[proofplan] We first write the expectation of the kernel estimator as a rescaled convolution and change variables so that the bias becomes an integral of $f(x-hu)-f(x)$ against $K(u)$. On the compact interior set, Taylor expansion through degree $m-1$ is valid uniformly for all translated points $x-hu$ in the support of $K$; the kernel moment conditions cancel every polynomial term of degree $1$ through $m-1$, leaving a uniformly bounded order-$m$ remainder. For the variance, we compute the second moment of one summand, change variables, and use compact support of $K$ together with boundedness and continuity of $f$ near $A$. The lower variance bound at a point with $f(x)>0$ follows by bounding the leading second moment from below by a positive multiple of $h^{-d}\int K^2\,d\mathcal L^d$ for small $h$, while the squared mean term stays bounded. [/proofplan]

[step:Choose a neighbourhood on which all translated kernel evaluations remain inside the differentiability region] Let $S:=\operatorname{supp}K\subset\mathbb R^d$. Since $S$ is compact, define \begin{align*} R:=\sup_{u\in S}|u|<\infty. \end{align*} Let $U\subset\mathbb R^d$ denote the neighbourhood of $A$ on which the stated derivatives of $f$ exist. Since $A$ is compact and $U$ is open with $A\subset U$, choose $\rho>0$ such that \begin{align*} A_\rho:=\{y\in\mathbb R^d:\operatorname{dist}(y,A)<\rho\}\subset U. \end{align*} If $R=0$, any $h_0>0$ small enough may be chosen. If $R>0$, choose \begin{align*} h_0:=\frac{\rho}{2R}. \end{align*} Then for every $x\in A$, every $u\in S$, and every $0<h<h_0$, \begin{align*} |x-hu-x|\le hR<\rho, \end{align*} so $x-hu\in A_\rho\subset U$. Therefore all derivatives of $f$ used below are evaluated inside $U$. [/step]

[step:Rewrite the expectation as an integral against translated values of $f$] Fix $x\in A$ and $0<h<h_0$. Let $X_1,\dots,X_n:(\Omega,\mathcal F,\mathbb P)\to(\mathbb R^d,\mathcal B(\mathbb R^d))$ be the i.i.d. random vectors from the theorem statement, each with density $f$ with respect to $\mathcal L^d$. The kernel density estimator at bandwidth $h$ is the map $\hat f_{n,h}:\mathbb R^d\times\Omega\to\mathbb R$ defined by \begin{align*} \hat f_{n,h}(x,\omega):=\frac{1}{nh^d}\sum_{i=1}^n K\left(\frac{x-X_i(\omega)}{h}\right). \end{align*} For each $i\in\{1,\dots,n\}$, define the real-valued [random variable](/page/Random%20Variable) $Z_{i,h,x}:\Omega\to\mathbb R$ by \begin{align*} Z_{i,h,x}(\omega):=h^{-d}K\left(\frac{x-X_i(\omega)}{h}\right) \end{align*} for $\omega\in\Omega$. Suppressing the sample variable in the standard way, this gives \begin{align*} \hat f_{n,h}(x)=\frac{1}{n}\sum_{i=1}^n Z_{i,h,x}. \end{align*} Since $X_i$ has density $f$ with respect to $\mathcal L^d$, \begin{align*} \mathbb E[Z_{i,h,x}] &=h^{-d}\int_{\mathbb R^d}K\left(\frac{x-y}{h}\right)f(y)\,d\mathcal L^d(y). \end{align*} Use the substitution $u=(x-y)/h$, equivalently $y=x-hu$. The [Lebesgue measure](/page/Lebesgue%20Measure) transforms as $d\mathcal L^d(y)=h^d\,d\mathcal L^d(u)$, and the domain $\mathbb R^d$ maps onto $\mathbb R^d$. Thus \begin{align*} \mathbb E[\hat f_{n,h}(x)] =\mathbb E[Z_{1,h,x}] =\int_{\mathbb R^d}K(u)f(x-hu)\,d\mathcal L^d(u). \end{align*} Using $\int_{\mathbb R^d}K(u)\,d\mathcal L^d(u)=1$, the bias is \begin{align*} \mathbb E[\hat f_{n,h}(x)]-f(x) =\int_{\mathbb R^d}K(u)\bigl(f(x-hu)-f(x)\bigr)\,d\mathcal L^d(u). \end{align*} [/step]

