Let$^1$


* $T>0$
* $U,H$ be separable $\mathbb R$-Hilbert spaces
* $Q\in\mathfrak L(U)$ be nonnegative and self-adjoint operator with finite trace $\operatorname{tr}Q$
* $(e^n)\_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe^n=\lambda\_ne^n\;\;\;\text{for all }n\in\mathbb N\tag1$$ for some $(\lambda\_n)\_{n\in\mathbb N}\subseteq(0,\infty)$ and $$e\_0^n:=Q^{1/2}e^n=\sqrt{\lambda\_n}e^n\;\;\;\text{for }n\in\mathbb N$$
* $U\_0:=Q^{1/2}U$ be equipped with $$\langle u\_0,v\_0\rangle\_{U\_0}:=\langle Q^{-1/2}u\_0,Q^{-1/2}v\_0\rangle\_U\;\;\;\text{for }u\_0,v\_0\in U\_0$$
* $(\Omega,\mathcal A,\operatorname P)$ be a probability space
* $\Phi:\Omega\times[0,T]\to\operatorname{HS}(U\_0,H)$ be predictable with $$\int\_0^T\operatorname E\left[\left\|\Phi\_t\right\|\_{\operatorname{HS}(U\_0,\:H)}^2\right]{\rm d}t<\infty\tag2$$



> 
> Let $u\in U$ and $h\in H$. Then, $$\langle\Phi\_s(\omega)Qu,h\rangle\_H=\sum\_{n\in\mathbb N}\lambda\_n\langle u,e^n\rangle\_U\langle\Phi\_s(\omega)e^n,h\rangle\_H\;\;\;\text{for all }(\omega,s)\in\Omega\times[0,T]\tag3\;.$$ I want to conclude that $$\operatorname E\left[\int\_0^t\langle\Phi\_sQu,h\rangle\_H\:{\rm d}s\mid\mathcal F\_r\right]=\operatorname E\left[\sum\_{n\in\mathbb N}\lambda\_n\langle u,e^n\rangle\_U\int\_0^t\langle\Phi\_se^n,h\rangle\_H\:{\rm d}s\mid\mathcal F\_r\right]\tag4$$ for all $r,t\in[0,T]$ with $r\le t$.
> 
> 
> 


My problem is that I'm not even able to show that, for fixed $s\in[0,T]$, $$S\_N:=\sum\_{n=1}^N\lambda\_n\langle u,e^n\rangle\_U\langle\Phi\_se^n,h\rangle\_H\;\;\;\text{for }N\in\mathbb N$$ converges for $N\to\infty$ to $\langle\Phi\_sQu,h\rangle\_H$ in $L^1(\operatorname P)$. This would follow from $(3)$ by *Lebesgue's dominated convergence theorem*, but the best bound I obtain is $$\left|S\_N(\omega)\right|\le\left\|u\right\|\_U\left\|h\right\|\_H\left\|\Phi\_s(\omega)\right\|\_{\operatorname{HS}(U\_0,\:H)}\sum\_{n=1}^N\sqrt{\lambda\_n}\;\;\;\text{for all }\omega\in\Omega\text{ and }N\in\mathbb N\;.\tag5$$ This is not sufficient, since I don't see how we should bound $\sum\_{n=1}^N\sqrt{\lambda\_n}$. Note that $$\sum\_{n\in\mathbb N}\lambda\_n=\operatorname{tr}Q<\infty\tag6\;.$$ 




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$^1$ Let $\mathfrak L(A,B)$ and $\operatorname{HS}(A,B)$ denote the space of bounded linear Operators and Hilbert-Schmidt operators, respectively. Moreover, let $\mathfrak L(A):=\mathfrak L(A,A)$.



Covariation of the stochastic integral and the Wiener process
Since $$\lambda\_n\left\|\Phi\_s(\omega)e^n\right\|\_H\le\begin{cases}\lambda\_n&\text{, if }\left\|\Phi\_s(\omega)e^n\right\|\_H\le1\\\left\|\Phi\_s(\omega)e\_0^n\right\|\_H^2&\text{, if }\left\|\Phi\_s(\omega)e^n\right\|\_H\ge1\end{cases}\tag7$$ for all $n\in\mathbb N$, $\sum\_{n\in\mathbb N}\lambda\_n=\operatorname{tr}Q$ and $\sum\_{n\in\mathbb N}\left\|\Phi\_s(\omega)e\_0^n\right\|\_H^2=\left\|\Phi\_s(\omega)\right\|\_{\operatorname{HS}(U\_0,\:H)}^2$, we obtain $$\sum\_{n\in\mathbb N}\lambda\_n\left\|\Phi\_s(\omega)e^n\right\|\_H<\infty\tag8$$ by the comparison test for $(\operatorname P\otimes\lambda^1)$-almost all $(\omega,s)\in\Omega\times[0,T]$.


