d-d_{2}(\tau+1)}{\gamma}}}{\hat{N}^{\kappa+\beta}}\leq\frac{1}{C_{1}}\right\\}.$
(B.3)

Then,

$\displaystyle
E\left[\exp\left(\frac{\lambda\sum_{l=1}^{|I_{j}|}G_{lN}^{r}}{|I_{j}|}\right)\right]\leq
2\exp\left(\frac{8\lambda^{2}(C_{0}\sigma)^{2}}{(|I_{j}|-\frac{4\lambda
C_{0}A}{3})}\right).$

Define $h(\lambda)=\frac{8\lambda^{2}(C_{0}\sigma)^{2}}{|I_{j}|-\frac{4\lambda
C_{0}A}{3})}-\frac{\lambda t}{2}$. We thus obtain Chernoff-type upper bound as
follow,

$\displaystyle\mathrm{P}\left(\begin{matrix}\sum_{j=1}^{P_{\hat{N}}^{d_{2}}}\end{matrix}p_{j}\begin{matrix}\sum_{l=1}^{|I_{j}|}\end{matrix}\frac{G_{lN}^{r}}{|I_{j}|}\geq\frac{t}{2}\right)\leq
2\exp\\{h(\lambda)\\}.$

Here, let
$\lambda^{*}=\frac{|I_{j}|t}{\xi((C_{0}\sigma)^{2}+\frac{4C_{0}A}{3}t)}$ for
some $\xi\geq 1$. Thus $\lambda^{*}\leq\frac{3|I_{j}|t}{4\xi
C_{0}At}<\frac{3|I_{j}|}{4C_{0}A}$, which coincides with the previous
requirement. Then

$\displaystyle h(\lambda^{*})$
$\displaystyle=\frac{8(C_{0}\sigma|I_{j}|t)^{2}}{\xi^{2}(|I_{j}|-\frac{4C_{0}A|I_{j}|t}{3\xi((C_{0}\sigma)})^{2}+\frac{4C_{0}At}{3})((C_{0}\sigma)^{2}+\frac{4C_{0}A}{3}t)^{2}}-\frac{|I_{j}|t^{2}}{2\xi((C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t}$
$\displaystyle=\frac{(8/\xi^{2})(C_{0}\sigma)|I_{j}|t)^{2}}{(|I_{j}|(C_{0}\sigma))^{2}+\frac{4}{3}C_{0}A|I_{j}|t(1-\frac{1}{\xi}))((C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t)}-\frac{(1/2\xi)|I_{j}|t^{2}}{(C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t}$
$\displaystyle\left(\because\xi\geq 1\ \textrm{and}\ C_{0}A>0,\therefore
1-\frac{1}{\xi}\geq 0,\ \frac{4}{3}C_{0}A|I_{j}|t(1-\frac{1}{\xi})\geq
0\right)$
$\displaystyle\leq\frac{(8/\xi^{2})(C_{0}\sigma)^{2}|I_{j}|t^{2}}{(C_{0}\sigma))^{2}((C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t)}-\frac{(1/2\xi)|I_{j}|t^{2}}{(C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t}$
$\displaystyle=\frac{(8/\xi^{2}-1/2\xi)|I_{j}|t}{C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t}$
$\displaystyle(\textrm{let}\ \xi=32)$
$\displaystyle=2\exp\left(-\frac{|I_{j}|t}{128(C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t)}\right).$

Due to sub-Gaussianity and definition of $P_{\hat{N}}$, we manage to prove,
for $\forall\ t>0$,

$\displaystyle Q_{2}\leq
4\exp\left(-\frac{|I_{j}|t}{128(C_{0}\sigma))^{2}+\frac{4C_{0}A}{3}t)}\right),$
(B.4) $\displaystyle Q_{1}\leq
2\left(\frac{N^{\kappa}}{\hat{N}^{2(\kappa+\beta)+\tau+1}t}\right)^{p}.$ (B.5)

The last job is to specify the relationship between $N$ and $\hat{N}$. Recall
Assumptions 3 actually indicates that $\Gamma_{N}\in R_{N}$ while $R_{N}$ can
be decomposed into $\prod_{k=d_{1}+1}^{d}m_{k}$-many cubes. Each cube contains
at most $C_{0}$ locations. Thus, we have

$\displaystyle N\leq C_{0}\begin{matrix}\prod_{k=d_{1}+1}^{d}\end{matrix}$
$\displaystyle
m_{k}=\frac{C_{0}}{H_{0}^{d_{2}}}\begin{matrix}\prod_{k=d_{1}+1}^{d}\end{matrix}H_{k}^{N}=\frac{C_{0}\hat{N}}{H_{0}^{d_{2}}}.$
$\displaystyle\Rightarrow$ $\displaystyle\
\frac{H_{0}^{d_{2}}N}{C_{0}}\leq\hat{N}.$ (B.6)

Conversely, according to Assumption 4,

$\displaystyle Vol(R_{N})$
$\displaystyle=H_{0}^{d_{1}}\hat{N}\leq(KN^{\frac{1}{d_{2}}})^{d_{2}}$
$\displaystyle\Rightarrow\hat{N}\leq\frac{K^{d_{2}}}{H_{0}^{d_{1}}}N=\frac{H_{0}^{d_{2}-d_{1}}}{2^{d_{2}}}N.$
(B.7)

Therefore, we obtain

$\displaystyle|I_{j}|$
$\displaystyle\geq\begin{matrix}\prod_{k=d_{1}+1}^{d}L_{\hat{N}}^{k}\end{matrix}=\begin{matrix}\prod_{k=d_{1}+1}^{d}\left[\frac{m_{k}}{P_{\hat{N}}}\right]\end{matrix}\geq\left(\frac{1}{P_{\hat{N}}}\right)^{d_{2}}\begin{matrix}\prod_{k=d_{1}+1}^{d}m_{k}\end{matrix}$
$\displaystyle=\left(\frac{1}{2P_{\hat{N}}H_{0}}\right)^{d_{2}}\hat{N}=\frac{\hat{N}}{\left(2H_{0}\left(\frac{[2(\kappa+\beta)+\tau+1]}{b}\right)^{\frac{1}{\gamma}}\right)^{d_{2}}(\log_{\nu}\hat{N})^{\frac{d_{2}}{\gamma}}}$
$\displaystyle\left(\textrm{let}\
N\geq\frac{H_{0}^{d_{2}-d_{1}}}{2^{d_{2}}}\right)$ (B.8)
$\displaystyle\geq\frac{N}{2\left(2C_{0}\left(\frac{[2(\kappa+\beta)+\tau+1]}{b}\right)^{\frac{1}{\gamma}}\right)^{d_{2}}(\log_{\nu}N)^{\frac{d_{2}}{\gamma}}}$
(B.9)

and

$\displaystyle\left(\frac{1}{\hat{N}}\right)^{2(\kappa+\beta)+\tau+1}\leq\left(\frac{C_{0}/H_{0}^{d_{2}}}{N}\right)^{2(\kappa+\beta)+\tau+1}.$
(B.10)

Hence, by combining (B.9) and (B.10) with (B.4) and (B.5) respectively, we
obtain the result shown in Theorem 1. To specify the $N_{0}$, we firstly apply
(B.6) and (B.7) to $\hat{N}_{0}$. We thus obtain a sufficient condition for
(B.3) as follow,

$\displaystyle N\geq
N_{1}:=\min\left\\{N\in\mathbf{N}:\frac{\left(\log_{\nu}\frac{H_{0}^{d_{2}-d_{1}}N}{2^{d_{2}}}\right)^{\frac{\tau
d-d_{2}(\tau+1)}{\gamma}}}{N^{\kappa+\beta}}\right\\}.$

Recall we also require $\frac{P_{\hat{N}}}{3}\geq H_{0}$, which, dut to (B.6),
yields

$\displaystyle
N\geq\frac{C_{0}}{H_{0}^{d_{2}}}\nu^{\frac{bH^{\gamma}}{2(\kappa+\beta)+\tau+1}}:=N_{2}.$

Together (B.8), we specify $N_{0}$ as
$N_{0}:=\max\\{N_{1},N_{2},\frac{H_{0}^{d_{2}-d_{1}}}{2^{d_{2}}}\\}$.

Proof of Corollary 1

Similar to Step 1 of the proof of Theorem 1, we can rewrite the averaged
partial sum as follow,

$\displaystyle\begin{matrix}\sum_{i=1}^{N}\end{matrix}\epsilon_{s_{iN}}/N^{-1}=\begin{matrix}\sum_{j=1}^{P_{\hat{N}}^{d_{2}}}\end{matrix}p_{j}\begin{matrix}\sum_{l=1}^{|I_{j}|}\end{matrix}G_{lN}/|I_{j}|.$

Here $G_{lN}=\sum_{i=1}^{N}\textbf{1}[s_{iN}\in B_{l}]\epsilon_{s_{iN}}$.
$p_{j}$ and $|I_{j}|$ are directly inherited from proof of Theorem 1. Then, by
using Chernoff approach, we obtain

$\displaystyle\mathrm{P}\left\\{\left|\frac{\begin{matrix}\sum_{i=1}^{N}\end{matrix}\epsilon_{s_{iN}}}{N}\right|\geq
t\right\\}\leq\exp\left(-\lambda
t\right)\begin{matrix}\sum_{j=1}^{P_{\hat{N}}^{d_{2}}}\end{matrix}\frac{1}{P_{\hat{N}}^{d_{2}}}E\left[\exp\left(\lambda\frac{\begin{matrix}\sum_{l=1}^{|I_{j}|}\end{matrix}G_{lN}}{|I_{j}|}\right)\right]$

Since process $\\{\epsilon_{s_{iN}}\\}_{s_{iN}\in\Gamma_{N}}$ is an
$\alpha$-mixing random field with geometric decay, from now on, we only need
to follow exactly the same procedures demonstrated in Step 3 of proof of
Theorem 1.

###### Lemma 3

(Bradley’s Lemma) Let $(X,Y)$ be a $\mathbf{R}^{n}\times\mathbf{R}$-valued
random vector such that $Y\in L^{p}(P)$ for some $p\in[1,+\infty)$. Let $c$ be
a real number such that $||Y+c||_{p}>0$, and $\xi\in[0,||Y+c||_{p}]$. Then,
there exists a random variable $Y^{*}$ such that

  1. 1.

$P_{Y^{*}}=P_{Y}$ and $Y^{*}$ is independent of $X$.

  2. 2.

$P(|Y-Y^{*}|>\xi)\leq
11\left(\frac{||Y+c||_{p}}{\xi}\right)^{\frac{p}{2p+1}}[\alpha(\sigma(X),\sigma(Y))]^{\frac{2p}{2p+1}}$.

Proof of Theorem 2

Step 1(Blocking)

Similar to the proof of Theorem 1, we let
$R_{N}=\begin{matrix}\prod_{k=d_{1}+1}^{d}\end{matrix}(0,H_{k}^{N}]\otimes(0,H_{o}]^{d_{1}}$,
where $\otimes$ denotes the Cartesian product among intervals. Without loss of
generality, set $\forall d_{1}+1\leq k\leq d$, $\min_{d_{1}+1\leq k\leq
d}H_{k}^{N}\geq H_{0}$. For $\forall\ d_{1}+1\leq k\leq d$, $1\leq
q\leq\hat{N}/2$, let $p_{k}=H_{k}^{N}/2q$. Then $R_{N}$ is partitioned into
$2^{d_{2}}$ groups of rectangles. Within each group, the quantity of rectangle
is $q^{d_{2}}$ and the volume of each of these rectangles is
$\prod_{k=d_{1}+1}^{d}p_{k}:=p$. Here we name the j-th rectangle belonging to
the m-th group as $A_{jm}$. Hence for $\forall\ j,m$,
$Vol(A_{jm})=H_{0}^{d_{1}}\prod_{k=d_{1}+1}^{d}p_{k}:=H_{0}^{d_{1}}p$.
Additionally, due to Proposition 1, we know that $|\Gamma_{N}\cap A_{jm}|\leq
C_{0}\prod_{k=d_{1}+1}^{d}([p_{k}/H_{0}]+1):=C_{p}$, where $C_{0}$ is
inherited from Theorem 1. Thus, we can write the partial sum process as
follow,

$\displaystyle S_{N}=\begin{matrix}\sum_{i=1}^{N}\end{matrix}Z_{s_{iN}}$
$\displaystyle=\begin{matrix}\sum_{j=1}^{q^{d_{2}}}\sum_{m=1}^{2^{d_{2}}}\end{matrix}(\begin{matrix}\sum_{i=1}^{N}\end{matrix}\textbf{1}[s_{iN}\in
A_{jm}]Z_{s_{iN}})$
$\displaystyle:=\begin{matrix}\sum_{m=1}^{2^{d_{2}}}\sum_{j=1}^{q^{d_{2}}}\end{matrix}G_{jm}.$

Define $G_{jm}^{r}=\sum_{i=1}^{N}\textbf{1}[s_{iN}\in
A_{jm}]E[Z_{s_{iN}}|\mathcal{F}_{iN}(r)]$. Then, for any $t>0$,

$\displaystyle\mathrm{P}\left(\left|S_{N}\right|\geq
Nt\right)\leq\mathrm{P}\left(\begin{matrix}\sum_{m=1}^{2^{d_{2}}}\end{matrix}\left|\begin{matrix}\sum_{j=1}^{q^{d_{2}}}\end{matrix}G_{jm}\right|\geq
Nt\right)\leq\begin{matrix}\sum_{m=1}^{2^{d_{2}}}\end{matrix}\mathrm{P}\left(\left|\begin{matrix}\sum_{j=1}^{q^{d_{2}}}\end{matrix}G_{jm}\right|\geq\frac{Nt}{2^{d_{2}}}\right)$
$\displaystyle\leq\begin{matrix}\sum_{m=1}^{2^{d_{2}}}\end{matrix}\mathrm{P}\left(\left|\begin{matrix}\sum_{j=1}^{q^{d_{2}}}\end{matrix}(G_{jm}-G_{jm}^{r})\right|\geq\frac{Nt}{2^{d_{2}+1}}\right)+\begin{matrix}\sum_{m=1}^{2^{d_{2}}}\end{matrix}\mathrm{P}\left(\left|\begin{matrix}\sum_{j=1}^{q^{d_{2}}}\end{matrix}G_{jm}^{r}\right|\geq\frac{Nt}{2^{d_{2}+1}}\right):=Q_{3}+Q_{4}$

Hereafter, denote $t/2^{d_{2}+1}$ as $\eta$. Hence the result can be obtained
if we find probabilistic upper bounds for $Q_{3}$ and $Q_{4}$ uniformly over
$m$.

Step 2(Projection Error)

Firstly, we have the following argument existed.

$\displaystyle
Q_{3}\leq\sum_{m=1}^{2^{d_{2}}}\mathrm{P}\left(\left|\sum_{j=1}^{q^{d_{2}}}(G_{jm}-G_{jm}^{r})\right|\geq\hat{N}\eta\frac{N}{\hat{N}}\right)\leq\sum_{m=1}^{2^{d_{2}}}\mathrm{P}\left(\left|\sum_{j=1}^{q^{d_{2}}}(G_{jm}-G_{jm}^{r})\right|\geq\frac{2^{d_{2}}\hat{N}\eta}{H_{0}^{d_{2}-d_{1}}}\right),$

Let
$\bar{\eta}=\frac{2^{d_{2}}\eta}{H_{0}^{d_{2}-d_{1}}}=\frac{t}{2H_{0}^{d_{2}-d_{1}}}$.
According to the same procedures as Step 2 in the proof of Theorem 1, we
obtain, for $\forall\ t>0$,

$\displaystyle
Q_{3}\leq\left(\frac{C_{0}2^{d_{2}+1}}{H_{0}^{d_{1}}}\right)^{s}\left(\frac{\alpha(N)r^{-\nu_{1}}}{t}\right)^{s}.$

Step 3(Moment Upper Bound for r.v $G_{jm}^{r}$)

Due to the definition of $G_{jm}^{r}$, it’s obvious that, for any $j$ and $m$,
$E[G_{jm}^{r}]=0$, $||G_{jm}^{r}||_{+\infty}\leq C_{p}A$. Now we focus on
deriving an upper bound for $Var(G_{jm}^{r})$. Firstly, according to the basic
property of variance operator, for some $r>0$,

$\displaystyle Var(G_{jm}^{r})$
$\displaystyle=Var\left(\sum_{i=1}^{N}\textbf{1}[s_{iN}\in
A_{jm}]E[Z_{s_{iN}}|\mathcal{F}_{iN}(r)]\right):=Var\left(\sum_{i=1}^{N}\textbf{1}[s_{iN}\in
A_{jm}]Z_{iN}^{r}\right)$ $\displaystyle\leq\sum_{s_{iN}\in
A_{jm}}Var(Z_{iN}^{r})+\sum_{i\neq j}|Cov(Z_{iN}^{r},Z_{jN}^{r})|$
$\displaystyle:=\Xi_{1}+\Xi_{2}.$

Obviously, $\Xi_{1}\leq C_{p}\sigma^{2}$. We only need to concentrate on
$\Xi_{2}$, which can be decomposed as follow,

$\displaystyle\Xi_{2}$
$\displaystyle=\sum_{s_{iN}}\sum_{s_{jN}:\rho(s_{iN}),s_{jN})\leq
2r}|Cov(Z_{iN}^{r},Z_{iN}^{r})|+\sum_{s_{iN}}\sum_{s_{jN}:\rho(s_{iN}),s_{jN})>2r}|Cov(Z_{iN}^{r},Z_{iN}^{r})|$
$\displaystyle:=\Xi_{21}+\Xi_{22}.$

