following for the Gaussian ensembles.

######  Proposition 5.12.

Let $G$ be a matrix from the Gaussian ensemble. Then for any $1\geq\eta\geq
N^{-2}$ and $\varepsilon>0$ and $x,y\in(-2+\kappa,2-\kappa)$ we have,

$N^{2}\left|\mathrm{Cov}_{G}(\mathrm{Im}[m_{N}(x+\mathrm{i}\eta)],\mathrm{Im}[m_{N}(y+\mathrm{i}\eta)])\right|\leq
CN^{\varepsilon}\frac{1}{\eta(|x-y|+\eta)}$ (5.79)

Moreover,

$N\left|\mathbb{E}^{G}[\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(x+\mathrm{i}\eta)]\right|\leq
C$ (5.80)

Proof. The first estimate follows from Theorem 3.6. The second follows from
writing the Poisson kernel as

$P_{\eta}(x)=\chi_{1}(x)P_{\eta}(x)+(1-\chi_{1}(x))P_{\eta}(x)=f_{1}(x)+f_{2}(x).$
(5.81)

where $\chi_{1}$ is a function that is $1$ on $(-2+\kappa/2,2-\kappa/2)$ and
$0$ outside of $(-2-\kappa/4,2-\kappa/4)$. From Theorem 2.14 we have

$\left|\mathbb{E}[\mathrm{tr}f_{1}(G)]-N\int
f_{1}(x)\rho_{\mathrm{sc}}(x)\right|\leq C.$ (5.82)

From Proposition 2.15 we have,

$\left|\mathbb{E}[\mathrm{tr}f_{2}(G)]-N\int
f_{2}(x)\rho_{\mathrm{sc}}(x)\right|\leq C.$ (5.83)

The claim follows. ∎

We can now conclude the main technical estimate of this section.

######  Theorem 5.13.

For any $\varepsilon>0$ the following holds. Let $\eta=2^{-k}$ and $k$
satisfy,

$\mathfrak{b}\log_{2}(N)\leq k\leq\log_{2}(N\mathfrak{b}^{-1}).$ (5.84)

We have

$\displaystyle\left|\mathbb{E}^{(\tilde{Z})}[N^{2}\mathbb{E}^{(H)}\big{[}\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(x+Z_{1}+\mathrm{i}\eta)]\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(y+Z_{2}+\mathrm{i}\eta)]\big{]}\right|$
$\displaystyle\leq$ $\displaystyle
CN^{\varepsilon}\left(\frac{1}{\eta(|x-y|+\eta)}+\frac{1}{\eta T}\right).$
(5.85)

Proof. Thanks to Proposition 5.9, it suffices to estimate the function
$F(z_{1},z_{2})$. For $|z_{1}|+|z_{2}|\leq N^{\varepsilon/1000}$ we have, by
Proposition 5.11,

$\displaystyle\left|N^{2}F(z_{1},z_{2})\right|$ $\displaystyle\leq
N^{2}\left|\mathrm{Cov}_{G}(\mathrm{Im}[m_{N}(x+N^{-1}z_{1}+\mathrm{i}\eta)],\mathrm{Im}[m_{N}(y+N^{-1}z_{2}+\mathrm{i}\eta)])\right|$
$\displaystyle+N^{2}\bigg{|}\mathbb{E}^{G}\big{[}\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(x+N^{-1}z_{1}+\mathrm{i}\eta)]\big{]}$
$\displaystyle\times\mathbb{E}^{G}\big{[}\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(y+N^{-1}z_{2}+\mathrm{i}\eta)]\big{]}\bigg{|}$
$\displaystyle+N^{\varepsilon}\frac{1}{\eta T}.$ (5.86)

The claim now follows from Proposition 5.12, as well as the estimates in
Proposition 5.7. ∎

Proof of Theorem 3.3. By (5) and Proposition 5.2 we have,

$\displaystyle\mathrm{Var}(\mathrm{tr}\varphi_{k}(H))$ $\displaystyle\leq$
$\displaystyle\int_{I_{\kappa/2}^{2}}\left|f_{k}(x)f_{k}(y)N^{2}\mathbb{E}^{(\tilde{Z})}\big{[}\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(x+Z_{1}+\mathrm{i}\eta)]\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(y+Z_{2}+\mathrm{i}\eta)]\big{]}\right|\mathrm{d}x\mathrm{d}y$
$\displaystyle+$ $\displaystyle CN^{-100}\|\varphi\|_{2}^{2}$ (5.87)

where

$I_{a}=(-2+a,2-a)$ (5.88)

and the support of $\varphi$ is in $(-2+\kappa,2-\kappa)$. For $x,y\in
I_{\kappa/2}$ we see from Theorem 5.13 that

$\displaystyle\int_{I_{\kappa/2}^{2}}\left|f_{k}(x)f_{k}(y)N^{2}\mathbb{E}^{(\tilde{Z})}[\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(x+Z_{1}+\mathrm{i}\eta)](\mathrm{Im}[(m_{N}-m_{\mathrm{sc}})(y+Z_{2}+\mathrm{i}\eta)]\right|\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle
N^{\varepsilon}\int_{I_{\kappa/2}^{2}}|f(x)f(y)|\left(\frac{1}{\eta(|x-y|+\eta)}+\frac{1}{\eta
T}\right)\mathrm{d}x\mathrm{d}y$ (5.89)

Clearly,

$\displaystyle\frac{1}{\eta
T}\int_{I_{\kappa/2}^{2}}|f_{k}(x)f_{k}(y)|\mathrm{d}x\mathrm{d}y\leq$
$\displaystyle\frac{1}{\eta
T}\int_{I_{\kappa/2}^{2}}|f_{k}(x)|^{2}+|f_{k}(y)|^{2}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle\frac{C}{\eta
T}\|f_{k}\|_{2}^{2}\leq\frac{C}{\eta T}\|\varphi_{k}\|_{2}^{2}$ (5.90)

where in the final inequality we used Proposition 5.1. Since,

$\int_{|x|<5}\frac{1}{\eta(|x-y|+\eta)}\mathrm{d}x\leq
C\frac{|\log(\eta)|}{\eta}$ (5.91)

for any $y$, we see that

$\displaystyle\int_{I_{\kappa/2}^{2}}\frac{|f_{k}(x)f_{k}(y)|}{(x-y)^{2}+\eta^{2}}\mathrm{d}x\mathrm{d}y\leq$
$\displaystyle\int_{I_{\kappa/2}^{2}}\frac{|f_{k}(x)|^{2}}{(x-y)^{2}+\eta^{2}}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq CN^{\varepsilon}\eta^{-1}\|f_{k}\|_{2}^{2}\leq
CN^{\varepsilon}\eta^{-1}\|\varphi_{k}\|_{2}^{2}.$ (5.92)

The result follows. ∎

##  6 Large frequencies

In this section we prove Theorems 3.1 and the variance and expectation
estimates of Theorem 3.2 via resolvent expansions. Fix $\tau>0$ and let us
define,

$\mathcal{D}=\mathcal{D}_{\tau,1}\cup\bar{\mathcal{D}}_{\tau,1},$ (6.1)

where $\mathcal{D}_{\tau,1}$ is defined in (2.9). Recall the definition of the
control parameter $\Psi(z)$ in (2.14).

In the rest of this section we will use the notation,

$z=x\pm\mathrm{i}y,\qquad w=u\pm\mathrm{i}v$ (6.2)

where $y,v>0$.

The main tool is the following self-consistent equation for expectations of
products of Stieltjes transforms.

######  Proposition 6.1.

For $H$ a real symmetric Wigner matrix and $z,w\in\mathcal{D}$ we have for any
$\varepsilon>0$,

$\displaystyle(z+2m_{\mathrm{sc}}(z))\mathbb{E}[m_{N}(z)(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$
$\displaystyle-\frac{2}{N^{2}}\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}+\frac{4s_{3}(m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)+m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}^{\prime}(w))}{N^{5/2}}$
$\displaystyle-$
$\displaystyle\frac{2s_{4}m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}}{N^{2}}$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-3}(y^{-2}v^{-1}+(y+v)^{-1}v^{-2}))$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-3/2}v^{-1}(\Psi(z)^{2}\Psi(w)+\Psi(z)\Psi(w)^{2})+N^{-2}\Psi(z)\Psi(w)v^{-1})$
(6.3)

The proof of this proposition is via cumulant expansions, and is given over
the course of Section 6.1.

###  6.1 Self-consistent equation for covariance

By the cumulant expansion we have for any $\varepsilon>0$ that,

$\displaystyle\frac{1}{N^{2}}\sum_{i,j}z\mathbb{E}[G_{ii}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{2}}\sum_{i,j,a}\mathbb{E}[H_{ia}G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{2}}\sum_{i,j,a}\sum_{k=2}^{7}\frac{(1+\delta_{ia})^{k-1}s_{k}}{(k-1)!N^{k/2}}\mathbb{E}[\partial_{ia}^{k-1}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)]))]$
$\displaystyle+$ $\displaystyle\mathcal{O}(N^{\varepsilon-3}).$ (6.4)

Here, the notation $\partial_{ia}$ means the derivative with respect to the
matrix entry $(i,a)$. Since we have symmetric matrices, note that this
involves the derivative both with respect to the entry $(i,a)$ and $(a,i)$.

For a treatment of the error term, see the discussion in [45] (the expansion
with error term appears there as Lemma 3.2); there is no substantial
difference from the treatment appearing there and here. We will refer to the
$k$th term in the sum (6.1) as the $k$th-order term. Each of these terms is
expanded and dealt with in each of the next few subsubsections. We will also
require expressions for derivatives of resolvent entries with respect to
matrix entries. These are collected in Appendix A.

####  6.1.1 Second order term

######  Lemma 6.2.

We have for any $\varepsilon>0$,

$\displaystyle\frac{1}{N^{3}}\sum_{ija}(1+\delta_{ia})\mathbb{E}[\partial_{ia}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)]))]$
$\displaystyle=$
$\displaystyle-2m_{\mathrm{sc}}(z)\mathbb{E}[(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
(6.5) $\displaystyle-$
$\displaystyle\frac{2}{N^{2}}\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}+\mathcal{O}(N^{\varepsilon-3}(y^{-2}v^{-1}+(y+v)^{-1}v^{-2})$
(6.6)

Proof. We have, using (A.2) and (A.3),

$\displaystyle\frac{1}{N^{3}}\sum_{ija}(1+\delta_{ia})\mathbb{E}[\partial_{ia}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)]))]$
$\displaystyle=$
$\displaystyle-\frac{1}{N^{3}}\sum_{ija}\mathbb{E}[G_{ii}(z)G_{aa}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
(6.7) $\displaystyle-$
$\displaystyle\frac{1}{N^{3}}\sum_{ija}\mathbb{E}[G_{ia}(z)^{2}(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
(6.8) $\displaystyle-$
$\displaystyle\frac{2}{N^{3}}\sum_{ija}\mathbb{E}[G_{ia}(z)G_{ja}(w)G_{ji}(w)]$
(6.9)

The first term (6.7) equals,

$\displaystyle-\frac{1}{N^{3}}\sum_{ija}\mathbb{E}[G_{ii}(z)G_{aa}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$
$\displaystyle-\mathbb{E}[m_{N}(z)^{2}(m_{N}(w)-\mathbb{E}[m_{N}(w))]$
$\displaystyle=$
$\displaystyle-2\mathbb{E}[m_{N}(z)]\mathbb{E}[(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w)])]+\mathcal{O}(N^{\varepsilon-3}y^{-2}v^{-1})$
$\displaystyle=$
$\displaystyle-2m_{\mathrm{sc}}(z)\mathbb{E}[(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w)])]+\mathcal{O}(N^{\varepsilon-3}y^{-2}v^{-1}).$
(6.10)

In the second to last line we wrote
$m_{N}(z)=(m_{N}(z)-\mathbb{E}[m_{N}(z)])+\mathbb{E}[m_{N}(z)]$, expanded the
square and used the local law (2.11). In the last line we used again the local
law (2.11).

The term (6.8) equals,

$\displaystyle-\frac{1}{N^{3}}\sum_{ija}\mathbb{E}[G_{ia}(z)^{2}(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]=-\frac{1}{N}\mathbb{E}[\partial_{z}m_{N}(z)(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$
$\displaystyle-\frac{1}{N}\mathbb{E}[\partial_{z}(m_{N}(z)-m_{\mathrm{sc}}(z))(m_{N}(w)-\mathbb{E}[m_{N}(w)])]=\mathcal{O}(N^{\varepsilon}N^{-3}y^{-2}v^{-1})$
(6.11)

where we used the local law (2.11) and the Cauchy integral formula to obtain
the last estimate. For the term (6.9) we have,

$\displaystyle-$
$\displaystyle\frac{2}{N^{3}}\sum_{ija}\mathbb{E}[G_{ia}(z)G_{ja}(w)G_{ji}(w)]=-2\partial_{w}\frac{1}{N^{3}}\mathbb{E}[\mathrm{tr}G(z)G(w)].$
(6.12)

We have the identity,

$\frac{1}{N}\mathrm{tr}G(z)G(w)=\frac{m_{N}(z)-m_{N}(w)}{z-w}.$ (6.13)

If $z$ and $w$ are in opposite half-planes, then since $|z-w|\geq y+v$ we have
by the local law (2.11) with overwhelming probability,

$\displaystyle\partial_{w}\frac{m_{N}(z)-m_{N}(w)}{z-w}=\frac{m_{N}(z)-m_{N}(w)}{(z-w)^{2}}-\frac{\partial_{w}m_{N}(w)}{(z-w)}$
$\displaystyle=$
$\displaystyle\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{(z-w)^{2}}-\frac{\partial_{w}m_{\mathrm{sc}}(w)}{z-w}+\mathcal{O}(N^{\varepsilon-1}(y^{-2}v^{-1}+(y+v)^{-1}v^{-2})$
(6.14)

and the main term in the last line equals
$\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}$. If $z$ and
$w$ are in the same half-plane we consider first the case
$10|z-w|\geq\max\\{y,v\\}$. In this case we use the same method as above to
derive the same estimate. In the case that $10|z-w|\leq\max\\{y,v\\}$ we see
that $v\leq 2y\leq 4v$. We have the identity,

$\frac{m_{N}(z)-m_{N}(w)}{z-w}=-\int_{0}^{1}m_{N}^{\prime}(z+s(w-z))\mathrm{d}s.$
(6.15)

For the integrand we have,
$m_{N}^{\prime}(z+s(w-z))=m_{\mathrm{sc}}^{\prime}(z+s(w-z))+\mathcal{O}(N^{\varepsilon-1}y^{-2})$
by the local law with overwhelming probability. Summarizing for (6.9) we have
derived that with overwhelming probability,

$\displaystyle-\frac{2}{N^{3}}\sum_{ija}G_{ia}(z)G_{ja}(w)G_{ji}(w)=$
$\displaystyle-2\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}$
$\displaystyle+$
$\displaystyle\mathcal{O}(N^{\varepsilon-3}(y^{-2}v^{-1}+(y+v)^{-1}v^{-2})).$
(6.16)

This yields the claim. ∎

####  6.1.2 Third order term

In the proof below, where possible, we will simplify errors that arise using

$\frac{1}{N}\leq\Psi(E\pm\mathrm{i}\eta)\leq C\frac{1}{(N\eta)^{1/2}}$ (6.17)

for $E\pm\mathrm{i}\eta\in\mathcal{D}$, to absorb them into error terms that
have arisen already in the proof.

######  Lemma 6.3.

We have,

$\displaystyle\frac{1}{N^{3+1/2}}\sum_{ija}(1+\delta_{ia})^{2}\mathbb{E}[\partial_{ia}^{2}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)]))]$
$\displaystyle=$ $\displaystyle
8\frac{m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)+m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}^{\prime}(w)}{N^{5/2}}$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-3}y^{-2}v^{-1}+N^{-3}v^{-2}(y+v)^{-1})$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-3/2}v^{-1}(\Psi(z)^{2}\Psi(w)+\Psi(z)\Psi(w)^{2})+N^{-2}\Psi(z)\Psi(w)v^{-1}).$
(6.18)

Proof. We have, using (A.4), (A.5) and (A.6),

$\displaystyle\frac{1}{N^{3+1/2}}\sum_{ija}(1+\delta_{ia})^{2}\mathbb{E}[\partial_{ia}^{2}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)]))]$
$\displaystyle=$
$\displaystyle\frac{2}{N^{2+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)^{3}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
(6.19) $\displaystyle+$
$\displaystyle\frac{6}{N^{2+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ii}(z)G_{aa}(z)(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
(6.20) $\displaystyle+$
$\displaystyle\frac{4}{N^{3+1/2}}\sum_{ia}\mathbb{E}[G_{ii}(z)G_{aa}(z)\partial_{w}G_{ia}(w)]$
(6.21) $\displaystyle+$
$\displaystyle\frac{4}{N^{3+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)^{2}\partial_{w}G_{ia}(w)]$
(6.22) $\displaystyle+$
$\displaystyle\frac{4}{N^{3+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ai}(w)\partial_{w}G_{ai}(w)]$
(6.23) $\displaystyle+$
$\displaystyle\frac{4}{N^{3+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ii}(w)\partial_{w}G_{aa}(w)].$
(6.24)

The first term (6.19) is,

$\displaystyle\frac{2}{N^{2+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)^{3}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$
$\displaystyle\frac{2}{N^{2+1/2}}\sum_{i}\mathbb{E}[(G_{ii}(z)^{3}-m_{\mathrm{sc}}(z)^{3})(m_{N}(w)-\mathbb{E}[m_{N}(w)])]+\mathcal{O}(N^{\varepsilon}N^{-3/2}\Psi(z)^{3}v^{-1})$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-7/2}y^{-1}v^{-1}+N^{-3/2}\Psi(z)^{3}v^{-1})$
(6.25)

In the second line we used (2.12) to estimate the off-diagonal Green’s
function elements and also applied the local law (2.11). In the last line we
expanded out $(G_{ii}-m_{\mathrm{sc}}+m_{\mathrm{sc}})^{3}$ and used the local
law (2.11) to estimate the term linear in $G_{ii}$ and then the entry-wise law
(2.12) to estimate the higher powers.

For the term (6.20) we first introduce the notation,

$\mathbf{e}=N^{-1/2}(1,\dots,1)^{T}\in\mathbb{R}^{N},$ (6.26)

i.e., the constant unit vector in $\mathbb{R}^{N}$. Then, writing
$G_{ii}=(G_{ii}-m_{\mathrm{sc}})+m_{\mathrm{sc}}$ and expanding (and doing the
same for $G_{aa}$) we find for (6.20),

$\displaystyle\frac{1}{N^{2+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ii}(z)G_{aa}(z)(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{3/2}}m_{\mathrm{sc}}(z)^{2}\mathbb{E}[\mathbf{e}^{T}G(z)\mathbf{e}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle+$
$\displaystyle\frac{2m_{\mathrm{sc}}(z)}{N^{2}}\sum_{i}\mathbb{E}[(G_{ii}(z)-m_{\mathrm{sc}}(z))(G(z)\mathbf{e})_{i}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle+$
$\displaystyle\mathcal{O}(N^{\varepsilon}N^{-3/2}\Psi(z)^{3}v^{-1}).$ (6.27)

By the isotropic local law (2.13) we see

$\frac{1}{N^{3/2}}m_{\mathrm{sc}}(z)^{2}\mathbb{E}[\mathbf{e}^{T}G(z)\mathbf{e}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]=N^{\varepsilon}\mathcal{O}(N^{-5/2}\Psi(z)v^{-1}).$
(6.28)

By the isotropic local law (2.13) we see with overwhelming probability that,

$(G(z)\mathbf{e})_{i}=N^{-1/2}m_{\mathrm{sc}}(z)+N^{\varepsilon}\mathcal{O}(\Psi(z))$
(6.29)

and so,

$\displaystyle\frac{2m_{\mathrm{sc}}(z)}{N^{2}}\sum_{i}\mathbb{E}[(G_{ii}(z)-m_{\mathrm{sc}}(z))(G(z)\mathbf{e})_{i}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$
$\displaystyle\frac{2m_{\mathrm{sc}}(z)^{2}}{N^{3/2}}\mathbb{E}[(m_{N}(z)-m_{\mathrm{sc}}(z))(m_{N}(w)-\mathbb{E}[m_{N}(w)])]+N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)^{2}v^{-1})$
$\displaystyle=$
$\displaystyle\mathcal{O}(N^{-7/2}y^{-1}v^{-1}+N^{-2}\Psi(z)^{2}v^{-1}).$
(6.30)

This completes estimating the term (6.20). We write (6.21) as,

$\displaystyle\frac{1}{N^{3+1/2}}\sum_{ia}\mathbb{E}[G_{ii}(z)G_{aa}(z)\partial_{w}G_{ia}(w)]$
$\displaystyle=$
$\displaystyle\frac{m_{\mathrm{sc}}(z)^{2}}{N^{5/2}}\mathbb{E}[\partial_{w}\mathbf{e}^{T}G(w)\mathbf{e}]+\frac{2m_{\mathrm{sc}}(z)}{N^{3}}\sum_{i}\mathbb{E}[(G_{ii}(z)-m_{\mathrm{sc}}(z))\partial_{w}(G(w)\mathbf{e})_{i}]$
$\displaystyle+$
$\displaystyle\frac{1}{N^{7/2}}\sum_{ia}\mathbb{E}[(G_{ii}(z)-m_{\mathrm{sc}}(z))(G_{aa}(z)-m_{\mathrm{sc}}(z))\partial_{w}G_{ia}(w)].$
(6.31)

By the isotropic local law (2.13) and the Cauchy integral formula, we have

$\frac{m_{\mathrm{sc}}(z)^{2}}{N^{5/2}}\mathbb{E}[\partial_{w}\mathbf{e}^{T}G(w)\mathbf{e}]=\frac{m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)}{N^{5/2}}+N^{\varepsilon}\mathcal{O}(N^{-5/2}\Psi(w)v^{-1}).$
(6.32)

By (6.29) with $z$ replaced by $w$ and the entry-wise local law (2.12) we
have,

$\frac{2m_{\mathrm{sc}}(z)}{N^{3}}\sum_{i}\mathbb{E}[(G_{ii}(z)-m_{\mathrm{sc}}(z))\partial_{w}(G(w)\mathbf{e})_{i}]=N^{\varepsilon}\mathcal{O}(N^{-3}y^{-1}v^{-1/2}+N^{-2}\Psi(z)\Psi(w)v^{-1}),$
(6.33)

where we used a crude bound $|m_{\mathrm{sc}}^{\prime}(w)|\leq Cv^{-1/2}$.
Finally, after separating into diagonal and off-diagonal terms we find,

$\displaystyle\frac{1}{N^{7/2}}\sum_{ia}\mathbb{E}[(G_{ii}(z)-m_{\mathrm{sc}}(z))(G_{aa}(z)-m_{\mathrm{sc}}(z))\partial_{w}G_{ia}(w)]$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-3/2}\Psi(z)^{2}\Psi(w)v^{-1}+N^{-5/2}\Psi(z)^{2}(\Psi(w)v^{-1}+v^{-1/2})),$
(6.34)

again using a crude bound $|m_{\mathrm{sc}}^{\prime}(w)|\leq Cv^{-1/2}$. This
completes (6.21). We now turn to (6.22). Clearly,

$\displaystyle\frac{4}{N^{3+1/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)^{2}\partial_{w}G_{ia}(w)]=\frac{4}{N^{7/2}}\sum_{i}\mathbb{E}[G_{ii}(z)^{2}\partial_{w}G_{ii}(w)]+N^{\varepsilon}\mathcal{O}(N^{-3/2}\Psi(z)^{2}\Psi(w)v^{-1}),$
(6.35)

and

$\displaystyle\frac{1}{N^{7/2}}\sum_{i}\mathbb{E}[G_{ii}(z)^{2}\partial_{w}G_{ii}(w)]$
$\displaystyle=$
$\displaystyle\frac{m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)}{N^{5/2}}+N^{\varepsilon}\mathcal{O}(N^{-7/2}(y^{-1}v^{-1/2}+v^{-2})+N^{-5/2}(\Psi(z)\Psi(w)v^{-1}),$
(6.36)

by expanding $G_{ii}(z)=(G_{ii}(z)-m_{\mathrm{sc}}(z))+m_{\mathrm{sc}}(z)$
(and similarly for $G_{ii}(w)$), using the local law (2.11) to estimate the
terms linear in $G_{ii}$, and the entrywise local law (2.12) for the terms
quadratic and cubic in $G_{ii}$. We also used the crude bound
$|m_{\mathrm{sc}}^{\prime}(w)|\leq v^{-1/2}$.

