text:
\section{Introduction}

The Wigner-Dyson-Mehta conjecture asserts that the local eigenvalue statistics of large random matrices are universal in the sense that they depend only on the symmetry class of the model - real symmetric or complex Hermitian - but are otherwise independent of the underlying details of the model. There are two types of universality results. Bulk universality involves the spacing distribution eigenvalues that lie well within the support of the limiting spectral distribution, while edge universality involves the extreme eigenvalues.

There has recently been a lot of progress made in proving the Wigner-Dyson-Mehta conjecture in a increasingly large class of models.  In \cite{Erdos2012a,Erdos2013a,Erdos2009b,Erdos2012c,Erdos2012b,Erdos2010universality}, universality was proved for Wigner matrices whose entries are independent and have identical variance; parallel results are obtained independently in various cases in \cite{Tao2010,Tao2011a}. In \cite{Ajanki2015,ziliang}, this type of result was extended to more general variance patterns, while still maintaining the independence of matrix entries.

Most of the previous works rely heavily on the independence between matrix entries, and  deal with bulk universality. Only recently have people proved results on models with general correlation structure. In \cite{Che2016, Erdos2015, Ajanki2016}, bulk universality is proved for matrices where the correlation decays fast enough. In a recent paper \cite{Erdos2017d}, Erd{\"o}s et al. consider a model where the correlation between matrix entries has a power law decay of order $d\geq 12$ in the long range and $d\geq 2$ in the short range. They use a combinatorial expansion to get optimal local law, then prove bulk universality. They remark in Example 2.12 that in the Gaussian case, $d\geq 2$ for both long range and short range correlation is sufficient to satisfy the assumptions of their main theorem.

In this paper, we prove edge universality for Gaussian matrices with a correlation structure that decays as a power law of order $d>2$, namely $|\EE{h_{ij}h_{kl}}|\leq\frac{1}{|i-l|^{d} + |j-k|^{d}}$ where $h_{ij}$ are the entries of the random matrix $H$. Our proof avoids the expansion of Greens function, but relies on a decomposition of Gaussian random variables into a sum of short range interactions.

Recent proofs of universality have followed a robust three step strategy:
\begin{enumerate}
    \item Prove a local law for the empirical eigenvalue distribution at small scales.
    \item  Study the convergence of the DBM (Dyson-Brownian motion) in short time scales to local equilibrium.
    \item  Prove that the eigenvalue spacing distribution does not change too much during the short time evolution of DBM.
\end{enumerate}
Step 1, finding the local law, is generally the most difficult and model dependent.
The strategy in proving this local law is deriving a self-consistent equation for the Green's function $G=(H-z)^{-1}$.

One can heuristically derive a self consistent equation by taking expectation and performing integration by parts on $G(H-z)=I$. One notices that there is a linear operator $S$ such that $\EE{G(-S(G)-z)} =1$. Removing the expectation creates some error term. The goal is to show that a small error exists with high probability on our matrix ensemble, as is done in \cite{Erdos2015,Che2016}.

From \cite{Che2016}, it is known that the self-consistent equation for correlated matrix entries is of the form $G(- S(G)-z ) = I$ that can be transformed into the following vector equation via local Fourier transform.
\begin{gather}
    g(x)(-\Psi(g)(x)-z)=1,\quad x\in L^\infty([0,1]^2)
\end{gather}
where $\Psi:L^\infty([0,1]^2)\to L^\infty([0,1]^2)$ is an integral operator, which is the continuous version of $S$.
There are two difficulties in our case: getting a small error for our self-consistent equation and proving the stability of the equation near the edge.

In order to get a small error for the self-consistent equation, we avoid the procedure of removing blocks of elements, which requires combinatorial expansion, but instead applied integration by parts and concentration results along a careful decomposition of the probability space. This gives us a weak local law which can be bootstrapped to give an even better bound for the expected value of the Green's function. Once we have bounds on the expected value, we use the concentration of eigenvalues about its mean value in order to show a version of upper bound for the top eigenvalue along the edge.

In order to prove the stability, we first embed the matrix space into the continuous space $C^\infty([0,1]^2)$, up to small errors. However, entry-wise error is not small enough to allow this embedding. We noticed the fact that the operator $S$ has a smoothing effect and will reduce the error; thus, a double iteration of the operator $F(G)=(-S(G)-z)^{-1}$ created a matrix $F(F(G))$ that satisfies
\begin{equation}
    F(F(G)) = F(F(F(G))) + R,
\end{equation}
where $R$ has sufficiently fast decay on off-diagonal entries. A similar strategy based on the smoothing effect of $F$ is also used in \cite{Ajanki2016}.  Then we can embed and apply stability of the continuous solution. In order to prove the decay properties of the double iteration, we applied a preturbation around a fixed matrix that is known to have decay of matrix entries. With sufficiently strong upper bounds on the top eigenvalue and lower bounds on the bottom eigenvalue, we are able to use the result of \cite{Yau2017} to get universality for the extreme eigenvalues. The result of \cite{Erdos2017d} is sufficient to locate the extremal eigenvalues but we have an approach that allows us to get optimal correlation decay without  a combinatorial expansion.

The structure of this paper is as follows. The second section is devoted to proving a self-consistent equation with sufficiently small error. The third section of this paper involves proving stability of the self-consistent equation to get a local law to prove an upper bound on eigenvalues. The final section uses this upper bound in order to prove universality.

\textbf{Acknowledgements}: We thank J. Huang and H-T Yau for useful discussions. The argument of Sec 3.3. came from a private communication with the two.

\section{Derivation of self-consistent equation}

\subsection{The Model and Assumptions}

For $N \in\mathbb{N}$, we consider a symmetric matrix $H = (h_{ij}^{(N)})_{1\leq i,j\leq N}$ whose entries are centered Gaussian random variables. For simplicity of notation we omit the dependence of $h_{ij}$ on $N$.  Let $\xi_{ijkl}:= N\EE{ h_{ij}h_{kl}}$. Assume there is a Lipschitz function $\phi: \mathbb{T} \times \mathbb{Z} \to {\mathbb R}$ such that
\begin{equation}\label{def:xi}
    \xi_{ijkl} = \phi(i/N, j/N, k-i,l-j)+O(N^{-1}),\forall i\leq k, j\leq l.
\end{equation}
Let $\mathbb{Z}_N = \mathbb{Z}/N\mathbb{Z}$, and from now on we view the indices $i,j,k,l$ as elements in $\mathbb{Z}/N\mathbb{Z}$. On  $\mathbb{Z}/N\mathbb{Z}$ we define the natural distance $\dist_{\mathbb{Z}/N\mathbb{Z}} (i,j):= \min\{\abs{i-j+kN}|k\in\mathbb{Z}\}$, which for simplicity of notation we still denote by $\abs{i-j}$ unless there is danger of confusion. Assume that there are universal constants  $d > 2$ and $\mathsf{c}_1>0$ such that
\begin{equation}\label{def:xidecay}
    \abs{\xi_{ijkl}} \leq \mathsf{c}_1^2 \max\left\{ \inv{(\abs{i-k}+\abs{j-l}+1)^d}, \inv{(\abs{i-l}+\abs{j-k} + 1)^d}\right\}, \quad \forall i,j,k,l \in \mathbb{Z}/N\mathbb{Z}.
\end{equation}
In this paper we fix an arbitrary  $\alpha\in(2,d)$ and consider it as a universal constant.  Assume that there is a constant $\mathsf{c}_2>0$, such that $H$ allows a decomposition
\begin{equation}\label{def:c2}
    H=\mathsf{c}_2 X + Y,
\end{equation}
where $X$ is a GOE matrix independent from $Y$.

We say that a constant is universal if it only depends on $\mathsf{c}_1,\mathsf{c}_2,d$ and $\phi$. In this paper we denote $a\lesssim b$ if there is a universal constant $c>0$ such that $a\leq cb$. We also denote $a\sim1$ if $a\lesssim 1$ and $1\lesssim a$.

For $\beta>0$ and any matrix $A$ (finite square or infinite) we define the following norms,
\begin{equation}
    \norm{A}_\beta : = \sup_{i,j} \left(\abs{A_{ij}}(1+\abs{i-j})^\beta\right),\quad \abs{A}_\infty : = \max_{i,j} \abs{A_{ij}}.
\end{equation}

Let $\lambda_1\leq \cdots \leq \lambda_N$ be the eigenvalues of $H$. Let $\hat{\lambda}_1\leq \cdots\leq \gamma_N$ be the eigenvalues of an $N$ by $N$ GOE matrix (i.e. a matrix $A=(Z_{ij}+Z_{ji})_{1\leq i,j\leq N}$ where $(Z_{ij})$ are i.i.d. copies of an $N(0,1/N)$ random variable). The main result we will prove is the following.
\begin{theorem}

There exists a universal constant $\gamma$  such that for any $f\in C^1( \mathbb{R}^{k-1})$, the following inequality holds for $N$ large enough.
\begin{gather}
 |\mathbb{E}_{H}[f(\gamma N^{2/3}(\lambda_{2} -\lambda_{1}), ...\gamma N^{2/3}(\lambda_{k} - \lambda_1))] -
\mathbb{E}_{\text{GOE}}[f(N^{2/3}(\hat{\lambda}_{2}-\hat{\lambda}_1),...N^{2/3}(\hat{\lambda}_{k} -\hat{\lambda}_1))] \le N^{-c}
\end{gather}
\end{theorem}

\subsection{The Loop Equation}\label{sec:loopequation}
We will use the following lemma.
\begin{lemma}\label{lem:stein}
Let $Z=(Z_k)_{k=1}^p$ be a centered Gaussian random vector in ${\mathbb R}^p$ with covariance matrix $\Sigma\in{\mathbb R}^{p\times p}$. Let $f\in C^1({\mathbb R}^p)$.Then,
\[
    \EE{f(Z)Z_l} = \sum_{k=1}^p \EE{\partial_kf(Z)}\Sigma_{kl}, \forall 1\leq l\leq p.
\]
\end{lemma}
\begin{proof}
By a linear change of variable, we may assume without loss of generality that $\Sigma=I$ and $l=1$. It is sufficient to show that $\mathbb{E}[f(Z)Z_1]=\mathbb{E}[\partial_1f(Z)]$. Let $\mathcal{F}:=\sigma(Z_2,\cdots,Z_p)$, it is sufficient to show $\mathbb{E}[f(Z)Z_1\vert \mathcal{F}] = \mathbb{E}[\partial_1f(Z)\vert \mathcal{F}]$. This directly follows from an identity known as Stein's lemma, which says that if $X\sim N(0,1)$ and $h\in C^1({\mathbb R})$, then $\EE{h(X)X}=\EE{h'(X)}$.
\end{proof}
We also use the following decomposition lemma.
\begin{lemma}\label{lem:decomp}
Let $Z=(Z_k)_{k=1}^p$ be a centered Gaussian random vector. Let $1\leq q< p$. Then, there is a constant matrix $(a_{kl})_{1\leq l\leq q,q+1\leq k\leq p}$ such that
\[
    Z_k = \sum_{l=1}^q a_{kl}Z_l + \tilde{Z}_k,
\]
where $(\tilde{Z}_k)_{k=q+1}^p$ are Gaussian random variables independent from $(Z_l)_{l=1}^q$.
\end{lemma}
\begin{proof}
Up to a linear transform, we may assume without loss of generality that $(Z_l)_{l=1}^q$ has covariance matrix $I_{q\times q}$. Let $\tilde{Z}_k:= Z_k - \sum_{l=1}^1 \EE{Z_kZ_l}Z_l$. It is easy to check that $(\tilde{Z}_k)_{k=q+1}^p$ are uncorrelated with $(Z_l)_{l=1}^q$. Being linear combinations of Gaussian random variables, $(\tilde{Z}_k)_{k=q+1}^p$ are still Gaussian.  Therefore $\tilde{Z}_k$ are independent from $(Z_l)_{l=1}^q$, since zero correlation is equivalent to independence for Gaussian random variables.
\end{proof}

We start with the trivial matrix identity $G(H-z) = I$, which can be written as follows
\begin{gather} \label{r:loopequation}
\sum_{k} G_{ik} h_{kj} - z G_{ij} = \delta_{ij}, \quad i,j\in\mathbb{Z}/N\mathbb{Z}.
\end{gather}
Without loss of generality, fix $j=1$. According to Lemma \ref{lem:decomp} we may write,
\begin{equation}\label{r:decomp}
	h_{ab} = \sum_{k=1}^N \gamma_{abk1} h_{k1} + \tilde h_{ab},
\end{equation}
 where $\tilde h_{ab}$ is a Gaussian random variable that is independent from $(h_{k1})_{1\leq k \leq N}$. In particular, $\gamma_{a1k1} = \delta_{ak}$, $\tilde{h}_{a1} = 0$, $\forall a\in\mathbb{Z}/N\mathbb{Z}$. In order to apply Lemma \ref{lem:stein} on \eqref{r:loopequation}, let  $\mathcal{F}_{1}$ be the $\sigma$-algebra generated by $(\tilde h_{ab})_{a\neq 1,b\neq 1}$.  Define conditional expectation operator
\[
	\mathbb{E}_1 [ \cdot ] : = \EE{ \cdot \vert \mathcal{F}_1}.
\]
We will then be able to apply Lemma \ref{lem:stein} to get the following
\begin{gather}\label{r:condexp1}
	 \delta_{i1} = \sum_k \mathbb{E}_1[G_{ik} h_{k1}] - z \mathbb{E}_1[G_{i1}] = -\sum_{k,a,b}\mathbb{E}_1[ G_{ia}G_{bk}] \xi_{abk1} - z \mathbb{E}_1[G_{i1}] .
\end{gather}
For technical reasons define the cut-off version of $\xi$ as follows, $\tilde{\xi}_{iklj}= \min\{\max\{\xi_{iklj},-\mathsf{c}_1^2\abs{i-j}^{-d}\},\mathsf{c}_1^2\abs{i-j}^{-d}\}$, so that $\tilde{\xi}_{iklj}$ has a power-law decay as $i$ and $j$ gets farther. Define a linear map $S:{\mathbb R}^{N\times N} \to {\mathbb R}^{N\times N}$ by.
\begin{equation}\label{def:S}
    	(S(M))_{pq} : = \inv{N}\sum_{\alpha,\beta} \tilde{\xi}_{p\alpha\beta q}  M_{\alpha\beta}.
\end{equation}
Therefore, \eqref{r:condexp1} is equivalent to
\begin{equation}\label{E1}
	 - \mathbb{E}_1 \left[[G S(G)]_{i1}\right]- z \mathbb{E}_1\left[G_{i1} \right] = \delta_{i1} + O(N^{-1}\max_{k,l}\abs{G_{kl}}).
\end{equation}
Notice that the expectation operator $\mathbb{E}_1$ is equivalent to integrating over $N$ weakly dependent Gaussian random variables, we may remove the expectation up to the cost of some small error terms, after which, we would get a self-consistent equation in the following form.
\begin{equation}
    G(-S(G)-z) = I + \textrm{error}.
\end{equation}
Define a map $F:{\mathbb R}^{N\times N} \to {\mathbb R}^{N\times N}$ via
\begin{equation}\label{def:F}
    F(M) = (-z- S(M))^{-1}.
\end{equation}
Then the above equation can be written as the perturbation of a fixed point equation
\begin{equation}
    G=F(G) +  \textrm{error}.
\end{equation}
Here the error is entry-wise bounded by roughly $O((N\eta)^{-\inv{2}})$. However, this entry-wise bound is not strong enough to use the stability of the equation $G=F(G)$. Therefore, we iterate the map $F$ on $G$ to get
\begin{equation}
    F(F(G)) = F(F(F(G))) + \textrm{new error}.
\end{equation}
The new error term has a power-law decay on the off-diagonal entries, hence is much smaller than the original error. This allows us to get an estimate on $F(F(G))$. Using $F(F(G))$ we can recover $G$ and get a bound on $\abs{G-G_0}$ where $G_0$ is some deterministic matrix.

