# Variance of the number of zeros of dependent Gaussian trigonometric
polynomials

Louis Gass

###### Abstract

We compute the variance asymptotics for the number of real zeros of
trigonometric polynomials with random dependent Gaussian coefficients and show
that under mild conditions, the asymptotic behavior is the same as in the
independent framework. In fact our proof goes beyond this framework and makes
explicit the variance asymptotics of various models of random Gaussian
polynomials. Though we use the Kac–Rice formula, we do not use the explicit
closed formula for the second moment of the number of zeros, but we rather
rely on intrinsic properties of the Kac–Rice density.

00footnotetext: Univ Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes,
France.00footnotetext: This work was supported by the ANR grant UNIRANDOM,
ANR-17-CE40-0008.00footnotetext: Email: louis.gass(at)ens-rennes.fr

###### Contents

  1. 1 Introduction
  2. 2 Kac–Rice formula for the second moment
    1. 2.1 Determinant formulas
    2. 2.2 Kac–Rice formula for the variance
    3. 2.3 Estimates for the singularity along the diagonal
    4. 2.4 Decay estimates far from the diagonal
  3. 3 Proof of the main theorems
    1. 3.1 Asymptotics of the Kac density
    2. 3.2 Proof of Theorem 1.2
    3. 3.3 Proof of Theorem 1.1

##  1 Introduction

The asymptotic behavior of the variance of the number of zeros of random
trigonometric polynomials $\sum a_{k}\cos(kt)+b_{k}\sin(kt)$ with independent
Gaussian coefficients has been established in [8]. Since then, the variances
of numerous models have been studied: for instance, see [4] for the analogous
model $\sum a_{k}\cos(kt)$, see [6, 7] in the independent framework with non
Gaussian coefficients, or more recently [10] for random orthogonal polynomials
on the real line. In greater dimension, the asymptotic behavior of the
variance of random nodal volume has been established in several kinds of
random waves models, see for instance the survey [12] and the references
therein.

Roughly speaking, two distinct methods are used in the literature to study the
asymptotic behavior of the variance: the Wiener chaos expansion of the number
of zeros (see for instance [4, 11]), and the Kac–Rice formula (see [8, 10]).
In this paper we adopt the second method to make explicit the asymptotics of
the variance of the number of zeros of random trigonometric polynomials with
dependent coefficients.

This paper can be viewed as the natural continuation of [2] and [3], in which
the asymptotic behavior of the mean number of zeros of this model has been
established. To the best of our knowledge, the question of variance for
dependent trigonometric polynomials has not been considered until now and, the
aforementioned examples do not seem to cover this situation.

Let us now detail our model. Let $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$ be the
one-dimensional torus, which can be identified with a segment of $\mathbb{R}$
of length $2\pi$. For $s,t\in\mathbb{T}$ we define the distance

$\operatorname{dist}(s,t)=d(s-t,2\pi\mathbb{Z}),$

For $s\in\mathbb{T}$ we define

$X_{n}(s)=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}a_{k}\cos(ks)+b_{k}\sin(ks),$

with $(a_{k})_{k},(b_{k})_{k}$ two independent stationary centered Gaussian
processes with correlation function $\rho:\mathbb{Z}\rightarrow\mathbb{R}$.
That is,

$\forall k,l\geq
0,\quad\mathbb{E}[a_{k}a_{l}]=\mathbb{E}[b_{k}b_{l}]=\rho(k-l).$

Thanks to Bochner Theorem, the correlation function $\rho$ is associated with
a spectral measure $\mu$ on the torus $\mathbb{T}$ via the relation

$\rho(k)=\frac{1}{2\pi}\int_{\mathbb{T}}e^{-iku}\mathrm{d}\mu(u).$

We denote by $Z_{X_{n}}(I)$ the number of zeros of $X_{n}$ on a subinterval
$I$ of the torus $\mathbb{T}$. Under suitable conditions on the spectral
measure $\mu$, it has been shown in [2] and [3] that the expectation of the
number of zeros of the process $X_{n}$ on $\mathbb{T}$ behaves like
$\frac{2}{\sqrt{3}}n$, as in the independent framework.

The first main theorem of this article is the following, and makes explicit
the variance asymptotics for the number of zeros of the process $X_{n}$ on
$\mathbb{T}$, as $n$ grows to infinity, under the assumption that the spectral
measure $\mu$ has a positive continuous density.

######  Theorem 1.1.

We suppose that the spectral measure $\mu$ has a positive continuous density
$\psi$ with respect to the Lebesgue measure on the torus $\mathbb{T}$. Then
there is an explicit positive constant $C_{\infty}$ that does not depend on
$\psi$, such that for any subinterval $J$ of the torus,
$\lim_{n\rightarrow+\infty}\frac{\operatorname{Var}(Z_{X_{n}}(J))}{n}=\mathrm{length}(J)\,C_{\infty}.$

The constant $C_{\infty}\simeq 0.089$ is thus the same constant computed in
[8] in the particular framework of independent Gaussian random variables. It
is quite remarkable that the limiting value is universal with respect to the
dependency of the Gaussian coefficients, as long as the spectral measure $\mu$
has a positive continuous density.

Theorem 1.1 is in fact an application of the more general next Theorem 1.2,
that puts forwards the principal ingredients necessary to obtain such a
universal asymptotics for the variance. Let $(X_{n})_{n}$ be a sequence of
centered Gaussian random processes defined on an open subinterval $I$ of the
torus $\mathbb{T}$, with covariance functions $(r_{n})_{n}$ defined for $n\geq
0$ and $s,t\in I$ by

$r_{n}(s,t)=\mathbb{E}[X_{n}(s)X_{n}(t)].$

######  Theorem 1.2.

Let $J$ be a subinterval of $I$ such that $\overline{J}\subset I$. We suppose
that the sequence of random processes $(X_{n})_{n}$ satisfies the two
following conditions $(A1)$ Let $K$ be a compact subset of $\mathbb{R}$. There
exists a continuous positive function $\psi$ on $I$ such that uniformly on
$x\in J$, $u,v\in K$, and $a,b\in\\{0,1,2,3,4\\}$,
$\lim_{n\rightarrow+\infty}\frac{1}{n^{a+b}}r_{n}^{(a,b)}\left(x+\frac{u}{n},x+\frac{v}{n}\right)=(-1)^{b}\psi(x)\operatorname{sinc}^{(a+b)}(u-v).$
$(A2)$ There is a constant $C$ and an exponent $\alpha>\frac{1}{2}$ such that
for $a,b\in\\{0,1\\}$, $\displaystyle\forall s,t\in
J,\quad|r_{n}^{(a,b)}(s,t)|\leq
C\frac{n^{a+b}}{(n\operatorname{dist}(s,t))^{\alpha}}.$ Then there is an
explicit positive constant $C_{\infty}$ independent of the function $\psi$,
such that
$\lim_{n\rightarrow+\infty}\frac{\operatorname{Var}(Z_{X_{n}}(J))}{n}=\mathrm{length}(J)\,C_{\infty}.$

This last Theorem 1.2 covers Theorem 1.1, as we will prove that the
assumptions on the spectral measure $\mu$ in Theorem 1.1 ensure that the
associated sequence of processes $(X_{n})_{n}$ satisfies hypotheses $(A1)$ and
$(A2)$ of Theorem 1.2. In this particular case, we can take advantage of the
fact that the covariance function $r_{n}$ of the model is a trigonometric
polynomial of degree $n$, to show that the general statements of hypotheses
$(A1)$ and $(A2)$ roughly follow from the case $a=b=0$.

In fact, Theorem 1.2 covers various previously known results about the
asymptotics of variance of the number of zeros of general random trigonometric
polynomials. Indeed, Theorem it also allows to make explicit the variance
asymptotics for the number of zeros on any compact subinterval of $]0,\pi[$ of
the process

$\widetilde{X}_{n}(s)=\frac{1}{\sqrt{n}}\sum_{k=0}^{n}a_{k}\cos(ks+\theta),\quad\quad\theta\in\mathbb{R},$

with $(a_{k})_{k\geq 0}$ a stationary sequence of centered random Gaussian
variables whose spectral measure has a positive continuous density. Note that
the variance asymptotics for the number of zeros on $[0,\pi]$ for this process
was established in the case of Gaussian iid $(a_{k})_{k\geq 0}$ in [4]. In
fact, a slight modification of our proof would prove the asymptotics for the
variance of the number of zeros on the whole interval $[0,\pi]$.

More recently, the authors in [10] established the variance asymptotics for
the number of zeros of sums of real orthogonal polynomials (with respect to
some compactly supported measure $\nu$) with independent Gaussian
coefficients. The one to one mapping between $[-1,1]$ and the half-torus
$[0,\pi]$ given by the relation $x=\cos(\theta)$ allows us to partially
connect their results and Theorem 1.2. Let $(P_{n})_{n}$ be a sequence of real
orthogonal polynomials with respect to a measure $\nu$ supported on $[-1,1]$.
We suppose that $\nu$ has a continuous positive density $\phi$ on $]-1,1[$
which satisfies,

$\int_{0}^{\pi}\log(\phi(\cos\theta))\mathrm{d}\theta>-\infty\quad\text{and}\quad\int_{0}^{1}\frac{\omega_{\phi}(x)}{x}\mathrm{d}\delta<+\infty.$

We define

$\overline{X}_{n}(s)=\frac{1}{\sqrt{n}}\sum_{k=0}^{n}a_{k}P_{k}(\cos(s)),$

where $(a_{n})_{n\geq 0}$ are iid centered Gaussian random variables. Then the
hypotheses $(A1)$ and $(A2)$ of Theorem 1.2 are satisfied with $\psi=1/\phi$
(see [9, 14]) and Theorem 1.2 implies [10, Cor. 1.3].

Section 2 of the paper is devoted to the study of the Kac density, that is the
integrand in the Kac–Rice formula, for the second moment of the number of
zeros of a general one-dimensional Gaussian process. We first recall standard
formulas for Gaussian conditional expectation, and the integral expression for
the second moment of the number of zeros of a smooth Gaussian process. Under
suitable regularity assumptions on the covariance function, we then deduce the
behavior of the Kac density near the diagonal and we give a non-singular
formula for this integrand, valid in a neighborhood of the diagonal. At last
we give an explicit bound for the Kac density far away from the diagonal.

In Section 3, we make explicit the asymptotics of the Kac density associated
with a sequence of processes satisfying the hypotheses of Theorem 1.2.
Hypothesis $(A1)$ implies in particular that as $n$ grows to infinity, the
sequence $(X_{n})_{n}$ has a local limit proportional to the
$\operatorname{sinc}$ process on $\mathbb{R}$. Together with the results of
Section 2, we deduce after a suitable scaling that the Kac density associated
to the process $X_{n}$ uniformly converges towards the Kac density of the
$\operatorname{sinc}$ process on $\mathbb{R}$, as $n$ grows to infinity. This
fact leads to the proof of Theorem 1.2. We then check that the model of
trigonometric polynomials with dependent coefficients satisfying the
hypotheses of Theorem 1.1, also satisfies the hypotheses of Theorem 1.2, from
which immediately follows the conclusion of Theorem 1.1.

