How to find the self-adjoint extension of an unbounded symmetric operator?

Asked Mar 23 '21 at 15:20

Active Mar 23 '21 at 15:20

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I have an unbounded symmetric operator, and I would like to find its self-adjoint extension if possible. First off, what properties does such an operator need in order to have a self-adjoint extension, and what are the steps I need to take to get this extension?

For concreteness, the operator I am looking at is an infinite dimensional matrix $M_{nm}$ acting on $l^2$ which reads

$$M_{nm}=\frac{(2n+1)!!}{(2n)!!}\Big((4n+3)\delta_{nm}+2n\delta_{n-1m}+(2n+3)\delta_{n+1m}\Big)$$

This is symmetric with respect to the inner product $(u,v)=\sum_{n=0}^{\infty}u^*_nv_n$.

asked Mar 23 '21 at 15:20

fewfew4

There is Carleman's test. How fast does the part with the !!s grow? – Keith McClary Mar 23 '21 at 18:02
@KeithMcClary The double factorials grow like $\sim \sqrt{\pi n}$. – fewfew4 Mar 23 '21 at 18:13
Carleman won't work then. There is another test for semibounded operators with growth $~n^2$. – Keith McClary Mar 23 '21 at 18:19
@KeithMcClary Where would I go to find such tests? – fewfew4 Mar 23 '21 at 19:00
Eqn. 6 on p. 26 here has the corresponding condition for semibounded Jacobi matrices. – Keith McClary Mar 23 '21 at 19:01
It is stated on p.25 #4 of Akhiezer's book "The Classical Moment Problem ...". – Keith McClary Mar 23 '21 at 19:22
Thank you so much, I had no clue these matrices are called Jacobi matrices, and that there is already a wealth of research on them. I have a lot of reading to do. – fewfew4 Mar 23 '21 at 20:49
Let us continue this discussion in chat. – fewfew4 Mar 24 '21 at 15:45

How to find the self-adjoint extension of an unbounded symmetric operator?

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