= Solution
Put $p_i=\langle\psi_i|\rho|\psi_i\rangle$. The <rank-one dephasing> is $\rho'=\sum_ip_i|\psi_i\rangle\langle\psi_i|$. If $p_i=0$, positivity gives
$$
0=p_i=\|\sqrt\rho\,|\psi_i\rangle\|^2,
$$
so $\rho|\psi_i\rangle=0$. The kernel of $\rho'$ 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>:
$$
\boxed{\operatorname{supp}\rho\subseteq\operatorname{supp}\rho'.}
$$
Here the <support of a positive operator> is the orthogonal complement of its kernel.
On that support, $\log_2\rho'$ is diagonal in the measurement basis, giving
$$
\operatorname{Tr}(\rho\log_2\rho')=\sum_{i:p_i>0}p_i\log_2p_i=\operatorname{Tr}(\rho'\log_2\rho').
$$
Consequently the <relative-entropy identity for rank-one dephasing> is
$$
D_{\mathrm{rel}}(\rho\|\rho')=\operatorname{Tr}\bigl[\rho(\log_2\rho-\log_2\rho')\bigr]=S(\rho')-S(\rho).
$$
<Klein's inequality> gives $D_{\mathrm{rel}}(\rho\|\rho')\geq0$, since both <density operators> have trace one. For singular $\rho$, first restrict to $\operatorname{supp}\rho'$ where $\rho'$ is positive definite, replace $\rho$ by $(\rho+\eta I)/(1+\eta\dim\operatorname{supp}\rho')$, and let $\eta\downarrow0$. The support inclusion ensures that the limit is finite. Thus
$$
\boxed{S(\rho')\geq S(\rho).}
$$
This proves <entropy increase under nonselective projective measurement> using <Klein's inequality>. Equality holds exactly when $\rho=\rho'$, so the original <density operator> was already diagonal in the chosen basis.
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