Let$^1$
- $U_0$ be a separable $\mathbb R$-Hilbert space
- $d\in\mathbb N$
- $G:\mathbb R^d\to\operatorname{HS}(U_0,\mathbb R^d)$
- $\Lambda\subseteq\mathbb R^d$ be bounded and open
- $V:=H_0^1(\Lambda,\mathbb R^d)$
- $H:=L^2(\Lambda,\mathbb R^d)$
Can we show that $$\left(\tilde G(v)u_0\right)(x):=G\left(v\left(x\right)\right)u_0\;\;\;\text{for }v\in V\text{, }u_0\in U\text{ and }x\in\Lambda\tag 1$$ is a well-defined mapping $V\to\operatorname{HS}(U_0,H)$?
I guess the answer is no in general. If $(e_n)_{n\in\mathbb N}$ is an orthonormal basis of $U_0$, then we know that $$\left\|G(x)\right\|_{\operatorname{HS}(U_0,\:\mathbb R^d)}^2=\sum_{n\in\mathbb N}\left|G(x)e_n\right|^2<\infty\;\;\;\text{for all }x\in\Lambda\tag 2\;.$$ But that's not enough to conclude that$^2$ $$\left\|\tilde G(v)\right\|_{\operatorname{HS}(U_0,\:H)}^2=\sum_{n\in\mathbb N}\int_\Lambda\left|G\left(v\left(x\right)\right)e_n\right|^2\:{\rm d}\lambda(x)=\int_\Lambda\left\|G\left(v\left(x\right)\right)\right\|_{\operatorname{HS}(U_0,\:\mathbb R^d)}^2\:{\rm d}\lambda(x)\tag 3$$ is finite, for any $v\in V$.
So, if we can't show the desired statement in general, is a proof possible under suitable further assumptions? In either case, how is the situation in the case where $\tilde G$ is given and we want to prove that $G$ defined in the sense of $(1)$ is a well-defined mapping $\mathbb R^d\to\operatorname{HS}(U_0,\mathbb R^d)$?
Since each $v\in V$ has a continuous representative, it should be sufficient to assume that $G$ is continuous. Since $\Lambda$ is bounded, we obtain that $\Lambda\ni x\mapsto G(v(x))$ is bounded and that $\lambda(\Lambda)<\infty$. Thus, $(3)$ should be finite.
$^1$ Let $\operatorname{HS}(A,B)$ denote the space of Hilbert-Schmidt operators from $A$ to $B$.
$^2$ Let $\lambda$ denote the Lebesgue measure on $\mathbb R^d$.