$\displaystyle\frac{1}{N^{5/2}}\sum_{ia}(1+\delta_{ia})^{4}\mathbb{E}[\partial_{ia}^{4}(e_{\mathfrak{a}}(G_{ia}-\mathbb{E}[\partial_{ia}^{4}G_{ia}])]$
$\displaystyle=$
$\displaystyle\frac{1}{N^{5/2}}\sum_{ia}(1+\delta_{ia})^{4}\mathbb{E}[e_{\mathfrak{a}}(\partial_{ia}^{4}G_{ia}-\mathbb{E}[\partial_{ia}^{4}G_{ia}])]$
$\displaystyle+$
$\displaystyle\sum_{n=1}^{4}\frac{1}{N^{5/2}}\sum_{ia}(1+\delta_{ia})^{4}c_{n}\mathbb{E}[(\partial_{ia}^{n}e_{\mathfrak{a}})\partial_{ia}^{4-n}G_{ia}].$
(10.48)

for some combinatorial factor $c_{n}$. From the entry-wise local law (2.12) we
easily see that,

$\frac{1}{N^{5/2}}\sum_{ia}(\partial_{ia}^{4}G_{ia}-\mathbb{E}[\partial_{ia}^{4}G_{ia}])=N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1/2})$
(10.49)

with overwhelming probability. Consider now each of the terms in the last line
of (10.48). Note that for $m=2k$ even we have,

$|\partial_{ia}^{2k}G_{ia}(z)|\leq C(\delta_{ia}+N^{\varepsilon}(Ny)^{-1/2})$
(10.50)

with overwhelming probability. Together with the estimate (10.5) we see that
the terms with $n=2,4$ of the last line of (10.48) are
$N^{\varepsilon}\mathcal{O}(N^{-1}y^{-1/2}(1+|\xi|)^{4}(1+\|f^{\prime}\|_{1})^{4})$.
For the terms with $n=1,3$ note that the estimate (10.46) holds also with LHS
$|\partial_{ia}e_{\mathfrak{a}}|$. Therefore, these terms are
$N^{\varepsilon}(1+|\xi|^{3})(1+\|f^{\prime}\|_{1}^{3})\mathcal{O}(N^{-1/2}(1+\|f^{\prime\prime}\|_{1})^{1/2})$.
This completes the proof. ∎

####  10.1.5 Proof of Proposition 10.1

This follows immediately from the lemmas of the previous subsections.

###  10.2 Proof of Theorem 3.2

The estimates for the expectation and variance are Lemma 6.11 and Theorem 6.9,
respectively. We turn to the esimate of the characteristic function. We have

$\frac{\mathrm{d}}{\mathrm{d}\xi}\psi_{\mathfrak{a}}(\xi)=\frac{\mathrm{i}}{\pi}\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{f}(z)N\mathbb{E}[e_{\mathfrak{a}}(m_{N}(z)-\mathbb{E}[m_{N}(z)])]\mathrm{d}x\mathrm{d}y.$
(10.51)

From Proposition 10.1 we then see that,

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\xi}\psi_{\mathfrak{a}}(\xi)$
$\displaystyle=(-\xi
V_{1}(f)+\frac{\mathrm{i}3\xi^{2}}{N^{1/2}}B_{1}(f))\psi_{\mathfrak{a}}(\xi)$
$\displaystyle+N^{\varepsilon}(1+|\xi|^{6})(1+\|f^{\prime}\|_{1,w}^{7})\mathcal{O}(N^{-1}(1+\|f^{\prime\prime}\|_{1,w}))$
(10.52)

where

$B_{1}(f):=\frac{2}{3\pi^{3}}\left(\int_{\Omega_{\mathfrak{a}}}\partial_{\bar{z}}\tilde{f}(z)m_{\mathrm{sc}}^{\prime}(z)\mathrm{d}x\mathrm{d}y\right)^{3}$
(10.53)

and

$V_{1}(f):=\int_{\Omega_{\mathfrak{a}}^{2}}\partial_{\bar{z}}\tilde{f}(z)\partial_{\bar{w}}\tilde{f}(w)F(z,w)\mathrm{d}x\mathrm{d}y\mathrm{d}u\mathrm{d}v$
(10.54)

where $F(z,w)$ is defined by the function on the second and third lines of
(6.2). Now, with $V(f)$ defined by (6.9) and $B(f)$ defined by,