\subsection{Limiting Version of self-consistent equation}
Consider $\mathcal{K}:=C(\mathbb{T}^2)$ and $\mathcal{K}_+ :=\{ g \in\mathcal{K}\vert \Im g(s,u)>0,\forall s,u\in\mathbb{T}\}$. Recall the function $\phi$ in \eqref{def:xi}. Define
\begin{equation}
    \varphi(s,t,u,v):= \sum_{k,l}\phi(s,t,k,l) \mathrm{e}^{-2\pi\mathrm{i}(uk-vl)}.
\end{equation}
The argument in Lemma 4.15 of \cite{Che2016} can be modified to show that $\varphi\sim 1$. Also, the decay condition \eqref{def:xidecay} guarantees that $\varphi$ is Lipschitz. Define $\Psi:\mathcal{K}_+\to\mathcal{K}_+$ via $\Psi(h)(s,u):=\iint_{\mathbb{T}^2}\varphi(s,t,u,v)h(t,v)\mathrm{d} t\mathrm{d} v$ and $\Phi:\mathcal{K}_+\to\mathcal{K}_+$ via $\Phi(h):=(-\Psi(h)-z)^{-1}$. Consider the fixed point equation $g = \Phi(g)$, or equivalently,
\begin{equation}\label{sceinfty}
    g(-\Psi(g)-z)=1.
\end{equation}
If we think of $\hat{g}$ as an infinite matrix, we may write the above equation as
\begin{equation}\label{sceinfty1}
    \hat{g}(-S(\hat{g})-z)=I.
\end{equation}
Equations like \eqref{sceinfty} are studied in detail in \cite{Ajanki2015a}. Since the function $\varphi$ is bounded above and below away from $0$, the function $\Phi$ satisfies conditions A1-A3 and is block fully indecomposible in Definition 2.9 of \cite{Ajanki2015a}. Also, since $\varphi$ is Lipschitz, it satisfies (2.22) in that article. Therefore,  their Theorem 2.6 says that the above equation has a unique solution $g\in\mathcal{K}_+$, and there is a universal constant $\mathsf{c}_3<+\infty $ such that
\begin{equation}
    \sup_{z\in\mathbb{C}^+} \norm{g}_\infty \leq \mathsf{c}_3.
\end{equation}
Let $m(z): = \iint_{\mathbb{T}^2} {g}(s,u)\mathrm{d} s \mathrm{d} u$. Then $m$ is the Stieltjes transform of a compactly supported probability measure $\nu$ on ${\mathbb R}$, i.e.,
\begin{equation}
    m(z) = \int_{\mathbb R} \frac{\nu(\mathrm{d} x)}{x- z},\quad\forall z\in\mathbb{C}^+.
\end{equation}
Then Theorem 2.6 in \cite{Ajanki2015a} says that $\nu$ has a $\frac{1}{3}$-H{\"o}lder continuous density $\rho \in C_c({\mathbb R})$ such that it has square-root behavior at the left and right edges, i.e., let
\begin{equation}
    E_L:= \inf \supp \nu, \quad E_R:= \sup \supp \nu.
\end{equation}
Then, there are $c_L, c_R>0$ s.t.
\begin{equation}
    \rho(E_L + t) = c_L\sqrt{t} + O(t),\quad \rho(E_R-t) = c_R\sqrt{t}+O(t),\text{ as } t\to 0_+.
\end{equation}
For $h\in\mathcal{K}$, define the Fourier coefficients $\hat{h}(s,k):= \int_\mathbb{T} h(s,u) \mathrm{e}^{-2\pi\mathrm{i} k u} \mathrm{d} u$. On $\mathcal{K}$ we may define a norm $\norm{\cdot}_\beta$ for $\beta\geq 0$:
\begin{equation}
    \norm{h}_\beta:= \sup_{s\in\mathbb{T},k\in\mathbb{Z}} \abs{\hat{h}(s,k)}(1+\abs{k})^\beta.
\end{equation}
In view of Theorem \ref{Jaffard}, it is easy to see that $\norm{g}_\alpha\vee \norm{g^{-1}}_\alpha \lesssim 1$ on any bounded subdomain of $\mathbb{C}^+$. For any $N\in\mathbb{N}$, define a discretization operator $D^{(N)}: \mathcal{K} \to \bC^{N\times N}$ by
\begin{equation}
    D(h)_{ij}: = \hat{h}(i/N, j-i).
\end{equation}
We have the following lemma concerning the discretization $D(g)$:
\begin{lemma}\label{D}
Let $a,b\in\mathcal{K}$. Assume that $b$ is Lipschitz in the first variable in the sense that $\abs{b(s,u)-b(s',u)}\leq L\abs{s-s'},\forall s,s'\in\mathbb{T},u\in\mathbb{T}$. Then, $\norm{D(a)D(b)-D(ab)}\lesssim N^{-\inv{2}}(L+\norm{b}_\alpha)\norm{a}_\alpha$, also, $\norm{D(a)D(b)^*-D(a\bar{b})}\lesssim N^{-\inv{2}}(L+\norm{b}_\alpha)\norm{a}_\alpha$.
\end{lemma}
\begin{proof}
By definition, $(D(a)D(b)-D(ab))_{ij}= \sum_k \hat{a}(i/N,k-i)(\hat{b}(k/N,j-k)-\hat{b}(i/N,j-k))$, therefore, using the decay of $\hat{a}$ and the Liptchitz continuity of $\hat{b}$, we have
\begin{equation}\label{D1}
    \abs{(D(a)D(b)-D(ab))_{ij}} \leq \sum_k \frac{\norm{a}_\alpha}{\abs{k-i}^\alpha}\frac{L\abs{k-i}}{N}\lesssim N^{-1}L\norm{a}_\alpha.
\end{equation}
On the other hand, $\norm{D(a)D(b)-D(ab)}_\alpha\lesssim \norm{a}_\alpha \norm{b}_\alpha$, hence
\begin{equation}\label{D2}
        \abs{(D(a)D(b)-D(ab))_{ij}} \leq \norm{a}_\alpha\norm{b}_\alpha(1+\abs{i-j})^{-\alpha}.
\end{equation}
Therefore $\norm{D(a)D(b)-D(ab)}_{l^\infty\to l^\infty} \lesssim (L+\norm{b}_\alpha)\norm{a}_\alpha\sum_k(N^{-1}\wedge  \abs{k}^{-2} ) \lesssim N^{-\inv{2}}(L+\norm{b}_\alpha)\norm{a}_\alpha$. Similarly, the $l^1\to l^1$ norm is bounded by the same quantity, hence the operator norm has the same bound by interpolation. The second estimate follows from a similar argument.
\end{proof}

Let $Z(z):= \{ \abs{g(s,t)}\vert s,t\in\mathbb{T}\}$. From equation \eqref{sceinfty} we know that $Z$ is bounded away from $0$ and $+\infty$. For $K>0$ let $\mathcal{D}_K=\{z\in\mathbb{C}^+\vert \abs{z}\leq K\}$.
\begin{corollary} \label{col:DisSV}
There is an $N(K)>0$ such that for any $ N>N(K)$ and $z\in \mathcal{D}_K$, the singular spectrum of $D(g)$ is in the $N^{-\inv{3}}(\log N)^{-1}$-neighborhood of $Z(z)$.
\end{corollary}
\begin{proof}
Let $\theta\ll 1 $ be some parameter to be chosen. Let $x\in{\mathbb R}_+$ s.t. $\dist(x,Z)\geq \theta$. Let $h:= \frac{1}{\abs{g}^2 -x^2}$. Then
\[
    \norm{(D(g)D(g)^* -x^2) D(h) -I}\leq \norm{D(g)D(g)^*-D(\abs{g}^2)}\norm{D(h)} + \norm{D(\abs{g}^2-x^2)D(h)-D(1)}.
\]
According to Lemma \ref{D}, we have $\norm{(D(g)D(g)^* -x^2) D(h) -I}\lesssim \norm{h}_2+L$, where $L$ is the Lipschitz constant of $h$ with respect to the first variable. By chain rule we know that $\norm{h''}_\infty \lesssim \theta^{-3}$ and $L\lesssim \theta^{-2}$. Therefore, $\norm{h}_2\lesssim \theta^{-3}$ and hence $\norm{(D(g)D(g)^* -x^2) D(h) -I}\lesssim \theta^{-3}$. Choose $\theta = N^{-\inv{3}}(\log N)^{-1}$. Then $D(g)D(g)^*-x^2$ is invertible for $N$ large enough. That means $x$ is not in the singular spectrum of $D(g)$.
\end{proof}

\begin{corollary}\label{col:DisSV1}
Let $R:= D(g)(-S(D(g))-z)-I$. Then, for any $z\in \mathcal{D}_K$,
\begin{equation}
    \abs{R_{ij}}\leq C(K) N^{-1} \wedge \abs{i-j}^{-2}.
\end{equation}
In particular, $\norm{R}\leq C(K) N^{-\inv{2}}$.
\end{corollary}
\begin{proof}
According to Lemma \ref{D} and equation \eqref{sceinfty}, we know
\[
    \abs{(D(g)(-D(\Psi(g))-z)-I)_{ij}}\lesssim N^{-1} \wedge \abs{i-j}^{-2}.
\]
By definition, $(D(\Psi(g))_{kl} = \sum_p \int_\mathbb{T} \phi(k/N,t,l-k,p) \hat{g}(t,p)\mathrm{d} t$, $(S(D(g)))_{kl}=\inv{N}\sum_{p,q} \phi(k/N,q/N,l-k,p)\hat{g}(t,p)$. Using the Lipschitz-ness of $\phi$ and $g$, we have $\abs{(D(\Psi(g))_{kl}-(S(D(g)))_{kl})}\lesssim N^{-1} \wedge \abs{k-l}^{-2}$. Therefore,
\[
    \abs{(D(g)(-S(D(g))-z)-I)_{ij}}\lesssim N^{-1} \wedge \abs{i-j}^{-2},
\]
as desired.
\end{proof}

\begin{corollary} \label{col:Nrmbndit}
Recall the definition \eqref{def:F} of $F$. For all sufficiently large N, there exists a constant $c>0$ such that $\norm{F(D(g))-D(g)}\vee\norm{F(F(D(g)))-D(g)}\leq cN^{-\inv{2}}$. In particular, the singular spectrum of $F(D(g))$ and $F(F(D(g)))$ are contained in a compact subset of ${\mathbb R}_+$.
\end{corollary}
\begin{proof}
Using the notation from the previous corollary, if $D(g)(-S(D(g))-z)-I=R$, then
\[
    F(D(g)) = (I+R)^{-1}D(g).
\]
Since $\norm{R}\lesssim N^{-\inv{2}}$ and $\norm{D(g)}\lesssim 1$, we know $\norm{(I+R)^{-1}D(g)-D(g)}\lesssim N^{-\inv{2}}$. From perturbation theory we know that the singular spectrum of $F(D(g))$ is within the $N^{-\inv{2}}$ of that of $D(g)$, therefore it is a compact subset of ${\mathbb R}_+$. On the other hand, a simple algebraic calculation yields
\[
    F(F(D(g)))= \left(I+F(D(g))S(F(D(g))R)\right)^{-1}F(D(g)).
\]
Note that $\norm{F(D(g))S(F(D(g))R)}\lesssim N^{-\inv{2}}$, so the singular spectrum of $F(F(D(g)))$ is within the $O(N^{-\inv{2}}$ neighborhood of that of $F(F(D))$, hence is a compact subset of ${\mathbb R}_+$.
\end{proof}

