= Solution
Put $v_p(u)=(1,u,\ldots,u^p)^T$ and $w_i(x)=K((x_i-x)/h)$, where $p$ is a nonnegative integer and $h$ is the <smoothing bandwidth>. The <local polynomial estimator> is the intercept of the <weighted least squares> fit
$$
\widehat\beta(x)=\mathop{\operatorname{argmin}}_{\beta\in\mathbb R^{p+1}}
\sum_{i=1}^nw_i(x)\{Y_i-\beta^Tv_p((x_i-x)/h)\}^2,
\qquad \widehat m_h(x;p)=\widehat\beta_0(x).
$$
When the <local polynomial Gram matrix> $G_p(x)=\sum_iw_i(x)v_p((x_i-x)/h)v_p((x_i-x)/h)^T$ is invertible, the <normal equations> give
$$
\boxed{\widehat m_h(x;p)=e_0^TG_p(x)^{-1}\sum_{i=1}^nw_i(x)v_p((x_i-x)/h)Y_i.}
$$
A singular <local polynomial Gram matrix> requires a selection convention and may leave the intercept unidentified; an arbitrary nonnegative <regression kernel> alone does not guarantee uniqueness. At $p=0$, provided $\sum_iw_i(x)>0$, the <Nadaraya–Watson estimator> is
$$
\boxed{\widehat m_h(x)=\frac{\sum_iK((x_i-x)/h)Y_i}{\sum_iK((x_i-x)/h)}.}
$$
Multiplying every weight by $1/h$ changes neither fit.
For the <uniform kernel>, let $I_x=\{i:|i/n-x|\leq h\}$ and $N_x=|I_x|$. Then $\widehat m_h(x)=N_x^{-1}\sum_{i\in I_x}Y_i$. For $0<h<1/2$ and $x\in[h,1-h]$, the interval $[x-h,x+h]$ is inside $[0,1]$, and the regular design has
$$
N_x\geq\lfloor2nh\rfloor\geq2nh-1\geq nh\qquad(nh\geq1).
$$
The same lower bound holds when the interval starts at zero, despite the absence of the design point zero. In particular the <Nadaraya–Watson estimator> is defined throughout the integration interval.
For every <Lipschitz continuous> mean function in $\Theta_L$, the <bias of an estimator> and <variance> satisfy
$$
\left|\mathbb E_m\widehat m_h(x)-m(x)\right|
\leq\frac L{N_x}\sum_{i\in I_x}|i/n-x|\leq Lh,
\qquad \operatorname{Var}_m\widehat m_h(x)=\frac{\sigma^2}{N_x}\leq\frac{\sigma^2}{nh}.
$$
Here the <variance> uses the <independence> of the errors in the <fixed-design nonparametric regression> model. The <bias-variance decomposition of mean squared error> and <Tonelli theorem> give a slightly stronger <integrated mean squared error> bound than required:
$$
\begin{aligned}
\sup_{m\in\Theta_L}\mathbb E_m\int_h^{1-h}\{\widehat m_h(x)-m(x)\}^2\,dx
&\leq(1-2h)\left(L^2h^2+\frac{\sigma^2}{nh}\right)\\
&\leq L^2h^2+\frac{3\sigma^2}{2nh}.
\end{aligned}
$$
Thus
$$
\boxed{\sup_{m\in\Theta_L}\mathbb E_m\int_h^{1-h}\{\widehat m_h(x)-m(x)\}^2\,dx\leq L^2h^2+\frac{3\sigma^2}{2nh}.}
$$
The printed integration interval represents a nonnegative <risk function> for $h\leq1/2$. For $1/2<h\leq1$, the written integral has reversed limits and is nonpositive, making the displayed upper bound trivial; it should not be interpreted as an <integrated mean squared error> in that range. For $h>1$, evaluation outside $[0,1]$ would additionally require a specified extension of the mean function. The subsequent optimization restricts to $h<1/3$.
