where in the last step we used a discussion similar to (B.30). Together with
(A.21), we can conclude that $\mathtt{R}_{11}\prec n^{-1/2-1/\alpha}.$ For
$\mathtt{R}_{12},$ using (B.19), (B.24) and (B.27)

$m_{2}-m^{(i)}_{2}\prec\frac{1}{n}\sum_{j\neq
i}\frac{\xi^{2}_{i}n^{-1-2/\alpha}}{n^{-1/\alpha}}+\frac{1}{n}\prec n^{-1}.$

Then by an argument similar to (B.20), we can conclude that
$\mathtt{R}_{12}\prec n^{-1-1/\alpha}.$ Similarly, for $R_{2},$ we have that

$\begin{split}&\frac{1}{n}\left|{\rm
tr}\left(\frac{(G^{(i)}-G)\Sigma(I+m_{2}\Sigma)^{-1}\Sigma}{1+\mathbf{y}_{i}^{*}G^{(i)}\mathbf{y}_{i}}\right)\right|=\frac{1}{n}\left|\frac{\mathbf{y}_{i}^{*}G^{(i)}\Sigma(I+m_{2}\Sigma)^{-1}\Sigma
G\mathbf{y}_{i}}{1+\mathbf{y}_{i}^{*}G^{(i)}\mathbf{y}_{i}}\right|\\\
&\asymp\frac{1}{n}\left|\mathbf{y}_{i}^{*}G^{(i)}\Sigma(I+m_{2}\Sigma)^{-1}\Sigma
G^{(i)}\mathbf{y}_{i}\right|\prec\frac{\xi_{i}^{2}}{n^{2}}\|G^{(i)}\|_{F}\|G\|_{F}\prec\frac{\xi_{i}^{2}}{n^{1+2/\alpha}}.\end{split}$

Consequently, we have that

$\frac{z}{n}\left|{\rm
tr}(R_{2}\Sigma)\right|\prec\frac{1}{n}\sum_{i}\frac{\xi_{i}^{2}}{n^{1+1/\alpha}}\prec
n^{-1-1/\alpha}.$

Combining all the above arguments, we find that (B.21) still holds true. Armed
with all the above controls, using an argument similar to the discussions
between (B.22) and (B.23), we can conclude the proof.

∎

Combining the above two lemmas, we now proceed to the proof of Proposition
B.1. We will use a continuity argument as in [27, Lemma A.12] or [14, Section
4.1]. In fact, our discussion is easier since the real part in the spectral
domain $\widetilde{\mathbf{D}}_{u}$ is divergent so that the rate is
independent of $\eta$. Due to similarity, we focus on explaining the key
ingredients.

###### Proof of Proposition B.1.

As before, due to similarity, we focus on the study of the separable
covariance i.i.d. model as in Case (2) of Assumption 2.1 when $\xi^{2}$
satisfies (2.5).

For each $z=E+\mathrm{i}\eta\in\widetilde{\mathbf{D}}_{u}$, we fix the real
part and construct a sequence $\\{\eta_{j}\\}$ by setting
$\eta_{j}=\mathtt{C}\mu_{1}-jn^{-3}$. Then it is clear that $\eta$ falls in an
interval $[\eta_{j-1},\eta_{j}]$ for some $0\leq j\leq Cn^{1/\alpha+3},C>0$ is
some constant.

In Lemma B.3, we have proved that the results hold for $\eta_{0}$. Now we
assume (B.2) holds for some $\eta_{j}$. Then according to Lemma B.4, we have
that

$|m_{1}(z_{j})-m_{1n}(z_{j})|+|m_{Q}(z_{j})-m_{n}(z_{j})|\prec
n^{-1/2-2/\alpha},\ |m_{2}(z_{j})-m_{2n}(z_{j})|\prec n^{-1/2-1/\alpha}.$

For any $\eta^{\prime}$ lying in the interval $[\eta_{k-1},\eta_{k}]$, denote
$z^{\prime}=E+\mathrm{i}\eta^{\prime}$ and $z_{j}=E+\mathrm{i}\eta_{j}$.
According to the first resolvent identity, we have that

$\|\mathcal{G}(z^{\prime})-\mathcal{G}(z_{j})\|\leqslant
n^{-3}\|\mathcal{G}(z^{\prime})\|\|\mathcal{G}(z_{j})\|\prec
n^{-11/6-1/\alpha},$ (B.35)

where in the second step we used the basic bound
$\|\mathcal{G}(z^{\prime})\|\leqslant n^{2/3},$ (B.24), (B.27) and Gershgorin
circle theorem.

On the one hand, according to the definitions in (A.20), using the first
resolvent identity, we have that

$\begin{split}m_{1}(z^{\prime})-m_{1}(z_{j})&=\frac{1}{n}{\rm
tr}[(G(z^{\prime})-G(z_{j}))\Sigma]=\frac{1}{n^{4}}{\rm
tr}(G(z^{\prime})G(z_{j})\Sigma)\prec\frac{1}{n^{4}\eta_{j}}\|G(z_{j})\|_{F}\prec
n^{-17/6-1/\alpha},\end{split}$

where in the last step we used a discussion similar to (B.30). Similarly,
combining (B.35) and (A.21), we have that

$m_{2}(z^{\prime})-m_{2}(z_{j})=\frac{1}{n}\sum_{i=1}^{n}\xi^{2}_{i}(\mathcal{G}_{ii}(z^{\prime})-\mathcal{G}_{ii}(z_{j}))\prec
n^{-11/6-1/\alpha},$

and by a discussion similar to (B.30)

$\displaystyle|m_{Q}(z^{\prime})-m_{Q}(z_{j})|$
$\displaystyle=\frac{1}{p}\left|\operatorname{tr}\left(G(z^{\prime})-G(z_{j})\right)\right|\leqslant
n^{-4}\|G(z^{\prime})\|_{F}\|G(z_{j})\|_{F}$ $\displaystyle\prec
n^{-4}(n^{1-2/\alpha})^{1/2}n^{1/2+2/3}=n^{-7/3-1/\alpha}.$

On the other hand, using the definitions in (2.16), we decompose that

$\begin{split}m_{1n}(z^{\prime})-m_{1n}(z_{j})&=\frac{1}{n}\sum_{i}\left(\frac{\sigma_{i}}{-z^{\prime}(1+\sigma_{i}m_{2n}(z^{\prime}))}-\frac{\sigma_{i}}{-z^{\prime}(1+\sigma_{i}m_{2n}(z_{j}))}\right)\\\
&+\frac{1}{n}\sum_{i}\left(\frac{\sigma_{i}}{-z^{\prime}(1+\sigma_{i}m_{2n}(z_{j}))}-\frac{\sigma_{i}}{-z_{j}(1+\sigma_{i}m_{2n}(z_{j}))}\right)\\\
&:=\mathcal{M}_{11}+\mathcal{M}_{12}.\end{split}$

For $\mathcal{M}_{11}$, according to Lemma A.1 and (2.16), we readily obtain
that

$\begin{split}\mathcal{M}_{11}&=\frac{1}{n}\sum_{i}\frac{\sigma^{2}_{i}}{-z^{\prime}(1+\sigma_{i}m_{2n}(z^{\prime}))(1+\sigma_{i}m_{2n}(z_{j}))}\big{(}m_{2n}(z_{j})-m_{2n}(z^{\prime})\big{)}\\\
&=\mathrm{O}(|z^{\prime}|^{-1})\times\frac{1}{n}\sum_{i}\Big{(}\frac{\xi^{2}_{i}}{-z_{j}(1+\xi^{2}_{i}m_{1n}(z_{j}))}-\frac{\xi^{2}_{i}}{-z_{j}(1+\xi^{2}_{i}m_{1n}(z^{\prime}))}\Big{)}\\\
&+\mathrm{O}(|z^{\prime}|^{-1})\times\frac{1}{n}\sum_{i}\left(\frac{\xi^{2}_{i}}{-z_{j}(1+\xi^{2}_{i}m_{1n}(z^{\prime}))}-\frac{\xi^{2}_{i}}{-z^{\prime}(1+\xi^{2}_{i}m_{1n}(z^{\prime}))}\right)\\\
&=\mathrm{O}(|z^{\prime}|^{-1})\times\left(\mathtt{M}_{11,1}+\mathtt{M}_{11,2}\right).\end{split}$

For $\mathtt{M}_{11,1},$ by a discussion similar to (B.26) and (B.27), we find
that

$\begin{split}\mathtt{M}_{11,1}&=\frac{1}{n}\sum_{i=1}^{n}\frac{\xi_{i}^{4}}{-z_{j}(1+\xi^{2}_{i}m_{1n}(z^{\prime}))(1+\xi^{2}_{i}m_{1n}(z_{j}))}(m_{1n}(z^{\prime})-m_{1n}(z_{j}))\\\
&\prec\left(\frac{\xi_{1}^{4}}{-nz_{j}d^{2}_{1}|m_{1n}(z_{j})||m_{1n}(z^{\prime})|}+\frac{1}{n}\sum_{i\geq
2}\frac{\xi_{i}^{4}}{|z^{\prime}(1+\xi^{2}_{1}m_{1n}(z^{\prime}))(1+\xi^{2}_{i}m_{1n}(z_{j}))|}\right)\times|m_{1n}(z_{j})-m_{1n}(z^{\prime})|\\\
&\prec\mathrm{o}(1)\times|m_{1n}(z_{j})-m_{1n}(z^{\prime})|.\end{split}$

Similarly, for $\mathtt{M}_{11,2},$ we have that

$\begin{split}\mathtt{M}_{11,2}=\frac{1}{n}\sum_{i}\frac{\xi_{i}^{2}(z_{j}-z^{\prime})}{z^{\prime}z_{j}(1+\xi_{i}^{2}m_{1n}(z^{\prime}))}\prec\frac{z^{\prime}-z_{j}}{z^{\prime}z_{j}}\prec
n^{-3-2/\alpha}.\end{split}$

Analogously, we can prove that $\mathcal{M}_{12}\prec n^{-3-4/\alpha}$.
Therefore, combining the above bounds with (A.21), we see that

$|m_{1n}(z^{\prime})-m_{1n}(z_{j})|\prec n^{-3-2/\alpha}.$

By similar procedures and arguments, we can also prove that

$|m_{2n}(z^{\prime})-m_{2n}(z_{j})|\prec n^{-3-2/\alpha},\
|m_{n}(z^{\prime})-m_{n}(z_{j})|\prec n^{-3-2/\alpha}.$

and

$\displaystyle\left\|(z^{\prime})^{-1}(I+m_{1n}(z^{\prime})D^{2})^{-1}-(z_{j})^{-1}(I+m_{1n}(z_{j})D^{2})^{-1}\right\|\prec
n^{-3-2/\alpha}.$

Therefore, combining all the above bounds with triangle inequality, we see
that the results of part 1 of Theorem A.7 hold for $z^{\prime}.$ Using an
induction procedure and a standard lattice argument (for example, see [14,
27]), we find that the results hold for all $z\in\widetilde{\mathbf{D}}_{u}$
and conclude the proof of Proposition B.1.

∎

####  B.1.2 Proof of Theorem A.7

Once Proposition B.1 is proved, we can roughly locate the edge eigenvalues of
$Q$ as in Lemma B.2 so that we can expand the spectral domain from
$\widetilde{\mathbf{D}}_{u}$ to $\mathbf{D}_{u}$ for $Q^{(1)}$ and conclude
the proof of Theorem A.7.

As before, we focus on the study of the separable covariance i.i.d. model as
in Case (2) of Assumption 2.1 when $\xi^{2}$ satisfies (2.5). Recall the
definitions of $\mu_{2}$ and $\lambda_{1}^{(1)}$ around (C.1) and in Figure 7.
By Lemma B.2 and an analogous argument, as well as Weyl’s inequality, we find
that conditional on the event $\Omega,$ with high probability,

$\mu_{1}>\lambda_{1}>\mu_{2}>\lambda_{1}^{(1)}.$ (B.36)

By (C.7) and a similar argument, we see that $\mu_{k}\asymp\xi_{k}^{2},k=1,2.$
Together with (A.21), we have that $\mu_{1}-\mu_{2}\geq
C_{1}n^{1/\alpha}\log^{-1}n$ for some constant $C_{1}>0$ on the event
$\Omega.$ This implies for some constant $C>0,$ for $z\in\mathbf{D}_{u},$

$\displaystyle|\lambda_{1}^{(1)}-z|\geq Cn^{1/\alpha}\log^{-1}n,$ (B.37)

Now we proceed to the proof of Theorem A.7. Recall (A.3) and (A.4).

###### Proof of Theorem A.7.

Again, we focus on the study of the separable covariance i.i.d. model as in
Case (2) of Assumption 2.1 when $\xi^{2}$ satisfies (2.5).

Observe by (B.37) that it holds uniformly for $z\in\mathbf{D}_{u}$ and
$\mathcal{T}\subset\\{2,\dots,n\\}$, for some constant $C_{1}>0$

$\displaystyle\|G^{(1\mathcal{T})}\|\leqslant C_{1}n^{-1/\alpha}\log n.$
(B.38)

By the definition of $m_{2}^{(1)}$ and a decomposition similar to (B.9), we
have that

$\displaystyle
m_{2}^{(1)}=\frac{1}{n}\sum_{i=2}^{n}\frac{\xi^{2}_{i}}{-z-z\mathbf{y}_{i}^{*}G^{(1i)}\mathbf{y}_{i}}=\frac{1}{n}\sum_{i=2}^{n}\frac{\xi^{2}_{i}}{-z(1+\xi^{2}_{i}n^{-1}\operatorname{tr}G^{(1i)}\Sigma+Z_{i}^{(1)})},$
$\displaystyle
Z_{i}^{(1)}=\mathbf{y}_{i}^{*}G^{(1i)}\mathbf{y}_{i}-\xi^{2}_{i}n^{-1}\operatorname{tr}G^{(1i)}\Sigma.$

By arguments similar to (B.10) and (B.11) but with (B.38), we obtain that

$\displaystyle
Z_{i}^{(1)}\prec\frac{\xi^{2}_{i}}{n}\|G^{(1i)}\Sigma\|_{F}\leq\frac{\xi^{2}_{i}}{n}\|G^{(1i)}\|\|\Sigma\|_{F}\prec\frac{\xi^{2}_{i}}{n^{1/2}}n^{-1/\alpha},$
$\displaystyle\frac{1}{n}\operatorname{tr}(G^{(1i)}\Sigma)-m_{1}^{(1)}(z)=\frac{1}{n}\mathbf{y}^{*}G^{(1)}\Sigma
G^{(1i)}\mathbf{y}_{i}\prec\frac{\xi^{2}_{i}}{n}n^{-2/\alpha}.$

In addition, using (B.38) and a discussion similar to (B.32)–(B.34), we
readily see that

$\displaystyle
m^{(1)}_{2}=\frac{1}{n}\sum_{i=2}^{n}\frac{\xi^{2}_{i}}{-z(1+\xi^{2}_{i}m^{(1)}_{1})}+\mathrm{O}_{\prec}\left(n^{-1/2-1/\alpha}\right).$

Using the decomposition

$\displaystyle
Q^{(1)}-zI=\sum_{i=2}^{n}\mathbf{y}_{i}\mathbf{y}_{i}^{*}+zm^{(1)}_{2}(z)\Sigma-z(I+m_{2}^{(1)}(z)\Sigma),$

by arguments similar to (B.14)–(B.21) with
$\|\mathcal{G}^{(1\mathcal{T})}\|=\|G^{(1\mathcal{T})}\|\prec n^{-1/\alpha}$,
we conclude that

$\displaystyle\begin{split}m_{1}^{(1)}&=-z^{-1}\frac{1}{n}\operatorname{tr}((I+m_{2}^{(1)}(z)\Sigma)^{-1}\Sigma)+\mathrm{O}_{\prec}\left(n^{-1/2-2/\alpha}\right)\\\
&=-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{z(1+m_{2}^{(1)}\sigma_{i})}+\mathrm{O}_{\prec}\left(n^{-1/2-2/\alpha}\right).\end{split}$

Combining with the definitions in (2.16), we see that

$\displaystyle\begin{split}m_{1}^{(1)}(z)-m_{1n}^{(1)}(z)&=-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{z(1+\sigma_{i}m_{2}^{(1)}(z))}+\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{z(1+\sigma_{i}m_{2n}^{(1)}(z))}+\mathrm{O}_{\prec}(n^{-1/2-2/\alpha})\\\
&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}(m_{2}^{(1)}(z)-m_{2n}^{(1)}(z))}{z(1+\sigma_{i}m_{2n}^{(1)}(z))(1+\sigma_{i}m_{2}^{(1)}(z))}+\mathrm{O}_{\prec}(n^{-1/2-2/\alpha})\\\
&=\left(\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}}{z(1+\sigma_{i}m_{2n}^{(1)}(z))(1+\sigma_{i}m_{2}^{(1)}(z))}\right)\left(\frac{1}{n}\sum_{i=2}^{n}\frac{\xi^{4}_{i}(m_{1}^{(1)}(z)-m_{1n}^{(1)}(z))}{z(1+\xi^{2}_{i}m_{1}^{(1)}(z))(1+\xi^{2}_{i}m_{1n}^{(1)}(z))}\right)+\mathrm{O}_{\prec}(n^{-1/2-2/\alpha})\\\
&=\mathrm{o}(1)(m_{1}^{(1)}(z)-m_{1n}^{(1)}(z))+\mathrm{O}_{\prec}(n^{-1/2-2/\alpha}),\end{split}$

where in the third step we used a discussion similar to (B.33) and (A.21).
This completes our proof.

∎

###  B.2 Bounded support setting: proof of Theorem A.8

In this section, we will prove Theorem A.8. In Section B.2.1, we study
$m_{1n,c}(z)-m_{1n}(z)$ and $m_{n,c}-m_{n},$ which is a counterpart of Lemma
4.4 of [60]. Then in Section B.2.2, we study $m_{1n}(z)-m_{1}(z)$ and
$m_{Q}-m_{n},$ which is a counterpart of Proposition 5.1 of [60]. Due to
similarity, we only provide the details for the separable covariance i.i.d
data as in (2) of Assumption 2.1. The elliptical data can be handled
similarly.

####  B.2.1 Control of $m_{1n,c}(z)-m_{1n}(z)$ and $m_{n,c}(z)-m_{n}(z)$

Due to similarity, we focus on $|m_{1n,c}-m_{1n}|$ and briefly discuss
$|m_{n,c}-m_{n}|$ in the end. The proof ideas follow Lemma 4.5 of [62] or
Lemma 4.4 of [60]. We focus on explaining the parts deviates the most.

