Given a linear Hilbert-Schmidt embedding $ι$ between Hilbert spaces, prove that $ιι^*$ is a bounded, linear operator with finite trace Let


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*$(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a separable Hilbert space

*$Q$ be a bounded, linear, nonnegative and symmetric operator on $U$

*$U_0:=Q^{\frac 12}(U)$, $$\langle u,v\rangle_0:=\langle Q^{-\frac 12}u,Q^{-\frac 12}v\rangle\;\;\;\text{for }u,v\in U_0$$ where $Q^{-\frac 12}$ is the pseudo inverse of $Q^{\frac 12}$ and $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U_0$

*$(U_1,\langle\;\cdot\;,\;\cdot\;\rangle_1)$ be a separable Hilbert space and $$\iota:(U_0,\langle\;\cdot\;,\;\cdot\;\rangle_0)\to(U,\langle\;\cdot\;,\;\cdot\;\rangle)$$ be a linear Hilbert-Schmidt embedding



How can we show that $$Q_1:=\iota\iota^\ast$$ is a bounded, linear, symmetric and nonnegative operator on $U_1$ with finite trace?

Clearly, $$\langle Q_1u,v\rangle_1\stackrel{\text{def}}=\langle\iota\color{blue}(\iota^\ast u\color{blue}),v\rangle_1\stackrel{\text{def}}=\langle\iota^\ast u,\iota^\ast v\rangle_0\;\;\;\text{for all }u,v\in U_1\tag 1\;.$$ Thus, $\iota$ is nonnegative (since $\langle\iota^\ast u,\iota^\ast u\rangle_0\ge 0$ for all $u\in U_1$) and symmetric, since $$\langle u,Q_1v\rangle_1\stackrel{\text{Hermitian symmetry}}=\overline{\langle Q_1v,u\rangle_1}\stackrel{\text{(1)}}=\overline{\langle\iota^\ast v,\iota^\ast u\rangle_0}\stackrel{\text{Hermitian symmetry}+(1)}=\langle Q_1u,v\rangle_1\;.$$

However, I fail to prove that $Q_1$ has finite trace, i.e. $$\operatorname{tr}Q_1:=\sum_{n\in\mathbb N}\langle Q_1e_n,e_n\rangle_1<\infty\tag 2$$ for any orthonormal basis $(e_n)_{n\in\mathbb N}$ of $U_1$. How can we show $(2)$ and that $\iota$ is bounded and linear?

 A: You have probably figured it out already, but in case someone else is faced with the same problem, here is an answer to your question:
Since the embedding operator $\iota$ is linear and Hilbert-Schmidt (in particular, a bounded linear operator), the same holds for its adjoint $\iota^{*}$. Thus, $Q_{1} = \iota \iota^{*}$ is bounded and linear as the composition of two such operators.
That it is also trace-class (an important property for defining cylindrical Wiener processes) follows from the fact that the operator $\iota$ is assumed to be Hilbert-Schmidt.
As you seem to be working with the book of Röckner and Prevot (or Röckner/Liu), Proposition B.0.8 in there ensures that $Q_{1}$, as the composition of two Hilbert-Schmidt operators is a nuclear operator, i.e. in $L_{1}(U_{1})$. Since every nuclear operator is trace-class (Remark B.0.4 in the book), you get that $\mathrm{tr}~ Q_{1} < \infty$.
If you prefer a concrete calculation over abstract results, you can write
\begin{align*}
\mathrm{tr}~ Q_{1} &= \sum_{n \in \mathbb{N}} \langle Q_{1} e_{n}, e_{n} \rangle_{1} =  \sum_{n \in \mathbb{N}} \langle \iota \iota^{*} e_{n}, e_{n} \rangle_{1} =  \sum_{n \in \mathbb{N}} \langle \iota^{*} e_{n}, \iota^{*} e_{n} \rangle_{0} \\
&=  \sum_{n \in \mathbb{N}} \| \iota^{*} e_{n} \|_{0}^{2} = \| \iota^{*} \|_{L_{2}(U_{1},U_{0})}^{2} = \| \iota \|_{L_{2}(U_{0},U_{1})}^{2} < \infty
\end{align*}
where in the penultimate step we used that the adjoint of a Hilbert-Schmidt operator has the same norm as the operator itself (Remark B.0.6 (i)), and that $\iota$ is a Hilbert-Schmidt embedding.
Hope that helps,
Andre
