[proofplan]
We prove the two claims separately. For claim 1, we argue by contradiction: if a unique Bayes rule were inadmissible, some dominating rule would have strictly lower risk at some $\theta$ and no higher risk elsewhere, which upon integrating against the prior would produce a strictly lower Bayes risk, contradicting uniqueness of the Bayes rule. For claim 2, if an admissible rule $\delta$ has constant risk $c$ and is not minimax, then a minimax rule $\delta^*$ would have maximal risk $c^* < c$, which would dominate $\delta$, contradicting admissibility.
[/proofplan]
[step:Prove that a unique Bayes rule is admissible by contradiction]
Let $\delta_\pi$ be the unique Bayes rule under prior $\pi$ and loss $L$, meaning $\delta_\pi$ is the unique minimizer of the Bayes risk
\begin{align*}
R_\pi(\delta) = \int_\Theta R(\delta, \theta) \, \pi(\theta) \, d\theta,
\end{align*}
where $R(\delta, \theta) = \mathbb{E}_\theta[L(\delta(X), \theta)]$ is the frequentist risk.
Suppose for contradiction that $\delta_\pi$ is inadmissible. Then there exists a decision rule $\delta'$ that dominates $\delta_\pi$: $R(\delta', \theta) \leq R(\delta_\pi, \theta)$ for all $\theta \in \Theta$, with strict inequality for at least one $\theta_1 \in \Theta$.
Integrating against $\pi$:
\begin{align*}
R_\pi(\delta') = \int_\Theta R(\delta', \theta) \, \pi(\theta) \, d\theta \leq \int_\Theta R(\delta_\pi, \theta) \, \pi(\theta) \, d\theta = R_\pi(\delta_\pi).
\end{align*}
If $\pi(\theta_1) > 0$ (which holds on a set of positive $\pi$-measure near $\theta_1$ since $R(\delta', \cdot)$ is measurable and strictly below $R(\delta_\pi, \cdot)$ on a set containing $\theta_1$), the inequality is strict: $R_\pi(\delta') < R_\pi(\delta_\pi)$. But this contradicts $\delta_\pi$ being a Bayes rule (a minimizer of the Bayes risk).
If $\pi(\theta_1) = 0$, we need the following refinement. Since $R(\delta', \theta) \leq R(\delta_\pi, \theta)$ for all $\theta$ with equality $\pi$-a.e., we have $R_\pi(\delta') = R_\pi(\delta_\pi)$, so $\delta'$ is also a Bayes rule. But $\delta'$ dominates $\delta_\pi$ with $R(\delta', \theta_1) < R(\delta_\pi, \theta_1)$, so $\delta' \neq \delta_\pi$. This contradicts the uniqueness of the Bayes rule $\delta_\pi$.
In either case we reach a contradiction. Therefore $\delta_\pi$ is admissible.
[guided]
The intuition is that a Bayes rule is already optimal on average with respect to $\pi$. If some rule $\delta'$ were uniformly at least as good and sometimes strictly better, then integrating against $\pi$ would make $\delta'$ at least as good on average, and the strict improvement at $\theta_1$ would either lower the Bayes risk (if $\pi$ assigns positive mass near $\theta_1$) or produce a second Bayes-optimal rule (if $\pi$ assigns zero mass near $\theta_1$). The uniqueness hypothesis rules out the second possibility.
Formally, suppose $\delta'$ dominates $\delta_\pi$ with $R(\delta', \theta_1) < R(\delta_\pi, \theta_1)$.
**Case 1:** $\pi$ assigns positive measure to any open set containing $\theta_1$. Since $R(\delta_\pi, \cdot) - R(\delta', \cdot) \geq 0$ everywhere and is strictly positive at $\theta_1$, and the risk functions are measurable, by continuity or measurability the set $\{\theta : R(\delta', \theta) < R(\delta_\pi, \theta)\}$ has positive $\pi$-measure. Then
\begin{align*}
R_\pi(\delta') = \int_\Theta R(\delta', \theta) \, \pi(\theta) \, d\theta < \int_\Theta R(\delta_\pi, \theta) \, \pi(\theta) \, d\theta = R_\pi(\delta_\pi),
\end{align*}
contradicting $\delta_\pi$ being a Bayes rule.
**Case 2:** The set $\{\theta : R(\delta', \theta) < R(\delta_\pi, \theta)\}$ has $\pi$-measure zero. Then $R_\pi(\delta') = R_\pi(\delta_\pi)$, so $\delta'$ also minimizes the Bayes risk. But $\delta' \neq \delta_\pi$ (they differ in risk at $\theta_1$), contradicting the uniqueness of the Bayes rule.
In both cases the assumption of inadmissibility leads to a contradiction.
[/guided]
[/step]
[step:Prove that an admissible rule with constant risk is minimax]
Suppose $\delta$ is admissible with constant risk: $R(\delta, \theta) = c$ for all $\theta \in \Theta$. Then the maximal risk of $\delta$ is
\begin{align*}
R_m(\delta, \Theta) := \sup_{\theta \in \Theta} R(\delta, \theta) = c.
\end{align*}
Suppose for contradiction that $\delta$ is not minimax. Then there exists a rule $\delta^*$ with
\begin{align*}
\sup_{\theta \in \Theta} R(\delta^*, \theta) =: c^* < c.
\end{align*}
Since $c^* < c$, we have $R(\delta^*, \theta) \leq c^* < c = R(\delta, \theta)$ for every $\theta \in \Theta$. In particular, $\delta^*$ dominates $\delta$: $R(\delta^*, \theta) \leq R(\delta, \theta)$ for all $\theta$, with strict inequality everywhere. This contradicts the admissibility of $\delta$.
Therefore $\delta$ is minimax.
[guided]
Why does constant risk play such a crucial role? If $\delta$ had non-constant risk, say $R(\delta, \theta_1) = c_1 < c_2 = R(\delta, \theta_2)$, then a competing rule could have lower maximal risk $c^* < c_2$ while still satisfying $R(\delta^*, \theta_1) > R(\delta, \theta_1) = c_1$. In that case $\delta^*$ would not dominate $\delta$, so admissibility would not be contradicted.
But when the risk is constant at $c$, any rule with maximal risk $c^* < c$ must satisfy $R(\delta^*, \theta) \leq c^* < c = R(\delta, \theta)$ for every single $\theta$. This means $\delta^*$ strictly dominates $\delta$ everywhere, directly contradicting admissibility.
Concretely: suppose $\delta^*$ has $\sup_\theta R(\delta^*, \theta) = c^* < c$. Then for every $\theta \in \Theta$:
\begin{align*}
R(\delta^*, \theta) \leq c^* < c = R(\delta, \theta).
\end{align*}
So $\delta^*$ dominates $\delta$, contradicting admissibility. Hence no such $\delta^*$ can exist, and $\delta$ is minimax.
[/guided]
[/step]
