[proofplan] We first write the expectation of the reflected estimator as a one-sided kernel integral over the support of the density. At the boundary point $0$, symmetry of $K$ makes the two reflected kernel terms identical, and the normalization of $K$ gives \begin{align*} \int_0^1 K(u)\,d\mathcal L^1(u)=\frac12. \end{align*} We then expand $f(hu)$ at $0$ with a right-sided Taylor estimate, integrate the resulting expression against $K(u)$ on $[0,1]$, and identify the first-order coefficient as $2\mu_{1,+}(K)f'(0+)$. If $f'(0+)=0$, this coefficient is zero, leaving only the quadratic remainder. [/proofplan]

[step:Rewrite the expectation as a one-sided kernel integral]Let $(\Omega,\mathcal F,\mathbb P)$ be the underlying probability space. Let $n\in\mathbb N$ be the sample size, and let $X_1,\dots,X_n:(\Omega,\mathcal F)\to([0,\infty),\mathcal B([0,\infty)))$ denote identically distributed random variables with density $f$ with respect to $\mathcal L^1$. The reflection estimator at the boundary is the [random variable](/page/Random%20Variable) \begin{align*} \hat f_{h,\mathrm{ref}}(0)=\frac{1}{n}\sum_{i=1}^n\frac{1}{h}\left[K\left(\frac{-X_i}{h}\right)+K\left(\frac{X_i}{h}\right)\right]. \end{align*} Choose $h_0\in(0,\delta]$, where $\delta>0$ is the neighbourhood from the hypotheses on $f$. For each $h\in(0,h_0)$, define the single-observation reflected kernel contribution $Y_h:[0,\infty)\to\mathbb R$ by \begin{align*} Y_h(x)=\frac{1}{h}\left[K\left(\frac{-x}{h}\right)+K\left(\frac{x}{h}\right)\right]. \end{align*} The hypothesis $K\in L^1(\mathbb R)$ makes the normalization integral a genuine [Lebesgue integral](/page/Lebesgue%20Integral) and implies that $x\mapsto K(x/h)$ is integrable on $[0,h]$ by the change of variables $x=hu$. The random variable $Y_h(X_1)$ is integrable: by the support condition on $K$, the function $Y_h(x)f(x)$ vanishes for $x>h$, while $f$ is bounded on $[0,h_0]$ because $f'_+$ is absolutely continuous there. Since $X_1,\dots,X_n$ are identically distributed with density $f$ with respect to $\mathcal L^1$, linearity of expectation gives \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]=\mathbb E[Y_h(X_1)]. \end{align*} The density formula for $X_1$ gives \begin{align*} \mathbb E[Y_h(X_1)]=\int_0^\infty \frac{1}{h}\left[K\left(\frac{-x}{h}\right)+K\left(\frac{x}{h}\right)\right]f(x)\,d\mathcal L^1(x). \end{align*} Because $K$ is symmetric, $K(-x/h)=K(x/h)$. Because $\operatorname{supp}K\subset[-1,1]$, the integrand vanishes for $x>h$. Therefore \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)] =2\int_0^h \frac{1}{h}K\left(\frac{x}{h}\right)f(x)\,d\mathcal L^1(x). \end{align*} Using the change of variables $x=hu$, so that $d\mathcal L^1(x)=h\,d\mathcal L^1(u)$ and $x\in[0,h]$ corresponds to $u\in[0,1]$, we obtain \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)] =2\int_0^1 K(u)f(hu)\,d\mathcal L^1(u). \end{align*}[/step]