This completes the treatment of (6.22). For (6.23), the diagonal terms are
handled similarly to (6.1.2) and we get

$\displaystyle\frac{1}{N^{7/2}}\sum_{i}\mathbb{E}[G_{ii}(z)G_{ii}(w)\partial_{w}G_{ii}(w)]$
$\displaystyle=$
$\displaystyle\frac{m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}^{\prime}(w)}{N^{5/2}}+N^{\varepsilon}\mathcal{O}(N^{-7/2}(y^{-1}v^{-1/2}+v^{-2})+N^{-5/2}\Psi(z)\Psi(w)v^{-1}).$
(6.37)

The off-diagonal terms are
$N^{\varepsilon}\mathcal{O}(N^{-3/2}\Psi(z)\Psi(w)^{2}v^{-1})$. We write
(6.24) as,

$\displaystyle\frac{1}{N^{7/2}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ii}(w)\partial_{w}G_{aa}(w)]$
$\displaystyle=$
$\displaystyle\frac{m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)}{N^{5/2}}\mathbb{E}[\mathbf{e}^{T}G(z)\mathbf{e}]$
$\displaystyle+$
$\displaystyle\frac{m_{\mathrm{sc}}^{\prime}(w)}{N^{3}}\sum_{i}\mathbb{E}[(G(z)\mathbf{e})_{i}(G_{ii}(z)-m_{\mathrm{sc}}(z))]+\frac{m_{\mathrm{sc}}(z)}{N^{3}}\mathbb{E}[(G(z)\mathbf{e})_{i}\partial_{w}(G_{ii}(w)-m_{\mathrm{sc}}(w))]$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-3/2}v^{-1}\Psi(w)^{2}\Psi(z)).$ (6.38)

By the isotropic local law (2.13) the first term is,

$\frac{m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)}{N^{5/2}}\mathbb{E}[\mathbf{e}^{T}G(z)\mathbf{e}]=\frac{m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}{N^{5/2}}+N^{\varepsilon}\mathcal{O}(N^{-5/2}\Psi(z)v^{-1/2}).$
(6.39)

Using (6.29) as well as the local law (2.11) we obtain easily that,

$\displaystyle\frac{m_{\mathrm{sc}}^{\prime}(w)}{N^{3}}\sum_{i}\mathbb{E}[(G(z)\mathbf{e})_{i}(G_{ii}(z)-m_{\mathrm{sc}}(z))]+\sum_{i}\frac{m_{\mathrm{sc}}(z)}{N^{3}}\mathbb{E}[(G(z)\mathbf{e})_{i}\partial_{w}(G_{ii}(w)-m_{\mathrm{sc}}(w))]$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-7/2}y^{-1}v^{-1/2}+N^{-2}\Psi(z)^{2}v^{-1/2}+N^{-7/2}v^{-2}+N^{-2}\Psi(z)\Psi(w)v^{-1})$
(6.40)

Adding up all the errors and simplifying yields the claim. ∎

####  6.1.3 Fourth order term

######  Lemma 6.4.

We have

$\displaystyle\frac{1}{N^{4}}\sum_{ija}(1+\delta_{ia})^{3}\mathbb{E}[\partial_{ia}^{3}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)]))]$
$\displaystyle=$
$\displaystyle-\frac{12m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}}{N^{2}}+N^{\varepsilon}\mathcal{O}(N^{-3}(v^{-1}y^{-1}+v^{-2})+N^{-2}\Psi(z)\Psi(w)v^{-1})$
(6.41)

Proof. We break up the left side of (6.4) into the four terms based on how the
three derivatives are distributed amongst the $G_{ia}(z)$ and $G_{jj}(w)$. We
first calculate, using (A.7),

$\displaystyle-\frac{1}{N^{3}}\sum_{ia}(1+\delta_{ia})^{3}\mathbb{E}[(\partial_{ia}^{3}G_{ai}(z))(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{3}}\sum_{ia}\mathbb{E}[(6G_{ia}^{4}(z)+36G_{ia}(z)^{2}G_{ii}(z)G_{aa}(z))(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle+$
$\displaystyle\frac{6}{N^{3}}\sum_{ia}\mathbb{E}[G_{ii}^{2}(z)G_{aa}(z)^{2}(m_{N}(w)-\mathbb{E}[m_{N}(w)])].$
(6.42)

Clearly for the second line,

$\displaystyle\frac{1}{N^{3}}\sum_{ia}\mathbb{E}[(6G_{ia}^{4}(z)+36G_{ia}(z)^{2}G_{ii}(z)G_{aa}(z))(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-2}v^{-1}\Psi(z)^{2}+N^{-3}v^{-1}).$ (6.43)

For the last term of (6.42) we have,

$\frac{1}{N^{3}}\sum_{ia}\mathbb{E}[G_{ii}^{2}(z)G_{aa}(z)^{2}(m_{N}(w)-\mathbb{E}[m_{N}(w)])]=N^{\varepsilon}\mathcal{O}(N^{-3}y^{-1}v^{-1})$
(6.44)

where we again expanded the resolvent entries around $m_{\mathrm{sc}}$ and
used the local law (2.11) to estimate the terms linear in $G_{ii}$ or
$G_{aa}$.

The next term is, using (A.4) and (A.3),

$\displaystyle-$
$\displaystyle\frac{1}{N^{4}}\sum_{ija}(1+\delta_{ia})^{3}\mathbb{E}[(\partial_{ia}^{2}G_{ia}(z))(\partial_{ia}G_{jj}(w))]$
$\displaystyle=$
$\displaystyle\frac{4}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(z)^{3}]$
$\displaystyle+$
$\displaystyle\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ii}(z)G_{aa}(z)G_{ia}(z)]$
(6.45)

For the first term, we have

$\displaystyle\frac{1}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(z)^{3}]=N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)^{3}\Psi(w)v^{-1}+N^{-3}v^{-1})$
(6.46)

where we used a crude bound $|\partial_{w}G_{ii}(w)|\leq Cv^{-1}$ for the
diagonal entries, which holds with overwhelming probability. For the term on
the last line of (6.1.3) we have by the same argument,

$\frac{1}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ii}(z)G_{aa}(z)G_{ia}(z)]=N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)\Psi(w)v^{-1}+N^{-3}v^{-1}).$
(6.47)

For the next term contributing to the LHS of (6.4) we have, using (A.2) and
(A.6),

$\displaystyle\frac{3}{N^{4}}\sum_{ija}(1+\delta_{ia})^{3}\mathbb{E}[(\partial_{ia}G_{ia}(z))(\partial_{ia}^{2}G_{jj}(w))]$
$\displaystyle=$
$\displaystyle-\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ii}(w))G_{aa}(w)G_{ii}(z)G_{aa}(z)]$
$\displaystyle-$
$\displaystyle\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ii}(w))G_{aa}(w)G_{ia}(z)^{2}]$
$\displaystyle-$
$\displaystyle\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(w)G_{ii}(z)G_{aa}(z)]$
$\displaystyle-$
$\displaystyle\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(w)G_{ia}(z)^{2}]$
(6.48)

We begin with the first term on the right side of (6.1.3). By expanding each
resolvent entry around $m_{\mathrm{sc}}(z)$ or $m_{\mathrm{sc}}(w)$ and using
the local law (2.11) to estimate the terms that are linear in the $G_{ii}$ and
the entrywise law (2.12) to estimate the terms quadratic or higher in the
$G_{ii}$ we obtain,

$\displaystyle-\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ii}(w))G_{aa}(w)G_{ii}(z)G_{aa}(z)]$
$\displaystyle=$
$\displaystyle-\frac{12m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}}{N^{2}}+N^{\varepsilon}\mathcal{O}(N^{-3}(y^{-1}v^{-1/2}+v^{-2})+N^{-2}(v^{-1}\Psi(w)\Psi(z))).$
(6.49)

For the second term on the right side of (6.1.3) we have,

$\frac{1}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ii}(w))G_{aa}(w)G_{ia}(z)^{2}]=N^{\varepsilon}\mathcal{O}(N^{-3}y^{-1}v^{-1}).$
(6.50)

The third term is,

$\displaystyle\frac{1}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(w)G_{ii}(z)G_{aa}(z)]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{3}}m_{\mathrm{sc}}(z)^{2}\mathbb{E}[\partial_{w}^{2}m_{N}(w)]+\frac{2m_{\mathrm{sc}}(z)}{N^{4}}\sum_{i}\mathbb{E}[(\partial_{w}^{2}G_{ii}(w))(G_{ii}(z)-m_{\mathrm{sc}}(z))]$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)^{2}\Psi(w)^{2}v^{-1}+N^{-3}v^{-1}).$
(6.51)

Clearly,

$\frac{1}{N^{3}}m_{\mathrm{sc}}(z)^{2}\mathbb{E}[\partial_{w}^{2}m_{N}(w)]=N^{\varepsilon}\mathcal{O}(N^{-3}v^{-2}),$
(6.52)

and

$\displaystyle\frac{1}{N^{4}}\sum_{i}\mathbb{E}[(\partial_{w}^{2}G_{ii}(w))(G_{ii}(z)-m_{\mathrm{sc}}(z))]=N^{\varepsilon}\mathcal{O}(N^{-3}v^{-2}).$
(6.53)

The last term on the right side of (6.1.3) is,

$\displaystyle\frac{1}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(w)G_{ia}(z)^{2}]=N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)^{2}\Psi(w)^{2}v^{-1}+N^{-3}v^{-1})$
(6.54)

The final contribution to the left side of (6.4) is, using (A.8),

$\displaystyle-$
$\displaystyle\frac{1}{N^{4}}\sum_{ija}\mathbb{E}[G_{ia}(z)(\partial_{ia}^{3}G_{jj}(w))]$
$\displaystyle=$
$\displaystyle\frac{24}{N^{4}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ai}(w)G_{ii}(w)\partial_{w}G_{aa}(w)]$
$\displaystyle+$
$\displaystyle\frac{12}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(z)(G_{aa}(w)G_{ii}(w)+G_{ia}(w)^{2})].$
(6.55)

By splitting into off-diagonal and diagonal terms, we see that the first term
is

$\frac{24}{N^{4}}\sum_{ia}\mathbb{E}[G_{ia}(z)G_{ai}(w)G_{ii}(w)\partial_{w}G_{aa}(w)]=N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)\Psi(w)v^{-1}+N^{-3}v^{-1})$
(6.56)

and the second term is

$\frac{1}{N^{4}}\sum_{ia}\mathbb{E}[(\partial_{w}G_{ia}(w))G_{ia}(z)(G_{aa}(w)G_{ii}(w)+G_{ia}(w)^{2})]=N^{\varepsilon}\mathcal{O}(N^{-2}\Psi(z)\Psi(w)v^{-1}+N^{-3}v^{-1}).$
(6.57)

This completes the proof. ∎

####  6.1.4 Fifth order

######  Lemma 6.5.

We have,

$\displaystyle\frac{1}{N^{9/2}}\sum_{ija}(1+\delta_{ia})^{4}\mathbb{E}[\partial_{ia}^{4}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-5/2}\Psi(z)v^{-1}+N^{-5/2}\Psi(w)v^{-1/2}).$
(6.58)

Proof. We write the term as,

$\displaystyle\frac{1}{N^{9/2}}\sum_{ija}(1+\delta_{ia})^{4}\mathbb{E}[\partial_{ia}^{4}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{7/2}}\sum_{ia}(1+\delta_{ia})^{4}\mathbb{E}[(\partial^{4}_{ia}G_{ai}(z)(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle-$
$\displaystyle\frac{1}{N^{9/2}}\sum_{n=0}^{3}c_{n}\sum_{ia}(1+\delta_{ia})^{4}\mathbb{E}[(\partial^{3-n}_{ia}G_{ai}(z))(\partial_{w}\partial_{ia}^{n}G_{ia}(w))]$
(6.59)

for some combinatorial constants $c_{n}$. Consider the first term on the right
side of (6.1.4). The expression $(1+\delta_{ia})^{4}\partial_{ia}^{4}G_{ia}$
is a linear combination of products of five resolvent entries. Among five
resolvent entries there are 10 indices, 5 of which must be $i$ and 5 of which
must be $a$. Therefore, each term contains at least one off-diagonal entry
$G_{ia}$. Therefore, with overwhelming probability we have,

$|\partial_{ia}^{4}G_{ia}|\leq C\delta_{ia}+N^{\varepsilon}\Psi(z).$ (6.60)

It follows that the second line of (6.1.4) is
$\mathcal{O}(N^{-5/2}\Psi(z)v^{-1})$. By the same parity argument, we find
that for any derivative of even order $n=2m$ there exists $C$ such that

$|\partial_{ia}^{2m}G_{ia}|\leq C\delta_{ia}+N^{\varepsilon}\Psi(z),$ (6.61)

with overwhelming probability. Therefore, for any $n=0,1,2,3$ we have

$|\partial^{3-n}_{ia}G_{ai}(z))(\partial_{w}\partial_{ia}^{n}G_{ia}(w)|\leq
N^{\varepsilon}v^{-1}(\delta_{ia}+\Psi(z)+\Psi(w)),$ (6.62)

with overwhelming probability. It follows that the last line of (6.1.4) is
$N^{\varepsilon}\mathcal{O}(N^{-5/2}\Psi(z)v^{-1}+N^{-5/2}\Psi(w)v^{-1/2})$.
This yields the claim. ∎

####  6.1.5 Sixth and seventh order

######  Lemma 6.6.

We have,

$\displaystyle\frac{1}{N^{5}}\sum_{ija}(1+\delta_{ia})^{5}\mathbb{E}[\partial_{ia}^{5}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$ $\displaystyle N^{\varepsilon}\mathcal{O}(N^{-3}v^{-1}).$
(6.63)

and

$\displaystyle\frac{1}{N^{11/2}}\sum_{ija}(1+\delta_{ia})^{6}\mathbb{E}[\partial_{ia}^{6}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$ $\displaystyle N^{\varepsilon}\mathcal{O}(N^{-3}v^{-1}).$
(6.64)

Proof. We consider,

$\displaystyle\frac{1}{N^{5}}\sum_{ija}(1+\delta_{ia})^{5}\mathbb{E}[\partial_{ia}^{5}(G_{ai}(z)(G_{jj}(w)-\mathbb{E}[G_{jj}(w)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{4}}\sum_{ia}(1+\delta_{ia})^{5}\mathbb{E}[(\partial^{5}_{ia}G_{ai}(z)(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle-$
$\displaystyle\frac{1}{N^{5}}\sum_{n=0}^{4}c_{n}\sum_{ia}(1+\delta_{ia})^{5}\mathbb{E}[(\partial^{4-n}_{ia}G_{ai}(z))(\partial_{w}\partial_{ia}^{n}G_{ia}(w))]$
(6.65)

for some combinatorial constants $c_{n}$. We easily conclude the proof from,
the fact that for any $n$ we have

$|\partial_{ia}^{k}G_{ia}|\leq
C,\qquad|\partial_{ia}^{k}\partial_{w}G_{ia}(w)|\leq Cv^{-1},$ (6.66)

with overwhelming probability. The result for the seventh order term is proven
in the same way. ∎

####  6.1.6 Proof of Proposition 6.1

This follows from the expansion (6.1); in each of the previous few
subsections, the lemmas provide the required estimates for each of the terms
in the expansion. ∎

###  6.2 Covariance estimates

Recall the notation,

$\kappa(E)=||E|-2|.$ (6.67)

For $z=x+\mathrm{i}y$ we define $\kappa(z)=\kappa(x)$. Since

$|z+2m_{\mathrm{sc}}(z)|\asymp\sqrt{\kappa+y},$ (6.68)

we see from Proposition 6.1 and Lemma B.2 that,

$\displaystyle
N^{2}\mathbb{E}[(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w)])]$
$\displaystyle=$ $\displaystyle
2\frac{m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}^{\prime}(w)}{(1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w))^{2}}-4s_{3}N^{-1/2}m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}^{\prime}(w)(m_{\mathrm{sc}}(z)+m_{\mathrm{sc}}(w))$
$\displaystyle+$ $\displaystyle
2s_{4}m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}(z)$
$\displaystyle+$
$\displaystyle\frac{N^{\varepsilon}}{\sqrt{\kappa(x)+y}}\mathcal{O}(N^{-3}(y^{-2}v^{-1}+(y+v)^{-1}v^{-2}))$
$\displaystyle+$
$\displaystyle\frac{N^{\varepsilon}}{\sqrt{\kappa(x)+y}}\mathcal{O}(N^{-3/2}v^{-1}(\Psi(z)^{2}\Psi(w)+\Psi(z)\Psi(w)^{2})+N^{-2}\Psi(z)\Psi(w)v^{-1})$
(6.69)

From this, we conclude the following proof.

Proof of Theorem 3.1. We apply (6.2) assuming that
$\sqrt{\kappa(x)+y}\geq\sqrt{\kappa(u)+v}$. Otherwise, we reverse the roles of
$z$ and $w$. Since $v,y\geq N^{-1/2}$ we have,

$\Psi(z)\leq C\frac{(\kappa(x)+y)^{1/4}}{(Ny)^{1/2}},$ (6.70)

and a similar estimate for $\Psi(w)$. This yields the claim. ∎

Let $\varphi$ be a function supported in $[-5,5]$ and recall the following
quasi-analytic extension of $\varphi$,

$\tilde{\varphi}(x+\mathrm{i}y)=(\varphi(x)+\mathrm{i}y\varphi^{\prime}(x))\chi(y),$
(6.71)

where $\chi$ is smooth even bump function that is on $[-1,1]$ and $0$ outside
of $[-2,2]$. Recall the formula

$\partial_{\bar{z}}\tilde{\varphi}(x+\mathrm{i}y)=\frac{1}{2}\mathrm{i}y\chi(y)\varphi^{\prime\prime}(x)+\mathrm{i}(\varphi(x)+\mathrm{i}\varphi^{\prime}(x)y)\chi^{\prime}(y),$
(6.72)

as well as the Helffer-Sjöstrand (HS) formula [7, Appendix C]

$\mathrm{tr}\varphi(H)-\mathbb{E}[\varphi(H)]=\frac{1}{\pi}\int_{\mathbb{R}^{2}}\partial_{\bar{z}}\tilde{\varphi}(x+\mathrm{i}y)N(m_{N}(z)-\mathbb{E}[m_{N}(z)])\mathrm{d}x\mathrm{d}y.$
(6.73)

Recall the definition of the norm $\|\varphi\|_{1,w}$ in (3.2). The following
is a simple consequence of the HS formula.