For $\Xi_{21}$, since
$Z_{iN}^{r}=Z_{s_{iN}}^{r}-Z_{s_{iN}}+Z_{s_{iN}}:=\Delta_{iN}^{r}+Z_{s_{iN}}$,
by using definition of NED and Cauchy-Schwarz inequality, we obtain

$\displaystyle|Cov(Z_{iN}^{r},Z_{jN}^{r})|$
$\displaystyle\leq|E[Z_{s_{iN}}Z_{s_{jN}}]|+||\Delta_{iN}^{r}||_{2}||\Delta_{jN}^{r}||_{2}+2\sigma\max_{i}||\Delta_{iN}||_{2}$
$\displaystyle\leq|E[Z_{s_{iN}}Z_{s_{jN}}]|+(\alpha(N)r^{-\nu_{1}})^{2}+2\sigma\alpha(N)r^{-\nu_{1}}$
$\displaystyle\leq\overline{\sigma}_{ij}+(\alpha(N)r^{-\nu_{1}})^{2}+2\sigma\alpha(N)r^{-\nu_{1}}.$

Here $\overline{\sigma}_{ij}=\max_{i,j}|E[Z_{s_{iN}}Z_{s_{jN}}]|$. Therefore,
together with Proposition 1, provided that
$\alpha(N)r^{-\nu_{1}}\leq\overline{\sigma}_{ij}$, we obtain

$\displaystyle\Xi_{21}\leq
C_{p}C_{0}\left(\left[\frac{r}{H_{0}}\right]+1\right)^{d_{2}}\max_{i,j}|Cov(Z_{iN}^{r},Z_{jN}^{r})|\leq
3C_{p}\left(C_{0}\left(\frac{2}{3H_{0}}\right)^{d_{2}}\right)r^{d_{2}}(\overline{\sigma}_{ij}\lor\overline{\sigma}_{ij}^{2}\lor
2\sigma\overline{\sigma}_{ij}).$

For $\Xi_{22}$, recall $Z_{iN}^{r}$ is a measurable function of random
variable $\epsilon_{s_{jN}}$ that $s_{jN}\in B(s_{iN},r)$ and
$\\{\epsilon_{s_{iN}}\\}_{s_{iN}\in\Gamma_{N}}$ is an $\alpha$-mixing random
field. Then, based on Lemma 2, by denoting $\max_{d_{1}+1\leq k\leq
d}p_{k}=\bar{p}$, $\min_{d_{1}+1\leq k\leq d}p_{k}=\underline{p}$ and letting
$r=\underline{p}/3$, we have,

$\displaystyle\Xi_{22}$ $\displaystyle=\sum_{s_{iN}\in
A_{jm}}\sum_{s_{jN}:\rho(s_{jN},s_{iN})\in(2r,\bar{p}]}|Cov(Z_{iN}^{r},Z_{jN}^{r})|$
$\displaystyle\leq\sum_{s_{iN}\in
A_{jm}}\sum_{s_{jN}:\rho(s_{jN},s_{iN})\in(\frac{2\underline{p}}{3},\bar{p}]}2^{\tau}\left(\frac{2\underline{p}}{3}\right)^{-\frac{\nu_{2}\delta}{2+\delta}}\sigma_{2+\delta}^{2}$
$\displaystyle\leq
C_{p}\sigma_{2+\delta}^{2}2^{\tau}\left(\frac{2}{3}\right)^{-\frac{\nu_{2}\delta}{2+\delta}}(C_{p}\underline{p}^{-\frac{\nu_{2}\delta}{2+\delta}})$
$\displaystyle\leq
C_{p}\sigma_{2+\delta}^{2}2^{\tau}\left(\frac{2}{3}\right)^{-\frac{\nu_{2}\delta}{2+\delta}}\left(\frac{C_{0}}{2^{d_{2}}H_{0}^{d_{1}}}\frac{N}{q^{d_{2}}}\underline{p}^{-\frac{\nu_{2}\delta}{2+\delta}}\right).$

According to Assumption 4, $\min_{d_{1}+1\leq k\leq
d}H_{k}^{N}\leq\max_{d_{1}+1\leq k\leq d}H_{k}^{N}\leq KN^{1/d_{2}}.$
Furthermore, according to geometric-mean inequality, we know
$\min_{d_{1}+1\leq k\leq d}H_{k}^{N}=O(\max_{d_{1}+1\leq k\leq d}H_{k}^{N})$
and $\underline{p}=O(p^{1/d_{2}})$. This implies, $\exists\ C^{*}>0$ such
that,

$\displaystyle\Xi_{2}2\leq
C_{p}\sigma_{2+\delta}^{2}\underbrace{C^{*}\left(\frac{N}{q^{d_{2}}}\right)^{\frac{-\nu_{2}\delta}{d_{2}(2+\delta)}+1}}_{:=B(N,q)}.$

Therefore,

$\displaystyle\Xi_{2}$ $\displaystyle\leq
3C_{p}\left(C_{0}\left(\frac{2}{H_{0}}\right)^{d_{2}}\right)r^{d_{2}}(\overline{\sigma}_{ij}\lor\overline{\sigma}_{ij}^{2}\lor
2\sigma\overline{\sigma}_{ij})+C_{p}\sigma_{2+\delta}^{2}B(N,q).$

Based on some simple algebra,

$\displaystyle Var(G_{jm}^{r})$ $\displaystyle\leq
C_{p}\sigma^{2}+3C_{p}\left(C_{0}\left(\frac{2}{H_{0}}\right)^{d_{2}}\right)r^{d_{2}}(\overline{\sigma}_{ij}\lor\overline{\sigma}_{ij}^{2}\lor
2\sigma\overline{\sigma}_{ij})+C_{p}\sigma_{2+\delta}^{2}B(N,q)$
$\displaystyle\leq 2C_{p}(\sigma^{2}+\sigma_{2+\delta}^{2}B(N,q)\lor
3\left(C_{0}\left(\frac{2}{H_{0}}\right)^{d_{2}}\right)r^{d_{2}}(\overline{\sigma}_{ij}\lor\overline{\sigma}_{ij}^{2}\lor
2\sigma\overline{\sigma}_{ij})).$

Recall $r=\frac{\underline{p}}{3}$, we obtain

$\displaystyle Var(G_{jm}^{r})$ $\displaystyle\leq
2C_{p}(\sigma^{2}+\sigma_{2+\delta}^{2}B(N,q)\lor
3\left(C_{0}\left(\frac{2}{3H_{0}}\right)^{d_{2}}\right)\underline{p}^{d_{2}}(\overline{\sigma}_{ij}\lor\overline{\sigma}_{ij}^{2}\lor
2\sigma\overline{\sigma}_{ij})).$

We thus get, there exists some $C^{**}>0$, such that

$\displaystyle Var(G_{jm}^{r})$ $\displaystyle\leq
2C_{p}\max\left\\{\sigma^{2}+\sigma_{2+\delta}^{2}B(N,q),\
3\left(C_{0}C^{**}\left(\frac{2}{3H_{0}}\right)^{d_{2}}\right)\frac{N}{q^{d_{2}}}(\overline{\sigma}_{ij}\lor\overline{\sigma}_{ij}^{2}\lor
2\sigma\overline{\sigma}_{ij})\right\\}$ $\displaystyle:=2C_{p}v(q).$ (B.11)

Step 4(Bernstein-type Inequality for $\\{G_{jm}^{r}\\}$)

Define $O_{jm}^{r}=\\{s_{jN}:s_{jN}\in\cap_{s_{iN}\in A_{jm}}B(s_{iN},r)\\}$,
where $B(s_{iN},r)$ denotes the closed ball centered at $s_{iN}$ at the radius
of $r$. For any given $m$, according the Step 1, we know $G_{jm}^{r}$ is a
real-valued measurable function of $\epsilon_{s_{iN}}$’s whose $s_{iN}\in
O_{jm}^{r}$, hence $\sigma(O_{jm}^{r})$-measurable. Moreover, the distance
between any $O_{jm}^{r}$ and $O_{im}^{r}$ is at least $\min_{d_{1}+1\leq k\leq
d}p_{k}-2r$, provided that $0<r<\min_{d_{1}+1\leq k\leq d}p_{k}/2$. Then, for
any $i\neq j$, according to Definition 2, the $\alpha$-mixing coefficient
between sigma fields $\sigma(G_{im}^{r})$ and $\sigma(G_{jm}^{r})$ can bounded
by inequality,

$\displaystyle\alpha(\sigma(G_{im}^{r}),\sigma(G_{jm}^{r}))\leq\alpha(\sigma(O_{im}^{r}),\sigma(O_{jm}^{r}))\leq\phi(|O_{im}^{r}|,|O_{jm}^{r}|)\psi_{\epsilon}(\rho(O_{im}^{r},O_{jm}^{r})).$

Based on Lemma 3, for any given $m$, there exists a sequence of independent
random variables, say $\\{W_{jm}\\}_{j=1}^{q^{d_{2}}}$, such that, for each
$j$, $W_{jm}$ is identically distributed as $G_{jm}^{r}$. Furthermore, for any
$s\geq 1$,

$\displaystyle P(|W_{jm}-G_{jm}^{r}|>\xi)$ $\displaystyle\leq
11\left(\frac{||G_{jm}^{r}+c||_{s}}{\xi}\right)^{\frac{s}{2s+1}}[\alpha(\sigma(G_{im}^{r}),\sigma(G_{jm}^{r}))]^{\frac{2s}{2s+1}}$
$\displaystyle\leq
11\left(\frac{||G_{jm}^{r}+c||_{s}}{\xi}\right)^{\frac{s}{2s+1}}[\phi(|O_{im}^{r}|,|O_{jm}^{r}|)\psi_{\epsilon}(\rho(O_{im}^{r},O_{jm}^{r}))]^{\frac{2s}{2s+1}}$
$\displaystyle(s=\infty)$ $\displaystyle\leq
11\left(\frac{||G_{jm}^{r}+c||_{\infty}}{\xi}\right)^{\frac{1}{2}}\phi(|O_{im}^{r}|,|O_{jm}^{r}|)\psi_{\epsilon}(\rho(O_{im}^{r},O_{jm}^{r}))$
$\displaystyle\leq
11\left(\frac{||G_{jm}^{r}+c||_{\infty}}{\xi}\right)^{\frac{1}{2}}\phi(|O_{im}^{r}|,|O_{jm}^{r}|)(\frac{\underline{p}}{3})^{-\nu_{2}}$
$\displaystyle\leq
11\left(\frac{||G_{jm}^{r}+c||_{\infty}}{\xi}\right)^{\frac{1}{2}}\left(C_{0}\left(\frac{4}{H_{0}}\right)^{d_{2}}p\right)^{\tau}(\frac{\underline{p}}{3})^{-\nu_{2}}.$

Now let $p=\infty$, $c=\delta C_{p}A$, $\delta=1+\frac{\bar{\eta}}{2A}$ and
$\xi=\min\\{\hat{N}\bar{\eta}/2q^{d_{2}},(\delta-1)C_{p}A\\}$. Since
$||G_{jm}^{r}+c||_{\infty}\geq\delta
C_{p}A-||G_{jm}^{r}||_{\infty}\geq(\delta-1)C_{p}A\geq\xi$, our setting
coincides with the preconditions of Lemma 3. Therefore, we obtain

$\displaystyle P(|W_{jm}-G_{jm}^{r}|>\xi)$ $\displaystyle\leq
11\left(\frac{(\delta+1)C_{p}A}{\min\\{\frac{\hat{N}\bar{\eta}}{2q^{d_{2}}},(\delta-1)C_{p}A\\}}\right)^{\frac{1}{2}}\left(C_{0}\left(\frac{4}{H_{0}}\right)^{d_{2}}p\right)^{\tau}(\frac{\underline{p}}{3})^{-\nu_{2}}$
$\displaystyle\leq 11\left(C_{0}\left(\frac{2}{H_{0}}\right)^{d_{2}}\lor
1\right)\left(1+\frac{4A}{\bar{\eta}}\right)^{\frac{1}{2}}\left(C_{0}\left(\frac{4}{H_{0}}\right)^{d_{2}}\right)^{\tau}p^{\tau}(\frac{\underline{p}}{3})^{-\nu_{2}}$
$\displaystyle\leq
11\left(1+\frac{4A}{\bar{\eta}}\right)^{\frac{1}{2}}p^{\tau}(\frac{\underline{p}}{3})^{-\nu_{2}}.$
(B.12)

The last inequality is due to Assumption 5. Furthermore, similar to the proof
of Theorem 1.3 in bosq2012nonparametric, we have

$\displaystyle Q_{4}$
$\displaystyle\leq\mathrm{P}\left(\left|\sum_{j=1}^{q^{d_{2}}}G_{jm}^{r}\right|\geq\hat{N}\bar{\eta}\right)$
$\displaystyle\leq\mathrm{P}\left(\left|\sum_{j=1}^{q^{d_{2}}}G_{jm}^{r}\right|\geq\hat{N}\bar{\eta};|W_{jm}-G_{jm}^{r}|\leq\xi\right)+\sum_{j=1}^{q^{d_{2}}}\mathrm{P}\left(|W_{jm}-G_{jm}^{r}|>\xi\right)$
$\displaystyle\leq\mathrm{P}\left(\left|\sum_{j=1}^{q^{d_{2}}}W_{jm}\right|\geq\hat{N}\bar{\eta}-q^{d_{2}}\xi\right)+\sum_{j=1}^{q^{d_{2}}}\mathrm{P}\left(|W_{jm}-G_{jm}^{r}|>\xi\right)$
$\displaystyle:=Q_{41}+Q_{42}.$

By applying (B.12), we instantly obtain

$\displaystyle Q_{42}\leq
11q^{d_{2}}\left(1+\frac{4A}{\bar{\eta}}\right)^{\frac{1}{2}}p(\frac{\underline{p}}{3})^{-\nu_{2}}=11q^{d_{2}}\left(1+\frac{8AH^{d_{2}-d_{1}}}{t}\right)^{\frac{1}{2}}p(\frac{\underline{p}}{3})^{-\nu_{2}}.$

For $Q_{41}$, note $\\{W_{jm}\\}_{j=1}^{q^{d_{2}}}$ is a mutually independent
sequence of real-valued random variables. Thus, Bernstein inequality for sum
of independent random variables could be applied, which yields

$\displaystyle Q_{41}$ $\displaystyle\leq
2\exp\left(-\frac{(\hat{N}\bar{\eta}/2)^{2}}{2(\sum_{j=1}^{q^{d_{2}}}\max_{j}Var(W_{jm})+\frac{||W_{jm}||_{\infty}\hat{N}\bar{\eta}}{6})}\right)$
$\displaystyle\leq
2\exp\left(-\frac{(\hat{N}\bar{\eta}/2)^{2}}{2(2q^{d_{2}}C_{p}v(q)+\frac{C_{p}A\hat{N}\bar{\eta}}{6})}\right)$
$\displaystyle\leq
2\exp\left(-\frac{(\hat{N}\bar{\eta}/2)^{2}}{2(2q^{d_{2}}pv(q)+\frac{pA\hat{N}\bar{\eta}}{6})}\right)$
$\displaystyle=2\exp\left(-\frac{\hat{N}t^{2}}{32H_{0}^{d_{2}-d_{1}}\left(2^{d_{2}-1}v(q)+\frac{Apt}{12H_{0}^{d_{2}-d_{1}}}\right)}\right)$

The second and third inequalities are due to (B.11) and $C_{p}\leq
C_{0}(2/H_{0})^{d_{2}}p\leq p$ respectively. The last equality is according to
$\hat{N}=(2q)^{d_{2}}p$ and $\bar{\eta}=\frac{t}{2H_{0}^{d_{2}-d_{1}}}$.
Together all the results above, we get

$\displaystyle\mathrm{P}\left(\left|\frac{\sum_{i=1}^{N}Z_{s_{iN}}}{N}\right|\geq
t\right)\leq
2\exp\left(-\frac{\hat{N}t^{2}}{32H_{0}^{d_{2}-d_{1}}\left(2^{d_{2}-1}v(q)+\frac{Apt}{12H_{0}^{d_{2}-d_{1}}}\right)}\right)$
$\displaystyle+$ $\displaystyle
11q^{d_{2}}\left(1+\frac{8AH^{d_{2}-d_{1}}}{t}\right)^{\frac{1}{2}}p^{\tau}(\frac{\underline{p}}{3})^{-\nu_{2}}+\left(\frac{C_{0}2^{d_{2}+1}}{H_{0}^{d_{1}}}\right)^{s}\left(\frac{\alpha(N)\underline{p}^{-\nu_{1}}}{3t}\right)^{s}$
$\displaystyle:=U_{1}+U_{2}+U_{3}.$

Together with Proposition 5, we have

$\displaystyle U_{1}\leq
2\exp\left(-\frac{H_{0}^{d_{1}}Nt^{2}}{32C_{0}\left(2^{d_{2}-1}v(q)+\frac{AH_{0}^{d_{1}}Nt}{12(2q)^{d_{2}}C_{0}}\right)}\right).$

For $U_{2}$ and $U_{3}$, the key is to specify $p$ and $\underline{p}$.
According to Assumption 3 and Assumption 4, we have $\min_{d_{1}+1\leq k\leq
d}H_{k}^{N}\leq\max_{d_{1}+1\leq k\leq d}H_{k}^{N}\leq KN^{1/d_{2}}.$ Then,
according to geometric-mean inequality, we know $\min_{d_{1}+1\leq k\leq
d}H_{k}^{N}=O(\max_{d_{1}+1\leq k\leq d}H_{k}^{N})$ and $\underline{p}\leq
p^{1/d_{2}}$. Thus, there exists some constant $K_{3}$ and $K_{4}>0$, such
that,

$\displaystyle U_{1}+U_{2}\leq
11K_{3}\left(1+\frac{8AH_{0}^{d_{2}-d_{1}}}{t}\right)^{\frac{1}{2}}q^{d_{2}}\left(\frac{N}{q^{d_{2}}}\right)^{-\frac{\nu_{2}}{d_{2}}+\tau}+K_{4}\left(\frac{N}{q^{d_{2}}}\right)^{-\frac{s\nu_{1}}{d_{2}}}\left(\frac{\alpha(N)}{t}\right)^{s}.$

Thus we finish the proof.

