Appendix
Proof of Theorem 1
The proof is based on the well-known chaining argument of Dehling and Taqqu (1989) [see also Giraitis et al. (2012, Theorem 10.2.3 and Lemma 13.2.3)]. Hence we shall only briefly indicate the extra steps that are needed for us to achieve our goal. Let \(h_3(x)=\) E \([-\sigma (X_0)I(X_0\le x)f_X'(X_0)/f_X(X_0)]\), and \(\xi _i(x)=\sigma (X_i)I(X_i \le x)-h_1(x)-h_3(x)(X_i-m_X)\). Then we define \(W_n(x)=\sum _{i=1}^n \xi _i(x) \varepsilon _i\), and \(W_n(x, y) = W_n(x)-W_n(y)\), \(-\infty<y \le x <\infty \).
Under the null hypothesis, by (1) and (6),
$$\begin{aligned} V_n(x)=W_n(x)+ h_1(x) S_{n1}+h_3(x)\sum _{i=1}^n (X_i-m_X)\varepsilon _i. \end{aligned}$$
(12)
Let \(H_1(x,y)=\int _y^x \sigma (u)|f_X'(u)| du\). Note that, by Assumptions (A3) and (A4), we obtain \(\sup _{x\in {\mathbb {R}}} h_1(x)<\infty \), and
$$\begin{aligned} \sup _{x\in {\mathbb {R}}}|h_3(x)|\le \sup _{x\in {\mathbb {R}}}\Big \{\text{ E }[\sigma ^2 (X_0)I(X_0\le x)] \text{ E }\big [|f_X'(X_0)|^2/f^2_X(X_0)\big ]\Big \}^{1/2}<\infty , \end{aligned}$$
and similarly, \(\sup _{x, y\in {\mathbb {R}}}|H_1(x, y)|<\infty \). In addition,
$$\begin{aligned} \int |u f_X'(u)| du\le \Big \{\text{ E }X_0^2 \text{ E }\big [|f_X'(X_0)|^2/f^2_X(X_0)\big ]\Big \}^{1/2}<\infty . \end{aligned}$$
Therefore, by (3) and (A2\('\)), for some large enough N,
$$\begin{aligned}&\text{ E } W_n^2(x, y) \nonumber \\&\quad =n\text{ E } (\xi _0 (x)-\xi _0(y))^2\text{ E }\varepsilon _0^2+\sum _{i\not =j}\gamma (i-j)\text{ Cov } \Big ((\xi _i(x)-\xi _i(y)), (\xi _j(x)-\xi _j(y))\Big )\nonumber \\&\quad \le C n +C\Big (\sum _{0<|i-j|\le N}+\sum _{|i-j|>N}\Big )\gamma (i-j)\int \prod _{k=1}^2\Big (\sigma (u_k) I(y<u_k\le x)\nonumber \\&\qquad -h_1(x, y)-h_3(x, y)(u_k-m_X)\Big )\big (f_{X, i,j}(u_1,u_2 )-f_X(u_1) f_X(u_2)\big ) du_1 du_2\nonumber \\\nonumber \\&\quad \le Cn+C n\sum _{|k|>N} |k|^{2d+2d_X-2}\nonumber \\&\quad \quad \times \,\Big (\int |\sigma (u) I(y<u\le x)-h_1(x, y)-h_3(x, y)(u-m_X)| |f_X'(u)|du\Big )^2 \nonumber \\&\quad \le Cn +C n^{2d+2d_X}\nonumber \\&\quad \quad \times \,\Big (C h_1(x, y)+\int _y^x \sigma (u)|f_X'(u)| du+|h_3(x, y)|\int |(u-m_X) f_X'(u)| du\Big )^2\nonumber \\&\quad \le Cn +C n^{2d+2d_X}\Big (h_1(x, y)+H_1(x,y)+|h_3(x, y)|\Big )^2\nonumber \\&\quad \le Cn +C n^{2d+2d_X} (h_1(x, y)+H_1(x, y))^2\nonumber \\&\quad \le C n^{2d+1-\kappa } (h_1(x, y)+H_1(x, y))^2, \end{aligned}$$
(13)
where \(\kappa =\min (2d, 1-2d_X)>0\). Now using (13) instead and along the same lines as those of the proofs of Theorem 10.2.3 and Lemma 13.2.3 of Giraitis et al. (2012), we obtain
$$\begin{aligned} \sup _{x\in {\mathbb {R}}}n^{-1/2-d}|W_n(x)|=o_P(1). \end{aligned}$$
(14)
Moreover, by (3) and (5),
$$\begin{aligned} \text{ E }\Big (n^{-1/2-d}\sum _{i=1}^n (X_i-m_X)\varepsilon _i\Big )^2\le & {} Cn^{-2d}+n^{-1-2d}\sum _{i\not =j}\gamma _X(i-j) \gamma (i-j)\\= & {} O(n^{-2d})+O (n^{2d_X-1})=o(1). \end{aligned}$$
This implies that
$$\begin{aligned} n^{-1/2-d}\sum _{i=1}^n (X_i-m_X)\varepsilon _i=o_P(1). \end{aligned}$$
(15)
Then combining (8), (12), (14) and (15) concludes the proof of (a).
