$\displaystyle=-\frac{\eta_{0}^{2}}{4\pi^{2}}\int\int{g}\left(\tilde{x}_{1}\right){g}\left(\tilde{x}_{2}\right)\left(\widehat{\alpha}\left(E+\tilde{x}_{1}{\eta_{0}}+\mathrm{i}\underline{\eta}_{1},E+\tilde{x}_{2}{\eta_{0}}+\mathrm{i}\underline{\eta}_{2}\right)+\widehat{\beta}\left(E+\tilde{x}_{1}{\eta_{0}}+\mathrm{i}\underline{\eta}_{1},E+\tilde{x}_{2}{\eta_{0}}+\mathrm{i}\underline{\eta}_{2}\right)\right)\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}$
$\displaystyle:=\left(\mathcal{K}_{1}\right)_{++}+\left(\mathcal{K}_{2}\right)_{++}.$

For $(\mathcal{K}_{1})_{++},$ we see that

$\displaystyle\left(\mathcal{K}_{1}\right)_{++}$
$\displaystyle=-\frac{{\eta_{0}}^{2}}{4\pi^{2}}\iint
g\left(\tilde{x}_{1}\right)g\left(\tilde{x}_{2}\right)\widehat{\alpha}_{++}\left(E+\tilde{x}_{1}{\eta_{0}},E+\tilde{x}_{2}{\eta_{0}}\right)\mathrm{d}\tilde{y}_{1}\mathrm{~{}d}\tilde{y}_{2}+\mathrm{O}(\sqrt{\tilde{{\eta}}})$
$\displaystyle=-\frac{{\eta_{0}}^{2}}{4\pi^{2}}\iint
g\left(\tilde{x}_{1}\right)g\left(\tilde{x}_{2}\right)\widehat{\alpha}_{++}\left(E+\tilde{x}_{1}{\eta_{0}},E+\tilde{x}_{2}{\eta_{0}}\right)\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}+\mathrm{O}\left(n^{-\varepsilon_{0}/2}\right)$
$\displaystyle=\mathrm{O}\left(\frac{1}{\sqrt{\kappa_{1}+\underline{\eta}_{1}}\sqrt{\kappa_{2}+\underline{\eta}_{2}}}\eta_{0}^{2}\right)=\mathrm{O}\left(\frac{1}{\sqrt{\underline{\eta}_{1}}\sqrt{\underline{\eta}_{2}}}\eta_{0}^{2}\right).$
(7.21)

This indicates that when $\eta_{0}=\mathrm{o}(1)$, only the $\beta$ terms will
have non-vanishing contributions. Similarly, we can calculate the integral in
$\mathcal{K}_{1}$ over the other three regions:
$\left(\mathcal{K}_{1}\right)_{+-}$for
$\mathcal{R}_{+-}:=\left\\{y_{1}=\underline{{\eta}}_{1},y_{2}=-\underline{{\eta}}_{2}\right\\}$,
$\left(\mathcal{K}_{1}\right)_{-+}$for
$\mathcal{R}_{-+}:=\left\\{y_{1}=-\underline{{\eta}}_{1},y_{2}=\underline{{\eta}}_{2}\right\\}$,
and $\left(\mathcal{K}_{1}\right)_{--}$for
$\mathcal{R}_{--}:=\left\\{y_{1}=-\underline{{\eta}}_{1},y_{2}=-\underline{{\eta}}_{2}\right\\}$
and obtain the same result. This completes the proof for
$\eta_{0}=\mathrm{o}(1)$ and hence that of $\mathbb{E}[\mathcal{Z}(f)^{l}].$

Finally, we discuss how to generalize the results to multiple functions.
Similar to (7.3), we can write

$\mathbb{E}\left[\mathcal{Z}_{{\eta_{0}},E}\left(f_{i_{1}}\right)\cdots\mathcal{Z}_{{\eta_{0}},E}\left(f_{i_{l}}\right)\right]=\int\theta_{f_{i_{1}}}\left(z_{1}\right)\cdots\theta_{f_{i_{l}}}\left(z_{l}\right)\mathbb{E}\left[\mathcal{Y}\left(z_{1}\right)\cdots\mathcal{Y}\left(z_{l}\right)\right]\mathrm{d}^{2}z_{1}\cdots\mathrm{d}^{2}z_{l}.$

Following the proof for a single function, it is not hard to see that

$\mathbb{E}\left[\mathcal{Z}_{{\eta_{0}},E}\left(f_{i_{1}}\right)\cdots\mathcal{Z}_{{\eta_{0}},E}\left(f_{i_{l}}\right)\right]=\sum_{s=2}^{l}\varpi\left(f_{i_{1}},f_{i_{s}}\right)\mathbb{E}\prod_{t\notin\\{1,s\\}}\mathcal{Z}_{{\eta_{0}},E}\left(f_{i_{t}}\right)+\mathrm{O}_{\prec}\left(n^{-c}\right)$

for some constant $c>0$. Then the proof follows from an induction argument.

∎

###  7.2 Proof of Theorems 3.2 and 3.7

Control of $\mathcal{M}_{\eta_{0},E}(f_{i})$. For simplicity, we first rewrite
the quantity a little bit. Using the fact that $Q$ and $\mathcal{Q}$ share the
same non-zero eigenvalues and $E\sim\sqrt{\phi}$ and
$g_{i}\in\mathcal{C}^{2}_{c}\left(\mathbb{R}\right),$ we have that
$f_{i}(0)=g_{i}(\frac{0-E}{\eta_{0}})$=0. Combining (5.5), we have that
$n\int_{\mathbb{R}}f_{i}(x)\mathrm{d}\varrho(x)=p\int_{\mathbb{R}}f_{i}(x)\mathrm{d}\varrho_{p}(x).$
Together with the definition of Stieltjes transform and Fubini’s theorem,
using Lemma 5.8, we have that

$n\int_{\mathbb{R}}f_{i}(x)\mathrm{d}\varrho(x)=\frac{1}{\pi}\int_{\mathbb{C}}\frac{\partial}{\partial\bar{z}}\tilde{f}_{i}(z)pm_{p}(z)\mathrm{d}^{2}z.$
(7.22)

Using Lemma 5.8 again, we have that

$\mathbb{E}\operatorname{Tr}f_{i}(Q)=\frac{1}{\pi}\int_{\mathbb{C}}\frac{\partial}{\partial\bar{z}}\tilde{f}_{i}(z)\mathbb{E}\operatorname{Tr}R_{1}(z)\mathrm{d}^{2}z.$
(7.23)

Therefore, due to the regularity properties of $f_{i},$ in order to study the
quantity $\mathcal{M}_{\eta_{0},E}$, it suffices to study
$\mathbb{E}\operatorname{Tr}{R_{1}}(z)-pm_{p}(z)$ and then apply a complex
integral. The control of $\mathbb{E}\operatorname{Tr}{R_{1}}(z)-pm_{p}(z)$ is
similar to the discussions in the proof of Lemma 6.2 using the cumulant
expansion formula Lemma 5.7. Due to similarity, we only sketch the main ideas
of the proof. We start with $\mathbb{E}({R_{1}})_{ii},1\leq i\leq p$. Using

$\frac{\partial({R_{1}})_{ab}}{\partial
y_{ij}}=-({R_{1}})_{ai}\left(Y^{*}{R_{1}}\right)_{jb}-({R_{1}}Y)_{aj}({R_{1}})_{ib},\quad\frac{\partial({R_{1}})_{ab}}{\partial
x_{ij}}=\frac{\partial({R_{1}})_{ab}}{\partial y_{ij}}\sqrt{\sigma_{i}},$

the relation $z{R_{1}}(z)=QR_{1}(z)-I$, Lemma 5.7 and Theorem 5.5, we have
that

$\displaystyle
z\mathbb{E}({R_{1}})_{ii}=\sqrt{\sigma_{i}}\sum_{j=1}^{n}\mathbb{E}X_{ij}({R_{1}}Y)_{ij}-1$
(7.24) $\displaystyle=$
$\displaystyle\frac{\sqrt{\sigma_{i}}}{\sqrt{np}}\mathbb{E}\sum_{j=1}^{n}\frac{\partial({R_{1}}Y)_{ij}}{\partial
x_{ij}}-1+\frac{\sqrt{\sigma_{i}}}{2!(np)^{\frac{3}{4}}}\sum_{j=1}^{n}\kappa_{3}\mathbb{E}\frac{\partial^{2}({R_{1}}Y)_{ij}}{\partial^{2}x_{ij}}+\frac{\sqrt{\sigma_{i}}}{3!np}\sum_{j=1}^{n}\kappa_{4}\mathbb{E}\frac{\partial^{3}({R_{1}}Y)_{ij}}{\partial^{3}x_{ij}}+\mathrm{O}_{\prec}\left(n^{-\frac{1}{4}}p^{-\frac{5}{4}}\right)$
$\displaystyle=$
$\displaystyle\frac{\sigma_{i}}{\sqrt{np}}\sum_{j=1}^{n}\mathbb{E}({R_{1}})_{ii}-\frac{\sigma_{i}}{\sqrt{np}}\sum_{j=1}^{n}\mathbb{E}\left(Y^{*}({R_{1}})Y\right)_{jj}({R_{1}})_{ii}-\frac{\sigma_{i}}{\sqrt{np}}\sum_{j=1}^{n}\mathbb{E}\left((({R_{1}})Y)_{ij}\right)^{2}-1$
$\displaystyle+\frac{\sigma_{i}^{3/2}}{2(np)^{\frac{3}{4}}}\sum_{j=1}^{n}\kappa_{3}\mathbb{E}\left(-6({R_{1}})_{ii}(({R_{1}})Y)_{ij}+6({R_{1}})_{ii}(({R_{1}})Y)_{ij}\left(Y^{*}({R_{1}})Y\right)_{jj}+2\left((({R_{1}})Y)_{ij}\right)^{3}\right)$
$\displaystyle+\frac{\sigma_{i}^{2}}{6np}\sum_{j=1}^{n}\kappa_{4}\mathbb{E}\left(-6\left(({R_{1}})_{ii}\right)^{2}+12\left(({R_{1}})_{ii}\right)^{2}\left(Y^{*}({R_{1}})Y\right)_{jj}-6\left(({R_{1}})_{ii}\right)^{2}\left(\left(Y^{*}({R_{1}})Y\right)_{jj}\right)^{2}\right)+\mathrm{O}_{\prec}\left(p^{-1}\Psi(z)\right)$
$\displaystyle:$
$\displaystyle=\mathsf{S}_{i,1}+\mathsf{S}_{i,2}+\mathsf{S}_{i,3}+\mathrm{O}_{\prec}\left(p^{-1}\Psi(z)\right).$

Then we deal with the terms $\mathsf{S}_{i,k},k=1,2,3.$ First, by a discussion
similar to the proof of Lemma 6.3, using Theorem 5.5, we have that

$\displaystyle\mathsf{S}_{i,1}=$
$\displaystyle\phi^{-1/2}\sigma_{i}\mathbb{E}({R_{1}})_{ii}-\frac{\sigma_{i}}{\sqrt{np}}\mathbb{E}\left[\operatorname{Tr}\left(Y^{*}{R_{1}}Y\right)({R_{1}})_{ii}\right]-\frac{\sigma_{i}}{\sqrt{np}}\mathbb{E}(R_{1}QR_{1})_{ii}-1$
(7.25) $\displaystyle=$
$\displaystyle\phi^{-1/2}\sigma_{i}\mathbb{E}({R_{1}})_{ii}-\frac{\sigma_{i}}{\sqrt{np}}\mathbb{E}\left[\operatorname{Tr}\left((Q-z+z){R_{1}}\right)({R_{1}})_{ii}\right]-\frac{\sigma_{i}}{\sqrt{np}}\mathbb{E}(R_{1}(Q-z+z)R_{1})_{ii}-1$
$\displaystyle=$
$\displaystyle\sigma_{i}(\phi^{-1/2}-\phi^{1/2})\mathbb{E}({R_{1}})_{ii}-z\sigma_{i}\phi^{1/2}\mathbb{E}\left[p^{-1}\operatorname{Tr}({R_{1}})({R_{1}})_{ii}\right]-\frac{\sigma_{i}}{\sqrt{np}}\mathbb{E}\left(\frac{\partial}{\partial
z}z{R_{1}}(z)\right)_{ii}-1$ $\displaystyle=$
$\displaystyle\sigma_{i}(\phi^{-1/2}-\phi^{1/2})\mathbb{E}({R_{1}})_{ii}-z\sigma_{i}\phi^{1/2}m_{p}\mathbb{E}\left(({R_{1}})_{ii}+\frac{1}{z\left(1+\phi^{-1/2}m\sigma_{i}\right)}\right)$
$\displaystyle+\frac{\sigma_{i}\phi^{1/2}}{p\left(1+\phi^{-1/2}m\sigma_{i}\right)}\mathbb{E}\operatorname{Tr}({R_{1}})-\frac{\sigma_{i}^{2}}{p}\frac{m^{\prime}}{\left(1+\phi^{-1/2}\sigma_{i}m\right)^{2}}-1+\tilde{\mathcal{E}}_{i,1},$

where we denote

$\displaystyle\tilde{\mathcal{E}}_{i,1}:=$
$\displaystyle\frac{z\sigma_{i}\phi^{1/2}}{p}\mathbb{E}\left[\left(\operatorname{Tr}({R_{1}(z)})-pm_{p}\right)\left(({R_{1}})_{ii}-\left(-\frac{1}{z(1+\phi^{-1/2}m\sigma_{i})}\right)\right)\right]$
$\displaystyle-\frac{\sigma_{i}}{\sqrt{np}}\left(\mathbb{E}\left(\frac{\partial}{\partial
z}z{R_{1}}(z)\right)_{ii}-\frac{\phi^{-1/2}\sigma_{i}m^{\prime}}{(1+\phi^{-1/2}m\sigma_{i})^{2}}\right).$

Here we used the identity $\frac{\partial(zR(z))}{\partial
z}={R_{1}}(z)+z{R_{1}}^{2}(z)$ to estimate
$\frac{\sigma_{i}}{\sqrt{np}}\mathbb{E}[({R_{1}}Q{R_{1}})_{ii}]$ in the second
step, and used the identity
$\mathbb{E}[AB]=\mathbb{E}[(A-a)(B-b)]+\mathbb{E}[aB-ab+Ab]$ to control
$-z\sigma_{i}\phi^{1/2}\mathbb{E}\left[p^{-1}\operatorname{Tr}({R_{1}})({R_{1}})_{ii}\right]$
in the last step. Plugging (7.25) into (7.24), we have

$\displaystyle z\mathbb{E}({R_{1}})_{ii}=$
$\displaystyle\sigma_{i}(\phi^{-1/2}-\phi^{1/2})\mathbb{E}({R_{1}})_{ii}-z\sigma_{i}\phi^{1/2}m_{p}\mathbb{E}\left(({R_{1}})_{ii}+\frac{1}{z\left(1+\phi^{-1/2}m\sigma_{i}\right)}\right)$
$\displaystyle+\frac{\sigma_{i}\phi^{1/2}}{p\left(1+\phi^{-1/2}m\sigma_{i}\right)}\mathbb{E}\operatorname{Tr}({R_{1}})-\frac{\sigma_{i}^{2}}{p}\frac{m^{\prime}}{\left(1+\phi^{-1/2}\sigma_{i}m\right)^{2}}-1+\tilde{\mathcal{E}}_{i,1}+\mathsf{S}_{i,2}+\mathsf{S}_{i,3}+\mathrm{O}_{\prec}\left(p^{-1}\Psi(z)\right).$

Using (5.5), we can rewrite the above equation as follows

$\displaystyle z(1+\sigma_{i}\phi^{-1/2}m)\mathbb{E}({R_{1}})_{ii}$ (7.26)
$\displaystyle=-\frac{\sigma_{i}\phi^{1/2}m_{p}}{\left(1+\phi^{-1/2}m\sigma_{i}\right)}+\frac{\sigma_{i}\phi^{1/2}}{p\left(1+\phi^{-1/2}m\sigma_{i}\right)}\mathbb{E}\operatorname{Tr}({R_{1}})-\frac{\sigma_{i}^{2}}{p}\frac{m^{\prime}}{\left(1+\phi^{-1/2}\sigma_{i}m\right)^{2}}-1$
$\displaystyle+\tilde{\mathcal{E}}_{i,1}+\mathsf{S}_{i,2}+\mathsf{S}_{i,3}+\mathrm{O}_{\prec}\left(p^{-1}\Psi(z)\right).$

This yields that

$\displaystyle\mathbb{E}\operatorname{Tr}({R_{1}})$ (7.27) $\displaystyle=$
$\displaystyle\sum_{i=1}^{p}\frac{1}{z(1+\sigma_{i}\phi^{-1/2}m)}\left(\frac{\sigma_{i}\phi^{1/2}}{p\left(1+\phi^{-1/2}m\sigma_{i}\right)}\mathbb{E}[\operatorname{Tr}({R_{1}})-pm_{p}]-\frac{\sigma_{i}^{2}}{p}\frac{m^{\prime}}{\left(1+\phi^{-1/2}\sigma_{i}m\right)^{2}}-1\right.$
$\displaystyle\left.+\tilde{\mathcal{E}}_{i,1}+\mathsf{S}_{i,2}+\mathsf{S}_{i,3}+\mathrm{O}_{\prec}\left(p^{-1}\Psi(z)\right)\right).$

