Thanks for the answer. I have two questions. First, I understand that we are able to assume WLOG $A$ and $B$ are Hermitian, but why can we assume that they are positive at the same time. Second, could you elaborate more on how Cauchy-Schwarz inequality implies that $|\mathrm{Tr}(UAV^*B)|$ attains the supremum when $V$ and $U$ are equal?

Thanks. It would be great if you could refer me to a proof of the last equality. To prove the inequality we can assume that $\hat{U}$ is the partial isometry that maximizes the LHS sup. Then, we need to show $$ |\mathrm{Tr}(B\hat{U}A)| \le \sup_{V}|\mathrm{Tr}(B\hat{U}AV)|. $$ Now, let $B\hat{U}A$ have the singular value decomposition $R\Sigma Q^*$, where $\Sigma$ has at most $k$ non-zero elements, then there exists a rank (at most) $k$ partial isometry $V=Q\Lambda Q^*$ s.t. $|\mathop{\mathrm{Tr}}(B\hat{U}A)|=|\mathop{\mathrm{Tr}}(B\hat{U}AV)|$. The inequality follows by taking the sup.