Let $(B_n)$ be a an increasing sequence such that (i) $B_n\rightarrow \infty$ and (ii) $r^{n,B_n}\rightarrow 0$. We remark that under <Ref> such a sequence always exists.
For example set $B_1=1$ and $L_1=1$; then for all $n$ if $r^{n, 2\, B_n}\le 2^{-L_n}$ then set $L_{n+1}=L_n+1$ and $B_{n+1}=2\, B_n$ (by <Ref> the latter will occur at some finite $n$); otherwise keep $B_{n+1}=B_n$ and $L_{n+1}=L_n$.

We note that:

\begin{align}
\II_2 \le~& \bn{g_n(Y^n)-\mathbb{E}[ g_n(Y^n)] - \Big(g_n(\tilde  Y^n)-\mathbb{E}[ g_n(\tilde Y^n)\mid X^n]\Big)}_{L_1}
\\
\overset{(a)}{\le}~& \Big\|\mathbb{I}(\sqrt{n}\big\|\bar{X}^n-\E\bb{X_1^n}\|_{2}\le B_n) \\
& \times\sup_{x\in B_{d_n}(0,B_n)}\ba{ g_n\bp{Y^n+\frac{x}{\sqrt{n}}} -\mathbb{E}\Big[g_n\bp{Y^n+\frac{x}{\sqrt{n}}} \mid X^n\Big]- g_n( Y^n)+\mathbb{E}\big[g_n\bp{Y^n}\big] }\Big\|_{L_1}
\\
& +\Big\|\mathbb{I}(\sqrt{n}\big\|\bar{X}^n-\E\bb{X_1^n}\|_{2}\ge B_n)  \times \Big[g_n(Y^n)-\mathbb{E}[ g_n(Y^n)] - \big(g_n(\tilde  Y^n)-\mathbb{E}[ g_n(\tilde Y^n)\mid X^n]\big)\Big]\Big\|_{L_1}
\end{align}

where (a) is a consequence of the triangle inequality. The first term is bounded by
$$\bn{\sup_{x\in B_{d_n}(0,B_n)}\ba{ g_n\bp{Y^n+\frac{x}{\sqrt{n}}} -\mathbb{E}\Big[g_n\bp{Y^n+\frac{x}{\sqrt{n}}} \mid X^n\Big]- g_n( Y^n)+\mathbb{E}\big[g_n\bp{Y^n}\big] }}_{L_1} \le r^{n,B_n}.$$
The second term can be bounded by the use of the Cauchy-Swartz inequality:

\begin{multline}
\Big\|\mathbb{I}(\sqrt{n}\big\|\bar{X}^n-\E\bb{X_1^n}\|_{2}\ge B_n)  \times \Big[g_n(Y^n)-\mathbb{E}[ g_n(Y^n)] - \big(g_n(\tilde  Y^n)-\mathbb{E}[ g_n(\tilde Y^n)\mid X^n]\big)\Big]\Big\|_{L_1}
\\
\le\Big\|\mathbb{I}\bc{\sqrt{n}\big\|\bar{X}^n-\E\bb{X_1^n}\|_{2}\ge B_n} \Big\|_{L_2}   \bp{ \bn{g_n(Y^n)}_{L_2}+ \bn{g_n(\tilde Y^n)}_{L_2}}
\end{multline}

Observe that:

\begin{align}
& \E\bb{\bn{\bar{X}^n-\E\bb{X_{1}^n}}_{2}^2} = \E\bb{ \sum_{k\leq d_n} \bp{\bar{X}_k^n-\E\bb{X_{1,k}^n}}^2}\\
&= \sum_{k\leq d_n} \Var\bp{\bar{X}_k^n} \leq \frac{4\sum_{k\le d_n}\|X_{1,k}^n\|_{L_2}^2}{n}
\end{align}

Hence, by Chebyshev's inequality:

\begin{align}
P\bp{\bn{\bar{X}^n-\E\bb{X_{1}^n}}_{2} \geq B_n/\sqrt{n}} \le \frac{n \Var\bp{\bn{\bar{X}^n-\E\bb{X_{1}^n}}_{2}}}{B_n^2}\leq \frac{4\sum_{k\le d_n}\|X_{1,k}^n\|_{L_2}^2}{B_n^2}
\end{align}

Thus we have:

\begin{equation}
\bn{\mathbb{I}\bc{ \sqrt{n}\bn{\bar{X}^n-\E\bb{X_{1}^n}}_2\ge B_n} }_{L_2} \leq \sqrt{P\bp{\bn{\bar{X}^n-\E\bb{X_{1}^n}}_2 \geq B_n/\sqrt{n}}} \leq \frac{2\sqrt{\sum_{k\le k_n}\|X_{1,k}^n\|_{L_2}^2}}{B_n}
\end{equation}

Thus we conclude that:

\begin{multline}
\bn{d_{\mcF}\bp{ g_n(Y^n)-\mathbb{E}[ g_n(Y^n)], g_n(\tilde  Y^n)-\mathbb{E}[ g_n(\tilde Y^n)\mid X^n]\mid X^n }}_{L_1}
\\
\le \frac{2\sqrt{\sum_{k\le k_n}\|X_{1,k}^n\|_{L_2}^2}}{B_n} \bp{ \bn{g_n(Y^n)}_{L_2}+ \bn{g_n(\tilde Y^n)}_{L_2}}+ r^{n,B_n}.
\end{multline}

§.§ Proof of <Ref>

We remark that both $g_{n,1}$ and $g_{n,2}$ are invariant under uniform perturbations. Indeed for all $x\in \RR$ we have:

\begin{equation}
\begin{split}
g_{n,1}(X_{1:n}+x/\sqrt{n})&=\bp{\frac{1}{\sqrt{n}}\sum_{i\le \lfloor n/2\rfloor}X_{i}-\frac{x}{\sqrt{n}}-\bp{x_{i+\lfloor n/2\rfloor}-\frac{x}{\sqrt{n}}}}^2
\end{split}
\end{equation}

\begin{equation}
\begin{split}
g_{n,2}(X_{1:n}+x/\sqrt{n})&=\sqrt{n}\bp{ \prod_{i=1}^n\bp{ 1+\frac{x_i-{x/\sqrt{n}}-\big(\bar{x}^n-x/\sqrt{n}\big)}{n}}-1}
\end{split}
\end{equation}

Therefore both functions $(g_{n,1})$ and $(g_{n,2})$ satisfy (<ref>). Moreover by a direct application of the chain rule we can verify that both $(g_{n,1})$ and $(g_{n,2})$ verify conditions $(H_0)$ and $(H_1)$. Hence <Ref> implies that:
$$d_{\mcF}\bp{g_{n,1}( Z^n)-\E(g_{n,1}(Z^n)|X),g_{n,1}( Y_{1:n})-\E(g_{n,1}(Y_{1:n}))}\rightarrow 0;$$

$$d_{\mcF}\bp{g_{n,2}( Z^n)-\E(g_{n,2}(Z^n)|X),g_{n,2}( Y_{1:n})-\E(g_{n,2}(Y_{1:n}))}\rightarrow 0.$$

§.§ Proof of <Ref>

The proof works by contradiction. Suppose that there is a measurable function $\mathcal{Q}_n:x_1,\dots,x_n\rightarrow \mathcal{P}_M(\RR)$ such that for all sequence of measures $(\nu_n)\in \prod_{n=1}^{\infty}\mathcal{P}_n$ the following holds:
$$\Big\|d_{\mcF}\Big(\mathcal{Q}_n(X^n),g_n(Y^n)-\E(g_n(Y^n))\mid X^n\Big)\Big\|_{L_1}\xrightarrow{n\rightarrow \infty }0.$$

By hypothesis we know that there is a sequence of measures $(p_{\theta_n}^n)\in\times_{l=1}^{\infty} \mathcal{P}_n$, an $\epsilon>0$ and and a sequence of vectors $(z_n)$ such that

* $\limsup \sup_{\tilde \theta_n\in[\theta_n,\theta_n+\frac{z_n}{\sqrt{n}}]}~~\|\mathcal{I}_n(\tilde \theta_n)^{1/2}\frac{z_n}{\sqrt{n}}\|_{v,d_n}<\infty$;

* The following holds for $(X^n)\overset{i.i.d}{\sim}{p_{\theta_n}^n}$

\begin{equation}\label{contrac}
d_{\mcF}\bp{ g_n\bp{X^n+\frac{z_n}{\sqrt{n}}}-\E\bb{g_n\bp{X^n+\frac{z_n}{\sqrt{n}}}},g_n(X^n)-\E\bb{g_n(X^n)} }>\epsilon.
\end{equation}

* $\theta_n+\frac{z_n}{\sqrt{n}}\in \Omega_n$

By abuse of notations we denote $p_{\theta_n}^n+\frac{z_n}{\sqrt{n}}$ the distribution of $X+\frac{z_n}{\sqrt{n}}$ for $X\sim p_{\theta_n}^n$. We define the following subset of distributions $\mathcal{P}_n^*:=\{p_{\theta_n}^n,p_{\theta_n}^n+\frac{z_n}{\sqrt{n}}\}$ and let $X^n$ be i.i.d random variables distributed according to $\mu_n\in \mathcal{P}^*_n$.
We want to test if $H_0: \E(X_1^n)=\theta_n$ against the alternative hypothesis $H_1: \E(X_1^n)=\theta_n+\frac{z_n}{\sqrt{n}}$. Using <ref> we know that  it is possible to find a rejection region $R_n$ such that:
$$P(X^n\in R_n|H_0)+P(X^n\not \in R_n|H_1)\rightarrow 0.$$

However by hypothesis the Kullback-Leibler divergence is smooth and according to the Taylor expansion we know that

$$KL\bp{ p_{\theta_n}^n+\frac{z_n}{\sqrt{n}},p_{\theta_n}^n }=\frac{1}{2n}z_n^{\top}\mathcal{I}_n(\tilde \theta_n)z_n,$$

where $\tilde\theta_n\in [\theta_n,\theta_n+\frac{z_n}{\sqrt{n}}]$. Moreover by definition of the total variation distance and the inequality: $\|p(\cdot)-q(\cdot)\|_{TV}\ge 1-\frac{1}{2}e^{-KL(p,q)}$ (see <Ref>) we know that for all rejection region $R_n$ we have
\begin{equation}
\begin{split}&
P(X^n\in R_n|H_0)+P(X^n\not \in R_n|H_1)\\&\ge \frac{1}{2}e^{-nKL(p_{\theta_n}^n+\frac{z_n}{\sqrt{n}},p_{\theta_n}^n)}\\&\overset{}{\ge} \frac{1}{2}e^{-\frac{1}{2}\sqrt{z_n^{\top}\mathcal{I}_n(\tilde \theta_n)z_n}}.
\end{split}
\end{equation}
Therefore using (i) we obtain that $\liminf_n  P(X^n\in R_n|H_0)+P(X^n\not \in R_n|H_1)>0$ and note there is a contradiction. Hence we have successfully showed that the desired result holds.

§.§ Proof of <Ref>

This is a direct consequence of using <ref> for the random functions
$$\tilde g_n:x_{1:n}\rightarrow g_n\Big(x_1+\mathbb{E}(X^n_1)-\bar{X^n},\dots,x_n+\mathbb{E}(X^n_1)-\bar{X^n}\Big);$$
which can be approximated by the following smooth functions:
$$\tilde f_n:x_{1:n}\rightarrow f_n\Big(x_1+\mathbb{E}(X^n_1)-\bar{X^n},\dots,x_n+\mathbb{E}(X^n_1)-\bar{X^n}\Big).$$

§.§ Proof of <Ref>

By the triangle inequality we know that:

\begin{align}
\label{yp}\ba{ g_n( Z^n)-\E\bb{g_n(Y^n)\mid X^n}}
\le~&  \ba{ \E\bb{g_n(Y^n)- g_n(\tilde{Y}^n)\mid X^n} }+  \ba{ g_n(\tilde Y^n)-\E\bb{g_n(\tilde Y^n)\mid X^n}}.
\end{align}

We bound successively each terms of the right-hand side. Firstly using <ref> we remark that for all $\delta>0$ we have
$$P\bb{\ba{\E\bb{g_n(\tilde Y^n)\mid X^n}-\E\bb{g_n(Z^n)\mid X^n}}\ge \delta}\rightarrow 0.$$
Hence by combining this with the definition of $t_{b,n}^{\beta/2}$ we know that

\begin{align}
\limsup_{\delta\downarrow 0}P\big(\ba{ g_n(Z^n)-\E\bb{g_n(\tilde Y^n)\mid X^n}}\ge t_{{\rm b},n}^{\beta/2}(X^n)+\delta \big)\le \beta/2.
\end{align}

Let $N^n\sim N(0,\Sigma_n)$ be a Gaussian vector with variance-covariance $\Sigma_n$.
Using [16] we know that there is a constant $C$ that does not depend on $n$ such that the following holds:

\begin{align}&\label{eqn:nn}
\sup_t \Big|P\bb{\sqrt{n}\max_k\big|\bar{X^n_k}-\mathbb{E}(X_{1,k}^n)\big|\ge t}-P(\max_k|N^{n,k}|\ge t)\Big|\le \frac{C\log(d_n)^{7/6}\max(\|\max_j|X^n_{1,j}|\|_{L_4}^4,1)}{n^{1/6}}
\end{align}

Combining this with hypothesis $(H_3)$ we obtain that:

\begin{align}&
P\bp{\big|  \E\bb{g_n(Y^n+\bar{X^n}-\mathbb{E}(X^n_1))-g_n(Y^n)\mid X^n} \big|\ge t_{g,n}^{\beta/2}}
\\&\le P\bp{\|\sqrt{n}[\bar{X^n}-\mathbb{E}(X^n_1)]\|_{\infty}^{\alpha}\ge \frac{t_{g,n}^{\beta/2}}{C_n}}
\\&\le\frac{C\log(d_n)^{7/6}\max(\|\max_j|X^n_{1,j}|\|_{L_4}^4,1)}{n^{1/6}}+   P\bp{\max_k|N_{k}^n|\ge \big[\frac{t_{g,n}^{\beta/2}}{C_n}\big]^{\frac{1}{\alpha}}}
\\&\le \beta/2+o(1)
\end{align}

Therefore  for all $\delta>0$ we have

\begin{align}&
P\bp{\big|g_n(Z^n)-\E(g_n(Y^n))\big|\le t_{{\rm b},n}^{\beta/2}(X^n)+ t_{g,n}^{\beta/2}+\delta}
\\&\le P\bp{\big|g_n(Z^n)-\E(g_n(\tilde Y^n))\big|\le t_{{\rm b},n}^{\beta/2}(X^n)+\delta}
\\&~+ P\bp{\big|\E(g_n(\tilde Y^n)-g_n(Y^n)\mid X^n)\big|\le  t_{g,n}^{\beta/2}}
\\&\le \beta+o(1)
\end{align}

§.§ Proof of <Ref>

Define the function: $g_n:x_{1:n}\rightarrow \bp{\frac{1}{\sqrt{n}}\sum_{i\le n}x_i}^2$. It is straightforward to check that $(g_n)$ and $(X^n)$ satisfy conditions $(H_0)$ and $(H_1)$. Moreover for all $x$ we note that:

\begin{align}
\ba{\mathbb{E}\bp{g_n(Y^n+\frac{x}{\sqrt{n}})-g_n(Y^n)}}&=  x^2.
\end{align}
Therefore $(g_n)$ also satisfy conditions $(H_3)$ with $C_n=1$ and $\alpha=2$. The result is a direct consequence of <Ref>.

§.§ Proof of <Ref>

Firstly we prove that with high-probability $\bar{X}^n-\mathbb{E}(X_1^n)$ is in $B_{d_n}(\gamma_n)$. Indeed, using <Ref> and Chebystchev inequality we know that there is a constant $C$ such that

\begin{align}
P\bb{\max_k\ba{\bar{X^n_k}-\mathbb{E}(X_{1,k}^n)}\ge \gamma_n }
&\le  \frac{\E\bp{\max_k\ba{\bar{X^n_k}-\mathbb{E}(X_{1,k}^n)}^2}}{\gamma^2_n}
\\&\le \frac{C\log(d_n)^2\bn{\max_{k\le k_n}\ba{X^n_{1,k}}}_{L_2}^2}{\gamma_n^2n}\rightarrow 0.
\end{align}

Therefore with high-probability $\tilde\mu:=\bar{X}^n-\mathbb{E}(X_1^n)$ is in $B_{d_n}(\gamma)$. Moreover, we note that:
\begin{align}&P\bp{\ba{g_n( Z^n-\tilde \mu)-\mathbb{E}(g_n( Z^n-\tilde \mu)\mid X^n)}\ge t^{\alpha}_{*}(X^n)~\mid X^n}\le \alpha.
\end{align}

We remark that $Z^n-\tilde \mu=\tilde Z^n$ therefore according to <Ref> for all $\delta>0$ we have:

\begin{align}& {P\bp{\ba{g_n(Y^n)-\mathbb{E}(g_n(Y^n)}\ge t^{\alpha}_{*}(X^n)+\delta}\mid X^n}\le \alpha+o_n(1).
\end{align}

Which implies that asymptotically the confidence intervals $\big[t^{\beta}_{*}(X^n)-\delta, ~t^{\alpha}_{*}(X^n)+\delta\big]$ has an asymptotic coverage of at least $1-\beta$.

