\big ) ^2
+ \big ( {18 \delta_m ^{1/6}\gamma }\big )^6
\big \}  \to (1 + \sqrt{c})^2$ as $m\to \infty$.
Furthermore, if $m = o ( \sqrt{n} )$ we use $\gamma=1/(1 + \varepsilon /2 )$ for $\varepsilon > $
to obtain

\begin{eqnarray*}
\big ( 2mq \big )^{1/k} \to   e^{2 +   \varepsilon}
\end{eqnarray*}

which proves (<ref>) for any $z > z(\gamma ) =e^2$
in the case (a) of Theorem <ref>.

If $ m = o(\log n)$, we have
$ { m ^{1+\gamma} \log m \over  n }  \to 0 $ for any $\gamma >0$, and it follows

\begin{eqnarray*}
\big ( 2mq \big )^{1/k} \to   e^{2 / \gamma }
\end{eqnarray*}

for any $\gamma $. Therefore (<ref>) holds for any $ z > (1 + \sqrt{c})^2$, which completes the proof of Theorem <ref>(a) and (c).

For a proof of part (b), note that it
follows from the arguments in  baiyin1993,

\begin{align*}
\P&\Big(\lambda_{\min}\big(\widehat{\Sigma}_n^*\big)<K\Big)\\
&\leq \P\Big(\big\Arrowvert \widehat{\Sigma}_n^*-(1+c)I_q\big\Arrowvert_{S_{\infty}}>2\sqrt{c}+\big(1-\sqrt{c}\big)^2-K\Big)\\
&\leq \P\Big(\big\Arrowvert \widehat{\Sigma}_n^*-cI_q-\diag\big(\widehat{\Sigma}_n^*\big)\big\Arrowvert_{S_{\infty}}>2\sqrt{c}+\frac{1}{2}\big(\big(1-\sqrt{c}\big)^2-K\big)\Big)\\
& \ \ \ +\P\Big(\big\Arrowvert \diag\big(\widehat{\Sigma}_n^*\big)-I_q\big\Arrowvert_{S_{\infty}}>\frac{1}{2}\big(\big(1-\sqrt{c}\big)^2-K\big)\Big).
\end{align*}

Hence, it remains to show that for any $\varepsilon>0$ and any $l\in\N$,

\begin{align}
\P\Big(\big\Arrowvert \diag\big(\widehat{\Sigma}_n^*\big)-I_q\big\Arrowvert_{S_{\infty}}>\varepsilon\Big)=o(m^{-l})\label{eq: app2}
\intertext{and}
\P\Big(\big\Arrowvert \widehat{\Sigma}_n^*-\diag\big(\widehat{\Sigma}_n^*\big)-cI_q\big\Arrowvert_{S_{\infty}}>2\sqrt{c}+\varepsilon\Big)=o(m^{-l}). \label{eq: app1}
\end{align}

Proof of (<ref>). Since $q=\mathcal{O}(m)$, it is sufficient to  show

\begin{align}
\label{ang16}
\P \bigg(\bigg\arrowvert \frac{1}{m}\sum_{i=1}^m \Big(\arrowvert X_{i1}^*\arrowvert^{2}-1\Big)\bigg\arrowvert>\varepsilon\bigg)=o(m^{-l}).
\end{align}

For the  sequence $k=k_m = \gamma \log m$,

an application of Markov's inequality yields an upper bound on the left-hand side
(recall that $| X_{ij} | \leq \delta_m \sqrt{m})$

\begin{align*}
m^{-2k}&\varepsilon^{-2k}\E\bigg[ \sum_{i=1}^m \Big(\arrowvert X_i^*\arrowvert^{2}-\E \arrowvert X_1\arrowvert^{2}\Big)\bigg]^{2k}\\
&=m^{-2k}\varepsilon^{-2k}\E\bigg[ \sum_{i=1}^n W_i\Big(\arrowvert X_i\arrowvert^{2}-\E \arrowvert X_1\arrowvert^{2}\Big)\bigg]^{2k}\\
&= m^{-2k}\varepsilon^{-2k}
{l}\sum_{\substack{i_1\geq 0,\dots, i_n\geq 0\\ i_1+\dots +i_n=2k}}\binom{2k}{i_1\dots i_n}\E\bigg[\prod_{t=1}^n W_t^{i_t}\Big(\arrowvert X_i\arrowvert^{2}-\E \arrowvert X_1\arrowvert^{2}\Big)^{i_t}\bigg]\\
&\leq 2^{2k}m^{-2k}\varepsilon^{-2k}\sum_{l=1}^k\binom{n}{l}\sum_{\substack{i_1\geq 2,\dots, i_l\geq 2\\ i_1+\dots +i_l=2k}}\binom{2k}{i_1\dots i_l}\E\bigg[\prod_{t=1}^l W_t^{i_t}\bigg]\prod_{t=1}^l \E\arrowvert X_1\arrowvert^{2i_t}\\
&\leq K 2^{2k}m^{-2k}\varepsilon^{-2k}\sum_{l=1}^k n^l \sum_{\substack{i_1\geq 2,\dots, i_l\geq 2\\ i_1+\dots +i_l=2k}}\binom{2k}{i_1\dots i_l}\Big(\frac{m}{n}\Big)^l
\Big  ( \
1 +   c_\gamma  {m^{1+\gamma}   \over n }
\Big )  ^{k}
\prod_{t=1}^l \E\arrowvert X_1\arrowvert^{2i_t}\\
&=K 2^{2k}m^{-2k}\varepsilon^{-2k}
\Big  ( \
1 +   c_\gamma  {m^{1+\gamma}   \over n }
\Big )  ^{k}
\sum_{l=1}^k m^l \sum_{\substack{i_1\geq 2,\dots, i_l\geq 2\\ i_1+\dots +i_l=2k}}\binom{2k}{i_1\dots i_l}\prod_{t=1}^l \E\arrowvert X_1\arrowvert^{2i_t},
\end{align*}

where Lemma <ref> has been applied in the last inequality.
Arguing as in the proof of Lemma 2' in
baisil2004, page 602 (where $f=1$ and
their  $m$ corresponds to  $k$), we obtain the upper  bound

\begin{align}
\label{ang17}
\gamma \log m \Big (
{ 16  \gamma \delta_m^2  \log m
\over
\varepsilon \log ( 4 \delta_m^4 m /
\E | X_{11}|^4) )
\Big )^{2 \gamma \log m} &\leq
\gamma (\log m)\Big (
{ 32   \gamma \delta_m^2
\over
\varepsilon
\Big )^{2 \gamma \log m} \nonumber\\
%&=\gamma(\log m)\exp\bigg(2\gamma(\log m)\log \Big({ 32   \gamma \delta_m^2
&=\gamma(\log m) m^{2 \gamma \log \big({ 32   \gamma \delta_m^2
\over
\varepsilon
\end{align}

if $m$ is sufficiently large  such that
$\log ( 4 \delta_m^4 m / \E | X_{11}|^4  )  \geq {1 \over 2} \log m$ (note that by (<ref>) eventually,  $\delta_m \geq m^{-1/8}$  as $m \to \infty $).
Because of
$\delta_m \to 0 $ as $m \to \infty$,  it follows that for  any $a \in (0,1)$, there exists  an integer
$m_0=m_0(a)$ such that  the expression in (<ref>) is bounded by  $m^{2\gamma\log a} $ for all
$m  \geq m_0$, which proves  (<ref>) and completes the proof of (<ref>).

Proof of (<ref>). With the notation $\hat{T}_n^*=\widehat{\Sigma}_n^*-\diag\big(\widehat{\Sigma}_n^*\big)$, it is sufficient to prove the following result. There exists a positive constant $C>0$, such that for every $r\in\N$ and positive $\varepsilon$ and $l$,

\begin{align}
\P\Big(\big\Arrowvert \hat{T}_n^*-cI_q\big\Arrowvert_{S_{\infty}}^r> Cr^42^rc^{r/2}+\varepsilon\Big)=o(m^{-l}).
\end{align}

To this aim, we need to establish the bootstrap analogs of lemmata 1' – 8' in the appendix of baisil2004. Since all of them can be deduced by our manipulation technique and Lemma <ref> in a straightforward manner, we omit them at this point.

We begin part (b). By the discussion in Section <ref> we can assume that $R_n=0$, which

\begin{align*}
{x^\top  \widehat \Sigma_n^* x \over  \|x \|^2 }
\lambda_{\rm min} \Big ( {1 \over m} \sum_{i=1}^m X_i^*X_i^{*\top} \Big)
{x^\top  L_nL_n^\top  x \over  \|x \|^2 }
\geq
\lambda_{\rm min} \Big ( {1 \over m} \sum_{i=1}^m X_i^*X_i^{*\top} \Big)
\lambda_{\rm min} ( \Sigma_n).
\end{align*}

The assertion follows applying Theorem <ref> for $q' \times q'$-matrix ${1 \over m} \sum_{i=1}^m X_i^*X_i^{*\top}$.
Part (a) is an immediate consequence of the fact that the spectral norm is a matrix norm.

§ PROOF OF THE RESULTS IN SECTION <REF>

Throughout this section, we assume that Assumptions (A1) – (A3+) are satisfied.
All proofs have in common the truncation steps in (<ref>),  (<ref>) and (<ref>) discussed in the following  Section <ref> and <ref>, respectively.  The proofs of Theorem <ref> and Proposition <ref> require additionally  (<ref>) and

\begin{align}
\label{det206}
\E \hat{T}_n (f_m)- \E \check {T}_n (f_m)\rightarrow_{\P} 0
\end{align}

§.§ Reduction to $L_n$ and $f_m$

Recalling the notation from
Section <ref>  we will will first prove that  we can replace the matrix $\Pi_n A_n$ in the decomposition (<ref>)
by the matrix $L_n$, that is

\begin{align}
\label{ar100}
&  \hat{T}_n^*(f)-\hat {T}_{n,L_n}^*(f_m)\rightarrow_{\P}0  \\
&  \E^*\hat{T}_n^*(f_m)-\E^*\hat {T}_{n,L_n}^*(f_m)\rightarrow_{\P}0, \label{ar101}
\end{align}

where  $f_m:=f \mathds{1}_{ \{ |f| \leq m^\ell\} }$ and
$ \hat {T}_{n,L_n}^*(f)$ and
$ \hat {T}_{n,L_n}^*(f_m)$
denotes the linear spectral statistics corresponding to the matrix

\widehat \Sigma_{n,L_n}^*:=\frac{1}{m} L_n X^*X^{*\top} L_n^\top,

and   $ L_n X^* = (L_nX^*_1, \ldots ,  L_nX^*_m) \in \mathbb{R}^{q \times m} $. Define

\begin{align*}
\mathcal{D}_n =\Big \{
\lambda_{\min }  (\widehat \Sigma_{n}^* ) > K_l ,~
\lambda_{\min }  (\widehat \Sigma_{n,L_n}^* ) > K_l ,~
\| \widehat \Sigma_{n,L_n}^*   \|_{S_\infty } <  K_r ,~
\| \widehat \Sigma_{n}^*   \|_{S_\infty } <  K_r \Big \}
\end{align*}

where constants $K_l$ and $K_r$ come from Corollary  <ref>. By this result we have $\P ( \mathcal{D}_n^c) = o(m^{-\ell}) $  for any $\ell \in \N $ (note that due to the  Representative Subpopulation Condition  <ref>, it follows that
$ \big |   \| \widehat \Sigma_{n,L_n}^* )  \|_{S_\infty } -
\| \widehat \Sigma_{n}^*\|_{S_\infty }  )
\big |  = o_{\P } (1))$.
By the Lipschitz continuity of $f$, the $1$-Wielandt-Hoffman inequality,  and Assumption (A1), it follows that  (note that on the set $[K_l, K_r] $ we have $f=f_m$ if $m$ is sufficiently large)

\begin{align*}
\big\arrowvert
\hat {T}_{n,L_n}^* (f_m)
- \hat{T}_n^*(f)
\big\arrowvert
\leq {\max}_{ \lambda \in [K_l, K_r] }  | f'(\lambda ) |
\sum_{j=1}^q
\big\arrowvert {\hat \lambda}_{j,L_n}^*-\hat{\lambda}_j^*\arrowvert  + o_\mathbb{P}  (1)\\
\leq {\max}_{ \lambda \in [K_l, K_r] }  | f'(\lambda ) |
\big\Arrowvert \widehat \Sigma_{n,L_n}^{*}-\widehat \Sigma_n^*\big\Arrowvert_{S_1}
+ o_\mathbb{P}  (1) ~,
\end{align*}

$\hat \lambda_{j,L_n}^*$ is the $j$th eigenvalue of the matrix $ \widehat \Sigma_{n,L_n}^{*}$. By the discussion in Section <ref>, the right-hand side is of order  $ o_\mathbb{P}  (1)$ if
\big [ ||  R_n   ||_{S_{2}}^{2} \big ] = o (1)$, which proves (<ref>).
we obtain

\begin{align*}
\E^* \Big [ \big\arrowvert
\hat {T}_{n,L_n}^* (f_m)
- \hat{T}_n^*(f_m)\big\arrowvert
\mathds{1}_{\mathcal{D}_n} \Big ]
&  \leq
{\max}_{ \lambda \in [K_l, K_r] }  | f'(\lambda ) |
\E^* \big\Arrowvert \widehat \Sigma_{n,L_n}^{*}-\widehat \Sigma_n^*\big\Arrowvert_{S_1} = o_{\P} (1).
%& \leq {\sup}_{\lambda \in G}|f(\lambda ) | \cdot  \E^*
%\big\Arrowvert \widehat \Sigma_{n,L_n}^{*}-\widehat \Sigma_n^*\big\Arrowvert_{S_1}+\E^*\Big[\big \arrowvert \hat {T}_{n,L_n}^* (f)
%- \hat{T}_n^*(f)\big\arrowvert\mathds{1}_{A_n^c}\Big].
\end{align*}

On the other hand, since
$\P (\mathcal{D}_n^c) =o(  m^{-\ell}) $    for any  $\ell \in \N$,  it follows   for the remaining part

\begin{align*}
\E \E^* \big [  \arrowvert
\hat {T}_{n,L_n}^* (f_m)
- \hat{T}_n^*(f_m) \arrowvert
\mathds{1}_{\mathcal{D}_n^c } \big ]  \leq 2q m^\ell
\P ( \mathcal{D}_n^c) = o(1)
\end{align*}

which proves (<ref>).

Therefore, we will assume in the following discussion
that given the random projection $\Pi_n$
the matrix $\widehat{\Sigma}_{n}^{*} $ can be represented as

\widehat{\Sigma}_{n}^{*}  =
{1 \over m } L_nX^*X^{*\top} L_n^\top ~,

where $L_n$  is a $q \times q'$ matrix
satisfying $\| L_n \|_{S_\infty} \leq \alpha < \infty $ (for all $n \in \mathbb{N}$)
and $X^* $ is an $  q'\times m$ matrix and $q' =  O(q)$.
Furthermore, $f$ is replaced by $f_m$.
Note that these arguments only require the existence of moments of order $4$.

§.§ Reduction to truncated components

We will continue truncating the random variables  $X_{ij}$. For this purpose we formulate the following lemma.

There exists a sequence $(\delta_m)_{n\in\N}$ converging decreasingly to zero such that

\begin{align} \label{hol101}
\delta_m^{-4}\E\Big(I_{\{\arrowvert X_{11}\arrowvert\geq\delta_mm^{1/2}\} }X_{11}^4\Big)\longrightarrow 0\ \ \text{as}\ \ n\rightarrow\infty.
\end{align}

If  in addition $\E\arrowvert X_{11}\arrowvert^{6}<\infty$, then the sequence can be chosen such that

\begin{align} \label{hol1011}
\delta_m^{-6}\E\Big(I_{\{\arrowvert X_{11}\arrowvert\geq\delta_mm^{1/2}\} }X_{11}^6\Big)\longrightarrow 0\ \ \text{as}\ \ n\rightarrow\infty.
\end{align}

If $\E\arrowvert X_{11}\arrowvert^{4+\varepsilon}<\infty$ for some $\varepsilon>0$, then (<ref>) holds true with $\delta_m^{-1}=o\big(m^{\frac{\varepsilon}{2(4+\varepsilon)}}\big).$

The proof of (<ref>) is given on page 559 in baisil2004, that we repeat here for the reader's convenience. First observe that for any $k\in\N$, there exists a strictly  increasing sequence $(n_k)_{k\in\N}$ with

k^4\E\Big( I\big\{\arrowvert X_{11}\arrowvert\geq m_{n_k}^{1/2}/k\big\}X_{11}^4\Big)<\frac{1}{2^k}

by monotone convergence, because $\E X_{11}^4<\infty$. Choose $\delta_m=1/k$ for $n\in[n_k,n_{k+1})$, $\delta_m=1$ for $n<n_1$. Then, $\delta_m\searrow 0$ and

\delta_m^{-4}\E\Big(I_{\{\arrowvert X_{11}\arrowvert\geq\delta_mm^{1/2}\}}  X_{11}^4\Big)\longrightarrow 0\ \ \text{as}\ \ n\rightarrow\infty.

The statement (<ref>) under the assumption $\E \arrowvert X_{11}\arrowvert^{6}<\infty$ follows in the same way.
If $\E \arrowvert X_{11}\arrowvert^{4+\varepsilon}<\infty$,

\delta_m^{-4}\E\Big(I_{\{\arrowvert X_{11}\arrowvert\geq\delta_mm^{1/2}\} }X_{11}^4\Big)\leq \delta_m^{-(4+\varepsilon)}m^{-\varepsilon/2}\E \arrowvert X_{11}\arrowvert^{4+\varepsilon},

which converges to zero if $\delta_m^{-1}=o\big(m^{\frac{\varepsilon}{2(4+\varepsilon)}}\big)$.

We choose the sequence  $(\delta_m)$
such that  Lemma  <ref> holds and additionally
such that

\begin{align}
\label{ar7}
\delta_m m^{1/8}\rightarrow\infty ~
\end{align}

With this sequence, we
show that it is sufficient
to consider random variables which satisfy

\begin{align}
\arrowvert  X_{ij}\arrowvert&\leq \delta_m\sqrt{m}\ \ i=1,\dots n,\ j=1,\dots, p\label{eq: trunk1}\\
\E \ X_{11}&=0\\
\Var ( X_{11})&=1\\
\E   X_{11}^4 &\rightarrow  3\label{eq: trunk4}.
\end{align}

For this purpose, we introduce the notation

\begin{eqnarray*}
\check{X}_{ij} =
\frac{
X_{ij} I \big \{ |X_{ij}| \leq \delta_m \sqrt{m} \big \} - \mathbb{E} \big[ X_{ij} I \big  \{|X_{ij}| \leq  \delta_m \sqrt{m} \big \} \big ] }
{ \sqrt{{\rm Var } \big (  X_{ij} I \big \{ |X_{ij}| \leq  \delta_m \sqrt{m} \big \}
\big )  } }.
\end{eqnarray*}

We write

\check {\Sigma}_n^{*}:=\frac{1}{m}\sum_{i=1}^m L_n \check  X_i^{*}{\check {X}^{*\top}_i} L_n^\top

and denote by
$\check {\lambda}_i^*$  ($i=1,\dots, q$)   its eigenvalues in decreasing order and by

\check {T}_n^*(f)=\sum_{i=1}^{q}f(\check {\lambda}_i^*)

the corresponding linear spectral statistic.

Grant Assumptions (A1)–(A3+). Then

\begin{align}
& \hat{T}_n^*(f_m)-\check {T}_n^*(f_m)\rightarrow_{\P}0.\label{eq: trunk0}
\end{align}

If additionally $\E | X_{11} |^6 < \infty $, we have

\begin{align}
& \E^* \hat{T}_n^*(f_m)- \E^* \check {T}_n^*(f_m)\rightarrow_{\P}0.\label{eq: trunk01}
\end{align}

For the sequence  $(\delta_m)$ specified  above we

\tilde{X}_{ij}:=X_{ij}I\{\arrowvert X_{ij}\arrowvert <\delta_m\sqrt{m}\}

and denote by $\tilde{X}_1^*,\dots, \tilde {X}_n^*$ the corresponding iid sample from $n^{-1}\sum_{i=1}^n\delta_{\hat{X}_i}$. With

\begin{align}
\label{hol201}
\widehat \Sigma_n^*&=\frac{1}{m}\sum_{i=1}^mY_i^*Y_i^{* \top }=\frac{1}{m}\sum_{i=1}^m
L_n {X}_i^*{X}_i^{* \top }L_n^\top
\end{align}

\begin{align}
\label{hol202}
\tilde{B}_n^*&=\frac{1}{m}\sum_{i=1}^m L_n \tilde {X}_i^*{\tilde {X}_i^{*\top}} L_n^\top
\end{align}

we get by the union bound and Lemma <ref> that

\begin{align}
\label{det112222}
\P(\widehat \Sigma_n^*\not=\tilde {B}_n^*)&\leq\P\Big (\tilde{X}_{ij}^*\not= X_{ij}^* \text{ for some } (i,j)\Big ) \leq mq'\,\P\big(\tilde {X}_{11}^*\not={X}_{11}^*\big)\\
&= mq' \,\E\Big (\frac{1}{n}\sum_{i=1}^nI\{\arrowvert X_{i1}\arrowvert\geq\delta_m\sqrt{m}\}\Big )
= mq' \,\P(\arrowvert X_{11}\arrowvert>\delta_m\sqrt{m})\nonumber \\
&\leq K\delta_m^{-4}\int_{\{\arrowvert X_{11}\arrowvert\geq\delta_m\sqrt{m}\}}\arrowvert X_{11}\arrowvert^4 d \mathbb{P} =o(1). \nonumber
\end{align}

Now, we are passing over
from $\tilde {X}_1,\dots, \tilde {X}_n$
to  the centered and standardized modifications
$\check {X}_1,\dots, \check{X}_n$, where

\check{X}_{ij}:=\frac{\tilde {X}_{ij} -\E\tilde{X}_{ij}}{\sigma_n}\ \ \text{ with }\sigma_n:=\sqrt{\Var(\tilde{X}_{ij})},

and denote by $\check {X}_1^*,\dots,\check {X}_m^*$ the corresponding iid sample from $n^{-1}\sum_{i=1}^n\delta_{\tilde{X}_i}$.
Further, we introduce
$\check {\X}^*=(\check X_1^*, \dots , \check  {X}_m^*)$ and $ {\tilde\X}^*=(\tilde X_1^* , \dots , \tilde X_m^*)$,

\check {\Sigma}_n^*=\frac{1}{m}\sum_{i=1}^mL_n \check {X}_i^*\check {X}_i^{*\top}L_n^\top
= L_n \tilde \X ^*  \tilde \X ^{*\top} L_n^\top ,

and denote by   $\check {\lambda}_1^*\geq \check {\lambda}_2^*\geq \dots\geq \check{\lambda}_q^*$  its eigenvalues
with corresponding linear spectral statistic

\check{T}_n^*(f_m)=\sum_{j=1}^q f_m(\check\lambda _j).

