Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
so you are basically saying its obvious? –  user88576 Jan 21 at 0:50
    
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 at 13:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.