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Let $\Delta =\rho -\sigma ={\Delta }_{+}-{\Delta }_{-}$ be the positive-negative decomposition of a Hermitian operator. Since $\mathrm{Tr}\Delta =0$,For $0\le P\le I$,Equality is attained by the projector onto the positive eigenspace of $\Delta$. This proves the variational characterization of trace distance

If $\epsilon =0$, then ${\rho }_{AB}={\sigma }_{AB}$ and the desired continuity bound is immediate, so assume $\epsilon >0$. Apply the positive-negative decomposition of a Hermitian operator toBecause $\mathrm{Tr}X=0$ and $\Vert X{\Vert }_1=2\epsilon$, one hasThus ${\Delta }_{AB}=X_{+}/\epsilon$ is positive with trace one, hence is a density operator. DefineIt is a convex combination of states. The equation $\epsilon {\Delta }_{AB}^{\prime }=(1+\epsilon ){\omega }_{AB}-{\rho }_{AB}$ giveswhich is likewise positive and has trace one.