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


1 Answer 1


Roughly, the trick is not to view $L$ as an operator on $L^2$, but on $C^0$

I will use the following version of Krein-Rutmann which is proven in "Du, Yihong: Order Structure and Topological Methods in Nonlinear Partial Differential Equations, Vol. 1: Maximum Principles and Applications.":

Let $X$ be a Banach space, $C \subset X$ a solid cone (i.e. a cone with nonempty interior) and $T : X \longrightarrow X$ a compact linear operator which is strongly positive, i.e. $Tu \in C$ if $u \in C$. Then the spectral radius $r(T)$ fulfills $r(T) > 0$ and is a simple eigenvalue admitting an eigenvector $\psi \in \mathrm{int} C$ and there is no other eigenvalue that admits an eigenvector in $C$. Furthermore, all other eigenvalues $\lambda$ fulfill $|\lambda| < r(T)$.

By partial integration, you easily show that the operator is bounded from below, so $L + \alpha$ is strongly positive from $C^2$ to $C^0$ for some $\alpha$ big enough. It is also well-known that the inverse $T := (L + \alpha)^{-1}$ exists and is a compact operator on $C^0$. By the strong maximum principle, $Lu>0$ implies $u>0$, so $T$ is a strongly positive operator.

For the strong maximum principle, you can consult Evans: Partial Differential Equations, for example. To get the statement on a manifold instead of an area in $\mathbb{R}^n$, use a partition of unity.

Now it is easy to show that the set of positive functions is actually a solid cone in $C^0$ (even though it has empty interior in $L^2$), so we can apply the Krein-Rutmann theorem.

By elliptic regularity, every $L^2$ eigenfunction is $C^\infty$ and because the manifold is compact, is bounded, hence in $C^0$. Conversely, every $C^0$ Eigenfunction is in $L^2$, again because the domain is bounded. Hence the eigenvectors and eigenfunctions of $L$ are the same, whether viewed as operator on $L^2$ or on $C^0$

  • $\begingroup$ You are using the elliptic regularity. What happens if the potenital V is not smooth but just bounded? I think the eigenfunctions will not be smooth anymore. Does the statement remeins true for this case or are there counterexamples? $\endgroup$
    – supersnail
    May 20, 2013 at 13:37
  • $\begingroup$ The argument in the last paragraph also works with $C^2$ instead of $C^\infty$, and the solutions are always $C^2$. You should check out some book about PDE (e.g. the mentioned one by Evans or Gilbarg-Trudinger) to get the exact statements about regularity (especially at the border), but in principle, the same argument should work in most related cases. $\endgroup$ May 20, 2013 at 14:27

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