$B(f):=\frac{2}{3\pi^{3}}\left(\int_{\mathbb{R}^{2}}\partial_{\bar{z}}\tilde{f}(z)m_{\mathrm{sc}}^{\prime}(z)\mathrm{d}x\mathrm{d}y\right)^{3}=-\frac{1}{12\pi^{3}}\left(\int_{-2}^{2}f(x)\frac{x}{\sqrt{4-x^{2}}}\mathrm{d}x\right)^{3},$
(10.55)

it is not hard to derive (see the proof of Theorem 6.9 for the estimate of
$V_{1}(f)-V(f)$),

$|V_{1}(f)-V(f)|+|B_{1}(f)-B(f)|\leq
N^{\varepsilon}(1+\|f^{\prime}\|_{1})^{2}N^{\mathfrak{a}-1}(1+\|f^{\prime\prime}\|_{1,w}).$
(10.56)

Therefore,

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\xi}\psi_{\mathfrak{a}}(\xi)$
$\displaystyle=(-\xi
V(f)+\frac{3\mathrm{i}\xi^{2}s_{3}}{N^{1/2}}B(f))\psi_{\mathfrak{a}}(\xi)$
$\displaystyle+N^{\mathfrak{a}+\varepsilon}(1+|\xi|^{6})(1+\|f^{\prime}\|_{1,w}^{7})\mathcal{O}(N^{-1}(1+\|f^{\prime\prime}\|_{1,w}))$
(10.57)

Therefore, using the fact that $B(f)$ is real we have,

$\displaystyle\left|\psi_{\mathfrak{a}}(\xi)-\exp\left[-\xi^{2}V(f)/2+\mathrm{i}\xi^{3}N^{-1/2}B(f)\right]\right|$
$\displaystyle\leq$ $\displaystyle
CN^{\mathfrak{a}+\varepsilon}(1+|\xi|^{6})(1+\|f^{\prime}\|_{1,w}^{7})N^{-1}(1+\|f^{\prime\prime}\|_{1,w})\int_{0}^{\xi}\mathrm{e}^{(s^{2}-\xi^{2})V(f)}\mathrm{d}s.$
(10.58)

By Lemma E.1 we have that $V(f)\geq 0$ for $N\geq 4$. Therefore, the integral
on the last line of (10.2) is bounded by $|\xi|$. Therefore, by Lemma 6.8 we
have the following estimate,

$\displaystyle\left|\mathbb{E}[\exp(\mathrm{i}\xi(\mathrm{tr}f(H)-\mathbb{E}[\mathrm{tr}f(H)]))]-\exp\left[-\xi^{2}V(f)/2+\mathrm{i}\xi^{3}N^{-1/2}B(f)\right]\right|$
$\displaystyle\leq$ $\displaystyle
N^{\varepsilon}(1+|\xi|^{6}+\|f^{\prime}\|_{1,w}^{7})N^{-1}(1+\|f^{\prime\prime}\|_{1,w}),$
(10.59)

and the claim follows. ∎

##  Appendix A Derivatives of resolvent entries with respect to matrix
elements

In this section, we collect some identities for derivatives of resolvent
entries with respect to matrix entries. We have the general formula

$(1+\delta_{ia})\partial_{ia}G_{jk}=-G_{ij}G_{ka}-G_{aj}G_{ki}.$ (A.1)

For first derivatives, this specializes to

$(1+\delta_{ia})\partial_{ia}G_{ia}=-G_{ii}G_{aa}-G_{ia}^{2}$ (A.2)

and

$(1+\delta_{ia})\partial_{ia}G_{jj}=-2G_{ij}G_{ja}.$ (A.3)

For second derivatives, we get

$(1+\delta_{ia})^{2}\partial^{2}_{ia}(G_{ia})=2G_{ia}^{3}+6G_{ii}G_{aa}G_{ia}.$
(A.4)

and

$(1+\delta_{ia})^{2}(\partial_{ia}G_{ia}(z))(\partial_{ia}G_{jj}(w))=2G_{ii}(z)G_{aa}(z)G_{ji}(w)G_{ja}(w)+2G_{ia}(z)^{2}G_{ji}(w)G_{ja}(w)$
(A.5)

and

$(1+\delta_{ia})^{2}\partial^{2}_{ia}G_{jj}=4G_{ij}G_{ja}G_{ai}+2G_{ij}G_{ji}G_{aa}+2G_{aj}G_{ja}G_{ii}.$
(A.6)

