Theorem 21 of these notes prove that finite rank operators are dense in the space of trace class operators (under the trace norm). I am unable to get through a certain step of the proof.
Let $H$ be a Hilbert space.
For a bounded linear operator $T$ on $H$, we will write $|T|$ to mean tha unique positive square root of $T^*T$.
The trace norm of $T$ will be denote by $\|T\|_1$.
Here is how the proof goes in the notes.
Fix a trace class operator $A$ on $H$ and let $\varepsilon >0$. We want to show that there is a finite rank bounded linear operator $B$ on $H$ such that $B$ is $\varepsilon$-close to $A$ in the trace norm.
Let $\mathcal E$ be an orthonormal basis of $H$.
So $\sum_{e\in \mathcal E}\langle |A|e, e \rangle$ is finite.
Therefore there is a finite subset $F$ of $\mathcal E$ such that $\sum_{e\in \mathcal E\setminus F}\langle |A|e, e\rangle<\varepsilon$.
Define $P:H\to H$ as the othogonal projection onto the span of $F$ and let $B=AP$.
(I follow so far).
Then
$$\|A-B\|_1 = \sum_{e\in\mathcal E}\langle |A-B|e, e \rangle = \sum_{e\in \mathcal E\setminus F}\langle |A|e, e \rangle<\varepsilon$$
I am unable to see how the second equality comes.
Suppose $e\in F$, then $|A-B|e=0$ since $|A-B|^2e=0$ and $|A-B|$ is self adjoint.
So we want to show that $\sum_{e\in \mathcal E\setminus F}\langle |A-B|e, e\rangle = \sum_{e\in \mathcal E\setminus F} \langle |A|e, e\rangle$.
But I am stuck here.
Best Answer
I haven't manage to decide whether the equality holds. In any case, here is how I would prove the claim in a similar way. Let $A \in S^1$. Decompose $A$ in two self-adjoin operators by $A = \Re(A) + i \, \Im(A)$ with $\Re(A) = (A + A^\ast)/2$ and $\Im(A) = (A - A^\ast)/2i$. Each of the self-adjoints can be decomposed as sum of two positives. By an $\epsilon/4$-argument, it is enough to prove it for $A$ positive. But for $A$ positive your calculation works fine since you don't have the $| \cdot |$ in your formula.