Minimizing the stated right side gives
$$
h_* =\left(\frac{3\sigma^2}{4nL^2}\right)^{1/3},\qquad
L^2h_*^2+\frac{3\sigma^2}{2nh_*}
=3\left(\frac34\right)^{2/3}\left(\frac{L\sigma^2}{n}\right)^{2/3}.
$$
For sufficiently large $n$, $nh_*\geq1$ and $h_*<1/3$. For example take $n_0=1+\lceil\max\{1,2L/(\sqrt3\sigma),81\sigma^2/(4L^2)\}\rceil$. Therefore the upper bound holds with
$$
\boxed{C=3\left(\frac34\right)^{2/3}.}
$$
For the lower bound, the PDF has the <mean squared error> $\mathbb E_m[(\widetilde m(x_0)-m(x_0))^2]$; the TeX erroneously moves the square outside the <expected value>. The PDF's suggested <smooth bump function> is $\exp(-1/(1-t^2))\mathbf1_{\{|t|\leq1\}}$, with value zero at the endpoints, rather than the corrupted TeX formula. A triangular alternative suffices because only <Lipschitz continuity> is needed.
The <Le Cam two-point lemma> says that for two data laws $P_0,P_1$ with parameter separation $\Delta=|\theta_1-\theta_0|$, every estimator $T$ satisfies
$$
\boxed{\max_{j=0,1}\mathbb E_j(T-\theta_j)^2\geq\frac{\Delta^2}{8}\{1-\|P_0-P_1\|_{\mathrm{TV}}\}.}
$$
Indeed, classify according to the nearer parameter value. Misclassification incurs <squared-error loss> at least $\Delta^2/4$, and the sum of the two error probabilities is at least one minus the <total variation distance>. Averaging the two <risk functions> proves the stated form.
Let $d=\min(x_0,1-x_0)>0$, $m_0=0$, and use the <two-point lower bound with a triangular bump>
$$
m_1(x)=L(h-|x-x_0|)_+,
\qquad h=\left(\frac{\sigma^2}{8nL^2}\right)^{1/3}.
$$
Both functions are in $\Theta_L$, and $\Delta=Lh$. Assume first that $h\leq d$ and $nh\geq1$. At most $2nh+1$ design points meet the bump's support. The <Kullback-Leibler divergence between normal distributions> and <chain rule for relative entropy> give
$$
D_{\mathrm{KL}}(P_1\Vert P_0)
=\frac1{2\sigma^2}\sum_{i=1}^nm_1(i/n)^2
\leq\frac{L^2h^2(2nh+1)}{2\sigma^2}
\leq\frac{3nL^2h^3}{2\sigma^2}=\frac3{16}.
$$
The <Pinsker's inequality> therefore gives $\|P_1-P_0\|_{\mathrm{TV}}\leq\sqrt{3/32}<1/2$. The <Le Cam two-point lemma> yields the <pointwise minimax rate for Lipschitz regression>
$$
\boxed{\inf_{\widetilde m}\sup_{m\in\Theta_L}\mathbb E_m[(\widetilde m(x_0)-m(x_0))^2]
\geq\frac1{64}\left(\frac{L\sigma^2}{n}\right)^{2/3}.}
$$
This universal constant is valid for $n\geq N:=\lceil\max\{1,\sigma^2/(8L^2d^3),2\sqrt2L/\sigma\}\rceil$.
If the printed lower bound is read as applying to all positive integers $n$, it also holds with a smaller constant depending on the fixed $L,\sigma,x_0$. For $n<N$, use the constant mean functions $m_0=0$ and $m_1=\sigma/\sqrt n$. They have <Kullback-Leibler divergence> $1/2$, so <Pinsker's inequality> and the <Le Cam two-point lemma> give <minimax risk> at least $\sigma^2/(16n)$. Combining the two ranges gives the explicit all-$n$ choice
$$
\boxed{c=\min\left\{\frac1{64},\ \frac1{16}\left(\frac{\sigma^2}{NL^2}\right)^{1/3}\right\}>0.}
$$
The universal asymptotic constant and this parameter-dependent all-$n$ constant are distinct conventions; the latter supplies the assertion without inserting an unstated large-$n$ restriction.
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