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 operator on $H$ 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 $$\left\|C(x)\right\|_{H_1}^2+\operatorname{tr}B(x)Q{B(x)}^\ast\le K(1+\left\|x\right\|_{H_1}^2)\;\;\;\text{or all }x\in H_1\tag 5$$ 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.