Timeline for Is there a good notion of "random bounded linear map" on a separable Hilbert space?
Current License: CC BY-SA 3.0
10 events
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| May 23, 2014 at 0:49 | comment | added | tmh | Although as has been pointed out, an operator as you've described cannot be bounded, the usual way around this is to define the `operator' formally as a Gaussian process, getting a so-called isonormal process on the space - there are ways of interpreting, for example, Brownian motion as such a process on a suitably chosen Hilbert space. The book "Gaussian Hilbert Spaces" by Janson might be a good place to look. | |
| May 15, 2014 at 4:02 | answer | added | Nate Eldredge | timeline score: 5 | |
| May 15, 2014 at 3:44 | comment | added | Nate Eldredge | Actually, without the "identically distributed" requirement, shouldn't it be pretty easy to let the entries of the "matrix" be independent Gaussians with variances chosen to guarantee that $T$ is bounded almost surely? | |
| May 15, 2014 at 3:36 | comment | added | Nate Eldredge | I guess you want some kind of non-degeneracy condition as well. Otherwise let $A$ be the operator that maps $e_1$ to $e_1$ and $e_2, e_3, \dots$ to 0, and let $T = \xi A$ where $\xi$ is real-valued Gaussian. | |
| May 15, 2014 at 2:49 | history | edited | Paul Siegel | CC BY-SA 3.0 |
added 396 characters in body
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| May 14, 2014 at 17:37 | comment | added | Abdelmalek Abdesselam | looks like you want to define an $N\times N$ GUE random matrix directly in the $N=\infty$ situation. You may look at free probability theory which does something of this kind although it is not what you had in mind, i.e., a probability measure on $B(H)$. | |
| May 14, 2014 at 16:09 | comment | added | Christian Remling | Let $X_{ij}$ be iid Gaussians. Why don't you just put $T_{ij}=X_{ij}$ now? As pointed out by Martin, that of course doesn't look like it could be bounded with positive probability. | |
| May 14, 2014 at 14:35 | comment | added | zhoraster | I don't quite get what do you mean by "chosen unifromly randomly". But if one multiplies a matrix with iid standard Gaussian entries by a non-random vector $h$ from $H$, the result is a well defined vector of iid centered Gaussians with variance $\|h\|^2$. Not sure if this helps you. | |
| May 14, 2014 at 12:07 | comment | added | Martin Hairer | How could your $T$ possibly be bounded (assuming $H$ is infinite-dimensional of course)? | |
| May 14, 2014 at 11:17 | history | asked | Paul Siegel | CC BY-SA 3.0 |