The definition of trace-class operators A trace-class operators are defined on a separable Hilbert space. However, is it possible to define trace or trace-class operators on an arbitrary Hilbert spaces? I am just curious...
 A: In Chapter 16 of "Introduction to Functional Analysis" by Meise & Vogt (1997), the trace class gets introduced on arbitrary infinite-dimensional Hilbert spaces ($\mathcal G,\mathcal H,\ldots)$. Basically, every compact operator $A:\mathcal G\to\mathcal H$ has a Schmidt representation
$$
Ax=\sum_{n=0}^\infty s_n(A)\langle e_n,x\rangle f_n
$$
with a non-negative, decreasing null sequence $(s_n(A))_{n\in\mathbb N_0}$ (singular values) and orthonormal systems $(e_n)_{n\in\mathbb N_0}$ in $\mathcal G$, $(f_n)_{n\in\mathbb N_0}$ in $\mathcal H$. Then the trace class is defined as
$$
\mathcal B_1(\mathcal G,\mathcal H):=\Big\lbrace A:\mathcal G\to\mathcal H\text{ compact}\,|\,\Vert A\Vert_1:=\sum_{n=0}^\infty s_n(A)<\infty\Big\rbrace.\tag1
$$
One then can define a trace on $\mathcal B_1(\mathcal H)$ as usual. Now


*

*the trace norm $\Vert\cdot\Vert_1$ actually is a norm. (Corollary 16.15 in Meise/Vogt)

*$\mathcal B_1(\mathcal G,\mathcal H)$ is a Banach space with respect to the trace norm. (Corollary 16.24)

*The dual space $\mathcal B_1(\mathcal G,\mathcal H)'$ is isometrically isomorphic to $\mathcal B(\mathcal H,\mathcal G)$ by means of the map $\Psi:B\mapsto\operatorname{tr}(B(\cdot))$. (Proposition 16.26)
One can also show that in the case of separable Hilbert spaces, (1) is equivalent to the usual definition
$$
\mathcal B_1(\mathcal H)=\big\lbrace A\in\mathcal B(\mathcal H)\,|\,\operatorname{tr}\sqrt{A^\dagger A}<\infty \big\rbrace
$$
see e.g. Theorem VI.21 in "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
