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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 $$\sum_{n\in\mathbb N}\lambda_n=\operatorname{tr}Q<\infty\tag6\;.$$


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

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  • $\begingroup$ Please note that I know that this is not a "research level" question. However, I've asked several stochastic analysis questions on MSE and almost never got an answer. $\endgroup$ – 0xbadf00d May 2 '17 at 14:36
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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 $(\omega,s)\in\Omega\times[0,T]$.

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