##  Appendix C Proof Related to Section 4

###  C.1 Section 4.1

Proof of Theorem 3

Proof of Theorem 3 is almost the same as the proof of Theorem 3 in
jenish2012nonparametric. Hence we only highlight the difference here. A basic
decomposition of $\hat{\theta}(x)-\theta(x)$ is as follow,

$\displaystyle\hat{\theta}(x)-\theta(x)=\begin{pmatrix}\hat{m}(x)-m(x)\\\
(\triangledown\hat{m}(x)-\triangledown
m(x))h\end{pmatrix}=S_{N}^{-1}(x)\underbrace{\left\\{T_{N}(x)-S_{N}(x)\begin{pmatrix}m(x)\\\
\triangledown m(x)h\end{pmatrix}\right\\}}_{:=W_{N}(x)},$

where, for $k=1,\dots,D+1$, by denoting $v^{k}$ as the k-th element of vector
$v$ and location $s_{iN}$ as $iN$,

$\displaystyle
W_{N}^{k}(x)=\frac{1}{Nh^{D}}\sum_{i=1}^{N}K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{k}(x)\omega_{iN}$
$\displaystyle\omega_{iN}=Y_{iN}-m(x)-\triangledown m(x)^{T}(X_{iN}-x).$

$\triangledown m(x)$ denotes the gratitude of $m(x)$. Hence, it suffice to
finish the proof if we demonstrate the following three things,

  1. (1)

$\sup_{x\in\mathcal{X}}||U^{-1}(x)||=O_{p}(1)$,

  2. (2)

$\sup_{x\in\mathcal{X}}|E[W_{N}^{1}(x)]|=O_{p}(h^{\beta})$,

  3. (3)

$\sup_{x\in\mathcal{X}}|W_{N}^{1}(x)-E[W_{N}^{1}(x)]|=\sqrt{\frac{\log
N}{Nh^{D}}}$,

where $||A||$ means Euclidean norm of matrix A. Furthermore, according to
Assumption 8, we have

$\displaystyle\sup_{x\in\mathcal{X}}|W_{N}^{1}(x)-E[W_{N}^{1}(x)]|$
$\displaystyle\leq$
$\displaystyle\sup_{x\in\mathcal{X}}\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}\left(K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{1}(x)Y_{iN}-E\left[K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{1}(x)Y_{iN}\right]\right)\right|$
$\displaystyle+$
$\displaystyle\sup_{x\in\mathcal{X}}||m||_{\infty}\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}\left(K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{1}(x)-E\left[K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{1}(x)\right]\right)\right|$
$\displaystyle+$ $\displaystyle\sup_{x\in\mathcal{X}}||\triangledown
m||\left\|\frac{1}{Nh^{D}}\sum_{i=1}^{N}\left((X_{iN}-x)K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{1}(x)-E\left[(X_{iN}-x)K\left(\frac{X_{iN}-x}{h}\right)U_{i}^{1}(x)\right]\right)\right\|$
$\displaystyle:=$ $\displaystyle\triangle_{1}+\triangle_{2}+\triangle_{3}$

Step 1(Union Upper Bound)

Define $Z_{iN}=(X_{iN},Y_{iN})$,

$\displaystyle\mathcal{K}(\mathcal{X})=\\{g_{x}(v)y:=K(v-x/h)y:v\in\prod_{l=1}^{D}[x^{(l)}-1,x^{(l)}+1],y\in[-L_{N},L_{N}],x\in\mathcal{X}\\}.$

Since $\mathcal{X}$ is compact in $\mathbf{R}^{D}$, here we assume $\exists\
M>0$, such that $\mathcal{X}\subset[-M,M]^{D}$. In this step, we focus on
deriving the convergence rate of

$\displaystyle\sup_{g\in\mathcal{K}(\mathcal{X})}\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}(g_{x}(X_{iN})Y_{iN}-E[g_{x}(X_{iN})Y_{iN}])\right|:=\sup_{g\in\mathcal{K}(\mathcal{X})}\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}h_{x}(Z_{iN})\right|.$

Therefore, according to some basic union bound argument based on covering
number, for any $t>0$,

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{K}(\mathcal{X})}\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}h_{x}(Z_{iN})\right|>t\right)\leq\mathcal{N}(th^{D}/3,\mathcal{K},||\cdot||_{\infty})\mathrm{P}\left(\left|\frac{1}{N}\sum_{i=1}^{N}h_{x}(Z_{iN})\right|>\frac{th^{D}}{3}\right),$

where $\mathcal{N}(th^{D}/3,\mathcal{K},||\cdot||_{\infty})$ denotes the
$th^{D}/3$-covering number of set $\mathcal{K}(\mathcal{X})$ with respect to
$||\cdot||_{\infty}$. Note that, based on Assumption 9 and some simple
algebra, for any $z\neq z^{\prime}$,

$\displaystyle|h_{x}(z)-h_{x}(z^{\prime})|\leq
2\sqrt{2}M\frac{L_{N}\lor||K||_{\infty}}{h}||z-z^{\prime}||_{2},$

which implies $Lip(h_{x})=O(L_{N}/h)$. Then, according to Proposition 2, for
any $x$ and $l$, $\\{h_{x}^{l}(Z_{iN})\\}_{s_{iN}\in\Gamma_{N}}$ is also a
centered and real-valued $L^{s}$-NED random field on process $\epsilon$ and
satisfies

$\displaystyle||h_{x}(Z_{iN})-E[h_{x}(Z_{iN})|\mathcal{F}_{iN}(r)]||_{s}\leq\underbrace{4\sqrt{2}M\frac{L_{N}\lor||K||_{\infty}}{h}}_{:=\alpha(N)}\psi_{Z}(r).$

Furthermore, due to Assumption 8, for any $x$ and the given $p_{0}>2$,

$\displaystyle||h_{x}||_{\infty}\leq||K||_{\infty}L_{N},\ Var(h_{x}(Z))\leq
h^{D}\sigma_{Y}^{2},$
$\displaystyle||h_{x}||_{p_{0}}^{2}=O(h^{\frac{2D}{p_{0}}}),\ \max_{i\neq
j}Cov(h_{x}(Z_{iN}),h_{x}(Z_{jN}))=O(h^{2D}).$

Therefore, by using Theorem 2, for any $q\in\mathbf{N}$,

$\displaystyle\mathrm{P}\left(\left|\frac{1}{N}\sum_{i=1}^{N}h_{x}(Z_{iN})\right|>th^{D}\right)$
$\displaystyle\lesssim\exp\left(-\frac{Nt^{2}h^{2D}}{v(q)+\frac{NL_{N}th^{D}}{q^{d_{2}}}}\right)$
$\displaystyle+\sqrt{\frac{1}{th^{D}}}\left(\frac{N}{q^{d_{2}}}\right)^{-\frac{\nu_{2}}{d_{2}}+\tau}q^{d_{2}}+\left(\frac{N}{q^{d_{2}}}\right)^{-\frac{sv_{1}}{d_{2}}}\left(\frac{L_{N}}{th^{D+1}}\right)^{s}$
$\displaystyle:=U_{1}+U_{2}+U_{3}.$ (C.2)

Step 2(Covering Number)

Define mapping $T(x):x\rightarrow K(v-x/h)y$, $x\in\mathcal{X}$. Then,
according to Assumption 9, it can be shown that, $T(x)$ is a Lipschitz mapping
with Lipschitz modulo $2\sqrt{2}M\frac{L_{N}\lor||K||_{\infty}}{h}$. Thus,
according to Proposition 13 of Hang ., we have

$\displaystyle\mathcal{N}\left(\frac{th^{D}}{3},\mathcal{K}(\mathcal{X}),||\cdot||_{\infty}\right)\lesssim\frac{L_{N}}{th^{D+1}}.$
(C.3)

Step 3(Generic Uniform)

Let $q=(N^{1+\frac{1}{p_{0}}}t)^{\frac{1}{d_{2}}}$, $t=\sqrt{\log N/Nh^{D}}$
and $L_{N}=N^{\frac{1}{p_{0}}}$. Firstly, we calculate $v(q)$. To calculate
$v(q)$, we only need to calculate

$\displaystyle B(N,q)h^{\frac{2D}{p_{0}}}\ \mbox{and}\
\frac{N}{q^{d_{2}}}h^{2D}.$

According to the (1)’s in Assumptions 10, 11 and 12, based on some simple
calculation, we directly obtain

$\displaystyle B(N,q)h^{\frac{2D}{P_{0}}}=O(h^{D}),\
\frac{N}{q^{d_{2}}}h^{2D}=O(h^{D}),$

which implies $v(q)=O(h^{D})$. Therefore,

$\displaystyle U_{1}$
$\displaystyle\lesssim\exp\left(-\frac{Nt^{2}h^{2D}}{h^{D}+\frac{N^{1+\frac{1}{p_{0}}}th^{D}}{q^{d_{2}}}}\right)\lesssim\exp(-Nt^{2}h^{D})\lesssim\exp(-\log
N)=\frac{1}{N}.$ (C.4)

Similarly, according to Assumption 10, we have

$\displaystyle U_{2}\lesssim
h^{\frac{\alpha\nu_{2}}{2d_{2}}-(\frac{\alpha\tau}{2}+\frac{5D}{4}+1)}(N\log
N)^{\frac{1}{4}}.$ (C.5) $\displaystyle
U_{3}\lesssim\left(h^{\frac{\alpha\nu_{1}}{2d_{2}}}\frac{N}{\sqrt{\log
N}}\right)^{s}.$ (C.6)

Combine (C.3) to (C.6) and Assumptions 10 and 11 and set $t=\sqrt{\log
N/Nh^{D}}$, we obtain

$\displaystyle\mathcal{N}\left(\frac{th^{D}}{3},\mathcal{K}(\mathcal{X}),||\cdot||_{\infty}\right)(U_{1}+U_{3}+U_{2})$
$\displaystyle\lesssim\frac{1}{\sqrt{\log
N}}+h^{\frac{\alpha\nu_{2}}{2d_{2}}-(\frac{\alpha\tau}{2}+\frac{5D}{4}+1)}(N^{5}\log
N)^{\frac{1}{4}}+\left(h^{\frac{\alpha\nu_{1}}{2d_{2}}}\frac{N^{2}}{\log
N}\right)^{s}=o(1).$

Furthermore, similar to (A.7) in hansen2008uniform, under Assumption 8, we can
also obtain

$\displaystyle
E\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}K\left(\frac{X_{iN}-x}{h}\right)Y_{iN}\textbf{1}[|Y_{iN}|>N^{\frac{1}{p_{0}}}]\right|^{p_{0}}=O\left(\frac{1}{N}\right).$

Hence, we manage to prove,

$\displaystyle\sup_{x\in\mathcal{X}}\left|\frac{1}{Nh^{D}}\sum_{i=1}^{N}\left(K\left(\frac{X_{iN}-x}{h}\right)Y_{iN}-E\left[K\left(\frac{X_{iN}-x}{h}\right)Y_{iN}\right]\right)\right|=O_{p}\left(\sqrt{\frac{\log
N}{Nh^{D}}}\right).$ (C.7)

Step 4

Based on (C.7), Assumption 8 and compactness of $\mathcal{X}$, it can be shown
that

$\displaystyle\sup_{x\in\mathcal{X}}|W_{N}^{1}(x)-E[W_{N}^{1}(x)]|=O_{p}\left(\sqrt{\frac{\log
N}{Nh^{D}}}\right).$

(2) can also be obtained by using standard argument of convolution kernel and
Hölder continuity of $m(x)$. Thus it is omitted here. To prove (1), it is
equivalent to prove matrix $U(x)$ is asymptotically invertable, which can be
done by using Proposition 4. Since procedure of this is almost the same as the
proof of Lemma 1 in Jenish, it is omitted as well. Then, we finish the proof.

###  C.2 Section 4.2

We firstly introduce the following to lemmas which are useful within our
proof.

###### Lemma 4

(Lemma 2.3.7 van1996weak) Let $=Z_{i}(f):=f(X_{i})-Ef(X_{i})$, where
$f\in\mathcal{F}$, for some real valued function class $\mathcal{F}$. For
arbitrary process $(Z_{i}(f))_{i=1}^{N}$ and arbitrary functions
$\mu_{1},...,\mu_{N}:\mathcal{F}\rightarrow\mathbf{R}$,

$\displaystyle\beta_{N}(\delta)\mathrm{P}^{*}\left(\sup_{f\in\mathcal{F}}|\sum_{i=1}^{N}Z_{i}(f)|>\delta\right)\leq
2\mathrm{P}^{*}\left(4\sup_{f\in\mathcal{F}}|\sum_{i=1}^{N}\epsilon_{i}(Z_{i}(f)-\mu_{i})|>\delta\right),$

for every $\delta>0$ and
$\beta_{N}(\delta)\leq\inf_{f}\mathrm{P}(|\sum_{i=1}^{N}Z_{i}(f)|<\delta/2)$.
Here $\mathrm{P}^{*}$ is outer measure and $\epsilon_{i}$ denotes Rademacher
random variable.

###### Lemma 5

Let $\mathcal{G}$ be a class of functions $g:\mathbf{R}^{m}\rightarrow[-A,A]$
with $2\leq V_{\mathcal{G}^{+}}<+\infty$. Let $Q$ be any probability measure
on $\mathbf{R}^{m}$ and let $0<\epsilon<A/2$. Then for any $p\geq 1$, we have

$\displaystyle\mathcal{M}(\epsilon,\mathcal{G},||\cdot||_{L^{p}(Q)})\leq
3\left(\frac{2^{p+1}eA^{p}}{\epsilon^{p}}\log\frac{3e(2A)^{p}}{\epsilon^{p}}\right)^{V_{\mathcal{G}}^{+}},$

where $\mathcal{M}(\epsilon,\mathcal{G},||\cdot||_{L^{p}(Q)})$ denotes the
$\epsilon$-packing number of set $\mathcal{G}$ with respect to
$||\cdot||_{L^{p}(Q)}$.

Please note, as we mentioned previously, we have postponed the problem of
measurability. Thus the outer probability in Lemma 4 is directly considered as
usual probabilistic measure when we apply it. As for Lemma 5, this result is a
direct consequence of Theorem 9.4 in gyorfi2002distribution. To prove this, we
only need to note that, this is equal to prove the the same argument works
fine on the following set,
$\\{g:\mathbf{R}^{m}\rightarrow[0,2A]\\}:=\mathcal{G}^{\prime}$ on condition
that $VC_{\mathcal{G}^{\prime}}^{+}\geq 2$. Hence we omit the proof here.

