1

Are there theorems similar to the following: If $T$ is symmetric and $D(T)$ contains an ONB of eigenvectors of $T$, then $T$ is essentially self adjoint and the spectrum of $\bar{T}$ is the closure of the point spectrum $\sigma_p(T)$?

I am interested in these theorems as in physics they do the following to deduce that the spectrum of $A:=-(1/2)(d^2/dx^2+x^2)$ is $1/2,1,3/2,\ldots$ find some of the eigenvalues and show that the associated eigenvectors are an ONB. Why does it follow that $\sigma(A)=\sigma_p(A)$?

Everything happens on Schwarz space

As a related question: Reason for Continuous Spectrum of Laplacian " has a discrete spectrum consisting of the eigenvalues ... as can be seen from the eigenfunction basis" how come?

Community
  • 1

1 Answers1

0

If/when the Hilbert-space closure of the algebraic span of whatever eigenfunctions/eigenvectors there happen to be is the whole Hilbert space... we're just done. In particular, if one wants to look at official definitions of "other" parts of putative spectra, there's just no room for anything else to happen.

In the case of that nicest Schroedinger operator, the way that we realize that the eigenvectors will span is by proving that the resolvent is compact... otherwise there's no general way to know what's going on. That is, one proves that the resolvent ... of a given unbounded/diff-op is (nevertheless...) compact (self-adjoint...) to produce an orthonormal basis of eigenfunctions.

paul garrett
  • 55,317
  • so you are basically saying its obvious? –  Jan 21 '14 at 00:50
  • Under the hypotheses in your first paragraph, I think the conclusions follow. The thing that requires work in general is the "completeness", that is, proving that the eigenvectors give an orthonormal basis. Orthogonality is often easy. – paul garrett Jan 21 '14 at 13:33