On the other hand, under the alternative,
$$\begin{aligned} V_n(x)= & {} W_n(x)+ h_1(x) S_{n1}+h_3(x)\sum _{i=1}^n (X_i-m_X)\varepsilon _i\nonumber \\&+nh_2(x)+h_4(x)S_{n2}+W_n^*(x), \end{aligned}$$
(16)
where \(h_4(x)=\) E \([-(\mu (X_0) -\mu _0(X_0))I(X_0\le x)f_X'(X_0)/f_X(X_0)]\), \(W_n^*(x)=\sum _{i=1}^n \xi _i^*(x)\) and \(\xi _i^*(x)=(\mu (X_i)-\mu _0(X_i))I(X_i\le x)-h_2(x)-h_4(x)(X_i-m_X)\). Let \(H_2(x,y)=\int _y^x |(\mu (u)-\mu _0(u))f_X'(u)| du\). Similarly to (13) and (14), we obtain
$$\begin{aligned}&\text{ E } (W_n^*(x, y))^2\\&\quad =n\text{ E } (\xi _0^* (x)-\xi _0^*(y))^2+\sum _{i\not =j}\text{ Cov } \Big ((\xi _i^*(x)-\xi _i^*(y)), (\xi _j^*(x)-\xi _j^*(y))\Big )\\&\quad \le C n +C\left( \sum _{0<|i-j|\le N}+\sum _{|i-j|>N}\right) \int \prod _{k=1}^2\Big ((\mu (u_k)-\mu _0 (u_k)) I(y<u_k\le x)\\&\qquad -h_2(x, y)-h_4(x, y)(u_k-m_X)\Big )\big (f_{X,i,j}(u_1,u_2 )-f_X(u_1) f_X(u_2)\big ) du_1 du_2\\&\quad \le Cn+C n\sum _{|k|>N} |k|^{2d_X-1}\\&\quad \left( \int |(\mu (u)-\mu _0 (u)) I(y<u\le x)-h_2(x, y)-h_4(x, y)(u-m_X)| |f_X'(u)|du\right) ^2 \\&\quad \le Cn +C n^{2d_X+1}(|h_2(x, y)|+H_2(x, y))^2\\&\quad \le C n^{2d+1-(2d-2d_X)} (|h_2(x, y)|+H_2(x, y))^2. \end{aligned}$$
By \(d_X<d\), we arrive at
$$\begin{aligned} \sup _{x\in {\mathbb {R}}}n^{-1/2-d}|W_n^*(x)|=o_P(1). \end{aligned}$$
(17)
Since \(\text{ E }S^2_{n2}=O(n^{2d_X+1})=o (n^{2d+1})\), the proof of part (b) is completed upon combining (8) and (14)–(17). \(\square \)
Proof of Theorem 2
Let \(\phi (X_i):=\dot{\mu }_0 (X_i, \beta )\sigma (X_i)\) and \(\phi ^{(l)}(X_i)\) be the l-th element of \(\phi (X_i)\), \(l=1,\ldots , q\). Then, under the null hypothesis,
$$\begin{aligned} M(\beta )=\sum _{i=1}^n (\phi (X_i)-\text{ E }\phi (X_i))\varepsilon _i+\text{ E }[\phi (X_0)]\sum _{i=1}^n\varepsilon _i. \end{aligned}$$
Similarly to the proof of (13), by (3), Assumptions (A2\('\)) and (B1), for some large enough N, and any \(1\le l, r\le q\),
$$\begin{aligned}&\text{ Cov } \left( \sum _{i=1}^n (\phi ^{(l)}(X_i)-\text{ E }\phi ^{(l)}(X_i))\varepsilon _i, \sum _{i=1}^n (\phi ^{(r)}(X_i)-\text{ E }\phi ^{(r)}(X_i))\varepsilon _i\right) \\&\quad \le n\text{ E } \Vert \phi (X_0)\Vert ^2\text{ E }\varepsilon _0^2+\sum _{i\not =j}\gamma (i-j)\text{ Cov } (\phi ^{(l)}(X_i), \phi ^{(r)}(X_j) )\\&\quad \le C n +C\sum _{|i-j|>N}\gamma (i-j)\int \phi ^{(l)}(u_1)\phi ^{(r)}(u_2) (f_{X, i,j}(u_1,u_2 )\\&\qquad -f_X(u_1) f_X(u_2)) du_1 du_2\\&\quad \le Cn+C n\sum _{|k|>N} |k|^{2d+2d_X-2}\left( \int \Vert \dot{\mu }_0 (u, \beta )\sigma (u)\Vert |f_X'(u)|du\right) ^2 \\&\quad \le Cn +C n^{2d+2d_X}. \end{aligned}$$
Since \(\sum _{i=1}^n\varepsilon _i=O_P(\rho _n)\), this implies that
