Gaussian Fluctuations and Moderate Deviations of Eigenvalues in Unitary Invariant Ensembles

Abstract

We study the limiting behavior of the k-th eigenvalue $x_k$ of unitary invariant ensembles with Freud-type and uniform convex potentials. As both k and $n-k$ tend to infinity, we obtain Gaussian fluctuations for $x_k$ in the bulk and soft edge cases, respectively. Multi-dimensional central limit theorems, as well as moderate deviations, are also proved. This work generalizes earlier results in the GUE and unitary invariant ensembles with monomial potentials of even degree. In particular, we obtain the precise asymptotics of corresponding Christoffel–Darboux kernels as well.

Appendix

Proof of Lemma 2.3

First, using Theorem 2.2, we have for all $n\ge N$ and $x\in [-1+\delta ,1-\delta ]$,

$$\begin{aligned} \rho _{V_n}^{-1}(x)=2\pi \frac{1}{\sqrt{1-x^2}}\frac{1}{h_n(x)}<\frac{2\pi }{h_0} \frac{1}{\sqrt{1-(1-\delta )^2}} <\infty . \end{aligned}$$

As regards $\rho '_V$, we have

$$\begin{aligned} 2\pi |\rho '_{V_n}|\le \frac{|x|}{\sqrt{1-x^2}}|h_n(x)|+\sqrt{1-x^2}|h_n'(x)|. \end{aligned}$$

(7.1)

Note that by (2.2) and (2.4),

$$\begin{aligned} h_n(x)&=\sum _{k=0}^{m-1}2\frac{A_{m-k-1}}{A_m}x^{2k}+\mathcal {O}\left( n^{-\frac{1}{2m}}\right) , \end{aligned}$$

(7.2)

$$\begin{aligned} h'_n(x)&=\sum _{k=0}^{m-1}4k\frac{A_{m-k-1}}{A_m}x^{2k-1} +\mathcal {O}\left( n^{-\frac{1}{2m}}\right) , \end{aligned}$$

(7.3)

which implies that $|h_n(x)|$ and $|h'_n(x)|$ are uniformly bounded for all $n\ge N$ and $x\in [-1+\delta ,1-\delta ]$, thereby completing the proof. $\square$

Proof of (3.11)

First consider $x,y \in (1-\delta ,1)$. As in the case where $x,y\in (-1+\delta , 1-\delta )$, (3.7) still holds, i.e.,

$$\begin{aligned} 2\pi i(x-y)\mathscr {K}_n(x,y) =(e^{-n\varphi _n(x)},e^{n\varphi _n(x)})S(x)^TS(y)^{-T}(-e^{n\varphi _n(y)},e^{-n\varphi _n(y)})^T. \end{aligned}$$

(7.4)

Since for $x\in (1-\delta ,1)$, $f_n(x+i\epsilon )$ lies in the region II in (2.24), taking $\epsilon \rightarrow 0$ we obtain

$$\begin{aligned} \Psi ^{\sigma }(f_n(x))=[AI(f_n(x))]e^{-\frac{\pi i}{6}\sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ -1 &{} 1 \\ \end{array} \right) , \end{aligned}$$

which along with (2.26) and (2.23) yields that

$$\begin{aligned} S=RE_n[AI(f_n)]e^{-\frac{\pi i}{6}\sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ -1 &{} 1 \\ \end{array} \right) e^{n\varphi _n\sigma _3}. \end{aligned}$$

(7.5)

Consequently, plugging (7.5) into (7.4), we obtain (3.11) for $x,y\in (1-\delta ,1)$.