[step:Cancel the Taylor polynomial by the integer moment conditions]Let $\mathbb N_0:=\{0,1,2,\dots\}$ denote the set of non-negative integers. For a multi-index $\alpha=(\alpha_1,\dots,\alpha_d)\in\mathbb N_0^d$, define $\alpha!:=\alpha_1!\cdots\alpha_d!$ and $u^\alpha:=u_1^{\alpha_1}\cdots u_d^{\alpha_d}$ for $u=(u_1,\dots,u_d)\in\mathbb R^d$. For a function $\varphi:U\to\mathbb R$ with continuous partial derivatives through order $|\alpha|$, define \begin{align*} D^\alpha \varphi:=\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}\varphi. \end{align*} Let $\mathcal L^1$ denote one-dimensional Lebesgue measure. For $x\in A$, $u\in S$, and $0<h<h_0$, the line segment $\{x-thu:0\le t\le1\}$ lies in $A_\rho\subset U$. Define the one-dimensional function $g_{x,u,h}:[0,1]\to\mathbb R$ by $g_{x,u,h}(t)=f(x-thu)$. Since $f$ has continuous partial derivatives through order $m$ on $U$ and the segment lies in $U$, the chain rule gives $g_{x,u,h}\in C^m([0,1])$. For each $k\in\{1,\dots,m\}$, repeated application of the chain rule gives \begin{align*} g_{x,u,h}^{(k)}(0)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}(-h)^k u^\alpha D^\alpha f(x). \end{align*} We apply the Taylor theorem with integral remainder to $g_{x,u,h}$ at $0$; its hypothesis $g_{x,u,h}\in C^m([0,1])$ has just been verified. This gives \begin{align*} f(x-hu)-f(x)=\sum_{1\le|\alpha|<m}\frac{(-h)^{|\alpha|}}{\alpha!}u^\alpha D^\alpha f(x)+R_m(x,u,h). \end{align*} Here the remainder is \begin{align*} R_m(x,u,h)=\sum_{|\alpha|=m}\frac{(-h)^m u^\alpha}{\alpha!}\int_0^1 m(1-t)^{m-1}D^\alpha f(x-thu)\,d\mathcal L^1(t). \end{align*} Taking absolute values and using $\int_0^1 m(1-t)^{m-1}\,d\mathcal L^1(t)=1$ gives \begin{align*} |R_m(x,u,h)|\le h^m |u|^m\sum_{|\alpha|=m}\frac{1}{\alpha!}\sup_{y\in A_\rho}|D^\alpha f(y)|. \end{align*} Define the finite constant \begin{align*} M_m:= \sum_{|\alpha|=m}\frac{1}{\alpha!} \sup_{y\in A_\rho}|D^\alpha f(y)|. \end{align*} The boundedness of each $D^\alpha f$ on $U$ gives $M_m<\infty$. Substituting the Taylor expansion into the bias identity gives the sum of the cancelled polynomial contribution and the remainder contribution. The polynomial contribution is \begin{align*} \sum_{1\le|\alpha|<m}\frac{(-h)^{|\alpha|}}{\alpha!