$$\begin{aligned} \Vert \rho _n^{-1}M(\beta )\Vert =\Big \Vert \rho _n^{-1}\text{ E }[\dot{\mu }_0 (X_0, \beta )\sigma (X_0)]\sum _{i=1}^n\varepsilon _i\Big \Vert +o_P(1)=O_P(1), \end{aligned}$$
(18)
where Assumption (B1) ensure that \(\text{ E }\Vert \dot{\mu }_0 (X_0, \beta )\sigma (X_0)\Vert <\infty \). In addition, from Assumptions (A4) and (B2),
$$\begin{aligned}&\text{ E }\big \Vert (\dot{\mu }_0 \big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta ))\sigma (X_i)\big \Vert \\&\quad \le \big \{\text{ E }\big \Vert \dot{\mu }_0 \big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta )\big \Vert ^2 \text{ E }\sigma ^2(X_i)\big \}^{1/2} =o(1). \end{aligned}$$
In a similar way as in (18), we obtain, for any \(u\in {\mathbb {R}}^q\),
$$\begin{aligned} \rho _n^{-1}\left\| \sum _{i=1}^n \big (\dot{\mu }_0 \big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta )\big )\sigma (X_i)\varepsilon _i\right\| =o_P(1). \end{aligned}$$
(19)
Moreover, Assumptions (B1) and (B2), (9) and the ergodic theorem imply that, for any \(\Vert u\Vert \le k\) and \(0<k<\infty \),
$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^n\big (\dot{\mu }_0 \big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta )\big )\big (\mu _0 (X_i, \beta )-\mu _0 \big (X_i, \beta +a_n^{-1}u\big )\big )\nonumber \\&\quad =\frac{1}{n}\sum _{i=1}^n\big (\dot{\mu }_0\big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta )\big )\big (-a_n^{-1}u'\dot{\mu }_0 (X_i, \beta )+o_P\big (a_n^{-1}\Vert u\Vert \big )\big )\nonumber \\&\quad \le \frac{1}{n}\sum _{i=1}^n\Vert \dot{\mu }_0 \big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta )\Vert \big (a_n^{-1}k\Vert \dot{\mu }_0 (X_i, \beta )\Vert +o_P\big (a_n^{-1}\big )\big )\nonumber \\&\quad =o_P\big (a_n^{-1}\big ). \end{aligned}$$
(20)
Thus (9), (19) and (20) yield
$$\begin{aligned}&M\big (\beta +a_n^{-1}u\big )-M(\beta )\\&\quad = \sum _{i=1}^n \dot{\mu }_0 (X_i, \beta )(\mu _0(X_i, \beta )-\mu _0 (X_i, \beta +a_n^{-1}u))\\&\qquad +\sum _{i=1}^n \big (\dot{\mu }_0 (X_i, \beta +a_n^{-1}u)-\dot{\mu }_0 (X_i, \beta ))(\mu _0 (X_i, \beta )-\mu _0 \big (X_i, \beta +a_n^{-1}u\big )\big )\\&\qquad +\sum _{i=1}^n\big (\dot{\mu }_0 \big (X_i, \beta +a_n^{-1}u\big )-\dot{\mu }_0 (X_i, \beta )\big )\sigma (X_i)\varepsilon _i.\\&\quad =-\rho _n n^{-1}\sum _{i=1}^n \dot{\mu }_0 (X_i, \beta )\big (\dot{\mu }_0' (X_i, \beta )u+o_P(\Vert u\Vert )\big )+o_P(\rho _n). \end{aligned}$$
Hence Assumption (B4) implies that
$$\begin{aligned} \big \Vert \rho _n^{-1}\big [M\big (\beta +a_n^{-1}u\big )-M(\beta )\big ]+Hu\big \Vert =o_P(1) \end{aligned}$$
(21)
for any \(\Vert u\Vert \le k\) and \(0<k<\infty \).
