Let$^1$
- $U$ be a separable $\mathbb R$-Hilbert space
- $d\in\mathbb N$
- $\lambda$ denote the Lebesgue measure on $\mathbb R^d$
- $\Omega\subseteq\mathbb R^d$ be a bounded domain
- $H:=L^2(\Omega,\mathbb R^d)$
- $Q:\mathbb R^d\to\operatorname{HS}(U,\mathbb R^d)$
I want to define some $\tilde Q\in\operatorname{HS}(U,H)$ with $$Q(x)u=(\tilde Qu)(x)\;\;\;\text{for all }x\in\Omega\text{ and > }u\in U\;.\tag 1$$
Is that possible? And if it is, how? Maybe something like a Nemytskii operator can do the job.
However, if we use $(1)$ as the defining equation for $\tilde Q$, we obtain
\begin{equation} \begin{split} \left\|\tilde Q\right\|_{\operatorname{HS}(U,\;H)}^2&\stackrel{\text{def}}=\sum_{n\in\mathbb N}\left\|\tilde Qe_n\right\|_H^2\\ &\stackrel{\text{def}}=\sum_{n\in\mathbb N}\int_\Omega\left\|Q(\omega)e_n\right\|_2^2\;{\rm d}\lambda(\omega)\\ &\color{red}{=}\int_\Omega\sum_{n\in\mathbb N}\left\|Q(\omega)e_n\right\|_2^2\;{\rm d}\lambda(\omega)\\ &\stackrel{\text{def}}=\int_\Omega\left\|Q(\omega)\right\|_{\operatorname{HS}(U,\;\mathbb R^d)}^2\;{\rm d}\lambda(\omega) \end{split}\;,\tag 2 \end{equation}
where $(e_n)_{n\in\mathbb N}$ is an arbitrary orthonormal basis of $U$. So, we need to impose the assumption that $$\int_\Omega\left\|Q\right\|_{\operatorname{HS}(U,\;\mathbb R^d)}^2\;{\rm d}\lambda<\infty\;.\tag 3$$
I'm unsure whether or not the $\color{red}{\text{red}}$ equality sign is justified. And for some reason I'm unsure whether or not $\tilde Q$ is even well-defined by $(1)$.
$^1$ Let $\operatorname{HS}(A,B)$ denote the space of Hilbert-Schmidt operators from $A$ to $B$.