}D^\alpha f(x)\int_{\mathbb R^d}u^\alpha K(u)\,d\mathcal L^d(u). \end{align*} The remainder contribution is \begin{align*} \int_{\mathbb R^d}K(u)R_m(x,u,h)\,d\mathcal L^d(u). \end{align*} Every polynomial term in the displayed sum vanishes by the order-$m$ moment assumptions on $K$. Hence, by the triangle inequality, \begin{align*} \left|\mathbb E[\hat f_{n,h}(x)]-f(x)\right|\le \int_{\mathbb R^d}|K(u)|\,|R_m(x,u,h)|\,d\mathcal L^d(u). \end{align*} Using the remainder estimate gives \begin{align*} \left|\mathbb E[\hat f_{n,h}(x)]-f(x)\right|\le h^m M_m\int_{\mathbb R^d}|u|^m|K(u)|\,d\mathcal L^d(u). \end{align*} Define \begin{align*} C_{\mathrm b}:= M_m\int_{\mathbb R^d}|u|^m|K(u)|\,d\mathcal L^d(u). \end{align*} The finite absolute moment assumption gives $C_{\mathrm b}<\infty$, and therefore \begin{align*} \left|\mathbb E[\hat f_{n,h}(x)]-f(x)\right|\le C_{\mathrm b}h^m \end{align*} for every $x\in A$ and $0<h<h_0$.[/step]

[guided]The goal is to isolate the part of $f(x-hu)-f(x)$ that the kernel moments are designed to kill. Let $\mathbb N_0:=\{0,1,2,\dots\}$ denote the set of non-negative integers. For each multi-index $\alpha=(\alpha_1,\dots,\alpha_d)\in\mathbb N_0^d$, define $\alpha!:=\alpha_1!\cdots\alpha_d!$ and $u^\alpha:=u_1^{\alpha_1}\cdots u_d^{\alpha_d}$ for $u=(u_1,\dots,u_d)\in\mathbb R^d$. For a function $\varphi:U\to\mathbb R$ with continuous partial derivatives through order $|\alpha|$, define \begin{align*} D^\alpha \varphi:=\partial_{x_1}^{\alpha_1}\cdots\partial_{x_d}^{\alpha_d}\varphi. \end{align*} Let $\mathcal L^1$ denote one-dimensional Lebesgue measure. Because $u$ lies in $S=\operatorname{supp}K$ and $0<h<h_0$, the point $x-hu$ lies in $A_\rho\subset U$. Thus the whole line segment $\{x-thu:0\le t\le1\}$ is contained in $A_\rho\subset U$, so all derivatives that appear are evaluated in $U$. For $x\in A$, $u\in S$, and $0<h<h_0$, define the one-dimensional function $g_{x,u,h}:[0,1]\to\mathbb R$ by $g_{x,u,h}(t)=f(x-thu)$. Since $f$ has continuous partial derivatives up to order $m$ on $U$ and the segment lies in $U$, the chain rule gives $g_{x,u,h}\in C^m([0,1])$. For each $k\in\{1,\dots,m\}$, repeated application of the chain rule gives \begin{align*} g_{x,u,h}^{(k)}(0)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}(-h)^k u^\alpha D^\alpha f(x). \end{align*} We apply the Taylor theorem with integral remainder through degree $m-1$ to $g_{x,u,h}$ at $0$; the required regularity hypothesis is exactly $g_{x,u,h}\in C^m([0,1])$, verified above. This gives \begin{align*} f(x-hu)-f(x)=\sum_{1\le|\alpha|<m}\frac{(-h)^{|\alpha|}}{\alpha!}u^\alpha D^\alpha f(x)+R_m(x,u,h). \end{align*} The integral remainder is \begin{align*} R_m(x,u,h)=\sum_{|\alpha|=m}\frac{(-h)^m u^\alpha}{\alpha!}\int_0^1 m(1-t)^{m-1}D^\alpha f(x-thu)\,d\mathcal L^1(t). \end{align*} Taking absolute values and using $\int_0^1 m(1-t)^{m-1}\,d\mathcal L^1(t)=1$ gives \begin{align*} |R_m(x,u,h)|\le h^m |u|^m\sum_{|\alpha|=m}\frac{1}{\alpha!}\sup_{y\in A_\rho}|D^\alpha f(y)|. \end{align*} The important point is that this bound is uniform in $x\in A$, because the same neighbourhood $A_\rho$ works for all $x\in A$ and all $u\in S$. Define \begin{align*} M_m:= \sum_{|\alpha|=m}\frac{1}{\alpha!