= Solution
The quantity $\chi^*(\Lambda)$ is the one-use Holevo information of the channel: by the Holevo--Schumacher--Westmoreland theorem, its regularization gives the classical capacity, and without entangled inputs across uses it is the asymptotically achievable classical rate using collective output measurements.
If an ensemble contains a mixed $\rho_x=\sum_jq_{j|x}|\psi_{xj}\rangle\langle\psi_{xj}|$, refine its label to $(x,j)$ with probability $p_xq_{j|x}$. The average channel output is unchanged, while <Concavity of Von Neumann entropy> gives
$$
S(\Lambda(\rho_x))\geq\sum_jq_{j|x}S(\Lambda(|\psi_{xj}\rangle\langle\psi_{xj}|)).
$$
Thus refinement cannot decrease the <Holevo quantity>, so the maximum may be restricted to pure input states.
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