###### Proof.

According to the definitions of $m_{1n,c}$ and $m_{1n}$ in (2.17) and (2.16),
we observe that

$\displaystyle|m_{1n,c}(z)-m_{1n}(z)|$
$\displaystyle\leqslant\left|\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z+\sigma_{i}\int\frac{s}{1+sm_{1n,c}(z)}\mathrm{d}F(s)}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(z)}}\right|$
$\displaystyle+|m_{1n,c}(z)-m_{1n}(z)|\left|\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n}(z))(1+\xi^{2}_{j}m_{1n,c}(z))}}{(-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)})(-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(z)})}\right|$
(B.39) $\displaystyle:=\mathsf{P}_{1}+\mathsf{P}_{2}.$

On the one hand, for $\mathsf{P}_{1},$ we have that

$\mathsf{P}_{1}=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}|n^{-1}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(z)}-\int\frac{s}{1+sm_{1n,c}(z)}\mathrm{d}F(s)|}{|(-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(z)})(-z+\sigma_{i}\int\frac{s}{1+sm_{1n,c}(z)}\mathrm{d}F(s))|}.$

Since $z\in\mathbf{D}_{b}^{\prime}\subset\mathbf{D}_{b},$ according to
Assumption 2.10 and the continuity of $m_{1n,c}$, we conclude that
$|-z+\sigma_{i}\int\frac{s}{1+sm_{1n,c}(z)}\mathrm{d}F(s))|\geqslant c$ for
some constant $c>0.$ Together with (A.23), we show that
$|-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(z)}|\geqslant
c^{\prime}$ for some $c^{\prime}>0$ when $n$ is sufficiently large. Using
(A.23) again, we conclude that on $\Omega,$ for some small constant
$\epsilon>0$ and some constant $C>0$

$\mathsf{P}_{1}\leqslant Cn^{-1/2+\epsilon}.$ (B.40)

On the other hand, for $\mathsf{P}_{2},$ for notional convenience, we further
write it as $\mathsf{P}_{2}=|m_{1n,c}(z)-m_{1n}(z)|\times|\mathsf{T}|.$ For
$\mathsf{T},$ by Cauchy-Schwarz inequality, we have that

$|\mathsf{T}|\leqslant\mathsf{E}_{1}\mathsf{E}_{2},$ (B.41)

where $\mathsf{E}_{k},k=1,2,$ are defined as

$\displaystyle\begin{split}&\mathsf{E}_{1}:=\left(\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma^{2}_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n}(z))^{2}}}{|-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}\right)^{1/2},\\\
&\mathsf{E}_{2}:=\left(\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma^{2}_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n,c}(z))^{2}}}{|-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(z)}|^{2}}\right)^{1/2}.\end{split}$
(B.42)

Together with the identity (D.21) below and the fact $m_{1n}(z)\asymp 1$ for
$z\in\mathbf{D}_{b}^{\prime}$, we find that $\mathsf{E}_{1}\leqslant 1.$ For
the term $\mathsf{E}_{2},$ we first consider a closely related quantity
$\mathsf{W}(z)$ defined as

$\mathsf{W}(z):=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}\int\frac{s^{2}}{|1+sm_{1n,c}(z)|^{2}}\mathrm{d}F(s)}{|-z+\sigma_{i}\int\frac{s^{2}}{(1+sm_{1n,c}(z))^{2}}\mathrm{d}F(s)|^{2}}=1-\eta\frac{|m_{1n,c}(z)|^{2}}{\operatorname{Im}m_{1n,c}(z)}.$

By assumption that $\phi^{-1}>\varsigma_{3},$ (3.7) that
$m_{1n,c}(L_{+})=-l^{-1}$ and recall the notations in (3.5), we see that

$\mathsf{W}(L_{+})<1.$ (B.43)

Armed with (B.43), using (A.23) and Assumption 2.10, we can apply an argument
similar to Lemma A.6 of [62] or Lemma A.7 of [60] to conclude that when $n$ is
sufficiently large, for $z\in\mathbf{D}_{b},$

$\mathsf{E}^{2}_{2}=\mathsf{W}(L_{+})+\mathrm{o}(1)<\mathfrak{c}^{\prime},$
(B.44)

for some constant $0<\mathfrak{c}^{\prime}<1.$ Consequently, we find that when
$n$ is sufficiently large, $\mathsf{E}_{2}<1.$ Together with (B.41), we can
conclude that $|\mathsf{T}|<1$. This yields that

$\mathsf{P}_{2}=\mathfrak{c}|m_{1n,c}-m_{1n}|,$

for some constant $0<\mathfrak{c}<1.$

Inserting the above control back into (B.2.1), using (B.40), we can conclude
our proof that

$m_{1n,c}=m_{1n}(z)+\mathrm{O}(n^{-1/2+\epsilon}).$ (B.45)

The proof of $m_{n,c}-m_{n}$ follows from an argument similar to (B.2.1) using
(2.16) and (2.17) that

$m_{n}(z)=\frac{1}{p}\sum_{i=1}^{p}\frac{1}{-z+\sigma_{i}n^{-1}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n}}},\
m_{n,c}(z)=\frac{1}{p}\sum_{i=1}^{p}\frac{1}{-z+\sigma_{i}\int_{0}^{l}\frac{s}{1+sm_{1n,c}(z)}\mathrm{d}F(s)},$

the results of $m_{1n,c}-m_{1n}$ and (A.23). We omit the details. ∎

####  B.2.2 Control of $m_{1n}(z)-m_{1}(z)$ and $m_{n}(z)-m_{Q}(z)$

Due to similarity, we focus on $|m_{1n,c}-m_{1n}|$ and will briefly discuss
how to study $|m_{Q}-m_{n}|$ from line to line. The proof ideas follow
Proposition 5.1 of [62] or Proposition 5.1 of [60]. We focus on explaining the
parts deviates the most. The proof relies on the following two lemmas.

######  Lemma B.5.

Conditional on the event $\Omega$ in Theorem A.8, for all
$z=E+\mathrm{i}\eta\in\mathbf{D}_{b}^{\prime}$ with
$n^{-1/2+\epsilon_{d}}\leq\eta\leq n^{-1/(d+1)+\epsilon_{d}}$, we have

$\displaystyle|m_{1n}(z)-m_{1}(z)|\prec\frac{1}{n\eta_{0}},\
|m_{n}(z)-m_{Q}(z)|\prec\frac{1}{n\eta_{0}}.$

######  Lemma B.6.

Assuming that $|m_{1n}(z)-m_{1}(z)|\prec n^{\epsilon_{d}}(n\eta_{0})^{-1}$,
then conditional on the event $\Omega,$ we have that for all
$z\in\mathbf{D}_{b}^{\prime}$

$\displaystyle|m_{1n}(z)-m_{1}(z)|\prec\frac{1}{n\eta_{0}},\
|m_{n}(z)-m_{Q}(z)|\prec\frac{1}{n\eta_{0}}.$

Armed with the above two lemmas, we now proceed to the control of
$m_{1n}(z)-m_{1}(z).$

###### Proof: control of $m_{1n}(z)-m_{1}(z)$ and $m_{n}(z)-m_{Q}(z)$.

Due to similarity, we only prove $m_{1n}(z)-m_{1}(z).$ We prove this by
mathematical induction as that of Proposition 5.1 of [62]. Fix $E$ such that
$z=E+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime}$, we consider a sequence
$(\eta_{j})$ defined by $\eta_{j}=\eta_{0}+jn^{-2}$. Let $K$ be the smallest
positive integer such that $\eta_{K}\geq n^{-1/2+\epsilon_{d}}$. Note that for
$j=K$, by Lemma B.5, we have that

$\displaystyle|m_{1n}(z_{j})-m_{1}(z_{j})|\prec\frac{1}{n\eta_{0}}.$

Then for any $z=E+\mathrm{i}\eta$ with $\eta_{j-1}\leq\eta\leq\eta_{j}$, we
have that for some constant $C>0$

$\displaystyle|m_{1}(z_{j})-m(z)|=\frac{1}{n}\operatorname{tr}\left[(G(z_{j})-G(z))\Sigma\right]=\frac{|z_{j}-z|}{n}\operatorname{tr}(G(z_{j})G(z)\Sigma)\leq
C\frac{|z_{j}-z|}{\eta_{j-1}^{2}}\leq C\frac{n^{2\epsilon_{d}}}{n},$

where we used the first resolvent identity and the trivial bound
$|G(z)|\leqslant\eta^{-1}$, and similarly

$\displaystyle|m_{1n}(z_{j})-m_{1n}(z)|=\left|\int\left[\frac{1}{x-z_{j}}-\frac{1}{x-z}\right]\rho(x)\mathrm{d}x\right|\leq\frac{|z_{j}-z|}{\eta_{j-1}^{2}}\leq\frac{n^{2\epsilon_{d}}}{n}.$

Thus we find that if $|m_{1n}(z_{j})-m_{1}(z_{j})|\prec\frac{1}{n\eta_{0}}$,
then by Lemma B.6, for some constant $C^{\prime}>0$

$\displaystyle|m_{1n}(z)-m_{1}(z)|\leq|m_{1n}(z_{j})-m_{1}(z_{j})|+\frac{C^{\prime}n^{2\epsilon_{d}}}{n}\prec\frac{n^{\epsilon_{d}}}{n\eta_{0}}.$
(B.46)

This gives the result that $|m_{1n}(z)-m_{1}(z)|\prec(n\eta_{0})^{-1}$ for
$z=E+\mathrm{i}\eta$ with $\eta_{j-1}\leq\eta\leq\eta_{j}$. The proof for each
$z$ can be completed by an induction on $j.$ Finally, using an induction
procedure and a standard lattice argument (for example, see [14, 27]), we find
that the results hold for all $z\in\mathbf{D}_{b}^{\prime}$. More
specifically, we construct a lattice $\mathcal{L}$ from
$z^{\prime}=E^{\prime}+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime}$ with
$|z-z^{\prime}|\leq n^{-3}$. It is obvious that the bound holds uniformly on
$\mathcal{L}$. For any $z=E+\mathrm{i}\eta_{0}\notin\mathcal{L}$, we find a
$z^{\prime}\in\mathcal{L}$ and then $|z-z^{\prime}|\leq n^{-3}$. Moreover,
using resolvent identity, we can conclude that
$|m_{1}(z)-m_{1}(z^{\prime})|\leq\eta^{-2}_{0}|z-z^{\prime}|\ll(n\eta_{0})^{-1}$.
Therefore, we conclude the proof.

∎

In what follows, we prove lemmas B.5 and B.6. The proofs are similar to those
of Lemmas 5.6 and 5.7 of [62], except that we will need a weak local law as
follows.

######  Proposition B.7 (Weak averaged local law).

Suppose the assumptions of Theorem A.8 hold. We have that for
$z\in\mathbf{D}_{b}^{\prime}$

$|m_{Q}(z)-m_{n}(z)|+|m_{1}(z)-m_{1n}(z)|+|m_{2}(z)-m_{2n}(z)|=\mathrm{O}_{\prec}\left((n\eta)^{-1/4}\right).$

###### Proof.

The proof of Proposition B.7 is relatively standard in the random matrix
literature, for example, see Section 4.1 of [14] or Section 3.6 of [37] or
Appendix A.2 of [27] or Section 5.2 of [92]. Due to similarity, as in Lemma
5.12 of [92], we only provide the key ingredients. Define the $z$-dependent
parameter

$\displaystyle\Psi(z):=\sqrt{\frac{\operatorname{Im}m_{1}(z)}{n\eta}}+\frac{1}{n\eta}.$
(B.47)

Recall (B.9). By Lemma A.14 and (A.24), we find that

$\displaystyle
Z_{i}\prec\frac{\xi^{2}_{i}}{n}\|G^{(i)}\Sigma^{1/2}\|_{F}\leqslant
l\sqrt{\frac{\operatorname{Im}m^{(i)}_{1}(z)}{n\eta}}\asymp\Psi,$ (B.48)

where in the last step we used (A.25). Together with (A.25) and the first
equation of (B.9), we conclude that

$\displaystyle
m_{2}=\frac{1}{n}\sum_{i=1}^{n}\frac{\xi^{2}_{i}}{-z(1+\xi^{2}_{i}m_{1}(z)+\mathrm{O}_{\prec}(\Psi))}.$
(B.49)

For $m_{1}(z),$ recall (B.14). According to the definition of $m_{1}(z)$ in
(A.20), we have that

$\displaystyle
m_{1}(z)=\frac{1}{n}\operatorname{tr}(G(z)\Sigma)=-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{z(1+m_{2}(z)\sigma_{i})}+\frac{1}{n}\operatorname{tr}(R_{1}\Sigma)+\frac{1}{n}\operatorname{tr}(R_{2}\Sigma).$
(B.50)

Similarly, for $m_{Q}(z)$ in (2.14), we have that

$m_{Q}(z)=\frac{1}{p}\operatorname{tr}(G(z))=-\frac{1}{p}\sum_{i=1}^{p}\frac{1}{z(1+m_{2}(z)\sigma_{i})}+\frac{1}{p}\operatorname{tr}(R_{1})+\frac{1}{p}\operatorname{tr}(R_{2}).$
(B.51)

On the one hand, when $\eta\asymp 1,$ by a discussion similar to (5.45) of
[92] , we find that $\|(I+m_{2}^{(i)}\Sigma)^{-1}\|<\infty.$ Then using
(B.48), by a discussion similar to the equations between (B.15) and (B.21), we
find that

$\displaystyle
m_{1}(z)=-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{z(1+m_{2}(z)\sigma_{i})}+\mathrm{O}_{\prec}\Big{(}\frac{1}{n}\sum_{i}\frac{\xi^{2}_{i}\Psi}{z(1+\xi^{2}_{i}m_{1}(z)+\mathrm{O}_{\prec}(\Psi))}\Big{)}.$
(B.52)

Similarly, we have

$m_{Q}(z)=-\frac{1}{p}\sum_{i=1}^{p}\frac{1}{z(1+m_{2}(z)\sigma_{i})}+\mathrm{O}_{\prec}\Big{(}\frac{1}{p}\sum_{i}\frac{\xi^{2}_{i}\Psi}{z(1+\xi^{2}_{i}m_{1}(z)+\mathrm{O}_{\prec}(\Psi))}\Big{)}.$

On the other hand, denote
$\Xi:=\\{|\mathcal{G}_{ij}(z)+\delta_{ij}(z(1+m_{1n}(z)\xi_{i}^{2}))^{-1}|+|m_{2}(z)-m_{2n}(z)|\leqslant(\log
n)^{-1}\\}.$ When restricted on $\Xi,$ by Assumption 2.10, we also have that
$\|(I+m_{2}^{(i)}\Sigma)^{-1}\|<\infty.$ By an analogous argument, we find
that (B.52) also holds true. By an argument similar to (B.29) using Lemma
A.12, we have that for $i\neq j,$ $\mathbf{1}(\eta\geqslant
1)\mathcal{G}_{ij}\prec\Psi,\ \mathbf{1}(\Xi)\mathcal{G}_{ij}\prec\Psi.$

The above arguments show that the counterparts of Lemmas 5.9 and 5.10 of [92]
hold. Therefore, by the same arguments as in Lemma 5.12 of [92], we can
conclude the proof.

∎

Next we provide some useful controls whose proofs and results will be used in
the proof of Lemmas B.5 and B.6. The results and arguments are analogous to
Lemma 5.4 of [62]. We only point out the key ingredients in our proof and
refer the readers to [62] for more details.

######  Lemma B.8.

Suppose the assumptions of Theorem A.8 hold. Then we have that on $\Omega$ and
for all $z=E+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime}$

$\displaystyle\operatorname{Im}m_{1}(z)\prec\frac{1}{n\eta_{0}},\ \
\operatorname{Im}m_{Q}(z)\prec\frac{1}{n\eta_{0}}.$

###### Proof.

Due to similarity, we focus our proof on $\operatorname{Im}m_{Q}(z).$ We prove
by contradiction. Given some $\epsilon>0,$ conditional on $\Omega,$ for some
sufficiently small constants $0<c_{1},c_{2}<1,$ we first introduce a
probability event $\Xi_{1}\equiv\Xi_{1}(\epsilon)$ so that the followings
holds:

  1. 1.

For $z\in\mathbf{D}_{b}^{\prime},$

$|m_{Q}(z)-m_{n}(z)|+|m_{1}(z)-m_{1n}(z)|+|m_{2}(z)-m_{2n}(z)|\leqslant(n\eta)^{-1/4+c_{1}\epsilon}.$

  2. 2.

For $z\in\mathbf{D}_{b}^{\prime}$ and $Z_{i}$ in (B.9),

$\max_{i}Z_{i}\leqslant n^{c_{2}\epsilon}\Psi.$

According to Proposition B.7 and (B.48), we find that there exists some large
constant $D>0$ so that $\mathbb{P}(\Xi_{1})=1-n^{-D}.$ In what follows, we
restrict ourselves on $\Xi_{1}$ so that the discussions are purely
deterministic.

Assuming that

$\operatorname{Im}m_{1}(z)>n^{\epsilon}\frac{1}{n\eta_{0}}.$

We then conclude from the definition of $\Psi$ in (B.47) that

$\Psi=\mathrm{o}(\operatorname{Im}m_{1}(z)).$ (B.53)

Moreover, by (A.15) and (A.16), we readily see that
$\operatorname{Im}m_{1n}(z)\ll\operatorname{Im}m_{1}(z)$. This implies that
for some constant $C>0$

$\displaystyle|m_{1n}(z)-m_{1}(z)|\geq|\operatorname{Im}m_{1n}-\operatorname{Im}m_{1}|>Cn^{\epsilon}\frac{1}{n\eta_{0}}.$
(B.54)

On the other hand, by Proposition B.7 and Assumption 2.10, we see that (B.52)
still holds. Together with $m_{1n}$ in (2.16), using (B.9), we find that

$\displaystyle m_{1n}-m_{1}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}(m_{1n}-m_{1})+\xi^{2}_{j}\mathrm{O}(n^{c_{2}\epsilon}\Psi)}{(1+\xi^{2}_{j}m_{1n})(1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi))}}{(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}})(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi)})}+\mathrm{O}(n^{c_{2}\epsilon}\Psi)$
$\displaystyle=\mathsf{C}_{1}(m_{1n}-m_{1})+\mathsf{C}_{2}+\mathrm{O}(n^{c_{2}\epsilon}\Psi),$
(B.55)

where $\mathsf{C}_{1},\mathsf{C}_{2}$ are defined as

$\displaystyle\mathsf{C}_{1}:=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n})(1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi))}}{(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}})(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi)})},$
$\displaystyle\mathsf{C}_{2}:=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{2}_{j}\mathrm{O}(n^{c_{2}\epsilon}\Psi)}{(1+\xi^{2}_{j}m_{1n})(1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi))}}{(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}})(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi)})}.$

We first control $\mathsf{C}_{2}.$ It is easy to see that $m_{1}\sim 1$ by
contradiction. If $m_{1}\ll 1$, one can observe from (B.49) that $m_{2}\sim 1$
which yields $m_{1}\asymp 1$ by (B.52). If $m_{1}\gg 1$, we have $m_{2}\ll 1$
from (B.49), and then it gives that $m_{1}\asymp 1$ by (B.52). Similarly, we
can show that $m_{2}\asymp 1.$ Together with Proposition B.7, we find that
$m_{1n},m_{2n}\asymp 1.$ Since $z\asymp 1,$ using the definition of $m_{2n}$
in (2.16) and Proposition B.7, we find that

$\frac{1}{n}\sum_{j}\frac{\xi^{2}_{j}\mathrm{O}(n^{c_{2}\epsilon}\Psi)}{(1+\xi^{2}_{j}m_{1n})(1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi))}=\mathrm{O}(n^{c_{2}\epsilon}\Psi).$

Moreover, by Proposition B.7, (B.45) and Assumption 2.10, we find that

$\frac{1}{n}\sum_{i}\frac{1}{(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}})(z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1}+\mathrm{O}(n^{c_{2}\epsilon}\Psi)})}\asymp
1.$

This yields that

$\mathsf{C}_{2}=\mathrm{O}(n^{c_{2}\epsilon}\Psi).$ (B.56)

For $\mathsf{C}_{1},$ using Proposition B.7 and Assumption 2.10, by an
argument similar to (B.41), we can conclude that when $n$ is sufficiently
large, for some constant $0<\mathfrak{c}<1,$

$\mathsf{C}_{1}\leqslant\mathfrak{c}.$ (B.57)

Combining (B.55), (B.56) and (B.57), we conclude that

$\displaystyle|m_{1n}-m_{1}|=\mathrm{O}(n^{c_{2}\epsilon}\Psi),$

which contradicts (B.54) since $c_{2}<1$ is sufficiently small. This completes
our proof for each fixed $z.$ For uniformity in $z,$ we can follow a standard
lattice argument as discussed below (B.46). This finishes the proof. The
discussion for $m_{Q}$ follows from an analogous discussion with the help of
(B.49) and (B.52). ∎

######  Remark B.9.

Two remarks are in order. First, it is easy to see that repeating the proof of
Lemma B.8, we can prove the results for all $\eta$ as specified in (A.2).
Second, we remark that combining (B.48) and Lemma B.8, when
$z=E+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime},$ we have that conditional
on $\Omega$

$Z_{i}\prec\frac{1}{n\eta_{0}}.$

Armed with the above discussions and results, following the strategies of
Lemmas 5.6 and 5.7 of [62] or [60], we prove Lemmas B.5 and B.6 using similar
arguments as in Lemma B.8. Due to similarity, we only provide the key
ingredients.

###### Proof of Lemmas B.5 and B.6.

Due to similarity, we focus our proof on Lemma B.5 and briefly mention that of
Lemma B.6 in the end. Due to similarity, we only explain
$|m_{1n}(z)-m_{1}(z)|.$

The proof is similar to that of Lemma B.8 and we prove by contradiction. We
also restrict ourselves on the event $\Xi_{1}$ in Lemma B.8. We assume that
$|m_{1n}(z)-m_{1}(z)|>n^{\epsilon}(n\eta_{0})^{-1}$. To see a contraction, in
addition to the the arguments of Lemma B.8, we need to provide a finer control
for $\Psi$ since in the current proof it depends on $\eta$ instead of
$\eta_{0}.$ Note that

$\displaystyle\begin{split}n^{c_{2}\epsilon}\Psi&=n^{c_{2}\epsilon}\sqrt{\frac{|\operatorname{Im}m_{1}-\operatorname{Im}m_{1n}+\operatorname{Im}m_{1n}|}{n\eta}}+n^{c_{2}\epsilon}\frac{1}{n\eta}\\\
&\leq
n^{c_{2}\epsilon}\sqrt{\frac{|\operatorname{Im}m_{1}-\operatorname{Im}m_{1n}|}{n\eta}}+n^{c_{2}\epsilon}\sqrt{\frac{\operatorname{Im}m_{1n}}{n\eta}}+n^{c_{2}\epsilon}\frac{1}{n\eta}\\\
&=\mathrm{o}(|m_{1}-m_{1n}|),\end{split}$ (B.58)

where in the last step we used (A.14) and the assumption
$|m_{1n}(z)-m_{1}(z)|>n^{\epsilon}(n\eta_{0})^{-1}\gg(n\eta)^{-1}$ when
$n^{-1/2+\epsilon_{d}}\leqslant\eta\leqslant n^{-1/(d+1)+\epsilon_{d}}.$
Replacing $n^{c_{2}\epsilon}\Psi$ with $\mathrm{o}(|m_{1}-m_{1n}|)$ in the
arguments between (B.55) and (B.57), we find that
$|m_{1n}-m_{1}|=\mathrm{o}(|m_{1n}-m_{1}|)$ which is a contraction. This
proves the result for each fixed $z.$ The uniformity follows from the same
lattice argument as mentioned in the end of the proof of Lemma B.8.