##  2 Kac–Rice formula for the second moment

In this section we give the two main lemmas that quantify the behavior of the
Kac density along the diagonal and far from the diagonal, for a general
centered Gaussian process satisfying some natural regularity assumptions.
First we recall some formula for the conditional density of Gaussian vectors.
For a Gaussian stochastic process $(Y(s))_{s\in I}$, where $I$ is a
subinterval of $\mathbb{R}$, we denote by $r$ its covariance function defined
for $s,t\in I$ by

$r(s,t)=\mathbb{E}[Y(s)Y(t)],$

and for $a,b>0$, we denote by $r^{(a,b)}(s,t)$ the derivatives

$r^{(a,b)}(s,t)=\partial_{1}^{a}\partial_{2}^{b}r(s,t)=\mathbb{E}[Y^{(a)}(s)Y^{(b)}(t)],$

provided that $Y$ is sufficiently regular.

###  2.1 Determinant formulas

The Kac–Rice formula for the expectation and variance of the number of zeros
of a centered Gaussian stochastic process $(Y(s))_{s\in I}$ involves the
conditional Gaussian distributions

$\gamma_{0}(s)\sim\mathrm{Law}\left((Y^{\prime}(s)|Y(s)=0)\right)\;\,\text{and}\;\,(\gamma_{1}(s,t),\gamma_{2}(s,t))\sim\mathrm{Law}\left((Y^{\prime}(s),Y^{\prime}(t)|Y(s)=Y(t)=0)\right).$

We will need formulas for the distribution of $\gamma_{0}$, $\gamma_{1}$ and
$\gamma_{2}$. For now we assume that the process $Y$ has $\mathcal{C}^{2}$
sample paths and that the joint distribution of the Gaussian vector
$(Y(s),Y(t))$ does not degenerate for $s\neq t$. Let

$\Omega(s):=\operatorname{Cov}(Y(s),Y^{\prime}(s)),\quad\text{and}\quad\Sigma(s,t):=\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s),Y^{\prime}(t)).$

Then we have

$\Omega(s)=\begin{pmatrix}r(s,s)&r^{(1,0)}(s,s)\\\
r^{(1,0)}(s,s)&r^{(1,1)}(s,s)\end{pmatrix},$

and

$\Sigma(s)=\begin{pmatrix}[cc|cc]r(s,s)&r(s,t)&r^{(1,0)}(s,s)&r^{(0,1)}(s,t)\\\
r(s,t)&r(t,t)&r^{(1,0)}(s,t)&r^{(1,0)}(t,t)\\\ \hline\cr
r^{(1,0)}(s,s)&r^{(1,0)}(s,t)&r^{(1,1)}(s,s)&r^{(1,1)}(s,t)\\\
r^{(0,1)}(s,t)&r^{(1,0)}(t,t)&r^{(1,1)}(s,t)&r^{(1,1)}(t,t)\end{pmatrix}=\begin{pmatrix}[c|c]\Sigma_{11}&\Sigma_{12}\\\
\hline\cr{}^{T}\Sigma_{12}&\vphantom{\int^{\int}}\Sigma_{22}\quad\end{pmatrix}.$

Now, set

$\omega(s):=r^{(1,1)}(s,s)-\frac{r^{(1,0)}(s,s)^{2}}{r(s,s)}\quad\text{and}\quad\Gamma:=\Sigma_{22}-^{T}\Sigma_{12}\Sigma_{11}^{-1}\Sigma_{12}.$

The matrix $\Gamma$ is the Schur complement of $\Sigma_{11}$ in $\Sigma$. Let
$x\in\mathbb{R}^{4}$ with $x=(x_{1},x_{2})$ and
$x_{1},x_{2}\in\mathbb{R}^{2}$. Let $y=(y_{1},y_{2})\in\mathbb{R}^{2}$. We
define

$m=\frac{r^{(1,0)}(s,s)}{r(s,s)}y_{1}\quad\text{and}\quad
M=\Sigma_{12}\Sigma_{11}^{-1}x_{1}.$

The identities

${}^{T}y\Omega^{-1}y=\frac{y_{1}^{2}}{r(s,s)}+\frac{(y_{2}-m)^{2}}{\omega(s)}\quad\text{and}\quad\,^{T}x\Sigma^{-1}x=\,^{T}x_{1}\Sigma_{11}^{-1}x_{1}+\,^{T}(x_{2}-M)\Gamma^{-1}(x_{2}-M)$

imply, according to the standard formula for the conditional density, the
following lemma.

######  Lemma 2.1.

$\mathbb{E}[\gamma_{0}]=0,\quad\mathbb{E}[\gamma_{0}^{2}]=\omega(s),\quad\text{and}\quad\mathbb{E}[\gamma_{1}]=0,\;\mathbb{E}[\gamma_{2}]=0,\quad\operatorname{Cov}(\gamma_{1},\gamma_{2})=\Gamma(s,t).$

By row reduction we have

$\det(\Sigma)=\det(\Sigma_{11})\det(\Gamma)\quad\text{and}\quad\Sigma^{-1}=\begin{pmatrix}*&*\\\
*&\Gamma^{-1}\end{pmatrix},$

and thus,

$\det[\operatorname{Cov}(\gamma_{1},\gamma_{2})]=\frac{\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s),Y^{\prime}(t))\right]}{\det\left[\operatorname{Cov}(Y(s),Y(t))\right]}.$

The relation between the inverse of a matrix and its adjugate matrix implies

$\mathbb{E}[\gamma_{1}^{2}]=\frac{\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s))\right]}{\det\left[\operatorname{Cov}(Y(s),Y(t))\right]},\quad\mathbb{E}[\gamma_{2}^{2}]=\frac{\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(t))\right]}{\det\left[\operatorname{Cov}(Y(s),Y(t))\right]},$
$\mathbb{E}[\gamma_{1}\gamma_{2}]=\frac{\det\left[\Sigma_{3,4}\right]}{\det\left[\operatorname{Cov}(Y(s),Y(t))\right]},\quad\text{and
also}\quad\mathbb{E}[\gamma_{0}^{2}]=\frac{\det\left[\operatorname{Cov}(Y(s),Y^{\prime}(s))\right]}{\mathbb{E}[Y(s)^{2}]},$

where $\Sigma_{3,4}$ is the matrix $\Sigma$ with the third row and the fourth
line removed.

From Definition of $\Gamma$ in Equation (2.1) and the identity
$\det(A)\leq\det(A+B)$ valid for any two positive semi-definite matrices $A$
and $B$, we also have the inequality

$\det(\Gamma)\leq\det(\Sigma_{22}).$

Note that when $s=t$, the formulas for the distribution of $\gamma_{0}$ and
$(\gamma_{1},\gamma_{2})$ are singular. We will make explicit in Lemma 3.4
below the behavior of these distributions near the diagonal, provided that the
process $Y$ is sufficiently regular.

###  2.2 Kac–Rice formula for the variance

Let $(Y(t))_{t\in I}$ be a centered Gaussian process such that the joint
distribution of the Gaussian vector $(Y(s),Y(t),Y^{\prime}(s),Y^{\prime}(t))$
does not degenerate as $s\neq t$. Denote by $p_{s}$ the density of $Y(s)$ and
by $p_{s,t}$ the density of $(Y(s),Y(t))$. We define

$Z_{Y}(I)=\mathrm{Card}\left\\{s\in I\leavevmode\nobreak\
\middle|\leavevmode\nobreak\ Y(s)=0\right\\}$

the number of zeros of $Y$ in $I$. Kac–Rice formula (see [5, Thm. 3.2]) then
asserts that

$\displaystyle\mathbb{E}[Z_{Y}]$
$\displaystyle=\int_{I}\rho_{1}(s)\mathrm{d}s,$

with

$\displaystyle\rho_{1}(s)$
$\displaystyle=\mathbb{E}\left[|Y^{\prime}(s)|\;\middle|\;Y(s)=0\right]p_{s}(0)\;=\frac{1}{2\pi\sqrt{\det(\Omega)}}\int_{\mathbb{R}}|y|\exp\left(-\frac{y^{2}}{2w(s)}\right)\mathrm{d}y,$

and

$\mathbb{E}[Z_{Y}^{2}]-\mathbb{E}[Z_{Y}]=\iint_{I^{2}}\rho_{2}(s,t)\mathrm{d}s\mathrm{d}t,$

with

$\displaystyle\rho_{2}(s,t)$
$\displaystyle=\mathbb{E}\left[|Y^{\prime}(s)||Y^{\prime}(t)|\;\middle|\;Y(s)=0,Y(t)=0\right]p_{s,t}(0,0)$
$\displaystyle=\frac{1}{(2\pi)^{2}\sqrt{\det(\Sigma(s,t))}}\iint_{\mathbb{R}^{2}}|y_{1}||y_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!y\Gamma^{-1}(s,t)y\right)\mathrm{d}y_{1}\mathrm{d}y_{2}.$

Hence we deduce the integral representation

$\displaystyle\operatorname{Var}(Z_{Y})$
$\displaystyle=\left(\mathbb{E}[Z_{Y}^{2}]-\mathbb{E}[Z_{Y}]-\mathbb{E}[Z_{Y}]^{2}\right)+\mathbb{E}[Z_{Y}]$
$\displaystyle=\left(\iint_{I^{2}}\left(\rho_{2}(s,t)-\rho_{1}(s)\rho_{1}(t)\right)\mathrm{d}s\mathrm{d}t\right)+\int_{I}\rho_{1}(t)\mathrm{d}t.$
(9)

######  Remark 2.2.

$\rho_{1}$ and $\rho_{2}$ have explicit values given by

$\rho_{1}(s)=\frac{1}{\pi}\frac{\sqrt{\vphantom{\int}\det(\Omega)}}{r(s,s)}\quad\text{and}\quad\rho_{2}(s,t)=\frac{1}{\pi^{2}\sqrt{\det(\Sigma_{11})}}\left(\sqrt{\det(\Gamma)}+\Gamma_{12}\arcsin\left(\frac{\Gamma_{12}}{\sqrt{\Gamma_{11}\Gamma_{22}}}\right)\right),$

but these explicit formulas will not be used throughout the proof.

###  2.3 Estimates for the singularity along the diagonal

Our aim now is to derive the asymptotics of formulas (2.1) and (2.1) as
$\tau=(t-s)$ goes to zero. In the case where the process $Y$ is sufficiently
regular, a Taylor development around the point $s$ with sufficiently high
order allows us to remove the singularity at $\tau=0$. This procedure is
standard, see for instance [1] for a general treatment, or [5, Prop. 5.8].

In the following, $(Y(s))_{s\in I}$ is a Gaussian process such that its
covariance function is of class $\mathcal{C}^{4}$ in each variable. We define
the following quantities.

$\xi(s)=\frac{1}{4}\det\left[\operatorname{Cov}(Y(s),Y^{\prime}(s),Y^{(2)}(s))\right],$
$\Delta(s)=\frac{1}{144}\det\left[\operatorname{Cov}(Y(s),Y^{\prime}(s),Y^{(2)}(s),Y^{(3)}(s))\right],$
$\zeta(s)=\frac{1}{24}\det\left[\operatorname{Cov}(Y(s),Y^{\prime}(s),Y^{(3)}(s))\right].$

The following lemma details the exact asymptotics for the distributions of
$\gamma_{0},\gamma_{1}$ and $\gamma_{2}$ as the parameter $\tau=(t-s)$ goes to
zero.

