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If a density operator $\sigma$ on an $m\ge 2$ dimensional Hilbert space has overlap $t=⟨\psi |\sigma |\psi ⟩$ with a fixed pure state, then $S(\sigma )\le h(t)+(1-t){\log }_2(m-1)$. Dephase in a basis containing $|\psi ⟩$ and apply the entropy bound with one prescribed probability. The abstract state bound is attained by $\sigma =t|\psi ⟩⟨\psi |+(1-t)(I-|\psi ⟩⟨\psi |)/(m-1)$.

Choose an orthonormal basis of the joint $d^2$-dimensional Hilbert space whose first vector is the given purification of a density operator $|\rho {⟩}_{QR}$. Apply rank-one dephasing to ${\sigma }_{QR}$ in this basis, and denote the resulting diagonal probabilities by $p_1,\dots ,p_{d^2}$. Their first entry isThe preceding entropy increase under nonselective projective measurement and entropy bound with one prescribed probability yieldTherefore the quantum Fano inequality isThe quantity $f^2$ is the entanglement fidelity of $\mathcal{N}$ on ${\rho }_Q$; it is already a squared overlap, so it is $f^2$, rather than $f$, that enters the binary entropy. The argument is an instance of the entropy bound from overlap with a pure state in dimension $d^2$. At $f^2=1$ the output is the original pure state and its Von Neumann entropy is zero. For $d=1$ the system is trivial and the same zero-entropy conclusion holds without evaluating ${\log }_2(d^2-1)$.

For $t<1$, normalize the remaining probabilities by $q_j=P_X(j)/(1-t)$, $j=2,\dots ,k$. Directly splitting the Shannon entropy sum givesHere $h$ is the binary entropy. The Shannon entropy of a distribution on $k-1$ points is at most ${\log }_2(k-1)$. For example, nonnegativity of its Kullback-Leibler divergence from the uniform distribution gives ${\log }_2(k-1)-H(q)\ge 0$. Thus the entropy bound with one prescribed probability isFor $t<1$, equality holds precisely when the remaining probabilities are all $(1-t)/(k-1)$. For $t=1$, the distribution is deterministic and both sides are zero. The displayed formula is for $k\ge 2$; a one-point alphabet simply has zero Shannon entropy.