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Dyson Brownian motion for general \(\beta \) and potential at the edge

Abstract

In this paper, we compare the solutions of the Dyson Brownian motion for general \(\beta \) and potential V and the associated McKean–Vlasov equation near the edge. Under suitable conditions on the initial data and the potential V, we obtain optimal rigidity estimates of particle locations near the edge after a short time \(t={{\,\mathrm{o}\,}}(1)\). Our argument uses the method of characteristics along with a careful estimate involving an equation of the edge. With the rigidity estimates as an input, we prove a central limit theorem for mesoscopic statistics near the edge, which, as far as we know, has been done for the first time in this paper. Additionally, combining our results with Landon and Yau (Edge statistics of Dyson Brownian motion. arXiv:1712.03881, 2017), we give a proof of the local ergodicity of the Dyson Brownian motion for general \(\beta \) and potential at the edge, i.e., we show the distribution of extreme particles converges to the Tracy–Widom \(\beta \) distribution in a short time.

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Acknowledgements

We want to thank the anonymous reviewers for their comments and suggestions, which significantly improved the readability of this manuscript.

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Correspondence to Arka Adhikari.

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Appendices

A Proof of Proposition 3.4

The proof of Proposition 3.4 is based on performing a power series expansion of the analytic functions \(A_0\) and \(B_0\) in a neighborhood around \(E_0\) and solving the McKean–Vlasov equation (2.2) term by term. \(A_0(z)\) and \(B_0(z)\) have power series representations

$$\begin{aligned} A_0(z)&=a_0+a_1(z-E_0)+a_2(z-E_0)^2+a_3(z-E_0)^3+\cdots ,\\ B_0(z)&=b_0+b_1(z-E_0)+b_2(z-E_0)^2+b_3(z-E_0)^3+\cdots , \end{aligned}$$

such that

$$\begin{aligned} |a_i|, |b_i|\leqslant \frac{C_0M^{i}}{(i+1)^2},\quad i=0,1,2,\ldots . \end{aligned}$$

The following proposition states that the Stieltjes transform of \(\hat{\mu }_t\) has the form \(A_t+\sqrt{B_t}\), and \(A_t, B_t\) have power series representations in a neighborhood of \(E_0\). Proposition 3.4 is a natural consequence of the proceeding proposition.

Proposition A.1

We assume (3.3) holds. We fix a sufficiently small \(T>0\) and let \(L= 1/T\). Then for \(t\in [0,T]\), the solution of (2.2) is given by

$$\begin{aligned} \hat{m}_t(z)=A_t(z)+\sqrt{B_t(z)}, \end{aligned}$$
(A.1)

with \(A_t(z)\) and \(B_t(z)\) analytic functions in a small neighborhood of \(E_0\). In this neighborhood, \(A_t(z)\) and \(B_t(z)\) have power series representations,

$$\begin{aligned} A_t(z)&=a_0(t)+a_1(t)(z-E_0)+a_2(t)(z-E_0)^2+a_3(t)(z-E_0)^3+\cdots ,\nonumber \\ B_t(z)&=b_0(t)+b_1(t)(z-E_0)+b_2(t)(z-E_0)^2+b_3(t)(z-E_0)^3+\cdots , \end{aligned}$$
(A.2)

where the coefficients satisfy

$$\begin{aligned} |a_k(t)|, |b_k(t)|\leqslant \frac{CM^ke^{Ltk}}{(k+1)^2}. \end{aligned}$$
(A.3)

Moreover, in a small neighborhood of \(E_0\), \(B_t(z)\) has a unique simple root at \(z=E_t\), which satisfies the differential equation

$$\begin{aligned} \partial _t E_t=-\hat{m}_t(E_t)-\frac{V'(E_t)}{2}. \end{aligned}$$
(A.4)

Proof

We suppose that the solution of (2.2) is given by (A.1), and \(A_t(z)\), \(B_t(z)\) have power series representations given by (A.2). We plug (A.1) into (2.2) to get

$$\begin{aligned} \partial _t A_t +\frac{\partial _t B_t}{2\sqrt{B_t}} =D_t\partial _zD_t+\frac{\partial _z B_t}{2} +R_t(z)+\frac{1}{\sqrt{B_t}}\left( B_t\partial _zD_t+\frac{1}{2}D_t\partial _z B_t\right) , \end{aligned}$$
(A.5)

where

$$\begin{aligned} D_t(z)=A_t+\frac{V'(z)}{2},\quad R_t(z)=\int _{\mathbb {R}}g(z,x)\mathrm{d}\hat{\mu }_t(x)-\frac{V'(z)V''(z)}{4}, \end{aligned}$$
(A.6)

are analytic in a neighborhood of \(E_0\). One should note here that our goal in this section is not to show the existence of solutions to the original McKean–Vlasov equation, but only that said solution is analytic in a small neighborhood of \(E_0\). Thus, the fact that we use \(\hat{\mu }_t\) in our expression of \(R_t(z)\) is not an issue.