For $z\in\mathbb{C}^+$, define
\begin{equation}\label{def:kappa}
    \kappa(z):= \dist(z,\supp \nu),\quad \rho(z):= \rho(\Re z),\quad \omega(z):=\kappa(z)^{\frac{2}{3}} + \rho(z)^2.
\end{equation}
Theorem 2.8 in \cite{Ajanki2015a} implies the following stability result:
\begin{lemma}\label{lem:stab}
    There is a universal constant $\mathsf{c}_6$ such that if $\tilde{g} \in \mathcal{K}$ satisfies
    \begin{equation}
        \tilde{g}(-{\Psi}(\tilde{g}) -z) = 1 + r
    \end{equation}
    and $\norm{\tilde{g}-g}_\infty\leq \mathsf{c}_6 ( \kappa^{\frac{2}{3}} + \rho)$, then $\norm{\tilde{g}-g}_\infty\leq \mathsf{c}_6^{-1} \omega^{-1}$.
\end{lemma}
\subsection{Concentration lemmas}
The following lemma says that a Lipschitz function of weakly dependent Gaussian random variables concentrates around its expectation.
\begin{lemma}\label{lem:gaussianconcentration}
Let $X = (X_1,\cdots,X_n)$ be an array of centered Gaussian random variables with covariance matrix $\Sigma$. Let $f:{\mathbb R}^N\to{\mathbb R} $ be a Lipschitz function, such that $\abs{f(x)-f(y)} \leq L\abs{x-y},\forall x,y\in{\mathbb R}^N$. Then
\[
		\mathbb{P}\left[ \left| f(X) - \mathbb{E} f(X)\right|\geq t \right] \leq 2 \mathrm{e} ^{-\frac{t^2}{2L^2 \norm{\Sigma}}} ,\quad\forall t> 0.
\]
\end{lemma}
\begin{proof}
Let $Y=\Sigma^{-1/2} X $ so that $Y$ is an $n$-dimensional random vector with independent $N(0,1)$ components. In \cite{Wainwright},
\[
	\mathbb{P}\left[ \left| f(\Sigma^{\frac{1}{2}}Y ) - \mathbb{E} f(\Sigma^{\frac{1}{2}}Y ) \right|\geq t \right] \leq 2 \mathrm{e} ^{-\frac{t^2}{2L_1^2}} \text{ for all } t> 0.
\]
Here $L_1$ is the Lipschitz constant for the function $y\mapsto f(\Sigma^{\frac{1}{2}}y)$. It is easy to see that $L_1 \leq L \norm{\Sigma}^{\frac{1}{2}}$, which concludes the proof.
\end{proof}
In the future, we will frequently use the following lemma.
\begin{lemma}\label{lem:decaymatrix}
Let $A\in\mathbb{C}^{N\times N}$. Assume that there are $\beta>0, \theta >1$,  s.t. $\abs{A_{ij}}\leq \beta(\left( \abs{i-j} + 1\right)^{-\theta} + N^{-1}).\forall 1\leq i,j \leq n$. Then $ \norm{A} \leq \frac{\beta\theta}{\theta-1}$. More generally, for any $p\in[1,+\infty]$, we have $\norm{A}_{l^p\to l^p} \leq \frac{\beta\theta}{\theta-1}$.
\end{lemma}
\begin{proof}
	Without loss of generality let $\beta=1$. For any vector $v\in{\mathbb R}^n$,
	\[
		\norm{A v}_\infty = \max_k\abs{\sum_i A_{ki}v_i}\leq \norm{v}_\infty \max_k \left(\sum_i ((\abs{i-k}+1)^{-\theta} + N^{-1})\right) \leq\norm{v}_\infty \left(\int_{1}^{+\infty}x^{-\theta}\mathrm{d} x + 1\right).
	\]
	Therefore, $\norm{A}_{l^\infty \to l^\infty} \leq \frac{\beta\theta}{\theta-1}$. Similarly, $\norm{A}_{l^1\to l^1} = \norm{A^*}_{l^\infty\to l^\infty} \leq \frac{\beta\theta}{\theta-1}$. By interpolation,
	\[
		\norm{A}_{l^p\to l^p}\leq \norm{A}_{l^\infty \to l^\infty}^ { \inv{p}}\norm{A}_{l^1\to l^1}^{1-\inv{p}} \leq \frac{\beta\theta }{\theta-1},\quad \forall p\in[1,+\infty].
	\]
\end{proof}

Recall that in Section \ref{sec:loopequation} we defined a map $S$ (see \eqref{def:S}). Thanks to the decay condition \eqref{def:xidecay}, the operator $S$ is a bounded operator, as will be seen in the following lemma.
\begin{lemma}\label{lem:S}
Let $A\in\mathbb{C}^{N\times N} $. Then there is a universal constant $c>0$ such that the following inequalities hold.
\begin{enumerate}
    \item $\norm{S(A)}_{d-1} \leq c \abs{A}_\infty$.
    \item $\norm{S(A)}_{l^p\to l^p}\leq c\abs{A}_\infty, \forall p\in[1,+\infty]$.
    \item $\norm{S(A)}_d \leq c\norm{A}_{d-1}$.
    \item $\norm{S(A)}_{d-\inv{2}}\leq c\norm{A}.$
\end{enumerate}
\end{lemma}
\begin{proof}
By definition $\abs{S(A)_{ij} }= \abs{\inv{N} \sum_{k,l} \xi_{iklj} A_{kl}}\leq \frac{\abs{A}_\infty}{N}\sum_{k,l} \abs{\xi_{iklj}} $. According to \eqref{def:xidecay}, $\inv{N}\sum_{k,l} \abs{\xi_{iklj}}\lesssim \inv{(1+\abs{i-j})^{d-1}} $. Hence $\abs{S(A)_{ij} } \lesssim  \frac{\abs{A}_\infty}{(1+\abs{i-j})^{d-1}} $, which implies the first inequality.  Setting $\theta=d-1$ in Lemma \ref{lem:decaymatrix}, we see that
$\norm{S(A)}_{l^p\to l^p} \lesssim  \abs{A}_\infty, \forall p\in[1,+\infty]$, which implies the second inequality. If $\norm{A}_{d-1}<+\infty$, then $\abs{S(A)_{ij} }= \abs{\inv{N} \sum_{k,l} \xi_{iklj} A_{kl}}\leq (1+\abs{i-j})^{-d}\frac{1}{N}\sum_{k,l}\abs{A_{kl}}\lesssim (1+\abs{i-j})^{-d}$. This proves the third inequality. As for the fourth inequality, we use Cauchy-Schwarz inequality to see that
$ \abs{S(A)_{ij} }\leq \inv{N} \left(\sum_{k,l}\abs{\xi_{iklj}}^2\right)^{\inv{2}} \left(\sum_{k,l}\abs{A_{kl}}^2\right)^{\inv{2}} \lesssim (1+\abs{i-j})^{\inv{2}-d}\norm{A}$.
\end{proof}

\subsection{Error estimate}\label{sec:errors}
Recall the decomposition \eqref{r:decomp}
\begin{equation}\label{r:decomp1}
	h_{ab} = \sum_{k=1}^N \gamma_{abk1} h_{k1} + \tilde h_{ab}, \quad \forall a,b\in\mathbb{Z}/N\mathbb{Z}.
\end{equation}
Taking the co-variance with $h_{l1}$ for any $l\in\mathbb{Z}/N\mathbb{Z}$, we see that
\[
    \xi_{abl1} = \sum_{k=1}^N \gamma_{abk1} \xi_{l1k1},\quad \forall l\in\mathbb{Z}/N\mathbb{Z}.
\]
Note that by assumption \eqref{def:xidecay} the matrix $\Sigma_1:= (\xi_{l1k1})_{l,k\in\mathbb{Z}/N\mathbb{Z}}$ satisfies $\abs{\xi_{l1k1}}\lesssim \frac{1}{(1+\abs{l-k})^\alpha}$ and by \eqref{def:c2}, $\norm{\Sigma_1^{-1}} \leq \mathsf{c}_2^{-1}$. Therefore Lemma \ref{Jaffard} implies that $\abs{(\Sigma^{-1}_1)_{ij}} \lesssim (1+\abs{i-j})^{-\alpha}$ and hence by Lemma \ref{lem:decaymatrix} we have $\norm{\Sigma_1^{-1}} \leq c$. Let $\nabla_1$ denote the partial gradient with respect to the first column $(h_{k1})_{1\leq k\leq N}$. Use the fact that $\frac{\partial G_{i1}}{\partial h_{ab}} = - G_{ia} G_{bj}$ and the chain rule, we have
\begin{equation}\nonumber
\norm{\nabla_1 G_{ij}}^2 \leq \sum_k \abs{-\sum_{a,b} G_{ia}G_{bj}\gamma_{abk1}}^2\lesssim \sum_k \abs{-\sum_{a,b} G_{ia}G_{bj}\xi_{abk1}}^2.
\end{equation}
In the second inequality above we have used the boundedness of $\norm{\Sigma_1^{-1}}$. Let
\begin{equation}\label{def:gamma}
    \Gamma = \max_{i,j}\abs{G_{ij}}\vee 1,\quad \gamma:=\max_{i} \Im G_{ii} \vee \eta.
\end{equation}
Use the decay rate \eqref{def:xidecay},
    \begin{equation}
        \norm{\nabla_1 G_{ij}}^2 \leq C\Gamma^2 \sum_k  \left( \sum_a \frac{\abs{G_{ia}}^2}{(\abs{a-k}+1)^{\alpha-1}} + \sum_b \frac{\abs{G_{bj}}^2}{(\abs{b-k}+1)^{\alpha-1}}\right)^2.
    \end{equation}
Since $\alpha-1 >1$, the operator norm of the matrix $\left( \frac{1}{(|a-k| + 1 )^{\alpha-1}}\right) _ { 1\leq a, k \leq N}$ is bounded by $C(\alpha-2)^{-1}$, according to Lemma \ref{lem:decaymatrix}. Therefore,
\begin{equation}\label{nablag}
\norm{\nabla_1 G_{ij}}^2 \leq C\Gamma^2 \left( \sum_a \abs{G_{ia}}^2 +\sum_b \abs{G_{bj}}^2\right) \leq C\Gamma^2 \gamma\eta^{-1}.
\end{equation}
In the second inequality we used Ward Identity.
Similarly,
\begin{equation}
    \norm{\nabla_1 (GS(G))_{ij}}\leq \norm{\sum_p \nabla_1 G_{ip} (S(G))_{pj}} + \norm{\sum_p  G_{ip} \nabla_1(S(G))_{pj}}.
\end{equation}
Define a short-hand notation $Q_{kl}:= \norm{\nabla_1 G_{kl}}$. By \eqref{nablag} we have $\abs{Q}_\infty^2 \leq C\Gamma^2 \gamma \eta^{-1}$. Then
\begin{equation}
\begin{aligned}
      \norm{\nabla_1 (GS(G))_{ij}} &\leq \sum_p Q_{ip} \abs{(S(G))_{pj}} + \sum_p \abs{G_{ip}}\inv{N}\sum_{k,l}\abs{\xi_{pklj}}Q_{kl}\\
        &\leq \abs{Q}_\infty \norm{S(G)}_{l^\infty \to l^\infty } + \abs{Q}_\infty \frac{\Gamma}{N}\sum_{k,l,p}\abs{\xi_{pklj}}.
\end{aligned}
\end{equation}
Now we use the bound \eqref{nablag}, and use the decay \eqref{def:xidecay} as well as Lemma \ref{lem:S} to see,
\begin{equation}\label{nablagsg}
    \norm{\nabla_1 (GS(G))_{ij}}^2 \leq C\Gamma^4\gamma\eta^{-1}.
\end{equation}

The observation above yields the following lemma.
\begin{lemma} \label{lem:ibp} Let $z=E+\mathrm{i}\eta\in\mathbb{C}^+$ and $K\geq 1$, then there is a universal constant $c>0$ such that
    \[
        -GS(G)-Gz = I + R,
    \]
    where $
	\PP{\abs{R}_\infty \geq t \sqrt{\frac{K^4\gamma}{N\eta}} , \Gamma \leq K}\leq 2N^2\mathrm{e}^{-ct^2},\forall t\geq 1$.
\end{lemma}
\begin{proof}
	For any $K>0$ let $\chi:{\mathbb R} \to [0,1]$ be a smooth function s.t. $\abs{\chi'} \leq 1$ and $\chi = 1$ on $[-K, K]$ and $\chi=0$ outside $[-3K,3K]$. Define
	\[
	\tilde{G} = \chi(\Gamma) G.
	\]
	Then
	$\norm{	\nabla \tilde{G}_{ij}}^2\lesssim K^2\gamma\eta^{-1}.$ According to Lemma \ref{lem:gaussianconcentration},
	\[
	\PP{\abs{ \tilde G_{ij} - \mathbb{E}_j\tilde {G}_{ij}} \geq t \sqrt{\frac{K^2\gamma}{N\eta}}}\leq 2\mathrm{e}^{-ct^2}.
	\]
	Note that $\tilde{G}=G$ on the event  $\{ \Gamma\leq K \}$. Therefore,
	\[
	\PP{\max_{i,j}\abs{ G_{ij} - \mathbb{E}_j{G}_{ij}} \geq t \sqrt{\frac{K^2\gamma}{N\eta}}, \Gamma \leq K}\leq 2N^2\mathrm{e}^{-ct^2}.
	\]
    On the other hand, in view of \eqref{nablagsg}, a similar argument yields,
	\[
	\PP{\max_{i,j}\abs{ (GS(G))_{ij}- \mathbb{E}_j(GS(G))_{ij}} \geq t \sqrt{\frac{K^4\gamma}{N\eta}} , \Gamma \leq K}\leq 2N^2\mathrm{e}^{-ct^2} .
	\]
    Now we go back to the identity \eqref{E1}, removing $\mathbb{E}_1$ at the cost of some error term, and replacing $1$ with a generic $j$, to see
    \[
        -GS(G)-Gz = I + R,
    \]
    where $
	\PP{\abs{R}_\infty \geq t \sqrt{\frac{K^4\gamma}{N\eta}} , \Gamma \leq K}\leq 2N^2\mathrm{e}^{-ct^2}$.
\end{proof}
In particular, for a crude bound, we may take $t= \log N$ and take $K = 2/\eta$ so that $\PP{\Gamma>K} = 0$. The lemma above yields,
\begin{corollary}\label{cor:globalerror}
Let $R$ satisfy
\[
    G(-S(G)-z) = I + R.
\]
Then $\abs{R}_\infty \leq \frac{8\log N }{\sqrt{N\eta^{6}}}$ with probability $1-N^{-c\log N}$.
\end{corollary}

\section{The Local Law for Correlated Gaussian Ensembles}

\subsection{Power Law Decay of Inverse Matrices}

\begin{lemma}\label{ProdofDecay}
Let $A,B\in\bC^{N\times N}$, $\beta_{1,2}>1$, then $\norm{AB}_{\min\{\beta_1,\beta_2\}} \leq C_{\min{\beta_1,\beta_2}}\norm{A}_{\beta_1}\norm{B}_{\beta_2}$.
\end{lemma}
\begin{proof}
Note that by definition, $\norm{A}_{\min\{\beta_1,\beta_2\}}\norm{B}_{\min\{\beta_1,\beta_2\}}\leq \norm{A}_{\beta_1}\norm{B}_{\beta_2}$, so it is sufficient to prove the case where $\beta_1=\beta_2=\beta$. Without loss of generality assume $\norm{A}_\beta=\norm{B}_\beta=1$, then,
\[
    \abs{(AB)_{ik}} \leq \sum_{j} \frac{1}{(1+\abs{i-j})^\beta} \frac{1}{(1+\abs{j-k})^\beta}.
\]
Since either $\abs{i-j}$ or $\abs{j-k}$ is $\geq \abs{i-k}/2$, the above quantity is bounded by
\[
     \abs{(AB)_{ik}}\leq 2 \sum_{l\in\mathbb{Z}} \frac{1}{(1+\frac{\abs{i-k}}{2})^\beta} \frac{1}{(1+\abs{l})^\beta} \leq \frac{2}{(1+\frac{\abs{i-k}}{2})^\beta} \left( 1+ 2\int_{1}^{+\infty} \frac{\mathrm{d} x}{x^\beta}\right),
\]
which is bounded by $2^{\beta +1} \frac{\beta+1}{\beta-1} (1+\abs{i-k})^{-\beta}$.
\end{proof}

The following argument is based off a similar argument of Jaffard \cite{Jaffard}.
\begin{theorem}\label{Jaffard}
Let $d>\frac{3}{2}$ and assume that a matrix $A=I+B$ (finite or infinite) satisfies $\norm{B}<1$ and $\norm{A}_d <+\infty$. Then, for any $\delta>0$, there exists a polynomial dependent on $d$ and $\delta>0$ such that $\norm{A^{-1}}_{d-1/2-\delta}\leq P_{d,\delta}(\norm{A}_d,\frac{1}{1-\norm{B}})$.