######  Lemma 6.7.

Assume that $\|\varphi^{\prime\prime}\|_{1}\leq N$ and let $\varepsilon>0$.
With overwhelming probability we have,

$\left|\mathrm{tr}\varphi(H)-\mathbb{E}[\mathrm{tr}\varphi(H)]\right|\leq
N^{\varepsilon}(1+\|\varphi^{\prime}\|_{1}+\|\varphi^{\prime\prime}\|_{1}N^{-1})$
(6.74)

Proof. We have by a direct application of the HS formula and the local law
(2.11),

$\displaystyle\left|\mathrm{tr}\varphi(H)-\mathbb{E}[\mathrm{tr}\varphi(H)]\right|$
$\displaystyle\leq
C\left|\int_{\mathbb{R}^{2}}y\varphi^{\prime\prime}(x)\chi(y)N(m_{N}(x+\mathrm{i}y)-\mathbb{E}[m_{N}(x+\mathrm{i}y)])\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle+CN^{\varepsilon}(\|\varphi\|_{1}+\|\varphi^{\prime}\|_{1}).$
(6.75)

For the term on the first line, by the symmetry of $\chi(y)$ we have,

$\displaystyle\int_{\mathbb{R}^{2}}y\varphi^{\prime\prime}(x)\chi(y)N(m_{N}(x+\mathrm{i}y)-\mathbb{E}[m_{N}(x+\mathrm{i}y)])\mathrm{d}x\mathrm{d}y$
$\displaystyle=$ $\displaystyle
2\int_{y>0}y\varphi^{\prime\prime}(x)\chi(y)N\mathrm{Im}[m_{N}(x+\mathrm{i}y)-\mathbb{E}[m_{N}(x+\mathrm{i}y)]]\mathrm{d}x\mathrm{d}y$

We have that $y\to y\mathrm{Im}[m_{N}(x+\mathrm{i}y)]$ is increasing and so
using the local law (2.11) it follows that for $0<y<N^{\varepsilon-1}$ that,

$yN\mathrm{Im}[m_{N}(x+\mathrm{i}y)]\leq N^{\varepsilon}$ (6.76)

with overwhelming probability and so,

$\left|\int_{y<N^{\varepsilon-1}}y\varphi^{\prime\prime}(x)\chi(y)N\mathrm{Im}[m_{N}(x+\mathrm{i}y)-\mathbb{E}[m_{N}(x+\mathrm{i}y)]]\mathrm{d}x\mathrm{d}y\right|\leq
CN^{2\varepsilon-1}\|\varphi^{\prime\prime}\|_{1}$ (6.77)

Let
$t=N^{\varepsilon}(1+\|\varphi^{\prime}\|_{1})/(\|\varphi^{\prime\prime}\|)$.
Then,

$\displaystyle\left|\int_{N^{\varepsilon-1}<y<t}y\varphi^{\prime\prime}(x)\chi(y)N\mathrm{Im}[m_{N}(x+\mathrm{i}y)-\mathbb{E}[m_{N}(x+\mathrm{i}y)]]\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$
$\displaystyle\int_{y<t}CN^{\varepsilon}\|\varphi^{\prime\prime}\|_{1}\mathrm{d}y\leq
CN^{\varepsilon}(1+\|\varphi^{\prime}\|_{1}).$ (6.78)

For larger $y$, we have by integration by parts and the compact support of
$\varphi$ that

$\displaystyle\left|\int_{y>t}y\varphi^{\prime\prime}(x)\chi(y)N\mathrm{Im}[m_{N}(x+\mathrm{i}y)-\mathbb{E}[m_{N}(x+\mathrm{i}y)]]\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle=$
$\displaystyle\left|\int_{y>t}y\varphi^{\prime}(x)\chi(y)N\mathrm{Im}[m^{\prime}_{N}(x+\mathrm{i}y)-\mathbb{E}[m^{\prime}_{N}(x+\mathrm{i}y)]]\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle
N^{\varepsilon}\int_{3>|y|>t}\|\varphi^{\prime}\|_{1}y^{-1}\mathrm{d}y\leq
CN^{2\varepsilon}\|\varphi^{\prime}\|_{1}$ (6.79)

where we used the Cauchy integral formula. ∎

Following [44], define

$\Omega_{\mathfrak{a}}:=\\{(x,y)\in\mathbb{R}^{2}:|y|>N^{\mathfrak{a}-1}\\}$
(6.80)

where $0<\mathfrak{a}<1$.

######  Lemma 6.8.

Let $\varepsilon>0,\mathfrak{a}>0$ and $H$ be a Wigner matrix. With
overwhelming probability,

$\displaystyle\left|(\mathrm{tr}\varphi(H)-\mathbb{E}[\mathrm{tr}\varphi(H)])-\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{\varphi}(z)N(m_{N}(z)-\mathbb{E}[m_{N}(z)])\mathrm{d}x\mathrm{d}y\right|\leq
CN^{\varepsilon}N^{\mathfrak{a}-1}\|\varphi^{\prime\prime}\|_{1}.$ (6.81)

Proof. This is a straightforward conseqeunce of the local law. The argument of
the previous lemma shows that the contribution from $|y|<N^{\varepsilon-1}$ is
$\mathcal{O}(N^{2\varepsilon-1}\|\varphi^{\prime\prime}\|_{1})$. The
contribution from $N^{\varepsilon-1}<|y|<N^{\mathfrak{a}-1}$ is
$\mathcal{O}(N^{\varepsilon+\mathfrak{a}-1}\|\varphi^{\prime\prime}\|_{1})$
due to the local law (2.11) and the explicit form of the quasi-analytic
extension $\varphi$. ∎

######  Theorem 6.9.

Assume $\|\varphi^{\prime\prime}\|_{1,w}\leq N$. We have for any
$\varepsilon>0$,

$\displaystyle\left|\mathrm{Var}(\mathrm{tr}\varphi(H))-V(\varphi)\right|\leq
N^{\varepsilon-2}\|\varphi^{\prime\prime}\|_{1,w}^{2}+N^{\varepsilon-1}(1+\|\varphi^{\prime}\|_{1,w})\|\varphi^{\prime\prime}\|_{1,w}$
(6.82)

where,

$\displaystyle V(\varphi)$
$\displaystyle=\frac{1}{2\pi^{2}}\int_{-2}^{2}\int_{-2}^{2}\frac{(\varphi(x)-\varphi(u))^{2}}{(x-u)^{2}}\frac{4-xu}{\sqrt{4-x^{2}}\sqrt{4-u^{2}}}\mathrm{d}x\mathrm{d}y$
$\displaystyle+\frac{s_{4}}{2\pi^{2}}\left(\int_{-2}^{2}\varphi(x)\frac{2-x^{2}}{\sqrt{4-x^{2}}}\mathrm{d}x\right)^{2}$
$\displaystyle-\frac{2s_{3}}{\pi^{2}N^{1/2}}\left(\int_{-2}^{2}\varphi(x)\frac{2-x^{2}}{\sqrt{4-x^{2}}}\mathrm{d}x\right)\left(\int_{-2}^{2}\varphi(x)\frac{x}{\sqrt{4-x^{2}}}\mathrm{d}x\right)$
(6.83)

Proof. By the previous two lemmas we have,

$\mathrm{Var}(\mathrm{tr}\varphi(H))=\frac{1}{\pi^{2}}\mathbb{E}\left[\left(\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{\varphi}N(m_{N}(z)-\mathbb{E}[m_{N}(z)]\mathrm{d}x\mathrm{d}y)\right)^{2}\right]+\mathcal{O}(N^{2\mathfrak{a}+\varepsilon-1}\|\varphi^{\prime\prime}\|_{1}(1+\|\varphi^{\prime}\|_{1}))$
(6.84)

The first term on the right side is,

$\displaystyle\mathbb{E}\left[\left(\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{\varphi}N(m_{N}(z)-\mathbb{E}[m_{N}(z)])\mathrm{d}x\mathrm{d}y)\right)^{2}\right]$
$\displaystyle=$
$\displaystyle\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)N^{2}\mathbb{E}\big{[}(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w)])\big{]}\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v$
(6.85)

where we use the notation $z=x+\mathrm{i}y$ and $w=u+\mathrm{i}v$. Let
$F(z,w)$ be the function defined by the second and third lines of (6.2) so
that,

$\displaystyle|H_{N}(z,w)|:=$
$\displaystyle\left|N^{2}\mathbb{E}[(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w)])]-F(z,w)\right|$
$\displaystyle\leq$
$\displaystyle\frac{N^{\varepsilon}}{\sqrt{\kappa(x)+|y|}}\left(\frac{1}{Ny^{2}|v|}+\frac{1}{Nv^{2}|y|}\right).$
(6.86)

We simplified the errors in (6.2) using the upper bound of (6.17). We now use
this estimate to replace the terms $m_{N}(z),m_{N}(w)$ with $F(z,w)$ in the
above integral expression for the variance. Clearly,

$\int_{\Omega_{\mathfrak{a}}^{2}}|\chi^{\prime}(y)\chi^{\prime}(v)(\varphi(x)+\mathrm{i}y\varphi^{\prime}(x))(\varphi(u)+\mathrm{i}v\varphi^{\prime}(u))H_{N}(z,w)|\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v\leq
CN^{\varepsilon-1}(1+\|\varphi^{\prime}\|_{1,w})^{2}$ (6.87)

and

$\displaystyle\int_{\Omega_{\mathfrak{a}}^{2}}|\chi(y)\chi^{\prime}(v)y\varphi^{\prime\prime}(x)(\varphi(u)+\mathrm{i}v\varphi^{\prime}(u))H_{N}(z,w)|\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v$
$\displaystyle\leq$ $\displaystyle
CN^{\varepsilon-1}\|\varphi^{\prime\prime}(x)\|_{1,w}(1+\|\varphi^{\prime}\|_{1,w})\int_{N^{-1}<|y|<1}|y|^{-1}\mathrm{d}y\leq
CN^{2\varepsilon-1}\|\varphi^{\prime\prime}\|_{1,w}(1+\|\varphi^{\prime}\|_{1,w}).$
(6.88)

Define
$t=N^{\varepsilon}(1+\|\varphi^{\prime}\|_{1,w})\|\varphi^{\prime\prime}\|_{1,w}^{-1}\wedge
1$. We have,

$\displaystyle\int_{|y|,|v|<t}|\varphi^{\prime\prime}(x)\varphi^{\prime\prime}(v)yvH_{N}(z,w)|\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v\leq$
$\displaystyle\|\varphi^{\prime\prime}\|_{1,w}^{2}N^{\varepsilon-1}\int_{|y|,|v|<t}|y|^{-1}\mathrm{d}v\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle
CN^{3\varepsilon-1}\|\varphi^{\prime\prime}\|_{1,w}(1+\|\varphi^{\prime}\|_{1,w}).$
(6.89)

By integration by parts in the $x$ variable, and the fact that $z\to
H_{N}(z,w)$ is holomorphic away from the real axis and the Cauchy integral
formula,

$\displaystyle\left|\int_{|y|>t,|v|<t}\varphi^{\prime\prime}(x)y\chi^{\prime}(y)\varphi^{\prime\prime}(u)v\chi^{\prime}(v)H_{N}(z,w)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v\right|$
$\displaystyle=$
$\displaystyle\left|\int_{|y|>t,|v|<t}\varphi^{\prime}(x)y\chi^{\prime}(y)\varphi^{\prime\prime}(u)v\chi^{\prime}(v)\partial_{z}H_{N}(z,w)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v\right|$
$\displaystyle\leq$ $\displaystyle
N^{\varepsilon}\|\varphi^{\prime}\|_{1,w}\|\varphi^{\prime\prime}\|_{1,w}\int_{|y|>t,|v|<t}(y^{-2}+|yv|^{-1})\mathrm{d}y\mathrm{d}v\leq
CN^{2\varepsilon}\|\varphi^{\prime}\|_{1,w}\|\varphi^{\prime\prime}\|_{1,w}.$
(6.90)

Finally,

$\displaystyle\left|\int_{|y|>t,|v|>t}\varphi^{\prime\prime}(x)y\chi^{\prime}(y)\varphi^{\prime\prime}(u)v\chi^{\prime}(v)H_{N}(z,w)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v\right|$
$\displaystyle=$
$\displaystyle\left|\int_{|y|>t,|v|>t}\varphi^{\prime}(x)y\chi^{\prime}(y)\varphi^{\prime}(u)v\chi^{\prime}(v)\partial_{z}\partial_{w}H_{N}(z,w)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v\right|$
$\displaystyle\leq$ $\displaystyle
N^{\varepsilon-1}\|\varphi^{\prime}\|_{1,w}^{2}\int_{|y|,|v|>t}y^{-2}|v|^{-1}\mathrm{d}v\mathrm{d}y\leq
N^{2\varepsilon-1}\|\varphi^{\prime}\|_{1,w}\|\varphi^{\prime\prime}\|_{1,w}$
(6.91)

The above estimates show that,

$\displaystyle\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)N^{2}\mathbb{E}[(m_{N}(z)-\mathbb{E}[m_{N}(z)])(m_{N}(w)-\mathbb{E}[m_{N}(w))]$
$\displaystyle=$
$\displaystyle\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)F(z,w)+N^{3\varepsilon-1}\mathcal{O}(\|\varphi^{\prime\prime}\|_{1,w}(1+\|\varphi^{\prime}\|_{1,w})).$
(6.92)

Using the estimate,

$|F(z,w)|\leq
C\frac{1}{\sqrt{\kappa(x)+y}\sqrt{\kappa(u)+v}}\frac{1}{y^{2}+v^{2}}$ (6.93)

it is straightforward to conclude that,

$\displaystyle\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)F(z,w)$
$\displaystyle=\int_{\mathbb{C}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)F(z,w)$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{2\mathfrak{a}-2}\|\varphi^{\prime\prime}\|_{1,w}^{2}+N^{\mathfrak{a}-1}\|\varphi^{\prime}\|_{1,w}\|\varphi^{\prime\prime}\|_{1,w})$
(6.94)

The integral on the right side may then be calculated by Green’s theorem which
in complex notation states that

$\int_{\Omega}\partial_{\bar{z}}F(z)\mathrm{d}x\mathrm{d}y=-\frac{\mathrm{i}}{2}\int_{\partial\Omega}F(z)\mathrm{d}z.$
(6.95)

Therefore,

$\displaystyle\frac{1}{\pi^{2}}\int_{\mathbb{C}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)F(z,w)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v$
$\displaystyle=$
$\displaystyle-\frac{1}{4\pi^{2}}\int_{\mathbb{R}^{2}}\varphi(x)\varphi(y)(F(x+\mathrm{i}0,y+\mathrm{i}0)+F(x-\mathrm{i}0,y-\mathrm{i}0)-2F(x+\mathrm{i}0,y-\mathrm{i}0))\mathrm{d}x\mathrm{d}y.$
(6.96)

The contributions from the terms that do not involve $s_{3}$ were calculated
in the proof of [44, Lemma 4.2]. The term that involves $s_{3}$ may be
calculated as

$\displaystyle-\frac{8s_{3}}{N^{1/2}\pi^{2}}\int_{\mathbb{C}^{2}}\partial_{\bar{z}}\tilde{\varphi}(z)\partial_{\bar{w}}\tilde{\varphi}(w)m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(z)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v$
$\displaystyle=$
$\displaystyle-\frac{8s_{3}}{N^{1/2}\pi^{2}}\left(\int_{\mathbb{R}}\varphi(x)\mathrm{Im}[m_{\mathrm{sc}}^{\prime}(x+\mathrm{i}0)m_{\mathrm{sc}}(x+\mathrm{i}0)]\mathrm{d}x\right)\left(\int_{\mathbb{R}}\varphi(y)\mathrm{Im}[m_{\mathrm{sc}}^{\prime}(y+\mathrm{i}0)]\mathrm{d}y\right)$
$\displaystyle=$
$\displaystyle-\frac{8s_{3}}{N^{1/2}\pi^{2}}\left(\int_{-2}^{2}\varphi(x)\frac{x^{2}-2}{2\sqrt{4-x^{2}}}\mathrm{d}x\right)\left(\int_{-2}^{2}\varphi(y)\frac{-y}{2\sqrt{4-y^{2}}}\mathrm{d}y\right)$
(6.97)

where the last line follows from the explicit formula of $m_{\mathrm{sc}}$. We
conclude the proof. ∎

###  6.3 Expectation

######  Lemma 6.10.

Let $H$ be a Wigner matrix. Let $z=x+\mathrm{i}y$. Then,

$\displaystyle(z+2m_{\mathrm{sc}}(z))(\mathbb{E}[m_{N}(z)]-m_{\mathrm{sc}}(z))$
$\displaystyle=-\frac{1}{N}m_{\mathrm{sc}}^{\prime}(z)$ $\displaystyle+$
$\displaystyle\frac{4s_{3}}{N^{3/2}}m_{\mathrm{sc}}(z)^{3}-\frac{s_{4}}{N}m_{\mathrm{sc}}(z)^{4}$
(6.98) $\displaystyle+$ $\displaystyle N^{\varepsilon}\mathcal{O}((Ny)^{-2}).$
(6.99)

Proof. By the cumulant expansion,

$\displaystyle\frac{1}{N}\sum_{i}z\mathbb{E}[G_{ii}(z)]+1=$
$\displaystyle\frac{1}{N}\sum_{ia}\mathbb{E}[H_{ia}G_{ia}(z)]$
$\displaystyle=$
$\displaystyle-\mathbb{E}[m_{N}(z)^{2}]-\partial_{z}\mathbb{E}[m_{N}(z)]$
$\displaystyle+$
$\displaystyle\sum_{n=3}^{5}\frac{1}{N^{1+n/2}}\sum_{ia}\frac{(1+\delta_{ia})^{n-1}s_{n}}{(n-1)!}\mathbb{E}[\partial_{ia}^{n}G_{ia}]$
$\displaystyle+$ $\displaystyle N^{\varepsilon}\mathcal{O}(N^{-2}).$ (6.100)

By the local law (2.11) we have,

$\displaystyle-\mathbb{E}[m_{N}(z)^{2}]-\frac{1}{N}\partial_{z}\mathbb{E}[m_{N}(z)]$
$\displaystyle=-m_{\mathrm{sc}}(z)^{2}-2m_{\mathrm{sc}}(z)\mathbb{E}[m_{N}(z)-m_{\mathrm{sc}}(z)]-\frac{1}{N}\partial_{z}m_{\mathrm{sc}}(z)$
$\displaystyle+N^{\varepsilon}\mathcal{O}((Ny)^{-2}).$ (6.101)

By (A.4) and similar arguments as in the proof of Lemma 6.3, we have that

$\frac{1}{N^{5/2}}\sum_{ia}(1+\delta_{ia})^{2}\mathbb{E}[\partial_{ia}^{2}G_{ia}]=\frac{8}{N^{3/2}}m_{\mathrm{sc}}(z)^{3}+N^{\varepsilon}\mathcal{O}((Ny)^{-2}).$
(6.102)

From (A.7) and the entrywise local law (2.12) we see,

$\displaystyle\frac{1}{N^{3}}\sum_{ia}(1+\delta_{ia})^{3}\mathbb{E}[\partial_{ia}^{3}G_{ia}]=-\frac{6}{N^{3}}\sum_{ia}\mathbb{E}[G_{ii}^{2}G_{aa}^{2}]+N^{\varepsilon}\mathcal{O}((Ny)^{-2}).$
(6.103)

Expanding $G_{ii}=(G_{ii}-m_{\mathrm{sc}})+m_{\mathrm{sc}}$ (and similarly for
$G_{aa}$) and estimating the terms linear in $G_{ii}$ using the local law
(2.11) and the higher order terms using the entry-wise local law (2.12) we see
that,

$\displaystyle\frac{1}{N^{3}}\sum_{ia}\mathbb{E}[G_{ii}^{2}G_{aa}^{2}]=\frac{m_{\mathrm{sc}}(z)^{4}}{N}+N^{\varepsilon}\mathcal{O}((Ny)^{-2}).$
(6.104)

Finally, the term with $n=5$ in (6.3) is estimate using the parity estimate
(6.60). ∎

######  Lemma 6.11.

Let $f$ be a function, and $H$ a Wigner matrix. Then,

$\displaystyle\mathbb{E}[\mathrm{tr}f(H)]$ $\displaystyle=N\int
f(x)\rho_{\mathrm{sc}}(x)\mathrm{d}x-\frac{1}{2\pi}\int
f(x)\frac{1}{\sqrt{4-x^{2}}}\mathrm{d}x$
$\displaystyle+\frac{f(2)+f(-2)}{4}+\frac{s_{4}}{2\pi}\int_{-2}^{2}f(x)\frac{x^{4}-4x^{2}+2}{\sqrt{4-x^{2}}}\mathrm{d}x$
$\displaystyle+\frac{2s_{3}}{N^{1/2}}\frac{1}{\pi}\int_{-2}^{2}f(x)\frac{3x-x^{3}}{\sqrt{4-x^{2}}}+N^{\varepsilon}\mathcal{O}(N^{-1}\|f^{\prime\prime}\|_{1,w})$
(6.105)

Proof. The following is proven in a straightforward manner from the HS formula
and Lemma 6.10, similar to the argument proving Theorem 6.9,

$\displaystyle\mathbb{E}[\mathrm{tr}f(H)]-N\int
f(x)\rho_{\mathrm{sc}}(x)\mathrm{d}x$
$\displaystyle=\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{f}(z)\frac{-m_{\mathrm{sc}}^{\prime}(z)-s_{4}m_{\mathrm{sc}}(z)^{4}+4s_{3}N^{-1/2}m_{\mathrm{sc}}(z)^{3}}{z+2m_{\mathrm{sc}}(z)}\mathrm{d}x\mathrm{d}y$
$\displaystyle+N^{\varepsilon}\mathcal{O}(N^{\mathfrak{a}-1}\|f^{\prime\prime}\|_{1,w})$
(6.106)

where $\tilde{f}$ is as in (6.71) and $\Omega_{\mathfrak{a}}$ is as in (6.80).
From the estimate,

$\left|\frac{-m_{\mathrm{sc}}^{\prime}(z)}{z+2m_{\mathrm{sc}}(z)}\right|C\leq\frac{1}{\kappa(x)+y},$
(6.107)

we see that

$\displaystyle\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{f}(z)\frac{-m_{\mathrm{sc}}^{\prime}(z)-s_{4}m_{\mathrm{sc}}(z)^{4}+4s_{3}N^{-1/2}m_{\mathrm{sc}}(z)^{3}}{z+2m_{\mathrm{sc}}(z)}\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\int_{\mathbb{C}}\partial_{\bar{z}}\tilde{f}(z)\frac{-m_{\mathrm{sc}}^{\prime}(z)-s_{4}m_{\mathrm{sc}}(z)^{4}+4s_{3}N^{-1/2}m_{\mathrm{sc}}(z)^{3}}{z+2m_{\mathrm{sc}}(z)}\mathrm{d}x\mathrm{d}y+N^{\varepsilon}\mathcal{O}(N^{\mathfrak{a}-1}\|f^{\prime\prime}\|_{1,w}).$
(6.108)

The integral without the $s_{3}N^{-1/2}$ term as calculated in [44, Lemma
4.3]. Calculating the additional term is similar to what appears there, and
the result follows. ∎

##  7 Mesoscopic estimates for Dyson Brownian Motion

In this section, we will consider Dyson Brownian Motion (DBM),

$\mathrm{d}\lambda_{i}(t)=\sqrt{\frac{2}{N\beta}}\mathrm{d}B_{i}(t)+\frac{1}{N}\sum_{j\neq
i}\frac{1}{\lambda_{i}(t)-\lambda_{j}(t)}\mathrm{d}t-\frac{\lambda_{i}(t)}{2}\mathrm{d}t$
(7.1)

where the initial data are the eigenvalues of a Wigner matrix. Compared to
(1.33) we introduced the parameter $\beta\in\mathbb{R}$ above which equals
$\beta=1$ in the real symmetric case and $\beta=2$ in the complex Hermitian
case. The parameter does not play a substantial role but we introduce it for
completeness.