For (A.5) to hold, it is sufficient that

$$\begin{aligned} \partial _tD_t&=D_t\partial _zD_t+\frac{\partial _z B_t}{2}+R_t(z),\nonumber \\ \frac{\partial _t B_t}{2}&=B_t\partial _zD_t+\frac{1}{2}D_t\partial _z B_t. \end{aligned}$$
(A.7)

We solve (A.7) using the power series representations. Let

$$\begin{aligned} B_t(z)&=b_0(t)+b_1(t)(z-E_0)+b_2(t)(z-E_0)^2+b_3(t)(z-E_0)^3+\cdots \\ D_t(z)&=d_0(t)+d_1(t)(z-E_0)+d_2(t)(z-E_0)^2+d_3(t)(z-E_0)^3+\cdots \\ R_t(z)&=r_0(t)+r_1(t)(z-E_0)+r_2(t)(z-E_0)^2+r_3(t)(z-E_0)^3+\cdots . \end{aligned}$$

Then, (A.7) is equivalent to the following infinite system of ordinary differential equations. For \(i=0,1,2,\ldots \),

$$\begin{aligned} \partial _t d_i(t)&=\sum _{j=0}^i(j+1)d_{i-j}(t)d_{j+1}(t) +\frac{1}{2}(i+1)b_{i+1}(t)+r_i(t)\nonumber \\ \partial _t b_i(t)&=2\sum _{j=0}^i(j+1)b_{i-j}(t)d_{j+1}(t) +\sum _{j=0}^i(j+1)d_{i-j}(t)b_{j+1}(t). \end{aligned}$$
(A.8)

Although (A.8) is not Lipschitz, we can still solve this system by the Picard iteration. Let

$$\begin{aligned} d_i^{(0)}(t)=d_i(0), \quad b_i^{(0)}(t)=b_i(0), \end{aligned}$$

and define recursively

$$\begin{aligned} d_i^{(n+1)}(t)&=d_i(0)+\int _0^t\left( \sum _{j=0}^i(j+1)d^{(n)}_{i-j}(t)d^{(n)}_{j+1}(t) +\frac{1}{2}(i+1)b^{(n)}_{i+1}(t)+r_i(t)\right) \mathrm{d}t,\nonumber \\ b_i^{(n+1)}(t)&= b_i(0)+\int _0^t\left( \sum _{j=0}^i 2(j+1)b^{(n)}_{i-j} (t)d^{(n)}_{j+1}(t)+\sum _{j=0}^i(j+1)d^{(n)}_{i-j}(t)b^{(n)}_{j+1}(t)\right) \mathrm{d}t. \end{aligned}$$
(A.9)

As we have noted before, the existence of the measure \(\hat{\mu }_t\) is not in question and, thus, we do not need to perform an iteration of the \(r_i(t)\) terms in n.

We take large constants \(C>0\), \(L>0\) and a small constant \(T=1/L\). We first prove by induction that uniformly for \(t\in [0,T]\),

$$\begin{aligned} |d_i^{(n)}(t)|, |b_i^{(n)}(t)|\leqslant \frac{CM^i e^{Lti}}{(i+1)^2}. \end{aligned}$$
(A.10)

Since \(R_t(z)\) is analytic in a neighborhood of \(E_0\), we can take \(C_0, M>0\) large enough such that

$$\begin{aligned} |r_i(t)|\leqslant \frac{C_0M^i }{(i+1)^2}. \end{aligned}$$

We assume that (A.10) holds for n. Using (A.9), we have

$$\begin{aligned} |d_i^{(n+1)}(t)|&\leqslant \frac{C_0M^i}{(i+1)^2}+\int _0^t \left( \sum _{j=0}^i \frac{C^2M^{i+1} e^{Lt(i+1)}}{(i-j+1)^2(j+2)} +\frac{CM^{i+1} e^{Lt(i+1)}}{2(i+2)}+\frac{C_0M^i }{(i+1)^2}\right) \mathrm{d}t\\&\leqslant \frac{(1+t)C_0M^i}{(i+1)^2}+\sum _{j=0}^i \frac{C^2M^{i+1} e^{Lt(i+1)}}{L(i+1)(j+2)(i-j+1)^2} +\frac{CM^{i+1} e^{Lt(i+1)}}{2L(i+1)(i+2)}\\&\leqslant \frac{CM^i e^{Lti}}{(i+1)^2}\left( \frac{(1+t)C_0}{C} +\frac{6CMe^{LT}}{L}+\frac{Me^{TL}}{2L}\right) \leqslant \frac{CM^i e^{Lti}}{(i+1)^2}, \end{aligned}$$

provided \(C>4C_0\) and \(L>24eCM\). Similarly, for \(b_i^{(n+1)}(t)\),

$$\begin{aligned} |b_i^{(n+1)}(t)|&\leqslant \frac{C_0M^i}{(i+1)^2}+\int _0^t 3\sum _{j=0}^i \frac{C^2M^{i+1} e^{Lt(i+1)}}{(i-j+1)^2(j+2)}\mathrm{d}t\nonumber \\&\leqslant \frac{CM^i e^{Lti}}{(i+1)^2}\left( \frac{C_0}{C} +\frac{18CMe^{LT}}{L}\right) \leqslant \frac{CM^i e^{Lti}}{(i+1)^2}, \end{aligned}$$
(A.11)

provided \(C>2C_0\) and \(L>36eCM\). This finishes the proof of claim (A.10).

In the following, we prove that \(d_i^{(n)}(t), b_i^{(n)}(t)\) converge uniformly as n goes to infinity. This follows from

$$\begin{aligned} |d_i^{(n)}(t)-d_i^{(n-1)}(t)|, |b_i^{(n)}(t)-b_i^{(n-1)}(t)| \leqslant \frac{1}{2^n}\frac{CM^ie^{Lti}}{(i+1)^2}. \end{aligned}$$
(A.12)

