3
$\begingroup$

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 $\mathfrak L(A):=\mathfrak L(A,A)$.

$\endgroup$
1
  • $\begingroup$ 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, 2017 at 14:36

1 Answer 1

0
$\begingroup$

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 $(\omega,s)\in\Omega\times[0,T]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.