[guided]Let $(\Omega,\mathcal F,\mathbb P)$ be the underlying probability space. Let $n\in\mathbb N$ be the sample size, and let $X_1,\dots,X_n:(\Omega,\mathcal F)\to([0,\infty),\mathcal B([0,\infty)))$ denote identically distributed random variables with density $f$ with respect to $\mathcal L^1$. The reflection estimator at the boundary is \begin{align*} \hat f_{h,\mathrm{ref}}(0)=\frac{1}{n}\sum_{i=1}^n\frac{1}{h}\left[K\left(\frac{-X_i}{h}\right)+K\left(\frac{X_i}{h}\right)\right]. \end{align*} The estimator is an average of $n$ identically distributed terms, so its expectation is the expectation of one term; independence is not needed for this calculation. Choose $h_0\in(0,\delta]$, where $\delta>0$ is the neighbourhood from the hypotheses on $f$, and let $h\in(0,h_0)$. We introduce the function $Y_h:[0,\infty)\to\mathbb R$ by \begin{align*} Y_h(x)=\frac{1}{h}\left[K\left(\frac{-x}{h}\right)+K\left(\frac{x}{h}\right)\right]. \end{align*} Before integrating, we check that $Y_h(X_1)$ is integrable. The hypothesis $K\in L^1(\mathbb R)$ makes the normalization integral a genuine Lebesgue integral and implies that $x\mapsto K(x/h)$ is integrable on $[0,h]$, since the substitution $x=hu$ gives an integral over $[0,1]$ of $K(u)$. The support condition $\operatorname{supp}K\subset[-1,1]$ implies $Y_h(x)f(x)=0$ for $x>h$. On $[0,h]$, the function $f$ is bounded because $f'_+$ is absolutely continuous on $[0,\delta]$. Hence $Y_h(X_1)$ is integrable. Since $X_1$ has density $f$ on $[0,\infty)$ with respect to $\mathcal L^1$, the expectation is computed in two parts: \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]=\mathbb E[Y_h(X_1)]. \end{align*} The density formula for $X_1$ gives \begin{align*} \mathbb E[Y_h(X_1)]=\int_0^\infty \frac{1}{h}\left[K\left(\frac{-x}{h}\right)+K\left(\frac{x}{h}\right)\right]f(x)\,d\mathcal L^1(x). \end{align*} The reflection is useful exactly here: since $K$ is symmetric, the two kernel values are equal: \begin{align*} K\left(\frac{-x}{h}\right)=K\left(\frac{x}{h}\right). \end{align*} Thus the reflected estimator doubles the one-sided contribution: \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)] =2\int_0^\infty \frac{1}{h}K\left(\frac{x}{h}\right)f(x)\,d\mathcal L^1(x). \end{align*} The support condition $\operatorname{supp}K\subset[-1,1]$ now restricts the integration region. For $x>h$, we have $x/h>1$, hence $K(x/h)=0$. Therefore the integral over $[0,\infty)$ reduces to the integral over $[0,h]$: \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)] =2\int_0^h \frac{1}{h}K\left(\frac{x}{h}\right)f(x)\,d\mathcal L^1(x). \end{align*} Finally, use the substitution $x=hu$. Under this substitution, $d\mathcal L^1(x)=h\,d\mathcal L^1(u)$, and the interval $x\in[0,h]$ becomes $u\in[0,1]$. Hence \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)] =2\int_0^1 K(u)f(hu)\,d\mathcal L^1(u). \end{align*}[/guided]

[step:Use symmetry to isolate the bias] Since $K$ is symmetric and $\int_{\mathbb R}K(u)\,d\mathcal L^1(u)=1$, the two halves of the integral over $[-1,1]$ are equal: \begin{align*} \int_0^1 K(u)\,d\mathcal L^1(u)=\frac{1}{2}\int_{-1}^1 K(u)\,d\mathcal L^1(u)=\frac12. \end{align*} Therefore \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0)=2\int_0^1 K(u)f(hu)\,d\mathcal L^1(u)-2f(0)\int_0^1K(u)\,d\mathcal L^1(u). \end{align*} Therefore \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0)=2\int_0^1 K(u)\bigl(f(hu)-f(0)\bigr)\,d\mathcal L^1(u). \end{align*} [/step]

[step:Apply the right-sided Taylor estimate uniformly on the kernel support]Let $M:=\operatorname{ess\,sup}_{t\in[0,\delta]}|f''_+(t)|$, which is finite by hypothesis. For $0<h<h_0$ and $u\in[0,1]$, the point $hu$ lies in $[0,\delta]$. Define the remainder function $R_h:[0,1]\to\mathbb R$ by \begin{align*} R_h(u)=f(hu)-f(0)-hu\,f'(0+). \end{align*} Because $f$ has a right derivative $f'_+$ on $[0,\delta]$ and $f'_+$ is absolutely continuous there, $f'_+$ is integrable on $[0,\delta]$ and $f$ is absolutely continuous on $[0,\delta]$. Therefore the [Fundamental Theorem of Calculus](/theorems/632) gives, for $s\in[0,\delta]$, \begin{align*} f(s)-f(0)=\int_0^s f'_+(r)\,d\mathcal L^1(r) \end{align*} and \begin{align*} f'_+(r)-f'(0+)=\int_0^r f''_+(t)\,d\mathcal L^1(t) \end{align*} for a.e. $r\in[0,s]$. Applying [Fubini's theorem](/theorems/2961) to the integrable signed function $(r,t)\mapsto \mathbb{1}_{\{0\le t\le r\le s\}}f''_+(t)$ on $[0,s]^2$, we obtain \begin{align*} f(s)-f(0)-s f'(0+)=\int_0^s (s-t)f''_+(t)\,d\mathcal L^1(t). \end{align*} With $s=hu$, this gives \begin{align*} R_h(u)=\int_0^{hu}(hu-t)f''_+(t)\,d\mathcal L^1(t). \end{align*} Therefore \begin{align*} |R_h(u)| \le M\int_0^{hu}(hu-t)\,d\mathcal L^1(t) =\frac{M}{2}h^2u^2. \end{align*}[/step]