By employing the Lipschitz continuity of $f$, the $1$-Wielandt-Hoffman inequality and the reasoning of the proof of  Lemma 2.7 in
bai1999, we deduce the upper bound

\begin{align}
\nonumber
\big\arrowvert\check  {T}_n^*(f_m)- \tilde {T}_n^*(f_m)\big\arrowvert&\leq
{\max}_{\lambda \in  [K_l, K_r]}
| f_m'(\lambda) | \cdot
\sum_{j=1}^q \big\arrowvert \check {\lambda}_j^*-\tilde {\lambda}_j^*\arrowvert + o_\P (1) \\
&\leq {2}
{\max}_{\lambda \in  [K_l, K_r] }
| f_m'(\lambda) | \cdot    \Big(\frac{1}{m}\tr\Big[L_n ({\tilde  \X}-\check{\X})({\X}-\check{\X})^\top L_n ^\top\Big]\Big)^{1/2}
\nonumber \\
&\hspace{50mm}\times\Big(\tr\big(\tilde{B}_n^* \big)+\tr\big(\check {\Sigma}_n^*\big)\Big)^{1/2} + o_\P (1)
\label{eq: b1}
\end{align}

where  $
\tilde {T}_n^*(f_m)=\sum_{j=1}^q f_m(\tilde \lambda _j) $ denotes the linear spectral statistic  corresponding to the matrix  $\tilde{B}_n^* = L_n \check {\X}^* \check {\X}^{* \top } L_n^\top $.
In order to bound the latter expression, observe that

\begin{align*}
\frac{1}{m}& \tr\Big[L_n (\tilde{\X}-\check{\X})(\tilde{\X}-\check{\X})^\top  L_n ^\top \Big]\\
&\leq \frac{2}{m}\Big(1-\frac{1}{\sigma_n}\Big)^2 q \Arrowvert \tilde {B}_n^*\Arrowvert_{S_\infty} + \frac{2}{m}\frac{1}{\sigma_n^2}\tr\left[L_n (\E \tilde  {X}_{1})(\E \tilde  {X}_1)^\top L_n^\top  \right]\\
&\leq 2c \frac{(\sigma_n^2-1)^2}{\sigma_n^2(1+\sigma_n)^2}\Arrowvert \tilde {B}_n^*\Arrowvert_{S_\infty} + \frac{2c}{\sigma_n^2}(\E \tilde {X}_{11})^2 \Arrowvert L_n\Arrowvert_{S_\infty} ^2.
\end{align*}

\begin{align}
\arrowvert \sigma_n^2-1\arrowvert&\leq 2\E\Big(I\{\arrowvert X_{11}\arrowvert\geq \delta_m\sqrt{m}\}\arrowvert X_{11}\arrowvert^2\Big)\nonumber =o(\delta_m^{2}m^{-1})\nonumber
\intertext{and}
\arrowvert \E\tilde {X}_{11}\arrowvert&=o(\delta_mm ^{-3/2}),\label{eq: Edelta}
\end{align}

such that

\begin{align*}
\frac{1}{m}& \tr\Big[L_n (\tilde{\X}-\check{\X})(\tilde{\X}-\check{\X})^\top L_n ^\top \Big]
=o(\delta_m^4m^{-2})\Arrowvert \tilde {B}_n^*\Arrowvert
+o(\delta_m^2m^{-3})\Arrowvert L_n\Arrowvert_{S_\infty}^2 .
\end{align*}

Plugging this bound into (<ref>), we find

\begin{align}
\big\arrowvert\check  {T}_n^*(f_m)- \tilde {T}_n^*(f_m)\big\arrowvert&
\leq \mathcal{O}_{\P}
(1)  o \big (  \delta_m^2m^{-1}\sqrt{\Arrowvert \tilde{B}_n^*\Arrowvert_{S_\infty}}+o(\delta_m m^{-3/2})\big) \Big )
O_{\P}(\sqrt{q} ) ,
%\sqrt{\Arrowvert \check {\Sigma}_n^*\Arrowvert_{S_\infty} +\Arrowvert \tilde{B}_n^*\Arrowvert_{S_\infty}},
\label{eq: tro}
\end{align}

where we used the fact that, by  Theorem <ref> and (<ref>),
{\max}_{\lambda \in [K_l, K_r]}
| f'(\lambda) |
= \mathcal{O}_{\P} (1) $  and

\begin{align*}
\tr\big(\tilde {B}_n^* \big )
+ \tr\big(\check{\Sigma}_n^*\big)  =\mathcal{O}_{\P}({q}).
\end{align*}

This proves (<ref>). Note that these arguments only require the existence of the moments  order $4$.

For a proof of (<ref>) we note that we need a substitute for (<ref>)  which allows to transfer the $\E^*$-expectations. For this purpose  recall that
$\hat T_n^* (f_m) $ and $  \tilde T_n^* (f_m) $  are the linear spectral statistics corresponding to the matrices $\widehat \Sigma_n^*$
and $\tilde B_n^*$ in (<ref>) and (<ref>), respectively. By a similar reasoning as in (<ref>) we have

\begin{align}
\nonumber
\E \big\arrowvert\hat   {T}_n^*(f_m)- \tilde {T}_n^*(f_m)\big\arrowvert
&\leq {2}
{\max}_{\lambda \in  [K_l, K_r] }
| f_m'(\lambda) | \cdot    \Big( \E  \frac{1}{m}\tr\Big[L_n ({\tilde  \X}-\check{\X})({\X}-\check{\X})^\top L_n ^\top\Big]\Big)^{1/2}
\nonumber \\
&\hspace{30mm}\times \E
\big (\tr\big(\tilde{B}_n^* \big)+\tr\big(\check {\Sigma}_n^*\big)\big )^{1/2} +  2q m^\ell \P (\mathcal{D}_n^c) .
\label{eq: b1a}
\end{align}

By Theorem <ref> we have
$2q m^\ell \P (\mathcal{D}_n^c) = o(1)$. Moreover,
\big (\tr\big(\tilde{B}_n^* \big)+\tr\big(\check {\Sigma}_n^*\big)\big )^{1/2} = O(\sqrt{q} ) $ while  we obtain for the other factor on the right-hand side of (<ref>)

\begin{align*}
\E  \frac{1}{m}\tr\Big[L_n ({\tilde  \X}-\check{\X})({\X}-\check{\X})^\top L_n ^\top\Big] & =\frac{1}{m} \E \| L_n ({\tilde  \X}-\check{\X}) \|^2_{S_2} \leq \frac{1}{m} \|L_n \|^2_{S_\infty}
\E \|  {\tilde  \X}-\check{\X} \|^2_{S_2}.
\end{align*}

Using (<ref>) in Lemma <ref>  it follows

\begin{align*}
\frac{1}{m}
\E \|  {\tilde  \X}-\check{\X} \|^2_{S_2} = \frac{1}{m}
\sum_{i=1}^m  \sum_{j=1}^{q'} \E \big [ |X_{ij}|^2 \mathds{1} \{ |X_{ij} | \geq \delta_m \sqrt{m} \} \big ] \leq  {\delta_m^2 \over m }  \E |X_{11}|^6 ~,
\end{align*}

which implies  $\E \big\arrowvert\hat   {T}_n^*(f_m)- \tilde {T}_n^*(f_m)\big\arrowvert = o(1)$.  The  bound
$\E^* \big\arrowvert\check   {T}_n^*(f_m)- \tilde {T}_n^*(f_m)\big\arrowvert  = o_{\P}(1)$ follows by similar arguments as used in the proof of the estimate (<ref>).

Summarizing the discussion of Section <ref> and <ref>, we will from now assume that the random variables $X_{ij}$
satisfy (<ref>) - (<ref>), that the vectors $X_i$ have $q'=O(q)$ components and that  the matrix $L_n$ is of dimension $q \times q' $.
Note that the matrix $L_n$ can be a random matrix which is independent of $X_1, \ldots X_n$.

§.§ Passing over to   the  bootstrapped  process of Stieltjes transforms

\begin{align*}
{M}_n^*(z):= q \Big(m_{\mu^{\widehat \Sigma_n^*}}(z)-m_{\E^*\mu^{\widehat \Sigma_n^*}}(z)\Big) = q \Big(m_{\mu^{\widehat \Sigma_n^*}}(z)- \E^* \Big[m_{\mu^{\widehat \Sigma_n^*}}(z)\Big] \Big)
\intertext{and}
{M}_{n,\mathcal{D}_n}^*(z): = q \Big (m_{\mu^{\widehat \Sigma_n^*}}(z)\mathds{1}_{\mathcal{D}_n}- \E^* \Big[m_{\mu^{\widehat \Sigma_n^*}}(z)\mathds{1}_{\mathcal{D}_n}\Big] \Big ),
\end{align*}

where $
\mathcal{D}_n =\big \{
\lambda_{\min }  (\widehat \Sigma_{n}^* ) > K_l ~,~
\| \widehat \Sigma_{n}^*   \|_{S_\infty } <  K_r \big \} ~.

\begin{equation}\label{eq: red 1}\widehat{T}_n(f_m)-\E^*\widehat{T}_n(f_m)
=\widehat{T}_n(f_m)\mathds{1}_{\mathcal{D}_n}-\E^*\Big(\widehat{T}_n(f_m)\mathds{1}_{\mathcal{D}_n}\Big) + o_{\P}(1)\end{equation}

by Corollary <ref> and the relation

\hat{T}_n(f_m)^*\mathds{1}_{\mathcal{D}_n}- \E^*\Big(\hat{T}_n(f_m)^*\mathds{1}_{\mathcal{D}_n}\Big)= -\frac{1}{2\pi i}\ointctrclockwise f_m(z){M}_{n,\mathcal{D}_n}^*(z)\dd z

provided by the Cauchy integral formula, the result is deduced from the corresponding limit theorem for

\ointctrclockwise f_m(z){M}_{n,\mathcal{D}_n}^*(z)\dd z,

where the curve integral is along any closed curve within a
region on which $f$ is analytic and which encloses the interval $[K_l,K_r]$. The latter indeed boils down to proving a conditional Donsker-type theorem of a truncated version of the bootstrapped process $M_n^*(.)$, denoted by $\widehat{M}_n^*(.)$,
see Section <ref>. For a precise definition of $\widehat{M}_n^*(.)$ let $x_l$, $x_r$ be two real numbers with

\begin{align*}
(0, K_l ) & \text{if }c'\in(0,1)\\
(-\infty,~0 ) & \text{if } c'\geq 1
\end{cases}
\end{align*}

and $
x_r> K_r
$, where $K_l$ and $K_r$ are the constants introduced in Corollary <ref> and
$$c' = \limsup_{n\to \infty}\frac{ q'}{m}  .

Moreover, define
\mathcal{C}_u=\{ x+iv_0:~x\in [x_l,x_r] \}

\begin{align*}
\mathcal{C}=\{x_l+iv:v\in[0,v_0]\}\cup \CC_u \cup \{x_r+iv: v\in [0,v_0]\}
\end{align*}

such that the closed curve $\CC \cup \bar \CC$ is contained in a region where $f$ is analytic. Further, for some null sequence $(\varepsilon_n)$ satisfying

\begin{align}
\label{det105}
%(\log n)\varepsilon_n\rightarrow 0\ \ \text{and}\ \
\varepsilon_n\geq m^{-\alpha}
\end{align}

for some $\alpha\in (0,1)$,  we introduce

\begin{align}
\CC_{l,n}&:=\begin{cases}
\{x_l+iv:~v\in[m^{-1}\varepsilon_n,v_0]\}, & \text{if }x_l>0\\
\{x_l+iv:~v\in[0,v_0]\} & \text{if }x_l\leq 0,
\end{cases}\\
\nonumber
\CC_{r,n}&:=\{x_r+iv:~v\in[m^{-1}\varepsilon_n,v_0]\}
\intertext{and}
\label{det11}
\CC_n&:=\CC_{l,n}\cup\CC_u\cup\CC_{r,n}.
\end{align}

Lastly, we define

\begin{align*}
\widehat M_n^*(z)=\begin{cases}
M_n^*(z), &\text{for }z\in\CC_n,\\
M_n^*(x_r+im^{-1}\varepsilon_n), & \text{for }z=x_r+iv \text{ with }v\in [0,m^{-1}\varepsilon_n] \\
M_n^*(x_l+im^{-1}\varepsilon_n), & \text{for }z=x_l+iv\text{ with }x_l>0 \text{ and }v \in[0,m^{-1}\varepsilon_n] ,
\end{cases}
\end{align*}

and because of $m(\bar{z})=\overline{m(z)}$ for any Stieltjes transform $m$, we have
\widehat{M}_n^*(\bar{z}):=\overline{\widehat M_n^*(z)}$ for $z\in\mathcal{C}$.
Analogously, we define $\widehat M_{n,\mathcal{D}_n}^*$ for the   truncated version $M_{n,\mathcal{D}_n}^*$.

\begin{align}
\widehat{T}_n(f_m)\mathds{1}_{\mathcal{D}_n}&-\E^*\Big(\widehat{T}_n(f_m)\mathds{1}_{\mathcal{D}_n}\Big)\nonumber \\
=&\ointctrclockwise_{\CC \cup \bar\CC} f_m(z) M_{n,\mathcal{D}_n}^*(z)\dd z \nonumber\\
=& \ointctrclockwise_{\CC \cup \bar\CC} f_m(z) \widehat M_{n,\mathcal{D}_n}^*(z)\dd z
+\mathcal{O}\bigg(8 \varepsilon_n {q \over m}   \| f_m \|_{\CC\cup\bar\CC} \Big( \big |
%\max  \{ \lambda_{\max} ( \widehat{\Sigma}_{n}),
-x_r \big |^{-1}+
\big |
%\min \big\{ \lambda_{\min}(\widehat{\Sigma}_{n}),
%\big ) \}
-x_l  \big |^{-1} \Big)\bigg)\nonumber\\
=&\ointctrclockwise_{\CC \cup \bar\CC} f_m(z) \widehat M_n^*(z)\dd z + o_{\P}(1)\label{eq: AR1005}
\end{align}

by Corollary <ref>,  is sufficient to consider $\widehat M_n^*$ in what follows.
The essential part of the proof of Theorem <ref> consists of verifying the following Donsker-type result. As the restrictions of $f_m$ and $f$  to $\CC \cup \bar\CC$ coincide for $m$ sufficiently large,  Gaussianity of  (<ref>)
then follows with (<ref>) from the continuous mapping Theorem.

Grant the conditions of either Theorem <ref> or Proposition <ref>, then

\begin{align*}
d_{BL}\bigg\{\LL\Big(\big( \widehat {M}_n^{*}(z)\big)_{z\in\CC}\Big\arrowvert X_1,\dots,X_n, \Pi_n \Big),\, \LL(Z)\bigg\}\longrightarrow_{\P} 0
\end{align*}

with a  centered Gaussian process $(Z)$ on $\CC$ satisfying $Z(\bar z)=\overline{Z(z)}$ and

\begin{align*}
\E\big(Z(z_1), Z(z_2)\big)= 2 \frac{(\underline{m}_{{H}}^0)'(z_1)(\underline{m}_{{H}}^0)'(z_2)}{\big(\underline{m}_{H}^0(z_1)-\underline{m}_{{H}}^0
(z_2)\big)^2}-\frac{2}{(z_1-z_2)^2} \ \text{ for }z_1,z_2\in\mathcal{C}
\end{align*}

(understood as its continuous extrapolation for the removable singularities at $z_1=z_2$).

A central  tool in the proof of this result is  an
analog of Proposition <ref> for the truncation in Section <ref> with a bound which is not depending on $z\in\mathcal{C}_n$.
These results will be presented first in the following section. The proof of Proposition <ref>
is given in Section <ref>.

§.§ A non-standard result on quadratic forms

For the statement of  an
analog of Proposition <ref> for the truncation in Section <ref> we study the following matrices  in the quadratic form

\begin{eqnarray}
\label{eq: C_i0}
C^*  &=&
C^*  (z) =
L_n^\top  D_1^*(z)^{-1} L_n  ,   \\
\label{eq: Cende}
C^*  &=&
C^*  (z) =
L_n^\top  D_1^*(z)^{-1}M L_n  ,   \\     \label{eq: C_i}
C^*  &=&
C^*  (z) =
L_n^\top  D_1^*(z)^{-2} L_n  ,   \\
\label{eq: C_ia}
C^*  & =&
C^*  (z_1,z_2) =
L_n ^\top D_{12}^*(z_1)^{-1}
L_nL_n^\top D_{12}^*(z_2)^{-1}
L_n  ~,
\\
\label{eq: C_ib}
C^*  & =&
C^*  (z_1,z_2) = L_n ^\top  \E_j^* \big [  D_{1j}^*(z_1)^{-1} \big ]
L_n L_n^\top D_{1j}^*(z_2)^{-1}
L_n  ~, \\
\label{eq: C_ic}
C^*  &=&   C^*  (z) =
L_n^\top  D_1^*(z)^{-2} L_n  ,   \\
%\HDB{C^* } &=&
%     \label{eq: C_idalt}
%  \HDB{  C^*  (z_1,z_2) =
%   L_n^\top  D_1^*(z_1)^{-2} L_n  L_n^\top  D_1^*(z_2)^{-1} L_n ,  }
%    \\
C^*  &=&
\label{eq: C_id}
C^*  (z_1,z_2) =
L_n^\top  D_1^*(z_1)^{-2}   D_1^*(z_2)^{-1} L_n ,  \\
C^*  &=&
\label{eq: C_ie}
C^*  (z_1,z_2) =
L_n^\top  D_1^*(z_1)^{-1}   D_1^*(z_2)^{-1} L_n
\end{eqnarray}

for $z, z_1,z_2\in\mathcal{C}_n$.  $M\in \C^{q\times q}$ is deterministic and of bounded spectral norm, uniformly in $n$. Recall that the notation $\E_X$ means integration with respect to $X=(X_1, \ldots , X_n)$. In other words (as the projection is independent of  $X$), the expectation is taken conditional on $\Pi_n $.

For any
$p\geq 2$, there exists some constant $K_p>0$, such that for any $n\in \N$,
\begin{align*}
\E_X
\big\arrowvert X_1^{*\top}
C^*X_1^*-\tr C^* \big\arrowvert^p  \leq
\Big( m^{p-1} \delta_m^{2p-4} +
\frac{m^{p+1}}{n}
\Big)
\end{align*}

where $K_p$ is a constant depending
only on $p$ and the matrix
$C^*$ is   given by  one of the matrices in   (<ref>) - (<ref>).

denote by
$\widehat \Sigma_{n, 1, \ldots ,j}^* $ the matrix.    which  is obtained  from $\widehat \Sigma_{n}^*$ by omitting the terms involving $X_1^* , \ldots , X_j^*$
(for $j=0$ this is $\Sigma_n^*$) and the matrix
$\tilde \Sigma_{n, 1}^*$
is the empirical covariance matrix of the  vectors $\tilde{X}_2^*,\dots, \tilde{X}_m^*$ with normalizing  factor $1/m$ which  are defined in (<ref>).  By Corollary <ref>  (with an adaptation of its proof to the matrices $\widehat \Sigma_{n, 1}  ^*, \widehat \Sigma_{n, 1,2}  ^*$

