mathbb{R}^n)$ with respect to the semi-norm $\|\nabla u\|^2_{L^2}.$
