# Correction term in the relation between the Itō and Stratonovich integrals in Hilbert spaces

I'm reading the paper On the relation between the Itō and Stratonovich integrals in Hilbert spaces and there is something I don't understand.

In the notation of the paper, let

• $$H,H_1$$ be separable $$\mathbb R$$-Hilbert spaces
• $$Q\in\mathfrak L(H)$$ be nonnegative and self-adjoint with finite trace
• $$(e_n)_{n\in\mathbb N}$$ be an orthonormal basis of $$H$$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N$$ for some $$(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$$
• $$B:H_1\to\mathfrak L(H,H_1)$$

They show (and that is easy to see) that for any $$y\in H_1$$ and $$L\in\mathfrak L(H,\mathfrak L(H,H_1))$$, there is a unique $$\tilde L(y)\in\mathfrak L(U)$$ with $$\langle y,(Lu)v\rangle_H=\langle v,\tilde L(y)u\rangle_U\;\;\;\text{for all }u,v\in H\tag 1$$ and a unique $$\tilde{\operatorname{tr}}\:QL\in H_1$$ with $$\operatorname{tr}Q\tilde L(y)=\langle y,\tilde{\operatorname{tr}}\:QL\rangle_H\;\;\;\text{for all }y\tag 2$$ given by $$\tilde{\operatorname{tr}}\:QL=\sum_{n\in\mathbb N}(Le_n)(\lambda_ne_n)\;.\tag 3$$

In particular, $$\tilde{\operatorname{tr}}\:Q{\rm D}B(x)B(x)=\sum_{n\in\mathbb N}({\rm D}B(x)B(x)e_n)(\lambda_ne_n)\;\;\;\text{for all }x\in H_1\;.\tag 4$$

In the paper, they consider a SDE $${\rm d}z_t=[Az_t+C(z_t)]{\rm d}t+B(z_t){\rm d}W_t$$ where only the operator $$C$$ on $$H_1$$ is of (marginal) interest in the following. They state the following assumptions:

(A3): $$\exists K>0$$ with $$\left\|C(x)\right\|_{H_1}^2+\operatorname{tr}B(x)Q{B(x)}^\ast\le K(1+\left\|x\right\|_{H_1}^2)\;\;\;\text{for all }x\in H_1\tag 5$$ and $$\left\|C(x)-C(y)\right\|_{H_1}^2+\operatorname{tr}(B(x)-B(y))Q(B(x)-B(y))^\ast\le K\left\|x-y\right\|_{H_1}^2\tag 6\;.$$

(A4): $$B$$ is continuously Fréchet differentiable with bounded, Lipschitz-continuous derivative.

Question: Now they state that (A4) and the convergence of the series in $$(4)$$ would imply (A3). Why? I absolutely don't understand this.

This latter conclusion sounds fishy at first glance, because the Ito-Stratonovich correction term does not appear to be globally Lipschitz (i.e., satisfies (A3)). To see what the issue is, suppose $B$ was in $C^2$, then the Frechet derivative of the correction term involves $B$ itself, which is not necessarily bounded by their assumptions. In much simpler terms, if a $C^2$ function $f$ has bounded first derivative then $f'(x) f(x)$ (the correction term) does not always have bounded first derivative; take for instance $f(x) = x + \cos(x)$. However, I think their point is that even though the term $D^2B(x) B(x)$ is not necessarily bounded, its convolution with $Q$ as defined by that series is bounded.
• No. They do not mean (5'') and (6''). They mean that the corrected nonlinear part of the drift still satisfies (A3). To be sure, if we label the correction term as $C_1(x)$, they mean: $\| C_1(x) \|_{H_1}^2 \le K (1+ \|x\|_{H_1}^2)$ (linear growth condition) and $\| C_1(x) - C_1(y) \|_{H_1}^2 \le K \| x - y \|_{H_1}^2$ (globally Lipschitz). With the caveat given in my answer, the proof of this result seems math.stackexchange level. – Nawaf Bou-Rabee Aug 25 '16 at 13:09
• Indeed, since the convolution defining the correction term is bounded and $C^1$, it is straightforward to prove that it is globally Lipschitz. Moreover, math.stackexchange.com/questions/1227314/… – Nawaf Bou-Rabee Aug 25 '16 at 13:18
• I'm not sure what you mean with "corrected nonlinear part of the drift", but I guess $$C_1(x)=C(x)+\frac12\tilde{\operatorname{tr}}\:Q{\rm D}B(x)B(x)\;,$$ right? – 0xbadf00d Aug 26 '16 at 19:26
• They really mean that $w$ is a $Q$-Wiener process: please note that the left-hand-side of (2.1) involves the eigenvalues of $Q$. I think (2.4) is right, since the role of this term is to eliminate the Ito correction to the standard chain rule -- that is the essence of the Stratonovich description. – Nawaf Bou-Rabee Aug 27 '16 at 13:39