Show that the product of two Hilbert-Schmidt operators is Hilbert-Schmidt

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Let

  • $U,H,\tilde H$ be infinite-dimensional separable $\mathbb R$-Hilbert spaces
  • $Q$ be a nonnegative, nuclear and self-adjoint operator on $U$
  • $\Psi$ be a Hilbert-Schmidt operator from $Q^{1/2}U$ to $H$, where $Q^{1/2}U$ is equipped with $$\langle u,v\rangle_{Q^{1/2}U}:=\langle Q^{-1/2}u,Q^{-1/2}v\rangle_U\;\;\;\text{for }u,v\in Q^{1/2}U$$ and $Q^{-1/2}$ denotes the pseudo inverse of $Q^{1/2}$ (i.e. $Q^{-1/2}:=\left(\left.Q^{1/2}\right|_{(\ker Q^{1/2})^\perp}\right)^{-1}$)
  • $\tilde Q:=\left(\Psi Q^{1/2}\right)\left(\Psi Q^{1/2}\right)^\ast$
  • $\Phi$ be a Hilbert-Schmidt operator from $\tilde Q^{1/2}H$ to $\tilde H$, where $\tilde Q^{1/2}H$ is equipped with $$\langle x,y\rangle_{\tilde Q^{1/2}H}:=\langle\tilde Q^{-1/2}x,\tilde Q^{-1/2}y\rangle_H\;\;\;\text{for }x,y\in\tilde Q^{1/2}H$$

Note that $\tilde Q$ is a nonnegative, nuclear and self-adjoint operator on $H$.

I want to show that $$\left\|\Phi\Psi\right\|_{\operatorname{HS}\left(Q^{1/2}U,\:H\right)}^2\le\left\|\Phi\right\|_{\operatorname{HS}\left(\tilde Q^{1/2}H,\:\tilde H\right)}^2\operatorname{tr}\tilde Q\;,\tag1$$ where $\left\|\;\cdot\;\right\|_{\operatorname{HS}(A,B)}$ denotes the Hilbert-Schmidt norm and $\operatorname{tr}$ denotes the trace functional.

I've tried the following: Let $(e_i)_{i\in I}$ ($I\subseteq\mathbb N$) be an orthonormal basis of $(\ker Q)^\perp$. Then, $\left(Q^{1/2}e_i\right)_{i\in I}$ is an orthonormal basis of $Q^{1/2}U$. Supplement $(e_i)_{i\in I}$ to an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $U$ by an orthonormal basis of $\ker Q$.

I don't know if it helps or not, but there are $\tilde\Psi\in\operatorname{HS}(U,H)$ and $\tilde\Phi\in\operatorname{HS}(H,\tilde H)$ with $\Psi=\tilde\Psi Q^{-1/2}$ and $\Phi=\tilde\Phi\tilde Q^{-1/2}$.

We obtain (since $Q^{-1/2}Q^{1/2}=\operatorname P_{(\ker Q)^\perp}$ is the orthogonal projection from $U$ to $(\ker Q)^\perp=(\ker Q^{1/2})^\perp$) $$\left\|\Phi\Psi\right\|_{\operatorname{HS}\left(Q^{1/2}U,\:H\right)}^2=\sum_{i\in I}\left\|\Phi\tilde\Psi e_i\right\|_{\tilde H}^2\tag2\;,$$ but I don't know how we can proceed from here.

So, how can we show $(1)$?