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For any orthonormal basis and Hermitian operator, ${\sum }_i|⟨{\psi }_i|X|{\psi }_i⟩|\le \Vert X{\Vert }_1$. In the trace-norm variational principle for Hermitian operators, choose the diagonal $T$ whose entries are the signs of these diagonal entries. This bounds the total distinguishability visible in one fixed measurement basis.

The constraints $-I\le T\le I$ are in the order of Hermitian operators. They imply $-1\le ⟨{\alpha }_i|T|{\alpha }_i⟩\le 1$. Using the spectral decomposition of $X$,Choose $T=\mathrm{sgn}(X)$, with eigenvalues $+1$ on the positive spectral subspace, $-1$ on the negative spectral subspace, and zero on the kernel. It satisfies the constraints and attains equality. The trace-norm variational principle for Hermitian operators is therefore

To prove the Holevo–Helstrom theorem, write a binary POVM as $\{E,I-E\}$, where $0\le E\le I$, and associate the first outcome with $\rho$. Any larger outcome set followed by a binary decision can be grouped into this form. Put $\Delta =p\rho -(1-p)\sigma$. Its success probability isThe substitution $T=2E-I$ bijects the allowed effects with the interval $-I\le T\le I$. Since $\mathrm{Tr}\Delta =2p-1$,Choosing $E$ as the positive spectral projection of $\Delta$ attains this value, with an arbitrary decision on its kernel. This proves the optimal success probability and constructs an optimal measurement.

Write $x_i=⟨{\psi }_i|X|{\psi }_i⟩$, which is real because $X$ is a Hermitian operator. Setwith $\mathrm{sgn}(0)=0$. This Hermitian operator satisfies $-I\le T\le I$. The trace-norm variational principle for Hermitian operators gives the diagonal absolute-sum bound for the trace norm:

For $X=\rho -P$, with $P=|\psi ⟩⟨\psi |$, extend $|\psi ⟩$ to an orthonormal basis $|{\psi }_1⟩=|\psi ⟩,|{\psi }_2⟩,\dots$. Put $r=⟨\psi |\rho |\psi ⟩$. The first diagonal entry is $r-1\le 0$, and all the others are nonnegative. Their sum is $1-r$, because $\mathrm{Tr}\rho =1$. The bound therefore yields $\Vert \rho -P{\Vert }_1\ge 2(1-r)$. Using the definitions of trace distance and quantum fidelity,This pure-target lower bound on trace distance is attained whenever $\rho$ has no coherence between $|\psi ⟩$ and its orthogonal complement. The proof used the trace-norm variational principle for Hermitian operators, together with positivity and normalization of a density operator.