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Use Klein's inequality, or equivalently nonnegativity of quantum relative entropy. First check the support needed for the logarithm. If $q_y=0$, positivity gives $|{\rho }_{yz}{|}^2\le {\rho }_{yy}{\rho }_{zz}=0$ for every $z$; the corresponding row and column vanish. Thus support inclusion under rank-one dephasing gives $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$, and the logarithms may be evaluated on this support.

Since $\log \sigma$ is diagonal in the dephasing basis,The relative-entropy identity for rank-one dephasing follows:Klein's inequality gives $D(\rho \Vert \sigma )\ge (\mathrm{Tr}\rho -\mathrm{Tr}\sigma )/\ln 2=0$. ThereforeEquality holds precisely when $\rho =\sigma$, meaning that the input was already diagonal in the chosen basis. This quantifies why rank-one dephasing removes coherence without reducing the entropy.

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 ${\log }_2(d^2-1)$.

Put $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩$. The rank-one dephasing is ${\rho }^{\prime }={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$. If $p_i=0$, positivity givesso $\rho |{\psi }_i⟩=0$. The kernel of ${\rho }^{\prime }$ is exactly the span of these zero-probability basis vectors and is therefore contained in the kernel of $\rho$. Taking orthogonal complements proves support inclusion under rank-one dephasing:Here the support of a positive operator is the orthogonal complement of its kernel.

On that support, ${\log }_2{\rho }^{\prime }$ is diagonal in the measurement basis, givingConsequently the relative-entropy identity for rank-one dephasing isKlein's inequality gives $D_{\mathrm{rel}}(\rho \Vert {\rho }^{\prime })\ge 0$, since both density operators have trace one. For singular $\rho$, first restrict to $\mathrm{supp}{\rho }^{\prime }$ where ${\rho }^{\prime }$ is positive definite, replace $\rho$ by $(\rho +\eta I)/(1+\eta \dim \mathrm{supp}{\rho }^{\prime })$, and let $\eta \downarrow 0$. The support inclusion ensures that the limit is finite. ThusThis proves entropy increase under nonselective projective measurement using Klein's inequality. Equality holds exactly when $\rho ={\rho }^{\prime }$, so the original density operator was already diagonal in the chosen basis.

If $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩=0$, positivity gives $\sqrt{\rho }|{\psi }_i⟩=0$, so the corresponding basis vector is in $\ker \rho$. Therefore $\ker \Delta (\rho )\subseteq \ker \rho$, and $\mathrm{supp}\rho \subseteq \mathrm{supp}\Delta (\rho )$. This makes the quantum relative entropy of $\rho$ relative to its rank-one dephasing finite.