We prove (A.12) by induction. We assume that (A.12) holds for n. By using (A.9), the difference \(|d_i^{(n+1)}(t)-d_i^{(n)}(t)|\) is bounded by

$$\begin{aligned}&\int _0^t\sum _{j=0}^i(j+1)\left( \left| d^{(n)}_{i-j}(t) -d^{(n-1)}_{i-j}(t)\right| d^{(n)}_{j+1}(t)+d_{i-j}^{(n-1)}(t) \left| d^{(n)}_{j+1}(t)-d^{(n-1)}_{j+1}(t)\right| \right) \\&+\frac{1}{2}(i+1)\left| b^{(n)}_{i+1}(t)-b^{(n-1)}_{i+1}(t)\right| \mathrm{d}t \leqslant \frac{1}{2^n}\int _0^t \left( \sum _{j=0}^i \frac{2C^2M^{i+1} e^{Lt(i+1)}}{(i-j+1)^2(j+2)} +\frac{CM^{i+1} e^{Lt(i+1)}}{2(i+2)}\right) \mathrm{d}t\\&\leqslant \frac{1}{2^n}\frac{CM^ie^{Lti}}{(i+1)^2} \left( \frac{12CMe^{LT}}{L}+\frac{Me^{TL}}{2L}\right) \leqslant \frac{1}{2^{n+1}}\frac{CM^ie^{Lti}}{(i+1)^2}, \end{aligned}$$

provided that \(L\geqslant 24eCM\). Similarly, the difference \(|b_i^{(n+1)}(t)-b_i^{(n)}(t)|\) is bounded by

$$\begin{aligned}&\int _0^t 2\sum _{j=0}^i(j+1)\left( \left| b^{(n)}_{i-j}(t) -b^{(n-1)}_{i-j}(t)\right| d^{(n)}_{j+1}(t)+b^{(n-1)}_{i-j}(t) \left| d^{(n)}_{j+1}(t)-d_{j+1}^{(n-1)}(t)\right| \right) \nonumber \\&\qquad +\sum _{j=0}^i(j+1)\left( \left| d^{(n)}_{i-j}(t) -d^{(n-1)}_{i-j}(t)\right| b^{(n)}_{j+1}(t)+d^{(n-1)}_{i-j}(t) \left| b^{(n)}_{j+1}(t)-b_{j+1}^{(n-1)}(t)\right| \right) \mathrm{d}t\nonumber \\&\quad \leqslant \frac{1}{2^n}\int _0^t \left( \sum _{j=0}^i \frac{6C^2M^{i+1} e^{Lt(i+1)}}{(i-j+1)^2(j+2)}\right) \mathrm{d}t \leqslant \frac{1}{2^n}\frac{CM^ie^{Lti}}{(i+1)^2}\frac{36CMe^{LT}}{L} \leqslant \frac{1}{2^{n+1}}\frac{CM^ie^{Lti}}{(i+1)^2}, \end{aligned}$$
(A.13)

provided \(L\geqslant 72eCM\).

We denote for \(i=0,1,2,\ldots \),

$$\begin{aligned} d_i(t)=\lim _{n\rightarrow \infty }d^{(n)}_i(t), \quad b_i(t)=\lim _{n\rightarrow \infty }b_i^{(n)}(t). \end{aligned}$$

They satisfy the system of differential equations (A.9), and (A.3) holds.

Using the same argument we used for (A.11) gives us

$$\begin{aligned} |b_i(t)-b_i(0)|\leqslant \frac{3C^2M^{i+1}(e^{Lt(i+1)}-1)}{(i+1)^2L}. \end{aligned}$$

And, thus, for \(T>0\) small enough, times \(0\leqslant t\leqslant T\), and z on a small circle centered at \(E_0\),

$$\begin{aligned} |B_t(z)-B_0(z)|\leqslant \sum _{i\geqslant 0}\frac{3C^2(Mz)^{i+1} (e^{Lt(i+1)}-1)}{(i+1)^2L}< |B_0(z)|. \end{aligned}$$

Thus, by Rouché’s theorem, \(B_t(z)\) has a unique simple root at \(z=E_t\) inside a small neighborhood of \(E_0\). Moreover, from our construction, \(\bar{B}_t(z)=B_t(\bar{z})\). Thus, \(E_t\) is real. By taking the derivative of \(B_t(E_t)=0\) with respect to t and using (A.7), we get

$$\begin{aligned} \partial _t E_t=-\frac{\partial _t B_t(E_t)}{\partial _z B_t(E_t)} =-D_t(E_t)=-\hat{m}_t(E_t)-\frac{V'(E_t)}{2}. \end{aligned}$$

This finishes the proof of Proposition A.1. \(\square \)

Proof of Proposition 3.4

It follows from Proposition A.1 that \(\hat{\mu }_t\) has square root behavior. In addition, \(\hat{m}_t(E_t) = A_t(E_t) +\sqrt{B_t(E_t)}\) is uniformly bounded for \(0\leqslant t\leqslant T\). The claim that \(E_t\) is Lipschitz in time simply follows from integrating (A.4).

Next, we prove that \({\mathfrak {C}}_t\) is Lipschitz in time. We notice that \({\mathfrak {C}}_t^2=\partial _z B_t(E_t)\). To prove \(|{\mathfrak {C}}_t-{\mathfrak {C}}_s|={{\,\mathrm{O}\,}}(t-s)\), it suffices to prove this for \(s=0\). Using the series expansion, as in the notation of (A.2), we see that

$$\begin{aligned} {\mathfrak {C}}_t^2-{\mathfrak {C}}_0^2=\partial _z B_t(E_t) -\partial _z B_0(E_0) = b_1(t) -b_1(0) + (E_t-E_0)\sum _{i=2}^{\infty } i b_i(t)(E_t-E_0)^{i-2}. \end{aligned}$$

Using the differential equation (A.8), we see that \(b_1(t) -b_1(0)={{\,\mathrm{O}\,}}(t)\). Earlier, we showed that \((E_t - E_0)\) is of \({{\,\mathrm{O}\,}}(t)\). The infinite sum converges provided we take T sufficiently small. Therefore, it follows that \({\mathfrak {C}}_t^2 - {\mathfrak {C}}_0^2 ={{\,\mathrm{O}\,}}(t)\). Since \({\mathfrak {C}}_0\) is bounded away from 0, this would imply that \({\mathfrak {C}}_t -{\mathfrak {C}}_0 = {{\,\mathrm{O}\,}}(t)\) as desired. \(\square \)

B Rigidity bounds in the Proof of the Mesoscopic Central Limit Theorem

We start this section by collecting some lemmas that analyze the movement of characteristic flows that terminate in a radius \(\eta \) (\(N^{-2/3} \ll \eta \ll 1\)) neighborhood of the edge \(E_t\). Knowing the precise values of \(\kappa _s\) and \(\eta _s\) along a characteristic of the form \(z_s = E_s + \kappa _s + \mathrm{i}\eta _s\) will allow us to apply many of the improved rigidity bounds from Theorem 4.8 and the following remarks.

