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.



Can be this operator extended to an unbounded self-adjoint operator ?  
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 
$$ 
\bigl\{ f\in L^2(X,\mu) \, | \, af \in L^2(X,\mu) \bigr\}.
$$


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.


