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Let $\cal{H}$ be a Hilbert space, $T$ a bounded linear operator on $\cal{H}$, $S$ a trace class operator, then can one verify that $$|Tr(TS)|\leq\|T\|\cdot|Tr(S)|?$$

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  • $\begingroup$ Are you perhaps confusing the absolute value of the trace of a trace operator with the trace norm? They are completely different things in general. The inequality you state holds true if you replase $|Tr(TS)|$ with $\|TS\|_{tr}$ where the last norm is the trace norm. For instance, you can see P. Lax, Functional Analysis, Ch. 30 Section 2, Thm 2. $\endgroup$ – guest Jun 19 '15 at 4:53
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No. Let $$ S=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ T=\begin{bmatrix}0&0\\1&0\end{bmatrix}. $$ Then $$ \text{Tr}(TS)=1,\ \ \text{ and } \|T\|\,|\text{Tr}(S)|=0. $$

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It is correct for positive operators, for more information see Gohberg, Krein, Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space page 27

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