For the third derivatives, we have

$\displaystyle-(1+\delta_{ia})^{3}\partial_{ia}^{3}G_{ia}$
$\displaystyle=6G_{ia}^{4}+36G_{ia}^{2}G_{ii}G_{aa}+6G_{ii}^{2}G_{aa}^{2},$
(A.7)

as well as

$\displaystyle-(1+\delta_{ia})^{3}G_{jj}=12G_{ai}^{2}G_{ij}G_{ja}+12G_{ja}^{2}G_{ii}G_{ai}+12G_{ij}^{2}G_{aa}G_{ai}+12G_{aj}G_{ij}G_{aa}G_{ii}$
(A.8)

##  Appendix B Semicircle calculations

######  Lemma B.1.

We have,

$\displaystyle\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}=\frac{m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}{1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}.$
(B.1)

Proof. Recall,

$m_{\mathrm{sc}}(z)=\frac{-z+\sqrt{z^{2}-4}}{2},\qquad
m_{\mathrm{sc}}(z)+z=-\frac{1}{m_{\mathrm{sc}}(z)}.$ (B.2)

Consider,

$\displaystyle m_{\mathrm{sc}}(z)-\frac{1}{m_{\mathrm{sc}}(w)}$
$\displaystyle=m_{\mathrm{sc}}(z)+w+m_{\mathrm{sc}}(w)=\frac{1}{2}\left(-z+\sqrt{z^{2}-4}+w+\sqrt{w^{2}-4}\right)$
(B.3)
$\displaystyle=\frac{1}{2}\left(w-z+\frac{w^{2}-z^{2}}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}\right)=\frac{1}{2}\left(w-z+\frac{(w-z)(w+z)}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}\right)$
(B.4)
$\displaystyle=\frac{w-z}{2}\left(\frac{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}+w+z}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}\right)$
(B.5)
$\displaystyle=(w-z)\left(\frac{m_{\mathrm{sc}}(w)+w-m_{\mathrm{sc}}(z)}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}\right)$
(B.6)
$\displaystyle=(w-z)\left(\frac{-m_{\mathrm{sc}}(z)-(m_{\mathrm{sc}}(w))^{-1}}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}\right)$
(B.7)

where in the second line we used difference of squares factoring
$(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$. Hence,

$\displaystyle 1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)$
$\displaystyle=-m_{\mathrm{sc}}(w)\left(m_{\mathrm{sc}}(z)-\frac{1}{m_{\mathrm{sc}}(w)}\right)$
$\displaystyle=$
$\displaystyle(-m_{\mathrm{sc}}(w))(w-z)\left(\frac{-m_{\mathrm{sc}}(z)-(m_{\mathrm{sc}}(w))^{-1}}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}\right)$
$\displaystyle=\frac{w-z}{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}(1+m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)).$
(B.8)

Therefore:

$\frac{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}{w-z}=\frac{1+m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}{1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}.$
(B.9)

Continuing, we have:

$\displaystyle\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}$
$\displaystyle=\frac{-z+w+\sqrt{z^{2}-4}-\sqrt{w^{2}-4}}{2(z-w)}$ (B.10)
$\displaystyle=-\frac{1}{2}+\frac{1}{2}\frac{\sqrt{w^{2}-4}-\sqrt{z^{2}-4}}{w-z}$
(B.11)
$\displaystyle=-\frac{1}{2}+\frac{1}{2}\left(\frac{1+m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}{1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}\right)$
(B.12)
$\displaystyle=\frac{m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}{1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)}$
(B.13)

This yields the claim. ∎

######  Lemma B.2.

We have,

$\displaystyle\frac{1}{z+2m_{\mathrm{sc}}(z)}\partial_{w}\frac{m_{\mathrm{sc}}(z)-m_{\mathrm{sc}}(w)}{z-w}=-\frac{m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}^{\prime}(w)}{(1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w))^{2}}$
(B.14)

Proof. This follows from the previous lemma and the fact that,

$\frac{m_{\mathrm{sc}}(z)}{z+2m_{\mathrm{sc}}(z)}=-m_{\mathrm{sc}}^{\prime}(z),$
(B.15)

which follows from differentiating the quadratic equation satisfied by
$m_{\mathrm{sc}}$. ∎

