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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.

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    $\begingroup$ Can't you just use Friedrichs extension? define $A$ on the linear subspace of finite sequences, which is dense. $A$ is clearly a symmetric operator, bounded from below, thus it admits a selfadjoint extension, and the $e_k$ are eigenvectors as you require $\endgroup$ – Piero D'Ancona May 4 '11 at 22:35
  • $\begingroup$ Just to clarify: are you asking if there is a choice of Schauder basis for which things work, or if things work for every choice of Schauder basis? $\endgroup$ – Yemon Choi May 5 '11 at 1:00
  • $\begingroup$ Hi Yemon, If this works for one basis it is good for me. $\endgroup$ – Leandro May 5 '11 at 1:22
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    $\begingroup$ Thanks for the remark Yemon, I added inthe question that I was thinking about orthogonal basis. $\endgroup$ – Leandro May 5 '11 at 1:32
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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.

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  • $\begingroup$ doesn't this assume that the Schauder basis mentioned in the question consists of orthogonal vectors? or have I misunderstood? $\endgroup$ – Yemon Choi May 5 '11 at 1:01
  • $\begingroup$ @Christian: You would be right if $X$ was a subset of $\mathbb R$, and the function $a(x)$ was given by the formula $a(x)=x$. But I'm allowing more general $X$, and more general $a(x)$. So there is no need to take orthogonal sums. $\endgroup$ – André Henriques Apr 24 '16 at 23:24

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