For the error term $\tilde{\mathcal{E}}_{i,1}$, using
$\frac{1}{z(1+\sigma_{i}\phi^{-1/2}m)}\asymp\phi^{-1/2}$ and a fluctuation
averaging argument as in the proof of (5.17) (c.f. Lemma A.9), we can obtain
that

$\sum_{i=1}^{p}\frac{1}{z(1+\sigma_{i}\phi^{-1/2}m)}\tilde{\mathcal{E}}_{i,1}=\mathrm{O}_{\prec}\left(n^{-\frac{1}{2}}p^{-\frac{1}{2}}\eta^{-2}\right).$
(7.28)

For the summation involving $\mathsf{S}_{i,2},$ among all the terms, we
discuss the following representative term

$(np)^{-3/4}\sum_{i=1}^{p}\sum_{j=1}^{n}({R_{1}})_{ii}(({R_{1}})Y)_{ij}\left(Y^{*}({R_{1}})Y\right)_{jj}\prec\phi^{1/2}(np)^{-3/4}(np)^{1/2}\phi^{-1/2}\Psi(z)=(np)^{-1/4}\Psi(z),$
(7.29)

where we used the identities ${R_{1}}Y=Y{R_{2}}$, $Y^{*}{R_{1}}={R_{2}}Y^{*}$
and (5.18). The other terms can be handled analogously (and more easily). This
results in

$\sum_{i=1}^{p}\frac{1}{z(1+\sigma_{i}\phi^{-1/2}m)}\mathsf{S}_{i,2}=\mathrm{O}_{\prec}\left(n^{\frac{1}{4}}p^{-\frac{3}{4}}\Psi(z)\right).$
(7.30)

Finally, using Theorem 5.5 and similar arguments as in (7.29), we see that

$\displaystyle\mathsf{S}_{i,3}=-\kappa_{4}\frac{\sigma_{i}^{2}}{p}\left(\frac{m^{2}}{\left(1+\phi^{-1/2}m\sigma_{i}\right)^{2}}\right)+\mathrm{O}_{\prec}\left(p^{-1}\Psi(z)\right).$
(7.31)

Inserting (7.28) (7.30) (7.31) back into (7.27) and use
$z\left(1+\phi^{-1/2}m\sigma_{i}\right)\asymp\sqrt{\phi}$ again, we get

$\displaystyle\mathbb{E}\operatorname{Tr}({R_{1}})-\sum_{i=1}^{p}\left(-\frac{1}{z\left(1+\phi^{-1/2}m\sigma_{i}\right)}\right)$
(7.32) $\displaystyle=$
$\displaystyle\frac{1}{z\left(1+\phi^{-1/2}m\sigma_{i}\right)}\left[\frac{\sigma_{i}\phi^{1/2}}{1+\phi^{-1/2}m\sigma_{i}}\mathbb{E}\left[p^{-1}\operatorname{Tr}({R_{1}})-m_{p}\right]-{\frac{1}{p}\frac{\sigma_{i}^{2}m^{\prime}}{\left(1+\phi^{-1/2}\sigma_{i}m\right)^{2}}}-\kappa_{4}\frac{1}{p}\frac{\sigma_{i}^{2}m^{2}}{\left(1+\phi^{-1/2}m\sigma_{i}\right)^{2}}\right]$
$\displaystyle+\mathrm{O}_{\prec}\left(n^{-\frac{1}{2}}p^{-\frac{1}{2}}\eta^{-2}+n^{\frac{1}{4}}p^{-\frac{3}{4}}\Psi(z)+\phi^{-1/2}\Psi(z)\right),$

Using Lemma 5.2, we can rewrite the above equation as follows

$\displaystyle\left(1-\frac{1}{p}\sum_{i=1}^{p}\frac{\phi^{1/2}\sigma_{i}}{z\left(1+\phi^{-1/2}m(z)\sigma_{i}\right)^{2}}\right)\mathbb{E}\left(\operatorname{Tr}({R_{1}})-pm_{p}\right)$
(7.33) $\displaystyle=$
$\displaystyle{-\frac{1}{p}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}m^{\prime}}{z\left(1+\phi^{-1/2}m\sigma_{i}\right)^{3}}}-\kappa_{4}\frac{1}{p}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}m^{2}}{z\left(1+\phi^{-1/2}m\sigma_{i}\right)^{3}}+\mathrm{O}_{\prec}\left(n^{-\frac{1}{2}}p^{-\frac{1}{2}}\eta^{-2}+n^{\frac{1}{4}}p^{-\frac{3}{4}}\Psi(z)+\phi^{-1/2}\Psi(z)\right).$

Combining (5.7) and (5.9) with a straightforward calculation, we obtain that

$\mathbb{E}(\operatorname{Tr}({R_{1}})-pm_{p})=b(z)+\mathrm{O}_{\prec}\left(\frac{n^{-1}\eta^{-2}+n^{-1/4}p^{-1/4}\Psi(z)+\Psi(z)}{\sqrt{\kappa+\eta}}\right),$
(7.34)

where we used the control $zm^{\prime}/m\asymp\phi^{1/2}/\sqrt{\kappa+\eta}.$
Recall (7.16) and let $\Omega$ be defined exactly in the same way as
$\Omega_{2}^{0}$ for $z.$ Combining (7.22), (7.23) and (7.34), by a discussion
similar to (7.6), we readily obtain that

$\mathcal{M}_{\eta_{0},E}=\frac{1}{\pi}\int_{\Omega}\frac{\partial}{\partial\bar{z}}\tilde{f}_{i}(z)b(z)\mathrm{d}^{2}z+\mathrm{O}_{\prec}\left(\frac{1}{\sqrt{n\eta_{0}\sqrt{\kappa+\eta_{0}}}}\right),$

where we used Lemma 5.1. By Lemma 5.9, we have that

$\mathcal{M}_{\eta_{0},E}=\frac{1}{2\pi\mathrm{i}}\int_{\partial\Omega}\tilde{f}_{i}(z)b(z)\mathrm{d}z+\mathrm{O}_{\prec}\left(\frac{1}{\sqrt{n\eta_{0}\sqrt{\kappa+\eta_{0}}}}\right).$

Finally, we can follow lines of the arguments between (7.17) and (7.1) to
conclude the proof of ((1).) and (3.15).

Distribution of $\mathcal{Z}_{\eta_{0},E}(f_{i})$. Using Lemma 7.1, by a
discussion similar to the proof of Theorem 6.1, we can use Lemma 5.11 to
conclude that $\\{\mathcal{Z}_{\eta_{0},E}(f_{i})\\}$ follows a multivariate
Gaussian distribution asymptotically whose mean is $\mathbf{0}$ and covariance
matrices satisfy (3.8) for $\eta_{0}\asymp 1$ and (3.16) for
$\eta_{0}=\mathrm{o}(1).$ This completes the proof of Theorems 3.2 and 3.7
using (3.17).

##  8 Proof of the corollaries

###  8.1 Proof of Corollary 3.5

In this subsection, we prove Corollary 3.5 using Theorem 3.2. Note that when
$\Sigma=I,$ (2.4) becomes

$z=-\frac{1}{m(z)}+\frac{\phi^{1/2}}{1+\phi^{-1/2}m(z)},$ (8.1)

or equivalently,

$m(z):=\frac{\phi^{1/2}-\phi^{-1/2}-z+\mathrm{i}\sqrt{\left(z-\gamma_{-}\right)\left(\gamma_{+}-z\right)}}{2\phi^{-1/2}z}.$
(8.2)

Proof of Part (1). Following [62, Section 3], for some $a,r$ and $\xi,$ we
consider a change of variable that

$m=-\frac{1}{a+r\xi},\ |\xi|=1.$ (8.3)

Together with (8.1), we see that

$\frac{\phi^{1/2}}{1+\phi^{-1/2}m}=\frac{1}{1+\frac{-\phi^{-1/2}}{a+r\xi}}=\frac{\phi^{1/2}(a+r\xi)}{-\phi^{-1/2}+a+r\xi}.$

Set $a=\phi^{-1/2}$, we further have

$\frac{\phi^{1/2}}{1+\phi^{-1/2}m}=r^{-1}\bar{\xi}+\phi^{1/2}.$

This yields that

$z=-\frac{1}{m}+\frac{\phi^{1/2}}{1+\phi^{-1/2}m}=\phi^{1/2}+\phi^{-1/2}+r\xi+r^{-1}\bar{\xi},\
\mathrm{d}z=(r-r^{-1}\xi^{-2})\mathrm{d}\xi,$ (8.4)

and

$\begin{aligned}
m^{\prime}(z)=-\frac{m^{2}+\phi^{1/2}m}{2zm+z\phi^{1/2}+1-\phi}=\frac{r^{2}\xi^{2}}{(\phi^{-1/2}+r\xi)^{2}(r^{2}\xi^{2}-1)}\end{aligned}.$

Together with the chain rule that
$m^{\prime\prime}(z)\mathrm{d}z=\left(\frac{\partial
m^{\prime}}{\partial\xi}\right)\mathrm{d}\xi,$ it is easy to see that

$\frac{m^{\prime\prime}(z)}{2m^{\prime}(z)}\mathrm{d}z=\left[\frac{2}{\xi}-\frac{2r}{\phi^{-1/2}+r\xi}-\frac{2r^{2}\xi}{r^{2}\xi^{2}-1}\right]\mathrm{d}\xi.$
(8.5)

We first with the mean part. According to Remark 3.3 and (3.4), we decompose
that

$\mathsf{M}(f)\equiv\mathbb{E}\mathscr{G}:=\mathsf{M}_{1}(f)+\mathsf{M}_{2}(f),$
(8.6)

where $\mathsf{M}_{k},k=1,2,$ correspond to the $b_{k}$ part in (3.4). On the
one hand, using (8.4), (8.5) and Remark 3.3 with calculations similar to [62,
Section 3.2], we see that

$\displaystyle\mathsf{M}_{1}(f)=$
$\displaystyle\frac{-1}{2\pi\mathrm{i}}\oint_{\Gamma}f(z)\left(\frac{m^{\prime\prime}(z)}{2m^{\prime}(z)}-\frac{m^{\prime}(z)}{m(z)}\right)\mathrm{d}z$
$\displaystyle=$
$\displaystyle\frac{-1}{2\pi\mathrm{i}}\oint_{|\xi|=1}f({\phi^{1/2}+\phi^{-1/2}}+r\xi+\frac{1}{r\xi})\left(\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right)\mathrm{d}\xi$
$\displaystyle=$ $\displaystyle\lim_{r\downarrow
1}\frac{-1}{2\pi\mathrm{i}}\oint_{|\xi|=1}f({\phi^{1/2}+\phi^{-1/2}}+\xi+\frac{1}{\xi})\left(\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right)\mathrm{d}\xi.$
(8.7)

On the other hand, for $\mathsf{M}_{2}(f),$ with a discussion similar to
(8.5), we have that

$\displaystyle\left[\frac{m^{2}(z)m^{\prime\prime}(z)}{2{m^{\prime}(z)}^{2}}-m(z)\right]\mathrm{d}z=$
$\displaystyle\left[\frac{1}{\xi}-\frac{1}{r^{2}\xi^{3}}-\frac{1}{\frac{\phi^{-1/2}}{r}+\xi}-\frac{1}{\xi}+\frac{1}{r^{2}\xi^{2}}\frac{1}{\xi+\frac{\phi^{-1/2}}{r}}+\frac{1}{\xi+\frac{\phi^{-1/2}}{r}}-\frac{1}{r^{2}\xi^{2}}\frac{1}{\xi+\frac{\phi^{-1/2}}{r}}\right]\mathrm{d}\xi$
$\displaystyle=$ $\displaystyle-\frac{1}{r^{2}\xi^{3}}\mathrm{d}\xi.$

Together with a discussion similar to the $\mathsf{M}_{1}(f)$ part, we see
that

$\displaystyle\mathsf{M}_{2}(f)$
$\displaystyle=\kappa_{4}\frac{-1}{2\pi\mathrm{i}}\oint_{|\xi|=1}f({\phi^{1/2}+\phi^{-1/2}}+r\xi+\frac{1}{r\xi})\left(-\frac{1}{r^{2}\xi^{3}}\right)\mathrm{d}\xi$
$\displaystyle=\kappa_{4}\frac{-1}{2\pi\mathrm{i}}\oint_{|\xi|=1}f({\phi^{1/2}+\phi^{-1/2}}+\xi+\frac{1}{\xi})\left(-\frac{1}{\xi^{3}}\right)\mathrm{d}\xi.$
(8.8)

We then work with the variance part. Similar to (8.6), we also decompose that

$\mathsf{V}(f)\equiv\operatorname{Var}(\mathscr{G}):=\mathsf{V}_{1}(f)+\mathsf{V}_{2}(f).$
(8.9)

We again follow the ideas of [62, Section 3] and consider two non-overlapping
contours defined by
$z_{j}=\phi^{1/2}+\phi^{-1/2}+r_{j}\xi_{j}+r_{j}^{-1}\bar{\xi}_{j},j=1,2,$
with $r_{2}>r_{1}>1$. According to Remark 3.3 and (8.3) that
$m^{\prime}(z_{j})\mathrm{d}z_{j}=\left(\frac{\partial}{\partial\xi_{j}}m(z_{j})\right)\mathrm{d}\xi_{j}=\frac{r_{j}}{(\phi^{-1/2}+r_{j}\xi_{j})^{2}}\mathrm{d}\xi_{j},$
by a discussion similar to (8.7), we have that

$\displaystyle\mathsf{V}_{1}(f)=$
$\displaystyle-\frac{1}{2\pi^{2}}\oint_{|\xi_{1}|=1}\oint_{|\xi_{2}|=1}\frac{f(z_{1})f(z_{2})}{(m(z_{1})-m(z_{2}))^{2}}\frac{r_{1}}{(\phi^{-1/2}+r_{1}\xi_{1})^{2}}\frac{r_{2}}{(\phi^{-1/2}+r_{2}\xi_{2})^{2}}\mathrm{d}\xi_{1}\mathrm{d}\xi_{2}$
$\displaystyle=$
$\displaystyle-\frac{1}{2\pi^{2}}\oint_{|\xi_{1}|=1}\oint_{|\xi_{2}|=1}\frac{f(z_{1})f(z_{2})}{(r_{1}\xi_{1}-r_{2}\xi_{2})^{2}}r_{1}r_{2}\mathrm{d}\xi_{1}\mathrm{d}\xi_{2}$
(8.10) $\displaystyle=$
$\displaystyle-\frac{1}{2\pi^{2}}\oint_{|\xi_{1}|=1}\oint_{|\xi_{2}|=1}\frac{\frac{r_{2}}{r_{1}}f(\phi^{1/2}+\phi^{-1/2}+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1})f(\phi^{1/2}+\phi^{-1/2}+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}\mathrm{d}\xi_{2}$
$\displaystyle=$
$\displaystyle-\frac{1}{2\pi^{2}}\lim_{\begin{subarray}{c}r_{2}>r_{1}\\\
r_{1},r_{2}\downarrow
1\end{subarray}}\oint_{|\xi_{1}|=1}\oint_{|\xi_{2}|=1}\frac{f(\phi^{1/2}+\phi^{-1/2}+\xi_{1}+\xi_{1}^{-1})f(\phi^{1/2}+\phi^{-1/2}+\xi_{2}+\xi_{2}^{-1})}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}\mathrm{d}\xi_{2}.$

Similarly, using Remark 3.3 and the fact

$\frac{\partial}{\partial
z_{j}}\left(\frac{1}{1+\phi^{-1/2}m(z_{j})}\right)\mathrm{d}z_{j}=\frac{\partial}{\partial\xi_{j}}\left(\frac{1}{1+\phi^{-1/2}m(z_{j})}(\xi_{j})\right)\mathrm{d}\xi_{j}=-\frac{\phi^{-1/2}r_{j}^{-1}}{\xi_{j}^{2}}\mathrm{d}\xi_{j},\quad
j=1,2,$

we readily obtain that

$\displaystyle\mathsf{V}_{2}(f)=$
$\displaystyle-\frac{\kappa_{4}}{4\pi^{2}}\oint_{|\xi_{1}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1})\frac{r_{1}^{-1}}{\xi_{1}^{2}}\mathrm{d}\xi_{1}\oint_{|\xi_{2}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})\frac{r_{2}^{-1}}{\xi_{2}^{2}}\mathrm{d}\xi_{2}$
$\displaystyle=$
$\displaystyle-\frac{\kappa_{4}}{4\pi^{2}}\left(\oint_{|\xi|=1}f(\phi^{1/2}+\phi^{-1/2}+\xi+\xi^{-1})\frac{1}{\xi^{2}}\mathrm{d}\xi\right)^{2}.$
(8.11)

This completes the proof of part (1).