Regarding the case where $x,y\in (1,1+\delta )$, by (2.14), (2.15), (2.16) and (3.5),

$$\begin{aligned} U=e^{n\frac{l_n}{2}\sigma _3}Se^{n\left( g_n-\frac{l_n}{2}\right) \sigma _3}, \end{aligned}$$

and

$$\begin{aligned} 2\pi i(x-y)\mathscr {K}_n(x,y) =(e^{-n\varphi _n(x)},0)S^T(x)S^{-T}(y)(0,e^{-n\varphi _n(y)})^T. \end{aligned}$$

(7.6)

Since for $x\in (1,1+\delta )$, $f_n(x+i\epsilon )$ is in the region I in (2.24), by (2.26) and (2.23),

$$\begin{aligned} S=RE_n[AI(f_n)]e^{-\frac{\pi i}{6}\sigma _3}e^{n\varphi _n\sigma _3}. \end{aligned}$$

(7.7)

Hence, combining (7.6) and (7.7) we get (3.11) for $x,y\in (1,1+\delta )$, thereby completing the proof of (3.11). $\square$

Proof of (3.14)

We first show that $I_2(x,y) = {\mathcal {O}}(n^{-\frac{5}{6}})$. Indeed, by (3.13), expressions of AI, $E_n$ and that $\det [AI(z)]= \frac{-1}{2\pi i} e^{-\frac{\pi i}{3}}$ (see [6, p. 890]),

$$\begin{aligned} I_2(x,y)&=-2\pi i e^{\frac{\pi i}{3}}(H_n(x)Ai(f_n(x)),H_n^{-1}(x)Ai'(f_n(x)))\left( \begin{array}{cc} 1 &{} -1 \\ -i &{} -i \\ \end{array} \right) ^T\\&\cdot \Delta _{R}(x,y)R^{-T}(y)\left( \begin{array}{cc} 1 &{} -1 \\ -i &{} -i \\ \end{array} \right) ^{-T}\\&\quad \times (-H_n^{-1}(y)Ai'(f_n(y)),H_n(y)Ai(f_n(y)))^T. \end{aligned}$$

Note that since $H_n=f_n^{\frac{1}{4}}a^{-1}$, by (2.19),

$$\begin{aligned} H_n=n^{\frac{1}{6}}(x-1)^{\frac{1}{4}}(\widehat{\phi }_n)^{\frac{1}{6}} \left( \frac{x-1}{x+1}\right) ^{-\frac{1}{4}} =n^{\frac{1}{6}}(x+1)^{\frac{1}{4}}(\widehat{\phi }_n)^{\frac{1}{6}} =\mathcal {O}\left( n^{\frac{1}{6}}\right) . \end{aligned}$$

Moreover, for $x\in \mathbb {R}$, $|Ai(x)|=\mathcal {O}(1)$ and $|Ai'(f_n(x))|=\mathcal {O}(|f_n(x)|^{\frac{1}{4}})=\mathcal {O}(n^{\frac{1}{6}})$, and by (2.29), $\Delta {R}(x,y)R^{-T}(y)=\mathcal {O}(n^{-1})$. Thus, we conclude that $I_2(x,y)$ is of order $\mathcal {O}(n^{\frac{1}{6}})\mathcal {O}(n^{-1})=\mathcal {O}(n^{-\frac{5}{6}})$.

It remains to check the first term on the right-hand side of (3.12). To this end, it follows from (3.12) and the computations as above that

$$\begin{aligned}&e^{-\frac{\pi i}{3}}(1,0)[AI(f_n(x))]^TE_n^T(x)E_n^{-T}(y)[AI(f_n(y))]^{-T}(0,1)^T\\&\quad =(-2\pi i)\left[ -Ai(f_n(x))Ai'(f_n(y))\frac{f_n^{\frac{1}{4}}(x)}{f_n^{\frac{1}{4}}(y)}\frac{a(y)}{a(x)} +Ai'(f_n(x))Ai(f_n(y))\frac{f_n^{\frac{1}{4}}(y)}{f_n^{\frac{1}{4}}(x)}\frac{a(x)}{a(y)}\right] . \end{aligned}$$

which yields the first term on the right-hand side of (3.14). $\square$

Proof of (3.15)

The proofs are similar to those of (3.11). First consider $x,y\in (-1,-1+\delta )$. As in the case where $x,y\in (1-\delta ,1)$, we have

$$\begin{aligned} 2\pi i(x-y)\mathscr {K}_n(x,y) =(e^{-n\varphi _n(x)},e^{n\varphi _n(x)})S(x)^TS(y)^{-T}(-e^{n\varphi _n(y)},e^{-n\varphi _n(y)})^T. \end{aligned}$$

