Timeline for Spectral gap of a Markov operator on $L^2$ with a symmetric $L^\infty$ kernel
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 12 at 21:03 | comment | added | Jochen Glueck | @GiuseppeNegro: Yes, there's an infinite-dimensional version of irreducibility that makes sense on general Banach lattices. See for instance this question and my answer to it. | |
| Feb 10 at 16:52 | comment | added | Giuseppe Negro | @JochenGlueck: Thanks, I didn't really know that, I learned something. So, just out of curiosity: is there a "continuum" version of the "discrete" concept of irreducibility? That is, a necessary and sufficient condition on the kernel $k(x, y)$ such that $Kf(x)=\int k(x, y)f(y)\, dy$ has $1$ as simple dominant eigenvalue. I think this is essentially what this question by JohnyB is asking. | |
| Feb 9 at 8:40 | comment | added | Johny B | Indeed, trivial:) Thank you Jochen Glueck! | |
| Feb 8 at 21:52 | comment | added | Mikael de la Salle | More generally, $1$ has higher mutliplicity if and only if there is a Borel subset $J \subset I$ of measure $\neq 0,1$ such that $k=0$ on $J \times J^{c}$. One direction is clear (the indicator function of $J$ is another eigenvector), and the converse is by some kind of maximum principle. | |
| Feb 8 at 14:19 | history | edited | gmvh | CC BY-SA 4.0 |
Added MathJax to title, corrected punctuation
|
| Feb 8 at 14:14 | history | edited | Johny B | CC BY-SA 4.0 |
added 15 characters in body
|
| Feb 8 at 13:29 | comment | added | Jochen Glueck | JohnyB: Let $k$ be twice the indicator function of $[0,1/2]^2 \cup [1/2,1]^2$. Then $1$ is a double eigenvalue. | |
| Feb 8 at 13:26 | comment | added | Jochen Glueck | @GiuseppeNegro: Strict positivity is an unnecessarily strong assumption, though. In the symmetric case, simplicity of the eigenvalue $1$ is equivalent to irreducibility (and irreducibility is sufficient for simplicity even in the non-symmetric case). | |
| Feb 8 at 12:44 | comment | added | Giuseppe Negro | Assuming $k>0$ at all points, and in the discrete case $k=k_{ij}$, the answer would be affirmative by Perron--Frobenius, as you might already know: en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem | |
| S Feb 8 at 12:29 | review | First questions | |||
| Feb 8 at 13:02 | |||||
| S Feb 8 at 12:29 | history | asked | Johny B | CC BY-SA 4.0 |