= Solution
Since the MLE has constant risk $p$, its worst-case risk is $p$, so the minimax risk is at most $p$. On the other hand, the worst-case risk of every estimator is at least its average risk under any prior, and therefore at least the optimal Bayes risk for that prior. Using $\pi_c=N_p(0,c^2I_p)$ gives
$$
\inf_\delta\sup_\theta R(\theta,\delta)
\geq r(\pi_c)=\frac{pc^2}{1+c^2}.
$$
Letting $c\to\infty$ yields a lower bound $p$. The bounds agree, proving the <minimaxity of the usual multivariate normal mean estimator>:
$$
\boxed{\inf_\delta\sup_\theta R(\theta,\delta)=p,
\qquad\widehat\theta_{\rm MLE}=X\text{ is minimax}.}
$$
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