} \sup_{y\in A_\rho}|D^\alpha f(y)|. \end{align*} Since the order-$m$ derivatives are bounded on $U$, this constant is finite. Substituting the Taylor expansion into the bias formula gives the sum of two contributions. The polynomial contribution is \begin{align*} \sum_{1\le|\alpha|<m}\frac{(-h)^{|\alpha|}}{\alpha!}D^\alpha f(x)\int_{\mathbb R^d}u^\alpha K(u)\,d\mathcal L^d(u). \end{align*} The remainder contribution is \begin{align*} \int_{\mathbb R^d}K(u)R_m(x,u,h)\,d\mathcal L^d(u). \end{align*} Now the reason for the order-$m$ kernel assumption appears: for every multi-index $\alpha$ with $1\le|\alpha|<m$, \begin{align*} \int_{\mathbb R^d}u^\alpha K(u)\,d\mathcal L^d(u)=0. \end{align*} Therefore every Taylor polynomial term of degree $1$ through $m-1$ cancels after integration against $K$. Only the remainder remains: \begin{align*} \mathbb E[\hat f_{n,h}(x)]-f(x) = \int_{\mathbb R^d}K(u)R_m(x,u,h)\,d\mathcal L^d(u). \end{align*} Taking absolute values and using the triangle inequality gives \begin{align*} \left|\mathbb E[\hat f_{n,h}(x)]-f(x)\right|\le \int_{\mathbb R^d}|K(u)|\,|R_m(x,u,h)|\,d\mathcal L^d(u). \end{align*} Using the remainder estimate gives \begin{align*} \left|\mathbb E[\hat f_{n,h}(x)]-f(x)\right|\le h^m M_m\int_{\mathbb R^d}|u|^m|K(u)|\,d\mathcal L^d(u). \end{align*} Finally define \begin{align*} C_{\mathrm b}:= M_m\int_{\mathbb R^d}|u|^m|K(u)|\,d\mathcal L^d(u). \end{align*} The assumed finite absolute moment of order $m$ makes $C_{\mathrm b}$ finite. Hence \begin{align*} \left|\mathbb E[\hat f_{n,h}(x)]-f(x)\right|\le C_{\mathrm b}h^m, \end{align*} uniformly for $x\in A$ and $0<h<h_0$.[/guided]

[step:Bound the variance from above by the rescaled second moment] Since $Z_{1,h,x},\dots,Z_{n,h,x}$ are i.i.d., \begin{align*} \operatorname{Var}(\hat f_{n,h}(x))=\frac{1}{n}\operatorname{Var}(Z_{1,h,x})\le\frac{1}{n}\mathbb E[Z_{1,h,x}^2]. \end{align*} Using the density of $X_1$, \begin{align*} \mathbb E[Z_{1,h,x}^2] &= h^{-2d}\int_{\mathbb R^d}K\left(\frac{x-y}{h}\right)^2f(y)\,d\mathcal L^d(y). \end{align*} With the substitution $u=(x-y)/h$, so that $y=x-hu$ and $d\mathcal L^d(y)=h^d\,d\mathcal L^d(u)$, \begin{align*} \mathbb E[Z_{1,h,x}^2] = h^{-d}\int_{\mathbb R^d}K(u)^2f(x-hu)\,d\mathcal L^d(u). \end{align*} Define \begin{align*} M_0:=\sup_{y\in A_\rho}|f(y)|. \end{align*} Since $f$ is bounded on $U$, $M_0<\infty$. For $u\notin S$, $K(u)=0$, and for $u\in S$ the point $x-hu$ lies in $A_\rho$. Therefore \begin{align*} \mathbb E[Z_{1,h,x}^2] &\le h^{-d}M_0\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u). \end{align*} Define \begin{align*} C_{\mathrm v}:= M_0\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u). \end{align*} Then $C_{\mathrm v}<\infty$, and \begin{align*} \operatorname{Var}(\hat f_{n,h}(x)) \le \frac{C_{\mathrm v}}{nh^d} \end{align*} for every $x\in A$ and $0<h<h_0$. [/step]