we find constants  $K_l$ and $K_r$
such that
the event

\begin{align}
\label{det21}
\mathcal{A}_n:=
\Big \{
%  \lambda_{\min} ( \widehat B_{m}) \geq K_l,
%  ~  \| \widehat B_{m}  \|_{S_\infty }\leq K_r,  ~
%  \\
&   \lambda_{\min} (\tilde \Sigma_{n, 1}^*) \geq K_l,
~  \| \tilde \Sigma_{n, 1}  ^* \|_{S_\infty }\leq K_r,  ~
\\
& \nonumber
~~~~~~~~  ~~~~~~~~  ~~~~~~~~
\lambda_{\min} (\widehat \Sigma_{n, 1, \ldots ,j}  ^*) \geq K_l,~
\| \widehat \Sigma_{n, 1, \ldots ,j}  ^* \|_{S_\infty}   \leq K_r ~\text{for } j=0,1,2
\Big \} ,
\end{align}

satisfies    for all $\ell \in \mathbb{N}$

\begin{align}
\label{det33}
\mathbb{P}( \mathcal{A}_n^c)  =  o ( m^{-\ell}) ~.
\end{align}

We will begin proving the statement for matrices of the form (<ref>). Observing the arguments as given in the proof of Proposition <ref> we obtain

\begin{align}
\E_X
\big\arrowvert X_1^{*\top}
C^*X_1^*-\tr C^* \big\arrowvert^p  & \leq
2^{p-1} \Big (  \E_X \big\arrowvert X_1^\top
\tilde  C^* X_1-\tr  \tilde  C^*
\big\arrowvert^p  \label{det35} \\
& ~~~~~~~~~~~
\E_X \big\arrowvert X_1^\top (
\tilde C^* -   C^*  ) X_1-\tr ( \tilde
C^* -    C^*  )
\big\arrowvert^p
\Big )
~,  \nonumber
\end{align}

where matrix $\tilde C^*$ is defined by (<ref>) with $M=I_q$.
By the same reasoning leading to equation (3.2) in baisil2004 with $a(v) =1 $ and  $B(v) =
\tilde C^*$ it follows that

$$ \E_X \big\arrowvert X_1^\top
\tilde  C^* X_1-\tr  \tilde  C^*
\big\arrowvert^p
\leq c {\delta_m^{2p-4 \vee 0} \over m^{1 \wedge p} } =  c {\delta_m^{2p-4 } \over m }~.

Note that the matrix $\tilde  C^*$ satisfies the corresponding assumption for such an  estimate, that is

\begin{align}
\label{det32}
\| \tilde  C^*    \|_{S_\infty}  \leq  \| L_n
\|_{S_\infty}^2
\| ( \tilde  D^*_1 (z))^{-1}   \|_{S_\infty}  \leq  c(
\mathds{1}_{\mathcal{A}_n} +
\mathds{1}_{\mathcal{A}_n^c}
m^{1+ \alpha}  )
\leq  c(
1  +
\mathds{1}_{\mathcal{A}_n^c}m^{1+ \alpha}  ) ,
\end{align}

where $\tilde{D}_1^*(z)$ is defined as $D_1^*(z)$ with $X_2^*,\dots, X_m^*$ replaced by $\tilde{X}_2^*,\dots, \tilde{X}_m^*$.

Now we turn to  the second term in (<ref>) and consider

\begin{align*}
\E_X \big | \tr ( \tilde
C^* -    C^*  )
\big | ^p       & \leq  m^p   \E_X \big\|  \tilde
C^* -    C^*
\big\|_{S_\infty} ^p    \leq
m^p  \| L_n
\|_{S_\infty}^{2p}  \E_X \big\|
D_1^* (z)^{-1} -    \tilde D_1^* (z)^{-1}
\big\|_{S_\infty} ^p  \\
& \leq  c  m^p  \E_X \big [ \|
D_1^* (z)^{-1} \|_{S_\infty} ^p   \|    \tilde D_1^* (z)^{-1}
\|_{S_\infty} ^p  \|
\widehat \Sigma_{n,1}^* -  \tilde \Sigma_{n,1}^*
\|_{S_\infty} ^p \big ] ,
\end{align*}

where the last inequality follows from the formula $B^{-1} - A^{-1}  = A^{-1} (A-B) B^{-1}$.
Note that we have

\begin{align}
\label{ang21}
\|
D_1^* (z)^{-1} \|_{S_\infty}  &\leq
\max_{i=1}^q {1\over | z - \lambda_i ( \widehat \Sigma_{n,1}^* ) | }
\leq
\max_{i=1}^q {1\over | \Re ( z - \lambda_i ( \widehat \Sigma_{n,1}^* ) )| }\mathds{1}_{\mathcal{A}_n}+ \frac{\arrowvert z\arrowvert}{\Im(z)}\mathds{1}_{\mathcal{A}_n^c}
\\
\nonumber
& \leq \max \Big \{ {1\over | x_r - K_r| },  {1\over | x_l - K_l| }
\Big \}  + \frac{\arrowvert z\arrowvert}{\Im(z)}\mathds{1}_{\mathcal{A}_n^c}
\end{align}

where $K_l$ and $K_r$ are the constants from Corollary  <ref>. Moreover,

\|
\widehat \Sigma_{n,1}^* -  \tilde \Sigma_{n,1}^*
\|_{S_\infty} \leq   \| \Sigma_n \|_{S_\infty} \|X_1\|^2 {\Delta_n^* \over m } ,
% \leq c \Delta_n,

\begin{align}
\label{det37a}
{1 \over m^{p/2} } \E_X   \|X_1\|^p   & \leq c ,  \\
\label{det37b}
\E_X   \| \Delta_n^*\|^p    & \leq c_p {m \over n }
\end{align}

(note that (<ref>) follows from the fact that here the random variable $X_1$ is of dimension $q'$ and has independent components  bounded by $\delta_m \sqrt{m}$).
Combining these estimates and

and using the corresponding  bound for the quantity  $\|   \tilde
D_1^* (z)^{-1} \|_{S_\infty}$
and  obtain  (observing (<ref>)), we arrive at

\begin{align}
\nonumber
\E_X \big | & \tr ( \tilde
C^* -    C^*  )
\big | ^p \\
& \leq  c  m^p  \Big \{  \E_X  \big [ \mathds{1}_{\mathcal{A}_n} \|
\widehat \Sigma_{n,1}^* -  \tilde \Sigma_{n,1}^*
\|_{S_\infty} ^p \big ]
+  \big |{ z \over  \Im (z)  } \big |^{2p} \E_X  \big [ \mathds{1}_{\mathcal{A}_n^c} \|
\widehat \Sigma_{n,1}^* -  \tilde \Sigma_{n,1}^*
\|_{S_\infty} ^p \big ]
\big \}  \nonumber
\\
\nonumber
& \leq  c  m^p  \Big \{ {1 \over m^p}  \E_X
\E_X \|X_1 \|^{2p}
+  \big | { z \over  \Im (z)  } \big |^{2p} \big ( \mathbb{P} ( {\mathcal{A}_n^c} ) \big )^{1/2}
\big (  \E_X \|
\widehat \Sigma_{n,1}^* -  \tilde \Sigma_{n,1}^*
\|_{S_\infty} ^{2p}  \big )^{1/2}
\big \}
\nonumber
\\
\nonumber
& \leq  c  m^p  \Big \{ {m \over n}
+  m^{2p(1+\alpha)} (\mathbb{P} ( {\mathcal{A}_n^c} ) \big )^{1/2} \sqrt{m\over n}
\Big \}  \\
& \leq c {m^{p+1}\over n}~.
\label{det38}
\end{align}

Similarly, we obtain

\begin{align*}
\E_X \big\| X_1^\top (
\tilde C^* -   C^*  ) X_1
\big| ^p & \leq  \E_X \big [ \| X_1\|1^{2p}  \| C^* -   C^*  \|^p_{S_\infty } \big ] \\
& \leq c \E_X \Big [ \| X_1\|^{2p}
\|
D_1^* (z)^{-1} \|_{S_\infty} ^p   \|    \tilde D_1^* (z)^{-1}
\|_{S_\infty} ^p   \| X_1\|^{2p}
\big ( {\Delta_n^* \over m }\big )^p
\Big  ] \\
& \leq {c  \over  m^p}
\Big \{  \E_X  \big [ \mathds{1}_{\mathcal{A}_n} |\Delta_n^*|^p
\|X_1 \|^{4p} \big ]
+  \big |{ z \over  \Im (z)  } \big |^{2p} \E_X  \big [ \mathds{1}_{\mathcal{A}_n^c}   |\Delta_n^*|^p
\|X_1 \|^{4p}  \big ]
\Big \}  \\
& \leq  {c  \over  m^p} \Big \{ {m^{2p+1} \over n } + m^{(1+\alpha)2p}
\big ( \mathbb{P} ( {\mathcal{A}_n^c}) \big )^{1/2} \big ( \E_X [ |\Delta_n^*|^{2p}
\|X_1 \|^{8p} ]
\big )^{1/2}
\Big \} \\
& \leq c  {  m^{p+1} \over  n},
\end{align*}

by (<ref>). Combining this estimate with (<ref>) and (<ref>) yields the statement of Proposition <ref> for the matrix (<ref>).

The statement for the other matrices follow by similar arguments, which are omitted for the sake of brevity. For, example, for the matrix (<ref>) we use the identities

\begin{align*}
\tilde D_1^* (z)^{-2}  - D_1^* (z)^{-2} &=  D_1^* (z)^{-2} \big (
D_1^* (z)^{2} -\tilde D_1^* (z)^{2}
\big ) \tilde D_1^* (z)^{-2} \\
D_1^* (z)^{2} -\tilde D_1^* (z)^{2}    & = \big (  D_1^* (z) -  \tilde D_1^* (z) \big )\tilde D_1^* (z) + D_1^* (z) \big ( D_1^* (z) - \tilde D_1^* (z)\big )
\end{align*}

which gives

\begin{align*}
\|  \tilde D_1^* (z)^{-2}  - D_1^* (z)^{-2} \|_{S_\infty} & \leq c
\|  \tilde D_1^* (z)^{-2}   \|_{S_\infty}  \|    D_1^* (z)^{-2} \|_{S_\infty}  \\
& ~~~~~~~~~~~~~~~ \times
\big ( 2 |z| +
\|   \widehat \Sigma_{n,1}^*   \|_{S_\infty}
+   \|   \tilde  \Sigma_{n,1}^*    \|_{S_\infty} \big )
\|  \widehat \Sigma_{n,1}^*  -\tilde  \Sigma_{n,1}^*  \|_{S_\infty} .
\end{align*}

We now proceed in the same way as before
multiplying with $( \mathds 1_{\mathcal{A}_n} +\mathds 1_{\mathcal{A}_n}^c)$, where we use

\|   \widehat \Sigma_{n,1}^*   \|^2_{S_\infty}  \leq  \|   \widehat \Sigma_{n,1}^*   \|^2_{S_2} \leq
{1 \over m^2 } \sum_{i,j=1}^m \|X_i\|^2 \| X_j\|^2

on $\mathcal{A}_n^c$.

Note that Proposition <ref> will replace Proposition <ref> in the following discussion. Moreover,  in the case $p=2$ both results yield the same estimate. This fact will be of importance as we will use some of the
estimates for Section <ref>  - <ref> in the following discussion which also hold under Assumption (A3+) (instead of (A3)) and the truncation scheme considered in this section.

§.§ Proof of Proposition <ref>

§.§.§ Weak convergence of finite dimensional distributions in probability

Recall  that $\E_j^*$ denotes
the conditional expectation operator corresponding to $\hat{\P}_n$ with respect to the $\sigma$-field generated by $X_1^*,\dots, X_j^*$ (conditional on $\Pi_n$)
Similar calculations as in Section 2 (p. 569-570) of  baisil2004
and an application of Proposition <ref>  for the matrices $L_n^\top  D_1^*(z)^{-1} L_n$ and $L_n^\top  D_1^*(z)^{-2} L_n$
yield for $ z\in \CC_n $

\begin{align*}
\widehat M_n^{*}(z) :=
q\big(m_{\mu^{\widehat \Sigma_n^*}}(z)-\E^* m_{\mu^{\widehat \Sigma_n^*}}(z)\big) = \sum_{j=1}^m Y_j^*+O_{\P} \Big ( \delta_m^2  + \sqrt{m^2 \over n} \Big )
= \sum_{j=1}^m Y_j^*+o_{\P} (1))~,
\end{align*}

\begin{align}
Y_j^*&=-(\E_j^*-\E_{j-1}^*)\Big(\bar{\beta}_j^*(z)\delta_j^*(z)-\bar{\beta}_j^*(z)^2\varepsilon_j^*(z)\frac{1}{m}\tr\big(L_nL_n^\top  D_j^*(z)^{-2}\big)\Big)\nonumber\\
&=-(\E_j^*-\E_{j-1}^*)\Big(\frac{\dd}{\dd z}\bar{\beta}_j^*(z)\varepsilon_j^*(z)\Big).\nonumber
\end{align}

For example, by the algebraic manipulations on page 569 in this reference we obtain

\widehat M_n^{*}(z)  = \sum_{i=1}^m Y_j^*
+  \sum_{i=1}^m Z_j^*

Z_j^* = (\E_j^*-\E_{j-1}^*) \big [
\bar{\beta}_j^*(z) (\varepsilon_j^*(z)
\delta_j^*(z)- {\beta}_j^*(z) r_j^{*\top} D_j^{-2} (z) r_j^* \varepsilon_j^*(z)^2)  \big].

The $L^2$-norm of the first term is now  estimated as follows

\begin{align}
\nonumber
\E \Big [ \E^* \Big | \sum_{i=1}^m
(\E_j^*-\E_{j-1}^*) (
\bar{\beta}_j^*(z) \varepsilon_j^*(z)
\delta_j^*(z) ) \Big |^2
\Big ] & =
\sum_{i=1}^m
\E  \big |
(\E_j^*-\E_{j-1}^*) (
\bar{\beta}_j^*(z) \varepsilon_j^*(z)
\delta_j^*(z) ) \big |^2
\Big ]  \\
& \leq  4 \sum_{i=1}^m
\E  \big |
\bar{\beta}_j^*(z) \varepsilon_j^*(z)
\delta_j^*(z)  \big |^2
\Big ]\nonumber\\
&= O_{\P} \Big ( \delta_m^2  + \sqrt{m^2 \over n} \Big )~,
\label{det111}
\end{align}

where the last estimate follows from Proposition <ref>
and the bounds $\arrowvert\bar{\beta}_j^*(z)\arrowvert\leq \arrowvert z\arrowvert/\Im(z)$.
The second term can be estimated by similar arguments and is of order
$O_{\P}  ( \delta_m^4  + m^2/ n  )$.
Note that one crucial difference to the analysis of baisil2004 is now caused by the fact that the random variables $\varepsilon_j^*$ and $\delta_j^*$ are not centered anymore with respect to $\E_{X_j^*}^*$:

\begin{align*}
\E_{X_j^*}^*\big(r_j^{*\top}D_j^*(z)^{-k}r_j^*\big)&= \frac{1}{m}\tr\Big( L_nL_n^\top   D_j^*(z)^{-k}\Big)\not=  \frac{1}{m}\tr\Big(L_nL_n^\top   D_j^*(z)^{-k}\Big),\ \ k=1,2.
\end{align*}

In view of the limiting distribution result, it is therefore sufficient to study linear combinations

\begin{align*}
\sum_{i=1}^r\sum_{j=1}^m\alpha_iY_j^*(z_i)\ \ \text{with }\alpha_1,\dots, \alpha_r\in\C,\ r\in\N,
\end{align*}

due to the Cramér-Wold device (since the real parts of these linear combinations are running over all real linear combinations of $\Re(\sum_{j=1}^mY_j^*(z_i)),\Im(\sum_{j=1}^mY_j(z_i))$, $i=1,\dots, r$, as $\alpha_1,\dots, \alpha_r$ varies over $\C$).
Note furthermore that it is sufficient to consider the case $\Im (z_i) > 0$ ($i=1, \ldots  , r$), because the distribution of any $C ( \CC , \R^2) $-valued random variable $Z$ is uniquely determined by its finite dimensional distributions $\mathcal{L}(Z(z_1),\dots, Z(z_k))$ with $z_1,\dots, z_k$ belonging to a dense subset of $\CC$ and $k\in\N$.
For this purpose, we shall prove that the conditions of
Theorem <ref> in the online supplement
satisfied for
\Re\Big (\sum_{i=1}^r\sum_{j=1}^m\alpha_iY_j^*(z_i)\Big ).

the bounds $\arrowvert\bar{\beta}_j^*(z)\arrowvert\leq \arrowvert z\arrowvert/\Im(z)$ and
the estimate

\Big\arrowvert\frac{1}{m}\tr\big(L_nL_n^\top   D_j^*(z)^{-2}\big)\Big\arrowvert\leq \frac{q}{m}
\Arrowvert L_n \Arrowvert^2_{S_{\infty}}\cdot\Arrowvert D_j^*(z)^{-2}\Arrowvert_{S_{\infty}}\leq  c(1+o(1))\frac{
\Arrowvert L_n \Arrowvert^2_{S_{\infty}}

we find by Proposition <ref>
(with $p=2$) that

\begin{align*}
\E\arrowvert Y_j^*(z)\arrowvert^4&\leq K\Big (\frac{\arrowvert z\arrowvert^4}{\Im(z)^4}\E\arrowvert\delta_j^*(z)\arrowvert^4+\frac{\arrowvert z\arrowvert^8}{\Im(z)^{16}}c^4(1+o(1))\E\arrowvert\varepsilon_j^*(z)\arrowvert^4\Big)  =\mathcal{O}\Big(\frac{\delta_m^4}{m}+\frac{m}{n}\Big)
\end{align*}

as $n\rightarrow\infty$. Consequently,

\begin{align}
\nonumber
\E \bigg [
\sum_{j=1}^m\E^*\bigg(\bigg\arrowvert \sum_{i=1}^r\alpha_iY_j^*(z_i)\bigg\arrowvert^2 I\bigg\{\sum_{i=1}^r\alpha_iY_j^*(z_i)\arrowvert\geq\varepsilon\bigg\}\bigg)
\bigg ] & \leq\frac{1}{\varepsilon^2}\sum_{j=1}^m\E\bigg\arrowvert\sum_{j=1}^r\alpha_iY_j^*(z_i)\bigg\arrowvert^4 \\
=  \mathcal{O}\Big(\delta_m^4
= o(1 )
\label{det112}
\end{align}

as $n\rightarrow\infty$, and condition (<ref>) of Theorem <ref> is fulfilled.

In order to verify  condition (<ref>), it is sufficient to show that for $z_1,z_2$ with $\Im(z_1)\not=0$, $\Im(z_2)\not=0$,

\begin{equation}\label{eq: 3.10}
\sum_{j=1}^m\E^*_{j-1}Y_j^*(z_1)Y_j^*(z_2)
\longrightarrow_\mathbb{P}
\begin{cases}
\frac{1}{3}\frac{(\underline{m}_H^0)'''(z_1)}{(\underline{m}_H^0)'(z_1)}
-\frac{1}{2}\Big(\frac{(\underline{m}_H^0)''(z_1)}{(\underline{m}_H^0)'(z_1)}\Big)^2 &\text{ if }z_1=z_2\\
(z_2)\big)^2}-\frac{2}{(z_1-z_2)^2}&\text{ if }z_1\not=z_2
\end{cases}
\end{equation}

(note that $\overline{Y_j^*(z)}=Y_j^*(\bar{z})$).
By the theorem of dominated convergence,

\begin{align*}
\frac{\partial^2}{\partial z_2\partial z_1}\eqref{eq: 3.10}=\eqref{eq: 3.11}
\end{align*}

\begin{align}\label{eq: 3.11}
\sum_{j=1}^m\E_{j-1}^*\Big[ (\E_j^*-\E_{j-1}^*)\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big) (\E_j^*-\E_{j-1}^*)\big( \bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big)\Big].
\end{align}

As for the classical CLT of linear spectral statistics, it follows from Vitali's convergence theorem that the convergence  of (<ref>) in probability  follows from the corresponding stochastic convergence of (<ref>).
For analyzing (<ref>)  we shall prove the following claims.

Claim I.

\begin{align}
\sum_{j=1}^m&\,\E_{j-1}^*\Big[(\E_j^*-\E_{j-1}^*)\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big)(\E_j^*-\E_{j-1}^*)\big(\bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big)\Big]\nonumber\\
\end{align}

Claim II.

\begin{align}
\sum_{j=1}^m&\, \E_{j-1}^*\Big[\E_j^*\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big)\E_j^*\big(\bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big)\Big]\nonumber\\
&=\sum_{j=1}^mb_n^*(z_1)b_n^*(z_2)\E_{j-1}^*\Big[\E_j^*\big(\varepsilon_j^*(z_1)\big)\E_j^*\big(\varepsilon_j^*(z_2)\big)\Big] + o_{\P}(1).
\end{align}

Claim III.

\begin{align*}
\sum_{j=1}^m&\, b_n^*(z_1)b_n^*(z_2)\E_{j-1}^*\big[\E_j^*\big(\varepsilon_j^*(z_1)\big)\E_j^*\big(\varepsilon_j^*(z_2)\big)\big]\nonumber\\
&=2 b_n^*(z_1)b_n^*(z_2)\frac{1}{m^2}\sum_{j=1}^m\tr \Big(L_n^\top  \E_j^*\big(D_j^*(z_1)^{-1}\big)
\E_j^*\big(D_j^*(z_2)^{-1}\big)L_n
\Big)
\end{align*}

In order to prove stochastic convergence and to determine the limit in probability of the right-hand side in Claim III, we shall prove the representation

\begin{align*}
\frac{1}{m^2}&\sum_{j=1}^m\tr \Big(L_n^\top
\E_j^*\big(D_j^*(z_1)^{-1}\big)
L_n L_n^\top
\E_j^*\big(D_j^*(z_2)^{-1}\big)L_n
\Big)\\
&= a_n(z_1,z_2)\frac{1}{m}\sum_{j=1}^m \Big(1-\frac{j-1}{m}a_n(z_1,z_2)\Big)^{-1} + o_{\P}(1)
\end{align*}

for some function $a_n(z_1,z_2)$ that will be specified in Claim VI. This is the most involved part of the proof. Claims IV and V are intermediate steps on this way. For this purpose, we
recall the notation
of $D_{ij}^*(z)$, $\beta_{ij}^*(z)$ and $b_1^*(z)$  in (<ref>),
(<ref>), and (<ref>), respectively, which will be used intensively in the following discussion.

Claim IV. There exists some constant $K>0$, such that for all $n\in\N$ and any $j\leq m$

\begin{align}
\tr&\Big(\E_j^*\big(D_j^*(z_1)^{-1}\big)
L_n L_n^\top
\E_j^*\big(D_j^*(z_2)^{-1}\big)
L_n L_n^\top
\Big)\nonumber\\
&\ \ \ \times \bigg[1{-}\frac{j-1}{m^2}b_1^*(z_1)b_1^*(z_2)\tr\bigg(
\Big(z_2 I-\frac{m-1}{m}b_1^*(z_2) L_n L_n^\top  \Big)^{-1}
L_n L_n^\top
\nonumber\\
&\hspace{45mm}\times\Big(z_1 I-\frac{m-1}{m}b_1^*(z_1)
L_n L_n^\top
\Big)^{-1}
L_n L_n^\top
\bigg)\bigg]\nonumber\\
&= \tr\bigg[\Big(z_1 I-\frac{m-1}{m}b_1^*(z_1) L_n L_n^\top  \Big)^{-1}
L_n L_n^\top
\Big(z_2 I-\frac{m-1}{m}b_1^*(z_2) L_n L_n^\top  \Big)^{-1}
L_n L_n^\top
\bigg] \label{eq: IV}\\
\end{align}

with $\E\arrowvert R(z_1,z_2)\arrowvert\leq K\sqrt{m}$.

Claim V. Recall the notation
of $\underline{\tilde m}_n^0$
in (<ref>). For  any $j\leq m$,

\begin{align}
\tr\Big(\E_j^*\big(D_j^*(z_1)^{-1}\big)
\E_j^*\big(D_j^*(z_2)^{-1}\big)L_n L_n^\top  \Big)\nonumber\\
\times\bigg[1-\frac{j-1}{m^2}\underline{\tilde m}_n^0(z_1)\underline{\tilde m}_n^0(z_2)\nonumber\\
&\hspace{40mm}\cdot\tr\Big((I+ \underline{\tilde m}_n^0(z_2)
L_nL_n^\top  )^{-1}
(I+\underline{\tilde m}_n^0(z_1)L_nL_n^\top  )^{-1}L_nL_n^\top  \Big)\bigg]\nonumber\\
&=\frac{1}{z_1z_2}\tr\Big((I+ \underline{\tilde m}_n^0(z_2)L_nL_n^\top  )^{-1}L_nL_n^\top  (I+\underline{\tilde m}_n^0(z_1)L_nL_n^\top  )^{-1}L_nL_n^\top  \Big)+R'(z_1,z_2)\label{eq: ClaimV}
\end{align}

with $\arrowvert R'(z_1,z_2)\arrowvert =\mathcal{O}_{\P}(\sqrt{m})$.

Claim VI: We shall conclude the stochastic convergence in (<ref>):