######  Lemma 2.3.

Suppose that there is a constant $C$ such that for $a,b\in\\{0,1,2,3,4\\}$ and
$s,t\in I$, $|r^{(a,b)}(s,t)|\leq
C\quad\text{and}\quad\det[\Omega(s)]\geq\frac{1}{C}.$ Then there is an
explicit constant $T_{\varepsilon}$ depending only on the constant $C$, such
that for all $s\in I$ and $\tau\in[-T_{\varepsilon},T_{\varepsilon}]$,
$\det[\Sigma_{11}(s,s+\tau)]=\tau^{2}\det[\Omega(s)]+\tau^{3}R_{\Sigma_{11}}(s,\tau),$
$\Gamma(s,s+\tau)=\frac{\tau^{2}}{\det[\Omega(s)]}\begin{pmatrix}\xi(s)&-\xi(s)\\\
-\xi(s)&\xi(s)\end{pmatrix}+\tau^{3}R_{\Gamma}(s,\tau)$
$\det\left[\Sigma(s,t)\right]=\tau^{8}\Delta(s)+\tau^{9}R_{\Sigma}(s,\tau),$
$\mathbb{E}[(\gamma_{1}+\gamma_{2})^{2}]=\tau^{4}\frac{\zeta(s)}{\det[\Omega(s)]}+\tau^{5}R_{\gamma}(s,\tau).$
Moreover, the functions
$\Omega,\gamma,\Delta,\zeta,R_{\Sigma_{11}},R_{\Gamma},R_{\Sigma},R_{\gamma}$
are continuous functionals with respect to the $\|.\|_{\infty}$ norm, of the
covariance function $r$ and its derivatives up to order $4$.

###### Proof.

By Taylor expansion with integral remainder, we have

$Y(s+\tau)=\sum_{k=0}^{n}\frac{Y^{(k)}(s)}{k!}\tau^{k}+\frac{\tau^{n+1}}{n!}\int_{0}^{1}(1-u)^{n}Y^{(n+1)}(s+\tau
u)\mathrm{d}u.$

We will use this expression with $n=1,2,3$. For instance, by applying a
unitary transformation on the covariance matrix and the above Taylor formula
with $n=1$, we get

$\displaystyle\det[\Sigma_{11}(s,t)]=\det\left[\operatorname{Cov}(Y(s),Y(t))\right]$
$\displaystyle=\det\left[\operatorname{Cov}(Y(s),Y(t)-Y(s))\right]$
$\displaystyle=\tau^{2}\det\left[\operatorname{Cov}(Y(s),Y^{\prime}(s))\right]+\tau^{3}R_{0}(s,\tau)$
$\displaystyle=\tau^{2}\det\left[\Omega(s)\right]+\tau^{3}R_{0}(s,\tau).$

The remainder $R_{0}(s,\tau)$ is explicit and follows from the expansion of
the determinant. Here we have

$\displaystyle R_{0}(s,\tau)=ae-2bc-c^{2},$

with

$a=\mathbb{E}[Y(s)^{2}]=r(s,s),\quad
b=\mathbb{E}[Y(s)Y^{\prime}(s)]=r^{(0,1)}(s,s)\quad
d=\mathbb{E}[Y^{\prime}(s)^{2}]=r^{(1,1)}(s,s),$
$c=\int_{0}^{1}(1-u)r^{(0,2)}(s,s+\tau u)\mathrm{d}u,$
$e=2\int_{0}^{1}(1-u)r^{(1,2)}(s,s+\tau
u)\mathrm{d}u+\tau\int_{0}^{1}\int_{0}^{1}(1-u)(1-v)r^{(2,2)}(s+\tau u,s+\tau
v)\mathrm{d}u\mathrm{d}v.$

By our assumptions the remainder $R_{0}(s,\tau)$ is a continuous functional of
the covariance function of $r$ and its partial derivatives up to order $2$. We
deduce that there are explicit positive constants $C^{\prime}$ and
$T_{\varepsilon}$ depending only on $C$, such that for all $s\in I$ and
$\tau\in[-T_{\varepsilon},T_{\varepsilon}]$,

$\frac{1}{\tau^{2}}\det[\Sigma_{11}(s,s+\tau)]\geq C^{\prime}.$

We can do similar computations for the others determinants. In particular, we
have

$\displaystyle\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s))\right]$
$\displaystyle=\det\left[\operatorname{Cov}(Y(s),Y^{\prime}(s),Y(t)-Y(s)-\tau
Y^{\prime}(s))\right]$ $\displaystyle=\tau^{4}\xi(s)+\tau^{5}R_{11}(s,\tau),$
$\displaystyle\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(t))\right]=\tau^{4}\xi(s)+\tau^{5}R_{22}(s,\tau),$
$\displaystyle\det\left[\Sigma_{3,4}(s,t)\right]=-\tau^{4}\xi(s)+\tau^{5}R_{12}(s,\tau),$
$\displaystyle\det\left[\Sigma(s,t)\right]=\tau^{8}\Delta(s)+\tau^{9}R_{\Sigma}(s,\tau),$
$\displaystyle\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(t)+Y^{\prime}(s))\right]=\tau^{6}\zeta(s)+\tau^{7}R_{3}(s,\tau).$

As for the computation of $R_{0}(s,\tau)$, the remainders $R_{11}$, $R_{12}$,
$R_{22}$, $R_{\Sigma}$ and $R_{3}$ can be explicitly computed and are defined
as a sum of integrals (on a compact interval) of quantities that depends
polynomially on $r^{(a,b)}(s+\tau_{1},s+\tau_{2})$ for $a,b\in\\{0,1,2,3,4\\}$
and $\tau_{1},\tau_{2}\in[s,s+T_{0}]$. In particular, they are continuous
functionals of the covariance function $r$ and its derivatives up to order
$4$.

The asymptotics for $\Gamma$ and $(\gamma_{1}+\gamma_{2})$ follows from these
identities, and the fact that the common denominator of
$\gamma_{1},\gamma_{2}$, $\gamma_{12}$ and $(\gamma_{1}+\gamma_{2})$ in their
expression (2.1), which is $\det(\Sigma_{11})$, is bounded from below by a
positive constant on $[-T_{\varepsilon},T_{\varepsilon}]$, according to (2.3).
∎

######  Remark 2.4.

From Lemma 2.3, we have the following behavior near the diagonal for $\Gamma$.

$\Gamma(s,s+\tau)\simeq\frac{\tau^{2}}{\det[\Omega(s)]}\begin{pmatrix}\xi(s)&-\xi(s)\\\
-\xi(s)&\xi(s)\end{pmatrix}.$

It means that at a scaling limit, the vector $(\gamma_{1},\gamma_{2})$
degenerates. If $X\simeq\mathcal{N}(0,\xi)$, then we have approximately

$\mathbb{E}[|\gamma_{1}||\gamma_{2}|]\simeq\tau^{2}\mathbb{E}[|X|\,|\\!-\\!X|]=\mathbb{E}[X^{2}]=\tau^{2}\frac{\xi(s)}{\det[\Omega(s)]}.$

It will be convenient to define

$(\widetilde{\gamma}_{1},\widetilde{\gamma}_{2})=\left(\frac{\gamma_{1}+\gamma_{2}}{\tau^{2}}\frac{\gamma_{2}}{\tau}\right),\quad\text{and}\quad\widetilde{\Gamma}\,(s,s+\tau)=\operatorname{Cov}(\widetilde{\gamma}_{1},\widetilde{\gamma}_{2}).$

We have then

$\det[\operatorname{Cov}(\widetilde{\gamma}_{1},\widetilde{\gamma}_{2})]=\frac{\det[\operatorname{Cov}(\gamma_{1},\gamma_{2})]}{\tau^{6}}=\frac{\Delta(s)}{\det\left[\Omega(s)\right]}+\tau
R_{\Sigma}(s,\tau).$

This time, we have the following asymptotics

$\widetilde{\Gamma}(s,s+\tau)\simeq\begin{pmatrix}\zeta(s)&\eta(s)\\\
\eta(s)&\xi(s)\end{pmatrix},$

with $\eta(s)$ some explicit function of $s$. The limit matrix is non
degenerate, provided that $\Delta(s)>0$.

The following lemma precises the behavior of $\rho_{2}(s,t)$ near the
diagonal.

######  Lemma 2.5.

Suppose that there is a constant $C$ such that for $a,b\in\\{0,1,2,3,4\\}$ and
$s,t\in I$, $|r^{(a,b)}(s,t)|\leq
C,\quad\det[\Omega(s)]\geq\frac{1}{C}\quad\text{and}\quad\Delta(s)\geq\frac{1}{C}.$
Then there exists a positive constant $T_{\varepsilon}$ depending only on the
constant $C$, and continuous bounded functions $\kappa$ and $R$ such that
$\forall\tau\in[-T_{\varepsilon},T_{\varepsilon}],\quad\rho_{2}(s,s+\tau)=\tau\kappa(s)+\tau^{2}R(s,\tau).$
Moreover, the functions $\kappa$ and $R$ are continuous functionals of the
covariance function $r$ and its derivatives up to order $4$.

###### Proof.

We use the results established in Lemma 2.3. In order to obtain a relevant
scaling, we follow Remark 2.4 and we make the change of variable
$(\gamma_{1},\gamma_{2})\rightarrow(\widetilde{\gamma_{1}},\widetilde{\gamma_{2}})$.
By the previous asymptotics, we deduce

$\displaystyle\rho_{2}(s,s+\tau)$
$\displaystyle=p_{s,s+\tau}(0,0)\mathbb{E}\left[|\gamma_{1}||\gamma_{2}|\right]$
$\displaystyle=\frac{\tau^{2}}{2\pi\det[\Sigma_{11}(s)]}\mathbb{E}\left[|\tau\widetilde{\gamma_{1}}-\widetilde{\gamma_{2}}||\widetilde{\gamma_{2}}|\right]$
$\displaystyle=\frac{\tau^{2}}{(2\pi)^{2}\sqrt{\det(\Sigma_{11})\det(\widetilde{\Gamma})}}\iint_{\mathbb{R}^{2}}|\tau
x_{1}-x_{2}||x_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!x\widetilde{\Gamma}^{-1}x\right)\mathrm{d}x_{1}\mathrm{d}x_{2}$
$\displaystyle=\frac{\tau^{2}}{(2\pi)^{2}\sqrt{\frac{1}{\tau^{6}}\det(\Sigma)}}\iint_{\mathbb{R}^{2}}|\tau
x_{1}-x_{2}||x_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!x\widetilde{\Gamma}^{-1}x\right)\mathrm{d}x_{1}\mathrm{d}x_{2}$
$\displaystyle=\frac{\tau}{(2\pi)^{2}\sqrt{\Delta(s)+\tau\mathbb{R}_{\Sigma}(s)}}\int_{\mathbb{R}}(I_{\tau}(x_{1})+J_{\tau}(x_{1}))\mathrm{d}x_{1},$

where

$I_{\tau}(x_{1})=\int_{\tau x_{1}}^{\infty}(\tau
x_{1}-x_{2})|x_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!x\widetilde{\Gamma}^{-1}x\right)\mathrm{d}x_{1}\mathrm{d}x_{2},$

and

$J_{\tau}(x_{1})=\int_{-\infty}^{\tau x_{1}}(x_{2}-\tau
x_{1})|x_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!x\widetilde{\Gamma}^{-1}x\right)\mathrm{d}x_{1}\mathrm{d}x_{2}.$

