Let
- $U$ be a Hilbert space
- $Q\in\mathfrak L(U)$ be nonnegative and symmetric
- $U_0:=Q^{1/2}U$ be equipped with $$\langle u,v\rangle_{U_0}:=\langle Q^{-1/2}u,Q^{-1/2}v\rangle_U\;\;\;\text{for }u,v\in U_0$$
- $V$ be a Hilbert space and $\iota\in\operatorname{HS}(U_0,V)$ be an embedding
- $C:=\iota\iota^\ast$
Let $$V_0:=C^{1/2}V\tag 1$$ be equipped with $$\langle u,v\rangle_{V_0}:=\langle C^{-1/2}u,C^{-1/2}v\rangle_V\;\;\;\text{for }u,v\in V_0\;.$$ We can show (see Röckner, Proposition 2.5.2) that $\iota$ is an isometry between $U_0$ and $$V_0=\iota U_0\;.\tag 2$$
However, I think I've found a special case which let me doubt on $(1)$ and $(2)$.
Choose $Q=\text{id}_U$ and $V$ such that $U_0\subseteq V$ with the inclusion $\iota$ being Hilbert-Schmidt. Then, $$V_0\stackrel{(2)}=U_0=U\tag 3$$ and $$C=\text{id}_V\tag 4$$ (compare with my other question).
The problem is that $(4)$ yields $C^{1/2}=\text{id}_V$ and hence $$V_0=V\tag 5$$ by $(1)$. In particular, $$V=V_0=U_0=U\;.\tag 6$$
So, either I made some mistake or the only way to choose a superspace $V$ of $U_0$ such that the inclusion $\iota$ is Hilbert-Schmidt is the trivial choice $U_0=V$.
$\mathfrak L(A,B)$ and $\operatorname{HS}(A,B)$ denote the space of bounded, linear operators and Hilbert-Schmidt operators from $A$ to $B$, respectively.