By (18), Assumption (B3) and Lemma 11.2.1 of Giraitis et al. (2012), we obtain
$$\begin{aligned} a_n\Vert \hat{\beta }-\beta \Vert =O_P(1). \end{aligned}$$
(22)
This, together with (21), yields
$$\begin{aligned} \beta -\hat{\beta }=-H^{-1}n^{-1}M(\beta )+o_P(1). \end{aligned}$$
Hence, under the null hypothesis and by (7) and (9) ,
$$\begin{aligned} \rho _n^{-1}\tilde{V}_n(x)= & {} \rho _n^{-1}V_n(x)-\rho _n^{-1}M'(\beta )H^{-1}n^{-1}\sum _{i=1}^n\dot{\mu }_0(X_i, \beta )I(X_i\le x)\\&+o_P(a_n\Vert \hat{\beta }-\beta \Vert ). \end{aligned}$$
The ergodic theorem implies that
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \beta )I(X_i\le x)=\text{ E }[\dot{\mu }_0 (X_0, \beta )I(X_0\le x)]+o_P(1). \end{aligned}$$
Then, by (18) and the proof of Theorem 1, we arrive at
$$\begin{aligned} \rho _n^{-1}\tilde{V}_n(x)= & {} \rho _n^{-1}\tilde{h}_1(x) S_{n1}+\rho _n^{-1}W_n(x)+\rho _n^{-1}h_3(x)\sum _{i=1}^n (X_i-m_X)\varepsilon _i+o_P(1)\\= & {} \tilde{h}_1(x)\rho _n^{-1} S_{n1}+o_P(1). \end{aligned}$$
This yields the first part of Theorem 2.
Next, under the alternative, similarly to the proof of Theorem 1, we obtain
$$\begin{aligned} \tilde{V}_n(x)= & {} nh_2(x)+\tilde{h}_1(x) S_{n1}\\&+ W_n(x)+h_3(x)\sum _{i=1}^n (X_i-m_X)\varepsilon _i+h_4(x)S_{n2}+W_n^*(x)+o_P(1)\\= & {} nh_2(x)+\tilde{h}_1(x) S_{n1}+o_P(1). \end{aligned}$$
This implies the part (b) of Theorem 2. \(\square \)
The proof of Theorem 3 is facilitated by the following lemmas. Lemmas 1 and 2 follow immediately from Lemmas 1 and 2 of Koul and Wang (2017).
Lemma 1
Under the Assumptions (A2) and (C1),
$$\begin{aligned} \sup _{x\in {\mathbb {R}}} | f_n(x)-f_X(x) |=O_P\big (\max \big \{n^{d_X-1/2}b_n^{-1}, b_n\big \}\big ). \end{aligned}$$
To state the next result, recall the definition of \(S_{n1}\) from (8), let \(\mu _\sigma =\) E\(\sigma (X_0)\), and let
$$\begin{aligned} \hat{Z}_i= & {} \hat{\mu }_n(X_i)-\mu (X_i), \,\, i=1,\ldots , n, \quad Z_n=\mu _\sigma n^{-1}S_{n1},\\ \xi _n= & {} \max (n^{1/2}h_n^2\log n,\,\, h_n^{-1/2}\log n,\,\, n^{d_X}\log n,\,\, n^{d} h_n^{1/2}, n^{d+d_X-1/2}h_n^{-1}\log n). \end{aligned}$$
Lemma 2
Assume (A2), (A3) and (C2)–(C6). Then
$$\begin{aligned} \sum _{i=1}^n \big (\hat{Z}_i-Z_n\big )^2w_i=O_P\big (\xi _n^2\big ). \end{aligned}$$
Since \(Z_n=O_P(n^{d-1/2})\) and \(\max _{1\le i\le n} |w_i|\le 1\), by Assumption (C6) and Lemma 2, we obtain
$$\begin{aligned} \sum _{i=1}^n \hat{Z}_i^2w_i=O_P\big (\xi _n^2\big )+O_P(n^{2d})=O_P(n^{2d}). \end{aligned}$$
(23)
Let
$$\begin{aligned} \tilde{\sigma }_{-i}^2 (x)=\frac{1}{(n-1) f_n(x)}\sum _{j\not =i} K_{b_n} (x-X_j) \varepsilon _{j}^2 \sigma ^2 (X_j)w_j. \end{aligned}$$
Lemma 3
Under the Assumptions (A2), (A3), (C1), (C4)–(C5) and (C7),
$$\begin{aligned} \sum _{i=1}^n \big |\tilde{\sigma }_{-i} (X_i)-w_i^{1/2}\sigma (X_i)\big |^2=O_P\big (\max \big \{n^{1/2}b_n^{-1/2}\log n, n^{2d}\log n, nb_n^{1/2}(\log n)^{1/2}\big \}\big ). \end{aligned}$$