If $d>1$ and there exists an $\epsilon>0$ such that $\norm{B} \le 1- \epsilon$, then  $\norm{A^{-1}}_{d-\delta} \le C(\delta,\epsilon, \norm{A}_d)$.
\end{theorem}

We will show matrix element decay of the solution to the self-consistent equation. Though we will only really apply this to the solution of the limiting equation \eqref{sceinfty1}, the following theorem will phrase the result in terms of Matrices for convenience of notation.

\begin{proposition}\label{DecayofSolution}
    Let $M$ be the solution to the following equation
    \begin{gather*}
        M(-z -S(M))=I.
    \end{gather*}
    If there exists a constant $c>0$ such that $\norm{M}, \norm{M^{-1}} \leq c$, then we have that $\norm{M}_{\alpha} \leq C(c,\alpha)$.
\end{proposition}
\begin{proof}
Notice that we are able to write
\begin{gather*}\label{ProdForm}
M=(M^{-1})^*((M^{-1})^*M^{-1})^{-1}.
\end{gather*}
By the equation of $M$, we have $M^{-1}=-z-S(M)$.  Let us first estimate the decay of $M^{-1}$. By Lemma \ref{lem:S} we have $\norm{M^{-1}}_{d-\inv{2}}\lesssim \norm{M}$. By Lemma \ref{lem:decaymatrix} we have $\norm{M^{-1}(M^{-1})^*}_{d-\inv{2}}\lesssim \norm{M}^2$. We would now like to apply theorem \ref{Jaffard}  to $(M^{-1}(M^{-1})^*)^{-1}$.

For any general positive semi-definite matrix, $A$, we will be able to write it as $A = \frac{\lambda_1 + \lambda_n}{2}[I + B]$ where $\lambda_1$ and $\lambda_n$ are respectively the largest and smallest eigenvalues of A.

 Theorem \ref{Jaffard} is applied  to the matrix $I+B$. The operator norm bound on $B$ will be $\frac{\lambda_1 -\lambda_n}{\lambda_1 + \lambda_n}$.
The important factor $r= 1- \|B\|$ will be $\frac{2 \lambda_n}{\lambda_1 + \lambda_n}$.  \ref{Jaffard}  now shows that the matrix decay of $A^{-1}$ will be the same matrix decay of $A^{-1}$.

Applying this logic to the positive semidefinite matrix $(-z-S(M))(-z-S(M))^*$, one will obtain that $\lambda_1$ and $\lambda_n$ are both of some bounded constant order. Thus, we see we have matrix decay of order $1$.

Finally applying the multiplication lemma \eqref{ProdofDecay} to \eqref{ProdForm}, we will be able to get a matrix decay of $M$ of order $1$. We use this decay of $M$ to argue that $-z-S(M)$ has a matrix decay of order $\alpha$. We can then apply the same logic as above to argue that $M$ has matrix decay of order $\alpha$.
\end{proof}

\begin{remark}
The solution to the limiting self-consistent equation, though ostensibly a vector, can be written as an infinite Toeplitz matrix and the above result can be applied.
\end{remark}

Now we define $J:\bC^{N\times N}\to \mathcal{K}$, such that for any $A\in \bC^{N\times N}$ and $i\in\mathbb{Z}/N\mathbb{Z}$, $u\in\mathbb{T}$,
\begin{equation}\label{def:J}
    J(A)(i/N,u):= \sum_{k=i-\floor{N/2}}^{i+\floor{N/2}} A_{i,i+k}\mathrm{e}^{2\pi\mathrm{i} ku}.
\end{equation}
and $J(A)(s,u)$ is linear in $s$ for $s\in[i/N,(i+1)/N]$. It is easy to check that
\begin{equation}\label{DJ}
    D(J(A)) = A,\quad \forall A\in\mathbb{C}^{N\times N}.
\end{equation}

\begin{proposition}\label{prop:iterate}
    Consider a fixed bounded subset $U\subset \mathbb{C}^+$. There are constants $\epsilon, C>0$ such that if $\abs{J(M)-g}_\infty\leq \epsilon$ , then $\norm{F(M)-F(D(g))}_{\alpha-1}\vee \norm{F(F(M))-F(F(D(g)))}_\alpha \leq C\abs{J(M)-g}_\infty$ and $\abs{F(M)-D(g)}_\infty\vee \abs{F(F(M))-D(g)}_\infty\leq C(\abs{J(M)-g}_\infty+N^{-\inv{2}}),\forall z\in U$.
\end{proposition}
\begin{proof} Let $A:=F(D(g))$ and $R:=S(M-D(g))$. Then
\[
    F(M)-A= \sum_{k=1}^\infty A(RA)^k.
\]
Hence $\norm{F(M)-A}_{\alpha-1}\leq \sum_{k=1}^\infty \norm{A(RA)^k}_{\alpha-1}$. It is easy to see that $\norm{R}_{\alpha-1}\leq  c\abs{J(M)-g}_\infty $ for some universal constant $c>0$. By Lemma  \ref{ProdofDecay} we have $\norm{A(RA)^k}_{\alpha-1}\lesssim (c\abs{J(M)-g}_\infty)^k$. Therefore, taking $\epsilon$ small enough, we have $\norm{F(M)-F(D(g))}_{\alpha-1} \leq C\abs{J(M)-g}_\infty,$.

Next, we define $R'=S(F(M)-F(D(g)))$, $A' = F(F(D(g)))$. Then $\norm{R'}_\alpha \leq c'\epsilon$ according to the above argument. We have
\[
    F(F(M))-A' = \sum_{k=1}^\infty A'(R'A')^k.
\]
By Lemma  \ref{ProdofDecay} we have $\norm{A'(R'A')^k}_\alpha \lesssim (c\abs{J(M)-g}_\infty)^k$. Therefore, taking $\epsilon$ small enough, we have $\norm{F(F(M))-F(F(D(g)))}_\alpha \leq C\abs{J(M)-g}_\infty$.

The last claim follows from the estimates above and Corollary \ref{col:Nrmbndit}.
\end{proof}

\subsection{Local Law}

Recall definition \eqref{def:kappa} and \eqref{def:gamma}, for a constant $T>0$ to be chosen, define
\begin{equation}
    \mathcal{D} : = \{z\in \mathbb{C}^+\vert \abs{z}\leq T, \Im z \geq (\log N)^{10}N^{-1} \omega^{-4}\}.
\end{equation}
\begin{theorem}[Local law]\label{thm:bulklaw}
Define $\Lambda(z):=\abs{D(g)-G}_\infty$. For $N$ large enough, we have
\[
    \sup_{z\in \mathcal{D}} \Lambda(z) \leq (\log N)^4 \left(\sqrt{\frac{\gamma}{N\eta}}\right)\omega^{-1},
\]
with probability $1- \mathrm{e}^{-\mathsf{a}_3(\log N)^2}$. If $\kappa>\rho$,
\[
    \sup_{z\in \mathcal{D}^\theta_N}\Lambda(z) \leq (\log N)^8 \left(\sqrt{\frac{\Im m}{N\eta}}\omega^{-1}+(N\eta)^{-1}\omega^{-2}\right) ,
\]
with probability $1- \mathrm{e}^{-\mathsf{a}_3(\log N)^2}$.
\end{theorem}

\begin{proof}
Take $K:=\log N$, and let $\{z_k\}$ be an $N^{-4}$-net of $\mathcal{D}$. Define
\[
    \Omega:= \bigcup_{k=1}^{N^{10}}\{ \Lambda(z)\in (K^4\sqrt{\gamma}(N\eta)^{-\inv{2}}\omega^{-1}, K^{-1}\omega\}
\]
Then by Proposition \ref{prop:iterate}, on $\Omega$ we have
\[
    F(F(G))(-S(F(F(G))-z) = I + \tilde{R},
\]
where $\tilde{R}\lesssim \abs{R}_\infty$. Then $J(F(F(G))(-\Psi(J(F(F(G))))-z) = 1 + O(\abs{R}_\infty+N^{-1})$. By Proposition \ref{prop:iterate}, on $\Omega$ we have $\abs{F(F(G))-D(g)}_\infty\lesssim K^{-1} \omega  + N^{-\inv{2}}$, which is $\ll \omega$. By stability Lemma \ref{lem:stab} we know $\Lambda(z) \lesssim (\abs{R}_\infty+N^{-1})\omega^{-1} $, which implies $\norm{J(F(F(G)))-g}_\infty\lesssim (\abs{R}_\infty+N^{-1})\omega^{-1}$, hence $\abs{G-D(g)}_\infty \lesssim (\abs{R}_\infty+N^{-1})\omega^{-1}$. Therefore, on $\Omega$ we have $\abs{R}_\infty \gtrsim K^4\sqrt{\gamma}(N\eta)^{-\frac{1}{2}}$. By Lemma \ref{lem:ibp} we know $\PP{\Omega}\leq 2N^{12}\mathrm{e}^{-c(\log N)^2}$. On $\Omega^c$, we either have $\inf_{z\in\mathcal{D}}\abs{G-D(g)}_\infty\geq K^{-1} \omega /2$ or $\sup_{z\in\mathcal{D}}\abs{G-D(g)}\leq 2K^4\sqrt{\gamma}(N\eta)^{-\inv{2}}$. The latter is true with probability $1- \mathrm{e}^{-c(\log N)^2}$, since if we take the $T$ in the definition of $\mathcal{D}$ to be a large enough constant, then the former case holds with $O(\mathrm{e}^{-c(\log N)^2})$ probability.
\end{proof}

\begin{corollary} \label{cor:Averaged Error}
Let $a>0$ be a small constant. Then on
\[
    \mathcal{D}' : = \{ z\in\mathcal{D}\vert \kappa\geq N^{-a}\}.
\]
we have
\[
    \abs{\EE{G}-D(g)}_\infty\lesssim (\log N)^{16}\left(\frac{1}{N\kappa \omega^3} + \frac{1}{(N\eta)^2 \omega^5}\right).
\]
\end{corollary}

\begin{proof}
    By integration by parts,
    \[
        -\EE{GS(G)} -\EE{G}z = I.
    \]
    Let $R= (\abs{G_{ij}-D(g)_{ij}})_{1\leq i,j\leq N}$.
\[
    -\EE{G}S(\EE{G}) - \EE{G}z = I + \EE{ (R) S(R)} = I + \EE{O(\abs{D(g)-G}_\infty^2)}.
\]
Repeating the argument in the proof of Theorem \ref{thm:bulklaw} on $\EE{G}$ instead of $G$, we have
\[
    \abs{\EE{G}-D(g)}_\infty\lesssim \EE{O(\abs{D(g)-G}_\infty^2)}\omega^{-1}.
\]
We use Theorem \ref{thm:bulklaw} and the crude bound $\Im m \leq \eta \kappa^{-2}$ to get the conclusion.
\end{proof}

\begin{remark} When we proved this local law, the only error estimates that depended strongly on the particular model we are considering are the stability results for the limiting vector equation. When considering the case of sample covariance matrices, though they are not exactly considered in the context of our proof, the stability results and the square root behavior at the right edge hold for sample covariance matrices. Thus, we will be able to prove a local law for sample covariance matrices.

\end{remark}

\subsection{Upper Bound of Top Eigenvalue}

Here we first show a lemma that combines our estimates on the average empirical spectral density with Gaussian concentration to prove upper bound for the top eigenvalue.
\begin{lemma} \label{lem:AverageComp}
For $N\in\mathbb{N}$, consider a family of random measures $\mu_N=\frac{1}{N}\sum_{k=1}^N \delta_{\lambda_k}$ where $\lambda_1\geq \cdots\geq \lambda_N$ such that there is a deterministic $\hat{\lambda}_1$ satisfying $\lambda_1 = \hat{\lambda}_1+ o(N^{\varepsilon-\inv{2}})$ for any $\varepsilon>0$. Assume that there exists a deterministic measure $\nu$ whose Green's function satisfies
\begin{gather}\label{squareRootBehavior}
\text{Im}[m_{\nu}](x + i \eta) \le C \frac{\eta}{\sqrt{\kappa+\eta}}.
\end{gather}
where $\kappa:=\dist(\supp(\nu),x)$ and that
\begin{gather}\label{Green's function Comparison}
    |\EE{m_{\mu_N}(z)} - m_{\nu}(z)| =o(N^{-\frac{1}{2}-\gamma}).
\end{gather}
for some $\gamma>0$ and all $z= E +i\eta$ with $\dist(E,\supp(\nu)) \ge N^{-\epsilon}$ and $\eta \ge N^{-\delta-\frac{1}{2}}$ for some $\delta,\epsilon>0$.

Then, $\dist(\lambda_1,\supp(\nu))\leq N^{-\epsilon'}$ for some $\epsilon'>0$.
\end{lemma}
\begin{proof}
Assume for contradiction that $\lambda_1$ lies outside a distance $N^{-\epsilon'}$ of $\supp(\nu)$ where $\epsilon'$ is smaller than the $\epsilon$ in the condition for \eqref{Green's function Comparison}.