Lemma B.1

Suppose V satisfies Assumption 2.1 and let \({\mathfrak {d}}\) be a small parameter. Consider a characteristic \(z_t=E_t+\kappa _t+\mathrm{i}\eta _t\) as in (2.4) with terminal time \(t=o(1)\). If \(z_t\) satisfies

$$\begin{aligned} - N^{{\mathfrak {d}}}\eta _t \leqslant \kappa _t \lesssim \eta _t,\quad N^{-2/3+20{\mathfrak {d}}} \leqslant \eta _t\leqslant N^{-20 {\mathfrak {d}}}. \end{aligned}$$
(B.1)

then for any \(0\leqslant s\leqslant t\), we have:

$$\begin{aligned}&N^{-2/3+20{\mathfrak {d}}}(1-C(t-s))\leqslant \eta _s \leqslant N^{-20{\mathfrak {d}}}+C(t-s),\\&\quad -CN^{{\mathfrak {d}}}\eta _s\leqslant \kappa _s\leqslant C(1+(t-s)/\eta _t)\eta _s. \end{aligned}$$

Proof

From the equation determining the motion of the characteristics (2.4), we know that each characteristic moves at a rate \({{\,\mathrm{O}\,}}(1)\) in time. Thus, if \(\eta _t \leqslant N^{-20{\mathfrak {d}}}\), we have \(\eta _s \leqslant N^{-20{\mathfrak {d}}}+C(t-s)\). Thanks to the estimates (2.7), if \(\eta _t \geqslant N^{-2/3+ 20{\mathfrak {d}}}\), we have \(\eta _s \geqslant N^{-2/3+ 20{\mathfrak {d}}}(1-C(t-s))\).

In the following, we estimate the ratio \( \kappa _s/\eta _s\). Let \(z_t - E_t =\kappa _t+\mathrm{i}\eta _t=(R_t+\mathrm{i}a_t)^2\), and \(z_s-E_s= \kappa _s+\mathrm{i}\eta _s=(R_s+\mathrm{i}a_s)^2\), where \(R_t, a_t, R_s\), and \(a_s>0\). By our assumption on \(z_t\), we have

$$\begin{aligned} -N^{{\mathfrak {d}}}\leqslant \frac{\kappa _t}{\eta _t}=\frac{1}{2}\left( \frac{R_t}{a_t} -\frac{a_t}{R_t}\right) \leqslant C, \end{aligned}$$

and it follows that \(N^{-{\mathfrak {d}}}/C\leqslant \frac{R_t}{a_t}\leqslant C\). Thanks to (2.7) and the assumption of square root behavior, we have \(a_t\asymp \mathrm{Im}[\hat{m}_t(z_t)]\asymp \mathrm{Im}[\hat{m}_s(z_s)]\asymp a_s\). Moreover, using (2.7) again, we have \(2a_sR_s=\eta _s\asymp (\eta _t+(t-s)a_t)\asymp (a_s R_t+(t-s)a_s)\). We conclude that \(R_s\asymp R_t+(t-s)\), and

$$\begin{aligned} N^{-{\mathfrak {d}}}\lesssim \frac{R_s}{a_s}\asymp \frac{R_t}{a_t}+\frac{t-s}{a_t} \lesssim \left( 1+\frac{t-s}{a_t}\right) . \end{aligned}$$

It follows, by rearranging, that

$$\begin{aligned} -N^{{\mathfrak {d}}}\lesssim \frac{\kappa _s}{\eta _s}=\frac{1}{2}\left( \frac{R_s}{a_s} -\frac{a_s}{R_s}\right) \lesssim 1+\frac{t-s}{a_t}. \end{aligned}$$

This finishes the proof of Lemma B.1. \(\square \)

Lemma B.2

Suppose V satisfies Assumption 2.1. Consider a characteristic \(z_t=E_t+\kappa _t+\mathrm{i}\eta _t\) as in (2.4) with terminal time \(t={{\,\mathrm{o}\,}}(1)\). If its terminal point \(z_t = E_t +\kappa _t + \mathrm{i}\eta _t\) satisfies \(-\eta _t \leqslant \kappa _t \leqslant C \eta _t \), then we have, for \(t-q \geqslant \sqrt{\eta _t}\), that \(\sqrt{|\kappa _q| +\eta _q} \asymp (t-q)\), and, for \(t-q \leqslant \sqrt{\eta _t}\), it holds that \(\sqrt{|\kappa _q|+ \eta _q} \asymp \sqrt{\eta _t}\). We always have \(\eta _q \asymp \eta _t + (t-q) \sqrt{\eta _t}\).

If we consider a characteristic whose terminal point instead satisfies \(-N^{{\mathfrak {d}}} \eta _t \leqslant \kappa _t \leqslant -\eta _t \), we instead find that \(\sqrt{\kappa _q + \eta _q} \asymp (t-q)\) for \(t-q \geqslant \sqrt{|\kappa _t| + \eta _t}\). For \(t-q \leqslant \sqrt{|\kappa _t| +\eta _t}\), we have \(\sqrt{|\kappa _q| + \eta _q} \asymp \sqrt{|\kappa _t| + \eta _t}\). We always have \(\eta _q \asymp \eta _t +(t-q) \sqrt{|\kappa _t| +\eta _t}\).

