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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.

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