Notice that we have the following inequality
\begin{gather} \label{ConcenGreen}
    \frac{1}{N}
    \le C \int_I \int_{-\infty}^{\infty} \frac{1}{N} \frac{\eta  \delta_{x=\lambda_1}}{(x-E)^2 + \eta^2} \mathrm{d} x \mathrm{d} E \le C \int_I \text{Im}[m_{\mu_N}](E+\mathrm{i} \eta) \mathrm{d} E,
\end{gather}
letting $I=[\hat{\lambda}_1-\frac{N^{\gamma'}}{\sqrt{N}}, \hat{\lambda}_1+ \frac{N^{\gamma'}}{\sqrt{N}}]$ with $\gamma'<\gamma \wedge \delta/2$.  We need to choose $\eta$ to be smaller than $\frac{1}{N^{1/2 + \gamma'}}$. This will ensure that at least a one sided $\eta$ neighborhood of $\lambda_1$ will always lie the region $[\hat{\lambda}_1 - \frac{1}{N^{1/2 +\gamma'}}, \hat{\lambda}_1 + \frac{1}{N^{1/2 +\gamma'}}]$. The integral of $\frac{\eta}{(x-z)^2 + \eta^2}$ inside this half interval of size $\eta$ will certainly be greater than $\frac{\eta^2}{2 \eta^2}= \frac{1}{2}$.

We can take the expectation of \eqref{ConcenGreen} to get,
\begin{gather} \label{GfuncCompar}
    \frac{1}{N} \le C' \int_I \Im[\EE{m_{\mu_N}}](E+\mathrm{i} \eta) \mathrm{d} E
    \le C' \int_I \frac{o(1)}{N^{\frac{1}{2} +\gamma}} + C' \int_I \text{Im}[m_{\nu}](E + \mathrm{i} \eta) \mathrm{d} E
    \\ \label{SquareRoot}
    \le  C' \frac{o(1)}{N} + C'' \int_I \frac{\eta}{\sqrt{\kappa}}
    \le C' \frac{o(1)}{N} + C'' \frac{\eta N^{\gamma'}}{\sqrt{N} \sqrt{\kappa}}
\end{gather}
In $\eqref{GfuncCompar}$ we used the assumption \eqref{Green's function Comparison}(since $\hat{\lambda}_1$ lies in the region this assumption is valid)
while in \eqref{SquareRoot}, we used the fact that $\nu$ satisfies \eqref{squareRootBehavior}.

Notice that we can set $\eta = N^{-1/2 - \delta}$ for $\delta$ positive and $\kappa = N^{-\min{(\epsilon, \delta/4)}}$ and see that the error of \eqref{SquareRoot} will be $\frac{o(1)}{N}$. This contradiction implies that for large $N$, $\hat{\lambda}_1$ must necessarily be less than $N^{-\epsilon'}$. By concentration of $\lambda_1$ around $\hat{\lambda}_1$, we would know that all $\lambda_1$ will be less than $N^{-\epsilon'}$.
\end{proof}

\begin{theorem} \label{thm:Edge upper bound for the top eigenvalue}
For the Gaussian Ensemble that we are considering, there exists an $\epsilon>0$ such that all eigenvalues lie within distance $N^{-\epsilon}$ from the edge.
\end{theorem}

\begin{proof}
We would like to apply Lemma \ref{lem:AverageComp}. First notice that by Gaussian concentration, we are able to prove that the distance of $|\lambda_1 - \EE{\lambda_1}| \le \frac{(\log N)^2}{\sqrt{N}}$ with probability $1-O(N^{-c\log N})$. We thus put $\hat{\lambda}_1 = \EE{\lambda_1}$ in the assumption of Lemma \ref{lem:AverageComp}.

Then we check that the error bounds in Corollary \ref{cor:Averaged Error} are sufficient for our purposes. The error that appears there is $\abs{\EE{G}-D(g)}_\infty\lesssim (\log N)^{16}\left(\frac{1}{N\kappa \omega^3} + \frac{1}{(N\eta)^2 \omega^5}\right)$. By the definition of $D$ and the Lipschitz continuity of $g$, we have $|\EE{\inv{N}{\Tr G}} - m_\nu |= O(N^{-\inv{2}-\gamma})$ for some $\gamma>0$  as long as we have $\eta\gg N^{-3/4+\delta}$ and $\kappa \sim N^{-\epsilon}$ for $\epsilon$ very small and $\delta>0$. Since $\delta$ can be arbitrarily small, we may choose $\eta$ such that $N^{-3/4+\delta}\ll \eta \ll N^{-1/2}$ and we can apply Lemma \ref{lem:AverageComp}.
\end{proof}

\section{Universality}

In the previous section, we proved a local law for $m_N$ as well as an improved local law for $\EE{m_N}$, and combining it with the concentration of the top eigenvalue to prove an upper bound on the top eigenvalue. According to a recent result by Landon and Yau \cite{Yau2017} below, the local law with upper bound on the top eigenvalue is sufficient to prove universality near the edge.

\begin{theorem} \label{YauEdge}
Let $\eta* = N^{-\phi^*}$ for some $0<\phi^*<\frac{2}{3}$. We call a deterministic matrix $V$ $\eta^*$-regular if it satisfies the following properties.

\begin{enumerate}
\item There exists a constant $C_{V} \ge 0$ such that
\begin{gather*}
\frac{1}{C_V} \frac{\eta}{\sqrt{|E|+\eta}} \le \text{Im}[m_V(E + i \eta)] \le C_{V} \frac{\eta}{\sqrt{|E| + \eta}}, -1 \le E \le 0, \eta^*\le \eta \le 10,
\end{gather*}
and
\begin{gather*}
\frac{1}{C_V} \sqrt{|E|+\eta} \le \text{Im}[m_V(E+ i\eta)] \le C_{V} \sqrt{|E| + \eta},  0 \le E \le 1 , (\eta^*)^{1/2}|E| + \eta^* \le \eta \le 10.
\end{gather*}
\item There exists no eigenvalue of V in the region $[-\eta^*,0]$.
\item We have $\|V\|\le N^{C_V}$ for some $C_V >0$.
\end{enumerate}

Consider the ensemble $V_t =V + \sqrt{t} G$. Where $G$ is an independent GOE ensemble. Let $t$ satisfy $N^{-\epsilon} \ge t \ge N^{\epsilon} \eta^*$ and let $F:\mathbb{R}^{k+1} \rightarrow \mathbb{R}$  be a test function such that $\|F\|_{\infty} \le C $ and $\|F\|_{\infty} \le C$. Then there are deterministic parameters $\gamma_0\sim 1$ and $E_{-}$ such that
\begin{gather*}
|\mathbb{E}[F(\gamma_{0} N^{2/3}(\lambda_{i_0} -E_{-}), ...\gamma_{0}N^{2/3}(\lambda_{i_k} - E_{-}))] -
\mathbb{E}_{GOE}[F(N^{2/3}(\hat{\lambda}_{1} +2),...N^{2/3}(\hat{\lambda}_{k} +2))] \le N^{-c}
\end{gather*}
The first expectation is with respect to the eigenvalues of the ensemble $V_t$. The latter expectation is taken with respect to the eigenvalues $\hat{\lambda}_{i}$ of a GOE. $i_0$ is the first index i such that $i$th smallest eigenvalue of $V$ is greater than $-\frac{1}{2}$.
\end{theorem}

Call $H$ the ensemble with correlation structure $\xi_{ijkl} $.
Theorem \ref{thm:Edge upper bound for the top eigenvalue} combined with \ref{thm:bulklaw} shows that there exists a parameter $\Phi>0$ such that with high probability a matrix $M$ produced by $H$ would be $\eta^*$ regular for any $N^{-\phi}$ such that $\phi<\Phi$. Now fix some $\phi$ sufficiently small and $\phi<\Phi$ and $t=N^{-\phi}$; we would like to write the ensemble $H$ as $H'+\sqrt{t}G$ for $G$ an independent GOE ensemble. We will use the fact that $\phi$ is sufficiently small in the following section.

When N is large enough, $H'$ is the ensemble with correlation structure given by $\xi_{ijkl}- t \delta_{ij=kl}$. With t sufficiently small, the covariance matrix is positive and one can construct the ensemble. Also note that a matrix produce from $H'$ would satisfy the regularity estimates with parameter $N^{-\phi}$ as well due to our proof of the local law and upper bound for the top eigenvalue of the edge.

We will apply \eqref{YauEdge} as follows. Any matrix, $M$, in $H$ can be written in the form $M'+ t \text{GOE}$ where $M'$ is a matrix produced from the ensemble $H'$.$M'$ can be diagonalized into the form $V'$ by some unitary transformation $U$, which will leave invariant the GOE part. We will then condition on this matrix $M'$ and apply theorem \eqref{YauEdge}.We will thus get the following statement, the matrices of the form $M'+t \text{GOE}$ will satisfy a universality statement of the form.

\begin{gather} \label{BadScale}
    |\mathbb{E}_{M'}[F(\gamma_{0} N^{2/3}(\lambda_{1} -E_{-}), ...\gamma_{0}N^{2/3}(\lambda_{k} - E_{-}))] -
\mathbb{E}_{GOE}[F(N^{2/3}(\hat{\lambda}_{1} +2),...N^{2/3}(\hat{\lambda}_{k} +2))] \le N^{-c}
\end{gather}

 where $\lambda_{1}$ are the eigenvalues of the considered matrix $M' + t \text{GOE}$.$\mathbb{E}_{M'}$ denotes the conditional expectation over the matrices of the form $M'+ t \text{GOE}$. We used for $N$ large enough, the largest eigenvalue of $M'$ is of distance less than $1/2$ from the edge, so the index $i_0$ is 1. The only issue with \eqref{BadScale} is that $\gamma_0$ is a function of the initial data, we will make this a universal constant in the next section.

 \subsection{Changing the scaling factor}

 Let $a^{t}$ be the edge of the ensemble corresponding to $H' + t \text{GOE}$ . Let us denote the Green's function of this ensemble as $m_{(H')^t}$; let us also write the density of this ensemble as $\rho_{(H')^t}$

As in Thm 2.2, let $E^{t}_{-}$ be the edge corresponding to the model $V + t \text{GOE}$ where $V$ is the deterministic diagonal matrix and the $\text{GOE}$ is an independent ensemble. Let us denote the Green's function of this ensemble as $m_{V^t}$; let us also write the density of this ensemble as $\rho_{V^t}$. We will be considering the case that $V$ is a fixed matrix coming from the ensemble $H$. From now on, we will assume that $V$ is $\eta^*$ regular so that the conditions of Thm 2.2. hold.

From the results of Thm 2.2. we can write $\rho_{V^t}(E) = \gamma_{V^t}^{-1/2} \sqrt{E- E^{t}_{-}}(1+ t^{-2} O(|E-E^{t}_{-}|))$
and $\rho_{H^t}(E) = \gamma_{H^t}^{1/2} \sqrt{E- a^{t}}(1+ t^{-2} O(|E-E^{t}_{-}|))$.

We will show the following bound on the sacling factors.
\begin{lemma} \label{ScalingFactor}
For sufficiently large N, we have that $\gamma_{H} - \gamma_{V^t} =\gamma_{(H')^{t}} - \gamma_{V^t} = O(t)$
\end{lemma}

\begin{proof}
Define $z_1$ to be the solution of $z_1 + t m_{(H')^0}(z_1) = a^t + \kappa$ and $z_2$ to be the solution of $z_2 + t m_{V^0}(z_1) = E_{-}^{t} + \kappa$. We would need to compare the values of $\text{Im} m_{(H')^0}[z_1]$ and $\text{Im} m_{V^0}[z_2]$ in order to compare the values of $\rho_{(H')^t}$ and $\rho_{V^t}$ at a distance $\kappa$ away from their edges.

Namely, we know that $\rho_{(H')^t}(a^t + \kappa) = \text{Im}[m_{(H')^{0}}[z_1]]$ and similarly for $z_2$.

Indeed, we have

\begin{gather}
\pi[\rho_{(H')^t}(a^t + \kappa) - \rho_{V^t}(E^t_{-} + \kappa)] = \text{Im}[m_{(H')^{0}}[z_1]] - \text{Im}[m_{V^{0}}[z_2]]\\ \label{Differences}
= \text{Im}[m_{(H')^{0}}[z_1]] - \text{Im}[m_{V^{0}}[z_1]]  + \text{Im}[m_{V^{0}}[z_1]] - \text{Im}[m_{V^{0}}[z_2]]
\end{gather}

In \eqref{Differences}, the first term can be bounded by a sufficiently good local law. The second term can be bounded by a Lipschitz condition provided $|z_1 - z_2|$ are sufficiently close to each other.

We will now attempt to bound the quantity $|z_1 -z_2|$

\begin{lemma}\label{Distance Lemma}
Assume that we are considering a matrix model $H$ that is $\eta^*$ regular for all $\eta^* = N^{-\phi}$, for $\Phi>\phi>0$.

Consider the time scale $t= N^{-\phi/2}$ and choose $\kappa$ to be the almost optimal $t^{2+\epsilon}$ for the edge expansion. Then there exists a small parameter $\delta$ such that for N large enough we can ensure that $|z_1 -z_2| \le t^{2+\delta}$
\end{lemma}

\begin{proof}

We have that
\begin{gather}
z_1 + t m_{(H')^0}(z_1) - (z_2 + t m_{V^{0}}(z_2)) = (a^t - E^{t}_{-}) \\ \label{Rouche}
(z_1 -z_2) + t(m_{(H')^{0}}(z_1) - m_{(H')^{0}}(z_2)) = (a^t -E^{t}_{-}) + t(m_{V^{0}}(z_2) - m_{(H')^{0}}(z_2))
\end{gather}

We will try to prove that $|z_1 - z_2|$ is sufficiently small. We will do this by appealing to Rouche's Theorem and a Local Law bound to the second term on the RHS of \eqref{Rouche}.

We will now address the Local Law portion of the above estimate. Recall the formula that $\Im[z_1]= t \text{Im}[m_{(H')^{t}}(a^t + \kappa)]$. From the earlier expansion of the density around $\kappa$, we know that for $\kappa \le ct^2$, we have that $\text{Im}[m(z_1)]$ is up to a constant factor equal to $\gamma_{(H')^{t}}\sqrt{\kappa}$ where the $\gamma$ scaling factor is of order 1.