Proof

Consider a characteristic whose terminal point \(z_t\) satisfies \(-\eta _t\leqslant \kappa _t \leqslant C \eta _t\). By square root behavior, this will imply \(\mathrm{Im}[\hat{m}_t(z_t)] \asymp \sqrt{\eta _t}\). Finally, by the final line of (2.7), we have that \(\eta _q \asymp \eta _t + (t-q) \sqrt{\eta _t}\). We see that, for \(t-q \leqslant \sqrt{\eta _t}\), we have that \(\eta _q \asymp \eta _t\), while, for \((t-q) \geqslant \sqrt{\eta _t}\), we have that \(\eta _q \asymp (t-q) \sqrt{\eta _t}\).

We additionally have the relation \(\mathrm{Im}[\hat{m}_q(z_q)] \asymp \mathrm{Im}[\hat{m}_t(z_t)]\) by the second line of (2.7). This will certainly imply that \(z_q\) cannot move far inside the spectrum, and, instead, we must have

$$\begin{aligned} \frac{\eta _q}{\sqrt{|\kappa _q| + \eta _q}} \asymp \frac{\eta _t}{\sqrt{|\kappa _t| + \eta _t}}. \end{aligned}$$

This will imply

$$\begin{aligned} \sqrt{|\kappa _q| + \eta _q} \asymp \frac{\eta _q}{\sqrt{\eta _t}} \asymp \sqrt{\eta _t} + (t-q). \end{aligned}$$

Thus, when \(t-q \geqslant \sqrt{\eta _t}\), we have that \(\sqrt{|\kappa _q|}\asymp (t-q)\).

Now let us consider, instead, the case that \(- N^{{\mathfrak {d}}} \eta _t \leqslant \kappa _t \leqslant - \eta _t\). We will have the relation \(\eta _q \asymp \eta _t + (t-q) \sqrt{|\kappa _t|}\). For \(t-q \geqslant \eta _t/\sqrt{\kappa _t}\), this will be comparable to \((t-q)\sqrt{|\kappa _t|}\). To maintain \(\mathrm{Im}[m_q(z_q)]\asymp \mathrm{Im}[m_t(z_t)]\), when \(t-q \leqslant \sqrt{\kappa _t}\), would necessitate \(|\kappa _q| + \eta _q \asymp |\kappa _t| + \eta _t\). When \((t-q) \geqslant \sqrt{\kappa _t}\), we would instead have the relation

$$\begin{aligned} \frac{\eta _q}{\sqrt{\kappa _q + \eta _q}} \asymp \sqrt{|\kappa _t| + \eta _t}. \end{aligned}$$

This would then imply that \(\sqrt{\kappa _q + \eta _q}\asymp (t-q)\). \(\square \)

The improved rigidity bounds in Theorem 4.8 allow us to get sharp estimates on the derivatives of the Stieltjes transform as in the following lemma.

Lemma B.3

Recall the times s and t from Eq. (5.1) as well as the event \({\mathcal {S}}\) from Sect. 5. Under the assumptions of Theorem 5.1, consider a characteristic \(z_t\) such that \(\eta _t\asymp \eta \) and \(-N^{{\mathfrak {d}}}\eta \leqslant \kappa _t \lesssim \eta \). On \({\mathcal {S}}\), we have the following estimate uniformly for times q satisfying \(s\leqslant q\leqslant t\).

$$\begin{aligned} \left| \partial _z^p m_q(z_q) - \partial _z^p\hat{m}_q(z_q)\right| \leqslant \frac{N^{\tilde{{\mathfrak {d}}}}}{(|\kappa _q| + \eta _q)\eta _q^p}, \end{aligned}$$
(B.2)

where \(\tilde{{\mathfrak {d}}}\) is any parameter strictly greater than \({\mathfrak {d}}\).

Proof

Since both \(m_q\) and \(\hat{m}_q\) are analytic on the upper half-plane, by Cauchy’s integral formula,

$$\begin{aligned} \partial _z^p m_q(z_q)-\partial _z^p\hat{m}_q(z_q) =\frac{p!}{2\pi \mathrm{i}}\oint _{{\mathcal {C}}} \frac{m_q(w)-\hat{m}_q(w)}{(w-z_q)^{p+1}}\mathrm{d}w , \end{aligned}$$

where \({\mathcal {C}}\) is a small contour in the upper half-plane centered at \(z_q\) with radius \(\eta _q/2\). We take the absolute value of both sides to get

$$\begin{aligned} \left| \frac{p!}{2\pi \text {i}}\oint _{{\mathcal {C}}} \frac{m_q(w) -\hat{m}_q(w)}{(w-z_q)^{p+1}}\mathrm{d}w\right| \leqslant \frac{p!}{2\pi } \oint _{{\mathcal {C}}} \frac{|m_q(w)-\hat{m}_q(w)|}{|w-z_q|^{p+1}}\mathrm{d}w. \end{aligned}$$
(B.3)

If we have that \(\kappa _q \geqslant \eta _q \gg N^{-2/3}\), we will be in a region in which we can apply the improved error estimates (4.18) to argue that \(|m_q(w) - \hat{m}_q(w)| \leqslant \frac{M(\log N)^2}{N (\kappa _q + \eta _q)}\). Otherwise, we have, instead, that \(\eta _q \geqslant \kappa _q \geqslant - N^{{\mathfrak {d}}} \eta _q\) by applying Lemma B.1. At this point, we can apply the error estimate from (4.13) to conclude that \(|m_q(w) -\hat{m}_q(w)| \leqslant \frac{N^{{\mathfrak {d}}} (\log N) M}{N(|\kappa _q|+\eta _q)}\).