[guided]The goal of this step is to obtain a Taylor expansion whose error bound is uniform for every $u\in[0,1]$, because this is exactly the interval over which the kernel will be integrated. Let \begin{align*} M:=\operatorname{ess\,sup}_{t\in[0,\delta]}|f''_+(t)|. \end{align*} This number is finite by the boundedness hypothesis on the second right derivative. Since $0<h<h_0\le\delta$ and $u\in[0,1]$, the point $hu$ belongs to $[0,\delta]$, so all right-sided derivative information is available at $hu$. Define the remainder function $R_h:[0,1]\to\mathbb R$ by \begin{align*} R_h(u)=f(hu)-f(0)-hu\,f'(0+). \end{align*} We now justify the integral formula for this remainder. Because $f$ has a right derivative $f'_+$ on $[0,\delta]$ and $f'_+$ is absolutely continuous there, $f'_+$ is integrable on $[0,\delta]$ and $f$ is absolutely continuous on $[0,\delta]$. Hence the [Fundamental Theorem of Calculus](/theorems/632) gives, for each $s\in[0,\delta]$, \begin{align*} f(s)-f(0)=\int_0^s f'_+(r)\,d\mathcal L^1(r). \end{align*} Applying the same theorem to the absolutely [continuous function](/page/Continuous%20Function) $f'_+$ gives, for a.e. $r\in[0,s]$, \begin{align*} f'_+(r)-f'(0+)=\int_0^r f''_+(t)\,d\mathcal L^1(t). \end{align*} Substituting this identity into the previous integral yields an iterated integral over the triangle $0\le t\le r\le s$. The integrand $(r,t)\mapsto \mathbb{1}_{\{0\le t\le r\le s\}}f''_+(t)$ is integrable on $[0,s]^2$, since $|f''_+(t)|\le M$ a.e. and $[0,s]^2$ has finite $\mathcal L^2$ measure. Therefore Fubini's theorem permits the order of integration to be exchanged for this signed integrable function, and we get \begin{align*} f(s)-f(0)-s f'(0+)=\int_0^s (s-t)f''_+(t)\,d\mathcal L^1(t). \end{align*} Taking $s=hu$ gives \begin{align*} R_h(u)=\int_0^{hu}(hu-t)f''_+(t)\,d\mathcal L^1(t). \end{align*} Now use the essential bound on $f''_+$ and the non-negativity of $hu-t$ on $[0,hu]$: \begin{align*} |R_h(u)| \le M\int_0^{hu}(hu-t)\,d\mathcal L^1(t) =\frac{M}{2}h^2u^2. \end{align*} This is the needed uniform second-order remainder estimate on the kernel support.[/guided]

[step:Integrate the Taylor expansion and identify the boundary term] Substituting the expansion from the previous step into the bias identity gives \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0)=2\int_0^1K(u)\left(hu\,f'(0+)+R_h(u)\right)\,d\mathcal L^1(u). \end{align*} By linearity of the Lebesgue integral, \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0)=2h f'(0+)\int_0^1uK(u)\,d\mathcal L^1(u)+2\int_0^1K(u)R_h(u)\,d\mathcal L^1(u). \end{align*} Using the definition of $\mu_{1,+}(K)$, this becomes \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0)=2h f'(0+)\mu_{1,+}(K)+2\int_0^1K(u)R_h(u)\,d\mathcal L^1(u). \end{align*} It remains to bound the remainder. Since $|R_h(u)|\le \frac{M}{2}h^2u^2$ and $0\le u\le1$, \begin{align*} \left|2\int_0^1K(u)R_h(u)\,d\mathcal L^1(u)\right|\le 2\int_0^1 |K(u)|\,|R_h(u)|\,d\mathcal L^1(u). \end{align*} Using $|R_h(u)|\le \frac{M}{2}h^2u^2$ gives \begin{align*} \left|2\int_0^1K(u)R_h(u)\,d\mathcal L^1(u)\right|\le Mh^2\int_0^1u^2|K(u)|\,d\mathcal L^1(u). \end{align*} Since $0\le u\le 1$ on the domain of integration, \begin{align*} \left|2\int_0^1K(u)R_h(u)\,d\mathcal L^1(u)\right|\le Mh^2\int_0^1|K(u)|\,d\mathcal L^1(u). \end{align*} The last integral is finite because $K\in L^1(\mathbb R)$. Thus \begin{align*} 2\int_0^1K(u)R_h(u)\,d\mathcal L^1(u)=O(h^2), \end{align*} and consequently \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0) =2h f'(0+)\mu_{1,+}(K)+O(h^2). \end{align*} [/step]

[step:Deduce the compatibility conclusion] If the even-extension compatibility condition $f'(0+)=0$ holds, then the first-order coefficient in the expansion is zero: \begin{align*} 2f'(0+)\mu_{1,+}(K)=0. \end{align*} The expansion therefore reduces to \begin{align*} \mathbb E[\hat f_{h,\mathrm{ref}}(0)]-f(0)=O(h^2), \end{align*} which is exactly the asserted vanishing of the leading first-order boundary bias term. [/step]