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 $K>0$?