We will consider the evolution of

$s_{N}(z,t):=\frac{1}{N}\sum_{i}\log(\lambda_{i}(t)-z)$ (7.2)

for $z$ in the upper half-plane. By the Itô formula,

$\displaystyle\mathrm{d}s_{N}=\frac{-m_{N}^{2}}{2}+\frac{1}{2N}\left(1-\frac{2}{\beta}\right)m_{N}^{\prime}-\frac{1}{2}-\frac{zm_{N}(z)}{2}+\frac{1}{N^{3/2}}\sqrt{\frac{2}{\beta}}\sum_{i}\frac{\mathrm{d}B_{i}}{\lambda_{i}-z}.$
(7.3)

Define now,

$\displaystyle s(z):=\int\log(x-z)\rho_{\mathrm{sc}}(x)\mathrm{d}x.$ (7.4)

The characteristics as defined in (2.2) go backwards in time. We need the
characteristics that go forward in time. For this, we fix a final time $T$ and
define

$\tilde{z}_{t}:=z_{T-t}$ (7.5)

for some final condition $\tilde{z}_{T}=z_{0}=z$ which we suppress from the
notation. Note that due to (2.24), $\tilde{z}_{s}$ satisfies

$\partial_{s}\tilde{z}_{s}=-m_{\mathrm{sc}}(\tilde{z}_{s})-\frac{\tilde{z}_{s}}{2}.$
(7.6)

Then, with $s(z)$ as in (7.4),

$\partial_{s}s(\tilde{z}_{s})=m_{\mathrm{sc}}(\tilde{z}_{s})^{2}+\tilde{z}_{s}\frac{m_{\mathrm{sc}}(\tilde{z}_{s})}{2}=\frac{m_{\mathrm{sc}}(\tilde{z}_{s})^{2}-1}{2},$
(7.7)

where we used $m_{\mathrm{sc}}^{2}+zm_{\mathrm{sc}}+1=0$. Plugging in the
characteristic, we see that

$\displaystyle\mathrm{d}s_{N}(\tilde{z}_{s})$
$\displaystyle=-m_{N}^{2}/2+\frac{1}{2N}\left(1-\frac{2}{\beta}\right)m_{N}^{\prime}-\frac{1}{2}-\tilde{z}_{s}m_{N}/2+\frac{1}{N^{3/2}}\sqrt{\frac{2}{\beta}}\sum_{i}\frac{\mathrm{d}B_{i}}{\lambda_{i}-\tilde{z}_{s}}$
(7.8) $\displaystyle+m_{N}m_{\mathrm{sc}}+m_{N}\tilde{z}_{s}/2.$ (7.9)

In Section 7.1 we estimate the difference of
$\mathrm{d}s_{N}(\tilde{z}_{s})-\partial_{s}s(\tilde{z}_{s})$ and use this to
find an estimate of $s_{N}(\tilde{z}_{T},T)$ in terms of
$s_{N}(\tilde{z}_{0},0)$ and an explicit Gaussian random variable. In Section
7.2 we calculate the variance of this Gaussian random variable as well as the
covariance between two of these variables for different characteristics that
are well-separated.

Before beginning, we record the following easy consequence of rigidity.

######  Proposition 7.1.

Let $\varepsilon>0$ and $\tau>0$. With overwhelming probability we have
simultaneously for all $t\in[0,1]$ and $z\in\mathcal{D}_{\tau,1}$ that,

$|m_{N}(z)-m_{\mathrm{sc}}(z)|\leq\frac{N^{\varepsilon}}{N\mathrm{Im}[z]}$
(7.10)

as well as

$|\lambda_{i}(t)-\gamma_{i}|\leq\frac{N^{\varepsilon}}{N^{2/3}\min\\{i^{1/3},(N+1-i)^{1/3}\\}}.$
(7.11)

###  7.1 Estimates along characteristics

######  Proposition 7.2.

Let $\tilde{z}_{s}$, $s\in[0,T]$ be a characteristic, and assume that the
final condition $z=\tilde{z}_{T}$ satisfies
$\mathrm{Re}[z]\in(-2+\kappa,2-\kappa)$ and $\mathrm{Im}[z]\geq N^{\tau-1}$
for some $\tau,\kappa>0$. Assume $T<(\log(N))^{-1}$. Let $\varepsilon>0$. With
overwhelming probability,

$\displaystyle s_{N}(\tilde{z}_{T},T)-s(\tilde{z}_{T})$
$\displaystyle=s_{N}(\tilde{z}_{0},0)-s(\tilde{z}_{0})$
$\displaystyle+\frac{1}{N^{3/2}}\sqrt{\frac{2}{\beta}}\sum_{i}\int_{0}^{T}\frac{\mathrm{d}B_{i}(s)}{\gamma_{i}-\tilde{z}_{s}}$
$\displaystyle+\frac{1}{2N}\left(1-\frac{2}{\beta}\right)\int_{0}^{T}m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s})\mathrm{d}s$
$\displaystyle+N^{\varepsilon}\mathcal{O}\left(\frac{1}{N^{2}\mathrm{Im}[z_{T}]}\right)$
(7.12)

Proof. Using (7.6) and (7.8) we have,

$\displaystyle\mathrm{d}(s_{N}(\tilde{z}_{s})-s(\tilde{z}_{s}))$
$\displaystyle=-\frac{1}{2}(m_{\mathrm{sc}}(\tilde{z}_{s})-m_{N}(\tilde{z}_{s}))^{2}$
(7.13)
$\displaystyle+\frac{1}{2N}\left(1-\frac{2}{\beta}\right)(m_{N}^{\prime}(\tilde{z}_{s})-m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s}))$
(7.14)
$\displaystyle+\frac{1}{2N}\left(1-\frac{2}{\beta}\right)m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s})$
(7.15)
$\displaystyle\frac{1}{N^{3/2}}\sqrt{\frac{2}{\beta}}\sum_{i}\mathrm{d}B_{i}\left(\frac{1}{\lambda_{i}-\tilde{z}_{s}}-\frac{1}{\gamma_{i}-\tilde{z}_{s}}\right)$
(7.16)
$\displaystyle+\frac{1}{N^{3/2}}\sqrt{\frac{2}{\beta}}\sum_{i}\frac{\mathrm{d}B_{i}}{\gamma_{i}-\tilde{z}_{s}}.$
(7.17)

We now estimate the integrals in $s$ of the various terms on the right. By the
local law estimates of Proposition 7.1 we have, with overwhelming probability,

$\displaystyle\int_{0}^{T}\left|m_{N}(\tilde{z}_{s})-m_{\mathrm{sc}}(\tilde{z}_{s})\right|^{2}\mathrm{d}s\leq
N^{\varepsilon}\int_{0}^{T}\frac{1}{N^{2}(\mathrm{Im}[\tilde{z}_{s}])^{2}}\mathrm{d}s.$
(7.18)

From the explicit form of the characteristics we see that,

$c\leq-\partial_{s}\mathrm{Im}[\tilde{z}_{s}]\leq c^{-1}$ (7.19)

for some $c>0$ depending on $\kappa>0$ and our assumption that
$T<(\log(N))^{-1}$. Therefore, for the term (7.13) we have the estimate,

$\int_{0}^{T}\left|m_{N}(\tilde{z}_{s})-m_{\mathrm{sc}}(\tilde{z}_{s})\right|^{2}\mathrm{d}s\leq
N^{\varepsilon}\frac{1}{N^{2}\mathrm{Im}[\tilde{z}_{T}]}.$ (7.20)

By the Cauchy integral formula we have for all $s\in[0,T]$,

$\left|m_{N}^{\prime}(\tilde{z}_{s})-m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s})\right|\leq\frac{N^{\varepsilon}}{N\mathrm{Im}[\tilde{z}_{s}]^{2}}$
(7.21)

and so we arrive at the same estimate for the time integral of (7.14) as we
did for (7.13). For the term (7.16), we have by the rigidity estimates of
Proposition 7.1 that,

$\sum_{i}\left|\frac{1}{\lambda_{i}-\tilde{z}_{s}}-\frac{1}{\gamma_{i}-\tilde{z}_{s}}\right|^{2}\leq\frac{N^{\varepsilon}}{N(\mathrm{Im}[\tilde{z}_{s}])^{3}}$
(7.22)

with overwhelming probability for all $s\in[0,T]$. Therefore, by the
Burkholder-Davis-Gundy inequality:

$\displaystyle\left|\int_{0}^{T}\frac{1}{N^{3/2}}\sqrt{\frac{2}{\beta}}\sum_{i}\mathrm{d}B_{i}\left(\frac{1}{\lambda_{i}-\tilde{z}_{s}}-\frac{1}{\gamma_{i}-\tilde{z}_{s}}\right)\right|\leq\frac{N^{\varepsilon}}{N^{2}\mathrm{Im}[\tilde{z}_{T}]}$
(7.23)

with overwhelming probability. The claim follows. ∎

###  7.2 Covariance calculation

We now calculate the variance of the Gaussian random variables appearing in
Proposition 7.2.

######  Lemma 7.3.

Let $\tilde{z}_{s}$ be a characteristic whose final condition $\tilde{z}_{T}$
satisfies $\mathrm{Re}[\tilde{z}_{T}]\in(-2+\kappa,2-\kappa)$ and
$\mathrm{Im}[\tilde{z}_{T}]\geq N^{\tau-1}$. Assume that $T<\log(N)^{-1}$.
Then,

$\displaystyle
2\frac{1}{N}\sum_{i}\int_{0}^{T}\frac{\mathrm{Im}[\tilde{z}_{s}]^{2}}{|\gamma_{i}-\tilde{z}_{s}|^{4}}\mathrm{d}s=$
$\displaystyle\log(\mathrm{Im}[\tilde{z}_{0}]/\mathrm{Im}[\tilde{z}_{T}])$
$\displaystyle-$
$\displaystyle\frac{T}{2}+\mathrm{Re}[\log(1-m_{\mathrm{sc}}(\tilde{z}_{T})^{2})-\log(1-m_{\mathrm{sc}}(\tilde{z}_{0})^{2})]$
$\displaystyle+$
$\displaystyle\mathcal{O}((N\mathrm{Im}[\tilde{z}_{T}])^{-1}).$ (7.24)

Remark. Note that as long as $z$ is separated from $\pm 2$ by a constant $c$,
then $|1-m_{\mathrm{sc}}^{2}(z)|\geq c^{\prime}$ for some $c^{\prime}>0$.

Proof. First, it is easy to see that,

$\left|\frac{1}{N}\sum_{i}\frac{1}{|\gamma_{i}-\tilde{z}_{s}|^{4}}-\int\frac{1}{|x-\tilde{z}_{s}|^{4}}\rho_{\mathrm{sc}}(x)\mathrm{d}x\right|\leq\frac{C}{N\mathrm{Im}[\tilde{z}_{s}]^{4}}.$
(7.25)

Therefore,

$\left|\frac{1}{N}\sum_{i}\int_{0}^{T}\frac{\mathrm{Im}[\tilde{z}_{s}]^{2}}{|\gamma_{i}-\tilde{z}_{s}|^{4}}\mathrm{d}s-\int_{0}^{T}\int\frac{\mathrm{Im}[\tilde{z}_{s}]^{2}}{|x-\tilde{z}_{s}|^{4}}\rho_{\mathrm{sc}}(x)\mathrm{d}x\mathrm{d}s\right|\leq\frac{C}{N\mathrm{Im}[\tilde{z}_{T}]}.$
(7.26)

The integral against $\rho_{\mathrm{sc}}$ admits an exact calculation. First,
observe that

$2\int\frac{\mathrm{Im}[\tilde{z}_{s}]^{2}}{|x-\tilde{z}_{s}|^{4}}\rho_{\mathrm{sc}}(x)\mathrm{d}x=-\mathrm{Re}[m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s})]+\frac{\mathrm{Im}[m_{\mathrm{sc}}(\tilde{z}_{s})]}{\mathrm{Im}[\tilde{z}_{s}]}.$
(7.27)

Since,

$\partial_{s}\tilde{z}_{s}=-m_{\mathrm{sc}}(\tilde{z}_{s})-\frac{\tilde{z}_{s}}{2}$
(7.28)

we have

$\frac{\mathrm{Im}[m_{\mathrm{sc}}(\tilde{z}_{s})]}{\mathrm{Im}[\tilde{z}_{s}]}=\frac{-\partial_{s}\mathrm{Im}[\tilde{z}_{s}]}{\mathrm{Im}[z_{s}]}-\frac{1}{2}.$
(7.29)

Next,

$\partial_{s}\log(1-m_{\mathrm{sc}}^{2}(\tilde{z}_{s}))=\frac{-2m_{\mathrm{sc}}m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s})}{1-m_{\mathrm{sc}}^{2}(\tilde{z}_{s})}\partial_{s}\tilde{z}_{s}=-m_{\mathrm{sc}}^{\prime}(\tilde{z}_{s})$
(7.30)

where we used,

$\partial_{s}\tilde{z}_{s}=-m_{\mathrm{sc}}(\tilde{z}_{s})-\tilde{z}_{s}/2=\frac{1-m_{\mathrm{sc}}^{2}(\tilde{z}_{s})}{2m_{\mathrm{sc}}(\tilde{z}_{s})}$
(7.31)

the second identity following from $m_{\mathrm{sc}}^{2}+zm_{\mathrm{sc}}+1=0$.
Therefore,

$2\int\frac{\mathrm{Im}[\tilde{z}_{s}]^{2}}{|x-\tilde{z}_{s}|^{4}}\rho_{\mathrm{sc}}(x)\mathrm{d}x=\partial_{s}\mathrm{Re}[\log(1-m_{\mathrm{sc}}^{2}(\tilde{z}_{s}))]-\frac{\partial_{s}\mathrm{Im}[\tilde{z}_{s}]}{\mathrm{Im}[\tilde{z}_{s}]}-\frac{1}{2}.$
(7.32)

The claim follows after integration in $s$. ∎

######  Lemma 7.4.

Let $\tilde{z}_{s}$ and $\tilde{w}_{s}$ be two characteristics such that the
final conditions satisfy
$\mathrm{Re}[\tilde{z}_{T}],\mathrm{Re}[\tilde{w}_{T}]\in(-2+\kappa,2-\kappa)$
and $\mathrm{Im}[\tilde{z}_{T}],\mathrm{Im}[\tilde{w}_{T}]\geq N^{\tau-1}$ for
some $\tau,\kappa>0$. Assume also that,

$\kappa\mathrm{Im}[\tilde{z}_{T}]\leq\mathrm{Im}[\tilde{w}_{T}]\leq\frac{1}{\kappa}\mathrm{Im}[\tilde{z}_{T}].$
(7.33)

Assume that

$T<\frac{|\mathrm{Re}[\tilde{z}_{T}]-\mathrm{Re}[\tilde{w}_{T}]|}{\log(N)}.$
(7.34)

Then,

$\displaystyle\left|\frac{1}{N}\int_{0}^{T}\sum_{i}\frac{\mathrm{Im}[\tilde{z}_{s}]\mathrm{Im}[\tilde{w}_{s}]}{|\gamma_{i}-\tilde{z}_{s}|^{2}|\gamma_{i}-\tilde{w}_{s}|^{2}}\mathrm{d}s\right|\leq
C\frac{T^{2}+(\mathrm{Im}[\tilde{z}_{T}])^{2}}{(\mathrm{Re}[\tilde{z}_{T}]-\mathrm{Re}[\tilde{w}_{T}])^{2}}+C\frac{1}{N\mathrm{Im}[\tilde{z}_{T}]}$
(7.35)

Proof. Due to the explicit form of the characteristics, we see that

$c\mathrm{Im}[\tilde{z}_{s}]\leq\mathrm{Im}[\tilde{w}_{s}]\leq\frac{1}{c}\mathrm{Im}[\tilde{z}_{s}].$
(7.36)

for some $c>0$ for all $s\in[0,T]$. Similar to the proof of the previous lemma
we have,

$\left|\frac{1}{N}\int_{0}^{T}\sum_{i}\frac{\mathrm{Im}[\tilde{z}_{s}]\mathrm{Im}[\tilde{w}_{s}]}{|\gamma_{i}-\tilde{z}_{s}|^{2}|\gamma_{i}-\tilde{w}_{s}|^{2}}\mathrm{d}s-\int_{0}^{T}\int\frac{\mathrm{Im}[\tilde{z}_{s}]\mathrm{Im}[\tilde{w}_{s}]}{|x-\tilde{z}_{s}|^{2}|x-\tilde{w}_{s}|^{2}}\rho_{\mathrm{sc}}(x)\mathrm{d}x\mathrm{d}s\right|\leq\frac{C}{N\mathrm{Im}[\tilde{z}_{T}]}.$
(7.37)

The integral in $x$ is bounded via convolution of two Poisson kernels,

$\int\frac{\mathrm{Im}[\tilde{z}_{s}]\mathrm{Im}[\tilde{w}_{s}]}{|x-\tilde{z}_{s}|^{2}|x-\tilde{w}_{s}|^{2}}\rho_{\mathrm{sc}}(x)\mathrm{d}x\leq
C\frac{\mathrm{Im}[\tilde{z}_{s}]}{(\mathrm{Re}[\tilde{z}_{s}]-\mathrm{Re}[\tilde{w}_{s}])^{2}+(\mathrm{Im}[\tilde{z}_{s}])^{2}}.$
(7.38)

Under our assumptions we have,

$|\mathrm{Re}[\tilde{z}_{s}]-\mathrm{Re}[\tilde{w}_{s}]|\geq
c|\mathrm{Re}[\tilde{z}_{T}]-\mathrm{Re}[\tilde{w}_{T}]|$ (7.39)

for all $s\in[0,T]$, using the explicit form of the characteristics. Hence,
using the explicit form of the characteristics to estimate
$\mathrm{Im}[\tilde{z}_{s}]\leq C(\mathrm{Im}[\tilde{z}_{T}]+T)$,

$\displaystyle\int_{0}^{T}\int\frac{\mathrm{Im}[\tilde{z}_{s}]\mathrm{Im}[\tilde{w}_{s}]}{|x-\tilde{z}_{s}|^{2}|x-\tilde{w}_{s}|^{2}}\rho_{\mathrm{sc}}(x)\mathrm{d}x\mathrm{d}s\leq$
$\displaystyle
C\int_{0}^{T}\frac{\mathrm{Im}[\tilde{z}_{T}]+T}{|\mathrm{Re}[\tilde{z}_{T}]-\mathrm{Re}[\tilde{w}_{T}]|^{2}}\mathrm{d}s$
$\displaystyle\leq$ $\displaystyle
C\frac{T^{2}+(\mathrm{Im}[\tilde{z}_{T}])^{2}}{(\mathrm{Re}[\tilde{z}_{T}]-\mathrm{Re}[\tilde{w}_{T}])^{2}}$
(7.40)

This yields the claim. ∎

The above two lemmas immediately imply the following.