§.§ Proof of <Ref>

$$\epsilon_n:=\sup_{\nu\in \mathcal{P}_n}\E_{X_1^n\in \mathcal{P}_n}\bp{\ba{ f_{n}(\tilde Z^n)-g_n(\tilde Z^n) }} + \E\bp{ \ba{f_{n}({Y}^n)-g_n({Y}^n) }}+\sup_{\nu\in \mathcal{P}_n}\max\bp{R_{n,1}^{c,\nu}, R_{n,2}^{c,\nu}, R_{n,3}^{c,\nu}}.$$

Choose $\nu\in \mathcal{Q}_n$, and let $X^n\sim \nu$ if $t^{\alpha}_{n}(X^n)$ is chosen such that:

\begin{equation}
{P\bp{ \ba{ g_n(Z^{n}+\E(X^n_1)-\bar{X^n} )-\E\bb{g_n(Z^{n}-\E(X^n_1)+\bar{X^n} )\mid X^n} }\ge t_{n}^{\alpha} (X^n)\mid X^n}} \le \alpha.
\end{equation}
then according to <Ref> and <Ref> for all $\delta>0$ we have

\begin{equation}
P\bp{ \ba{ g_n(X^{n})-\E\bb{g_n(X^n)} }\ge t^{\alpha}_{n}+\delta}\le \alpha+O(\epsilon_n).
\end{equation}
But as  $\mathbb{E}(X_1^n)\in A_n$ by  definition of $t_{n}^{\alpha}(X^n)$ we know that: $t_{n}^{\alpha}(X^n)\le t_{\alpha}^{*,n}$. This directly implies that

\begin{equation}
{{P_{X^n,Y^n\overset{i.i.d}{\sim} \nu}\bp{ \ba{ g_n(Y^{n})-\E\bb{g_n(Y^n)} }\ge t_{\alpha}^{
*,n} (X^n)+\delta}}} \le \alpha+O(\epsilon_n).
\end{equation}

As this holds for all $\nu \in A_n$ we get the desired result.

§ PROOFS FROM <REF>

§.§ Proof of <ref>

We will use <Ref>.
We note that the condition $(H_0)$ holds automatically as the functions  $(g_n)$ are three-times differentiable. Therefore to we only need to verify that $(H_1)$  holds. By the chain rule we have:
$$\partial_i g_n( Z^{n, i, \bar{X}^n})=\frac{p}{\sqrt{n}}\big(\frac{1}{\sqrt{n}}\sum_{j\ne i}^{n}X_j+\frac{1}{\sqrt{n}}\bar{X^n}\big)^{p-1}; $$
$$ \partial_i^2 g_n( Z^{n,i,\bar{X^n}} )=\frac{p(p-1)\mathbb{I}(p\ge 2)}{n}\big(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_i+\frac{1}{\sqrt{n}}\bar{X^n}\big)^{p-2};$$

$$\partial_i^3 g_n( Z^{n,i,x} )=\frac{p(p-1)(p-2)\mathbb{I}(p\ge 3)}{\sqrt{n}^3}\big(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_i+\frac{1}{\sqrt{n}}x\big)^{p-3}.$$

Moreover, according to the  Rosenthal inequality for martingales [30], there is a constant $C$ such that

\begin{equation}
\begin{split}
&\bn{\frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_i}_{12p}\le C\|X_1\|_{12p}
\end{split}
\end{equation}
This implies that: $R_{n,1}=O(n^{-1/2}),\quad R_{n,2}=O(n),\quad \&\quad R_{n,3}=O(n^{-3/2}).$
<Ref> concludes the proof.

§.§ Proof of <Ref>

We define $g_n(X):= \sqrt{n}\prod_{l=1}^n (1+\frac{X_l}{{n}}).$ Condition $(H_0)$ holds automatically as $g_n$ is smooth. To obtain the desired result we only need to verify that $(H_1)$  holds.
By the chain rule we have:
$$\partial_i g_n( Z^{n,i,\bar{X^n}})=\frac{1}{\sqrt{n}}\prod_{l\ne i}(1+\frac{ Z^n_l}{{n}})\times(1+\frac{\bar{X^n}}{n}); \quad \rm{and}\quad \partial_i^2 g_n( Z^{n,i} )=0.$$
Therefore we note that for condition $(H_1)$ to hold we only need to upper bound $\partial_i g_n(Z^{n,i,\bar{X^n}})$. Using the fact that the observations $(X_i)$ are assumed to be bounded we know that there is a constant $C$ such that

\begin{equation}
\begin{split}
&\big\|\prod_{l\ne i}\Big(1+\frac{ Z^n_l}{{n}}\Big)\big\|_{L_{12}}^{12}
\\&\overset{(a)}{\le} \mathbb{E}\Big[ \prod_{l\ne i}\mathbb{E}\Big((1+\frac{ Z^n_l}{{n}})^{12}\Big|X\Big)\Big]
\\& \overset{(b)} {\le} \prod_{l\ne i}(1+\frac{\max(C,1)}{n})
\sim e^{\max(C,1)}
\end{split}
\end{equation}
where to get (a) we exploited the conditional independence of the observations $(Z^n_i)$ and to obtain (b) we used the fact that the observations $(X_i)$ are assumed to be bounded. This implies that $(H_1)$ holds.

§.§ Proof of <Ref>

We want to used <Ref> and <Ref>. Let $(\beta_n)\in \mathbb{R}^{\mathbb{N}}$ be a sequence satisfying: (i) $\beta_n\rightarrow \infty$ and (ii) $\beta_n=o(n^{1/4})$

We define $(f_n:\mathbb{R}^n\rightarrow\mathbb{R})$ as the following sequence of functions:
$$f_n(x_{1:n}):=\frac{1}{\beta_n}\log\bp{1+e^{\beta_n\frac{1}{\sqrt{n}}\sum_{i\le n}x_i}}.$$
We note that the functions $(f_n)$ are three times differentiable functions and that the following holds:

$$\bn{f_n(Z^n)-g_n(Z^n)}\le \frac{\log(2)}{\beta_n}\quad \rm{and}\quad \bn{f_n(Z^n)-g_n(Z^n)}\le \frac{\log(2)}{\beta_n}.$$
Therefore the functions $(f_n)$ and $(g_n)$ satisfy conditions $(H_0)$. Moreover by the chain rule we have:

$$\partial_i f_n(x_{1:n})=\frac{1}{\sqrt{n}},\quad \partial^2_i f_n(x_{1:n})=\frac{\beta_n}{{n}},\quad \partial^3_i f_n(x_{1:n})=\frac{\beta_n^2}{{n}^{3/2}}. $$

This implies that condition $(H_1)$ also holds as $R_{n,1}=O(n^{-1/2}),\quad R_{n,2}=o(n^{-1/2}),\quad \&\quad R_{n,3}=0(n^{-1}).$
<Ref> and <Ref> conclude the proof.

§.§ Proof of <Ref>

For ease of notations  for all element $\z\in\mathbb{Z}^2$ we denote by $(\z_1,\z_2)$ its coordinates. Moreover we remark that the i.i.d random variables $(X^n_{i,j})$ are indexed by $\dbracket{n}^2$ rather than $\dbracket{n^2}.$ As there is a one-to-one mapping between those two sets, we note that the random variables $(X^n_{i,j})$ could be indexed by $\dbracket{n^2}.$ Therefore <Ref> applies. However for ease of notations we keep the original indexing.

We note that $(g_n)$ is three times differentiable so $(H_0)$ holds. Moreover by the chain rule we have:

\begin{equation}
\partial_{\z}g_n(ZZ^{n,\z,\overline{X^n}}):=\frac{1}{n\sqrt{n}}\frac{\sum_{m\in \{-1,1\}^n}m_{\z_1}m_{\z_2}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}{\sum_{m\in \{-1,1\}^n}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}.
\end{equation}

This implies that
\begin{equation}
\big\| \partial_{\z}g_n(ZZ^{n,\z,\overline{X^n}})\big\|_{L_{12}}\le \frac{1}{n\sqrt{n}}.
\end{equation}

Using the chain rule we have:

\begin{equation}
\partial^2_{\z}g_n(ZZ^{n,\z,\overline{X^n}}):=\frac{1}{n^2}\frac{\sum_{m\in \{-1,1\}^n}(m_{\z_1}m_{\z_2})^2e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}{\sum_{m\in \{-1,1\}^n}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}
-\frac{1}{n^2}\Big[\frac{\sum_{m\in \{-1,1\}^n}m_{\z_1}m_{\z_2}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}{\sum_{m\in \{-1,1\}^n}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}\Big]^2 .
\end{equation}

This implies that
\begin{equation}
\big\| \partial^2_{\z}g_n(ZZ^{n,\z,\overline{X^n}})\big\|_{L_{12}}\le \frac{1}{n^2}.
\end{equation}

Finally by a last use of the chain rule we have

\begin{equation}\begin{split}
\partial^3_{\z}g_n(ZZ^{n,\z,x}):=&\frac{1}{n^{5/2}}\frac{\sum_{m\in \{-1,1\}^n}m_{\z_1}m_{\z_2}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}{\sum_{m\in \{-1,1\}^n}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}\Big[1-
\Big[\frac{\sum_{m\in \{-1,1\}^n}m_{\z_1}m_{\z_2}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}{\sum_{m\in \{-1,1\}^n}e^{\frac{1}{\sqrt{n}}m^{\top}ZZ^{n,\z,\overline{X^n}}m}}\Big]^2\Big]
\end{split}
\end{equation}

This implies that
\begin{equation}
\bn{ \sup_{x\in [Z_{\z}^n,\tilde X^n]\cup  [\tilde Y_{\z}^n,\tilde X^n]}\ba{\partial^3_{\z}g_n(ZZ^{n,\z,x})}}_{L_{12}}\le \frac{1}{n^{5/2}}.
\end{equation}

This implies that the desired result holds. The same can be proved for the centered bootstrap $(\tilde Z^n_{\z})$ and $(Y_{\z}^n)$.

§ PROOFS FROM <REF>

§.§ Proof of <Ref>

We note that we can suppose without loss of generality that $\E\bp{X_1^n}=0$. Let $(\beta_n)$ be a sequence of  reals satisfying: $(i)$ $\beta_n\rightarrow \infty$ and (ii) $\beta_n=o\big(\frac{n^{1/8}}{\sqrt{\log(p_n)}}\big)$.

To get the desired result we use <Ref>. In this goal we define $(f_n)$ to be the following sequence of function:
$$ f_n(x_{1:n}):= \frac{\log\Big(\sum_{l\le p_n}e^{\beta_n\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x'_{i,l}}\Big)}{\beta_n\log(p_n)}.$$
We remark that the functions $(f_n)$ are three times differentiable, and that they satisfy:
\begin{equation}
\begin{split}\label{fr}&
\Big\|f_n(\tilde Z^n)-g_n( \tilde Z^n)\Big\|_{L_1}\le\frac{1}{\beta_n}
\\&\Big\|f_n( Y^n)-g_n( Y^n)\Big\|_{L_1}\le\frac{1}{\beta_n}.
\end{split}
\end{equation}

We prove that the conditions of <Ref> hold. Using <ref> we know that hypothesis $(H_0)$ is holding; and we only need to prove that  $(H_1^*)$ also holds. For ease of notations we write:
$$\omega_k(x_{1:n}):=\frac{e^{\beta\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x
_{i,k}}}{\sum_{l\le p_n}e^{\beta\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x_{i,l}}}.$$
By the chain rule we remark that  for all $k_1\le p_n$ and all $x_{1:n}\in \RR^{p_n}$  we have

\begin{equation}
\begin{split}
\Big|\partial_{i,k_1}f_n(x_{1:n})\Big|
&\le \frac{1}{\sqrt{n}}\omega_{k_1}(x'_{1:n})
\end{split}\end{equation}

As $\sum_{k\le p_n}\omega_k(x_{1:n})=1$ we obtain for all $i\le n$ that

\begin{equation}
\begin{split}
\Big\|\big\|\partial_{i}f_n(\tilde Z^{n, i, \bar{X}^n})\big\|_{v,p_n}\Big\|_{L_{12}}
&\le \frac{1}{\sqrt{n}}\Big\|\sum_{j\le p_n}\omega_j({ \tilde Z^{n, i, \bar{X}^n}_{1:n}})\Big\|_{L_{12}}
\le  \frac{1}{\sqrt{n}}.\end{split}\end{equation}

This directly implies that $R_{n,1}^*=O(\frac{1}{n^{1/6}})$.
Moreover by using the chain rule we obtain that:

\begin{equation}\begin{split}&
\Big| \partial_{i,k_1,k_2}^2f_n(x_{1:n})\Big|
\le \frac{\beta_n\log(p_n)}{n}\omega_{k_1}(x_{1:n})\omega_{k_2}(x_{1:n})+\mathbb{I}(k_1=k_2)\frac{\beta_n\log(p_n)}{n}\omega_{k_1}(x_{1:n}).
\end{split}
\end{equation}

Therefore  as $\sum_{k\le p_n}\omega_{k}(x_{1:n})=1$ we have for all $i\le n$

\begin{equation}
\begin{split}
\Big\|\big\|\partial_{i}^2f_n(\tilde Z^{n, i, \bar{X}^n})\big\|_{m,p_n}\Big\|_{L_{12}}
\le  \frac{2\beta_n \log(p_n)}{n}.\end{split}\end{equation}

this implies that $R_{n,2}^*=O(\beta_n\log(p_n)/\sqrt{n})$.
Finally, by another application of the chain rule we have:

\begin{equation}\begin{split}&\label{landing}
\Big| \partial_{i,k_{1:3}}^3f_n(x_{1:n})\Big|
\\&\le2 \frac{\beta_n^2\log(p_n)^2}{n^{\frac{3}{2}}}\omega_{k_1}(x_{1:n})\omega_{k_2}(x_{1:n})\omega_{k_3}(x_{1:n})+[\mathbb{I}(k_2=k_3)+\mathbb{I}(k_1=k_3)\\&+\mathbb{I}(k_1=k_2)] \frac{\beta_n^2\log(p_n)^2}{n^{\frac{3}{2}}}\omega_{k_1}(x_{1:n})\omega_{k_2}(x_{1:n})+\mathbb{I}(k_1=k_2=k_3) \frac{\beta_n^2\log(p_n)^2}{n^{\frac{3}{2}}}\omega_{k_1}(x_{1:n}).
\end{split}
\end{equation}

Therefore we obtain for all $i\le n$ that:

\begin{equation}\begin{split}&
\Big\| \max_{x\in [\bar{X}^n,Z^{n}_i]\bigcup [0,\tilde Y^n_i]}\big\|\partial_{i}^3f_n(Z^{n,i,x}_{1:n})\big\|_{t,p_n}\Big\|_{L_{12}}
\le \frac{6\beta^2_n\log(p_n)^2}{{n^{\frac{3}{2}}}}
\end{split}
\end{equation}
This implies that $R_{3,n}^*=O(\frac{\beta_n^3\log(p_n)^2}{n^{1/2}})$
Finally by assumption we know that $\|\sup_{l\le p_n}X^n_{1,l}\|_{L_{12}}<\infty$, hence the assumption $(H_1^*)$ holds as well as the desired result.

§ PROOFS FROM <REF>

§.§ Proof of <Ref>

To get the desired results we use <Ref>. Firstly we remark that the observations $X_1^n$ take value in $\RR^{p_n\times p_n}$. As there is a one to one mapping between $\RR^{p_n\times p_n} $ and $\RR^{p_n^2}$ we could take $X_1^n$ to take value in $\RR^{p_n^2}$. For ease of notations we do not do so, but note that the conditions of <Ref> can alternatively be expressed for observations taking value in $\RR^{p_n\times p_n}$.

We denote $g_n(x_{1:n}):=\min_{i\le p_n}\max_{j\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}X^n_{l,i,j}$;
and let $(\beta_n)$ be a sequence of positive reals satisfying (i) $\beta_n\rightarrow \infty$ and (ii) $\beta_n=o(\log(p_n)/n^{-1/4})$.  We set $(f_n)$ to be the following sequence of functions:
$$f_n(x_{1:n}):=\frac{1}{\beta_n log(p_n)}\log\Big(\sum_{i\le p_n}\big(\sum_{j\le p_n}e^{\frac{\beta_n\log(p_n)}{\sqrt{n}}\sum_{l\le n}x_{l,i,j}}\big)^{-1}\Big).$$
We remark that they are three times differentiable and that following holds:

\begin{equation}
\begin{split}
&\big\|g_n(Z^n)-f_n(Z^n)\|_{L_1}\le \frac{2}{\beta_n}
\\&\big\|g_n(\tilde Y^n)-f_n(\tilde Y^n)\|_{L_1}\le \frac{2}{\beta_n}.
\end{split}
\end{equation}

Therefore hypothesis $(H_0)$ holds and we only need to check that hypothesis $(H_1^*)$ also holds to be able to use <Ref>. For ease of notations we write $\hat{R}_{i,j}:=\frac{1}{\sqrt{n}}\sum_{l\le n}X_{l,i,j}$ and denote
$$\omega_{i,j}(x_{1:n}):=\frac{e^{\beta_n\log(p_n)\hat{R}_{i,j}}}{\sum_{k\le p_n}e^{\beta_n\log(p_n)\hat{R}_{i,k}}};\qquad \overline{\omega}_{i}(x_{1:n}):=\frac{\big[\sum_{k\le p_n}e^{\beta_n\log(p_n)\hat{R}_{i,k}}\big]^{-1}}{\sum_{l\le p_n}\big[\sum_{k\le p_n}e^{\beta_n\log(p_n)\hat{R}_{i,k}}\big]^{-1}}.$$

We remark that the following holds:
\begin{equation}\label{eq:s}\sum_{k,j\le p_n}\omega_{k,j}(x_{1:n})\overline{\omega}_k(x_{1:n})=1,\quad \sum_{j\le p_n}\omega_{k,j}(x_{1:n})=1\end{equation}
By the chain rule, we remark that for all $k,j\le p_n$ we have:

Therefore as $\sum_{k,j\le p_n}\omega_{k,j}(x_{1:n})\overline{\omega}_k(x_{1:n})=1$, this implies that:

$$\max_{l\le n} \Big\|\|\partial_{l,i,j}f_n(Z^{n, i, \bar{X}^n})\|_{v,p_n^2}\Big\|_{L_{12}}\le \frac{1}{\sqrt{n}}.
This implies that $R_{n,1}^*=O(\frac{1}{n^{1/6}}).$
Moreover by another application of the chain rule we have for all $i_1,j_1,i_2,j_2\le p_n$ and all $l\le  n$:

\begin{equation}
\begin{split}&
\partial_{l,i_1,j_1,i_2,j_2}^2f_n(x_{1:n})\\&=\frac{\beta_n\log(p_n)\mathbb{I}(i_1=i_2) }{\sqrt{n}} \partial_{l,i_1,j_1}f_n(x_{1:n})\Big[\mathbb{I}(j_1=j_2)-2 \omega_{i_2,j_2}(x_{1:n})\Big]
\\&-\partial_{l,i_1,j_1}f_n(x_{1:n})\frac{\log(p_n)\beta_n }{\sqrt{n}}\omega_{i_2,j_2}(x_{1:n})\overline{\omega}_{i_2}(x_{1:n}).
\end{split}
\end{equation}

Therefore we have by combining this with <ref> and the fact that $\sum_{k,j\le p_n}|\partial_{l,k,j}f_n(x_{1:n})|\le 1/\sqrt{n}$  we have

$$\max_{l\le n}\Big\|\big\|  \partial_{l,i_1,j_1,i_2,j_2}^2f_n(Z^{n, i, \bar{X}^n})\big\|_{m,p_n}\Big\|\le \frac{4\beta_n\log(p_n)}{n}.

