6
$\begingroup$

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

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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 $C^0$

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.