Thus, we see that $\Im[z_1]$ is of the order of $t\sqrt{\kappa}$. Notice that if we take $\kappa$ near the limit scale of $t^{2+\epsilon}$, as we will do later, then we will have that $\text{Im}[z_1]$ is of the size $t^{2+\epsilon/2}$ Using the fact that we are dealing with time scales of the order $t= N^{-\phi/2}$, we see that $\Im[z_1] = N^{-\phi(1+\epsilon/4)}$. This is in a regime where we can apply the local law \ref{thm:bulklaw}.

To confirm this carefully, note  that
$\dist(z_1,\supp \nu) \ge \text{Im}[z_1]$
so the following should hold for
\begin{gather*}
\text{Im}[z_1] = N^{-\phi(1+\epsilon/4)}  \ge (\log N)^{10} N^{-1}(N^{2/3 (-\phi)(1+\epsilon)})^{-4} \gg (\log N)^{\log\log N} N^{-1/2}
\end{gather*}
so the point $z_1$ is in the region $\mathcal{D}$ when  when we have that $\phi$ is sufficiently small.
Clearly, we would also have that a circle of radius $t^{2+\delta}$ around $z_1$ for $\delta > \epsilon/2$ would also lie in the region $\mathcal{D}$ .

Applying \ref{thm:bulklaw} for z in a circle of radius $t^{2+\delta}$ around $z_1$ will give us that the error of $|m_{V^{0}}(z) -m_{(H')^{0}}(z)| \le (\log N)^4(\sqrt{1 /(N\text{Im}(z))}) (\text{Im}(z))^{- 2/3}$. This can be seen to be much less than $t^3$ given that we set $\phi$ to be sufficiently small. Thus, we have a good local law bound on the second term of \eqref{Rouche} once $\phi$ is set to be sufficiently small.

We know that since we assumed $V$ is $\eta^*$ regular for $\eta^* = N^{-\Phi}$ for $\phi<\Phi$ from the local law on the ensemble $H$, we also know that with high probability $|a^t - E^t|$ should be less than $N^{-\Phi}$. Again choosing $\phi$ small enough, this will imply that $|a^t - E^t| \le t^{3}$

Consider a circle of radius equal to $R= \frac{t^{2+\delta}}{1- t K}$ where $K$ is such that we have  $|m_{(H')^{0}}(z_1) - m_{(H')^{0}}(z_2)| \le K |z_1 -z_2|$.
around the point $z_1$. Notice that t decreases as N increases; thus for very large N, we will have that $tK \le \frac{1}{2}$. Therefore, we have that R is a circle of radius less than $2 t^{2+\delta}$ for large enough N.

On this circle of radius R, we have by the local law and estimates on $|a^{t}- E^{t}_{-}|$ that the right hand side of \eqref{Rouche} will be less than the left hand side of \eqref{Rouche} in absolute value on the boundary. If the left hand side of $\eqref{Rouche}$ were 0, then we would clearly have the unique solution $z_2=z_1$. Rouche's theorem then shows that there is a solution such that $|z_2-z_1| \le R= t^{2+\delta}$\end{proof}. Putting this content back into \eqref{Differences} with $\kappa = t^{2+\epsilon}$.

\begin{gather*}\label{KappaCompare}
\gamma_{H^t}^{1/2} t^{1+\epsilon/2} (1+ t^{-2} O(t^{2+\epsilon}))-\gamma_{V^t}^{-1/2} t^{1+\epsilon} (1+ t^{-2} O(t^{2+\epsilon}) \le \\ \label{LLandLip}
\text{Im}[m_{(H')^{0}}[z_1]] - \text{Im}[m_{V^{0}}[z_1]]  + \text{Im}[m_{V^{0}}[z_1]] - \text{Im}[m_{V^{0}}[z_2]] \le
t^3 + K t^{2+\delta}
\end{gather*}

For the first term in \eqref{LLandLip}, we used the local law around $z_1$ to bound the quantity by $t^3$ for the second quantity we used Lipschitz continuity of $m_{V^{0}}$ combined with the estimate on $|z_1-z_2|$ coming from \eqref{Distance Lemma}
Notice that if we now have that $|\gamma_{V^t}^{1/2}- \gamma_{H^t}^{1/2}| \ge t $, then it would clearly be impossible for the inequality in $\eqref{KappaCompare}$ to hold.
Thus, we have proved a bound on
$|\gamma_{V^t}^{1/2}- \gamma_{H^t}^{1/2}| \le t$, which can be turned into an o(1) bound on $\gamma_{V^t} -\gamma_{H^t}$ by squaring and using the fact that $\gamma_{V^t}$ is of constant order.
\end{proof}

\subsection{Final universality Result}

Using the scaling results coming from the previous section we can translate \eqref{BadScale} as follows.

\begin{theorem}\label{AlmostUniversality}

There exists a scaling factor $\gamma$ that depends only on the matrix ensemble H such that the following inequality holds for functions $G: \mathbb{R}^{k} \rightarrow \mathbb{R}$ such that $\|G\|_{\infty}, \|\nabla G\|_{\infty} \le C$

\begin{gather}\label{AlUeq}
 |\mathbb{E}_{H}[G(\gamma N^{2/3}(\lambda_{2} -\lambda_{1}), ...\gamma N^{2/3}(\lambda_{k} - \lambda_1))] -
\mathbb{E}_{\text{GOE}}[G(N^{2/3}(\hat{\lambda}_{2}-\hat{\lambda}_1),...N^{2/3}(\hat{\lambda}_{k} -\hat{\lambda}_1))] \le N^{-c}
\end{gather}
\end{theorem}
\begin{proof}

First, notice that we can find a function $F:{\mathbb R}^{k+1} \rightarrow {\mathbb R}$ such that $\|F\|_{\infty}$ and $\|\nabla F\|_{\infty}$ are bounded  and
\begin{gather*}
    F(x_1,..x_{k+1}) = G(x_1-x_2,...x_1-x_{k+1})
\end{gather*}

Recall from earlier discussion that we can write any matrix from the ensemble $H$ as $M' + t GOE$ where M' is generated from the ensemble $H'$ with correlation structure $\xi_{abcd} - t^2 \delta_{ab=cd}$
Let $\Omega$ be the set in which we know that $M'$ has sufficiently good regularity so that $\eqref{BadScale}$ holds for the function F.
On $\Omega$, we would like to change the scaling factor $\gamma_0$ to $\gamma$, which is the scaling factor at the edge for the limiting spectral density.
As before, with high probability $M'$ has sufficient regularity so we can ensure that \eqref{BadScale} holds. We only need to change the $\gamma_0$ factor to $\gamma$ which the edge scaling coefficient of the ensemble $H$.

From \eqref{ScalingFactor}, we know that the difference between the $\gamma_0$ appearing in \eqref{BadScale} and the $\gamma$ appearing here is of the order $t = N^{-\phi/2}$.
Finally, one can appeal to the Lipschitz nature of $F$ as well as the fact that the $N^{2/3}(\lambda_{i_k} - E_{-})$
are bounded to say that

\begin{gather*}
    |F(\gamma N^{2/3}(\lambda_{1} -E_{-}), ...\gamma N^{2/3}(\lambda_{k} - E_{-}))-
    F(\gamma_0 N^{2/3}(\lambda_{3} -E_{-}), ...\gamma_0 N^{2/3}(\lambda_{k} - E_{-}))| \le  C k N^{-\phi/2}
\end{gather*}

One can then take expectation with respect to the ensemble $M' + t \text{GOE}$ with $M'$ fixed and then apply the triangle inequality with respect \eqref{BadScale} to prove

\begin{gather*}
    |\mathbb{E}_{M'}[F(\gamma N^{2/3}(\lambda_1 - E^{M}_{-}) ,...,\gamma N^{2/3}(\lambda_k -E^{M}_{-})] - \mathbb{E}_{\text{GOE}}[F(N^{2/3}(\hat{\lambda}_1+2),...N^{2/3}(\hat{\lambda}_k +2))]| \le N^{-c}
\end{gather*}

Translating this statement to $G$, we get for matrices $M'$ in $\Omega$

\begin{gather}\label{CondUniv}
    |\mathbb{E}_{M'}[G(\gamma N^{2/3}(\lambda_1 - \lambda_2) ,...,\gamma N^{2/3}(\lambda_1 -
    \lambda_k)] - \mathbb{E}_{\text{GOE}}[G(N^{2/3}(\hat{\lambda}_1-\hat{\lambda}_2),...N^{2/3}(\hat{\lambda}_1 -\hat{\lambda}_k))]| \le N^{-c}
\end{gather}

One would now like to remove the conditional expectation in the above expression. Namely, we would like to integrate \eqref{CondUniv} over the matrices $M'$ found in $\Omega$ while using the trivial bound that $|E_{H'}[G] - E_{GOE}[G]|$ is bounded by a constant for all matrices $M'$ not found in  $\Omega$.
We thus get the full universality statement

\begin{gather}\label{FullUniv}
    |\mathbb{E}_{H}[G(\gamma N^{2/3}(\lambda_1 - \lambda_2) ,...,\gamma N^{2/3}(\lambda_1 -
    \lambda_k))] - \mathbb{E}_{\text{GOE}}[G(N^{2/3}(\hat{\lambda}_1-\hat{\lambda}_2),...N^{2/3}(\hat{\lambda}_1 -\hat{\lambda}_k))]| \le N^{-c}
\end{gather}

as desired.

\begin{remark}
As long as we know that a version of the Dyson-Brownian Motion result holds for sample covariance matrices, then we will be able prove edge universality using the local law and edge upper bound for the top eigenvalue results from the previous section.
\end{remark}
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
The Outer Worlds to launch on 25 October
Stream of the Day
Hero? Villain? Lunatic? Your decisions drive the outcome in The Outer Worlds, an immersive single-player sci-fi RPG from Obsidian Entertainment and Private Division. Along with the release date, a new trailer has also been released.
Click below to watch the trailer and to read more about the game.
From the renowned RPG (role-playing game) developer, Obsidian Entertainment, comes The Outer Worlds, a dark and witty player-driven story set in a colony at the farthest reaches of the galaxy. As the cold, firm grasp of corporate bureaucracy starts to unravel due to one unknown visitor – the player – the character the player decides to become – will determine how the story unfolds. Their choices will affect the way the story develops, as well as their character progression, companion stories, and endgame scenarios.
"When I found myself contractually obligated to work with Leonard, we decided to make the game we always wanted," said Tim Cain, co-game director at Obsidian. "A game where player agency matters, and allows you to approach any situation however you'd like. Because it's your story. And, because we haven't figured out how to contractually oblige you to play how we want you to. Yet."
The Outer Worlds presents its players with a variety of options to play through the game. With an array of ranged and melee sci-fi weaponry at a player's disposal, they can also solve issues with good, old-fashioned spoken diplomacy, or even leadership-driven delegation to their companions.
"All kidding aside," said Leonard Boyarsky, co-game director at Obsidian, "Tim is truly honored to be working with me again on a game that focuses not only on a player-driven story, but also on establishing a new universe that is fun for players to explore with depth, humor, and corporate approved action."
"Obsidian has such a great history when it comes to creating exceptional single-player RPGs," said Allen Murray, VP of Production. "The Outer Worlds is unmistakably an Obsidian game, delivering an experience rich with dark humor, interesting characters, and plenty of player agency."
The game is set to launch on PlayStation 4, Xbox One and PC on 25 October.
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Tom Clancy's Rainbox Six Siege reveals Operation Phantom Sight
Microsoft Teams goes big with ad at NFL
New legendary items in Borderlands 3
Tokyo Mirage Sessions #FE Encore available tomorrow
Microsoft will be showcasing a 30-second ad for its Slack competitor, Microsoft Teams, during the NFL Playoffs.
The NFL playoffs are among the biggest events in US sporting history and will also be shown in the UK, France and Germany. That explains why Microsoft has spent an undisclosed, but clearly massive, amount of money on screening a 30-second ad for Microsoft Teams.
One would think that Microsoft would showcase one of its bigger products, like Windows or Office, during such a big event, especially when there's so much money being spent on the ad. It suggests that Teams may very well be a money-maker for Microsoft, which is why they'll be pushing it to such a wide audience. Of course, the metaphor of teamwork ties in strongly, too, during a sporting event.
At the launch of Teams nearly four years ago, Satya Nadella, CEO of Microsoft, said: "Just like Outlook brought together email, contacts, calendar, into this one magical user experience scaffolding that changed how we work, Microsoft Teams will bring together chat, meetings, notes, Office, Planner, Power BI and a host of other ecosystem-developed extensions and applications to help teams get work done. "
The ad showcases instant team meeting organisation, background blur on video chats, a digital whiteboard that's shared during virtual meetings, and group chat functionality.
The best gear of Borderlands 3 will be much easier to acquire during a time-limited 'Farming Frenzy' mini-event, which is currently running, through to 30 January.
Hoping to snag a specific Legendary drop? Still looking to cross some of those rare and exotic enemies off your list? Well, good news, Vault Hunter: for the next two weeks, the Borderlands 3 farming frenzy is on.
Not only will rare enemies be guaranteed to appear, their specific Legendary gear will drop at a higher than normal rate when players defeat them. Additionally, Takedown at the Maliwan Blacksite will be more manageable for small squads and solo players, meaning its bosses (and their Legendary loot) will be highly accessible.
2K Games provided these specifics of this limited-time mini-event:
Rare spawns are guaranteed to appear;
All enemies with dedicated loot pools (e.g. rare spawns, named bosses, etc.) have a higher chance of dropping Legendary gear from their dedicated loot pool;
The Anointment rate for that gear has been increased;
Takedown at the Maliwan Blacksite's difficulty will now scale to match the size of the player party rather than defaulting to four-player difficulty (available on PC, PS4, and Xbox One only).
These changes went live on 16 January and will be reversed on 30 January. That means players have two weeks of prime loot hunting to kick off the new year, so get out there and level up your arsenal.

redpajama_set_name: RedPajamaCommonCrawl


Daniel J. Rea, Mahdi Rahmani, Neil Bruce, James E. Young. "Tortoise and the Hare Robot: Slow and steady almost wins the race, but finishes more safely," IEEE International Symposium on Robot and Human Interactive Communication, RO-MAN, 2017. Lisbon, Portugal.


