Timeline for How to diagonalize this tridiagonal difference operator with unbounded coefficients?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2023 at 21:09 | history | edited | Christian Remling | CC BY-SA 4.0 |
added 13 characters in body
|
| Dec 18, 2023 at 21:06 | comment | added | Christian Remling | corresponding to boundary conditions imposed on elements of $D(T_0^*)$. | |
| Dec 18, 2023 at 21:05 | comment | added | Christian Remling | I was assuming some material that is basic in this subject but of course not necessarily well known generally. An attempt at a summary is as follows: your operator is unbounded, so needs a domain. One can define it on finitely supported $g$'s and then take the closure $T_0$ of this operator. $T_0$ is symmetric, but not necessarily self-adjoint. Whether or not this is the case depends on the lc/lp alternative that I discuss in my answer. In your case, it has deficiency $(1,1)$ (one endpoint lp, the other lc), and there is a one parameter family of self-adjoint extensions (cont'd) | |
| Dec 18, 2023 at 20:59 | comment | added | Christian Remling | @LeonidPetrov: I consider the two half lines separately and show that all solutions of $Tg=0$ are in $\ell^2(\mathbb Z_+)$, for $n\ge 1$, so we have limit point case at $+\infty$ (the limit circle/limit point alternative refers to endpoints, so no other set-up makes sense if this is what I want to discuss). | |
| Dec 18, 2023 at 20:53 | comment | added | Leonid Petrov | I don't think the function $q^{n/2}$ is exponentially decaying for $n\in \mathbb{Z}$, so it is not summable. I do not think I understand the construction | |
| Dec 18, 2023 at 20:35 | history | edited | Christian Remling | CC BY-SA 4.0 |
added 154 characters in body
|
| Dec 18, 2023 at 20:23 | history | edited | Christian Remling | CC BY-SA 4.0 |
added 93 characters in body
|
| Dec 18, 2023 at 20:09 | history | answered | Christian Remling | CC BY-SA 4.0 |