How can we prove that $\langle\int_0^t\Phi_s{\rm d}W_s,x\rangle_H=\sum_{n\in\mathbb N}\int_0^t\langle\sqrt{λ_n}\Phi_se_n,x\rangle_H{\rm d}B_s^{(n)}$? Let$^1$


*

*$U$ and $H$ be separable $\mathbb R$-Hilbert spaces

*$Q\in\mathfrak L(U,H)$ be nonnegative and symmetric operator on $U$ with finite trace

*$(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space and $(\mathcal F_t)_{t\ge 0}$ be a filtration of $\mathcal A$

*$W$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with respect to $\mathcal F$, $$B^{(n)}:=\begin{cases}\frac{\langle W,e_n\rangle_U}{\sqrt{\lambda_n}}&\text{, if }n\in\mathbb N\text{ with }\lambda_n>0\\0&\text{, else}\end{cases}$$ and $$W^{(n)}:=\sum_{i=1}^n\sqrt{\lambda_i}B^{(i)}e_i\;\;\;\text{for }n\in\mathbb N$$



Note that the $B^{(n)}$ are independent $\mathcal F$-Brownian motions on $(\Omega,\mathcal A,\operatorname P)$, $$\left\|W_t^{(n)}-W_t\right\|_{L^2(\operatorname P,\;U)}\stackrel{n\to\infty}\to 0\;\;\;\text{for all }t\ge 0\tag 1$$ and $$\operatorname P\left[\left\|W^{(n)}-W\right\|_{C^0([0,\;t],\;U)}\stackrel{n\to\infty}\to 0\right]=1\;\;\;\text{for all }t\ge 0\tag 2$$

Now, let $(\Phi_t)_{t\ge 0}$ be a $\mathfrak L(U,H)$-valued $\mathcal F$-adapted and locally bounded$^2$ stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $$\Phi_t=\sum_{i=1}^n\xi_{i-1}1_{(t_{i-1},t_i]}\;\;\;\text{for all }t\ge 0\tag 3$$ for some $\mathfrak L(U,H)$-valued random variables $\xi_0,\ldots,\xi_{n-1}$ on $(\Omega,\mathcal A,\operatorname P)$ and some $0=t_0<\cdots<t_n$. Moreover, let $$\int_0^t\Phi_s{\rm d}W_s:=\sum_{i=1}^n\xi_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right)\;\;\;\text{for }t\ge 0\;.$$

Let $x\in H$. How can we prove that $$\langle\int_0^t\Phi_s{\rm d}W_s,x\rangle_H=\sum_{n\in\mathbb N}\int_0^t\langle\sqrt{\lambda_n}\Phi_se_n,x\rangle_H{\rm d}B_s^{(n)}\tag 4$$ in $L^2(\operatorname P)$, for all $t\ge 0$?

Let $$X^{(n)}_t:=\langle\sqrt{\lambda_n}\Phi_te_n,x\rangle_H\;\;\;\text{for }n\in\mathbb N\text{ and }t\ge 0\;.$$ By definition of the Itō integral with respect to a Brownian motion, $$\int_0^tX^{(n)}_s{\rm d}B_s=\sqrt{\lambda_n}\sum_{i=1}^n\langle\xi_{i-1}e_n,x\rangle_H\left(B^{(n)}_{t_i\wedge t}-B^{(n)}_{t_{i-1}\wedge t}\right)\;\;\;\text{for all }n\in\mathbb N\text{ and }t\ge 0\;.\tag 5$$
$\color{red}{\text{Now comes the crucial part!}}$ In the following equation chain, why do $(6)$ and $(7)$ hold?
\begin{equation}
\begin{split}
\langle\int_0^t\Phi_s{\rm d}W_s,x\rangle_H&=\sum_{i=1}^n\langle\xi_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right),x\rangle_H\\
&\stackrel{(6)}{\color{red}=}\sum_{i=1}^n\langle\xi_{i-1}\sum_{k\in\mathbb N}\sqrt{\lambda_k}\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)e_k,x\rangle_H\\
&=\sum_{i=1}^n\sum_{k\in\mathbb N}\sqrt{\lambda_k}\langle\xi_{i-1}e_k,x\rangle_H\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)\\
&\stackrel{(7)}{\color{red}=}\sum_{k\in\mathbb N}\sum_{i=1}^n\sqrt{\lambda_k}\langle\xi_{i-1}e_k,x\rangle_H\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)\\
&=\sum_{k\in\mathbb N}\int_0^tX^{(k)}_s{\rm d}B^{(k)}_s
\end{split}\tag 8
\end{equation}
for all $t\ge 0$. In particular, I don't really understand in which sense the equation chain $(8)$ holds. In the sense of $L^2(\operatorname P)$-convergence or in the sense of $\operatorname P$-almost sure convergence?

