Can be this operator extended to an unbounded self-adjoint operator ?

2011-05-04T22:19:00Z

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 by André Henriques for Can be this operator extended to an unbounded self-adjoint operator ?

2011-05-04T23:15:09Z

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.

Can be this operator extended to an unbounded self-adjoint operator ? - MathOverflow

Can be this operator extended to an unbounded self-adjoint operator ?

Answer by André Henriques for Can be this operator extended to an unbounded self-adjoint operator ?