Hello,

let $(M,g)$ be a compact and connected Riemannian manifold (possibly with $\partial M\neq \emptyset$). We consider the Friedrichs extension of $L=-\Delta +V: C^{\infty}(M,\mathbb{R})\subset L^2(M)\rightarrow L^2(M)$ with bounded potential $V:M\rightarrow \mathbb{R}$ (and Dirichlet/Neumann boundary conditions for nonempty boundary). Then the spectrum of L is a discrete sequence $\lambda_0 < \lambda_1 \leq \lambda_2\leq ...$

I'm trying to understand why the first eigenvalue is always simple. I've found the same question here: First eigenvalue of Schrödinger operator is simple

The answer given there is very helpful, but I still couldn't understand the situation on the whole.

First, the answer remarks the operator $L$ satisfies the maximum principle, i.e. for $f\geq 0$ and $f\neq 0$ there is a unique solution $u$, s.t. $Lu=f$. The solution $u$ is positive. My questions are, whether this is true for all $f\in L^2_{+}:=\lbrace f\in L^2(M) : f\geq 0 \text{ }a.e.\rbrace$ and do I need M to be connecet? Where can I find a proof?

Second, the given answer suggest to use the Krein-Rutman thm. for $L^{-1}$. But the Krein-Rutman version I've found, doesn't state that the largest eigenvalue of $L^{-1}$ must be simple:

Krein-Rutman-thm: "Let $X$ be a Banach space, $K\subset X$ a total cone and $T\in L(X)$ compact positive with $r(T)>0$. Then $r(T)$ is an eigenvalue with a positive eigenvector." (r(T) is the spectral radius of T)

In our situation we can put $X=L^2(M)$, $K=L^2_{+}$ (?) and $T=L^{-1}$.

How to get the simplicity out of the above Krein-Rutman-thm?

I will remark something else, maybe it will help you to answer my questions: On the other hand, I've found a stronger result which provides that the eigenvalue $r(T)$ is simple. But then the cone $K$ must have non empty interior and T must be strongly positive. The problem is that $L^2_{+}$ has empty interior. Therefore I can't use this result for the choice of cone I've made above.

I hope you can help me. Regards