[proofplan] The equivalent kernel is built from the population normal matrix, so its first $p$ algebraic moments are exact identities: multiplying $S_{p,h}(x)$ by the coordinate vector selects the corresponding monomial moment. After these identities remove the Taylor polynomial of $m$ through degree $p$, the bias is controlled by the moments of $L_{p,h,x}$ at orders $p+1$ and $p+2$. The parity of the symmetric kernel implies an additional cancellation when $p$ is even, and the first-order perturbation of the design density gives the stated density-dependent correction. [/proofplan]

[step:Choose a small bandwidth range on which all integrals are local and the moment matrix is invertible] Because $K$ is compactly supported, choose $R>0$ such that $\operatorname{supp}K\subset[-R,R]$. Since $x$ is an interior point and $f_X,m\in C^{p+2}(U)$ on an open neighbourhood $U$ of $x$, choose $\eta>0$ such that \begin{align*} [x-\eta,x+\eta]\subset U. \end{align*} For $0<h<\eta/R$ when $R>0$, every point $x+hu$ with $u\in\operatorname{supp}K$ lies in $U$. Let $f_0=f_X(x)>0$. For each pair of indices $j,k\in\{0,\dots,p\}$, the integrand $u^{j+k}K(u)f_X(x+hu)$ converges pointwise to $u^{j+k}K(u)f_0$ as $h\downarrow0$. Since $f_X$ is continuous on $[x-\eta,x+\eta]$, define $A_f=\sup_{y\in[x-\eta,x+\eta]}|f_X(y)|<\infty$; the absolute value of the integrand is bounded by $A_f|u|^{j+k}K(u)$ on $\operatorname{supp}K$, and this dominating function is integrable with respect to $\mathcal{L}^1$ because $K$ is bounded and compactly supported. The [Dominated Convergence Theorem](/theorems/4) applies because the integrands converge pointwise and are dominated by the integrable functions just described. It gives entrywise convergence to $f_0M_p$ as $h\downarrow0$. The determinant is continuous on the space of real $(p+1)\times(p+1)$ matrices, and \begin{align*} \det(f_0M_p)=f_0^{p+1}\det(M_p)\ne0. \end{align*} Hence there exists $h_0>0$ such that $S_{p,h}(x)$ is nonsingular for every $0<h<h_0$. Shrinking $h_0$ if necessary, the convergence $S_{p,h}(x)\to f_0M_p$ and continuity of the matrix inversion map on $GL_{p+1}(\mathbb{R})$ also give a constant $C_S<\infty$ such that \begin{align*} \|S_{p,h}(x)^{-1}\|_{\mathrm{op}}\le C_S \end{align*} for every $0<h<h_0$. All definitions below are taken for such $h$. [/step]

[step:Derive the exact moment identities from the population normal matrix]For each $j\in\{0,\dots,p\}$, let $e_j\in\mathbb{R}^{p+1}$ denote the coordinate vector whose $j$th coordinate, using the indexing $0,\dots,p$, is $1$ and whose other coordinates are $0$. Then \begin{align*} r_p(u)^\top e_j=u^j. \end{align*} Therefore the definition of $S_{p,h}(x)$ gives \begin{align*} S_{p,h}(x)e_j= \int_{\mathbb{R}}r_p(u)r_p(u)^\top e_jK(u)f_X(x+hu)\,d\mathcal{L}^1(u). \end{align*} Using $r_p(u)^\top e_j=u^j$, this becomes \begin{align*} S_{p,h}(x)e_j= \int_{\mathbb{R}}r_p(u)u^jK(u)f_X(x+hu)\,d\mathcal{L}^1(u). \end{align*} Using the definition of $L_{p,h,x}$, we obtain \begin{align*} \int_{\mathbb{R}}u^jL_{p,h,x}(u)\,d\mathcal{L}^1(u) = e_0^\top S_{p,h}(x)^{-1} \int_{\mathbb{R}}r_p(u)u^jK(u)f_X(x+hu)\,d\mathcal{L}^1(u). \end{align*} Substituting the identity for $S_{p,h}(x)e_j$ gives \begin{align*} \int_{\mathbb{R}}u^jL_{p,h,x}(u)\,d\mathcal{L}^1(u) = e_0^\top S_{p,h}(x)^{-1}S_{p,h}(x)e_j. \end{align*} Since $S_{p,h}(x)$ is nonsingular, this reduces to \begin{align*} \int_{\mathbb{R}}u^jL_{p,h,x}(u)\,d\mathcal{L}^1(u)=e_0^\top e_j. \end{align*} Thus the value is $1$ when $j=0$ and $0$ when $1\le j\le p$.[/step]

