[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 $1\le j\le p$.
[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]
[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]