Animal Shelter 1Kittens wait at the Missoula Animal Control Shelter for new homes and owners on Friday. The shelter has reached capacity and now looks to Missoulians to adopt some of the 31 dogs and 49 cats there.
Animal Shelter 2From his perch in the shelter lobby, the Big Lebowski glances at visitors in between naps. According to shelter attendant Elaine Sehnert, Lebowski was returned to the shelter five years ago but because of his relaxed behavior, he is the only pet not up for adoption.
Dog Show 1Mary Ellen Guy of Littleton, Colo., sits next to her Bernese mountain dog Lily at the 58th annual Five Valley Kennel Club dog show on Tuesday. Although tired, Lily won Best in Breed as an 8-year-old veteran show dog.
Dog Show 2Edward Boyes and Taylor, a smooth fox terrier from Sacramento, Calif., relax while waiting to compete in the terrier group during the Five Valley Kennel Club Dog Show on Tuesday. Taylor went on to win the group and took second place in Best of Show.
Dog Show 3Virginia Bair of Klamath Falls, Ore., removes hair products from Glee's coat on Tuesday. Bair explained the importance of caring for a show dog's coat between events and commented that product is only in the dog's hair for a short time before and after shows.


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President Joe Biden and first lady Jill Biden arrive at the North Portico of the White House, Wednesday, Jan. 20, 2021, in Washington.
President Joe Biden signed three executive actions Wednesday. The first mandated masks and social distancing on federal property, the second s…
In one of the inauguration's most talked about moments, poet Amanda Gorman summoned images dire and triumphant Wednesday as she called out to …


text:
The Duckworth Family Tennis Facility
Live Stats/Video
Men's Tennis Season Preview
Under the direction of first-year head coach Robbie Weiss, the Clemson men's tennis team looks to begin a new era as it kicks off its 2020 championship campaign. Last season, the Tigers went 10-17 overall and 2-10 in conference action. Nine tigers return to the roster for the 2019-20 season after competing in a combined 130 singles and 77 doubles matches in 2018-19.
"The much-anticipated Spring season has officially arrived, and I could not be more excited to get started," stated Coach Weiss. "The team has done a fantastic job preparing themselves throughout the fall for the upcoming team competition. I am extremely proud of them for the improvements they have made and for their commitment to excellence. Playing in the Atlantic Coast Conference is as tough as it gets, but the team is ready to compete and start going toe-to-toe with the best in the country. Leading the Clemson men's tennis program is an honor and a privilege, and I will continue to give everything I have to this amazing university and program."
The Tigers will field a young team as six of its 11 athletes hold underclassman status. Of the nine student-athletes returning to the roster this season, three are seniors, one is a junior and five are sophomores. In addition, the program welcomed two walk-ons to the roster, junior Jaxon Wiley and freshman Tim Pitts. Seeing action in six tournaments during the fall schedule, Clemson athletes went a combined 66-45 in singles and 27-23 in doubles.
At the conclusion of the 2019 fall season, Jacob Jahn posted a team-best record of 13-2 in singles while going 7-5 in doubles. In singles, the South Carolina native finished the season on a 6-match win streak, dropping just two sets during that span. The sophomore also earned a singles draw title at the Wake Forest Invitational. During the 2019 spring campaign, Jahn played predominantly in the No. 6 singles position, where he went 6-5 in addition to going 7-10 overall. Jahn paired with senior Pierre-Louis Dodens of Geneva, Switzerland, for doubles last season, where they earned a team-best record of 10-3 at line two.
Dodens, who saw limited action during the fall, maintained a 5-4 record when partnered with Jahn in doubles as well as a 5-2 record in singles. Though the senior competed with Jahn in doubles for the majority of last season, Dodens was also undefeated when he played with sophomore Mike Agee, going 5-0 in the third position and 1-0 at line two. In singles last season, Dodens played a majority of his matches in the No. 5 slot, where he went 5-4.
Senior Carlos Kelaidis also had an impressive fall season after tallying an 11-3 record in singles. Native to Glyfada, Greece, Kelaidis enjoyed a 7-match win streak which spanned nearly an entire month from September to October. In addition, the senior went 5-5 in doubles matches during the fall. Last spring, Kelaidis went 7-15 overall in singles and 3-4 in the No. 4 slot.
Sophomore Teodor Giusca and senior Gabriel Diaz Freire each tallied 8-5 singles marks during the fall season. In doubles, Giusca posted 5-7 marks after utilizing three different partners. Diaz Freire and junior Yizhou Liu had an outstanding fall season as a doubles pair, earning a team-best record of 10-4. Last season, Giusca went 9-13 overall in singles and 11-13 in doubles while Diaz Freire was sidelined most of last season due to injury.
Liu, who hails from Nanjing, China, posted a 10-11 record last season, including a 4-4 mark at line two. In addition to his accomplishments in doubles during the fall, Liu added 5-8 marks in singles play.
From Atlanta, Ga., Agee tallied a 6-9 singles record during the fall and 4-7 combined in doubles. Last season, the sophomore went 4-10 in singles and 6-4 overall in doubles. Sophomore Cole Gordon, who played in only three matches last season, earned a 3-1 record in doubles during the fall season with partner Pitts while also going 1-1 in singles. Pitts, a South Carolina native, went 3-2 in singles and 5-3 in doubles in his first fall of action.
Sophomore Matteo Vialmin, native to Montreux, Switzerland, made four appearances in the fall, posting a 3-1 record in singles. In 2019, Vialmin finished the season 5-4 in singles and 3-2 in doubles. Meanwhile, Wiley, from Asheville, N.C., played to 3-7 marks in singles and a 3-3 record in doubles.
The Tigers face a competitive schedule this spring, matching up with five of the ITA's top-25 programs in addition to eight NCAA Tournament teams from 2019. Much of Clemson's strength of schedule comes from conference play as four of their five top-25 opponents hail from the ACC.
Clemson will enjoy an 11-match homestand to start the season and will open the schedule on Jan. 18, when they host Furman and Wofford for the Duckworth Family Classic. During their 11-match home stretch, the Tigers will host No. 16 South Carolina for Palmetto series action on Feb. 14 and host Miami to open ACC action on Feb. 21. Clemson hits the road for the first time in 2020 on Feb. 28, when they head to Georgia Tech. The Tigers are set to close out the regular season at home against Boston College on April 9 and will take part in the ACC Tournament, which begins on April 15 in Rome, Ga.


text:
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NR-643 has been confirmed as subspecies tularensis (Type A) by PCR amplification of a subspecies-specific sequence of approximately 390 base pairs from extracted DNA, however, it was found to be highly attenuated in the Fischer 344 rat model of pneumonic tularemia. Genetic analyses of NR-643 showed nine regions of difference in the genomic sequence when compared to the published reference SCHU S4 sequence. Given the genetic differences and decreased virulence, it is recommended that NR-643 should not be used in efficacy studies to evaluate potential vaccines and therapeutics against tularemia (Lovchik, JA, et al. (2021) Pathogens. 10: 638).
Each vial contains approximately 0.5 mL of bacterial culture in Mueller Hinton broth supplemented with 10% glycerol.
Citations: Acknowledgment for publications should read "The following reagent was obtained through BEI Resources, NIAID, NIH: Francisella tularensis subsp. tularensis, Strain SCHU S4 (FSC237), NR-643."
Publications Citing
this Reagent: Barrigan, L. M., et al. "Infection with Francisella tularensis LVS clpB Leads to an Altered yet Protective Immune Response." Infect. Immun. 81 (2013): 2028-2042. PubMed: 23529616.
Barrow, E. W., et al. "High-Throughput Screening of a Diversity Collection Using Biodefense Category A and B Priority Pathogens." J. Biomolec. Screen. 17 (2012): 946-956. PubMed: 22653912.
Brown, M. J., et al. "The Contribution of the Glycine Cleavage System to the Pathogenesis of Francisella tularensis." Microbes Infect. 16 (2013): 300-309. PubMed: 24374051.
Conlan, J. W., et al. "Infection of Mice with Francisella as an Immunological Model." Curr. Protoc. Immunol. 19 (2011). PubMed: 21462168.
Dotson, R. J., et al. "Repression of Inflammasome by Francisella tularensis During Early Stages of Infection." Biol. Chem. 288 (2013): 23844-23857. PubMed: 23821549.
Duncan, D., et al. "Identification and Typing of Francisella tularensis with a Highly Automated Genotyping Assay." Lett. Appl. Microbiol. 56 (2013): 128-134. PubMed: 23121644.
Emery, F. D., et al. "Efficient Inactivation of Burkholderia pseudomallei or Francisella tularensis in Infected Cells for Safe Removal from Biosafety Level 3 Containment Laboratories." Pathog. Dis. (2014): doi: 10.1111/2049-632X.12138. PubMed: 24449562.
Faith, S. A., et al. "Growth Conditions and Environmental Factors Impact Aerosolization but Not Virulence of Francisella tularensis Infection in Mice." Front. Cell. Infect. Microbiol. 2 (2012): doi: 10.3389. PubMed: 23087911.
Gregory, S. H., et al. "Detoxified Endotoxin Vaccine (J5dLPS/OMP) Protects Mice Against Lethal Respiratory Challenge with Francisella tularensis SchuS4." Vaccine 28 (2010): 2908-2915. PubMed: 20170768.
Li, Y., et al. "A Rapid One-Step Method for the Characterization of Membrane Lipid Remodeling in Francisella Using Matrix-Assisted Laser Desorption Ionization Time-of-Flight Tandem Mass Spectrometry." Rapid Commun. Mass Spectrom. 25 (2011): 2641-2648. PubMed: 23657959.
Lovchik, J. A., et al. "Identification of an Attenuated Substrain of Francisella tularensis SCHU S4 by Phenotypic and Genotypic Analyses." Pathogens 10 (2021): 638. doi: 10.3390/pathogens10060638.
Lu, Z., et al. "Protective B-Cell Epitopes of Francisella tularensis O-Polysaccharide in a Mouse Model of Respiratory Tularaemia." Immunol. 136 (2012): 352-360. PubMed: 22486311.
Lu, Z., et al. "B-Cell Epitopes in GroEL of Francisella tularensis." PloS One 9 (2014): e99847. PubMed: 24968190.
Mahawar, M., et al. "Identification of a Live Attenuated Vaccine Candidate for Tularemia Prophylaxis." PloS One 8 (2013): e61539. PubMed: 23613871.
Mortensen, B. L., et al. "Effects of the Putative Transcriptional Regulator IclR on Francisella tularensis Pathogenesis." Infect. Immun. 78 (2010): 5022-5032. PubMed: 20921148.
Petersen, J. M. and C. R. Molins. "Subpopulations of Francisella tularensis ssp. tularensis and holarctica: Identification and Associated Epidemiology." Future Microbiol. 5 (2010): 649-661. PubMed: 20353304.
Pflughoeft, K. (2010). "The Immune Inhibitor A1 Protease of Bacillus anthracis."
Riojas, M.A., et al. "Multiplex PCR for Species-Level Identification of Bacillus anthracis and Detection of pXO1, pXO2, and Related Plasmids." Health Secur. 13 (2015): 122-129. PubMed: 25813976.
Roberts, L. M., et al. "Identification of Early Interactions Between Francisella and the Host." Inf. Imm. 82 (2014): 2504-2510. PubMed: 24686053.
Roche, M. I., et al. "Characterization of Monoclonal Antibodies to Terminal and Internal O-Antigen Epitopes of Francisella tularensis Lipopolysaccharide." Hybridoma 30 (2011): 19-28. PubMed: 21466282.
Stark, J. C., et al. "On-Demand Biomanufacturing of Protective Conjugate Vaccines." Sci. Adv. 7 (2021): eabe9444. PubMed: 33536221.
Valentino, M. D., et al. "Identification of T-Cell Epitopes in Francisella tularensis Using an Ordered Protein Array of Serological Targets." Immunol. 132 (2011): 348-360. PubMed: 21214540.
Wang, Q., et al. "A Typical Preparation of Francisella tularensis O-Antigen Yields a Mixture of Three Types of Saccharides." Biochem. 50 (2011): 10941-10950. PubMed: 22091710.
Zinman, G., et al. "Large Scale Comparison of Innate Responses to Viral and Bacterial Pathogens in Mouse and Macaque." PloS One 6 (2011): e22401. PubMed: 21789257.
Select Agents require facility registration for specific agents with CDC and completion of APHIS/CDC Form 2. For information about toxin amounts excluded from CDC Select Agent regulations, click here.
Genomic DNA from Francisella tularensis subsp. tularensis, Strain SCHU S4 (FSC237)


text:
using System.Linq;
using System.Reflection;
using System.Text;
using System.Threading.Tasks;
using Wyam.Common;

namespace Wyam.Modules.TextGeneration
{
    public class GenerateContent : RantModule
    {
        public GenerateContent(object template) : base(template)
        {
        }

        public GenerateContent(ContextConfig template) : base(template)
        {
        }

        public GenerateContent(DocumentConfig template) : base(template)
        {
        }

        public GenerateContent(params IModule[] modules) : base(modules)
        {
        }

        protected override IDocument Execute(string content, IDocument input)
        {
            return input.Clone(content);
        }
    }
}