[guided]The point of the equivalent kernel is that it preserves the intercept of every polynomial of degree at most $p$. We verify this directly from the matrix defining the population fit. For each $j\in\{0,\dots,p\}$, define $e_j\in\mathbb{R}^{p+1}$ to be the coordinate vector corresponding to the monomial $u^j$. Since \begin{align*} r_p(u)=(1,u,\dots,u^p)^\top, \end{align*} we have \begin{align*} r_p(u)^\top e_j=u^j. \end{align*} Multiplying the matrix $S_{p,h}(x)$ by $e_j$ therefore selects the $j$th monomial column. First, \begin{align*} S_{p,h}(x)e_j= \int_{\mathbb{R}}r_p(u)r_p(u)^\top e_jK(u)f_X(x+hu)\,d\mathcal{L}^1(u). \end{align*} Using $r_p(u)^\top e_j=u^j$, we get \begin{align*} S_{p,h}(x)e_j= \int_{\mathbb{R}}r_p(u)u^jK(u)f_X(x+hu)\,d\mathcal{L}^1(u). \end{align*} Now insert this identity into the $j$th moment of $L_{p,h,x}$. By definition, \begin{align*} L_{p,h,x}(u)=e_0^\top S_{p,h}(x)^{-1}r_p(u)K(u)f_X(x+hu), \end{align*} so \begin{align*} \int_{\mathbb{R}}u^jL_{p,h,x}(u)\,d\mathcal{L}^1(u) = e_0^\top S_{p,h}(x)^{-1} \int_{\mathbb{R}}r_p(u)u^jK(u)f_X(x+hu)\,d\mathcal{L}^1(u). \end{align*} Using the displayed identity for $S_{p,h}(x)e_j$, this is \begin{align*} \int_{\mathbb{R}}u^jL_{p,h,x}(u)\,d\mathcal{L}^1(u) = e_0^\top S_{p,h}(x)^{-1}S_{p,h}(x)e_j. \end{align*} Since $S_{p,h}(x)$ is nonsingular, we conclude \begin{align*} \int_{\mathbb{R}}u^jL_{p,h,x}(u)\,d\mathcal{L}^1(u)=e_0^\top e_j. \end{align*} The final scalar product is $1$ if $j=0$ and $0$ if $1\le j\le p$. This is the exact algebraic reason the local polynomial target reproduces constants and kills the first $p$ centered monomials in the bias calculation.[/guided]