\begin{equation*}
\sum_{j=1}^m\E^*_{j-1}Y_j^*(z_1)Y_j^*(z_2)
\longrightarrow_\mathbb{P}
\begin{cases}
\frac{1}{3}\frac{(\underline{m}_H^0)'''(z_1)}{(\underline{m}_H^0)'(z_1)}
-\frac{1}{2}\Big(\frac{(\underline{m}_H^0)''(z_1)}{(\underline{m}_H^0)'(z_1)}\Big)^2 &\text{ if }z_1=z_2\\
(z_2)\big)^2}-\frac{2}{(z_1-z_2)^2}&\text{ if }z_1\not=z_2.
\end{cases}
\end{equation*}

Note that the expression for $z_2=z_2$ is the continuous extrapolation of the one for $z_1\not=z_2$ for the removable singularities at $z_1=z_2$.

§.§.§ Proofs of Claim I – Claim VI

$ $

Proof of Claim I.
Due to the identity

\begin{align*}
\E_{j-1}^*&\,\Big[(\E_j^*-\E_{j-1}^*)\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big)(\E_j^*-\E_{j-1}^*)\big(\bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big)\Big]\\
&\quad\quad\quad\quad- \E_{j-1}^*\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big)\E_{j-1}^*\big(\bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big),
\end{align*}

the claim follows if

\begin{align*}
\sum_{j=1}^m \E_{j-1}^*\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big)\E_{j-1}^*\big(\bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big)\longrightarrow_{\P}0.
\end{align*}

Employing the operator identity $\E_{j-1}^*=\E_{j-1}^*\E_{X_j^*}^*$, the independence of $\bar{\beta}_j^*(z)$ and $D_j^*(z)$ from $X_j^*$ and the bound

\arrowvert \bar{\beta}_j^*(z)\arrowvert  \leq \frac{\arrowvert z\arrowvert}{\Im(z)},

we deduce
with $A_j(z)=\E_{j-1}^*\big(\bar\beta_j^*(z)D_j^*(z)^{-1}\big)$ and
Lemma <ref> that

\begin{align}
\nonumber
\E\bigg\arrowvert &\sum_{j=1}^m \E_{j-1}^*\big(\bar{\beta}_j^*(z_1)\varepsilon_j^*(z_1)\big)\E_{j-1}^*\big(\bar{\beta}_j^*(z_2)\varepsilon_j^*(z_2)\big)\bigg\arrowvert\\
\nonumber
&\leq \sum_{j=1}^m\E\bigg\arrowvert  \E_{j-1}^*\big(\bar{\beta}_j^*(z_1)\E_{X_j^*}^*\varepsilon_j^*(z_1)\big)\E_{j-1}^*\big(\bar{\beta}_j^*(z_2)\E_{X_j^*}^*\varepsilon_j^*(z_2)\big)\bigg\arrowvert\\
\nonumber
= \sum_{j=1}^m\E\Big\arrowvert \E_{X_j^*}^*\big(r_j^{*\top}A_j(z_1)r_j^*-{1 \over m} \tr A_j(z_1)\big)\E_{X_j^*}^*\big(r_j^{*\top}A_j(z_2)r_j^*-{1 \over m}  \tr A_j(z_2)\big)\Big\arrowvert
\\
&\leq K(z_1,z_2) \Big({\delta_m^2 m  \over \sqrt{n}} +\frac{m^2}{n}\Big)
%\frac{\arrowvert z_1\arrowvert \arrowvert z_2\arrowvert}{\Im(z_1)\Im(z_2)}K \frac{m^2}{n}.
\label{det113}
\end{align}

Proof of Claim II.
Inserting the conditional expectation operator $\E^*$, the proof follows by representing the difference as a martingale difference sum and Burkholder's inequality with the exponent $2$ and Lemma <ref>.

Proof of Claim III.
Since $\arrowvert b_n^*(z)\arrowvert \leq \arrowvert z\arrowvert/\Im(z)$, $\E_{j-1}^*D_j^*(z)^{-1}=\E_{j}^*D_j^*(z)^{-1}$ and

\begin{align*}
\E\bigg\arrowvert  \sum_{j=1}^m&\bigg\{\E_{j-1}^*\Big[\E_j^*\big(\varepsilon_j^*(z_1)\big)\E_j^*\big(\varepsilon_j^*(z_2)\big)\Big]\\
&\quad\quad\quad -\frac{2}{m^2}\tr \Big(L_n^\top  \E_j^*\big(D_j^*(z_1)^{-1}\big)
L_n L_n^\top
\E_j^*\big(D_j^*(z_2)^{-1}\big)L_n^\top  \Big)\bigg\}\bigg\arrowvert \\
&\leq \sum_{j=1}^m\E\bigg\arrowvert \E_{j-1}^*\Big[\E_j^*\big(\varepsilon_j^*(z_1)\big)\E_j^*\big(\varepsilon_j^*(z_2)\big)\Big]\\
&\quad\quad\quad\quad -\frac{2}{m^2}\tr \Big(L_n^\top  \E_j^*\big(D_j^*(z_1)^{-1}L_n L_n^\top  \E_j^*\big(D_j^*(z_2)^{-1}\big) L_n^\top
\Big)\bigg\arrowvert\\
&= \sum_{j=1}^m \E \bigg\arrowvert \E_{X_j^*}^*\bigg[\big(X_j^{*\top}A_{j} (z_1)X_j^*-\tr A_{j}(z_1) )\big(X_j^{*\top}A_{j}(z_2) X_j^*- \tr A_{j}(z_2)) \\
& ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-2 \tr(A_{j} (z_1) A_{j} (z_2) )\bigg]\bigg\arrowvert
\end{align*}

A_{j} (z_k)= \frac{1}{m}L_n^\top  \E_{j-1}^*\big(D_j^*(z_k)^{-1}\big)L_n
,\ \ k=1,2,

the proof is an immediate consequence of Lemma <ref>.

Proof of Claim IV.
The algebraic manipulations in [4] on page 572 provide the representation

\begin{align}\label{eq: Dj}
D_j^*(z_1)^{-1}&= -\Big(z_1 I-  \frac{m-1}{m}b_1^*(z_1) L_n L_n^\top  \Big)^{-1} + b_1^*(z_1)A^*(z_1)+ B^*(z_1)+C^*(z_1)
\end{align}

\begin{align*}
A^*(z_1)&=\sum_{\substack{1\leq i\leq m\\i\not=j}}\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)
L_n L_n^\top  \Big)^{-1}
\Big(r_i^* {r_i^{*}}^\top -\frac{1}{m} L_n L_n^\top  \Big)D_{ij}^*(z_1)^{-1},\\
B^*(z_1)&= \sum_{\substack{1\leq i\leq m\\i\not=j}}\big(\beta_{ij}^*(z_1)-b_1^*(z_1)\big)\Big(z_1I -\frac{m-1}{m}b_1^*(z_1)
L_n L_n^\top  \Big)^{-1} r_i^* {r_i^{*}}^\top D_{ij}^*(z_1)^{-1}
\intertext{and}
C^*(z_1)&=\frac{1}{m}b_1^*(z_1)\Big(z_1 I -\frac{m-1}{m}b_1^*(z_1)
L_n L_n^\top  \Big)^{-1}
L_nL_n^\top \\
&\hspace{40mm}\times \sum_{\substack{1\leq i\leq m\\i\not=j}}\Big(D_{ij}^*(z_1)^{-1}-D_j^*(z_1)^{-1}\Big).
\end{align*}

In order to prove Claim IV, we establish the following steps.

If a possibly random $q\times q$-matrix $M$ satisfies $\Arrowvert M\Arrowvert_{S_{\infty}}\leq c$, then

\begin{align*}
\E\big\arrowvert \tr\big(B^*(z_1)M\big)\big\arrowvert\leq Kc\frac{\arrowvert z_1\arrowvert^2(1+m/(q\Im(z_1)))}{\Im(z_1)^5}\sqrt{m}.
\end{align*}

(ii)If a possibly random $q\times q$-matrix $M$ satisfies $\Arrowvert M\Arrowvert_{S_{\infty}}\leq c$, then

\begin{align*}
\E\big\arrowvert \tr\big(C^*(z_1)M\big)\big\arrowvert\leq Kc\frac{\arrowvert z_1\arrowvert(1+m/(q\Im(z_1)))}{\Im(z_1)^3}.
\end{align*}

(iii) We have

\begin{align}
\tr \Big(\E_j^*&\big(A^*(z_1)\big) L_n L_n^\top
L_n L_n^\top
\Big)\nonumber\\
&= \tr \Big(\E_j^*\big(\check{A}^*(z_1)\big)
L_n L_n^\top
L_n L_n^\top  \Big)+\check{R}(z_1,z_2),\nonumber
\end{align}

\begin{align*}
\check{A}^*(z_1)&=\sum_{1\leq i < j}\Big(z_1I  -\frac{m-1}{m}b_1^*(z_1) L_nL_n^\top  \Big)^{-1} \Big(r_i^* {r_i^{*}}^\top -\frac{1}{m}
L_n L_n^\top  \Big)D_{ij}^*(z_1)^{-1}
\end{align*}

and $\E\arrowvert\check{R}(z_1,z_2)\arrowvert\leq K(z_1,z_2)\sqrt{m}$.

(iv) Moreover,

\begin{align*}
\tr \Big(\E_j^*\big(\check{A}^*(z_1)\big)L_nL_n^\top D_j^*(z_2)^{-1}L_nL_n^\top \Big)= A_1^*(z_1,z_2)+ R''(z_1,z_2),
\end{align*}

where $\E\arrowvert R''(z_1,z_2)\arrowvert\leq K\sqrt{m}$ and

\begin{align*}
&=-\sum_{1\leq i<j}\beta_{ij}^*(z_2) {r_i^{*}}^\top \E_j^*\Big(D_{ij}^*(z_1)^{-1}\Big)L_nL_n^\top D_{ij}^*(z_2)^{-1}r_i^* {r_i^{*}}^\top D_{ij}^*(z_2)^{-1}\\
&\hspace{30mm}\times L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}r_i^*.
%\tr\bigg\{\sum_{\substack{1\leq i,j\leq m\\ i<j}}\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}r_i^* {r_i^{*}}^\top \E_j^*\Big(D_{ij}^*(z_1)^{-1}\Big)\\
%&\hspace{30mm}\times L_nL_n^\top \beta_{ij}^*(z_2)D_{ij}^*(z_2)^{-1}r_i^* {r_i^{*}}^\top L_nL_n^\top
\end{align*}

(v) Each summand of $A_1^*(z_1,z_2)$ satisfies the approximation

\begin{align*}
\beta_{ij}^*(z_2) &{r_i^{*}}^\top \E_j^*\Big(D_{ij}^*(z_1)^{-1}\Big)L_nL_n^\top D_{ij}^*(z_2)^{-1}r_i^* {r_i^{*}}^\top D_{ij}^*(z_2)^{-1}\\
&\hspace{25mm}\times L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}r_i^*\\
& =
b_1^*(z_2)\frac{1}{m^2}\tr\Big[\E_j^*\Big(D_{ij}^*(z_1)^{-1}\Big)L_nL_n^\top D_{ij}^*(z_2)^{-1} L_nL_n^\top \Big]\\
&\hspace{25mm}\times\tr\bigg[D_{ij}^*(z_2)^{-1} L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^\top \bigg]\\
&\quad + R'''(z_1,z_2)
\end{align*}

with $\E\arrowvert R'''(z_1,z_2)\arrowvert\leq Km^{-1/2}$.

Having established these five steps, the proof of Claim IV is conducted as follows. By the triangle and Jensen inequality,

\begin{align*}
\E&\bigg\arrowvert \tr\Big(\E_j^*\big(D_j^*(z_1)^{-1}\big)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\\
&\quad - \tr \bigg[\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^{\top}\E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \bigg]\\
&\quad - b_1^*(z_1)\tr\Big(\E_j^*\big(A^*(z_1)\big)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\bigg\arrowvert\\
&= \E\bigg\arrowvert \E_j^*\Big[\tr \Big(B^*(z_1)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\Big]\\
&\quad \quad\quad+  \E_j^*\Big[\tr \Big(C^*(z_1)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\Big]\bigg\arrowvert\\
&\leq \E\Big\arrowvert\tr \Big(B^*(z_1)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\Big\arrowvert\\
&\quad +   \E\Big\arrowvert\tr \Big(C^*(z_1)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\Big\arrowvert\\
&\leq K(z_1,z_2)\sqrt{m},
\end{align*}

where the last inequality is established in steps (i) and (ii). By Jensen's inequality, the bound $\arrowvert b_1^*(z_1)\arrowvert\leq \arrowvert z_1\arrowvert/\Im(z_1)$ and steps (iii) and  (iv),

\begin{align*}
\E&\bigg\arrowvert  \tr\Big(b_1^*(z_1)\E_j^*\big(A^*(z_1)\big)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)- b_1^*(z_1)\E_j^*\big(A_1^*(z_1,z_2)\big)
\bigg\arrowvert\\
&\leq K(z_1,z_2)\sqrt{m}.
\end{align*}

It remains to prove the approximation

\begin{align}
\E&\bigg\arrowvert b_1^*(z_1)\E_j^*\big(A_1^*(z_1,z_2)\big) +\nonumber \tr\Big(\E_j^*\big(D_j^*(z_1)^{-1}\big)L_nL_n^\top \E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top \Big)\nonumber\\
&\hspace{3mm} \times \frac{j-1}{m^2}b_1^*(z_1)b_1^*(z_2)\tr\bigg(\E_j^*\big(D_j^*(z_2)^{-1}\big)L_nL_n^\top
\nonumber \Big(z_1 I_q-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^\top \bigg)\bigg\arrowvert\nonumber\\
&\leq K(z_1,z_2)\sqrt{m}.\label{eq: intermed}
\end{align}

Applying the approximation of step (v), we deduce

\begin{align*}
\E\Big\arrowvert b_1^*(z_1)\E_j^*\big(A_1^*(z_1,z_2)\big)-b_1^*(z_1)\E_j^*\big(\tilde{A}_1^*(z_1,z_2)\big)\Big\arrowvert\leq K(z_1,z_2)\sqrt{m},
\end{align*}

\begin{align*}
\tilde{A}_1^*&(z_1,z_2)\\
&=-\sum_{1\leq i<j\leq m}b_1^*(z_2)\frac{1}{m^2}\tr\Big[\E_j^*\Big(D_{ij}^*(z_1)^{-1}\Big)L_nL_n^\top D_{ij}^*(z_2)^{-1} L_nL_n^\top \Big]\\
&\hspace{5mm}\times\tr\bigg[D_{ij}^*(z_2)^{-1} L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^\top \bigg].
\end{align*}

By Lemma 2.10 of [3], we may successively replace $\E_j^*\big(D_{ij}^*(z)^{-1}\big)$ by $\E_j^*\big(D_j^*(z)^{-1}\big)$ within the traces

\begin{align*}
&\bigg\arrowvert \tr\Big[\E_j^*\Big(D_{ij}^*(z_1)^{-1}\Big)L_nL_n^\top D_{ij}^*(z_2)^{-1} L_nL_n^\top \Big]\\
&\hspace{5mm}\times\tr\bigg[D_{ij}^*(z_2)^{-1} L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^\top \bigg]\\
&\ \  -  \tr\Big[\E_j^*\Big(D_{j}^*(z_1)^{-1}\Big)L_nL_n^\top D_{j}^*(z_2)^{-1} L_nL_n^\top \Big]\\
&\hspace{5mm}\times\tr\bigg[D_{j}^*(z_2)^{-1} L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^\top \bigg]\bigg\arrowvert\\
&\leq K(z_1,z_2)\cdot m,
\end{align*}

which proves (<ref>) and therefore the equation

\begin{align}\label{eq: D.5200}
\tr&\Big(\E_j^*\big(D_j^*(z_1)^{-1}\big)
L_n L_n^\top
\E_j^*\big(D_j^*(z_2)^{-1}\big)
L_n L_n^\top
\Big)\nonumber\\
&\ \ \ \times \bigg[1+\frac{j-1}{m^2}b_1^*(z_1)\tr\bigg(
\E_j^*\big(D_{j}^*(z_2)^{-1}\big) L_n L_n^\top
\Big(z_1 I-\frac{m-1}{m}b_1^*(z_1)
L_n L_n^\top
\Big)^{-1}
L_n L_n^\top
\bigg)\bigg]\nonumber\\
&= -\tr\bigg[\Big(z_1 I-\frac{m-1}{m}b_1^*(z_1) L_n L_n^\top  \Big)^{-1}
L_n L_n^\top
\E_j^*\big(D_{j}^*(z_2)^{-1}\big)
L_n L_n^\top
\bigg]
~+\tilde R(z_1,z_2)
\end{align}

with $\E\arrowvert \tilde R(z_1,z_2)\arrowvert\leq K\sqrt{m}$. Now, inserting   the representation (<ref>) into (<ref>), this time for $D_j^*(z_2)^{-1}$, and using (i) and (ii) together with the bound

\begin{align*}
\E\big\arrowvert \tr\big(A(z_2)M\big)\big\arrowvert&\leq \sum_{i\not=j}\E^{1/2}\bigg\arrowvert {r_i^*}^{\top}D_{ij}^*(z_2)^{-1}M\Big(z_2I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}r_i^*\\
&\ \ \ \ \ \ -\frac{1}{m}\tr\Big(D_{ij}^*(z_2)^{-1}M\Big(z_2I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}\Big)\bigg\arrowvert^2\\
&\leq K(z_2,\Arrowvert M\Arrowvert_{S_{\infty}})\sqrt{m}\Big(1+\frac{m}{\sqrt{n}}\Big)
\end{align*}

for non-random matrices $M$ of uniformly bounded spectral norm by   Proposition <ref>, Claim IV is verified.

* Proof of (i). First, applying the same reasoning as for inequality (2.10) in Bai and Silverstein (2004), we obtain the spectral norm bound

\begin{align}\label{eq: snb}
\Big\Arrowvert\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}\Big\Arrowvert_{S_{\infty}}\leq \frac{1+m/(q\Im(z_1))}{\Im(z_1)}.
\end{align}

By the Cauchy-Schwarz inequality, the upper bounds $\arrowvert \beta_{12}^*(z)\arrowvert,\arrowvert b_1^*(z)\arrowvert\leq \arrowvert z\arrowvert/\Im(z)$ and  $\Arrowvert D_{ij}^*(z)^{-1}\Arrowvert_{S_{\infty}}\leq 1/\Im(z)$, (<ref>),
and Lemma <ref>,

\begin{align*}
\E\big\arrowvert \tr\big(B^*(z_1)M\big)\big\arrowvert
& \leq m\E^{1/2}\big\arrowvert\beta_{12}^*(z_1)-b_1^*(z_1)\big\arrowvert^2\\
&\hspace{10mm}\times \E^{1/2}\Big\arrowvert  {r_i^{*}}^\top D_{ij}^*(z_1)^{-1}M\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}r_i^*\Big\arrowvert^2\\
\leq \frac{\arrowvert z_1\arrowvert^2}{(\Im(z_1))^2}
\E^{1/2}\bigg[\big\arrowvert r_1^{*\top}r_1^*\big\arrowvert^2\big\Arrowvert D_{ij}^*(z_1)^{-1}\big\Arrowvert_{S_{\infty}}^2\Arrowvert M\Arrowvert_{S_{\infty}}^2\\
&\hspace{30mm}\times\Big\Arrowvert \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}\Big\Arrowvert_{S_{\infty}}^2\bigg]\\
&\leq   K  \frac{\arrowvert z_1\arrowvert^2(1+m/(q\Im(z_1)))}{\Im(z_1)^5}\sqrt{m}.
\end{align*}

* Proof of (ii). By the cyclic invariance of the trace, the bound $\arrowvert b_1^*(z)\arrowvert\leq\arrowvert z\arrowvert/\Im(z)$, the submultiplicativity of $\Arrowvert\cdot\Arrowvert_{S_{\infty}}$, (<ref>), and Lemma 2.6 in silversteinbai1995,

\begin{align*}
\E\big\arrowvert \tr\big(C^*(z_1)M\big)\big\arrowvert
& \leq \sum_{\substack{1\leq i\leq m\\i\not=j}}\E\bigg[\frac{1}{m}\big\arrowvert b_1^*(z_1)\big\arrowvert  \bigg\arrowvert  \tr\bigg( \Big(D_{ij}^*(z_1)^{-1}-D_j^*(z_1)^{-1}\Big)M\\
&\hspace{40mm}\times  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}L_nL_n^\top \bigg)\bigg\arrowvert\bigg]\\
&\leq Kc\frac{\arrowvert z_1\arrowvert(1+m/(q\Im(z_1)))}{\Im(z_1)^3}.
\end{align*}

* Proof of (iii). Because of $\E_j^*=\E_j^*\E_{X_i^*}^*$ for $i>j$, the triangle and Cauchy-Schwarz inequality,

\begin{align*}
\E\bigg\arrowvert &\tr \Big(\E_j^*\big(A^*(z_1)\big)L_nL_n^\top D_j^*(z_2)^{-1}L_nL_n^\top \Big) -  \tr \Big(\E_j^*\big(\check{A}^*(z_1)\big)L_nL_n^\top D_j^*(z_2)^{-1}L_nL_n^\top \Big)\bigg\arrowvert\\
& =\E\bigg\arrowvert \sum_{i>j}\tr\Big[\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}\E_{X_i^*}^*\Big(r_i^* {r_i^{*}}^\top -\frac{1}{m}L_nL_n^\top \Big)\\
&\hspace{8mm}\times \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top D_j^*(z_2)^{-1}L_nL_n^\top \Big]\bigg\arrowvert\\
& \leq \frac{1}{m}\sum_{i>j}\E^{1/2}\Big\Arrowvert \frac{1}{n}\sum_{i=1}^n Y_iY_i^\top-L_nL_n^\top\Big\Arrowvert_{S_2}^2\\
\E^{1/2}\Big\Arrowvert
\E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top D_j^*(z_2)^{-1}L_nL_n^\top \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}\Big\Arrowvert_{S_2}^2.
\end{align*}

Applying (<ref>) and the estimates $\Arrowvert D_{ij}^*(z)\Arrowvert_{S_{\infty}}$, $ \Arrowvert D_{j}^*(z)\Arrowvert_{S_{\infty}}\leq 1/\Im(z)$,

\begin{align*}
\E^{1/2}&\Big\Arrowvert
\E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top D_j^*(z_2)^{-1}L_nL_n^\top  \times\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}\Big\Arrowvert_{S_2}^2\\
&\leq K(z_1,z_2)\sqrt{m},
\end{align*}

\begin{align}
\label{det114}
\E^{1/2}\Big\Arrowvert \frac{1}{n}\sum_{i=1}^n Y_iY_i^\top - L_nL_n^\top \Big\Arrowvert_{S_2}^2=\mathcal{O}\Big(\frac{m}{\sqrt{n}}\Big),
\end{align}

since both matrices are of dimension $q \times q$.
This  proves (iii).

* Proof of (iv).
Adding and subtracting $D_{ij}^*(z_2)$ and applying the Sherman-Morrison formula to the difference $D_j^*(z_2)^{-1}-D_{ij}^*(z_2)^{-1}$ in the subsequent expression $A_1^*(z_1,z_2)$, we obtain the decomposition

\begin{align*}
\tr \Big(\E_j^*\big(\check{A}^*(z_1)\big)&L_nL_n^\top  D_j^*(z_2)^{-1}L_nL_n^\top  \Big) =A_1^*(z_1,z_2) + A_2^*(z_1,z_2)+A_3^*(z_1,z_2)
\end{align*}

\begin{align*}
A_1^*(z_1,z_2)&=-\sum_{1\le i<j} \beta_{ij}^*(z_2) {r_i^{*}}^\top \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1}r_i^*\\
{r_i^{*}}^\top D_{ij}^*(z_2)^{-1}L_nL_n^\top  \Big(z_1I_q-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}r_i^*\\
A_2^*(z_1,z_2)&= -\tr\bigg\{\sum_{1\leq i<j}\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}\frac{1}{m}L_nL_n^\top  \\
&\hspace{15mm}\times\E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  \Big(D_j^*(z_2)^{-1}-D_{ij}^*(z_2)^{-1}\Big)L_nL_n^\top  \bigg\}\\
A_3^*(z_1,z_2)&=\tr\bigg\{\sum_{1\leq i<j}\Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}\Big(r_i^* {r_i^{*}}^\top -\frac{1}{m}L_nL_n^\top  \Big)\\
&\hspace{10mm}\times \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1}L_nL_n^\top  \bigg\}.
\end{align*}

By Lemma 2.6 in silversteinbai1995 and (<ref>),

\begin{align*}
\big\arrowvert A_2^*(z_1,z_2)\big\arrowvert \leq K\frac{1+m/(q\Im(z_1))}{(\Im(z_1))^2}.
\end{align*}