redpajama_set_name: RedPajamaGithub


text:
You Have the Same Chance of Winning a Lottery Whether You Play Or Not
Fran Lebowitz? Herb Caen? Don Bleu? Rob Morse? Rebecca Blagrave? Liz Smith? William Deresiewicz?
Dear Quote Investigator: The probability that you will purchase a lottery ticket worth millions of dollars is miniscule. Here are two comically exaggerated quips based on this observation:
I figure your odds of winning the lottery are the same, whether you buy a ticket or whether you don't.
I've done the calculation and your chances of winning the lottery are identical whether you play or not.
Commentator Fran Lebowitz has received credit for this saying. Would you please explore its provenance?
Quote Investigator: The earliest evidence located by QI occurs in a video segment dated September 18, 1985 from the television show "Late Night with David Letterman" during which Fran Lebowitz spoke about gambling to the host Letterman. The segment is available via YouTube. Boldface added to excerpts by QI:[1]YouTube video, Title: Fran Lebowitz Collection on Letterman, 1980-2010, Uploaded on January 11, 2021, Uploaded by: Don Giller, Airdate of television episode: September 18, 1985, (Date is shown at 1 … Continue reading
We have a lottery here in New York. I feel you have the same chance of winning a lottery whether you play or not.
Currently, Lebowitz is the leading candidate for originator of this humorous observation. The statement's phrasing is highly variable which makes it difficult to trace. QI has not independently verified the date of the video segment.
Interestingly, another person received credit for the saying in the October 4, 1985 column of famous San Francisco journalist Herb Caen:[2] 1985 October 4, San Francisco Chronicle, Keep in Cool Dry Place by Herb Caen, Quote Page 45, Column 1, San Francisco, California. (GenealogyBank)
Don Bleu, the KYUU discjock, figures that your chances of winning $2 million in the state lottery are about equal whether you buy a ticket or not. Keep cool and move slowly through the weekend.
On October 8, 1985 columnist Rob Morse of "The San Francisco Examiner" contradicted Caen and stated that Lebowitz employed the joke during an appearance on Letterman's late-night television show. Morse spelled "Lebowitz" as "Liebowitz":[3] 1985 October 8, The San Francisco Examiner, Me and my chateau by Rob Morse, Quote Page D1, Column 1, San Francisco, California. (Newspapers_com)
Add this to your list of lines that were not originally spoken by a KYUU disc jockey. It was said by Fran Liebowitz on a recent David Letterman show: "I figure your odds of winning the lottery are the same, whether you buy a ticket or whether you don't."
On October 13, 1985 Caen responded, and he conceded that Fran Lebowitz probably deserved credit although he spelled her name as "Liebowitz":[4] 1985 October 13, San Francisco Chronicle, Section: Sunday Punch, Out Of My Mind by Herb Caen, Quote Page 1, Column 1, San Francisco, California. (GenealogyBank)
A few columns ago, I credited a local disc jockey with the crack that your chances of winning $2 million are about the same whether you buy a ticket or not. Several people tell me Fran Liebowitz originated the line, but I am not accusing the local fellow of plagiarism; however, I'll concede the odds are not in his favor. Fran has a helluva batting average.
In September 1988 physical fitness advocate Rebecca Blagrave wrote a piece in "Vincennes Sun-Commercial" newspaper of Indiana. She used a version of the jest without attribution:[5]1988 September 4, Vincennes Sun-Commercial, Do you feel lucky in the automobile? by Rebecca Blagrave, (Knox County Coordinating Council on Physical Fitness), Quote Page 29, Column 1, Vincennes, … Continue reading
Each week, thousands of Hoosiers cross the state line to buy lottery tickets, knowing full well the chance of winning the lottery is about the same whether they buy a ticket or not.
In January 1989 science magazine "Omni" published the following instance as a filler item:[6] 1989 January, OMNI, Volume 11, Number 4, Section: Continuum, (Filler item), Quote Page 35, Column 2, OMNI Publications, New York. (Verified with scans)
"I figure you have the same chance of winning the lottery whether you play or not."
A week later prominent gossip columnist Liz Smith repeated the information in "Omni" magazine:[7] 1989 January 8, Daily News, Mere five Ferrari fans to get lucky by Liz Smith, Quote Page 10, Column 2, New York, New York. (ProQuest)
"I figure you have the same chance of winning the lottery whether you play or not," Fran Lebowitz told Omni.
In April 1989 the "Pittsburgh Post-Gazette" published an article about the Pennsylvania lottery's $100 million dollar jackpot. The newspaper quoted Timothy Means of Penn Hills who decided not to buy a ticket:[8] 1989 April 26, Pittsburgh Post-Gazette, Buying frenzy swells pot by Barbara White Stack (Post-Gazette Staff Writer), Quote Page 1, Column 5, Pittsburgh, Pennsylvania. (ProQuest)
"You stand the same chance of winning the lottery, whether you play or not," he said as he waited in line at Somma Pizza in the food court at PPG Place.
In September 1989 a newspaper in Kingston, Ontario, Canada printed an anonymous instance:[9] 1989 September 22, The Whig-Standard, Lotteries: So Little for the Mind by Gerald Walton Paul, Quote Page 1, Kingston, Ontario, Canada. (ProQuest)
So remote are the chances of winning a lottery million that there's truth in the joke that's making the rounds: "I figure you have the same chance of winning the lottery whether you buy a ticket or not!"
In 1997 "The Forbes Book of Business Quotations" included the following entry:[10]1997, The Forbes Book of Business Quotations: 14,173 Thoughts on the Business of Life, Edited by Ted Goodman, Topic: Chance, Quote Page 125, Column 2, Black Dog & Leventhal Publishers, New York. … Continue reading
In 2006 a columnist in a Tucson, Arizona newspaper credited Lebowitz with another variant:[11] 2006 June 20, Arizona Daily Star, Guest Opinion: Willing to bet we can ax another scam — lotteries by Gilbert D. Shapiro, Quote Page A7, Column 5, Tucson, Arizona. (Newspapers_com)
I agree with author Fran Lebowitz, who insightfully commented, "I've done the calculation and your chances of winning the lottery are identical whether you play or not."
In 2020 cultural critic William Deresiewicz published "The Death of the Artist" about the precarious existence of many modern artists. He included another instance of the saying:[12] 2020, The Death of the Artist by William Deresiewicz, Chapter 4: The New Conditions, Unnumbered Page, Henry Holt and Company, New York. (Google Books Preview)
Of the two million artists on Spotify, less than 4 percent account for over 95 percent of streams. Of the roughly fifty million songs available, at last count, on the site, 20 percent have never been streamed, not even once. (Let that one sink in for a moment.) I'm reminded of something that the writer Fran Lebowitz has said: your chances of winning the lottery are the same whether you buy a ticket or not.
QI believes Fran Lebowitz should receive credit for this remark. She used it during an appearance on the television show "Late Night with David Letterman" in 1985. Variant phrasings have proliferated over time.
Image Notes: Public domain image of the painting titled "The State Lottery" by Vincent van Gogh circa 1882. The image has been cropped and resized.
(Great thanks to the anonymous person who told QI about the instance in William Deresiewicz's book. This led QI to formulate this question and perform this exploration.)
YouTube video, Title: Fran Lebowitz Collection on Letterman, 1980-2010, Uploaded on January 11, 2021, Uploaded by: Don Giller, Airdate of television episode: September 18, 1985, (Date is shown at 1 hour, 45 minutes, 18 seconds), (Quotation starts at 1 hour, 51 minutes, 50 seconds of 2 hours, 30 minutes, 44 seconds) (Accessed on youtube.com on April 29, 2021) link
1985 October 4, San Francisco Chronicle, Keep in Cool Dry Place by Herb Caen, Quote Page 45, Column 1, San Francisco, California. (GenealogyBank)
1985 October 8, The San Francisco Examiner, Me and my chateau by Rob Morse, Quote Page D1, Column 1, San Francisco, California. (Newspapers_com)
1985 October 13, San Francisco Chronicle, Section: Sunday Punch, Out Of My Mind by Herb Caen, Quote Page 1, Column 1, San Francisco, California. (GenealogyBank)
1988 September 4, Vincennes Sun-Commercial, Do you feel lucky in the automobile? by Rebecca Blagrave, (Knox County Coordinating Council on Physical Fitness), Quote Page 29, Column 1, Vincennes, Indiana. (Newspapers_com)
1989 January, OMNI, Volume 11, Number 4, Section: Continuum, (Filler item), Quote Page 35, Column 2, OMNI Publications, New York. (Verified with scans)
1989 January 8, Daily News, Mere five Ferrari fans to get lucky by Liz Smith, Quote Page 10, Column 2, New York, New York. (ProQuest)
1989 April 26, Pittsburgh Post-Gazette, Buying frenzy swells pot by Barbara White Stack (Post-Gazette Staff Writer), Quote Page 1, Column 5, Pittsburgh, Pennsylvania. (ProQuest)
1989 September 22, The Whig-Standard, Lotteries: So Little for the Mind by Gerald Walton Paul, Quote Page 1, Kingston, Ontario, Canada. (ProQuest)
1997, The Forbes Book of Business Quotations: 14,173 Thoughts on the Business of Life, Edited by Ted Goodman, Topic: Chance, Quote Page 125, Column 2, Black Dog & Leventhal Publishers, New York. (Verified with scans)
2006 June 20, Arizona Daily Star, Guest Opinion: Willing to bet we can ax another scam — lotteries by Gilbert D. Shapiro, Quote Page A7, Column 5, Tucson, Arizona. (Newspapers_com)
2020, The Death of the Artist by William Deresiewicz, Chapter 4: The New Conditions, Unnumbered Page, Henry Holt and Company, New York. (Google Books Preview)
Posted on April 29, 2021 April 30, 2021 Author quoteresearchCategories Fran Lebowitz, Herb Caen, Liz SmithTags Don Bleu, Fran Lebowitz, Herb Caen, Liz Smith, Rebecca Blagrave, Rob Morse, William Deresiewicz
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text:
Intercourse ist ein zu Statistikzwecken definiertes Siedlungsgebiet, ein Census-designated place (CDP) mit 1274 Einwohnern im Lancaster County des US-Bundesstaats Pennsylvania.

Geographie
Intercourse liegt 18 Kilometer östlich von Lancaster und rund 80 Kilometer westlich der Großstadt Philadelphia. Der U.S. Highway 30 tangiert Intercourse im Süden.

Geschichte
Das Dorf wurde 1754 unter dem Namen Cross Keys, wie auch eine lokale Taverne hieß, gegründet. Es gibt viele Spekulationen hinsichtlich der Herkunft des späteren Namens dieses kleinen Ortes. So werden die folgenden verschiedenen Erklärungen diskutiert, aber keine kann wirklich begründet werden:
 Das Zentrum um eine alte Rennstrecke, die auf dem Old Philadelphia Pike existierte wurde als Entercourse bezeichnet. Es wird angenommen, dass sich Entercourse allmählich zu Intercourse entwickelte und 1814 der Name des Ortes wurde.
 Eine andere Theorie betrifft zwei berühmte Straßen, die sich hier kreuzten. Der Old King's Highway von Philadelphia nach Pittsburgh (später Old Philadelphia Pike) verlief nach Osten und Westen durch das Zentrum des Ortes und kreuzte sich in der Mitte mit der Straße von Wilmington nach Erie. Die Kreuzung dieser beiden Straßen wird von einigen als Grundlage für den Namen Cross Keys genannt. Später wurde die Kreuzung möglicherweise als Intercourse bezeichnet.
 Eine weitere Idee basiert auf dem Sprachgebrauch in den frühen Tagen des Dorfes. Das Wort Intercourse wurde damals häufig verwendet, um die "Gemeinschaft" und "soziale Interaktion und Unterstützung" zu beschreiben, die in der Glaubensgemeinschaft eines kleinen ländlichen Dorfes geteilt wurde.

Die Einwohner von Intercourse bestreiten ihren Lebensunterhalt zu Beginn des 21. Jahrhunderts in erster Linie durch touristische und landwirtschaftliche Betriebe. Intercourse ist wegen seiner Deitscherei-Lage sowie seines sexuell suggestiven Namens (Intercourse = Geschlechtsverkehr) ein beliebter Ort für Touristen.

Demografische Daten
Im Jahr 2010 wurde eine Einwohnerzahl von 1274 Personen ermittelt. Das Durchschnittsalter lag zu diesem Zeitpunkt mit 35,0 Jahren unterhalb des Wertes von Pennsylvania, der 40,6 Jahre betrug. 25,3 % der heutigen Bewohner gehen auf Einwanderer aus Deutschland zurück. Weitere maßgebliche Zuwanderungsgruppen während der Anfänge des Ortes kamen zu 4,2 % aus der Schweiz und zu 10,2 % aus weiteren europäischen Ländern.

Einzelnachweise

Weblinks

 discoverlancaster.com – American Military Edged Weaponry Museum
 pacamping.com/attractions – People's Place Quilt Museum

Census-designated place in Pennsylvania

redpajama_set_name: RedPajamaWikipedia


Shop.org, a division of the National Retail Federation, is the world's leading community for digital retail. The renowned judges of the group's Digital Commerce Startup of the Year competition named Reflektion the winner. After being selected as finalist in September, Reflektion was able to prevail over two competitor. The company's technology aims to capture and respond to individual shopper's intent in real time.
Reflektion will receive a free one-year membership with the National Retail Federation, a complimentary registration to Shop.org's 2016 Digital Summit in Dallas and $5,000 dollars. Sean Moran, president and CEO of Reflektion decided to donate the prize money to Disney supported charities. A great gesture!


text:
and Personal Assistant
Jodi Wareing
jodiwareing@parryfield.com
Parry Field Lawyers
Alex Summerlee
LLB (Hons), BA – University of Canterbury
Email: alexsummerlee@parryfield.com
Alex specialises in civil litigation and dispute resolution, with experience resolving complex legal issues for clients both in and outside of the Courts. Alex has particular expertise acting in disputes involving insurance, construction, solvency, commercial leases, trusts & estates, child custody and relationship property. He has represented clients in arbitrations, a variety of tribunals, the District Court, the High Court, and in the Court of Appeal.
Alex is active in the community, serving as Vice Chancellor of the Anglican Diocese of Christchurch. He also sits on the governance board of his local church and is a member of the Executive Committee of the Alliance Française de Christchurch.
Outside of the office, you may find him enjoying a glass of vintage red with a good book, running along the Port Hills, doing some DIY renovations, on the piste at his local fencing club, or pulling out some tunes on piano. Alex is also fluent in French and conversant in Spanish.
Alex graduated from the University of Canterbury in 2015 with a LLB(Hons) and a BA majoring in French and Political Science.
Cases:
Hortiventure Ltd v Alpine Sun Ltd [2016] NZHC 2130;
A successful strike out application against a landlord arising out of commercial lease dispute.
Gull v Smart [2016] NZHC 2566;
In this custody proceeding, the father successfully opposed the mother's application to relocate the parties' son to Australia. The result was upheld on appeal in the High Court.
Lee v IAG New Zealand Ltd [2017] NZHC 2626; [2018] Lloyd's Rep. IR 345;
Litigation concerning the interpretation of an insurance policy, and an insurer's liability for interest on a payment obligation in an insurance contract.
Black Rock Administration Limited [2018] NZHC 3450;
The insured obtained costs against IAG in litigation relating to its material damage claims caused by the 2010-2011 Canterbury earthquakes, after Black Rock was able to successfully settle its claims.
Redline NZ Ltd v Conveyor Technology Ltd [2016] NZHC 1083;
The respondent company succeeded in opposing an application to set aside a statutory demand, and had the debtor company immediately placed into liquidation by deploying a rarely used procedure in the Companies Act.


text:
package multihost

import "testing"

func TestServe(t *testing.T) {

} //TestServe