Proof
Note that
$$\begin{aligned} \left( \sum _{i=1}^n \big |\tilde{\sigma }_{-i} (X_i)-w_i^{1/2}\sigma (X_i)\big |^2\right) ^2\le & {} n \sum _{i=1}^n \big |\tilde{\sigma }_{-i} (X_i)-w_i^{1/2}\sigma (X_i)\big |^4\\\le & {} n \sum _{i=1}^n \big |\tilde{\sigma }_{-i}^2 (X_i)-w_i\sigma ^2 (X_i)\big |^2. \end{aligned}$$
Therefore it suffices to show
$$\begin{aligned} \sum _{i=1}^n \big |\tilde{\sigma }_{-i}^2 (X_i)-w_i\sigma ^2 (X_i)\big |^2=O_P\big (\max \big \{b_n^{-1}\log ^2 n, n^{4d-1}\log ^2 n, nb_n\log n\big \}\big ).\nonumber \\ \end{aligned}$$
(24)
To prove (24), by Assumption (C4) and Lemma 1, it suffices to show
$$\begin{aligned}&\frac{1}{(n-1)^2}\sum _{i=1}^n \text{ E } \Big (\sum _{j\not =i}K_{b_n} (X_i-X_j)\sigma ^2 (X_j)w_j (\varepsilon _j^2-1)\Big )^2\nonumber \\&\quad =O(b_n^{-1})+O(n^{4d-1}) \end{aligned}$$
(25)
and
$$\begin{aligned}&\sum _{i=1}^n \text{ E }\Big (\frac{1}{f_X(X_i)}\sum _{j\not =i}K_{b_n} (X_i-X_j)\sigma ^2 (X_j)w_j-(n-1)w_i\sigma ^2 (X_i)\Big )^2\nonumber \\&\quad =O(n^2b_n^{-1}\log ^2 n)+O(n^3b_n\log n). \end{aligned}$$
(26)
Note that the left hand side of (25) is less than or equal to \((n-1)^{-2}\sum _{i=1}^n (A_{1i}+A_{2i})\), where
$$\begin{aligned} A_{1i}= \sum _{j\not =i}\text{ E } \Big ( K_{b_n} (X_i-X_j)\sigma ^2 (X_j)w_j\Big )^2\text{ E }(\varepsilon _j^2-1)^2, \end{aligned}$$
and
$$\begin{aligned} A_{2i}= & {} \sum _{j\not =k, j, k\not =i}\text{ E } \Big ( K_{b_n} (X_i-X_j)K_{b_n} (X_i-X_k)\sigma ^2 (X_j)\sigma ^2(X_k)w_jw_k\Big ) \text{ E }\\&\big [\big (\varepsilon _j^2-1\big )\big (\varepsilon _k^2-1\big )\big ]. \end{aligned}$$
Since E\(\sigma ^4(X_0)<\infty \) and E\(\varepsilon _0^4<\infty \), we obtain
$$\begin{aligned} \frac{1}{(n-1)^2}\sum _{i=1}^n A_{1i}=O\big ( b_n^{-1}\big ). \end{aligned}$$
(27)
Moreover, by the result from Guo and Koul (2008),
$$\begin{aligned} \text{ E }\big [\big (\varepsilon _j^2-1\big )\big (\varepsilon _k^2-1\big )\big ]\le C |j-k|^{4d-2}, \ \ |j-k|\rightarrow \infty . \end{aligned}$$
This yields
$$\begin{aligned} A_{2i}= & {} \sum _{j\not =k, j, k\not =i} \text{ E }\big [\big (\varepsilon _j^2-1\big )\big (\varepsilon _k^2-1\big )\big ]\int K(u) K(v) \sigma ^2(x-b_nu)\sigma ^2 (x-b_n v) \\&\cdot w(x-b_nu) w(x-b_n v) f_{X,i ,j,k}(x, x-b_n u, x-b_nv)dx du dv \\\le & {} C n+C\sum _{|j-k|>N}|j-k|^{4d-2}\\= & {} O(n)+O(n^{4d}), \,\,\,\text{ not } \text{ depending } \text{ on } i. \end{aligned}$$
This implies that
$$\begin{aligned} \frac{1}{(n-1)^2}\sum _{i=1}^n A_{2i}=O(1)+O(n^{4d-1}). \end{aligned}$$
(28)
The claim (25) follows from the bounds (27) and (28). Similarly, the left hand side of (26) equals to \(\sum _{i=1}^n B_i:=\sum _{i=1}^n (B_{1i}+B_{2i}-2B_{3i}+(n-1)^2\text{ E } [w^2(X_0)\sigma ^4(X_0)])\), where