The proof of Lemma B.6 is similar. We also prove by contradiction and assume
that $n^{\epsilon}(n\eta_{0})^{-1}<|m_{1}(z)-m_{1n}(z)|\leq
n^{\epsilon+\epsilon_{d}}(n\eta_{0})^{-1}.$ Under this assumption, together
with (A.15) and (A.16), we find that (B.58) still holds true. The rest of the
arguments are similar and we omit the details.

∎

##  Appendix C Locations for extreme eigenvalues and proof of the main results

In this section, we study the first order convergent limits of the largest
eigenvalues of $Q,$ i.e., $\lambda_{1}(Q).$ In Section C.1, we investigate the
case when $\\{\xi_{i}^{2}\\}$ have unbounded support as in (i) of Assumption
2.4. In Section C.2, we study the bounded support case as in (ii) of
Assumption 2.4.

###  C.1 The unbounded support case

In order to quantify the location of $\lambda_{1}\equiv\lambda_{1}(Q),$ we
need to introduce several auxiliary quantities. Recall $\mu_{1}$ defined in
(3.14). Similarly, we denote $\mu_{2}$ by replacing $\xi^{2}_{(1)}$ with
$\xi^{2}_{(2)}$ in (3.14). Moreover, for $d_{1}$ and the sufficiently small
constant $\epsilon>0$ in (3.15), we denote

$\mu_{1}^{\pm}:=\mu_{1}\pm n^{-1/2+2\epsilon}d_{1},$ (C.1)

and recall that

$Q^{(1)}:=Q-\mathbf{y}_{(1)}\mathbf{y}^{*}_{(1)},$ (C.2)

where $\mathbf{y}_{(1)}$ is the column of $Y$ in (2.1) associated with
$\xi^{2}_{(1)}$. Accordingly, we denote the largest eigenvalue of $Q^{(1)}$ as
$\lambda_{1}^{(1)}\equiv\lambda_{1}(Q^{(1)}).$ Throughout this section, we
shall prove Figure 7 so that the location of $\lambda_{1}$ can be quantified
with high probability on the event $\Omega$.

$\lambda_{2}$$\lambda_{1}^{(1)}$$\mu_{2}$$\mu_{1}^{-}$$\mu_{1}$$\mu_{1}^{+}$$\lambda_{1}$
is here Figure 7: Location of the largest eigenvalue of $Q$.

More formally, the main result is summarized in Proposition C.1 below.

######  Proposition C.1.

Suppose Assumptions 2.1, 2.6 and (i) of Assumption 2.4 hold. For some
sufficiently small constant $\epsilon>0$ and $\mu_{1}^{\pm}$ defined in (C.1),
condition on the probability event $\Omega$ in Lemma A.11, with high
probability, we have that

$\lambda_{1}\in[\mu_{1}^{-},\mu_{1}^{+}].$

We now proceed to the proof of Proposition C.1 following the structure
described in Figure 7.

###### Proof of Proposition C.1.

Due to similarity, we focus on the study of the separable covariance i.i.d.
model as in Case (2) of Assumption 2.1 when $\xi^{2}$ satisfies (2.5). The
main differences from the other cases will be explained in the end of the
proof.

In what follows, we restrict the discussion on the probability event $\Omega$
in Lemma A.11. By Weyl’s inequality, we have that
$\lambda_{2}\leqslant\lambda_{1}^{(1)}.$ Moreover, by (B.36), we see that with
high probability $\lambda_{1}^{(1)}<\mu_{2}.$ The rest of the proof leaves to
prove that the following two claims:

$\mu_{1}-\mu_{2}\geqslant n^{1/\alpha}\log^{-1}n,$ (C.3)

and for $Q^{(1)}$ in (C.2) and

$M(\lambda)=1+\mathbf{y}^{*}_{(1)}G_{1}^{(1)}(\lambda)\mathbf{y}_{(1)},\
G^{(1)}_{1}(\lambda):=(Q^{(1)}-\lambda I)^{-1},$ (C.4)

$M(\lambda)$ changes sign with high probability at $\mu_{1}^{-}$ and
$\mu_{1}^{+}.$ In fact, for $\lambda_{1},$ it should satisfy the following
equation with high probability

${\rm
det}(\lambda_{1}I-\mathbf{y}_{(1)}\mathbf{y}^{*}_{(1)}-Q^{(1)})=0\Rightarrow
M(\lambda_{1})=0,$ (C.5)

as long as $\lambda_{1}>\lambda_{1}^{(1)}.$ On the other hand, if $M(\lambda)$
changes sign at $\mu_{1}^{\pm},$ by continuity, there must at least be an
eigenvalue of $Q$ in the interval $[\mu_{1}^{-},\mu_{1}^{+}].$ If (C.3) holds,
combining the above arguments, we see that the only possibility is
$\lambda_{1}$ and it is also true that $\lambda_{1}>\lambda_{1}^{(1)}.$

We first justify (C.3). Recall that $\mu_{1}$ is defined in (3.14) according
to

$1+(\xi^{2}_{(1)}+d_{1})m_{1n}(\mu_{1})=0.$

Together with (2.19) and (2.21), we readily obtain that

$1=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{\frac{\mu_{1}}{\xi^{2}_{(1)}+d_{1}}-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi_{(j)}^{2}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}}.$

Recall (A.5). Using the definition of $d_{1}$ and (A.21), we see that on
$\Omega,$ for some constant $C>0$

$\begin{split}\frac{1}{n}\sum_{j=1}^{n}\frac{\xi_{(j)}^{2}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}&=\frac{1}{n}\frac{\xi_{(1)}^{2}}{d_{1}}+\frac{1}{n}\sum_{j=2}^{n}\frac{\xi_{(j)}^{2}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}\\\
&\leq\frac{Cn^{\epsilon}\log
n}{n}+\frac{1}{n}\sum_{j=2}^{n}\frac{\xi_{(j)}^{2}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(2)}}\\\
&\leq\frac{Cn^{\epsilon}\log n}{n}+\frac{C\log
n}{n^{1/\alpha}}=Ce.\end{split}$ (C.6)

Using the definition of $\varphi$ in (3.2), the above arguments imply that on
$\Omega$

$\frac{\mu_{1}}{\xi_{(1)}^{2}+d_{1}}=\varphi+\mathrm{O}(e).$ (C.7)

By an analogous argument, we have that for some constants $C_{k}>0,k=1,2,3,$

$\begin{split}&\frac{1}{n}\left(\sum_{j=1}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}-\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(2)}+d_{1}-\xi^{2}_{(j)}}\right)\\\
&=\frac{1}{n}\left(\frac{\xi^{2}_{(1)}}{d_{1}}+\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}-\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(2)}+d_{1}-\xi^{2}_{(j)}}\right)\\\
&=C_{1}e-\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}(\xi^{2}_{(1)}-\xi^{2}_{(2)})}{(\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)})(\xi^{2}_{(2)}+d_{1}-\xi^{2}_{(j)})}\\\
&\geq
C_{1}e-\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(2)}+d_{1}-\xi^{2}_{(j)}}\\\
&\geqslant C_{2}e,\end{split}$

and on the other hand

$\begin{split}&\frac{1}{n}\left(\sum_{j=1}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}-\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(2)}+d_{1}-\xi^{2}_{(j)}}\right)\\\
&\leq\frac{1}{n}\left(\frac{\xi^{2}_{(1)}}{d_{1}}+\sum_{j=2}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(1)}+d_{1}-\xi^{2}_{(j)}}+\sum_{j=2}\frac{\xi^{2}_{(j)}}{\xi^{2}_{(2)}+d_{1}-\xi^{2}_{(j)}}\right)\\\
&\leqslant C_{3}e.\end{split}$

Using the above control, the definition of $\mu_{2}$ and a discussion similar
to (C.7), we can prove that

$\frac{\mu_{2}}{\xi_{(2)}^{2}+d_{1}}=\phi\bar{\sigma}+\mathrm{O}(e).$ (C.8)

Combining (C.7) and (C.8), we immediately see that

$\frac{\mu_{1}}{\xi^{2}_{(1)}+d_{1}}-\frac{\mu_{2}}{\xi^{2}_{(2)}+d_{1}}=\mathrm{O}(e).$
(C.9)

This implies that

$\begin{split}\mu_{1}-\mu_{2}&=(\xi^{2}_{(1)}+d_{1})\left(\frac{\mu_{1}}{\xi^{2}_{(1)}+d_{1}}-\frac{\mu_{2}}{\xi^{2}_{(2)}+d_{1}}\right)+\mu_{2}\left(\frac{\xi^{2}_{(1)}+d_{1}}{\xi^{2}_{(2)}+d_{1}}-1\right)\\\
&=(\xi^{2}_{(1)}+d_{1})\left(\frac{\mu_{1}}{\xi^{2}_{(1)}+d_{1}}-\frac{\mu_{2}}{\xi^{2}_{(2)}+d_{1}}\right)+\frac{\mu_{2}}{\xi^{2}_{(2)}+d_{1}}\left(\xi^{2}_{(1)}-\xi^{2}_{(2)}\right)\\\
&\geq n^{2/\alpha}\log^{-1}n,\end{split}$

where in the third step we used (C.9), (A.21) and the definition of $e_{2}$ in
(A.5). This completes the proof of (C.3).

Next, we will show that

$M(\mu_{1}^{-})<0,\ M(\mu_{1}^{+})>0.$

Due to similarity, in what follows, we focus on the first inequality. Note
that

$\mathbf{y}_{1}^{*}G_{1}^{(1)}(\mu_{1}^{-})\mathbf{y}_{1}=\xi_{(1)}^{2}\mathbf{u}_{1}^{*}\Sigma^{1/2}G_{1}^{(1)}(\mu_{1}^{-})\Sigma^{1/2}\mathbf{u}_{1}.$

Moreover, recall that
$m_{1}^{(1)}(\mu_{1}^{-})=n^{-1}\operatorname{tr}(\Sigma^{1/2}G_{1}^{(1)}(\mu_{1}^{-})\Sigma^{1/2}).$
Then according to Lemma A.14, we have that

$\displaystyle\mathbf{y}_{1}^{*}G_{1}^{(1)}(\mu_{1}^{-})\mathbf{y}_{1}$
$\displaystyle=\xi^{2}_{(1)}m_{1}^{(1)}(\mu_{1}^{-})+\mathrm{O}_{\prec}\left(\frac{\xi^{2}_{(1)}}{n}\|G_{1}^{(1)}(\mu_{1}^{-})\|_{F}\right)$
$\displaystyle=\xi^{2}_{(1)}m_{1}^{(1)}(\mu_{1}^{-})+\mathrm{O}_{\prec}\left(\frac{\xi^{2}_{(1)}}{n^{1/2+1/\alpha}}\right),$
(C.10)

where in the second step we used (C.3) and the fact
$\mu_{2}>\lambda_{1}^{(1)}$. Moreover, for some sufficiently small constant
$\epsilon_{0}>0$ and $z_{0}=\mu_{1}^{-}+\mathrm{i}n^{-1/2-\epsilon_{0}},$ we
can decompose that

$\displaystyle m_{1}^{(1)}(\mu_{1}^{-})$
$\displaystyle=\left[m_{1}^{(1)}(\mu_{1}^{-})-m_{1}^{(1)}(z_{0})\right]+\left[m_{1}^{(1)}(z_{0})-m_{1n}^{(1)}(z_{0})\right]+\left[m_{1n}^{(1)}(z_{0})-m_{1n}^{(1)}(\mu_{1}^{-})\right]+m_{1n}^{(1)}(\mu_{1}^{-})$
$\displaystyle=\mathsf{P}_{1}+\mathsf{P}_{2}+\mathsf{P}_{3}+m_{1n}^{(1)}(\mu_{1}^{-}).$
(C.11)

First, by Theorem A.7, we have that $\mathsf{P}_{2}\prec n^{-1/2-2/\alpha}$.
Second, let $\\{\mathbf{v}^{(1)}_{i}\\}$ be the eigenvectors of $Q^{(1)}$
associated with the eigenvalues $\\{\lambda_{i}^{(1)}\\},$ then we have that

$\begin{split}|\mathsf{P}_{1}|&\leq\frac{1}{n}\sum_{i=1}^{p}|T\mathbf{v}^{(1)}_{i}|^{2}\left|\frac{1}{\lambda^{(1)}_{i}-\mu_{1}^{-}}-\frac{1}{\lambda^{(1)}_{i}-z_{0}}\right|\\\
&=\frac{1}{n}\sum_{i=1}^{p}|T\mathbf{v}^{(1)}_{i}|^{2}\left|\frac{\mathrm{i}n^{-1/2-\epsilon_{0}}}{(\lambda^{(1)}_{i}-\mu_{1}^{-})(\lambda^{(1)}_{i}-z_{0})}\right|\\\
&\leq\frac{1}{n}\sum_{i=1}^{p}|T\mathbf{v}^{(1)}_{i}|^{2}\Big{|}\frac{\mathrm{i}n^{-1/2-\epsilon_{0}}+\mathrm{O}_{\prec}(n^{-2/\alpha-1/2}\log^{2}n)}{|\lambda^{(1)}_{i}-z_{0}|^{2}}\Big{|}\\\
&\prec\operatorname{Im}(m_{1}^{(1)}(z_{0}))\times\mathrm{O}_{\prec}(n^{-1/2-\epsilon_{0}})\prec
n^{-1-1/\alpha},\end{split}$

where in the third step we used (C.3) and the fact $\mu_{2}>\lambda_{1}^{(1)}$
and in the last step we used Lemma A.1, (C.7) and (A.21). Third, according to
Lemma 2.8, we can decompose that

$\begin{split}\mathsf{P}_{3}&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z_{0}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-z_{0}(1+m^{(1)}_{1n}(z_{0})\xi^{2}_{(j)})})}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}^{-}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+m^{(1)}_{1n}(\mu_{1}^{-})\xi^{2}_{(j)})})}\\\
&=\left[\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z_{0}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-z_{0}(1+m^{(1)}_{1n}(z_{0})\xi^{2}_{(j)})})}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z_{0}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+m^{(1)}_{1n}(\mu_{1}^{-})\xi^{2}_{(j)})})}\right]\\\
&+\left[\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z_{0}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+m^{(1)}_{1n}(\mu_{1}^{-})\xi^{2}_{(j)})})}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}^{-}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+m^{(1)}_{1n}(\mu_{1}^{-})\xi^{2}_{(j)})})}\right]\\\
&:=\mathcal{M}^{(1)}_{31}+\mathcal{M}^{(1)}_{32}.\end{split}$

Note that according to (C.7), (C.3), (A.21) and Lemma A.1, we conclude that
with high probability
$|1+\sigma_{i}m_{2n}^{(1)}(z_{0})|,|1+\sigma_{i}m_{2n}^{(1)}(\mu_{1}^{-})|\sim
1.$ For $\mathcal{M}^{(1)}_{31}$, using the definitions in (2.16) and the
above bounds, we have that with high probability

$\displaystyle\mathcal{M}^{(1)}_{31}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma^{2}_{i}}{-z_{0}(1+\sigma_{i}m^{(1)}_{2n}(z))(1+\sigma_{i}m^{(1)}_{2n}(\mu_{1}^{-}))}\big{(}m^{(1)}_{2n}(\mu_{1}^{-})-m^{(1)}_{2n}(z)\big{)}$
(C.12)
$\displaystyle=\mathrm{O}(|z_{0}|^{-1})\times\frac{1}{n}\sum_{j=2}^{n}\left(\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))}-\frac{\xi^{2}_{(j)}}{-z(1+\xi^{2}_{(j)}m_{1n}^{(1)}(z_{0}))}\right)$
$\displaystyle=\mathrm{O}(|z_{0}|^{-1})\times\frac{1}{n}\sum_{j=2}^{n}\left(\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m^{(1)}_{1n}(\mu_{1}^{-}))}-\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m^{(1)}_{1n}(z_{0}))}\right)$
$\displaystyle+\mathrm{O}(|z_{0}|^{-1})\times\frac{1}{n}\sum_{j=2}^{n}\left(\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m^{(1)}_{1n}(z))}-\frac{\xi^{2}_{(j)}}{-z(1+\xi^{2}_{(j)}m_{1n}^{(1)}(z_{0}))}\right)$
$\displaystyle=\mathrm{O}(|z_{0}|^{-1})\times\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{4}_{(j)}(m^{(1)}_{1n}(z_{0})-m_{1n}^{(1)}(\mu_{1}^{-}))}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m^{(1)}_{1n}(\mu_{1}^{-}))(1+\xi^{2}_{(j)}m^{(1)}_{1n}(z_{0}))}+\mathrm{O}(|z_{0}|^{-1})\times\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}(\mu_{1}^{-}-z_{0})}{z\mu_{1}^{-}(1+\xi^{2}_{(j)}m^{(1)}_{1n}(z_{0}))}$
$\displaystyle=\mathrm{o}(1)\times(m^{(1)}_{1n}(z_{0})-m^{(1)}_{1n}(\mu_{1}^{-}))+\mathrm{O}(n^{-2/\alpha-1/2-\epsilon_{0}}).$

where in the second to last step we used Lemma A.1 and in the last step we
used Lemma A.1 and (A.21). This implies with high probability

$\mathcal{M}_{31}^{(1)}=\mathrm{O}\left(n^{-2/\alpha-1/2-\epsilon_{0}}\right).$

Similarly, for $\mathcal{M}^{(1)}_{32}$, we have that

$\begin{split}\mathcal{M}^{(1)}_{32}&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}(z_{0}-\mu_{1}^{-})}{z_{0}\mu_{1}^{-}(1+\sigma_{i}m^{(1)}_{2n}(\mu_{1}^{-}))}\\\
&=\mathrm{O}(n^{-2/\alpha-1/2-\epsilon_{0}}).\end{split}$ (C.13)

Combining the above arguments, we have that
$\mathsf{P}_{3}=\mathrm{O}\left(n^{-2/\alpha-1/2-\epsilon_{0}}\right).$

Inserting the bounds of $\mathsf{P}_{k},1\leqslant k\leqslant 3$ into (C.1),
we conclude that

$|m^{(1)}_{1}(\mu_{1}^{-})-m^{(1)}_{1n}(\mu_{1}^{-})|\prec
n^{-1/\alpha-1/2-\epsilon_{0}}.$

Together with (C.1) and (A.21), it yields that

$M(\mu_{1}^{-})=1+(\xi^{2}_{(1)}+d_{1})m^{(1)}_{1n}(\mu_{1}^{-})+\mathrm{O}_{\prec}(n^{-1/2-\epsilon_{0}}).$
(C.14)

In what follows, we study $1+(\xi^{2}_{(1)}+d_{1})m^{(1)}_{1n}(\mu_{1}^{-})$.
We rewrite that

$\begin{split}1+(\xi^{2}_{(1)}+d_{1})m^{(1)}_{1n}(\mu_{1}^{-})&=1+(\xi^{2}_{(1)}+d_{1})m^{(1)}_{1n}(\mu_{1})-(\xi^{2}_{(1)}+d_{1})\big{(}m^{(1)}_{1n}(\mu_{1})-m^{(1)}_{1n}(\mu_{1}^{-})\big{)}.\end{split}$
(C.15)

Using the definition for $\mu_{1}$ that
$1+(\xi^{2}_{(1)}+d_{1})m_{1n}(\mu_{1})=0$ and the definitions in (2.16), by a
discussion similar to (C.12), we have that for some constant $C>0$

$\begin{split}&1+(\xi^{2}_{(1)}+d_{1})m^{(1)}_{1n}(\mu_{1})\\\
&=1+(\xi^{2}_{(1)}+d_{1})m_{1n}(\mu_{1})+(\xi^{2}_{(1)}+d_{1})\big{(}m_{1n}^{(1)}(\mu_{1})-m_{1n}(\mu_{1})\big{)}\\\
&=(\xi^{2}_{(1)}+d_{1})\frac{1}{n}\sum_{i=1}^{p}\left(\frac{\sigma_{i}}{-\mu_{1}(1+\sigma_{i}m_{2n}^{(1)}(\mu_{1}))}-\frac{\sigma_{i}}{-\mu_{1}(1+\sigma_{i}m_{2n}(\mu_{1}))}\right)\\\
&=(\xi^{2}_{(1)}+d_{1})\left[\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma^{2}_{i}}{-\mu_{1}(1+\sigma_{i}m_{2n}^{(1)}(\mu_{1}))(1+\sigma_{i}m_{2n}(\mu_{1}))}\right]\big{(}m_{2n}(\mu_{1})-m_{2n}^{(1)}(\mu_{1})\big{)}\\\
&\leqslant
C\frac{(\xi^{2}_{(1)}+d_{1})}{\mu_{1}}\times\left|\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}(\mu_{1}))}-\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m^{(1)}_{1n}(\mu_{1}))}\right|\\\
&\leqslant
C\frac{(\xi^{2}_{(1)}+d_{1})}{\mu_{1}}\times\xi^{2}_{(2)}|m_{1n}^{(1)}(\mu_{1})-m_{1n}(\mu_{1})|+n^{-1}.\end{split}$
(C.16)

where in the last step we again used (A.21). This yields that

$|(\xi^{2}_{(1)}+d_{1})\big{(}m_{1n}^{(1)}(\mu_{1})-m_{1n}(\mu_{1})\big{)}|\leqslant
C\frac{(\xi^{2}_{(1)}+d_{1})}{\mu_{1}}\times\xi^{2}_{(2)}|m_{1n}^{(1)}(\mu_{1})-m_{1n}(\mu_{1})|+n^{-1},$

which implies that
$(\xi^{2}_{(1)}+d_{1})\big{(}m^{(1)}_{1n}(\mu_{1})-m_{1n}(\mu_{1})\big{)}=\mathrm{O}(n^{-1})$.
Together with (C.15), we have

