Every compact subset of real line is spectrum of self-adjoint operator

Question

It is known that if $H$ is Hilbert space and $T$ is self-adjoint operator on $H$, then the spectrum is real and closed. But is every closed or compact subset of real numbers a spectrum of some self-adjoint operator on some $H?$ For interval I think we can find the construction but how if the set is a mess, for example, Cantor set?

Answer 1

There are bits and pieces in the comments that I put together to an answer.

Empty set: The empty set is compact and does only arise as the spectrum of a self-adjoint operator if the Hilbert space in question is the trivial Hilbert space (see for example here Self-adjoint operator has non-empty spectrum.).

Finite set: It's instructive to see how things play out for a finite set, say $\Omega=\{\lambda_1, \dots \lambda_n\}$. Then we can pick $H=\mathbb{C}^n$ and $$ T: \mathbb{C}^n \rightarrow \mathbb{C}^n, x \mapsto \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix} x. $$

General case: Let $\Omega \subset \mathbb{R}$ compact and non-empty. Then there exists a dense, countable subset $\{ \lambda_n \in \Omega \ : \ n\in \mathbb{N} \}$ in $\Omega$ (see Prove if $X$ is a compact metric space, then $X$ is separable.). Now we can consider a similar operator as in the finite case, namely,

$$ T: \ell^2(\mathbb{N}, \mathbb{C}) \rightarrow \ell^2(\mathbb{N}, \mathbb{C}), (x_n)_{n\in \mathbb{N}} \mapsto (\lambda_n x_n)_{n\in \mathbb{N}}. $$

With a bit of work (see Spectrum of $l^p$ multiplication operator (Brezis 6.17)) one can show that for such operators one has $$ \sigma(T) = \overline{ \{ \lambda_n \ : \ n \in \mathbb{N} \} }. $$

By construction we have that $\{ \lambda_n \ : \ n \in \mathbb{N}\}$ is dense in $\Omega$. Thus, as $\Omega$ is closed, we get $\sigma(T)= \Omega$.

If you like bigger Hilbert spaces: Before Mariano Suárez-Álvarez's comment I had a similiar, but uglier construction in mind. Namely, I would have taken a Hilbert space with an orthonormal bases that is large enough to associate to every point in $\Omega$ an orthonormal basis vector and then do a similar construction as above. Namely, one can consider the Hilbert space $$ \ell^2(\Omega, \mathbb{C}) = \left\{ (x_s)_{s\in \Omega} \subseteq \mathbb{C}^\Omega \ : \ \sum_{s\in \Omega} \vert x_s \vert^2 < \infty \right\} $$ and the scalar product $$ \langle (x_s)_{s\in \Omega}, (y_s)_{s\in \Omega} \rangle = \sum_{s\in \Omega} \overline{x_s} y_s. $$ We can define the operator $$ T: \ell^2(\Omega, \mathbb{C}) \rightarrow \ell^2(\Omega, \mathbb{C}), (x_s)_{s\in \Omega} \mapsto (s x_s)_{s\in \Omega}. $$ One checks that this operator is bounded and symmetric, thus self-adjoint. Furthermore, $(\delta_{s_0,s})_{s\in \Omega}$ is an eigenvector to the eigenvalue $s_0$. Furthermore, if $\lambda\notin \Omega$, then there exists (as $\Omega$ is closed) $\varepsilon>0$ such that for all $s\in \Omega$ holds $\vert \lambda -\lambda_s\vert>\varepsilon$. Thus, the operator $T-\lambda Id$ admits the inverse $$ A: \ell^2(\Omega, \mathbb{C})\rightarrow \ell^2(\Omega, \mathbb{C}), (x_s)_{s\in \Omega} \mapsto \left( \frac{1}{s-\lambda} x_s \right)_{s\in \Omega} $$ which is bounded (with operator norm bounded by $\varepsilon^{-1}$). Thus, if $\lambda\notin \Omega,$ then $\lambda$ is not in the spectrum of $T$ and hence $\Omega = \sigma(T)$.