Proof of Proposition 6

Step 1(Symmetrization by a ghost sample)

Let $X_{N}=(X_{s_{iN}})_{i=1}^{N}$ and introduce a shadow sample vector
$X_{N}^{{}^{\prime}}:=(X_{s_{iN}}^{{}^{\prime}})_{i=1}^{N}$ such that
$X_{N}^{{}^{\prime}}$ is an independent copy of $X_{N}$. Then, we introduce a
Rademacher sequence, $\\{\epsilon_{i}\\}_{i=1}^{N}$, i.e. $\epsilon_{i}$’s are
mutually independent and for each $i$,
$P(\epsilon_{i}=1)=P(\epsilon_{i}=-1)=\frac{1}{2}$. Let vector
$(\epsilon_{i})_{i=1}^{N}$ is also independent from vector $X_{N}$. Due to
Corollary 1, we know, for $\forall\ g\in\mathcal{G}$ and some constants
$C_{0}$ and $K_{2}$ inherited from Theorem 1, for $\forall\ t>0$

$\displaystyle\mathrm{P}\left(\left|N^{-1}\sum_{i=1}^{N}(g(X_{s_{iN}})-E[g(X_{s_{iN}})])\right|>t\right)<4\exp\left(-\frac{Nt^{2}}{K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}((C_{0}A)^{2}+\frac{8C_{0}At}{3})}\right),$

where the upper bound holds uniformly over function class $\mathcal{G}$. Thus,
according to Lemma 4, by letting $\delta=Nt$,

$\displaystyle\beta_{N}(Nt)$
$\displaystyle=\inf_{g\in\mathcal{G}}\mathrm{P}\left(\left|N^{-1}\sum_{i=1}^{N}(g(X_{s_{iN}})-E[g(X_{s_{iN}})])\right|\leq\frac{t}{2}\right)$
$\displaystyle=\inf_{g\in\mathcal{G}}1-\mathrm{P}\left(\left|N^{-1}\sum_{i=1}^{N}(g(X_{s_{iN}})-E[g(X_{s_{iN}})])\right|>\frac{t}{2}\right)$
$\displaystyle=1-\sup_{g\in\mathcal{G}}\mathrm{P}\left(\left|N^{-1}\sum_{i=1}^{N}(g(X_{s_{iN}})-E[g(X_{s_{iN}})])\right|>\frac{t}{2}\right)$
$\displaystyle\geq 1-4\exp\left(-\frac{Nt^{2}}{4K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}((C_{0}A)^{2}+\frac{8C_{0}At}{3})}\right).$

Then for any $N\geq\min\\{N\in\mathbf{N}:\frac{N}{(\log N)^{d_{2}/\gamma}}\geq
4K_{2}\log 8C_{0}A((C_{0}A)+8t/3)/t^{2}\\}$, we have
$\beta_{N}(Nt)\leq\frac{1}{2}$. Therefore, according to Lemma 4, by letting
$\mu_{i}=g(X_{s_{iN}}^{{}^{\prime}})-E[g(X_{s_{iN}}^{{}^{\prime}})]$,

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|N^{-1}\sum_{i=1}^{N}(g(X_{s_{iN}})-E[g(X_{s_{iN}}])\right|>t\right)$
$\displaystyle\leq
4\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|N^{-1}\sum_{i=1}^{N}\epsilon_{i}(g(X_{s_{iN}})-g(X_{s_{iN}}^{{}^{\prime}}))\right|>t\right)$
$\displaystyle\leq
4\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|N^{-1}\sum_{i=1}^{N}\epsilon_{i}g(X_{s_{iN}})\right|>\frac{t}{2}\right)+4\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|N^{-1}\sum_{i=1}^{N}\epsilon_{i}g(X_{s_{iN}}^{{}^{\prime}})\right|>\frac{t}{2}\right)$
$\displaystyle\leq
8\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|N^{-1}\sum_{i=1}^{N}\epsilon_{i}g(X_{s_{iN}})\right|>\frac{t}{2}\right).$

Step 2(Chaining)

This step is similar to the proof of Theorem 19.1 in gyorfi2002distribution.
But in order to maintain the completeness of our proof, we still demonstrate
the detail here. Let $||\cdot||_{2}$ denotes the empirical $L^{2}$ norm on
condition that $X_{N}=x_{N}$. Since due to Cauchy-Schwarz inequality, for any
$g\in\mathcal{G}$,

$\displaystyle\sup_{g\in\mathcal{G}}\left|\frac{1}{N}\sum_{i=1}^{N}g(x_{s_{iN}})\epsilon_{i}\right|\leq||g||_{2}\sqrt{\frac{1}{N}\sum_{i=1}^{N}\epsilon_{i}^{2}}\leq
A.$

Hence we only need to consider the situation $t<2A$. For any $s\in\mathbf{N}$,
let $\\{g_{1}^{s},...,g_{N_{s}}^{s}\\}:=\mathcal{G}_{s}$ be a
$||\cdot||_{N}$-cover of $\mathcal{G}$ of radius $A/2^{s}$ on condition that
$X_{N}=x_{N}$. Here $N_{s}=\mathcal{N}_{2}(A/2^{s},\mathcal{G},||\cdot||_{N})$
is the $||\cdot||_{N}$\- covering number of $\mathcal{G}$ of radius $A/2^{s}$.
Thus, given $X_{N}=x_{N}$, by using chaining argument, for $\forall\
g\in\mathcal{G}$, we can rewrite it as

$\displaystyle g=g-g^{0}=g-g^{S}+\sum_{s=1}^{S}(g_{s}-g_{s-1}),$

where for each $s$, $g^{s}\in\mathcal{G}_{s}$ and obviously we can define
$g^{0}=0$. For simplicity, denote $g(x_{s_{iN}})$ as $g_{i}$. Let
$S=\min\\{s\geq 1:A/2^{s}\leq t/4\\}$. Thus, due to the chaining, we have

$\displaystyle\left|\frac{1}{N}\sum_{i=1}^{N}g_{i}\epsilon_{i}\right|$
$\displaystyle=\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}-g_{i}^{S})\epsilon_{i}+\sum_{s=1}^{S}\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|$
$\displaystyle\leq\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}-g_{i}^{S})\epsilon_{i}\right|+\sum_{s=1}^{S}\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|$
$\displaystyle\leq||g-g^{S}||\cdot\sqrt{\frac{1}{N}\sum_{i=1}^{N}\epsilon_{i}^{2}}+\sum_{s=1}^{S}\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|$
$\displaystyle\leq\frac{t}{4}+\sum_{s=1}^{S}\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|.$

Therefore, for any $\eta_{s}\geq 0,s=1,...,S$ and $\sum_{s=1}^{S}\eta_{s}\leq
1$, we have,by denoting $P_{X_{N}}(\cdot)=P(\cdot|X_{N}=x_{N})$,

$\displaystyle\mathrm{P}_{X_{N}}\left(\sup_{g\in\mathcal{G}}\left|\frac{1}{N}\sum_{i=1}^{N}g_{i}\epsilon_{i}\right|>\frac{t}{2}\right)$
$\displaystyle\leq\mathrm{P}_{X_{N}}\left(\exists
g\in\mathcal{G}:\frac{t}{4}+\sum_{s=1}^{S}\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|>\frac{t}{4}+\sum_{s=1}^{S}\eta_{s}\frac{t}{4}\right)$
$\displaystyle\leq\sum_{s=1}^{S}\mathrm{P}_{X_{N}}\left((\exists
g\in\mathcal{G}:\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|>\eta_{s}\frac{t}{4}\right)$
$\displaystyle\leq\sum_{s=1}^{S}N_{s}N_{s-1}\max_{g\in\mathcal{G}}\mathrm{P}_{X_{N}}\left(\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|>\eta_{s}\frac{t}{4}\right).$

Step 3(Conditional Concentration)

Note that, on condition of $X_{N}=x_{N}$,
$\\{(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\\}_{i=1}^{N}$ is an iid sequence. For
each i, the support is $[-g_{i}^{s}-g_{i}^{s-1},g_{i}^{s}-g_{i}^{s-1}]$. Thus,
by using Hoeffing inequality under iid setting, we have

$\displaystyle\mathrm{P}_{X_{N}}\left(\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|>\eta_{s}\frac{t}{4}\right)$
$\displaystyle\leq
2\exp\left(-\frac{Nt^{2}}{8||g^{s}-g^{s-1}||_{N}^{2}}\right)\leq
2\exp\left(-\frac{N(\eta_{s}t)^{2}}{72\frac{A^{2}}{2^{2s}}}\right),$

where the second inequality above ensures an upper bound independent of the
value of sample vector. Therefore,

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|\frac{1}{N}\sum_{i=1}^{N}g(X_{s_{iN}})\epsilon_{i}\right|>\frac{t}{2}\right)$
$\displaystyle=E\left[\mathrm{P}_{X_{N}}\left(\sup_{g\in\mathcal{G}}\left|\frac{1}{N}\sum_{i=1}^{N}g_{i}\epsilon_{i}\right|>\frac{t}{2}\right)\right]$
$\displaystyle\leq
2\sum_{s=1}^{S}E\left[N_{s}^{2}\max_{g\in\mathcal{G}}\mathrm{P}_{X_{N}}\left(\left|\frac{1}{N}\sum_{i=1}^{N}(g_{i}^{s}-g_{i}^{s-1})\epsilon_{i}\right|>\eta_{s}\frac{t}{4}\right).\right]$
$\displaystyle\leq 2\sum_{s=1}^{S}\exp\left(2\log
E[N_{s}]-\frac{N(\eta_{s}t)^{2}}{72\frac{A^{2}}{2^{2s}}}\right).$

.

Step 4(Proper design of $\eta_{s}$)

In order to get rid of the covering number, we firstly let $\eta_{s}$ satisfy

$\displaystyle 2\log
E[N_{s}]\leq\frac{Nt^{2}\eta_{s}^{2}}{144\frac{A^{2}}{2^{2s}}},$

which yields

$\displaystyle\eta_{s}\geq\frac{24\frac{A}{2^{s}}\sqrt{\log
E[\mathcal{N}(A/2^{s},\mathcal{G},||\cdot||_{N})]}}{\sqrt{N}t}.$

Combing Lemma 5 and Lemma 9.2 in gyorfi2002distribution, we can show that

$\displaystyle E[\mathcal{N}(A/2^{s},\mathcal{G},||\cdot||_{N})]\leq
3\left(\frac{8e}{(1/2^{s})^{2}}\log\frac{12e}{(1/2^{s})^{2}}\right)^{V_{\mathcal{G}}^{+}}.$

Hence, we define

$\displaystyle\frac{24A\frac{1}{2^{s}}\sqrt{\log
3\left(\frac{8e}{(1/2^{s})^{2}}\log\frac{12e}{(1/2^{s})^{2}}\right)^{V_{\mathcal{G}}^{+}}}}{\sqrt{N}t}:=\bar{\eta}_{s}$

and let

$\displaystyle\eta_{s}=\max\left\\{\bar{\eta}_{s},\frac{\sqrt{s}}{2^{s}4}\right\\}.$

Thus, due to the facts

$\displaystyle\sum_{s=1}^{S}\frac{\sqrt{s}}{2^{s}4}\leq\sum_{s=1}^{+\infty}\frac{\sqrt{s}}{2^{s}4}\leq\frac{1}{2}$

and

$\displaystyle\sum_{s=1}^{S}\bar{\eta}_{s}$
$\displaystyle\leq\sum_{s=1}^{S}\frac{24A\frac{1}{2^{s}}\sqrt{V_{\mathcal{G}}^{+}\log
3\left(\frac{8e}{(1/2^{s})^{2}}\log\frac{12e}{(1/2^{s})^{2}}\right)}}{\sqrt{N}t}$
$\displaystyle\leq
24A\frac{\sum_{s=1}^{S}\int_{\frac{1}{2^{s+1}}}^{\frac{1}{2^{s}}}\sqrt{V_{\mathcal{G}}^{+}\log
3\left(\frac{8e}{u^{2}}\log\frac{12e}{u^{2}}\right)}du}{\sqrt{N}t}$
$\displaystyle=24A\frac{\int_{\frac{1}{2^{S+1}}}^{\frac{1}{2}}\sqrt{V_{\mathcal{G}}^{+}\log
3\left(\frac{8e}{u^{2}}\log\frac{12e}{u^{2}}\right)}du}{\sqrt{N}t}$
$\displaystyle\leq
24A\frac{\int_{\frac{t}{16A}}^{\frac{1}{2}}\sqrt{V_{\mathcal{G}}^{+}\log
3\left(\frac{8e}{u^{p}}\log\frac{12e}{u^{p}}\right)}du}{\sqrt{N}t}.$

Note that for any $t>0$, there exists some constant $c>0$ such that,

$\displaystyle\int_{\frac{t}{16A}}^{\frac{1}{2}}\sqrt{V_{\mathcal{G}}^{+}\log
3\left(\frac{8e}{u^{p}}\log\frac{12e}{u^{p}}\right)}du\leq
c\int_{\frac{1}{2}}^{\frac{t}{16A}}\sqrt{\log\frac{1}{u^{2}}}du$
$\displaystyle=\sqrt{2}c\int_{\frac{1}{2}}^{\frac{t}{16A}}\sqrt{\log\frac{1}{u}}du\leq\sqrt{2}c\int_{\frac{1}{2}}^{\frac{t}{16A}}u^{-\frac{1}{2}}du\leq
2\sqrt{2}c<+\infty.$

Therefore, for any $t>0$,
$\int_{\frac{t}{16A}}^{\frac{1}{2}}\sqrt{V_{\mathcal{G}}^{+}\log
3\left(\frac{8e}{u^{p}}\log\frac{12e}{u^{p}}\right)}du<+\infty$. This ensures
the $N_{0}$ is always existed. Then, for any $N\geq N_{0}$, we have
$\sum_{s=1}^{S}\eta_{s}\leq 1$, which coincides with the requirement of
$\eta_{s}$’s.

Step 5(Plug in $\eta_{s}$)

Since $\eta_{s}\geq\frac{\sqrt{s}}{2^{s}4}\lor\bar{\eta}_{s}$, we obtain

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|\frac{1}{N}\sum_{i=1}^{N}g(X_{s_{iN}})\epsilon_{i}\right|>\frac{t}{2}\right)$
$\displaystyle\leq
2\sum_{s=1}^{S}\exp\left(-\frac{Nt^{2}\eta_{s}^{2}}{144\frac{A^{2}}{2^{2s}}}\right)$
$\displaystyle\leq 2\sum_{s=1}^{S}\exp\left(-\frac{sNt^{2}}{16\cdot
144A^{2}}\right)$ $\displaystyle\leq
5\exp\left(-\frac{Nt^{2}}{(48A)^{2}}\right),$

for any $\sqrt{N}t\geq 36A$. Hence we finish the proof.

Proof of Proposition 7

Similar to the proof of Proposition 6, the major tool here is still chaining.
Hence, we will skip some details and highlight the difference.

Step 1(Chaining)

Let $\\{g^{s}_{1},\dots,g^{s}_{N_{s}}\\}$ be the minimum $\sigma/2^{2s}$-net
of $\mathcal{G}$ with respect to infinite norm $||\cdot||_{\infty}$. Then, for
any $g\in\mathcal{G}$, we can construct chaining

$g=g-g^{0}+\sum_{s=1}^{S}g^{s}-g^{s-1},$

where $g^{0}=0$ and
$S=\min\\{s\in\mathbf{N}:\frac{\sigma}{2^{2s}}\leq\frac{t}{4L}\\}$. Then, we
have

$\displaystyle\left|P_{N}(gY)-P(gY)\right|$ $\displaystyle\leq$
$\displaystyle\left|P_{N}((g-g^{S})Y)-P((g-g^{S})Y)\right|+\sum_{s=1}^{S}\left|P_{N}((g^{s}-g^{s-1})Y)-P((g^{s}-g^{s-1})Y)\right|$
$\displaystyle\leq$ $\displaystyle
2L||g-g^{S}||_{\infty}+\sum_{s=1}^{S}\left|P_{N}((g^{s}-g^{s-1})Y)-P((g^{s}-g^{s-1})Y)\right|$
$\displaystyle\leq$
$\displaystyle\frac{t}{2}+\sum_{s=1}^{S}\left|P_{N}((g^{s}-g^{s-1})Y)-P((g^{s}-g^{s-1})Y)\right|.$

Therefore, similar to Step 2 in the proof of Proposition 6, for any
$\eta_{s},\ s=1,\dots,S$ such that $\sum_{s=1}^{S}\eta_{s}\leq 1$,

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|P_{N}(gY)-P(gY)\right|\geq
t\right)$ $\displaystyle\leq$
$\displaystyle\sum_{s=1}^{S}\mathrm{P}\left(\exists\
g:\left|P_{N}((g^{s}-g^{s-1})Y)-P((g^{s}-g^{s-1})Y)\right|\geq\frac{\eta_{s}t}{2}\right)$
$\displaystyle\leq$
$\displaystyle\sum_{s=1}^{S}N_{s}N_{s-1}\max_{g\in\mathcal{G}}\mathrm{P}\left(\left|P_{N}((g^{s}-g^{s-1})Y)-P((g^{s}-g^{s-1})Y)\right|\geq\frac{\eta_{s}t}{2}\right)$

Step 2(Application of Corollary 1)

Note that, for each $s=1,\dots,S$, by letting
$Z^{s}_{iN}=(g^{s}(X_{s_{iN}})-g^{s-1}(X_{s_{iN}}))Y_{s_{iN}}-E[(g^{s}(X_{s_{iN}})-g^{s-1}(X_{s_{iN}}))Y_{s_{iN}}]$,
due to the measurability of function $g^{s}$, we know process
$\\{Z^{s}_{iN}\\}_{s_{iN}\in\Gamma_{N}}$ is also a mean-zero geometric
irregularly-spaced $\alpha$-mixing random field whose mixing coefficient is
smaller than $\\{(X_{s_{iN}},Y_{s_{iN}})\\}_{s_{iN}\in\Gamma_{N}}$’s. In order
to apply Corollary1, we firstly need to investigate the upper bound of
$\sup_{iN}Var(Z^{s}_{iN})$ and $\sup_{iN}||Z^{s}_{iN}||_{\infty}$ for each s.