$1+(\xi^{2}_{(1)}+d_{1})m^{(1)}_{1n}(\mu_{1}^{-})=-(\xi^{2}_{(1)}+d_{1})\big{(}m^{(1)}_{1n}(\mu_{1})-m^{(1)}_{1n}(\mu_{1}^{-})\big{)}+\mathrm{O}(n^{-1}).$
(C.17)

Recall that we have proved the facts that
$\mu_{1},\mu_{1}^{-}>\lambda_{1}^{(1)},$ by Theorem A.8 and the monotonicity
of $m_{1}^{(1)}$ outside the bulk, the first term on the right-hand side of
(C.17) is negative. In order to show $M(\mu_{1}^{-})<0,$ in light of (C.14),
it suffices to show that its magnitude is much larger than
$\mathrm{O}(n^{-1/2-\epsilon_{0}}).$ To see this, we decompose that

$\begin{split}&m^{(1)}_{1n}(\mu_{1})-m^{(1)}_{1n}(\mu_{1}^{-})\\\
&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}(1+\sigma_{i}m^{(1)}_{2n}(\mu_{1}))}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}^{-}(1+\sigma_{i}m^{(1)}_{2n}(\mu_{1}^{-}))}\\\
&=\left[\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1})})}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))})}\right]\\\
&+\left[\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))})}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\mu_{1}^{-}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))})}\right]\\\
&:=\tilde{\mathcal{M}}_{11}^{(1)}+\tilde{\mathcal{M}}_{12}^{(1)}.\end{split}$

Similar to the discussion of (C.12), we have that
$\tilde{\mathcal{M}}_{11}^{(1)}$,

$\begin{split}\tilde{\mathcal{M}}_{11}^{(1)}&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma^{2}_{i}\Big{(}\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\lambda))}-\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}))}\Big{)}}{-\mu_{1}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}))})(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))})}\\\
&=\mathrm{O}(\frac{1}{\mu_{1}})\times\frac{1}{n}\sum_{j=2}^{n}\frac{\xi^{4}_{(j)}(m^{(1)}_{1n}(\mu_{1})-m^{(1)}_{1n}(\mu_{1}^{-}))}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}))}\\\
&=\mathrm{o}(1)\times(m^{(1)}_{1n}(\mu_{1})-m^{(1)}_{1n}(\mu_{1}^{-})).\end{split}$

Moreover, similar to (C.13), for $\tilde{\mathcal{M}}_{12}^{(1)}$ we have that
with high probability

$\begin{split}\tilde{\mathcal{M}}_{12}^{(1)}&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}(\mu_{1}-\mu_{1}^{-})}{\mu_{1}\mu_{1}^{-}(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))})(1+\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi^{2}_{(j)}}{-\mu_{1}^{-}(1+\xi^{2}_{(j)}m_{1n}^{(1)}(\mu_{1}^{-}))})}\\\
&\asymp\frac{\mu_{1}-\mu_{1}^{-}}{\mu_{1}^{-}\mu_{1}}\asymp
n^{-1/2-1/\alpha+\epsilon}.\end{split}$

This implies that

$m^{(1)}_{1n}(\mu_{1})-m^{(1)}_{1n}(\lambda)\asymp
n^{-1/2-1/\alpha+\epsilon}.$

Together with (C.17), the definition of $d_{2}$ and (A.21), we readily see
that

$1+(\xi^{2}_{(1)}+d_{2})m^{(1)}_{1n}(\mu_{1}^{-}))\asymp-n^{-1/2+\epsilon},$
(C.18)

which concludes the proof of $M(\mu_{1}^{-})<0$ when $n$ is sufficiently
large. Similarly, we can prove that $M(\mu_{1}^{+})>0.$

Before concluding the proof, we briefly discuss the proof of the other cases.
For the case (2.6), the main difference is to utilize the second part of
Theorem A.7. For elliptical model, we repeat the proof verbatim with some
minor modification. For example, in (C.6), we need to apply (A.22) and in
(C.7) we need to use the definition in (2.20) and for (C.1) we need to use (2)
of Lemma A.14. We omit further details.

∎

###  C.2 The bounded support case

In this section, we study the first order convergence of $\lambda_{1}$ under
the assumptions of part (1) of Theorem 3.3. The other parts will be discussed
in Section C.3. The main result of this section can be summarized in the
following proposition. As before, due to similarity, we only prove for the
separable covariance i.i.d. data as in (2) of Assumption 2.1.

######  Proposition C.2.

Suppose the assumptions of part (1) of Theorem 3.3 hold, then conditional on
the probability event as in Lemma A.4, we have that

$\displaystyle\left|\lambda_{1}-\left(\widehat{L}_{+}-\frac{1-\phi\widehat{\varsigma}_{3}}{\widehat{\varsigma}_{4}}\frac{l-\xi^{2}_{(1)}}{l\xi^{2}_{(1)}}\right)\right|=\mathrm{O}_{\mathbb{P}}\left[\frac{1}{n^{1/(d+1)}}\left(\frac{n^{3\epsilon_{d}}}{n^{-1/(d+1)+1/2}}+\frac{\log
n}{n^{1/(d+1)}}\right)\right].$

The proof of Proposition C.2 relies crucially on the following lemma whose
justification will be provided in the end of this section.

######  Lemma C.3.

Suppose the assumptions of Proposition C.2 hold. Recall $E_{0}$ defined in
(3.13). Conditional on the probability event $\Omega$ as in Lemma A.4, we have
that

$\lambda_{1}=E_{0}+\mathrm{O}_{\mathbb{P}}\left(n^{-1/2+3\epsilon_{d}}\right),$
(C.19)

and

$\displaystyle\operatorname{Re}m_{1n}(\lambda_{1}+\mathrm{i}\eta_{0})=-\frac{1}{\xi^{2}_{(1)}}+\mathrm{O}_{\mathbb{P}}(n^{-1/2+3\epsilon_{d}}).$
(C.20)

Armed with Lemma C.3, we proceed to the proof of Proposition C.2.

###### Proof of Proposition C.2.

Recall (A.9), we have that conditional on $\Omega,$
$m_{1n}(\widehat{L}_{+})=-l^{-1}.$ Together with (A.12), we conclude that

$\operatorname{Re}m_{1n}(\widehat{L}_{+}+\mathrm{i}n^{-1/2-\epsilon_{d}})=-l^{-1}+\mathrm{O}(n^{-1/2-\epsilon_{d}}).$

Moreover, according to the definition in (3.13), we have that

$\operatorname{Re}m_{1n}(E_{0}+\mathrm{i}n^{-1/2-\epsilon_{d}})=-\xi_{(1)}^{2}.$

By (A.23), we obtain that conditional on $\Omega$

$\operatorname{Re}m_{1n}(\widehat{L}_{+}+\mathrm{i}n^{-1/2-\epsilon_{d}})=\operatorname{Re}m_{1n}(E_{0}+\mathrm{i}n^{-1/2-\epsilon_{d}})+\mathrm{O}\left(\frac{\log
n}{n^{d+1}}\right),$

which implies that $\widehat{L}_{+}=E_{0}+\mathrm{O}(\log n/n^{d+1}).$ Let
$\Xi$ be the probability event that (C.19) holds. We therefore conclude from
(A.11) that when restricted on $\Xi$ and $n$ is sufficiently large,
$\lambda_{1}+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}.$ Consequently, by part I of
Lemma A.4, we find the following holds on $\Xi$

$\displaystyle
m_{1n}(\widehat{L}_{+})-m_{1n}(\lambda_{1}+\mathrm{i}\eta_{0})=\frac{\widehat{\varsigma}_{4}}{(1-\phi\widehat{\varsigma}_{3})}\left(\widehat{L}_{+}-\lambda_{1}-\mathrm{i}\eta_{0}\right)+\mathrm{O}\left((\log
n)(n^{-1/(d+1)})^{\min\\{d,2\\}}\right).$ (C.21)

Again by $m_{1n}(\widehat{L}_{+})=-l^{-1},$ Together with the second part of
Lemma C.3, considering the real parts of both sides of (C.21), we obtain that
on $\Xi$

$-l^{-1}+\xi_{(1)}^{-2}+\mathrm{O}_{\mathbb{P}}(n^{-1/2+3\epsilon_{d}})=\frac{\widehat{\varsigma}_{4}}{(1-\phi\widehat{\varsigma}_{3})}\left(\widehat{L}_{+}-\lambda_{1}\right)+\mathrm{O}\left((\log
n)(n^{-1/(d+1)})^{\min\\{d,2\\}}\right).$

This completes our proof.

∎

The rest of this section leaves to the proof of Lemma C.3. We first prove the
following lemma which will be used in the proof of Lemma C.3. It essentially
locates the points in $\mathbf{D}_{b}^{\prime}$ for which
$\operatorname{Im}m_{Q}(z)\gg\eta_{0}$ near the edge. It is a counterpart of
[62, Lemmas 5.12, 5.13 and 5.15] and [60, Lemmas 5.13, 5.14 and 5.16]. Due to
similarity, we only sketch the key points of the proof. Recall $z_{0}$ defined
in (3.13).

######  Lemma C.4.

Suppose the assumptions of Lemma C.3, we have that the followings holds with
high probability

  1. (1).

For any $z=E+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime}$ satisfying that
$|z-z_{0}|\geqslant n^{-1/2+3\epsilon_{d}},$

$\operatorname{Im}m_{1}(z)\asymp\eta_{0},\
\operatorname{Im}m_{Q}(z)\asymp\eta_{0}.$ (C.22)

  2. (2).

For $m_{1}^{(1)}(z)$ and $m_{Q}^{(1)}(z)$ defined around (A.4), we have that
for all $z=E+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime},$

$\operatorname{Im}m_{1}^{(1)}(z)\asymp\eta_{0},\
\operatorname{Im}m_{Q}^{(1)}(z)\asymp\eta_{0}.$ (C.23)

  3. (3).

There exists some $E_{0}^{\prime}\in\mathbb{R}$ such that for
$z_{0}^{\prime}=E_{0}^{\prime}+\mathrm{i}\eta_{0},$ the followings holds
simultaneously

$|z_{0}^{\prime}-z_{0}|\leqslant n^{-1/2+3\epsilon_{d}},\ \text{and}\
\operatorname{Im}m(z_{0}^{\prime})\gg\eta_{0}.$ (C.24)

###### Proof.

Due to similarity, we focus our discussion on $m_{1}(z)$ and will explain the
minor differences for $m_{Q}(z)$ from line to line. Recall (B.9). Our proof
relies on the following fluctuation average which provides a stronger control
on $n^{-1}\sum_{i=1}^{p}Z_{i}$ than the one in Remark B.9. They are
counterparts of Lemmas 5.8 and 5.9 and Corollary 5.10 of [62]. We deter its
proof to Appendix D.3.

######  Lemma C.5.

Suppose the assumptions of Lemma C.4 hold, we have that the followings holds
on $\Omega$

  1. (1).

For all $i\neq 1$ and all $z=E+\mathrm{i}\eta_{0}\in\mathbf{D}_{b}^{\prime},$
we have that

$|m_{2}-m_{2}^{(1)}|\prec\frac{1}{n\eta_{0}},\ \
|m_{2}-m_{2}^{(i)}|+|m_{2}^{(i)}-m_{2}^{(i1)}|\prec
n^{-1+1/(d+1)+4\epsilon_{d}}.$ (C.25)

  2. (2).

For all $z\in\mathbf{D}_{b}^{\prime},$

$\left|\frac{1}{n}\sum_{i=2}^{n}Z_{i}\right|+\left|\frac{1}{n}\sum_{i=2}^{n}Z_{i}^{(1)}\right|\prec
n^{-1/2-\frac{1}{2}(\frac{1}{2}-\frac{1}{d+1})+2\epsilon_{d}}.$

  3. (3).

For all $z\in\mathbf{D}_{b}^{\prime},$

$\left|\frac{1}{n}\sum_{i=2}^{n}\frac{(\xi_{i}^{2}+\xi_{i}^{4})Z_{i}}{(1+\xi_{i}^{2}m_{1n}(z))^{2}}\right|\prec
n^{-1/2-\frac{1}{2}(\frac{1}{2}-\frac{1}{d+1})+2\epsilon_{d}}.$

Armed with Lemma C.5, we proceed to finish our proof. The proof of part (1) is
similar to that of (A.16) by using the local law Theorem A.8. We only provide
the key arguments. Using (B.50), (B.48), and (B.49), we see that

$\displaystyle
m_{1}=-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi_{j}^{2}m_{1}+Z_{j}}}+\frac{1}{n}\operatorname{tr}(R_{1}\Sigma)+\frac{1}{n}\operatorname{tr}(R_{2}\Sigma).$

According to Theorem A.8, by a discussion similar to (B.52) using Remark B.9,
we have that

$\displaystyle\begin{split}\operatorname{Im}m_{1}(z)&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}\eta_{0}}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}+Z_{j}}|^{2}}+\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1}+\xi^{2}_{j}\operatorname{Im}Z_{j}}{|1+\xi^{2}_{j}m_{1n}+Z_{j}|^{2}}}{|z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}+Z_{j}}|^{2}}\\\
&+\mathrm{O}_{\prec}\left(\frac{1}{n}\sum_{j=1}^{n}\frac{Z_{j}}{|1+\xi^{2}_{j}m_{1n}+Z_{j}|^{2}}+\frac{1}{(n\eta_{0})^{2}}\right)\\\
&:=\mathsf{R}_{1}+\mathsf{R}_{2}+\mathsf{R}_{3}.\end{split}$ (C.26)

Together with Assumption 2.10, (A.23) and Remark B.9, we find that for some
small constant $c^{\prime}>0,$ when $n$ is sufficiently large,

$\left|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n}(z)+Z_{j}}\right|\geqslant
c^{\prime}.$ (C.27)

This implies that

$\mathsf{R}_{1}\asymp\eta_{0}.$ (C.28)

For $\mathsf{R}_{2},$ on the one hand, by Theorem A.8 and (B.44), we can
conclude that there exists some constant $0<\mathsf{c}^{\prime}<1,$

$\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}}{|1+\xi^{2}_{j}m_{1n}+Z_{j}|^{2}}}{|z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}+Z_{j}}|^{2}}\leqslant\mathsf{c}^{\prime}<1.$

Moreover, as $|z-z_{0}|\geqslant n^{-1/2+3\epsilon_{d}},$ according to (A.13)
and Remark B.9, we find that $1+\xi_{j}^{2}m_{1n}(z)+Z_{j}\geqslant
Cn^{-1/2+3\epsilon_{d}}$ for some constant $C>0.$ Together with (C.27) and (3)
of Lemma C.5, we find that with high probability

$\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}\operatorname{Im}Z_{j}}{|1+\xi^{2}_{j}m_{1n}+Z_{j}|^{2}}}{|z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}+Z_{j}}|^{2}}\ll\eta_{0}.$

Consequently, we have that with high probability

$\mathsf{R}_{2}=\mathsf{c}^{\prime}\operatorname{Im}m_{1}+\mathrm{o}(\eta_{0}).$
(C.29)

Similarly, we can prove that with high probability
$\mathsf{R}_{3}=\mathrm{o}(\eta_{0}).$ Together with (C.28), (C.29) and
(C.26), we can conclude the prove of $m_{1}(z).$ The discussion for $m_{Q}(z)$
is similar except we need to use (B.49) and (B.51).

The proof of part (2) is similar to that of part (1) using Lemma C.5, Theorem
A.8 and Remark B.9. The idea is analogous to the proof of Lemma 5.13 of [62]
or Lemma 5.14 of [60]. We omit further details.

Finally, for part (3), we find from Lemma A.14, (C.23) and (A.24) that for
some small constant $\epsilon^{\prime}<\epsilon_{d}/2,$ with high probability,

$|Z_{1}|\leqslant n^{-1/2+\epsilon^{\prime}}.$ (C.30)

Without loss of generality, we assume that
$\xi_{1}^{2}\geqslant\xi_{2}^{2}\geqslant\cdots\geqslant\xi_{n}^{2}.$ On the
one hand, by the definition of $m_{2}(z)$ and (B.9), we have

$m_{2}=\frac{\xi_{1}^{2}\mathcal{G}_{11}}{n}+\frac{1}{n}\sum_{i=2}^{n}\frac{\xi^{2}_{i}}{-z(1+\xi^{2}_{i}m_{1}^{(i)}+Z_{i})}.$

Together with Lemma A.12, we see that

$\displaystyle\frac{1}{\xi_{1}^{2}\mathcal{G}_{11}}=-z(1+\xi^{2}_{1}m_{1}^{(1)}+Z_{1}).$
(C.31)

Denote

$z^{\pm}_{0}=z_{0}\pm n^{-1/2+3\epsilon_{d}}.$

Recall (3.13) that
$1+\xi_{1}^{2}\operatorname{Re}m_{1n}(E_{0}+\mathrm{i}\eta_{0})=0.$ Using
Theorem A.8, (A.25) and (A.13), together with (C.23) and Remark B.9, we
conclude that for some constant $C>0$

$\frac{1}{\xi_{1}^{2}\mathcal{G}_{11}(z_{0}^{-})}\geqslant
Cn^{-1/2+3\epsilon_{d}},\ \text{and}\ \
\frac{1}{\xi_{1}^{2}\mathcal{G}_{11}(z_{0}^{+})}\leqslant-
Cn^{-1/2+3\epsilon_{d}}.$

Consequently, by continuity, we find that there exists
$z_{1}=E_{1}+\mathrm{i}\eta_{0}$ with
$E_{1}\in(E_{0}-n^{-1/2+3\epsilon_{d}},E_{0}+n^{-1/2+3\epsilon_{d}})$ that
$\operatorname{Re}\mathcal{G}_{11}(z_{1})=0$. For the choice of $z_{1},$
together with (C.31), we find that

$\displaystyle|\operatorname{Im}(z_{1}\xi_{1}^{2}\mathcal{G}_{11}(z_{1}))|=\frac{1}{|\operatorname{Im}m^{(1)}_{1}(z_{1})+\operatorname{Im}Z_{1}|}\geq
n^{1/2-\epsilon_{d}/2},$ (C.32)

where we used (C.30) with the assumption $\epsilon^{\prime}<\epsilon_{d}/2$
and (C.23). On the other hand, following lines of the proof of [62, Lemma
5.15], by a decomposition similar to (B.21) and a discussion similar to
(C.26), using Lemma C.5, we find that

$\displaystyle\operatorname{Im}m_{1}(z_{1})\asymp\eta_{0}+\frac{\operatorname{Im}(\xi_{1}^{2}z_{1}\mathcal{G}_{11}(z_{1}))}{n}.$

Together with (C.32), we conclude that

$\operatorname{Im}m_{1}(z_{1})\gg\eta_{0}.$

The discussion for $m_{Q}$ is similar and we omit the details. This completes
our proof.

∎

Finally, armed with Lemma C.4, we proceed to the proof of Lemma C.3. Since the
details are similar to those of Proposition 4.6 of [62] or Proposition 4.7 of
[60], we only provide the key ingredients.

###### Proof of Lemma C.3.

We first prove (C.19). Using the spectral decomposition of $Q,$ for $m_{Q}(z)$
in (2.14), we find that

$\displaystyle\operatorname{Im}m_{Q}(E+\mathrm{i}\eta_{0})=\frac{1}{n}\sum_{i=1}^{n}\frac{\eta_{0}}{(\lambda_{i}-E)^{2}+\eta_{0}^{2}}.$
(C.33)

This yields that

$\operatorname{Im}m_{Q}(\lambda_{1}+\mathrm{i}\eta_{0})\geq(n\eta_{0})^{-1}\gg\eta_{0},$

where we used the definition of $\eta_{0}$ in (3.12). It is clear that
$\lambda_{1}=\mathrm{O}_{\mathbb{P}}(1).$ First, if
$\lambda_{1}\in\mathbf{D}_{b}^{\prime},$ then the proof follows directly from
(C.22). Second, if $\lambda_{1}\notin\mathbf{D}_{b}^{\prime},$ on the other
hand, for the upper bound, by (3) of Lemma C.4 and (A.15), using the
definition of $\mathbf{D}_{b}^{\prime}$ in (A.7), with an argument similar to
Proposition 4.7 of [62], we have that on $\Omega,$
$\lambda_{1}<E_{0}+n^{-1/2+3\epsilon_{d}}$ holds with $1-\mathrm{o}(1)$
probability. On the other hand, for the lower bound, we prove by
contradiction. We assume that $\lambda_{1}<E_{0}(1)-n^{-1/2+3\epsilon_{d}}$.
Then we see from (C.33) that $\operatorname{Im}m_{Q}(E+\mathrm{i}\eta_{0})$ is
a decreasing function of $E$ on the interval
$(E_{0}-n^{-1/2+3\epsilon_{d}},E_{0}+n^{-1/2+3\epsilon_{d}})$. However, from
Lemma C.4 and its proof (recall that
$z_{0}^{-}=E_{0}-n^{-1/2+3\epsilon_{d}}$), we have seen that,
$\operatorname{Im}m_{Q}(z_{0})\gg\eta_{0}$,
$\operatorname{Im}m_{Q}(z_{0}^{-})\sim\eta_{0},$ which is a contradiction. It
implies that $\lambda_{1}\geq E_{0}(1)-n^{-1/2+3\epsilon_{d}}$ and completes
the proof of (C.19).

