= Solution
For <fixed-design nonparametric regression>, let $x_i=i/n$ and $K_h(u)=h^{-1}K(u/h)$. The degree-$p$ <local polynomial estimator> takes the fitted intercept $\widehat m_h(x;p)=\widehat\beta_0$, where
$$
(\widehat\beta_0,\ldots,\widehat\beta_p)
\in\operatorname*{argmin}_{\beta\in\mathbb R^{p+1}}
\sum_{i=1}^n K_h(x_i-x)\left[Y_i-\sum_{r=0}^p\beta_r(x_i-x)^r\right]^2.
$$
Use a nonnegative <regression kernel> and an invertible <local polynomial Gram matrix> to obtain a unique fit. For $p=0$, solving the single <weighted least squares> <normal equation> gives the <local constant estimator>:
$$
\boxed{\widehat m_h(x;0)=\frac{\sum_i K_h(x_i-x)Y_i}{\sum_i K_h(x_i-x)}.}
$$
At the interior point $x_0$, use the standard unit-integral <regression kernel> convention $\mu_0(K)=1$. Otherwise the leading <variance> below is $R(K)/(nh\mu_0(K)^2)$; unlike a <kernel density estimator>, the <local constant estimator> itself is unchanged by rescaling $K$.
The <local constant estimator> is a <linear estimator in nonparametric regression> with weights $W_i=K_h(x_i-x_0)/(ns_{0,h})$. The independent unit-variance errors imply
$$
\operatorname{Var}\{\widehat m_h(x_0;0)\}
=\sum_i W_i^2=\frac{t_{0,h}}{n s_{0,h}^2}
=\frac{R(K)}{nh}+o((nh)^{-1}).
$$
To control the <bias of an estimator>, <differentiability> at this one point is enough. Write
$$
m(x_0+u)-m(x_0)=m'(x_0)u+r(u),\qquad
\eta(h):=\sup_{0<|u|\leq h}\frac{|r(u)|}{|u|}\longrightarrow0.
$$
Symmetry makes $\mu_1(K)=0$, and the supplied moment approximation gives $s_{1,h}=o(h)$. Therefore the linear contribution to the <bias of an estimator> is $m'(x_0)s_{1,h}/s_{0,h}=o(h)$. The remaining contribution has absolute value at most $\eta(h)h$, because only $|x_i-x_0|\leq h$ receive positive weight. This is the <interior first-order bias cancellation for local constant regression>, and gives squared <bias of an estimator> $o(h^2)$. Combining with the <variance> proves
$$
\boxed{\operatorname{MSE}\{\widehat m_h(x_0;0)\}
=\frac{R(K)}{nh}+o\left(h^2+\frac1{nh}\right).}
$$
For fixed $\varepsilon>0$, take $h_\varepsilon=\varepsilon^{-1/3}n^{-1/3}$. It satisfies both asymptotic <smoothing bandwidth> conditions, and the displayed remainder, multiplied by $n^{2/3}$, tends to zero for this fixed $\varepsilon$. Thus
$$
0\leq\limsup_{n\to\infty}n^{2/3}\inf_{h>0}\operatorname{MSE}\{\widehat m_h(x_0;0)\}
\leq R(K)\varepsilon^{1/3}.
$$
Letting $\varepsilon\downarrow0$ proves \b[the optimized scaled pointwise error tends to zero]. The order of these limits matters: no remainder uniform in $\varepsilon$ has been assumed.
For the lower bound over the whole function class, the absence of a uniform bound on derivatives permits a particularly direct argument. It also avoids assuming a <normal distribution> for errors when the model only supplies their first two moments. Fix $n$ and $x_0$. Choose a <smooth bump function> $b$ on the interval, with $b(x_0)=1$, whose support contains no design point except possibly $x_0$. Such a bump exists because the design is finite; at an endpoint use the restriction of a smooth bump on the real line. Every $m_a=ab$, $a\in\mathbb R$, belongs to the <differentiability> class.
If $x_0$ is not a design point, all observation means are zero for every $a$. The <Le Cam two-point lemma> applied to $a=\pm A$ has <total variation distance> zero and gives a lower bound tending to infinity with $A$. In fact, the <supremum> of the <mean squared error> is infinite at every such $x_0$.
If $x_0$ is a design point, only its response depends on $a$. The resulting <location family> is the single observation $Y=a+\epsilon$; all remaining observations are independent of $a$ and give no extra information. The <single-observation location minimax bound> is
$$
\inf_T\sup_{a\in\mathbb R}\mathbb E\{T(a+\epsilon)-a\}^2=\operatorname{Var}(\epsilon)=1.
$$
Here is a proof covering discrete as well as continuous errors. Give $a$ the <prior distribution> $U[-A,A]$, independently of the noise. Put $p_M=\mathbb P(|\epsilon|\leq M)$ and $v_M=\operatorname{Var}(\epsilon\mid |\epsilon|\leq M)$. Consider an easier experiment in which an oracle also reveals whether $|\epsilon|\leq M$, and reveals $a$ itself on the complementary event. On the event $|\epsilon|\leq M$ and $|Y|\leq A-M$, every possible truncated noise value places $a=Y-\epsilon$ within $[-A,A]$. The flat <prior distribution> then makes the <Bayesian posterior> noise law exactly its truncated original law. The smallest conditional <squared-error loss> is its <conditional variance> $v_M$, attained by the <posterior mean>.
The probability of that event is at least $p_M(1-2M/A)$ for $A>2M$, by restricting further to $|a|\leq A-2M$. Thus every rule's <supremum> <risk function> is at least the original <Bayes risk>, which is at least the easier experiment's <Bayes risk>, and therefore at least $p_M(1-2M/A)v_M$. Let $A\to\infty$ and then $M\to\infty$. Finite second moment and zero mean give $p_Mv_M\to1$. The rule $T(Y)=Y$ has constant <mean squared error> one, proving the equality.
Consequently, at every $x_0\in[0,1]$, the unrestricted <differentiability> class satisfies the stronger conclusion
$$
\boxed{\sup_{m\in\mathcal F}\mathbb E\{\widetilde m(x_0)-m(x_0)\}^2\geq1\geq n^{-2/3}.}
$$
Thus $c=1$ works for every $n\geq1$. There is no conflict with the preceding fixed-function limit: the worst functions, including arbitrarily narrow bumps, can vary with $n$. This is the <pointwise versus uniform risk distinction>. A smoothness ball with a common derivative bound would pose a different <minimax risk> problem.
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