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For a finite-dimensional Hermitian operator $A$ with decreasing eigenvalues ${\lambda }_j$ and integer $0\le r\le d$, maximizing $\mathrm{Tr}(AB)$ over positive contractions $B$ with trace $r$ gives the sum of its leading $r$ eigenvalues. In an orthonormal eigenbasis, the diagonal entries of $B$ lie in $[0,1]$ and sum to $r$. Moving their weight to the largest eigenvalues can only increase the objective. An orthogonal projection onto their eigenvectors attains the maximum.

The allowed operators are exactly the positive contractions $0\le T\le I$, or binary measurement effects. As in (ii),because $\mathrm{Tr}(TR)\ge 0$ and $\mathrm{Tr}[(I-T)Q]\ge 0$. The positive-spectral orthogonal projector $P_{+}$ is among these effects and attains the bound. ThusAllowing general binary effects instead of projectors does not improve the optimal trace distance test.

Use an orthonormal eigenbasis of the Hermitian operator $A$ and set $b_j=⟨{\alpha }_j|B|{\alpha }_j⟩$. Since $B$ is a positive contraction,For $0<r<d$, the missing weight among the first $r$ entries equals the weight in the remaining entries:The ordered eigenvalues satisfy ${\lambda }_j\ge {\lambda }_{r-1}$ for $j<r$ and ${\lambda }_j\le {\lambda }_{r-1}$ for $j\ge r$. ThereforeThis argument does not require the eigenvalues of $A$ to be positive. For $r=0$, positivity and zero trace give $B=0$; for $r=d$, the same argument applied to $I-B$ gives $B=I$. Thus in every caseThe orthogonal projection onto the leading $r$ eigenvectors attains equality. This is the Hermitian effect variational principle, extending the Ky Fan maximum principle to all positive contractions with the prescribed trace.

Choose $P_X$ uniform on ${\mathcal{A}}_M$ and zero outside it, so the marginal density operator of $Q$ is ${\bar{\rho }}_Q=k^{-1}{\sum }_m\rho (m{)}_Q$. Define the binary test for quantum decoding success byEach $E(m)$ is a positive contraction. The computational basis blocks of $F(0)$ are these effects on ${\mathcal{A}}_M$ and zero elsewhere. Thus $0\le F(0)\le I$, making $F$ a POVM on the whole space, including labels outside ${\mathcal{A}}_M$.

On the correlated classical-quantum state, its acceptance probability isOn the product of the marginal density operators,since ${\sum }_{m}E(m)=I_Q$. The two requested probabilities are therefore $1-ϵ$ and $1/k$. In the notation for binary output density operators, the measurement channel sends the joint input to $\omega [ϵ]$ and the product input to $\omega [1-1/k]$; their parameters describe outcome one, not outcome zero.

Put $p=⟨\psi |M_1^2|\psi ⟩\ge 1-\epsilon$. Since $M_1$ is a positive contraction, the successful post-measurement state is the pure state represented by $|{\psi }^{\prime }⟩=M_1|\psi ⟩/\sqrt{p}$. Its squared quantum fidelity with the input isThe finite-dimensional spectral theorem and $0\le M_1\le I$ give $M_1^2\le M_1$, so $⟨M_1⟩\ge p$. Therefore the pure-state gentle measurement bound isThe positive-operator assumption matters: an arbitrary unitary measurement operator can have success probability one while rotating the state to an orthogonal vector.

Let $\Delta =\rho -\sigma ={\Delta }_{+}-{\Delta }_{-}$ be its positive-negative decomposition. Since $\mathrm{Tr}\Delta =0$, the positive and negative parts have equal trace, namely $D(\rho ,\sigma )$. For every positive contraction $M$,Here both products have nonnegative trace, and $M\le I$ 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

If $M$ is a positive contraction and $p=⟨\psi |M^2|\psi ⟩>0$, the normalized successful post-measurement state is $M|\psi ⟩/\sqrt{p}$. Since $M^2\le M$,A likely outcome therefore preserves a pure input well in quantum fidelity when its measurement operator is positive. Positivity excludes a hidden unitary rotation that could otherwise change the state even for an outcome of probability one.