Therefore we have $R_{2,n}^*=O(\frac{\log(p_n)\beta_n}{\sqrt{n}})$. Finally, by a last application of  the chain rule  for all $i_{1:3},j_{1:3}\le p_n$ we have:

\begin{equation}
\begin{split}&
\partial_{l,i_1,j_1,i_2,j_2,i_3,j_3}^3f_n(x_{1:n})\\&=\frac{\beta_n\log(p_n)\mathbb{I}(i_1=i_2) }{\sqrt{n}} \partial_{l,i_1,j_1,i_3,j_3}^2f_n(x_{1:n})\Big[\mathbb{I}(j_1=j_2)-2 \omega_{i_2,j_2}(x_{1:n})\Big]
\\&-\partial_{l,i_1,j_1,i_3,j_3}^2f_n(x_{1:n})\frac{\log(p_n)\beta_n }{\sqrt{n}}\omega_{i_2,j_2}(x_{1:n})\overline{\omega_{i_2}}(x_{1:n})
\\&-2\frac{\log(p_n)^2\beta_n^2\mathbb{I}(i_1=i_2=i_3) }{n} \partial_{l,i_1,j_1}f_n(x_{1:n})\omega_{i_2,j_2}(x_{1:n}) \Big[\mathbb{I}(j_2=j_3)-\omega_{i_3,j_3}(x_{1:n})\Big]
\\&-\partial_{l,i_1,j_1}   \mathbb{I}(i_2=i_3)f_n(x_{1:n})\frac{\log(p_n)^2\beta^2  }{n}\omega_{i_2,j_2}(x_{1:n})\overline{\omega_{i_2}}(x_{1:n})
\times\Big[\mathbb{I}(j_2=j_3)-2\omega_{i_3,j_3}(x_{1:n})\Big]
\\&+\partial_{l,i_1,j_1}f_n(x_{1:n}){\omega_{i_2,j_2}}(x_{1:n})\overline{\omega_{i_2}}(x_{1:n})
\end{split}
\end{equation}

Therefore by combining this with <ref>  and the fact that $\sum_{i_{1:2}\le p_n,j_{1:2}\le p_n}\ba{ \partial_{l,i_1,j_1,i_2,j_2}^2f_n(Z^{n, i, \bar{X}^n})}\le \frac{4\beta_n\log(p_n)}{n}$ we have

\begin{equation}
\begin{split}&
\max_{l\le n}   \bn{\max_{z\in [\bar{X}^n,\tilde Z_1^n]\bigcup [\bar{X}^n,Y_1^n] }  \bn{\partial_{l,i_1,j_1,i_2,j_2,i_3,j_3}^3f_n(Z^{i,z})}_{t,p_n}}_{L_{12}}
\le \frac{10\log(p_n)^2\beta_n^2}{n^{3/2}}.
\end{split}
\end{equation}
This implies that $R_{3,n}^*=O(\frac{\log(p_n)^2\beta_n^2}{\sqrt{n}})$.
Therefore if we choose $\beta_n=\frac{n^{1/6}}{{\log(p_n)}^{2/3}}$ then
by using <Ref> guarantees that:

\begin{equation}
\begin{split}&
\Big\|d_{\mcF}(g_n(Z^n),g_n(\tilde Y^n)|X^n)\Big\|_{L_1}=O\bp{\frac{{\log(p_n)}^{2/3}}{n^{1/6}}}
\end{split}
\end{equation}

Moreover by definition of $\tilde Y^n$ we have $\E\bp{Z_{1,i,j}^n\mid X^n}=\E\bp{\tilde Y_{1,i,j}^n\mid X^n}$ therefore $\theta_{Z^n}=\theta_{\tilde Y^n}$. By combining this with <ref> we conclude that:

\begin{equation}
\begin{split}&
\Big\|d_{\mcF}(g_n(Z^n)- \sqrt{n}\theta_{Z^n},~g_n(\tilde Y^n)-\sqrt{n}\theta_{\tilde Y^n}|X^n)\Big\|_{L_1}=O\bp{\frac{{\log(p_n)}^{2/3}}{n^{1/6}}}
\end{split}
\end{equation}

Moreover by following the exact same line of reasoning we can prove that:

\begin{equation}
\begin{split}&
\bn{d_{\mcF}\bp{g_n(\tilde Z^n),g_n( Y^n)\mid X^n}}_{L_1}=O\bp{\frac{{\log(p_n)}^{2/3}}{n^{1/6}}}
\\&\bn{d_{\mcF}\bp{g_n(\tilde Z^n)-\sqrt{n}\theta_{\tilde Z^n},~g_n( Y^n)-\sqrt{n}\theta_0\mid X^n}}=O\bp{\frac{{\log(p_n)}^{2/3}}{n^{1/6}}}
\end{split}
\end{equation}

Therefore according to <Ref> if $(t_{n,\alpha})$ is a sequence chosen such that:
$$P\bp{\ba{g_n(\tilde Z^n)-\sqrt{n}\theta_{\tilde Z^n}}\ge t_{n,\alpha}\mid  X^n}=\alpha$$
then for all sequence $(\epsilon_n)$ satisfying $\epsilon_n\downarrow 0$ and $\bp{\frac{\log(p_n)^2}{\sqrt{n}}}^{1/3}=o(\epsilon_n)$ we have
$$\limsup_{n\rightarrow 0}P\bp{\ba{g_n(X_n)-\sqrt{n}\theta_0}\ge t_{n,\alpha}+\epsilon_n\mid  X^n}\le\alpha.$$

Let $(p_n)$ be a sequence of integers satisfying $\log(p_n)=o(n^{1/4})$. Define $(X^n_i)$ to be a triangular array of sequences of i.i.d random variables taking value in $\RR^{p_n\times p_n}$. Suppose that $\|\sup_{l_1,l_2\le p_n}|X_{1,l_1,l_2}^n|\|_{L_{12}}<\infty$ and that $\frac{\log(p_n)}{\sqrt{n}}=o\Big[\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|\Big]$. We denote: $\hat \theta_n(x_{1:n})=\frac{1}{n}\min_{i\le p_n}\max_{j\le p_n}\sum_{l\le n}X_{l,i,j}^n$.
Then the following holds
$$\Big\|d_{\mcF}\bp{\sqrt{n}\Big[\hat \theta_n( Z^n)-\theta_{Z^n}\Big],~\sqrt{n}\Big[\hat \theta_n( Y^n)-\theta_{Y^n}\Big]\mid X^n}\Big\|_{L_1}=o(1);$$
where  for a process $(Y'_{1:n})$ we have written, by abuse of notations,
$\theta_{Y'}:=\min_{i\le p_{n}}\max_{j\le p_{n}} ~\E\big[Y^{'}_{l,i,j}|X^n\big].$

Firstly, we remark that <Ref> imply that:

$$\Big\|d_{\mcF}\bp{\sqrt{n}\Big[\hat \theta_n( Z^n)-\theta_{Z^n}\Big],~\sqrt{n}\Big[\hat \theta_n(\tilde Y^n)-\theta_{\tilde Y^n}\Big]\mid X^n}\Big\|_{L_1}=o(1).$$
Therefore to get the desired result we only need to prove that:

$$\Big\|d_{\mcF}\bp{\sqrt{n}\Big[\hat \theta_n( Y^n)-\theta_{Y^n}\Big],~\sqrt{n}\Big[\hat \theta_n(\tilde Y^n)-\theta_{\tilde Y^n}\Big]\mid X^n}\Big\|_{L_1}=o(1).$$

We denote $I,J\le p_n$ the unique pair of indices satisfying $\E\bp{X^n_{1,I,J}}=\theta_0$; and define:
$$\epsilon_n:=\frac{\log(p_n)\bn{\max_{i,j}\ba{X^n_{1,i,j}}}_{L_2}}{\sqrt{n}\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|}.$$

Using <Ref> we know that there is a constant $C$ such that

\begin{equation}
\begin{split}&
P\bp{\max_{i,j\le p_n} \ba{\frac{1}{{n}}\sum_{l\le n}Y^n_{l,i,j}-\E(Y^n_{l,i,j})}\ge \frac{\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|}{4}}
\\&=  P\bp{\max_{i,j\le p_n} \ba{\frac{1}{{n}}\sum_{l\le n}Y^n_{l,i,j}-\E(Y^n_{l,i,j})}\ge \frac{\log(p_n)\bn{\max_{i,j}\ba{X^n_{1,i,j}}}_{L_2}}{4\epsilon_n\sqrt{n}}}
\\&\overset{}{\le} {C\epsilon_n^2}\rightarrow 0.
\end{split}
\end{equation}

We remark that if $\hat{\theta}_n(Y^n)$ is distinct from  $\frac{1}{{n}}\sum_{l\le  p_n}Y^n_{l,I,J}$ this means that there are indexes $i,j\le p_n$ such that  $\ba{\frac{1}{{n}}\sum_{l\le n}Y^n_{l,i,j}-\E(Y^n_{l,i,j})}$ is bigger than $ \frac{\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|}{2}$. Therefore we have:
\begin{equation}
\begin{split}
& P\bp{\hat{\theta}_n(Y^n)\ne\frac{1}{{n}}\sum_{l\le  p_n}Y^n_{l,I,J}}
\\&\le P\bp{\max_{i,j\le p_n} \ba{\frac{1}{{n}}\sum_{l\le n}Y^n_{l,i,j}-\E(Y^n_{l,i,j})}\ge \frac{\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|}{2}}
\\& \le {C\epsilon_n^2}\to 0.
\end{split}
\end{equation}
Similarly we have

\begin{equation}
\begin{split}
& P\bp{\hat\theta_n(\tilde Y^n)=\frac{1}{{n}}\sum_{l\le  p_n}\tilde Y^n_{l,I,J}}
\\&\le P\bp{\max_{i,j\le p_n} \ba{\frac{1}{{n}}\sum_{l\le n}X^n_{l,i,j}-\E(X^n_{l,i,j})}\ge \frac{\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|}{4}}\\&+P\bp{\max_{i,j\le p_n} \ba{\frac{1}{{n}}\sum_{l\le n}Y^n_{l,i,j}-\E(Y^n_{l,i,j})}\ge \frac{\inf_{(i,j)\ne (l,k)}\big|\E\big[X^{n}_{1,l,k}\big]-\E\big[X^{n}_{1,i,j}\big]\big|}{4}}
\\& \le {2C\epsilon_n^2}\to 0.
\end{split}
\end{equation}

Therefore with high-probability we have:
$$\hat\theta_n(Y^n)=\hat\theta_n(\tilde Y^n)-\theta_{\tilde Y^n}+\theta_{Y^n}.$$

Using the dominated convergence theorem we thus obtain that

$$\Big\|d_{\mcF}\Big[~\sqrt{n}\big[{\hat\theta_n(\tilde Y^n)}-\theta_{\tilde Z^n}\big],\sqrt{n}\big[{g_n(Y^n)}-\theta_{Y^n}\big]\mid X^n\Big]\Big\|_{L_1}\rightarrow 0.$$

Let $(p_n)$ be a sequence of integers satisfying $\log(p_n)=o(n^{1/7})$. Define $(X^n_i)$ to be a triangular array of i.i.d random variables taking value in $\RR^{p_n\times p_n}$. Suppose that $\sup_n\|\sup_{l_1,l_2\le p_n}|X_{1,l_1,l_2}^n|\|_{L_{12}}<\infty$. We denote $ t_{g,n}^{\beta/4}$ a constant such that:
$$P\bp{\max_{i,j\le p_n}\big|N^n_{i,j}|\ge t_{g,n}^{\beta/4}}\le \beta/4 ,$$
where $(N^n)$ is a sequence of gaussian matrices satisfying $\rm{Cov}(N^n_{i,j},N^n_{k,l})=\rm{Cov}(X^n_{i,j},X^n_{k,l})$ and write $\hat \theta_n(x_{1:n})=\frac{1}{n}\min_{i\le p_n}\max_{j\le p_n}\sum_{l\le n}X_{l,i,j}^n$. Choose $t_{\beta/4}(X^n)$ such that the following holds:

$$P\bp{\sqrt{n}\ba{\hat \theta_n( Z^n)-\mathbb{E}(\hat{\theta}_n(Z^n)\mid X^n)}\ge t_{\beta/4}(X^n)\mid X^n}\le \beta/4.$$

The following holds
$$\lim_{\delta \downarrow 0}\limsup_{n\rightarrow \infty} P\bp{\sqrt{n}\ba{\hat \theta_n( X^n)-\theta_{X^n}}\ge t_{\beta/4}(X^n)+3t_{g,n}^{\beta/4}+\delta}\le \beta;$$
where  for a process $(Y'_{1:n})$ we have written, by abuse of notations,
$\theta_{Y'}:=\min_{i\le p_{n}}\max_{j\le p_{n}} ~\E\big[Y^{'}_{l,i,j}|X^n\big].$

We want to use <Ref> to prove this result.  By the triangle inequality we know that:

\begin{align}
\sqrt{n}\ba{\hat\theta_n(Y^n)-\theta_{Y^n}}
\le  \sqrt{n}\ba{\hat\theta_n(Y^n)-\mathbb{E}(\hat\theta_n({Y^n}))}+\sqrt{n}\ba{\mathbb{E}(\hat\theta_n({Y^n}))-\theta_{Y^n}}.
\end{align}

This implies that:

\begin{align}&
P\bp{\sqrt{n}\ba{\hat\theta_n(Y^n)-\theta_{Y^n}}\le t_{\beta/2}(X^n)+3t_{g,n}^{\beta/4}+\delta }
\\&\le P\bp{ \sqrt{n}\ba{\hat\theta_n(Y^n)-\mathbb{E}(\hat\theta_n({Y^n}))}\le t_{\beta/4}(X^n)+2t_{g,n}^{\beta/4}+\delta/2 }\\&+P\bp{\sqrt{n}\ba{\mathbb{E}(\hat\theta_n({Y^n}))-\theta_{Y^n}}\le t_{g,n}^{\beta/4}+\delta/2}.
\end{align}

We bound the first term using <Ref> and use arguments that are inspired by the proof of <Ref> to bound the second.

Firstly, we remark that in <Ref> we proved that $\sqrt{n}\hat{\theta}_n$ satisfies <Ref>. Moreover we note that for all $x\in \mathbb{R}^{p_n\times p_n}$ we have

$$\sqrt{n}\ba{\mathbb{E}\bp{\hat\theta_n(Y^n+\frac{x}{\sqrt{n}})-\hat\theta_n(Y^n)}}\le 2\max_{i,j\le p_n}|x_{i,j}| .$$
Therefore according to <Ref> we know that:

\begin{align}\label{sdcp_1}\lim_{\delta\downarrow 0}\limsup_{n\rightarrow \infty}P\bp{\sqrt{n}\ba{\hat \theta_n( Y^n)-\mathbb{E}(\hat{\theta}_n(Y^n))}\ge t_{\beta/4}(Y^n)+2 t_{g,n}^{\beta/4}+\delta}\le \beta/2.\end{align}

Moreover we note that:
$$\sqrt{n}\ba{\mathbb{E}(\hat\theta_n(Y^n))-\theta_{X^n}}\le \sup_{i,j}\ba{\frac{1}{\sqrt{n}}\sum_{l\le n}X_{l,i,j}^n-\mathbb{E}(X_{1,i,j}^n)}.$$

Using [16] we know that there is a constant $C$ that does not depend on $n$ such that the following holds:

\begin{align}&\label{sdcp_2}
\sup_t \Big|P\bb{\sup_{i,j}\ba{\frac{1}{\sqrt{n}}\sum_{l\le n}X_{l,i,j}^n-\mathbb{E}(X_{1,i,j}^n)}\ge t}-P(\max_{i,j}|N^{n}_{i,j}|\ge t)\Big|\\&\le \frac{C\log(p_n)^{7/6}\max(\|\max_{i,j\le p_n}|X^n_{1,i,j}|\|_{L_4}^4,1)}{n^{1/6}}\to 0.
\end{align}

Therefore by combining <ref> and <ref> we obtain that:
\begin{align}\label{sdcp_1}\lim_{\delta\downarrow 0}\limsup_{n\rightarrow \infty}P\bp{\sqrt{n}\ba{\hat \theta_n( Y^n)-\theta_{X^n}}\ge t_{\beta/4}(Y^n)+3 t_{g,n}^{\beta/4}+\delta}\le \beta.\end{align}

§.§ Proof of <Ref>

We separately prove each statement. Firstly we examine the case where both  $(H_0^{\rm{minmax}})$ and $(H_1^{\rm{minmax}})$ are assumed to hold.

The first step of the proof is to approximate $\sqrt{n}\hat{\theta}_n(\cdot)$ by a discrete version and use <Ref> to prove the desired results. In this goal, choose $(\epsilon_n)$  and $(\delta_n)$ to be two  sequences satisfying:
$$\epsilon_n=\frac{1}{n^{1/8}\sqrt{\log(nC_n)d'_nC_nc_n}};\qquad \delta_n=\frac{1}{(C_nn)^{3/2}}.$$

We denote $\Omega^{\mathcal{P}}_n:=\{\omega_i^*,~i\le p_n\}$ a maximal $\epsilon_n$-packing of $B_{d'_n}^1$, it is well known that $p_n\le \Big(\frac{2}{ \epsilon_n}+1\Big)^{d'_n}$. Moreover we remark that for all $\theta\in B_{d'_n}^1$ there is an $\omega^*\in\Omega^{\mathcal{P}}_n $ such that $\bn{\omega^*-\omega}\le \epsilon_n$. Indeed if this would not be the case then $\{\omega\}\bigcup \Omega^{\mathcal{P}}_n$ would also be a packing set of $B_{d'_n}^1$ contradicting the fact that we assumed it to be maximal. Similarly we can find a $\delta_n$-covering net of $B_{d'_n}^1$:  $\Omega^{\delta_n}:=\{\omega'_i,~i\le q_n\}$  with $q_n\le \bp{\frac{2}{\delta_n}+1}^{d'_n}$.