(7.8)

Since for $x\in (-1,-1+\delta )$, $-\widetilde{f}_n(x+i\epsilon )$ lies in the region III in (2.24), letting $\epsilon \rightarrow 0$ we have

$$\begin{aligned} \widetilde{\Psi }^{\sigma }(-\widetilde{f}_n(x))=[\widetilde{AI}(-\widetilde{f}_n(x))]e^{-\frac{\pi i}{6}\sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ 1 &{} 1 \\ \end{array} \right) , \end{aligned}$$

which along with (2.26) and (2.25) yields

$$\begin{aligned} S=R\widetilde{E}_n\sigma _3[\widetilde{AI}(-\widetilde{f}_n)]e^{-\frac{\pi i}{6}\sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ 1 &{} 1 \\ \end{array} \right) \sigma _3e^{n\widetilde{\varphi }_n\sigma _3}. \end{aligned}$$

(7.9)

Thus, plugging (7.9) into (7.8), since $\widetilde{\varphi }_n(z)=\varphi _n(z)+\pi i$, $z\in \mathbb {C}^+$, we get (3.15).

Regarding the case where $x,y\in (-1-\delta ,-1)$. As in the case where $x,y\in (1,1+\delta )$ in the proof of (3.11), we have

$$\begin{aligned} 2\pi i(x-y)\mathscr {K}_n(x,y)=(e^{-n\varphi _n(x)},0)S^T(x)S^{-T}(y)(0,e^{-n\varphi _n(y)})^T. \end{aligned}$$

(7.10)

Since for $x\in (-1-\delta ,-1)$, $-\widetilde{f}_n(x+i\epsilon )$ is in the region IV in (2.24), taking ${\varepsilon }\rightarrow 0$, we obtain from (2.25) and (2.26) that

$$\begin{aligned} S=R\widetilde{E}_n\sigma _3 [\widetilde{AI}(-\widetilde{f}_n)] e^{-\frac{\pi i}{6}\sigma _3}\sigma _3e^{n\widetilde{\varphi }_n\sigma _3}. \end{aligned}$$

(7.11)

Therefore, plugging (7.11) into (7.10) yields (3.15). $\square$

Proof of Lemma 6.1

Define the Hilbert transform $\mathscr {H}$ and the Borel transform $\mathscr {B}$ by

$$\begin{aligned} \mathscr {H} \rho _V(x)&=\frac{1}{\pi }P.V.\int \frac{\rho _V(y)}{x-y}\mathrm{d}y,\\ \mathscr {B}\rho _V(z)&=\frac{1}{\pi i}\int \frac{\rho _V(s)}{s-z}\mathrm{d}s, \ \ z\in \mathbb {C}/\mathbb {R}. \end{aligned}$$

In view of [7, (3.10), (3.12)], for $x\in \mathbb {R}$ we have

$$\begin{aligned} (\mathscr {B}\rho _V)_{\pm }(x)&=\pm \rho _V(x)+i\mathscr {H} \rho _V(x)=\pm \rho _V(x)-\frac{1}{2\pi i}V'(x). \end{aligned}$$

Moreover, by virtue of [7, (3.17), (3.18)],

$$\begin{aligned} \mathscr {B}\rho _V(z)=-\frac{1}{2\pi i}V'(z)-\frac{\sqrt{R(z)}}{4\pi ^2 }\oint \limits _{\Gamma _z}\frac{V'(s)}{\sqrt{R(s)}}\frac{\mathrm{d}s}{s-z}, \end{aligned}$$

where $\sqrt{R(z)}$ is as in Sect. 6, and $\Gamma _z$ is a counterclockwise contour with z and $[-1,1]$ in its interior. Note that due to the analytic branch of $\sqrt{R(z)}$, we have

$$\begin{aligned} (\sqrt{R(x)})_+=i\sqrt{(x+1)(1-x)}=-(\sqrt{R(x)})_-. \end{aligned}$$

(7.12)

Hence, it follows that for $x\in (-1,1)$,

$$\begin{aligned} \rho _V(x)=\frac{1}{2\pi }\sqrt{(1-x)(x+1)} \left( \frac{1}{2\pi i}\oint \limits _{\Gamma _z}\frac{V'(s)}{\sqrt{R(s)}}\frac{\mathrm{d}s}{s-z}\right) _+ , \end{aligned}$$

and so

$$\begin{aligned} h(z)=\frac{1}{2\pi i}\oint \limits _{\Gamma _z}\frac{V'(s)}{\sqrt{R(s)}}\frac{\mathrm{d}s}{s-z}, \end{aligned}$$

(7.13)

which is an analytic function.