Analogously to the proof of (i), we find

\begin{align*}
\E\big\arrowvert A_3^*(z_1,z_2)\big\arrowvert \leq K\frac{1+m/(q\Im(z_1))}{(\Im(z_1))^3}\sqrt{m}.
\end{align*}

* Proof of (v). By the Cauchy-Schwarz inequality and
Lemma   <ref>, we have

\begin{align*}
\hspace{-5mm}\E\bigg\arrowvert &\big(\beta_{ij}^*(z_2)-b_1^*(z_2)\big) {r_i^{*}}^\top \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_n L_n^\top  D_{ij}^*(z_2)^{-1}r_i^*\\
&\hspace{20mm}\times  {r_i^{*}}^\top D_{ij}^*(z_2)^{-1}L_n L_n^\top  \Big(z_1I  -\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}r_i^*\bigg\arrowvert\\
\leq \E^{1/2}\big\arrowvert \beta_{ij}^*(z_2)-b_1^*(z_2)\big\arrowvert^2\cdot \E^{1/2}\bigg\arrowvert  {r_i^{*}}^\top \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1}r_i^*
\\
\ \times  {r_i^{*}}^\top D_{ij}^*(z_2)^{-1}L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top \Big)^{-1}r_i^*\bigg\arrowvert^2
\\
&\leq K(z_1,z_2)\frac{1}{\sqrt{m}},
\end{align*}

where the last inequality follows by a further application of the Cauchy-Schwarz inequality,
(<ref>), the bound
$|{r_1^{*}}^\top  C r^*_1| \leq \| C\|_{S_\infty}  \| r_1^* \|^2$  and  the fact that $\E  \| r_1^* \|^p \leq  c $, which follows from (<ref>).

\begin{align*}
\hspace{-5mm}\E\bigg\arrowvert &b_1^*(z_2) {r_i^{*}}^\top \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1}r_i^*\\
&\hspace{20mm}\times  {r_i^{*}}^\top D_{ij}^*(z_2)^{-1}L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}r_i^*\\
&- b_1^*(z_2)\frac{1}{m^2}\tr\Big[\E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1} L_nL_n^\top  \Big]\\
&\hspace{5mm}\times\tr\bigg[D_{ij}^*(z_2)^{-1} L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}L_nL_n^\top  \bigg]\bigg\arrowvert\\
&\leq \E\bigg\arrowvert b_1^*(z_2)\frac{1}{m}\tr\Big[\E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1} L_nL_n^\top  \Big]\\
&\hspace{10mm}\times\bigg(  {r_i^{*}}^\top D_{ij}^*(z_2)^{-1}L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}r_i^*\\
&\hspace{15mm}-\frac{1}{m}\tr\bigg[D_{ij}^*(z_2)^{-1} L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}L_nL_n^\top  \bigg]\bigg)\bigg\arrowvert\\
+ \E\bigg\arrowvert b_1^*(z_2)\bigg( {r_i^{*}}^\top \E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1}r_i^*
\\
-\frac{1}{m}\tr\Big[\E_j^*\big(D_{ij}^*(z_1)^{-1}\big)L_nL_n^\top  D_{ij}^*(z_2)^{-1} L_nL_n^\top  \Big]\bigg)
\\
\times
{r_i^{*}}^\top D_{ij}^*(z_2)^{-1}L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}r_i^*
%\tr\bigg[D_{ij}^*(z_2)^{-1} L_nL_n^\top  \Big(z_1I-\frac{m-1}{m}b_1^*(z_1)L_nL_n^\top  \Big)^{-1}L_nL_n^\top  \bigg]
\bigg\arrowvert.
\end{align*}

By an application of the Cauchy-Schwarz inequality to the first term,  (<ref>), the bounds $\arrowvert b_1^*(z)\arrowvert\leq \arrowvert z\arrowvert/\Im(z)$ and $\Arrowvert D_{ij}^*(z)^{-1}\Arrowvert_{S_{\infty}}\leq 1/\Im(z)$,  and  Proposition  <ref>, the right-hand side of the last inequality is bounded by

Proof of Claim V.
We shall prove

\begin{equation}
\big\arrowvert b_1^*(z) + z\underline{\tilde m}_n^0(z)\big\arrowvert=\mathcal{O}_{\P}(m^{-1/2}).\label{eq: VHb}
\end{equation}

\begin{align*}
\big\arrowvert b_1^*(z)-b_n^*(z)\big\arrowvert &=\Big\arrowvert b_1^*(z)b_n^*(z)\frac{1}{m}\E^*\tr\Big(L_n L_n^\top \big(D_1^*(z)^{-1}-D_{12}^*(z)^{-1}\big)\Big)\Big\arrowvert\\
&\leq \frac{\arrowvert z\arrowvert^2}{\big(\Im(z)\big)^3}\frac{\Arrowvert L_n \Arrowvert_{S_{\infty}}^2}{m}
\end{align*}

by the inequalities $\arrowvert b_1^*(z)\arrowvert, \arrowvert b_n^*(z)\arrowvert\leq \arrowvert z\arrowvert/\Im(z)$ and Lemma 2.6 in silversteinbai1995. Next, note that the estimate
(<ref>) is still valid under the truncation scheme used in section. This follows by showing the inequality
$\E \arrowvert \gamma_1^* (z) \arrowvert^2=\mathcal{O}
( { 1 \over m }
+ \frac{m}{n} )$
by an application of Proposition <ref> with $p=2$ and using the same arguments following equation (<ref>). Therefore, it follows that

\E\arrowvert b_n^*(z)-\E^*\beta_1^*(z)\arrowvert=\mathcal{O}(m^{-1/2}),

and by  (<ref>)
we have
$\E^*\beta_1^*(z)= - z  \E^* \underline{m}_n^*(z)$. Together with (<ref>),
which is also  valid under the truncation scheme used in section (see Remark <ref>),
we conclude (<ref>). Replacing successively $b_1^*(z_i)$ in the left-hand side of (<ref>) by $ z_i  \underline{\tilde{m}}_n^0(z_i)$, $i=1,2$, and employing Lemma 2.6 in silversteinbai1995, the proof of Claim V is completed.

Proof of Claim VI.
Note that we may rewrite (<ref>) as

\begin{align*}
{1 \over q}
\tr\Big(\E_j^*&\big(D_j^*(z_1)^{-1}\big)
L_n L_n^\top
\E_j^*\big(D_j^*(z_2)^{-1}\big)L_n L_n^\top  \Big)\\
&\quad\times\bigg[1+\frac{j-1}{m^2}\underline{\tilde m}_n^0(z_1)\underline{\tilde m}_n^0(z_2)\int \frac{t^2}{\big(1+t\underline{\tilde m}_n^0(z_1)\big)\big(1+t \underline{\tilde m}_n^0(z_2)\big)}d\mu^{\tilde \Sigma_n}(t)\bigg]\\
&= {q \over m } \frac{1}{z_1z_2}\int  \frac{t^2}{\big(1+t\underline{\tilde m}_n^0(z_1)\big)\big(1+ t\underline{\tilde m}_n^0(z_2)\big)}d\mu^{\tilde \Sigma_n}(t) + \mathcal{O}_{\P}(m^{1/2}).
\end{align*}

As for (2.19) in [4],

\begin{align*}
\limsup_n\bigg\arrowvert {q \over m} \frac{1}{z_1z_2}\int  \frac{t^2}{\big(1+ t\underline{\tilde m}_n^0(z_1)\big)\big(1+ t \underline{\tilde m}_n^0(z_2)\big)}d\mu^{\tilde \Sigma_n}(t)  \bigg\arrowvert <1.
\end{align*}

a_n(z_1,z_2)={q \over m}
\frac{1}{z_1z_2}\int  \frac{t^2}{\big(1+t\underline{\tilde m}_n^0(z_1)\big)\big(1+  t \underline{\tilde m}_n^0(z_2)\big)}d\mu^{\tilde \Sigma_n}(t),

can be written as

a_n(z_1,z_2)\frac{1}{m}\sum_{j=1}^m \Big(1-\frac{j-1}{m}a_n(z_1,z_2)\Big)^{-1} + O_{\P}(m^{-1/2}).

We now use the Representative  Subpopulation Condition <ref> to
conclude that

\begin{equation}\label{AR10000}
a_n(z_1,z_2  ) \longrightarrow
\frac{c}{z_1z_2}\int  \frac{t^2}{\big(1+t\underline{ m}^{0}_H (z_1)\big)\big(1+  t \underline{ m}^{0}_H(z_2)\big)}d H(t),
\end{equation}

in probability. For this purpose,  we note that it follows from (<ref>), (<ref>)  and Condition <ref>
that $ \mu^{\tilde \Sigma_n}   \Rightarrow   H  $
in probability and  $\underline{\tilde m}_n^0(z) \to  \underline{ m}^{0}
_{H}(z)$ in probability, see equation (<ref>).  Together with the fact that the sequence of  bounded continuous functions

t\mapsto \frac{t^2}{\big(1+t\underline{ \tilde  m}_{n}^0 (z_1)\big)\big(1+  t \underline{\tilde  m}_n^0(z_2)\big)}

converges in probability uniformly on compacts to its limit, this implies (<ref>).
The final steps in the proof of (<ref>) are then the same as on page 578 in baisil2004 and omitted for the sake of brevity.

§.§.§ Proof of Proposition <ref>
under the unconditional  tightness assumption on $M_n^{*} $

We start with the following lemma whose assumption can be deduced from the unconditional tightness of the sequence $\mathcal{L} ({\widehat M_n^*} )$
by an  application of Markov's inequality.
Subsequently, we use the abbreviation $U:=(X_1,, X_2, \dots  \Pi_1, \Pi_2, \dots  )$ and denote by $\P^{\widehat{M}_n^*}(\cdot\,\arrowvert u)$ the conditional distribution $\mathcal{L}\big((\widehat{M}_n^*(z))_{z\in\mathcal{C}}\arrowvert  U=u  \big)$.

Assume that for every $\varepsilon>0$ and $\eta>0$, there exists some compact set $K$ such that

\begin{equation}\label{eq: tt}
\sup_n\P\Big(\P^{\widehat{M}_n^*}(K^c\arrowvert U )>\varepsilon\Big)< \eta.
\end{equation}

Then there exists some array of measurable sets $(A_{m,n})_{m,n\in\N}$ with  $\sup_n\P(U\in A_{m,n}^c)\rightarrow 0$ as $ m\rightarrow\infty$,
such that the family

\Big(\P^{\widehat{M}_n^*}(\cdot\arrowvert u): u\in A_{m,n},n\in\N \Big)

is tight for every $m\in\N$.

$\varepsilon\searrow 0$ be some null sequence and $\eta_k=2^{-k}$, $k\in\N$. Let $(K_k)$ be some increasing sequence of compacts such that $K_k$ satisfies (<ref>) for $\varepsilon_k$ and $\eta_k$. Define

A_{m,n}^c:=\bigcup_{k\geq m}\Big\{u: \P^{\widehat{M}_n^*}(K_k^c\arrowvert u)>\varepsilon_k\Big\}.

Every measure $\P^{\widehat{M}_n^*}(\cdot\arrowvert u)$ with $u\in A_{m,n}$  satisfies $\P^{\widehat{M}_n^*}(K_k^c\arrowvert u)\leq\varepsilon_k$, $k\geq m$. Hence, the family

\Big(\P^{\widehat{M}_n^*}(\cdot\arrowvert u): u\in A_{m,n},n\in\N\Big)

is tight for every $m\in\N$ and by the sigma-subadditivity,

\sup_n\P(A_{n,m}^c )\leq\sum_{k\geq m}\frac{1}{2^k}\longrightarrow 0\ \ \text{as }m\rightarrow\infty.

Assume now that the condition of Lemma <ref> holds. Then the weak convergence of the finite dimensional distribution in probability implies
the assertion of Proposition <ref>, which can be seen as follows.
We define for any measure $R$ on the Borel $\sigma$-field on the continuous function on $\CC$ the operation

R\cdot \mathds{1}_A (u)  = \begin{cases}
R & \text{ if } u \in A \\
\delta_0 & \text{  otherwise. }
\end{cases}

\begin{align}
\mathbb{P} \big ( d_{\rm BL}  &\big ( \mathbb{P}^{\widehat{M}_n^*}(\cdot | U ), \mathcal{L} (Z) \big ) > \varepsilon \big )\nonumber\\
&  \leq
\mathbb{P} \big ( d_{\rm BL}  \big ( \mathbb{P}^{\widehat{M}_n^*}(\cdot | U ), \mathcal{L} (Z) \big ) > \varepsilon , ~ U \in  A_{m.n}  \big ) +
\mathbb{P} \big ( U \in  A_{m.n}^c\big ) \nonumber \\
&  =
\mathbb{P} \big ( d_{\rm BL}  \big ( \mathbb{P}^{\widehat{M}_n^*}(\cdot | U ) \mathds{1}_{A_{m,n} }(U)  , \mathcal{L} (Z) \mathds{1}_{A_{m,n} }(U) \big ) > \varepsilon , ~  U \in  A_{m.n}  \big ) +
\mathbb{P} \big ( U \in  A_{m.n}^c\big ) \nonumber \\
&  \leq
\mathbb{P} \big ( d_{\rm BL}  \big ( \mathbb{P}^{\widehat{M}_n^*}(\cdot | U ) \mathds{1}_{A_{m,n} }(U)  , \mathcal{L} (Z) \mathds{1}_{A_{m,n} }(U) \big ) > \varepsilon  \big ) +
\mathbb{P} \big (   U \in  A_{m.n}
^c\big )
\label{det104}
\end{align}

By Lemma <ref>, the family $ \big \{ \mathbb{P}^{\widehat{M}_n^*}(\cdot | u ) \mathds{1}_{A_{m,n} }(u) \big \}_{n \in \mathbb{N}}$
is tight uniformly in $u$ for every $m \in \mathbb{N}$. The same holds true for the sequence $\big \{ \mathcal{L} (Z) \mathds{1}_{A_{m,n} }(u) \}_{n \in \mathbb{N}} $ which attains only the measures  $\mathcal{L} (Z)$ and $\delta_0$.
Now, let $(n_k)$ denote an arbitrary subsequence of $(n)$. Then for every $u$, there exists a further subsequence $(n_k') = (n_k' (u)) $ of $(n_k)$   such that

\begin{align*}
\mathbb{P}^{\widehat{M}_{n_k'}^*}(\cdot | u ) \mathds{1}_{A_{m,n_k'} }(u) & \Rightarrow \nu (u) \\
\mathcal{L} (Z) \mathds{1}_{A_{m,n_k'} }(u)
& \Rightarrow \bar \nu (u)
\end{align*}

for some measures $\nu (u)  $ and $\bar \nu (u) $. Moreover, by the weak convergence of the finite dimensional distributions in probability established in Section <ref>, we can choose this  subsequence such that additionally,

\mathcal{L} \big ( (\widehat M_n^* (z_1), \ldots , \widehat M_n^* (z_\ell ) ) |   u \big )  \Rightarrow
\mathcal{L} \big ( (Z (z_1), \ldots , Z (z_\ell ) ) |   u \big )

for all $z_1, \ldots , z_\ell   \in (\mathbb{Q} + i\mathbb{Q}^+ ) \cap \CC$,  $\ell \in \mathbb{N}$ and for all $u \in A $, where $A $  is a measurable set with $\mathbb{P} (A) =1 $.
If $\bar \nu (u)  = \delta_0$,
then there exists a $k_0 = k_0(u)  $ such that $u \in A_{m,n_k'}^c$ for all $k \geq k_0$. In this case,
$$d_{\rm BL}  \bigg ( \mathbb{P}^{\widehat{M}_{n_k'}^*}(\cdot | u ) \mathds{1}_{A_{m,n_k'} }(u)  , \ \mathcal{L} (Z) \mathds{1}_{A_{m,n_k'} }(u) \bigg ) =0 \ \ \text{for all $k \geq k_0$.}$$

Otherwise, if $\bar \nu (u) = \mathcal{L} (Z)$, there exists a $k_1=k_1(u) $ such that
$u \in A_{m,n_k'} $ for all $k \geq k_1$. Consequently it follows  that $ \nu (u) = \bar \nu (u)$
which implies
$$d_{\rm BL}  \bigg ( \mathbb{P}^{\widehat{M}_{n_k'}^*}(\cdot | u ) \mathds{1}_{A_{m,n_k'} }(u)  , \ \mathcal{L} (Z) \mathds{1}_{A_{m,n_k'} }(u) \bigg)
\mathds{1}_{A} (u)  \rightarrow 0 .$$

Summarizing, we have shown that for any subsequence $(n_k)$ and any $u$, there exists some further subsubsequence $(n_k'(u))$ such that

d_{\rm BL}  \bigg ( \mathbb{P}^{\widehat{M}_{n_k'}^*}(\cdot | u )   ,\  \mathcal{L} (Z)  \bigg)
\mathds{1}_{A\cap A_{m,n_k'(u)}} (u)  \rightarrow 0.

d_{\rm BL}  \bigg ( \mathbb{P}^{\widehat{M}_{n}^*}(\cdot | u )   ,\  \mathcal{L} (Z)  \bigg)\mathds{1}_{A\cap A_{m,n}}(u)    \rightarrow 0
~~\text { for every } u ~.

By dominated convergence and $\P(A)=1$, this in turn implies

d_{\rm BL}  \bigg ( \mathbb{P}^{\widehat{M}_{n}^*}(\cdot | U )   ,\  \mathcal{L} (Z)  \bigg)\mathds{1}_{A_{m,n}}(U)    \rightarrow_{\P} 0.  $$

Thus, it follows from (<ref>) that

\limsup_{n\to \infty}
\mathbb{P} \big ( d_{\rm BL}  \big ( \mathbb{P}^{\widehat{M}_n^*}(\cdot | U ), \mathcal{L} (Z) \big ) > \varepsilon \big )
\leq  \sup_{n \in \mathbb{N}}
\mathbb{P} \big ( U \in  A_{m.n}^c\big )

As the left-hand side does not depend on $m$, the assertion of Proposition <ref>  now follows by taking the limit $ m \to \infty$.

§.§.§ Unconditional tightness of the process $M_n^{*} $

We only need to verify that assumption (<ref>) in Lemma  <ref> is satisfied. But this can easily be deduced from the unconditional tightness of the sequence $\mathcal{L} ({\widehat M_n^*} )$
by an  application of Markov's inequality.

Therefore, it is  sufficient to prove the moment condition  (12.51) in
billingsley1999, that is

\begin{align}
\label{det66}
\sup_{n}
\sup_{ z_1,z_2}    \mathbb{E}  \Big [
\mathbb{E}^*   \Big [
{ |  \widehat M_n^{*}  (z_1)  - \widehat M_n^{*}  (z_2) |^{1+r}   \over |z_1 - z_2 |^{1+r}  }
\Big ] \Big ] \leq K
\end{align}

for a sufficiently small number $r>0$, which will be specified below.
Using similar arguments as in baisil2004, p.582 we obtain

\begin{align}
\label{det1}
{   \widehat M_n^{*}  (z_1)  - \widehat M_n^{*}  (z_2)    \over z_1 - z_2   }
=  H_{1n}  + H_{2n}  + H_{3n}
\end{align}

\begin{align*}
H_{1n}  &=
\sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_1) \beta_j^* (z_2) \big (r_j{^*}^\top (D_j^* (z_1))^{-1} (D_j^* (z_2))^{-1} r_j^* \big ) ^2
% \label{det1a}
\\
H_{2n}  &=
-   \sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_1)  r_j{^*}^\top (D_j^* (z_1))^{-2} (D_j^* (z_2))^{-1} r_j^*
% \label{det1b}
\\
H_{3n}  &= -
\sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_2)  r_j{^*}^\top (D_j^* (z_2))^{-2} (D_j^* (z_1))^{-1} r_j^*
% \label{detc}
\end{align*}

We  will show that  the absolute moments of $1+r$ of all  three terms are bounded,
where $r$ will be specified below and
we have to use different arguments
for the two different assumptions in  Theorem <ref> and Proposition <ref>.
In particular,  from  now on the arguments differ from those given in the tightness proof of
baisil2004 as we have to address the specific properties of  the bootstrap.

Proof of tightness  under the assumptions of Proposition <ref>

begin deriving a bound for the quantity $H_{2n}$
the identity

\begin{equation*}
%    \label{det61}
b_n^*(z) = \beta_1^*(z) +  \beta_1^*(z)  b_n^*(z) \gamma_1^*(z) ,
\end{equation*}

where $\beta_1^*(z)$
and $\gamma_1^*(z)$ are defined in (<ref>) and (<ref>), respectively. It  then follows by a
straightforward calculation that

H_{2n}  = b_n^*(z_1)W_1^* - b_n^*(z_1)W_2^*

\begin{align*}
W_1^*  &=   \sum_{j=1}^m  \big (  \mathbb{E}^*_j
- \mathbb{E}^*_{j-1} \big)
r_j{^*}^\top D_j^* (z_1)^{-2} D_j^* (z_2)^{-1} r_j^* ~,
\\
W_2^* &=
\sum_{j=1}^m  \big (  \mathbb{E}^*_j
- \mathbb{E}^*_{j-1} \big)  \Big [
\beta_j^*(z_1)
r_j{^*}^\top D_j^* (z_1)^{-2} D_j^* (z_2)^{-1} r_j^*  \gamma_j^*(z_1)
\Big   ] ~.
\end{align*}

Note that

\begin{align}
\label{det41}
\E  | H_{2n} |^{1+r} \leq  C \big (
\E  | b_n^*(z_1)W_1^*  |^{1+r}  +  \E  |  b_n^*(z_1)W_2^* |^{1+r}\big ) ~,
\end{align}

and that
Hölder's inequality gives for both terms on the right-hand side

\begin{align}
\label{det42}
\mathbb{E} \big [ | b_n^*(z_1)  W_j^* |^{1+r}
\big ] \leq \big (  \mathbb{E} \big [ | b_n^*(z_1) |^{2(1+r)/(1-r)} \big ]  \big )^{(1-r)/2} \big (
\E \big [ | W_j^* |^{2}
\big ] \big )^{(1+r)/2} ~~~~~~~~~(j=1,2).
\end{align}

We will show at the end of this section that  both factors s on the
right-hand side are bounded, that is

\begin{align}
\label{det43a}
\mathbb{E} \big [ | b_n^*(z) |^{q(r)}\big ]  &\leq C
\\
\E | W_j^* |^{2}
&\leq C  ~~~~~~~~~(j=1,2)
\label{det43b}
\end{align}

for $z \in  \mathcal{C}_n$ and

\begin{align}
\label{det63}
q(r) = {2(1+r) \over 1-r}  ~~\text{  with  ~} r \in  \Big ( 0 ,  {\eta \over 8 + \eta } \Big )
\end{align}

and $\eta $ is the constant in the assumptions of Proposition <ref>. Therefore, observing (<ref>) it follows that $\E |  H_{2n} |^{1+r} \leq c
$  for some constant $c$. The  term $H_{3n}$ in the decomposition (<ref>)
can be bounded in the same way, which means

\begin{align}
\label{det65a}
\E |  H_{jn} |^{1+r} \leq c~~~j=2,3 ~.
\end{align}

To derive a bound for  the first term $H_{1n}$ we use the same arguments as in baisil2004 and obtain the representation

