Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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 –  Piero D'Ancona May 4 '11 at 22:35
    
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? –  Yemon Choi May 5 '11 at 1:00
    
Hi Yemon, If this works for one basis it is good for me. –  Leandro May 5 '11 at 1:22
1  
Thanks for the remark Yemon, I added inthe question that I was thinking about orthogonal basis. –  Leandro May 5 '11 at 1:32
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
doesn't this assume that the Schauder basis mentioned in the question consists of orthogonal vectors? or have I misunderstood? –  Yemon Choi May 5 '11 at 1:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.