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 $U$).
  • $\begingroup$ 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 agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.