\begin{align}
\label{det65b}
H_{1n} =
b_n^*(z_1)b_n^*(z_2)Y_1^* -b_n^*(z_2)Y_2^* -b_n^*(z_1)b_n^*(z_2)Y_3^*
\end{align}

\begin{align*}
Y_1^* &= \sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big )
\Big [ \big (r_j{^*}^\top (D_j^* (z_1))^{-1} (D_j^* (z_2))^{-1} r_j^* \big ) ^2
\\
& ~~~~~~~~~~~~~~~~~~~~~
- \Big ( {1 \over m }  {\rm tr } \big ( L_n (D_j^* (z_1))^ {-1} (D_j^* (z_2) )^{-1}
L_n^\top  \big ) \Big )^2 \Big ]
\\
Y_2^* &=   \sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_1) \beta_j^* (z_2) \big (r_j{^*}^\top (D_j^* (z_1))^{-1} (D_j^* (z_2))^{-1} r_j^* \big ) ^2 \gamma_j^* (z_2)  \\
Y_3^* &=   \sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_1) \beta_j^* (z_2) \big (r_j{^*}^\top (D_j^* (z_1))^{-1} (D_j^* (z_2))^{-1} r_j^* \big ) ^2 \gamma_j^* (z_1) \\
\end{align*}

This yields

\begin{align}
\label{det65c}
\mathbb{E} [ |H _{1n} |^{1+r} ]
& \leq C (\big (  \mathbb{E} \big [ | b_n^*(z_1) b_n^*(z_2) |^{q(r) } \big ]  \big )^{(1-r)/2} \big (
\E \big [ | Y_1^* |^{2}
\big ] \big )^{(1+r)/2}   \\
\nonumber
& + (\big (  \mathbb{E} \big [ |  b_n^*(z_2) |^{q(r) } \big ]  \big )^{(1-r)/2} \big (
\E \big [ | Y_2^* |^{2}
\big ] \big )^{(1+r)/2}  \\
\nonumber
& +
(\big (  \mathbb{E} \big [ | b_n^*(z_1) b_n^*(z_2) |^{q(r) } \big ]  \big )^{(1-r)/2} \big (
\E \big [ | Y_3^* |^{2}
\big ] \big )^{(1+r)/2}  ,
\end{align}

where $q(r)$ is defined in (<ref>).
At the end of this proof  we will show
that all expectations in this expression are bounded, that

\begin{align}
\label{det65d}
\mathbb{E} \big [ |  b_n^*(z_1)b_n^*(z_2) |^{q(r)} \big ] &\leq C  \\
\mathbb{E} \big [ |Y_j^* |^2 \big ]  & \leq C. ~~~~(j=1,2,3)
\label{det65e}
\end{align}

for some constant $C>0$.
It then follows that $\mathbb{E} [ |H _{1n} |^{1+r} ] \leq c $ and combining this estimate with (<ref>) and (<ref>)
yields (<ref>).

This proves tightness under the assumptions  of Proposition  <ref>,
if we are able to show the estimates (<ref>), (<ref>), (<ref>) and (<ref>), which will be done next.

Proof of (<ref>):
We recall the definition of the set $\mathcal{A}_n$ in (<ref>), the estimate
in (<ref>) and use the decomposition
| b_n^*(z) | = | b_n^*(z) | ( \mathds 1_{\mathcal{A}_n}+\mathds 1_{\mathcal{A}_n^c}).
For the second  term we have

\begin{align}
\label{det50a}
\E \big [ \mathds 1_{\mathcal{A}_n^c} | b_n^*(z) |^{q(r)} \big ] & \leq C \Big  | {z \over \Im (z)} \Big |^{q(r)}
\mathbb{P} ( \mathcal{A}_n^c ) \\
& \nonumber
\leq C m^{q(r)} \varepsilon_n^{-q(r)}  \mathbb{P} ( \mathcal{A}_n^c )
\leq C m^{q(r)(1+\alpha )} \mathbb{P} ( \mathcal{A}_n^c )  = o(1),
\end{align}

by (<ref>). For the first term we note that on the set
$\mathcal{A}_n$ we have $ \|  D^*(z)^{-1} \|_{S_\infty} < C$
for some constant  by the arguments given  in the derivation of (<ref>). Moreover, on the set $\mathcal{A}_n$ we have

\begin{align}
\|r_1^* \|^2  & =  \|  r_1^* {r_1^*}^\top  \|_{S_\infty}
\leq  \Big \|  \sum_{j=1}^m r_j^* {r_j^*}^\top  \Big \|_{S_\infty}  +\Big \|  \sum_{j=2}^m r_j^* {r_j^*}^\top  \Big \|_{S_\infty}  \leq 2 K_r,
\end{align}

and,  recalling the definition of $\beta_1^*(z) $ in (<ref>) yields

\begin{equation}
\label{det62}
|  \beta_1^*(z) | = | 1 -{r_j^*}^\top  D^*(z)^{-1}{r_j^*} |  \leq 1  + \|  D^*(z)^{-1} \|_{S_\infty}
\| {r_j^*} \|^2 < C
\end{equation}

on $\mathcal{A}_n$ for some positive constant $C$.
This  gives

\begin{align}
\label{det44}
\E \big [ \mathds 1_{\mathcal{A}_n} | b_n^*(z) |^{q(r)} \big ]  \leq C  \Big (  1    +
\E \big [
\mathds 1_{\mathcal{A}_n}
| b_n^*(z) |^{q(r)}
|   \gamma_1^*(z) |^{q(r)}
\big ]
\Big ).
\end{align}

Next, we show that $ \E \big [ \mathds 1_{\mathcal{A}_n}  |  \gamma_1^*(z) |^{q(r)} \big ]$ converges to $0$.
For this purpose we recall the definition of $\varepsilon_1^* (z) $ in (<ref>) and obtain,  the bound

\begin{align*}
\E \big [ &\mathds 1_{\mathcal{A}_n}   |  \gamma_1^*(z)  - \varepsilon_1^* (z)   |^{q(r)} \big ]\\
&  \leq {1 \over m^{q(r)} } \E \Big | \sum_{j=2}^m (\E^*_{j} -\E^*_{j-1} ) \beta_{1j}^*(z) {r_j^*}^\top  D_{1j}^*(z)^{-1}L_nL_n^\top
\Big |^{q(r)}
\\
\leq {C \over m^{q(r) /2} }
\E \big [ \big |
\beta_{12}^*(z) {r_2^*}^\top  D_{12}^*(z)^{-1}L_nL_n^\top
\big  |^{q(r)} \big ]
\end{align*}

where $\beta_{12}^*(z) $ is defined in (<ref>) and the last inequality follows from Lemma 2.1 in [23]. Note that on the set ${\mathcal{A}_n} $  we have $ \|  D^*_{12} (z)^{-1} \|_{S_\infty} < C$ and $  |
\beta_{12}^*(z) | \leq C $   (which follows by similar calculations as used for the derivation of (<ref>)).
Observing (<ref>), this yields

\begin{align}
\E \big [ \mathds 1_{\mathcal{A}_n}   |  \gamma_1^*(z)  - \varepsilon_1^* (z)   |^{q(r)} \big ]
& \leq {C \over m^{q(r)/2} }\Big (  \mathbb{P} ( \mathcal{A}_n ) \nonumber \\
& ~~~~~
+  \E \big [ \mathds 1_{\mathcal{A}_n^c}  \big |
\beta_{12}^*(z) {r_2^*}^\top  D_{12}^*(z)^{-1}L_nL_n^\top
\big  |^{q(r)} \big ] \Big )
\nonumber \\
&  \leq  {C \over m^{q(r)/2} } ~,
\label{det50aa}
\end{align}

where the last inequality follows by (<ref>) and similar arguments as given before.

An application of Proposition <ref>
with the matrix (<ref>) then gives for any $v \geq 2$

\begin{align}
\label{det50}
\E \big [ \mathds 1_{\mathcal{A}_n}   |  \gamma_1^*(z)  |^{\nu } \big ]
& \leq  c \Big (  \E \big [ \mathds 1_{\mathcal{A}_n}   |  \varepsilon_1^* (z)   |^{\nu } \big ] +
{1 \over m^{\nu /2} }  \Big )  \leq c \Big (
{\delta_m^{2\nu -4} \over m } +
{m \over n } +  {1 \over m^{\nu /2} }
\Big )  = o (1) .
\end{align}

Using this bound with $\nu =q(r)$
we continue with the estimate (<ref>), and obtain by. Hölder's inequality  (with $q=1/a$ and $q'=1/(1-a)$ for any $a \in (0,1)$)

\begin{align}
\nonumber
\E &\big [ \mathds 1_{\mathcal{A}_n} | b_n^*(z) |^{q(r)} \big ]\nonumber\\
& \leq C \Big  \{
\Big | { z \over \Im (z) } \Big |^{{q(r)}(1-a)}
\E \big [
\mathds 1_{\mathcal{A}_n}
| b_n^*(z) |^{{q(r)}a}
|   \gamma_1^*(z) |^{{q(r)}} \big ]
\Big \} \nonumber \\
\nonumber
& \leq
C \Big \{
1 +
m^{{q(r)}(1-a)} \varepsilon_n^{-{q(r)}(1-a)}
\big (   \E \big [ \mathds 1_{\mathcal{A}_n} | b_n^*(z) |^{q(r)} \big ]  \big )^{a}
( \E \big [ \mathds 1_{\mathcal{A}_n}   |  \gamma_1^*(z)  |^{{q(r)}/(1-a)} )^{1-a}
\Big \}
\\
\label{det51}
& \leq
C \Big \{
1 +  \big (   \E \big [ \mathds 1_{\mathcal{A}_n} | b_n^*(z) |^{q(r)} \big ]  \big )^{a} \kappa_n
\Big \}  ~,
\end{align}

where we use (<ref>) with $\nu = q(r)/(1-a)$ and the sequence $\kappa_n$ is defined by

\begin{align}
\label{det64}
\kappa_n & =
\varepsilon_n^{-{q(r)}(1-a)}
m^{{q(r)}(1-a) }
\Big (  {\delta_m^{2{q(r)}-4(1-a)} \over m^{1-a}} + \Big ( {m \over n } \Big )^{1-a}
+{1 \over m^{{q(r)}/2}}
\Big )
\end{align}

We now discuss  the three terms separately noting that we can choose the sequence $\varepsilon_n$ in (<ref>)
as a sequence converging to $0$ at an arbitrarily slow rate as long as it satisfies $\varepsilon_n \geq m^{-\alpha}$ for some $\alpha\in (0,1)$. Therefore, we choose
$\varepsilon_n = 1/\log m$ and show that
each of the three terms without the logarithmic factor
$ \varepsilon_n^{-{q(r)}(1-a)}
m^{{q(r)}(1-a) } $ converges to $0$ at a polynomial rate. We begin with the second  one

m^{{q(r)}(1-a) } \Big ( {m \over n } \Big )^{1-a} =  \Big ( {m^{q(r) +1 } \over n } \Big )^{1-a}
=   O \bigg ( \Big ( {m^{q(r) +1 } \over m^{5+\eta}  } \Big )^{1-a} \bigg )  = O \Big ({1 \over m^{2 + \eta/2}} \Big )

where we use  (<ref>) and $n=m^{5 + \eta}$ for the last equality.
Note that the assumption
$m^{3 + \eta } = O(n) $ would also yield the polynomial bound $O \big ({1 \over m^{ \eta/2}} \big )$, but we will need the stronger assumption to derive a corresponding bound in the investigation of the term $H_{1n}$.

Turning to the third term in (<ref>), it is easy to see that this term converges to $0$ at a polynomial rate in $m$ whenever $a >1/2$, and therefore it remains to consider the first term.
Here  we use the fact  that we can choose
$\delta_m=m^{-  1/12} $,

by Lemma <ref> (with $\varepsilon =2 $),

which gives for the third term

m^{{q(r)}(1-a) } {\delta_m^{2{q(r)}-4(1-a)} \over m^{1-a}} = m^{ (1-a)  {4(2r+1) \over 3(1-r)}   - {r+1 \over 3(1-r)} } .

Therefore, by choosing $a \in (0,1)$ larger than

$ 1 - {r+1 \over 4 (2r+1) }  $
it follows that this expression converges to $0$ at a  polynomial rate as well.
Consequently, (<ref>) holds with a sequence $\kappa_n =o(1)$ and
a constant $a \in (0,1)$, which implies that the sequence
$\E \big [ \mathds 1_{\mathcal{A}_n} | b_n^*(z) |^{q(r)} \big ]$ is bounded. Combining this result with (<ref>) proves the assertion  (<ref>).

Proof of (<ref>): Note that

W_1^*  =   \sum_{j=1}^m
\big (  \mathbb{E}^*_j
- \mathbb{E}^*_{j-1} \big)
\Big \{
r_j{^*}^\top D_j^* (z_1)^{-2} D_j^* (z_2)^{-1} r_j^*
-  {1 \over m} \tr  \big (
L_n L_n^\top  D_j^* (z_1)^{-2}
D_j^* (z_2)^{-1} \big )
\Big
\} ~,

and Proposition <ref>  with the matrix (<ref>)  gives

\begin{align}
\nonumber
&\mathbb{E} \big [  | W_1^* |^2
\big ] \\
&\ = \mathbb{E} \big [  \mathbb{E}^* [
| W_1^* |^2] \big ] \nonumber \\
&\ = \sum_{j=1}^m\mathbb{E}
\Big [
\Big |
\big (  \mathbb{E}^*_j
- \mathbb{E}^*_{j-1} \big) \Big \{ r_j{^*}^\top D_j^* (z_1)^{-2} D_j^* (z_2)^{-1} r_j^*
- {1 \over m } {\rm tr}  \big (
L_n^\top D_j^* (z_1)^{-2} D_j^* (z_2)^{-1}L_n
\big ) \Big ) \Big |^2
\Big ]
\nonumber \\
\nonumber
% \label{det2}
\ \leq  4 m
\mathbb{E}
\Big [ \Big |
r_1{^*}^\top D_1^* (z_1)^{-2} D_1^* (z_2)^{-1} r_1^*
- {1 \over m } {\rm tr}  \big (
L_n^\top D_1^* (z_1)^{-2} D_1^* (z_2)^{-1}L_n
\big ) \Big |^2
\Big ]   \\
&\  \leq 4 m K_2 \Big ( {1\over m}  + {m \over n } \big )
\leq 4 K_2 (1 + o(1))
\label{det115}
\end{align}

For  the term $W_2^*$ we have

\begin{align*}
\mathbb{E}[ | W_2^*||^2 ]  &=     \mathbb{E}\big [ \mathbb{E}^* [| W_2^*|^2 ]  \big ] \\
&\ =   \sum_{j=1}^m \mathbb{E} \Big [ \Big |
\big (  \mathbb{E}^*_j
- \mathbb{E}^*_{j-1} \big)  \big [
\beta_j^*(z_1)
r_j{^*}^\top D_j^* (z_1)^{-2} D_j^* (z_2)^{-1} r_j^*  \gamma_j^*(z_1)
\big   ] \Big |^2 \Big ] \\
&\  \leq  4m \mathbb{E} \big |
\beta_1^*(z_1)
r_1{^*}^\top D_1^* (z_1)^{-2} D_1^* (z_2)^{-1} r_1^*  \gamma_1^*(z_1)
\big |^2 ~.
\end{align*}

We now use the decomposition $\mathds{1} = \mathds{1}_{\mathcal{A}_n} +  \mathds{1}_{\mathcal{A}_n^c}$,   (<ref>) and (<ref>)
to obtain

\begin{align*}
\mathbb{E}[ | W_2^*||^2 ]  & \leq  4m \mathbb{E} \big |
\mathds{1}_{\mathcal{A}_n}
\gamma_1^*(z_1)
\big |^2   + o(1)  =O(1),
\end{align*}

where the last equality follows from (<ref>)  with $ \nu =2$.

Proof of (<ref>): By the
Cauchy-Schwarz inequality the assertion follows from the bound

\begin{align*}
\mathbb{E} \big [ |  b_n^*(z_1) |^{2q(r)} \big ]  \leq c
\end {align*}

where $q(r)= {2(1+r)/(1-r)}$.  For this purpose we proceed as in the proof of
(<ref>) where $q(r)$ is replaced by $2q(r)$. Consequently, we have to show that the sequence

\varepsilon_n^{-{2q(r)}(1-a)}
m^{{2q(r)}(1-a) }
\Big (  {\delta_m^{4{q(r)}-4(1-a)} \over m^{1-a}} + \Big ( {m \over n } \Big )^{1-a}
+{1 \over m^{{q(r)}}}
\Big )

converges to $0$. We use again  $\varepsilon_n  = \log m $ and show that each term without the logarithmic factor $\varepsilon_n^{-{2q(r)}(1-a)}$
converges at a polynomial rate, say $m^{-\rho}$ for some small
$\rho >0$. For the third this follows again for any $a \in (1/2,1)$. For the second we now use the assumption $m^{5+\eta} = O(n)  $  which gives

\begin{align}
m^{{2q(r)} }   \Big ( {m \over n } \Big ) =
O \Big ( m^{{8r \over 1-r} - \eta} \Big ) ,
\end{align}

where the exponent is negative by
(<ref>). Finally, using
$\delta_m=m^{-1/ 12 }  $
we have for the third term

m^{{2q(r)}(1-a) } {\delta_m^{{4q(r)}-4(1-a)} \over m^{1-a}} = m^{ (1-a)  {2(7r+5) \over 3(1-r)}   - {r+1 \over 3(1-r)} }

which converges to $0$ at a polynomial rate in $m$, if

$ a>  1 -  {r+1 \over 2(7r+5)}   $.

Proof of (<ref>):  For the sake of brevity we
restrict ourselves to the term $ \mathbb{E} |Y_1^*|^2$. Corresponding bounds for $ \mathbb{E} |Y_2^*|^2$ and $ \mathbb{E} |Y_3^*|^2$ are derived in a similar way.
Note that

\begin{align*}
\mathbb{E}& |Y_1^*|^2  =
\sum_{j=1}^{m} \E
\Big |  \big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big )
\Big [ \big (r_j{^*}^\top (D_j^* (z_1))^{-1} (D_j^* (z_2))^{-1} r_j^* \big ) ^2  \\
& ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
- \Big ( {1 \over m }  {\rm tr } \big ( L_n (D_j^* (z_1))^ {-1} (D_j^* (z_2) )^{-1}
L_n^\top  \big ) \Big )^2 \Big ]
\Big |^2 \\
&  \leq
4 m \E
\Big |
\big (r_1{^*}^\top (D_1^* (z_1))^{-1} (D_1^* (z_2))^{-1} r_1^* \big ) ^2\\
&\hspace{60mm}- \big ( {1 \over m }  {\rm tr } \big ( L_n (D_1^* (z_1))^ {-1} (D_1^* (z_2) )^{-1}
L_n^\top  \big ) \big )^2
\Big |^2
\end{align*}

We recall the definition of the set $\mathcal{A}_n$ defined in (<ref>), the decomposition
$\mathds{1} = \mathds{1}_{\mathcal{A}_n} +  \mathds{1}_{\mathcal{A}_n^c}$
and the identity
$|a^2-b^2|^2 = |a-b|^4 + 2( \bar a b + a\bar  b)
|a-b|^2$, which give  (observing (<ref>))

\begin{align*}
m  \E  \mathds{1}_{\mathcal{A}_n^c}&
\Big |
\big (r_1{^*}^\top (D_1^* (z_1))^{-1} (D_1^* (z_2))^{-1} r_1^* \big ) ^2   - \big ( {1 \over m }  {\rm tr } \big ( L_n (D_1^* (z_1))^ {-1} (D_1^* (z_2) )^{-1}
L_n^\top  \big ) \big )^2  \Big |^2\\
& \leq  C \big (1 + o(1) \big )
\end{align*}

for some positive constant $C$

\begin{align*}
&  \E  \mathds{1}_{\mathcal{A}_n}
\Big |
\big (r_1{^*}^\top (D_1^* (z_1))^{-1} (D_1^* (z_2))^{-1} r_1^* \big ) ^2   - \big ( {1 \over m }  {\rm tr } \big ( L_n (D_1^* (z_1))^ {-1} (D_1^* (z_2) )^{-1}
L_n^\top  \big ) \big )^2  \Big |^2 \\
& \leq    \mathbb{E}  \Big |
r_1{^*}^\top (D_1^* (z_1))^{-1} (D_1^* (z_2))^{-1} r_1^*    -  {1 \over m }  {\rm tr } \big ( L_n (D_1^* (z_1))^ {-1} (D_1^* (z_2) )^{-1}
L_n^\top  \big ) \Big |^4 \\
& + 4  \mathbb{E}  \Big [ \Big |
r_1{^*}^\top (D_1^* (z_1))^{-1} (D_1^* (z_2))^{-1} r_1^*    -  {1 \over m }  {\rm tr } \big ( L_n (D_1^* (z_1))^ {-1} (D_1^* (z_2) )^{-1}
L_n^\top  \big ) \Big |^2
\\
& \times
\mathds{1}_{\mathcal{A}_n}
\| (D_1^* (z_1))^ {-1} (D_1^* (z_2) )^{-1} \|_{{S}_\infty}^2  {q \over m} \|r_1^* \|^2
\Big ]  \\
&  = O \Big (  {1 \over m} \Big ),
\end{align*}

where we have used Proposition <ref> for the matrix   (<ref>) and the estimate (<ref>) on the set $\mathcal{A}_n$.  The bound on $\E |Y_1^*|^2$ now follows.

Proof of tightness under the assumptions of  Theorem <ref>
the function  $f$ is analytic  in a region of
the complex plane containing the interval $I$ defined in (<ref>) we can choose the contour in (<ref>) such that

x_l < K_l -  \limsup_{n \to \infty}  \| \Sigma_n \|_{S_\infty} ~~\text{ and }~x_r >  K_r +  \limsup_{n \to \infty}  \| \Sigma_n \|_{S_\infty} ,

it follows from (<ref>) that

\| D_1^*(z)^{-1}\|_{S_\infty}
\leq  \max \Big \{
{1 \over | x_l -K_l| },  {1 \over | x_r -K_r| }
\Big \}  <  {1  \over \limsup_{n \to \infty}  \| \Sigma_n \|_{S_\infty} }

on the set $\mathcal{A}_n$.
This implies that

\limsup_{n \to \infty}
%{1 \over m }
\bigg |   \mathbb{E}^* \Big [ \frac{1}{m}
\tr \big ( L_nL_n^\top D_1^*(z)^{-1} \big)  \Big ] \bigg |
\leq
\limsup_{n \to \infty}
%{ 1 \over m}
\|  L_nL_n^\top \|_{S_{\infty}}
\| D_1^*(z)^{-1}\|_{S_\infty} \leq c < 1

on the set $ \mathcal{A}_n$.
Consequently, we obtain the on the set $\mathcal{A}_n$ that

\limsup_{n \to \infty} | b_n^*(z) |
\leq
\limsup_{n \to \infty} {1 \over 1 - {1 \over m}  \|  L_nL_n^\top \|_{S_\infty}
\| D_1^*(z)^{-1}\|_{S_\infty} } \leq C < \infty ,

and it follows

\begin{align}
\label{det207}
| b_n^*(z) | \leq   \mathds{1}_{\mathcal{A}_n} | b_n^*(z) | +  \mathds{1}_{\mathcal{A}_n^c} | b_n^*(z) |  \leq  C +  \mathds{1}_{\mathcal{A}_n^c} | b_n^*(z) |
\leq C +  \mathds{1}_{\mathcal{A}_n^c}m^{1+\alpha}
\end{align}

for some positive constant $C$. We now prove  (<ref>)  observing that

\E \Big |
{   \widehat M_n^{*}  (z_1)  - \widehat M_n^{*}  (z_2)    \over z_1 - z_2   }
\Big |  ^{1+r}   \leq C \big ( \E  |   H_{1n} | ^{1+r}   +  \E  |  H_{2n} | ^{1+r}    +  \E  |  H_{3n} | ^{1+r}   \big ) ,

where the moments  $\E  |  H_{jn} | ^{1+r} $ can estimated by similar but simpler arguments as given before. For example,

\E  | H_{2n} | ^{1+r}  \leq  C  \big (
\E  | b_n^*(z_1)W_1^*  | ^{1+r}   +  \E  |  b_n^*(z_1)W_2^* | ^{1+r} \big )

where, by (<ref>),

\begin{align*}
\mathbb{E} \big [ | b_n^*(z)  W_j^* |^{1+r}
\big ] \leq \big (  \mathbb{E} \big [ | b_n^*(z) |^{2(1+r)/(1-r)} \big ]  \big )^{(1-r)/2} \big (
\E \big [ | W_j^* |^{2}
\big ] \big )^{(1+r)/2} ~~~~~~~~~(j=1,2).
\end{align*}

The term
$ \E \big [ | W_j^* |^{2}
\big ]$ can be bounded by the same arguments as given in Section <ref>.  Moreover, by (<ref>)

and Corollary <ref>
we have

\mathbb{E} \big [ | b_n^*(z) |^{2(1+r)/(1-r)} \big ]  \leq