By exploiting conditions $(H_0^{\rm{minmax}})$  for all $\omega_1,\omega_2\in B_{d'_n}^1$ we have

\begin{equation}\label{gold35}\sup_{x\in \RR^{d_n}}\inf_{\omega_1^*,\omega_2^*\in \Omega^{\mathcal{P}}_n}\|f_n(x,\omega_1,\omega_2)-f_n(x,\omega^*_1,\omega^*_2)\big\|_{L_{\infty}}< C_n \epsilon_n;
\end{equation}

and  by using $(H_1^{\rm{minmax}})$ we remark that

\begin{equation}\label{gold2}\inf_{\substack{(i,j)\ne (k,l)}}\Big|\E\Big[f_n(Y_1^n,\omega^*_k,\omega^*_l)\Big]-\E\Big[f_n(Y_1^n,\omega^*_i,\omega^*_j)\Big]\Big|>c_n\epsilon_n.
\end{equation}

We define $g_n(x_{1:n}):=\min_{i\le p_n}\max_{j\le p_n}\frac{1}{\sqrt{n}}\sum_{l\le n}f_n(x_l,\omega^*_i,\omega^*_j)$ and remark that it approximates $\sqrt{n}\hat{\theta}_n(\cdot).$ Indeed by combining <ref> and <Ref> we know that there is a constant $K$ such that :
\begin{align}
\label{gold3}&\bn{g_n(\xi^n)-\sqrt{n}\hat{\theta}_n(\xi^n)}_{L_{1}}
\\&\le \bn{\sup_{\omega_1,\omega_2\in B_{d'_n}^1}\sup_{\omega^*_1,\omega^*_2\in \Omega^{\mathcal{P}}_n}\ba{\frac{1}{\sqrt{n}}\sum_{i\le n}  f_n(\xi^n_{i},\omega_1,\omega_2)-f_n(\xi^n_{i},\omega^*_1,\omega^*_2)}}_{L_{1}}
\\&\le  C_n\delta_n\sqrt{n}+\bn{\sup_{\omega'_1,\omega'_2\in \Omega^{\delta_n}}\sup_{\omega^*_1,\omega^*_2\in \Omega^{\mathcal{P}}_n}\ba{\frac{1}{\sqrt{n}}\sum_{i\le n}  f_n(\xi^n_{i},\omega'
\\&\le C_n^{-1/6} n^{-1/6}+K\log(q_n)C_n\epsilon_n
\\&\le C_n^{-1/6} n^{-1/6}+K\log(\frac{2}{(C_nn)^{3/2}}+1)d'_n C_n\epsilon_n\rightarrow 0.
\end{align}

Therefore if we establish that the bootstrap method is consistent for $(g_n)$ we will have demonstrated that it is for $\sqrt{n}\hat\theta_n(X_{1:n}^n)$.
For ease of notations we denote
$$\theta^{dis}_{Z^n}:=\min_{i\le p_n}\max_{j\le p_n}\E\big[f_n(Z^n_1, \omega^*_i,\omega^*_j)|\xi^n\big],\qquad \theta^{dis}_{\xi^n}:=\min_{i\le p_n}\max_{j\le p_n}\E\big[f_n(\xi^n_1, \omega^*_i,\omega^*_j)\big].$$

We remark that as $\log(p_n)=o(n^{1/4})$
and as <ref> guarantees that
$$\log(p_n)/\sqrt{n}=o(\min_{(i,j)\ne (k,l)}\ba{\E\bp{f_n(\xi_1^n,\omega_i,\omega_j)}-\E\bp{f_n(\xi_1^n,\omega_l,\omega_k)}})$$
then <Ref> guarantees that
$$\bn{d_{\mcF}\Big[g_n(\xi^n)-\sqrt{n}\theta^{dis}_{\xi^n},~{g_n(Z^n)-\sqrt{n}\theta^{dis}_{Z^n}}\Big|\xi^n\Big]}_{L_1}\rightarrow 0.$$

Hence by a last application of <ref>  we get the desired result.

We now examine the case where we only assume that $(H_0^{\rm{minmax}})$ holds.

The first step is to approximate $\sqrt{n}\hat{\theta}_n(\cdot)$ by a discrete version and use <Ref> to prove the desired results. In this goal, choose $(\epsilon_n)$  and $(\delta_n)$ to be two  sequences satisfying:
$$\epsilon_n=\frac{1}{n^{1/8}\sqrt{\log(nC_n)d'_nC_n}};\qquad \delta_n=\frac{1}{(C_nn)^{3/2}}.$$

Similarly as previously, we denote $\Omega^{\mathcal{P}}_n:=\{\omega_i^*,~i\le p_n\}$ a maximal $\epsilon_n$-packing of $B_{d'_n}^1$ with $p_n\le \Big(\frac{2}{ \epsilon_n}+1\Big)^{d'_n}$. Similarly we can find a $\delta_n$-covering net of $B_{d'_n}^1$:  $\Omega^{\delta_n}:=\{\omega'_i,~i\le q_n\}$  with $q_n\le \bp{\frac{2}{\delta_n}+1}^{d'_n}$.

As above, we remark that for all $\theta\in B_{d'_n}^1$ there is an $\omega^*\in\Omega^{\mathcal{P}}_n $ such that $\bn{\omega^*-\omega}\le \epsilon_n$.
By exploiting conditions $(H_0^{\rm{minmax}})$ we obtain for all $\omega_1,\omega_2\in B_{d'_n}^1$ that:

\begin{equation}\label{gold352}\sup_{x\in \RR^{d_n}}\inf_{\omega_1^*,\omega_2^*\in \Omega^{\mathcal{P}}_n}\|f_n(x,\omega_1,\omega_2)-f_n(x,\omega^*_1,\omega^*_2)\big\|_{L_{\infty}}<  C_n \epsilon_n.
\end{equation}

We define $g_n(x_{1:n}):=\min_{i\le p_n}\max_{j\le p_n}\frac{1}{\sqrt{n}}\sum_{l\le n}f_n(x_l,\omega^*_i,\omega^*_j)$ and remark that it approximates $\sqrt{n}\hat{\theta}_n(\cdot).$ Indeed by combining <ref> and <Ref> we know that there is a constant $K$ such that :
\begin{align}
\label{gold32}&\bn{g_n(\xi^n)-\sqrt{n}\hat{\theta}_n(\xi^n)}_{L_{1}}
\\&\le \bn{\sup_{\omega_1,\omega_2\in B_{d'_n}^1}\sup_{\omega^*_1,\omega^*_2\in \Omega^{\mathcal{P}}_n}\ba{\frac{1}{\sqrt{n}}\sum_{i\le n}  f_n(\xi^n_{i},\omega_1,\omega_2)-f_n(\xi^n_{i},\omega^*_1,\omega^*_2)}}_{L_{1}}
\\&\le  C_n\delta_n\sqrt{n}+\bn{\sup_{\omega'_1,\omega'_2\in \Omega^{\delta_n}}\sup_{\omega^*_1,\omega^*_2\in \Omega^{\mathcal{P}}_n}\ba{\frac{1}{\sqrt{n}}\sum_{i\le n}  f_n(\xi^n_{i},\omega'
\\&\le C_n^{-1/6} n^{-1/6}+K\log(q_n)C_n\epsilon_n
\\&\le C_n^{-1/6} n^{-1/6}+K\log(\frac{2}{(C_nn)^{3/2}}+1)d'_n C_n\epsilon_n\rightarrow 0.
\end{align}

Therefore if we establish the desired result for $(g_n)$ it will also hold for $\sqrt{n}\hat\theta_n(X_{1:n}^n)$.
We write $X_1^n:=(f_n(\xi_1^n,\omega_i,\omega_i))_{i,j\le p_n}$ and denote $ t_{g,n}^{\beta/4}$ a constant such that:
$$P\bp{\max_{i,j\le p_n}\big|N^n_{i,j}|\ge t_{g,n}^{\beta/4}}\le \beta/4 ,$$
where $N^n$ is a gaussian matrix with $\rm{Cov}\bp{N_{i,j}^n,N_{k,l}^n}=\rm{Cov}\bp{X_{i,j}^n,X_{k,l}^n}$. Choose $t_{\beta/4}(X^n)$ such that the following holds:

$$P\bp{\ba{g_n(Z^n)-\E(g_n(Z^n)\mid X^n)}\ge t_{\beta/4}(X^n)\mid X^n}\le \beta/4.$$

We remark that $\log(p_n)=o(n^{1/7})$
and that $g_n$ satisfies all of the conditions of <Ref>. Therefore according to <Ref> we know that:
$$\lim_{\delta\downarrow 0}\limsup_{n\rightarrow \infty}P\bp{\ba{g_n(Z^n)-\E(g_n(Z^n)\mid X^n)}\ge t_{\beta/4}(X^n)+3t_{g,n}^{\beta/4}+\delta\mid X^n}\le \beta.$$
We note that as $\|g_n(X^n)-\sqrt{n}\theta_n(X^n)\|_{L_1}$ we therefore have that
$$\lim_{\delta\downarrow 0}\limsup_{n\rightarrow \infty}P\bp{\ba{g_n(Y^n)-\E(g_n(Y^n))}\ge t_{\beta/4}(X^n)+3t_{g,n}^{\beta/4}+\delta\mid X^n}\le \beta.$$
Finally we note that $t^{*,\beta/4}_{g,n}\ge t^{\beta/4}_{g,n}$ which implies the desired result.

§ PROOFS FROM <REF>

§.§ Proof of <Ref>

For simplicity we write $\Theta_n:=\{\theta_k,~k\le p_n\}$ and for all $\theta\in \Theta_n$ we denote

$$D_n:=4\max\bp{\big\|\sup_{k\le p_n}K_{\theta_k}(X_{1,1}^{M},X_{1,1}^{M})\big\|_{L_{120}}, \sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k},~1}.$$

Using Mercer's theorem we note that there are orthonormal eigenfunctions $(\psi_{l,k})$ and  positive eigenvalues $(\lambda_{l,k})$ such that:

As we assumed that $\max_{k}\sum_{l}\lambda_{l,k}<\infty$ then there is a sequence $(l_n)$ such that:
$$\max_{k}\sum_{l\ge l_n}\lambda_{l,k}=o\bp{\min(\lambda_n,1)^2(\beta_nnp_nD_n^3)^{-1}}.$$
We show that we can suppose without loss of generality that the kernels $(K_{\theta_k})$ are of rank $l_n$ or less. In this goal, we denote $K_{\theta_k}^*(\cdot,\cdot)$ the following kernels $K_{\theta_k}^*(\cdot,\cdot):=\sum_{l\le l_n}\lambda_{l,k}\psi_{l,k}(\cdot)\psi_{l,k}(\cdot)$ and shorthand

$$p^*_{\theta}(X^{M}):=\frac{\frac{1}{n^2}\sum_{i,j\le n}H_{i,j}^{*\theta}(X^M)}{\frac{4}{|n|^3}\sum_{i\le n}\big[\sum_{j\le n} H_{i,j}^{*,\theta}(X^M)]^2-\frac{4}{n^4}\Big[\sum_{i,j\le n} H_{i,j}^{*,\theta}(X^M)\Big]^2+\lambda_n}.$$

$$ \omega^*_k(x_{1:n}):=\frac{e^{\beta_np^*_{\theta_k}(X^M)}}{\sum_{k'\le p_n}e^{\beta_np^*_{\theta_k'}(X^M)}}$$

We prove that:
$$\bn{ nT_n(X^M)-\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(X^M)\omega^*_k(X^M)}_{L_1}\to 0$$
and that
$$\bn{ nT_n(Z^M)-\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(Z^M)\omega^*_k(Z^M)}_{L_1}\to 0.$$
In this goal, we remark that:

\begin{equation}
\begin{split}\label{dm_0}
\sup_{i,j\le n} \bn{\sup_{k\le p_n}\ba{ \bp{ H_{i,j}^{*\theta_k}(X^M)-H^{\theta_k}_{i,j}}}}_{L_4}
&\le\sup_{i,j\le n}\sum_{k\le p_n} \bn{{ \bp{ H_{i,j}^{*\theta_k}(X^M)-H^{\theta_k}_{i,j}}}}_{L_4}
\\&\overset{(a)}{\le}4\sup_{i,j\le n}\bn{{ \sup_{k\le p_n}\ba{ K^{*}_{\theta_k}(X^M_{i,1},X^M_{j,2})-K_{\theta_k}(X^M_{i,1},X^M_{j,2})}}}_{L_4}
\\&\le 4\sup_{i,j\le n}\bn{{ \sup_{k\le p_n}\sum_{l\ge l_n}\lambda_{l,k}\ba{ \psi_{l,k}(X_{i,1}^M)\psi(X_{j,2}^M)}}}_{L_4}
\\&\overset{}{\le}4p_n\sup_{k\le p_n}\sum_{l\ge l_n} \lambda_{l,k} \sup_{i\le n}\sup_{k\le p_n, l\in \mathbb{N}}\bn{{ { \psi_{l,k}(X_{i,1}^M)^2}}}_{L_4}
\\&\overset{}{\le}4p_nD_n\sup_{k\le p_n}\sum_{l\ge l_n} \lambda_{l,k}
\end{split}
\end{equation}
where to get (a) we used the fact that $X_{i,1}^M$ has the same distribution than $X_{i,2}^M$.

We remark that the function $\sigma_n:(x_1,\dots,x_{p_n})\rightarrow \sum_{k\le p_n}H_{i,j}^{*\theta_k}(X^M) \frac{e^{\beta_nx_k}}{\sum_{k'\le p_n} e^{\beta_nx_{k'}}}$ is Lipchitz in the max norm: $\ba{\sigma_n(x)-\sigma_n(y)}\le \beta_n \sup_{k\le p_n}\ba{H_{i,j}^{*\theta_k}(X^M)} \max_{k\le p_n}\ba{x_k-y_k}.$ Therefore coupling this with  the triangle inequality we get that

\begin{equation}
\begin{split}&~\label{dm_1}
\bn{ nT_n(X^M)-\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(X^M)\omega^*_k(X^M)}_{L_1}
\\\le& \bn{\sum_{\substack{k\le p_n\\i,j\le n}}\frac{1}{n} \bp{ H_{i,j}^{*\theta_k}(X^M)-H^{\theta_k}_{i,j}}\omega_k(X^M)}_{L_1}+ \bn{\frac{1}{n}\sum_{\substack{k\le p_n\\i,j\le n}}  H_{i,j}^{*\theta_k}(X^M)\bp{\omega_k(X^M)-\omega^*_k(X^M)}}_{L_1}
\\\overset{(a)}{\le}&n \bn{\sup_{k\le p_n}\ba{\frac{1}{n^2}\sum_{i,j\le n} \bp{ H_{i,j}^{*\theta_k}(X^M)-H^{\theta_k}_{i,j}}}}_{L_1}+2n\beta_n\sup_{i,j\le n}\bn{\sup_{k\le p_n}\ba{H_{i,j}^{*\theta_k}(X^M)}}_{L_2}
\bn{\sup_{k\le p_n}\ba{p_{\theta_k}^*(X^n)-p_{\theta_k}(X^n)}}_{L_2}.
\end{split}
\end{equation}

The first term of <ref> is bounded by:

$$n \bn{\sup_{k\le p_n}\ba{\frac{1}{n^2}\sum_{i,j\le n} \bp{ H_{i,j}^{*\theta_k}(X^M)-H^{\theta_k}_{i,j}}}}_{L_1}\le np_n\sup_{k\le p_n}\sum_{l\ge l_n}\lambda_{l,k}\rightarrow 0.$$

Moreover we can bound the second term of <ref> using the triangular inequality, <ref> and the fact that
$$\bp{\frac{4}{|n|^3}\sum_{i\le n}\big[\sum_{j\le n} H_{i,j}^{*,\theta}]^2-\frac{4}{n^4}\Big[\sum_{i,j\le n} H_{i,j}^{*,\theta}\Big]^2+\lambda_n}^{-1}\ge \lambda_n,\qquad \bn{\sup_{k\le p_n}\ba{H_{i,j}^{*\theta_k}(X^M)}}_{L_4}\le D_n.$$
Indeed we have,

\begin{equation}
\begin{split}\label{dm_2}&~
2n\beta_n\sup_{i,j\le n}\bn{\sup_{k\le p_n}\ba{H_{i,j}^{*\theta_k}(X^M)}}_{L_2}
\bn{\sup_{k\le p_n}\ba{p_{\theta_k}^*(X^n)-p_{\theta_k}(X^n)}}_{L_2}
\\\overset{}{\le}& \frac{2n}{\lambda_n}\beta_nD_n
\bn{\sup_{k\le p_n}\ba{\frac{1}{n^2}\sum_{i,j\le n} \bp{ H_{i,j}^{*\theta_k}(X^M)-H^{\theta_k}_{i,j}}}}_{L_2}
+\frac{2n\beta_n}{\lambda_n^2}D_n^2\sup_{k\le p_n}
\Big\|\frac{4}{|n|^3}\sum_{i\le n}\Big(\big[\sum_{j\le n} H_{i,j}^{*\theta_k}(X^M)]^2\\&-\big[\sum_{j\le n} H_{i,j}^{\theta_k}]^2\Big)-\frac{4}{n^4}\bp{\Big[\sum_{i,j\le n} H_{i,j}^{*\theta_k}(X^M)\Big]^2-\Big[\sum_{i,j\le n} H_{i,j}^{\theta_k}\Big]^2}\Big\|_{L_2}
\\\le & 4np_n\sup_{k\le p_n}\sum_{l\ge l_n}\lambda_{l,k}\big[1+\frac{2D_n\beta_n}{\lambda_n}+\frac{16D_n^3}{\lambda_n^2}\big]\rightarrow 0.
\end{split}
\end{equation}

This implies directly that: $ \bn{nT_n(X^M)-\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(X^M)\omega^*_k(X^M)}_{L_1}\rightarrow 0$.
Following the exact same road map we can show that:$ \bn{nT_n(Z^M)-\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(Z^M)\omega^*_k(Z^M_{1:n})}_{L_1}\rightarrow 0$.