Then we prove (C.20). By (A.13), (A.15) and (A.16), we find that

$\operatorname{Re}m_{1n}(\lambda_{1}+\mathrm{i}\eta_{0})=\operatorname{Re}m_{1n}(z_{0})+\mathrm{O}_{\mathbb{P}}(n^{-1/2+3\epsilon_{d}})=-\frac{1}{\xi_{(1)}^{2}}+\mathrm{O}_{\mathbb{P}}(n^{-1/2+3\epsilon_{d}}),$

where in the last step we used (3.13). This completes our proof. ∎

###  C.3 Proof of main results of Section 3

In this section, we prove Theorems 3.1 and 3.3 using the results in Sections
C.1 and C.2.

###### Proof of Theorem 3.1.

For the first part, according to (C.7), by Lemma A.11, when $n$ is
sufficiently large,

$\mu_{1}=(\xi_{(1)}^{2}+d_{1})\left(\phi\bar{\sigma}+\mathrm{O}_{\mathbb{P}}(e)\right).$

Together with Proposition C.1, we find that

$\lambda_{1}=(\xi_{(1)}^{2}+d_{1})\left(\phi\bar{\sigma}+\mathrm{O}_{\mathbb{P}}(e)\right)+\mathrm{O}_{\mathbb{P}}\left(n^{-1/2+2\epsilon}d_{1}\right).$

Using (A.21) and (A.22) and the definition of $d_{1}$ in (3.15), we can
complete the proof for the first part.

For the second part, it follows directly from the results in Lemma A.15,
(A.21) and (A.22). This completes our proof. ∎

###### Proof of Theorem 3.3.

Due to similarity, we focus our discussion on the separable covariance i.i.d.
data as in case (2) of Assumption 2.1. The elliptical case can be handled
similarly. For part (1), (3.7) has been proved in (II) of Lemma A.4. For
(3.8), the proofs follow from Proposition C.2, II of Lemma A.4 with the fact
$d>1$ and (A.23). Then (3.9) follows from (3.8) and Lemma A.15.

Then we proceed to the proof of parts (2) and (3). Following [28, Lemma 2.5],
we see that conditional on $\Omega$ in Lemma A.11,
$(\widehat{L}_{+},m_{1n}(\widehat{L}_{+}))$ should satisfy the following
systems of equations

$\displaystyle
m_{1n}(\widehat{L}_{+})=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-\widehat{L}_{+}+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})}},\quad
1=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}}{|1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})|^{2}}}{\left|\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})}\right|^{2}}.$
(C.34)

Similarly, $(L_{+},m_{1n,c}(L_{+}))$ should satisfy the following equations

$\displaystyle
m_{1n,c}(L_{+})=\frac{1}{n}\sum_{i}\frac{\sigma_{i}}{-L_{+}+\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s)},\quad
1=\frac{1}{n}\sum_{i}\frac{\sigma^{2}_{i}\int\frac{s^{2}}{|1+sm_{1n,c}(L_{+})|^{2}}\mathrm{d}F(s)}{|L_{+}-\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s)|^{2}}.$
(C.35)

Using the definitions of $\varsigma_{k}$ and
$\widehat{\varsigma}_{k},1\leqslant k\leqslant 3,$ by (A.23) and an argument
similar to II of Lemma A.4, when $n$ is sufficiently large, we see that our
assumption $\phi^{-1}<\varsigma_{3}$ implies
$\phi^{-1}<\widehat{\varsigma}_{3}$ on $\Omega.$ This yields that for some
constant $\delta>0$

$\displaystyle\frac{1}{n}\sum_{i}\frac{\sigma_{i}^{2}\widehat{\varsigma}_{1}}{(\widehat{L}_{+}-\sigma_{i}\widehat{\varsigma}_{2})^{2}}>1+\delta,\quad\frac{1}{n}\sum_{i}\frac{\sigma_{i}^{2}\varsigma_{1}}{(L_{+}-\sigma_{i}\varsigma_{2})^{2}}>1+\delta.$
(C.36)

where the first inequality is restricted on the event $\Omega$. From now on,
for notional simplicity, we always restrict ourselves on $\Omega$ so that the
discussion is purely deterministic. Recall (3.5) and (A.6). Combining the
second equations in (C.34) and (C.35) with (C.36), we readily obtain that

$m_{1n}(\widehat{L}_{+})>-l^{-1},\ \ m_{1n,c}(L_{+})>-l^{-1}.$ (C.37)

Together with (A.23), we have that for all $1\leqslant j\leqslant n$ and some
constant $\delta^{\prime}>0$

$\frac{1}{\left|1+\xi_{j}^{2}m_{1n}(\widehat{L}_{+})\right|}\geqslant\delta^{\prime},\
\frac{1}{\left|1+sm_{1n,c}(L_{+})\right|}\geqslant\delta^{\prime}\ \text{for
any}\ 0<s\leqslant l.$ (C.38)

We now proceed to the proof. The proof consists of two steps. In the first
step, we prove the results assuming that

$\left|m_{1n,c}(L_{+})-m_{1n}(\widehat{L}_{+})\right|=\mathrm{O}_{\mathbb{P}}(n^{-1/2}),\
|L_{+}-\widehat{L}_{+}|=\mathrm{O}_{\mathbb{P}}(n^{-1/2}).$ (C.39)

In the second step, we justify (C.39). We start with step one.

Step one: Under the assumption C.39, the key component of the proof is the
following lemma. Denote

$\mathsf{C}_{1}:=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{\left(L_{+}-\sigma_{i}\int\frac{s}{1+sm_{1n,c}}\mathrm{d}F(s)\right)^{2}},$

and

$\mathcal{X}:=\frac{1}{n}\sum_{j=1}^{n}\left(\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n,c}(L_{+})}-\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s)\right).$

According to Assumption 2.10, we have that

$\mathsf{C}_{1}\asymp 1.$

######  Lemma C.6.

Under the assumptions of Theorem 3.3 and (C.39), we have that

$\mathsf{C}_{1}(\widehat{L}_{+}-L_{+})=\mathsf{C}_{1}\mathcal{X}+\mathrm{O}_{\mathbb{P}}(n^{-1}).$

Armed with Lemma C.6, we can easily prove parts (2) and (3). Recall (3.6). It
is clear from (A.23) $\mathcal{X}=\mathrm{O}_{\mathbb{P}}(n^{-1/2}),$ and from
central limit theorem that $\mathcal{X}$ is asymptotically Gaussian with
variance $n^{-1}\vartheta$. We decompose that

$\lambda_{1}-L_{+}=\lambda_{1}-\widehat{L}_{+}+\widehat{L}_{+}-L_{+}.$

According to [25, 29] and Lemma A.4, we find that on $\Omega,$
$|\lambda_{1}-\widehat{L}_{+}|\prec n^{-2/3}$ and
$n^{2/3}\gamma(\lambda_{1}-\widehat{L}_{+})$ follows type-1 Tracy-Widom law.
This concludes the general results in part (3). Moreover, for part (2), it is
easy to see that when $d>1,$ by Cauchy-Schwarz inequality, $\vartheta$ is
bounded from blow so that the Gaussian part dominates the Tracy-Widom part and
we hence conclude the proof.

To complete Step one, we now prove Lemma C.6.

###### Proof of Lemma C.6.

Using the first parts in equations (C.34) and (C.35), we see that

$\displaystyle m_{1n,c}(L_{+})-m_{1n}(\widehat{L}_{+})$
$\displaystyle=\frac{1}{n}\sum_{i}\frac{\sigma_{i}}{\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})}}-\frac{1}{n}\sum_{i}\frac{\sigma_{i}}{L_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})}}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{-\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s)+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})}}{(L_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})(L_{+}-\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s))}$
$\displaystyle=\frac{1}{n}\sum_{i}\frac{\sigma_{i}(L_{+}-\widehat{L}_{+})}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})(L_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{\sigma_{i}}{\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})}}-\frac{1}{n}\sum_{i}\frac{\sigma_{i}}{\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})}}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s)-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})}}{(L_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})(L_{+}-\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s))}$
$\displaystyle=\frac{1}{n}\sum_{i}\frac{\sigma_{i}(L_{+}-\widehat{L}_{+})}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})(L_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{-\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))(1+\xi^{2}_{j}m_{1n,c}(L_{+}))}}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{-\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s)+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})}}{(L_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})(L_{+}-\sigma_{i}\int\frac{s}{1+sm_{1n,c}(L_{+})}\mathrm{d}F(s))}$
$\displaystyle:=\mathsf{T}_{1}+\mathsf{T}_{2}+\mathsf{T}_{3}.$ (C.40)

For the term $\mathsf{T}_{1},$ by (C.39), Assumption 2.10 and (A.23), we can
see that

$\mathsf{T}_{1}=\mathsf{C}_{1}(L_{+}-\widehat{L}_{+})+\mathrm{O}_{\mathbb{P}}(n^{-1}).$
(C.41)

For the term $\mathsf{T}_{2},$ we see that

$\displaystyle\mathsf{T}_{2}=\frac{1}{n}\sum_{i}\frac{-\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))(1+\xi^{2}_{j}m_{1n,c}(L_{+}))}}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})^{2}}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{\big{(}\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))(1+\xi^{2}_{j}m_{1n,c}(L_{+}))}\big{)}^{2}(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))^{2}}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})^{2}(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})}$
$\displaystyle=\frac{1}{n}\sum_{i}\frac{-\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))^{2}}}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})^{2}}+\frac{1}{n}\sum_{i}\frac{-\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))^{2}}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))^{2}(1+\xi^{2}_{j}m_{1n,c}(L_{+}))}}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})^{2}}$
$\displaystyle+\frac{1}{n}\sum_{i}\frac{\big{(}\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))(1+\xi^{2}_{j}m_{1n,c}(L_{+}))}\big{)}^{2}(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))^{2}}{(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n,c}(L_{+})})^{2}}$
(C.42)
$\displaystyle=-(m_{1n}(\widehat{L}_{+})-m_{1n,c}(L_{+}))+\mathrm{O}_{\mathbb{P}}(n^{-1}),$

where in the last step we used the second equation of (C.34) for the first
term of (C.3), and (C.39), Theorem A.8, (A.23) and Assumption 2.10 for the
second and third terms. Similarly, for $\mathsf{T}_{3},$ we have that

$\mathsf{T}_{3}=\mathsf{C}_{1}\mathcal{X}+\mathrm{O}_{\mathbb{P}}(n^{-1}).$
(C.43)

Insert (C.41), (C.3) and (C.43) into (C.3), we can conclude the proof. ∎

Then we prove (C.39) to complete step two and the proof of the theorem.

Step two: To prove (C.39), we first rewrite (C.34) and (C.35) a little bit.
Recall (2.21). We find that (C.34) can be rewritten as

$\displaystyle
F_{n}(m_{1n}(\widehat{L}_{+}),\widehat{L}_{+})=0,\quad\frac{\partial
F_{n}}{\partial x}(m_{1n}(\widehat{L}_{+}),\widehat{L}_{+})=0,$

where we denote

$\displaystyle
F_{n}(x,y)=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-y+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+x\xi^{2}_{j}}}-x.$
(C.44)

Similarly, (C.35) can be rewritten as

$\displaystyle F_{n,c}(m_{1n,c}(L_{+}),L_{+})=0,\quad\frac{\partial
F_{n,c}}{\partial x}(m_{1n,c}(L_{+}),L_{+})=0,$

where we denote

$\displaystyle
F_{n,c}(x,y)=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-y+\sigma_{i}\int\frac{s}{1+xs}dF(s)}-x.$

For pair $(\widetilde{x},\widetilde{y})$ so that $\widetilde{x}>-l^{-1}$
(recall (C.37)), as long as they satisfy Assumption 2.10 in the sense that
$\min_{1\leqslant i\leqslant
p}|\tilde{y}-\sigma_{i}\int\frac{s}{1+\tilde{x}s}dF(s)|\geqslant\tau,$ by
(A.23), we find that

$\displaystyle\left|F_{n,c}(\tilde{x},\tilde{y})-F_{n}(\tilde{x},\tilde{y})\right|+\left|\frac{\partial
F_{n,c}}{\partial x}(\tilde{x},\tilde{y})-\frac{\partial F_{n}}{\partial
x}(\tilde{x},\tilde{y})\right|+\left|\frac{\partial F_{n,c}}{\partial
y}(\tilde{x},\tilde{y})-\frac{\partial F_{n}}{\partial
y}(\tilde{x},\tilde{y})\right|=\mathrm{O}_{\mathbb{P}}(n^{-1/2}).$ (C.45)

Set $(x_{0},y_{0})=(m_{1n,c}(L_{+}),L_{+}).$ Then we have that

$\displaystyle F_{n,c}(x_{0},y_{0})=0,\quad\frac{\partial F_{n,c}}{\partial
x}(x_{0},y_{0})=0,\quad 0<\frac{\partial F_{n,c}}{\partial
y}(x_{0},y_{0})<\infty,\quad\frac{\partial^{2}F_{n}}{\partial
y^{2}}(x_{0},y_{0})<0.$ (C.46)

It suffices to prove the following lemma.

######  Lemma C.7.

There exists a pair $(x_{1},y_{1})$ with condition
$|x_{1}-x_{0}|+|y_{1}-y_{0}|=\mathrm{O}_{\mathbb{P}}(n^{-1/2})$ such that with
probability $1-\mathrm{o}(1)$

$\displaystyle F_{n}(x_{1},y_{1})=0,\quad\frac{\partial F_{n}}{\partial
x}(x_{1},y_{1})=0.$ (C.47)

With Lemma C.7, according to (C.34) and Theorem 2.8, we see that (C.39) holds.
In the rest, we prove Lemma C.7 using (C.45).

###### Proof of Lemma C.7.

For some small $\epsilon>0,$ we consider the probability event $\Xi$ so that
(A.23) holds and (C.45) reads as

$\left|F_{n,c}(\tilde{x},\tilde{y})-F_{n}(\tilde{x},\tilde{y})\right|+\left|\frac{\partial
F_{n,c}}{\partial x}(\tilde{x},\tilde{y})-\frac{\partial F_{n}}{\partial
x}(\tilde{x},\tilde{y})\right|+\left|\frac{\partial F_{n,c}}{\partial
y}(\tilde{x},\tilde{y})-\frac{\partial F_{n}}{\partial
y}(\tilde{x},\tilde{y})\right|=\mathrm{O}(n^{-1/2+\epsilon}).$ (C.48)

We have seen that $\mathbb{P}(\Xi)=1-\mathrm{o}(1).$ Now we fix a realization
$\\{\xi_{i}^{2}\\}\in\Xi$ so that the discussions below are purely
deterministic.

For the above fixed constant $\epsilon>0,$ we set the region

$\displaystyle\mathcal{N}(x,y):=\\{(x,y):|x-x_{0}|+|y-y_{0}|\leqslant
n^{-1/2+\epsilon}\\},$

To prove the first part of (C.47), it suffices to prove that there exists a
solution of $F_{n}(x,y)=0$ in the region $\mathcal{N}(x,y).$ By Bolzano’s
theorem, we see that for sufficiently large $n,$ we can find two points
$(x_{11},y_{11})$ and $(x_{12},y_{12})$ on $\mathcal{N}(x,y)$ so that
$F_{n,c}(x_{11},y_{11})<0,\ F_{n,c}(x_{12},y_{12})>0.$ Together with (C.48),
we see that $F_{n}(x_{11},y_{11})<0,\ F_{n}(x_{12},y_{12})>0.$ Therefore, by
continuity, we can find some point $(x^{\prime},y^{\prime})$ so that
$F_{n}(x^{\prime},y^{\prime})=0.$ Repeating the above procedure, by implicit
function theorem, we find that there exists a curve $x\equiv x(y)$ on
$\mathcal{N}(x,y)$ so that $F_{n}(x,y)=0.$ Similarly, we can show that there
exists another curve $\widehat{x}\equiv\widehat{x}(\widehat{y})$ on
$\mathcal{N}(\widehat{x},\widehat{y})$ so that the second part of (C.47) holds
in the sense that $\partial
F_{n}(\widehat{x},\widehat{y})/\partial\widehat{x}=0.$

In order to show (C.47), we need to prove that the curves $(x,y)$ and
$(\widehat{x},\widehat{y})$ must have at least one intersection in the region
$\mathcal{N}(x,y).$ We prove by contradiction. Otherwise, the curve $(x,y)$
will lie in one of the areas separated by $(\widehat{x},\widehat{y})$ with
strictly $\partial F_{n}(x,y)/\partial x<0$ or $\partial F_{n}(x,y)/\partial
x>0$. By (C.46), we see that $\mathcal{N}(x,y),$ $\partial F_{n}(x,y)/\partial
y>0$. Without loss of generality, we assume $\partial F_{n}(x,y)/\partial
x<0$. On the one hand, as $F_{n}(x,y)=0$, one may conclude that for small
neighbor around the points on $(x,y)$, it holds that
$\mathrm{d}x/\mathrm{d}y>0$. On the other hand, taking the derivative
$F_{n}(x,y)$ with respect to $y$, we obtain that

$\displaystyle\frac{\mathrm{d}x}{\mathrm{d}y}\times\left(\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}}{(1+x\xi^{2}_{j})^{2}}}{(-y+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+x\xi^{2}_{j}})^{2}}-1\right)=0,$

which implies $\partial F_{n}(x,y)/\partial x=0$ and gives the contradiction.
This concludes our proof.

∎

∎

###  C.4 Proof of the results of Section 4

In this section, we prove the results of Section 4 which are related to our
statistical applications.