C (1 + \mathbb{E} \big [ \mathds{1}_{\mathcal{A}_n^c} | b_n^*(z) |^{2(1+r)/(1-r)} \big ]  \leq C(1+ o(1)) ,

which shows that $ \E  | H_{2n} | ^{1+r} $ is bounded.
The arguments for $  \E  | H_{3n} |^{1+r}$ and $  \E  | H_{1n} |^{1+r}$ are similar and omitted for the sake of brevity.

§.§ Auxiliary results

Recall that in the next two Lemmata we consider
truncated, centered and normalized $q'$-dimensional random vector $X_1 , \ldots, X_n$
(see the discussion in Section <ref> and <ref>).

For $k=1, \ldots , m $  let

\begin{align*}
A_{k} (z)=L_n^\top  \E_{k-1}^*\big( \bar\beta_k^*(z) D_k^* (z)^{-1}\big) L_n^\top  \ .
\end{align*}

\begin{align}
\label{det107}
&  \E \bigg\arrowvert \E_{X_k^*}^*\Big(X_k^{*\top}A_k (z_1) X_k^*-\tr(A_k(z_1))\Big)\E_{X_k^*}^*\Big(X_k^{*\top}A_k(z_2)X_k^*-\tr(A_k(z_2))\Big)\bigg\arrowvert \\
\nonumber
&   ~~~~~~~~~~~~~ ~~~~~~~~~~~~~ ~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~  \leq K (z_1,z_2)
\Big({\delta_m^2m^{2} \over \sqrt{n}} +\frac{m^3}{n}\Big)
\end{align}

Evaluating $E_{X_k^*}^*$ provides the upper bound

\begin{align}
\E\Big\arrowvert &\E_{X_k^*}^*\Big(X_k^{*\top}A_k (z_1) X_k^*-\tr(A_k  (z_1) )\Big)\E_{X_k^*}^*\Big(X_k^{*\top}A_k  (z_2)X_k^*-\tr(A_k  (z_2) )\Big)\Big\arrowvert \nonumber\\
&\leq  \E\bigg\arrowvert \frac{1}{n^2}\sum_{i=1}^n\big(X_i^\top A_k  (z_1) X_i-\tr(A_k  (z_1))\big)\big(X_{i}^\top A_k  (z_2) X_{i}-\tr(A_k  (z_2))\big)\bigg\arrowvert \nonumber\\
&\quad\quad +  \E\bigg\arrowvert \frac{1}{n^2}\sum_{\substack{i,i'\in\{1,\dots n\} :\\ i\not= i'}}\big(X_i^\top A_k  (z_1)X_i-\tr(A_k  (z_1))\big)\big(X_{i'}\top A_k  (z_2) X_{i'}-\tr(A_k  (z_2))\big)\bigg\arrowvert.\label{eq: term2!}
\end{align}

By the Cauchy-Schwarz inequality, we have

\begin{align}
\label{det106}
\E\bigg\arrowvert &\frac{1}{n^2}\sum_{i=1}^n\big(X_i^\top A_k  (z_1) X_i-\tr(A_k  (z_1))\big)\big(X_{i}^\top A_k  (z_2) X_{i}-\tr(A_k  (z_2))\big)
\bigg\arrowvert\\
\nonumber
&\leq \frac{1}{n}\E^{1/2}\big\arrowvert X_1^\top A_k  (z_1)X_1-\tr(A_k  (z_1))\big\arrowvert^2\E^{1/2}\big\arrowvert X_1^\top A_k  (z_2)X_1-\tr(A_k  (z_2))\big\arrowvert.
\end{align}

We now investigate one factor  using Jensen's inequality for the conditional expectation and  the bound
| \bar\beta_j^*(z)| \leq \frac{\arrowvert z \arrowvert }{\Im(z ) }

\begin{align*}
\E \big\arrowvert &X_1^\top A_k  (z_1)X_1-\tr(A_k  (z_1))\big\arrowvert^2\\
& \leq
\Big (  \frac{\arrowvert z_1\arrowvert }{\Im(z_1)} \Big )^2
\E \big\arrowvert X_1^\top  L_n^\top D_k^* (z_1)^{-1}  (z_1) L_n X_1-\tr( L_n^\top  D_k^* (z_1)^{-1}L_n )\big\arrowvert^2
\\
& \leq K \Big ( m + {m^3 \over n} \Big ),
\end{align*}

where last inequality follows by the same arguments as given after formula (<ref>) in the proof Proposition <ref>. Therefore, the right-hand side of (<ref>) is bounded by

K \frac{\arrowvert z_1\arrowvert \arrowvert z_2\arrowvert}{\Im(z_1)\Im(z_2)} \Big(1+\frac{m^2}{n}\Big)\frac{m}{n}.

We define $i^*_\ell $  as the index $j \in \{ 1, \ldots , n \} $ with $X_\ell ^* = X_j$
For $k=1, \ldots , m $, we denote $\hat{I}_k=\{1,\dots,n\}\cap\{i_1^*,\ldots,  i_{k-1}^*,  i_{k+1}^*, \ldots , i_m^*\}$ and observe  that $\sharp\hat{I}_k\leq m$  and that  $\hat{I}_k, X_1,\dots,X_n$ are jointly independent. With the notation

R_{ki} (z) = X_i^\top A_k (z) X_i-\tr(A_k (z) ),\ \ i=1,\dots, n ,µ k=1,  \ldots , m,

we decompose the expression in (<ref>) into

\begin{align*}
\E\bigg\arrowvert &\frac{1}{n^2}\sum_{\substack{i,i'\in\{1,\dots n\} :\\ i\not= i'}} R_{ki  }(z_1) R_{ki' }  (z_2) \bigg\arrowvert\\
&\leq \E\bigg\arrowvert \frac{1}{n^2}\sum_{i,i'\in\hat{I}_k, i\not=i'}
R_{ki  }(z_1) R_{ki' }  (z_2)
\bigg\arrowvert+\ \E\bigg\arrowvert \frac{1}{n^2}\sum_{i,i'\in\hat{I}_k^c, i\not=i'}  R_{ki  }(z_1) R_{ki' }  (z_2)
\bigg\arrowvert\\
&\quad\quad +  \E\bigg\arrowvert \frac{1}{n^2}\sum_{i\in\hat{I}_k^c, i'\in\hat{I}_k} R_{ki  }(z_1) R_{ki' }  (z_2)  \bigg\arrowvert
+ \E\bigg\arrowvert \frac{1}{n^2}\sum_{i\in\hat{I}_k, i'\in\hat{I}_k^c} R_{ki  }(z_1) R_{ki' }  (z_2) \bigg\arrowvert
\end{align*}

and bound each summand separately.
By the Cauchy-Schwarz inequality, we have

\begin{align*}
\E\bigg\arrowvert \frac{1}{n^2}\sum_{i,i'\in\hat{I}_k, i\not=i'}R_{1i}R_{2i'}\bigg\arrowvert&\leq \frac{1}{n^2}\sum_{i\not=i'}\E^{1/4} ( I {\{ i\in\hat{I}_k \}} )\E^{1/4} ( I {\{ i'\in\hat{I}_k \}} )\E^{1/4} \arrowvert R_{1i} \arrowvert^4 \E^{1/4} \arrowvert R_{2i'}\arrowvert^4 \\
\leq \sqrt{K} \Big({\delta_m^2m^{2} \over \sqrt{n}} +\frac{m^3}{n}\Big).
\end{align*}

For the last  inequality,  we use  Jensen's inequality for the conditional expectation,  the same arguments as given after formula (<ref>) in the proof Proposition <ref> and  the fact that $\E ( I {\{ i\in\hat{I}_k \}} ) \leq {m\over n}$, which follows because $\E ( I {\{ i\in\hat{I}_k \}} )$ is independent of $i$ and $\sum_{i=1}^n\E ( I {\{ i\in\hat{I}_k \}} ) \leq m$.

Observing that, by  Jensen's inequality, the  conditional

$$\E [ R_{ki}  (z_1) \arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\} ] $$
vanish for $i \in \hat I^c_k$, the
conditional independence of $R_{ki (z) }$ and $R_{ki'}(z)$ for $i\not=i'$, $i,i'\in\hat{I}^c_k$ given $\hat{I}_k$ and $\{X_j:j\in\hat{I}_k\}$, and Lemma 2.7 of baisil1998,

\begin{align*}
\E^2&\bigg\arrowvert \frac{1}{n^2}\sum_{i,i'\in\hat{I}_k^c, i\not=i'}R_{ki} (z_1) R_{ki'} (z_2) \bigg\arrowvert\\
&\leq \E\frac{1}{n^4}\sum_{i,i'\in\hat{I}_k^c, i\not=i'}\bigg\{\E\Big[ R_{ki} (z_1) \overline{R}_{ki} (z_1) R_{ki'}(z_2) \overline{R}_{ki'}(z_2) \Big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\Big]\\
&\quad \quad\quad \quad\quad\quad\quad\quad\quad+ \E\Big[R_{ki}
(z_1) R_{ki}(z_2) \overline{R}_{ki'} (z_1) \overline{R}_{ki'} (z_2) \Big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\Big]\bigg\}\\
&=  \E\frac{1}{n^4}\sum_{i,i'\in\hat{I}k^c, i\not=i'}\bigg\{\E\big[ \arrowvert R_{ki} (z_1) \arrowvert^2\big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\big]\E\big[ \arrowvert R_{ki'} (z_2)\arrowvert^2\big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\big]\\
& \quad \quad\quad  + \E\big[R_{ki}(z_1) R_{ki} (z_2)\big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\big]\E\big[ \overline{R}_{ki'} (z_1)\overline{R}_{2i'} (z_2)\big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\big]\bigg\}\\
&\leq K \frac{m^2}{n^2}.
\end{align*}

Next, by the same arguments as above and the same arguments as given after formula (<ref>) in the proof of Proposition <ref>,

\begin{align*}
\E^2\bigg\arrowvert &\frac{1}{n^2}\sum_{i\in\hat{I}_k^c, i'\in\hat{I}_k}R_{ki} (z_1) R_{ki'} (z_2) \bigg\arrowvert\\
&\leq \E\frac{1}{n^4}\sum_{\substack{i,l\in \hat{I}_k^c\\ i',l'\in\hat{I}_k}}R_{ki'} (z_2) \overline{R}_{kl'} (z_2)\E\Big[R_{ki} (z_1) \overline{R}_{kl} (z_1) \Big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\Big]\\
&=\E\frac{1}{n^4}\sum_{\substack{i\in \hat{I}_k^c\\ i',l'\in\hat{I}_k}}R_{ki'} (z_2 )\overline{R}_{kl'} (z_2 )\E\Big[R_{ki}(z_1) \overline{R}_{ki} (z_1) \Big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\Big]\\
&\leq  K (z_1)  \frac{m}{n^3}\sum_{i,i', i\not=i'}\E^{1/2}\arrowvert R_{ki'} (z_2) \arrowvert^2 \E^{1/2}\arrowvert R_{kl'} (z_2) \arrowvert^2 \\
&\leq K (z_1, z_2)   \frac{m^2}{n}.
\end{align*}

Here, we   applied Lemma B.26 of baisilverstein2010 to obtain

\E\Big[R_{ki}(z_1) \overline{R}_{ki} (z_1) \Big\arrowvert \hat{I}_k, \{X_j:j\in\hat{I}_k\}\Big] \leq  K(z_1 ) m ,

which yields the  second inequality.
The last expression

\E\bigg\arrowvert \frac{1}{n^2}\sum_{i\in\hat{I}_k, i'\in\hat{I}_k^c}R_{ki} (z_1) R_{ki'} (z_2)\bigg\arrowvert

in the decomposition of (<ref>) can be bounded analogously. Summarizing these  calculations yield (<ref>), which completes the proof.

For any $k=1, \ldots , m $  define

\begin{align*}
&A_{k}(z)= \frac{1}{m}
L_n^\top  \E_{k-1}^*\big(D_k^*(z)^{-1}\big)L_n  = \frac{1}{m}L_n^\top  \E_{k}^*\big(D_k^*(z)^{-1}\big)L_n ~.
\end{align*}

Then there exists some positive constant $K(z_1,z_2)$ (independent of $k$)  such that

\begin{align*}
\E&\Big\arrowvert \E_{X_k^*}\Big[\big(X_k^{*\top}A_k (z_1) X_k^*-\tr A_k (z_1) \big)\big(X_k^{*\top}A_k (z_2) X_k^*-\tr A_k (z_2) \big)\\
&\hspace{80mm}- 2 \tr(A_k (z_1) A_k (z_2 ))\Big]\Big\arrowvert\\
&\leq  K (z_1,z_2) \Big ( \frac{\delta_m^2}{\sqrt{n}}+\frac{m}{n} +  o \Big(\frac{1}{m}\Big) \Big )  .
\end{align*}

Recall the notation of $\hat I_k$ in the proof of Lemma <ref>.
Evaluating $\E_{X_1^*}^*$ provides the upper bound

\begin{align}
\E&\Big\arrowvert \E_{X_k^*}\Big[\big(X_k^{*\top}A_k (z_1) X_k^*-\tr A_k (z_1) \big)\big(X_k ^{*\top}A_k (z_2) X_k^*-\tr A_k (z_2)\big)\nonumber\\
&\hspace{80mm}-  2\tr(A_k (z_1) A_k (z_2) )\Big]\Big\arrowvert\nonumber\\
&=\E\Big\arrowvert \frac{1}{n}\sum_{i=1}^n \big(X_i^\top A_k (z_1) X_i-\tr A_k (z_1) \big)\big(X_i^\top A_k (z_2 ) X_i-\tr A_k (z_2) \big)
\nonumber \\
&\hspace{80mm}- 2\tr(A_k (z_1) A_k (z_2))\Big\arrowvert\nonumber\\
&\leq \E\Big\arrowvert \frac{1}{n}\sum_{i\in\hat{I}_k} \big(X_i^\top A_k (z_1) X_i-\tr A_k (z_1) \big)\big(X_i^\top A_k (z_2) X_i-\tr A_k (z_2 )\big)\nonumber\\
&\hspace{80mm}- 2 \tr(A_k (z_1)A_k (z_2) )\Big\arrowvert\label{eq: sum1}\\
&\quad\quad +  \E\Big\arrowvert \frac{1}{n}\sum_{i\in\hat{I}_k^c} \big(X_i^\top A_k (z_1) X_i-\tr A_k (z_1) \big)\big(X_i^\top A_k (z_2 )X_i-\tr A_k (z_2) \big)\nonumber\\
&\hspace{80mm}- 2 \tr(A_k (z_1) A_k (z_2) )\Big\arrowvert\label{eq: sum2}
\end{align}

We start with bounding (<ref>). Since   $\arrowvert \tr (A_k (z_1) A_k (z_2) )\arrowvert  \leq  m \| A_k (z_1) \|_{S_\infty} \| A_k (z_2) \|_{S_\infty} $,  $\| A_k (z_1) \|_{S_\infty}\leq K(z_1)/m$ and  $\sharp \hat{I}_k\leq m$, we have

\E\Big\arrowvert \frac{\sharp\hat{I}_k}{n}\tr(A_1A_2)\Big\arrowvert\leq K (z_1, z_2) \frac{1}{n}.

Next, by the Cauchy-Schwarz inequality and
by the same arguments as given after formula (<ref>) in the proof Proposition <ref>,  it follows that

\begin{align*}
\E&\Big\arrowvert \frac{1}{n}\sum_{i\in\hat{I}_k} \big(X_i^\top A_k (z_1) X_i-\tr A_k (z_1)  \big)\big(X_i^\top A_k (z_2) X_i-\tr A_k (z_2)  \big)\Big\arrowvert\\
&\leq \frac{1}{n}\sum_{i=1}^n\E^{1/2} \big(\mathds{1}_{\{i\in\hat{I}_k\}} \big) \E^{1/4}\big\arrowvert  X_i^\top A_k (z_1)  X_i-\tr A_k (z_1)  \big\arrowvert^4\\
&\hspace{60mm}\times \E^{1/4}\big\arrowvert  X_i^\top A_k (z_2) X_i-\tr A_k (z_2) \big\arrowvert^4\\
\leq K(z_1,z_2) \frac{1}{\sqrt{n}}\Big(\delta_m^2+\frac{m}{\sqrt{n}}\Big).
\end{align*}

As concerns (<ref>),
we use the same arguments as in the proof of Lemma <ref> and obtain

\begin{align*}
&  \E^2\bigg\arrowvert \frac{1}{n}\sum_{i\in\hat{I}_k^c} \big(X_i^\top A_k (z_1)  X_i-\tr A_k (z_1)  \big)\big(X_i^\top A_k (z_2) X_i-\tr A_k (z_2)  \big)- 2 \tr(A_k (z_1)  A_k (z_2) )\bigg\arrowvert\\
&\leq \E \bigg \{ \E\bigg[\frac{1}{n^2}\sum_{i,i'\in\hat{I}_k^c}\Big[\big(X_i^\top A_k (z_1)  X_i-\tr A_k (z_1)  \big)\big(X_i^\top A_k (z_2) X_i-\tr A_k (z_2) \big)\\
&\hspace{90mm}- 2\tr(A_k (z_1) A_k (z_2) )\Big]\\  &\hspace{20mm}\times \Big[\big(X_{i'}^\top \overline{A_k (z_1) }X_{i'}-\tr \overline{A_k (z_1)  }\big)\big(X_{i'}^\top \overline{A_k (z_2)  }X_{i'}-\tr \overline{A_k (z_2) }\big)\\
&\hspace{90mm}-2 \tr(\overline{A_k (z_1) A_k (z_2) })\Big]\\
\bigg\arrowvert \hat{I}_k,\{X_j: j\in\hat{I}_k\}\bigg] \bigg \}
\\
&= \E\frac{1}{n^2}\sum_{i\in\hat{I}_k^c}\E\bigg[\Big\arrowvert \big(X_i^\top A_k (z_1)  X_i-\tr A_k (z_1)  \big)\big(X_i^\top A_k (z_2) X_i-\tr A_k (z_2) \big)\\
&\hspace{40mm}- 2 \tr(A_l (z_1)  A_k (z_2) )\Big\arrowvert^2\bigg\arrowvert \hat{I}_k,\{X_j: j\in\hat{I}_k\}\bigg] ~+ ~ K(z_1,z_2)o\Big(\frac{1}{m^2}\Big) \\
&\leq\frac{2}{n}\E\bigg\{\E^{1/2} \Big[\big\arrowvert X_i^\top A_k (z_1)  X_i-\tr A_k (z_1)  \big\arrowvert^4\Big\arrowvert \hat{I}_k,\{X_j: j\in\hat{I}_k\}\Big]\\
&\quad\quad\quad\quad\quad\quad\quad\quad\times\E^{1/2}\Big[\big\arrowvert \big\arrowvert X_i^\top A_k(z_2)  X_i-\tr A_k (z_2) \big\arrowvert^4\Big\arrowvert \hat{I}_k,\{X_j: j\in\hat{I}_k\}\Big]\bigg\}\\
&\hspace{66mm} + \frac{8}{n}\E\arrowvert \tr(A_k (z_1)  A_k (z_2) )\arrowvert^2  ~+ ~K(z_1,z_2)o\Big(\frac{1}{m^2}\Big) \\
&\leq K(z_1,z_2) \Big( \frac{\delta_m^4}{mn} + \frac{m}{n^2}  +  o\Big(\frac{1}{m^2}\Big) \Big),
\end{align*}

where the last line follows from Lemma B.26 in
in baisilverstein2010 and the truncation at level $\delta_m\sqrt{m}$. In this estimate, the term  $K(z_1,z_2)o(1/m^2)$ is caused by the sum over the mixed products with indices $i\not=i'$ where the conditional expectation factorizes. Its bound is due to  (<ref>)  and  formula (1.15) in baisil2004, using that

\Big | \sum_{i=1}^{q'} (A_k (z_1))_{ii} (A_k (z_2))_{ii}  \Big | \leq  \|A_k (z_1) \|_{S_2}
\|A_k (z_2) \|_{S_2} \leq \frac{K(z_1,z_2)}{m}.

There exists some constant $K(z) >0$, such that

\begin{align*}
\E\arrowvert \bar{\beta}_1^*(z)-b_n^*(z)\arrowvert^2\leq K(z) \frac{1}{m} ~, \\
\E\arrowvert {\beta}_{12}^*(z)-b_1^*(z)\arrowvert^2\leq K(z) \frac{1}{m}~.
\end{align*}

Note that by Proposition <ref>, we have

\begin{align*}
\E \big | \bar{\beta}_j^*(z) -{\beta}_j^*(z)   \big |^2
%  &
%\leq 2 \Big | \frac{1}{1+r_j^{*^\top}{D_j^*}(z)^{-1}r_j^*} -     \frac{1}{1+m^{-1}\tr\big(L_n L_n^{^\top}{D_j^*}(z)^{-1}\big)} \Big |   \\
& \leq \E  \bigg |
\frac{r_j^{*^\top}{D_j^*}(z)^{-1}r_j^* -
{1\over m } \tr\big(L_n L_n^{^\top}{D_j^*}(z)^{-1}) }{(1+r_j^{*^\top}{D_j^*}(z)^{-1}r_j^*) (1+{1\over m }\tr\big(L_n L_n^{^\top}{D_j^*}(z)^{-1}\big)} \bigg |^2    \\
& \leq \Big ( { | z | \over \Im (z)} \Big )^2
\E  \Big |
{r_j^{*^\top}{D_j^*}(z)^{-1}r_j^* -
{1\over m } \tr\big(L_n L_n^{^\top}{D_j^*}(z)^{-1}) } \Big |^2 \\
& \leq  K(z) \Big ( {1 \over m  }  + {m \over n  } \Big )  \leq  K(z)  {1 \over m  } ~.
\end{align*}

\begin{align*}
\E\arrowvert {\beta}_1^*(z)-b_n^*(z)\arrowvert^2   & = \E |  {\beta}_1^*(z) b_n^*(z)) \gamma_1^*(z) |^2  \\
& \leq
\Big ( { | z | \over \Im (z)} \Big )^2
\E |   \gamma_1^*(z) |^2 \leq K(z) \Big ( {1 \over m  }  + {m \over n  } \Big ) \leq  K(z)  {1 \over m  },
\end{align*}

which follows as in (<ref>), where use Proposition <ref> instead of Proposition <ref> because of the different truncation scheme. The first assertion now follows by Minkowskii's inequality.  The second inequality is obtained analogously.

§.§ Bootstrap version of the Martingale-CLT

Let $(X_j)_{j\in\N}$ be an iid-sequence, $(\Pi_n)_{n\in\N}$ be some  further sequence of  random variables independent of $(X_j)_{j\in\N}$, and  $m\leq n$ with $m=m(n)\rightarrow\infty$. Conditional on  $X_1,\dots, X_n$,  let

X_1^*,\dots, X_m^*\overset{iid}{\sim}\hat{\P}_n=\frac{1}{n}\sum_{k=1}^n\delta_{X_i}
denote the'$m$ out of $n$' bootstrap sample. Suppose that conditional on $X_1,\dots, X_n$ and $\Pi_n$,
$$Y_{n,1}^*,\dots, Y_{n,m}^*$$
is a real square integrable martingale difference sequence with respect to the bootstrap canonical filtration $(\FF_{n,k}^*)_{k=1}^m$ with $\FF_{n,k}^*=\sigma(X_1^*,\dots, X_k^*\arrowvert X_1,\dots, X_n, \Pi_n)$.
Assume that, as $n\rightarrow\infty$,

\begin{align}\label{eq: cond 1}
\sum_{k=1}^{m}\E^*\big(\arrowvert Y_{n,k}^*\arrowvert^2\big\arrowvert \FF_{n,k-1}^*\big)\overset{\P}{\longrightarrow}\sigma^2
\end{align}

for some constant $\sigma^2>0$ and

\begin{align}\label{eq: cond 2}
\sum_{k=1}^m\E^*\Big[\arrowvert Y_{n,k}^*\arrowvert^2\ind_{\{\arrowvert Y_{n,k}^*\arrowvert\geq\varepsilon\}}\Big]\overset{\P}{\longrightarrow}0
\end{align}

for each $\varepsilon>0$. Then, as $n\rightarrow\infty $,

\begin{align}
\LL\bigg(\sum_{k=1}^mY_{n,k}^*\bigg\arrowvert X_1,\dots,X_n,\Pi_n\bigg)\Longrightarrow\NN(0,\sigma^2)\ \ \text{in probability}.
\end{align}

Preliminary, we assume that there exists some constant $c>0$ with

\begin{align}\label{eq: step 11}
\sup_{n}\sum_{k=1}^m \E^*\big(Y_{n,k}^{*2}\big\arrowvert Y_1^*,\dots,Y_{k-1}^*\big) \leq c.
\end{align}

Claim I. With

\begin{align}\label{eq: step 1}
\sup_{t\in K}\Big\arrowvert \E^*\exp(it Z_n^*)-\exp\Big(-\frac{1}{2}t^2\sigma^2\Big)\Big\arrowvert \overset{\P}{\longrightarrow}0
\end{align}

for any compact subset  $K\subset\R$.

Proof of Claim I.
The proof follows the lines in the proof of the classical martingale CLT, replacing all expectations  by conditional expectations $\E^*$ and the canonical filtration correspondingly. However, we state here locally uniform stochastic convergence rather than pointwise  stochastic convergence of the characteristic functions, which requires some extra care with the transfer of arguments.

\begin{align*}
\sigma_{n,l}^{*2}&=\E^*(Y_{n,l}^{*2}\arrowvert Y_1^*,\dots, Y_{l-1}^*)\ \ \text{for }2\leq l\leq m,\ \ \sigma_{n,1}^{*2}=\E^*Y_{n,1}^{*2},\\
\Sigma_{n,l}&=\sum_{k=1}^l\sigma_{n,k}^{*2}\ \ \text{ for } 0\leq l\leq m, \ \text{and}\\
Z_{n,l}^*&=\sum_{k=1}^lY_{n,k}^* \ \ \text{for } 0\leq l\leq m.