[step:Obtain the matching lower variance bound at positive density points]Fix $x\in A$ with $f(x)>0$, and suppose \begin{align*} \int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u)>0. \end{align*} Since $f$ is continuous at $x$ and $f(x)>0$, choose $\delta_x>0$ such that $|f(y)-f(x)|<f(x)/2$ whenever $|y-x|<\delta_x$. Since $\sup_{u\in S}|hu|\le hR\to0$ as $h\downarrow0$, choose $h_1\in(0,h_0)$ such that $hR<\delta_x$ for every $0<h<h_1$. Then for every $u\in S$ and every $0<h<h_1$, \begin{align*} f(x-hu)\ge \frac{f(x)}{2}. \end{align*} Thus \begin{align*} \mathbb E[Z_{1,h,x}^2]=h^{-d}\int_{\mathbb R^d}K(u)^2f(x-hu)\,d\mathcal L^d(u). \end{align*} Using $f(x-hu)\ge f(x)/2$ on $S$ and $K(u)=0$ off $S$, we get \begin{align*} \mathbb E[Z_{1,h,x}^2]\ge h^{-d}\frac{f(x)}{2}\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u). \end{align*} Define \begin{align*} a_x:=\frac{f(x)}{2}\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u)>0. \end{align*} Also define \begin{align*} B_x:= \sup_{0<h<h_1}\left|\mathbb E[Z_{1,h,x}]\right|. \end{align*} This quantity is finite because \begin{align*} \left|\mathbb E[Z_{1,h,x}]\right|\le\int_{\mathbb R^d}|K(u)|\,f(x-hu)\,d\mathcal L^d(u). \end{align*} Since $x-hu\in A_\rho$ on $S$ and $K(u)=0$ off $S$, \begin{align*} \left|\mathbb E[Z_{1,h,x}]\right|\le M_0\int_{\mathbb R^d}|K(u)|\,d\mathcal L^d(u). \end{align*} Therefore \begin{align*} \operatorname{Var}(Z_{1,h,x})=\mathbb E[Z_{1,h,x}^2]-\left(\mathbb E[Z_{1,h,x}]\right)^2. \end{align*} Combining the lower second-moment bound with the definition of $B_x$ gives \begin{align*} \operatorname{Var}(Z_{1,h,x})\ge a_xh^{-d}-B_x^2. \end{align*} Choose $h_x\in(0,h_1)$ such that $B_x^2\le \frac{a_x}{2}h^{-d}$ whenever $0<h<h_x$. Then \begin{align*} \operatorname{Var}(Z_{1,h,x})\ge \frac{a_x}{2}h^{-d}. \end{align*} Since $\operatorname{Var}(\hat f_{n,h}(x))=n^{-1}\operatorname{Var}(Z_{1,h,x})$, we obtain \begin{align*} \operatorname{Var}(\hat f_{n,h}(x)) \ge \frac{a_x}{2nh^d}. \end{align*} Define the pointwise constants \begin{align*} c_x:=\frac{a_x}{2},\qquad C_x:=C_{\mathrm v}. \end{align*} Both constants are positive, and the already proved upper bound gives \begin{align*} \frac{c_x}{nh^d}\le\operatorname{Var}(\hat f_{n,h}(x))\le\frac{C_x}{nh^d} \end{align*} for every $0<h<h_x$.[/step]

[guided]This step turns the second-moment asymptotic into a genuine variance lower bound. Fix $x\in A$ with $f(x)>0$ and assume \begin{align*} \int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u)>0. \end{align*} For the fixed bandwidth $h>0$, define the real-valued random variable $Z_{1,h,x}:\Omega\to\mathbb R$ by \begin{align*} Z_{1,h,x}(\omega):=h^{-d}K\left(\frac{x-X_1(\omega)}{h}\right) \end{align*} for $\omega\in\Omega$. Also define \begin{align*} M_0:=\sup_{y\in A_\rho}|f(y)|. \end{align*} Since $f$ is bounded on $U$ and $A_\rho\subset U$, this constant is finite. By continuity of $f$ at $x$, choose $\delta_x>0$ such that $|f(y)-f(x)|<f(x)/2$ whenever $|y-x|<\delta_x$. Because $S=\operatorname{supp}K$ is bounded and $\sup_{u\in S}|hu|\le hR\to0$, choose $h_1\in(0,h_0)$ such that $hR<\delta_x$ for $0<h<h_1$. Then $f(x-hu)\ge f(x)/2$ for every $u\in S$ and every $0<h<h_1$. We