= Solution
Let $\Delta=\rho-\sigma=\Delta_+-\Delta_-$ be its <positive-negative decomposition>. Since $\operatorname{Tr}\Delta=0$, the positive and negative parts have equal trace, namely $D(\rho,\sigma)$. For every <positive contraction> $M$,
$$
\operatorname{Tr}(M\Delta)
=\operatorname{Tr}(M\Delta_+)-\operatorname{Tr}(M\Delta_-)
\leq\operatorname{Tr}\Delta_+=D(\rho,\sigma).
$$
Here both products have nonnegative trace, and $M\leq I$ bounds the first term. Equality is attained by the <orthogonal projection> $P_+$ onto the positive spectral subspace of $\Delta$, since $P_+\Delta_+=\Delta_+$ and $P_+\Delta_-=0$. Thus the <variational characterization of trace distance> is
$$
\boxed{D(\rho,\sigma)=\max_{0\leq M\leq I}\operatorname{Tr}[M(\rho-\sigma)]}.
$$
Back to article page