Therefore we have
$$ \bn{ d_{\mcF}\bp{nT_n(Z^M),\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(Z^M)\omega^*_k(Z^M)\mid X^n}}_{L_1}\rightarrow 0;$$

$$ \bn{ d_{\mcF}\bp{nT_n(Y^M),\sum_{k\le p_n} \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(Y^M)\omega^*_k(Y^M)\mid X^n}}_{L_1}\rightarrow 0.$$
Hence using <ref> we know that to establish the desired result it is sufficient to study the asymptotics of $ \frac{1}{n}\sum_{i,j\le n}  H_{i,j}^{*\theta_k}(Z^M)\omega^*_k(Z^M)$. Hence we can suppose without loss of generality that all the Kernels $(K_{\theta_i})$ have ranks of $l_n$ or less, and do so

As the distribution of $X_{i,1}^M$ is the same than the one of $X_{i,2}^M$ they also have the same mean embedding. Moreover we note that for all $c\in\mathbb{R}$ we have :
\begin{equation}
\begin{split}
H_{i,j}^{\theta_k}=&\sum_{l\le l_n}\lambda_{l,k}\big[\psi_{l,k}(X^M_{i,1})-\psi_{l,k}(X^M_{i,2})\big]\big[\psi_{l,k}(X_{j,1}^n)-\psi_{l,k}(X_{j,2}^n)\big]
\\&=\sum_{l\le l_n}\lambda_{l,k}\big[\psi_{l,k}(X^M_{i,1})+c-\big(\psi_{l,k}(X^M_{i,2})+c\big)\big]\big[\big(\psi_{l,k}(X_{j,1}^n)+c\big)-\big(\psi_{l,k}(X_{j,2}^n)+c\big)\big] .
\end{split}
\end{equation}
Therefore as the test statistics $\hat{T_n}$ depends only on $(H_{i,j}^{\theta_k})$ we can suppose without loss of generality that
$$\E(\psi_{l,k}(X^{M}_{1,1}))= \E(\psi_{l,k}(X^{M}_{1,2}))=0,\quad \forall\le l_n,~ k\le p_n.$$

We define $X_i^{*,n}:=(X_{i,l,k}^{*})_{l,k}$ to be random variables, defined as
\begin{equation}\begin{split}X_{i,l,k}^{*}:=&\psi_{l,k}(X^{n}_{i,1})-\psi_{l,k}(X^{n}_{i,2}).\end{split}\end{equation}

We define the process $X^{*,n}:=\big(X_i^{*,n}\big)$ and note that the observations $(X^{*,n}_i)$ take value in $M_{p_n\times l_n}(\RR)$. We remark that we could have taken the observations $(X^{*,n}_i)$ to take value in $\RR^{p_nl_n}$ as there is a one to one mapping from $M_{p_n\times l_n}(\RR)$ to $\RR^{p_nl_n}$. However for ease of notations we keep them as defined.
Let $( Z^{*}_i)$  and $(Y^{*}_i)$ be defined in the following way:

$$Z^*_i:=\big(\psi_{l,k}(Z^M_{i,1})- \psi_{l,k}(Z^M_{i,2})\big) \quad {\rm and}\quad Y^*_i:=\big(\psi_{l,k}(Y^M_{i,1})-\psi_{l,k}(Y^M_{i,2})\big).$$
We note that they form
respectively a bootstrap sample and  an independent copy  of $(X^{*}_i)$. Moreover we remark that $Z_{i,1}^M|X^n\overset{d}{=}Z_{i,2}^M|X^n$ therefore we can also assume with out loss of generality that $\E(Z^*_{i,l,k}|X^n)=0$. We choose $(g_n)$ to be the following sequence of functions:

$$g_n(x_{1:n}):= \sum_{k}\omega_{k}(x_{1:n})\sum_{l}\lambda_{l,k}\Big(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k}\Big)^2$$

where we have set $\omega_k(x_{1:n})= \frac{e^{\beta_nh_{n,k}(x_{1:n})}}{\sum_{k}e^{\beta_n h_{n,k}(x_{1:n})}}$ and
$$h_{n,k}(x_{1:n}):=\frac{\sum_{l}\lambda_{l,k}\Big(\frac{1}{n}\sum_{j\le n}x_{j,l,k}\Big)^2}{\frac{4}{n^3}\sum_l\lambda_{l,k} (\sum_ix_{i,l,k})^3-\frac{4}{n^4}(\sum_l\lambda_{l,k}(\sum_ix_{i,l,k})^2)^2+\lambda_n}.$$

We remark that we have
$$g_n(Z^{*}_{1:n})=n T_n( Z^M),\quad g_n(\tilde Y^{*}_{1:n})=n T_n( Y^M),\quad g_n(X^{*}_{1:n})=n T_n( X^M).$$
It is therefore enough to study $g_n$ and $(X^{*}_{{1:n}})$.
For ease of notations we write:
\begin{equation}\label{cvd}\widehat{\sigma_{n,k}}(x_{1:n}):=\frac{4}{n^3}\sum_l\lambda_{l,k} (\sum_ix_{i,l,k})^3-\frac{4}{n^4}(\sum_l\lambda_{l,k}(\sum_ix_{i,l,j})^2)^2+\lambda_n;\end{equation}

\begin{equation}\label{cvd2}\overline{x_{l,k_1}}:=\frac{1}{{n}}\sum_{j\le n}x_{j,l,k_1},\quad\hat{H}_{n,k}(x_{1:n}):=\sum_{l}\lambda_{l,k}\Big(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k}\Big)^2.\end{equation}

Moreover we note that we have
\begin{equation}
\begin{split}\label{eq:ker:1}
\big\| \sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}[X^{*}_{1,l,k}]^2\|_{L_{120}}&\overset{(a)}{\le} 2  \big\| \sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}\Big([\psi_{l,k}(X^M_{i,1})]^2+[\psi_{l,k}(X^M_{i,2})]^2\Big)\|_{L_{120}}
\\&\overset{(b)}{\le}4\big\|\sup_{k\le p_n}K_{\theta_k}(X_{1,1}^{M},X_{1,1}^{M})\big\|_{L_{120}}
\end{split}
\end{equation}
where (a) and (b) come from the Cauchy-Schwarz inequality. Similarly we have:
\begin{equation}
\begin{split}\label{eq:ker:2}
\bn{ \sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}\ba{X^{*}_{1,l,k}}}_{L_{120}}
&\overset{(a)}{\le}  \sqrt{\sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}}\sqrt{ \bn{ \sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}\ba{X^{*}_{1,l,k}}^2}_{L_{120}}}
\\&\le 2\sqrt{\sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}}\sqrt{\bn{\sup_{k\le p_n}K_{\theta_k}(X_{1,1}^{M},X_{1,1}^{M})}_{L_{120}}}
\end{split}
\end{equation}
where (a) comes from the Cauchy-Schwarz inequality. Therefore using <Ref> we note that there is a constant $C$ such that:

\begin{align}&
\Big\|\sup_{k\le p_n}\sup_{z\in [\bar{X}^n, Z_1^{*}]\cup[\bar{X}^n,\tilde Y_1^*]} \frac{1}{\sqrt{n}}\hat{H}_{n,k}(Z^{*,i,z})\Big\|_{L_{120}}
\\&\overset{(a)}{\le}\frac{2}{n^{3/2}}\big\| \sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}[X^{*}_{1,l,k}]^2\|_{L_{120}}
+2\Big\|\sup_{k\le p_n} \frac{1}{\sqrt{n}}\sum_{l\le l_n}\lambda_{l,k}(\frac{1}{\sqrt{n}}\sum_{l\le n}Z_l^{*,n,i})^2\Big\|_{L_{120}}
\\&\overset{(b)}{\le} C\log(p_n) \|\sup_{k\le p_n}K_{\theta_k}(X_{1,1}^{M},X_{1,1}^{M})\big\|_{L_{120}}.
\end{align}

where (a) is a consequence of the triangle inequality and (b) of <Ref> and <ref>. Similarly using <Ref> and <ref> we can establish that there is a constant $C'$ such that:

\begin{align}&\Big\|\sup_{k\le p_n}\sup_{z\in [\bar{X}^n, Z_1^{*}]\cup [\bar{X}^n,\tilde Y_1^*]} \sum_{l\le l_n} \lambda_{l,k}\frac{1}{\sqrt{n}}\Big|z_{l,k}+\sum_{i\ge 2} Z_{i,l,k}^{*}\Big|\Big\|_{L_{120}}\\&\le C'\log(p_n) \sqrt{\sup_{k\le p_n}\sum_{l\le l_n}\lambda_{l,k}}\sqrt{\bn{\sup_{k\le p_n}K_{\theta_k}(X_{1,1}^{M},X_{1,1}^{M})}_{L_{120}}}\end{align}

To prove the desired result we use <Ref>. We remark that the functions $(h_{n,k})$ and $(g_n)$ are three times differentiable. This implies that $(H_0)$ hold. We check that $(H_1^*)$ also holds.
In this goal we first check that the partial derivatives of $(h_{n,k})$ are bounded. For ease of notations for a function $f_n$ we shorthand:
$$\partial_{i,k,l}f_n(x_{1:n}):=\partial_{x_{i,k,l}}f_n(x_{1:n}),\quad \partial_{i,k_{1:2},l_{1:2}}^2f_n(x_{1:n}):=\partial_{x_{i,k_1,l_1},x_{i,k_2,l_2}}^2f_n(x_{1:n})$$

In this goal, using the chain rule we note that
for all $k\le p_n$ and all $l\le l_n$ we have:

\begin{align}
\partial_{{i,k,l}}\hat{\sigma}_{n,k}(x_{1:n}):=
\frac{\lambda_{l,k}}{n}\bp{12\overline{x_{l,k}}^2
-16\overline{x_{l,k}} \frac{\hat{H}_{n,k}(x_{1:n})}{n}}
\end{align}

\begin{equation}
\begin{split}
\partial_{{i,k,l}}h_{n,k}(x_{1:n})=&
\frac{2}{{n}}\frac{\lambda_{l,k}\overline{x_{l,k}}}{\widehat{\sigma_{n,k}}(x_{1:n})}
-\frac{\hat{H}_{n,k}(x_{1:n}) \partial_{{i,k,l}}\widehat{\sigma}_{n,k}(x_{1:n})}{\widehat{\sigma_{n,k}}(x_{1:n})^2}.
\end{split}
\end{equation}

Therefore we obtain that there is a constant $K$ such that:

\begin{equation}
\begin{split}
{n}  \bn{\sup_{k\le p_n}\sum_{l\le l_n}  \ba{ \partial_{{i,k,l}}h_{n,k}(Z^{n, i, \bar{X}^n})}}_{L_{12}}
\le
K\Big[\frac{D_n}{\lambda_n}+ \frac{D_n^2+D_n^3}{\lambda_n^2}\Big]
\end{split}
\end{equation}

Using once again the chain rule we have that:

\begin{equation}
\begin{split}
\partial_{{i,k,l}}\omega_{k'}(x_{1:n})=& \beta_n  \partial_{{i,k,l}}h_{n,k}(x_{1:n})\omega_{k'}(x_{1:n})\bp{\mathbb{I}(k'=k)-\omega_{k}(x_{1:n})}
\end{split}
\end{equation}

as well as:

\begin{equation}
\begin{split}
\partial_{{i,k,l}}g_n(x_{1:n})=& \frac{2}{\sqrt{n}}\lambda_{l,k}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k}) \omega_k(x_{1:n})+\sum_{k'\le p_n}\hat{H}_{n,k'}(x_{1:n}) \partial_{{i,k,l}}\omega_{k'}(x_{1:n}).
\end{split}
\end{equation}

Therefore, using <ref> and <ref> and as $\sum_{k\le p_n}\omega_k(x_{1:n})=1$
we obtain that there is a constant $K_2$ that do not depend on $n$ such that

\begin{equation}
\begin{split}&
\sup_{i\le n}  \Big\| \big\| \partial_{{i,k,l}}g_n( Z^{*,i,\bar{X}^n}_{1:n})\big\|_{v,p_n\times l_n} \Big\|_{L_{12}}
\\&\le  \beta_n\sup_{i\le n}  \Big\| \sup_{k\le p_n}\big|\frac{1}{\sqrt{n}}\hat{H}_{n,k}(Z^{*,i,\bar{X^n}})\big|\Big\|_{L_{24}}\Big\| \sup_{k\le p_n}\sum_{l\le l_n}\sqrt{n}\big| \partial_{{i,k,l}}h_{n,k}(Z^{*,i,\bar{X^n}})\big| \Big\|_{L_{24}}
\\&\quad+ \sup_{i\le n}  \Big\| \sup_{k\le p_n}\sum_{l\le l_n}\big|\lambda_{l,k}(\frac{1}{\sqrt{n}}\sum_{j\le n}Z^{*,i,\bar{X^n}}_{j,l,k})\big| \Big\|_{L_{12}}
\\&\le \frac{K_2\beta_n\log(p_n)D_n^4}{\sqrt{n}\min(\lambda_n^2,1)} .
\end{split}
\end{equation}

This implies that $R^*_{n,1}:=O(\frac{\beta_n\log(p_n)D_n^4}{n^{1/6}\min(\lambda_n^2,1)}).$
Moreover all $k_1,l_1,k_2,l_2$ using the chain rule we have: $\partial_{{i,k_{1:2},l_{1:2}}}^2h_{n,k_1}(x_{1:n})=0$ if $k_1$ is distinct from $k_2$. Moreover if $k_1=k_2$ we have:

\begin{align}
\partial_{i,k_{1:2},l_{1:2}}\hat{\sigma}_{n,k}(x_{1:n}):=
\frac{\lambda_{k_1,l_1}}{n^2}\bp{24\overline{x_{l_1,k_1}}\mathbb{I}(l_1=l_2)
-16 \frac{\hat{H}_{n,k_1}(x_{1:n})}{n}\mathbb{I}(l_1=l_2)
\end{align}

and by the Chain rule we have:

\begin{equation}
\begin{split}
\partial_{i,k_{1:2},l_{1:2}}^2h_{n,k_1}(x_{1:n})
\\&-\frac{\partial_{{i,k_2,l_2}}h_{n,k_2}(x_{1:n})}{n}\frac{\partial_{{i,k_1,l_1}}\widehat{\sigma_{n,k_1}}(x_{1:n})}{\widehat{\sigma_{n,k_1}}(x_{1:n})}
\\&+ 2\frac{\hat{H}_{n,k_1}(x_{1:n})}{n}\frac{\partial_{{i,k_1,l_1}}\widehat{\sigma_{n,k_1}}(x_{1:n})\partial_{{i,k_2,l_2}}\widehat{\sigma_{n,k_1}}(x_{1:n})}{\widehat{\sigma_{n,k_1}}(x_{1:n})^3}.
\end{split}
\end{equation}

Therefore there is a constant $K_3$ such that:

\begin{equation}
\begin{split}&
\Big\| \sup_{k_{1:2}\le p_n} \sum_{l_{1:2}\le l_n}\ba{ \partial_{i,k_{1:2},l_{1:2}}^2h_{n,k_1}(Z^{*, i, \bar{X}^n}_{1:n}) }\Big\|_{L_{12}}
\le\frac{ K_3D_n^5}{\max(\lambda_n^3,1)n^2} .
\end{split}\end{equation}

By another application of the chain rule we have:

\begin{equation}
\begin{split}
\partial_{i,k_{1:2},l_{1:2}}\omega_{k'}(x_{1:n})=& \beta_n \bp{\mathbb{I}(k'=k_1)-\omega_{k_1}(x_{1:n})}
\\&\times \bp{\partial_{i,k_{1:2},l_{1:2}}^2h_{n,k_1}(x_{1:n})\omega_{k'}(x_{1:n})+ \partial_{{i,k_1,l_1}}h_{n,k_1}(x_{1:n})\partial_{{i,k_2,l_2}}\omega_{k'}(x_{1:n})}
\\-&    \beta_n \partial_{{i,k_1,l_1}}h_{n,k_1}(x_{1:n})\omega_{k'}(x_{1:n})\partial_{{i,k_2,l_2}}\omega_{k_1}(x_{1:n})
\end{split}
\end{equation}

by the chain rule we also have that:

\begin{equation}
\begin{split}
\partial_{i,k_{1:2},l_{1:2}}g_n(x_{1:n})=& \frac{2}{{n}}\lambda_{k_1,l_1}\mathbb{I}(k_1=k_2) \omega_{k_1}(x_{1:n})+\frac{2}{\sqrt{n}}\lambda_{k_1,l_1}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k_1}) \partial_{{i,k_2,l_2}}\omega_{k_1}(x_{1:n})
\\&+\frac{2}{\sqrt{n}}\lambda_{k_2,l_2}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,k_2,l_2})\partial_{{i,k_1,l_1}}\omega_{k_2}(x_{1:n})
\\&+\sum_{k'\le p_n}\hat{H}_{n,k'}(x_{1:n})\partial_{i,k_{1:2},l_{1:2}}^2\omega_{k'}(x_{1:n})
\end{split}
\end{equation}

Therefore we know that there are constants $K_3$ such that:

\begin{equation}
\begin{split}&
\Big\| \big\|\partial^2_{x_{1}}g_n( Z^{*,i,\bar{X^n}}_{1:n})\big\|_{m,p_n\times l_n} \Big\|_{L_{12}}
\le \frac{K_3D_n^7}{\min(\lambda_n^4,1)n}\beta_n^2.
\end{split}\end{equation}

