Properties of the trace term in the Itō formula

Question

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 $K>0$?

Answer 1

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).