[step:Expand the equivalent-kernel moments through the required orders] For each $s\in\{0,\dots,p+2\}$, define the equivalent-kernel moment map $a_s:(0,h_0)\to\mathbb{R}$ by \begin{align*} a_s(h)=\int_{\mathbb{R}}u^sL_{p,h,x}(u)\,d\mathcal{L}^1(u). \end{align*} For $s\in\{0,\dots,p+2\}$, the [Taylor Theorem](/theorems/631) applied to $f_X$ at $x$ gives, uniformly for $u\in\operatorname{supp}K$, \begin{align*} f_X(x+hu)=f_0+hf_1u+h^2\rho_h(u), \end{align*} where $f_1=f_X'(x)$ and the functions $\rho_h:\operatorname{supp}K\to\mathbb{R}$ satisfy $\sup_{u\in\operatorname{supp}K}|\rho_h(u)|\le C_f$ for a constant $C_f<\infty$ and all sufficiently small $h$. This uniform bound follows because $f_X\in C^{p+2}(U)$ and all points $x+hu$ with $u\in\operatorname{supp}K$ lie in the compact interval $[x-\eta,x+\eta]$. Since $K$ is bounded and compactly supported, every componentwise constant \begin{align*} \int_{\mathbb{R}} |u|^qK(u)\,d\mathcal{L}^1(u) \end{align*} is finite for $0\le q\le 2p+2$. Hence, entrywise, \begin{align*} S_{p,h}(x)=f_0M_p+hf_1N_p+O(h^2), \end{align*} where $N_p=(\mu_{j+k+1})_{0\le j,k\le p}$. Apply the [Matrix Inverse Perturbation Formula](/theorems/1126), namely $(A+hB+O(h^2))^{-1}=A^{-1}-hA^{-1}BA^{-1}+O(h^2)$ for an invertible finite-dimensional matrix $A$. With $A=f_0M_p$ and $B=f_1N_p$, the hypothesis holds because $f_0>0$ and $M_p$ is nonsingular, so \begin{align*} S_{p,h}(x)^{-1} = \frac{1}{f_0}M_p^{-1} - h\frac{f_1}{f_0^2}M_p^{-1}N_pM_p^{-1} + O(h^2). \end{align*} The same uniform remainder estimate, applied component by component and using the same finite moment bounds up to order $2p+2$, also gives, for each $s\in\{0,\dots,p+1\}$, \begin{align*} \int_{\mathbb{R}}r_p(u)u^sK(u)f_X(x+hu)\,d\mathcal{L}^1(u) = f_0c_s+hf_1c_{s+1}+O(h^2). \end{align*} This range restriction is needed because the largest moment used here is $\mu_{s+p+1}$, which is at most $\mu_{2p+2}$ when $s\le p+1$. Multiplying the two expansions yields, for each $s\in\{0,\dots,p+1\}$, \begin{align*} a_s(h) = e_0^\top M_p^{-1}c_s + h\frac{f_1}{f_0} e_0^\top\left(M_p^{-1}c_{s+1}-M_p^{-1}N_pM_p^{-1}c_s\right) + O(h^2). \end{align*} For $s=p+2$, we only use the zeroth-order limit. By dominated convergence, applied component by component with dominating functions $A_f|u|^{p+2+j}K(u)$ for $0\le j\le p$, the vector integral satisfies \begin{align*} \int_{\mathbb{R}}r_p(u)u^{p+2}K(u)f_X(x+hu)\,d\mathcal{L}^1(u) = f_0c_{p+2}+o(1). \end{align*} Together with $S_{p,h}(x)^{-1}=f_0^{-1}M_p^{-1}+o(1)$, this gives \begin{align*} a_{p+2}(h)=e_0^\top M_p^{-1}c_{p+2}+o(1). \end{align*} [/step]

[step:Use symmetry to identify the parity cancellation] Since $K$ is symmetric, $\mu_q=0$ for every odd integer $q$ with $0\le q\le 2p+2$. Suppose first that $p$ is even. Then $M_p$ has nonzero entries only when its row and column indices have the same parity, so $M_p^{-1}$ has the same even-odd block structure. The vector $c_{p+1}$ has entries \begin{align*} (c_{p+1})_j=\mu_{p+1+j},\qquad 0\le j\le p. \end{align*} Because $p+1$ is odd, this entry is nonzero only when $j$ is odd. Thus $M_p^{-1}c_{p+1}$ has zero even-index coordinates, and in particular \begin{align*} e_0^\top M_p^{-1}c_{p+1}=0. \end{align*} If $p$ is odd, the theorem does not require this cancellation. The leading moment is \begin{align*} a_{p+1}(h)=e_0^\top M_p^{-1}c_{p+1}+O(h), \end{align*} which follows from the expansion of $a_s(h)$ with $s=p+1$. [/step]