For $Var(Z^{s}_{iN})$, when $s\geq 2$, according to Step 1,

$\displaystyle Var(Z^{s}_{iN})$
$\displaystyle=||(g^{s}-g^{s-1})(X_{s_{iN}})Y_{s_{iN}}||_{2}^{2}\leq
L^{2}\left(||g^{s}-g^{s-1}||_{\infty}\right)^{2}$ $\displaystyle\leq
L^{2}\left(||g^{s}-g||_{\infty}+||g^{s-1}-g||_{\infty}\right)^{2}\leq
L^{2}\left(\frac{\sigma}{2^{2s}}+\frac{\sigma}{2^{2(s-1)}}\right)^{2}$
$\displaystyle\leq
L^{2}\left(\frac{\sigma}{2^{s}}+\frac{\sigma}{2^{(s-1)}}\right)^{2}=\frac{9L^{2}\sigma^{2}}{2^{2s}}.$

When $s=1$, since $g^{0}=0$,

$\displaystyle Var(Z^{1}_{iN})$
$\displaystyle=||g^{s}(X_{s_{iN}})Y_{s_{iN}}||_{2}^{2}\leq
L^{2}\sigma^{2}<\frac{9L^{2}\sigma^{2}}{4}.$

Therefore, for any $s$, we have
$\sup_{iN}Var(Z^{1}_{iN})\leq\frac{9L^{2}\sigma^{2}}{2^{2s}}$. Based on the
same type of argument, we can also show that, for any s,
$\sup_{iN}||Z^{s}_{iN}||_{\infty}\leq\frac{10AL}{2^{2s}}$. Then, according to
Corollary 1, we obtain

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|P_{N}(gY)-P(gY)\right|\geq
t\right)\leq 4\sum_{s=1}^{S}\exp\left(2\log
N_{s}-\frac{Nt^{2}\eta_{s}^{2}}{4K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}(C_{0}^{2}\frac{9L^{2}\sigma^{2}}{2^{2s}}+\frac{40C_{0}ALt}{3\cdot
2^{2s}})}\right).$

Together with the fact that $C_{0}\geq 1$ and condition C1,

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|P_{N}(gY)-P(gY)\right|\geq
t\right)$ $\displaystyle\leq 4\sum_{s=1}^{S}\exp\left(2\log
N_{s}-\frac{Nt^{2}\eta_{s}^{2}}{40C_{0}^{2}K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}\frac{L^{2}\sigma^{2}}{2^{2s}}}\right)$
$\displaystyle:=4\sum_{s=1}^{S}\exp(2\log N_{s}-H(N)).$

Finally, by the following the same procedure as Step 4 and 5 in the proof of
Proposition 6, we let

$\displaystyle\eta_{s}=\max\left\\{\frac{\sqrt{s}}{2^{s}4},\bar{\eta}_{s}\right\\},$
$\displaystyle\bar{\eta}_{s}=4\sqrt{10K_{2}\sigma}C_{0}L\sqrt{\frac{\sigma}{2^{2s}}}\sqrt{\log\mathcal{N}(\frac{\sigma}{2^{2s}},\mathcal{G},||\cdot||_{\infty})}\sqrt{\frac{(\log
N)^{\frac{d_{2}}{\gamma}}}{Nt^{2}}}.$

Obviously, it’s obvious that, based on this definition of $\eta_{s}$, we
immediately get $2\log N_{s}\leq H(N)/2$. The only condition needs to be
specified is $\sum_{s=1}^{S}\eta_{s}\leq 1$. It has already been shown that
$\sum_{s=1}^{S}\frac{\sqrt{s}}{2^{s}4}\leq\frac{1}{2}$. We only need to
calculate $\sum_{s=1}^{S}\bar{\eta}_{s}$.

$\displaystyle\sum_{s=1}^{S}\bar{\eta}_{s}=4\sqrt{10K_{2}\sigma}C_{0}L\sqrt{\frac{(\log
N)^{\frac{d_{2}}{\gamma}}}{Nt^{2}}}\sum_{s=1}^{S}\sqrt{\frac{\sigma}{2^{2s}}}\sqrt{\log\mathcal{N}(\frac{\sigma}{2^{2s}},\mathcal{G},||\cdot||_{\infty})}.$

Note
$\sqrt{\frac{\sigma}{2^{2s}}}=\sqrt{\frac{\sigma}{2^{2(s-1)}}}-\sqrt{\frac{\sigma}{2^{2s}}}$,
then based on the definition of Stieltjes integral, we know

$\displaystyle\sum_{s=1}^{S}\bar{\eta}_{s}\leq\sqrt{10K_{2}\sigma}C_{0}L\sqrt{\frac{(\log
N)^{\frac{d_{2}}{\gamma}}}{Nt^{2}}}\int_{\frac{\sqrt{\sigma}}{2^{S}}}^{\sqrt{\sigma}}\sqrt{\log\mathcal{N}(u,\mathcal{G},||\cdot||_{\infty})}d\sqrt{u}.$

Even though the integral above is a Stieltjes integral, considering that for
any interval $[a,b]\subset\mathbf{R}$, function $\sqrt{u}$ is continuous and
differentialble, Stieltjes integral is equal to Riemann integral. Thus, we
have

$\displaystyle\int_{\frac{\sqrt{\sigma}}{2^{S}}}^{\sqrt{\sigma}}\sqrt{\log\mathcal{N}(u,\mathcal{G},||\cdot||_{\infty})}d\sqrt{u}$
$\displaystyle=\int_{\frac{\sigma^{1/4}}{2^{S/2}}}^{\sigma^{1/4}}\sqrt{\log\mathcal{N}(u^{2},\mathcal{G},||\cdot||_{\infty})}du$
$\displaystyle\leq\int_{\frac{1}{2}(\frac{t}{L})^{\frac{1}{4}}}^{\sigma^{1/4}}\sqrt{\log\mathcal{N}(u^{2},\mathcal{G},||\cdot||_{\infty})}du,$

where the last inequality is due to the definition of $S$. Thus, according to
C2, we know $\sum_{s=1}^{S}\bar{\eta}_{s}\leq\frac{1}{2}$. Therefore, we
ensure the $\eta_{s}$’s are well defined. Finally, we obtain

$\displaystyle\mathrm{P}\left(\sup_{g\in\mathcal{G}}\left|P_{N}(gY)-P(gY)\right|\geq
t\right)\leq 4\sum_{s=1}^{S}\exp\left(-\frac{sNt^{2}}{1440C_{0}^{2}K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}L^{2}\sigma^{2}}\right)$
$\displaystyle\leq\frac{4}{1-\exp\left(-\frac{Nt^{2}}{1440C_{0}^{2}K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}L^{2}\sigma^{2}}\right)}\exp\left(-\frac{Nt^{2}}{1440C_{0}^{2}K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}L^{2}\sigma^{2}}\right)$ $\displaystyle\leq
8\exp\left(-\frac{Nt^{2}}{1440C_{0}^{2}K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}L^{2}\sigma^{2}}\right),$

where the last inequality is due to condition C3.

Proof of Proposition 8

Denote $X_{s_{iN}}=X_{i}$. For any $l=0,1,\dots,p$, let

$\displaystyle
g^{l}_{h}(z)=\left(\frac{z-x}{h}\right)^{l}K_{h}\left(\frac{z-x}{h}\right),\
K_{h}\left(\frac{z-x}{h}\right)=\frac{1}{h}K\left(\frac{z-x}{h}\right),$
$\displaystyle\mathcal{G}_{l}=\\{g^{l}(z):[a,b]\rightarrow\mathbf{R},K\left(\frac{z-x}{h}\right)\
\textrm{satisfies Assumption 7},x\in[a,b]\\}.$

Step 1

Obviously, we have the following simple decomposition

$\displaystyle\frac{1}{N}\sum_{i=1}^{N}g_{h}^{l}(X_{i})F_{N}(X_{i})$
$\displaystyle=$
$\displaystyle\frac{1}{N}\sum_{i=1}^{N}g_{h}^{l}(X_{i})\underbrace{(F_{N}(X_{i})-F(X_{i}))}_{\triangle_{N}(X_{i})}+\frac{1}{N}\sum_{i=1}^{N}g^{l}_{h}(X_{i})F(X_{i}),$

and similarly,

$\displaystyle
E\left[\left(\frac{X_{i}-x}{h}\right)^{l}K_{h}\left(\frac{X_{i}-x}{h}\right)F_{N}(X_{i})\right]=E\left[g^{l}_{h}(X_{i})\triangle_{N}(X_{i})\right]+E\left[g^{l}_{h}(X_{i})F(X_{i})\right].$

This implies,

$\displaystyle\left|\frac{1}{N}\sum_{i=1}^{N}\left(g^{l}_{h}(X_{i})F_{N}(X_{i})-E\left[g^{l}_{h}(X_{i})F_{N}(X_{i})\right]\right)\right|$
$\displaystyle\leq$
$\displaystyle\left|\frac{1}{N}\sum_{i=1}^{N}\left(g^{l}_{h}(X_{i})\triangle_{N}(X_{i})-E\left[g^{l}_{h}(X_{i})\triangle_{N}(X_{i})\right]\right)\right|$
$\displaystyle+$
$\displaystyle\left|\frac{1}{N}\sum_{i=1}^{N}\left(g^{l}_{h}(X_{i})F(X_{i})-E\left[g^{l}_{h}(X_{i})F(X_{i})\right]\right)\right|:=S_{1}(x)+S_{2}(x).$

Therefore,

$\displaystyle\sup_{x\in[a,b]}\left|\frac{1}{N}\sum_{i=1}^{N}\left(g^{l}_{h}(X_{i})F_{N}(X_{i})-E\left[g^{l}_{h}(X_{i})F_{N}(X_{i})\right]\right)\right|\leq\sup_{x\in[a,b]}S_{1}(x)+\sup_{x\in[a,b]}S_{2}(x)$

Step 2($S_{2}(x)$)

In order to measure the capacity of set $\mathcal{G}_{l}$, we firstly
investigate the Lipschitz property of $g^{l}\in\mathcal{G}_{l}$. Considering
the support is $[a,b]$, due to Assumption 13, we have, for any $z\in[a,b]$,

$\displaystyle|g_{h}^{l}(z-x)-g^{l}_{h}(z-x^{\prime})|$ $\displaystyle\leq$
$\displaystyle\left|K_{h}\left(\frac{z-x}{h}\right)\right|\left|\left(\frac{z-x}{h}\right)^{l}-\left(\frac{z-x^{\prime}}{h}\right)^{l}\right|+\left|\frac{z-x^{\prime}}{h}\right|^{l}\left|K_{h}\left(\frac{z-x}{h}\right)-K_{h}\left(\frac{z-x^{\prime}}{h}\right)\right|$
$\displaystyle\leq$ $\displaystyle(\overline{K}\lor
l(b-a)^{l-1})h^{-(l+1)}|x-x^{\prime}|+(b-a)^{l}h^{-(l+2)}|x-x^{\prime}|$
$\displaystyle\leq$ $\displaystyle ch^{-(l+2)}|x-x^{\prime}|,\ \textrm{for
some}\ c>0.$

Hence, define a mapping
$T:([a,b],|\cdot|)\rightarrow(\mathcal{G}^{l},||\cdot||_{\infty})$. The
Lipschitz modulo of $T$ is $O(1/h^{l+2})$. Under Assumption 13, according to
Proposition 13 of hang2018kernel, we have, there exists some $c^{\prime}>0$
such that

$\mathcal{N}(u^{2},\mathcal{G}_{l},||\cdot||_{\infty})\leq\frac{c^{\prime}}{u^{2}h^{l+2}}.$

Also note that for any $g^{l}\in\mathcal{G}_{l}$, $Var[g^{l}(X)]\leq O(1/h)$,
$l=0,1,\dots,p$. Then, to apply Proposition 7, we only need to ensure
conditions C1-C3 hold in an asymptotic level. Let $t=\sqrt{\log N/Nh}$.
Actually, under Assumption 14 and16, C1 and C3 hold obviously. Hence, we only
need to check C2.

By letting ”$A\preceq B$ ”denote that A is asymptotically smaller than B, we
have

$\displaystyle\sqrt{\sigma}\int_{\frac{1}{2}(\frac{t}{L})^{1/4}}^{\sigma^{1/4}}\sqrt{\log\mathcal{N}(u^{2},\mathcal{K},||\cdot||_{\infty})}du\preceq\left(\frac{1}{h}\right)^{\frac{1}{4}}\int_{(\frac{\log
N}{Nh})^{1/8}}^{\frac{1}{h}^{1/8}}\sqrt{\log\frac{1}{u^{2}h^{l+2}}}du$
$\displaystyle=\left(\frac{1}{h}\right)^{\frac{1}{4}}\int_{(\frac{\log
N}{Nh})^{1/8}}^{\frac{1}{h}^{1/8}}\sqrt{2\log\frac{1}{u}+(l+2)\log\frac{1}{h}}du\preceq\left(\frac{1}{h}\right)^{\frac{1}{4}}\int_{(\frac{\log
N}{Nh})^{1/8}}^{\frac{1}{h}^{1/8}}\left(\sqrt{2\log\frac{1}{u}}+(l+2)\log\frac{1}{h}\right)du$
$\displaystyle\preceq\left(\frac{1}{h}\right)^{\frac{3}{8}}\log\frac{1}{h}+\left(\frac{1}{h}\right)^{\frac{1}{4}}\int_{(\frac{\log
N}{Nh})^{1/8}}^{\frac{1}{h}^{1/8}}u^{-\frac{1}{2}}du\preceq\left(\frac{1}{h}\right)^{\frac{3}{8}}\log\frac{1}{h}.$

On the other hand, by setting $t=\sqrt{\frac{\log N}{Nh}}$,
$\sqrt{\frac{Nt^{2}}{(\log N})^{d_{2}/\gamma}}=O(\sqrt{(\log
N)^{1-d_{2}/\gamma}/h})$. C2 is easily satisfied. According to Proposition 7,
and $F(X_{i})=Y_{i}$, for sufficiently large N,

$\displaystyle\mathrm{P}\left(\sup_{x\in[a,b]}S_{2}(x)>\sqrt{\frac{\log
N}{Nh}}\right)\leq 8\exp\left(-\frac{(\log
N)^{1-\frac{d_{2}}{\gamma}}}{1440C_{0}^{2}K_{2}}\right),$

which implies $\sup_{x\in[a,b]}S_{2}(x)=O_{p}\left(\sqrt{\frac{\log
N}{Nh}}\right)$.

Step 3($S_{1}(x)$)

Firstly, due to Corollary 2, an immediate consequence is

$\displaystyle\sup_{x\in[a,b]}|F_{N}(x)-F(x)|=\sup_{x\in[a,b]}\triangle_{N}(x)=O_{a.s.}\left(\sqrt{\frac{\log
N}{N}}\right).$

Then, according to some simple algebra, we almost surely have

$\displaystyle\sup_{x\in[a,b]}S_{1}(x)\leq\sup_{x\in[a,b]}\left(\left|\frac{1}{N}\sum_{i=1}^{N}g^{l}_{h}(X_{i})\right|+\left|E\left[g^{l}_{h}(X)\right]\right|\right)\sqrt{\frac{\log
N}{N}}$
$\displaystyle\leq\sup_{x\in[a,b]}\left(\underbrace{\left|\frac{1}{N}\sum_{i=1}^{N}g^{l}_{h}(X_{i})-E\left[g^{l}_{h}(X)\right]\right|}_{T^{1}_{N}(x)}+\underbrace{\left|E\left[g^{l}_{h}(X)\right]\right|}_{T^{2}_{N}(x)}\right)\sqrt{\frac{\log
N}{N}}.$

Please note the calculation of $\sup_{x\in[a,b]}T_{N}^{1}(x)$ is exactly the
same as Step 2 by considering $Y_{i}=1$. Thus, according to Corollary 2, we
obtain

$\displaystyle\sup_{x\in[a,b]}T_{N}^{1}(x)=O_{p}\left(\sqrt{\frac{\log
N}{Nh}}\right).$

As for $T_{N}^{2}(x)$, according to Assumption 15, we know the density
function $f$ at least has continuously derivative on support $[a,b]$ and is
uniformly bounded. Moreover, due to Assumption 13, it can be shown that

$\displaystyle|E[g^{l}_{h}(X)]|\leq\int_{\mathbf{R}}\left|\left(\frac{z-x}{h}\right)^{l}K_{h}\left(\frac{z-x}{h}\right)f(z)\right|dz\leq||f||_{\infty}\int_{-1}^{1}|v^{l}K(v)|dv<+\infty$

Therefore, we obtain

$\displaystyle\sup_{x\in[a,b]}S_{1}(x)=O_{p}\left(\sqrt{\frac{\log
N}{N}}\right)+o_{p}(1).$

Above all, we have shown that, for any $l=0,1,\dots,p$,

$\displaystyle\sup_{x\in[a,b]}\left|\frac{1}{N}\sum_{i=1}^{N}\left(g^{l}_{h}(X_{i})-E[g^{l}_{h}(X)]\right)\right|=O_{p}\left(\sqrt{\frac{\log
N}{Nh}}\right).$

Proof of Theorem 4

Firstly, we have the following decomposition existed,

$\displaystyle\hat{\theta}(x)-\theta(x)$ $\displaystyle=$
$\displaystyle\hat{\theta}(x)-U_{N}^{-1}(x)E(U_{N}(x)\theta(x))+U_{N}^{-1}(x)E(U_{N}(x)\theta(x))-\theta(x)$
$\displaystyle=$ $\displaystyle
U_{N}^{-1}(x)[V_{N}(x)-E(V_{N}(x))]+U_{N}^{-1}(x)E[V_{N}(x)-U_{N}(x)\theta(x)]+U_{N}^{-1}(x)[E(U_{N}(x))-U_{N}(x)]\theta(x)$
$\displaystyle:=$ $\displaystyle
U_{N}^{-1}(x)(W_{N}^{1}(x)+W_{N}^{2}(x)+W_{N}^{3}(x))$ .

