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A compact linear operator $A$ over a separable Hilbert space $H$ is said to be in the trace class or in the nuclear operator space if for some (and hence all ) orthonormal bases $(e_{n})$ of $H$ the trace norm $$\Vert A \Vert_{tr} = Tr| A |= \sum_{n}\langle ( A ^{\ast} A ) ^{\frac{1}{2}} e_{ n} , e_{ n} \rangle$$ is finite. In this case $$TrA=\sum_{n}\langle A e_{ n} , e_{ n} \rangle$$ converges absolutely and is independent of the choice of the orthonormal basis.

Moreover,we have $$\Vert A \Vert_{tr} \leq \sum_{n=1}^{\infty}\Vert A e_{n} \Vert<\infty$$

The trace norm also can be characterized as follows

$$\Vert A\Vert_{tr}= sup \sum_{n=1}^{\infty}|\langle A e_{ n} , f_{ n} \rangle|$$

Where the supremum is taken over any pair of orthonormal bases $\{e_n\}$ and $\{f_n\}$

My question is : Can we characterize the trace norm to be $$\Vert A \Vert_{tr} =sup \sum_{n=1}^{\infty}\Vert A e_{n} \Vert$$ Where the supremum is taken over any orthonormal basis $\{e_n\}$?

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It is very easy to show that, if $A$ is trace-class, $$\tag{1} \|A\|_{\rm tr}=\min\left\{\sum_n\|Af_n\|:\ \{f_n\}\ \text{ orthonormal basis}\right\}. $$ Note that since $\|Ax\|=\|\,|A|x\,\|$, we may assume that $A$ is positive. So $A=\sum_n\lambda_j\,e_n\otimes e_n$ for a decreasing sequence $\{\lambda_n\}$ with $\lambda_n\searrow0$, and $\{e_n\}$ an orthonormal basis. Thus $$ \|A\|_{\rm tr}=\sum_n\langle |A|e_n,e_n\rangle=\sum_n\langle Ae_n,e_n\rangle=\sum_n\lambda_n=\sum_n\|Ae_n\|. $$ For an arbitrary orthonormal basis $\{f_n\}$, we have $$ \|A\|_{\rm tr}=\sum_n\langle Af_n,f_n\rangle\leq\sum_n\|Af_n\|. $$ So $(1)$ is established.

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