This implies that $R_{n,2}^*=O(\frac{\beta_n^2D_n^7}{\min(\lambda_n^4,1)\sqrt{n}})$.    Finally using a similar line of reasoning we note that Moreover all $k_1,l_1,k_2,l_2,l_3,k_3$ using the chain rule we have: $\partial_{{i,k_{1:3},l_{1:3}}}^3h_{n,k_1}(x_{1:n})=0$ if $k_1$ is distinct from $k_2$ or from $k_3$. Moreover if $k_1=k_2=k_3$ we have:

\begin{align}
\partial_{i,l_{1:3},k_{1:3}}^3\hat{\sigma}_{n,k}(x_{1:n})=
\frac{\lambda_{k_1,l_1}}{n^3}\Big(&24\mathbb{I}(l_1=l_2=l_3)-32\lambda_{k_1,l_3}\overline{x_{l_3,k_1}}\mathbb{I}(l_1=l_2)
\\&-16\lambda_{k_2,l_2}(\overline{x_{l_1,k_1}}\mathbb{I}(l_3=l_2)+\overline{x_{l_2,k_1}}\mathbb{I}(l_1=l_3) )\Big)
\end{align}

and by the Chain rule we have:

\begin{equation}
\begin{split}&
\partial_{i,k_{1:3},l_{1:3}}^3h_{n,k_1}(x_{1:n})
\\=&-\frac{\partial_{l_3,k_3}\widehat{\sigma_{n,k_1}}(x_{1:n})}{n\widehat{\sigma_{n,k_1}}(x_{1:n})^2}\Big(\frac{\overline{x_{l_1,k_1}}\partial_{{i,k_2,l_2}}\widehat{\sigma_{n,k_1}}(x_{1:n})}{\widehat{\sigma_{n,k_1}}(x_{1:n})}\big[2\lambda_{k_1,l_1}+3\frac{\hat{H}_{n,k_1}(x_{1:n})\partial_{{i,k_1,l_1}}\widehat{\sigma_{n,k_1}}(x_{1:n})}{\widehat{\sigma_{n,k_1}}(x_{1:n})}\Big]\\-&\partial_{{i,k_2,l_2}}\hat{H}_{n,k_2}(x_{1:n}) \partial_{{i,k_1,l_1}}\widehat{\sigma_{n,k_1}(x_{1:n})}
\\+&\frac{2\lambda_{k_1,l_1}}{n\widehat{\sigma_{n,k_1}}(x_{1:n})^2}\bp{\frac{\partial_{{i,k_2,l_2}}\widehat{\sigma_{n,k_1}}(x_{1:n})\mathbb{I}(l_2=l_3)}{n}+\overline{x_{l_3,k_1}}\partial^2_{{i,k_{2:3},l_{2:3}}}\widehat{\sigma_{n,k_1}}(x_{1:n})}\\-&\frac{\partial_{{i,k_{2:3},l_{2:3}}}h_{n,k_2}(x_{1:n})}{n}\frac{\partial_{{i,k_1,l_1}}\widehat{\sigma_{n,k_1}}(x_{1:n})}{\widehat{\sigma_{n,k_1}}(x_{1:n})}
\\& -\frac{\hat{H}_{n,k_1}(x_{1:n})}{n\widehat{\sigma_{n,k_1}}(x_{1:n})}\bp{\partial_{i,k_{1:3},l_{1:3}}^3\widehat{\sigma_{n,k_1}}(x_{1:n})-\frac{\partial_{{i,k_1,l_1}}\widehat{\sigma_{n,k_1}}(x_{1:n})\partial_{{i,k_{2:3},l_{2:3}}}^2\widehat{\sigma_{n,k_1}}(x_{1:n})}{\widehat{\sigma_{n,k_1}}(x_{1:n})}}
\end{split}
\end{equation}

Therefore there is a constant $K_3$ such that:

\begin{equation}
\begin{split}&
\Big\| \sup_{k_{1:3}\le p_n}\sup_{z\in [\bar{X}^n,Z_i^*]\cup[\bar{X}^n,\tilde Y^*_i]} \sum_{l_{1:3}\le l_n} \ba{\partial_{i,k_{1:3},l_{1:3}}^3h_{n,k_1}(Z^{*, i, z}_{1:n}) }\Big\|_{L_{12}}
\le\frac{ K_3D_n^7}{\max(\lambda_n^4,1)n^3} .
\end{split}\end{equation}

By another application of the chain rule we have:

\begin{equation}
\begin{split}
\partial_{i,k_{1:3},l_{1:3}}\omega_{k'}(x_{1:n})=& \beta_n\bp{\mathbb{I}(k'=k_1)-\omega_{k_1}(x_{1:n})}
\\&\times \Big({\partial_{i,k_{1:3},l_{1:3}}^3h_{n,k_1}(x_{1:n})\omega_{k'}(x_{1:n})+\partial_{i,k_{1:2},l_{1:2}}^2h_{n,k_1}(x_{1:n})\partial_{i,k_3,l_3}\omega_{k'}(x_{1:n})}%
\\& \quad +{ \partial_{{i,k_1,l_1}}h_{n,k_1}(x_{1:n})\partial^2_{{i,k_{2:3},l_{2:3}}}\omega_{k'}(x_{1:n})+\partial^2_{{i,k_1,l_1},i,k_3,l_3}h_{n,k_1}(x_{1:n})\partial_{{i,k_2,l_2}}\omega_{k'}(x_{1:n})
\\-&\beta_n\partial_{i,k_3,l_3}\omega_{k_1}(x_{1:n})\\&\times \bp{ \partial_{i,k_{1:2},l_{1:2}}^2h_{n,k_1}(x_{1:n})\omega_{k'}(x_{1:n})- \partial_{{i,k_1,l_1}}h_{n,k_1}(x_{1:n})\partial_{{i,k_{2},l_{2}}}\omega_{k'}(x_{1:n})}
\\-&    \beta_n\omega_{k'}(x_{1:n})\\&\times\Big( \partial^2_{{i,k_1,l_1},i,k_3,l_3}h_{n,k_1}(x_{1:n})\partial_{{i,k_2,l_2}}\omega_{k_1}(x_{1:n})+ \partial_{{i,k_1,l_1}}h_{n,k_1}(x_{1:n})\partial^2_{{i,k_{2:3},l_{2:3}}}\omega_{k_1}(x_{1:n})\Big)
\end{split}
\end{equation}

as well as:

\begin{equation}
\begin{split}&
\partial_{i,k_{1:3},l_{1:3}}^3g_n(x_{1:n})\\=& \frac{2}{{n}}\lambda_{k_1,l_1}\mathbb{I}(k_1=k_2=k_3) \partial_{i,k_3,l_3}\omega_{k_1}(x_{1:n})+\frac{2}{\sqrt{n}}\lambda_{k_1,l_1}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k_1}) \partial_{{i,k_{2:3},l_{2:3}}}\omega_{k_1}(x_{1:n})
\\&+\frac{2\mathbb{I}(k_1=k_3)}{n}\lambda_{k_1,l_1} \partial_{{i,k_{2},l_{2}}}\omega_{k_1}(x_{1:n})+\frac{2\mathbb{I}(k_2=k_3)}{{n}}\lambda_{k_2,l_2}\partial_{{i,k_1,l_1}}\omega_{k_2}(x_{1:n})
\\&+\frac{2}{\sqrt{n}}\lambda_{k_2,l_2}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,k_2,l_2})\partial_{{i,k_1,l_1},i,k_3,l_3}^2\omega_{k_2}(x_{1:n})\\&+\frac{2}{\sqrt{n}}\lambda_{k_3,l_3} (\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,k_3,l})\partial_{i,k_{1:2},l_{1:2}}^2\omega_{k_3}(x_{1:n})
+\sum_{k'\le p_n}\hat{H}_{n,k'}(x_{1:n})\partial_{i,k_{1:3},l_{1:3}}^3\omega_{k'}(x_{1:n})
%   +{\beta_n} \partial_{{i,k,l}}h_{n,k}(x_{1:n})\hat{H}_{n,k}(x_{1:n})\omega_k(x_{1:n})
%     \\&-{\beta_n}\omega_k(x_{1:n}) \partial_{{i,k,l}}h_{n,k}(x_{1:n})\sum_{k''\le p_n}\hat{H}_{n,k''}(x_{1:n})\omega_{k''}(x_{1:n})
%\\&+\frac{2\beta_n}{{n}^{3/2}}\frac{\lambda_{l,k}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k}) e^{\beta_nh_{n,k}(x_{1:n})}}{\widehat{\sigma_{n,k}}(x_{1:n})\sum_{k'} e^{\beta_n h_{n,k'}(x_{1:n})}}\hat{H}_{n,k}(x_{1:n})\Big[1-\frac{e^{\beta_nh_{n,k}(x_{1:n})}}{\sum_{k'} e^{\beta_n h_{n,k'}(x_{1:n})}}\Big]
%  \\&-\frac{12\beta_n}{n^3}\frac{\lambda_{l,k}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k})^2 e^{\beta_nh_{n,k}(x_{1:n})}}{\widehat{\sigma_{n,k}}(x_{1:n})^2\sum_{k'} e^{\beta_n h_{n,k'}(x_{1:n})}} \hat{H}_{n,k}(x_{1:n})^2\Big[1-\frac{e^{\beta_nh_{n,k}(x_{1:n})}}{\sum_{k'} e^{\beta_n h_{n,k'}(x_{1:n})}}\Big]
% \\&+\frac{16\beta_n}{n^{7/2}}\frac{\lambda_{l,k}(\frac{1}{\sqrt{n}}\sum_{j\le n}x_{j,l,k}) e^{\beta_nh_{n,k}(x_{1:n})}}{\widehat{\sigma_{n,k}}(x_{1:n})^2\sum_{k'} e^{\beta_n h_{n,k'}(x_{1:n})}} \hat{H}_{n,k}(x_{1:n})^3\Big[1-\frac{e^{\beta_nh_{n,k}(x_{1:n})}}{\sum_{k'} e^{\beta_n h_{n,k'}(x_{1:n})}}\Big]
\end{split}
\end{equation}

Therefore we know that there are constants $K_4$ such that:

\begin{equation}
\begin{split}&
\max_i  \Big\|\max_{z\in [\bar{X^n},Z^*_{i}]\cup[\bar{X^n},\tilde Y^*_i]} \big\|\partial^2_{x_{1}}g_n( Z^{*,i,z}_{1:n})\big\|_{t,p_n\times l_n} \Big\|_{L_{12}}
%  \\&\le\frac{ K_4D_n}{n}\Big[1+\beta_n\frac{D_n^4}{\lambda_n^3}+\beta^2_n\frac{D_n^6}{\lambda_n^4}\Big]
\le \frac{K_4D_n^{10}}{\min(\lambda_n^6,1)n^2}\beta_n^3
%\\&\Big[1+\frac{\beta_n}{\lambda_n^2}\big[1+\frac{\beta_n}{\lambda_n^2}\big]\Big]\max\Big(\big\|\sup_{k\le p_n}K_{\theta_k}(X_{1,1}^{n},X_{1,1}^{n})\big\|^{5}_{120},1\Big)
% \\&\le\frac{K_4\beta_nD_n}{\lambda_nn}\Big[
%\Big[1+\frac{D_n+D_n^2}{\lambda_n}\Big]+\beta_n \frac{D_n}{\lambda_n}
%D_n\sup_{k_1}\sum_{l}\lambda_{l,k_1} \Big[1+\frac{D_n^2}{n^4}\Big]\\&~~+D_n
%  +\frac{1}{n\lambda_n}\Big[D_n+D_n^2+D_n^3\big]\Big[D_n+D_n^2\Big]
%          +\big[D_n^2+D_n^3\big]\Big]\Big]
%  \\&~~+\sup_{k_1}\sum_{l}\lambda_{l,k_1}\Big]
%   \\&\le\frac{\beta_n K_4}{\lambda_nn}
\end{split}\end{equation}

Therefore we have $R_{n,3}^*=O(\frac{D_n^{10}\beta_n^3}{\min(\lambda_n^6,1)n})$.

This implies that both $(H_0)$ and $(H_1^*)$ hold and <Ref> guarantees that
$$\Big\|d_{\mcF}\Big(n\hat{T}_n( Z^M),g_n( Y^M)\mid X^n\Big)\Big\|_{L_1}\rightarrow 0.$$

§ PROOF OF <REF>

We divide the proof of the two statements of <Ref> in two. Firstly we prove that the bootstrap method is consistent if $\beta_n=o(\sqrt{n-m_n})$.

Let $(Y^n_i)$ be an independent copy of $(X^n_i)$.
We shorthand
\frac{e^{-\beta_n\mcR^k_n(x_{1:n})}}{\sum_{k'\le p_n}e^{-\beta_n\mcR^{k'}_n(x_{1:n})}},\qquad \hat\theta^p_n:=\hat\theta^p_n(X_{1:m_n}^n);$$

$$\hat\Theta_n:=\hat\Theta_n(X_{1:n}^n),\qquad \hat\Theta^Y_n:=\hat\Theta_n(X_{1:m_n}^nY^{n}_{m_n+1:n}). $$

The first step of the proof is to realize that as conditionally on $\hat\Theta_n$ and  $\hat\Theta_n^Y$  the observations $\bp{\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta^Y_n(X^n_{i+m_n})}-\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta_n(X^n_{i+m_n})}}_{i\ge 0}$ are independent and identically distributed we have

\begin{equation}
\begin{split}&\label{fin_1}
\bn{ \sqrt{{\rm var}\Big[\frac{1}{\sqrt{n-m_n}}\sum_{i\le n- m_n}\mathcal{L}_n\bp{X_{i+m_n}^n,~\hat\Theta_n^Y(X^n_{i+m_n})}-\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta_n(X^n_{i+m_n})} \Big|\hat\Theta_n^Y,\hat\Theta_n \Big]}}_{L_1}
\\&=\bn{  \sqrt{{\rm var}\Big[\mathcal{L}_n\bp{X_{m_n+1}^n,~\hat\Theta_n^Y(X^n_{m_n+1})}-\mathcal{L}_n\Big(X^n_{m_n+1},\hat\Theta_n(X^n_{m_n+1})\Big)\Big|\hat\Theta_n,\hat\Theta_n^Y\Big]}}_{L_1}
% \\&\le 2\sum_{l\le d'_n}\Big\|\sup_{\theta \in \Omega(\{\hat\theta^p_n,~p\le p_n\})}\partial_{2,l}\mathcal{L}_n(X^n_{m_n+1},\theta)\Big\|_{L_{\infty}} \times\Big\|\hat\Theta_{n,l} -\hat\Theta_{n,l}(X_{1:m_n}^nY^{n}_{m_n+1:n}\Big)\Big\|_{L_2}
% \\&\le L_nd'_n\sup_{l\le d'_n}\Big\|\hat\Theta_{n,l} -\hat\Theta_{n,l}(X_{1:m_n}^nY^{n}_{m_n+1:n}\Big)\Big\|_{L_2}
\end{split}\end{equation}

Moreover by exploiting Taylor expansions we know that
\begin{equation}
\begin{split}&
\bn{  \sqrt{{\rm var}\Big[\mathcal{L}_n\bp{X_{m_n+1}^n,~\hat\Theta_n^Y(X^n_{m_n+1})}-\mathcal{L}_n\Big(X^n_{m_n+1},\hat\Theta_n(X^n_{m_n+1})\Big)\Big|\hat\Theta_n,\hat\Theta_n^Y\Big]}}_{L_1}
\\&\le \sum_{l\le d'_n}\Big\|\sup_{\theta \in \Omega(\{\hat\theta^p_n,~p\le p_n\})}\ba{\partial_{2,l}\mathcal{L}_n(X^n_{m_n+1},\theta(X^n_{m_n+1}))}\Big\|_{L_{\infty}} \times\Big\|\hat\Theta_{n,l}(X^n_{m_n+1}) -\hat\Theta_{n,l}^Y(X^n_{m_n+1})\Big\|_{L_2}
\\&\le L_nd'_n\sup_{l\le d'_n}\Big\|\hat\Theta_{n,l}(X^n_{m_n+1}) -\hat\Theta_{n,l}^Y(X^n_{m_n+1})\Big\|_{L_2}
\end{split}\end{equation}
We define $(X_{i}^{n,j})$ as the following interpolating process: $X_{i}^{n,j}:=\begin{cases}X_i^n~{\rm if}~i\le j\\Y_j^{n}~{\rm otherwise}\end{cases}$ and shorthand $\hat\Theta_{n,l}^j:=\hat\Theta_{n,l}(X_{1:n}^{n,j})$. By the Effron-Stein inequality we have:

\begin{equation}
\begin{split}\label{fin_2}
\bn{\hat\Theta_{n,l}(X^n_{m_n+1}) -\hat\Theta_{n,l}^Y(X^n_{m_n+1})}^2_{L_2}
%    \\&=\bn{\sum_{j\ge m_n+1}\hat\Theta_{n,l} -\hat\Theta_{n,l}^Y}^2_{L_2}
&\overset{}{\le} \sum_{j\ge m_n+1}\bn{\hat\Theta_{n,l}^j(X^n_{m_n+1})-\hat\Theta_{n,l}^{j-1}(X^n_{m_n+1})}^2_{L_2}        \\&\le (n-m_n)\bn{\hat\Theta_{n,l}^{n}(X^n_{m_n+1})-\hat\Theta_{n,l}^{n-1}(X^n_{m_n+1})}_{L_2}^2
\end{split}\end{equation}

Moreover using the triangle inequality  and the inequality $(a+b)^2\le 2(a^2+b^2)$ we can upper-bound the right-hand side as

\begin{equation}
\begin{split}&\label{fin_3}
\bn{\hat\Theta_{n,l}^n(X^n_{m_n+1})-\hat\Theta_{n,l}^{n-1}(X^n_{m_n+1})}^2_{L_2}
\\&= \bn{\sum_{p\le p_n}\hat\theta^p_{n,l}\bp{\omega_n^p(X^n_{m_n+1:n})-\omega_n^p(X^n_{m_n+1:n-1}Y^n_n)}}^2_{L_2} %  \\&\le \bn{\hat\Theta_{n,l}-\hat\Theta_{n,l}(X_{1:n}^{n,n})}_{L_2}^2
\\\le&  2\bn{\sum_{p\le p_n}\hat\theta^p_{n,l}\omega_n^p(X^n_{m_n+1:n})\Big[e^{\frac{\beta_n}{n-m_n}[\mathcal{L}_n(X_n^n,\hat\theta_n^p)-\mathcal{L}_n(Y_n^n,\hat\theta_n^p)]}-1\Big]}_{L_2}^2
\\&~+2\bn{\sum_{p\le p_n}\hat\theta^p_{n,l}\omega_n^p(X^n_{m_n+1:n})\sum_{p'\le p_n}\omega_n^{p'}(X^n_{m_n+1:n})\Big[e^{\frac{\beta_n}{n-m_n}[\mathcal{L}_n(X_n^n,\hat\theta_n^{p'})-\mathcal{L}_n(Y_n^n,\hat\theta_n^{p'})]}-1\Big]}_{L_2}^2
\\&\le4\frac{\beta_n^2}{(n-m_n)} \|\sup_{p\le p_n}|\hat{\theta}_{n,l}^p|\|^2_{L_{\infty}}\bn{\sup_{p\le p_n}\ba{\mathcal{L}_n(X_1^n,\hat\theta_n^p)}}^2_{L_{\infty}}e^{\frac{2\beta_n}{n-m_n}\bn{\sup_{p\le p_n}\mathcal{L}_n(X_1^n,\hat\theta_n^p)}_{L_{\infty}}}% 4(n-m_n)^2\bn{\sum_{p\le p_n}\hat\theta^p_n\omega_n^p(X^n_{m_n+1:n})e^{\frac{\beta_n}{n-m_n}[\mathcal{L}_n(X_1^n,\hat\theta_n^p)-\mathcal{L}_n(Y_1^n,\hat\theta_n^p)]}
\\&\le 4\frac{\beta_n^2}{(n-m_n)} T^2_{n}L^2_n e^{\frac{2\beta_n}{n-m_n}L_n}
\end{split}\end{equation}

By combining <ref>
and <ref>  we therefore obtain that:

\begin{equation}
\begin{split}&
\bn{ \sqrt{{\rm var}\Big[\frac{1}{\sqrt{n-m_n}}\sum_{i\le n- m_n}\mathcal{L}_n\bp{X_{i+m_n}^n,~\hat\Theta_n^Y(X^n_{i+m_n})}-\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta_n(X^n_{i+m_n})} \Big|\hat\Theta_n^Y,\hat\Theta_n \Big]}}_{L_1}
\\&\longrightarrow 0.
\end{split}\end{equation}

Hence by using the fact that $d_{\mcF}$ satisfies the triangle inequality (see <Ref>) we observe that it is enough to prove that the distribution of
$\frac{1}{\sqrt{n}}\sum_{i\le n}\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta^Y_n(X^n_{i+m_n})}$ can be correctly approximated by the bootstrap method. We note that the process $\bp{\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta^Y_n(X^n_{i+m_n})}}$ is exchangeable and that if we define the function: $g_n(x_1,\dots,x_n)=\frac{1}{\sqrt{n}}\sum_{i\le n}x_i$ then:
$$g_n\bp{\bp{\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta^Y_n(X^n_{i+m_n})}}_{1:n}}:=\frac{1}{\sqrt{n}}\sum_{i\le n}\mathcal{L}_n\bp{X^n_{i+m_n},\hat\Theta^Y_n(X^n_{i+m_n})}.$$
We remark that $(g_n)$ satisfy all the conditions of <Ref> which implies that the desired result holds.