It remains to prove that for some $c>0$, $h(x)\ge c$$\forall x\in \mathbb {R}$. We first claim that

$$\begin{aligned} \frac{1}{2\pi i}\oint \limits _{\Gamma _z}\frac{1}{\sqrt{R(s)}}\frac{\mathrm{d}s}{s-z}\equiv 0 \end{aligned}$$

(7.14)

To this end, we have for r large enough that

$$\begin{aligned} \frac{1}{2\pi i}\oint \limits _{\Gamma _z}\frac{1}{\sqrt{R(s)}}\frac{\mathrm{d}s}{s-z}&=\frac{1}{2\pi i}\oint \limits _{|s|=r}\frac{1}{\sqrt{(s+1)(s-1)}}\frac{\mathrm{d}s}{s-z}. \end{aligned}$$

(7.15)

Using Taylor’s extension, we see that

$$\begin{aligned} \frac{1}{\sqrt{s^2-1}}\frac{1}{s-z} =\left( A_0\frac{1}{s}+A_1\frac{1}{s^3}+A_2\frac{1}{s^5}+\cdots \right) \left( \frac{1}{s}+\frac{z}{s^2}+\frac{z^2}{s^3}+\cdots \right) , \end{aligned}$$

which by Cauchy’s theorem implies that the right-hand side of (7.15) equals to the coefficient of $\frac{1}{s}$ which is exactly zero, thereby yielding (7.14), as claimed.

Now, it follows from (7.13) and (7.14) that

$$\begin{aligned} h(z) =\frac{1}{2\pi i}\oint \limits _{\Gamma _z}\frac{V'(s)-V'(z)}{s-z}\frac{1}{\sqrt{R(s)}}\mathrm{d}s =\frac{1}{2\pi i}\oint \limits _{\Gamma _1}\frac{V'(s)-V'(z)}{s-z}\frac{1}{\sqrt{R(s)}}\mathrm{d}s, \end{aligned}$$

(7.16)

where $\Gamma _1$ is a counterclockwise contour with $[-1,1]$, but not z, in the interior.

Therefore, in view of (7.16) we have that for $x\in (-1,1)$,

$$\begin{aligned} h(x)&=\lim \limits _{z\in \mathbb {C}^+\rightarrow x}h(z) =\frac{1}{\pi i}\int _1^{-1}\frac{V'(s)-V'(x)}{s-x}\frac{1}{(\sqrt{R(s)})_+}\mathrm{d}s \\&= \frac{1}{\pi } \int _{-1}^1\frac{V'(s)-V'(x)}{s-x}\frac{1}{\sqrt{(s+1)(1-s)}}\mathrm{d}s, \end{aligned}$$

which implies by the mean value theorem and the uniform convexity of V that

$$\begin{aligned} h(x) =\frac{1}{\pi }\int _{-1}^1\frac{V''(\xi )}{\sqrt{(s+1)(1-s)}}\mathrm{d}s \ge \frac{c}{\pi }\int _{-1}^1\frac{\mathrm{d}s}{\sqrt{(s+1)(1-s)}}&=c>0. \end{aligned}$$

where $\xi \in (-1,1)$. The proof is complete. $\square$

Before proving Lemma 6.4, we recall that

Theorem 7.1

([7, (1.62)–(1.64)]) Consider the uniform convex potential as in (1.4). For the leading coefficients of orthogonal polynomials, we have

$$\begin{aligned} (\gamma _{n-1}^{(n)})^2&=e^{-nl}\left[ \frac{1}{4\pi } + \mathcal {O}(n^{-1})\right] , \ (\gamma _{n}^{(n)})^{-2}=e^{nl} \left[ \pi + \mathcal {O}(n^{-1})\right] , \\ \frac{\gamma _{n-1}^{(n)}}{\gamma _{n}^{(n)}}&=\frac{1}{2}+\mathcal {O}(n^{-1/2}). \end{aligned}$$