###### Proof of Theorem 4.1.

We start with the proof of part (1). For the data matrix $\widetilde{Y}$
defined using (4.2) around (4.3), we denote
$\widetilde{Q}:=\widetilde{Y}\widetilde{Y}^{*}$ and
$\widetilde{\mathcal{Q}}:=\widetilde{Y}^{*}\widetilde{Y}.$ Since these two
matrices have the same non-zero eigenvalues, we focus on the later one for
convenience. For the spiked covariance matrix model in (4.1), we decompose it
as follows

$\widetilde{\Sigma}:=\Sigma_{s}+\Sigma_{o},$

where we denote the two $p\times p$ matrices as

$\Sigma_{s}:=\sum_{i=1}^{r}\tilde{\sigma}_{i}\mathbf{v}_{i}\mathbf{v}_{i}^{*}\equiv
V_{1}\Lambda_{s}V_{1}^{*},\quad\Sigma_{o}:=\sum_{i=r+1}^{p}\sigma_{i}\mathbf{v}_{i}\mathbf{v}_{i}^{*}\equiv
V_{2}\Lambda_{o}V_{2}^{*}.$ (C.49)

Consequently, we can decompose $\widetilde{\mathcal{Q}}$ as follows

$\widetilde{\mathcal{Q}}=DX^{*}\tilde{\Sigma}XD=DX^{*}\Sigma_{s}XD+DX^{*}\Sigma_{o}XD.$

Note that with high probability

$\|DX^{*}\Sigma_{o}XD\|=\|D^{2}X^{*}\Sigma_{0}X\|\leqslant\|D^{2}\|\|X^{*}\Sigma_{0}X\|\leq\sigma_{r}\xi_{(1)}^{2}\|X^{*}X\|\sim\xi^{2}_{(1)},$

where in the last step we used [90] that $\|X^{*}X\|$ is bounded from above
with high probability. Using (A.21) and (A.22) as well as Weyl’s inequality,
we see that from the assumption of (4.6) that, for $1\leqslant i\leqslant r,$

$\frac{\mu_{i}-\lambda_{i}(DX^{*}\Sigma_{s}XD)}{\widetilde{\sigma}_{i}}=\mathrm{o}_{\mathbb{P}}(1).$
(C.50)

Then we consider the first few largest eigenvalues of $DX^{*}\Sigma_{s}XD,$ or
equivalently those of $\Sigma_{s}^{1/2}XD^{2}X^{*}\Sigma_{s}^{1/2}.$ By a
discussion similar to Lemma D.1 of [28], we find that if $\lambda$ is an
eigenvalues of $\Sigma_{s}^{1/2}XD^{2}X^{*}\Sigma_{s}^{1/2},$ recalling
(C.49), we have that

$\displaystyle\operatorname{det}(V_{1}^{*}XD^{2}X^{*}V_{1}-\lambda\Lambda_{s}^{-1})=0.$
(C.51)

Moreover, due to the rotational invariant property of $X$, without loss of
generality, we can assume the columns of $V_{1}$ are standard basis in
$\mathbb{R}^{p}.$ Consequently, we observe that

$\displaystyle
V_{1}^{*}XD^{2}X^{*}V_{1}=\begin{pmatrix}\sum_{i}\xi^{2}_{i}u_{i1}^{2}&\sum_{i}\xi^{2}_{i}u_{i1}u_{i2}&\cdots&\sum_{i}\xi^{2}_{i}u_{i1}u_{ir}\\\
\sum_{i}\xi^{2}_{i}u_{i2}u_{i1}&\sum_{i}\xi^{2}_{i}u_{i2}^{2}&\cdots&\sum_{i}\xi^{2}_{i}u_{i2}u_{ir}\\\
\vdots&\vdots&\ddots&\vdots\\\
\sum_{i}\xi^{2}_{i}u_{ir}u_{i1}&\sum_{i}\xi^{2}_{i}u_{ir}u_{i2}&\cdots&\sum_{i}\xi^{2}_{i}u_{ir}^{2}\end{pmatrix}_{r\times
r},$

where we used the fact that the $i$-th column of $X$ is
$\mathbf{u}_{i}=(u_{i1},u_{i2},\cdots,u_{ip})^{*}.$ Recall that $u_{i1}$ has
the same distribution as the self-normalized random variable
$\mathsf{g}_{i1}:=g_{i1}/\sqrt{\sum_{j=1}^{p}g_{ij}^{2}},$ where
$\\{g_{ij}\\}$ are i.i.d. standard Gaussian random variables. Let
$\operatorname{Kol}$ be the Kolmogorov distance. According to the discussions
in [81, Section 4.2.1], we find that for some standard Gaussian random
variable $\mathsf{g}$ independent of $u_{i1},$ we have that

$\operatorname{Kol}(\mathsf{g},\sqrt{p}u_{i1})=\mathrm{O}(p^{-1}).$ (C.52)

Since $r$ is finite and $\\{\xi_{i}^{2}\\}$ and $X$ are independent, using the
assumptions in (i) of Assumption 2.4, by straightforward calculations using
Markov inequality, we conclude that

$V_{1}^{*}XD^{2}X^{*}V_{1}=\phi^{-1}\mathbb{E}\xi^{2}I_{r}+\mathrm{o}_{\mathbb{P}}(1),$

where $I_{r}$ is a $r\times r$ identity matrix. Together with (C.51), we
conclude that for $1\leqslant i\leqslant r$

$\frac{\lambda_{i}(DX^{*}\Sigma_{s}XD)}{\widetilde{\sigma}_{i}}=\phi^{-1}\mathbb{E}\xi^{2}+\mathrm{o}_{\mathbb{P}}(1).$
(C.53)

Combining (C.50), we have completed the proof of part (1).

Then we proceed with part (2). The proof follows closely from a discussion
similar to the proof of [28, Theorem 3.7], or [57, Theorem 2.7], or [15,
Theorem 2.7], or [24, Theorem 3.6]. Due to similarity, we only sketch the
proof strategies, provide the key ingredients and point out the main
differences. In fact, our proof will be easier since the spikes are much
larger than the edges and we only consider the first few extremal non-outlier
eigenvalues. As discussed in [28, Appendix D], or [57, Section 6], or [15,
Section 4], the proof consists of the following three steps.

  1. (i).

We first find the permissible regions in which contain the eigenvalues of
$\widetilde{\mathcal{Q}}$ with high probability.

  2. (ii).

Then we apply a counting argument to a special case (where all the spikes are
well-separate), and show that the results hold under this special case.

  3. (iii).

Finally we use a continuity argument to extend the results in (ii) to the
general case using the gaps in the permissible regions.

In what follows, we choose a realization $\\{\xi_{i}^{2}\\}\in\Omega$ so that
(A.23) holds with $1-\mathrm{o}(1)$ probability as in Lemma A.11. With this
restriction, $m_{1n}$ and $\mu_{1}$ in (3.14) are purely deterministic. Recall
$d_{1}$ in (3.15) and the $\epsilon$ used therein. Due to similarity, we focus
on the polynomial decay setting (2.5). The exponential decay case can be
handled similarly.

For Step (i), to find the permissible region, for some large constant
$\mathsf{C}>0,$ we denote the set for $1\leqslant i\leqslant k$

$\Gamma_{i}:=\left\\{x\in[\lambda_{i},\mu_{1}+n^{-1/2+2\epsilon}d_{1}]:\
\operatorname{dist}(x,\operatorname{spec}({Q}))>\mathsf{C}n^{-1/2+2\epsilon}d_{1}\right\\},$
(C.54)

where $\text{spec}(\widetilde{Q})$ stands for the spectrum of ${Q}.$ The
results of Step (i) can be summarized as follows.

######  Lemma C.8.

There exists some constant $\mathsf{C}>0$ so that the set $\cup_{i}\Gamma_{i}$
contains no eigenvalue of $\widetilde{{Q}}.$

###### Proof.

The proof is similar to that of Lemma D.4 of [28] or Lemma 5.4 of [24] and we
only sketch the key points here. Due to the rational invariance of $X,$ we can
without of generality assume that $\widetilde{\Sigma}$ is diagonal and
decompose that $\widetilde{\Sigma}=\Sigma_{1}+\Sigma$ with
$\Sigma_{1}=\widetilde{\Sigma}-\Sigma=U^{*}DU$, where we recall that
$\widetilde{\Sigma}$ is constructed based on $\Sigma.$ Here $D$ is an $r\times
r$ matrix containing the nonzero eigenvalues of $\Sigma_{1}$ and $U^{*}$ is
the $p\times r$ matrix containing the first $r$ standard basis in
$\mathbb{R}^{p}.$ Under the assumption of (4.6) and (2.12), we see that $D$ is
invertible when $n$ is sufficiently large. By a discussion similar to (C.51),
we see that $x$ is an eigenvalue of $\widetilde{\mathcal{Q}}$ but not
$\mathcal{Q}$ if and only if

$\displaystyle\operatorname{det}(I-D^{1/2}U^{*}XD(xI-DX^{*}\Sigma
XD)^{-1}DX^{*}UD^{1/2})=0.$ (C.55)

Moreover, for $x\in\Gamma_{i},1\leqslant i\leqslant k$ and $\eta:=n^{-1/2},$
we define $z_{x}=x+\mathrm{i}\eta.$ According to Proposition B.1, we have that
with high probability

$\left\|(DX^{*}\Sigma XD-
z_{x}I)^{-1}+z^{-1}(I+m_{1n}(z_{x})D^{2})^{-1}\right\|_{\infty}=\mathrm{O}(n^{-1/2-1/\alpha+\epsilon}).$

Together with the arguments around (C.52), using the fact $r$ is finite, $U$
contains the standard basis and the definition of $m_{2n}(z)$ in (2.15), we
find that

$\displaystyle\left\|z^{-1}U^{*}XD(I+m_{1n}(z_{x})D^{2})^{-1}DX^{*}U-m_{2n}(z_{x})I\right\|_{\infty}=\mathrm{O}_{\mathbb{P}}(n^{-1/2-1/\alpha+2\epsilon}),$

where we used the fact that $|z_{x}|\asymp\mu_{1}$ and (B.6). According to a
discussion similar to equation (D.29) of [28] and Lemma A.1, we see that for
$t=1,2,$

$m_{tn}(z_{x})-m_{tn}(x)\asymp\operatorname{Im}m_{tn}(z)=\mathrm{O}(n^{-1/2-1/\alpha+\epsilon}).$

Combining all the above controls, we find that for some constant $C>0$

$\left\|I-D^{1/2}U^{*}XD(xI-DX^{*}\Sigma
XD)^{-1}DX^{*}UD^{1/2}\right\|_{\infty}=C\max_{1\leqslant j\leqslant
r}|m_{2n}(z_{x})+\widetilde{\sigma}_{j}^{-1}|+\mathrm{O}_{\mathbb{P}}\left(n^{-1/2-1/\alpha+\epsilon}\right).$

Since $x\in\Gamma_{i},$ together with Lemma A.1, we see that
$|m_{2n}(z_{x})+\widetilde{\sigma}_{j}^{-1}|\gg n^{-1/2-1/\alpha+\epsilon}.$
This implies that $x$ is not an eigenvalue of $\widetilde{\mathcal{Q}}$ and
completes the proof. ∎

As mentioned in the proof of Theorem 2.7 of [15], once Step (i) is done, Steps
(ii) and (iii) are more standard. For Step (ii), together with the interlacing
results as in Lemma C.3 of [28], we perform the counting argument to prove
(4.7) for a special case assuming
$\widetilde{\sigma}_{1}>\widetilde{\sigma}_{2}>\cdots>\widetilde{\sigma}_{r}.$
The details can be found in Lemma D.5 of [28] or Lemma 5.5 of [24]. For Step
(iii), we use a continuity argument for all possible configurations
$\\{\widetilde{\sigma}_{i}\\}_{1\leqslant i\leqslant r}.$ The details can be
found in the proof of Theorem 3.7 of [28]. Since most of the arguments can be
made verbatim following lines of the counterparts of [28] or [24] or [15] or
[57], we omit further details. This completes the proof.

∎

###### Proof of Corollary 4.3.

(4.11) follows directly from (4.7), Theorem 3.1, and the discussions in Remark
4.2. On the other hand, when $\mathbf{H}_{a}$ and the assumption that
$r_{*}>r$ hold, we have that

$\mathbb{T}\geqslant\frac{\mu_{r}-\mu_{r+1}}{\mu_{r+1}-\mu_{r+2}}.$

Moreover, according to part (2) of Theorem 4.1 and (A.21) or (A.22), we see
that with $1-\mathrm{o}(1)$ probability, for some constant $C>0$

$(\mu_{r+1}-\mu_{r+2})^{-1}\geqslant C\mathsf{T}^{-1}.$

In addition, according to part (1) of Theorem 4.1, we see from the assumption
of (4.6) and (A.21) or (A.22) that

$\mu_{r}-\mu_{r+1}\geqslant C\widetilde{\sigma}_{r}.$

Combining the above two controls, we can prove (4.12) and (4.13) for the
statistic $\mathbb{T}.$ The other two equations can be proved similarly for
the statistic $\mathbb{T}_{r_{0}}$. This completes our proof. ∎

###### Proof of Theorem 4.5.

The proof follows from strategies similar to Theorem 2.4 of [16] or Theorem
3.5 of [96]. We focus on explaining the main ideas and omit the details. The
core of the proof is to introduce the auxiliary quantities
$\theta_{i},1\leqslant i\leqslant r,$ where for each $1\leqslant i\leqslant
r,$ $\theta_{i}$ satisfies the equation

$\displaystyle\frac{\theta_{i}}{\tilde{\sigma}_{i}}=\Big{(}1-\frac{1}{n\theta_{i}}\sum_{j=r+1}^{p}\frac{\sigma_{j}}{1-\tilde{\sigma}_{i}^{-1}\sigma_{j}}\Big{)}^{-1}.$

With the restriction that
$\theta_{i}\in[\widetilde{\sigma}_{i},2\widetilde{\sigma}_{i}],$ the existence
of uniqueness of $\theta_{i}$ have been justified in [16, 96]. Furthermore,
under the assumption of (4.6), we can conclude from equation (2.10) of [16]
that

$\theta_{i}/\widetilde{\sigma}_{i}=1+\mathrm{o}(1),1\leqslant i\leqslant r.$
(C.56)

Moreover, following lines of the proof of Theorem 2.2 of [16] or Lemma 3.4 of
[96], we can conclude that for $1\leqslant i\leqslant r,$ we can obtain that

$\displaystyle\frac{\widehat{\lambda}_{i}}{\theta_{i}}=1+\mathbf{k}_{i}^{*}(V_{1}^{*}XX^{*}V_{1}-I_{r})\mathbf{k}_{i}+\mathrm{o}_{\mathbb{P}}(\frac{1}{\sqrt{n}}),$

where we recall (C.49) for the definition of $V_{1}$ and
$\mathbf{k}_{i},1\leqslant i\leqslant r,$ are the standard basis in
$\mathbb{R}^{r}.$ Together with Lemma A.14, we see that

$\frac{\mu_{i}}{\widehat{\lambda}_{i}}=\frac{\mu_{i}}{\theta_{i}}\left(1+\mathrm{O}_{\mathbb{P}}(n^{-1/2})\right).$
(C.57)

The rest of the proof leaves to establish the asymptotics of
$\mu_{i}/\theta_{i}.$ In the actual proof, it is more convenient to work with
random quantities $\widehat{\theta}_{i},1\leqslant i\leqslant r,$ defined
according to

$\displaystyle\frac{\widehat{\theta}_{i}}{\tilde{\sigma}_{i}}=\left(1-\frac{1}{n\widehat{\theta}_{i}}\sum_{j=r+1}^{p}\frac{\sigma_{j}}{1-\tilde{\sigma}_{i}^{-1}\sigma_{j}}+\frac{1}{n\widehat{\theta}_{i}}\sum_{k=r+1}^{p}\frac{\sigma_{k}}{1-\frac{\sigma_{k}}{n\widehat{\theta}_{i}}\sum_{j=1}^{n}\xi^{2}_{j}/\mathbb{E}\xi^{2}}\right)^{-1}.$
(C.58)

Similar to $\theta_{i},$ with the restriction
$\widehat{\theta}_{i}\in[\widetilde{\sigma}_{i},2\widetilde{\sigma}_{i}],$ we
can obtain the uniqueness and existence of the solutions with high
probability.

We first summarize some important properties of $\widehat{\theta}_{i}.$ On the
one hand, it is easy to use the definitions of
$\theta_{i},\widehat{\theta}_{i},$ the assumption of (4.6) and (C.56), we can
obtain that

$\frac{\theta_{i}}{\widehat{\theta}_{i}}=1+\mathrm{o}_{\mathbb{P}}(1).$

In light of (C.57), we find that it suffices to work with
$\mu_{i}/\widehat{\theta}_{i}.$ On the other hand, using the above controls
with (C.58) and (A.21) or (A.22), we find that

$\displaystyle\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{\mathbb{E}\xi^{2}}-\frac{\widehat{\theta}_{i}}{\tilde{\sigma}_{i}}$
$\displaystyle=\frac{\widehat{\theta}_{i}}{\tilde{\sigma}_{i}}\times\left(\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}-\mathbb{E}\xi^{2}-\frac{\xi^{2}_{j}}{n\widehat{\theta}_{i}}\sum_{k=r+1}^{p}\frac{\sigma_{k}}{1-\tilde{\sigma}_{1}^{-1}\sigma_{k}}+\frac{\xi^{2}_{i}}{n\widehat{\theta}_{i}}\sum_{k=r+1}^{p}\frac{\sigma_{k}}{1-\frac{\sigma_{k}}{n\widehat{\theta}_{i}}\sum_{j=1}^{n}\xi^{2}_{j}/\mathbb{E}\xi^{2}}}{\mathbb{E}\xi^{2}}\right)$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}-\mathbb{E}\xi^{2}}{\mathbb{E}\xi^{2}}-\frac{\frac{1}{n}\sum_{j=1}^{n}\xi^{2}_{j}/\mathbb{E}\xi^{2}-\frac{\widehat{\theta}_{i}}{\tilde{\sigma}_{i}}}{\tilde{\sigma}_{i}}\times\mathrm{o}_{\mathbb{P}}(1)+\mathrm{o}_{\mathbb{P}}(n^{-1/2}),$

which implies that

$\frac{\widehat{\theta}_{i}}{\widetilde{\sigma}_{i}}=1+\mathrm{o}_{\mathbb{P}}(n^{-1/2}).$
(C.59)

Now we proceed to complete the proof following that of Theorem 2.4 of [16].
For notational convenience, we now work with the rescaled matrix

$\check{\mathcal{Q}}:=\check{D}X^{*}\widetilde{\Sigma}X\check{D},\
\check{D}^{2}:=(\mathbb{E}\xi^{2})^{-1}D^{2},$

whose eigenvalues are denoted as
$\check{\lambda}_{1}\geqslant\check{\lambda}_{2}\geqslant\cdots\geqslant\check{\lambda}_{\\{p\wedge
n\\}}>0.$ Note that $\mu_{i}=\mathbb{E}\xi^{2}\check{\lambda}_{i}.$ By a
discussion similar to (C.51) and (C.55), using (C.49), we find that
$\check{\lambda}_{i},1\leqslant i\leqslant r$ satisfy the equation

$\displaystyle\operatorname{det}(\Lambda_{s}^{-1}-V_{1}^{*}X\check{D}(\check{\lambda}_{i}I-\check{D}X^{*}\Sigma_{o}X\check{D})^{-1}\check{D}X^{*}V_{1})=0.$

Denote $\mathbf{B}(x):=xI-\check{D}X^{*}\Sigma_{o}X\check{D}$ and
$\delta_{i}=(\check{\lambda}_{i}-\widehat{\theta}_{i})/\widehat{\theta}_{i}$,
the above determinant can be rewritten into

$\displaystyle\operatorname{det}(\widehat{\theta}_{i}\Lambda_{s}^{-1}-\widehat{\theta}_{i}V_{1}^{*}X\check{D}\mathbf{B}^{-1}(\widehat{\theta}_{i})\check{D}X^{*}V_{1}+\delta_{i}\widehat{\theta}_{i}^{2}V_{1}^{*}X\check{D}\mathbf{B}^{-1}(\check{\lambda}_{i})\mathbf{B}^{-1}(\widehat{\theta}_{i})\check{D}X^{*}V_{1})=0.$
(C.60)

Following the procedure in Section 7.1 of [16] or Lemma C.5 of [96], we find
that for $1\leqslant i,l\leqslant r$ (recall that $\widetilde{\Sigma}$ is
assumed to be diagonal)

$\displaystyle\widehat{\theta}_{i}\mathbf{e}_{i}^{*}V_{1}^{*}X\check{D}\mathbf{B}^{-1}(\widehat{\theta}_{i})\check{D}X^{*}V_{1}\mathbf{e}_{l}=\mathbf{1}(l=i)\sum_{j=1}^{n}x_{kj}^{2}\xi^{2}_{j}/\mathbb{E}\xi^{2}+\mathrm{o}_{\mathbb{P}}(n^{-1/2}).$

Similarly, by a discussion similar to Lemma C.6 of [96], we conclude that

$\displaystyle\delta_{i}\widehat{\theta}_{i}^{2}[V_{1}^{*}X\check{D}\mathbf{B}^{-1}(\check{\lambda}_{i})\mathbf{B}^{-1}(\widehat{\theta}_{i})\check{D}X^{*}V_{1}]_{il}=\delta_{i}(\mathbf{1}(l=i)+\mathrm{o}_{\mathbb{P}}(1)).$

Inserting the above two controls into (C.60), by the assumption of (4.17),
using Leibniz’s formula for determinant, one has that

$\displaystyle\delta_{i}(1+\mathrm{o}_{\mathbb{P}}(1))=\sum_{j=1}^{n}x_{ij}^{2}\xi^{2}_{j}/\mathbb{E}\xi^{2}-\frac{\widehat{\theta}_{i}}{\tilde{\sigma}_{i}}+\mathrm{o}_{\mathbb{P}}(n^{-1/2}).$
(C.61)

Combining (C.59), we conclude that

$\frac{\check{\lambda}_{i}-\widehat{\theta}_{i}}{\widehat{\theta}_{i}}=\sum_{j=1}^{n}x_{ij}^{2}\xi^{2}_{j}/\mathbb{E}\xi^{2}-1+\mathrm{o}_{\mathbb{P}}(n^{-1/2}).$

Together with (C.57) and (C.59), using central limit theorem, we can conclude
the proof.

∎

##  Appendix D Proof of some auxiliary lemmas

###  D.1 Preliminary estimates: Proof of Lemmas A.1 and A.4

####  D.1.1 Proof of Lemma A.1

Due to similarity, we only prove the results for the separable covariance
i.i.d. data model when $\xi^{2}$ decays polynomially, i.e., when (2.21) and
(2.5) hold. The other cases can be proved analogously and we omit the details.