§ PROOF OF  <REF>
For simplicity we write

\begin{equation}\begin{split}&T_{l,n}:=\bn{\sup_{p\le p_n}\ba{\hat\theta^p_{n,l}(X_{1:m_n}^n)(Y_1^{n,1})}}_{L_{\infty}}\lor 1,\\&L^{*}_{l_{1:i},n}:=\big\|\max_{\substack{p\le p_n}}\big|\partial^i_{2,l_{1:i}}\mathcal{L}_n(X_n^n,\hat\theta^p_{n}(X^n_{1:m_n})(X_n^n))\big|\big\|_{L_{\infty}}\lor 1.\end{split}\end{equation}

For ease of notations, we shorthand:

$$ \hat\Theta_n^{Z^{n,2}}:=\hat\Theta_n(X^n_{1:m_n}Z^{n,2}_{1:n-m_n}),\qquad \hat\theta^p_n:=\hat\theta^p_n(X_{1:m_n}^n).$$
We first notice using <ref> and <ref> that:

\begin{equation}
\begin{split}\label{eq:sm_last} &\Big\|d_{\mcF}\Big(\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Z_{1:n-m_n}^{n,1})-\E\bp{\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Z_{1:n-m_n}^{n,1})\mid\hat\Theta_n^{Z^{n,2}},X^n},\\&\qquad \quad\mcR^{{\rm s}}_{\hat\Theta_n^{'}}(Y_{1:n-m_n}^{n,1})-\E\bp{\mcR^{{\rm s}}_{\hat\Theta_n^{'}}(Y_{1:n-m_n}^{n,1})\mid\hat\Theta_n^{'}}\mid X^n\Big)\Big\|_{L_1}
\\&\le \Big\|d_{\mcF}\Big(\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Z_{1:n-m_n}^{n,1})-\mathbb{E}\big(\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Z_{1:n-m_n}^{n,1})\mid\hat\Theta_n^{Z^{n,2}},X^n\big) \\&~\qquad\qquad ,\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Y_{1:n-m_n}^{n,1})-\mathbb{E}\big(\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Y_{1:n-m_n}^{n,1})\mid\hat\Theta_n^{Z^{n,2}}\big)\Big)\Big\|_{L_1}
\\&+\Big\|d_{\mcF}\Big(\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Y_{1:n-m_n}^{n,1})-\E\bp{\mcR^{{\rm s}}_{\hat\Theta_n^{Z^{n,2}}}(Y_{1:n-m_n}^{n,1})\mid\hat\Theta_n^{Z^{n,2}}},\\&\qquad\qquad  \mcR^{{\rm s}}_{\hat\Theta_n^{'}}(Y_{1:n-m_n}^{n,1})-\E\bp{\mcR^{{\rm s}}_{\hat\Theta_n^{'}}(Y_{1:n-m_n}^{n,1})\mid\hat\Theta_n^{'}}\mid X^n\Big)\Big\|_{L_1}
\end{split}
\end{equation}

We upper-bound each term seperately. In this goal, we define as $(X'_i)$ the process defined as $X'_i=\mathcal{L}_n(X_{i+m_n}^n,\hat\Theta^{Z^{n,2}}_n).$ We remark that the sequence $(X'_i)$ is an exchangeable sequence and we define:
$$Z'_i:=\mathcal{L}_n(Z^{n,1}_i,\hat\Theta_n^{Z,n2}),\quad Y'_i:=\mathcal{L}_n(Y^{n,1}_i,\hat\Theta_n^{Z,n2}). $$
We note that  $(Z'_i)$ and $(Y'_i)$ respectively form a bootstrap sample of $(X'_i)_{i\le n-m_n}$ and a copy of  $(X'_i)$ that is conditionally on $\hat \Theta_n^{Z^{n,2}}$ independent.
We define $F_{1,n}$ to be the following functions:

$$F_{1,n}(x_{1:n-m_n}):=\frac{1}{\sqrt{n-m_n}}\sum_{i\le n-m_n}x_i.$$

Moreover we denote $(X^{*}_i)$ the following random vectors $X^{*}_i=\Big(\mathcal{L}_n(X_{i+m_n}^{n},\hat\theta^p_n\Big)_{p\le p_n}.$ We also define:
$$Z^{*}_i=\Big(\mathcal{L}_n(Z_{i+m_n}^{n,2},\hat\theta^p_n\Big)_{p\le p_n},\quad Z^{*}_i=\Big(\mathcal{L}_n(Y_{i+m_n}^{n,2},\hat\theta^p_n\Big)_{p\le p_n}.$$
We remark that $(Z^{*}_i)$ and $(Y^{*}_i)$ are respectively bootstrap samples and (conditionally) independent copy of $(X^{*}_i)$.

We shorthand
$$ R_n^p:=\E(\mathcal{L}_n(X^n_p,\hat\theta^p_n)|\hat\theta^p_n);\qquad\bar{X'_p}:=\frac{1}{n-m_n}\sum_{i\le n-m_n}X'_{i,p}.$$

We define the following weight functions: $\omega_n^p:\times_{i=1}^{n-m_n}\RR\rightarrow \RR$ as:
\frac{e^{ \frac{-\beta_n}{n-m_n}\sum_{i\le n-m_n}x_{i,p}}}{\sum_{p'\le p_n}e^{\frac{-\beta_n}{n-m_n}\sum_{i\le n-m_n}x_{i,p'}}}.$$
We define
$F_{2,n}:\times_{l=1}^n\RR^{p_n}\rightarrow\RR$ as the following random function:
\begin{equation}\begin{split}F_{2,n}(x_{1:n-m_n}):=\frac{1}{\sqrt{n-m_n}}\sum_{i\le n-m_n}&\mathcal{L}_n\bp{Y^{n,1}_i,\sum_{p\le p_n}\hat\theta^p_n(Y_{i}^{n,1})\omega_n^p(x_{1:n-m_n})}\\&-\E\bp{\mathcal{L}_n\bp{Y^{n,1}_i,\sum_{p\le p_n}\hat\theta^p_n(Y_{i}^{n,1})\omega_n^p(x_{1:n-m_n})}\mid X^n}.
\end{split}\end{equation}
We remark that, conditionally on $X^n$ and $Y^{n,1}$, the functions $(F_{2,n})$ are independent from $(Z^*_i)$ and $(Y^*_i)$.
Using <ref> we note that if
the following hold then the desired result also holds:

\begin{equation}
\begin{split}
\bn{d_{\mcF}\bp{F_{2,n}(Z^{*}_{1:n-m_n}),F_{2,n}(\tilde Y^{*}_{1:n-m_n}) |X^n}}_{L_1}\rightarrow 0;
\end{split}\end{equation}

\begin{equation}
\begin{split}
\bn{d_{\mcF}\bp{F_{1,n}(Z'_{1:n-m_n}),F_{1,n}(\tilde Y'_{1:n-m_n}) \mid X^n}}_{L_1}\rightarrow 0.
\end{split}\end{equation}

To prove that those hold we use <Ref> extended to exchangeable sequences and random functions. We notice that the functions $(F_{2,n})$ and $(F_{1,n})$ are three times differentiable (random) functions. Therefore for both functions hypothesis $(H_0)$ holds and to get the desired results we only need to check that hypothesis $(H_1^*)$ also holds. For simplicity for all $i\le 3$ we write
\begin{equation}\begin{split}\hat{R}_{j}^{n,l_{1:i}}(x_{1:n}):=\frac{1}{\sqrt{n-m_n}}\Big(& \partial_{2,l_{1:i}}^i\mathcal{L}_n\Big({Y^{n,1}_j,\sum_{p\le p_n}\hat\theta^p_n(Y_j^{n,1})\omega_n^p(x_{1:n-m_n})}\Big)\\&-\E\Big({\partial_{2,l_{1:i}}^i\mathcal{L}_n\big({Y^{n,1}_j,\sum_{p\le p_n}\hat\theta^p_n(Y_j^{n,1})\omega_n^p(x_{1:n-m_n})}\big)\mid X^n}\Big)\Big).\end{split}\end{equation}
We define the following random functions: $f_{l_{1:3},x}(z):=\partial_{2,l_{1:3}}^3\mathcal{L}_n\bp{z,\sum_{p\le p_n}\hat{\theta}_n^p(z)\omega_p( xZ^{*}_{2:n-m_n}))};$
and remark that the set of functions $\{f_{l_{1:3},x},~l_{1:3}\le d'_n, x\in [0, Z^{*}_1]\bigcup [0,  Y^{*}_1]\}$  has an epsilon covering number proportional to $\big(\sum_{l\le n}L^*_{l,n}T_{l,n}\beta_n/((n-m_n)\epsilon)\big)^{p_n}p_n$. This implies that there is a constant $K$ such that  we have:

\begin{equation}\begin{split}\label{train}\max_{i\le n}\bn{\sup_{x\in [\bar{X}^{n,*}, Z^{*}_1]\bigcup [\bar{X}^{n,*}, \tilde Y^{*}_1]}\ba{\sum_{j\le m_n}\hat{R}_{j}^{n,l_{1:3}}(Z^{*,i,x}_{1:n})}}_{L_{72}}\le {K\log(p_n)}L^*_{n,l_{1:3}}.\end{split}\end{equation}

Moreover we note that for all $k\le 2$ there is a constant $C$ such that

\begin{equation}\begin{split}\label{train_233}\max_{i\le n}\bn{\ba{\sum_{j\le m_n}\hat{R}_{j}^{n,l_{1:k}}(Z^{*,i,\overline{Z^*}}_{1:n})}}_{L_{72}}\le {C}L^*_{n,l_{1:k}}.\end{split}\end{equation}

For ease of notations we write: $\overline{\theta^p_{n,l}}(x_{1:n-m_n},y):=\hat{\theta}_{n,l}^p(y)-
\sum_{p'\le p_n}\hat{\theta}_{n,l}^{p'}(y)\omega_{n}^{p'}(x_{1:n-m_n}).$
We remark that  for all $p\le p_n$ we have

\begin{equation}
\begin{split}&
\partial_{i,p}F_{2,n}(x_{1:n-m_n})\\=&-\frac{\beta_n}{n-m_n}
\omega^p_n(x_{1:n-m_n})\sum_{l\le d'_n}\sum_{j\le m_n}\hat{R}_{l}^{n,j}(x_{1:n})\Big[\hat{\theta}_{n,l}^p(Y_j^{n,1})-
\sum_{p'\le p_n}\sum_{j\le m_n}\hat{\theta}_{n,l}^{p'}(Y_j^{n,1})\omega_{n}^{p'}(x_{1:n-m_n})\Big]
\\=&-\frac{\beta_n}{n-m_n}
\omega^p_n(x_{1:n-m_n})\sum_{l\le d'_n}\sum_{j\le m_n}\hat{R}_{l}^{n,j}(x_{1:n})\overline{\theta^p_{n,l}}(x_{1:n-m_n},Y_j^{n,1})
\end{split}\end{equation}

Therefore as $\sum_{p\le p_n}\omega_n^p(x_{1:n})=1$, using <ref> we obtain that there is a constant $C$ that does not depend on $n$ such that
\begin{equation}
\begin{split}&
\max_{i\le n} \bn{ \bn{\partial_{i}F_{2,n}(Z^{*,i,\overline{X^*}}_{1:n-m_n})}_{v,p_n}}_{L_{12}}
\le \frac{C\beta_n}{n-m_n}\sum_{l\le d'_n}L^*_{l,n}
\Big\|\max_{p\le p_n}\big|\hat{\theta}_{n,l}^p(Y_1^{n,1})\big|\Big\|_{L_{12}}.
%\\&\le \frac{2\beta_n}{n-m_n}\sum_{l\le d'_n}L^*_{l,n}T_{l,n}
\end{split}\end{equation}

Moreover for all $p\le p_n$ we have $\partial_{i}F_{1,n}(x_{1:n-m_n})=\frac{1}{\sqrt{n-m_n}}.$
Therefore we obtain that
\begin{equation}\begin{split}&
\max_i \Big\| \partial_{i}F_{1,n}(Z^{',i,\overline{X'}}_{1:n-m_n})\Big\|_{L_{12}}
\le \frac{1}{\sqrt{n-m_n}}.
\end{split}\end{equation}
In addition, we remark that we have $\partial_{i}^2F_{1,n}(x_{1:n-m_n})=0$. Therefore the condition $(H_1^*)$ holds for $F_{1,n}$. For ease of notations we write $\omega^{p_1,p_2}_n(x_{1:n-m_n}):=\mathbb{I}(p_1=p_2)-\omega_{n}^{p_2}(x_{1:n-m_n}).$ We have

\begin{equation}
\begin{split}&
\partial_{i,p_1,p_2}^2F_{2,n}(x_{1:n-m_n})
\\&=\frac{\beta_n^2\omega_n^{p_1}(x_{1:n-m_n})}{(n-m_n)^2}
\omega^{p_1,p_2}_n(x_{1:n-m_n})\sum_{l\le d'_n}\sum_{j\le m_n}\hat{R}_{l}^{n,j}(x_{1:n})\overline{\theta^{p_1}_{n,l}}(x_{1:n},Y_j^{n,1})
\\&+\frac{\beta_n^2\omega_n^{p_1}(x_{1:n-m_n})\omega_n^{p_2}(x_{1:n-m_n})}{(n-m_n)^2}
\sum_{l_1,l_2\le d'_n}\sum_{j\le m_n}\hat{R}_{l_1,l_2}^{n,j}(x_{1:n})\overline{\theta^{p_1}_{n,l_1}}(x_{1:n},Y_j^{n,1})\overline{\theta^{p_2}_{n,l_2}}(x_{1:n},Y_j^{n,1})
\\&-\frac{\beta_n^2}{(n-m_n)^2}\omega_{n}^{p_1}(x_{1:n-m_n})\omega_{n}^{p_2}(x_{1:n-m_n})\sum_{l\le d'_n}\sum_{j\le m_n}\hat{R}^{n,j}_l(x_{1:n})\overline{\theta^{p_2}_{n,l}}(x_{1:n},Y_j^{n,1}).
\end{split}\end{equation}

This combined with <ref> implies that there is a constant $C_2$ such that:

\begin{equation}
\begin{split}&
\max_i \bn{ \bn{\partial_{i}^2F_{2,n}(Z^{*,i,\overline{X^*}}_{1:n-m_n})}_{m,p_n}}_{L_{12}}
%\\& \le \frac{C_2\beta_n^2}{(n-m_n)^2}
\le \frac{C_2\beta_n^2}{(n-m_n)^2}
\Big[\sum_{l\le d'_n}L^*_{l,n}T_{l,n}+\sum_{l\le d'_n}L^*_{l_1,l_2,n}T_{l_1,n}T_{l_2,n}\Big]
\end{split}\end{equation}