Theorem 7.2

([7, Theorems 1.1–1.3]) Consider the uniform convex potential as in (1.4). For the monic polynomials, we have

(i). For $x\in \mathbb {R}/(-1-\delta ,1+\delta )$,

$$\begin{aligned} \pi _n(x;n)&=e^{ng(x)}\left( M_1(x)+\mathcal {O}(n^{-1})\right) , \end{aligned}$$

(7.17)

$$\begin{aligned} -2\pi i(\gamma _{n-1}^{(n)})^2\pi _{n-1}(x;n)&=e^{n(g(x)-l)}\left( M_2(x)+\mathcal {O}(n^{-1})\right) , \end{aligned}$$

(7.18)

where $M_1=\frac{a+a ^{-1}}{2}$, $M_2=\frac{a^{-1}-a}{2i}$ with a as in (2.28) and l is as in (1.7).

(ii). For $x\in (-1+\delta ,1-\delta )$,

$$\begin{aligned} \pi _n(x;n)&=2e^{\frac{n}{2}(V(x)+l)}\left[ \hbox {Re}(M_1e^{i\pi n F(x)})+\mathcal {O}(n^{-1})\right] , \end{aligned}$$

(7.19)

$$\begin{aligned} -2\pi i(\gamma _{n-1}^{(n)})^2\pi _{n-1}(x;n)&=2e^{\frac{n}{2}(V(x)-l)} \left[ \hbox {Im}(M_2e^{i\pi n F(x)})+\mathcal {O}(n^{-1})\right] , \end{aligned}$$

(7.20)

where F is as in (6.2).

(iii). For $x\in (1-\delta ,1)\cup (-1,-1+\delta )$,

$$\begin{aligned} \pi _n(x;n)&=\left( e^{\frac{nl}{2}\sigma _3} (I+\mathcal {O}(n^{-1}))M_pe^{n\left( g(x)-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ e^{nV} &{} 1 \\ \end{array} \right) \right) _{11}, \end{aligned}$$

(7.21)

$$\begin{aligned} -2\pi i(\gamma _{n-1}^{(n)})^2\pi _{n-1}(x;n)&=\left( e^{\frac{nl}{2}\sigma _3}(1+\mathcal {O}(n^{-1}))M_pe^{n\left( g(x) -\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ e^{nV} &{} 1 \\ \end{array} \right) \right) _{21}, \end{aligned}$$

(7.22)

and for $x\in (1,1+\delta )\cup (-1-\delta ,-1)$,

$$\begin{aligned} \pi _n(x;n)&=\left( (I+\mathcal {O}(n^{-1}))M_p\right) _{11}e^{ng(x) }, \end{aligned}$$

(7.23)

$$\begin{aligned} -2\pi i(\gamma _{n-1}^{(n)})^2\pi _{n-1}(x;n)&=\left( (I+\mathcal {O}(n^{-1}))M_p\right) _{21}e^{n(g(x)-l)}, \end{aligned}$$

(7.24)

Proof of Lemma 6.4

We first note that by the analytic branch of $R^{1/2}(z)$,

$$\begin{aligned} (z-1)^{\frac{1}{2}}=i(1-z)^{\frac{1}{2}},~(z+1)^{\frac{1}{2}}=i(-1-z)^{\frac{1}{2}}, \ z\in \mathbb {C}_+. \end{aligned}$$

Then, using (6.5) we have

$$\begin{aligned} G(x)&=-\int _1^x\sqrt{(s-1)(s+1)}h(s)\mathrm{d}s=-2 \pi F(x), \ \ x>1, \end{aligned}$$

(7.25)

$$\begin{aligned} G(x)&=2\pi i F(x), \ \ x\in (-1,1), \end{aligned}$$

(7.26)

$$\begin{aligned} G(x)&=- 2\pi \widetilde{F}(x)+2\pi i, \ \ x<-1, \end{aligned}$$

(7.27)

where F(x) and $\widetilde{F}(x)$ are as in (6.2).