We can regard the M and \(\log N\) factors as unimportant relative to the \(N^{{\mathfrak {d}}}\) factor in the numerator for sufficiently large N. We can additionally bound \(|w-z_q|\) in the denominator from below by \(\eta _q\). Applying these considerations to the contour integral (B.3), we derive the bound

$$\begin{aligned} |\partial _z^p m_q(z_q) - \partial _z^p \hat{m}_q(z_q)| \leqslant \frac{N^{\tilde{{\mathfrak {d}}}}}{(|\kappa _q| + \eta _q) \eta _q^p}, \end{aligned}$$

where \(\tilde{{\mathfrak {d}}}\) is any parameter greater than \({\mathfrak {d}}\). \(\square \)

We will often have to perform the time integral of the error term coming from our rigidity estimates. This lemma explicitly performs the time integrals that frequently appear in our analysis.

Lemma B.4

Recall the times s and t from Eq. (5.1). Under the assumptions of Theorem 5.1, consider a characteristic \(z_\tau = E_\tau + \kappa _\tau + \mathrm{i}\eta _\tau \) that, at time t, satisfies the conditions \(\eta _t \asymp \eta \) and \(-N^{{\mathfrak {d}}}\eta \lesssim \kappa _t \lesssim \eta \). We have

$$\begin{aligned} \int _{s}^{t} \frac{\mathrm{d}q}{\eta _q^{p}}= {{\,\mathrm{O}\,}}\left( \frac{1}{\eta _t^{p-1/2}}\right) , \end{aligned}$$

where the implicit constant is universal.

Proof

We can perform explicit calculations thanks to (2.7)

$$\begin{aligned} \int _{s}^{t} \frac{1}{\eta _q^p} \mathrm{d}q ={{\,\mathrm{O}\,}}\left( \int _{s}^{t} \frac{1}{(\eta _t + (t-q)\mathrm{Im}[\hat{m}_t(z_t)])^p} \mathrm{d}q \right) = {{\,\mathrm{O}\,}}\left( \frac{1}{ \mathrm{Im}[\hat{m}_t(z_t)]\eta _t^{p-1}}\right) . \end{aligned}$$

The claim follows from our assumption of square root behavior. We chose the time difference \(t -s\) large enough so that the quantity \(\eta _t + (t-s) \mathrm{Im}[\hat{m}_t(z_t)]\) is much larger than \(\eta _t\). Finally, we used the fact that, at \(z_t\), we have \(\mathrm{Im}[\hat{m}_t(z_t)] \leqslant C \sqrt{\eta _t}\) for some large constant C. \(\square \)

The following lemma computes the main contribution of the quadratic variance terms from Proposition 5.3.

Lemma B.5

We recall \({\mathfrak {C}}_q\) from (3.5) and the times s and t from Eq. (5.1). Under the assumptions of Theorem 5.1 and for any time q satisfying \(s\leqslant q\leqslant t\), we have the following integral evaluation,

$$\begin{aligned} \int _{-\infty }^{E_q} \frac{\mathrm{d}\hat{\mu }_q(x)}{(x -z_q)^2(x-z_q')^2} =\frac{\pi }{2} \frac{(1+ {{\,\mathrm{O}\,}}(N^{- {\mathfrak {d}}^2})){\mathfrak {C}}_q}{\sqrt{\kappa _q +\mathrm{i}\eta _q}\sqrt{\kappa _q' + \mathrm{i}\eta _q'}(\sqrt{\kappa _q + \mathrm{i}\eta _q} +\sqrt{\kappa _q' + \mathrm{i}\eta _q'})^3}, \end{aligned}$$
(B.4)

where the characteristics \(z_t=E_t+\kappa _t+\mathrm{i}\eta _t,z_t'=E_t+\kappa _t'+\mathrm{i}\eta _t'\) satisfy \(\eta _t, \eta _t'\asymp \eta \), \(-N^{{\mathfrak {d}}}\eta \leqslant \kappa _t,\kappa _t'\leqslant \eta \).

Proof

We recall from Proposition 3.4 that the measure \(\hat{\mu }_q\) is supported on \((-\infty , E_q]\), and has a density in a small neighborhood of \(E_q\) of the form

$$\begin{aligned} \mathrm{d}\hat{\mu }_q(x)=(1+{{\,\mathrm{O}\,}}(|E_q-x|)){\mathfrak {C}}_q\sqrt{[E_q-x]_+}\mathrm{d}x. \end{aligned}$$

Now, we try to explicitly identify the main term found in (B.4). We have

$$\begin{aligned} \int _{-\infty }^{E_q} \frac{\text {d} \hat{\mu }_q(x)}{(x-z_q)^2(x-z_q')^2}&=\int _{E_q-(\log N)^{-1}}^{E_q} \frac{\hat{\rho }_q(x)\mathrm{d}x}{(x- z_q)^2 (x - z_q')^2} \nonumber \\&\quad + \int _{-\infty }^{E_q-(\log N)^{-1}} \frac{\mathrm{d}\hat{\mu }_q(x)}{(x- z_q)^2(x - z_q')^2}. \end{aligned}$$
(B.5)

The second term of the above equation can clearly be bounded by \({{\,\mathrm{O}\,}}((\log N)^4)\) if one takes into account the fact that \(z_q\) and \(z_q'\) are far away from \(E_q-(\log N)^{-1}\). This can easily be seen to be an \({{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2})\) factor of the main term in (B.4).