Proof of Part (2). Armed with the results from Part (1), we now proceed to the
calculations for the concrete examples. First, for
$f_{1}(x)=x-(\phi^{1/2}+\phi^{-1/2}-c)$ and its mean part, using (8.7) and
(8.1) with straightforward calculations and Cauchy’s integral theorem, we see
that

$\displaystyle\mathsf{M}_{1}(f_{1})$
$\displaystyle=\frac{-1}{2\pi\mathrm{i}}\oint_{|\xi|=1}(c+r\xi+\frac{1}{r\xi})\left(\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right)\mathrm{d}\xi=0,$
(8.12) $\displaystyle\mathsf{M}_{2}(f_{1})$
$\displaystyle=\frac{-1}{2\pi\mathrm{i}}\kappa_{4}\oint_{|\xi|=1}(c+r\xi+\frac{1}{r\xi})\left(-\frac{1}{r^{2}\xi^{3}}\right)\mathrm{d}\xi=0.$

For its variance part, according to (8.10), we see that

$\displaystyle\mathsf{V}_{1}(f_{1})=-\frac{1}{2\pi^{2}}\oint_{|\xi_{2}|=1}(c+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})\cdot\oint_{|\xi_{1}|=1}\frac{c+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1}}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}\mathrm{d}\xi_{2}.$

Moreover, it is easy to see that

$\displaystyle\frac{1}{2\pi\mathrm{i}}\oint_{|\xi_{1}|=1}\frac{c+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1}}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}$
$\displaystyle=$
$\displaystyle\frac{1}{2\pi\mathrm{i}}\oint_{|\xi_{1}|=1}\frac{c\xi_{1}+{r_{1}}\xi_{1}^{2}+\frac{1}{{r_{1}}}}{\xi_{1}(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}=$
$\displaystyle\frac{1}{2\pi\mathrm{i}}\left[\oint_{|\xi_{1}|=1}\frac{c}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}+\oint_{|\xi_{1}|=1}\frac{{r_{1}}\xi_{1}}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}+\oint_{|\xi_{1}|=1}\frac{\frac{1}{{r_{1}}}}{\xi_{1}(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{{r_{1}}(\frac{r_{2}}{r_{1}})^{2}\xi_{2}^{2}}.$

This yields that

$\displaystyle\mathsf{V}_{1}(f_{1})=$
$\displaystyle\frac{2\frac{r_{2}}{r_{1}}}{2\pi\mathrm{i}(\frac{r_{2}}{r_{1}})^{2}r_{1}}\oint_{|\xi_{2}|=1}\frac{c\xi_{2}+{r_{2}}\xi_{2}^{2}+\frac{1}{{r_{2}}}}{\xi_{2}^{3}}\mathrm{d}\xi_{2}={2}.$
(8.13)

Similarly, using (8.1), we can obtain that

$\displaystyle\mathsf{V}_{2}(f_{1})$ $\displaystyle=$
$\displaystyle-\frac{\kappa_{4}}{4\pi^{2}}\oint_{|\xi_{1}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1})\frac{r_{1}^{-1}}{\xi_{1}^{2}}\mathrm{d}\xi_{1}\oint_{|\xi_{2}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})\frac{r_{2}^{-1}}{\xi_{2}^{2}}\mathrm{d}\xi_{2}$
$\displaystyle=$ $\displaystyle\kappa_{4}.$

This concludes the proof for $f_{1}(x).$

Second, for $f_{2}(x)=[x-(\phi^{1/2}+\phi^{-1/2}-c)]^{2},$ by a discussion
similar to (8.12) using residual theorem, we see that

$\mathsf{M}_{1}(f_{2})=1,\ \mathsf{M}_{2}(f_{2})=\kappa_{4}.$

For its variance part, using an argument similar to (8.13), it is not hard to
see that

$\displaystyle\mathsf{V}_{1}(f_{2})$
$\displaystyle=-\frac{\frac{r_{2}}{r_{1}}}{2\pi^{2}}\oint_{|\xi_{2}|=1}((c+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})^{2})\cdot\oint_{|\xi_{1}|=1}\frac{(c+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1})^{2}}{(\xi_{1}-\frac{r_{2}}{r_{1}}\xi_{2})^{2}}\mathrm{d}\xi_{1}\mathrm{d}\xi_{2}$
$\displaystyle=2\frac{r_{2}}{r_{1}}\left[\frac{2c}{r_{1}(\frac{r_{2}}{r_{1}})^{2}}{2cr_{2}}+\frac{2}{r_{1}^{2}(\frac{r_{2}}{r_{1}})^{3}}{r_{2}^{2}}\right]=8c^{2}+4.$

Analogously, we have that

$\displaystyle\mathsf{V}_{2}(f_{2})=$
$\displaystyle-\frac{\kappa_{4}}{4\pi^{2}}\oint_{|\xi_{1}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1})\frac{r_{1}^{-1}}{\xi_{1}^{2}}\mathrm{d}\xi_{1}\oint_{|\xi_{2}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})\frac{r_{2}^{-1}}{\xi_{2}^{2}}\mathrm{d}\xi_{1}$
$\displaystyle=$ $\displaystyle{4c^{2}\kappa_{4}}.$

This concludes the proof for $f_{2}(x).$

Finally, for $f_{3}(x)=\log(x-\phi^{1/2}-\phi^{-1/2}+t+t^{-1}),$ for the mean
part, we first decompose that

$\displaystyle\mathsf{M}_{1}(f_{3})=$
$\displaystyle-\frac{1}{2\pi\mathrm{i}}\oint_{|\xi|=1}\log\left((t+{r}\xi)(1+\frac{1}{{r}t\xi})\right)\left[\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right]\mathrm{d}\xi$
$\displaystyle=$
$\displaystyle-\frac{1}{2\pi\mathrm{i}}\oint_{|\xi|=1}\left[\log(t+{r}\xi)+\log(1+\frac{1}{{r}t\xi})\right]\left[\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right]\mathrm{d}\xi$
$\displaystyle:=$
$\displaystyle\mathsf{M}_{11}(f_{3})+\mathsf{M}_{12}(f_{3}),$

where

$\displaystyle\mathsf{M}_{11}(f_{3})=$
$\displaystyle-\frac{1}{2\pi\mathrm{i}}\oint_{|\xi|=1}\log(t+{r}\xi)\left[\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right]\mathrm{d}\xi$
$\displaystyle=$
$\displaystyle-\left[\log(t)-\frac{1}{2}\log(t-1)-\frac{1}{2}\log(t+1)\right]=\frac{1}{2}\log(1-\frac{1}{t^{2}}),$

and

$\displaystyle\mathsf{M}_{12}(f_{3})=$
$\displaystyle-\frac{1}{2\pi\mathrm{i}}\oint_{|\xi|=1}\log(1+\frac{1}{{r}t\xi})\left[\frac{1}{\xi}-\frac{1}{2}\frac{1}{\xi+\frac{1}{r}}-\frac{1}{2}\frac{1}{\xi-\frac{1}{r}}\right]\mathrm{d}\xi$
$\displaystyle=$
$\displaystyle-\frac{1}{2\pi\mathrm{i}}\oint_{|z|=1}\log(1+\frac{z}{{r}t})\left[z-\frac{1}{2}\frac{1}{\frac{1}{z}+\frac{1}{r}}-\frac{1}{2}\frac{1}{\frac{1}{z}-\frac{1}{r}}\right]\frac{1}{z^{2}}\mathrm{d}z=0.$

This results in $\mathsf{M}_{1}(f_{3})=\frac{1}{2}\log(1-\frac{1}{t^{2}}).$
Similarly, one can show that

$\displaystyle\mathsf{M}_{2}(f_{3})=-\frac{\kappa_{4}}{2\pi\mathrm{i}}\oint_{|\xi|=1}\log(t+{r}\xi)\left[-\frac{1}{r^{2}\xi^{3}}\right]\mathrm{d}\xi-\frac{1}{2\pi\mathrm{i}}\oint_{|\xi|=1}\log(1+\frac{1}{{r}t\xi})\left[-\frac{1}{r^{2}\xi^{3}}\right]\mathrm{d}\xi=-\frac{\kappa_{4}}{2t^{2}}.$

For its variance part, using a straightforward calculation with Cauchy’s
differentiation formula and residual theorem, we have that

$\displaystyle\mathsf{V}_{1}(f_{3})$
$\displaystyle=2\left[\frac{\frac{r_{2}}{r_{1}}}{2\pi\mathrm{i}(\frac{r_{2}}{r_{1}})^{2}}\oint_{\left|\xi_{2}\right|=1}\frac{\log\left(t+{r_{2}}\xi_{2}\right)}{\xi_{2}\left(\frac{1}{\frac{r_{2}}{r_{1}}}+{r_{1}}t\xi_{2}\right)}\mathrm{d}\xi_{2}+\frac{\frac{r_{2}}{r_{1}}}{2\pi\mathrm{i}\frac{r_{2}}{r_{1}}^{2}}\oint_{\left|\xi_{2}\right|=1}\frac{\log\left(1+\xi_{2}^{-1}/({r_{2}}t)\right)}{\xi_{2}\left(\frac{1}{\frac{r_{2}}{r_{1}}}+{r_{1}}t\xi_{2}\right)}\mathrm{d}\xi_{2}\right]$
$\displaystyle=2\left(\log t-\log(t-t^{-1})\right).$

Similarly, we have that

$\displaystyle\mathsf{V}_{2}(f_{3})=$
$\displaystyle-\frac{\kappa_{4}}{4\pi^{2}}\oint_{|\xi_{1}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{1}}\xi_{1}+\frac{1}{{r_{1}}}\xi_{1}^{-1})\frac{r_{1}^{-1}}{\xi_{1}^{2}}\mathrm{d}\xi_{1}\oint_{|\xi_{2}|=1}f(\phi^{1/2}+\phi^{-1/2}+{r_{2}}\xi_{2}+\frac{1}{{r_{2}}}\xi_{2}^{-1})\frac{r_{2}^{-1}}{\xi_{2}^{2}}\mathrm{d}\xi_{1}$
$\displaystyle=$ $\displaystyle\kappa_{4}/t^{2}.$

This completes the proof.

###  8.2 Proof of Corollary 3.9

As in Section 7.1, we focus our discussion on a single function. The multiple
functions setting can be handled in the same ways as in the end of Section
7.1.

Simplification of variance. Similar to (7.18), we rewrite

$\operatorname{Var}(\mathcal{Z}_{\mathrm{l}})=(\mathcal{K}_{2})_{++}+(\mathcal{K}_{2})_{--}-(\mathcal{K}_{2})_{+-}-(\mathcal{K}_{2})_{-+}.$

Here we use $\mathcal{Z}_{\mathrm{l}}\equiv\mathcal{Z}_{\eta_{0},E}$ when
$\eta_{0}=\mathrm{o}(1).$ In view of (7.19) and (7.1), we can write

$\displaystyle(\mathcal{K}_{2})_{++}$
$\displaystyle=-\frac{\eta_{0}^{2}}{4\pi^{2}}\int\int{g}\left(\tilde{x}_{1}\right){g}\left(\tilde{x}_{2}\right)\widehat{\beta}\left(E+\tilde{x}_{1}{\eta_{0}}+\mathrm{i}\underline{\eta}_{1},E+\tilde{x}_{2}{\eta_{0}}+\mathrm{i}\underline{\eta}_{2}\right)\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}$
$\displaystyle=-\frac{{\eta_{0}}^{2}}{2\pi^{2}}\iint_{1\leq\tilde{y}_{1,2}\leq
2}g\left(\tilde{x}_{1}\right)g\left(\tilde{x}_{2}\right)\left[\frac{m^{\prime}\left(\left(E+\tilde{x}_{1}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{1}\right)m^{\prime}\left(\left(E+\tilde{x}_{2}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{2}\right)}{\left(m\left(\left(E+\tilde{x}_{1}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{1}\right)-m\left(\left(E+\tilde{x}_{2}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{2}\right)\right)^{2}}\right.$
$\displaystyle\left.-\frac{1}{\left(\left(\tilde{x}_{1}-\tilde{x}_{2}\right){\eta_{0}}+\mathrm{i}\frac{1}{2}{\eta_{0}}n^{-\varepsilon_{1}}\right)^{2}}\right]\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}$
$\displaystyle:=\left(\mathcal{K}_{2,1}\right)_{++}+\left(\mathcal{K}_{2,2}\right)_{++},$
(8.14)

where we used the definition of $\widehat{\beta}$ as in (3.6).

We first reformulate the case in the bulk
$E\in\left(\gamma_{-}+\tau^{\prime},\gamma_{+}-\tau^{\prime}\right)$ for some
fixed small constant $\tau^{\prime}>0$. Note that $m$ is holomorphic in
$\mathbb{C}\backslash[\gamma_{-},\gamma_{+}]$. Since we are dealing with
$z_{1},z_{2}$ in the same upper half plane, we can use the taylor expansion at
$z_{1}$ that

$\displaystyle\frac{m^{\prime}(z_{1})m^{\prime}(z_{2})}{(m(z_{1})-m(z_{2}))^{2}}-\frac{1}{(z_{1}-z_{2})^{2}}=$
$\displaystyle\frac{m^{\prime}(z_{1})(m^{\prime}(z_{1})+(z_{2}-z_{1})m^{\prime\prime}(z_{1})+\mathrm{O}(|z_{1}-z_{2}|^{2}))}{((z_{2}-z_{1})m^{\prime}(z_{1})+\mathrm{O}(|z_{1}-z_{2}|^{2}))^{2}}-\frac{1}{(z_{1}-z_{2})^{2}}$
(8.15) $\displaystyle=$
$\displaystyle\frac{m^{\prime\prime}(z_{1})}{(z_{2}-z_{1})m^{\prime}(z_{1})+\mathrm{O}(|z_{1}-z_{2}|^{2})}+\mathrm{O}(1).$

Similarly we have

$\displaystyle\frac{m^{\prime}(z_{1})m^{\prime}(z_{2})}{(m(z_{1})-m(z_{2}))^{2}}-\frac{1}{(z_{1}-z_{2})^{2}}=$
$\displaystyle\frac{m^{\prime}(z_{2})(m^{\prime}(z_{2})+(z_{1}-z_{2})m^{\prime\prime}(z_{2})+\mathrm{O}(|z_{1}-z_{2}|^{2}))}{((z_{1}-z_{2})m^{\prime}(z_{2})+\mathrm{O}(|z_{1}-z_{2}|^{2}))^{2}}-\frac{1}{(z_{1}-z_{2})^{2}}$
$\displaystyle=$
$\displaystyle\frac{m^{\prime\prime}(z_{2})}{(z_{1}-z_{2})m^{\prime}(z_{2})+\mathrm{O}(|z_{1}-z_{2}|^{2})}+\mathrm{O}(1).$

This implies that

$\widehat{\beta}(z_{1},z_{2})\sim\frac{1}{z_{2}-z_{1}}\left(\frac{m^{\prime\prime}(z_{1})}{m^{\prime}(z_{1})}-\frac{m^{\prime\prime}(z_{2})}{m^{\prime}(z_{2})}\right).$
(8.16)

Using the above control, we can conclude that

${\eta_{0}}^{2}\left[\frac{m^{\prime}\left(\left(E+\tilde{x}_{1}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{1}\right)m^{\prime}\left(\left(E+\tilde{x}_{2}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{2}\right)}{\left(m\left(\left(E+\tilde{x}_{1}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{1}\right)-m\left(\left(E+\tilde{x}_{2}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{2}\right)\right)^{2}}-\frac{1}{\left(\left(\tilde{x}_{1}-\tilde{x}_{2}\right){\eta_{0}}+\mathrm{i}\frac{1}{2}{\eta_{0}}n^{-\varepsilon_{1}}\right)^{2}}\right]=\mathrm{O}({\eta_{0}}^{2}).$

This shows that $(\mathcal{K}_{2})_{++}$ is negligible and similar discussions
and results apply to $(\mathcal{K}_{2})_{--}.$

When $z_{1}$ and $z_{2}$ are in different half planes, we have that

$\left|m^{\prime}\left(\left(E+\tilde{x}_{1}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{1}\right)\right|,\left|m^{\prime}\left(\left(E+\tilde{x}_{2}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{2}\right)\right|\lesssim\frac{1}{\sqrt{\tau^{\prime}}},$

and

$\left|\left(m\left(\left(E+\tilde{x}_{1}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{1}\right)-m\left(\left(E+\tilde{x}_{2}{\eta_{0}}\right)+\mathrm{i}\underline{\eta}_{2}\right)\right)\right|\sim\rho(E+\tilde{x}_{1}\eta_{0})+\rho(E+\tilde{x}_{2}\eta_{0})\sim
1.$

We can therefore conclude that
$(\mathcal{K}_{2,1})_{+-}=\mathrm{O}(\eta_{0}^{2}/\tau^{\prime})$ and
$(\mathcal{K}_{2,1})_{-+}=\mathrm{O}(\eta_{0}^{2}/\tau^{\prime}).$ For the
remaining term $(\mathcal{K}_{2,2})_{+-}+(\mathcal{K}_{2,2})_{-+}$, we have

$-\frac{1}{\pi^{2}}\iint
g\left(\tilde{x}_{1}\right)g\left(\tilde{x}_{2}\right)\frac{1}{\left(\tilde{x}_{1}-\tilde{x}_{2}\pm\mathrm{i}\frac{3}{2}\eta_{0}n^{-\varepsilon_{1}}\right)^{2}}\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}.$

Note that we can add additional
$\frac{1}{2\pi^{2}}\iint\frac{g\left(\tilde{x}_{1}\right)^{2}}{\left(\tilde{x}_{1}-\tilde{x}_{2}\pm\mathrm{i}\frac{3}{2}n^{-\varepsilon_{1}}\right)^{2}}\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}$
and
$\frac{1}{2\pi^{2}}\iint\frac{g\left(\tilde{x}_{2}\right)^{2}}{\left(\tilde{x}_{1}-\tilde{x}_{2}\pm\mathrm{i}\frac{3}{2}n^{-\varepsilon_{1}}\right)^{2}}\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}$
to the above terms since they are negligible. Then we can conclude the proof
using dominated convergence theorem.