[step:Apply Taylor expansion of $m$ and cancel the reproduced polynomial terms] The [Taylor Theorem](/theorems/631) applied to $m$ at $x$ gives, uniformly for $u\in\operatorname{supp}K$, \begin{align*} m(x+hu) = \sum_{j=0}^{p+2}\frac{m^{(j)}(x)}{j!}h^ju^j + h^{p+2}u^{p+2}\varepsilon_h(u), \end{align*} where $\varepsilon_h:\operatorname{supp}K\to\mathbb{R}$ satisfies \begin{align*} \sup_{u\in\operatorname{supp}K}|\varepsilon_h(u)|\to0 \end{align*} as $h\downarrow0$. Since $S_{p,h}(x)^{-1}$ remains bounded for $0<h<h_0$, choose $C_S<\infty$ with $\|S_{p,h}(x)^{-1}\|_{\mathrm{op}}\le C_S$. On $\operatorname{supp}K\subset[-R,R]$, the vector $r_p(u)$ satisfies $|r_p(u)|\le C_r$, where $C_r=\left(\sum_{j=0}^p R^{2j}\right)^{1/2}$. Since $f_X$ is bounded on $[x-\eta,x+\eta]$, with $A_f=\sup_{y\in[x-\eta,x+\eta]}|f_X(y)|$, we have \begin{align*} |L_{p,h,x}(u)|\le C_SC_rA_fK(u) \end{align*} for every $u\in\operatorname{supp}K$ and $0<h<h_0$. Therefore \begin{align*} \left|\int_{\mathbb{R}}h^{p+2}u^{p+2}\varepsilon_h(u)L_{p,h,x}(u)\,d\mathcal{L}^1(u)\right| \le h^{p+2}C_SC_rA_f\sup_{u\in\operatorname{supp}K}|\varepsilon_h(u)| \int_{\mathbb{R}}|u|^{p+2}K(u)\,d\mathcal{L}^1(u) = o(h^{p+2}). \end{align*} Using the moment identities $a_0(h)=1$ and $a_j(h)=0$ for $1\le j\le p$, we obtain \begin{align*} m_{p,h,\mathrm{pop}}(x)-m(x) = \frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1}a_{p+1}(h) + \frac{m^{(p+2)}(x)}{(p+2)!}h^{p+2}a_{p+2}(h) + o(h^{p+2}). \end{align*} [/step]

[step:Read off the odd-order and even-order bias expansions] If $p$ is odd, then \begin{align*} a_{p+1}(h)=e_0^\top M_p^{-1}c_{p+1}+O(h). \end{align*} Substituting this into the preceding bias identity gives \begin{align*} m_{p,h,\mathrm{pop}}(x)-m(x) = h^{p+1}\frac{m^{(p+1)}(x)}{(p+1)!}\,e_0^\top M_p^{-1}c_{p+1} + o(h^{p+1}). \end{align*} Now suppose $p$ is even. The parity cancellation gives \begin{align*} e_0^\top M_p^{-1}c_{p+1}=0. \end{align*} Hence the expansion for $a_s(h)$ with $s=p+1$ yields \begin{align*} a_{p+1}(h) = h\frac{f_X'(x)}{f_X(x)} e_0^\top\left(M_p^{-1}c_{p+2}-M_p^{-1}N_pM_p^{-1}c_{p+1}\right) + O(h^2). \end{align*} Also, \begin{align*} a_{p+2}(h)=e_0^\top M_p^{-1}c_{p+2}+o(1). \end{align*} Substituting both identities into the Taylor bias formula gives \begin{align*} m_{p,h,\mathrm{pop}}(x)-m(x) = h^{p+2} \frac{m^{(p+1)}(x)}{(p+1)!}\frac{f_X'(x)}{f_X(x)} e_0^\top\left(M_p^{-1}c_{p+2}-M_p^{-1}N_pM_p^{-1}c_{p+1}\right) + h^{p+2} \frac{m^{(p+2)}(x)}{(p+2)!}e_0^\top M_p^{-1}c_{p+2} + o(h^{p+2}). \end{align*} This is exactly \begin{align*} m_{p,h,\mathrm{pop}}(x)-m(x) = h^{p+2}B_{p+2}(x)+o(h^{p+2}), \end{align*} with the stated definitions of $B_{p+2,m}(x)$ and $B_{p+2,f}(x)$. The leading constants are explicit scalar quantities, so they may vanish for special choices of $m$ and $f_X$. [/step]