######  Proposition 7.5.

Let $\tilde{z}_{s}$ and $\tilde{w}_{s}$ be two characteristics such that the
final conditions satisfy
$\mathrm{Re}[\tilde{z}_{T}],\mathrm{Re}[\tilde{w}_{T}]\in(-2+\kappa,2-\kappa)$
and $\mathrm{Im}[\tilde{z}_{T}],\mathrm{Im}[\tilde{w}_{T}]\geq N^{\tau-1}$ for
some $\tau,\kappa>0$. Assume that there is a $t=N^{-\omega}$ such that

$t\asymp\mathrm{Im}[\tilde{z}_{T}]\asymp\mathrm{Im}[\tilde{w}_{T}],$ (7.41)

and a parameter $\delta>0$ such that,

$t\leq N^{-\delta}|\mathrm{Re}[\tilde{z}_{T}]-\mathrm{Re}[\tilde{w}_{T}]|.$
(7.42)

Assume that there is $0<\delta_{1}<\delta$ such that

$T\asymp N^{\delta_{1}}t.$ (7.43)

Then there is a $c>0$ so that the variances of the Gaussian random variables

$Z_{z}:=\frac{1}{N^{1/2}}\sum_{i}\int_{0}^{T}\frac{\mathrm{d}B_{i}}{\gamma_{i}-\tilde{z}_{s}}\mathrm{d}s,\qquad
Z_{w}:=\frac{1}{N^{1/2}}\sum_{i}\int_{0}^{T}\frac{\mathrm{d}B_{i}}{\gamma_{i}-\tilde{w}_{s}}\mathrm{d}s$
(7.44)

satisfy

$\mathrm{Var}(Z_{z})\geq c\delta_{1}\log(N),\qquad\mathrm{Var}(Z_{w})\geq
c\delta_{1}\log(N)$ (7.45)

and their covariance satisfies

$|\mathrm{Cov}(Z_{z},Z_{w})|\leq
CN^{2(\delta_{1}-\delta)}+(Nt)^{-1}\leq\frac{C}{\log(N)}.$ (7.46)

Proof. The estimate for the covariance follows immediately. For the variance,
note that $\mathrm{Im}[\tilde{z}_{0}]\asymp\mathrm{Im}[\tilde{z}_{T}]+T\asymp
T$ by the explicit form of the characteristics and the assumption that $T\gg
t$. The lower bound follows. ∎

##  8 Estimates for the Gaussian ensembles

If $\rho_{k}$ are the $k$-point functions of a point process (in our case the
eigenvalue of a random matrix), i.e.,

$\sum_{i}\mathbb{E}[f(\lambda_{i})]=\int
f(x)N\rho_{1}(x)\mathrm{d}x,\qquad\sum_{i\neq
j}\mathbb{E}[f(\lambda_{i},\lambda_{j})]=N(N-1)\int
f(x,y)\rho_{2}(x,y)\mathrm{d}x\mathrm{d}y$ (8.1)

then for two functions $f,g$ we have

$\mathrm{Cov}\left(\mathrm{tr}f(H),\mathrm{tr}g(H)\right)=\frac{1}{2}\int\left(f(x)-f(y)\right)\left(g(x)-g(y)\right)T_{N}(x,y)\mathrm{d}x\mathrm{d}y$
(8.2)

where the cluster function is,

$T_{N}(x,y)=N^{2}\rho_{1}(x)\rho_{1}(y)-N(N-1)\rho_{2}(x,y).$ (8.3)

Introduce the function,

$g_{E,\eta}(x):=\frac{\eta}{(x-E)^{2}+\eta^{2}}$ (8.4)

The un-normalized Hermite polynomials are

$H_{n}(x)=(-1)^{n}\mathrm{e}^{x^{2}}\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{n}\mathrm{e}^{-x^{2}}.$
(8.5)

They satisfy,

$\int_{\mathbb{R}}\mathrm{e}^{-x^{2}}H_{n}(x)H_{m}(x)=\delta_{nm}2^{n}n!\sqrt{\pi}.$
(8.6)

The normalized Hermite polyomials $h_{n}$ are,

$h_{n}(x):=\frac{1}{\sqrt{2^{n}n!\sqrt{\pi}}}H_{n}(x)$ (8.7)

which satisfy,

$\int_{\mathbb{R}}h_{i}(x)h_{j}(x)\mathrm{e}^{-x^{2}}\mathrm{d}x=\delta_{ij}.$
(8.8)

###  8.1 Covariance for GUE

With the above notation, define,

$\psi_{n}(x):=\mathrm{e}^{-x^{2}/4}h_{n}(x/\sqrt{2})2^{-1/4}.$ (8.9)

For the GUE, the Christoffel-Darboux formula is (see, e.g., Chapter 5 of
[51]),

$T_{N}(x,y)=N\left(\frac{\psi_{N}(\sqrt{N}x)\psi_{N-1}(\sqrt{N}y)-\psi_{N}(\sqrt{N}y)\psi_{N-1}(\sqrt{N}x)}{x-y}\right)^{2}=:\frac{\Psi_{N}(x,y)^{2}}{(x-y)^{2}}.$
(8.10)

The estimates in Appendix C imply that,

$|\Psi_{N}(x,y)|^{2}\leq
C\left(\frac{\mathrm{e}^{-cN(|x|-2)_{+}^{3/2}}}{|x-2|^{1/2}+N^{-1/3}}\right)\left(\frac{\mathrm{e}^{-cN(|y|-2)_{+}^{3/2}}}{|y-2|^{1/2}+N^{-1/3}}\right)$
(8.11)

######  Proposition 8.1.

Let $0<\eta,\delta<10^{-1}$. There is a $C>0$ depending on $\delta$ but not on
$\eta>0$ so that the following holds. For any $E_{1},E_{2}$ with
$|E_{1}|,|E_{2}|<2-2\delta$ we have,

$\left|\mathrm{Cov}(\mathrm{tr}g_{E_{1},\eta}(H),\mathrm{tr}g_{E_{2},\eta}(H))\right|\leq
C\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}$ (8.12)

Proof. We see from the exponential tails in (8.11) that,

$\displaystyle\mathrm{Cov}(\mathrm{tr}g_{E_{1},\eta}(H),\mathrm{tr}g_{E_{2},\eta}(H))$
$\displaystyle=\frac{1}{2}\int_{|x|,|y|<2+\delta}\frac{(g_{1}(x)-g_{1}(y))(g_{2}(x)-g_{2}(y))}{(x-y)^{2}}\Psi_{N}(x,y)^{2}\mathrm{d}x\mathrm{d}y$
$\displaystyle+\mathcal{O}\left(\eta^{2}\mathrm{e}^{-cN}\right),$ (8.13)

where we abbreviate $g_{i}=g_{E_{i},\eta}$. If $||x|-2|<\delta$ and
$|y|<2+\delta$ it is not hard to see that,

$\left|\frac{(g_{1}(x)-g_{1}(y))(g_{2}(x)-g_{2}(y))}{(x-y)^{2}}\right|\leq\frac{C\eta^{2}}{((y-E_{1})^{2}+\eta^{2})((y-E_{2})^{2}+\eta^{2})}$
(8.14)

Therefore,

$\displaystyle|\mathrm{Cov}(\mathrm{tr}g_{1}(H),\mathrm{tr}g_{2}(H))|\leq$
$\displaystyle\left|\frac{1}{2}\int_{|x|,|y|<2-\delta}\frac{(g_{1}(x)-g_{1}(y))(g_{2}(x)-g_{2}(y))}{(x-y)^{2}}\Psi_{N}(x,y)^{2}\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle+$ $\displaystyle C\frac{\eta}{((E_{1}-E_{2})^{2}+\eta^{2}}$
(8.15)

by the convolution property of Poisson kernels. Now,

$\displaystyle\frac{(g_{1}(x)-g_{1}(y))(g_{2}(x)-g_{2}(y))}{(x-y)^{2}}$
$\displaystyle=$
$\displaystyle\eta^{2}\frac{(x-E_{1}+y-E_{1})(x-E_{2}+y-E_{2})}{((x-E_{1})^{2}+\eta^{2})((y-E_{1})^{2}+\eta^{2})((x-E_{2})^{2}+\eta^{2})((y-E_{2})^{2}+\eta^{2})}.$
(8.16)

The claim then follows from Lemma 8.2 below and (8.11) above. ∎

######  Lemma 8.2.

For $0\leq E<5$ and $0<\eta<10^{-1}$ we have,

$\eta^{2}\int_{|x|,|y|<10}\frac{|(x+y)(x+y-2E)|}{(x^{2}+\eta^{2})(y^{2}+\eta^{2})((x-E)^{2}+\eta^{2})((y-E)^{2}+\eta^{2})}\mathrm{d}x\mathrm{d}y\leq
C\frac{1}{E^{2}+\eta^{2}}.$ (8.17)

Proof. We write the numerator as $(x+y)(x+y-2E)=x(x-E)+y(y-E)+x(y-E)+y(x-E)$.
By symmetry it suffices to bound the $x(x-E)$ term and the $x(y-E)$ term. For
the first we have,

$\displaystyle\int_{|x|,|y|<10}\frac{|x(x-E)|}{(x^{2}+\eta^{2})(y^{2}+\eta^{2})((x-E)^{2}+\eta^{2})((y-E)^{2}+\eta^{2})}\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\left(\int_{|x|<10}\frac{|x(x-E)|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x\right)\left(\int_{|y|<10}\frac{1}{(y^{2}+\eta^{2})((y-E)^{2}+\eta^{2})}\mathrm{d}y\right).$
(8.18)

The $y$ integration is bounded above by $\eta^{-1}(E^{2}+\eta^{2})^{-1}$ by
the convolution property of the Poisson kernel. The integrand in the $x$
integral can be bounded by $(x^{2}+\eta^{2})^{-1}+((x-E)^{2}+\eta^{2})^{-1}$
by Cauchy-Schwarz, and so yields $\mathcal{O}(\eta^{-1})$.

We now estimate the integral corresponding to $|x(y-E)|$. We have,

$\displaystyle\int_{|x|<10}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x$
$\displaystyle=\int_{|x|<E/2}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x$
$\displaystyle+\int_{2E>|x|>E/2}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x$
$\displaystyle+\int_{10>|x|>2E}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x.$
(8.19)

The last integral is bounded by,

$\int_{|x|>2E}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x\leq
C\int_{\mathbb{R}}\frac{1}{x^{3}+(E+\eta)^{3}}\mathrm{d}x\leq\frac{C}{E^{2}+\eta^{2}}.$
(8.20)

The second last integral is bounded by,

$\int_{2E>|x|>E/2}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x\leq\frac{CE}{E^{2}+\eta^{2}}\int_{\mathbb{R}}\frac{1}{(x-E)^{2}+\eta^{2}}\mathrm{d}x\leq
C\frac{E}{\eta(E^{2}+\eta^{2})}.$ (8.21)

The first integral is bounded by,

$\int_{|x|<E/2}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x\leq
C\frac{|E|}{E^{2}+\eta^{2}}\int_{\mathbb{R}}\frac{1}{x^{2}+\eta^{2}}\mathrm{d}x\leq
C\frac{E}{\eta(E^{2}+\eta^{2})}.$ (8.22)

Therefore,

$\int_{|x|<10}\frac{|x|}{(x^{2}+\eta^{2})((x-E)^{2}+\eta^{2})}\mathrm{d}x\leq
C\frac{1}{\eta(E+\eta)}.$ (8.23)

By making the change of variable $x=y-E$ we see by the same argument as above,

$\int_{|y|<10}\frac{|y-E|}{(y^{2}+\eta^{2})((y-E)^{2}+\eta^{2})}\mathrm{d}y\leq
C\frac{1}{\eta(E+\eta)}.$ (8.24)

This completes the proof. ∎

###  8.2 GOE preliminaries

We first make some simplifications that will allow us to avoid considering the
behavior of the cluster function near the spectral edges. Write $R_{N}(x,y)$
for twice the cluster function of the GOE, i.e.,

$\frac{1}{2}R_{N}(x,y)=N^{2}p_{1}(x)p_{1}(y)-N(N-1)p_{2}(x,y)$ (8.25)

Let now $g_{i}$ be as in the GUE case above, i.e., a Poisson kernel centered
at $|E_{i}|\leq 2-6\delta$, some $\delta>0$. Let $\chi$ be a bump function
that is $1$ on $[-2+4\delta,2-4\delta]$ and $0$ outside of
$[-2+3\delta,2-3\delta]$. Define $f_{i}=\chi g_{i}$.

Let $\chi_{1}$ be a bump function that is $0$ outside of
$[-2+\delta,2-\delta]$ and $1$ on $[-2+2\delta,2-2\delta]$.

######  Lemma 8.3.

Let $\varepsilon>0$ and $D>0$. Then, for $E_{i}$, $f_{i}$ as $0<\eta<1$ as
above we have

$\displaystyle\left|\mathrm{Cov}(\mathrm{tr}g_{1}(H),\mathrm{tr}g_{2}(H))\right|$
$\displaystyle\leq\left|\frac{1}{4}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}(x)\chi_{1}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle+CN^{\varepsilon}\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}+\eta^{-2}N^{-D}.$
(8.26)

Proof. We have, (writing $\mathrm{tr}h(H)=h$ for notational simplicity),

$\displaystyle\mathrm{Cov}(g_{1},g_{2})=$ $\displaystyle\mathrm{Cov}(\chi
g_{1},\chi g_{2})$ $\displaystyle+$
$\displaystyle\mathrm{Cov}((1-\chi)g_{1},g_{2})+\mathrm{Cov}(g_{1},(1-\chi)g_{2})-\mathrm{Cov}((1-\chi)g_{1},(1-\chi)g_{2}).$
(8.27)

Since $(1-\chi)g_{i}$ equals $\eta$ times a smooth function having derivatives
bounded uniformly in $N$ it is straightforward to conclude by Theorem 6.9 or
rigidity that,

$\mathrm{Var}((1-\chi)g_{i})\leq C\eta^{2}N^{\varepsilon}.$ (8.28)

Moreover, since the function $\eta\to\eta\mathrm{Im}[m_{N}(x+\mathrm{i}\eta)]$
is increasing, it follows that

$\mathrm{Var}(g_{i})\leq\eta^{-2}\mathbb{E}[N^{\varepsilon}\mathrm{Im}[m_{N}(E_{i}+\mathrm{i}N^{\varepsilon-1})]^{2}]\leq
CN^{3\varepsilon}\eta^{-2},$ (8.29)

where the last inequality follows by the local law (2.11). It follows by
Cauchy-Schwarz that,

$|\mathrm{Cov}(g_{1},g_{2})|\leq|\mathrm{Cov}(f_{1},f_{2})|+CN^{\varepsilon}$
(8.30)

where we defined $f_{i}=\chi g_{i}$.

We now show that we can restrict the domain of integration in the formula for
the covariance of the $f_{i}$’s. Let $\chi_{1}$ be a bump function as defined
before the statement of the lemma that is $0$ outside of
$[-2+\delta,2-\delta]$ and $1$ on $[-2+2\delta,2-2\delta]$. Recall that
$R_{N}$ is symmetric and so,

$\displaystyle\mathrm{Cov}(\mathrm{tr}f_{1}(H),\mathrm{tr}f_{2}(H))$
$\displaystyle=\frac{1}{4}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))R_{N}(x,y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}^{2}}f_{1}(x)f_{2}(x)R_{N}(x,y)\mathrm{d}x\mathrm{d}y-\frac{1}{2}\int_{\mathbb{R}^{2}}f_{1}(x)f_{2}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y.$
(8.31)

On the other hand, consider,

$\displaystyle\frac{1}{4}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}(x)\chi_{1}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{2}}f_{1}(x)f_{2}(x)\chi_{1}(x)\chi_{1}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y$
$\displaystyle-$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{2}}f_{1}(x)f_{2}(y)\chi_{1}(x)\chi_{1}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y.$
(8.32)

Since the $f_{i}$ are supported in $[-2+3\delta,2-3\delta]$ and $\chi_{1}$ is
$1$ on this interval, we see that
$f_{1}(x)f_{2}(y)\chi_{1}(x)\chi_{1}(y)=f_{1}(x)f_{2}(y)$ and
$f_{1}(x)f_{2}(x)\chi_{1}(x)\chi_{1}(y)=f_{1}(x)f_{2}(x)\chi_{1}(y)$.
Therefore,

$\displaystyle\mathrm{Cov}(\mathrm{tr}f_{1}(H),\mathrm{tr}f_{2}(H))-\frac{1}{4}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}(x)\chi_{1}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{2}}f_{1}(x)f_{2}(x)(1-\chi_{1}(y))R_{N}(x,y)\mathrm{d}x\mathrm{d}y.$
(8.33)

Recall that $R_{N}(x,y)=2(N^{2}p_{1}(x)p_{1}(y)-N(N-1)p_{2}(x,y))$ and so,
letting $F(x)=f_{1}(x)f_{2}(x)$ and $G(y)=1-\chi_{1}(y)$ we have,

$\displaystyle-$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{2}}f_{1}(x)f_{2}(x)(1-\chi_{1}(y))R_{N}(x,y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=$ $\displaystyle\sum_{i\neq
j}\mathbb{E}[F(\lambda_{i})G(\lambda_{j})]-\sum_{i,j}\mathbb{E}[F(\lambda_{i})]\mathbb{E}[G(\lambda_{j})]$
$\displaystyle=$
$\displaystyle\sum_{i,j}\mathbb{E}[F(\lambda_{i})G(\lambda_{j})]-\sum_{i,j}\mathbb{E}[F(\lambda_{i})]\mathbb{E}[G(\lambda_{j})]$
$\displaystyle=$
$\displaystyle\mathrm{Cov}\left(\mathrm{tr}F(H),\mathrm{tr}G(H)\right)$ (8.34)

where in the second equality we used that since $F$ and $G$ have disjoint
support, $F(\lambda_{i})G(\lambda_{i})=0$. To bound the last term we treat the
cases of $\eta\lesssim N^{-1}$ and $\eta\gtrsim N^{-1}$ separately.

Let $\varepsilon>0$, and consider $\eta\leq N^{\varepsilon-1}$. By rigidity we
have with overwhelming probability that
$\mathrm{tr}G(H)-\mathbb{E}[\mathrm{tr}G(H)]=\mathcal{O}(N^{\varepsilon})$.
Therefore, (using that $F$ is positive),

$|\mathrm{Cov}\left(\mathrm{tr}F(H),\mathrm{tr}G(H)\right)|=|\mathbb{E}[\mathrm{tr}F(H)(\mathrm{tr}G(H)-\mathbb{E}[\mathrm{tr}G(H)])]|\leq
N^{\varepsilon}\mathbb{E}[\mathrm{tr}F(H)]+CN^{-D}\eta^{-2}$ (8.35)

for any $D>0$. On the other hand, since the support of $F$ is restricted to
the bulk, we have that, (since the 1-point function or density of states of
the GOE is bounded in the bulk, see Theorem 2.14)

$\mathbb{E}[\mathrm{tr}F(H)]\leq
CN\int_{\mathbb{R}}f_{1}(x)f_{2}(x)\mathrm{d}x\leq
CN\frac{\eta}{(E_{1}-E_{2})^{2}+\eta^{2}}\leq\frac{CN^{\varepsilon}}{(E_{1}-E_{2})^{2}+\eta^{2}}$
(8.36)

where we used the convolution property for Poisson kernels as well as the fact
that we are assuming $N\eta\leq N^{\varepsilon}$.

Assume now that $\eta\geq N^{\varepsilon-1}$. Then a straightforward
calculation using rigidity shows that with overwhelming probability,

$|\mathrm{tr}F(H)-\mathbb{E}[\mathrm{tr}F(H)]|\leq
CN^{\varepsilon}\|F^{\prime}\|_{1}\leq
C\frac{N^{\varepsilon}}{\eta}\|F\|_{1}\leq
CN^{\varepsilon}\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}},$ (8.37)

again using the convolution property of the Poisson kernels. This yields the
claim. ∎

###  8.3 Covariance for GOE, $N$ even

In this section, we prove the following.

######  Proposition 8.4.

Let $g_{E,\eta}$ be as in (8.4). Let $\kappa>0$ and $\mathfrak{c}>0$ and
assume that $|E_{1}|,|E_{2}|\leq 2-\kappa$. Let $\varepsilon>0$. For any
$N^{-\mathfrak{c}}<\eta<\frac{1}{2}$ we have,

$\left|\mathrm{Cov}\left(\mathrm{tr}g_{E_{1},\eta}(H),\mathrm{tr}g_{E_{2},\eta}(H)\right)\right|\leq
CN^{\varepsilon}\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right)$
(8.38)

where $H$ is a GOE matrix of even dimension.

In order to prove the above Proposition, we have by Lemma 8.3 that it suffices
to estimate the first term on the right side of (8.26). We now state the exact
formula for the cluster function found in [51]. With $\psi_{n}$ as in (8.9),
introduce

$\varphi_{N}^{(N)}(x):=N^{1/4}\psi_{N}(\sqrt{N}x),\qquad\varphi_{N-1}^{(N)}(x):=N^{1/4}\psi_{N-1}(\sqrt{N}x)$
(8.39)

By the asymptotics in Appendix C, we have

$|\varphi_{N}^{(N)}(x)|+|\varphi_{N-1}^{(N)}(x)|\leq
C\frac{\mathrm{e}^{-cN(|x|-2)_{+}^{3/2}}}{||x|-2|^{1/4}+N^{-1/6}}$ (8.40)

Define the operator, for $x\in\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{C}$,

$\mathrm{sg}(x):=\frac{1}{2}\mathrm{sign}(x),\qquad(\mathrm{sg}f)(t):=\int_{\mathbb{R}}\mathrm{sg}(t-s)f(s)\mathrm{d}s.$
(8.41)

Note that for sufficiently well-behaved $f$ we have,

$\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{sg}f=f.$ (8.42)

From (6.1.19) and Theorem 6.1.6 and (6.2.10) of [51] we have for the cluster
function defined as in (8.25) and $N$ even,

$\displaystyle R_{N}(x,y)=$ $\displaystyle 2S_{N}(x,y)S_{N}(y,x)$
$\displaystyle+$ $\displaystyle
D_{N}(x,y)(I_{N}(y,x)-\mathrm{sg}(y-x))+D_{N}(y,x)(I_{N}(x,y)-\mathrm{sg}(x-y))$
(8.43)

and

$\displaystyle
S_{N}(x,y)=K_{N}(x,y)+\frac{N}{2}\varphi_{N-1}^{(N)}(x)\mathrm{sg}\varphi_{N}^{(N)}(y)$
(8.44)

where

$\displaystyle
K_{N}(x,y)=\frac{\varphi_{N}^{(N)}(x)\varphi_{N-1}^{(N)}(y)-\varphi_{N}^{(N)}(y)\varphi_{N-1}^{(N)}(x)}{(x-y)}$
(8.45)

(see [51, (4.2.17), (5.1.3)] for the Christoffel-Darboux formula in this
scaling) and,

$\displaystyle D_{N}(x,y)=-\partial_{y}S_{N}(x,y),\qquad
I_{N}(x,y)=\int_{y}^{x}S_{N}(s,y)\mathrm{d}s.$ (8.46)

We apply the exact formula above, and decompose the covariance of $f_{i}$ into
several terms corresponding to the various terms in $R_{N}$ above. To
facilitate this we further introduce,

$E_{N,1}(x,y)=\frac{N}{2}\varphi_{N-1}^{(N)}(x)\mathrm{sg}\varphi_{N}^{(N)}(y),\qquad\tilde{I}_{N}(x,y)=\int_{y}^{x}K_{N}(s,y)\mathrm{d}s$
(8.47)

and

$F_{N,1}(x,y)=\int_{y}^{x}E_{N,1}(s,y)\mathrm{d}s.$ (8.48)

Importantly, note that

$\partial_{x}\tilde{I}_{N}(x,y)=K_{N}(x,y),\qquad\partial_{x}F_{N,1}(x,y)=E_{N,1}(x,y).$
(8.49)

Proof of Proposition 8.4. By Lemma 8.3 it suffices to estimate the first term
on the right side of (8.26). After changing switching the roles of $x$ and $y$
in a few terms we find the following formula for the first term on the RHS of
(8.26).

