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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 ${\hat{m}}_h(x;p)={\hat{\beta }}_0$, whereUse 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: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 $W_i=K_h(x_i-x_0)/(n{s}_{0,h})$. The independent unit-variance errors implyTo control the bias of an estimator, differentiability at this one point is enough. WriteSymmetry 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 $m^{\prime }(x_0)s_{1,h}/s_{0,h}=o(h)$. The remaining contribution has absolute value at most $\eta (h)h$, because only $|x_i-x_0|\le h$ 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 provesFor fixed $\epsilon >0$, take $h_{\epsilon }={\epsilon }^{-1/3}{n}^{-1/3}$. It satisfies both asymptotic smoothing bandwidth conditions, and the displayed remainder, multiplied by $n^{2/3}$, tends to zero for this fixed $\epsilon$. ThusLetting $\epsilon \downarrow 0$ proves the optimized scaled pointwise error tends to zero. The order of these limits matters: no remainder uniform in $\epsilon$ 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 $Y=a+ϵ$; all remaining observations are independent of $a$ and give no extra information. The single-observation location minimax bound isHere is a proof covering discrete as well as continuous errors. Give $a$ the prior distribution $U[-A,A]$, independently of the noise. Put $p_M=\mathbb{P}(|ϵ|\le M)$ and $v_M=\mathrm{Var}(ϵ\mid |ϵ|\le M)$. Consider an easier experiment in which an oracle also reveals whether $|ϵ|\le M$, and reveals $a$ itself on the complementary event. On the event $|ϵ|\le M$ and $|Y|\le A-M$, every possible truncated noise value places $a=Y-ϵ$ 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 $|a|\le A-2M$. 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 $p_M{v}_M\to 1$. 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 conclusionThus $c=1$ works for every $n\ge 1$. 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.

At a target $x$, the degree-$p$ local polynomial regression estimate minimizesand ${\hat{m}}_n(x;p,h,K)={\hat{b}}_0$. Let $R_x$ have row
$(1,x_i-x,\dots ,(x_i-x{)}^p)$ and let $W_x$ be diagonal with the kernel weights. Assume the local polynomial Gram matrix $R_x^T{W}_x{R}_x$ is positive definite. The weighted normal equations giveso it is a linear estimator in nonparametric regression.

DefineInverting the two-by-two Gram matrix for $p=1$ yields

Now put $M=\min \{n,\lfloor nh\rfloor \}$. At $x=0$ and for the uniform kernel,For the usual bandwidth range $0<h\le 1$, $M=\lfloor nh\rfloor$. The elementary power-sum formulas and $|M-nh|\le 1$ show, for $r=0,1,2,3$, thatThe limiting moment determinant isFor $nh\ge 32$, the displayed errors can consume at most a fixed fraction of this value, sofor a universal $c>0$ and $0<h\le 1$.

As printed, the claim “for all $h>0$” cannot hold for this design: if $h>1$, all $n$ points receive weight $1/2$, each $s_r=O(h^{-1})$, and the determinant is $O(h^{-2})$, not bounded below by a positive multiple of $h^2$. The standard bandwidth restriction $h\le 1$ is therefore necessary for that intermediate assertion. The final bias bound remains harmless for $h>1$, since $h^2\ge 1$ and the finite design response means are uniformly bounded.

The polynomial reproduction property of local polynomial regression makes the local-linear weights reproduce both $1$ and $u$. Writefor $0\le x_i\le \min (h,1)$. The reproduced terms have zero bias. Using $s_r=O(h^r)$, the numerator contributed by the remainders is bounded byDivision by the determinant lower bound givesThe degree-zero Nadaraya–Watson estimator reproduces constants but not linear functions. The term $x_i$ therefore remains, and its boundary bias is bounded by $Ch$; this first-order boundary bias is the improvement that local linear fitting removes.

PutThe degree-$p$ local polynomial estimator is ${\hat{m}}_n(x)={\hat{b}}_0$, whereDefine the local polynomial Gram matrixAssume that it is positive definite, and writeThe normal equations then giveThus the estimator is a linear estimator in nonparametric regression, and the displayed $w_{p,i}$ are its effective kernel weights.

If $R$ is a polynomial of degree at most $p$, then $R(x+hu)=c^T{r}_p(u)$ for some $c$. Feeding $Y_i=R(x_i)$ into the weighted least-squares problem gives the exact zero-residual fit $c$, whose intercept is $R(x)$. Hence the polynomial reproduction property of local polynomial regression is

Let ${\beta }_0=\lceil \beta \rceil -1$. The Hölder class $\mathcal{H}(\beta ,L)$ consists of functions with derivatives through order ${\beta }_0$ bounded by $L$ andThe subclass $\tilde{\mathcal{H}}(\beta ,L)$ additionally makes every derivative of order below ${\beta }_0$ $L$-Lipschitz.

Suppose $p<{\beta }_0$. Taylor's theorem and that additional Lipschitz condition give, for the degree-$p$ Taylor polynomial $T_p$ at $x$,On the support of $K_i$, $|u_i|\le 1$, soFor the regular design $x_i=i/n$, at most $2nh+1\le 4nh$ points satisfy $|x_i-x|\le h$ when $h\ge 1/(2n)$. Polynomial reproduction cancels $T_p$, and thereforeThus the universal exponent in the question is $\gamma =1$, with the displayed choice of $C({\lambda }_0,p,L,K)$.

A linear estimator in nonparametric regression at $x_0$ has the form ${\hat{m}}_n(x_0)={\sum }_i{W}_i(x_0)Y_i$, where the weights may depend on the design, $x_0$, $p$, $h$, and $K$, but not on the responses. Putand define the local polynomial Gram matrixWhen ${\lambda }_0>0$, the weighted least squares normal equations have the unique solutionTherefore ${\hat{m}}_n(x_0)={\sum }_i{W}_i{Y}_i$, where the effective kernel weight is

The identityshows that these weights exactly reproduce at $x_0$ every multivariate polynomial of total degree at most $p$. This is the required polynomial reproduction property of local polynomial regression.

Only grid points with $\Vert x_i-x_0{\Vert }_{\infty }\le h$ have nonzero weight. There are at most $(2n_{1}h+1{)}^d\le (4n_{1}h{)}^d=4^{d}n{h}^d$ such points because $n_{1}h\ge 1/2$. On this support,so the operator norm bound $\Vert B_n^{-1}{\Vert }_{\mathrm{op}}={\lambda }_0^{-1}$ givesBecause the errors are independent random variables with the stated variance bounds,Thus ${\gamma }_1=d$.

Let $T_{x_0}$ be the Multivariate Taylor polynomial of $m$ at $x_0$ through total degree ${\beta }_0$. The assumed Hölder continuity of the derivatives andgive the Taylor remainder boundPolynomial reproduction cancels $T_{x_0}$ in the bias. The same support count and weight bound giveConsequentlyso ${\gamma }_2=\beta$.