= Solution
The <absolute value of an operator>, <trace norm>, and <operator norm> are respectively
$$
|L|=\sqrt{L^\dagger L},\qquad
\|L\|_1=\operatorname{Tr}|L|,\qquad
\|L\|_{\rm op}=\sup_{\|v\|=1}\|Lv\|.
$$
The <positive square root of an operator> makes $|L|$ well defined. Its <eigenvalues> $s_j\geq0$ are the <singular values> of $L$, so $\|L\|_1=\sum_js_j$ and $\|L\|_{\rm op}=\max_js_j$.
Use the <polar decomposition of a bounded operator> $L=U|L|$, with the given <unitary operator> $U$. In an <orthonormal eigenbasis> $e_j$ for $|L|$,
$$
\operatorname{Tr}(ZL)=\sum_js_j\langle e_j|ZU|e_j\rangle.
$$
The <operator norm> assumption and unitarity imply $|\langle e_j|ZU|e_j\rangle|\leq\|Z\|_{\rm op}\leq1$. The <triangle inequality> therefore yields \b[the required trace estimate]:
$$
\boxed{|\operatorname{Tr}(ZL)|\leq\sum_js_j=\|L\|_1.}
$$
Choosing $Z=U^\dagger$ attains equality, which also gives the finite-dimensional <trace duality> formula $\|L\|_1=\max_{\|Z\|_{\rm op}\leq1}|\operatorname{Tr}(ZL)|$. A singular $L$ causes no difficulty: in a finite-dimensional square space the polar factor can be extended to a <unitary operator> on the complementary subspaces.
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