By Remark 2.4 and Equation (2.4), there is a constant $T_{\varepsilon}$ and
$C^{\prime}$ such that the matrix $\widetilde{\Gamma}$ satisfies for all $s\in
I$ and $\tau\in[-T_{\varepsilon},T_{\varepsilon}]$ the inequality
$\det(\widetilde{\Gamma})\geq 1/C^{\prime}$. The formula for the inverse
matrix shows that that the functions $\tau\mapsto I_{\tau}(x_{1})$ and
$\tau\mapsto J_{\tau}(x_{2})$ are continuous functionals of the covariance
function $r$ and its derivatives up to order $4$, as well as their derivatives
with respect to the parameter $\tau$. It means that we can write

$\rho_{2}(s,s+\tau)=\tau\kappa(s)+\tau^{2}R(s,\tau),$

with $\kappa$ and $R$ some explicit continuous functional of the covariance
function $r$ and its derivatives up to order $4$. ∎

######  Remark 2.6.

As suggested by Remark 2.4 and Equation (2.4), the proof of Lemma 2.5 shows
that

$\kappa(s)=\frac{\xi(s)}{2\pi\det[\Omega(s)]^{2}}.$

###  2.4 Decay estimates far from the diagonal

In order to derive the asymptotics of the variance of the number of zeros, we
need to estimate the quantity $(\rho_{2}(s,t)-\rho_{1}(s)\rho_{1}(t))$ in the
Kac–Rice formula (9) for the variance. Suppose that as $s$ and $t$ are far
enough from each other, the random vector $(Y(s),Y^{\prime}(s))$ and
$(Y(t),Y^{\prime}(t))$ are then "almost" decorrelated. Heuristically we should
therefore have

$\displaystyle\mathbb{E}\left[|Y^{\prime}(s)||Y^{\prime}(t)|\;\middle|\;Y(s)=0,Y(t)=0\right]\simeq\mathbb{E}\left[|Y^{\prime}(s)|\;\middle|\;Y(s)=0\right]\;\mathbb{E}\left[|Y^{\prime}(t)|\;\middle|\;Y(t)=0\right],$

and

$p_{s,t}(0,0)\simeq p_{s}(0)p_{t}(0).$

The two following lemmas make the above heuristic rigorous and quantify the
error term. We define

$M(\tau):=\sup_{s\in I}\sup_{a,b\in\\{0,1\\}}|r^{(a,b)}(s,s+\tau)|.$

######  Lemma 2.7.

We assume the existence of a constant $C$ such that for $a,b\in\\{0,1\\}$ and
$s,t\in I$, $|r^{(a,b)}(s,t)|\leq C.$ Then
$|\det(\Sigma(s,s+\tau)-\det(\Omega(s)\det(\Omega(s,+\tau))|\leq
20C^{2}M(\tau)^{2}.$

###### Proof.

For a pair $(p,q)$ in $\\{1,2,3,4\\}$, with $p<q$ we denote by
$(\tilde{p},\tilde{q})$ the complementary pair of $(p,q)$ in $\\{1,2,3,4\\}$.
For instance, if $(p,q)=(2,3)$ then $(\tilde{p},\tilde{q})=(1,4)$. We set

$\Sigma_{p,q}=\begin{pmatrix}\Sigma_{1p}&\Sigma_{1q}\\\
\Sigma_{3p}&\Sigma_{3q}\end{pmatrix},\quad\text{and}\quad\widetilde{\Sigma}_{p,q}=\begin{pmatrix}\Sigma_{2\tilde{p}}&\Sigma_{2\tilde{q}}\\\
\Sigma_{4\tilde{q}}&\Sigma_{4\tilde{q}}\end{pmatrix}.$

The formula for Laplace expansion along rows $1$ and $3$ asserts that

$\det(\Sigma)=\sum_{p<q}(-1)^{p+q}\det(\Sigma_{p,q})\det(\widetilde{\Sigma}_{p,q}).$

Observe that

$\Sigma_{1,3}=\Omega(s),\quad\text{and}\quad\widetilde{\Sigma}_{1,3}=\Omega(t),$

and thus,

$\det(\Sigma)=\det(\Omega(s))\det(\Omega(t))+\sum_{\underset{(p,q)\neq(1,3)}{p<q}}(-1)^{p+q}\det(\Sigma_{p,q})\det(\widetilde{\Sigma}_{p,q}).$

For each pair $(p,q)\neq(1,3)$, there is at least a column of the form
${}^{T}(r^{(a,b)}(s,s+\tau),r^{(c,d)}(s,s+\tau))$ in both $\Sigma_{p,q}$ and
$\widetilde{\Sigma}_{p,q}$. For instance,

$\Sigma_{1,4}=\begin{pmatrix}r(s,s)&r^{(0,1)}(s,t)\\\
r^{(1,0)}(s,s)&r^{(1,1)}(s,t)\end{pmatrix},\quad\text{and}\quad\widetilde{\Sigma}_{1,4}=\begin{pmatrix}r(t,t)&r^{(0,1)}(s,t)\\\
r^{(0,1)}(t,t)&r^{(1,1)}(s,t)\end{pmatrix}.$

It means that for every pair $(p,q)\neq(1,3)$, we have the inequalities

$|\det(\Sigma_{p,q})|\leq
2CM(\tau),\quad\text{and}\quad|\det(\widetilde{\Sigma}_{p,q})|\leq 2CM(\tau),$

and thus

$\left|\det(\Sigma(s,t))-\det(\Omega(s))\det(\Omega(t))\right|\leq
20C^{2}M(\tau)^{2}.$

∎

######  Lemma 2.8.

We assume the existence of a constant $C$ such that for $a,b\in\\{0,1\\}$ and
$s,t\in I$, $|r^{(a,b)}(s,t)|\leq C,\quad\quad\det(\Omega(s))\geq 1/C.$ There
are positive constants $\varepsilon$ and $C^{\prime}$ depending only on the
constant $C$, such that for all $\tau$ satisfying $M(\tau)\leq\varepsilon$, we
have $|\rho_{2}(s,s+\tau)-\rho_{1}(s)\rho_{1}(s+\tau)|\leq
C^{\prime}M(\tau)^{2}.$

###### Proof.

The hypotheses of Lemma 2.8 imply that

$r(s,s)\geq\frac{\det(\Omega(s))}{r^{(1,1)}(s,s)}\geq\frac{1}{C^{2}},\quad\text{and}\quad\omega(s)=\frac{\det(\Omega(s))}{r(s,s)}\geq\frac{1}{C^{2}}.$

Set $\varepsilon=1/(\sqrt{40}C^{2})$. Lemma 2.7 above implies that for all
$\tau$ such that $M(\tau)\leq\varepsilon$, we have the following inequality

$\displaystyle\det(\Sigma(s,s+\tau))$
$\displaystyle\geq\det(\Omega(s))\det(\Omega(t))-20C^{2}M(\tau)^{2}\geq\frac{1}{2C^{2}}.$
(15)

Moreover, Equations (2.1) and (2.1) implies that

$\det(\Sigma)=\det(\Sigma_{11})\det(\Gamma)\leq
C^{2}\det(\Gamma),\quad\text{and}\quad\det(\Sigma)\leq\det(\Sigma_{11})\det(\Sigma_{22})\leq
C^{2}\det(\Sigma_{11}).$

Combining these inequalities with (15), we deduce the following inequalities
for all $\tau$ satisfying $M(\tau)\leq\varepsilon$

$\det(\Sigma_{11}(s,s+\tau))\geq\frac{1}{2C^{4}}\,,\quad\text{and}\quad\det(\Gamma(s,s+\tau))\geq\frac{1}{2C^{4}}.$

Now let $s,t\in I$ such that for $\tau=t-s$ we have $M(\tau)\leq\varepsilon$.
We can express the difference

$\displaystyle\rho_{2}(s,s+\tau)-\rho_{1}(s)\rho_{1}(s+\tau)=R_{1}+R_{2},$

with

$R_{1}=\left[\frac{1}{\sqrt{\det(\Sigma)}}\\!-\\!\frac{1}{\sqrt{\det(\Omega(s))\det(\Omega(t))}}\right]\\!\\!\left(\int_{\mathbb{R}}|x|\exp\left(-\frac{x^{2}}{2\omega(s)}\right)\mathrm{d}x\right)\\!\\!\left(\int_{\mathbb{R}}|y|\exp\left(-\frac{y^{2}}{2\omega(t)}\right)\mathrm{d}y\right),$
$R_{2}=\frac{1}{\sqrt{\det(\Sigma)}}\iint_{\mathbb{R}^{2}}|x||y|\left[\exp\left(-\frac{1}{2}\,^{T}\\!(x,y)\Gamma^{-1}\,(x,y)\right)-\exp\left(-\frac{1}{2}\left(\frac{x^{2}}{\omega(s)}+\frac{y^{2}}{\omega(t)}\right)\right)\right]\mathrm{d}x\mathrm{d}y.$

We treat first the term $R_{1}$. Let

$\mathrm{Denom}=\sqrt{\det(\Sigma)\det(\Omega(s))\det(\Omega(t))}\left(\sqrt{\det(\Sigma)}+\sqrt{\det(\Omega(s))\det(\Omega(t))}\right).$

From Lemma 2.7 and Equation (15), there is an explicit constant $C^{\prime}$
depending only on $C$ such that

$\displaystyle\left|\frac{1}{\sqrt{\det(\Sigma)}}-\frac{1}{\sqrt{\det(\Omega(s))\det(\Omega(t))}}\right|$
$\displaystyle=\frac{\left|\det(\Sigma)-\det(\Omega(s))\det(\Omega(t))\right|}{\mathrm{Denom}}$
$\displaystyle\leq C^{\prime}M(\tau)^{2}.$

From the estimate (2.4), the quantity $\omega(s)$ is bounded from below
$1/C^{2}$. We deduce the existence of a constant $C^{\prime\prime}$ depending
only on $C$ and such that

$|R_{1}|\leq C^{\prime\prime}M(\tau)^{2}.$

For the term $R_{2}$, we estimate the distance between $\Gamma$ and the
diagonal matrix $\mathrm{diag}(\omega(s),\omega(t))$. First, we have

$\left|\vphantom{\Sigma^{2}}\det\left[\operatorname{Cov}(Y(s),Y(t))\right]-r(s,s)r(t,t)\right|=r(s,t)^{2}\leq
M(\tau)^{2}.$

Developing the determinant $\det[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s))]$
along the second row, we also deduce

$\left|\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s))\right]-r(t,t)\det(\Omega(s))\right|\leq
6CM(\tau)^{2}.$

If the parameter $\tau$ satisfies the inequality $M(\tau)\leq\varepsilon$, the
inequalities (2.4) and (2.4) imply

$\displaystyle
r(s,s)r(t,t)\geq\det(\Sigma_{11}(s,s+\tau))-M(\tau)^{2}\geq\frac{1}{2C^{4}}-\frac{1}{40C^{4}}\geq\frac{1}{3C^{4}}.$
(19)

Using the formulas (2.1) for the coefficients of $\Gamma(s,t)$, we get

$\displaystyle\left|\mathbb{E}[\gamma^{2}_{1}]-\omega(s)\right|=\left|\frac{\det\left[\operatorname{Cov}(Y(s),Y(t),Y^{\prime}(s))\right]}{\det\left[\operatorname{Cov}(Y(s),Y(t))\right]}-\frac{r(t,t)\det(\Omega(s))}{r(s,s)r(t,t)}\right|\leq
C^{\prime}M(\tau)^{2},$ (20)

where the last inequality is justified by the inequalities (2.4) and (2.4),
and the lower bounds for the denominators given by inequalities (2.4) and
(19). Similarly,

