Question

Consider an enumeration $\{q_1,q_2,\ldots\}$ of $\mathbb{Q}\cap [1,\infty)$ and a orthogonal Schauder basis $\{e_1,e_2,\ldots\}$ of $\ell^2(\mathbb{N})$. Define $Ae_{2k-1}=e_{2k-1}$ and $Ae_{2k}=q_ke_{2k}$ for all $k\geq 1$.
Question 1: Is it possible to extend $A$ to a linear self-adjoint operator defined in some infinite dimensional subspace of $\ell^2(\mathbb{N})$ ?

If I am not wrong, this possible linear self-adjoint extension of $A$ can not be defined everywhere in $\ell^2(\mathbb{N})$ and I would like to know if the following set $$ \left\{ v\in \ell^2(\mathbb{N}); v=\lim_{n\to\infty} \sum_{i=1}^n\alpha_ie_i \ \text{and}\ \ \lim_{n\to\infty}\sum_{i=1}^n q_i\alpha_ie_i \in \ell^2(\mathbb{N}) \right\} $$ is a good candidate to be the domain of $A$ ?
Question 2: Is the point spectrum $\sigma_p(A)\supset \{q_1,\ldots,q_n,\ldots\}$ ?

Motivavation: I would like to know if there is an example of an unbounded self-adjoint operator such that the point spectrum is not composed only by isolated points in $\mathbb{R}$ and there is at least one eigenvalue with infinite dimensional eigenspace.

Answer 1

The spectral theorem for unbounded self-adjoint operators says the following:

Up to isomorphism, any unbounded self-adjoint operator $A$ on a Hilbert space $H$ can be written in the following form:

$$H=L^2(X,\mu)$$

$$Af(x)=a(x)f(x)$$

for some measure space $(X,\mu)$ and some $\mu$-measurable real valued function $a:X\to \mathbb R$.

The domain of the operator is $H$ iff the function $a$ is bounded. If $a$ is unbounded, then the domain is $\{ f\in L^2(X,\mu)\,|\,af\in L^2(X,\mu)\}$.

You operator is given in that form. So, yes, it is (i.e. extends to) a self adjoint operator.

The answer to your second question is also yes, and your construction indeed provides an example of what you're looking for.