Spectrum of an unbounded operator

Question

Consider a densely defined unbounded operator $A_0:D(A_0)(\subset{H})\to H$ which has the following properties:

1- Symmetric, $\langle A_0x,y\rangle=\langle x,A_0y \rangle$

2- Positive, $\langle A_0x,x\rangle \ge0$

3- $A_0$ is surjective and $A_0^{-1}$ is a compact operator

I am wondering whether these are sufficient ground for saying that:

$A_0$ admits an infinite set of eigenvalues which are positive and strictly increasing; furthermore, the corresponding eigenfunctions form an orthonormal basis of $H$.

why don't you simply apply the spectral theorem to $A_0^{-1}$ ? (note that for compact operators, you can directly prove the spectral theorem by generalizing the algorithm for the SVD of matrices) — reuns, Apr 22 '16 at 02:37

Disintegrating By Parts · Accepted Answer · 2016-04-22T03:16:53.417

Yes, it is enough. Because $A_0$ is positive, symmetric and surjective, then $A_0$ is densely-defined, injective and selfadjoint. Therefore, $A_0^{-1}$ is compact, selfadjoint with trivial null space. So $A_{0}^{-1}$ has an orthnormal basis of eigenfunctions $\{e_n \}$ with corresponding eigenvalue sequence of positive numbers $$ \lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \cdots \ge \lambda_n \ge \cdots \rightarrow 0 $$ Thus $A_0e_n = \lambda_n^{-1}e_n$ and $\lambda_n^{-1}\rightarrow\infty$.

score 1 · Answer 2 · answered Apr 22 '16 at 02:11

You are implicitly assuming that $A_0$ is injective.

Let $y\in H$. Then there exists $x\in D(A_0)$ with $y=A_0x$. Then $$ \langle A_0^{-1}y,y\rangle=\langle x,A_0x\rangle\geq0. $$ Thus $A_0^{-1}$ is positive, and so there exists an orthonormal basis $\{e_n\}$ of eigenvectors. Since $A_0^{-1}$ is compact, its eigenvalues $\lambda_n$ satisfy $\lambda_n\searrow0$.

Also, from $A_0^{-1}e_n=\lambda_ne_n$ and $A_0^{-1}$ bijective $H\to D(A_0)$, we get that $e_n\in D(A_0)$ and that $\lambda_n>0$ for all $n$. So $$ A_0e_n=\frac1{\lambda_n}\,e_n, $$ and $\lambda_n\nearrow\infty$.

Spectrum of an unbounded operator

2 Answers2