Timeline for When we integrate with respect to a $Q$-Wiener process on $U$, why do we restrict integrands to be operators on $Q^{1/2}U$ (instead of $U$)?
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| Oct 18, 2016 at 12:18 | comment | added | Nate Eldredge | There should be some relevant things in Kuo, Gaussian Measures on Banach Spaces. I'm at a bit of a handicap without seeing the definitions, but the proof should be something like this: Suppose $x \in U_0^\perp$. Since $Q$ is self-adjoint this implies $Q^{1/2}x = 0$. Now it should follow from the definition of $W_t$ that $0 = \|Q^{1/2}x\|^2 = \mathrm{E} |\langle W_1, x \rangle|^2$. Taking a countable dense subset of $U_0^\perp$ should show that $W_1 \in \overline{U_0}$ a.s. Now replace $1$ with any rational $t$ and take countable intersection. | |
| Oct 18, 2016 at 10:22 | comment | added | 0xbadf00d | @NateEldredge Do you have a reference for that? | |
| Oct 16, 2016 at 14:41 | comment | added | Nate Eldredge | For (2), I don't have the book, but if "$Q$-Wiener process" is defined the way I think it is, then it should be a theorem that $\mathrm{P}(W_t \in \overline{U_0} \,\forall t) = 1$. | |
| Oct 16, 2016 at 10:49 | comment | added | 0xbadf00d | @NateEldredge (1) I've written down all the assumptions as (for example) made in the linked book of Röckner. I don't think that they imply that $U_0$ is dense in $U$. (2) What do you mean by "your Brownian motion will always stay within the closure of $U_0$? | |
| Oct 15, 2016 at 17:56 | comment | added | Nate Eldredge | $U_0$ isn't dense in $U$? Are you sure? In this setting one almost always treats the case where $U_0$ is dense, i.e. where $Q$ has dense range. If $U_0$ isn't dense then $\mathfrak{L}(U,H)$ and $\mathfrak{L}(U_0, H)$ may simply be incomparable. But this case is rather degenerate; for instance, your Brownian motion will always stay within the closure of $U_0$. | |
| Oct 15, 2016 at 11:09 | comment | added | 0xbadf00d | @NateEldredge By definition, $\overline{\mathcal E}$ is an abstract object. So, it wouldn't be surprising for me, if $\mathcal E$ is not a subset of $\overline{\mathcal E}$ in the set-theoretic sense. However, as I said before, I don't see at which point of the proof of $(5)$ we really need that the processes which belong to the set on the right-hand side of $(5)$ are $\operatorname{HS}(U_0,H)$- instead of $\mathfrak L(U,H)$-valued. I've linked the proof. Maybe you can tell me where we need $\operatorname{HS}(U_0,H)$-valuedness. | |
| Oct 15, 2016 at 10:27 | comment | added | 0xbadf00d | @NateEldredge But the inclusion $\mathfrak{L}(U,H)\subset\operatorname{HS}(U_0, H)$ isn't really an inclusion, cause, as you say, only $\left\{\left.L\right|_{U_0}:L\in\mathfrak L(U,H)\right\}\subset\operatorname{HS}(U_0, H)$. Since $U_0$ is not dense in $U$, the $L\in\mathfrak L(U,H)$ are not uniquely determined by $\left.L\right|_{U_0}$. So, in that sense, $\mathfrak L(U,H)$ is richer than $\mathfrak L(U_0,H)$, isn't it? | |
| Oct 15, 2016 at 2:14 | comment | added | Nate Eldredge | By the way, $H$ and "Hilbert-Schmidt vs bounded" are red herrings here; you will have the same issue even if $H = \mathbb{R}$. At the root is the issue that the continuous dual of $U_0$ is larger than the continuous dual of $U$. | |
| Oct 15, 2016 at 0:40 | comment | added | Nate Eldredge | So if you replaced $HS(U_0,H)$ on the right side of (5) with $\mathfrak{L}(U,H)$, your space would not equal $\overline{\mathcal{E}}$; it would be too small. | |
| Oct 15, 2016 at 0:39 | comment | added | Nate Eldredge | If I'm reading the notation correctly, we are not "restricting ourselves"; quite the contrary. As you say, every bounded linear operator from $U$ to $H$ restricts to a Hilbert-Schmidt operator from $U_0$ to $H$. So we effectively have $\mathfrak{L}(U,H) \subset HS(U_0, H)$. Moreover the containment is strict; there are Hilbert-Schmidt operators from $U_0$ to $H$ that do not extend to a bounded operator from $U$ to $H$. | |
| Oct 15, 2016 at 0:34 | history | edited | 0xbadf00d | CC BY-SA 3.0 |
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| Oct 15, 2016 at 0:28 | history | asked | 0xbadf00d | CC BY-SA 3.0 |