Let
- $U,H,\mathcal H$ be separable $\mathbb R$-Hilbert spaces
- $Q\in\mathfrak L_1(U)$$^1$ be nonnegative and self-adjoint
- $\Psi\in\operatorname{HS}(Q^{1/2}U,H)$$^2$$^3$
- $\tilde Q:=\Psi Q\Psi^\ast$
I want to show that $$\Psi Q^{1/2}U\subseteq\tilde Q^{1/2}H\;.\tag 1$$
Maybe there is a useful identity for the square-root $\tilde Q^{1/2}$ of $\tilde Q$. I wasn't able to find one. So, how can we show $(1)$?
$^1$ $\mathfrak L_1(A,B)$ denotes the space of nuclear operators from $A$ to $B$.
$^2$ $\operatorname{HS}(A,B)$ denotes the space of Hilbert-Schmidt operators from $A$ to $B$.
$^3$ If $R\in\mathfrak L_1(A)$ is nonnegative and self-adjoint, then $R^{1/2}A$ is considered as being equipped with $$\langle x,y\rangle_{R^{1/2}A}:=\langle R^{-1/2}x,R^{-1/2}y\rangle_A\;\;\;\text{for }x,y\in R^{1/2}A\;,$$ where $R^{-1}$ denotes the pseudo inverse of $R$.