= Solution
Let $X$ have regular density $p_\theta$, let
$$
S_\theta(X)=\frac{\partial}{\partial\theta}\log p_\theta(X)
$$
be its <score function>, and let
$$
I(\theta)=\mathbb E_\theta[S_\theta(X)^2]
$$
be its <Fisher information>. Regularity gives the <mean-zero score identity> $\mathbb E_\theta S_\theta=0$.
Suppose an estimator $T$ has mean $\mathbb E_\theta T=\psi(\theta)$. Differentiating under the integral gives
$$
\psi'(\theta)
=\mathbb E_\theta[T S_\theta]
=\operatorname{Cov}_\theta(T,S_\theta).
$$
The <Cauchy-Schwarz inequality> therefore yields
$$
\psi'(\theta)^2
\le\operatorname{Var}_\theta(T)\,I(\theta),
$$
and hence the <Cramér-Rao bound>
$$
\boxed{\operatorname{Var}_\theta(T)
\ge\frac{\psi'(\theta)^2}{I(\theta)}}.
$$
In particular, an unbiased estimator of $\theta$ has variance at least $1/I(\theta)$. For an independent sample, the information is the sum of the individual informations.
In a decision problem with <risk of a decision rule> $R(\theta,\delta)$, a rule $\delta^*$ is <minimax decision rule>[minimax] when
$$
\boxed{
\sup_{\theta\in\Theta}R(\theta,\delta^*)
=\inf_\delta\sup_{\theta\in\Theta}R(\theta,\delta).
}
$$
Now let $X_1,\ldots,X_n$ be independent $N(\theta,1)$ variables with $\theta\ge0$, under <squared-error loss>. The <sample mean> is unbiased with variance $1/n$, so
$$
R(\theta,\overline X_n)
=\mathbb E_\theta(\overline X_n-\theta)^2
=\frac1n
$$
for every $\theta\ge0$. Thus the minimax value is at most $1/n$.
For the matching lower bound, choose a continuously differentiable density $\pi$ on $(0,1)$ that vanishes at both endpoints and has finite prior information
$$
J=\int_0^1\frac{\pi'(u)^2}{\pi(u)}\,du;
$$
for example, $\pi(u)=30u^2(1-u)^2$. For $L>0$, define the prior
$$
\pi_L(\theta)=\frac1L\pi(\theta/L),
\qquad 0<\theta<L.
$$
It is supported inside the parameter space, and its information is $J/L^2$.
Here the likelihood score is
$$
S_\theta(X)=\sum_{i=1}^n(X_i-\theta),
$$
whose Fisher information is $n$. For any decision rule $\delta(X)$, combine it with the prior score to form the joint score
$$
U=S_\theta(X)+\frac{\pi_L'(\theta)}{\pi_L(\theta)}.
$$
Integration by parts in $\theta$, with no boundary term because $\pi_L$ vanishes there, gives
$$
\mathbb E_{\pi_L}\bigl[(\delta(X)-\theta)U\bigr]=1.
$$
The likelihood score has conditional mean zero, so its cross term with the prior score vanishes and
$$
\mathbb E_{\pi_L}U^2=n+\frac{J}{L^2}.
$$
Cauchy--Schwarz now proves the <Van Trees inequality> in this case:
$$
\int_0^L R(\theta,\delta)\pi_L(\theta)\,d\theta
\ge\frac1{n+J/L^2}.
$$
The worst-case risk dominates every integrated risk, and therefore
$$
\sup_{\theta\ge0}R(\theta,\delta)
\ge\frac1{n+J/L^2}.
$$
Letting $L\to\infty$ shows that every rule has worst-case risk at least $1/n$. Since $\overline X_n$ attains that value, the result on the <minimax sample mean for a nonnegative normal location> is
$$
\boxed{\overline X_n\text{ is minimax on }[0,\infty)}.
$$
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