###### Proof.

We start with the first statement. We now abbreviate $F_{n}(m_{1n}(z))\equiv
F_{n}(m_{1n}(z),z)$ throughout the proof. For the real part, it suffices to
prove that with high probability for some $0<C_{2}<1<C_{1}$

$\operatorname{Re}m_{1n}(z)\in\left[-C_{1}\frac{\phi\bar{\sigma}E}{E^{2}+\eta^{2}},-C_{2}\frac{\phi\bar{\sigma}E}{E^{2}+\eta^{2}}\right].$
(D.1)

Moreover, by continuity and Theorem 2.8, it suffices to prove the following
inequalities

$\operatorname{Re}F_{n}(-C_{2}\phi\bar{\sigma}E(E^{2}+\eta^{2})^{-1}+\mathrm{i}\operatorname{Im}m_{1n}(z))<0,\
\operatorname{Re}F_{n}(-C_{1}\phi\bar{\sigma}E(E^{2}+\eta^{2})^{-1}+\mathrm{i}\operatorname{Im}m_{1n}(z))>0.$
(D.2)

We only focus on the first part. By definition, we have that

$\displaystyle\operatorname{Re}$ $\displaystyle
F_{n}(m_{1n}(z))=-\operatorname{Re}m_{1n}(z)$ (D.3)
$\displaystyle-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}\operatorname{Re}(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}(z)\xi^{2}_{j}})}{\operatorname{Re}^{2}(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}(z)\xi^{2}_{j}})+\operatorname{Im}^{2}(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}(z)\xi^{2}_{j}})}.$

Note that

$\displaystyle\operatorname{Re}\left(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}}\right)=E-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}(1+\xi^{2}_{j}\operatorname{Re}m_{1n})}{(1+\xi^{2}_{j}\operatorname{Re}m_{1n})^{2}+\xi^{4}_{j}\operatorname{Im}^{2}m_{1n}},$
$\displaystyle\operatorname{Im}\left(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}}\right)=\eta+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}}{(1+\xi^{2}_{j}\operatorname{Re}m_{1n})^{2}+\xi^{4}_{j}\operatorname{Im}^{2}m_{1n}}.$

By a discussion similar to (C.6), if
$\operatorname{Re}m_{1n}=-C_{2}(\phi\bar{\sigma}E)/(E^{2}+\eta^{2})$, we have
that

$\begin{split}\operatorname{Re}\left(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}}\right)&\geq
E(1-\mathrm{o}(1)),\\\
\operatorname{Im}\left(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}}\right)&\leq\eta+E\times\mathrm{o}(1).\end{split}$
(D.4)

Therefore, together with (D.3), we see that

$\begin{split}\operatorname{Re}F_{n}(-C_{2}\phi\bar{\sigma}E(E^{2}+\eta^{2})^{-1}+\mathrm{i}\operatorname{Im}m_{1n}(z))&\leq
C_{2}\phi\bar{\sigma}\frac{E}{(E^{2}+\eta^{2})}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}\times
E(1-\mathrm{o}(1))}{E^{2}+(\eta+E\times\mathrm{o}(1))^{2}}\\\
&\leq(C_{2}-1+\mathrm{o}(1))\frac{\phi\bar{\sigma}E}{(E^{2}+\eta^{2})}<0,\end{split}$
(D.5)

for sufficient large $n$. This completes the discussion for the real part. For
the complex part, the idea is similar and it suffices to prove that when
$z\in\mathbf{D}_{u},$ for some constants $C_{1},C_{2}>0$

$\operatorname{Im}m_{1n}(z)\in\left[C_{1}\frac{\eta\phi\bar{\sigma}}{E^{2}+\eta^{2}},C_{2}\eta\left|\operatorname{Re}m_{1n}(z)\right|\right].$
(D.6)

Equivalently, it suffices to prove that

$\operatorname{Im}F_{n}(\operatorname{Re}m_{1n}(z)+\mathrm{i}C_{2}\eta\left|\operatorname{Re}m_{1n}(z)\right|)<0,\
\operatorname{Im}F_{n}\left(\operatorname{Re}m_{1n}(z)+\mathrm{i}C_{1}\frac{\eta\phi\bar{\sigma}}{E^{2}+\eta^{2}}\right)>0,$

where by definition

$\operatorname{Im}F_{n}(m_{1n},z)=-\operatorname{Im}m_{1n}+\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}\operatorname{Im}(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}})}{\operatorname{Re}^{2}(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}})+\operatorname{Im}^{2}(z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+m_{1n}\xi^{2}_{j}})}.$

The proof of the above inequalities is similar to (D.2) using (D.1). We
briefly discuss the proof of the first inequality in which case by a
discussion similar to (D.4)

$\begin{split}\operatorname{Im}F_{n}(m_{1n},z)&=\eta\operatorname{Re}m_{1n}+\frac{\phi\bar{\sigma}\eta(1+E)}{E^{2}+\eta^{2}(1+E\times\mathrm{o}(1))^{2}}\\\
&\leq-C_{2}\frac{\phi\bar{\sigma}\eta
E}{(E^{2}+\eta^{2})}+\frac{\phi\bar{\sigma}\eta(1+E)}{E^{2}(1+\eta^{2}\times\mathrm{o}(1))+\eta^{2}(1+2E\times\mathrm{o}(1))}<0.\end{split}$

This completes our proof.

For the second statement, from the first statement, we see that it is valid to
write $m_{1n}(E).$ Since $\mu_{1}\gg d_{1}$ holds with high probability (see
(B.6)), it suffices to prove that for some constants $0<C_{2}<1<C_{1},$ when
$|E-\mu_{1}|\leqslant Cd_{1},$

$m_{1n}(E)\in\left[-C_{1}\frac{\phi\bar{\sigma}}{E},-C_{2}\frac{\phi\bar{\sigma}}{E}\right].$
(D.7)

Due to simplicity, we again only focus on the proof of the upper bound.
According to Theorem 2.8 and (2.19), we shall have that $F_{n}(m_{1n}(E))=0.$
Moreover, since $F_{n}(m_{1n}(\mu_{1}))=0,$ to prove (D.7), it suffices to
prove

$F_{n}(-C_{2}\phi\bar{\sigma}/E)<0,\ F_{n}(-C_{1}\phi\bar{\sigma}/E)>0.$ (D.8)

Due to similarity, we focus our discussion on the first inequality. By
definition, we have that

$\begin{split}F_{n}(-C_{2}\phi\bar{\sigma}/E)&=C_{2}\frac{\phi\bar{\sigma}}{E}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{E-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1-\xi^{2}_{j}(C_{2}\frac{\phi\bar{\sigma}}{E})}}\\\
&=C_{2}\frac{\phi\bar{\sigma}}{E}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{E(1-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{E-C_{2}\xi_{j}^{2}\phi\bar{\sigma}})}.\end{split}$

By a discussion similar to (C.6), we further have that

$F_{n}(-C_{2}\phi\bar{\sigma}/E)=C_{2}\frac{\phi\bar{\sigma}}{E}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{E(1-\mathrm{o}(1))}=(C_{2}-1+\mathrm{o}(1))\frac{\phi\bar{\sigma}}{E}<0.$
(D.9)

This completes the proof of the first statement.

Finally, we prove the third statement using the first two statements. For
$z\in\mathbf{D}_{u},$ by definition, we have that

$\begin{split}m_{2n}(z)&=\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{-E-\mathrm{i}\eta-(E+\mathrm{i}\eta)\xi^{2}_{j}(\operatorname{Re}m_{1n}+\mathrm{i}\operatorname{Im}m_{1n})}\\\
&=\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}\left[(-E-\xi^{2}_{j}(E\operatorname{Re}m_{1n}-\eta\operatorname{Im}m_{1n})+\mathrm{i}\eta+\mathrm{i}\xi^{2}_{j}(E\operatorname{Im}m_{1n}+\eta\operatorname{Re}m_{1n}))\right]}{(-E-\xi^{2}_{j}(E\operatorname{Re}m_{1n}-\eta\operatorname{Im}m_{1n}))^{2}+(-\eta-\xi^{2}_{j}(E\operatorname{Im}m_{1n}+\eta\operatorname{Re}m_{1n}))^{2}}.\end{split}$
(D.10)

According to the results in the first two statements, the definition of
$\mathbf{D}_{u}$ in (A.1) and the elementary relation that
$|m_{1n}(z)|=\mathrm{O}(1),$ we find that for some constants $C_{1},C_{2}>0,$
when $n$ is sufficiently large

$|(-E-\xi^{2}_{j}(E\operatorname{Re}m_{1n}-\eta\operatorname{Im}m_{1n})+\mathrm{i}\eta+\mathrm{i}\xi^{2}_{j}(E\operatorname{Im}m_{1n}+\eta\operatorname{Re}m_{1n}))|\leqslant
C_{1}E,$

and

$(-E-\xi^{2}_{j}(E\operatorname{Re}m_{1n}-\eta\operatorname{Im}m_{1n}))^{2}+(-\eta-\xi^{2}_{j}(E\operatorname{Im}m_{1n}+\eta\operatorname{Re}m_{1n}))^{2}\geqslant
C_{2}(E+\xi_{j}^{2})^{2}.$

Together with a discussion similar to (C.6), we readily see that for some
large constant $C>0$

$\begin{split}|m_{2n}(z)|&\leq\frac{C}{n}\sum_{j=1}^{n}\frac{E\xi^{2}_{j}}{(E+\xi_{j}^{2})^{2}}\leq\frac{C}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{E-\xi^{2}_{j}}=\mathrm{O}(e_{2}).\end{split}$
(D.11)

Then together with the definition of $m_{n}(z),$ we see that

$\begin{split}|m_{n}(z)|\leq\frac{1}{p|z|}\sum_{i=1}^{p}\frac{1}{|1+\sigma_{i}m_{2n}(z)|}\leq\frac{1}{p|z|}\sum_{i=1}^{p}\frac{1}{1+\sigma_{i}|m_{2n}(z)|}\leq\frac{1}{|z|}=\mathrm{O}(E^{-1}).\end{split}$

To control the imaginary part, by (D.10), we can write

$\operatorname{Im}m_{2n}(z)=\frac{1}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}(\eta+\xi^{2}_{j}(E\operatorname{Im}m_{1n}+\eta\operatorname{Re}m_{1n}))}{(-E-\xi^{2}_{j}(E\operatorname{Re}m_{1n}-\eta\operatorname{Im}m_{1n}))^{2}+(-\eta-\xi^{2}_{j}(E\operatorname{Im}m_{1n}+\eta\operatorname{Re}m_{1n}))^{2}}.$

Combining with (D.1) and (D.6), we see that for some constant $C>0$

$\begin{split}\operatorname{Im}m_{2n}(z)&\leqslant\frac{C}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}(\eta+\xi^{2}_{j}\eta)}{(E+\xi^{2}_{j}(\mathrm{O}(1)+\eta^{2}\times\mathrm{O}(E^{-1})))^{2}+(\eta+\xi^{2}_{j}\times\mathrm{O}(1)+\eta\times\mathrm{O}(E^{-1}))^{2}}\\\
&=\mathrm{O}\left(\frac{1}{n}\sum_{j=1}^{n}\frac{\eta\xi^{4}_{j}}{\mathrm{O}(E^{2})}\right)=\mathrm{O}\left(\frac{\eta}{E}\right),\end{split}$
(D.12)

where in the last step we used (A.21). Moreover, using the definition of
$m_{n}(z)$ in (2.16), we can write

$\operatorname{Im}m_{n}(z)=\frac{1}{p}\sum_{i=1}^{p}\frac{\eta+\sigma_{i}\eta\operatorname{Re}m_{2n}+\sigma_{i}E\operatorname{Im}m_{2n}}{(E+\sigma_{i}E\operatorname{Re}m_{2n}-\sigma_{i}\eta\operatorname{Im}m_{2n})^{2}+(\eta+\sigma_{i}\eta\operatorname{Re}m_{2n}+\sigma_{i}E\operatorname{Im}m_{2n})^{2}}.$

Together with (D.11) and (D.12), we can easily see that

$\operatorname{Im}m_{n}(z)=\mathrm{O}\left(\frac{\eta}{E^{2}}\right).$

This completes our proof. ∎

######  Remark D.1.

We may observe from the proof of Lemma A.1 that in many cases we can directly
write $m_{1n}(\mu_{1})$ without considering its imaginary part. For example,
when (2.5) holds, for $z_{0}=\mu_{1}+\mathrm{i}\eta$, using (2.21) and (3.14),
we see that

$\displaystyle\lim_{\eta\downarrow 0}\operatorname{Im}m_{1n}(z_{0})$
$\displaystyle=\lim_{\eta\downarrow
0}\frac{1}{n}\sum_{i}\frac{\sigma_{i}(\eta+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z_{0})}{|1+\xi^{2}_{j}m_{1n}(z_{0})|^{2}})}{|z_{0}-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z_{0})}|^{2}}$
$\displaystyle=\lim_{\eta\downarrow
0}\frac{1}{n}\sum_{i}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z_{0})}{|1+\xi^{2}_{j}m_{1n}(z_{0})|^{2}}}{|\mu_{1}-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z_{0})}|^{2}}$
$\displaystyle=\Big{(}\frac{1}{n}\sum_{i}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{(\xi^{2}_{(1)}+d_{2})^{2}\xi^{4}_{j}}{|\xi^{2}_{(1)}+d_{2}-\xi^{2}_{j}|^{2}}}{|\mu_{1}-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{(\xi^{2}_{(1)}+d_{2})\xi^{2}_{j}}{\xi^{2}_{(1)}+d_{2}-\xi^{2}_{j}}|^{2}}\Big{)}\times\lim_{\eta\downarrow
0}\operatorname{Im}m_{1n}(z_{0}).$

Then by a discussion similar to (D.12), using the fact that $\alpha\geqslant
2$, we find that for any $\mu_{1}\gtrsim\xi^{2}_{(1)}$

$\displaystyle\lim_{\eta\downarrow
0}\operatorname{Im}m_{1n}(z_{0})=\mathrm{o}(1)\times\lim_{\eta\downarrow
0}\operatorname{Im}m_{1n}(z_{0}),$

which holds true if and only if $\lim_{\eta\downarrow
0}\operatorname{Im}m_{1n}(z_{0})=0$. This shows that $m_{1n}(\mu_{1})$ is
well-defined for $\mu_{1}\gtrsim\xi^{2}_{(1)}$.

####  D.1.2 Proof of Lemma A.4

Due to similarity, we focus our discussion on the separable covariance i.i.d.
data model. The elliptical data model can be handled analogously using (2.20)
instead of (2.21) whenever it is necessary. We omit the details due to
similarity.

###### Proof.

We first prove the results when conditionally. Let $\Omega$ be the event
satisfying (c) of Definition A.9. According to Lemma A.11, we find that
$\mathbb{P}(\Omega)=1-\mathrm{o}(1).$ Now we choose a realization
$\\{\xi_{i}^{2}\\}\in\Omega$ so that the proofs of parts (a) and (b) are
purely deterministic.

###### Proof of (a).

We start with (A.9). According to (2.16), we have that

$\displaystyle
m_{1n}(z)=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}}.$

To characterize the bulk of the spectrum, we take the imaginary part on the
both sides of the above equation and let $\eta\downarrow 0$ to obtain that

$\displaystyle\begin{split}\operatorname{Im}m_{1n}(z)&=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}\left(\frac{1}{n}\sum_{j}\frac{\xi^{4}_{j}\operatorname{Im}(m_{1n})}{\operatorname{Re}^{2}(1+\xi^{2}_{j}m_{1n})+\xi^{4}_{j}\operatorname{Im}^{2}(m_{1n})}\right)}{(E-\operatorname{Re}(\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}}))^{2}+\operatorname{Im}^{2}(\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}})}.\end{split}$
(D.13)

The above equation can be further rewritten as

$\displaystyle 0=\operatorname{Im}m_{1n}(z)(1-g(m_{1n},E)),$ (D.14)

where $g(m_{1n},E)$ is denoted as

$\displaystyle
g(m_{1n},E):=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}\left(\frac{1}{n}\sum_{j}\frac{\xi^{4}_{j}}{\operatorname{Re}^{2}(1+\xi^{2}_{j}m_{1n})+\xi^{4}_{j}\operatorname{Im}^{2}(m_{1n})}\right)}{(E-\operatorname{Re}(\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}}))^{2}+\operatorname{Im}^{2}(\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}})}.$

Similar to the arguments used in [60, 62], it is easy to see that for any
fixed $\operatorname{Re}m_{1n}<-l^{-1}$ and $E$, $g(m_{1n},E)\rightarrow 0$
when $|\operatorname{Im}m_{1n}|\rightarrow\infty,$ and
$g(m_{1n},E)\rightarrow+\infty$ in order to satisfy (D.14) when
$|\operatorname{Im}m_{1n}|\rightarrow 0.$ Therefore, by monotonicity, there
exists a unique $\operatorname{Im}m_{1n}>0$ such that (D.14) holds, which
corresponds to the bulk of the spectrum.

Furthermore, for any fixed $\operatorname{Re}m_{1n}>-l^{-1}$ and fixed $E$ so
that Theorem 2.8 holds, we have that $g(m_{1n},E)$ is monotone decreasing in
terms of $|\operatorname{Im}m_{1n}|.$ Let $E_{+}$ be defined according to
$m_{1n}(E_{+})=-l^{-1}.$ In view of (2.21) and (2.19), we have that

$l^{-1}=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{E_{+}-\sigma_{i}\widehat{\varsigma}_{2}}.$

Let $\widetilde{\varsigma}_{3}$ be defined similarly as
$\widehat{\varsigma}_{3}$ in (A.6) by replacing $\widehat{L}_{+}$ with
$E_{+}.$ Based on the above arguments and definitions, it is easy to see that

$\begin{split}&\sup_{\operatorname{Re}m_{1n}\in(-l^{-1},\infty)}g(m_{1n},E)=g(-l^{-1},E_{+})=\phi\widetilde{\varsigma}_{3}.\\\
\end{split}$

Assuming that $\phi\widetilde{\varsigma}_{3}<1,$ we conclude that (D.14) holds
only if $\operatorname{Im}m_{1n}(z)=0$ which corresponds to the outside part
of the spectrum. This shows that $m_{1n}=-l^{-1}$ is at the right edge of the
spectrum and gives the expression of the end point $\widehat{L}_{+}$ as in
(A.9). Therefore, $\widehat{L}_{+}=E_{+}$ and
$\widehat{\varsigma}_{3}=\widetilde{\varsigma}_{3}.$ This completes the proof.

Second, the proof of (A.10) follows from an argument similar to Lemma 8.4 of
[62] utilizing the estimate (2.7), we omit the details.

Third, we prove (A.11). The closeness of $\varsigma_{k}$ and
$\widehat{\varsigma}_{k},k=1,2,3,4,$ follows from arguments similar to the
last equation of (A.23). Now we proceed to the proof of the second equation.
According to the proof of (A.9) and an analogous argument, we found that
$m_{1n}(\widehat{L}_{+})=-l^{-1}$ and $m_{1n,c}(L_{+})=-l^{-1}.$ Together with
the definitions of $\varsigma_{2},\widehat{\varsigma}_{2},$ $m_{2n}$ and
$m_{2n,c},$ we find that

$\varsigma_{2}=-m_{2n,c}(L_{+})L_{+},\
\widehat{\varsigma}_{2}=-m_{2n}(\widehat{L}_{+})\widehat{L}_{+}.$ (D.15)

Next, by (A.9) and an analogous argument for $L_{+}$ (see (3.7) and the proof
of part II below), we have that

$\displaystyle\begin{split}0&=\frac{1}{n}\sum_{i=1}^{p}\frac{-l\sigma_{i}}{(-L_{+}+\sigma_{i}\varsigma_{2})}+\frac{1}{n}\sum_{i=1}^{p}\frac{l\sigma_{i}}{(-\widehat{L}_{+}+\sigma_{i}\widehat{\varsigma}_{2})}\\\
&=\frac{1}{n}\sum_{i}\frac{-l\sigma_{i}}{(-L_{+}+\sigma_{i}\varsigma_{2})}+\frac{1}{n}\sum_{i}\frac{l\sigma_{i}}{(-L_{+}+\sigma_{i}\widehat{\varsigma}_{2})}+\frac{1}{n}\sum_{i}\frac{l\sigma_{i}(\widehat{L}_{+}-L_{+})}{(-\widehat{L}_{+}+\sigma_{i}\widehat{\varsigma}_{2})(-L_{+}+\sigma_{i}\widehat{\varsigma}_{2})}.\end{split}$
(D.16)

By first equation of (A.11), (D.15) and Assumption 2.10, we can conclude our
proof.