######  Lemma B.3.

We have, for $z=x+\pm\mathrm{i}y$ and $w=u\pm\mathrm{i}v$, where $v,y>0$

$\displaystyle\left|\frac{m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}^{\prime}(w)}{(1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w))^{2}}\right|\leq
C\frac{|m_{\mathrm{sc}}^{\prime}(z)m_{\mathrm{sc}}^{\prime}(w)|}{(x-u)^{2}+y^{2}+v^{2}}$
(B.16)

Proof. From the fact that $|m_{\mathrm{sc}}(z)|\leq 1-cy$ we see that

$|1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)|\geq c(y+v).$ (B.17)

From (B.1) we see that

$|1-m_{\mathrm{sc}}(z)m_{\mathrm{sc}}(w)|\geq c|z-w|\geq c|x-u|.$ (B.18)

This completes the proof. ∎

##  Appendix C Hermite polynomial asymptotics

We collect here some standard results on the Hermite polynomials $h_{n}(x)$,
all available from [20]. In the bulk we have that there exists a
$\delta_{a}>0$, we have

$h_{n}\left(\sqrt{\frac{n}{2}}x\right)\mathrm{e}^{-n\frac{x^{2}}{4}}=2\sqrt{\frac{1}{\pi\sqrt{2n}}}\frac{1}{(4-x^{2})^{1/4}}\left(\cos\left[\frac{n}{2}F(x)-\frac{1}{2}\arcsin\left(\frac{x}{2}\right)\right]+\mathcal{O}(n^{-1})\right)$
(C.1)

uniformly in $-2+\delta_{a}\leq x\leq 2-\delta_{a}$ where

$F(x)=\left|\int_{x}^{2}\sqrt{|4-y^{2}|}\mathrm{d}y\right|.$ (C.2)

In particular,

$n^{1/4}\left|h_{n}\left(\sqrt{\frac{n}{2}}x\right)\mathrm{e}^{-n\frac{x^{2}}{4}}\right|\leq
C$ (C.3)

for $|x|\leq 2-\delta_{a}$. Inside the spectrum, but near the edge we have

$\displaystyle
h_{n}\left(\sqrt{\frac{n}{2}}x\right)\mathrm{e}^{-n\frac{x^{2}}{4}}$
$\displaystyle=(2n)^{-1/4}$
$\displaystyle\times\bigg{\\{}\left(\frac{x+2}{x-2}\right)^{1/4}\left[\frac{3}{4}nF(x)\right]^{1/6}\mathrm{Ai}\left(-\left[\frac{3}{4}nF(x)\right]^{2/3}\right)(1+\mathcal{O}(n^{-1}))$
$\displaystyle-\left(\frac{x-2}{x+2}\right)^{1/4}\left[\frac{3}{4}nF(x)\right]^{-1/6}\mathrm{Ai}^{\prime}\left(-\left[\frac{3}{4}nF(x)\right]^{2/3}\right)(1+\mathcal{O}(n^{-1}))\bigg{\\}},$
(C.4)

for $2-\delta_{a}<x<2$. Above, $\mathrm{Ai}$ is the Airy function. It is
bounded on the real line and smooth. So for $r>1$, the following asymptotics
are useful, and can be found in Chapter 11, Section 1 of [49],

$\displaystyle\mathrm{Ai}(-r)$
$\displaystyle=\pi^{-1/2}r^{-1/4}\left\\{\cos\left[\frac{2}{3}r^{3/2}-\frac{\pi}{4}\right]+\mathcal{O}(r^{-3/2})\right\\}$
$\displaystyle\mathrm{Ai}^{\prime}(-r)$
$\displaystyle=\pi^{-1/2}r^{1/4}\left\\{\sin\left[\frac{2}{3}r^{3/2}-\frac{\pi}{4}\right]+\mathcal{O}(r^{-3/2})\right\\}.$
(C.5)

Letting $x=2-y$ for $\delta_{a}>y>0$ we see that

$nF(x)\asymp ny^{3/2}.$ (C.6)

So, using the above we see that

$\left|h_{n}\left(\sqrt{\frac{n}{2}}x\right)\mathrm{e}^{-n\frac{x^{2}}{4}}\right|\leq\frac{C}{y^{1/4}+n^{-1/6}}$
(C.7)

for $x=2-y$, $0<y<\delta_{a}$.