Then we reformulate the case at the edge. Due to similarity, we focus on the
right edge $\gamma_{+}.$ Note that near the edge, for some constants
$c_{1},c_{2}$ and $c_{3}=c_{1}/2,$ we have the following approximation

$m(z)=m(\gamma_{+})+c_{1}\sqrt{z-\gamma_{+}}+\mathrm{O}\left(|z-\gamma_{+}|\right),\
m^{\prime}(z)=c_{2}+\frac{c_{3}}{\sqrt{z-\gamma_{+}}}+\mathrm{O}(\sqrt{|z-\gamma_{+}|}).$
(8.17)

For the terms in (8.2) and counterparts in
$(\mathcal{K}_{2})_{--},(\mathcal{K}_{2})_{-+}$ and $(\mathcal{K}_{2})_{+-}$,
it is not hard to see that

$(\mathcal{K}_{2,2})_{++}+(\mathcal{K}_{2,2})_{--}+(\mathcal{K}_{2,2})_{+-}+(\mathcal{K}_{2,2})_{-+}=0.$

Then it suffices to control
$(\mathcal{K}_{2,1})_{++},(\mathcal{K}_{2,1})_{--},(\mathcal{K}_{2,1})_{-+}$
and $(\mathcal{K}_{2,1})_{+-}.$ Note that

$\displaystyle(\mathcal{K}_{2,1})_{++}=-\frac{1}{8\pi^{2}}\iint\frac{g(\tilde{x_{1}})g(\tilde{x}_{2})\left(\frac{1}{\sqrt{\tilde{x}_{1}+\mathrm{i}n^{-\varepsilon_{1}}}}+\mathrm{O}(\sqrt{{\eta_{0}}})\right)\left(\frac{1}{\sqrt{\tilde{x}_{2}+\mathrm{i}n^{-\varepsilon_{1}}/2}}+\mathrm{O}(\sqrt{{\eta_{0}}})\right)}{\left(\sqrt{\tilde{x}_{1}+\mathrm{i}n^{-\varepsilon_{1}}}-\sqrt{\tilde{x}_{2}+\mathrm{i}n^{-\varepsilon_{1}}/2}+\mathrm{O}(\sqrt{{\eta_{0}}})\right)^{2}}\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}$
$\displaystyle=-\frac{1}{8\pi^{2}}\iint\frac{g(\tilde{x_{1}})g(\tilde{x}_{2})}{\sqrt{\tilde{x}_{1}+\mathrm{i}n^{-\varepsilon_{1}}}\sqrt{\tilde{x}_{2}+\mathrm{i}n^{-\varepsilon_{1}}/2}\left(\sqrt{\tilde{x}_{1}+\mathrm{i}n^{-\varepsilon_{1}}}-\sqrt{\tilde{x}_{2}+\mathrm{i}n^{-\varepsilon_{1}}}\right)^{2}}\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}+\mathrm{O}(\sqrt{{\eta_{0}}}n^{3\varepsilon_{1}}).$

Similar to the discussion in the bulk, we could add vanishing
$g(\tilde{x}_{1})^{2}$ and $g(\tilde{x}_{1})^{2}$ terms by consider the
corresponding complex integral and use dominated convergence theorem to show
that

$\displaystyle(\mathcal{K}_{2,1})_{++}=$
$\displaystyle\frac{1}{16\pi^{2}}\iint\frac{(g(\tilde{x}_{1})-g(\tilde{x}_{2}))^{2}}{\sqrt{\tilde{x_{1}}+\mathrm{i}0}\sqrt{\tilde{x}_{2}+\mathrm{i}0}(\sqrt{\tilde{x_{1}}+\mathrm{i}0}-\sqrt{\tilde{x_{2}}+\mathrm{i}0})^{2}}\mathrm{d}\tilde{x}_{1}\mathrm{d}\tilde{x}_{2}+\mathrm{o}(1)$
(8.18) $\displaystyle=$
$\displaystyle\frac{1}{4\pi^{2}}\int_{\psi(\mathbb{R}+\mathrm{i}0)}\int_{\psi(\mathbb{R}+\mathrm{i}0)}\frac{\left(g\left(w_{1}^{2}\right)-g\left(w_{2}^{2}\right)\right)^{2}}{\left(w_{1}-w_{2}\right)^{2}}\mathrm{~{}d}w_{1}\mathrm{~{}d}w_{2}+\mathrm{o}(1),$

where we used the change the variable $\psi(z):=\sqrt{z}$ with branch cut such
that $\psi:\mathbb{C}^{+}\rightarrow\mathbb{C}^{+}$. Similarly, we also have

$\displaystyle(\mathcal{K}_{2,1})_{--}$
$\displaystyle=\frac{1}{4\pi^{2}}\int_{\psi(\mathbb{R}-\mathrm{i}0)}\int_{\psi(\mathbb{R}-\mathrm{i}0)}\frac{\left(g\left(w_{1}^{2}\right)-g\left(w_{2}^{2}\right)\right)^{2}}{\left(w_{1}-w_{2}\right)^{2}}\mathrm{d}w_{1}\mathrm{d}w_{2}+\mathrm{o}(1),$
$\displaystyle(\mathcal{K}_{2,1})_{+-}$
$\displaystyle=\frac{1}{4\pi^{2}}\int_{\psi(\mathbb{R}+\mathrm{i}0)}\int_{\psi(\mathbb{R}-\mathrm{i}0)}\frac{\left(g\left(w_{1}^{2}\right)-g\left(w_{2}^{2}\right)\right)^{2}}{\left(w_{1}-w_{2}\right)^{2}}\mathrm{d}w_{1}\mathrm{d}w_{2}+\mathrm{o}(1),$
$\displaystyle(\mathcal{K}_{2,1})_{-+}$
$\displaystyle=\frac{1}{4\pi^{2}}\int_{\psi(\mathbb{R}-\mathrm{i}0)}\int_{\psi(\mathbb{R}+\mathrm{i}0)}\frac{\left(g\left(w_{1}^{2}\right)-g\left(w_{2}^{2}\right)\right)^{2}}{\left(w_{1}-w_{2}\right)^{2}}\mathrm{d}w_{1}\mathrm{d}w_{2}+\mathrm{o}(1).$

This yields that

$\mathcal{K}_{2,1}=\frac{1}{4\pi^{2}}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty}\frac{\left(g\left(w_{1}^{2}\right)-g\left(w_{2}^{2}\right)\right)^{2}}{\left(w_{1}-w_{2}\right)^{2}}\mathrm{d}w_{1}\mathrm{d}w_{2}+\mathrm{o}(1)=\frac{1}{4\pi^{2}}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{\left(g\left(-w_{1}^{2}\right)-g\left(-w_{2}^{2}\right)\right)^{2}}{\left(w_{1}-w_{2}\right)^{2}}\mathrm{d}w_{1}\mathrm{d}w_{2}+\mathrm{o}(1).$

Simplification of mean. When $\operatorname{Re}z$ is in the bulk, we have that
$|m(z)|\sim 1,\ {m}^{\prime}(z)\sim(\kappa+{\eta})^{-1/2}.$ Using the change
of variable as in (7.20) and a discussion similar to (7.1), we have that

$\frac{1}{2\pi\mathrm{i}}\left(\int_{\mathbb{R}}f(x)b_{1}^{\mathfrak{+}}(x)\mathrm{d}x-\int_{\mathbb{R}}f(x)b_{1}^{\mathfrak{-}}(x)\mathrm{d}x\right)=\frac{\eta_{0}}{2\pi\mathrm{i}}\left(\int_{\mathbb{R}}g(\tilde{x})b_{1}^{\mathfrak{+}}(E+\tilde{x}\eta_{0})\mathrm{d}\tilde{x}-\int_{\mathbb{R}}g(\tilde{x})b_{1}^{\mathfrak{-}}(E+\tilde{x}\eta_{0})\mathrm{d}\tilde{x}\right).$

Furthermore, by definition and Lemma 5.1, we have that
$b_{1}(x)=\mathrm{O}\left(1+n^{\varepsilon_{1}/2}\eta_{0}^{-1/2}\right).$ This
shows that the mean is asymptotically $0$ in the bulk.

Next, at the edge, using (8.17) and

$m^{\prime\prime}(z)=-\frac{c_{3}}{2\left(\sqrt{z-\gamma_{+}}\right)^{3}}+\mathrm{O}\left(\left|z-\gamma_{+}\right|^{-1/2}\right).$

Consequently, we have that

$\left(\frac{m^{\prime\prime}(z)}{2m^{\prime}(z)}-\frac{m^{\prime}(z)}{m(z)}\right)+\kappa_{4}\left(\frac{m^{2}(z)m^{\prime\prime}(z)}{2({m^{\prime}}(z))^{2}}-m(z)\right)=\frac{1}{4(z-\gamma_{+})}+\mathrm{O}\left(\left|z-\gamma_{+}\right|^{-1/2}\right).$

Inserting the above estimate into (3.15), by Lemma 5.10, we have that the mean
is asymptotically $g(0)/4.$

###  8.3 Proof of Corollary 4.2

For the global statistics, (4.2) and (4.4) follow directly from (2) of
Corollary 3.5 and (3.14). For (4.3), we can prove it using Delta method, i.e.,
if $\mathfrak{g}^{\prime}(\mathbf{a})$ exists and
$n^{b}\left(\bm{x}_{n}-\mathbf{a}\right)\Rightarrow\bm{x}$ for some fixed
constant $b>0$, then
$n^{b}\left(\mathfrak{g}\left(\bm{x}_{n}\right)-\mathfrak{g}(\mathbf{a})\right)\Rightarrow[\mathfrak{g}^{\prime}(\mathbf{a})]^{*}\bm{x}.$
In light of (4.2) and the definition of $\mathcal{T}_{4}^{\mathrm{g}},$ we set

$\mathfrak{g}(y_{1},y_{2})=\frac{y_{2}}{y_{1}^{2}},\
\bm{x}_{n}=(\mathcal{T}_{1}^{\mathrm{g}},\mathcal{T}_{2}^{\mathrm{g}})^{\prime}.$

Therefore, in order to apply Delta method, it suffices to find the
distribution of $\bm{x}_{n}$ using Theorem 3.2 under $\Sigma=I.$ Recall (8.9).
Following lines of the arguments in Section 8.1, we have that

$\displaystyle\mathsf{V}_{1}(f_{1},f_{2})=$ $\displaystyle
4c,\quad\mathsf{V}_{2}(f_{1},f_{2})=2c\kappa_{4},$

so that $\mathsf{V}(f_{1},f_{2})=4c+2c\kappa_{4}.$ Then together with Theorem
3.2 and (2) of Corollary 3.5, we see from (4.2) that the distribution of
$\bm{x}_{n}$ follows

$\begin{pmatrix}\mathcal{T}_{1}^{\mathrm{g}}-n(c-\phi^{-1/2})\\\
\mathcal{T}_{2}^{\mathrm{g}}-n(1+c^{2}-2c\phi^{-1/2}+\phi^{-1})\end{pmatrix}\Rightarrow\mathcal{N}_{2}\left(\begin{pmatrix}0\\\
1+\kappa_{4}\end{pmatrix},\begin{pmatrix}2+\kappa_{4}&4c+2c\kappa_{4}\\\
4c+2c\kappa_{4}&4+4c^{2}(\kappa_{4}+2)\end{pmatrix}\right).$

Then the proof follows from Delta method with straightforward calculations.

For the local statistics, (4.5) follows directly from (2) of Corollary 3.9.
The proof of (4.6) follows from the Delta method as for (4.3). This completes
the proof.

###  8.4 Proof of Corollary 4.3

Note that (3.1), the centering depends on $\Sigma$ under $\mathbf{H}_{a}$ in
(4.7) so that the definitions in (3.2) need to be modified accordingly. In the
proof, we focus on a single test function and let $f_{0}$ be defined as in
(3.2) under $\mathbf{H}_{0}$ and $f_{a}$ be that under $\mathbf{H}_{a}.$
Moreover, under $\mathbf{H}_{a},$ we denote

$\mathrm{m}_{\mathfrak{d}}:=\mathbb{E}\left(Z_{\eta_{0},E}(f_{\mathfrak{d}})\right),\
\mathrm{v}_{\mathfrak{d}}=\operatorname{Var}\left(Z_{\eta_{0},E}(f_{\mathfrak{d}})\right),\
\mathfrak{d}=0,\ a.$

We also denote
$\mathrm{m}^{\prime}:=\mathbb{E}\left(Z_{\eta_{0},E}(f_{0})\right)$ under
$\mathbf{H}_{0}.$

###### Proof.

Note that under $\mathbf{H}_{a},$ we have that

$\frac{Z_{\eta_{0},E}(f_{a})-\mathrm{m}_{a}}{\sqrt{\mathrm{v}_{a}}}\Rightarrow\mathcal{N}(0,1).$

Moreover, under the nominal level $\alpha,$ we shall reject $\mathbf{H}_{a}$
if

$\frac{Z_{\eta_{0},E}(f_{0})-\mathrm{m}_{0}}{\sqrt{\mathrm{v}_{0}}}>z_{1-\alpha/2}\
\text{or}\
\frac{Z_{\eta_{0},E}(f_{0})-\mathrm{m}_{0}}{\sqrt{\mathrm{v}_{0}}}<z_{\alpha/2}.$

Due to similarity, we focus on the left tail and the power can be written as

$\mathsf{Po}:=\mathbb{P}\left(\frac{Z_{\eta_{0},E}(f_{0})-\mathrm{m}_{0}}{\sqrt{\mathrm{v}_{0}}}<z_{\alpha/2}|\mathbf{H}_{a}\right).$

After a straightforward calculation, we can rewrite

$\mathsf{Po}=\mathbb{P}\left(\frac{Z_{\eta_{0},E}(f_{a})-\mathrm{m}_{a}}{\sqrt{\mathrm{v}_{a}}}<T_{1}+T_{2}+T_{3}+T_{4}|\mathbf{H}_{a}\right),$

where $T_{i},1\leq i\leq 4$ are defined as

$T_{1}:=z_{\alpha/2}\frac{\sqrt{\mathrm{v}_{0}}}{\sqrt{\mathrm{v}_{a}}},\
T_{2}:=\frac{\mathrm{m}^{\prime}-\mathrm{m}_{0}}{\sqrt{\mathrm{v}_{a}}},\
T_{3}:=-\frac{\sum_{i=1}^{n}(f_{0}(\lambda_{i})-f_{a}(\lambda_{i}))+n(\int
f_{a}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{a})+\mathrm{m}_{a}-\mathrm{m}_{0}}{\sqrt{\mathrm{v}_{a}}},$
$T_{4}:=-\frac{n(\int f_{0}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{0})}{\sqrt{\mathrm{v}_{a}}}.$

It is not hard to see that $T_{i}=\mathrm{O}(1),1\leq i\leq 3.$ Consequently,
to have $\mathsf{Po}\rightarrow 1,$ we require $T_{4}\rightarrow\infty$ for
various test functions. Due to similarity, we only focus on the setting when
$\mathsf{sy}=\mathrm{g}$ and $\ell=1,2.$

When $\ell=1,$ if $(\mathfrak{m}_{1}(\pi_{a})-1)\sqrt{\phi}\gg 1,$ we a=have
that $\int f_{0}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{0}=\int(f_{0}-f_{a})\mathrm{d}\varrho_{a}+(\int
f_{a}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{0})=(\mathfrak{m}_{1}(\pi_{a})-1)\sqrt{\phi}+\mathrm{O}(1).$
This yields that

$T_{4}\asymp n\phi^{1/2}(\mathfrak{m}_{1}(\pi_{a})-1).$

Furthermore, $(\mathfrak{m}_{1}(\pi_{a})-1)\sqrt{\phi}=\mathrm{O}(1),$ we
still have $\int f_{0}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{0}=\int(f_{0}-f_{a})\mathrm{d}\varrho_{a}+(\int
f_{a}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{0})=(\mathfrak{m}_{1}(\pi_{a})-1)\sqrt{\phi}+\mathrm{O}((\mathfrak{m}_{1}(\pi_{a})-1)).$
This verifies (4.9) for $\ell=1$.