compute the second moment directly in this guided argument. Since $X_1$ has density $f$ with respect to $\mathcal L^d$, \begin{align*} \mathbb E[Z_{1,h,x}^2]=h^{-2d}\int_{\mathbb R^d}K\left(\frac{x-y}{h}\right)^2f(y)\,d\mathcal L^d(y). \end{align*} Use the substitution $u=(x-y)/h$, equivalently $y=x-hu$. The Lebesgue measure transforms as $d\mathcal L^d(y)=h^d\,d\mathcal L^d(u)$, and $\mathbb R^d$ maps onto $\mathbb R^d$. Therefore \begin{align*} \mathbb E[Z_{1,h,x}^2]=h^{-d}\int_{\mathbb R^d}K(u)^2f(x-hu)\,d\mathcal L^d(u). \end{align*} Since $K(u)=0$ off $S$, the lower bound on $f(x-hu)$ over $S$ gives \begin{align*} \mathbb E[Z_{1,h,x}^2]\ge h^{-d}\frac{f(x)}{2}\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u). \end{align*} Define \begin{align*} a_x:=\frac{f(x)}{2}\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u)>0. \end{align*} The mean term cannot cancel this leading $h^{-d}$ contribution because it remains bounded as $h\downarrow0$. Indeed, define \begin{align*} B_x:=\sup_{0<h<h_1}|\mathbb E[Z_{1,h,x}]|. \end{align*} For $u\in S$, the point $x-hu$ lies in $A_\rho$, and $K(u)=0$ off $S$; hence \begin{align*} |\mathbb E[Z_{1,h,x}]|\le M_0\int_{\mathbb R^d}|K(u)|\,d\mathcal L^d(u), \end{align*} so $B_x<\infty$. Therefore \begin{align*} \operatorname{Var}(Z_{1,h,x})=\mathbb E[Z_{1,h,x}^2]-(\mathbb E[Z_{1,h,x}])^2\ge a_xh^{-d}-B_x^2. \end{align*} Choose $h_x\in(0,h_1)$ such that $B_x^2\le (a_x/2)h^{-d}$ for $0<h<h_x$. Then \begin{align*} \operatorname{Var}(Z_{1,h,x})\ge \frac{a_x}{2}h^{-d}. \end{align*} Since $\operatorname{Var}(\hat f_{n,h}(x))=n^{-1}\operatorname{Var}(Z_{1,h,x})$, this gives the lower bound $\operatorname{Var}(\hat f_{n,h}(x))\ge a_x/(2nh^d)$. Define \begin{align*} c_x:=\frac{a_x}{2},\qquad C_x:=C_{\mathrm v}. \end{align*} Then $c_x>0$ and $C_x>0$, and combining the lower bound with the upper variance estimate yields \begin{align*} \frac{c_x}{nh^d}\le\operatorname{Var}(\hat f_{n,h}(x))\le\frac{C_x}{nh^d} \end{align*} for every $0<h<h_x$.[/guided]

[step:Translate the uniform bounds into the stated bandwidth asymptotics] Let $(h_n)_{n\in\mathbb N}$ be a sequence of positive [real numbers](/page/Real%20Numbers) such that $h_n\to0$ as $n\to\infty$ and $nh_n^d\to\infty$. Since $h_n<h_0$ for all sufficiently large $n$, the bias estimate gives \begin{align*} \sup_{x\in A}\left|\mathbb E[\hat f_{n,h_n}(x)]-f(x)\right| \le C_{\mathrm b}h_n^m, \end{align*} so the pointwise bias is $O(h_n^m)$ uniformly on $A$. Similarly, \begin{align*} \sup_{x\in A}\operatorname{Var}(\hat f_{n,h_n}(x)) \le \frac{C_{\mathrm v}}{nh_n^d}, \end{align*} so the pointwise variance is $O((nh_n^d)^{-1})$ uniformly on $A$. At any fixed $x\in A$ with $f(x)>0$ and $\int_{\mathbb R^d}K(u)^2\,d\mathcal L^d(u)>0$, the two-sided estimate from the preceding step gives variance of exact order $(nh_n^d)^{-1}$. In the additional standard regime $nh_n^d\to\infty$, this variance bound tends to zero. This proves all asserted bias and variance orders. [/step]