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Let $l^2$ denote the Hilbert space of all complex sequences $\phi = (\phi_j)_{j=0}^{\infty}$ such that $\sum_{j=0}^{\infty} |\phi_j|^2 < \infty$. Consider the linear subspace of $l^2$ defined by \begin{equation} D = \{ \phi \in l^2 : \sum_{j=0}^{\infty} j^2 |\phi_j|^2 < \infty \}, \end{equation} and let $A$ be the operator with domain $D$ which associates to each $\phi \in D$ the vector $A \phi$ whose j-th component (j=0,1,2,...) is \begin{equation} (A \phi)_j = \phi_{j+2} \sqrt{(j+2)(j+1)} - \phi_j - \phi_{j-2}\sqrt{j(j-1)}, \end{equation} where we have set $\phi_{-1}=\phi_{-2}=0$. $A$ is not a closed operator. To see this, consider the vector $\phi$ with components given by \begin{equation} \phi_j = (-1)^{\lfloor j/2 \rfloor} j^{-\beta}, \end{equation} with $1 < \beta < 3/2$. It is not difficult to see that $\phi \in l^2 \backslash D$, that the sequence $\xi = (\xi_j)_{j=0}^{\infty}$ defined by \begin{equation} \xi_j = \phi_{j+2} \sqrt{(j+2)(j+1)} - \phi_j - \phi_{j-2}\sqrt{j(j-1)}, \end{equation} belongs to $l^2$, and that if $\phi^{(n)}$ is the vector whose first $n$ components are equal to those of $\phi$ and all the remaining components are zero, then $A \phi^{(n)}$ converges to $\xi$. So $\phi$ belong to the domain of closure of $A$, but not to $D$. I conjectured that the domain of the closure of $A$ is the linear subspace defined by (again I mean $\phi_{-1}= \phi_{-2}=0$) \begin{equation} S = \{ \phi \in l^2 : \sum_{j=0}^{\infty} |\phi_{j+2} \sqrt{(j+2)(j+1)} - \phi_j - \phi_{j-2}\sqrt{j(j-1)}|^2 < \infty \}. \end{equation} Do you have any idea of how this fact could be proved? Thank you very much for your attention in advance.

Maurizio Barbato
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  • $A$ looks like a sum of products of these ladder operators. – Keith McClary Jan 30 '16 at 02:58
  • For sure it has. I met this operator in a course about mathematical methods of physics. Anyway, my question is purely mathematical. Since theory of operator is new for me, I do not have a sharp intuition about these objects, and cannot even assess the degree of difficulty of my question. Maybe it has a plain answer, or maybe one would need some deeper knowledge of operator theory to answer it. I simply don't know. – Maurizio Barbato Jan 31 '16 at 14:57
  • Is $i(A+1)$ a symmetric operator? Symmetric operators are closable. – Keith McClary Jan 31 '16 at 22:15
  • @Keith McClary: Clearly $A$ is closable. My question asks whether the characterization I conjectured for the closure is correct. – Maurizio Barbato Feb 01 '16 at 18:13
  • If it is symmetric then, by Nelson's analytic vector theorem, it is essentially self-adjoint on the finite component vectors. It follows that the closure of A restricted to that domain is the same as the closure of A (with domain D). – Keith McClary Feb 02 '16 at 02:39
  • As I understand it, you have made a conjecture about the closure of $a^a^ - aa$. We could make the corresponding conjecture about $a^* + a$. This is equivalent to multiplication by $x$ on the span of Hermite functions. We can characterize the closure there and then check whether it agrees with the conjecture. – Keith McClary Feb 05 '16 at 18:51
  • @Keith McClary: For sure, with $a$ and $a$ you meant the annihilation and creation operator quoted in the second answer to my [post] (http://math.stackexchange.com/questions/1629025/adjoint-of-an-operator-in-l2/1629234#1629234), in which I also made the corresponding conjecture about $X=a + a$, and as you said in a comment there, the answer is affirmative also in that case. So why should we use Hermite functions? – Maurizio Barbato Feb 09 '16 at 22:26

1 Answers1

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$i(A+1)$ is a symmetric tri-diagonal matrix on each of the subspaces spanned by the even and odd coordinates. These are equivalent to Jacobi matrices (real symmetric with positive off-diagonal elements). It is a classical result that these define essentially self-adjoint operators if the off-diagonal elements grow no more rapidly than $n$.

By Theorem 2.7 of Barry Simon's The Classical Moment Problem as a Self-Adjoint Finite Difference Operator, the conjecture characterizes $D(A^*)$, which is the same as the closure of $A$ if $i(A+1)$ is essentially self-adjoint.

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Keith McClary
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  • We have \begin{equation} S = { \phi \in l^2 : \sum_{j=0}^{\infty} |\phi_{j+2} \sqrt{(j+2)(j+1)} - \phi_j - \phi_{j-2}\sqrt{j(j-1)}|^2 < \infty } = { \phi \in l^2 : \sum_{j=0}^{\infty} |\phi_{j+2} \sqrt{(j+2)(j+1)} - \phi_{j-2}\sqrt{j(j-1)}|^2 < \infty } \end{equation} so that, by the same argument in your answer to my previous question http://math.stackexchange.com/questions/1629025/adjoint-of-an-operator-in-l2/1629234#1629234, it is easy to see that $S$ is the domain of the adjoint of $i(A+1)$. If $i(A+1)$ is essentially self-adjoint, my conjecture is then true. – Maurizio Barbato Feb 08 '16 at 21:30
  • So the question becomes: is $i(A+1)$ essentially self-adjoint? Unfortunately, I am not familiar with matrix representations and analytic vectors, so I can't check the conditions you quoted. – Maurizio Barbato Feb 08 '16 at 21:33
  • You don't need to know about analytic vectors. The simple proof in this answer shows that if $\Sigma_{n=1}^\infty \frac{1}{b_n} = \infty$ then there is no $l^2$ sequence orthogonal to the range of $A-\lambda$ for non-real $\lambda$ where $A$ is a Jacobi matrix (not the same as your $A$) and $b_n$ are the off-diagonal elements. – Keith McClary Feb 08 '16 at 22:24
  • Oops, the Jacobi matrix is called $J$, not $A$. – Keith McClary Feb 09 '16 at 01:59
  • After a little bit of thinking I have succeeded in understanding the argument you gave in the answer you quoted (a proof can also be found in Schmudgen, Unbounded Self-Adjoint Operators on Hilber Space, Example 7.6), which in turn implies (see Schmudgen, Unbounded Self-Adjoint Operators on Hilber Space, Proposition 3.8) that the restriction of $i(A+1)$ to the space of finite linear combinations of even (odd) coordinates is essentially self-adjoint. – Maurizio Barbato Feb 09 '16 at 22:08
  • We conclude that $i(A+1)$ is essentially self-adjoint on $D$ (see my last comment in my other my [post] (http://math.stackexchange.com/questions/1629025/adjoint-of-an-operator-in-l2/1629234#1629234>) ), and this proves my conjecture. – Maurizio Barbato Feb 09 '16 at 22:09
  • By the way, I made a slip in my comment above, where I quoted the answer to my other post http://math.stackexchange.com/questions/1629025 to Keith McClary, but it was actually given by TrialAndError. – Maurizio Barbato Feb 09 '16 at 22:34