Timeline for Representation of support of Gaussian measure by kernels of no-variance functionals
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2019 at 19:48 | comment | added | trying | @NateEldredge yes, I am trying to work out the paper exercise by exercise. Thank you for your response | |
| Apr 21, 2019 at 14:11 | comment | added | Nate Eldredge | Given a ball $B(x,\epsilon)$, separability guarantees there is some other ball $B(y, \epsilon/2)$ with positive measure. Now the non-degeneracy of $q$ implies the Cameron-Martin subspace $H \subset F$ is dense, so find some $h$ very close to $x-y$ and translate the measure by $h$, which by the Cameron-Martin theorem leaves it quasi-invariant. You can see Section 4.3 of my notes at arxiv.org/abs/1607.03591 for some more information. | |
| Apr 21, 2019 at 14:10 | comment | added | Nate Eldredge | @orange: It is clear that $F = \bigcap_{q(f,f)=0} \ker f$ is closed because it is an intersection of closed sets (recall that all the $f$ are continuous). To see it is the smallest takes more work. The best way I know to see it is to restrict everything to $F$, so that $q$ is now non-degenerate, and use the Cameron-Martin theorem to show that this would imply that $\mu$ has full support. [...] | |
| Apr 21, 2019 at 9:33 | comment | added | trying | Makes sense but why is this the smallest such set and why is it closed | |
| Apr 19, 2019 at 19:08 | comment | added | Nate Eldredge | @orange: Mercury Bench doesn't write it, but from context I assume that we are dealing here with a centered Gaussian measure. In this case we have $\int f\,d\mu = 0$ for any continuous linear functional $f$. | |
| Apr 19, 2019 at 19:04 | comment | added | trying | why not as $\int fg-\int f\int g $? | |
| Apr 19, 2019 at 18:52 | comment | added | Nate Eldredge | @orange: The covariance operator $q$ is defined by $q(f,g) = \int_X fg\,d\mu$. So if $q(f,f) = 0$ then $\int_X f^2\,d\mu = 0$, which is to say $f = 0$ $\mu$-a.e., which is to say $\mu(\ker f) = 1$. | |
| Apr 19, 2019 at 16:56 | comment | added | trying | Why do such kernels have $\mu$ measure 1? | |
| Jun 27, 2016 at 19:16 | comment | added | Philipp Wacker | Ah, thanks. I was so caught up with $X^*$ not being separable that I missed this argument. | |
| Jun 27, 2016 at 16:59 | comment | added | Nate Eldredge | @FasEtNefas: Well, a lazy way to see it is that by Alaoglu the unit ball is weak-* compact, and compact metric spaces are separable. | |
| Jun 27, 2016 at 12:39 | comment | added | Philipp Wacker | Sorry to bother again but after a careful look into Conway's book (and also Brezis) I could only find statements regarding the metrizability of the unit ball in X∗ with the weak- * -topology, but I couldn't find anything about the separability. Without that I don't see how we can choose a subsequence weak-*-dense in $B^*\cap F$ of $(f_n)_n$ in the first place. | |
| Jun 24, 2016 at 15:09 | comment | added | Nate Eldredge | The fact that $B^*$ is separable and metrizable in the weak-* topology (whenever $X$ is separable) is pretty standard; I think I first saw it in Conway's A Course in Functional Analysis. Given this, "weak-* sequentially dense" is easy: choose a weak-* dense sequence $\{f_n\}$ (by separability) and fix a compatible metric $d$. Given $f$, by density, for each $k$ there is an $f_{n_k}$ with $d(f,f_{n_k}) < 1/k$. So $f_{n_k} \to f$ weak-*. | |
| Jun 24, 2016 at 14:38 | comment | added | Philipp Wacker | Alright. Could you maybe provide me with a suggestion for further reading, especially on the statement "weak-* dense sequence + metrizable space = weak-* sequentially dense sequence"? I fear my textbook skips this. | |
| Jun 24, 2016 at 14:14 | comment | added | Nate Eldredge | @FasEtNefas: I mean that if $X^*$ is equipped with the weak-* topology, the subset $B^*$ is both separable and metrizable. And I am not using weak-* (sequential) compactness here, but simply the fact that $\{f_n\}$ is weak-* dense. Thanks to metrizability it is also sequentially dense. So it is possible to extract a subsequence which not only converges, but converges to the given $f$. | |
| Jun 24, 2016 at 14:08 | comment | added | Philipp Wacker | Thanks! So, just to make sure I got this: With "weak-* separable metrizable" you just mean that $B^*$, being in a separable space, is metrizable in the weak-* topology, right? And metrizability is needed for "weak-* sequential compactness" of $(f_n)_n$ in order to be able to extract a subsequence converging weak-*. | |
| Jun 24, 2016 at 13:12 | vote | accept | Philipp Wacker | ||
| Jun 24, 2016 at 1:47 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
added 15 characters in body
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| Jun 24, 2016 at 1:39 | history | answered | Nate Eldredge | CC BY-SA 3.0 |