When $\ell=2,$ we decompose that $\int f_{0}\mathrm{d}\varrho_{a}-\int
f_{0}\mathrm{d}\varrho_{0}=\int(f_{0}-f_{a})\mathrm{d}\varrho_{a}+(\int
f_{a}\mathrm{d}\varrho_{a}-\int f_{0}\mathrm{d}\varrho_{0}).$ Using

$\mathrm{d}\varrho_{a}(x)=\left(\frac{\sqrt{4\mathfrak{m}_{2}(\pi_{a})^{2}-(x-\mathfrak{m}_{1}(\pi_{a})\sqrt{\phi})^{2}}}{2\pi\mathfrak{m}_{2}(\pi_{a})}+\mathrm{O}(1/\sqrt{\phi})\right)\mathrm{d}x,$

we have that

$\int
x\mathrm{d}\varrho_{a}=\mathfrak{m}_{1}(\pi_{a})\sqrt{\phi}+\mathrm{O}(\mathfrak{m}_{1}(\pi_{a})),$
$\int
x^{2}\mathrm{d}\varrho_{a}=\mathfrak{m}_{2}(\pi_{a})^{2}+2(\mathfrak{m}_{1}(\pi_{a})\sqrt{\phi})^{2}-(\mathfrak{m}_{1}(\pi_{a})\sqrt{\phi})^{2}+\mathrm{O}(\mathfrak{m}_{1}(\pi_{a})^{2}\sqrt{\phi})=\mathfrak{m}_{2}(\pi_{a})^{2}+\mathfrak{m}_{1}(\pi_{a})^{2}\phi+\mathrm{O}(\mathfrak{m}_{1}(\pi_{a})^{2}\sqrt{\phi}),$
$\displaystyle\int(x-\sqrt{\phi}-1/\sqrt{\phi}+c)^{2}\mathrm{d}\varrho_{a}$
$\displaystyle=$
$\displaystyle\mathfrak{m}_{1}(\pi_{a})^{2}\phi-2\sqrt{\phi}\mathfrak{m}_{1}(\pi_{a})\sqrt{\phi}+\phi+(c-1/\sqrt{\phi})^{2}+{\mathfrak{m}_{2}(\pi_{a})^{2}}+\mathrm{O}((\mathfrak{m}_{1}(\pi_{a})-1)^{2}\sqrt{\phi})$
$\displaystyle=$
$\displaystyle(\mathfrak{m}_{1}(\pi_{a})-1)^{2}\phi+(c-1/\sqrt{\phi})^{2}+{\mathfrak{m}_{2}(\pi_{a})^{2}}+2c(\mathfrak{m}_{1}(\pi_{a})-1)\sqrt{\phi}+\mathrm{O}((\mathfrak{m}_{1}(\pi_{a})-1)^{2}\sqrt{\phi}).$

Meanwhile, we have for $\ell=2,$

$\int f_{0}\mathrm{d}\varrho_{0}=1+c^{2}-2c\phi^{-1/2}+\phi^{-1}.$

Combining all the above controls, we have that

$\displaystyle T_{4}\asymp
n\left((\mathfrak{m}_{1}(\pi_{a})-1)^{2}{\phi}+2c(\mathfrak{m}_{1}(\pi_{a})-1)\sqrt{\phi}+(\mathfrak{m}_{2}(\pi_{a})^{2}-1)+\mathrm{O}((\mathfrak{m}_{1}(\pi_{a})-1)^{2}\sqrt{\phi})\right).$

This verifies (4.10) for $\ell=2.$ The other cases can be handled similarly.
This completes our proof.

∎

##  Appendix A Proof of Theorem 5.5

In this section, we prove the main ingredient, Theorem 5.5 following [14, 35]
with necessary modifications. In Section A.1, we summarize the basic tools. In
Section A.2, we prove the entrywise local law when
$\mathbf{u}_{k},\mathbf{v}_{k}$ in (5.14) and (5.15) are standard basis. We
also prove (5.16), (5.17) and (5.18). In Section A.3, armed with the entrywise
local law, we prove (5.14) and (5.15) under the general setting.

###  A.1 Some tools

In this section, we provide some technical preparation. Recall (5.12). We
denote

$G_{1}(z):=z\Sigma^{1/2}R_{1}(z)\Sigma^{1/2},\ G_{2}(z):=R_{2}(z).$ (A.1)

It is not hard to see that by Schur’s complement,

$G=\left(\begin{array}[]{cc}\Sigma XG_{2}X^{*}\Sigma-\Sigma&\Sigma XG_{2}\\\
G_{2}X^{*}\Sigma&G_{2}\end{array}\right)=\left(\begin{array}[]{cc}G_{1}&z^{-1}G_{1}X\\\
z^{-1}X^{*}G_{1}&z^{-2}X^{*}G_{1}X-z^{-1}\end{array}\right).$ (A.2)

Following Definition 4.1 of [35], for matrices of the form $A=(A_{st}:s\in
l(A),t\in r(A)),$ whose entries are indexed by arbitrary finite subsets of
$l(A),r(A)\subset\mathbb{N},$ then the matrix multiplication $AB$ is defined
for $s\in l(A)$ and $t\in r(B)$ by

$(AB)_{st}:=\sum_{q\in l(A)\cap r(B)}A_{sq}B_{qt}.$

Recall (5.10). Let $H\equiv H(z):=G^{-1}(z).$ For $S\subset\mathcal{I}$ we
define the minor $H^{(S)}:=\left(H_{st}:s,t\in\mathcal{I}\backslash S\right).$
Note that $H^{(S)}$ is also an $(p+n)\times(p+n)$ matrix where the undefined
entries are zeros. We also write $G^{(S)}:=\left(H^{(S)}\right)^{-1}.$ The
matrices $G_{1}^{(S)}$ and $G_{2}^{(S)}$ can be defined similarly. Throughout
the paper, we abbreviate $(\\{s\\})\equiv(s)$ and $(\\{s,t\\})\equiv(st)$. The
following identities will be frequently used in our calculations.

######  Lemma A.1.

Recall Definition 5.3, (5.10), (A.1) and the assumption that $\Sigma$ is
diagonal. We have that

  1. (1).

For $\mu\in\mathcal{I}_{2}$ we have

$\frac{1}{G_{\mu\mu}}=-z-\left(X^{*}G^{(\mu)}X\right)_{\mu\mu},$

and for $\mu\neq\nu\in\mathcal{I}_{2}$

$G_{\mu\nu}=-G_{\mu\mu}\left(X^{*}G^{(\mu)}\right)_{\mu\nu}=-G_{\nu\nu}\left(G^{(\nu)}X\right)_{\mu\nu}=G_{\mu\mu}G_{\nu\nu}^{(\mu)}\left(X^{*}G^{(\mu\nu)}X\right)_{\mu\nu}.$

  2. (2).

For $i\in\mathcal{I}_{1}$ we have

$\frac{1}{G_{ii}}=-\frac{1}{\sigma_{i}}-\left(XG^{(i)}X^{*}\right)_{ii},$
(A.3)

and for $i\neq j\in\mathcal{I}_{1}$

$G_{ij}=-G_{ii}\left(XG^{(i)}\right)_{ij}=-G_{jj}\left(G^{(j)}X^{*}\right)_{ij}=G_{ii}G_{jj}^{(i)}\left(XG^{(ij)}X^{*}\right)_{ij}.$
(A.4)

  3. (3).

For $i\in\mathcal{I}_{1}$ and $\mu\in\mathcal{I}_{2}$ we have,

$G_{i\mu}=-G_{\mu\mu}\left(G^{(\mu)}X\right)_{i\mu},\quad G_{\mu
i}=-G_{\mu\mu}\left(X^{*}G^{(\mu)}\right)_{\mu i},$

and

$\displaystyle
G_{i\mu}=-G_{ii}\left(XG^{(i)}\right)_{i\mu}=G_{ii}G_{\mu\mu}^{(i)}\left(-x_{i\mu}+\left(XG^{(i\mu)}X\right)_{i\mu}\right)$
$\displaystyle G_{\mu i}=-G_{ii}\left(G^{(i)}X\right)_{\mu
i}=G_{\mu\mu}G_{ii}^{(\mu)}\left(-x_{\mu i}+\left(X^{*}G^{(\mu
i)}X^{*}\right)_{\mu i}\right).$

  4. (4).

For $r\in\mathcal{I}$ and $s,t\in\mathcal{I}\backslash\\{r\\}$ we have

$G_{st}^{(r)}=G_{st}-\frac{G_{sr}G_{rt}}{G_{rr}},\quad\frac{1}{G_{ss}}=\frac{1}{G_{ss}^{(r)}}-\frac{G_{sr}G_{rs}}{G_{ss}G_{ss}^{(r)}G_{rr}}.$
(A.5)

  5. (5).

All of the identities from (1)-(4) hold for $G^{(S)}$ if
$S\subset\mathcal{I}_{2}$ or $S\subset\mathcal{I}.$

###### Proof.

See Lemma 4.4 of [35]. ∎

Next, we introduce the spectral decomposition of $G$. Denote the singular
value decomposition of $\Sigma^{1/2}X$ as
$\Sigma^{1/2}X=\sum_{k=1}^{n}\sqrt{\lambda_{k}}\bm{\xi}_{k}\bm{\zeta}_{k}^{*},$
where$\sqrt{\lambda_{1}}\geqslant\sqrt{\lambda_{2}}\geqslant\cdots\geqslant\sqrt{\lambda_{n}}>0$
are the singular values and $\left\\{\bm{\xi}_{k}\right\\}$ and
$\left\\{\bm{\zeta}_{k}\right\\}$ the associated singular vectors. Then
according to (A.2), for $\mu,\nu\in\mathcal{I}_{2},$ we have

$G_{\mu\nu}(z)=\sum_{k=1}^{n}\frac{\bm{\zeta}_{k}(\mu)\overline{\bm{\zeta}_{k}(\nu)}}{\lambda_{k}-z}.$
(A.6)

Similarly, by defining

$\mathbf{u}_{k}:=\left(\begin{array}[]{c}\mathbf{1}(k\leqslant
p)\sqrt{\lambda_{k}}\bm{\xi}_{k}\\\ \mathbf{1}(k\leqslant
n)\zeta_{k}\end{array}\right)\in\mathbb{R}^{\mathcal{I}},\
\overline{\Sigma}:=\left(\begin{array}[]{cc}\Sigma&0\\\
0&I_{n}\end{array}\right)\in\mathbb{R}^{\mathcal{I}\times\mathcal{I}},$

we have (see equation (4.15) of [35])

$G=-\overline{\Sigma}+\overline{\Sigma}^{1/2}\sum_{k=1}^{p}\frac{\mathbf{u}_{k}\mathbf{u}_{k}^{*}}{\lambda_{k}-z}\overline{\Sigma}^{1/2}.$
(A.7)

Next, we provide some preliminary controls for the resolvent $G.$ For the
deterministic vectors $\mathbf{v}\in\mathbb{R}^{\mathcal{I}_{1}}$ and
$\mathbf{w}\in\mathbb{R}^{\mathcal{I}_{2}},$ we identity them with their
natural embeddings $(\mathbf{v}\ \ 0)^{*}$ and $(0\ \ \mathbf{w})^{*}$ in
$\mathbb{R}^{\mathcal{I}}.$ Moreover, for an $\mathcal{I}\times\mathcal{I}$
matrix $A,$ we use the following notations for their generalized entries

$A_{\mathbf{vw}}:=\mathbf{v}^{*}A\mathbf{w},\quad
A_{\mathbf{v}s}:=\mathbf{v}^{*}A\mathbf{e}_{s},\quad
A_{s\mathbf{v}}:=\mathbf{e}_{s}^{*}A\mathbf{v},$

where $\mathbf{e}_{s}$ denotes the standard unit vector in the coordinate
direction $s$ in $\mathbb{R}^{\mathcal{I}}$. The following deterministic
controls will be frequently used in our proofs.

######  Lemma A.2.

For $z\in\mathbf{S}$ uniformly in (5.1), we have that for some constant $C>0,$
with high probability

$\left\|\overline{\Sigma}^{-1/2}G\overline{\Sigma}^{-1/2}\right\|\leqslant
C{(\phi^{1/2}+\phi^{-1/2})}\eta^{-1},\quad\left\|\overline{\Sigma}^{-1/2}\partial_{z}G\overline{\Sigma}^{-1/2}\right\|\leqslant
C{(\phi^{1/2}+\phi^{-1/2})}\eta^{-2}.$ (A.8)

Furthermore, for unit vectors $\mathbf{v}\in\mathbb{R}^{\mathcal{I}_{1}}$ and
$\mathbf{w}\in\mathbb{R}^{\mathcal{I}_{2}}$, then for some constant $C>0$

$\sum_{\mu\in\mathcal{I}_{2}}\left|G_{\mathbf{w}\mu}\right|^{2}=\frac{\operatorname{Im}G_{\mathbf{w}\mathbf{w}}}{\eta},\\\
$ (A.9)
$\sum_{i\in\mathcal{I}_{1}}\left|G_{\mathbf{v}i}\right|^{2}\leqslant\frac{C(\phi^{1/2}+\phi^{-1/2})}{\eta}\operatorname{Im}G_{\mathbf{v}\mathbf{v}}+2\left(\Sigma^{2}\right)_{\mathbf{v}\mathbf{v}},\\\
$ (A.10) $\sum_{i\in\mathcal{I}_{1}}\left|G_{\mathbf{w}i}\right|^{2}\leqslant
C(\phi^{1/2}+\phi^{-1/2})\sum_{\mu\in\mathcal{I}_{N}}\left|G_{\mathbf{w}\mu}\right|^{2},\\\
$ (A.11)
$\sum_{\mu\in\mathcal{I}_{2}}\left|G_{\mathbf{v}\mu}\right|^{2}\leqslant
C(\phi^{1/2}+\phi^{-1/2})\sum_{i\in\mathcal{I}_{1}}\left|G_{\mathbf{v}i}\right|^{2}.$
(A.12)

Finally, the estimates (A.9)-(A.12) remain true for $G^{(S)}$ if
$S\subset\mathcal{I}_{2}$ or $S\subset\mathcal{I}$.

###### Proof.

Note that according to [5] and the scaling (2.1), we find that with high
probability, for some constant $C>0,$

$\left\|X^{*}X\right\|\leq C(\phi^{1/2}+\phi^{-1/2}).$

The estimates (A.8) follow from (A.7), using
$\|A\|=\sup\\{|\langle\mathbf{x},A\mathbf{y}\rangle|:|\mathbf{x}|,|\mathbf{y}|\leqslant
1\\}$, and since we are in the spectral domain $\mathbf{S}$ so
$|\operatorname{Re}z-\lambda_{k}|\asymp 1$,
$\left|\lambda_{k}\right|/\left|\lambda_{k}-z\right|\leqslant
C{(\phi^{1/2}+\phi^{-1/2})}\eta^{-1}$. Moreover, (A.9) directly follows from
(A.2) and (A.6). In order to prove (A.10), we use (A.2) to write

$\displaystyle\sum_{i\in\mathcal{I}_{1}}\left|G_{\mathbf{v}i}\right|^{2}=\sum_{i\in\mathcal{I}_{1}}\left|\left(\Sigma
XGX^{*}\Sigma\right)_{\mathbf{v}i}-\Sigma_{\mathbf{v}i}\right|^{2}\leqslant
2\left(\Sigma
XGX^{*}\Sigma^{2}XG^{*}X^{*}\Sigma\right)_{\mathbf{vv}}+2\left(\Sigma^{2}\right)_{\mathbf{vv}}$
$\displaystyle\leqslant C\left\|X^{*}X\right\|\left(\Sigma
XG_{N}G_{N}^{*}X^{*}\Sigma\right)_{\mathbf{vv}}+2\left(\Sigma^{2}\right)_{\mathbf{vv}}=\frac{C\left\|X^{*}X\right\|}{\eta}\operatorname{Im}\left(\Sigma
XGX^{*}\Sigma\right)_{\mathbf{vv}}+2\left(\Sigma^{2}\right)_{\mathbf{vv}}$
$\displaystyle=\frac{C\left\|X^{*}X\right\|}{\eta}\operatorname{Im}G_{\mathbf{vv}}+2\left(\Sigma^{2}\right)_{\mathbf{vv}}.$

In order to prove (A.11), we use (A.2) and (A.6) to get

$\sum_{i\in\mathcal{I}_{1}}\left|G_{\mathbf{w}i}\right|^{2}=\left(GX^{*}\Sigma^{2}XG^{*}\right)_{\mathbf{ww}}\leqslant
C\left(GX^{*}XG^{*}\right)_{\mathbf{ww}}\leqslant
C\left\|X^{*}X\right\|\left(G_{2}G_{2}^{*}\right)_{\mathbf{ww}}.$

The estimate (A.12) can be proved similarly in the sense that

$\sum_{\mu\in\mathcal{I}_{N}}\left|G_{\mathbf{v}\mu}\right|^{2}=|z|^{-2}\left(GXX^{*}G^{*}\right)_{\mathrm{vv}}\leqslant
C\left\|X^{*}X\right\|\left(G_{1}G_{1}^{*}\right)_{\mathbf{vv}}.$

Finally, the same estimates for $G^{(S)}$ instead of $G$ using minor
modifications of the above arguments. ∎

Finally, we provide some large deviation estimates.

######  Lemma A.3.

Let $(\xi_{i})$ and $(\zeta_{i})$ be two independent families satisfying that
$\mathbb{E}\xi_{i}=\mathbb{E}\zeta_{i}=0,\
\mathbb{E}|\xi_{i}|^{2}=\mathbb{E}|\zeta_{i}|^{2}=1$,
$(\mathbb{E}|\xi_{i}|^{q})^{1/q}\leq C$ and
$(\mathbb{E}|\zeta_{i}|^{q})^{1/q}\leq C$ for all $q\in\mathbb{N}$ and some
constant $C>0.$ Then for deterministic sequences $(a_{ij})$ and $(b_{j}),$ we
have that

$\displaystyle\sum_{i}b_{i}\xi_{i}\prec\left(\sum_{i}|b_{i}|^{2}\right)^{1/2},\
\sum_{i,j}a_{ij}\xi_{i}\zeta_{j}\prec\left(\sum_{i,j}|a_{ij}|^{2}\right)^{1/2},\
\sum_{i\neq j}a_{i}a_{j}\xi_{i}\xi_{j}\prec\left(\sum_{i\neq
j}|a_{ij}|^{2}\right)^{1/2}.$

###### Proof.

See Lemma 3.1 of [14]. ∎

###  A.2 Entrywise local law and proof of (5.16), (5.17) and (5.18)

In this section, we prove the entrywise local law and (5.16), (5.17) and
(5.18). More specifically, we will prove the following proposition.