$$\begin{aligned} B_{1i}= & {} \sum _{j\not =i}\text{ E } \left( \frac{ K^2_{b_n} (X_i-X_j)\sigma ^4 (X_j)w_j^2}{f_X^2(X_i)}\right) ,\\ B_{2i}= & {} \sum _{j\not =k, j, k\not =i}\text{ E } \left( \frac{ K_{b_n} (X_i-X_j)K_{b_n} (X_i-X_k)\sigma ^2 (X_j)\sigma ^2(X_k)w_jw_k}{f_X^2(X_i)}\right) , \end{aligned}$$
and
$$\begin{aligned} B_{3i}= (n-1)\sum _{j\not =i}\text{ E } \left( \frac{ K_{b_n} (X_i-X_j)\sigma ^2 (X_j)\sigma ^2(X_i)w_jw_i}{f_X(X_i)}\right) . \end{aligned}$$
Therefore, by (C4),
$$\begin{aligned} B_{1i}=O(nb_n^{-1}\log ^2 n). \end{aligned}$$
Again, by (C4), (C5), (A2\('\)) and (A3),
$$\begin{aligned} B_{2i}= & {} \sum _{j\not =k, j, k\not =i}\int \frac{K(u) K(v) \sigma ^2(x-b_nu)\sigma ^2 (x-b_n v)w(x-b_nu) w(x-b_n v)}{f_X^2(x)}\\&\quad \times \, f_{X,i,j, k}(x, x-b_nu, x-b_nv) dx du dv\\= & {} \sum _{j\not =k, j, k\not =i}\int \frac{K(u) K(v) \sigma ^2(x-b_nu)\sigma ^2 (x-b_n v)w(x-b_nu) w(x-b_n v)}{f_X^2(x)}\\&\times \, f_X(x)f_X(x-b_nu) f_X(x-b_nv) dx du dv\\&+\sum _{j\not =k, j, k\not =i}\int \frac{K(u) K(v) \sigma ^2(x-b_nu)\sigma ^2 (x-b_n v)w(x-b_nu) w(x-b_n v)}{f_X^2(x)}\\&\times \,\big (f_{X,i,j, k}(x, x-b_nu, x-b_nv)-f_X(x) f_{X,j, k}(x-b_nu, x-b_nv)\big ) dx du dv\\&+\sum _{j\not =k, j, k\not =i}\int \frac{K(u) K(v) \sigma ^2(x-b_nu)\sigma ^2 (x-b_n v)w(x-b_nu) w(x-b_n v)}{f_X^2(x)}f_X(x)\\&\times \, \big (f_{X,j, k}(x-b_nu, x-b_nv)-f_X(x-b_nu) f_X(x-b_nv)\big )dx du dv\\\le & {} (n-1)^2\text{ E }[w^2(X_0)\sigma ^4 (X_0)]+O(n^{2}b_n\log n)+O(n\log ^2 n)\\&+\,O(\log ^2 n)\Big [\sum _{|j-i|>N}|j-i|^{2d_X-1} +\sum _{|k-i|>N}|i-k|^{2d_X-1}+\sum _{|j-k|>N}|j-k|^{2d_X-1}\Big ]. \end{aligned}$$
This implies that
$$\begin{aligned}&B_{2i}-(n-1)^2\text{ E }\big [w^2(X_0)\sigma ^4 (X_0)\big ] = O(n^2b_n\log n)+O\big (n^{2d_X+1}\log ^2 n\big ). \end{aligned}$$
Furthermore, in a similar way,
$$\begin{aligned} B_{3i}-(n-1)^2\text{ E }\big [w^2(X_0)\sigma ^4 (X_0)\big ]=O(n^2b_n\log n)+O\big (n^{2d_X+1}\log ^2 n\big ). \end{aligned}$$
These bounds imply that
$$\begin{aligned} B_i=O\big (nb_n^{-1}\log ^2 n\big )+O\big (n^2b_n\log n\big )+O\big (n^{2d_X+1}\log ^2 n\big ),\ \ \text{ not } \text{ depending } \text{ on } i. \end{aligned}$$
Then, by Assumption (C7), we arrive at
$$\begin{aligned} \sum _{i=1}^n B_i=O\big (n^2b_n^{-1}\log ^2 n\big )+O\big (n^3 b_n\log n\big ). \end{aligned}$$
This completes the proof of Lemma 3. \(\square \)
Lemma 4
Under the Assumptions (A2), (A3) and (C),
$$\begin{aligned} \sum _{i=1}^n |\hat{\sigma }_{-i} (X_i)-\tilde{\sigma }_{-i} (X_i)|^2=O_P(n^{d+1/2}\log n). \end{aligned}$$
Proof
It suffices to show
$$\begin{aligned}&\sum _{i=1}^n \big |\hat{\sigma }^2_{-i} (X_i)-\tilde{\sigma }^2_{-i} (X_i)\big |=O_P(n^{d+1/2}\log n). \end{aligned}$$
Recall that \(\hat{Z}_j=\sigma (X_j)\varepsilon _j-\hat{\varepsilon }_j\). Then
$$\begin{aligned}&(n-1)\big (\hat{\sigma }^2_{-i} (X_i)-\tilde{\sigma }^2_{-i} (X_i)\big )\nonumber \\&\quad = \frac{1}{f_n(X_i)}\left\{ \sum _{j\not =i} K_{b_n}(X_i-X_j)\hat{Z}_j^2w_j-2\sum _{j\not =i} K_{b_n}(X_i-X_j)\sigma (X_j) \varepsilon _j\hat{Z}_jw_j\right\} \nonumber \\&\quad := I_{1i}+I_{2i}. \end{aligned}$$
