Let
- $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
- $T>0$
- $I:=(0,T]$
- $(\mathcal F_t)_{t\in\overline I}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A)$
- $U$ be a separable $\mathbb R$-Hilbert space
- $M$ be a $U$-valued continuous square-integrable $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$
- $\mathfrak L_1(U)$ denote the space of nuclear operators on $U$ and $\mathfrak L_1^+(U)$ denote the subspace of nuclear, nonnegative and self-adjoint operators on $U$
- $[\![M]\!]$ denote the unique (up to indistinguishability) $\mathfrak L_1(H)$-valued continuous $\mathcal F$-adapted process of locally bounded variation such that $M\otimes M-[\![M]\!]$ is an $\mathcal F$-martingale
- $\operatorname{Cov}(M)$ denote the unique (up to indistinguishability) $\mathfrak L_1^+(H)$-valued $\mathcal F$-predictable process on $(\Omega,\mathcal A,\operatorname P)$ with $$[\![M]\!]_t=\int_{(0,\:t]}\operatorname{Cov}(M)\:{\rm d}[M]\;\;\;\text{for all }t\in\overline I\text{ almost surely}\tag1\;,$$ where $[M]$ denotes the quadratic variation of $M$
There is a real-valued $\mathcal F$-predictable process $\lambda^n$ and a $U$-valued $\mathcal F$-predictable process $e^n$ on $(\Omega,\mathcal A,\operatorname P)$ for $n\in\mathbb N$ such that
- $(e^n_t(\omega))_{n\in\mathbb N}$ is an orthonormal basis of $\overline{Q_t(\omega)U}$ with $$Q_t(\omega)e^n_t(\omega)=\lambda^n_t(\omega)e_t^n(\omega)\;\;\;\text{for all }n\in\mathbb N\cap[0,\operatorname{rank}Q_t(\omega)]\tag2$$ for all $(\omega,t)\in\Omega\times\overline I$
- $(\lambda^n_t(\omega))_{n\in\mathbb N}\subseteq[0,\infty)$ is nondecreasing with $$\lambda^n_t(\omega)>0\;\;\;\text{for all }n\in\mathbb N\cap[0,\operatorname{rank}Q_t(\omega)]\tag3$$ and $$\lim_{n\to\infty}\lambda_t^n(\omega)=0\tag4$$ for all $(\omega,t)\in\Omega\times\overline I$
Now, let $H$ be a separable $\mathbb R$-Hilbert space and $\mathcal E_M(H)$ denote the space of elementary $\mathcal F$-predictable processes (defined as usual) with values in $\mathfrak L(U,H)$ equipped with the seminorm $$\left\|\Phi\right\|_{\mathcal E_M(H)}^2:=\operatorname E\left[\int\left\|\Phi_sQ^{1/2}_s\right\|_{\operatorname{HS}(U,\:H)}^2\:{\rm d}[M]_s\right]\;\;\;\text{for }\Phi\in\mathcal E_M(H)\;.$$ Let $\mathcal I^2_M(H)$ denote the completion of $\mathcal E_M(H)$ with respect to that seminorm. Then we can characterize $\mathcal I^2_M(H)$ as follows: If $\mathcal H$ is any separable $\mathbb R$-Hilbert space, $(\mathcal h_n)_{n\in\mathbb N}$ is an orthonormal basis of $\mathcal H$ and $$A_t(\omega):=\sum_{n\in\mathbb N}\sqrt{\lambda^n_t(\omega)}e_t^n(\omega)\otimes\mathcal h_n\in\mathfrak L(Q_t^{1/2}(\omega)U,\mathcal H)\;\;\;\text{for }(\omega,t)\in\Omega\times\overline I\;,$$ then $$\mathcal I^2_M(H)=\left\{\Psi A:\Psi\in\mathcal L^2(\operatorname P\otimes{\rm d}[M];\operatorname{HS}(\mathcal H,H))\text{ is }\mathcal F\text{-predictable}\right\}\tag5$$ and $$\left\|\Psi A\right\|_{\mathcal I^2_M}=\left\|\Psi\right\|_{\mathcal L^2(\operatorname P\otimes{\rm d}[M];\:\operatorname{HS}(\mathcal H,\:H))}\;\;\;\text{for all }\Psi\in\mathcal L^2(\operatorname P\otimes{\rm d}[M];\operatorname{HS}(\mathcal H,H))\;.\tag6$$
Now, assume $\Psi\in\mathcal I^2_M(H)$ and $\Phi\in\mathcal I^2_{\Psi\:\cdot\:M}(\tilde H)$, where $\Psi\cdot M$ denotes the integral process and $\tilde H$ is another separable $\mathbb R$-Hilbert space. In order to prove the associativity known from the real-valued case, i.e. $\Phi\Psi\in\mathcal I^2_M(\tilde H)$ and $$\Phi\Psi\cdot M=\Phi\cdot(\Psi\cdot M)\;\;\;\text{almost surely}\tag7\;,$$ we first need to show that the composition $\Phi\Psi$ is even well-defined. We know that $\Psi=\tilde \Psi A$ for some $\tilde\Psi\in\mathcal L^2(\operatorname P\otimes{\rm d}[M];\operatorname{HS}(\mathcal H,H))$. So, in order to show the well-definedness, we need at least to show that $$\tilde\Psi_t(\omega)\mathcal h\in\operatorname{Cov}(\Psi\cdot M)^{1/2}H\;\;\;\text{for all }\mathcal h\in\mathcal H\text{ and }(\omega,t)\in\Omega\times\overline I\tag8\;.$$
How can we do that?
