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Let $H$ be a separable Hilbert space and $L^{1}(H)$ be the space of trace class operators on $H$.

Let $f:\mathbb{R}\to L^{1}(H)$ be a measurable function that is $t\to \operatorname{tr}(f(t))$ is Borel measurable.

Q. Suppose that $\int \operatorname{tr}(f(t))d\mu$ is finite. Can we conclude that $\int \operatorname{tr}(|f(t)|)d\mu$ is finite too?

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    $\begingroup$ What does $\lvert f(t)\rvert$ mean? (That is, what is the absolute value of a trace-class operator?) $\endgroup$ Commented Aug 3, 2018 at 11:33

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What if $H$ is two dimensional, $\mu$ = Lebesgue measure and $$ f(t) = \begin{bmatrix} 1 & 0\\0 &-1\end{bmatrix} $$ so that $\mathrm{tr}(f(t))=0$

Then I guess $$ |f(t)| = \begin{bmatrix} 1 & 0\\0 &1\end{bmatrix} $$ so that $\mathrm{tr}\big(|f(t)|\big) = 2$ is not integrable.

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