Trace Class: Decomposition

Question

This is only Q&A.

Preview

Trace class operators decompose.
So proofs reduce to Hilbert-Schmidt!

Problem

Given a Hilbert Space $\mathcal{H}$.

For the trace class: $$\mathcal{B}_\textrm{Tr}(\mathcal{H})=\mathcal{B}_\textrm{HS}(\mathcal{H})^*\mathcal{B}_\textrm{HS}(\mathcal{H})\subsetneq\mathcal{B}_\textrm{HS}(\mathcal{H})$$ Reminding classes:* $$\mathcal{B}_\textrm{Tr}(\mathcal{H}):=\{A\in\mathcal{B}(\mathcal{H}):\operatorname{Tr}|A|<\infty\}$$ $$\mathcal{B}_\textrm{HS}(\mathcal{H}):=\{A\in\mathcal{B}(\mathcal{H}):\|A\|_\textrm{HS}<\infty\}$$ How can I check this?

*Hilbert-Schmidt Norm: $\|\cdot\|_\textrm{HS}$

Motivation

Relations on trace class operators are complex.
This series aims to enlighten these by reduction.
Hilbert-Schmidt operators become canonical.
Proofs do not require compact operators.

Answer 1

Inclusion

Suppose $A\in\mathcal{B}_\textrm{Tr}(\mathcal{H})$.

Polar decomposition: $$A=J|A|=J|A|^{1/2}\cdot|A|^{1/2}$$

But for square root: $$\left\||A|^{1/2}\right\|_\textrm{HS}^2=\operatorname{Tr}|A|<\infty$$

Concluding inclusion.

Converse

Suppose $A,B\in\mathcal{B}_\textrm{HS}(\mathcal{H})$.

Polar decomposition: $$B^*A=J|B^*A|\implies|B^*A|=J^*(B^*A)$$

Thus one obtains: $$\operatorname{Tr}|B^*A|=\langle A,BJ\rangle_\textrm{HS}<\infty$$

Concluding converse.

Strictness

Given the Hilbert spae $\ell^2(\mathbb{N})$.

Consider the operator: $$A\delta_n:=\frac{1}{n}\delta_n:_\quad\operatorname{Tr}|A|=\infty\quad\|A\|_\textrm{HS}<\infty$$ Concluding strictness.