Step 1($U_{N}(x)$)

Since the calculation of expectation has nothing to do with dependence, we
here directly borrow Lemma 1 in the supplementary of cattaneo2020simple which
yields,

$\displaystyle
E[U_{N}(x)]=\begin{cases}f(x)\int_{-1}^{1}r_{p}(u)r_{p}(u)^{T}K(u)du+O_{p}(1/\sqrt{Nh}),\
\mbox{if}\ x\ \mbox{is interior},\\\
f(x)\int_{-c}^{1}r_{p}(u)r_{p}(u)^{T}K(u)du+O_{p}(1/\sqrt{Nh}),\ \mbox{if}\ x\
\mbox{is lower boundary},c\in(0,1),\\\
f(x)\int_{-1}^{c}r_{p}(u)r_{p}(u)^{T}K(u)du+O_{p}(1/\sqrt{Nh}),\ \mbox{if}\ x\
\mbox{is upper boundary},c\in(0,1),\end{cases}$

where $r_{p}(u)=(1,u,u^{2},...,u^{p})$. By using $U_{N}(x)_{ij}$ to indicate
the (i,j)-th element of matrix $U_{N}(x)$, its variance is

$\displaystyle
Var(U_{N}(x)_{ij})=\frac{1}{N^{2}}Var\left(\sum_{i=1}^{N}g^{i+j-2}_{h}(X_{i})\right)$
$\displaystyle\leq\frac{1}{N^{2}}\sum_{i=1}^{N}Var(g^{i+j-2}_{h}(X_{i}))+\frac{1}{N^{2}}\sum_{i\neq
j}Cov(g^{i+j-2}_{h}(X_{i}),g^{i+j-2}_{h}(X_{j})):=V_{1}+V_{2}.$

According to Assumptions 13 and 15,

$\displaystyle
V_{1}\leq\frac{1}{Nh}||f||_{\infty}\int_{-1}^{1}u^{2i+2j-4}K^{2}(u)du\leq\frac{1}{Nh}||f||_{\infty}\int_{-1}^{1}K^{2}(u)du=O(\frac{1}{Nh}).$

Based on Lemma 3 and Assumption 14, we have, for some $\delta>0$,

$\displaystyle V_{2}$ $\displaystyle=\frac{1}{N^{2}}\sum_{i\neq
j}Cov(g^{i+j-2}_{h}(X_{i}),g^{i+j-2}_{h}(X_{j}))$
$\displaystyle=\frac{1}{N^{2}}\sum_{i=1}^{N}\sum_{m=1}^{+\infty}\sum_{s_{jN},\rho(s_{iN},s_{jN})\in[m,m+1)}Cov(g^{i+j-2}_{h}(X_{i}),g^{i+j-2}_{h}(X_{j}))$
$\displaystyle\leq\frac{1}{N^{2}}\sum_{i=1}^{N}\sum_{m=1}^{+\infty}m^{d-1}\exp\left(-\frac{\delta
bm^{\gamma}}{2+\delta}\right)||g^{i+j-2}||_{2+\delta}^{2}\leq\frac{M}{Nh},$

for some $M>0$, which yields, $V_{1}+V_{2}=O(\frac{1}{Nh})$. Then, $U_{N}(x)$
is asymptotically positive definite. Hence, the uniform convergence rate of
the k-th element of vector $\hat{\theta}(x)-\theta(x)$ is determined by
$\sup_{x\in[a,b]}|W_{N}^{i}(x)_{k}|$, i=1,2,3, where $W_{N}^{i}(x)_{k}$
denotes the k-th entry of vector $W_{N}^{i}(x)$.

Step 2($W_{N}^{3}(x)_{k}$)

Due to Assumption 15, CDF $F(x)$ is $p+1$ times continuously differentiable at
$x\in[a,b]$. Thus, for any $l=0,1,\dots,p$, we have
$||F^{l}(x)||_{\infty}<+\infty$. Then, for any $k=1,2,...,p+1$,

$\displaystyle|W_{N}^{3}(x)_{k}|$ $\displaystyle=$
$\displaystyle\left|\sum_{l=0}^{p}\frac{1}{N}\sum_{i=1}^{N}\left[\left(\frac{X_{i}-x}{h}\right)^{l+k-1}K_{h}\left(\frac{X_{i}-x}{h}\right)-E\left[\left(\frac{X_{i}-x}{h}\right)^{l+k-1}K_{h}\left(\frac{X_{i}-x}{h}\right)\right]\right]F^{(l)}(x)h^{l}\right|$
$\displaystyle\leq$
$\displaystyle\sum_{l=0}^{p}\left|\frac{1}{N}\sum_{i=1}^{N}\left[\left(\frac{X_{i}-x}{h}\right)^{l+k-1}K_{h}\left(\frac{X_{i}-x}{h}\right)-E\left[\left(\frac{X_{i}-x}{h}\right)^{l+k-1}K_{h}\left(\frac{X_{i}-x}{h}\right)\right]\right]\right|||F^{(l)}||_{\infty}h^{l}$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{p}\left|\frac{1}{N}\sum_{i=1}^{N}\left[g^{l+k-1}_{h}(X_{i})-E[g^{l+k-1}_{h}(X_{i})]\right]\right|||F^{(l)}||_{\infty}h^{l},$

which implies,

$\displaystyle\sup_{x\in[a,b]}|W_{N}^{3}(x)_{k}|$
$\displaystyle\leq\sum_{l=0}^{p}\sup_{g\in\mathcal{G}_{l+k}}\left|\frac{1}{N}\sum_{i=1}^{N}\left[g^{l+k-1}_{h}(X_{i})-E[g^{l+k-1}_{h}(X_{i})]\right]\right|||F^{(l)}||_{\infty}h^{l}$
$\displaystyle\leq p\max_{0\leq l\leq
p}||F^{(l)}||_{\infty}\sup_{g\in\mathcal{G}_{l+k}}\left|\frac{1}{N}\sum_{i=1}^{N}\left[g^{l+k-1}_{h}(X_{i})-E[g^{l+k-1}_{h}(X_{i})]\right]\right|$
$\displaystyle\leq p\max_{0\leq l\leq p}||F^{(l)}||_{\infty}\sqrt{\frac{\log
N}{Nh}},$

where the last inequality is due to ”$\sup_{x\in[a,b]}T_{N}^{1}(x)$” in Step 3
of the proof of Proposition 8.

Step 3($W_{N}^{1}(x)$ and $W_{N}^{2}(x)$)

For $W_{N}^{2}(x)$, Lemma 2 in cattaneo2020simple has already shown that
$\sup_{x\in[a,b]}|W_{N}^{2}(x)_{k}|=O(h^{p+1})$, for any $k=1,2,..,p+1$.

For $W_{N}^{1}(x)$, according to Proposition 8, we know,
$\sup_{x\in[a,b]}|W_{N}^{2}(x)_{k}|=O(\sqrt{\log N/Nh})$, for any
$k=1,2,..,p+1$.

Finally, based on all the results above, we prove, for any $v=0,1,...,p$

$\displaystyle\sup_{x\in[a,b]}|\hat{F}^{(v)}(x)-F^{(v)}(x)|=h^{-v}\sup_{x\in[a,b]}|\hat{\theta}_{v+1}(x)-\theta_{v+1}(x)|=O\left(\sqrt{\frac{\log
N}{Nh^{2v+1}}}+h^{p+1-v}\right).$

##  Appendix D Proof Related to Section 5

Proof of Theorem 5

Denote

$\displaystyle G_{h}=G_{h}(z^{\prime}-z/h)=h^{-(D+1)}G(z^{\prime}-z/h),$
$\displaystyle\mathcal{G}(\mathcal{Z})=\\{G_{h}(z^{\prime}-z/h):\mathcal{Z}\rightarrow[0,A];z\in\mathcal{Z}\\}$

Since density function is Holder continuous with exponent $\alpha$, it has
been widely know, the bias is smaller than $h^{\alpha}$. We only concentrate
on the stochastic error,
$\sup_{G_{h}\in\mathcal{G}(\mathcal{Z})}|P_{N}(G_{h})-P(G_{h})|$. Proof is
divided into four steps. The first three steps are very similar to the
previous proof, thus we only show the sketch here. Important difference lies
in Step 4. It is also the last step which demonstrate the advantage that
$\beta$ is allowed to be arbitrary large in Theorem 1. Additionally, in order
to simplify the notation, within the whole proof, we always admit
$\sigma=\sqrt{1/h^{D+1}}$. .

Step 1

By following Step 1 in the proof of Proposition 7, we introduce the same
chaining as follow,

$\displaystyle G_{h}=G-G_{h}^{S}+\sum_{s=1}^{S}G_{h}^{s}-G_{h}^{s-1},$

where $G_{h}^{0}=0$ and $S=\min\\{s\in\mathbf{N},\sigma/2^{2s}\leq t/4\\}$.
Also, introduce constants $\\{\eta_{s},s=1,\dots,S\\}$ such that
$\sum_{s=1}^{S}\eta_{s}\leq 1$. We can instantly obtain,

$\displaystyle\mathrm{P}\left(\sup_{G_{h}\in\mathcal{G}(\mathcal{Z})}\left|P_{N}(G_{h})-P(G_{h})\right|>t\right)$
$\displaystyle\leq$
$\displaystyle\sum_{s=1}^{S}N_{s}N_{s-1}\max_{G_{h}\in\mathcal{G}(\mathcal{Z})}\underbrace{\mathrm{P}\left(\left|P_{N}(G_{h}^{s}-G_{h}^{s-1})-P(G_{h}^{s}-G_{h}^{s-1})\right|>\frac{\eta_{s}t}{2}\right)}_{:=\triangle(s)}.$

Step 2

Firstly, by denoting

$\displaystyle
g_{iN}^{s}=G_{h}^{s}\left(\frac{Z_{iN}-z}{h}\right)-G_{h}^{s-1}\left(\frac{Z_{iN}-z}{h}\right)-P(G_{h}^{s}-G_{h}^{s-1}),$

we investigate array $\\{g_{iN}^{s}\\}_{s_{iN}\in\Gamma_{N}}$. Obviously, we
have, for any $z^{\prime}\neq z^{\prime\prime}$,

$\displaystyle\left|G_{h}\left(\frac{z^{\prime}-z}{h}\right)-G_{h}\left(\frac{z^{\prime\prime}-z}{h}\right)\right|\leq\frac{1}{h^{D+2}}||z^{\prime}-z^{\prime\prime}||,$

which, together with stationary, implies the Lipschitz modulo of function
$g_{iN}^{s}$ is $2/h^{D+2}$, for any $s=1,\dots,S$. Please note, for kernel
density estimator, a natural requirement is $1/Nh^{D+1}=o(1)$. Thus,

$\displaystyle\frac{1}{h^{D+2}}/N^{1+\frac{1}{D+1}}=o(1).$

According to Proposition 2, we know $\\{g_{iN}^{s}\\}_{s_{iN}\in\Gamma_{N}}$
is also a mean-zero and real-valued $L^{p}$ NED random field on $\epsilon$ and
satisfies, for some $\kappa>1+\frac{1}{D+1}$,

$\displaystyle||g_{iN}^{s}-E\left[g_{iN}^{s}|\mathcal{F}_{iN}(r)\right]||_{p}\leq\frac{4}{h^{D+2}}\psi_{Z}(r)\leq
N^{\kappa}\psi_{Z}(r).$

Furthermore, similar to Step 2 in the proof of Proposition 7, for any $i$ and
$N$,

$\displaystyle\sup_{iN}Var(g_{iN}^{s})\leq\frac{9}{2^{2s}h^{D+1}},\
\sup_{iN}||g_{iN}^{s}||_{\infty}\leq\frac{10A}{2^{2s}h^{D+1}}$

By applying Theorem 1, we obtain, for each $s$, when $t<3C_{0}/40A$,

$\displaystyle\triangle(s)$
$\displaystyle=\mathrm{P}\left(\left|\frac{1}{N}\sum_{i=1}^{N}g_{iN}^{s}\right|>\frac{\eta_{s}t}{2}\right)$
$\displaystyle\leq
2K_{1}\left(\frac{1}{N^{\kappa+2\beta+\tau+1}(\eta_{s}t/2)}\right)^{p}+4\exp\left(-\frac{Nt^{2}\eta_{s}^{2}}{4K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}(C_{0}^{2}\frac{9}{2^{2s}h^{D+1}}+\frac{40C_{0}At}{3\cdot
2^{2s}h^{D+1}})}\right)$ $\displaystyle\leq
2K_{1}\left(\frac{1}{N^{\kappa+2\beta+\tau+1}(\eta_{s}t/2)}\right)^{p}+4\exp\left(-\frac{Nt^{2}\eta_{s}^{2}}{40K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}\frac{C_{0}^{2}}{2^{2s}h^{D+1}}}\right).$

Therefore,

$\displaystyle\mathrm{P}\left(\sup_{G_{h}\in\mathcal{G}(\mathcal{Z})}\left|P_{N}(G_{h})-P(G_{h})\right|>t\right)$
$\displaystyle\leq$ $\displaystyle
2K_{1}\underbrace{\sum_{s=1}^{S}N_{s}^{2}\left(\frac{1}{N^{\kappa+2\beta+\tau+1}(\eta_{s}t/2)}\right)^{p}}_{:=M_{1}}+\underbrace{4\sum_{s=1}^{S}N_{s}^{2}\exp\left(-\frac{Nt^{2}\eta_{s}^{2}}{40K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}\frac{C_{0}^{2}}{2^{2s}h^{D+1}}}\right)}_{:=M_{2}}.$

Step 3($M_{2}$)

Please note that the treatment of $M_{2}$ is the same as Step 2 in the proof
of Proposition 7, which means we only need to specify conditions C2 and C3 in
Proposition 7. As for the covering number of set $\mathcal{G}(\mathcal{Z})$,
similar to the previous argument, the mapping
$T:(\mathcal{Z},||\cdot||)\rightarrow(\mathcal{G}(\mathcal{Z}),||\cdot||_{\infty})$
is Lipschitz continuous with Lipshitz modulo as $1/h^{-(D+2)}$. Again,
according to Proposition 13 in hang2018kernel, we obtain, for any $u>0$,
$\exists\ c^{\prime}>0$, such that,

$\displaystyle\mathcal{N}(u,\mathcal{G}(\mathcal{Z}),||\cdot||_{\infty})\leq\frac{c^{\prime}}{uh^{D+2}}.$

Additionally, we still design $\eta=\max\\{\sqrt{s}/2^{s}\cdot
4,\bar{\eta}_{s}\\}$, where $\bar{\eta}$ is directly inherited from Step 2 in
the proof of Proposition 7 except that $\sigma=\sqrt{1/h^{D+1}}$. Furthermore,
similar to Step 2 of the proof of Proposition 8, by letting
$t=\sqrt{\frac{\log N}{Nh^{D+1}}}$, for sufficiently large $N$, we have

$\displaystyle M_{2}\leq 8\exp\left(-\frac{(\log
N)^{1-\frac{d_{2}}{\gamma}}}{40K_{2}C_{0}^{2}}\right).$

Step 4($M_{1}$)

It suffices to only consider the case that $p=1$.

Note, for function $m(x)=\frac{\sqrt{x}}{2^{x}}$, $m^{\prime}(x)=\frac{1-2x\ln
2}{2^{x+1}\sqrt{x}}$, which is negative for any $x\geq 1$. Thus, for $\forall\
s=1,2,\dots,S$,

$\displaystyle\eta_{s}\geq\frac{\sqrt{s}}{2^{s}\cdot
4}\geq\frac{\sqrt{S}}{2^{S}\cdot
4}\Rightarrow\eta_{s}^{2}\leq\frac{S}{2^{2S}\cdot 16}.$

Note, according to the definition of $S$,

$\displaystyle\frac{\sigma}{2^{2S}}\leq\frac{t}{4}$ $\displaystyle\Rightarrow\
S\geq\frac{\log\frac{4L\sigma}{t}}{2\log 2},$
$\displaystyle\frac{1}{2^{2(S-1)}}>\frac{t}{4\sigma}$
$\displaystyle\Rightarrow\frac{1}{2^{2S}\cdot 16}>\frac{t}{256}\Rightarrow
S<\frac{\log\frac{16\sigma}{t}}{2\sigma}.$ (D.1)

Then,

$\displaystyle\frac{S}{2^{2S}\cdot
16}\geq\frac{t\log\frac{4\sigma}{t}}{1024\sigma}$ $\displaystyle\Rightarrow$
$\displaystyle\eta_{s}^{3}=(\eta_{s}^{2})^{\frac{3}{2}}\geq\frac{(t\log\frac{4\sigma}{t})^{\frac{3}{2}}}{2^{15}\sigma^{\frac{3}{2}}}.$

Therefore, we obtain

$\displaystyle\frac{1}{N^{\kappa+\tau+2\beta+1}(\eta_{s}t/2)}=\frac{2\eta_{s}^{2}}{N^{\kappa+\tau+2\beta+1}\eta_{s}^{3}}\leq\frac{2^{16}\sigma^{\frac{3}{2}}\eta_{s}^{2}}{N^{\kappa+\beta+\tau+1}\underbrace{(N^{\beta}t^{\frac{5}{2}}(\log\frac{4\sigma}{t})^{\frac{3}{2}})}_{:=R_{1}}}.$

Recall that $t=\sqrt{\frac{\log N}{Nh^{D+1}}}>\frac{1}{\sqrt{N}}$ and
$\sigma=\sqrt{\frac{1}{h^{D+1}}}$. Then,

$\displaystyle R_{1}\geq N^{\beta}t^{\frac{5}{2}}\geq N^{\beta-\frac{5}{4}}.$

Thus, by requiring $\beta>\frac{5}{4}$, we get $R_{1}\geq 1$ for $\forall\ N$,
which yields

$\displaystyle\frac{1}{N^{\kappa+\tau+2\beta+1}(\eta_{s}t/2)}\leq\frac{2^{16}\sigma^{\frac{3}{2}}\eta_{s}^{2}}{N^{\kappa+\tau+\beta+1}},\
\textrm{when}\ \beta>\frac{5}{4}.$ (D.2)

According to D.1, we have

$\displaystyle
M_{1}\leq\frac{2^{16}\sigma^{\frac{3}{2}}}{N^{\kappa+\beta+\tau+1}}\sum_{s=1}^{S}\eta_{s}^{2}N_{s}^{2}=\frac{2^{16}}{N^{\kappa+\beta+\tau+1}}\left(h^{-\frac{3(D+1)}{4}}\sum_{s=1}^{S}\eta_{s}^{2}N_{s}^{2}\right).$
(D.3)

