• $U$, $H$, $\tilde H$ 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?


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 $Q^{1/2}$.

  • $\begingroup$ Why is it important that it is infinite dimensional? Is the proof much easier in the finite dimensional case? $\endgroup$
    – username
    Feb 28, 2018 at 9:18
  • $\begingroup$ @username I don't know if it's easier to prove then, but I need the result for infinite-dimensional spaces. $\endgroup$
    – 0xbadf00d
    Feb 28, 2018 at 19:23

1 Answer 1


We'll need the following fact: Let $U_i,H$ be $\mathbb R$-Hilbert spaces and $A_i\in\mathfrak L(U_i,H)$ with $$A_1A_1^\ast=A_2A_2^\ast.\tag4$$ Then, $$\iota_{12}:=A_2^{-1}A_1\operatorname P_{(\ker A_1)^\perp}$$ and $$\iota_{21}:=A_1^{-1}A_2\operatorname P_{(\ker A_2)^\perp}$$ are well-defined partial isometries from $U_1$ to $U_2$ and $U_2$ to $U_1$, respectively, with $$

  1. $A_1=A_2\iota_{12}$
  2. $A_2=A_1\iota_{21}$
  3. $A_1^{-1}=\iota_{21}A_2^{-1}$
  4. $A_2^{-1}=\iota_{12}A_1^{-1}$
  5. $\iota_{12}^\ast=\iota_{21}$
  6. $\iota_{21}^\ast=\iota_{12}$
  7. $\iota_{21}\iota_{21}^\ast$ is the orthogonal projection from $U_1$ onto $(\ker A_1)^\perp$
  8. $\iota_{12}\iota_{12}^\ast$ is the orthogonal projection from $U_2$ onto $(\ker A_2)^\perp$

In particular, if $U$ is a $\mathbb R$-Hilbert space and $A\in\mathfrak L(U,H)$, then $$\iota:=(AA^\ast)^{-1/2}A\operatorname P_{(\ker A)^\perp}$$ is a well-defined isometry from $U$ to $H$ with

  1. $A=(AA^\ast)^{1/2}\iota$
  2. $(AA^\ast)^{1/2}=A\iota^\ast$
  3. $\iota^\ast=A^{-1}(AA^\ast)^{1/2}\operatorname P_{\overline{AU}}$

In the situation of the question, we choose $A=\Psi Q^{1/2}$ and obtain $$\left\|\Phi\Psi Q^{1/2}\right\|_{\operatorname{HS}(U,\:\tilde H)}\le\left\|\Phi\tilde Q^{1/2}\right\|_{\operatorname{HS}(U,\:H)}\left\|\iota\right\|_{\mathfrak L(U,\:H)}\tag5$$ and clearly $\left\|\iota\right\|_{\mathfrak L(U,\:H)}\le1$.


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