Now, near the edge, outside the spectrum we have, for $2<x<2+\delta_{a}$,

$\displaystyle
h_{n}\left(\sqrt{\frac{n}{2}}x\right)\mathrm{e}^{-n\frac{x^{2}}{4}}$
$\displaystyle=(2n)^{-1/4}$
$\displaystyle\times\bigg{\\{}\left(\frac{x+2}{x-2}\right)^{1/4}\left[\frac{3}{4}nF(x)\right]^{1/6}\mathrm{Ai}\left(\left[\frac{3}{4}nF(x)\right]^{2/3}\right)(1+\mathcal{O}(n^{-1}))$
$\displaystyle-\left(\frac{x-2}{x+2}\right)^{1/4}\left[\frac{3}{4}nF(x)\right]^{-1/6}\mathrm{Ai}^{\prime}\left(\left[\frac{3}{4}nF(x)\right]^{2/3}\right)(1+\mathcal{O}(n^{-1}))\bigg{\\}}.$
(C.8)

In this region, we have from [49] that,

$|\mathrm{Ai}(r)|\leq
C\frac{\mathrm{e}^{-\frac{2}{3}r^{3/2}}}{r^{1/4}},\qquad|\mathrm{Ai}^{\prime}(r)|\leq
C(1+r)^{1/4}\mathrm{e}^{-\frac{2}{3}r^{3/2}}.$ (C.9)

Therefore, for $x=2+y$ and $0<y<\delta_{a}$ we have

$\left|h_{n}\left(\sqrt{\frac{n}{2}}x\right)\mathrm{e}^{-n\frac{x^{2}}{4}}\right|\leq\frac{C\mathrm{e}^{-cny^{3/2}}}{y^{1/4}+n^{-1/6}}$
(C.10)

The asymptotics in [20] show that the above also extends to $y>\delta_{a}$.

It is a straightforward calculation to see that the above upper bounds extend
also to

$y\to\left|n^{1/4}h_{n-1}\left(\sqrt{\frac{n}{2}}(2+y)\right)\mathrm{e}^{-n\frac{(2+y)^{2}}{4}}\right|.$
(C.11)

##  Appendix D Some integrals of Hermite polynomials

Let $H_{n}(x)$ denote the physicist’s Hermite polynomials,

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

and the Hermite functions,

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

In this appendix we will use several results from Gradshteyn and Ryzhik [34].
There are many editions of this work, and all references are to the most
recent 2014 version (8th edition) of [34] unless noted.

######  Lemma D.1.

We have

$\int\varphi_{2m}(x)\mathrm{d}x=\frac{2^{1/2}}{m^{1/4}}\left(1+\mathcal{O}(m^{-1})\right)$
(D.3)

Proof. The integral equals,

$\int\varphi_{2m}(t)\mathrm{d}t=\frac{1}{2^{m}((2m)!)^{1/2}\pi^{1/4}}\int_{\mathbb{R}}\mathrm{e}^{-t^{2}/2}H_{2m}(t)\mathrm{d}t.$
(D.4)

Formula (7.376.1) of [34] is,

$\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}xy}\mathrm{e}^{-x^{2}/2}H_{n}(x)\mathrm{d}x=(2\pi)^{1/2}H_{n}(y)\mathrm{e}^{-y^{2}/2}\mathrm{i}^{n}.$
(D.5)

We need to evaluate this at $n=2m$ and $y=0$. Formula (8.956.6) of [34] is,

$H_{2m}(0)=(-1)^{m}2^{m}(2m-1)!!$ (D.6)

For the double factorial we have,

$(2m-1)!!=\frac{(2m-1)!}{2^{m-1}(m-1)!}$ (D.7)

Therefore,

$\displaystyle\int_{\mathbb{R}}\varphi_{2m}(t)\mathrm{d}t$
$\displaystyle=\left(\frac{1}{2^{m}((2m)!)^{1/2}\pi^{1/4}}\right)\left((2\pi)^{1/2}(-1)^{m}\right)\left((-1)^{m}2^{m}\frac{(2m-1)!}{2^{m-1}(m-1)!}\right)$
$\displaystyle=2^{1/2}\pi^{1/4}\frac{(2m-1)!}{2^{m-1}((2m)!)^{1/2}(m-1)!}$
$\displaystyle=2^{1/2}\pi^{1/4}\frac{((2m)!)^{1/2}}{2^{m}(m)!}.$ (D.8)

The first line is just plugging in the formulas (D.5), (D.6) and (D.7). The
second line is just a few cancellations. In the third line we simplified the
factorial terms. Stirling’s approximation is,

$n!=(2\pi n)^{1/2}n^{n}\mathrm{e}^{-n}\left(1+\mathcal{O}(n^{-1})\right)$
(D.9)