Now,we concentrate on $\sum_{s=1}^{S}\eta_{s}^{2}N_{s}^{2}$. Since
$\eta_{s}=\max\\{\frac{\sqrt{s}}{2^{s}\cdot 4},\bar{\eta}_{s}\\}$, we only
need to calculate them separately. On one hand,

$\displaystyle\sum_{s=1}^{S}\left(\frac{\sqrt{s}}{2^{s}\cdot
4}\right)^{2}N_{s}^{2}$
$\displaystyle=\frac{1}{16}\sum_{s=1}^{S}\frac{N_{s}^{2}s}{2^{2s}}\leq\frac{S}{16\sigma}\sum_{s=1}^{S}\frac{\sigma}{2^{2s}}N_{s}^{2}=\frac{S}{16\sigma}\sum_{s=1}^{S}\frac{\sigma}{2^{2s}}\mathcal{N}^{2}(\frac{\sigma}{2^{2s}},\mathcal{G},||\cdot||_{\infty})$
$\displaystyle\leq\frac{S}{16\sigma}\int_{\sigma/2^{2S}}^{\sigma/4}\mathcal{N}^{2}(u,\mathcal{G},||\cdot||_{\infty})du\leq\frac{S}{16\sigma}\int_{t/16}^{\sigma/4}\mathcal{N}^{2}(u,\mathcal{G},||\cdot||_{\infty})du$
$\displaystyle\leq\frac{\log\frac{16\sigma}{t}}{32\sigma}\int_{t/16}^{\sigma/4}\mathcal{N}^{2}(u,\mathcal{G},||\cdot||_{\infty})du\leq\frac{\log\frac{16\sigma}{t}}{32\sigma}\int_{t/16}^{\sigma/4}\frac{1}{(uh^{D+2})^{2}}du,$

where the third inequality is due to (D.1). Then, since
$\sigma=1/\sqrt{h^{D+1}}$ and $t=\sqrt{\frac{\log N}{Nh^{D+1}}}$, together
with some simple calculation, we get

$\displaystyle
h^{-\frac{3(D+1)}{4}}\sum_{s=1}^{S}\eta_{s}^{2}N_{s}^{2}\leq\frac{\sqrt{\frac{N}{\log
N}}\log 16\sqrt{\frac{N}{\log N}}}{(h^{D+1})^{\frac{7}{4}}}\leq
N^{\frac{7}{4}}\sqrt{\frac{N}{\log N}}\log 16\sqrt{\frac{N}{\log N}}\leq
N^{\frac{11}{4}}.$

Hence, we require $\beta>\frac{11}{4}$. Here the second $\preceq$ is due to
the fact that, for sufficiently large N, $Nh^{D+1}\leq 1$, which is a
mandatory precondition for kernel density estimator for
$\mathbf{R}^{D+1}$-valued data.

On the other hand, based on some similar calculation, we have

$\displaystyle\sum_{s=1}^{S}\bar{\eta}_{s}^{2}N_{s}^{2}$ $\displaystyle\leq
160K_{2}C_{0}^{2}\sigma\frac{(\log
N)^{\frac{d_{2}}{\gamma}}}{Nt^{2}}\int_{t/16}^{\sigma/4}\mathcal{N}^{3}(u,\mathcal{G},||\cdot||_{\infty})du$
$\displaystyle\leq
1440K_{2}C_{0}^{2}h^{-\frac{3(D+3)}{2}}t^{-2}\leq\frac{N}{(\log
N)^{\frac{d_{2}}{\gamma}}h^{\frac{D+7}{2}}}\leq
N^{\frac{3}{2}+\frac{3}{D+1}}.$

Thus, we also require $\beta>\frac{3}{2}+\frac{3}{D+1}$.

Above all, for any $\beta>\max\\{\frac{11}{4},\frac{3}{2}+\frac{3}{D+1}\\}$,
we obtain $M_{1}\leq\frac{1}{N^{\kappa+\tau+1}}$.

Proof of Theorem 6

According to Assumption 19, we immediately have, for any $\xi>0$, $\exists\
a>0$,

$\displaystyle\mathrm{P}\left(\sup_{x\in\mathcal{X}}|\hat{y}_{x}-y_{x}|>\xi\right)\leq\mathrm{P}\left(\sup_{x\in\mathcal{X}}|f(\hat{y}_{x}|x)-f(y_{x}|x)|>a\right).$

Note that

$\displaystyle\sup_{x\in\mathcal{X}}|f(\hat{y}_{x}|x)-f(y_{x}|x)|$
$\displaystyle\leq\sup_{x\in\mathcal{X}}|f(\hat{y}_{x}|x)-\hat{f}(\hat{y}_{x}|x)|+\sup_{x\in\mathcal{X}}|\hat{f}(\hat{y}_{x}|x)-f(y_{x}|x)|$
$\displaystyle\leq\sup_{x\in\mathcal{X}}\sup_{y\in\mathcal{Y}}|\hat{f}(y|x)-f(y|x)|+\sup_{x\in\mathcal{X}}|\hat{f}(\hat{y}_{x}|x)-f(y_{x}|x)|.$

Recall the definition of $\hat{y}_{x}$ and $y_{x}$, we have
$\hat{f}(\hat{y}_{x}|x)=\sup_{y\in\mathcal{Y}_{x}}\hat{f}(y|x)$,
$f(y_{x}|x)=\sup_{y\in\mathcal{Y}_{x}}f(y|x)$. Then,

$\displaystyle\sup_{x\in\mathcal{X}}|f(\hat{y}_{x}|x)-f(y_{x}|x)|$
$\displaystyle\leq$
$\displaystyle\sup_{x\in\mathcal{X}}\sup_{y\in\mathcal{Y}}|\hat{f}(y|x)-f(y|x)|+\sup_{x\in\mathcal{X}}|\sup_{y\in\mathcal{Y}_{x}}\hat{f}(y|x)-\sup_{y\in\mathcal{Y}_{x}}f(y|x)|$
$\displaystyle\leq$ $\displaystyle
sup_{x\in\mathcal{X}}\sup_{y\in\mathcal{Y}}|\hat{f}(y|x)-f(y|x)|+\sup_{x\in\mathcal{X}}\sup_{y\in\mathcal{Y}_{x}}|\hat{f}(y|x)-f(y|x)|$
$\displaystyle\leq$ $\displaystyle
2\sup_{x\in\mathcal{X}}\sup_{y\in\mathcal{Y}}|\hat{f}(y|x)-f(y|x)|.$

Therefore,

$\displaystyle\mathrm{P}\left(\sup_{x\in\mathcal{X}}|\hat{y}_{x}-y_{x}|>\xi\right)\leq\mathrm{P}\left(\sup_{x\in\mathcal{X}}\sup_{y\in\mathcal{Y}}|\hat{f}(y|x)-f(y|x)|>\frac{a}{2}\right),$

which shows that, to prove the uniform convergence rate of uni-modal
regression, it suffice to investigate the $L^{\infty}$ loss of conditional
density estimator. Furthermore, according to the construction of
$\hat{f}(y|x)$, we have

$\displaystyle|\hat{f}(y|x)-f(y|x)|$ $\displaystyle=$
$\displaystyle\left|\frac{\hat{f}(x,y)}{\hat{f}(x)}-f(y|x)\right|=\left|\left(\frac{\hat{f}(x,y)}{\hat{f}(x)}-f(y|x)\right)\left(\frac{\hat{f}(x)}{f(x)}-1+\frac{\hat{f}(x,y)}{\hat{f}(x)}-f(y|x)\right)\right|$
$\displaystyle\leq$
$\displaystyle\left|\frac{f(x)\hat{f}(x,y)-f(x,y)\hat{f}(x)}{f^{2}(x)}\right|+\left|\frac{(\hat{f}(y|x)-f(y|x))(\hat{f}(x)-f(x))}{f(x)}\right|$
$\displaystyle\leq$
$\displaystyle\left|\frac{f(x,y)-\hat{f}(x,y)}{f(x)}\right|+\left|\frac{(f(x)-\hat{f}(x))f(x,y)}{f^{2}(x)}\right|+\left|\frac{(\hat{f}(y|x)-f(y|x))(\hat{f}(x)-f(x))}{f(x)}\right|$
$\displaystyle:=$ $\displaystyle A_{1N}(x,y)+A_{2N}(x,y)+A_{3N}(x,y).$

Obviously, $\sup_{x,y}|\hat{f}(y|x)-f(y|x)|=O(\sup_{x,y}A_{1N}(x,y))$.
Together with assumption $\inf f(x)>0$ and Theorem 5, we finish the proof.

##  Appendix E Proof Related to Section 6

Generally speaking, the proof of Theorem 7 is essentially same as the proof of
Theormem 3.1 in rigollet2009optimal. Major differences are some technical
details, since we have additional remainder term
”$(N^{\theta}\delta)^{-p}$”(see Definition 5). Withiout loss of generality, we
here assume $B=1$.

Before we prove Theorem 7, we firstly introduce the following useful lemma.

###### Lemma 6

Under the conditions of Theorem 7, for any given $l_{N}$, the plug-in level
set estimator $\hat{\Gamma}_{f}(\lambda):=\\{z\in
Z:\hat{f}(z)\geq\lambda+l_{N}\\}$ satisfies that, for some $C^{\prime}>0$,

$\displaystyle\sup_{f\in\mathcal{P}}E[d_{H}(\hat{\Gamma},\Gamma)]\leq
C^{\prime}(\varphi_{N}\lor l_{N})^{1+\rho}.$ (23)

Proof of Lemma 6

Note that
$\hat{\Gamma}\triangle\Gamma=(\hat{\Gamma}\cap\Gamma^{c})\cup(\hat{\Gamma}^{c}\cap\Gamma)$.
Then, due to the definition of $d_{H}$, we have

$\displaystyle
d_{H}(\hat{\Gamma},\Gamma)=\int_{\hat{\Gamma}\cap\Gamma^{c}}|f(z)-\lambda|d\mu+\int_{\hat{\Gamma}^{c}\cap\Gamma}|f(z)-\lambda|d\mu:=I_{1}+I_{2}.$

Define $\alpha_{N}=C_{\alpha}(\varphi_{N}\lor l_{N})$ and
$\beta_{N}=C_{\beta}(\varphi_{N}\lor\phi_{N})\sqrt{\log N}$, where
$C_{\alpha}=2\max(c_{\varphi},c_{\phi},1)$ and $C_{\beta}\geq
C_{\alpha}\max(c_{l},2\nu(1+\rho)/c_{2},1)$. Let $N_{0}$ be the smallest
integer satisfying $\alpha_{N}<\beta_{N}<\eta\land\epsilon_{0}\land\triangle$.
Without loss of generality, we always assume $N\geq N_{0}$ in the remainder of
the proof.

Since $\hat{\Gamma}^{c}\cap\Gamma=\\{\hat{f}<\lambda+l_{N},f>\lambda\\}$, by
defining

$\displaystyle A_{1}=\\{\hat{f}<\lambda+l_{N},\lambda<f\leq f+\alpha_{N}\\},$
$\displaystyle
A_{2}=\\{\hat{f}<\lambda+l_{N},\lambda+\alpha_{N}<f\leq\lambda+\beta_{N}\\},$
$\displaystyle A_{3}=\\{\hat{f}<\lambda+l_{N},f>\lambda+\beta_{N}\\},$

we immediately have $\hat{\Gamma}^{c}\cap\Gamma\subset A_{1}\cup A_{2}\cup
A_{3}$. This yields

$\displaystyle
I_{1}\leq\sum_{i=1}^{3}\int_{A_{i}}|f(z)-\lambda|d\mu:=I_{11}+I_{12}+I_{13}.$

For $I_{11}$, since $A_{1}\subset\\{0<|f-\lambda|\leq\alpha_{N}\\}$, we have

$\displaystyle I_{11}\leq
E\left[\int_{\\{0<|f-\lambda|\leq\alpha_{N}\\}}|f-\lambda|d\mu\right]\leq\alpha_{N}\mu\\{|f-\lambda|\leq\alpha_{N}\\}.$

Since we have assumed that $N\geq N_{0}$ which indicates
$\alpha_{N}<\eta\land\epsilon_{0}$, together with the condition that f is
$\rho$-exponent at level $\lambda$, we have

$\displaystyle I_{11}\leq c_{0}\alpha_{N}^{1+\rho}.$ (24)

For $I_{12}$, we firstly partition $A_{2}$ as follow,

$\displaystyle A_{2}=\cup_{j=1}^{J_{N}}B_{j}\cap A_{2},$

where
$B_{j}=\\{\hat{f}<\lambda+l_{N},\lambda+2^{j-1}\alpha_{N}<f\leq\lambda+2^{j}\alpha_{N}\\}\cap
D(\eta\land\epsilon_{0})$ and $J_{N}=[\log(\beta_{N}/\alpha_{N})]+2$. Thus we
can rewrite

$\displaystyle I_{12}=\sum_{j=1}^{J_{N}}E\int_{B_{j}\cap
A_{2}}|f-\lambda|d\mu:=\sum_{j=1}^{J_{N}}I_{12}^{j}.$

For each $j=1,\dots,J_{N}$, since $l_{N}\leq\alpha_{N}/2$, we have

$\displaystyle
B_{j}\subset\\{|\hat{f}-f|>2^{j-2}\alpha_{N}\\}\cap\\{|f-\lambda|<2^{j}\alpha_{N}\\}\cap
D(\eta\land\epsilon_{0}).$

Then based on Fubini’s Theorem, we obtain, for each $j$,

$\displaystyle I_{12}^{j}$ $\displaystyle=\int\int_{B_{j}\cap
A_{2}}|f(z)-\lambda|d\mu d\mathrm{P}$
$\displaystyle\leq\int_{\\{|f-\lambda|<2^{j}\alpha_{N}\\}\cap
D(\eta\land\epsilon_{0})}|f(z)-\lambda|\int\mathbf{1}_{\\{|\hat{f}-f|>2^{j-2}\alpha_{N}\\}}(\omega)d\mathrm{P}d\mu$
$\displaystyle\leq
2^{j}\alpha_{N}\int_{D(\eta\land\epsilon_{0})\cap\\{|f-\lambda|<2^{j}\alpha_{N}\\}}\mathrm{P}(|\hat{f}(z)-f(z)|>2^{j-2}\alpha_{N})d\mu.$

Since for any $j$, $c_{\varphi}\varphi_{N}\leq
2^{j-2}\alpha_{N}\leq\beta_{N}<\triangle$ and $z\in D(\eta)$, we obtain

$\displaystyle\mathrm{P}(|\hat{f}(z)-f(z)|>2^{j-2}\alpha_{N})\leq
c_{3}\exp\left(-c_{4}\left(\frac{2^{j-2}\alpha_{N}}{\varphi_{N}}\right)^{2}\right)+\left(\frac{1}{N^{\theta}2^{j-2}\alpha_{N}}\right)^{p},\
\forall\ \theta>0,$

which yields

$\displaystyle I_{12}^{j}\leq
c_{0}c_{3}(2^{j}\alpha_{N})^{1+\rho}\left(\exp\left(-c_{4}\left(\frac{2^{j-2}\alpha_{N}}{\varphi_{N}}\right)^{2}\right)+\left(\frac{1}{N^{\theta}2^{j-2}\alpha_{N}}\right)^{p}\right).$
(25)

According to (25), we instantly have

$\displaystyle I_{12}$ $\displaystyle\leq
c_{0}c_{3}\sum_{j=1}^{+\infty}(2^{j}\alpha_{N})^{1+\rho}\left(\exp\left(-c_{4}\left(\frac{2^{j-2}\alpha_{N}}{\varphi_{N}}\right)^{2}\right)+\left(\frac{1}{N^{\theta}2^{j-2}\alpha_{N}}\right)^{p}\right)$
(26) $\displaystyle\leq
c_{0}c_{3}\alpha_{N}^{1+\rho}\sum_{j=1}^{+\infty}2^{j(1+\rho)}e^{-c_{4}2^{2j-4}}+\frac{c_{0}c_{3}\alpha_{N}^{1+\rho-p}}{4^{p}N^{\theta
p}}\sum_{j=1}^{+\infty}\left(\frac{1}{2^{p-1-\rho}}\right)^{j}.$ (27)

Due to Assumption 21, we know there exits some positive constants
$C_{1}^{{}^{\prime}}$ and $C_{2}^{{}^{\prime}}$ such that,

$\displaystyle(25)\leq
c_{o}c_{3}C_{1}^{{}^{\prime}}\alpha_{N}^{1+\rho}+c_{0}c_{3}C_{2}^{{}^{\prime}}\frac{1}{N^{\theta
p}\alpha_{N}^{p-1-\rho}}.$

Recall $\alpha_{N}=O(\varphi_{N}\lor l_{N})$. Thus $\alpha_{N}\geq CN^{-\nu}$.
Hence, since $\theta$ is allowed to be arbitrary large, there exits
sufficiently large $\theta_{1}>0$, such that, for any $\theta>\theta_{1}$, we
have

$\displaystyle I_{12}\leq c_{0}c_{3}(C_{1}^{\prime}\lor
C_{2}^{\prime})\alpha_{N}^{1+\rho}.$ (28)

For $I_{13}$, since $A_{3}\subset\\{f>\lambda\\}$, due to Markov inequality
with respect to measure $\mu$, we have $\mu(A_{3})\leq\frac{1}{\lambda}$.
Furthermore, since $\beta_{N}/2>l_{N}$,