$\displaystyle\frac{1}{4}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))R_{N}(x,y)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))K_{N}(x,y)^{2}\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.50) $\displaystyle-$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}K_{N}(x,y))\tilde{I}_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.51) $\displaystyle+$
$\displaystyle\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))E_{N,1}(x,y)K_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.52) $\displaystyle+$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))E_{N,1}(x,y)E_{N,1}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.53) $\displaystyle-$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}E_{N,1}(x,y))\tilde{I}_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.54) $\displaystyle-$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}K_{N}(x,y))F_{N,1}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.55) $\displaystyle-$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}E_{N,1}(x,y))F_{N,1}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.56) $\displaystyle+$
$\displaystyle\frac{1}{2}\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}S_{N}(x,y))\mathrm{sg}(y-x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.57)

Over the next sequence of lemmas, we bound each of the terms appearing above,
which will conclude the Proposition. ∎

######  Lemma 8.5.

For the term (8.50) we have,

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))K_{N}(x,y)^{2}\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y\right|\leq\frac{C}{(E_{1}-E_{2})^{2}+\eta^{2}}$
(8.58)

Proof. The region of integration is restricted to $|x|,|y|\leq 2-\delta$ on
which we have,

$|K_{N}(x,y)|\leq\frac{C}{|x-y|},$ (8.59)

which follows from the explicit form of $K_{N}$ above and the Hermite
polynomial asymptotics in Appendix C. When $|x|,|y|<2-4\delta$ we have that
$f_{i}=g_{i}$ and so by the proof of Proposition 8.1 (using Lemma 8.2) we
easily see that,

$\int_{|x|,|y|<2-4\delta}\left|\frac{(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))}{(x-y)^{2}}\right|\mathrm{d}x\mathrm{d}y\leq\frac{C}{(E_{1}-E_{2})^{2}+\eta^{2}}.$
(8.60)

When $|x|,|y|>2-5\delta$ we see from the explicit form of $f_{i}$ that,

$\left|\frac{(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))}{(x-y)^{2}}\right|\leq
C\eta^{2},$ (8.61)

and so this portion of the integral contributes $\mathcal{O}(1)$. This leaves
the cross-terms where $|x|>2-4\delta$ and $|y|<2-5\delta$, or vice versa. By
symmetry, it suffices to deal with the former region. When $|x|>2-4\delta$ and
$|y|<2-5\delta$ we have that,

$\left|\frac{(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))}{(x-y)^{2}}\right|\leq\frac{C\eta^{2}}{(y-E_{1})^{2}+\eta^{2}}\frac{1}{(y-E_{2})^{2}+\eta^{2}}$
(8.62)

and so this region contributes
$\mathcal{O}(\eta/((E_{1}-E_{2})^{2}+\eta^{2})$. The claim follows. ∎

######  Lemma 8.6.

We have for $|x|,|y|<2-\delta$ that

$|\tilde{I}_{N}(x,y)|\leq C\log(N)(1\wedge(N|x-y|)).$ (8.63)

Proof. We may assume $x>y$; the proof for $x<y$ is similar. We write, for some
$y<A<x$,

$\tilde{I}_{N}(x,y)=\int_{y}^{A}K_{N}(s,y)\mathrm{d}s+\int_{A}^{x}K_{N}(s,y)\mathrm{d}s.$
(8.64)

By (8.59) we have,

$\left|\int_{A}^{x}K_{N}(s,y)\mathrm{d}s\right|\leq
C\log\left(\frac{x-y}{A-y}\right).$ (8.65)

For the other term we have by the explicit form of $K_{N}$ and the fact that
the $\varphi_{N}^{(N)}$ and $\varphi_{N-1}^{(N)}$ are bounded in the domain of
integration,

$\left|\int_{y}^{A}K_{N}(s,y)\mathrm{d}s\right|\leq
C\int_{y}^{A}\frac{|\varphi_{N}^{(N)}(s)-\varphi_{N}^{(N)}(y)|+|\varphi_{N-1}^{(N)}(s)-\varphi_{N-1}^{(N)}(y)|}{s-y}\mathrm{d}s.$
(8.66)

By [51, (5.2.4)], we have that
$|\partial_{s}\varphi_{N}^{(N)}(s)|+|\partial_{s}\varphi_{N-1}^{(N)}(s)|\leq
CN$ for $|s|<2-\delta$ and so this integral is bounded by $CN|A-y|$. If
$|x-y|\leq N^{-1}$ we take $A=x$. If $|x-y|\geq N^{-1}$ we take $A=y+N^{-1}$.
The estimate follows. ∎

######  Lemma 8.7.

For the term (8.51) we have,

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}K_{N}(x,y))\tilde{I}_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle
C\log(N)\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right)$
(8.67)

Proof. We will treat the integral by integrating by parts in $y$. Note that
the integrand is of compact support so there is no boundary term. For the term
arising from the partial integration where the derivative hits $\tilde{I}_{N}$
we have, $\partial_{y}\tilde{I}_{N}(y,x)=K_{N}(y,x)$, and so this term was
dealt with in Lemma 8.5. Consider the term where where the partial derivative
hits $\chi_{1}(y)$. Note that

$(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}^{\prime}(y)=f_{1}(x)f_{2}(x)\chi_{1}^{\prime}(y)$
(8.68)

and that this term is non-zero only if $|x-y|\geq\delta$. Therefore,

$\displaystyle\int_{\mathbb{R}^{2}}|(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}^{\prime}(y)\chi_{1}(x)K_{N}(x,y)\tilde{I}_{N}(x,y)|\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle
C\log(N)\int_{\mathbb{R}}|f_{1}(x)f_{2}(x)\chi_{1}(x)|\mathrm{d}x\leq
C\log(N)\frac{\eta}{(E_{1}-E_{2})^{2}+\eta^{2}}.$ (8.69)

It remains to consider the term,

$\displaystyle\int_{\mathbb{R}^{2}}|(\partial_{y}[(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))])K_{N}(x,y)\tilde{I}_{N}(y,x)|\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle
C\log(N)\int_{\mathbb{R}^{2}}\left|\frac{\partial_{y}[(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))]}{x-y}\right|\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.70)

where we used (8.59) and (8.63). If both $|x|,|y|>2-5\delta$ then,

$\left|\frac{\partial_{y}[(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))]}{x-y}\right|\leq
C\eta^{2}$ (8.71)

by the explicit form of the $f_{i}$. If $|x|>2-4\delta$ and $|y|<2-5\delta$
then one can estimate easily,

$\displaystyle\left|\frac{\partial_{y}[(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))]}{x-y}\right|\leq
C\frac{\eta}{((y-E_{1})^{2}+\eta^{2})((y-E_{2})^{2}+\eta^{2})}.$ (8.72)

The integral over $x$ and $y$ contributes an acceptable error by the
convolution property of Poisson kernels. For the region $|x|<2-5\delta$ and
$|y|>2-4\delta$ we have,

$\left|\frac{\partial_{y}[(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))]}{x-y}\right|\leq
C\frac{\eta^{2}}{((x-E_{1})^{2}+\eta^{2})((x-E_{2})^{2}+\eta^{2})}$ (8.73)

and we obtain a similar (in fact, better) estimate. For $|x|,|y|<2-4\delta$ we
have that $f_{i}=g_{i}$. By symmetry it will suffice to consider the case that
the derivative hits $g_{1}(y)$. In this case, we can estimate,

$\displaystyle\left|(\partial_{y}g_{1}(y))\frac{g_{2}(x)-g_{2}(y)}{(x-y)}\right|\leq
C\eta^{2}\frac{|y-E_{1}|}{(y-E_{1})^{4}+\eta^{4}}\frac{|x-E_{2}|+|y-E_{2}|}{((x-E_{2})^{2}+\eta^{2})((y-E_{2})^{2}+\eta^{2})}.$
(8.74)

By estimating the integral of the right side over $|x|,|y|<10$ it suffices to
check the case $E_{1}=0$ and $E_{2}=E$ for some $0<E<5$. There are two terms,
corresponding to the two terms $|x-E_{2}|$ and $|y-E_{2}|$ on the right side
above. The $|x-E_{2}|$ term is bounded by,

$\displaystyle\eta^{2}\int_{|y|<10}\frac{|y|}{(y^{4}+\eta^{4})((y-E)^{2}+\eta^{2})}\mathrm{d}y\int_{|x|<10}\frac{|x-E|}{(x-E)^{2}+\eta^{2}}\mathrm{d}x$
$\displaystyle\leq$ $\displaystyle
C\log(N)\int_{|y|<10}\frac{\eta}{(y^{2}+\eta^{2})((y-E)^{2}+\eta^{2})}\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle C\log(N)\frac{1}{E^{2}+\eta^{2}}.$ (8.75)

In the first inequality we did the $x$ integration which contributes
$\mathcal{O}(|\log\eta|)=\mathcal{O}(\log(N))$ and cleared a power of
$y^{2}+\eta^{2}$ in the denominator using the $|y\eta|$ in the numerator. The
last estimate follows from the convolution property of Poisson kernels.

The final term to consider is,

$\displaystyle\eta^{2}\int_{|y|<10}\frac{|y||y-E|}{(y^{4}+\eta^{4})((y-E)^{2}+\eta^{2})}\mathrm{d}y\int_{|x|<10}\frac{1}{(x-E)^{2}+\eta^{2}}\mathrm{d}x$
$\displaystyle\leq$ $\displaystyle
C\eta\int_{|y|<10}\frac{|y||y-E|}{(y^{4}+\eta^{4})((y-E)^{2}+\eta^{2})}\mathrm{d}y$
(8.76)

For this integral, first,

$\eta\int_{|y|<E/2}\frac{|y||y-E|}{(y^{4}+\eta^{4})((y-E)^{2}+\eta^{2})}\mathrm{d}y\leq
C\frac{E}{E^{2}+\eta^{2}}\int\frac{\eta|y|}{y^{4}+\eta^{4}}\mathrm{d}y\leq
C\frac{E}{\eta(E^{2}+\eta^{2})}.$ (8.77)

Second,

$\eta\int_{|y|>E/2}\frac{|y||y-E|}{(y^{4}+\eta^{4})((y-E)^{2}+\eta^{2})}\mathrm{d}y\leq\frac{C}{E^{2}+\eta^{2}}\int_{|y|<10}\frac{|y-E|}{(y-E)^{2}+\eta^{2}}\mathrm{d}y\leq\frac{C\log(N)}{E^{2}+\eta^{2}}.$
(8.78)

Summarizing, we have shown

$\displaystyle\int_{|x|,|y|<2-4\delta}\left|(\partial_{y}g_{1}(y))\frac{g_{2}(x)-g_{2}(y)}{(x-y)}\right|\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle
C\log(N)\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right).$
(8.79)

As stated above, the same argument applies to when the derivative hits $g_{2}$
instead. This completes the proof. ∎

######  Lemma 8.8.

We have, for $|x|,|y|<2-\delta$,

$|E_{N,1}(x,y)|\leq C,\qquad|F_{N,1}(x,y)|\leq C.$ (8.80)

Proof. The estimate follows from [51, (6.2.9)]. ∎

######  Lemma 8.9.

We have for the terms (8.52) and (8.53),

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))E_{N,1}(x,y)K_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{(E_{1}-E_{2})^{2}+\eta^{2}}$ (8.81)

and

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))E_{N,1}(x,y)E_{N,1}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y\right|\leq\frac{C}{(E_{1}-E_{2})^{2}+\eta^{2}}$
(8.82)

Proof. By Lemma 8.8 we have that $|E_{N,1}(x,y)|\leq C$ for all $x$ and $y$ in
the region where the integrand is non-zero. Therefore, we can crudely
estimate,

$|E_{N,1}(s,t)|\leq\frac{C}{|s-t|}.$ (8.83)

The integral of $|(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(x-y)^{-2}|$ over
$|x|,|y|<2-\delta$ was estimated in the proof of Lemma 8.5. ∎

######  Lemma 8.10.

For the term (8.54) we have,

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}E_{N,1}(x,y))\tilde{I}_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle
C\log(N)\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right)$
(8.84)

Proof. We integrate by parts in $y$. There is no boundary term due to the
compact support of the integrand. When this derivative hits
$\tilde{I}_{N}(y,x)$ one gets $K_{N}(y,x)$. This term was dealt with in Lemma
8.9. When the derivative hits the $f_{i}$’s, one can, as in Lemma 8.9 simply
bound $|E_{N,1}(s,t)|\leq C|s-t|^{-1}$ and so one is left with estimating the
integral of

$\left|(\partial_{y}[(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y)])|x-y|^{-1}\tilde{I}_{N}(x,y)\right|$
(8.85)

over $|x|,|y|<2-\delta$. This was completed in Lemma 8.7. The final term is
when the derivative hits $\chi_{1}(y)$, which is treated similarly to the
analogous term in the proof of Lemma 8.7. That is, due to the support
properties of the $f_{i}$ we have that the integrand is bounded by,

$|(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}^{\prime}(y)E_{N,1}(x,y)\tilde{I}_{N}(x,y)|\leq
C\log(N)f_{1}(x)f_{2}(x)$ (8.86)

which contributes an acceptable error after integration in $x$ due the
convolution property of the Poisson kernel. ∎

######  Lemma 8.11.

For the terms (8.55) and (8.56) we have,

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}K_{N}(x,y))F_{N,1}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle
C\log(N)\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right)$
(8.87)

and

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))[\partial_{y}E_{N,1}(x,y)F_{N,1}(y,x)]\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle
C\log(N)\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right)$
(8.88)

Proof. We start with the first estimate. Partial integration in $y$ leads to a
term with $K_{N}(x,y)E_{N,1}(y,x)$ as well as one where the derivative hits
the $f_{i}$’s and $\chi_{1}$. The former term already appeared in Lemma 8.9
and was estimated there. The term where the derivative hits the $f_{i}$’s can
be estimated in the same manner as Lemma 8.7 using the estimate of Lemma 8.8
for $F_{N,1}$ (instead of the estimate $|\tilde{I}_{N}(x,y)|\leq C\log(N)$
which was used in that proof). The integrand that arises when the derivative
hits $\chi_{1}(y)$ is bounded by $Cf_{1}(x)f_{2}(x)$ which again contributes
an acceptable error after integration in $x$.

For the second estimate stated in the lemma, we again integrate by parts. The
term arising when the derivative hits $F_{N,1}$ is $E_{N,1}(x,y)E_{N,1}(y,x)$,
which was estimated in Lemma 8.9. When the derivative hits the $f_{i}$’s we
can proceed similarly to Lemma 8.7. ∎

######  Lemma 8.12.

For the term (8.57) we have,

$\displaystyle\left|\int(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))[\partial_{y}S_{N}(x,y)\mathrm{sg}(y-x)]\mathrm{d}x\mathrm{d}y\right|$
$\displaystyle\leq$ $\displaystyle
C\log(N)\left(\frac{|E_{1}-E_{2}|}{\eta((E_{1}-E_{2})^{2}+\eta^{2})}+\frac{1}{(E_{1}-E_{2})^{2}+\eta^{2}}\right)$
(8.89)

Proof. We again integrate by parts. Since we have $\mathrm{sg}(y-x)$ there is
a boundary term at $y=x$. However, this boundary term is $0$ due to the fact
that $S_{N}$ is a continuous function, and the factors $(f_{i}(x)-f_{i}(y))$
which vanish when $x=y$. There are then terms that arise when the derivative
hits the $f_{i}$’s and when it hits $\chi_{1}$. Since $|S_{N}(x,y)|\leq
C\log(N)|x-y|^{-1}$ and $|\mathrm{sg}(y-x)|\leq 1$, these kinds of terms have
been estimated in the previous lemmas. ∎

###  8.4 Covariance for GOE, $N$ odd

In this section we prove the following.

######  Proposition 8.13.

The conclusions of Proposition 8.4 hold also for GOE matrices of odd
dimension.

We again use Lemma 8.3 and so in this section we estimate the first term on
the right side of (8.26). The exact formula for $R_{N}(x,y)$ in the case of
odd dimension is similar to the case of even dimension, except for a few
terms. After introducing the formulas, we just show how to handle the extra
terms that arise, as the rest is the same as the case of even dimension.

The work [51] does not contain formulas for $N$ odd, so we use Mehta’s book
[48]. We will list the formulas there, rescale them to match our convention
and then relate them to the quantities listed in the $N$-even case. We will
need a few new estimates for some of the terms that appear.

The formulas in [48] are given in terms of the Hermite functions,

$\varphi_{n}(x)=\frac{1}{(2^{n}n!\sqrt{\pi})^{1/2}}\mathrm{e}^{-x^{2}/2}H_{n}(x)=\mathrm{e}^{-x^{2}/2}h_{n}(x),$
(8.90)

with $H_{n}$ and $h_{n}$ as in (8.5) and (8.7), and are orthonormal. Compared
with (8.9) we have,

$\psi_{n}(x)=2^{-1/4}\varphi_{n}(x2^{-1/2}).$ (8.91)

From [48, (7.2.20)-(7.2.27)] (which defines $\tilde{K}$ below), and [48,
(7.2.31)] (which defines the cluster function in terms of $\tilde{K}$, but see
also [48, (7.2.30)] for the correlation functions as well as Chapter 5.1 for
the definition of the determinant of quarternions, and the formula for the
covariance in terms of correlation functions above) and [48, (A.10.1)] for the
Christoffel-Darboux formula in this scaling, we have,

$\displaystyle\mathrm{Cov}(\mathrm{tr}f(H),\mathrm{tr}g(H))=\frac{1}{4}\int_{\mathbb{R}^{2}}$
$\displaystyle(f(x(2/N)^{1/2})-f(y(2/N)^{1/2}))(g(x(2/N)^{1/2})-g(y(2/N)^{1/2}))$
$\displaystyle\times\mathrm{tr}\left(\tilde{K}_{N}(x,y)\tilde{K}_{N}(y,x)\right)\mathrm{d}x\mathrm{d}y$
(8.92)

where,

$\displaystyle\tilde{K}_{N}(x,y):=\left(\begin{matrix}\tilde{S}_{N}(x,y)&\tilde{D}_{N}(x,y)\\\
\tilde{J}_{N}(x,y)&\tilde{S}_{N}(y,x)\end{matrix}\right)$ (8.93)

where

$\displaystyle\tilde{S}_{N}(x,y)$
$\displaystyle:=\left(\frac{N}{2}\right)^{1/2}\frac{\varphi_{N}(x)\varphi_{N-1}(y)-\varphi_{N}(y)\varphi_{N-1}(x)}{x-y}$
$\displaystyle+\varphi_{N-1}(x)\left(\left(\frac{N}{2}\right)^{1/2}\mathrm{sg}\varphi_{N}(y)+\frac{1}{\int\varphi_{N-1}(t)\mathrm{d}t}\right),$
$\displaystyle\tilde{D}_{N}(x,y)$ $\displaystyle:=-\frac{\partial}{\partial
y}\tilde{S}_{N}(x,y),$ $\displaystyle\tilde{J}_{N}(x,y)$
$\displaystyle:=\int\mathrm{sg}(x-t)\tilde{S}_{N}(t,y)\mathrm{d}t$
$\displaystyle-\mathrm{sg}(x-y)-\mathrm{sg}\varphi_{N-1}(y)\left(\int\varphi_{N-1}(t)\mathrm{d}t\right)^{-1}.$
(8.94)

So, rescaling, we obtain

$\mathrm{Cov}(\mathrm{tr}f(H),\mathrm{tr}g(H))=\frac{1}{4}\int_{\mathbb{R}^{2}}(f(x)-f(y))(g(x)-g(y))\mathrm{tr}\hat{K}_{N}(x,y)\hat{K}_{N}(y,x)\mathrm{d}x\mathrm{d}y$
(8.95)

where

$\hat{K}_{N}(x,y)=\left(\begin{matrix}\hat{S}_{N}(x,y)&(2/N)^{1/2}\hat{D}_{N}(x,y)\\\
(N/2)^{1/2}\hat{J}_{N}(x,y)&\hat{S}_{N}(y,x)\end{matrix}\right)$ (8.96)

and

$\displaystyle\hat{S}_{N}(x,y)$
$\displaystyle:=K_{N}(x,y)+\hat{E}_{N,1}(x,y),$
$\displaystyle\hat{E}_{N,1}(x,y)$
$\displaystyle:=\frac{N}{2}\varphi_{N-1}^{(N)}(x)\mathrm{sg}\varphi_{N}^{(N)}(y)+\varphi_{N-1}^{(N)}(x)(\int\varphi_{N-1}^{(N)}(t)\mathrm{d}t)^{-1},$
$\displaystyle\hat{D}_{N}(x,y)$ $\displaystyle:=-\frac{\partial}{\partial
y}\hat{S}_{N}(x,y),$ $\displaystyle\hat{J}_{N}(x,y)$
$\displaystyle:=\int\mathrm{sg}(x-t)\hat{S}_{N}(t,y)\mathrm{d}t$
$\displaystyle-\varepsilon(x-y)-\mathrm{sg}\varphi_{N-1}^{(N)}(y)\left(\int\varphi_{N-1}^{(N)}(t)\mathrm{d}t)\right)^{-1}.$
(8.97)

So we have in this case that twice the cluster function is given by,

$R_{N}(x,y)=2\hat{S}_{N}(x,y)\hat{S}_{N}(y,x)+\hat{J}_{N}(x,y)\hat{D}_{N}(y,x)+\hat{J}_{N}(y,x)\hat{D}_{N}(x,y).$
(8.98)

We now prove some estimates on the terms arising above.

######  Lemma 8.14.

Let $\delta>0$. For $x,y$ satisfying $|x|,|y|<2-\delta$ we have,

$|\hat{E}_{N,1}(x,y)|\leq C.$ (8.99)

Proof. Let $N=2m+1$. By [51, (6.2.8)] we have,

$|\hat{E}_{N,1}(x,y)-\hat{E}_{N,1}(x,0)|\leq C.$ (8.100)

For the latter term, we have

$\hat{E}_{N,1}(x,0)=\varphi_{N-1}^{(N)}(x)\left(\frac{N}{2}\mathrm{sg}\varphi_{N}^{(N)}(0)+(\int\varphi_{N-1}^{(N)}(t)\mathrm{d}t)^{-1}\right),$
(8.101)

by definition. We now estimate the factor appearing on the RHS, showing it is
$\mathcal{O}(1)$. We have,

$\displaystyle\frac{N}{2}\mathrm{sg}\varphi_{N}^{(N)}(0)$
$\displaystyle=-\frac{N}{2}\int_{0}^{\infty}\varphi_{N}^{(N)}(t)\mathrm{d}t$
$\displaystyle=-\left(\frac{N}{2}\right)^{3/4}\int_{0}^{\infty}\varphi_{2m+1}(t)\mathrm{d}t$
$\displaystyle=-\frac{m^{1/2}}{2^{1/2}}+\mathcal{O}\left(1\right).$ (8.102)

In the first line we used the fact that $\varphi_{N}^{(N)}$ is an odd
function. In the second line we used the definition of $\varphi_{N}^{(N)}$ in
terms of the Hermite functions. In the last line we used Lemma D.2. For the
other term we have,

$\displaystyle\int\varphi_{N-1}^{(N)}(t)\mathrm{d}t$
$\displaystyle=\left(\frac{N}{2}\right)^{-1/4}\int\varphi_{2m}(t)\mathrm{d}t$
$\displaystyle=\frac{2^{1/2}}{m^{1/2}}\left(1+\mathcal{O}(m^{-1})\right)$
(8.103)

We used Lemma D.1 in the second line. The claim follows after using the fact
that the Hermite functions are bounded in the domain in question. ∎

######  Lemma 8.15.

Let $\delta>0$. We have for $x,y$ satisfying $|x|,|y|<2-\delta$ that.