Therefore,

$\displaystyle\int_{\mathbb{R}}\varphi_{2m}(t)\mathrm{d}t$
$\displaystyle=2^{1/2}\pi^{1/4}\frac{((2m)!)^{1/2}}{2^{m}(m)!}$
$\displaystyle=2^{1/2}\pi^{1/4}\frac{(2\pi)^{1/4}(2m)^{1/4}(2m)^{m}\mathrm{e}^{-m}}{2^{m}(2\pi)^{1/2}m^{1/2}m^{m}\mathrm{e}^{-m}}\left(1+\mathcal{O}(m^{-1})\right)$
$\displaystyle=\frac{2^{1/2}}{m^{1/4}}\left(1+\mathcal{O}(m^{-1})\right)$
(D.10)

This yields the claim. ∎

######  Lemma D.2.

We have,

$\displaystyle\left(\frac{2m+1}{2}\right)^{1/2}\int_{0}^{\infty}\varphi_{2m+1}(x)\mathrm{d}x$
$\displaystyle=\frac{m^{1/4}}{2^{1/2}}\left(1+\mathcal{O}(m^{-1/2})\right)$
(D.11)

Proof. Clearly $(\frac{2m+1}{2})^{1/2}=m^{1/2}(1+\mathcal{O}(m^{-1}))$. By
definition,

$\displaystyle m^{1/2}\int_{0}^{\infty}\varphi_{2m+1}(t)\mathrm{d}t=$
$\displaystyle=\frac{m^{1/2}}{(2^{2m+1})^{1/2}((2m+1)!)^{1/2}\pi^{1/4}}\int_{0}^{\infty}\mathrm{e}^{-t^{2}/2}H_{2m+1}(t)\mathrm{d}t.$
(D.12)

We require the formula,

$\int_{0}^{\infty}\mathrm{e}^{-2\alpha
x^{2}}x^{\nu}H_{2n+1}(x)\mathrm{d}x=(-1)^{n}2^{2n-\nu/2}\frac{\Gamma(\nu/2+1)\Gamma(n+3/2)}{\pi^{1/2}\alpha^{\nu/2+1}}F(-n,\nu/2+1;3/2;(2\alpha)^{-1}),$
(D.13)

where $F$ is the (Gauss) hypergeometric function (also denoted ${}_{2}F_{1}$).
It is given in the 2007 version (7th edition) of Gradshteyn and Ryzhik [33] as
formula (7.376.3). It was then erroneously changed in the 2014 version [34],
and then corrected back to the 2007 (7th) edition formula in the erratum
[56]1.††footnotetext: $1$. The source of confusion seems to be that the
original source [14] itself contains an error; note that the formula (13.18b)
in that work is derived via substitution of (13.3b) into (10.4b). The $\mu$ in
(10.4b) equals $\mu=1/2$ and so the $\Gamma(n+1/2)$ in (13.18b) should be
$\Gamma(n+\mu+1/2)=\Gamma(n+3/2)$. In any case, it is also not hard to
directly check (D.13) from the relation
$H_{2n+1}(x)=(-1)^{n}(2n+1)!(n!)^{-1}(2x)_{1}F_{1}(-n,3/2;x^{2})$, where
${}_{1}F_{1}$ is the confluent hypergeometric function of the first kind, and
direct integration of the series term-by-term.

Evaluating this with $n=m$ and $\nu=0$ and $\alpha=1/4$ we see that,

$\displaystyle\frac{m^{1/2}}{(2^{2m+1})^{1/2}((2m+1)!)^{1/2}\pi^{1/4}}\int_{0}^{\infty}\mathrm{e}^{-t^{2}/2}H_{2m+1}(t)\mathrm{d}t$
$\displaystyle=$
$\displaystyle\frac{m^{1/2}}{(2^{2m+1})^{1/2}((2m+1)!)^{1/2}\pi^{1/4}}(-1)^{m}2^{2m}\frac{\Gamma(m+3/2)}{\pi^{1/2}4^{-1}}F(-m,1;3/2;2),$
(D.14)

where we simplified $\Gamma(1)=1$. From (8.339) of [34] and (D.7),

$\Gamma\left(\frac{1}{2}+n\right)=\frac{\sqrt{\pi}}{2^{n}}(2n-1)!!=\frac{(2n)!}{4^{n}n!}\sqrt{\pi}.$
(D.15)

