2
$\begingroup$

I am looking at the proof of this theorem which states that if $H$ is a separable Hilbert space and $A:H\rightarrow H$ a compact self-adjoint operator, then there exists a sequence of real eigenvalues $(\lambda_n)$ converging to $0$, and an orthonormal basis $\{v_n\}$ of eigenvectors with $Av_n=\lambda_nv_n$ for all $n\geq1$.

After he found that $V=\langle e_1,...\rangle$ and that $A\mid_{V^{\perp}}=0$, he uses the separability condition to choose an orthonormal basis for $V^{\perp}$ "(which might be zero, in which case the theorem is already proved, or might be finite dimensional)" and conclude saying that listing the orthonormal basis for $V^{\perp}$ with the one constructed of $V$ Proves the theorem.

I think that it should be fine till now. The point is that as a remark he states that separability is used for making use of orthonormal basis. How would be the idea for the conclusion without the separability condition?

$\endgroup$
3
$\begingroup$

The separability of $H$ has little influence, the only difference between the separable and the non-separable case is that in the non-separable case, $V^\perp$ has an uncountable Hilbert basis (of the same cardinality as any Hilbert basis of $H$, naturally). Some authors prefer to only treat countable Hilbert bases and therefore restrict to the separable case.

To see that, we look at the spectral theorem for compact operators on Banach spaces, as it is formulated as theorem 4.25 in Rudin's Functional Analysis.

Suppose $X$ is a Banach space, $T \in \mathscr{B}(X)$, and $T$ is compact.

  1. If $\lambda \neq 0$, then the four numbers \begin{align} \alpha &= \dim \mathscr{N}(T - \lambda I)\\ \beta &= \dim X/\mathscr{R}(T - \lambda I)\\ \alpha^\ast &= \dim \mathscr{N}(T^\ast - \lambda I)\\ \beta^\ast &= \dim X^\ast/\mathscr{R}(T^\ast - \lambda I)\end{align} are equal and finite.

  2. If $\lambda \neq 0$ and $\lambda \in \sigma(T)$, then $\lambda$ is an eigenvalue of $T$ and of $T^\ast$.

  3. $\sigma(T)$ is compact, at most countable, and has at most one limit point, namely, $0$.

Nothing in that requires separability, or reflexivity, or that the space is a Hilbert space.

In our case, where we have a self-adjoint compact operator $A\colon H \to H$, we know that the eigenspaces to the eigenvalues $\lambda \neq 0$ are mutually orthogonal, so choosing an orthonormal basis for each such eigenspace, we obtain a countable (finite or countably infinite) orthonormal set $\{ e_i : i \in I\}$ consisting of eigenvectors of $A$. Letting $V = \overline{\operatorname{span}}\: \{ e_i : i \in I\}$, by self-adjointness we have $A(V^\perp) \subset V^\perp$, and the restriction $B \colon V^\perp \to V^\perp$ of $A$ is a self-adjoint compact operator with $\sigma(B) = \{0\}$. It follows that $B = 0$, and hence $V^\perp = \ker A$. Then by extending $\{ e_i : i \in I\}$ by a Hilbert basis $\{ \nu_\kappa : \kappa \in K\}$ of $V^\perp$, we obtain a Hilbert basis of $H$ consisting of eigenvectors of $A$. The obtained Hilbert basis is countable if and only if $H$ is separable.

$\endgroup$
  • $\begingroup$ Do you mean to put a conjugate on the \lambda for the 3rd and 4th numbers? $\endgroup$ – Stephen Mar 11 '16 at 16:52
  • $\begingroup$ No, in that theorem $T^{\ast}$ denotes the transpose/Banach space adjoint/dual $T^{\ast} \colon X^{\ast} \to X^{\ast}$, and then $T^{\ast}$ has the same spectrum as $T$. It's unfortunate that the same notation is used for the transpose as for the Hilbert space adjoint, which is likely to lead to confusion sometimes, but since it's a quote, I wouldn't like to change the notation. $\endgroup$ – Daniel Fischer Mar 11 '16 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.