Can be this operator extended to an unbounded self-adjoint operator ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:35:41Z http://mathoverflow.net/feeds/question/63952 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63952/can-be-this-operator-extended-to-an-unbounded-self-adjoint-operator Can be this operator extended to an unbounded self-adjoint operator ? Leandro 2011-05-04T22:19:00Z 2011-05-05T01:25:08Z <p>Consider an enumeration <code>$\{q_1,q_2,\ldots\}$</code> of $\mathbb{Q}\cap [1,\infty)$ and a orthogonal Schauder basis <code>$\{e_1,e_2,\ldots\}$</code> of $\ell^2(\mathbb{N})$. Define $Ae_{2k-1}=e_{2k-1}$ and $Ae_{2k}=q_ke_{2k}$ for all $k\geq 1$. <br> <strong> Question 1: </strong> Is it possible to extend $A$ to a linear self-adjoint operator defined in some infinite dimensional subspace of $\ell^2(\mathbb{N})$ ?</p> <p>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 <code>$$\left\{ v\in \ell^2(\mathbb{N}); v=\lim_{n\to\infty} \sum_{i=1}^n\alpha_ie_i \ \text{and}\ \ \lim_{n\to\infty}\sum_{i=1}^n q_i\alpha_ie_i \in \ell^2(\mathbb{N}) \right\}$$</code> is a good candidate to be the domain of $A$ ? <br> <strong> Question 2:</strong> Is the point spectrum <code>$\sigma_p(A)\supset \{q_1,\ldots,q_n,\ldots\}$</code> ? <br></p> <p><strong> Motivavation: </strong> 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.</p> http://mathoverflow.net/questions/63952/can-be-this-operator-extended-to-an-unbounded-self-adjoint-operator/63956#63956 Answer by André Henriques for Can be this operator extended to an unbounded self-adjoint operator ? André Henriques 2011-05-04T23:15:09Z 2011-05-04T23:15:09Z <p>The spectral theorem for unbounded self-adjoint operators says the following:</p> <p>Up to isomorphism, any unbounded self-adjoint operator $A$ on a Hilbert space $H$ can be written in the following form:</p> <p>$$H=L^2(X,\mu)$$</p> <p>$$Af(x)=a(x)f(x)$$</p> <p>for some measure space $(X,\mu)$ and some $\mu$-measurable real valued function $a:X\to \mathbb R$.</p> <p>The domain of the operator is $H$ iff the function $a$ is bounded. If $a$ is unbounded, then the domain is $\{ f\in L^2(X,\mu)\,|\,af\in L^2(X,\mu)\}$.</p> <p>You operator is <i>given</i> in that form. So, yes, it is (i.e. extends to) a self adjoint operator.</p> <p>The answer to your second question is also yes, and your construction indeed provides an example of what you're looking for.</p>