######  Proposition A.4.

Suppose the assumptions of Theorem 5.5 hold. Then we have that (1) of Theorem
5.5 holds when $\mathbf{u}_{k}$ and $\mathbf{v}_{k}$ are the standard basis in
$\mathbb{R}^{p}$ and $\mathbb{R}^{n},$ respectively, and (2) and (3) of
Theorem 5.5 hold.

We first prepare some notations. For each $i\in\mathcal{I}_{1},$ we define

$m_{i}:=\frac{-\sigma_{i}}{1+\phi^{-1/2}m\sigma_{i}}.$ (A.13)

Recall (2.4). We find that the functions $m$ and $m_{i}$ satisfy

$\frac{1}{m}=-z-{\frac{\phi^{1/2}}{p}}\sum_{i\in\mathcal{I}_{1}}m_{i},\quad\frac{1}{m_{i}}=-\frac{1}{\sigma_{i}}-{\phi^{-1/2}}m.$

Using (1.2), (5.2) and (A.13), we readily obtain that

$\left|m_{i}\right|\asymp\sigma_{i}\quad\text{ for }z\in\mathbf{S}\text{ and
}i\in\mathcal{I}_{1}\text{. }$ (A.14)

For notational simplicity, we extend the definitions of $\sigma_{i}$ and
$m_{i}$ by setting $\sigma_{\mu}:=1$ and $m_{\mu}:=m$ for
$\mu\in\mathcal{I}_{2}$. Recall (5.10). We define the blockwise random control
parameters as follows

$\Lambda_{2}:=\max_{\mu,\nu\in\mathcal{I}_{2}}\frac{\left|G_{\mu\nu}-\delta_{\mu\nu}m_{\mu}\right|}{\sigma_{\mu}\sigma_{\nu}},\quad\Lambda_{o,2}=\max_{\mu\neq\nu\in\mathcal{I}_{2}}\frac{\left|G_{\mu\nu}\right|}{\sigma_{\mu}\sigma_{\nu}},\
\Lambda_{1}=\max_{i,j\in\mathcal{I}_{1}}\frac{\left|G_{ij}-\delta_{ij}m_{i}\right|}{\sigma_{i}\sigma_{j}},\quad\Lambda_{o,1}=\max_{i\neq
j\in\mathcal{I}_{1}}\frac{\left|G_{ij}\right|}{\sigma_{i}\sigma_{j}},$

and

$\Lambda_{o,12}=\max_{i\in\mathcal{I}_{1},\mu\in\mathcal{I}_{2}}\frac{\left|G_{i\mu}\right|}{\sigma_{i}\sigma_{\mu}}.$

Armed with the above notations, we further denote

$\Theta_{1}:=\left|\frac{1}{p}\sum_{i\in\mathcal{I}_{1}}\left(G_{ii}-m_{i}\right)\right|,\quad\Theta_{2}:=\left|\frac{1}{n}\sum_{\mu\in\mathcal{I}_{2}}\left(G_{\mu\mu}-m\right)\right|=\left|m_{2n}-m\right|.$

Moreover, we define

$\Lambda:=\phi^{1/2}\Lambda_{1}+\phi^{1/4}\Lambda_{o,12}+\Lambda_{2},\
\Lambda_{o}:=\phi^{1/2}\Lambda_{o,1}+\phi^{1/4}\Lambda_{o,12}+\Lambda_{o,2},\
\Theta:=\phi^{1/2}\Theta_{1}+\Theta_{2}.$ (A.15)

Note that all the above parameters depend on $z$ implicitly and we have the
trivial bound $\Theta=\mathrm{O}\left(\Lambda\right).$ With (A.15), we further
denote the parameters

$\Psi_{\Theta,2}:=\sqrt{\frac{\operatorname{Im}m+\Theta}{n\eta}},\
\Psi_{\Theta,1}:=\sqrt{\frac{\operatorname{Im}m+\Theta}{p\eta}},\
\Psi_{\Theta,12}:=\sqrt{\frac{\operatorname{Im}m+\Theta}{\sqrt{pn}\eta}},$

and the $z$-dependent probability event $\Xi\equiv\Xi(z)$

$\Xi=\left\\{\Lambda(z)\leqslant(\log n)^{-1}\right\\}.$ (A.16)

To prove Proposition A.4, we basically follow the approach of [35, Section 5]
or Section A.2 of [25] and focus on the details that departs significantly
from that of [25, 35].

####  A.2.1 Weak entrywise local law

In this section, we prove a weaker version of entrywise local law. It is an
analog of [35, Proposition 5.1] or [25, Lemma A.12].

######  Proposition A.5.

Suppose that the assumptions of Theorem 5.5 hold. Then
$\Lambda\prec(n\eta)^{-1/4}$ uniformly in $z\in\mathbf{S}$.

The rest of this subsection is devoted to the proof of Proposition A.5. For
$G$ in (5.10), recall that we set $H=G^{-1}.$ For $s\in\mathcal{I},$ we
introduce the conditional expectation

$\mathbb{E}_{s}[\cdot]:=\mathbb{E}\left[\cdot|H^{(s)}\right].$

Recall (A.1). According to Lemma A.1, we have that for $i\in\mathcal{I}_{1}$

$\frac{1}{G_{ii}}=-\frac{1}{\sigma_{i}}-{\frac{\phi^{-1/2}}{n}}\operatorname{Tr}G_{2}^{(i)}-Z_{i},\quad
Z_{i}:=\left(1-\mathbb{E}_{i}\right)\left(XG^{(i)}X^{*}\right)_{ii},$ (A.17)

and for $\mu\in\mathcal{I}_{2}$

$\frac{1}{G_{\mu\mu}}=-z-{\frac{\phi^{1/2}}{p}}\operatorname{Tr}G_{1}^{(\mu)}-Z_{\mu},\quad
Z_{\mu}:=\left(1-\mathbb{E}_{\mu}\right)\left(X^{*}G^{(\mu)}X\right)_{\mu\mu}.$
(A.18)

In order to prove Proposition A.5, we start with the following estimate which
is analogous to Lemma 5.2 of [35] or [25, Lemma A.9]. Recall (A.16).

######  Lemma A.6.

Suppose the assumptions of Proposition A.5 hold. Then for $s\in\mathcal{I}$
and $z\in\mathbf{S},$ we have

$\mathbf{1}(\Xi)\left(\left|Z_{s}\right|+\Lambda_{o}\right)\prec\Psi_{\Theta,2},$
(A.19)

and

$\mathbf{1}(\eta\geqslant
1)\left(\left|Z_{s}\right|+\Lambda_{o}\right)\prec\Psi_{\Theta,2}.$ (A.20)

###### Proof.

Due to similarity, we mainly focus on the proof of (A.19). Using (A.14), (A.5)
and a simple induction argument, it is not hard to see that

$\mathbf{1}(\Xi)\left|G_{tt}^{(S)}\right|\asymp\sigma_{t},$ (A.21)

for any $S\subset\mathcal{I}$ and $t\in\mathcal{I}\backslash S$ satisfying
$|S|\leqslant C$. We first work with $\Lambda_{o}$ in (A.15). It suffices to
prove that

$\displaystyle\mathbf{1}(\Xi)\left|G_{ij}\right|\prec\sigma_{i}\sigma_{j}\left(\frac{\operatorname{Im}m+\Theta_{2}+\Lambda_{o,12}^{2}}{p\eta}\right)^{1/2},$
(A.22)
$\displaystyle\mathbf{1}(\Xi)\left|G_{\mu\nu}\right|\prec\sigma_{\mu}\sigma_{\nu}\left(\frac{\operatorname{Im}m+\sqrt{\phi}\Theta_{1}+\Lambda_{o,12}^{2}}{n\eta}\right)^{1/2},$
$\displaystyle\mathbf{1}(\Xi)\left|G_{i\mu}\right|\prec\sigma_{i}\sigma_{\mu}\left(\frac{\operatorname{Im}m+\Theta_{2}+\Lambda_{o,2}^{2}}{\sqrt{pn}\eta}\right)^{1/2}.$

In fact, using the second estimate in (A.22), it is not hard to see that on
$\Xi$

$\Lambda_{o,2}\prec\Psi_{\Theta,2}+(n\eta)^{-1/2}\Lambda_{o,12}.$

Moreover, using the first estimate in (A.22), we see that on $\Xi$

$\phi^{1/2}\Lambda_{o,1}\prec\Psi_{\Theta,2}+(n\eta)^{-1/2}\Lambda_{o,12}.$

Similarly, using the last estimate in (A.22), we have that on $\Xi$

$\phi^{1/4}\Lambda_{o,12}\prec\Psi_{\Theta,2}+(n\eta)^{-1/2}\Lambda_{o,2}.$

Combining the above bounds with the definition $\Lambda_{o}$ in (A.15), we see
that $\mathbf{1}(\Xi)\Lambda_{o}\prec\Psi_{\Theta,2}.$ Now we prove (A.22).
Let us start with $G_{ij}$ for $i\neq j\in\mathcal{I}_{1}$. By (A.4), (A.21)
and Lemma A.3, we find on $\Xi$

$\displaystyle\left|G_{ij}\right|$
$\displaystyle\leqslant\left|G_{ii}G_{jj}^{(i)}\right|\left|\sum_{\mu,\nu\in\mathcal{I}_{2}}x_{i\mu}G_{\mu\nu}^{(ij)}x_{\nu
j}\right|\prec\sigma_{i}\sigma_{j}\left(\frac{1}{np}\sum_{\mu,\nu\in\mathcal{I}_{2}}\left|G_{\mu\nu}^{(ij)}\right|^{2}\right)^{1/2}$
$\displaystyle=\sigma_{i}\sigma_{j}\left(\frac{1}{np\eta}\sum_{\mu\in\mathcal{I}_{2}}\operatorname{Im}G_{\mu\mu}^{(ij)}\right)^{1/2}\prec\left(\frac{1}{np\eta}\sum_{\mu\in\mathcal{I}_{2}}\operatorname{Im}G_{\mu\mu}+\frac{\Lambda_{o,12}^{2}}{p\eta}\right)^{1/2}\prec\sqrt{\frac{\operatorname{Im}m+\Theta_{2}+\Lambda_{o,12}^{2}}{p\eta}},$

where in the third step we used (A.9) and the fourth step we used (A.5). This
completes the proof for the first estimate of (A.22). Similarly, for
$G_{\mu\nu}$ with $\mu\neq\nu\in\mathcal{I}_{2},$ using (1) of Lemma A.1 and
(A.10), as well as the bound
$\operatorname{Im}m_{i}=\mathrm{O}({\phi^{-1/2}}\sigma_{i}^{2}\operatorname{Im}m)$
for all $i\in\mathcal{I}_{1}$ which follows from (A.13), we have that on $\Xi$

$\displaystyle\left|G_{\mu\nu}\right|$
$\displaystyle\leqslant\left|G_{\mu\mu}G_{\nu\nu}^{(\mu)}\right|\left|\sum_{i,j\in\mathcal{I}_{1}}x_{\mu
i}G_{ij}^{(\mu\nu)}x_{j\nu}\right|\prec\sigma_{\mu}\sigma_{\nu}\left(\frac{1}{np}\sum_{i,j\in\mathcal{I}_{1}}\left|G_{ij}^{(\mu\nu)}\right|^{2}\right)^{1/2}$
$\displaystyle\prec\left(\frac{1}{np}\frac{\phi^{1/2}}{\eta}\sum_{k\in\mathcal{I}_{1}}\operatorname{Im}G_{kk}^{(\mu\nu)}+\frac{1}{np}\sum_{k\in\mathcal{I}_{1}}((\Sigma^{(\mu\nu)})^{2})_{kk}\right)^{1/2}\prec\sqrt{\frac{\operatorname{Im}m+\phi^{1/2}\Theta_{1}+\phi^{1/2}\Lambda_{o,12}^{2}}{n\eta}}.$

This completes the proof for the second estimate of (A.22). The last estimate
on $\mathbf{1}(\Xi)G_{i\mu}$ with $i\in\mathcal{I}_{1}$ and
$\mu\in\mathcal{I}_{N}$ can be estimated similarly using (3) of Lemma A.1 and
(A.11) in the sense that

$\displaystyle\left|G_{i\mu}\right|$
$\displaystyle=|G_{ii}G_{\mu\mu}^{(i)}|\left|-x_{i\mu}+\left(XG^{(i\mu)}X\right)_{i\mu}\right|\prec\left(\frac{1}{np}\sum_{\nu\in\mathcal{I}_{2},j\in\mathcal{I}_{1}}\left|G_{\nu
j}^{(i\mu)}\right|^{2}\right)^{1/2}.$ $\displaystyle\prec$
$\displaystyle\left(\frac{\phi^{1/2}}{np}\sum_{\nu\in\mathcal{I}_{2}}\frac{\operatorname{Im}G_{\nu\nu}^{(i\mu)}}{\eta}\right)^{1/2}\prec{\frac{\phi^{1/2}\operatorname{Im}m+\phi^{1/2}\Theta_{2}+\phi^{1/2}\Lambda_{o,2}^{2}}{p\eta}}=\frac{\operatorname{Im}m+\Theta_{2}+\Lambda_{o,2}^{2}}{\sqrt{np}\eta}.$

This completes the proof of (A.22).

Then we proceed to the control of $Z_{s}.$ On the one hand, when
$s=i\in\mathcal{I}_{1},$ we decompose that

$\left|Z_{i}\right|\leqslant\left|\sum_{\mu\in\mathcal{I}_{2}}^{(i)}\left(x_{i\mu}^{2}-\frac{1}{\sqrt{np}}\right)G_{\mu\mu}^{(i)}\right|+\left|\sum_{\mu\neq\nu}^{(i)}x_{i\mu}G_{\mu\nu}^{(i)}x_{\nu
i}\right|.$

By Lemma A.3 and (A.9), on $\Xi,$ we have that the first term of the above
equation can be stochastically dominated by
$\left(\sum_{\mu}^{(i)}(pn)^{-1}\left|G_{\mu\mu}^{(i)}\right|^{2}\right)^{1/2}\prec\frac{1}{\sqrt{p\eta}}$.
For the second term, we have

$\sum_{\mu,\nu}^{(i)}\frac{1}{np}\left|G_{\mu\nu}^{(i)}\right|^{2}\leqslant\frac{1}{np}\sum_{\mu,\nu}^{(i)}\left|G_{\mu\nu}^{(i)}\right|^{2}=\frac{1}{np\eta}\sum_{\mu}^{(i)}\operatorname{Im}G_{\mu\mu}^{(i)}\prec\frac{\operatorname{Im}m+\Theta_{2}+\Lambda_{o}^{2}}{p\eta}.$

Since $p\gg n,$ this results in

$\left|Z_{i}\right|\prec\sqrt{\frac{\operatorname{Im}m+\Theta+\Lambda_{o}^{2}}{p\eta}}\prec\Psi_{\Theta,2}.$

Similarly, for $s=\mu\in\mathcal{I}_{2},$ we can show that

$|Z_{\mu}|\prec\sqrt{\frac{\operatorname{Im}m+\phi^{1/2}\Theta_{1}+\Lambda_{o}^{2}}{n\eta}}\prec\Psi_{\Theta,2}.$

This completes the proof of (A.19).

Finally, for the proof of (A.20), the argument is similar except we need to
use the bound $\|G^{(S)}\|=\mathrm{O}(1)$ which follows from Lemma A.2 and
$\eta\geq 1$ whenever it is necessary instead of (A.21). This completes our
proof.

∎

The following estimate is an analog of [25, Lemma A.10] or [35, Lemma 5.3].