Finally using the chain rule for all $p_{1},p_2,p_3\le p_n$ we have

\begin{equation}
\begin{split}&
\partial_{i,p_{1:3}}^3F_{2,n}(x_{1:n-m_n})
\\&=\frac{\beta_n^3\sum_{l\le d'_n}\sum_{j\le m_n}\hat{R}_{l}^{n,j}(x_{1:n})\overline{\theta^{p_1}_{n,l_1}}(x_{1:n},Y_j^{n,1})\omega_n^{p_1}(x_{1:n-m_n})}{(n-m_n)^3} \Big[\omega_n^{p_2}(x_{1:n-m_n})\omega^{p_2,p_3}_n(x_{1:n-m_n})\\&-\omega_{n}^{p'}(x_{1:n-m_n}) \omega^{p_1,p_3}_n(x_{1:n-m_n})\Big]
%\\&=\frac{-\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega^{p_1,p_2}_n(x_{1:n-m_n}) \omega^{p_1,p_3}_n(x_{1:n-m_n})}{(n-m_n)^3}
%\sum_{l\le d'_n}\hat{R}_{l}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})
%\\&+\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega_n^{p_2}(x_{1:n-m_n})}{(n-m_n)^3} \omega^{p_2,p_3}_n(x_{1:n-m_n})
%\sum_{l\le d'_n}\hat{R}_{l}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})
\sum_{l_1,l_2\le d'_n}\sum_{j\le m_n}\hat{R}_{l_1,l_2}^{n,j}(x_{1:n})\overline{\theta^{p_1}_{n,l_1}}(x_{1:n},Y_j^{n,1})
\\&\quad \times\overline{\theta^{p_3}_{n,l_3}}(x_{1:n},Y_j^{n,1})\Big[\omega_n^{p_3}(x_{1:n-m_n})\omega_n^{p_1,p_2}(x_{1:n-m_n})+\omega_n^{p_3}(x_{1:n-m_n})\omega_n^{p_1,p_2}(x_{1:n-m_n})\Big]\\&-\frac{\beta_n^3\prod_{i=1}^3\omega_n^{p_i}(x_{1:n-m_n})}{(n-m_n)^3}
\Big[\sum_{l_{1:3}\le d'_n}\sum_{j\le p_n}\hat{R}_{l_1,l_2,l_3}^{n,j}(x_{1:n})\prod_{k=1}^3\overline{\theta^{p_k}_{n,l_k}}(x_{1:n},Y_j^{n,1})-\sum_{l_{1:2}\le d'_n}\hat{R}_{l_1,l_2}^{n,j}(x_{1:n},Y_j^{n,1})\\&\quad \times\overline{\theta^{p_2}_{n,l_1}}(x_{1:n},Y_j^{n,1})\overline{\theta^{p_3}_{n,l_2}}(x_{1:n},Y_j^{n,1})\Big]
%\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})\overline{\theta^{p_3}_{n,l_3}}(x_{1:n})
%\\&-\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega^{p_2}_n(x_{1:n}) \big[\omega^{p_1,p_3}_n(x_{1:n-m_n})+\omega^{p_2,p_3}_n(x_{1:n-m_n})\big]}{(n-m_n)^3}
%\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})\overline{\theta^{p_2}_{n,l_2}}(x_{1:n})
%\\&+\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega^{p_2}_n(x_{1:n}) \omega^{p_2,p_3}_n(x_{1:n-m_n})}{(n-m_n)^3}
%\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})\overline{\theta^{p_2}_{n,l_2}}(x_{1:n})
\\&\quad \times \sum_{l\le d'_n}\sum_{j\le m_n}\hat{R}_{l}^{n,j}(x_{1:n})\overline{\theta^{p_2}_{n,l_1}}(x_{1:n},Y_j^{n,1})
%\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_2}_{n,l_1}}(x_{1:n})\overline{\theta^{p_3}_{n,l_2}}(x_{1:n})
%\\&\\&MMM-\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})}{(n-m_n)^3}\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_2}_{n,l_1}}(x_{1:n})\overline{\theta^{p_3}_{n,l_1}}(x_{1:n}) \omega^{p_1,p_2}_n(x_{1:n-m_n}) \omega^{p_1,p_3}_n(x_{1:n-m_n})
%\\&-\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega_n^{p_2}(x_{1:n-m_n})}{(n-m_n)^3}\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})c\Big[ \omega^{p_1,p_3}_n(x_{1:n-m_n})+ \omega^{p_2,p_3}_n(x_{1:n-m_n})\Big]\\&
%-\frac{\beta_n^3\prod_{i=1}^3\omega_n^{p_i}(x_{1:n-m_n})}{(n-m_n)^3}\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})\overline{\theta^{p_3}_{n,l_2}}(x_{1:n})
%\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_2}_{n,l_1}}(x_{1:n})\overline{\theta^{p_3}_{n,l_3}}(x_{1:n})
%\sum_{l_1,l_2,l_3\le d'_n}\hat{R}_{l_1,l_2,l_3}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})\overline{\theta^{p_2}_{n,l_2}}(x_{1:n})\overline{\theta^{p_3}_{n,l_3}}(x_{1:n})
%\sum_{l\le d'_n}\hat{R}_{l}^n\overline{\theta^{p_2}_{n,l}}(x_{1:n})\omega^{p_1,p_3}(x_{1:n-m_n})
%\sum_{l\le d'_n}\hat{R}_{l}^n\Big[\hat{\theta}_{n,l}^p-
%\sum_{p'\le p_n}\hat{\theta}_{n,l}^{p'}\omega_{n}^{p'}(x_{1:n-m_n})\Big]
%\sum_{l\le d'_n}\hat{R}_{l}^n\overline{\theta^{p_3}_{n,l}}(x_{1:n})
%\sum_{l_1,l_2\le d'_n}\hat{R}_{l_1,l_2}^n\overline{\theta^{p_1}_{n,l_1}}(x_{1:n})\overline{\theta^{p_2}_{n,l_2}}(x_{1:n})\omega^{p_1,p_3}(x_{1:n-m_n})
%\\&+\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega_n^{p_2}(x_{1:n-m_n})}{(n-m_n)^3}\Big[\hat{R}_{p_3}^n(Z^{'}_{1:n-m_n})-\sum_{p'\le p_n}\hat{R}_{p'}^n(Z^{'}_{1:n-m_n})\omega_n^{p'}(x_{1:n-m_n})\Big]
%\\&\quad \times \Big[\mathbb{I}(p_1=p_3)+\mathbb{I}(p_1=p_2)-2\omega_{n}^{p_3}(x_{1:n-m_n})\Big]
%\\&+\frac{\beta_n^3\omega_n^{p_1}(x_{1:n-m_n})\omega_n^{p_3}(x_{1:n-m_n})}{(n-m_n)^3}\Big[\hat{R}_{p_3}^n(Z^{'}_{1:n-m_n})-\sum_{p'\le p_n}\hat{R}_{p'}^n(Z^{'}_{1:n-m_n})\omega_n^{p'}(x_{1:n-m_n})\Big]
%\\&\quad \times \Big[\mathbb{I}(p_1=p_3)-\omega_{n}^{p_3}(x_{1:n-m_n})\Big]
%\\&+\frac{\beta_n^3}{(n-m_n)^3}\omega_{n}^{p_1}(x_{1:n-m_n})\omega_{n}^{p_2}(x_{1:n-m_n})\omega_{n}^{p_3}(x_{1:n-m_n})\\&\qquad \times\Big[
%R_{p_3}^n(Z^{'}_{1:n-m_n})-\sum_{p'\le p_n}\hat{R}_{p'}^n(Z^{'}_{1:n-m_n})\omega_n^{p'}(x_{1:n})\Big].
\end{split}\end{equation}

Therefore using <ref> we establish that that there is a constant $C_3<\infty$ such that

\begin{equation}
\begin{split}&
\max_{i\le n} \bn{\max_{x\in [\bar{X}^n, Z^{*}_1]\bigcup [\bar{X}^n, \tilde Y^{*}_1]} \bn{\partial_{i}^3F_{2,n}(Z^{*,i,x}_{1:n-m_n})}_{t,p_n}}_{L_{12}}
%\\& \le \frac{C_2\beta_n^2}{(n-m_n)^2}
\\&\le \frac{C_3\beta_n^3\log(p_n)}{(n-m_n)^3}
\sum_{l_1\le d'_n}T_{l_1,n}\Big[L^*_{l_1,n}+\sum_{l_{2}\le d'_n}L^*_{l_{1:2},n}T_{l_2,n}+\sum_{l_{2:3}\le d'_n}L^*_{l_{1:3},n}T_{l_2,n}T_{l_3,n}\Big]
\end{split}\end{equation}

This implies that hypothesis $(H_1^*)$ holds for $F_{2,n}$; which means that

\begin{equation}
\begin{split}
\bn{d_{\mcF}\bp{F_{2,n}(Z^{*}_{1:n-m_n}),F_{2,n}(\tilde Y^{*}_{1:n-m_n}) |X^n}}_{L_1}\rightarrow 0;
\end{split}\end{equation}
We note that $F_{2,n}(\tilde Y^{*}_{1:n-m_n}) :=\mathcal{R}^s_{\hat\Theta'_n}(Y_{1:n-m_n}^{n,1}).$

Moreover, we remark that we have proved that the condition $(H_1^*)$ also holds for $F_{1,n}$. This implies that we have

\begin{equation}
\begin{split}
\bn{d_{\mcF}\bp{F_{1,n}(Z'_{1:n-m_n}),F_{1,n}(\tilde Y'_{1:n-m_n}) \mid X^n}}_{L_1}\rightarrow 0.
\end{split}\end{equation}

As $F_{1,n}$ is linear this gives us the desired result.

§ PROOFS FROM <REF>

§.§ Proof of <Ref>

The first step is to bound $\E\Big[\max_{k\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}\tilde X^n_{i,k}\Big]$.

Define $(f_n:\RR^d\rightarrow\RR)$ to be the following sequence of functions:

$$f_n(x_{1:n}):= \frac{\log\Big(\sum_{l\le p_n}e^{\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x_{i,l}}\Big)}{\log(p_n)}.$$
We observe that
$\|f_n(\tilde X^n)-\max_{k\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}\tilde X^n_{i,k}\|_{L_1}\le 1$; and that the functions $(f_n)$ are infinitely differentiable.
We denote,  for all $z\in \RR$, $(\tilde X^{i,n}_{l})$ and  $(\tilde X^{i,z}_{l})$ the processes respecting
$$\tilde X^{i,n}_l:=\begin{cases}\tilde X^n_l~{\rm if}~i\le l\\ 0~{\rm otherwise}.\end{cases}\qquad \tilde X^{i,z}_l:=\begin{cases}\tilde X^{i,n}_l~{\rm if}~i\ne l\\ z~{\rm otherwise}\end{cases}.$$

We first suppose that $\|\max_{k\le p_n}\tilde X^n_{1,k}\|_{L_p}\le 1$.
Using the Taylor expansion we have:

\begin{equation}
\begin{split}\label{idaho}
\Big| \E \Big[f_n(\tilde X^n)\Big]\Big|&\le \sum_{i\le n}  \Big| \E \Big[f_n(\tilde X^{i,n}_{1:n})-f_n(\tilde X^{i-1,n}_{1:n})\Big]\Big|
\\ &\le \sum_{i\le n}\Big|\E\Big[~(\tilde X^n_i)^{\top}~\partial_if_n(\tilde X^{i,0}_{1:n}) \Big]\Big|
+ \sum_{i\le n}\Big|\E\Big[(\tilde X^n_i)^{\top}\partial^2_if_n(\tilde X^{i,0}_{1:n})\tilde X^n_i\Big]\Big|
\\&+ \sum_{i\le n}\E\Big[\sup_{z\in [0,\tilde X^n_i]}\Big<\Big|\partial^3_if_n(\tilde X^{i,z}_{1:n})\Big|,\Big|\tilde X^{\otimes 3}_i\Big|\Big>\Big]
\\&\overset{(a)}{\le} n \Big|\sum_{d_1,d_2\le p_n}\E\Big[\partial^2_{i,d_1,d_2}F^{n}_{\beta}(\tilde X^{i,0}_{1:n})\Big]\E(\tilde X^n_{1,d_1}\tilde X^n_{1,d_2})\Big|
\\&+ n\sum_{d_1,d_2,d_2\le n}\E\Big[\sup_{z\in [0,\tilde X^n_i]}\Big|\partial^3_{i,d_{1:3}}F^{n}_{\beta}(\tilde X^{i,z}_{1:n})\Big|~\Big|\tilde X^n_{1,d_1}\tilde X^n_{1,d_2}\tilde X^n_{1,d_3}\Big|\Big]
\end{split}
\end{equation}
where to get (a) we used the fact that as $\sigma(\tilde X^{i,0}_{1:n})=\sigma(\tilde X_0^n,\dots, \tilde X_{i-1}^n)$ we have $\E(\tilde X^n_i|\tilde X^{i,0}_{1:n})=0$. The next step is to upper-hand the right-hand side of <Ref>. For ease of notations we write:
$$\omega_k(x_{1:n}):=\frac{e^{\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x_{i,k}}}{\sum_{l\le p_n}e^{\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x_{i,l}}}.$$

For all $k_1,k_2\le p_n$ we obtain by the chain rule that

\begin{equation}\begin{split}&
\Big| \partial_{i,k_1,k_2}^2f_n(x_{1:n})\Big|
\le \begin{cases}\frac{\log(p_n)}{n}\omega_{k_1}(x_{1:n})\omega_{k_2}(x_{1:n})\qquad{\rm if}~k_1\ne k_2\\\frac{\log(p_n)}{n}\Big[ \omega_{k_1}^2(x_{1:n})+\omega_{k_1}(x_{1:n})\Big]\qquad{\rm if}~k_1=k_2.\end{cases}
\end{split}
\end{equation}

As $\sum_{k\le p_n}\omega_k(x_{1:n})=1$, this implies that

\begin{equation}
\begin{split}&
n \Big|\sum_{d_1,d_2\le p_n}\E\Big[\partial^2_{i,d_1,d_2}F_{\beta}(\tilde X^{i,0})\Big]\E(\tilde X^n_{1,d_1}\tilde X^n_{1,d_2})\Big|
\le  {2 \log(p_n)}.\end{split}\end{equation}

We now bound the second term of the right hand side of <ref>. For all  integers $k_1,k_2,k_3\le p_n$, by the chain rule, we obtain that:

\begin{equation}\begin{split}\label{bj_1}&
\ba{ \partial_{i,k_{1:3}}^3F_{\beta}(x'_{1:n})}
\\&\le\quad \frac{\log(p_n)^{3/2}}{n^{3/2}} \omega_{k_1}(x_{1:n})\omega_{k_2}(x_{1:n})\omega_{k_3}(x_{1:n})\bp{1+\mathbb{I}(k_1=k_3)+\mathbb{I}(k_2=k_3)}
\\&+\frac{\log(p_n)^{3/2}}{n^{3/2}} \omega_{k_1}(x_{1:n})\omega_{k_2}(x_{1:n})\bp{\mathbb{I}(k_1=k_3)+\mathbb{I}(k_2=k_3)}\\&+\frac{\log(p_n)^{3/2}}{n^{3/2}} \omega_{k_1}(x_{1:n})\bp{\omega_{k_3}(x_{1:n})\mathbb{I}(k_1=k_2)
%\\&+\Big|\frac{\beta\log(p_n)}{nn!}\sum_{\pi\in \mathbb{S}_n}\frac{e^{2\beta\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x'_{i,\pi(1)}}}{\Big[\sum_{l\le p_n}e^{\beta\log(p_n)\frac{1}{\sqrt{n}}\sum_{i\le n} x'_{i,l}}\Big]^2}\Big|
% \\&\le \frac{\beta\log(p_n)}{n}
\end{split}
\end{equation}

Therefore we obtain that

\begin{equation}\begin{split}&
n\sum_{d_1,d_2,d_3\le n}\E\bp{\sup_{z\in [0,\tilde X^n_i]}\ba{\partial^3_{i,d_{1:3}}F^{\beta,n}(\tilde X^{i,z})}~\ba{\tilde X^n_{1,d_1}\tilde X^n_{1,d_2}\tilde X^n_{1,d_3}}}
\le \frac{7\beta^2\log(p_n)^2}{\sqrt{n}}.
\end{split}
\end{equation}

Hence using <ref> we establish that:

\begin{equation}\begin{split}
\ba{ \E\bp{\max_{k\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}\tilde X^n_{i,k}}}
%    &\le \inf_{\beta}\Big[\frac{1}{\beta}+  \Big[{\log(p_n)\beta} +\frac{6\beta^2\log(p_n)^2}{\sqrt{n}}\Big]\Big]
&\le 1+ \Big[{\log(p_n)} +\frac{7\log(p_n)^2}{6\sqrt{n}}\Big] . \end{split}
\end{equation}

By potentially renormalizing $\|\max_{k\le p_n}\ba{\tilde X_{i,k}^n}\|_{L_p}$ we obtain therefore than in general we have:
\begin{equation}\begin{split}
&\ba{ \E\bp{\max_{k\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}\tilde X^n_{i,k}}}
\\&\le \bn{\max_{k\le p_n}|X_{1,k}|}_{L_p}\bp{1+ {\log(p_n)} +\frac{6\log(p_n)^2}{\sqrt{n}}}. \end{split}
\end{equation}

Lastly, according to the  Rosenthal inequality for martingales [30], there are constants $(C_p)$ that do not depend on $d$ or $(\tilde X^n_i)$ such that

\begin{equation}\begin{split}&
\bn{\max_{k\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}\tilde X^n_{i,k}-\E\bp{\max_{k\le p_n}\frac{1}{\sqrt{n}}\sum_{i\le n}\tilde X^n_{i,k}}}_{L_p}
\le 2 C_p\bn{\sup_{d_1\le p_n}\ba{\tilde X^n_{i,d_1}}}_{L_p}.
\end{split}
\end{equation}

Therefore we get the desired results using the triangular inequality.

Finally, let $(\mathbb{F}_i)$ designates the filtration $\mathbb{F}_i:=\sigma\Big( X^n_1,\dots, X^n_i\Big)$ then the following is an array of martingale differences $\bp{\E(g_{k,n}( X^n)|\mathbb{F}_i)-\E(g_{k,n}( X^n)|\mathbb{F}_{i-1})}$. The last point of <Ref> follows directly from this observation.


explainers are also implemented as wrapper functions and currently provide
support for SHAP, LIME, LEMNA, and ELI5. The modules can be extended for added
support of custom parsers, models and explainers. For the sake of
reproducibility, the pipeline saves the model, predictions and explanations in
a file.

##  Appendix C LEMNA Implementation

We based our implementation of LEMNA on the code by Warnecke _et al._[23]. For
the explanation generation, we use the following settings: $N=500,K=6,S=10$.
The values of $N$ and $K$ are based on the original LEMNA paper. We do not
need fused Lasso since our features do not have a temporal structure.
Therefore we set $S$ to a high value, effectively turning off the fusing
effect. We expect LEMNA to perform better on tabular data when using feature
discretization. However, optimizing LEMNA is out of the scope of this work as
we are only using existing methods for model debugging.

##  Appendix D Explanations from the Tutorial

Figure 6 shows the global SHAP plot for the Gradient Boosting Machine (GBM)
learned from the experimental dataset. Figures 7 and 8 show the post-hoc
explanations for the false positive Netflow with and without the identified
spurious features, respectively. Figures 9 and 10 show the post-hoc
explanations for the false negative Netflow with and without the identified
spurious features, respectively.