Equivalent expressions of the trace class norm on infinite dimensional separable Hilbert spaces

Question

Let $H$ be an infinite dimensional separable Hilbert space. Suppose $A$ is a trace class operator on $H$, i.e., $A$ is a compact operator such that $$\Vert A\Vert_{tr}:=\sum_{n=0}^{\infty}s_n(A)<\infty$$ where $s_n(A)$ are singular values of $A$, i.e., eigenvalues of the square root of $A^*A$. My question is whether the following relation holds: $$\Vert A\Vert_{tr}=\sup\left\{\sum_{n=0}^{\infty}|(Ae_n,e_n)|\Big|\ \{e_n\}_{n=0}^{\infty} \ \mbox{is an orthonormal basis of $H$}\right\}.$$ Moreover, I have already verified that $$\Vert A\Vert_{tr}=\sup\left\{\sum_{n=0}^{\infty}|(Ae_n,f_n)|\Big|\ \{e_n\}_{n=0}^{\infty},\{f_n\}_{n=0}^{\infty} \ \mbox{are orthonormal bases of $H$}\right\}.$$ And I am also curious about if there are other equivalent expressions. Thanks in advance for your kind help!!

Answer 1

As for the first definition, I have discovered a counter-example. $$\Vert A-A^*\Vert=\sup\left\{\sum_n|(Ae_n,e_n)-(A^*e_n,e_n)|\right\}=0,$$ which is a contradiction, since the above relation means $A=A^*$ for any trace class operator. So I think the first equivalent definition does not hold at least in the real Hilbert space.

As for the second definition, i.e., $$\Vert A\Vert_{tr}=\sup\left\{\sum_{n=0}^{\infty}|(Ae_n,f_n)|\Big|\ \{e_n\}_{n=0}^{\infty},\{f_n\}_{n=0}^{\infty} \ \mbox{are orthonormal bases of $H$}\right\},$$ it could be obtained via the spectra decomposition of $A$. First, we take the polar decomposition of $A$, i.e., $$A=Q|A|$$ where $Q$ is the partial isometry on $\overline{\mathcal{R}|A|}$. Then, we obtain $$A=\sum_ns_n(\cdot, e_n)Qe_n.$$ where $\{e_n\}_{n=0}^{\infty}$ is the orthonormal basis induced by $A^*A$. Since $Q$ is partial isometry, then $\{Qe_n\}_{n=0}^{\infty}$ is an orthonormal family. Then $$\sum_n|(Ae_n,Qe_n)|=\sum_n s_n=\Vert A\Vert_{tr}.$$ And we are left to verify $$\Vert A\Vert_{tr}\ge \sum_{n}|(Ae_n^{'},f_n^{'})|$$ for any orthonormal bases $\{e_n^{'}\}_{n=0}^{\infty}$ and $\{f_n^{'}\}_{n=0}^{\infty}$. Again by applying the spectra decompostion and the Bessel's inequality, we get $$\sum_n|(Ae_n^{'},f^{'}_n)|=\sum_n\left|\left(\sum_ks_k(e_n^{'},e_k)Qe_k,f_n^{'}\right)\right|\le\sum_k\sum_n\left|s_k(e_n^{'},e_k)\left(Qe_k,f_n^{'}\right)\right|\le\sum_ks_k\left(\sum_n\left|(e_n^{'},e_k)\right|^2\right)^{\frac{1}{2}}\left(\sum_n\left|\left(Qe_k,f_n^{'}\right)\right|^2\right)^{\frac{1}{2}}\le\sum_ks_k=\Vert A\Vert_{tr}$$ as claimed.

And could anyone supply other equivalent definition of the trace norm? Thanks a lot!!!