Let $S^2$ be the 2-sphere equipped with some Riemannian metric. Let $\Delta$ denote the Laplace-Beltrami operator. Let $f\geq 0$ be a smooth, non-negative function such that $f(x)>0$ at some $x\in S^2$. Consider the linear second order differential operator $L=-\Delta+f$.

I am interested in the eigenvalues of this operator. Specifically, I would like to show that for all the eigenvalues $\lambda$ we have $\lambda \geq \lambda_0 >0$.

Assuming $u$ is an eigenfunction with eigenvalue $\lambda$, assuming appropriate regularity, and integrating by parts, we have

$\lambda ||u||_{L^2}=\int_{S^2} uLu=\int_{S^2}\left( |\nabla u|^2 + fu^2 \right) \geq 0$

showing $\lambda \geq 0$. I think I can use Theorem 8.38 in Gilbarg-Trudinger which says that the minimum eigenvalue is simple and has a positive eigenfunction. In our case this then shows the first eigenvalue is positive. However, this book deals with domains in $\mathbb{R}^n$ rather than Riemannian manifolds, but I assume that all the results can be translated to the manifold setting. I'd just like confirmation that this is indeed the case.

Many thanks!


1 Answer 1


Yes, it is the case. I think there are two different ways to approach this problem. I think the second one is more direct if you already know the behavior of the operator $-\Delta$ (without the potential $f\,$). Let's start by applying some spectral analysis (this is why I prefer the second approach).

First of all, note that $-\Delta+f$ is essentially self-adjoint in $C_0^\infty(S^2)$ and self-adjoint in $H^2(S^2)$. Thus, its spectrum is non-empty and it is contained in $\mathbb{R}$. Moreover, there is a basic result in Spectral Theory saying that: Let $T$ be a self-adjoint operator. Then, the spectrum of $T$ is contained in $[M,\infty)$ if and only if $T$ is semi-bounded from below by $$ \langle Tu,u\rangle \geq M\Vert u\Vert^2, \qquad \forall u\in D(T). $$ Thus, we can derive a first piece of information following your procedure. In fact, by integration by parts, it is easy to prove that spectrum of $-\Delta+f$ is contained in $[0,\infty)$.

Now, let's try to follow a different approach. First of all, let's recall Poincare's inequality (valid for $S^2$) $$ \int_M \vert \nabla g\vert^2 \geq \lambda_1\int_M \vert g\vert^2, $$ where $\lambda_1$ is the first eigenvalue of $-\Delta$. Thus, following you procedure again, integrating by parts and applying Poincare's inequality, we see that the first eigenvalue $\lambda_1^\star$ of $-\Delta+f$ satisfies $\lambda_1^\star\geq \lambda_1$ (due to the positivity of $f\,$). Thus, if you already know that $\lambda_1>0$, we conclude the property we are looking for.


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