Basis of eigenvectors of a self-adjoint operator with $\sigma(T) = \sigma_{p}(T)$

Question

Suppose $T$ is a self-adjoint operator on a Hilbert space $H$ with only point spectrum: $$\sigma(T) = \sigma_{p}(T) = \{x \in H: (T-\lambda)x = 0\}$$ Eigenvectors associated to different eigenvalues are orthogonal. In the above case, is it true that one can find a basis of $H$ consisting of eigenvectors of $T$?

Answer 1

A was observed by John Doe the claim is not true for nonseparable Hilbert space.

I assume an orthonormal basis is considered in OP. The claim can be proved without applying the spectral theorem.

Assume $\mathcal{H}$ is separable. Let $\mathcal{H}_{\lambda} =\ker (T-\lambda I).$ Every subspace admits an orthonormal system $B_{\lambda}$ and $B_\lambda\cap B_\mu=\emptyset$ for $\lambda\neq \mu.$ We claim that $$B=\bigcup_{\lambda\in \sigma_p(T)}B_\lambda $$ is an orthonormal basis. Indeed, the sum is countable as the set $\sigma(T)$ is countable. Assume that $V=B^\perp.$ Due to the self-adjointness, $V$ is a closed invariant subspace of $T.$ We have $\sigma(T\mid _V)\subset \sigma(T),$ hence $\sigma(T\mid_V)$ is a countable compact subset of the real line. Therefore it contains isolated point $\lambda_0.$ Thus $\lambda_0$ is an eigenvalue of $T\mid_V,$ i.e. $\ker (\lambda_0 I-T)\cap V\neq 0,$ which leads to a contradiction.

Remark A proof that isolated point in the spectrum of a self-adjoint operator is an eigenvalue can be proved by continuous functional calculus. Namely let $f$ be a continuous real valued function such that $f(x_0)=1$ and $f(x)=0$ for $x\in\sigma(T),$ $x\neq x_0.$ Then $\sigma(f(T))=f(\sigma(T))=\{0,\ 1\}.$ We have $Tf(T)=x_0f(T),$ hence $x_0$ is an eigenvalue.

Answer 2

Yes, because the spectral projection associated with a singleton of a point in the point spectrum is the projection on the eigenspace of that point, and the spectral projection of the entire spectrum is the identity operator. It all follows from the spectral theorem if you study its exact detail.

In fact, it follows from the spectral theorem for that a self-adjoint operator that has an orthonormal basis of eigenvectors - the closure of the point spectrum must be the whole spectrum. The converse need not be true because a (self-adjoint) operator can have an independent continuous spectrum part that overlaps the closure of its point spectrum.

Edit: the difference between the first and second cases is that the point spectrum is countable, and the sigma-additivity of the spectral measure is used to establish the first statement if the point spectrum is the whole spectrum. However, if you just know that that the closure of the point spectrum is the whole spectrum - it is insufficient to prove there's an eigenbasis because the whole spectrum may be uncountable!

Edit 2: I should mention that the discussion is for a separable Hilbert space. In a non-separable Hilbert space the claim in the original post is false. An operator can be constructed such that its point spectrum is the interval $[0,1]$, and it also has a continuous spectrum $[0, 1]$ so that there is no eigenbasis.