Let $H$ be a (separable) Hilbert space and $B(H)$ be the space of bounded linear operators defined on $H$. Suppose $\{ e_i \}$ be an orthonormal basis of $H$, we say $T \in B(H)$ is a trace-class operator if
$$ \|T\|_1 := \sum_i \langle |T| e_i, e_i \rangle < \infty, $$
where $|T|$ is the positive operator such that $|T|^2 = T^*T$ and $T^*$ is the adjoint operator of $T$.
If $T \in B(H)$ is a trace-class operator, we can define its trace as
$$ \newcommand{\Tr}{\mathrm{Tr}} \Tr(T) := \sum_i \langle Te_i, e_i \rangle. $$
From Wikipedia, or Conway's book A course in functional analysis, we know that trace-class operators have some nice properties:
- If $T \in B(H)$ is a trace-class operator, then for any $A \in B(H)$, we know both $AT$, $TA$ are trace-class, and $\Tr(AT) = \Tr(TA)$, and $\|AT\|_1 \leq \|A\| \|T\|_1$.
However, I'm curious about the definition of trace-class operators between two different Hilbert spaces $H_1$ and $H_2$. Let $B(H_1, H_2)$ be the space of bounded linear operators from $H_1$ to $H_2$.
It seems the above definition can be generalized to this case as follows: We say an operator $T \in B(H_1, H_2)$ is trace-class if $\sqrt{T^*T} : H_1 \to H_1$ is trace-class and define its trace-norm as $\|T\|_1 := \| \sqrt{T^*T} \|_1$. This generalization looks reasonable to me, but I am not sure whether there exist some subtleties.
I am trying to prove the aforementioned properties of trace-class operators under such definition. Suppose $H_3$ is another Hilbert space, and $A \in B(H_2, H_3)$. To show that $AT$ is trace-class, by definition it suffices to show that $\sqrt{T^* A^* AT}$ is trace-class. But I don't know how to proceed.
I guess that the definition of trace-class operators can be generalized and the aforementioned properties also hold. I would appreciate it if you could provide some references about these.
(BTW: It seems that the Hilbert-Schmidt operator can be easily generalized to different Hilbert spaces.)
Update: Thanks for the answer of Ruy. There is another question about these trace-class operators: When $A \in B(H_2, H_1)$, can we conclude that $\Tr(AT) = \Tr(TA)$?
Best Answer
Notice that $$ |AT|^2 = T^*A^*AT \leq \|A\|^2T^*T = \|A\|^2|T|^2, $$ so, by Proposition 1.3.8 in [1] (square-root is an operator monotone function), one has that $$ |AT| \leq \|A\||T|, $$ and it follows that $AT$ is trace class according to the OP's definition.
[1] Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London - New York -San Francisco: Academic Press. X, 416 p. $ 60.00 (1979). ZBL0416.46043.