$\left|\mathbb{E}[\gamma^{2}_{2}]-\omega(t)\right|\leq
C^{\prime}M(\tau)^{2}\quad\text{and}\quad|\mathbb{E}[\gamma_{1}\gamma_{2}]|\leq
C^{\prime}M(\tau)^{2}.$

From Equation (2.4), $\det(\Gamma)$ is bounded from below by an explicit
positive constant. Estimates (20) and (2.4) imply the existence of a constant
$C^{\prime}$ depending only on $C$ such that

$\left\|\Gamma-\mathrm{diag}(\omega(s),\omega(t))\right\|\leq
C^{\prime}M(\tau)^{2},$
$\left\|\Gamma^{-1}-\mathrm{diag}\left(\frac{1}{\omega(s)},\frac{1}{\omega(s)}\right)\right\|\leq
C^{\prime}M(\tau)^{2}.$

In order to recover the estimate on $R_{2}$ we use the following inequality
valid for $a,b\in\mathbb{R}$

$|e^{b}-e^{a}|\leq|b-a|\left(e^{a}+e^{b}\right).$

Since the quantities $\omega^{-1}$ and $\det(\Sigma)=\det(\Omega)\det(\Gamma)$
are uniformly bounded from below by an explicit positive constant, and
$\Gamma^{-1}$ is bounded, we get

$\displaystyle|R_{2}|$
$\displaystyle\leq\frac{C^{\prime}M(\tau)^{2}}{\sqrt{\det(\Sigma)}}\iint_{\mathbb{R}^{2}}|x||y||(|x|+|y|)\left[\exp\left(-\frac{1}{2}\,^{T}\\!(x,y)\Gamma^{-1}\,(x,y)\right)\right.$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\left.\exp\left(-\frac{1}{2}\left(\frac{x^{2}}{\omega(s)}+\frac{y^{2}}{\omega(t)}\right)\right)\right]\mathrm{d}x\mathrm{d}y.$

Since $\omega(s),\det(\Gamma(s,s+\tau)$ and $\det(\Sigma(s,s+\tau)$ are
bounded below by an explicit constant depending only on $C$, we deduce the
existence of a constant $C^{\prime\prime}$ (depending only on $C$) such that

$|R_{2}|\leq C^{\prime\prime}M(\tau)^{2}.$

Combining the estimates for $R_{1}$ and $R_{2}$ we obtain

$|\,\rho_{2}(s,s+\tau)-\rho_{1}(s)\rho_{1}(t)\,|\leq C^{\prime}M(\tau)^{2},$

where $C^{\prime}$ is a constant depending only on the constant $C$. ∎

##  3 Proof of the main theorems

###  3.1 Asymptotics of the Kac density

Let $(X_{n})_{n}$ be a sequence of processes of the torus $\mathbb{T}$
satisfying the hypotheses of Theorem 1.2 and let $J$ be a subinterval of $I$
with $\overline{J}\subset I$. We define for $x\in J$

$\forall
s\in\mathbb{R},\;Y_{n}(s):=X_{n}\left(\frac{s}{n}\right)\quad\text{and}\quad\forall\tau\in\mathbb{R},\;Y_{n}^{x}(\tau):=X_{n}\left(x+\frac{\tau}{n}\right).$

Let $Y_{\infty}$ be a centered stationary Gaussian process with
$\operatorname{sinc}$ covariance function. An explicit computation gives

$\mathbb{E}[(Y_{\infty}^{\prime}(0))^{2}]=\frac{1}{3},\quad\mathbb{E}[(Y_{\infty}^{(2)}(0))^{2}]=\frac{1}{5},\quad\text{and}\quad\mathbb{E}[(Y_{\infty}^{(3)}(0))^{2}]=\frac{1}{7}.$

Let

$q_{n}(s,t):=\mathbb{E}[Y_{n}(s)Y_{n}(t)]=r_{n}\left(\frac{s}{n},\frac{t}{n}\right),\quad\text{and
thus}\quad
q_{n}^{(a,b)}(s,t)=\frac{1}{n^{a+b}}r_{n}^{(a,b)}\left(\frac{s}{n},\frac{t}{n}\right).$

Hypothesis $(A1)$ of Theorem 1.2 implies that for $a,b\in\\{0,1,2,3,4\\}$, we
have the following uniform convergence on $x\in J$ and on $\tau$ in compact
subsets of $\mathbb{R}$

$\lim_{n\rightarrow+\infty}q_{n}^{(a,b)}(nx,nx+\tau)=\psi(x)(-1)^{b}\operatorname{sinc}^{(a+b)}(\tau).$

It means that the finite dimensional distributions of the process
$(Y_{n}^{x}(\tau))_{\tau\in\mathbb{R}}$ and its derivatives up to order $4$
converge in distribution towards the finite dimensional distributions of the
process $\psi(x)Y_{\infty}$ and its derivatives up to order $4$. Moreover, it
implies the quantity $q_{n}^{(a,b)}$ is uniformly bounded for all $x\in J$ and
$a,b\in\\{0,1,2,3,4\\}$ by some constant $C$. By Cauchy–Schwarz inequality for
covariance functions, we deduce then that for $s,t\in J$,

$q_{n}^{(a,b)}(s,t)\leq\sqrt{q_{n}^{(a,a)}(s,s)}\sqrt{q_{n}^{(b,b)}(t,t)}\leq
C.$

Denote by $\rho_{1,n}$ and $\rho_{2,n}$ (resp. $\rho_{1,\infty}$ and
$\rho_{2,\infty}$) the Kac densities of the process $Y_{n}$ (resp.
$Y_{\infty}$). Since the process $Y_{\infty}$ is stationary, the function
$\rho_{1,\infty}$ is a constant and the function $\rho_{2,\infty}$ depends
only on $t-s$. Similarly, we will use the notation
$\Omega_{n},\Omega_{\infty},\Gamma_{n},\Gamma_{\infty}$, etc. as defined in
Section 2. The following lemma give the asymptotics of the Kac densities as
$n$ grows to infinity.

######  Lemma 3.1.

We have the following uniform convergences, with $x\in J$ and $\tau$ in a
compact set of $\mathbb{R}$
$\lim_{n\rightarrow+\infty}\rho_{1,n}(nx)=\rho_{1,\infty}\quad\text{and}\quad\lim_{n\rightarrow+\infty}\rho_{2,n}(nx,nx+\tau)=\rho_{2,\infty}(\tau).$

###### Proof.

We begin with the convergence of $\rho_{1,n}$. From Equation (3.1) with
$\tau=0$ and the fact that the function $\psi$ is bounded from below by a
positive constant $C_{\psi}$, we deduce the following uniform convergence on
$x\in J$

$\lim_{n\rightarrow+\infty}\Omega_{n}(nx)=\psi(x)\Omega_{\infty},\quad\text{and}\quad\lim_{n\rightarrow+\infty}\omega_{n}(nx)=\psi(x)\omega_{\infty}.$

It follows from (3.1) that $\Omega_{\infty}=\mathrm{diag}(1,1/3)$ and
$\omega_{\infty}=1/3$. It implies the existence of rank $n_{0}$ independent of
$x$ such that

$\forall n\geq n_{0},\forall x\in
J,\quad\det[\Omega_{n}(nx)]\geq\frac{C_{\psi}^{2}}{4}\quad\text{and}\quad\omega_{n}(nx)\geq\frac{C_{\psi}}{4}.$

It implies the following uniform convergence on $x\in J$

$\displaystyle\lim_{n\rightarrow+\infty}\rho_{1,n}(nx)$
$\displaystyle=\lim_{n\rightarrow+\infty}\frac{1}{2\pi\sqrt{\det(\Omega_{n}(nx))}}\int_{\mathbb{R}}|y|\exp\left(-\frac{y^{2}}{2w_{n}(nx)}\right)\mathrm{d}y$
$\displaystyle=\frac{1}{2\pi\psi(x)\sqrt{\det(\Omega_{\infty})}}\int_{\mathbb{R}}|y|\exp\left(-\frac{y^{2}}{2\psi(x)\omega_{\infty}}\right)\mathrm{d}y$
$\displaystyle=\frac{1}{2\pi\sqrt{\det(\Omega_{\infty})}}\int_{\mathbb{R}}|u|\exp\left(-\frac{u^{2}}{2\omega_{\infty}}\right)\mathrm{d}u$
$\displaystyle=\rho_{1,\infty}.$

Now for the convergence of $\rho_{2,n}$, let $T_{\varepsilon}$ be some small
positive constant and $K$ a large compact set of $\mathbb{R}$. Observe that
the process $Y_{\infty}$ is stationary and the support of its spectral measure
has an accumulation point. It implies (see [5, Ex. 3.5]) that for $\tau\in
K\setminus[-T_{\varepsilon},T_{\varepsilon}]$, the covariance matrix of the
Gaussian vector
$(Y_{\infty}(s),Y_{\infty}(s+\tau),Y_{\infty}^{\prime}(s),Y_{\infty}^{\prime}(s+\tau))$
is nondegenerate and moreover, one can find an explicit positive constant $C$
such that

$\forall\tau\in K\setminus[-T_{\varepsilon},T_{\varepsilon}],\quad
1/C\leq\det(\Sigma_{\infty})\leq C\quad\text{and}\quad
1/C\leq\det(\Sigma_{11,\infty})\leq C.$

From Equation (2.1) we deduce that the matrices $\Gamma_{\infty}$ and
$\Sigma_{\infty}$ are nondegenerate on
$K\setminus[-T_{\varepsilon},T_{\varepsilon}]$ and we have the following
convergences, uniformly on $x\in J$ and $\tau\in
K\setminus[-T_{\varepsilon},T_{\varepsilon}]$

$\lim_{n\rightarrow+\infty}\Sigma_{n}(nx,\tau)=\psi(x)\Sigma_{\infty}(\tau),$
$\lim_{n\rightarrow+\infty}\Gamma_{n}(nx,\tau)=\psi(x)\Gamma_{\infty}(\tau).$

Moreover, there exists a rank $n_{0}$ depending only on $T_{\varepsilon}$ such
that

$\forall n\geq n_{0},\forall\tau\in
K\setminus[-T_{\varepsilon},T_{\varepsilon}],\quad\det(\Sigma_{n}(nx,nx+\tau))\geq\frac{C_{\psi}^{4}}{2C}\quad\text{and}\quad\det(\Gamma_{n}(nx,nx+\tau))\geq\frac{C_{\psi}^{2}}{2C^{2}}.$

We deduce the following convergence, uniform on $x\in J$ and $\tau\in
K\setminus[-T_{\varepsilon},T_{\varepsilon}]$

$\displaystyle\lim_{n\rightarrow+\infty}\rho_{2}(nx,nx+\tau)$
$\displaystyle=\lim_{n\rightarrow+\infty}\frac{1}{(2\pi)^{2}\sqrt{\det(\Sigma_{n})}}\iint_{\mathbb{R}^{2}}|y_{1}||y_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!y\Gamma_{n}^{-1}y\right)\mathrm{d}y_{1}\mathrm{d}y_{2}$
$\displaystyle=\frac{1}{(2\pi)^{2}\psi^{2}(x)\sqrt{\det(\Sigma_{\infty})}}\iint_{\mathbb{R}^{2}}|y_{1}||y_{2}|\exp\left(-\frac{1}{2\psi(x)}\,^{T}\\!y\Gamma_{\infty}^{-1}y\right)\mathrm{d}y_{1}\mathrm{d}y_{2}$
$\displaystyle=\frac{1}{(2\pi)^{2}\sqrt{\det(\Sigma_{\infty})}}\iint_{\mathbb{R}^{2}}|u_{1}||u_{2}|\exp\left(-\frac{1}{2}\,^{T}\\!u\Gamma_{\infty}^{-1}u\right)\mathrm{d}u_{1}\mathrm{d}u_{2}$
$\displaystyle=\rho_{2,\infty}(\tau).$