$^1$ Let $\mathfrak L(U,H)$ denote the space of bounded, linear operators from $U$ to $H$.
$^2$ i.e. $$\sup_\Omega\left\|\Phi_t\right\|_{\mathfrak L(U,\;H)}<\infty\;\;\;\text{for all }t\ge 0\;.$$
 A: This is not really an answer, (but too long as a comment) let us assume $\sup_{\omega\in\Omega}\|\xi_{i-1}(\omega)\|<\infty$ (otherwise we might have to use conditioning etc.).
We want to show
$$
\sum_{i=1}^n\langle\xi_{i-1}\sum_{k\in\mathbb N}\sqrt{\lambda_k}\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)e_k,x\rangle_H=\sum_{i=1}^n\sum_{k\in\mathbb N}\sqrt{\lambda_k}\langle\xi_{i-1}e_k,x\rangle_H\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right),\tag{*}
$$
where the convergence is in $L^2(\Omega;\mathbb{R})$.  
For $\xi(\omega)\in \mathfrak{L}(U,H)$ such that $\sup_{\omega\in\Omega}\|\xi(\omega)\|_{\mathfrak{L}(U,H)}<\infty$ we have 
\begin{align}
\|\sum_{k=1}^K\xi\sqrt{\lambda_k}B^{(k)}(t)e_k\|_{L^2(\Omega;{H})}^2
=\mathbb{E}\|\xi\sum_{k=1}^K\sqrt{\lambda_k}B^{(k)}(t)e_k\|_{H}^2
&\le 
\sup_{\omega\in\Omega}\|\xi(\omega)\|_{\mathfrak{L}(U,H)}^2
\mathbb{E}\|\sum_{k=1}^K\sqrt{\lambda_k}B^{(k)}(t)e_k\|_{U}^2\\
&=
\sup_{\omega\in\Omega}\|\xi(\omega)\|_{\mathfrak{L}(U,H)}^2
\Big\|\sum_{k=1}^K \sqrt{\lambda_k}B^{(k)}(t)e_k\Big\|_{L^2(\Omega;U)}^2,
\end{align}
from which we see 
$$
\sum_{k=1}^\infty \xi\sqrt{\lambda_k}B^{(k)}(t)e_k
=\xi\sum_{k=1}^\infty \sqrt{\lambda_k}B^{(k)}(t)e_k,
$$
and the series is $L^2(\Omega;H)$-convergent.
Further, noting $F_x:=\langle \cdot,x\rangle_H\colon H\to \mathbb{R}$ is bounded linear, similarly to the above, for a $L^2(\Omega;H)$ convergent sequence $\{\sum_{k}^{n}a_k\}_n \subset L^2(\Omega,H)$, we see $\{\sum_{k}^{n} F_x(a_k)\}\subset L^2(\Omega;\mathbb{R})$ congerves in $L^2(\Omega;\mathbb{R})$ and $$
\sum_{k}^{\infty} F_x(a_k)=F_x\Big(\sum_{k}^{\infty} a_k\Big)
\ \text{ in }\ L^2(\Omega;\mathbb{R}).
$$
Going back to (*), we have for each $i$
\begin{align}
F_x\Big(\sum_{k\in\mathbb N}\sqrt{\lambda_k}\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)\xi_{i-1}e_k\Big)
%&=
%\mathbb{E}\left[\bigg|
%F_x\Big(\sum_{k\in\mathbb N}\sqrt{\lambda_k}\left(B^{(k)}_{t_i\wedge t}-%B^{(k)}_{t_{i-1}\wedge t}\right)\xi_{i-1}e_k\Big)\bigg|^2\right]\\
&=
\sum_{k\in\mathbb N}
\sqrt{\lambda_k}\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)F_x(\xi_{i-1}e_k)\ \text{ in }\  L^2(\Omega;\mathbb{R}),
\end{align}
which is essentially (8) (swapping sum should be easier than acting by $\xi$).