$\left|\int\mathrm{sg}(x-t)\hat{E}_{N,1}(t,y)\right|\leq C$ (8.104)

and

$\left|\mathrm{sg}\varphi_{N-1}^{(N)}(y)\left(\int\varphi_{N-1}^{(N)}(t)\mathrm{d}t)\right)^{-1}\right|\leq
C$ (8.105)

Proof. By [51, (6.2.8)] we have

$|\mathrm{sg}\varphi_{N}^{(N)}(y)-\mathrm{sg}\varphi_{N}^{(N)}(0)|\leq
CN^{-1}$ (8.106)

and so it suffices to prove the first estimate with $y=0$ due to the fact that
scaled Hermite functions $\varphi_{N}^{(N)}$ and $\varphi_{N-1}^{(N)}$ have
bounded $L^{1}$ norms. For $y=0$, we have the explicit form (8.101) for which
the factor on the RHS is $\mathcal{O}(1)$, as shown in the previous proof. The
first estimate follows.

For the second estimate, note that the first estimate of [51, (6.2.9)] (this
applies since here $N$ is odd and there $N$ is even) implies that
$|\mathrm{sg}\varphi_{N-1}^{(N)}(y)|\leq CN^{-1}$. The previous proof shows
that the other factor is $\mathcal{O}(N^{1/2})$ and so we conclude the proof.
∎

######  Lemma 8.16.

Let $|x|,|y|<2-\delta$. Then,

$\left|\int\mathrm{sg}(x-t)\tilde{S}_{N}(t,y)\mathrm{d}t\right|\leq C\log(N),$
(8.107)

and so,

$|\hat{J}_{N}(x,y)|\leq C\log(N).$ (8.108)

Proof. First,

$\int_{t:|t-y|>\delta/2}|\tilde{S}_{N}(t,y)|\mathrm{d}t\leq
C(\|\varphi_{N}^{(N)}\|_{1}+\|\varphi_{N-1}^{(N)}\|_{1})\leq C,$ (8.109)

by the asymptotics (8.40). Next, the proof of Lemma 8.6 shows that,

$\int_{t:|t-y|\leq\delta/2}|K_{N}(t,y)|\mathrm{d}t\leq C\log(N),$ (8.110)

and the claim follows. ∎

Proof of Proposition 8.13. If suffices to estimate the first term on the right
side of (8.26). For this term we see from the above that we have the formula,

$\displaystyle\frac{1}{4}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\chi_{1}(x)\chi_{1}(y)R_{N}(x,y)\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))\hat{S}_{N}(x,y)\hat{S}_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y$
(8.111) $\displaystyle-$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{2}}(f_{1}(x)-f_{1}(y))(f_{2}(x)-f_{2}(y))(\partial_{y}\hat{S}_{N}(x,y))\hat{J}_{N}(y,x)\chi_{1}(x)\chi_{1}(y)\mathrm{d}x\mathrm{d}y.$
(8.112)

Given the estimates that we have established above, estimating the above terms
is very similar to the proof of Proposition 8.4. Given Lemma 8.14, the term
(8.111) is estimated in the exact same fashion as the terms (8.50), (8.52) and
(8.53). For the term (8.112), we integrate by parts in $y$. When the
derivative hits $\hat{J}_{N}(y,x)$ we use
$\partial_{y}\hat{J}_{N}(y,x)=\hat{S}_{N}(y,x)$ and see that this term is
identical to (8.111). When the derivative hits the other terms we see from
Lemma 8.16 that the integrand is bounded by

$C\log(N)^{2}|(\partial_{y}[\chi_{1}(y)(f_{1}(y)-f_{1}(x))(f_{2}(x)-f_{2}(y))])(x-y)^{-1}|.$
(8.113)

Estimation of the integral of this quantity was handled in the proof of Lemma
8.7. ∎

###  8.5 Expectation

######  Lemma 8.17.

Let $f$ be a function, and $H$ a matrix from the GOE. Then,

$\displaystyle\mathbb{E}[\mathrm{tr}f(H)]$ $\displaystyle=N\int
f(x)\rho_{\mathrm{sc}}(x)\mathrm{d}x-\frac{1}{2\pi}\int
f(x)\frac{1}{\sqrt{4-x^{2}}}\mathrm{d}x$
$\displaystyle+\frac{f(2)+f(-2)}{4}+N^{\varepsilon}\mathcal{O}(N^{-1}\|f^{\prime\prime}\|_{1,w})$
(8.114)

Proof. This follows immediately from Lemma 6.11. ∎

##  9 Proof of main theorems

######  Theorem 9.1.

Let $\varphi\in H^{1/2+s}$ for $s>0$. Let $H$ be a Gaussian divisible ensemble
of the form

$H=\mathrm{e}^{-T/2}W+\sqrt{1-\mathrm{e}^{-T}}G$ (9.1)

where $T=N^{-\omega}$ and $\omega>0$ satisfies,

$\omega<\frac{s}{100}.$ (9.2)

Then,

$\mathrm{Var}(\mathrm{tr}(\varphi(H)))\leq C\|\varphi\|_{H^{1/2+s}}^{2}.$
(9.3)

Proof. Fix a small $\mathfrak{a}>0$. Let $\chi$ be a smooth bump function that
is 1 on the interval $[-1,1]$ and $0$ outside of $[-2,2]$. Define

$\hat{\psi}_{1}(\xi)=\chi(\xi/N^{\mathfrak{a}})\hat{\varphi}(\xi),$ (9.4)

and

$\hat{\psi}_{2}(\xi)=(1-\chi(\xi/N^{\mathfrak{a}}))\sum_{|k-\mathfrak{a}\log_{2}(N)|<100}\hat{\varphi}_{k}(\xi)$
(9.5)

and

$\hat{\psi}_{3}(\xi)=\sum_{k\geq
50+\mathfrak{a}\log_{2}(N)}\hat{\varphi}_{k}(\xi).$ (9.6)

By the properties of the Littlewood-Paley decomposition,

$\varphi=\psi_{1}+\psi_{2}+\psi_{3}.$ (9.7)

By [3, Lemma 2.1, p. 52], we have that

$\|\psi_{1}\|_{C^{2}}+\|\psi_{2}\|_{C^{2}}\leq
CN^{3\mathfrak{a}/2}\|\varphi\|_{2}.$ (9.8)

We choose $\mathfrak{a}=1/10$. Now, if $p$ is the inverse Fourier transform of
$\chi$ we have,

$\psi_{1}(x)=N^{\mathfrak{a}}\int
p(N^{\mathfrak{a}}(x-y))\varphi(y)\mathrm{d}x.$ (9.9)

If $x\notin(-2+\kappa/2,2-\kappa/2)$ we see from the fact that,

$|p^{(m)}(x)|\leq\frac{C_{n,m}}{1+|x|^{n}}$ (9.10)

that,

$|\psi^{(m)}_{1}(x)|\leq CN^{-100}\frac{1}{1+|x|^{10}}\|\varphi\|_{2}.$ (9.11)

for $0\leq m\leq 5$ and $x\notin(-2+\kappa/2,2-\kappa/2)$. By the properties
of the Littlewood-Paley decomposition, we see that

$\hat{\psi}_{2}(\xi)=(1-\chi(\xi/N^{\mathfrak{a}}))q(\xi/N^{\mathfrak{a}})\hat{\varphi}(\xi)$
(9.12)

for some smooth $q$ that is of compact support. Explicitly,

$q(\xi)=\sum_{|k|\leq 100}\hat{\omega}(2^{-k}\xi).$ (9.13)

By the same argument as above, we see that

$|\psi^{(m)}_{2}(x)|\leq CN^{-100}\frac{1}{1+|x|^{10}}\|\varphi\|_{2}$ (9.14)

for $0\leq m\leq 5$ and $x\notin(-2+\kappa/2,2-\kappa/2)$ as well. Therefore
by Theorem 6.9 we have that,

$|\mathrm{Var}(\mathrm{tr}\psi_{1}(H))-V(\psi_{1})|+|\mathrm{Var}(\mathrm{tr}\psi_{2}(H))-V(\psi_{2})|\leq
N^{-1}N^{3\mathfrak{a}+\varepsilon}\|\varphi\|_{2}^{2}.$ (9.15)

Clearly

$\|\psi_{1}\|_{2}+\|\psi_{2}\|_{2}\leq C\|\varphi\|_{2}.$ (9.16)

We see from the above estimates that

$|V(\psi_{i})|\leq
C\|\varphi\|_{2}^{2}+C\int_{(-2+\kappa/4,2-\kappa/4)^{2}}\frac{(\psi_{i}(x)-\psi_{i}(y))^{2}}{(x-y)^{2}}\mathrm{d}x\mathrm{d}y.$
(9.17)

The last term is bounded above by the integral over $\mathbb{R}^{2}$ which is,
up to constants, the norm $\|\psi_{i}\|_{\dot{H}^{1/2}}$. Clearly,

$\|\psi_{i}\|_{\dot{H}^{1/2}}\leq C\|\varphi\|_{H^{1/2+s}}$ (9.18)

and so

$\mathrm{Var}(\mathrm{tr}(\psi_{1}(H)))+\mathrm{Var}(\mathrm{tr}(\psi_{2}(H)))\leq
C\|\varphi\|_{H^{1/2+s}}.$ (9.19)

Let

$M=10s^{-1}\log_{2}(N).$ (9.20)

and

$\psi_{4}:=\sum_{k\geq M}\varphi_{k}.$ (9.21)

By Theorem 3.5 we have,

$\mathrm{Var}(\mathrm{tr}(\psi_{4}(H)))\leq CN^{-1}\|\varphi\|_{H^{1/2+s}}.$
(9.22)

Consider now $k$ such that

$\mathfrak{a}\log_{2}(N)-100\leq k\leq M.$ (9.23)

By Theorems 3.3 and 3.4 we have,

$\mathrm{Var}(\mathrm{tr}\varphi_{k}(H))\leq
CN^{\varepsilon+\omega}2^{k}\|\varphi_{k}\|_{2}^{2}+N^{-100}\|\varphi\|_{2}^{2}$
(9.24)

for any $\varepsilon>0$. We may now choose $\varepsilon>0$ so that,

$\varepsilon+\omega<\frac{\mathfrak{a}s}{2}.$ (9.25)

Then

$N^{\varepsilon+\omega}2^{k}\|\varphi_{k}\|_{2}^{2}\leq
N^{-\mathfrak{a}s}\|\varphi\|^{2}_{H^{1/2+s}}.$ (9.26)

Hence

$\mathrm{Var}(\mathrm{tr}(\psi_{3}-\psi_{4})(H))\leq
C\log(N)N^{-\mathfrak{a}s}\|\varphi\|^{2}_{H^{1/2+s}}.$ (9.27)

The claim follows.∎

###  9.1 Proof of Theorem 1.4

We recall the following, [27, Proposition 4.1].

######  Proposition 9.2.

Let $u$ be a Wigner-smooth probability density as in Definition 1.2. For any
$K>0$ there is an $\alpha_{K}>0$ so that for any positive $t<\alpha_{K}$ there
exists a probability density $p_{t}$ with mean $0$ and variance $1$ so that

$\int|\mathrm{e}^{tA}p_{t}-u|\mathrm{d}x\leq Ct^{K},$ (9.28)

where $A=\frac{1}{2}\partial_{x}^{2}-\frac{x}{2}\partial-x$ is the generator
of the Ornstein-Uhlenbeck flow. Moreover, $g_{t}$ is Wigner-smooth with
uniform constants depending on those for $u$.

The only difference from the above and the construction in [27] is the
slightly weaker assumption on the decay, but that this holds can be checked by
the explicit form of the short construction given there.

Now consider a smooth Wigner matrix $H$ and a function $\varphi\in H^{1/2+s}$.
Denote the density of its off-diagonal entries by $u_{0}$ and of the diagonal
entries by $u_{d}$. Let,

$M=10s^{-1}\log_{2}(N),$ (9.29)

and

$T=N^{-\omega},\qquad\omega=\frac{s}{100}.$ (9.30)

Let,

$\psi_{1}=\sum_{k\geq-1}^{M}\varphi_{k},\qquad\psi_{2}=\sum_{k>M}\varphi_{k}.$
(9.31)

We see from Theorem 3.5 that

$\mathrm{Var}(\mathrm{tr}\psi_{2}(H))\leq CN^{-1}\|\varphi\|_{H^{1/2+s}}^{2}.$
(9.32)

We choose $K$ sufficiently large so that with $t=T$, Proposition 9.2
guarantees existence of a densities $p_{o},p_{d}$ so that,

$\int|\mathrm{e}^{tA}p_{o}-u_{o}|\mathrm{d}x+\int|\mathrm{e}^{tA}p_{d}-u_{d}|\mathrm{d}x\leq
C\frac{1}{N^{100}}\frac{1}{2^{10M}}.$ (9.33)

Let $W_{1}$ be the Wigner ensemble with off-diagonal densities $p_{0}$ and
diagonal densities $p_{d}$ and
$W=\mathrm{e}^{-t/2}W_{1}+\sqrt{1-\mathrm{e}^{-t}}G$ with $G$ GOE so that $W$
has densities $\mathrm{e}^{tA}p_{d}$ and $\mathrm{e}^{tA}p_{o}$. We apply
Theorem 9.1 to $W$.1††footnotetext: $1$. Note that strictly speaking, the
third and fourth cumulants of $W_{1}$ may not satisfy (1.19) but, by (9.33),
and since $H$ does satisfy (1.19), $W_{1}$ satisfies it up to, say, an error
of $\mathcal{O}(N^{-100})$ which does not affect the proof. For
$\eta_{1},\eta_{2}\geq 2^{-M}$ it follows from the Lipschitz continuity of the
resolvent that,

$\displaystyle\bigg{|}$
$\displaystyle\mathrm{Cov}\left(\mathrm{Im}\left[\mathrm{tr}(H-x-\mathrm{i}\eta_{1})^{-1}\right],\mathrm{Im}\left[\mathrm{tr}(H-y-\mathrm{i}\eta_{2})^{-1}\right]\right)$
$\displaystyle-\mathrm{Cov}\left(\mathrm{Im}\left[\mathrm{tr}(W-x-\mathrm{i}\eta_{1})^{-1}\right],\mathrm{Im}\left[\mathrm{tr}(W-y-\mathrm{i}\eta_{2})^{-1}\right]\right)\bigg{|}\leq
CN^{-10}.$ (9.34)

From the decomposition,

$\displaystyle\mathrm{Var}(\sum_{k\leq M}\mathrm{tr}\varphi_{k}(H))$
$\displaystyle=$ $\displaystyle\sum_{l,k\leq
M}\int_{\mathbb{R}^{2}}g_{l}(x)g_{k}(y)\mathrm{Cov}\left(\mathrm{Im}\left[\mathrm{tr}(H-x-\mathrm{i}\eta_{l})^{-1}\right],\mathrm{Im}\left[\mathrm{tr}(H-y-\mathrm{i}\eta_{k})^{-1}\right]\right)\mathrm{d}x\mathrm{d}y,$
(9.35)

the estimate

$\left|\mathrm{Var}\left(\sum_{k\leq
M}\mathrm{tr}\varphi_{k}(H)\right)-\mathrm{Var}\left(\sum_{k\leq
M}\mathrm{tr}\varphi_{k}(H)\right)\right|\leq N^{-2}\|\varphi\|^{2}_{2}$
(9.36)

follows. This yields the estimate for functions in $H^{1/2+s}$. ∎

##  10 Fine estimates of characteristic functions

Let $f$ be a smooth function supported in $[-5,5]$ and $H$ a Wigner matrix.
Recall the quasi-analytic extension of $f$ given by,

$\tilde{f}(x+\mathrm{i}y):=(f(x)+\mathrm{i}yf^{\prime}(x))\chi(y)$ (10.1)

where $\chi$ is an even bump function that is $1$ for $|y|<1$ and $0$ for
$|y|>2$. We fix $\mathfrak{a}>0$ for the rest of the section and define

$e_{\mathfrak{a}}(\xi):=\exp\left[\mathrm{i}\xi\left(\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{f}(z)N(m_{N}(z)-\mathbb{E}[m_{N}(z)])\mathrm{d}x\mathrm{d}y\right)\right],$
(10.2)

and

$\psi_{\mathfrak{a}}(\xi):=\mathbb{E}[e_{\mathfrak{a}}(\xi)],$ (10.3)

where $\Omega_{\mathfrak{a}}$ was defined in (6.80). Lemma 6.8 implies that
$\psi_{\mathfrak{a}}$ is a good approximation for the characteristic function
of the linear statistic.

###  10.1 Self-consistent equation for Stein’s method

We will apply Stein’s method to calculate $\psi_{\mathfrak{a}}(\xi)$. The main
tool technical estimate that we will use in our application of Stein’s method
is the following proposition.

######  Proposition 10.1.

Let $\varepsilon>0$ and $\tau>0$. For any $z=x+\mathrm{i}y$ with
$z\in\mathcal{D}_{\tau,1}$ we have,

$\displaystyle(z+2m_{\mathrm{sc}}(z))N\mathbb{E}[e_{\mathfrak{a}}(\xi)(m_{N}(z)-\mathbb{E}[m_{N}(z)])]$
$\displaystyle=$
$\displaystyle-\frac{2\mathrm{i}\xi}{N}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\psi_{a}\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\mathrm{d}u\mathrm{d}v$
$\displaystyle+$
$\displaystyle\psi_{\mathfrak{a}}\frac{4s_{3}m_{\mathrm{sc}}(z)\mathrm{i}\xi}{\pi
N^{1/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)(m_{\mathrm{sc}}(z)+m_{\mathrm{sc}}(w))m_{\mathrm{sc}}^{\prime}(w)\mathrm{d}u\mathrm{d}v$
$\displaystyle-$
$\displaystyle\frac{2\xi^{2}s_{3}}{\pi^{2}N^{1/2}}\psi_{\mathfrak{a}}m_{\mathrm{sc}}(z)\left(\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)\mathrm{d}u\mathrm{d}v\right)^{2}$
$\displaystyle-$
$\displaystyle\frac{-2\xi\mathrm{i}s_{4}}{\pi}\psi_{\mathfrak{a}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}\mathrm{d}u\mathrm{d}v$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}(1+|\xi|^{6})(1+\|f^{\prime}\|_{1}^{6})\mathcal{O}(N^{-1}(1+\|f^{\prime\prime}\|_{1})y^{-1}+N^{-1}y^{-2})$
(10.4)

Above, $w=u+\mathrm{i}v$.

The proof of the above proposition is largely similar to the proof of
Proposition 6.1, in that it will be seen to be a consequence of a cumulant
expansion. First, we have that whenever the entry-wise local semicircle law
holds,

$|\partial_{ia}^{k}e_{\mathfrak{a}}(\xi)|\leq
C(1+|\xi|)^{k}N^{\varepsilon}(1+\|f^{\prime}\|_{1})^{k}$ (10.5)

with overwhelming probability for any $k$ [44]. This allows for,

$\displaystyle
z\sum_{i}\mathbb{E}[e_{\mathfrak{a}}(G_{ii}(z)-\mathbb{E}[G_{ii}(z)])]$
$\displaystyle=$
$\displaystyle\sum_{ia}\mathbb{E}[e_{\mathfrak{a}}(G_{ia}(z)H_{ia}-\mathbb{E}[G_{ia}(z)H_{ia}])]$
$\displaystyle=$
$\displaystyle\sum_{n=2}^{5}\sum_{ia}\frac{(1+\delta_{ia})^{n-1}s_{n}}{N^{n/2}(n-1)!}\mathbb{E}[\partial_{ia}^{n-1}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial_{ia}^{n-1}G_{ia}(z)])]$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}((1+|\xi|)^{6}N^{-1}(1+\|f^{\prime}\|_{1})^{6})$
(10.6)

In the next few subsections we will deal with each of the terms on the second
last line of (10.1). We note that many of the terms that appear when the
derivatives $\partial_{ia}$ act on $e_{\mathfrak{a}}$ have been dealt with in
Section 6.1 in the proof of Proposition 6.1 and so we will refer to the
results in that section as appropriate. As in that proof, we will refer to the
term in the second line of (10.1) with $n=k$ as the $k$th order term.

Throughout the proof we will use the notation $z=x\pm\mathrm{i}y$ as well as
$w=u\pm\mathrm{i}v$ and $w_{i}=u_{i}\pm\mathrm{i}v_{i}$. We will often use the
fact that since $f$ is assumed to be supported in $[-5,5]$ that $\|f\|_{1}\leq
C\|f^{\prime}\|_{1}\leq C^{2}\|f^{\prime\prime}\|_{1}$ to simplify some error
terms. In applying the entrywise local law (2.12) we will often use that
$\Psi(z)\leq C(Ny)^{-1/2}$ when $y\geq N^{-1}$.