Fourth, we work with (A.12) and (A.13). Due to similarity, we only prove
(A.12). According to (2.8), we have that

$\displaystyle
m_{1n}(z)=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}}.$

Consequently, it is easy to see that for
$z=\widehat{L}_{+}-\kappa+\mathrm{i}\eta\in\mathbf{D}_{b},$

$\displaystyle\begin{split}m_{1n}(\widehat{L}_{+})-m_{1n}(z)=R_{1}(\widehat{L}_{+}-z)+R_{2}(m_{1n}(\widehat{L}_{+})-m_{1n}(z)),\end{split}$
(D.17)

where we denote

$\displaystyle
R_{1}:=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}}{(-\widehat{L}_{+}+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})(-z+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)})}$
$\displaystyle
R_{2}:=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}}{(1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+}))(1+\xi^{2}_{j}m_{1n}(z))}}{(-\widehat{L}_{+}+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(\widehat{L}_{+})})(-z+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)})}.$

To study the terms $R_{1}$ and $R_{2},$ we will need the following control
whose proof follows from equations (4.24)-(4.28) of [62]

$\frac{1}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{4}}{(1-\xi_{j}^{2}l^{-1})(1+\xi_{j}^{2}m_{1n}(z))}=\begin{cases}\mathrm{O}\left(\log
n\right),&d\geqslant 2;\\\ \mathrm{O}\left(|l^{-1}+m_{1n}(z)|^{d-2}\log
n\right),&1<d\leqslant 2.\end{cases}$ (D.18)

For the denominator of $R_{1}$, since $z\in\mathbf{D}_{b},$ by a discussion
similar to (D.16), we find they are bounded from below so that
$R_{1}=\mathrm{O}(1).$ Furthermore, since $m_{1n}(\widehat{L}_{+})=-l^{-1},$
by a straightforward calculation, using the definition of
$\widehat{\varsigma}_{2}$ in (A.6) and the control (D.18), we observe that

$\displaystyle\begin{split}R_{1}&=\widehat{\varsigma}_{4}-\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}(z-\widehat{L}_{+})+\frac{\sigma_{i}^{2}}{n}\sum_{j}\frac{\xi^{4}_{j}(m_{1n}(z)+l^{-1})}{(1-\xi^{2}_{j}l^{-1})(1+\xi^{2}_{j}m_{1n}(s))}}{(-\widehat{L}_{+}+\sigma_{i}\widehat{\varsigma}_{2})^{2}(-z+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)})}\\\
&=\widehat{\varsigma}_{4}+\mathrm{O}(|z-\widehat{L}_{+}|)+\mathrm{O}(|m_{1n}(z)+l^{-1}|^{\min\\{d-1,1\\}}\log
n),\end{split}$ (D.19)

where in the second step we again used an argument similar to (D.16).
Similarly, for $R_{2},$ we find that

$\displaystyle\begin{split}R_{2}&=\phi\widehat{\varsigma}_{3}+\mathrm{O}(|z-\widehat{L}_{+}|)+\mathrm{O}(|m_{1n}(z)+l^{-1}|^{\min\\{d-1,1\\}}\log
n).\end{split}$ (D.20)

We next provide a useful deterministic control. Using a discussion similar to
[60, Lemma A.4], we find from (2.16) that

$\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}\frac{1}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{4}}{(1+\xi_{j}^{2}m_{1n}(z))^{2}}}{|-z+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n}(z)}|^{2}}=1-\frac{1}{n}\sum_{i}\frac{\sigma_{i}\eta/\operatorname{Im}m_{1n}(z)}{|-z+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}=1-\eta\frac{|m_{1n}(z)|^{2}}{\operatorname{Im}m_{1n}(z)}.$
(D.21)

Since $\operatorname{Im}m_{1n}(z)>0,$ this implies that

$0\leqslant
1-\frac{1}{n}\sum_{i}\frac{\sigma_{i}\eta/\operatorname{Im}m_{1n}(z)}{|(-z+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)})|^{2}}\leqslant
1.$

Together with Cauchy-Schwarz inequality, we see that

$\displaystyle\begin{split}|R_{2}|\leqslant(\phi\widehat{\varsigma}_{3})^{1/2}\left(1-\frac{1}{n}\sum_{i}\frac{\sigma_{i}\eta/\operatorname{Im}m_{1n}(z)}{|(-z+\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)})|^{2}}\right)^{1/2}<1,\end{split}$

where we used the fact $\operatorname{Im}m_{1n}(z)>0$ and $\eta>0.$ Using
(D.17), we find that
$m_{1n}(\widehat{L}_{+})-m_{1n}(z)\asymp\widehat{L}_{+}-z.$ Then we can
conclude our proof using (D.17), (D.19) and (D.20).

Finally, we prove the controls for the imaginary parts. For (A.14), the
discussion is similar to that of Lemma 4.5 of [60]. According to (2.19), we
find see that

$\displaystyle\begin{split}-m_{1n}(z)&=\frac{1}{n}\frac{\sigma_{1}}{z-\frac{\sigma_{1}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}}+\frac{1}{n}\sum_{i=2}^{p}\frac{\sigma_{i}}{z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}}\\\
&=\mathrm{O}(\frac{1}{n\eta})+\frac{1}{n}\sum_{i=2}^{p}\frac{\sigma_{i}}{z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}},\end{split}$
(D.22)

where in the step we the fact that $\operatorname{Im}m_{1n}(z)\geqslant 0$ and
the trivial bound that

$\frac{1}{n}\left|\left(z-\frac{\sigma_{1}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}\right)^{-1}\right|\leqslant
n^{-1}\left(\eta+\frac{\sigma_{1}}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{4}\operatorname{Im}m_{1n}(z)}{|1+\xi_{j}^{2}m_{1n}(z)|^{2}}\right)^{-1}\leqslant(n\eta)^{-1}.$
(D.23)

Taking the imaginary part on both sides of (D.22), we see that for some
constant $0<c<1,$

$\displaystyle\begin{split}\operatorname{Im}m_{1n}(z)&=\frac{1}{n}\sum_{i=2}^{p}\frac{\sigma_{i}(\eta+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z)}{|1+\xi^{2}_{j}m_{1n}(z)|})}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}+\mathrm{O}(\frac{1}{n\eta})\\\
&=\frac{1}{n}\sum_{i=2}^{p}\frac{\sigma_{i}\eta}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}+\frac{1}{n}\sum_{i=2}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z)}{|1+\xi^{2}_{j}m_{1n}(z)|}}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}+\mathrm{O}(\frac{1}{n\eta})\\\
&=\mathrm{O}(\eta)+\mathrm{O}(\frac{1}{n\eta})+c\operatorname{Im}m_{1n}(z),\end{split}$

where in the last step we used discussions similar to (D.25) and (D.28) below.
This concludes the proof. Then we prove (A.16) and (A.15) following [62, Lemma
5.2]. Due to similarity, we focus our analysis on $m_{1n}(z)$ and discuss
$m_{n}(z)$ briefly in the end. In what follows, for notational simplicity,
without loss of generality, we assume that on $\Omega,$
$\xi_{(i)}^{2}=\xi_{i}^{2}.$ In what follows, we identify $\eta\equiv\eta_{0}$
till the end of the proof of the lemma. According to (2.16), we find that

$\displaystyle\operatorname{Im}m_{1n}(z)$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}(\eta+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z)}{|1+\xi^{2}_{j}m_{1n}(z)|^{2}})}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{p}\frac{\sigma_{i}\eta}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}+\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\frac{\xi^{4}_{1}\operatorname{Im}m_{1n}(z)}{|1+\xi^{2}_{1}m_{1n}(z)|^{2}}}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}+\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=2}^{n}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z)}{|1+\xi^{2}_{j}m_{1n}(z)|^{2}}}{|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}$
$\displaystyle=\mathsf{L}_{1}+\mathsf{L}_{2}+\mathsf{L}_{3}.$ (D.24)

For the denominator, by the results and arguments in Section B.2.1, we observe
that when $z\in\mathbf{D}_{b}^{\prime}$

$\displaystyle
z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n}(z)}$
$\displaystyle=z-\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n}(z)}+\frac{\sigma_{i}}{n}\frac{\xi_{1}^{2}}{1+\xi_{1}^{2}m_{1n}(z)}$
$\displaystyle=z-\frac{\sigma_{i}}{n}\sum_{j=2}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n,c}(z)}+\mathrm{O}((n\eta)^{-1}+n^{-1/2-1/(d+1)}).$

Together with Assumption 2.10 and (A.23), we find that for some small constant
$c^{\prime}>0,$ when $n$ is sufficiently large,

$\left|z-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi_{j}^{2}}{1+\xi_{j}^{2}m_{1n}(z)}\right|\geqslant
c^{\prime}.$

This implies that

$\mathsf{L}_{1}\asymp\eta.$ (D.25)

For $\mathsf{L}_{2},$ on the one hand, when $|z-z_{0}|\geqslant
Cn^{-1/2+3\epsilon_{d}},$ by (A.13), we conclude that on $\Omega,$ for some
constant $C>0$

$|\mathsf{L}_{2}|\leqslant n^{-1/2-3\epsilon_{d}}\operatorname{Im}m_{1n}(z).$
(D.26)

On the other hand, when $z=z_{0}$ so that
$\operatorname{Re}m_{1n}(z)=-\xi_{1}^{2},$ we can rewrite $\mathsf{L}_{2}$ as

$\mathsf{L}_{2}=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}}{n\operatorname{Im}m_{1n}(z)}}{|z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}.$
(D.27)

Next, for $\mathsf{L}_{3},$ by (B.44), (A.23) and the results and arguments in
Section B.2.1, using the trivial bound for $\mathsf{L}_{2}$ that
$|\mathsf{L}_{2}|=\mathrm{O}((n\eta)^{-1}),$ we conclude that when
$z\in\mathbf{D}_{b}^{\prime},$ for some constant $0<\mathfrak{c}<1$

$|\mathsf{L}_{3}|\leqslant\mathfrak{c}\operatorname{Im}m_{1n}(z).$ (D.28)

Consequently, we find that (A.16) follows from (D.1.2), (D.25), (D.26) and
(D.28). Moreover, (A.15) follows from (D.1.2), (D.25), (D.27) and (D.28) by
solving the associated quadratic equation. Finally, we mention that the
results for $\operatorname{Im}m_{n}(z)$ essentially follows from (2.16) that

$\displaystyle\operatorname{Im}m_{n}(z)=\frac{1}{n}\sum_{i=1}^{p}\frac{\eta}{|z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}}+\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{1}{n}\sum_{j}\frac{\xi^{4}_{j}\operatorname{Im}m_{1n}(z)}{|1+\xi^{2}_{j}m_{1n}(z)|^{2}}}{|z-\frac{\sigma_{i}}{n}\sum_{j}\frac{\xi^{2}_{j}}{1+\xi^{2}_{j}m_{1n}(z)}|^{2}},$

with the results for $\operatorname{Im}m_{1n}(z)$. This completes our proof.

∎

###### Proof of Part (b).

For (A.17), on the one hand, when when $d>1$ and
$\phi^{-1}<\widehat{\varsigma}_{3},$ the result has been proved in (C.37). On
the other hand, when $-1<d\leqslant 1,$ we employ the proof idea as in the
proof of Lemma A.3 of [61] using a continuity argument. Recall (C.44). Denote

$g(x,y)\equiv\frac{\partial F_{n}(x,y)}{\partial
x}+1=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{\xi^{4}_{j}}{(1+x\xi^{2}_{j})^{2}}}{(-y+\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{\xi^{2}_{j}}{1+x\xi^{2}_{j}})^{2}}.$

From our assumption that $-1<d\leqslant 1$ and (2.7), we find that there exist
constants $C,C_{0}>0$ such that $\mathrm{d}F(x)\geq C(l-x)^{d}\geq C_{0}(l-x)$
for $x\in(0,l)$. Let $D$ be a sufficiently large constant and choose a
sufficiently small constant $0<\epsilon<D^{-1}$, we have that when $n$ is
sufficiently large, there exists some constants $C_{1},C_{2},C_{3}>0$

$\displaystyle g(-(l+\epsilon)^{-1},\widehat{L}_{+})$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{p}\frac{\frac{\sigma_{i}^{2}}{n}\sum_{j=1}^{n}\frac{(l+\epsilon)^{2}\xi^{4}_{j}}{(l+\epsilon-\xi^{2}_{j})^{2}}}{\left(\widehat{L}_{+}-\frac{\sigma_{i}}{n}\sum_{j=1}^{n}\frac{(l+\epsilon)\xi^{2}_{j}}{l+\epsilon-\xi^{2}_{j}}\right)^{2}}$
$\displaystyle\geq\frac{C_{1}}{n}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}\int_{l-(D-1)\epsilon}^{l}\frac{(l+\epsilon)^{2}x^{2}}{(l+\epsilon-x)^{2}}\mathrm{d}F(x)}{(\widehat{L}_{+}-\sigma_{i}\mathrm{O}(1))^{2}}\geq\frac{1}{n}\sum_{i=1}^{p}\frac{C_{2}\int_{l-(D-1)\epsilon}^{l}\frac{(l-x)}{(l+\epsilon-x)^{2}}\mathrm{d}x}{(\widehat{L}_{+}-\sigma_{i}\mathrm{O}(1))^{2}}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{p}\frac{C_{2}\int_{\epsilon}^{D\epsilon}\frac{(t-\epsilon)}{t^{2}}\mathrm{d}t}{(\widehat{L}_{+}-\sigma_{i}\mathrm{O}(1))^{2}}\geqslant
C_{3}(\log D-1+\frac{1}{D})>1,$

for sufficiently large $D>0.$ Similar arguments apply to
$g(-(l-\epsilon),\widehat{L}_{+}).$ Consequently, by the continuity of
$g(x,y)$, we obtain that $\partial F_{n}(-l^{-1},\widehat{L}_{+})/\partial
x>0.$ Since $\partial F_{n}(m_{1n}(\widehat{L}_{+}),\widehat{L}_{+})/\partial
x=0,$ we can conclude that (C.37) still holds. That is,
$m_{1n}(\widehat{L}_{+})>-l^{-1}$. This finishes the proof of (A.17).

For (A.18) and (A.19), using (A.17), by a discussion similar to (C.46), we see
that

$\frac{\partial^{2}F_{n}(m_{1n}(\widehat{L}_{+}),\widehat{L}_{+})}{\partial
x^{2}}\asymp 1.$ (D.29)

Armed with this input, the square root behavior of $\rho$ at $\widehat{L}_{+}$
can be obtained in the same way as Lemma A.1 of [61]. Due to similarity, we
omit the details. This completes our proof of Part (b). ∎

Finally, it is easy to check that we can follow lines of the proofs of parts
(a) and (b) to prove the unconditional results by replacing the related
quantities verbatim. We omit further details.

∎

###  D.2 Control of some bad probability events: proof of Lemma A.11

In this subsection, we prove Lemma A.11 case by case. We first prove Case (a)
in Definition A.9.

###### Proof of Case (a)..

First, the last statement of (A.21) follows directly from strong law of large
number. In fact, the result holds almost surely.

Then, we prove the second statement of (A.21). For the upper bound, since
$\\{\xi_{i}^{2}\\}$ are independent, we readily see that when $n$ is
sufficiently large, for some constant $C^{\prime}>0,$

$\begin{split}\mathbb{P}(\xi_{(1)}^{2}\leqslant Cn^{1/\alpha}\log
n)&=\left(1-\mathbb{P}(\xi^{2}>Cn^{1/\alpha}\log
n)\right)^{n}\geq\left(1-\frac{L(Cn^{1/\alpha}\log n)}{(Cn^{1/\alpha}\log
n)^{\alpha}}\right)^{n}\\\
&\geq(1-C^{\prime}n^{-1}\log^{-\alpha}n)^{n}\asymp\exp\left(-1/(C^{\prime}\log^{\alpha}n)\right)\asymp
1-\mathrm{O}(\log^{-\alpha}n).\end{split}$

where in the second step we used the assumption (2.5) and in the third step we
used the assumption that $L(\cdot)$ is a slowly varying function. This proves
the upper bound. Similarly, for the lower bound, we can show that for some
large constant $C>0$

$\mathbb{P}(\xi^{2}_{(1)}\leq n^{1/\alpha}\log^{-1}n)=\mathrm{O}(\log
n/n^{C}).$ (D.30)

This concludes the proof of the second statement.

Next, we prove the first statement using the second one. Note that

$\begin{split}\mathbb{P}(\xi^{2}_{(1)}-\xi^{2}_{(2)}<n^{1/\alpha}\log^{-1}n)&=\mathbb{P}(\xi^{2}_{(1)}<n^{1/\alpha}\log^{-1}n+\xi^{2}_{(2)})=\mathbb{P}(\xi^{2}_{(1)}<Cn^{1/\alpha}\log^{-1}n)=\mathrm{O}(\log
n/n^{C}),\end{split}$

where the second and third steps we used the results of the second statement.

Then we justify the fourth statement. In what follows, without loss of
generality, we assume that $n^{b}$ is an integer. For $c>1$ and $b>1/2$, we
notice that for some large constant $C>0$

$\begin{split}\mathbb{P}(&\xi^{2}_{(1)}-\xi^{2}_{(n^{b})}<c^{-1}n^{1/\alpha}\log^{-1}n)\leq\mathbb{P}(\xi^{2}_{(n^{b})}\geq(1-c^{-1})n^{1/\alpha}\log^{-1}n)\\\
&=\sum_{k=n^{b}}^{n}\binom{n}{k}\left[\mathbb{P}(\xi^{2}\geq(1-c^{-1})n^{1/\alpha}\log^{-1}n)\right]^{k}\left[\mathbb{P}(\xi^{2}\leq(1-c^{-1})n^{1/\alpha}\log^{-1}n)\right]^{n-k}\\\
&=\sum_{k=n^{b}}^{n}\binom{n}{k}\left(\frac{\log^{\alpha}n}{n}\right)^{k}\left(1-\frac{\log^{\alpha}n}{n}\right)^{n-k}\\\
&\leq\sum_{k=n^{b}}^{n}\left(\frac{en}{k}\right)^{k}\left(\frac{\log^{\alpha}n}{n}\right)^{k}\left(1-\frac{\log^{\alpha}n}{n}\right)^{n-k}\\\
&\leq\sum_{k=n^{b}}^{n}\left(\frac{e}{k}\right)^{k}\log^{\alpha
k}ne^{-\log^{\alpha}n(1-k/n)}=\mathrm{O}(\log n/n^{C}),\end{split}$ (D.31)

where in the first step we used (D.30), in the third step we used (2.5) and in
the fourth step we used Stirling’s formula. This concludes the proof.

Finally, we proceed to the proof of the third statement. Define a sequence of
intervals
$I_{k}:=\\{Cn^{1/\alpha}\log^{-1}n+kn^{\epsilon},Cn^{1/\alpha}\log^{-1}n+(k+1)n^{\epsilon}\\},k=[\\![1,n^{1/\alpha-\epsilon}]\\!]$.
It is easy to see that if $\xi^{2}_{(i)}-\xi^{2}_{(i+1)}<n^{\epsilon}$ when
$\xi^{2}_{(i)},\xi^{2}_{(i+1)}\in I_{k}$ for some $k$. Setting
$\mathsf{p}_{k}:=\mathbb{P}(\xi^{2}\in I_{k})$, we see that

$\mathbb{P}(|j\in[\\![1,n]\\!]:\xi^{2}_{j}\in
I_{k}|=0)=(1-\mathsf{p}_{k})^{n},\quad\mathbb{P}(|j\in[\\![1,n]\\!]:\xi^{2}_{j}\in
I_{k}|=1)=n\mathsf{p}_{k}(1-\mathsf{p}_{k})^{n-1}.$

We now provide an estimate for $p_{k}$ using (2.5). Note that

$\displaystyle\mathsf{p}_{k}$
$\displaystyle=\mathbb{P}(Cn^{1/\alpha}\log^{-1}n+kn^{\epsilon}\leq\xi^{2}\leq
Cn^{1/\alpha}\log^{-1}n+(k+1)n^{\epsilon})$
$\displaystyle\leq\frac{1}{(Cn^{1/\alpha}\log^{-1}n+kn^{\epsilon})^{\alpha}}-\frac{1}{(Cn^{1/\alpha}\log^{-1}n+(k+1)n^{\epsilon})^{\alpha}}$
$\displaystyle\leq
Cn^{-1}\log^{\alpha}n\frac{(1+(k+1)n^{-1/\alpha+\epsilon}\log
n)^{\alpha}-(1+kn^{-1/\alpha+\epsilon}\log
n)^{\alpha}}{(1+kn^{-1/\alpha+\epsilon}\log n)^{\alpha}}$ $\displaystyle\leq
Cn^{-1}\log^{\alpha}n^{-1/\alpha+\epsilon}\log
n=Cn^{-(1+1/\alpha)+\epsilon}\log^{\alpha}n.$ (D.32)

Armed with the above estimate, we see that when $n$ is sufficiently large,

$\mathbb{P}(\xi^{2}_{(i)},\xi^{2}_{(i+1)}\in
I_{k})\leq\mathbb{P}(|j\in[\\![1,n]\\!]:\xi^{2}_{j}\in I_{k}|\geq
2)=1-(1-\mathsf{p}_{k})^{n}-n\mathsf{p}_{k}(1-\mathsf{p}_{k})^{n-1}\leq
n^{2}\mathsf{p}_{k}^{2}.$

Consequently, together with (D.2), we have that for some constant $C_{1}>0$

$\mathbb{P}(\xi^{2}_{(i)}-\xi^{2}_{(i+1)}\leq
n^{\epsilon})\leq\sum_{k=1}^{n^{1/\alpha-\epsilon}}n^{2}\mathsf{p}_{k}^{2}\leq
Cn^{1/\alpha+2-\epsilon}n^{-(2+2/\alpha)+2\epsilon}\log^{\alpha}n=n^{-1/\alpha+\epsilon}\log^{\alpha}n=\mathrm{o}(1),$
(D.33)

as long as $\epsilon<1/\alpha$. This finishes the proof of the third
statement.

∎

Then we prove Case (b) of Definition A.9.

###### Proof of Case (b)..

Due to similarity and for notational simplicity, we focus on the case
$\beta=1.$ The general setting can be proved analogously and we omit the
details.

We start with the second statement of (A.22). For the upper bound, for any