(i). For $x\in \mathbb {R}/(-1-\delta ,1+\delta )$, by (7.17) and (6.6),

$$\begin{aligned} \pi _n(x;n)=e^{\frac{n}{2}(V+l+G)}\left\{ \frac{1}{2} \left[ \left( \frac{x+1}{x-1}\right) ^{\frac{1}{4}} +\left( \frac{x-1}{x+1}\right) ^{\frac{1}{4}}\right] +\mathcal {O}(n^{-1})\right\} . \end{aligned}$$

Then, by Theorem 7.1,

$$\begin{aligned} p_n(x;n)=\gamma _n^{(n)}\pi _n(x;n)=\frac{1}{\sqrt{\pi }}e^{\frac{n}{2}(V+G)} \left\{ \frac{1}{2}\left[ \left( \frac{x+1}{x-1}\right) ^{\frac{1}{4}} +\left( \frac{x-1}{x+1}\right) ^{\frac{1}{4}}\right] +\mathcal {O}(n^{-1})\right\} . \end{aligned}$$

Thus, using (7.25) and (7.27) for $x > 1+\delta$ and $x< -1-\delta$, respectively, we obtain (6.26) and (6.28).

Similarly, by Theorem 7.1, (7.18) and (6.6),

$$\begin{aligned} p_{n-1}(x;n)e^{-\frac{n}{2}V}&=\frac{1}{\sqrt{4 \pi }} e^{\frac{ n G}{2}} \left\{ \left[ \left( \frac{x+1}{x-1}\right) ^{\frac{1}{4}}-\left( \frac{x-1}{x+1}\right) ^{\frac{1}{4}}\right] +\mathcal {O}(n^{-1})\right\} , \end{aligned}$$

which yields (6.27) and (6.29) by (7.25) and (7.27), respectively.

(ii). By the definitions of $M_1$ and $M_2$, we have that (cf. [8, (8.33), (8.34)])

$$\begin{aligned} M_1&=\frac{a+a^{-1}}{2}=\frac{\sqrt{2}}{2}\frac{1}{(1-x)^{\frac{1}{4}}(1+x)^{\frac{1}{4}}}e^{-\frac{i}{2}\arcsin x}\\ M_2&=\frac{a^{-1}-a}{2i}=-\frac{\sqrt{2}}{2}\frac{1}{(1-x)^{\frac{1}{4}}(1+x)^{\frac{1}{4}}}e^{\frac{i}{2}\arcsin x}. \end{aligned}$$

Then, in view of Theorems 7.1 and 7.2 (ii), we obtain (6.30) and (6.31).

(iii). We consider four cases $(iii.1)--(iii.4)$.

(iii.1). For $x\in (1-\delta ,1)$, by (6.18) and (6.6),

$$\begin{aligned} e^{\frac{nl}{2}\sigma _3}M_pe^{n\left( g-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ e^{nV} &{} 1 \\ \end{array} \right) =e^{\frac{nl}{2}\sigma _3}\widehat{E}_n \left[ AI(\Phi _1)\right] e^{-\frac{i\pi }{6}\sigma _3}e^{\frac{n}{2}V\sigma _3}. \end{aligned}$$

Then, since $\widehat{E}_n=\sqrt{\pi } e^{\frac{\pi i}{6}}\left( \begin{array}{cc} a^{-1}\Phi _1^{\frac{1}{4}} &{} -a\Phi _1^{-\frac{1}{4}} \\ -ia^{-1}\Phi _1^{\frac{1}{4}} &{} -ia\Phi _1^{-\frac{1}{4}} \\ \end{array} \right)$, direct calculations show that

$$\begin{aligned}&\left( e^{\frac{nl}{2}\sigma _3}M_pe^{n\left( g-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} o \\ e^{nV} &{} 1 \\ \end{array} \right) \right) _{11}\nonumber \\&\quad =\sqrt{\pi }e^{\frac{n}{2}(V+l)} \left[ a^{-1}\Phi _1^{\frac{1}{4}}Ai(\Phi _1)-a\Phi _1^{-\frac{1}{4}}Ai'(\Phi _1)\right] . \end{aligned}$$

