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We consider the Riemannian structure on the sphere $\mathbb{S}^n$ seen as a submanifold of $\mathbb{R}^{n+1}$ and the Laplace-Beltrami operator defined on $C^\infty(\mathbb{S}^n)$ by the equation

$$\Delta f= -\operatorname{div}\operatorname{grad} f = -\frac{1}{\sqrt{g}}\frac{\partial}{\partial u^i}\left(\sqrt{g}g^{ij}\frac{\partial f}{\partial u^j}\right).$$

We regard $C^{\infty}(\mathbb{S}^n)$ as a dense subspace of the Hilbert space $L^2(\mathbb{S}^n)$.

Question Is it true that $\Delta$ has compact resolvents, meaning that there exists $\lambda \in \mathbb{R}$ such that the closure of $\Delta-\lambda$ is invertible and its inverse operator is compact?

I think that we can easily work out the special case $n=1$: in this case the equation $\Delta u-\lambda u = v$ reduces to the standard Sturm-Liouville problem

$$\begin{cases} -\frac{d^2}{dt^2}u-\lambda u = v & t\in (-\pi, \pi) \\ {}\\ u(-\pi)=u(\pi) \\ u'(-\pi)=u'(\pi)\end{cases}$$

which admits Green's function for, say, $\lambda=-1$ (actually any $\lambda \notin \{0, 1, 4, 9 \ldots\}$ will do).

So the inverse of $-d^2/dt^2+1$ is an integral operator and in particular it is compact. I suspect that, similarly, the operator $\Delta_{\mathbb{S}^n}+1$ admits Green's function in any dimension $n$, but I am unable to prove (or disprove) this.

Thank you for reading.

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1 Answer 1

up vote 4 down vote accepted

This might be a stupid way to argue, but anyway.

  1. The solution of $-\mathrm{div}\ \mathrm{grad}\ f+f=v$ is the unique minimizer of the energy $E_v(f)=\int_{S^n}|\mathrm{grad}\ f|^2+|f-v|^2$.
  2. Since $0$ is admissible in minimization, the minimizer satisfies $E_v(f)\le E_v(0)=\int_{S^n}|v|^2$.
  3. Hence, every function $f\in L^2$ such that $-\mathrm{div}\ \mathrm{grad}\ f+f=v$ with $\|v\|_{L^2}\le 1$ satisfies $\int_{S^n}|\mathrm{grad}\ f|^2\le 1$.
  4. By the Rellich-Kondrachov theorem, the set of such $f$ is precompact in $L^2$.
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I find this approach most interesting, I'm thinking about it. –  Giuseppe Negro Jun 28 '12 at 16:51

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