####  10.1.1 Second order term

######  Lemma 10.2.

For any $\varepsilon>0$ we have,

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})}{N}\mathbb{E}[\partial_{ia}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial_{ia}G_{ia}(z)]$
$\displaystyle=$
$\displaystyle-2Nm_{\mathrm{sc}}(z)\mathbb{E}[e_{\mathfrak{a}}(m_{N}(z)-\mathbb{E}[m_{N}(z)])]$
$\displaystyle+$
$\displaystyle-\frac{2\mathrm{i}\xi}{N}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\psi_{a}\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\mathrm{d}u\mathrm{d}v$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}\mathcal{O}(N^{-1}(1+\|f^{\prime\prime}\|_{1})y^{-1}+N^{-1}y^{-2}(1+\|f^{\prime}\|_{1})).$
(10.7)

Proof. We write,

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})}{N}\mathbb{E}[\partial_{ia}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial_{ia}G_{ia}(z)]$
$\displaystyle=$
$\displaystyle-N\mathbb{E}[e_{\mathfrak{a}}(m_{N}(z)^{2}-\mathbb{E}[m_{N}(z)^{2}])]$
$\displaystyle-$
$\displaystyle\mathbb{E}[e_{\mathfrak{a}}(\partial_{z}m_{N}(z)-\mathbb{E}[\partial_{z}m_{N}(z)])]$
$\displaystyle+$
$\displaystyle\frac{1+\delta_{ia}}{N}\sum_{ia}\mathbb{E}[(\partial_{ia}e_{\mathfrak{a}})G_{ia}(z)]$
(10.8)

Using the local law (2.11) we have,

$\displaystyle-N\mathbb{E}[e_{\mathfrak{a}}(m_{N}(z)^{2}-\mathbb{E}[m_{N}(z)^{2}])]$
$\displaystyle=-2Nm_{\mathrm{sc}}(z)\mathbb{E}[e_{\mathfrak{a}}(m_{N}(z)-\mathbb{E}[m_{N}(z)])]$
$\displaystyle+N^{\varepsilon}\mathcal{O}(N^{-1}y^{-2}),$ (10.9)

and

$\mathbb{E}[e_{\mathfrak{a}}(\partial_{z}m_{N}(z)-\mathbb{E}[\partial_{z}m_{N}(z)])]=N^{\varepsilon}\mathcal{O}(N^{-1}y^{-2}).$
(10.10)

The last term is,

$\displaystyle\frac{1+\delta_{ia}}{N}\sum_{ia}\mathbb{E}[(\partial_{ia}e_{\mathfrak{a}})G_{ia}(z)]$
$\displaystyle=$
$\displaystyle-\frac{2\mathrm{i}\xi}{N}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}\mathrm{tr}G(z)G(w)]\mathrm{d}u\mathrm{d}v.$
(10.11)

In order to deal with this term, we apply the estimate (6.1.1) derived in the
proof of Lemma 6.2. We first have, denoting $w=u+\mathrm{i}v$,

$\displaystyle\frac{1}{N}\int_{\Omega_{\mathfrak{a}}}|(f(u)+\mathrm{i}f^{\prime}(u)v)\chi^{\prime}(v)|\left|\mathbb{E}\left[e_{\mathfrak{a}}\left(\partial_{w}\mathrm{tr}G(w)G(z)-\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\right)\right]\right|\mathrm{d}u\mathrm{d}v$
$\displaystyle\leq$ $\displaystyle
C\int_{\Omega_{\mathfrak{a}}}|(f(u)+\mathrm{i}f^{\prime}(u)v)\chi^{\prime}(v)|N^{-1}(y^{-2}v^{-1}+v^{-2}(y+v)^{-1})\mathrm{d}u\mathrm{d}v$
$\displaystyle\leq$ $\displaystyle
CN^{\varepsilon}(1+\|f^{\prime}\|_{1})N^{-1}y^{-2}$ (10.12)

Let
$t=N^{\varepsilon}(1+\|f^{\prime}\|_{1})\|f^{\prime\prime}\|_{1}^{-1}\wedge
1$. We have,

$\displaystyle\int_{N^{\mathfrak{a}-1}<|v|<t}|f^{\prime\prime}(u)v\chi(v)|\left|\mathbb{E}\left[e_{\mathfrak{a}}\left(\partial_{w}\mathrm{tr}G(w)G(z)-\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\right)\right]\right|\mathrm{d}u\mathrm{d}v\mathrm{d}u\mathrm{d}v$
$\displaystyle\leq$ $\displaystyle
CN^{\varepsilon}\|f^{\prime\prime}\|_{1}\int_{N^{-1}<v<t}vN^{-1}(y^{-2}v^{-1}+y^{-1}v^{-2})\mathrm{d}v\leq
CN^{3\varepsilon-1}((1+\|f^{\prime}\|_{1})y^{-2}+\|f^{\prime\prime}\|_{1}y^{-1}).$
(10.13)

By integration by parts, the Cauchy-Riemann equations (i.e.,
$\partial_{z}F=\partial_{x}F$ for holomorphic $F$) and the Cauchy integral
formula we have,

$\displaystyle\left|\int_{|v|>t}f^{\prime\prime}(u)v\chi(v)\mathbb{E}\left[e_{\mathfrak{a}}\left(\partial_{w}\mathrm{tr}G(w)G(z)-\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\right)\right]\mathrm{d}u\mathrm{d}v\right|$
$\displaystyle=$
$\displaystyle\left|\int_{|v|>t}f^{\prime}(u)v\chi(v)\mathbb{E}\left[e_{\mathfrak{a}}\left(\partial^{2}_{w}\mathrm{tr}G(w)G(z)-\partial^{2}_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\right)\right]\mathrm{d}u\mathrm{d}v\right|$
$\displaystyle\leq$ $\displaystyle
CN^{\varepsilon}\|f^{\prime}\|_{1}\int_{t<v<2}N^{-1}v(y^{-2}v^{-2}+y^{-1}v^{-3})\mathrm{d}v\leq
CN^{3\varepsilon-1}(1+\|f^{\prime}\|_{1})y^{-2}+(1+\|f^{\prime\prime}\|_{1})y^{-1}).$
(10.14)

Therefore,

$\displaystyle-$
$\displaystyle\frac{2\mathrm{i}\xi}{N}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}\mathrm{tr}G(w)G(w)]\mathrm{d}u\mathrm{d}v$
$\displaystyle=$
$\displaystyle-\frac{2\mathrm{i}\xi}{N}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\psi_{a}\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}\mathrm{d}u\mathrm{d}v$
$\displaystyle+$ $\displaystyle
N^{3\varepsilon}\mathcal{O}((1+\|f^{\prime}\|_{1})N^{-1}y^{-2}+N^{-1}(1+\|f^{\prime\prime}\|_{1})y^{-1}).$
(10.15)

This completes the proof. ∎

####  10.1.2 Third order term

######  Lemma 10.3.

We have,

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[\partial^{2}_{ia}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial^{2}_{ia}G_{ia}(z)]$
$\displaystyle=$
$\displaystyle\psi_{\mathfrak{a}}\frac{8m_{\mathrm{sc}}(z)\mathrm{i}\xi}{\pi
N^{1/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)(m_{\mathrm{sc}}(w)+m_{\mathrm{sc}}(z))m_{\mathrm{sc}}^{\prime}(w)\mathrm{d}u\mathrm{d}v$
$\displaystyle-$
$\displaystyle\frac{4\xi^{2}}{\pi^{2}N^{1/2}}\psi_{\mathfrak{a}}m_{\mathrm{sc}}(z)\left(\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)\right)^{2}\mathrm{d}u\mathrm{d}v$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}(1+|\xi|)\mathcal{O}(N^{-1}(\|f^{\prime\prime}\|_{1}+1)y^{-1})+N^{\varepsilon}|\xi|^{2}\mathcal{O}(N^{-1}y^{-1/2}(1+\|f^{\prime}\|_{1})(1+\|f^{\prime\prime}\|_{1}))$
$\displaystyle+$ $\displaystyle N^{\varepsilon}\mathcal{O}(N^{-1}y^{-3/2})$
(10.16)

Proof. We write,

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[\partial^{2}_{ia}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial^{2}_{ia}G_{ia}(z)])]$
$\displaystyle=$
$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[e_{\mathfrak{a}}(\partial_{ia}^{2}G_{ia}(z)-\mathbb{E}[\partial_{ia}^{2}G_{ia}(z)])]$
(10.17) $\displaystyle+$
$\displaystyle\sum_{ia}\frac{2(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[\partial_{ia}e_{\mathfrak{a}}\partial_{ia}G_{ia}(z)]$
(10.18) $\displaystyle+$
$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[(\partial^{2}_{ia}e_{\mathfrak{a}})G_{ia}(z)].$
(10.19)

We start with (10.17). Using (A.4) this is split into,

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[e_{\mathfrak{a}}(\partial_{ia}^{2}G_{ia}(z)-\mathbb{E}[\partial_{ia}^{2}G_{ia}(z)])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}2\mathbb{E}[e_{\mathfrak{a}}(G_{ia}(z)^{3}-\mathbb{E}[G_{ia}(z)^{3}])]$
$\displaystyle+$
$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}6\mathbb{E}[e_{\mathfrak{a}}(G_{ia}(z)G_{ii}(z)G_{aa}(z)-\mathbb{E}[G_{ia}(z)G_{ii}(z)G_{aa}(z)])].$
(10.20)

By similar logic as in the estimation of the term (6.19) we obtain,

$\frac{1}{N^{3/2}}\sum_{ia}2\mathbb{E}[e_{\mathfrak{a}}(G_{ia}(z)^{3}-\mathbb{E}[G_{ia}(z)^{3}])]=N^{\varepsilon}\mathcal{O}(N^{-1}y^{-3/2}).$
(10.21)

Similar logic as in the estimation of the term (6.20) yields,

$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}6\mathbb{E}[e_{\mathfrak{a}}(G_{ia}(z)G_{ii}(z)G_{aa}(z)-\mathbb{E}[G_{ia}(z)G_{ii}(z)G_{aa}(z)])]=N^{\varepsilon}\mathcal{O}(N^{-1}y^{-3/2}).$
(10.22)

This completes estimation of the term (10.17). We turn to (10.18). It is split
into,

$\displaystyle\sum_{ia}\frac{2(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[\partial_{ia}e_{\mathfrak{a}}\partial_{ia}G_{ia}(z)]$
$\displaystyle=$ $\displaystyle\mathrm{i}\xi\sum_{ia}\frac{4}{\pi
N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}G_{ia}(w)(G_{ii}(z)G_{aa}(z))]$
$\displaystyle+$ $\displaystyle\mathrm{i}\xi\sum_{ia}\frac{4}{\pi
N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}G_{ia}(w)G_{ia}(z)^{2}]$
(10.23)

Similar logic to the estimation of (6.21) yields that the following estimate
holds with overwhelming probability,

$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}\partial_{w}G_{ia}(w)G_{ii}(z)G_{aa}(z)=N^{-1/2}m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)+N^{\varepsilon}(N^{-1}v^{-3/2}y^{-1}).$
(10.24)

By direct integration, we obtain the estimate

$\displaystyle\mathrm{i}\xi\sum_{ia}\frac{4}{N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}G_{ia}(w)(G_{ii}(z)G_{aa}(z))]$
$\displaystyle=$ $\displaystyle\mathrm{i}\xi
4N^{-1/2}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)\psi_{\mathfrak{a}}+N^{\varepsilon}|\xi|\mathcal{O}(N^{-1}y^{-1}(1+\|f^{\prime\prime}\|_{1})).$
(10.25)

In a similar fashion, we use the logic used to estimate (6.22) to find,

$\displaystyle i\xi\sum_{ia}\frac{4}{\pi
N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}G_{ia}(w)G_{ia}(z)^{2}]$
$\displaystyle=$ $\displaystyle\mathrm{i}\xi
4N^{-1/2}\frac{1}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}(z)^{2}m_{\mathrm{sc}}^{\prime}(w)\psi_{\mathfrak{a}}+N^{\varepsilon}|\xi|\mathcal{O}(N^{-1}y^{-1}(1+\|f^{\prime\prime}\|_{1})).$
(10.26)

This completes the estimation of (10.18). We split the term (10.19) into,

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{2}}{N^{3/2}}\mathbb{E}[(\partial^{2}_{ia}e_{\mathfrak{a}})G_{ia}(z)]$
$\displaystyle=$ $\displaystyle\sum_{ia}\frac{2\mathrm{i}\xi}{\pi
N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}(G_{ii}(w)G_{aa}(w))G_{ia}(z)]$
$\displaystyle+$ $\displaystyle\sum_{ia}\frac{2\mathrm{i}\xi}{\pi
N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}(G_{ia}(w)^{2})G_{ia}(z)]$
$\displaystyle+$
$\displaystyle\sum_{ia}\frac{-4\xi^{2}}{\pi^{2}N^{3/2}}\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{w}_{1}}\partial_{\bar{w}_{2}}\tilde{f}(w_{1})\tilde{f}(w_{2})\mathbb{E}[e_{\mathfrak{a}}\partial_{w_{1}}G_{ia}(w_{1})\partial_{w_{2}}G_{ia}(w_{2})G_{ia}(z)]$
(10.27)

Similar logic to the estimation of (6.23) and (6.24) shows the estimates,

$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}G_{ia}(z)G_{ii}(w)\partial_{w}G_{aa}(w)$
$\displaystyle=$ $\displaystyle
m_{\mathrm{sc}}(w)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(z)+N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1/2}v^{-2})$
(10.28)

and

$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}G_{ia}(z)G_{ia}(w)\partial_{w}G_{ia}(w)$
$\displaystyle=$ $\displaystyle
m_{\mathrm{sc}}(w)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(z)+N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1/2}v^{-2})$
(10.29)

holds with overwhelming probability. Therefore, by direct integration, we
have:

$\displaystyle\sum_{ia}\frac{2\mathrm{i}\xi}{\pi
N^{3/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}(G_{ia}(w)^{2}+G_{ii}(w)G_{aa}(w))G_{ia}(z)]$
$\displaystyle=\frac{8\mathrm{i}\xi}{\pi
N^{1/2}}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)\psi_{\mathfrak{a}}+N^{\varepsilon}|\xi|\mathcal{O}(N^{-1}(1+\|f^{\prime\prime}\|_{1})y^{-1/2}).$
(10.30)

Denote $w_{i}=u_{i}\pm v_{i}$. Appying the entry-wise local law (2.12) we see
that with overwhelming probability,

$\displaystyle\frac{1}{N^{3/2}}\sum_{ia}\partial_{w_{1}}G_{ia}(w_{1})\partial_{w_{2}}G_{ia}(w_{2})G_{ia}(z)$
$\displaystyle=$
$\displaystyle\frac{m_{\mathrm{sc}}(z)m_{\mathrm{sc}}^{\prime}(w_{1})m_{\mathrm{sc}}^{\prime}(w_{2})}{N^{1/2}}+N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1/2}v_{1}^{-3/2}v_{2}^{-3/2}),$
(10.31)

and so by a similar argument to that leading to the estimate (10.1.1) we have,

$\displaystyle\sum_{ia}\frac{-4\xi^{2}}{\pi^{2}N^{3/2}}\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{w}_{1}}\partial_{\bar{w}_{2}}\tilde{f}(w_{1})\tilde{f}(w_{2})\mathbb{E}[e_{\mathfrak{a}}\partial_{w_{1}}G_{ia}(w_{1})\partial_{w_{2}}G_{ia}(w_{2})G_{ia}(z)]$
$\displaystyle=$
$\displaystyle\frac{-4\xi^{2}}{\pi^{2}N^{1/2}}\psi_{\mathfrak{a}}m_{\mathrm{sc}}(z)\left(\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)\mathrm{d}u\mathrm{d}v\right)^{2}+N^{\varepsilon}|\xi|^{2}\mathcal{O}(N^{-1}y^{-1/2}(1+\|f^{\prime}\|_{1})(1+\|f^{\prime\prime}\|_{1}))$
(10.32)

This yields the claim. ∎

####  10.1.3 Fourth order term

######  Lemma 10.4.

We have,

$\displaystyle\frac{1}{N^{2}}\sum_{ia}\frac{(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[\partial_{ia}^{3}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial_{ia}^{3}G_{ia}(z)])]$
$\displaystyle=$
$\displaystyle\frac{-12\xi\mathrm{i}}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}\mathrm{d}u\mathrm{d}v$
$\displaystyle+$ $\displaystyle
N^{\varepsilon}(1+|\xi|^{3})(1+\|f^{\prime}\|_{1}^{3})\mathcal{O}(N^{-1}y^{-1}(1+\|f^{\prime\prime}\|_{1}))$
(10.33)

Proof. We write,

$\displaystyle\frac{1}{N^{2}}\sum_{ia}\frac{(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[\partial_{ia}^{3}e_{\mathfrak{a}}(G_{ia}(z)-\mathbb{E}[\partial_{ia}^{3}G_{ia}(z)])]$
$\displaystyle=$
$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[e_{\mathfrak{a}}(\partial_{ia}^{3}G_{ia}(z)-\mathbb{E}[\partial_{ia}^{3}G_{ia}(z)])]$
(10.34) $\displaystyle+$
$\displaystyle\sum_{ia}\frac{3(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[(\partial_{ia}e_{\mathfrak{a}})(\partial_{ia}^{2}G_{ia}(z))]$
(10.35) $\displaystyle+$
$\displaystyle\sum_{ia}\frac{3(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[(\partial_{ia}^{2}e_{\mathfrak{a}})(\partial_{ia}G_{ia}(z))]$
(10.36) $\displaystyle+$
$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[(\partial_{ia}^{3}e_{\mathfrak{a}})(G_{ia}(z))]$
(10.37)

For (10.34), the same logic used to estimate (6.42) shows that with
overwhelming probability,

$\frac{1}{N^{2}}\sum_{ia}(1+\delta_{ia})^{3}(\partial_{ia}^{3}G_{ia}(z)-\mathbb{E}[\partial_{ia}^{3}G_{ia}(z)])=N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1}).$
(10.38)

For (10.35), we see from the entry-wise local law (2.12) that

$\frac{1}{N^{2}}\sum_{ia}\partial_{w}G_{ia}(w)(2G_{ia}^{3}(z)+6G_{ii}(z)G_{aa}(z)G_{ia}(z))=N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1/2}v^{-3/2})$
(10.39)

holds with overwhelming probability. Therefore by direct integration,

$\sum_{ia}\frac{3(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[(\partial_{ia}e_{\mathfrak{a}})(\partial_{ia}^{2}G_{ia}(z))]=N^{\varepsilon}\mathcal{O}(|\xi|N^{-1}y^{-1/2}\|f^{\prime\prime}\|_{1}).$
(10.40)

For (10.36), we first write it as

$\displaystyle\sum_{ia}\frac{(1+\delta_{ia})^{3}}{N^{2}}\mathbb{E}[(\partial_{ia}^{2}e_{\mathfrak{a}})(\partial_{ia}G_{ia}(z))]$
$\displaystyle=$ $\displaystyle\frac{-2\xi\mathrm{i}}{\pi
N^{2}}\sum_{ia}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}(G_{ii}(w)G_{aa}(w)+G_{ia}(w)^{2})(G_{ii}(z)G_{aa}(z)+G_{ia}(z)^{2})]$
$\displaystyle+$
$\displaystyle\frac{4\xi^{2}}{\pi^{2}N^{2}}\sum_{ia}\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{w}_{1}}\partial_{\bar{w}_{2}}\tilde{f}(w_{1})\tilde{f}(w_{2})\partial_{w_{1}}\partial_{w_{2}}\mathbb{E}[e_{\mathfrak{a}}G_{ia}(w_{1})G_{ia}(w_{2})(G_{ii}(z)G_{aa}(z)+G_{ia}(z)^{2})]$
(10.41)

Using the same logic as leading to the estimate of (6.1.3), we see that with
overwhelming probability,

$\displaystyle\frac{1}{N^{2}}\sum_{ia}\partial_{w}(G_{ii}(w)G_{aa}(w)+G_{ia}(w)^{2})(G_{ii}(z)G_{aa}(z)+G_{ia}(z)^{2})$
$\displaystyle=$ $\displaystyle
2m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}+N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1}v^{-2}),$
(10.42)

and so

$\displaystyle\frac{-2\xi\mathrm{i}}{\pi
N^{2}}\sum_{ia}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)\mathbb{E}[e_{\mathfrak{a}}\partial_{w}(G_{ii}(w)G_{aa}(w)+G_{ia}(w)^{2})(G_{ii}(z)G_{aa}(z)+G_{ia}(z)^{2})]$
$\displaystyle=$
$\displaystyle\frac{-4\xi\mathrm{i}}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{w}}\tilde{f}(w)m_{\mathrm{sc}}^{\prime}(w)m_{\mathrm{sc}}(w)m_{\mathrm{sc}}(z)^{2}\mathrm{d}u\mathrm{d}v+N^{\varepsilon}|\xi|\mathcal{O}(N^{-1}\|f^{\prime\prime}\|_{1}y^{-1}).$
(10.43)

By the entrywise local law (2.12) we have,

$\frac{1}{N^{2}}\sum_{ia}\partial_{w_{1}}\partial_{w_{2}}G_{ia}(w_{1})G_{ia}(w_{2})(G_{ii}(z)G_{aa}(z)+G_{ia}(z)^{2})=N^{\varepsilon}\mathcal{O}(N^{-1}v_{1}^{-3/2}v_{2}^{-3/2}),$
(10.44)

with overwhelming probability. Therefore, by a similar argument as to the one
that leads to (10.1.1) that uses integration by parts we find,

$\displaystyle\frac{\xi^{2}}{\pi^{2}N^{2}}\sum_{ia}\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{w}_{1}}\partial_{\bar{w}_{2}}\tilde{f}(w_{1})\tilde{f}(w_{2})\partial_{w_{1}}\partial_{w_{2}}\mathbb{E}[e_{\mathfrak{a}}G_{ia}(w_{1})G_{ia}(w_{2})(G_{ii}(z)G_{aa}(z)+G_{ia}(z)^{2})]$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}|\xi|^{2}\mathcal{O}(N^{-1}(1+\|f^{\prime\prime}\|_{1})(1+\|f^{\prime}\|_{1}))$
(10.45)

This completes (10.36). For the final term (10.37), we may simply use the
estimate,

$|\partial_{ia}^{3}e_{\mathfrak{a}}|\leq(1+|\xi|^{3})(1+\|f^{\prime}\|_{1}^{3})\left(\delta_{ia}+N^{-1/2}(1+\|f^{\prime\prime}\|_{1})^{1/2}\right)$
(10.46)

which follows from direct calculation, the entry-wise local law (2.12) and an
argument using integration by parts similar to the one leading to (10.1.1).
This completes the proof. ∎

####  10.1.4 Fifth order term

######  Lemma 10.5.

We have,

$\displaystyle\frac{1}{N^{5/2}}\sum_{ia}(1+\delta_{ia})^{4}\mathbb{E}[\partial_{ia}^{4}(e_{\mathfrak{a}}(G_{ia}-\mathbb{E}[\partial_{ia}^{4}G_{ia}])]$
$\displaystyle=$ $\displaystyle
N^{\varepsilon}(1+|\xi|^{4})(1+\|f^{\prime}\|_{1})^{4}\mathcal{O}(N^{-1}y^{-1}(1+\|f^{\prime\prime}\|_{1}))$
(10.47)

Proof. We write the fifth order term as,