Properties of the trace term in the Itō formula

Let's consider the SDE $${\rm d}X_t=u_t(X_t){\rm d}t+\xi_t(X_t){\rm d}W_t\;\;\;\text{for all }t\ge 0\tag 1$$ where

• $U,H$ are separable $\mathbb R$-Hilbert spaces
• $Q\in\mathfrak L(U)$ is nonnegative and self-adjoint with finite trac
• $W$ is a $Q$-Wiener process
• $u\in C^{1,\:2}(\mathbb R_{\ge 0}\times H,H)$
• $\xi:\mathbb R_{\ge 0}\times H\to\operatorname{HS}(U_0,H)$ with $U_0:=\sqrt QU$ being equipped with the usual inner product

By the Itō formula, we obtain $${\rm d}u^{(n)}(t,X_t)=\left[\frac{∂u^{(n)}}{∂t}(t,X_t)+{\rm D}u^{(n)}(t,X_t)u(t,X_t)+\frac12\text{tr}\left[{\rm D}^2u^{(n)}(t,X_t)\left(ξ(t,X_t)\sqrt Q\right)\left(ξ(t,X_t)\sqrt Q\right)^*\right]\right]{\rm d}t+{\rm D}u^{(n)}(t,X_t)ξ(t,X_t){\rm d}W_t\tag 2$$ for all $n\in\mathbb N$ and hence $${\rm d}u(t,X_t)=\left[\frac{∂u}{∂t}(t,X_t)+\underbrace{{\rm D}u(t,X_t)u(t,X_t)}_{=:\:C_1(t,X_t)}+\frac12\underbrace{\sum_{n∈ℕ}{\rm D}^2u(t,X_t)\left(ξ(t,X_t)\sqrt Qe_n\right)\left(ξ(t,X_t)\sqrt Qe_n\right)}_{=:\:C_2(t,X_t)}\right]{\rm d}t+{\rm D}u(t,X_t)ξ(t,X_t){\rm d}W_t\;,\tag 3$$ where $$u^{(n)}:=\langle u,f_n\rangle_H\;\;\;\text{for }n\in\mathbb N$$ for some orthonormal basis $(f_n)_{n\in\mathbb N}$ of $H$.

Let $C:=C_1+\frac12C_2$. The question is: In analogy to my question about the Itō-Stratonovich correction term, can we show that $$\left\|C(t,x)-C(t,y)\right\|_H\le K\left\|x-y\right\|_H\;\;\;\text{for all }t\ge 0\text{ and }x,y\in H\tag 4$$ and $$\left\|C(t,x)\right\|_H\le K(1+\left\|x\right\|_H)\;\;\;\text{for all }t\ge 0\text{ and }x\in H\tag 5$$ for some $K>0$?

• @NawafBou-Rabee I'm not sure how exactly I need to reformulate it. Can you share your thoughts? Aug 27, 2016 at 10:28
• @NawafBou-Rabee $u_t(X_t)$ is the drift of $(1)$. You can imagine that it is the determinstic part of the velocity of the perturbed particle trajectory $t\mapsto X_t$ at time $t$. In the application of the Itō formula, $u$ still denotes the $u$ from $(1)$, cause I want to obtain an expression for the differential velocity ${\rm d}u_t(X_t)$. I hope that answers your question. "I suggest to simply incorporate the Ito correction term to the usual chain rule into the drift": Could you please explain what you mean in more detail? Aug 27, 2016 at 12:20
• @NawafBou-Rabee (a) Are you sure? The Itō formula yields$${\rm d}u^{(n)}(t,X_t)=\left[\frac{∂u^{(n)}}{∂t}(t,X_t)+{\rm D}u^{(n)}(t,X_t)u(t,X_t)+\frac12\text{tr}\left[{\rm D}^2u^{(n)}(t,X_t)\left(ξ(t,X_t)\sqrt Q\right)\left(ξ(t,X_t)\sqrt Q\right)^*\right]\right]{\rm d}t+{\rm D}u^{(n)}(t,X_t)ξ(t,X_t){\rm d}W_t$$and hence$${\rm d}u(t,X_t)=\underbrace{\left[\frac{∂u}{∂t}(t,X_t)+{\rm D}u(t,X_t)u(t,X_t)+\frac12\sum_{n∈ℕ}{\rm D}^2u(t,X_t)\left(ξ(t,X_t)\sqrt Qe_n\right)\left(ξ(t,X_t)\sqrt Qe_n\right)\right]}_{=:\:C(t,X_t)}{\rm d}t+{\rm D}u(t,X_t)ξ(t,X_t){\rm d}W_t\;.$$ Aug 27, 2016 at 19:25
• @NawafBou-Rabee So, the drift of the SDE for $u$ should be $C$, shouldn't it? (b) In my real application, I know another expression for ${\rm d}u(t,X_t)$ and consider the SDE obtained by equating these two expressions. My problem is that I don't see that the nonlinear part of $C$ satisfies the usual growth and Lipschitz conditions as, for example, presented in Da Prato's book (Google books link) in (7.25) and (7.26). Aug 27, 2016 at 19:26
• @NawafBou-Rabee I've completely updated the question. Aug 28, 2016 at 13:16

1 Answer

Yes, under the following conditions:

For any $t>0$, the function $u(t,\cdot)$ is twice-differentiable and the function $\xi(t,\cdot)$ is differentiable, in agreement with the context set by the OP. Moreover, these functions satisfy:

(A1) For all $x \in H$, there exists a real constant $C>0$ such that $$\| D^3 u(t,x) \|_H \vee\| D^2 u(t,x) \|_H \vee \| Du(t,x)\|_H \vee \| u(t,x) \|_H \le C \;.$$ (A2) For all $x \in H$, there exists a real constant $C>0$ such that $$\| D\xi(t,x) \|_{L_0^2} \le C \;.$$ where $L_0^2$ is standard notation for the set of linear operators from the Cameron-Martin space $U_0$ to $H$.

These conditions are a bit restrictive since they assume more derivatives than one actually needs to obtain (4) and (5) given by the OP. However, the conditions are transparent.

Why do these conditions suffice?

Recall that a sufficient condition for a differentiable function to be globally Lipschitz is that its derivative is uniformly bounded. By differentiating $C_1(t,\cdot)$, and invoking (A1), its clear that $C_1(t,\cdot)$ is globally Lipschitz. Similarly, by differentiating $C_2(t,\cdot)$, invoking (A1) and (A2), and using the fact that $\xi(t,x)$ is a Hilbert-Schmidt operator, its clear that $C_2(t,\cdot)$ is globally Lipschitz. The linear growth condition follows directly from (A1) and (A2).