(29)
Since, for any \(i\not =j\),
$$\begin{aligned} \text{ E }|K_{b_n}(X_i-X_j)|=\int K(u) f_{X, i, j}(x, x-b_n u) dx du=O(1), \end{aligned}$$
it follows from Lemma 1 and (23) that
$$\begin{aligned} \sum _{i=1}^n |I_{1i}|\le O_P(n\log n)\sum _{j=1}^n \hat{Z}_j^2w_j=O_P(n^{2d+1}\log n). \end{aligned}$$
(30)
On the other hand,
$$\begin{aligned} \text{ E }|K_{b_n}(X_i-X_j)\sigma (X_j)\varepsilon _j|\le C\int K(u)\sigma (x-b_n u) f_{X, i, j}(x, x-b_n u) dx du=O(1). \end{aligned}$$
Hence
$$\begin{aligned} \sum _{i=1}^n |I_{2i}|\le & {} O_P(n\log n)\sum _{j=1}^n |\hat{Z}_jw_j |\nonumber \\\le & {} O_P( n^{3/2}\log n)\Big |\sum _{j=1}^n\hat{Z}_j^2w_j\Big |^{1/2}=O_P(n^{d+3/2}\log n). \end{aligned}$$
(31)
Now (29)-(31) concludes the proof of Lemma 4. \(\square \)
Proof of Theorem 3
First we observe that
$$\begin{aligned}&|\hat{h}_1(x)-h_1(x)|\nonumber \\&\quad \le \frac{1}{n}\sum _{i=1}^n|\hat{\sigma }_{-i}(X_i)-\sigma (X_i)|I(X_i\le x)+\Big |\frac{1}{n}\sum _{i=1}^n\sigma (X_i)I(X_i\le x)-h_1(x)\Big |.\qquad \quad \end{aligned}$$
(32)
Note that
$$\begin{aligned} E|w(X_0)-1|\le & {} \Big |\int _{\tau _{1n}+r}^{\tau _{2n}-r} f_X(x)dx -1\Big | + \int _{\tau _{1n}}^{\tau _{1n}+r} f_X(x)dx+ \int _{\tau _{2n}-r}^{\tau _{2n}} f_X(x)dx \\\longrightarrow & {} 0. \end{aligned}$$
Then, by Lemmas 3 and 4,
$$\begin{aligned}&\sum _{i=1}^n|\hat{\sigma }_{-i}(X_i)-\sigma (X_i)|^2\nonumber \\&\quad \le C\sum _{i=1}^n|\hat{\sigma }_{-i}(X_i)-\tilde{\sigma }_{-i} (X_i)|^2+C\sum _{i=1}^n\big |\tilde{\sigma }_{-i}(X_i)-w_i^{1/2}\sigma (X_i)\big |^2\nonumber \\&\qquad +C\sum _{i=1}^n\big |\big (w_i^{1/2}-1\big )\sigma (X_i)\big |^2\nonumber \\&\quad =O_P\big (n^{1/2}b_n^{-1/2}\log n\big )+O_P\big (nb_n^{1/2}\big (\log n)^{1/2}\big )+O_P\big (n^{d+1/2}\log n\big )+o_P(n)\nonumber \\&\quad =o_P(n). \end{aligned}$$
(33)
This implies that
$$\begin{aligned} \sup _{x\in {\mathbb {R}}}\frac{1}{n}\sum _{i=1}^n|\hat{\sigma }_{-i}(X_i)-\sigma (X_i)|I(X_i\le x)=o_P(1). \end{aligned}$$
(34)
Moreover, the ergodic theorem implies that
$$\begin{aligned} \sup _{x\in {\mathbb {R}}}\left| \frac{1}{n}\sum _{i=1}^n\sigma (X_i)I(X_i\le x)-h_1(x)\right| =o_P(1). \end{aligned}$$
(35)
The proof is completed upon combining (32), (34) and (35). \(\square \)
Proof of Theorem 4
By Assumption (B2) and (22),
$$\begin{aligned}&\left\| \frac{1}{n}\sum _{i=1}^n(\dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta ))I(X_i\le x)\right\| \\&\quad \le \frac{1}{n}\sum _{i=1}^n\Vert \dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta )\Vert =o_P(1). \end{aligned}$$
Thus