To estimate the other quantity, we do a Taylor expansion of \(\hat{\rho }_q(x)\) around the point 0 as \(|\hat{\rho }_q(x) -{\mathfrak {C}}_q\sqrt{x}| = {{\,\mathrm{O}\,}}(x^{3/2})\). Without loss of generality, we will assume that \(|\kappa _q| \geqslant |\kappa _q'|\). To illustrate the computation, we will only consider the case that \(-\kappa _q \geqslant \eta _q\), and \(-\kappa _q' \geqslant \eta _q'\). Similar techniques can be used in all other cases and are generally simpler. The above computation can be divided into two cases; the first case is when \(|\kappa _q - \kappa _q'| \geqslant |\kappa _q|/2\). We will perform a decomposition of the integral as follows:

$$\begin{aligned} \int _{-\infty }^{E_q} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x-z_q'|^2} \mathrm{d}x&= \int _{-\infty }^{E_q+2 \kappa _q} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x -z_q'|^2} \mathrm{d}x \\&\quad + \int _{E_q+2 \kappa _q}^{E_q+(\kappa _q + \kappa _q')/2} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x-z_q'|^2} \mathrm{d}x \\&\quad + \int _{ E_q+(\kappa _q + \kappa _q')/2}^{E_q} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x-z_q'|^2} \mathrm{d}x \\&\lesssim \int _{-\infty }^{E_q+2 \kappa _q} |E_q-x|^{-5/2} \mathrm{d}x +\frac{| \kappa _q|^{5/2}}{\eta _q^2 |\kappa _q - \kappa _q'|^2} +\frac{|\kappa _q' + \kappa _q|^{5/2}}{\eta _q'^2 |\kappa _q - \kappa _q'|^2}\\&\lesssim |\kappa _q|^{-3/2}+|\kappa _q|^{1/2}\eta _q^{-2}. \end{aligned}$$

We need to show that the above quantity will be less than \(N^{-{\mathfrak {d}}^2} |\kappa _q|^{-2} |\kappa _q'|^{-1/2}\). This is implied by the two inequalities \(\eta _q \leqslant N^{-20{\mathfrak {d}}}\) and \(|\kappa _q| \leqslant N^{{\mathfrak {d}}} \eta _q \) from Lemma B.1.

Next, we consider the case in which \(|\kappa _q - \kappa _q'| \leqslant |\kappa _q|/2\). We can divide the integral as

$$\begin{aligned} \int _{-\infty }^{E_q} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x-z_q'|^2} \mathrm{d}x&= \int _{-\infty }^{E_q+2 \kappa _q} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x-z_q'|^2} \mathrm{d}x +\int _{E_q+2 \kappa _q}^{E_q} \frac{|E_q-x|^{3/2}}{|x-z_q|^{2}|x-z_q'|^2} \mathrm{d}x\\&\lesssim |\kappa _q|^{-3/2}+|\kappa _q|^{5/2}\eta _q^{-2} \eta _q'^{-2}. \end{aligned}$$

To show that the above error is smaller than \(N^{-{\mathfrak {d}}^2} |\kappa _q|^{-2} |\kappa _q'|^{-1/2}\), we would only need to show that \(|\kappa _q|^{5} \leqslant N^{-{\mathfrak {d}}^2} \eta _q^2 \eta _q'^2\). Applying \(\eta _q \asymp \eta _q'\) and the fact that \(|\kappa _q| \leqslant N^{{\mathfrak {d}}} \eta _q \), we see that we would only need to prove that \(\eta _q \leqslant N^{- {\mathfrak {d}}^2 - 5 {\mathfrak {d}}}\). This is due to the fact that \(\eta _q \leqslant N^{-20{\mathfrak {d}}}\).

We now only need to compute the following integral

$$\begin{aligned}&\int _{E_q-(\log N)^{-1}}^{E_q} \frac{{\mathfrak {C}}_q \sqrt{E_q-x}}{(x- z_q)^2 (x - z_q')^2} \mathrm{d}x \nonumber \\&\quad = \int _{-\infty }^{0} \frac{{\mathfrak {C}}_q \sqrt{-x}}{(x - \kappa _q-\mathrm{i}\eta _q)^2 (x -\kappa _q'-\mathrm{i}\eta _q')^2} \mathrm{d}x + {{\,\mathrm{O}\,}}((\log N)^{5/2})\nonumber \\&\quad =\frac{\pi }{2} \frac{{\mathfrak {C}}_q}{\sqrt{\kappa _q+\mathrm{i}\eta _q} \sqrt{\kappa _q'+\mathrm{i}\eta _q'} (\sqrt{\kappa _q+\mathrm{i}\eta _q} +sqrt{\kappa _q'+\mathrm{i}\eta _q'})^3}+ {{\,\mathrm{O}\,}}((\log N)^{5/2}), \end{aligned}$$
(B.6)

where the square roots come from the branch with positive real part, and the error term \((\log N)^{5/2}\) can be absorbed into the \(N^{-{\mathfrak {d}}^2}\) factor of the main term. This finishes the proof of Lemma B.5. \(\square \)

We now compute the time integrals of the main terms in Proposition 5.3 weighted by the factors \({\mathcal {I}}_t ({\mathcal {I}}_s)^{-1}\). By using the equation of the characteristics, we can integrate this by a simple change of variables formula.

Lemma B.6

We recall \({\mathfrak {C}}_q\) from (3.5) and times s and t from Eq. (5.1). Under the assumptions of Theorem 5.1, take two characteristics \(z_t=E_t+\kappa _t+\mathrm{i}\eta _t\) and \(z_t'=E_t+\kappa _t'+\mathrm{i}\eta _t'\) such that \(\eta _t, \eta _t'\asymp \eta \), \(-N^{{\mathfrak {d}}}\eta \lesssim \kappa _t,\kappa _t'\lesssim \eta \). We have that

