**Part 1.** Let $\delta$ be any decision rule. By Proposition 4.3 and the defining hypothesis: \begin{align*} \sup_{\theta \in \Theta} R(\delta, \theta) \geq R_\lambda(\delta) = \mathbb{E}_\lambda[R(\delta, \theta)] \geq \mathbb{E}_\lambda[R(\delta_\lambda, \theta)] = R_\lambda(\delta_\lambda) = \sup_{\theta \in \Theta} R(\delta_\lambda, \theta). \end{align*} The first inequality is the Bayes–maximal risk bound. The second is because $\delta_\lambda$ minimises Bayes risk over $\lambda$. Taking the infimum over $\delta$ on the left shows $\inf_\delta R_m(\delta, \Theta) \geq R_m(\delta_\lambda, \Theta)$, so $\delta_\lambda$ is minimax. **Part 2.** If $\delta_\lambda$ is the unique Bayes rule, then the second inequality $\mathbb{E}_\lambda[R(\delta, \theta)] \geq \mathbb{E}_\lambda[R(\delta_\lambda, \theta)]$ is strict for any $\delta \neq \delta_\lambda$, making the whole chain strict. So no other rule can match $\delta_\lambda$'s maximal risk. **Part 3.** For any prior $\lambda'$ with Bayes rule $\delta_{\lambda'}$: \begin{align*} R_{\lambda'}(\delta_{\lambda'}) = \mathbb{E}_{\lambda'}[R(\delta_{\lambda'}, \theta)] \leq \mathbb{E}_{\lambda'}[R(\delta_\lambda, \theta)] \leq \sup_{\theta \in \Theta} R(\delta_\lambda, \theta) = \mathbb{E}_\lambda[R(\delta_\lambda, \theta)] = R_\lambda(\delta_\lambda). \end{align*} The first inequality uses that $\delta_{\lambda'}$ minimises Bayes risk under $\lambda'$. The second is the Bayes–maximal risk bound. The third uses the hypothesis. Hence $\lambda$ achieves the largest Bayes risk.