$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \hat{\beta })I(X_i\le x)-\text{ E }[\dot{\mu }_0(X_0, \beta )I(X_0\le x)]\\&\quad = \frac{1}{n}\sum _{i=1}^n(\dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta ))I(X_i\le x)+\frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \beta )I(X_i\le x)\\&\qquad -\text{ E }[\dot{\mu }_0(X_0, \beta )I(X_0\le x)]\\&\quad =\frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \beta )I(X_i\le x)-\text{ E }[\dot{\mu }_0(X_0, \beta )I(X_0\le x)]+o_P(1). \end{aligned}$$
Then it follows from the ergodic theorem that
$$\begin{aligned} \sup _{x\in {\mathbb {R}}}\left\| \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \hat{\beta })I(X_i\le x)-\text{ E }[\dot{\mu }_0(X_0, \beta )I(X_0\le x)]\right\| =o_P(1). \end{aligned}$$
(36)
Similarly, by Assumptions (B1), (B2) and (22),
$$\begin{aligned}&\left\| \frac{1}{n}\sum _{i=1}^n(\dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta ))\dot{\mu }_0'(X_i, \beta )\right\| \\&\quad \le \frac{1}{n}\sum _{i=1}^n\Vert \dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta )\Vert \Vert \dot{\mu }_0(X_i, \beta )\Vert \\&\quad =\text{ E }(\Vert \dot{\mu }_0(X_0, \hat{\beta })-\dot{\mu }_0(X_i, \beta )\Vert \Vert \dot{\mu }_0(X_0, \beta )\Vert )+o_P(1)\\&\quad \le \big \{\text{ E }\Vert \dot{\mu }_0(X_0, \hat{\beta })-\dot{\mu }_0(X_i, \beta )\Vert ^2 \text{ E }\Vert \dot{\mu }_0(X_0, \beta )\Vert ^2\big \}^{1/2}+o_P(1)=o_P(1). \end{aligned}$$
This implies that
$$\begin{aligned}&\left\| \frac{1}{n}\sum _{i=1}^n(\dot{\mu }_0(X_i, \hat{\beta })\dot{\mu }_0'(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta )\dot{\mu }_0'(X_i, \beta ))\right\| \\&\quad = \left\| \frac{1}{n}\sum _{i=1}^n(\dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta ))\dot{\mu }_0'(X_i, \beta )\right\| \\&\qquad +\,\left\| \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \hat{\beta })(\dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta ))'\right\| \\&\quad =o_P(1). \end{aligned}$$
Then the ergodic theorem yields that
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \hat{\beta })\dot{\mu }_0'(X_0, \hat{\beta })=H+o_P(1). \end{aligned}$$
(37)
Using similar arguments and by (33), we obtain
$$\begin{aligned}&\left\| \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \hat{\beta })(\hat{\sigma }_{-i}(X_i)-\sigma (X_i))\right\| \\&\quad \le \Big \{\frac{1}{n}\sum _{i=1}^n\Vert \dot{\mu }_0(X_i, \hat{\beta })\Vert ^2 \frac{1}{n}\sum _{i=1}^n|\hat{\sigma }_{-i}(X_i)-\sigma (X_i)|^2\Big \}^{1/2}=o_P(1) \end{aligned}$$
and
$$\begin{aligned} \left\| \frac{1}{n}\sum _{i=1}^n(\dot{\mu }_0(X_i, \hat{\beta })-\dot{\mu }_0(X_i, \beta ))\sigma (X_i)\right\| =o_P(1). \end{aligned}$$
This means that
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \hat{\beta })\hat{\sigma }_{-i}(X_i)= & {} \frac{1}{n}\sum _{i=1}^n\dot{\mu }_0(X_i, \beta )\sigma (X_i)+o_P(1)\nonumber \\= & {} \text{ E }[\dot{\mu }_0(X_0, \beta )\sigma (X_0)]+o_P(1). \end{aligned}$$
(38)
Combining (36)–(38), Theorem 4 follows from Theorem 3. \(\square \)
