3
$\begingroup$

When we integrate with respect to a $Q$-Wiener process $(W_t)_{t\ge 0}$ ($Q$ being a bounded, linear, nonnegative and self-adjoint operator on a separable $\mathbb R$-Hilbert space $U$ with finite trace) on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge 0},\operatorname P)$, the subspace $U_0:=Q^{1/2}U$ is considered as being important.

The question is: Why?

Let's take this step by step: If $(\Phi_t)_{t\in[0,\:T]}$ ($T>0$) is of the form $$\Phi_t=\sum_{i=1}^k\zeta_{i-1}1_{(t_{i-1},\:t_i]}(t)\;\;\;\text{for all }t\in[0,T]\tag 1$$ for some $k\in\mathbb N$, $0\le t_0<\cdots<t_k\le T$ and $\mathfrak L(U,\:H)$-valued ($H$ being another separable $\mathbb R$-Hilbert space) strongly (i.e. with respect to the strong operator topology; see Da Prato's book) $\mathcal F_{t_{i-1}}$-measurable random variables $\zeta_{i-1}$ on $(\Omega,\mathcal A,\operatorname P)$ (let $\mathcal E$ be the set of all such $\Phi$), then $$(\Phi\cdot W)_t:=\sum_{i=1}^k\zeta_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right)\tag 2$$ is called Itō integral of $\Phi$ with respect to $W$ up to $t\in[0,T]$. If $$\mathcal M^2:=\left\{(X_t)_{t\in[0,\:T]}\subseteq\mathcal L^2(\operatorname P,H):X\text{ is an almost surely right-continuous }\mathcal F\text{-martingale with }\operatorname P[X_0=0]=1\right\}$$ is equipped with $$\left\|X\right\|_{\mathcal M^2}^2:=\sup_{t\in[0,\:T]}\left\|X_t\right\|_{\mathcal L^2(\operatorname P,\:H)}$$ and $\mathcal E$ is equipped with $$\left\|\Phi\right\|_{\mathcal E}^2:=\int_0^T\operatorname E\left[\left\|\Phi_tQ^{1/2}\right\|_{\operatorname{HS}(U,\:H)}^2\right]{\rm d}t\;,\tag 3$$ then it's easy to see that $$\mathcal E\to\mathcal M^2\;,\;\;\;\Phi\mapsto\Phi\cdot W\tag 3$$ is a linear isometry ($\mathcal E$ and $\mathcal M^2$ can be considered as being normed $\mathbb R$-vector spaces under the usual identifications). At this point, we're ready to extend the Itō integral $(2)$ to the abstract completion $\overline{\mathcal E}$ of $\mathcal E$ with respect to $\left\|\;\cdot\;\right\|_{\mathcal E}$. It's clear that this extension is trivial by the isometric nature of $(3)$.

Da Prato and Röckner prove that $$\overline{\mathcal E}=\left\{\Psi:\Omega\times[0,T]\to\operatorname{HS}(U_0,H):\Psi\text{ is }\mathcal F\text{-predictable with }\left\|\Psi\right\|_{\mathcal E}<\infty\right\}\tag 5$$ (where $\operatorname{HS}(U_0,H)$ denotes the space of Hilbert-Schmidt operators from $U_0$ to $H$ and predictability is defined as usual). However, it's obvious that this equality has to be understood in some sense. In fact, they don't consider $\mathcal E$, but $$\left\{\left.\Phi\right|_{U_0}:\Phi\in\mathcal E\right\}\tag 6$$ (and it's clear that restrictions of bounded linear operators from $U$ to $H$ to the subspace $U_0$ are in fact Hilbert-Schmidt operators from $U_0$ to $H$).

Why do we need to restrict ourselves to processes with values in $\operatorname{HS}(U_0,H)$? Why can't we replace this operator space with $\operatorname{HS}(U,H)$ or even $\mathfrak L(U,H)$ (bounde linear operators from $U$ to $H$)?

I don't see anything in the proof linked above (Proposition 2.3.8 in Röckner's book), which wouldn't work for $\mathfrak L(U,H)$-valued $\mathcal F$-predictable processes as well.

Of course, there are two points we need to note:

  1. $\mathfrak L(U,H)$ is not separable (in contrast to $\operatorname{HS}(U_0,H)$) and hence the Borel $\sigma$-algebra on might be too rich. However, we should be able to fix that problem by considering measurability with respect to the strong Borel $\sigma$-algebra, again
  2. Obviously, $\left\|\Phi-\Psi\right\|_{\mathcal E}=0$, for two processes in $\mathcal E$, if and only if they are equal on $U_0$ (and not on the whole space $U$) $\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}$-almost everywhere ($\lambda^1$ denotes the Lebesgue measure on $\mathbb R)$, but that shouldn't be a problem (note in this context that the Itō integral in $(2)$ does depend on the values on the whole space $U$).
$\endgroup$
  • $\begingroup$ 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 $H$. $\endgroup$ – Nate Eldredge Oct 15 '16 at 0:39
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 15 '16 at 0:40
  • $\begingroup$ 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 $U$. $\endgroup$ – Nate Eldredge Oct 15 '16 at 2:14
  • $\begingroup$ @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? $\endgroup$ – 0xbadf00d Oct 15 '16 at 10:27
  • $\begingroup$ @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. $\endgroup$ – 0xbadf00d Oct 15 '16 at 11:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.