######  Lemma A.7.

Suppose the assumptions of Proposition A.5 hold. Denote

$[Z]_{2}:=\frac{1}{n}\sum_{\mu\in\mathcal{I}_{2}}Z_{\mu},\quad[Z]_{1}:=\frac{\phi^{1/2}}{p}\sum_{i\in\mathcal{I}_{1}}\frac{\sigma_{i}^{2}}{\left(1+\phi^{-1/2}m_{2n}\sigma_{i}\right)^{2}}Z_{i}.$
(A.23)

Then for $z\in\mathbf{S},$ we have for $f$ defined in (2.7)

$\mathbf{1}(\Xi)\left(f\left(m_{2n}\right)-z\right)=\mathbf{1}(\Xi)\left([Z]_{2}+[Z]_{1}+\mathrm{O}_{\prec}\left(\Psi_{\Theta,2}^{2}\right)\right),$
(A.24)

and

$\mathbf{1}(\eta\geq 1)\left(f\left(m_{2n}\right)-z\right)=\mathbf{1}(\eta\geq
1)\left([Z]_{2}+[Z]_{1}+\mathrm{O}_{\prec}\left(n^{-1}\right)\right).$ (A.25)

###### Proof.

As before, due to similarity, we focus our proof on (A.24). Recall that
$m_{2n}=n^{-1}\sum_{\mu\in\mathcal{I}_{2}}G_{\mu\mu}(z).$ Using
$G_{\mu\mu}=m_{2n}+\left(G_{\mu\mu}-m_{2n}\right),$ we see that

$\frac{1}{G_{\mu\mu}}=\frac{1}{m_{2n}}-\frac{1}{m_{2n}^{2}}\left(G_{\mu\mu}-m_{2n}\right)+\frac{1}{m_{2n}^{2}}\left(G_{\mu\mu}-m_{2n}\right)^{2}\frac{1}{G_{\mu\mu}}.$
(A.26)

Moreover, according to (A.18), using Lemma A.6 and (A.5), we see that

$\mathbf{1}(\Xi)\frac{1}{G_{\mu\mu}}=\mathbf{1}(\Xi)\left(-z-\frac{\phi^{1/2}}{p}\sum_{i\in\mathcal{I}_{1}}G_{ii}-Z_{\mu}+\mathrm{O}_{\prec}\left({\phi^{1/2}}\Psi_{\Theta,{12}}^{2}\right)\right),$
(A.27)

and

$\mathbf{1}(\Xi)\left|G_{\mu\mu}-m_{2n}\right|\prec\Psi_{\Theta,2}.$ (A.28)

Similarly, according to (A.17), we have that

$\mathbf{1}(\Xi)G_{ii}=\mathbf{1}(\Xi)\frac{-\sigma_{i}}{1+\phi^{-1/2}m_{2n}\sigma_{i}+\sigma_{i}Z_{i}+\mathrm{O}_{\prec}\left(\sigma_{i}\phi^{-1/2}\Psi_{\Theta,{12}}^{2}\right)}.$
(A.29)

On the one hand, combining (A.26) and (A.28), we see that

$\mathbf{1}(\Xi)\frac{1}{n}\sum_{\mu\in\mathcal{I}_{2}}\frac{1}{G_{\mu\mu}}=\mathbf{1}(\Xi)\frac{1}{m_{2n}}+\mathrm{O}_{\prec}\left(\Psi_{\Theta,2}^{2}\right).$
(A.30)

On the other hand, plugging (A.29) into (A.27), we see that

$\displaystyle\mathbf{1}(\Xi)\frac{1}{n}\sum_{\mu\in\mathcal{I}_{2}}\frac{1}{G_{\mu\mu}}$
$\displaystyle=$
$\displaystyle\mathbf{1}(\Xi)\left(-z+\frac{\phi^{1/2}}{p}\sum_{i\in\mathcal{I}_{1}}\frac{\sigma_{i}}{1+\phi^{-1/2}m_{2n}\sigma_{i}+\sigma_{i}Z_{i}+\mathrm{O}_{\prec}\left(\sigma_{i}\phi^{1/2}\Psi_{\Theta,12}^{2}\right)}-\frac{1}{n}\sum_{\mu\in\mathcal{I}_{2}}Z_{\mu}+\mathrm{O}_{\prec}\left(\Psi_{\Theta,2}^{2}\right)\right).$

Moreover, using (A.21) and the assumption $p\gg n$, we see that
$1+\phi^{-1/2}m_{2n}\sigma_{i}\asymp 1.$ Therefore, together with (A.30), we
can conclude the proof of (A.24).

The proof of (A.25) is similar as discussed in the end of the proof of Lemma
A.6. We omit further details.

∎

The following lemma provides the stability condition for (2.4). Roughly, it
says that if $f(m_{2n})-z$ is small and $m_{2n}(\tilde{z})-m(\tilde{z})$ is
small for $\tilde{z}:=z+\mathrm{i}{n^{-5}}$, then $m_{2n}(z)-m(z)$ is small.
For $z\in\mathbf{S},$ we introduce the discrete set

$L(z):=\\{z\\}\cup\left\\{w\in\mathbf{S}:\operatorname{Re}w=\operatorname{Re}z,\operatorname{Im}w\in[\operatorname{Im}z,1]\cap\left(n^{-5}{\phi^{-1/2}}\mathbb{N}\right)\right\\}.$

Thus, if $\operatorname{Im}z\geqslant 1$ then $L(z)=\\{z\\}$ and if
$\operatorname{Im}z\leqslant 1$ then $L(z)$ is a one-dimensional lattice with
spacing $n^{-5}{\phi^{-1/2}}$ plus the point $z$. Clearly, we have the bound
$|L(z)|\leqslant n^{5}{\phi^{1/2}}$.

######  Lemma A.8.

Suppose that $\delta:\mathbf{S}\rightarrow(0,\infty)$ satisfies
$n^{-2}\leqslant\delta(z)\leqslant(\log n)^{-1}$ for $z\in\mathbf{S}$ and that
$\delta$ is Lipschitz continuous with Lipschitz constant $n^{2}{\phi^{1/2}}$.
Suppose moreover that for each fixed $E$, the function
$\eta\mapsto\delta(E+\mathrm{i}\eta)$ is nonincreasing for $\eta>0$. Suppose
that $u:\mathbf{S}\rightarrow\mathbb{C}$ is the Stieltjes transform of a
probability measure supported in $[0,C]$ for some large enough constant $C>0$.
Let $w\in L(z)$ and suppose that

$|f(u(w))-w|\leqslant\delta(w).$

Then we have that

$|u(z)-m(z)|\leqslant\frac{C^{\prime}\delta(z)}{\operatorname{Im}m(z)+\sqrt{\delta(z)}},$
(A.31)

for some constant $C^{\prime}>0$ independent of $z$ and $n.$

###### Proof.

This lemma can proved with the same method as in e.g. [14, Lemma 4.5] and [35,
Appendix A.2]. The main inputs are Lemmas 2.5 and 5.1 and some estimates in
their proofs. We omit further details. ∎

Armed with the above results, we can follow [14, 25, 35] to complete the
proof.

###### Proof of Proposition A.5.

When $\eta\geq 1,$ using the above controls and Lemma 5.1, following the same
arguments as in [14, Lemma 4.6] or [25, (A.44)-(A.46)], we find that
$\Lambda\prec n^{-1/4}.$ It remains to deal with the small $\eta$ case. One
can prove this lemma using a continuity argument as in e.g. [14, Section 4.1].
The key inputs are Lemmas A.6–A.8, and the estimates in the $\eta\geq 1$ case.
All the other parts of the proof are essentially the same. We omit the
details. ∎

####  A.2.2 Proof of Proposition A.4

To get strong laws as in Proposition A.4, we need stronger bounds on $[Z]_{2}$
and $[Z]_{1}$ in (A.23). The following lemma is a counterpart of Lemma A.13 of
[25] or Lemma 5.6 of [35] or Lemma 4.9 of [14]. Due to similarity, we only
sketch the proof.

######  Lemma A.9.

Suppose the assumptions of Proposition A.4 hold. Moreover, if $\Upsilon$ is a
positive, $n$-dependent, deterministic function on $\mathbf{S}$ satisfying
$n^{-1/2}\leqslant\Upsilon\leqslant n^{-c}$ for some constant $c>0$. In
addition, if we assume that that $\Lambda_{2}\prec n^{-c}$ and
$\Lambda_{o,2}\prec\Upsilon$ on $\mathbf{S}$. Then on $\mathbf{S},$ we have

$\frac{1}{n}\sum_{\mu\in\mathcal{I}_{2}}\left(1-\mathbb{E}_{\mu}\right)\frac{1}{G_{\mu\mu}}=\mathrm{O}_{\prec}\left(\Upsilon^{2}\right).$

Similarly, for some other positive deterministic function on $\mathbf{S}$
satisfying $p^{-1/2}\leqslant\widetilde{\Upsilon}\leqslant p^{-c}$, suppose
that $\Lambda_{1}\prec p^{-c}$ and $\Lambda_{o,1}\prec\widetilde{\Upsilon}$ on
$\mathbf{S}$. Then we have

$\frac{1}{p}\sum_{i\in\mathcal{I}_{1}}\frac{\sigma_{i}^{2}}{\left(1+\phi^{-1/2}m_{2n}\sigma_{i}\right)^{2}}\left(1-\mathbb{E}_{i}\right)\frac{1}{G_{ii}}=\mathrm{O}_{\prec}\left(\widetilde{\Upsilon}^{2}\right).$

###### Proof.

As discussed in the proof of Lemma 5.6 of [35], the proof of the first
estimate follows from a straightforward extension of Lemma 4.9 of [14]. For
the second estimate, it can be proved similarly using the discussions as
explained in Appendix B of [14]. The only complication is the term

$\frac{\sigma_{i}^{2}}{\left(1+\phi^{-1/2}m_{2n}\sigma_{i}\right)^{2}}=\frac{\sigma_{i}^{2}}{\left(1+\phi^{-1/2}m\sigma_{i}\right)^{2}}-\frac{\sigma_{i}^{2}[\phi^{-1}(m_{2n}^{2}-m^{2})\sigma_{i}^{2}+2\phi^{-1/2}(m_{2n}-m)\sigma_{i}]}{(1+\phi^{-1/2}m_{2n}\sigma_{i})^{2}(1+\phi^{-1/2}m\sigma_{i})^{2}}.$

On the one hand, running the argument on
$\frac{1}{p}\sum_{i\in\mathcal{I}_{1}}\frac{\sigma_{i}^{2}}{\left(1+\phi^{-1/2}m\sigma_{i}\right)^{2}}\left(1-\mathbb{E}_{i}\right)\frac{1}{G_{ii}}=\mathrm{O}_{\prec}\left(\widetilde{\Upsilon}^{2}\right)$
is a straightforward extension of the proof of the first estimate. On the
other hand, for the second term, since we have $|m_{2n}-m|\prec(n\eta)^{-1/4}$
from Proposition A.5, we see that

$\frac{\sigma_{i}^{2}[\phi^{-1}(m_{2n}^{2}-m^{2})\sigma_{i}^{2}+2\phi^{-1/2}(m_{2n}-m)\sigma_{i}]}{(1+\phi^{-1/2}m_{2n}\sigma_{i})^{2}(1+\phi^{-1/2}m\sigma_{i})^{2}}\prec\phi^{-1/2}(n\eta)^{-1/4}.$

Using Proposition A.5, we see that

$\frac{1}{p}\sum_{i\in\mathcal{I}_{1}}\frac{\sigma_{i}^{2}[\phi^{-1}(m_{2n}^{2}-m^{2})\sigma_{i}^{2}+2\phi^{-1/2}(m_{2n}-m)\sigma_{i}]}{(1+\phi^{-1/2}m_{2n}\sigma_{i})^{2}(1+\phi^{-1/2}m\sigma_{i})^{2}}(1-\mathbb{E}_{i})\frac{1}{G_{ii}}\prec\phi^{-1/2}(n\eta)^{-1/2}.$

This completes the proof. For more details, we refer the readers to [14,
Appendix B], [35, Lemma 5.6] and [25, Lemma A.13]. ∎

Armed with Lemma A.9, we can prove the following iterative estimates which is
an analog of Lemma 4.8 of [14].

######  Lemma A.10.

Let $t\in(0,1)$ and suppose that $\Theta\prec(n\eta)^{-t}$ uniformly in
$z\in\mathbf{S}$. Then we have that uniformly in $z\in\mathbf{S}$

$|[Z]_{1}|+|[Z]_{2}|\prec\frac{\operatorname{Im}m+(n\eta)^{-t}}{n\eta}.$

###### Proof.

Due to similarity, we focus on $[Z]_{2}.$ According to (A.18) and (1) of Lemma
A.1, we see that

$-Z_{u}=(1-\mathbb{E}_{\mu})\frac{1}{G_{\mu\mu}}.$ (A.32)

We now apply Lemma A.9 to the above identity. According to Proposition A.5 and
its proof, we conclude that

$\Lambda_{o,2}\prec\Upsilon,\quad\Lambda\prec\Upsilon^{\prime},\quad\Upsilon:=\sqrt{\frac{\operatorname{Im}m+(n\eta)^{-t}}{n\eta}},\quad\Upsilon^{\prime}:=(n\eta)^{-1/4}.$

Using Lemma 5.1, it is easy to see that the conditions of $\Upsilon$ and
$\Upsilon^{\prime}$ satisfy the conditions of Lemma A.9 so that the result
follows from Lemma A.9 and (A.32). Similar discussions apply to $[Z]_{1}$
using

$\widetilde{\Upsilon}:=\sqrt{\frac{\operatorname{Im}m+(n\eta)^{-t}}{p\eta}},\quad\widetilde{\Upsilon}^{\prime}=(p\eta)^{-1/4}.$

This completes the proof. ∎

Armed with Lemma A.10 and Proposition A.5, we can prove Proposition A.4
following the arguments in [14, Section 4.2] verbatim. Due to similarity, we
only sketch the proof.

###### Proof of Proposition A.4.

We notice that Lemma A.10 implies that if $\Theta\prec(n\eta)^{-\tau},$ one
may invoke (A.31) to obtain that for some constant $C>0$

$\Theta_{2}\prec\frac{\operatorname{Im}m}{n\eta}\frac{1}{\sqrt{\kappa+\eta}}+\sqrt{\frac{(n\eta)^{-\tau}}{n\eta}}\leqslant
C(n\eta)^{-1/2-\tau/2}.$

According to Proposition A.5, we can iterate the above results and push the
rate all the way down to $\Theta_{2}\prec(n\eta)^{-1}$ on $\mathbf{S}$.
Similarly, we can obtain the control for $\phi^{1/2}\Theta_{1}$ and show that
$\Theta\prec(n\eta)^{-1}$ using the definition (A.15). This completes the
proof of (5.16) and (5.17). Once the averaged local law is proved, we can
prove the entrywise local law (5.18) and (5.14) and (5.15) with standard basis
using Proposition A.5 and Lemma A.1; see Section Section 4.2 of [14] for more
details. This concludes our proof. ∎

###  A.3 Proof of (5.14) and (5.15)

It remains to prove (5.14) and (5.15) for general projections $\mathbf{u}$ and
$\mathbf{v}.$ Our argument basically follows that of Section 5 of [14]. We
only sketch the proof and refer the readers to [14] for more details. Due to
similarity, we only provide the arguments for the proof of (5.14). By
linearity and polarization, we see that it suffices to prove that for any fix
deterministic unit vector $\mathbf{v}\in\mathbb{R}^{p},$
$\mathbf{v}^{*}R_{1}\mathbf{v}=\mathbf{v}^{*}\Pi\mathbf{v}+\mathrm{O}_{\prec}(\phi^{-1}\Psi).$
Furthermore, using the definitions in (5.11), (5.12), (A.1) and (A.13), it
suffices to prove

$\mathbf{v}^{*}G_{1}(z)\mathbf{v}=\mathbf{v}^{*}M\mathbf{v}+\mathrm{O}_{\prec}\left(\phi^{-1/2}\Psi\right),$
(A.33)

where for $m_{i}$ in (A.13) we denote

$M=\operatorname{diag}\\{m_{1},m_{2},\cdots,m_{p}\\}.$

The rest of this section is devoted to the proof of (A.33). Throughout the
proof of this section, to make more explicit connections with Section 5 of
[14], with a little abuse of notations, for (A.1), we use

$G\equiv G_{1}\in\mathbb{R}^{p\times p},\ R\equiv G_{2}\in\mathbb{R}^{n\times
n}.$ (A.34)

Moreover, we will use $a,b$ for indices valued in $\\{1,2,\ldots,p\\}$ and
$\mathbf{v}=(v_{a}).$ We decompose that

$\mathbf{v}^{*}G\mathbf{v}-\mathbf{v}^{*}M\mathbf{v}=\sum_{a}v_{a}^{2}(G_{aa}-m_{i})+\mathcal{Z},$

where

$\mathcal{Z}:=\sum_{a\neq b}v_{a}G_{ab}v_{b}.$ (A.35)

Note that the first part can be can be controlled using Proposition A.4 in the
following way

$|\sum_{a}v_{a}^{2}(G_{aa}-m_{a})|\leq\max_{a}|G_{aa}-m_{a}|\prec\phi^{-1/2}\Psi.$
(A.36)

It suffices to control $\mathcal{Z}.$ Throughout this section, we will follow
Section 5 of [14] to prove that the following result. Note that due to our
linearization as in (A.1) and (A.2), we do not need to rescale the resolvent
as in Section 5.1 of [14].

######  Lemma A.11.

Suppose the assumptions of Theorem 5.5 hold. Then we have that for all
$q\in\mathbb{N},$

$\mathbb{E}|\mathcal{Z}|^{q}\prec\left(\phi^{-1/2}\Psi\right)^{q}.$

###### Proof.

The proof is almost the same as Sections 5.5-5.15 of [14]. We omit the
details. ∎

Armed with Lemma A.11, we see from Chebyshev’s inequality that
$\mathcal{Z}\prec\phi^{-1/2}\Psi.$ Combining with (A.36), we can complete the
proof of (A.33) and hence (5.14).

##  Appendix B Additional technical proofs

In this section, we prove Lemmas 2.5 and 5.1.