It remains to prove the uniform convergence in the case where $\tau$ lives in
a neighborhood of the origin. To this end, we apply Lemma 2.5. We check first
that our process $Y_{n}$ fulfills the hypotheses of the lemma. From Equation
$\eqref{eq:35}$, there is a constant $C$ such that for $n\geq 0$, $x\in J$,
$\tau\in\mathbb{R}$ and $a,b\in\\{0,1,2,3,4\\}$,

$|q_{n}^{(a,b)}(nx,nx+\tau)|\leq C.$

Moreover, we have uniformly on $x\in J$

$\lim_{n\rightarrow+\infty}\det(\Omega_{n}(nx))=\psi^{2}(x)\Omega_{\infty}(0)=\frac{\psi^{2}(x)}{3}.$

Set

$\Delta_{n}(nx)=\det[\operatorname{Cov}(Y_{n}(nx),Y_{n}^{\prime}(nx),Y_{n}^{(2)}(nx),Y_{n}^{(3)}(nx))].$

From identities (3.1), we have

$\displaystyle\lim_{n\rightarrow+\infty}\Delta_{n}(nx)=\psi^{4}(x)\Delta_{\infty}=\psi^{4}(x)\begin{vmatrix}1&0&-\frac{1}{3}&0\\\
0&\frac{1}{3}&0&-\frac{1}{5}\\\ -\frac{1}{3}&0&\frac{1}{5}&0\\\
0&-\frac{1}{5}&0&\frac{1}{7}\end{vmatrix}=\frac{16}{23625}\psi^{4}(x).$

In particular there is a rank $n_{0}$ such that uniformly on $x\in J$

$\forall n\geq
n_{0},\quad\det(\Omega_{n}(nx))\geq\frac{C_{\psi}^{2}}{4}\quad\text{and}\quad\Delta_{n}(nx)\geq\frac{C_{\psi}^{4}}{2000}.$

Hence, the hypotheses of Lemma 2.5 are satisfied. There exists a positive
constant $T_{\varepsilon}$ such that for $n$ greater than $n_{0}$, there
exists continuous functions $R_{n}$ and $\kappa_{n}$ such that

$\forall\tau\in[-T_{\varepsilon},T_{\varepsilon}],\quad\rho_{2,n}(nx,nx+\tau)=\tau\kappa_{n}(nx)+\tau^{2}R_{n}(nx,\tau),$

and the functions $\kappa_{n}$ and $R_{n}$ are continuous functionals of
$q_{n}$ and its partial derivatives up to order $4$. From the uniform
convergence of $q_{n}(nx,nx+\tau)$ and its derivatives, we obtain the uniform
convergence of $\kappa_{n}(nx)$, $R_{n}(nx,nx+\tau)$ and thus
$\rho_{2,n}(nx,nx+\tau)$ towards $\kappa_{\infty}$, $R_{\infty}(\tau)$ and
$\rho_{2,\infty}(\tau)$ on $[-T_{\varepsilon},T_{\varepsilon}]$.

Gathering the uniform convergence of $\rho_{2,n}(nx,nx+\tau)$ towards
$\rho_{2,\infty}(\tau)$ on $K\setminus[-T_{\varepsilon},T_{\varepsilon}]$, and
on $[T_{\varepsilon},T_{\varepsilon}]$ we have proved the uniform convergence
for $x\in J$ and $\tau\in K$ of $\rho_{2,n}(nx,nx+\tau)$ towards its limit. ∎

The following lemma establishes a decay property for the Kac density.

######  Lemma 3.2.

Let $\alpha$ be the exponent in Assumption $(A2)$ of Theorem 1.2. There exists
a constant $C$ and a rank $n_{0}$ independent of $x\in J$ such that for all
$\tau$ with $x+\frac{\tau}{n}\in J$ and $n\geq n_{0}$, we have
$\left|\rho_{2,n}(nx,nx+\tau)-\rho_{1,n}(nx)\rho_{1,n}(nx+\tau)\right|\leq\frac{C}{\operatorname{dist}(\tau,2\pi
n\mathbb{Z})^{2\alpha}}.$

###### Proof.

We check that the hypotheses of Lemma 2.8 are satisfied with the process
$Y_{n}$ defined on the subinterval $nJ$ of $\mathbb{R}$. In virtue of
Hypothesis $(A2)$, there is a constant $C$ such that for $a,b\in\\{0,1\\}$,
$n\in\mathbb{N}$ and $x\in J$,

$\displaystyle
q_{n}^{(a,b)}(nx,nx+\tau)=\frac{1}{n^{a+b}}r_{n}^{(a,b)}\left(x,x+\frac{\tau}{n}\right)\leq\frac{C}{\operatorname{dist}(\tau,2\pi
n\mathbb{Z})^{\alpha}}.$

It implies that the function $M(\tau)$ defined in (2.4) satisfies,

$\forall\tau\in\mathbb{R},\quad
M(\tau)\leq\frac{C}{\operatorname{dist}(\tau,2\pi n\mathbb{Z})^{\alpha}}.$

From Equation (3.1), the function $q_{n}^{(a,b)}$ is uniformly bounded for
$a,b\in\\{0,1\\}$ by a constant $C$, and inequality (3.1) states that for $n$
greater than a rank $n_{0}$ independent of $x$, and uniformly in $x\in J$,
$\det[\Omega_{n}(nx)]\geq C_{\psi}^{2}/4$. For $n\geq n_{0}$, Lemma 2.8
implies the existence of positive constants $\varepsilon$ and $C^{\prime}$
independent of $n$ and $x$, such that for all $\tau$ with $x+\frac{\tau}{n}\in
J$ satisfying $M(\tau)\leq\varepsilon$, we have

$|\rho_{2,n}(nx,nx+\tau)-\rho_{1,n}(nx)\rho_{1,n}(nx+\tau)|\leq\frac{C^{\prime}}{\operatorname{dist}(\tau,2\pi
n\mathbb{Z})^{2\alpha}}.$

If $M(\tau)\geq\varepsilon$ then

$\operatorname{dist}(\tau,2\pi n\mathbb{Z})\leq\frac{C}{\varepsilon}.$

In that case, Lemma 3.1 implies that the left-hand side of Equation (3.1) is
bounded by a constant $C_{\varepsilon}$ independent of $n$ and $x$. To
conclude, note that Inequality (3.1) remains valid for all
$\tau\in\mathbb{R}$, with $C^{\prime}$ replaced by
$C^{\prime}+C_{\varepsilon}(C/\varepsilon)^{2\alpha}$. ∎

###  3.2 Proof of Theorem 1.2

Let $J$ be a subinterval of $I$ with $\overline{J}\subset I$. We identify $J$
with a segment $[a,b]$, such that $|b-a|\leq 2\pi$. Let $n\geq n_{0}$, where
$n_{0}$ is the rank defined in Lemma 3.2. We write

$\displaystyle\operatorname{Var}(Z_{X_{n}}(J))$
$\displaystyle=\operatorname{Var}(Z_{Y_{n}}(nJ))$
$\displaystyle=n\int_{a}^{b}\\!\rho_{1,n}(nx)\mathrm{d}x+n\int_{a}^{b}\\!\int_{-n(a-x)}^{n(b-x)}\left(\rho_{2,n}(nx,nx+\tau)-\rho_{1,n}(nx)\rho_{1,n}(nx+\tau)\right)\mathrm{d}\tau\mathrm{d}x.$

For the first term, Lemma 3.1 asserts that uniformly on $x\in J$, we have

$\lim_{n\rightarrow+\infty}\rho_{1,n}(nx)=\rho_{1,\infty},\quad\text{and
thus}\quad\lim_{n\rightarrow+\infty}\int_{a}^{b}\rho_{1,n}(nx)\mathrm{d}x=|b-a|\rho_{1,\infty}.$

For the second term, we define

$\displaystyle R_{n}=$
$\displaystyle\int_{a}^{b}\int_{n(a-x)}^{n(b-x)}\left(\rho_{2,n}(nx,nx+\tau)-\rho_{1,n}(nx)\rho_{1,n}(nx+\tau)\right)\mathrm{d}\tau\mathrm{d}x$
$\displaystyle\quad-\quad|b-a|\int_{\mathbb{R}}\left(\rho_{2,\infty}(\tau)-\rho_{1,\infty}^{2}\right)\mathrm{d}\tau.$

Let $A$ be a large constant. We split $R_{n}$ into four parts:

$R_{n}=R_{1,n}(A)-R_{2,n}(A)+R_{3,n}(A)-R_{3,\infty}(A),$

where,

$\displaystyle R_{1,n}(A)$
$\displaystyle=\int_{a}^{b}\int_{n(a-x)}^{n(b-x)}\mathds{1}_{|\tau|\leq
A}\left[\left(\rho_{2,n}(nx,nx+\tau)-\rho_{2,\infty}(\tau)\right)\right]\mathrm{d}\tau\mathrm{d}x,$
$\displaystyle R_{2,n}(A)$
$\displaystyle=\int_{a}^{b}\int_{n(a-x)}^{n(b-x)}\mathds{1}_{|\tau|\leq
A}\left[\left(\rho_{1,n}(nx)\rho_{1,n}(nx+\tau)-\rho_{1,\infty}^{2}\right)\right]\mathrm{d}\tau\mathrm{d}x,$
$\displaystyle R_{3,n}(A)$
$\displaystyle=\int_{a}^{b}\int_{n(a-x)}^{n(b-x)}\mathds{1}_{|\tau|\geq
A}\left(\rho_{2,n}(nx,nx+\tau)-\rho_{1,n}(nx)\rho_{1,n}(nx+\tau)\right)\mathrm{d}\tau\mathrm{d}x,$
$\displaystyle R_{3,\infty}(A)$
$\displaystyle=|b-a|\int_{\mathbb{R}\setminus[-A,A]}(\rho_{2,\infty}(\tau)-\rho_{1,\infty}^{2})\mathrm{d}\tau\mathrm{d}x.$

For the the terms $R_{1,n}(A)$ and $R_{2,n}(A)$, Lemma 3.1 shows that the two
integrands converge uniformly towards $0$ and thus,

$\lim_{n\rightarrow+\infty}R_{1,n}(A)=0\quad\text{and}\quad\lim_{n\rightarrow+\infty}R_{2,n}(A)=0.$

For the term $R_{3,n}(A)$, Lemma 3.2 and the inequality $\alpha>1/2$ gives

$\displaystyle|R_{3,n}(A)|\leq\int_{a}^{b}\int_{n(a-x)}^{n(b-x)}\mathds{1}_{|\tau|\geq
A}\frac{C}{\tau^{2\alpha}}\mathrm{d}\tau\,\leq\,2C|b-a|\int_{A}^{\infty}\frac{1}{\tau^{2\alpha}}\mathrm{d}\tau\,\leq\,\frac{2C|b-a|}{(2\alpha+1)A^{2\alpha-1}}.$
(28)