\end{align*}

Let $K\subset\R$ be compact. Then

\begin{align*}
\sup_{t\in K}\bigg\arrowvert\E^*&\bigg[\exp\Big(itZ_n^*\Big)-\exp\Big(-\frac{1}{2}t^2\sigma^2\Big)\bigg]\bigg\arrowvert\\
&\leq \sup_{t\in K}\E^*\bigg[\bigg\arrowvert 1-\exp\Big(\frac{1}{2}t^2\Sigma_{n,m}\Big)\exp\Big(-\frac{1}{2}t^2\sigma^2\Big)\bigg]\\
&\quad\quad + \sup_{t\in K}\bigg\arrowvert \E^*\bigg[\exp\Big(\frac{1}{2}t^2\Sigma_{n,m}\Big)\exp\Big(itZ_n^*\Big)-1\bigg]\bigg\arrowvert \\
&=: A_K+B_K.
\end{align*}

Choose some $0\leq t_K\in\R$ satisfying $-t_K\leq x\leq t_K$ for all $x\in K$.
As $\Sigma_{n,m}\rightarrow_{\P}\sigma^2$ by assumption and (<ref>),

\begin{align*}
\E A_K\leq \E\bigg(\exp\Big(\frac{1}{2}t_K^2\arrowvert \Sigma_{n,m}-\sigma^2\arrowvert\Big)-1\bigg)\rightarrow 0.
\end{align*}

As concerns $B_K$,

\begin{align*}
B_K&=\sup_{t\in K}\bigg\arrowvert \sum_{k=1}^m\E^*\bigg[\exp\Big(it Z_{n,{k-1}}^*\Big)\exp\Big(\frac{1}{2}t^2\Sigma_{n,k}\Big)\bigg(\exp(itY_{n,k}^*)-\exp\Big(-\frac{1}{2}t^2\sigma_{n,k}^{*2}\Big)\bigg)\bigg]\bigg\arrowvert\\
&\leq \exp\Big(\frac{1}{2}t_K^2 c\Big)\sup_{t\in K}\sum_{k=1}^m\E^*\bigg[\bigg\arrowvert\E^*\bigg(\exp\big(itY_{n,k}^*\big)-\exp\Big(-\frac{1}{2}t^2\sigma_{n,k}^{*2}\Big)\bigg\arrowvert \FF_{n,k-1}^*\bigg)\bigg\arrowvert\bigg].
\end{align*}

Still assuming the temporary condition (<ref>), it remains to prove that the latter expression converges to $0$ in probability. Taylor's approximation reveals

\begin{align}\label{eq: step 12}
\exp\Big(it Y_{n,k}^*\Big)=1+ it Y_{n,k}^*-\frac{1}{2}t^2Y_{n,k}^{*2} +\theta_{n,k}^*(t)
\end{align}

\begin{align*}
\sup_{t\in K}\arrowvert \theta_{n,k}^*(t)\arrowvert &\leq \sup_{t\in K}\min\Big\{\arrowvert t Y_{n,k}^*\arrowvert^3, \arrowvert tY_{n,k}^*\arrowvert^2\Big\}\\
&\leq (t_K^2+t_K^3)\Big(Y_{n,k}^{*2}\ind_{\{\arrowvert Y_{n,k}^{*}\arrowvert\geq\varepsilon\}}+\varepsilon Y_{n,k}^{*2}\Big)
\end{align*}

for any $\varepsilon>0$ as well as

\begin{align}\label{eq: step 13}
\exp\Big(\frac{1}{2}t^2\sigma_{n,k}^{*2}\Big)=1-\frac{1}{2}t^2\sigma_{n,k}^{*2}+\theta_{n,k}'(t)
\end{align}

\begin{align*}
\sup_{t\in K}\arrowvert\theta_{n,k}'(t)\arrowvert& \leq \sup_{t\in K}\Big(\frac{1}{2}t^2\sigma_{n,k}^{*2}\Big)^2\exp\Big(\frac{1}{2}t^2\sigma_{n,k}^{*2}\Big) \leq t_K^4\sigma_{n,k}^{*4}\exp\Big(\frac{1}{2}t_K^2c\Big).
\end{align*}

Therefore, with

we arrive at

\begin{align}
\sup_{t\in K}\sum_{k=1}^m&\E^*\bigg[\bigg\arrowvert\E^*\bigg(\exp\big(itY_{n,k}^*\big)-\exp\Big(-\frac{1}{2}t^2\sigma_{k,n}^{*2}\Big)\bigg\arrowvert \FF_{n,k-1}^*\bigg)\bigg\arrowvert\bigg]\nonumber\\
&\leq  c_K\sum_{k=1}^m\bigg[\E^*\Big(Y_{n,k}^{*2}\ind_{\{\arrowvert Y_{n,k}^{*}\arrowvert\geq\varepsilon\}}\Big)+\varepsilon\E^*(\sigma_{n,k}^{*2})+\E^*(\sigma_{n,k}^{*4})\bigg]\nonumber\\
&\leq c_K\bigg[\varepsilon c+c\E^*\Big(\max_{k=1,\dots, n}\sigma_{n,k}^{*2}\Big)+\sum_{k=1}^m \E^*\Big(Y_{n,k}^{*2}\ind_{\{\arrowvert Y_{n,j}^{*}\arrowvert\geq\varepsilon\}}\Big)\bigg].\label{eq: step 17}
\end{align}

Because of $\sigma_{n,l}^{*2}\leq \varepsilon^2+\sum_{k=1}^m\E^*(Y_{n,k}^{*2}\ind_{\{\arrowvert Y_{n,l}^{*}\arrowvert\geq\varepsilon\}})$ for any $1\leq l\leq m$, (<ref>) reveals that (<ref>) is upper bounded by

\begin{align*}
c_K(\varepsilon c+c\varepsilon^2)+ o_{\P}(1).
\end{align*}

Since $\varepsilon>0$ was chosen arbitrarily, this proves  (<ref>).

Claim II.  The sequence $(Z_n^*)$ is tight in probability, i.e. for any $\varepsilon>0$, there exists some compact subset $K_{\varepsilon}\subset\R$, such that

\begin{align}\label{eq: step 2}
\limsup_{n\to\infty}\P\Big(\P\big( Z_n^*\not\in K_{\varepsilon}\big\arrowvert X_1,\dots, X_n,\Pi_n\big)> \varepsilon\Big)=0.
\end{align}

Proof of Claim II. Using the identity

\begin{align*}
\int_{-u}^u\big(1-\exp(itx)\big)\dd t = 2u -\frac{\exp(iux)-\exp(-iux)}{ix}=2u-\frac{2\sin(ux)}{x},
\end{align*}

we obtain by the Theorem of Fubini

\begin{align*}
\frac{1}{u}&\int_{-u}^u\big(1-\E^*\exp(itZ_n^*)\big)\dd t\\
&=\int\Big(2-\frac{2\sin(ux)}{ux}\Big)\dd \P^{Z_n^*\arrowvert X_1,\dots, X_n,\Pi_n}(x)\\
&\geq 2\int_{\{\arrowvert x\arrowvert\geq 2/u\}}\Big(1-\Big\arrowvert\frac{\sin(ux)}{ux}\Big\arrowvert\Big)\dd \P^{Z_n^*\arrowvert X_1,\dots, X_n,\Pi_n}(x)\\
&\geq \P\Big(\arrowvert Z_n^*\arrowvert\geq\frac{2}{u}\Big\arrowvert X_1,\dots, X_n,\Pi_n\Big),
\end{align*}

where we used $\arrowvert \sin(v)/v\arrowvert\leq 1$ in the first inequality and $\arrowvert \sin(ux)\arrowvert\leq 1$ in the last line. Fix now $\varepsilon>0$ and choose $u>0$ such that

\begin{align*}
\frac{1}{u}\int_{-u}^u\bigg(1-\exp\Big(-\frac{1}{2}t^2\sigma^2\Big)\bigg)\dd t\leq\frac{\varepsilon}{2}.
\end{align*}

With $K_{\varepsilon}=[-2/u, 2/u]$, we obtain

\begin{align*}
\limsup_{n\rightarrow\infty}&\, \P \Big(\P\big( Z_n^*\not\in K_{\varepsilon}\big\arrowvert X_1,\dots, X_n,\Pi_n\big)> \varepsilon\Big)\\
&\leq \limsup_{n\rightarrow\infty}\P\bigg(\frac{1}{u}\int_{-u}^u\Big(1-\E^*\exp(itZ_n^*)\Big)\dd t>\varepsilon\bigg)\\
&\leq \limsup_{n\rightarrow\infty}\P\bigg(\frac{1}{u}\int_{-u}^u\Big(\E^*\exp(itZ_n^*)-\exp\Big(-\frac{1}{2}t^2\sigma^2\Big)\Big)\dd t>\frac{\varepsilon}{2}\bigg).
\end{align*}

The last expression is equal to zero by (<ref>), which proves (<ref>).

Claim III. For any bounded Lipschitz function $f$,

\begin{align}\label{eq: step 3}
\int f(Z_n^*)\dd \hat{\Q}_n  \overset{\P}{\longrightarrow}\int f\dd \NN(0,\sigma^2)\ \ \ (n\rightarrow\infty),
\end{align}

where $\hat{\Q}_n$ denotes the random distribution $\P(~\cdot ~\arrowvert X_1,\dots, X_n,\Pi_n)$.

Proof of Claim III.
For any $\varepsilon>0$, fix $K_{\varepsilon}=[-c_{\varepsilon},c_{\varepsilon}]$ which satisfies $\NN(0,\sigma^2)(K_{\varepsilon})\geq 1-\varepsilon$ and (<ref>).  Next, for any bounded Lipschitz function $f$  and any $\varepsilon>0$, there exists some bounded Lipschitz function $\tilde{f}_{\varepsilon}$ with

f_{\arrowvert K_{\varepsilon}}=\tilde{f}_{\varepsilon\arrowvert K_{\varepsilon}}\ \ \ \text{and}\ \ \ \tilde{f}_{\varepsilon}(x)=0\ \text{for all}\ x\in [- c_{\varepsilon}-\Arrowvert f\Arrowvert_{\sup}, c_{\varepsilon}+\Arrowvert f\Arrowvert_{\sup}]^c,

\begin{align*}
\bigg\arrowvert& \int f(Z_n^*)\dd \hat{\Q}_n -\int f\dd \NN(0,\sigma^2)\bigg\arrowvert\\
&\leq 2 \Arrowvert f\Arrowvert_{\sup} \hat{\Q}_n\big(Z_n^*\not\in K_{\varepsilon}\big)+ 2\Arrowvert f\Arrowvert_{\sup}\NN(0,\sigma^2)\big(K_{\varepsilon}^c\big)\\
&\quad\quad+\bigg\arrowvert \int \tilde{f}_{\varepsilon}(Z_n^*)\dd \hat{\Q}_n -\int \tilde{f}_{\varepsilon}\dd \NN(0,\sigma^2)\bigg\arrowvert\\
&\leq \bigg\arrowvert \int \tilde{f}_{\varepsilon}(Z_n^*)\dd \hat{\Q}_n -\int \tilde{f}_{\varepsilon}\dd \NN(0,\sigma^2)\bigg\arrowvert +4\Arrowvert f\Arrowvert_{\sup}\varepsilon+o_{\P}(1).
\end{align*}

Identifying the endpoint $-c_{\varepsilon}-\Arrowvert f\Arrowvert_{\sup}$  with $c_{\varepsilon}+\Arrowvert f\Arrowvert_{\sup}$ yields the torus $T_{\varepsilon}$ on which $f_{\varepsilon}$ is continuous. As the complex linear combinations of the monomials

m_j:x\mapsto \exp \left(i \frac{2\pi}{2c_{\varepsilon}+2}jx\right), \ j\in\Z,
restricted to $T_{\varepsilon}$ form a point-separating self-adjoint $\C$-algebra of functions on $T_{\varepsilon}$ which includes constants, the Stone-Weierstraß theorem reveals that they are dense in the space of continuous complex functions on $T_{\varepsilon}$ with respect to the topology of uniform convergence. Hence,  there exists some linear combination $P_{f,\varepsilon}=\sum_{j=1}^m\alpha_jm_j$ such that

\begin{align}\label{eq: step 3a}
\sup_{-c_{\varepsilon}-\Arrowvert f\Arrowvert_{\sup}\leq x\leq c_{\varepsilon}+\Arrowvert f\Arrowvert_{\sup}}\Big\arrowvert \tilde{f}_{\varepsilon}(x)-P_{f,\varepsilon}(x)\Big\arrowvert<\varepsilon.
\end{align}

Moreover, since $\tilde{f}_{\varepsilon}$ is bounded in absolute value by $1$ and $P_{f,\varepsilon}$ is $(2c_{\varepsilon}+2)$-periodic, (<ref>) reveals $\Arrowvert P_{f,\varepsilon}\Arrowvert_{\sup}\le \Arrowvert f\Arrowvert_{\sup} +\varepsilon$.
Claim I and Claim II then imply

\begin{align*}
\bigg\arrowvert& \int \tilde{f}_{\varepsilon}(Z_n^*)\dd \hat{\Q}_n -\int \tilde{f}_{\varepsilon}\dd \NN(0,\sigma^2)\bigg\arrowvert\\
&\leq \int_{-c_{\varepsilon}-\Arrowvert f\Arrowvert_{\sup}}^{c_{\varepsilon}+\Arrowvert f\Arrowvert_{\sup}}\Big\arrowvert \tilde{f}_{\varepsilon}(x)-P_{f,\varepsilon}(x)\Big\arrowvert\dd\Big( \hat{\Q}_n^{Z_n^*}+ \NN(0,\sigma^2)\Big)\\
&\quad\quad +(\Arrowvert f\Arrowvert_{\sup}+\varepsilon)\hat{\Q}_n\big(Z_n^*\not\in K_{\varepsilon}\big)+ (\Arrowvert f\Arrowvert_{\sup}+\varepsilon)\NN(0,\sigma^2)\big(K_{\varepsilon}^c\big)\\
&\quad\quad +\bigg\arrowvert \int P_{f,\varepsilon}(Z_n^*)\dd \hat{\Q}_n -\int P_{f,\varepsilon}\dd \NN(0,\sigma^2)\bigg\arrowvert\\
&\leq 2\varepsilon +2(\Arrowvert f\Arrowvert_{\sup}+\varepsilon)\varepsilon +o_{\P}(1).
\end{align*}

\begin{align*}
\bigg\arrowvert& \int f(Z_n^*)\dd \hat{\Q}_n -\int f\dd \NN(0,\sigma^2)\bigg\arrowvert\leq 6\Arrowvert f\Arrowvert_{\sup}\varepsilon+2\varepsilon(1+\varepsilon)+o_{\P}(1).
\end{align*}

Since $\varepsilon>0$ is arbitrary, this proves (<ref>).

Claim IV. As $n\rightarrow\infty$,

\begin{align}\label{eq: step 4}
d_{BL}\Big(\LL\big(Z_n^*\big\arrowvert X_1,\dots,X_n,\Pi_n\big),\,\NN(0,\sigma^2)\Big)\overset{\P}{\longrightarrow}0.
\end{align}

Proof of Claim IV. As in the proof of claim III, for any $\varepsilon>0$, fix $K_{\varepsilon}=[-c_{\varepsilon},c_{\varepsilon}]$ which satisfies $\NN(0,\sigma^2)(K_{\varepsilon})\geq 1-\varepsilon$ and (<ref>).
Denote the closed unit ball of bounded Lipschitz functions as introduced in Subsection <ref> by

B=\big\{f\in BL: \Arrowvert f\Arrowvert_{BL}\leq 1\big\}

and define

B_{K_{\varepsilon}}=\big\{g: K_{\varepsilon}\rightarrow\R: g=f_{\arrowvert K_{\varepsilon}}\ \text{for some}\ f\in B\big\}.

Then the set $B_{K_{\varepsilon}}$ is closed with respect to the topology of uniform convergence and therefore compact by the Arzelà-Ascoli theorem. Hence, for any $\varepsilon>0$, there exist $N\in N$ and $f_1,\dots, f_N\in B$, such that for any $f\in B$, there exists some $g_f\in\{f_1,\dots,f_N\}$ with

\begin{align*}
\sup_{x\in K_{\varepsilon}}\arrowvert f(x)-g_f(x)\arrowvert <\varepsilon.
\end{align*}

As a consequence,

\begin{align*}
\sup_{f\in B}\bigg\arrowvert \int &(f-g_f)(\dd \hat{\Q}_n^{Z_n^*}-\dd  \NN(0,\sigma^2))\bigg\arrowvert\\
&\leq \sup_{f\in B}\int_{K_{\varepsilon}}\arrowvert f-g_f\arrowvert\dd(\hat{\Q}_n^{Z_n^*} +  \NN(0,\sigma^2)) + 2\hat{\Q}_n^{Z_n^*}(K_{\varepsilon}^c) +  2\NN(0,\sigma^2)(K_{\varepsilon}^c)\\
&\leq 6\varepsilon+o_{\P}(1)
\end{align*}

and therefore,

\begin{align*}
d_{BL}&\Big(\LL\big(Z_n^*\big\arrowvert X_1,\dots,X_n,\Pi_n\big),\, \NN(0,\sigma^2)\Big)\\
&\leq \max_{j=1,\dots, N}\bigg\arrowvert \int f_j(\dd \hat{\Q}_n^{Z_n^*}-\dd  \NN(0,\sigma^2))\bigg\arrowvert+6\varepsilon + o_{\P}(1)\\
&=6\varepsilon + o_{\P}(1)
\end{align*}

by (<ref>). As $\varepsilon>0$ was arbitrary, this proves (<ref>).

To remove assumption (<ref>), the same argument as in the proof of the classical martingale-CLT can be applied.

ich bin nicht sicher, ob das wirklich gut in die Einleitung poasst. Eigentlich ziehen wir mit zuücklegen - oder ?Eventuell ans Ende von Kapitel 4! Es wäre schön, noch an einem Bsp. zu demonstrieren, dass unsere Vorausetzungen an $m$ scharf sind!
The challenges  in the construction of a consistent  ”$m$ out of $n$” bootstrap can easily be visualized in the example $T_n(f)={\rm tr} (\widehat \Sigma_n^2)$. If the population matrix is given by $\Sigma_n = I_p$, a simple calculation gives for the expectation of this statistic

\begin{equation}\label{ex4a}
\mathbb{E} \Big[{\rm tr} \big(\widehat \Sigma_n^2 \big)\Big] = \frac {p}{n} (p+1+n )= n (1+c) c \cdot (1+o(1))
\end{equation}

as $n,p \to \infty$ and $p/n \to c$. Now
let $\widehat \Sigma^*_n=m^{-1}q\sum_{i=1}^mZ_i^*(Z_i^*)^\top$  denote the corresponding matrix which is obtained by first sampling $m$ times with replacement from $Y_1, \ldots, Y_n$,
and then  resampling $q$ coordinates   as in Example <ref>  (i.e.
then drawing $q$ components randomly). Then it is easy to see

\widehat \Sigma^*_n  ~{\stackrel {\mathcal{D}}{=}} ~\frac {1}{m} \sum^n_{i=1} W_i \Pi_n Y_i (\Pi_n Y_i)^\top  ,

where $W=(W_1,\ldots,W_n)$ has a multinomial  $(m, \frac {1}{n}, \ldots, \frac {1}{n})$ distribution independent of ${\cal Y} =(Y_{1}, \ldots , Y_{n})$
and $\Pi_n$ denotes a random variable independent of  ${\cal Y} $ and $W$ and uniformly distributed on the set of all matrices $A \in \{ 0,1 \}^{q \times p}$ with exactly one entry equal to $1$ in each row and at most one entry equal to $1$ in each column korrekt?  Da müssen wir ganz genau aufpassen - ziehen mit oder ohne zurücklegen !  Vorne schrieben wir ohne - wie argumentieren wir hier ??. A straightforward calculation finally gives

\begin{eqnarray*}
\mathbb{E} \Big [ \mbox{tr} \big (  ( \widehat \Sigma^*_n )^2  \big ) \big |  {\cal Y} \Big ]
&=& \frac {1}{n} \big(1 - \frac {1}{n} + \frac {m}{n}\big) \frac{1}{m} \sum^n_{i=1} \| \Pi_n Y_i \|^4 + \frac {1}{n^2} \frac {m-1}{m} \sum_{i_1 \neq i_2} (\Pi_n Y_{i_1})^\top   (\Pi_n Y_{i_2} ) \\
& {\stackrel {\mathcal{D}}{=}}  & \frac {1}{nm} \big ( 1- \frac {1}{n} + \frac {m}{n}\big) \sum^n_{i=1} \| Y_i^{(1)} \|^4  +  \frac {1}{n^2} \frac {m-1}{m}  \sum_{i_1 \neq i_2} Y^{(1)^\top  }_{i_1} Y^{(1)}_{i_2}
\end{eqnarray*}

where $Y^{(1)}_1 , \ldots , Y^{(1)}_n  \in \mathbb{R}^{q}$ denote the vectors containing the first $q$ components of the vectors $Y_1 , \ldots , Y_{n} \in \mathbb{R}^p$, respectively  (note  that $\Sigma_{n} =I_{p}$, and by (<ref>)
the components of $Y_{i}$ are independent identically distributed).  Consequently, we obtain

\begin{eqnarray}
\label{h4}
\mathbb{E} \Big [ \mbox{tr} \big (  ( \widehat \Sigma^*_n )^2  \big )  ] &=& \frac {1}{m} \big ( 1 - \frac {1}{n} + \frac {m}{n} \big) (2 q + q^2) + \frac {n-1}{n} \frac {m-1}{m} q \\
&=& \frac {q}{m}\Big \{ \big( 1 - \frac {1}{n} + \frac {m}{n} \big) (2+q) + \frac {n-1}{n} (m-1) \Big \}~.
\nonumber
\end{eqnarray}

From this expression we already see that the classical  bootstrap  ”sampling with replacement”, which corresponds to $m=n$, $q=p$ yields to a bias,
that is $ \mathbb{E}  [  \mbox{tr} ( ( \widehat \Sigma^*_n  )^2)  ] =  nc (2c+1) \cdot (1+o(1))$.
On the other hand, if $m=o(1)$ and $q/m \to c$  the expectation in  (<ref>)
is of the order $mc (1+c) \cdot (1+o(1))$ indicating consistency of our approach.

On the other hand, for the variance additional conditions on $m \to \infty$ are required. It is well known (and relatively easy to verify) that

{\rm Var} \big( {\rm tr} \big ( \widehat \Sigma_n ^2\big)  \big) = (8c + 20c^2 +8c^3) \cdot  (1+o(1))

as $p/n \to \infty$. However, a similar statement holds for the bootstrap version only if $m$ is substantially smaller than $n$. More precisely, it can be shown

\mathbb{E} \Big [ {\rm Var} \Big ( {\rm tr} \big ( ( \widehat \Sigma^*_n)^2  \big) \mid {\cal Y} \Big) \Big ] = (8c + 20c^2 + 8c^3) (1+o(1))

if $m = o (\sqrt{n})$.

§.§.§ Tightness of the process $M_n^{1*} $

Angelika: abstraktes Argument, dass wir unbedingt arbeiten können!

It sufficient to prove the moment condition  (12.51) in
billingsley1999, that is
$\mathbb{E} $  oder  $\mathbb{E}_X $

\begin{align}
\sup_{n}
\sup_{ z_1,z_2}    \mathbb{E}  \Big [
\mathbb{E}^*   \Big [
{ |  M_n^{1*}  (z_1)  - M_n^{1*}  (z_2) |^{1+r}   \over |z_1 - z_2 |^{1+r}  }
\Big ] \Big ] \leq K
\end{align}

for some $r \in (0,1)$.
Using similar arguments as in baisil2004, p.582 we obtain

\begin{align}
\label{det1}
{   M_n^{1*}  (z_1)  - M_n^{1*}  (z_2)    \over z_1 - z_2   }
=  H_{1n}  + H_{2n}  + H_{3n}
\end{align}

\begin{align}
H_{1n}  &=
\sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_1) \beta_j^* (z_2) \big (r_j{^*}^\top (D_j^* (z_1))^{-1} (D_j^* (z_2))^{-1} r_j^* \big ) ^2
\label{det1a}
\\
H_{2n}  &=
-   \sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_1)  r_j{^*}^\top (D_j^* (z_1))^{-2} (D_j^* (z_2))^{-1} r_j^*
\label{det1b}
\\
H_{3n}  &= -
\sum_{j=1}^m
\big ( \mathbb{E}_{j}^* -\mathbb{E}_{j-1}^* \big ) \beta_j^* (z_2)  r_j{^*}^\top (D_j^* (z_2))^{-2} (D_j^* (z_1))^{-1} r_j^*
\label{detc}
\end{align}

We now bound the three terms separately, beginning with $ H_{2n}$.
From  now on the arguments differ from baisil2004 as we have to address the specific properties of  the bootstrap.
the identity

\begin{equation}
\label{det61}
b_n^*(z) = \beta_1^*(z) +  \beta_1^*(z)  b_n^*(z) \gamma_1^*(z)
\end{equation}

where $\beta_1^*(z)$
and $\gamma_1^*(z)$ are defined in (<ref>) and (<ref>), respectively. It  then follows by a
straightforward calculation that

H_{2n}  = b_n^*(z_1)W_1^* - b_n^*(z_1)W_2^*,

\begin{align*}