= Solution
The <density operators> are Hermitian, so their squared <Hilbert-Schmidt distance> is $2V=\operatorname{Tr}(\rho^2)+\operatorname{Tr}(\rho_d^2)-2\operatorname{Tr}(\rho\rho_d)$. Both <purity of a density operator> values are constant under their respective <unitary time evolution>: cyclicity gives $\operatorname{Tr}(\rho[H,\rho])=0$ for every Hermitian $H$. Consequently
$$
\dot V=-\operatorname{Tr}(\dot\rho\rho_d)-\operatorname{Tr}(\rho\dot\rho_d)=-f\operatorname{Tr}\bigl(\rho_d[-iH_1,\rho]\bigr).
$$
The terms containing $H_0$ cancel because $\operatorname{Tr}([-iH_0,\rho]\rho_d)=-\operatorname{Tr}(\rho[-iH_0,\rho_d])$. The quantity $F=\operatorname{Tr}(\rho_d[-iH_1,\rho])$ is real: $[-iH_1,\rho]$ is Hermitian, and the <trace> of the product of two <Hermitian matrices> equals its conjugate. The <Lyapunov quantum control> choice $f=F$ therefore gives $\boxed{\dot V=-f^2\le0}$. For $t_2\ge t_1$, integration yields $V(t_2)=V(t_1)-\int_{t_1}^{t_2}f(t)^2\,dt\le V(t_1)$. Since the distance is $\sqrt{2V}$, it is \b[monotonically nonincreasing]. The decrease need not be strict when $f=0$.
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