For the term $R_{3,\infty}$, we apply dominated convergence to $R_{3,n}(A)$.
Equation (28) implies in particular that the integrand is bounded by a
integrable function that does not depend on the parameter $n$, and 3.1 shows
that the integrand converges to the integrand of $R_{3,\infty}(A)$. Hence by
Equation (28),

$\displaystyle|R_{3,\infty}(A)|=\lim_{n\rightarrow+\infty}|R_{3,n}(A)|\leq\frac{2C|b-a|}{(2\alpha+1)A^{2\alpha-1}}.$

Gathering the estimates for $R_{1,n}(A)$, $R_{2,n}(A)$, $R_{3,n}(A)$ and
$R_{3,\infty}(A)$, we deduce that

$\limsup_{n\rightarrow+\infty}|R_{n}|\leq\frac{4C|b-a|}{(2\alpha+1)A^{2\alpha-1}}.$

Letting $A$ go to infinity, we deduce that
$\lim_{n\rightarrow+\infty}R_{n}=0$, from which follows the following
convergence

$\lim_{n\rightarrow+\infty}\frac{\operatorname{Var}(Z_{X_{n}}(J))}{n}=\mathrm{length}(J)\,C_{\infty},$

where

$C_{\infty}:=\int_{\mathbb{R}}\left(\rho_{2,\infty}(\tau)-\rho_{1,\infty}^{2}\right)\mathrm{d}\tau+\rho_{1,\infty}.$

It remains to show the positiveness of the constant $C_{\infty}$. A slight
modification of the above proof would show that

$C_{\infty}=\lim_{n\rightarrow+\infty}\frac{\operatorname{Var}(Z_{Y_{\infty}}[0,n])}{n}.$

It has been shown by several methods (see for example [1, 13]) that
$C_{\infty}>0$, not only for the process $Y_{\infty}$, but also for a large
class of stationary Gaussian processes satisfying mild conditions on their
covariance functions.

######  Remark 3.3.

Using the explicit formulas given by Remark 2.2, we have

$\rho_{1,\infty}=\frac{1}{\pi\sqrt{3}}\quad\rho_{2,\infty}(\tau)=\frac{1}{\pi^{2}\sqrt{\det(\Sigma_{11,\infty}(\tau))}}\left(\sqrt{\det(\Gamma_{\infty}(\tau))}+\Gamma_{12,\infty}\arcsin\left(\frac{\Gamma_{12,\infty}}{\sqrt{\Gamma_{11,\infty}\Gamma_{22,\infty}}}\right)\right).$

###  3.3 Proof of Theorem 1.1

In the following, we consider the sequence of trigonometric Gaussian
polynomials $(X_{n})_{n}$ defined in introduction. Assuming the hypotheses of
Theorem 1.1, we show that this sequence of processes satisfies hypotheses
$(A1)$ and $(A2)$ of Theorem 1.2 with $I=\mathbb{T}$, from which follows the
conclusion of Theorem 1.1. Following [3], the next computation gives an
integral expression for the covariance function of $X_{n}$.

$\displaystyle r_{n}(s,t):\\!$ $\displaystyle=\mathbb{E}[X_{n}(s)X_{n}(t)]$
$\displaystyle=\frac{1}{n}\sum_{k,l=1}^{n}\rho(k-l)\cos(ks-lt)$
$\displaystyle=\frac{1}{n}\sum_{k,l=1}^{n}\left(\frac{1}{2\pi}\int_{0}^{2\pi}e^{-i(k-l)u}\mathrm{d}\mu(u)\right)\cos(ks-
lt)$ $\displaystyle=\frac{1}{2\pi
n}\int_{0}^{2\pi}\mathrm{Re}\left(\sum_{k,l=1}^{n}e^{-i(k-l)u}e^{iks-
ilt}\right)\mathrm{d}\mu(u)$ $\displaystyle=\frac{1}{2\pi
n}\int_{0}^{2\pi}\mathrm{Re}\left(\left(\sum_{\,k=1}^{n}e^{ik(s-u)}\right)\left(\sum_{\,l=1}^{n}e^{il(t-u)}\right)\right)\mathrm{d}\mu(u)$
$\displaystyle=\cos\left(\frac{n+1}{2}(s-t)\right)\frac{1}{2\pi}\int_{0}^{2\pi}K_{n}(s-u,t-u)\mathrm{d}\mu(u),$
(29)

where $K_{n}$ is the two points Fejér kernel

$K_{n}(s,t)=\frac{1}{n}\frac{\sin\left(\frac{ns}{2}\right)}{\sin\left(\frac{s}{2}\right)}\frac{\sin\left(\frac{nt}{2}\right)}{\sin\left(\frac{t}{2}\right)}.$

In the case where $\rho(k-l)=\delta_{k,l}$, the measure $\mu$ is the
normalized Lebesgue measure on $[-\pi,\pi]$. In that case, we denote by
$r_{0,n}$ its covariance function. The function $r_{0,n}$ has the following
explicit expression.

$r_{0,n}(s,t)=\frac{1}{2n}\left[\frac{\sin\left(\left(n+\frac{1}{2}\right)(s-t)\right)}{\sin\left(\frac{s-t}{2}\right)}-1\right].$

Let us now assume that the spectral measure $\mu$ has a continuous and
positive density $\psi$ on $\mathbb{T}$. The two following lemmas show that
the covariance function $r_{n}$ satisfies hypotheses $(A1)$ and $(A2)$ of
Theorem 1.2 with any exponent $\alpha\in]1/2,1[$.

######  Lemma 3.4.

Let $a,b\in\\{0,1,2,3,4\\}$. Uniformly for $s\in\mathbb{T}$ and $u,v$ in any
compact subset of $\mathbb{R}$,
$\lim_{n\rightarrow+\infty}\frac{1}{n^{a+b}}r_{n}^{(a,b)}\left(s+\frac{u}{n},s+\frac{v}{n}\right)=\psi(s)(-1)^{b}\operatorname{sinc}^{(a+b)}(u-v).$

###### Proof.

Let us first remark that the covariance function $r_{n}$ is here a
trigonometric polynomial an can thus be extended to an analytic function on
$\mathbb{C}$. We will prove that the conclusion of Lemma 3.4 holds when $u$
and $v$ belong to a compact subset of $\mathbb{C}$. In that case, it suffices
to prove the lemma for $a=b=0$. The general case follows from the analyticity
of the covariance function $r_{n}$ with respect to the parameters $u$ and $v$,
and the uniform convergence on compact subsets of $\mathbb{C}$. We have

$\displaystyle r_{n}\left(s+\frac{u}{n},s+\frac{v}{n}\right)$
$\displaystyle=I_{n}+\psi(s)r_{0,n}\left(s+\frac{u}{n},s+\frac{v}{n}\right),$

where

$I_{n}=\cos\left(\frac{n+1}{2}\frac{u-v}{n}\right)\frac{1}{2\pi}\int_{-\pi}^{\pi}K_{n}\left(\frac{u}{n}-x,\frac{v}{n}-x\right)\left[\psi\left(x+s\right)-\psi(s)\right]\mathrm{d}x.$

Firstly, we have

$\displaystyle r_{0,n}\left(s+\frac{u}{n},s+\frac{v}{n}\right)$
$\displaystyle=\frac{1}{2n}\left[\frac{\sin\left(\left(n+\frac{1}{2}\right)\frac{u-v}{n}\right)}{\sin\left(\frac{u-v}{2n}\right)}-1\right]$
$\displaystyle=\operatorname{sinc}(u-v)+O\left(\frac{1}{n}\right),$

where the remainder in uniform on $s\in\mathbb{R}$ and $u,v$ in compact
subsets of $\mathbb{C}$. It remains to prove that the quantity $I_{n}$
converges towards $0$ uniformly on $s\in\mathbb{T}$ and $u,v$ in compact sets
of $\mathbb{C}$. Let $K$ be a compact subset of $\mathbb{C}$ and
$C(K)=1+\sup_{u\in K}|\mathrm{Re}(u)|$. We have

$\displaystyle|I_{n}|$ $\displaystyle\leq\frac{1}{2\pi
n}\int_{-\pi}^{\pi}\left|\frac{\sin\left(\frac{u-nx}{2}\right)}{\sin\left(\frac{1}{2}\left(\frac{u}{n}-x\right)\right)}\frac{\sin\left(\frac{v-nx}{2}\right)}{\sin\left(\frac{1}{2}\left(\frac{v}{n}-x\right)\right)}\right|\omega_{\psi}(x)\mathrm{d}x$
$\displaystyle\leq\frac{1}{2\pi
n^{2}}\int_{-n\pi}^{n\pi}\left|\frac{\sin\left(\frac{u-y}{2}\right)}{\sin\left(\frac{u-y}{2n}\right)}\frac{\sin\left(\frac{v-y}{2}\right)}{\sin\left(\frac{v-y}{2n}\right)}\right|\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y$
$\displaystyle\leq R_{1}+R_{2},$

where

$R_{1}=\frac{1}{2\pi
n^{2}}\int_{-C(K)}^{C(K)}\left|\frac{\sin\left(\frac{u-y}{2}\right)}{\sin\left(\frac{u-y}{2n}\right)}\frac{\sin\left(\frac{v-y}{2}\right)}{\sin\left(\frac{v-y}{2n}\right)}\right|\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y,$

and

$R_{2}=\frac{1}{2\pi n^{2}}\int_{-n\pi}^{n\pi}\mathds{1}_{\\{|y|\geq
C(K)\\}}\left|\frac{\sin\left(\frac{u-y}{2}\right)}{\sin\left(\frac{u-y}{2n}\right)}\frac{\sin\left(\frac{v-y}{2}\right)}{\sin\left(\frac{v-y}{2n}\right)}\right|\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y.$

For the term $R_{1}$, we have

$R_{1}\leq\frac{1}{2\pi
n^{2}}\int_{-C(K)}^{C(K)}\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y.$

Since the spectral density is continuous (hence uniformly continuous) on
$\mathbb{T}$, the quantity $R_{1}$ converges towards zero as $n$ goes to
infinity by dominated convergence, uniformly on $s\in\mathbb{T}$ and $u,v\in
K$. For the term $R_{2}$ we use the following inequalities, valid for
$\mathrm{Re}(z)\in[-5\pi/6,5\pi/6]$:

$\frac{3}{5\pi}|\mathrm{Re}(z)|\leq|\sin(\mathrm{Re}(z))|\leq|\sin(z)|.$

There is a rank $n_{0}$ depending only on the compact subset $K$ such that,
for all $n\geq n_{0}$, $u,v\in K$ and $y\in[-n\pi,n\pi]$,

$-\frac{5\pi}{6}\leq\frac{u-y}{2n}\leq\frac{5\pi}{6}\quad\text{and}\quad-\frac{5\pi}{6}\leq\frac{v-y}{2n}\leq\frac{5\pi}{6}.$

Define

$y_{K}:=\sup_{z\in K}|\mathrm{Im}(z)|\quad\text{and}\quad
I(K):=\left[\sup_{|\mathrm{Re}(z)|\leq y_{K}}|\sin(z)|\right]^{2}.$