Evaluating this with $n=m+1$ we get,

$\Gamma\left(\frac{3}{2}+m\right)=\frac{(2(m+1))!}{4^{m+1}(m+1)!}\sqrt{\pi}.$
(D.16)

We use the following integral representation of the hypergeometric function,
(9.111) of [34]

$F(a,b;c;z)=\frac{1}{B(b,c-b)}\int_{0}^{1}t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a}\mathrm{d}t,$
(D.17)

where the beta function is,

$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$ (D.18)

Therefore,

$\displaystyle F(-m,1,3/2,2)$
$\displaystyle=\frac{\Gamma(3/2)}{\Gamma(1)\Gamma(1/2)}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t$
$\displaystyle=\frac{1}{2}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t$
(D.19)

So,

$\displaystyle\frac{m^{1/2}}{(2^{2m+1})^{1/2}((2m+1)!)^{1/2}\pi^{1/4}}\int_{0}^{\infty}\mathrm{e}^{-t^{2}/2}H_{2m+1}(t)\mathrm{d}t$
$\displaystyle=$
$\displaystyle\left(\frac{m^{1/2}}{(2^{2m+1})^{1/2}((2m+1)!)^{1/2}\pi^{1/4}}\right)\times\left(\frac{(-1)^{m}2^{2m}}{\pi^{1/2}4^{-1}}\frac{(2(m+1))!}{4^{m+1}(m+1)!}\sqrt{\pi}\right)$
$\displaystyle\times$
$\displaystyle\left(\frac{1}{2}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t\right)$
$\displaystyle=$
$\displaystyle\pi^{-1/4}(-1)^{m}\frac{m^{1/2}(2(m+1))!}{2^{m}2^{1/2}(m+1)!((2m+1)!)^{1/2}}\left(\frac{1}{2}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t\right)$
$\displaystyle=$
$\displaystyle\pi^{-1/4}\frac{m^{1/2}(2m+2)(2m+1)((2m)!)^{1/2}}{2^{m}2^{3/2}(m+1)m!(2m+1)^{1/2}}\left((-1)^{m}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t\right)$
$\displaystyle=$
$\displaystyle\pi^{-1/4}m\left(\frac{((2m)!)^{1/2}}{2^{m}m!}\right)\left((-1)^{m}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t\right)\left(1+\mathcal{O}(m^{-1})\right)$
(D.20)

The first equality is direct substitution of (D.16) and (D) into (D). In the
second equality we cleaned up the powers of $2$ and $\pi$. In the third
equality we simplified the factorials to only have $(2m)!$ and $m!$. The third
asymptotic is using things like $2m+1\sim 2m$, etc., and simplifying the
powers of $2$. Via Stirling’s approximation,

$\frac{((2m)!)^{1/2}}{2^{m}m!}=\frac{(2\pi)^{1/4}(2m)^{1/4}(2m)^{m}\mathrm{e}^{-m}}{2^{m}(2\pi)^{1/2}m^{1/2}m^{m}\mathrm{e}^{-m}}\left(1+\mathcal{O}(m^{-1})\right)=\pi^{-1/4}m^{-1/4}\left(1+\mathcal{O}(m^{-1})\right).$
(D.21)

Now consider the integral,

$(-1)^{m}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t=\int_{0}^{1}(2t-1)^{m}(1-t)^{-1/2}\mathrm{d}t.$
(D.22)

It is easy to see that,

$\int_{0}^{1/2}|1-2t|^{m}(1-t)^{-1/2}\leq
2\int_{0}^{1/2}(1-2t)^{m}\leq\frac{C}{m}$ (D.23)

by exact calculation. For the other half of the integral,

$\displaystyle\int_{1/2}^{1}(2t-1)^{m}(1-t)^{-1/2}\mathrm{d}t$
$\displaystyle=\int_{0}^{1/2}(1-2u)^{m}u^{-1/2}\mathrm{d}u$
$\displaystyle=\int_{0}^{1/2}\mathrm{e}^{m\log(1-2u)}u^{-1/2}\mathrm{d}u$
$\displaystyle=\frac{1}{2^{1/2}m}\int_{0}^{\infty}\exp\left[-x(1+m^{-1})\right]\frac{1}{(1-\mathrm{e}^{-x/m})^{1/2}}\mathrm{d}x,$
(D.24)

via the change of variable $x=-m\log(1-2u)$. It is straightforward to check
that,