###  B.1 Proof of (5.2)

Taking imaginary part and real part on both sides of (2.4), we have

$\displaystyle\frac{\operatorname{Re}m}{|m|^{2}}=-E+\int\frac{\phi(\phi^{1/2}x^{-1}+\operatorname{Re}m)}{|\phi^{1/2}x^{-1}+m|^{2}}\pi(\mathrm{d}x),$
$\displaystyle\frac{-\operatorname{Im}m}{|m|^{2}}=-\eta+\int\frac{\phi(-\operatorname{Im}m)}{|\phi^{1/2}x^{-1}+m|^{2}}\pi(\mathrm{d}x).$
(B.1)

For simplicity, we denote $R:=\operatorname{Re}m$ and $I:=\operatorname{Im}m$.
We can further rewrite the above equations as

$\displaystyle
R\int\frac{(\phi^{1/2}x^{-1}+R)^{2}+I^{2}-\phi(R^{2}+I^{2})}{(R^{2}+I^{2})[(\phi^{1/2}x^{-1}+R)^{2}+I^{2}]}\pi(\mathrm{d}x)$
$\displaystyle=-E+\int\frac{\phi^{3/2}x^{-1}}{(\phi^{1/2}x^{-1}+R)^{2}+I^{2}}\pi(\mathrm{d}x),$
(B.2) $\displaystyle
I\int\frac{(\phi^{1/2}x^{-1}+R)^{2}+I^{2}-\phi(R^{2}+I^{2})}{(R^{2}+I^{2})[(\phi^{1/2}x^{-1}+R)^{2}+I^{2}]}\pi(\mathrm{d}x)$
$\displaystyle=\eta.$

Recall that the Stieltjes transform should map $\mathbb{C}_{+}$ to
$\mathbb{C}_{+}.$ Therefore, for all $z\in\mathbb{C}_{+},$ we see from the
second equation of (B.2) that

$\int\frac{(\phi^{1/2}x^{-1}+R)^{2}+I^{2}-\phi(R^{2}+I^{2})}{(R^{2}+I^{2})[(\phi^{1/2}x^{-1}+R)^{2}+I^{2}]}\pi(\mathrm{d}x)>0.$

This implies that there must exist $x_{0}\in\operatorname{supp}(\pi)$ so that

$(\phi^{1/2}x_{0}^{-1}+R)^{2}+I^{2}-\phi(R^{2}+I^{2})>0.$

Set $t:=|m|=\sqrt{R^{2}+I^{2}}$. We see that for all $\phi>0$

$\displaystyle(\phi-1)t^{2}-2\phi^{1/2}x_{0}^{-1}t-\phi x_{0}^{-2}$
$\displaystyle<0.$

Consequently, for $\phi\gg 1,$ we shall have that $\forall z\in\mathbb{C}_{+}$

$t<\frac{2\phi^{1/2}x_{0}^{-1}+2\sqrt{\phi
x_{0}^{-2}+\phi(\phi-1)x_{0}^{-2}}}{2(\phi-1)}\sim
x_{0}^{-1}(1+\phi^{-1/2})=\mathrm{O}(1).$

This proves the upper bound for $|m|.$ For the lower bound, equipped with the
upper bound, according to (2) of Assumption 2.1, it is easy to see from (B.2)
that when $\phi\gg 1$

$\displaystyle R\left(\frac{1}{R^{2}+I^{2}}-\int
x^{2}\pi(\mathrm{d}x)+\mathrm{O}(\phi^{-1/2})\right)$
$\displaystyle=-E+\int\frac{\phi^{3/2}x^{-1}}{(\phi^{1/2}x^{-1}+R)^{2}+I^{2}}\pi(\mathrm{d}x),$
$\displaystyle I\left(\frac{1}{R^{2}+I^{2}}-\int
x^{2}\pi(\mathrm{d}x)+\mathrm{O}(\phi^{-1/2})\right)$ $\displaystyle=\eta.$

This yields that

$\displaystyle(R^{2}+I^{2})\left(\frac{1}{R^{2}+I^{2}}-\int
x^{2}\pi(\mathrm{d}x)+\mathrm{O}(\phi^{-1/2})\right)^{2}=\left(-E+\int\frac{\phi^{3/2}x^{-1}}{(\phi^{1/2}x^{-1}+R)^{2}+I^{2}}\pi(\mathrm{d}x)\right)^{2}+\eta^{2}.$

Using the upper bound for $t,$ when $z\in\mathbf{S}$ in (5.1), we see that the
right-hand side of the above equation can be bounded by $\mathrm{O}(1).$
Moreover, the left-hand side of the above equation can be rewritten as
$\frac{(t^{-2}-\int x^{2}\pi(\mathrm{d}x))^{2}}{t^{-2}}.$ It shows that $t$ is
bounded from both above and below so that $m(z)\asymp 1$. Together with (B.1),
we see that $\operatorname{Im}m$ is bounded from below by $c\eta.$ This
concludes the proof.

###  B.2 Proof of Lemma 2.5 and (5.3)

The proofs are similar to those in [35]. Recall $f$ defined in (2.7) and the
intervals $I_{k},0\leq k\leq 2p,$ defined below it. We abbreviate
$r_{i}:=\phi\pi(\\{s_{i}\\})$. Then we can rewrite

$f(x)=-\frac{1}{x}+\sum_{i=1}^{p}\frac{r_{i}}{x+s_{i}^{-1}}.$

By definition of $f$ we have

$f^{\prime}(x)=\frac{1}{x^{2}}-\sum_{i=1}^{p}\frac{r_{i}}{\left(x+s_{i}^{-1}\right)^{2}},\quad
f^{\prime\prime}(x)=\frac{-2}{x^{3}}+2\sum_{i=1}^{p}\frac{r_{i}}{\left(x+s_{i}^{-1}\right)^{3}}.$
(B.3)

Denote the multiset $\mathcal{C}\subset\mathbb{R}$ of the critical points of
$f,$ using the conventions that a nondegenerate critical point is counted once
and a degenerate critical point twice. According to Lemma 2.4 of [35], we have
that $|\mathcal{C}\cap I_{0}|=|\mathcal{C}\cap I_{1}|=1$ and $|\mathcal{C}\cap
I_{i}|\in\\{0,2\\}$ for $i=2,\cdots,p.$ This implies that $|\mathcal{C}|=2q$
is even and we denote by $x_{1}\geq x_{2}\geq\cdots\geq x_{2q-1}\geq x_{2q}$
be these critical points. Moreover, when $\phi\gg 1,$ we find that for
$0<t\leq 1,$

$\partial_{t}\partial_{x}f^{t}(x)=-\sum_{i=1}^{p}\frac{\phi
t\pi(\\{\sigma_{i}\\})x}{(x+(\phi t)^{1/2}\sigma_{i}^{-1})^{3}}<0,\ x\in
I_{2}\cup I_{3}\cup\cdots\cup I_{p}.$

Following lines of the proof of Lemma 2.5 of [35], we have that for
$a_{k}=f(x_{k}),$ $a_{1}\geq a_{2}\geq\cdots\geq a_{2q}$ and $x_{k}=m(a_{k}).$

Note that by definition, $x_{k}\asymp\phi^{1/2}$ for
$k\in\\{2,\cdots,2q-1\\}.$ However, according to (5.2) that $m(z)$ is bounded
above and the relation that $f(m(z))=z$, this shows that $q=1$ and we only
have two critical points in $I_{1}$ and $I_{0},$ denoted as $x_{1}$ and
$x_{2}.$ Following lines of the proof of Lemma 2.6 of [35], we can complete
the first part of Lemma 2.5. To prove the second part of Lemma 2.5 and (5.3),
we need the following lemma which is an analog of Lemma A.3 of [35].

######  Lemma B.1.

There exists a $\tau>0,$ such that for $k=1,2$ and some constant $C>0$

$\left|x_{k}\right|\asymp
1,\quad\left|f^{\prime\prime}\left(x_{k}\right)\right|\asymp
1,\quad\left|f^{\prime\prime\prime}(\zeta)\right|\leqslant C\quad\text{ for
}\quad\left|\zeta-x_{k}\right|\leqslant\tau^{\prime}.$

###### Proof.

First, the above bounds for $x_{k}$ are clear. The lower bound
$\left|x_{k}\right|\geqslant{c}$ follows from $f^{\prime}\left(x_{k}\right)=0$
and (B.3). Second, using

$f^{\prime\prime}\left(x_{k}\right)=-\frac{2}{x_{k}}\sum_{i}\frac{s_{i}^{-1}r_{i}}{\left(x_{k}+s_{i}^{-1}\right)^{3}},$

and $|x_{k}|\asymp 1,$ we see that $|f^{\prime\prime}(x_{k})|\asymp 1.$
Finally, the last control follows from

$f^{\prime\prime\prime}(x)=\frac{6}{x^{4}}-6\sum_{i=1}^{p}\frac{\phi\pi(\\{\sigma_{i}\\})}{(x+\phi^{1/2}\sigma_{i}^{-1})^{4}}.$

∎

Using (2.7), $x_{1},x_{2}\asymp 1$ and
$\gamma_{+}=f(x_{1}),\gamma_{-}=f(x_{2}),$ when $\phi\geq 1,$ it is easy to
see that $\gamma_{+},\gamma_{-}\asymp\phi^{1/2}.$ Similarly, we can show that
$\gamma_{+}-\gamma_{-}=f(x_{1})-f(x_{2})=\mathrm{O}(1).$ This completes the
second part of Lemma 2.5. Finally, (5.3) follows from the following expansion

$z-a_{k}=\frac{f^{\prime\prime}\left(x_{k}\right)}{2}\left(m(z)-x_{k}\right)^{2}+\mathrm{O}\left(\left|m(z)-x_{k}\right|^{3}\right)\quad\text{
for }\quad\left|z-a_{k}\right|\leqslant\tau.$

Due to similarity, we omit more details and refer the readers to Section A.2
of [35] or Lemma C.5 of [26] for more details.

Finally, we prove (5.4) following that of Lemma 8.5 of [43]. Using (5.6), we
obtain that

$E\operatorname{Im}m+\eta\operatorname{Re}m=\frac{\phi^{1/2}}{p}\sum_{i=1}^{p}\frac{\sigma_{i}\operatorname{Im}m}{|1+\phi^{-1/2}m\sigma_{i}|^{2}},$

and

$E\operatorname{Re}m-\eta\operatorname{Im}m=\phi-1-\frac{\phi}{p}\sum_{i=1}^{p}\frac{1+\phi^{-1/2}\sigma_{i}\operatorname{Im}m}{|1+\phi^{-1/2}m\sigma_{i}|^{2}}.$

This yields that

$\left|z-\frac{\phi^{1/2}}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{|1+\phi^{-1/2}m\sigma_{i}|^{2}}\right|=\left|\mathrm{i}\eta-\frac{\operatorname{Re}m}{\operatorname{Im}m}\eta\right|=\frac{|m|\eta}{\operatorname{Im}m}.$

Together with (5.2) and (5.3), we obtain an uppper bound $C\sqrt{\kappa+\eta}$
for the above term. Note that

$\displaystyle\left|\frac{1}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{|1+\phi^{-1/2}m\sigma_{i}|^{2}}-\frac{1}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{(1+\phi^{-1/2}m\sigma_{i})^{2}}\right|$
$\displaystyle=$ $\displaystyle
2\left|\frac{1}{p}\sum_{i=1}^{p}\frac{\sigma_{i}[(\operatorname{Im}(1+\phi^{-1/2}m\sigma_{i}))^{2}+\mathrm{i}(\operatorname{Im}(1+\phi^{-1/2}m\sigma_{i}))(\operatorname{Re}(1+\phi^{-1/2}m\sigma_{i}))]}{|1+\phi^{-1/2}m\sigma|^{4}}\right|\leq
C\phi^{-1/2}\sqrt{\kappa+\eta}.$

Combining the above bounds and (5.7), we get the upper bound
$C\sqrt{\kappa+\eta}$ for $|\frac{m}{m^{\prime}}|$, it remains to derive a
lower bound. When $E\in[\gamma_{-},\gamma_{+}]$, we have

$\displaystyle\left|z-\frac{\phi^{1/2}}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{(1+\phi^{-1/2}m\sigma_{i})^{2}}\right|\geq\left|\operatorname{Im}\left(z-\frac{\phi^{1/2}}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{(1+\phi^{-1/2}m\sigma_{i})^{2}}\right)\right|$
$\displaystyle\geq$
$\displaystyle\left|\eta+\frac{2}{p}\sum_{i=1}^{p}\frac{\sigma_{i}(1+\phi^{-1/2}\operatorname{Re}m\sigma_{i})\sigma_{i}\operatorname{Im}m}{|1+\phi^{-1/2}m\sigma_{i}|^{4}}\right|\geq\frac{2}{p}\sum_{i=1}^{p}\frac{\sigma_{i}^{2}(1+\phi^{-1/2}\operatorname{Re}m\sigma_{i})\operatorname{Im}m}{|1+\phi^{-1/2}m\sigma_{i}|^{4}}\geq
C{\sqrt{\kappa+\eta}}.$

When $E\notin[\gamma_{-},\gamma_{+}]$, we have

$\displaystyle\left|z-\frac{\phi^{1/2}}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{(1+\phi^{-1/2}m\sigma_{i})^{2}}\right|\geq\left||z|-\frac{\phi^{1/2}}{p}\sum_{i=1}^{p}\frac{\sigma_{i}}{|1+\phi^{-1/2}m\sigma_{i}|^{2}}\right|$
$\displaystyle=$
$\displaystyle\left|\sqrt{E^{2}+\eta^{2}}-E-\eta\frac{\operatorname{Re}m}{\operatorname{Im}m}\right|=\left|\eta\frac{\operatorname{Re}m}{\operatorname{Im}m}-\frac{\eta}{E+\sqrt{E^{2}+\eta^{2}}}\right|\geq
C\sqrt{\kappa+\eta}.$

This completes our proof.

###  B.3 Proof of (7.7)

Let $z=E+\mathrm{i}\eta$ and $E\in\mathbf{R}.$ First, when
$\eta\geq\eta_{c}:=n^{-1+\tau},$ we see that the result follows from Theorem
5.5. Then we handle the case when $\eta\leq\eta_{c}.$ Denote
$z_{c}:=E+\mathrm{i}{\eta}_{c}$. Since
$\operatorname{Tr}R_{1}(z)=(p-n)/z+\operatorname{Tr}R_{2}(z),$ it is
equivalent to establishing a bound for $\operatorname{Tr}R_{2}(z)$. In view of
(A.2), it suffices to study
$(1-\mathbb{E})\sum_{\mu\in\mathcal{I}_{2}}G_{\mu\mu}$. Using (A.6), Lemma 5.1
and Theorem 5.5, we have that

$\displaystyle\left|{G}_{\mu\mu}(z)-{G}_{\mu\mu}\left(z_{c}\right)\right|\lesssim\sum_{j=1}^{n}\frac{{\eta}_{c}\left|\left\langle\mathbf{e}_{\mu},\bm{\zeta}_{j}\right\rangle\right|\left|\left\langle\mathbf{e}_{\mu},\bm{\zeta}_{j}\right\rangle\right|}{\left|\left(\lambda_{j}-E-\mathrm{i}{\eta}\right)\left(\lambda_{j}-E-\mathrm{i}{\eta}_{c}\right)\right|}$
$\displaystyle\leqslant$
$\displaystyle{\eta}_{c}\left(\sum_{j=1}^{n}\frac{\left|\left\langle\mathbf{e}_{\mu},\bm{\zeta}_{j}\right\rangle\right|^{2}}{\left|\left(\lambda_{j}-E-\mathrm{i}{\eta}\right)\right|^{2}}\right)^{1/2}\left(\sum_{j=1}^{n}\frac{\left|\left\langle\mathbf{e}_{\mu},\bm{\zeta}_{j}\right\rangle\right|^{2}}{\left|\left(\lambda_{j}-E-\mathrm{i}{\eta}_{c}\right)\right|^{2}}\right)^{1/2}$
$\displaystyle\leqslant$
$\displaystyle{\eta}_{c}\left(\frac{{\eta}_{c}^{2}}{{\eta}^{2}}\sum_{j=1}^{n}\frac{\left|\left\langle\mathbf{e}_{\mu},\bm{\zeta}_{j}\right\rangle\right|^{2}}{\left|\left(\lambda_{j}-E-\mathrm{i}{\eta}_{c}\right)\right|^{2}}\right)^{1/2}\sqrt{\frac{\operatorname{Im}\left[G_{\mu\mu}\left(z_{c}\right)\right]}{{\eta}_{c}}}$
$\displaystyle=$
$\displaystyle\frac{{\eta}_{c}}{{\eta}}\sqrt{\operatorname{Im}\left[G_{\mu\mu}\left(z_{c}\right)\right]\operatorname{Im}\left[G_{\mu\mu}\left(z_{c}\right)\right]}\prec\frac{1}{n{\eta}}.$

We readily obtain that

$\displaystyle|(1-\mathbb{E})\operatorname{Tr}{R_{1}}(z)|=|(1-\mathbb{E})\operatorname{Tr}{R_{2}}(z)|$
$\displaystyle=$
$\displaystyle|(1-\mathbb{E})\operatorname{Tr}({R_{2}}(z)-{R_{2}}(z_{c}))+(1-\mathbb{E})\operatorname{Tr}({R_{2}}(z_{c}))|$
$\displaystyle\leq$
$\displaystyle\sum_{\mu\in\mathcal{I}_{2}}\left|{G}_{\mu\mu}(z)-{G}_{\mu\mu}\left(z_{c}\right)\right|+|(1-\mathbb{E})\operatorname{Tr}({R_{2}}(z_{c}))|\prec\frac{1}{{\eta_{c}}}+\frac{1}{|\operatorname{Im}(z)|}\prec\frac{1}{|\operatorname{Im}(z)|}.$

This completes our proof.

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