While we know the definition for trace class is that $\sum_{n=0}^{\infty} \langle \mid A\mid x_n,x_n\rangle \leq +\infty$ holds for a orthogonal basis ${x_n}$, I wonder if it's true that $\sum_{n=0}^{\infty} \mid\langle A x_n,x_n\rangle\mid\leq +\infty$ holds for every orthogonal basis ${x_n}$ a sufficient condition for $A$ to be in the trace class.
Now I make some progress. We need prove the bounded self-disjoint operators with the property is compact.
Thanks in advance.
Ok, I found one by google.
Define $A^+=\frac{|A|+A}{2}$ and $A^-=\frac{|A|-A}{2}$.
Since we have $2|\langle Bx,x\rangle|\leq|\langle Ax,x\rangle|+|\langle A^*x,x\rangle|=2|\langle Ax,x\rangle|$ , where $B=\frac{A+A^*}{2}$ and similarly for $C=\frac{A-A^*}{2i}$, we can only deal with the case when $A$ self-adjoint operator.
It can be easily checked that since $A$ is self-adjoint, which means $\left\| A \right\|=\sup_{\left\| \phi \right\|=1} |(A\phi,\phi)|$, $A$ is a bounded operator. And it's a less obvious fact that $A|A|=|A|A$, where $|A|=\sqrt{A^*A}=\sqrt{A^2}$. The fact is often derived from $A|A|^2=A^3=|A|^2A$ and $|A|=\left\| A \right\|\sqrt{I-(I-\frac{A^2}{\left\| A \right\|^2})}$, a polynomial serie of $A$.
Then we have $A^+A^-=\frac{1}{4}(|A|A-A|A|)=0$, which implies $\mathrm{Im} A^-\subset \ker A^+$. The Hilbert space has a decomposition $H=\mathrm{Im} A^- \bigoplus \ker A^-$ . Choose a basis for $\ker A^-$ and , by Zorn's lemma or somehow, extend it to a basis of $H$. The basis chosen has the poperty that at least one of $ \langle A^+e,e\rangle$ and $ \langle A^-e,e\rangle$ is $0$ and therefore:
$$\sum_k|\langle Ae_k,e_k\rangle|\leq\sum_k|\langle A^+e_k,e_k\rangle-\langle A^-e_k,e_k\rangle|=\sum_k|\langle A^+e_k,e_k\rangle+\langle A^-e_k,e_k\rangle|=\sum_k\langle |A|e_k,e_k\rangle$$
Then the conclusion can be obtained.