$\left|\frac{1}{(1-\mathrm{e}^{-s})^{1/2}}-\frac{1}{s^{1/2}}\right|\leq C$
(D.25)

for all $s>0$. Therefore,

$\displaystyle\frac{1}{2^{1/2}m}\int_{0}^{\infty}\exp\left[-x(1+m^{-1})\right]\frac{1}{(1-\mathrm{e}^{-x/m})^{1/2}}\mathrm{d}x$
$\displaystyle=$
$\displaystyle\frac{1}{2^{1/2}m^{1/2}}\int_{0}^{\infty}\exp\left[-x(1+m^{-1})\right]x^{-1/2}\mathrm{d}x+\mathcal{O}(m^{-1})$
$\displaystyle=$
$\displaystyle\frac{1}{2^{1/2}m^{1/2}}\int_{0}^{\infty}\exp\left[-x\right]x^{-1/2}\mathrm{d}x+\mathcal{O}(m^{-1})$
(D.26)

The final integral can be calculated exactly by relating it to the Gaussian
integral,

$\frac{1}{2^{1/2}m^{1/2}}\int_{0}^{\infty}\exp\left[-x\right]x^{-1/2}\mathrm{d}x=\frac{\pi^{1/2}}{2^{1/2}m^{1/2}}.$
(D.27)

Summarizing,

$(-1)^{m}\int_{0}^{1}(1-t)^{-1/2}(1-2t)^{m}\mathrm{d}t=\frac{\sqrt{\pi}}{\sqrt{2m}}(1+\mathcal{O}(m^{-1/2})).$
(D.28)

Plugging (D.21) and (D.28) into (D.20) we obtain,

$\displaystyle\frac{m^{1/2}}{(2^{2m+1})^{1/2}((2m+1)!)^{1/2}\pi^{1/4}}\int_{0}^{\infty}\mathrm{e}^{-t^{2}/2}H_{2m+1}(t)\mathrm{d}t$
$\displaystyle=$
$\displaystyle\pi^{-1/4}m\left(\pi^{-1/4}m^{-1/4}\right)(\pi^{1/2}(2m)^{-1/2})\left(1+\mathcal{O}(m^{-1/2})\right)$
$\displaystyle=$
$\displaystyle\frac{m^{1/4}}{2^{1/2}}\left(1+\mathcal{O}(m^{-1/2})\right).$
(D.29)

This completes the proof. ∎

##  Appendix E Verification that the variance functional is positive

######  Lemma E.1.

The functional $V(\varphi)$ defined in (3) is positive for $N\geq 4$.

Proof. With the functional $V(\varphi)$ as in (3) we have, (see, e.g., [40,
Theorem 2.4])

$V(\varphi)=\frac{1}{2}\sum_{k=1}^{\infty}kc_{k}^{2}+\frac{s_{4}}{2}c_{2}^{2}-\frac{2s_{3}}{N^{1/2}}c_{1}c_{2}$
(E.1)

where $c_{i}$ is the $i$th coefficient of $\varphi$ in the Chebyshev
polynomial basis,

$c_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}\varphi(2\cos(\theta))\cos(k\theta)\mathrm{d}\theta=\frac{2}{\pi}\int_{-1}^{1}\varphi(2x)T_{k}(x)\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}$
(E.2)

and $T_{k}$ is the $k$th Chebyschev polynomial of the first kind,
$T_{k}(\cos(\theta))=\cos(k\theta)$.

Now for any centered random variable with variance $1$ we have the inequality,
(see, for example, the remark after [29, Theorem 2.2])

$s_{4}\geq s_{3}^{2}-2.$ (E.3)

Hence,

$\displaystyle V(\varphi)$
$\displaystyle\geq\frac{c_{1}^{2}}{2}+c_{2}^{2}+\frac{s_{4}}{2}c_{2}^{2}-2\frac{s_{3}}{N^{1/2}}c_{1}c_{2}$
$\displaystyle\geq\frac{c_{1}^{2}}{2}+\frac{s_{3}^{2}}{2}c_{2}^{2}-2\frac{s_{3}}{N^{1/2}}c_{1}c_{2}$
$\displaystyle\geq\frac{c_{1}^{2}}{2}+\frac{s_{3}^{2}}{2}c_{2}^{2}-\frac{s_{3}^{2}}{2}c_{2}^{2}-2\frac{c_{1}^{2}}{N}.$
(E.4)

We therefore conclude the proof. ∎

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