= Solution
Omit indices of probability zero. The <support of a positive operator> of every remaining $\rho_i$ lies within that of $\overline\rho$, because $\overline\rho$ is a positive sum containing $p_i\rho_i$. Thus $D(\rho_i\Vert\overline\rho)$ is finite.
First suppose $\operatorname{supp}\overline\rho\subseteq\operatorname{supp}\sigma$, so all quantities are finite. Expand the trace definitions of <quantum relative entropy>:
$$
\begin{aligned}
\sum_ip_iD(\rho_i\Vert\sigma)-\sum_ip_iD(\rho_i\Vert\overline\rho)
&=\sum_ip_i\operatorname{Tr}\rho_i(\log_2\overline\rho-\log_2\sigma)\\
&=\operatorname{Tr}\overline\rho(\log_2\overline\rho-\log_2\sigma)\\
&=D(\overline\rho\Vert\sigma).
\end{aligned}
$$
Therefore <Donald's identity> is
$$
\boxed{\sum_ip_iD(\rho_i\Vert\sigma)=\sum_ip_iD(\rho_i\Vert\overline\rho)+D(\overline\rho\Vert\sigma).}
$$
If the support condition on $\sigma$ fails, at least one positive-weight $\rho_i$ also fails it. The left side and the final term on the right are then $+\infty$, while the first right-hand sum remains finite. The equality is valid in this extended sense, without taking any undefined difference of infinities.
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