$$\begin{aligned}&\int _{s}^{t}\frac{\sqrt{\kappa _q + \mathrm{i}\eta _q}\sqrt{\kappa _q' + \mathrm{i}\eta _q'}}{\sqrt{\kappa _t + \mathrm{i}\eta _t}\sqrt{\kappa _t' + \mathrm{i}\eta _t'}}\int _{-\infty }^{\infty } \frac{\mathrm{d}\hat{\mu }_q(x)}{(x -z_q)^2(x-z_q')^2} \mathrm{d}q \nonumber \\&\quad = \frac{(1+ {{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2}))}{4 \sqrt{\kappa _t + \mathrm{i}\eta _t} \sqrt{\kappa _t' + \mathrm{i}\eta _t'} (\sqrt{\kappa _t + \mathrm{i}\eta _t} +\sqrt{\kappa _t' + \mathrm{i}\eta _t'})^2} \nonumber \\&\qquad - \frac{(1+ {{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2}))}{4 \sqrt{\kappa _t + \mathrm{i}\eta _t} \sqrt{\kappa _t' + \mathrm{i}\eta _t'} (\sqrt{\kappa _s + \mathrm{i}\eta _s} +\sqrt{\kappa _s' + \mathrm{i}\eta _s'})^2}. \end{aligned}$$
(B.7)

Proof

Using the results of Lemma B.5, we have

$$\begin{aligned}&\int _{s}^{t}\frac{\sqrt{\kappa _q + \mathrm{i}\eta _q}\sqrt{\kappa _q' + \mathrm{i}\eta _q'}}{\sqrt{\kappa _t + \mathrm{i}\eta _t}\sqrt{\kappa _t' + \mathrm{i}\eta _t'}}\int _{-\infty }^{E_q} \frac{\mathrm{d}\hat{\mu }_q(x)}{(x -z_q)^2(x-z_q')^2} \mathrm{d}q \\&\quad = (1+{{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2})) \int _{s}^{t} \pi /2 \frac{{\mathfrak {C}}_q}{(\sqrt{\kappa _q + \mathrm{i}\eta _q} +\sqrt{\kappa _q' + \mathrm{i}\eta _q'})^3} \mathrm{d}q. \end{aligned}$$

Let \(A= \sqrt{\kappa _s + \mathrm{i}\eta _s}\) , \(A' = \sqrt{\kappa _s' + \mathrm{i}\eta _s'}\) and \(B_q = \pi /2 \int _{q}^{t} {\mathfrak {C}}_\tau \mathrm{d}\tau \). By using our characteristic estimates from (5.27), we would like to replace \(\sqrt{\kappa _q + \mathrm{i}\eta _q}\) with the sum \(\sqrt{\kappa _t + \mathrm{i}\eta _t} + \pi /2 \int _q^t {\mathfrak {C}}_\tau \mathrm{d}\tau \) up to some small error. For times \((t-q) \ll (|\kappa _t| + \eta _t)^{1/4}\), this replacement is valid up to a multiplicative factor of \((1+ {{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2}))\).

$$\begin{aligned} \int _{s}^{t} \pi /2 \frac{{\mathfrak {C}}_q}{(\sqrt{\kappa _q + \mathrm{i}\eta _q} +\sqrt{\kappa _q' + \mathrm{i}\eta _q'})^3} \mathrm{d}q = (1+ {{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2})) \int _{s}^{t} \pi /2 \frac{{\mathfrak {C}}_q}{(A+A'- 2 B_q)^3} \mathrm{d}q. \end{aligned}$$

We will change the integration variable from \(q \rightarrow B_q\); the Jacobian of this transform will be \(\pi /2 {\mathfrak {C}}_q\). Upon this transformation, we see that we would have to integrate \(\int _{B_s=0}^{B_t} \frac{1}{(A+A' -2 x)^3} \mathrm{d}x\). This can be evaluated explicitly and will give us the last line of (B.7) up to a multiplicative factor of the form \(1+{{\,\mathrm{O}\,}}(N^{-{\mathfrak {d}}^2})\), due to the fact that we have to invert the substitution \(\sqrt{\kappa _t + \mathrm{i}\eta _t} + \pi /2 \int _q^t {\mathfrak {C}}_\tau \mathrm{d}\tau \rightarrow \sqrt{\kappa _q + \mathrm{i}\eta _q}\). We remark here that if we have that \(t-s\gg \sqrt{\eta }\), the last line of (B.7) can be subsumed as an \({{\,\mathrm{O}\,}}(N^{- {\mathfrak {d}}^2})\) error of the second term. \(\square \)

The following lemma collects the last computation appearing in Sect. 5. It follows from the arguments of Lemma B.5 and Lemma B.6. We will only mention the result here for reference.

Lemma B.7

We recall \({\mathfrak {C}}_q\) from (3.5) and times s and t from Eq. (5.1). Under the assumptions of Theorem 5.1, consider a characteristic \(z_t=E_t+\kappa _t+\mathrm{i}\eta _t\) such that \( \eta _t\asymp \eta \) and \(-N^{{\mathfrak {d}}}\eta \lesssim \kappa _t\lesssim \eta \). We have

$$\begin{aligned} \partial _z^2 \hat{m}_q(z_q) = \frac{\pi {\mathfrak {C}}_q}{4} \frac{(1+ {{\,\mathrm{O}\,}}(N^{- {\mathfrak {d}}^2}))}{(\sqrt{\kappa _q+ \mathrm{i}\eta _q})^3}, \end{aligned}$$
(B.8)

and

$$\begin{aligned} \int _{s}^t {\mathcal {I}}_t ({\mathcal {I}}_q)^{-1} \partial _z^2 \hat{m}_q(z_q) \mathrm{d}q= \frac{1}{2} \frac{(1+ {{\,\mathrm{O}\,}}(N^{- {\mathfrak {d}}^2}))}{\kappa _t + \mathrm{i}\eta _t} . \end{aligned}$$
(B.9)

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Adhikari, A., Huang, J. Dyson Brownian motion for general \(\beta \) and potential at the edge. Probab. Theory Relat. Fields 178, 893–950 (2020). https://doi.org/10.1007/s00440-020-00992-9

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  • 60F05
  • 60B20