Let

- $U$, $H$, $\tilde H$ be infinite-dimensional separable $\mathbb R$-Hilbert spaces
- $Q$ be a self-adjoint and nonnegative nuclear linear operator on $U$
- $\Psi$ be a Hilbert-Schmidt operator from$^1$ $Q^{1/2}U$ to $H$
- $\tilde Q:=\left(\Psi Q^{1/2}\right)\left(\Psi Q^{1/2}\right)^\ast$
- $\Phi$ be a Hilbert-Schmidt operator from $\tilde Q^{1/2}H$ to $\tilde H$

Note that $\tilde Q$ is a self-adjoint and nonnegative nuclear linear operator on $H$ and $$\Psi Q^{1/2}U=\tilde Q^{1/2}H.\tag1$$

By $(1)$, the composition $\Phi\Psi$ is well-defined. How can we show that $$\left\|\Phi\Psi\right\|_{\operatorname{HS}\left(Q^{1/2}U,\:\tilde H\right)}^2\le\left\|\Phi\right\|_{\operatorname{HS}\left(\tilde Q^{1/2}H,\:\tilde H\right)}^2\operatorname{tr}\tilde Q\tag2,$$ where $\left\|\;\cdot\;\right\|_{\operatorname{HS}}$ denotes the Hilbert-Schmidt norm and $\operatorname{tr}$ the trace functional?

**EDIT**:

Maybe it's useful to note that $$\operatorname{tr}\tilde Q=\left\|\Psi\right\|_{\operatorname{HS}(Q^{1/2}U,\:H)}^2=\left\|\Psi Q^{1/2}\right\|_{\operatorname{HS}(U,\:H)}^2\tag3.$$

$^1$ As usual, $Q^{1/2}U$ is equipped with $$\langle u,v\rangle_{Q^{1/2}U}:=\langle Q^{-1/2}u,Q^{-1/2}v\rangle_U\;\;\;\text{for }u,v\in Q^{1/2}U,$$ where $Q^{-1/2}$ denotes the pseudo-inverse of $Q^{1/2}$.