(7.28)

Similarly,

$$\begin{aligned}&\left( e^{\frac{nl}{2}\sigma _3}M_pe^{n\left( g-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} o \\ e^{nV} &{} 1 \\ \end{array} \right) \right) _{21}\nonumber \\&\quad =(-i) \sqrt{\pi }e^{\frac{n}{2}(V-l)} \left[ a^{-1}\Phi _1^{\frac{1}{4}}Ai(\Phi _1)+a\Phi _1^{-\frac{1}{4}}Ai'(\Phi _1)\right] . \end{aligned}$$

(7.29)

Plugging these into (7.21) and (7.22) and using Theorem 7.1, we obtain (6.32) and (6.33).

(iii.2). For $x\in (1,1+\delta )$, using (6.20) we note that in (7.23) and (7.24), $(M_p)_{11}e^{ng}$ and $(M_p)_{21}e^{n(g-l)}$ have the same formulations as (7.28) and (7.29). Thus, arguing as above we obtain (6.32) and (6.33).

(iii.3). For $x\in (-1,-1+\delta )$, it follows from (6.22) that

$$\begin{aligned} e^{\frac{nl}{2}\sigma _3} M_p e^{n\left( g-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ e^{nV} &{} 1 \\ \end{array} \right) =(-1)^n e^{\frac{nl}{2}\sigma _3} \widehat{\widetilde{E}}_n \sigma _3 [\widetilde{AI}(\Phi _{-1})] e^{-\frac{\pi i}{6}\sigma _3} e^{\frac{n}{2}V\sigma _3} \sigma _3. \end{aligned}$$

Then, since $\widehat{\widetilde{E}}_n =\sqrt{\pi } e^{\frac{\pi i}{6}}\left( \begin{array}{cc} a\Phi _{-1}^{\frac{1}{4}} &{} a^{-1}\Phi _{-1}^{-\frac{1}{4}} \\ ia\Phi _{-1}^{\frac{1}{4}} &{} -ia^{-1}\Phi _{-1}^{-\frac{1}{4}} \\ \end{array} \right)$, we have

$$\begin{aligned}&\left( e^{\frac{nl}{2}\sigma _3} M_p e^{n\left( g-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ e^{nV} &{} 1 \\ \end{array} \right) \right) _{11} \nonumber \\&\quad =(-1)^n \sqrt{\pi } e^{\frac{n(V+l)}{2}} \left[ a \Phi _{-1}^{\frac{1}{4}} Ai(\Phi _{-1}) - a^{-1} \Phi _{-1}^{-\frac{1}{4}} Ai'(\Phi _{-1})\right] . \end{aligned}$$

(7.30)

Similarly,

$$\begin{aligned}&\left( e^{\frac{nl}{2}\sigma _3} M_p e^{n\left( g-\frac{l}{2}\right) \sigma _3}\left( \begin{array}{cc} 1 &{} 0 \\ e^{nV} &{} 1 \\ \end{array} \right) \right) _{21}\nonumber \\&\quad =(-1)^{n+1} (-i) \sqrt{\pi } e^{\frac{n(V-l)}{2}} \left[ a \Phi _{-1}^{\frac{1}{4}} Ai(\Phi _{-1}) + a^{-1} \Phi _{-1}^{-\frac{1}{4}} Ai'(\Phi _{-1})\right] . \end{aligned}$$

(7.31)

Thus, (6.34) and (6.35) follow from (7.21), (7.22) and Theorem 7.1.

(iii.4) For $x\in (-1-\delta ,-1)$, by (6.24) we note that $(M_p)_{11}e^{ng}$ and $(M_p)_{11}e^{n(g-l)}$ in (7.23) and (7.24) have the same formulations as (7.30) and (7.31), which consequently implies (6.34) and (6.35). The proof of Lemma 6.4 is complete. $\square$