Timeline for Is the spectrum of a "self adjoint" operator real on $\ell^p$?
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14 events
| when toggle format | what | by | license | comment | |
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| Apr 6, 2020 at 15:43 | vote | accept | an_ordinary_mathematician | ||
| Apr 6, 2020 at 13:18 | answer | added | an_ordinary_mathematician | timeline score: 4 | |
| Apr 2, 2020 at 15:26 | comment | added | M.González | @Giorgio Metafune It is something I used in my Thesis, around 1983. I have tried to find the reference, but I failed. | |
| Apr 2, 2020 at 13:36 | comment | added | Bill Johnson | I guess the only correct part of my comment above is that the point spectrum is real when $p<2$ (because the operator is bounded on $\ell_2$ and $\ell_p \subset \ell_2$). | |
| Apr 2, 2020 at 10:38 | comment | added | Giorgio Metafune | @M. Gonzalez Very nice If you can recover that example (or give a more precise reference). I am also locked at home but I have online access to the library. | |
| Apr 2, 2020 at 10:19 | comment | added | M.González | I think I remember there is a paper by Gohberg and Krupnik in which they construct an operator on $\ell_p$ which is self-adjoint for $p=2$ and the spectrum is an ellipsoid por $p\neq 2$ which increases with $p>2$. Unfortunately, I have no access to my office. | |
| Apr 2, 2020 at 7:57 | comment | added | Giorgio Metafune | There are self-adjoint operators in $L^2$, generating (analytic) semigroups in $L^p$ for all $p$, such that the spectrum is not real for $p \neq 2$. One can take $L=r^2D_{rr}+2rD_r$ in the half-line. The spectrum is a parabola which degenerates into the negative half-line when $p=2$. The resolvents and the semigroups have similar bad properties, by the spectral mapping theorem. Maybe a discrete version can be contructed using them. | |
| Apr 2, 2020 at 3:52 | comment | added | Yemon Choi | Actually, it may be enough to work on ${\bf FS}_2$, the free monoid generated by 2 elements: for $p=1$ the construction in the proof of Theorem 5.1 of Jenkins's 1970 paper projecteuclid.org/euclid.pjm/1102977529 but as I said, I have not tried to see if the same ideas wor for $1<p<2$. | |
| Apr 2, 2020 at 3:46 | comment | added | Yemon Choi | Comment for the OP: I suspect the answer is negative for suitable convolution operators on $\ell^p({\bf F}_2)$ but right now I don't remember if the details for the $p=1$ case work for $1<p<2$. | |
| Apr 2, 2020 at 3:38 | comment | added | Yemon Choi | @BillJohnson Your argument doesn't seem to use the fact $p>1$; but there are known examples of "self-adjoint" (i.e. conjugate symmetric) convolution operators on $l^1$(free group) whose spectrum has non-empty interior | |
| Apr 1, 2020 at 23:36 | comment | added | Bill Johnson | Yes, Jochen, because the hypothesis implies that $A$ also is bounded on $\ell_q$, $1/p+1/q=1$, hence all eigenvectors are in $\ell_2$. Basically the same argument gives a positive answer to the OP's question (I think). | |
| Apr 1, 2020 at 22:49 | comment | added | Jochen Glueck | Do you know whether the point spectrum is always real? | |
| Apr 1, 2020 at 21:09 | history | edited | an_ordinary_mathematician | CC BY-SA 4.0 |
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| Apr 1, 2020 at 20:06 | history | asked | an_ordinary_mathematician | CC BY-SA 4.0 |