[proofplan] The lower bound already gives the left-hand inequality in the claimed minimax comparison. The upper bound is obtained by inserting the exhibited estimator into the infimum defining the minimax risk. Since the minimax risk is trapped between two fixed positive multiples of $r_n(\Theta)$, the sequence $r_n(\Theta)$ is exactly the minimax rate up to constants. If the estimator is generated by an explicit algorithm, the same risk calculation applies to the estimator returned by that algorithm. [/proofplan] [step:Use the testing lower bound as the lower minimax inequality] By hypothesis, there is a constant $c>0$ such that \begin{align*} R_n^*(\Theta,L) \ge c\,r_n(\Theta). \end{align*} This is precisely the lower half of the desired two-sided comparison. [/step] [step:Insert the constructed estimator into the minimax infimum] The estimator $\hat{\theta}_n$ belongs to the admissible class $\mathcal A_n$. Since $R_n^*(\Theta,L)$ is the infimum over all estimators in $\mathcal A_n$, it is bounded above by the risk of any particular admissible estimator. Therefore the definition of $R_n^*(\Theta,L)$ gives \begin{align*} R_n^*(\Theta,L) = \inf_{\tilde{\theta} \in \mathcal A_n} \sup_{\theta \in \Theta} \mathbb E_\theta[L(\tilde{\theta},\theta)]. \end{align*} Since an infimum is bounded above by the value at any admissible element and $\hat{\theta}_n \in \mathcal A_n$, we have \begin{align*} R_n^*(\Theta,L) \le \sup_{\theta \in \Theta} \mathbb E_\theta[L(\hat{\theta}_n,\theta)]. \end{align*} Using the assumed risk bound for $\hat{\theta}_n$ gives \begin{align*} R_n^*(\Theta,L) \le C\,r_n(\Theta). \end{align*} [guided] The minimax risk is defined by first measuring the worst-case risk of an estimator and then taking the infimum over all admissible estimators. Thus, for every fixed estimator $\tilde{\theta} \in \mathcal A_n$, the minimax risk is \begin{align*} R_n^*(\Theta,L) = \inf_{\eta \in \mathcal A_n} \sup_{\theta \in \Theta} \mathbb E_\theta[L(\eta,\theta)]. \end{align*} The defining property of an infimum then gives \begin{align*} R_n^*(\Theta,L) \le \sup_{\theta \in \Theta} \mathbb E_\theta[L(\tilde{\theta},\theta)]. \end{align*} We apply this observation to the specific estimator $\hat{\theta}_n \in \mathcal A_n$ supplied by the hypothesis. This gives \begin{align*} R_n^*(\Theta,L) \le \sup_{\theta \in \Theta} \mathbb E_\theta[L(\hat{\theta}_n,\theta)]. \end{align*} The assumed upper risk bound for this estimator then yields \begin{align*} R_n^*(\Theta,L) \le C\,r_n(\Theta). \end{align*} This step is the whole content of an upper bound argument: to prove that the minimax risk is no larger than a target quantity, it is enough to exhibit one admissible estimator whose worst-case risk is no larger than that quantity. [/guided] [/step] [step:Conclude the matching rate comparison] Combining the lower bound from the first step with the upper bound from the second step gives \begin{align*} c\,r_n(\Theta) \le R_n^*(\Theta,L) \le C\,r_n(\Theta). \end{align*} Since $0<c<C<\infty$, the minimax risk is bounded above and below by positive constant multiples of $r_n(\Theta)$. Hence $r_n(\Theta)$ is the minimax rate for $(\Theta,L)$. [/step] [step:Interpret an explicit construction as an algorithmic upper bound] If $\hat{\theta}_n$ is computed by a specified algorithm, then the estimator appearing in the upper bound is exactly the output of that algorithm. Therefore the bound \begin{align*} \sup_{\theta \in \Theta} \mathbb E_\theta[L(\hat{\theta}_n,\theta)] \le C\,r_n(\Theta) \end{align*} is not merely an existential statistical upper bound; it is a worst-case risk bound for the stated procedure. No further minimax argument is required for the algorithmic claim. [/step]