It follows from the series of inequalities (3.3) that

$\displaystyle R_{2}$ $\displaystyle\leq\frac{I(K)}{2\pi
n^{2}}\int_{-n\pi}^{n\pi}\mathds{1}_{\\{|y|\geq
C(K)\\}}\frac{1}{\left|\sin\left(\frac{u-y}{2n}\right)\right|}\frac{1}{\left|\sin\left(\frac{v-y}{2n}\right)\right|}\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y$
$\displaystyle\leq\frac{5}{3}I(K)\int_{-n\pi}^{n\pi}\mathds{1}_{\\{|y|\geq
C(K)\\}}\frac{1}{|\mathrm{Re}(u)-y||\mathrm{Re}(v)-y|}\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y$
$\displaystyle\leq\frac{5}{3}I(K)\int_{-\infty}^{\infty}\mathds{1}_{\\{|y|\geq
C(K)\\}}\frac{1}{\left(|y|-C(K)+1\right)^{2}}\,\omega_{\psi}\left(\frac{y}{n}\right)\mathrm{d}y.$

By dominated convergence, the quantity $R_{2}$ also converges towards $0$ when
$n$ goes to infinity, uniformly on $s\in\mathbb{T}$ and $u,v\in K$. ∎

######  Lemma 3.5.

Let $a,b\in\\{0,1\\}$ and $0<\alpha<1$. There is a constant $C$ such that
$\displaystyle\forall s,t\in\mathbb{T},\quad|r_{n}^{(a,b)}(s,t)|\leq
C\frac{n^{a+b}}{(n\operatorname{dist}(s,t))^{\alpha}}.$

###### Proof.

Let $s,t\in\mathbb{T}$. If $\operatorname{dist}(s,t)\leq 4/n$ then according
to the previous Lemma 3.4, there is a constant $C$ such that

$|r_{n}^{(a,b)}(s,t)|\leq Cn^{a+b}\leq
4^{\alpha}C\frac{n^{a+b}}{(n\operatorname{dist}(s,t))^{\alpha}},$

and the conclusion of Lemma 3.5 holds. It suffices then to prove Lemma 3.5 for
$s,t\in\mathbb{T}$ such that $\operatorname{dist}(s,t)\geq 4/n$. Let
$C(s,\varepsilon)$ denote the circle of center $s$ and radius $\varepsilon$.
By Cauchy integral formula,

$\displaystyle|r_{n}^{(a,b)}(s,t)|$
$\displaystyle=\left|\frac{1}{(2i\pi)^{a+b}}\int_{C\left(s,\frac{1}{n}\right)}\int_{C\left(t,\frac{1}{n}\right)}\frac{r_{n}(w,z)}{(s-w)^{a+1}(t-z)^{b+1}}\mathrm{d}w\mathrm{d}z\right|$
$\displaystyle\leq\vphantom{\int^{\int}}n^{a+b}\\!\\!\\!\sup_{w\in
C\left(s,\frac{1}{n}\right)}\sup_{z\in
C\left(t,\frac{1}{n}\right)}|r_{n}(w,z)|$ $\displaystyle\leq
n^{a+b}\sup_{|u|\leq 1}\sup_{|v|\leq
1}\left|r_{n}\left(s+\frac{u}{n},t+\frac{v}{n}\right)\right|.$

Let $u,v$ be complex numbers such that $|u|\leq 1$ and $|v|\leq 1$. Using the
explicit formula (29) for $r_{n}$ we obtain

$\displaystyle r_{n}\\!\left(s+\frac{v}{n},t+\frac{v}{n}\right)$
$\displaystyle=\frac{1}{2\pi
n}\cos\left(\frac{n+1}{2}\left(s-t+\frac{u-v}{n}\right)\right)\\!\\!\int_{-\pi}^{\pi}\\!\frac{\sin\left(\frac{n(s-x)+u}{2}\right)}{\sin\left(\frac{s-x+\frac{u}{n}}{2}\right)}\frac{\sin\left(\frac{n(t-x)+v}{2}\right)}{\sin\left(\frac{t-x+\frac{v}{n}}{2}\right)}\psi(x)\mathrm{d}x.$

Using the fact that the sine and cosine functions are bounded by some constant
$C$ on the complex strip $\left\\{z\in\mathbb{C}\leavevmode\nobreak\
\middle|\leavevmode\nobreak\ |\mathrm{Im}(z)|\leq 1\right\\}$ , we have

$\displaystyle\left|r_{n}\left(s+\frac{v}{n},t+\frac{v}{n}\right)\right|$
$\displaystyle\leq\frac{C\|\psi\|_{\infty}}{2\pi
n}\int_{\mathbb{T}}\left|\frac{\sin\left(\frac{n(s-x)+u}{2}\right)}{\sin\left(\frac{s-x+\frac{u}{n}}{2}\right)}\frac{\sin\left(\frac{n(t-x)+v}{2}\right)}{\sin\left(\frac{t-x+\frac{v}{n}}{2}\right)}\right|\mathrm{d}x.$

Let $\delta=\operatorname{dist}(s,t)/2$. Up to translating $s$ and $t$ by $\pm
2\pi$ and exchanging $s$ and $t$, we can assume that $\delta=\frac{t-s}{2}$.
We then make the change of variable $x=y+\frac{t+s}{2}$ to obtain

$\int_{\mathbb{T}}\left|\frac{\sin\left(\frac{n(s-x)+u}{2}\right)}{\sin\left(\frac{s-x+\frac{u}{n}}{2}\right)}\frac{\sin\left(\frac{n(t-x)+v}{2}\right)}{\sin\left(\frac{t-x+\frac{v}{n}}{2}\right)}\right|\mathrm{d}x=\int_{-\pi}^{\pi}\left|\frac{\sin\left(\frac{n(\delta+y)-u}{2}\right)}{\sin\left(\frac{\delta+y-\frac{u}{n}}{2}\right)}\frac{\sin\left(\frac{n(\delta-y)+v}{2}\right)}{\sin\left(\frac{\delta-y+\frac{v}{n}}{2}\right)}\right|\mathrm{d}y.$

This last integral splits into two integrals $I_{1}$ and $I_{2}$ defined by

$I_{1}=\int_{0}^{\pi}\left|\frac{\sin\left(\frac{n(\delta+y)-u}{2}\right)}{\sin\left(\frac{\delta+y-\frac{u}{n}}{2}\right)}\frac{\sin\left(\frac{n(\delta-y)+v}{2}\right)}{\sin\left(\frac{\delta-y+\frac{v}{n}}{2}\right)}\right|\mathrm{d}y\quad\text{and}\quad
I_{2}=\int_{-\pi}^{0}\left|\frac{\sin\left(\frac{n(\delta+y)-u}{2}\right)}{\sin\left(\frac{\delta+y-\frac{u}{n}}{2}\right)}\frac{\sin\left(\frac{n(\delta-y)+v}{2}\right)}{\sin\left(\frac{\delta-y+\frac{v}{n}}{2}\right)}\right|\mathrm{d}y.$

Both term can be treated the exact same way. We have by Hölder inequality with
$0<\alpha<1$,

$\displaystyle I_{1}$
$\displaystyle\leq\left(\int_{0}^{\pi}\left|\frac{\sin\left(\frac{n(\delta+y)-u}{2}\right)}{\sin\left(\frac{\delta+y-\frac{u}{n}}{2}\right)}\right|^{\frac{1}{1-\alpha}}\mathrm{d}y\right)^{1-\alpha}\left(\int_{0}^{\pi}\left|\frac{\sin\left(\frac{n(\delta-y)+v}{2}\right)}{\sin\left(\frac{\delta-y+\frac{v}{n}}{2}\right)}\right|^{\frac{1}{\alpha}}\mathrm{d}y\right)^{\alpha}.$
(32)

For the left integral in (32), we make use of the following inequalities,
which are consequences of inequalities (3.3), $|u|\leq 1$ and $\delta\geq
2/n$.

$\left|\sin\left(\frac{\delta+y-\frac{u}{n}}{2}\right)\right|\geq\frac{3}{10\pi}\left(\delta+y-\frac{\mathrm{Re}(u)}{n}\right)\geq\frac{3}{10\pi}\left(\frac{\delta}{2}+y\right),$

to get

$\displaystyle\left(\int_{0}^{\pi}\left|\frac{\sin\left(\frac{n(\delta+y)-u}{2}\right)}{\sin\left(\frac{\delta+y-\frac{u}{n}}{2}\right)}\right|^{\frac{1}{1-\alpha}}\mathrm{d}y\right)^{1-\alpha}$
$\displaystyle\leq
C\left(\int_{0}^{\infty}\frac{\mathrm{d}y}{\left(y+\frac{\delta}{2}\right)^{\frac{1}{1-\alpha}}}\right)^{1-\alpha}$
$\displaystyle\leq\frac{C^{\prime}}{\delta^{\alpha}}.$ (33)

For the right integral in (32), we make the change of variable
$x=n(\delta-y)+\mathrm{Re}(v)$. We also use the inequality

$|x+iy|\leq C|\sin(x+iy)|,$

valid for $x\in[-2\pi/3,2\pi/3]$ and $y\in[-1/4,1/4]$, and some positive
constant $C$.

$\displaystyle\left(\int_{0}^{\pi}\left|\frac{\sin\left(\frac{n(\delta-y)+v}{2}\right)}{\sin\left(\frac{\delta-y+\frac{v}{n}}{2}\right)}\right|^{\frac{1}{\alpha}}\mathrm{d}y\right)^{\alpha}$
$\displaystyle\leq
n^{-\alpha}\left(\int_{n(\delta-\pi)+\mathrm{Re}(v)}^{n\delta+\mathrm{Re}(v)}\left|\frac{\sin\left(\frac{x-i\,\mathrm{Im}(v)}{2}\right)}{\sin\left(\frac{x-i\,\mathrm{Im}(v)}{2n}\right)}\right|^{\frac{1}{\alpha}}\mathrm{d}y\right)^{\alpha}$
$\displaystyle\leq
2Cn^{1-\alpha}\left(\int_{-\infty}^{\infty}\left|\frac{\sin\left(\frac{x-i\,\mathrm{Im}(v)}{2}\right)}{x-i\,\mathrm{Im}(v)}\right|^{\frac{1}{\alpha}}\mathrm{d}y\right)^{\alpha}$
$\displaystyle\leq C^{\prime}n^{1-\alpha},$ (34)

where in the last inequality we used the fact that the integrand is uniformly
bounded in a neighborhood of the origin, and that $\frac{1}{\alpha}>1$ so the
integrand is also integrable near $\pm\infty$. Plugging estimates (33) and
(34) into inequality (32) we obtain

$\left|r_{n}\left(s+\frac{v}{n},t+\frac{v}{n}\right)\right|\leq\frac{C\|\psi\|_{\infty}}{2\pi
n}(I_{1}+I_{2})\leq\frac{2C\|\psi\|_{\infty}}{2\pi n}\left(C^{\prime
2}\frac{n^{1-\alpha}}{\delta^{\alpha}}\right)\leq\frac{C^{\prime\prime}}{(n\operatorname{dist}(s,t))^{\alpha}},$

where $C^{\prime\prime}$ is a positive constant independent of $s$, $t$ and
$n$. ∎

######  Remark 3.6.

If we assume the following Dini–Lipschitz condition on the spectral density
$\psi$, that is

$\int_{0}^{1}\frac{\omega_{\psi}(x)}{x}\mathrm{d}x<+\infty,$

then we can take $\alpha=1$ in Lemma 3.4, and we have the following inequality

$|r^{(a,b)}(s,t)|\leq
C\frac{n^{a+b-1}}{\operatorname{dist}(s,t)}\left[\|\psi\|_{\infty}+\int_{0}^{2\pi}\frac{\omega_{\psi}(x)}{x}\mathrm{d}x\right],$

Where $C$ is an explicit constant.

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