$A_{3}\subset\\{f>\lambda+\beta_{N}\\}\cap\\{|\hat{f}-f|\geq\beta_{N}/2\\}.$

By using Funibi’s Theorem, we have

$\displaystyle I_{13}$
$\displaystyle\leq\int\int_{\\{f>\lambda+\beta_{N}\\}\cap\\{|\hat{f}-f|\geq\beta_{N}/2\\}}|f-\lambda|d\mu
d\mathrm{P}$
$\displaystyle=\int_{\\{f>\lambda+\beta_{N}\\}}|f(z)-\lambda|\mathrm{P}(|\hat{f}(z)-f(z)|>\beta_{N}/2)d\mu$
$\displaystyle\leq\int_{\\{f>\lambda+\beta_{N}\\}}(f(z)-\lambda)\left(c_{1}\exp\left(-c_{2}\left(\frac{\beta_{N}}{2\phi_{N}}\right)^{2}\right)+\left(\frac{1}{N^{\theta}\beta_{N}}\right)^{p}\right)d\mu$
$\displaystyle\leq
c_{1}\exp\left(-c_{2}\left(\frac{\beta_{N}}{2\phi_{N}}\right)^{2}\right)+\left(\frac{1}{N^{\theta}\beta_{N}}\right)^{p}.$

Since $p>1$, $\beta_{N}=C_{\beta}(\varphi_{N}\lor\phi_{N})\sqrt{\log N}$ and
$C_{\beta}=2\max(c_{\varphi},c_{\phi},1)C_{\alpha}:=2C_{\beta}^{\prime}$ and
$C_{\beta}^{\prime 2}>C_{\beta}^{\prime}>2\nu(1+\rho)/c_{2}$,

$\displaystyle I_{13}\leq
c_{1}CN^{-\nu(1+\rho)}+\frac{1}{N^{\theta-\nu}\sqrt{\log N}}\leq
c_{1}\alpha_{N}^{1+\rho}+\frac{1}{N^{\theta-\nu}\sqrt{\log N}}.$

Obviously there exists sufficiently large $\theta_{2}>0$, such that, for any
$\theta>\theta_{2}$, we have $(N^{\theta-\nu}\sqrt{\log N})^{-1}\leq
c_{1}CN^{-\nu(1+\rho)}\leq c_{1}\alpha_{N}^{1+\rho}$. Above all, we obtain

$\displaystyle I_{13}\leq 2c_{1}\alpha_{N}^{1+\rho}.$ (29)

By combining (23), (28) and (29), we finish the proof. As for the proof of
$I_{2}$, in the same manner it can be shown that, for any $N\geq N_{0}$,
$I_{2}\lesssim\varphi_{N}^{1+\rho}$. The only difference with the part of the
proof detailed above is that in the step that corresponds to proving the upper
bound of $I_{13}$, we use the assumption that $\mu(\hat{f}>\lambda)\leq M$
almost surely with respect to $\mathrm{P}$, instead of Markov inequality.

Proof of Theorem 7

Obviously, (19) is a direct consequence of Lemma 6 applied with
$l_{N}\lesssim\varphi_{N}$. Hence, we concentrate on proving (20). In order to
apply (16, we need to show the existence of such a constant $L_{\mu}$
satisfying $\mu(\hat{\Gamma}\triangle\Gamma)$. This is indeed true, since we
have

$\displaystyle\mu(\hat{\Gamma}\triangle\Gamma)\leq\mu(\hat{f}\geq\lambda)+\mu(f\geq\lambda)\leq
M+\frac{1}{\lambda}.$

Therefore, we can choose $L_{\mu}=M+\frac{1}{\lambda}$. Thus, according to
(16), we instantly know, there exitst some positive constant $C$ such that

$\displaystyle E[d_{\triangle}(\hat{\Gamma},\Gamma)]$ $\displaystyle\leq
E[\mu(\hat{\Gamma}\triangle\Gamma\cap\\{f=\lambda\\})]+CE[(d_{H}(\hat{\Gamma},\Gamma))^{\frac{\rho}{1+\rho}}]$
$\displaystyle\lesssim
E[\mu(\hat{\Gamma}\triangle\Gamma\cap\\{f=\lambda\\})]+C(\varphi_{N}\sqrt{\log
N})^{\rho},$ (30)

where the second term of the right side of the second inequality above is due
to Jensen inequality and the Lemma 6. Additionally, please note that

$\displaystyle\hat{\Gamma}\triangle\Gamma\cap\\{f=\lambda\\}\subset\\{|\hat{f}-f|>l_{N}\\}\cap\\{f=\lambda\\}.$

Then, for any $\theta>0$ and $p>1+\rho$, by using Fubini’s Theorem we have

$\displaystyle E[\mu(\hat{\Gamma}\triangle\Gamma\cap\\{f=\lambda\\})]\leq
c_{1}\exp\left(-c_{2}\left(\frac{l_{N}}{2\varphi_{N}}\right)^{2}\right)+\left(\frac{1}{N^{\theta}l_{N}}\right)^{p}.$

Recall that $l_{N}=O(\varphi_{N}\sqrt{\log N})$ and $l_{N}\geq CN^{-\nu}$. We
thus obtain, there exists sufficiently large $\theta_{3}$, such that for any
$\theta>\theta_{1}\lor\theta_{2}\lor\theta_{3}$,

$\displaystyle
E[\mu(\hat{\Gamma}\triangle\Gamma\cap\\{f=\lambda\\})]\lesssim\varphi_{N}^{\rho}.$
(31)

Based on (E) and (31), the proof is finished.

Proof of Theorem 8

We prove Theorem 8 by following Theorem 7.

Denote $\bar{K}(u)=\prod_{k=1}^{D}K(u^{k})$, $u=(u^{1},\dots,u^{D})$, where
$K$ satisfies Assumption 23 with $\alpha=\beta^{*}$. Thus, it’s easy to
specify that $\bar{K}$ also satisfies (1) and (2) in Assumption 23.

Step 1(Specify $\exists\ M>0$ such that $\mu(\hat{f}\geq\lambda)\leq M\ a.s.\
w.r.t\ \mathrm{P}$)

In this step, instead of $d\mu$, we use $\mu(dz)$. Note that since $\mu$ is
the Lebesgue measure and $\bar{K}\in L^{1}_{\mu}(\mathbf{R}^{D})$, we
instantly have

$\displaystyle\infty>\int_{\mathbf{R}^{D}}|\bar{K}(z)|\mu(dz)\geq\int_{\\{\hat{f}\geq\lambda\\}}|\bar{K}(z)|\mu(dz).$

Recall that $\hat{f}=N^{-1}\sum_{i=1}^{N}\bar{K}_{h}((Z_{s_{iN}}-z)/h)$. For
arbitrary realization of sample, .i.e $Z_{s_{iN}}=z_{i}$, we have, for any
$i$,

$\displaystyle\int_{\\{\hat{f}\geq\lambda\\}}|\bar{K}(z)|\mu(dz)$
$\displaystyle=\int_{\\{\hat{f}\geq\lambda\\}}|\bar{K}((z_{i}-z)/h)|\mu(d(z_{i}-z/h))$
$\displaystyle=\int_{\\{\hat{f}\geq\lambda\\}}|\bar{K}_{h}((z_{i}-z)/h)|\mu(dz).$

This indicates, for any realization of sample $(z_{i})_{i=1}^{N}$

$\displaystyle\int_{\\{\hat{f}\geq\lambda\\}}|\bar{K}(z)|\mu(dz)=\frac{1}{N}\sum_{i=1}^{N}\int_{\\{\hat{f}\geq\lambda\\}}|\bar{K}_{h}((z_{i}-z)/h)|\mu(dz).$

Then based on simple triangle inequality and arbitrary of this realization, we
manage to obtain

$\displaystyle\infty>\int_{\mathbf{R}^{D}}|\bar{K}(z)|\mu(dz)\geq\int_{\\{\hat{f}\geq\lambda\\}}|\hat{f}(z)|\mu(dz)\geq\lambda\mu(\\{\hat{f}\geq\lambda\\}).$

Hence, we can set $M=\int|\bar{K}|/\lambda$.

Step 2($\hat{f}$ satisfies Definition 5)

Note that, for any given $z_{0}\in Z$,

$\displaystyle\left|\hat{f}(z_{0})-f(z_{0})\right|$
$\displaystyle=\left|\frac{1}{N}\sum_{i=1}^{N}\bar{K}_{h}\left(\frac{Z_{s_{iN}}-z_{0}}{h}\right)-f(z_{0})\right|=\left|\frac{1}{N}\sum_{i=1}^{N}W_{iN}(z_{0})\right|$
$\displaystyle\leq\left|\frac{1}{N}\sum_{i=1}^{N}(W_{iN}(z_{0})-E[W_{iN}(z_{0})])\right|+\max_{i}|E[W_{iN}(z_{0})]|$
$\displaystyle:=\left|\frac{1}{N}\sum_{i=1}^{N}\widetilde{W}_{iN}(z_{0})\right|+\max_{i}|E[W_{iN}(z_{0})]|,$
$\displaystyle W_{iN}(z_{0})$
$\displaystyle=\bar{K}_{h}\left(\frac{Z_{s_{iN}}-z_{0}}{h}\right)-f(z_{0}).$

Furthermore, due to the proof of Lemma 4.1 in rigollet2009optimal, under
Assumptions 22 and 23, it’s easy to obtain that, for any $i$,

$\displaystyle|E[W_{iN}(z_{0})]|\leq Lc_{5}h^{\beta},\ \mbox{when}\ z_{0}\in
D(\eta),$ $\displaystyle|E[W_{iN}(z_{0})]|\leq Lc_{5}h^{\beta^{{}^{\prime}}},\
\mbox{when}\ z_{0}\in D(\eta)^{c},$

where $c_{5}=\int_{\mathbf{R}^{D}}||t||^{\beta}|\bar{K}(t)|dt$. Then, provided
that $z_{0}\in D(\eta)$, for any $\delta>0$ such that
$Lc_{5}h^{\beta\land\beta^{{}^{\prime}}}\leq\delta/2$, we have

$\displaystyle\mathrm{P}\left(|\hat{f}(z_{0}))-f(z_{0})|\geq\delta\right)$
$\displaystyle\leq\mathrm{P}\left(\left|\frac{1}{N}\sum_{i=1}^{N}\widetilde{W}_{iN}(z_{0})\right|\geq\delta-
Lc_{5}h^{\beta}\right)$
$\displaystyle\leq\mathrm{P}\left(\left|\frac{1}{N}\sum_{i=1}^{N}\widetilde{W}_{iN}(z_{0})\right|\geq\frac{\delta}{2}\right).$
(32)

Similarly, when $z_{0}\in D(\eta)^{c}$, under condition
$Lc_{5}h^{\beta\land\beta^{{}^{\prime}}}\leq\delta/2$, we can obtain (E) as
well. Since the following argument is nearly the same no matter $z_{0}\in
D(\eta)$ or not, we concentrate solely on the case that $z_{0}\in D(\eta)$.

In order to apply Theorem 1, we need to ensure that
$(W_{iN}(z_{0}))_{s_{iN}\in\Gamma_{N}}$ is a centered real-valued and
geometric $L^{p}$-NED random field on process $\epsilon$.

Due to the definition of $\widetilde{W}_{iN}(z_{0})$, it is certainly
centered. Note that $\widetilde{W}_{iN}(z_{0})$ can be regarded as a function
of random vector $Z_{s_{iN}}$, for each i. According to stationarity of
$(Z_{s_{iN}})_{s_{iN}\in\Gamma_{N}}$, we know $E[W_{iN}(z_{0})]\equiv C$, for
some constant $C$. Now considering function

$\widetilde{W}(z,z_{0})=\bar{K}_{h}\left(\frac{z-z_{0}}{h}\right)-f(z_{0})-C,z\in\mathbf{R}^{D},$

we have, for any $z_{1}\neq z_{2}$ and some positive constant $E$,

$\displaystyle|\widetilde{W}(z_{1},z_{0})-\widetilde{W}(z_{2},z_{0})|$
$\displaystyle=\left|\bar{K}_{h}\left(\frac{z_{1}-z_{0}}{h}\right)-\bar{K}_{h}\left(\frac{z_{2}-z_{0}}{h}\right)\right|$
$\displaystyle=h^{-D}\left|\bar{K}\left(\frac{z_{1}-z_{0}}{h}\right)-\bar{K}\left(\frac{z_{2}-z_{0}}{h}\right)\right|$
$\displaystyle\leq
h^{-D}||K||_{\infty}^{D}\sum_{k=1}^{D}\left|K\left(\frac{z_{1}-z_{0}}{h}\right)-K\left(\frac{z_{2}-z_{0}}{h}\right)\right|$
$\displaystyle\leq
h^{-(D+1)}||K||_{\infty}^{D}\sum_{k=1}^{D}|z_{1}^{k}-z_{2}^{k}|$
$\displaystyle\leq h^{-(D+1)}||K||_{\infty}^{D}E||z_{1}-z_{2}||_{2}.$

Thus,
$Lip(\widetilde{W}(z,z_{0}))\leq||K||_{\infty}^{D}MEh^{-(D+1)}:=m_{1}h^{-(D+1)}$.
Then according to Proposition 2 and Assumption 20, we know
$\\{\widetilde{W}_{iN}(z_{0})\\}$ is a centered $L^{p}$-NED random field on
$\epsilon$ such that

$\displaystyle\left\|\widetilde{W}_{iN}(z_{0})-E[\widetilde{W}_{iN}(z_{0})|\mathcal{F}_{iN}(r)]\right\|_{p}\leq
2m_{1}h^{-(D+1)}e^{-br^{\gamma}}\leq 2N^{\kappa_{0}}e^{-br^{\gamma}},$ (33)

for some $\kappa_{0}>0$. Meanwhile, due to some simple calculation, we can
easily obtain that, for any $i$,

$\displaystyle||\widetilde{W}_{iN}(z_{0})||_{\infty}\leq(||K||_{\infty}/h)^{D}+L^{*}+Lc_{5}h^{\beta}\leq
c_{7}h^{-D},\ c_{7}=||K||_{\infty}^{D}+L^{*}+Lc_{5};$ (34) $\displaystyle
Var(\widetilde{W}_{iN}(z_{0}))\leq h^{-D}\int\bar{K}(u)^{2}f(z_{0}+hu)du\leq
L^{*}||\bar{K}||_{2}^{2}h^{-D}:=c_{8}h^{-D}.$ (35)

Based on (31)-(33), by applying Theorem 1, provided that $\delta\leq\triangle$
and

$\triangle=\frac{3C_{0}c_{8}}{4c_{7}}=\frac{3C_{0}L^{*}||\bar{K}||_{2}^{2}}{4(||K||_{\infty}^{D}+L^{*}+L\int_{\mathbf{R}^{D}}||t||^{\beta}|\bar{K}(t)|dt)},$

we have

$\displaystyle\mathrm{P}\left(\left|\frac{1}{N}\sum_{i=1}^{N}\widetilde{W}_{iN}(z_{0})\right|\geq\delta\right)$
$\displaystyle\leq$ $\displaystyle 2\exp\left(-\frac{N\delta^{2}}{4K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}(C_{0}^{2}c_{8}h^{-D}+\frac{4C_{0}c_{7}\delta}{3h^{D}})}\right)+2K_{1}\left(\frac{2}{N^{\kappa_{0}+2\vartheta+\tau+1}\delta}\right)^{p}$
$\displaystyle=$ $\displaystyle
2\exp\left(-\frac{N\delta^{2}h^{D}}{8K_{2}(\log
N)^{\frac{d_{2}}{\gamma}}}C_{0}^{2}c_{8}\right)+2K_{1}\left(\frac{2}{N^{\kappa_{0}+2\vartheta+\tau+1}\delta}\right)^{p}$
$\displaystyle=$ $\displaystyle
2\exp\left(-c_{6}\left(\frac{\delta}{\sqrt{(\log
N)^{d_{2}/\gamma}/Nh^{D}}}\right)^{2}\right)+\left(\frac{B}{N^{R(\vartheta)}\delta}\right)^{p},$
$\displaystyle c_{6}=8K_{2}C_{0}^{2}c_{8},\
R(\vartheta)=\kappa_{0}+2\vartheta+\tau+1,\
B=2^{1+\frac{1}{p}}K_{1}^{\frac{1}{p}}.$

Here $\vartheta$ is used to replace the $beta$ used in Theorem 1 so that we
won’t confuse it with the smoothness of density function. $K_{1}$ and $K_{2}$
are defined the same as Theorem 1 by setting $\kappa=\kappa_{0}$. Note that
$R(\vartheta)$ here can be arbitrary large. Therefore, we prove that, when
$z_{0}\in D(\eta)$, $\hat{f}$ satisfies Definition 5. Similarly, when
$z_{0}\in D(\eta)$, the arguments are the same.

Step 3

Set $\varphi_{N}=\sqrt{\log N/Nh^{D}}$. To prove (21) and (22), let
”$h=N^{-\frac{1}{2\beta+D}}$, $l_{N}=N^{-\frac{\beta}{2\beta+D}}$” and
”$h=(\log N/N)^{\frac{1}{2\beta+D}}$,
$l_{N}=N^{-\frac{\beta}{2\beta+D}}\sqrt{\log N}$” respectively. Hence, it’s
easy to specify that the $\nu$ defined in Theorem 7 existed. Then, by
following Theorem 7, we can instantly obtain the results.