# CLT for real $\beta$-ensembles at high temperature††thanks: This project has
received funding from the European Research Council (ERC) under the European
Union Horizon 2020 research and innovation program (grant agreement No.
884584).

Charlie Dworaczek Guera111Université de Lyon, ENSL, CNRS, France
email<EMAIL_ADDRESS>Ronan Memin222Université de Lyon, ENSL,
CNRS, France
email<EMAIL_ADDRESS>
###### Abstract

We establish a central limit theorem for the fluctuations of the empirical
measure in the $\beta$-ensemble of dimension $N$ at a temperature proportional
to $N$ and with confining smooth potential. The space of test functions for
which the CLT holds includes $\mathcal{C}^{1}$, vanishing functions at
infinity. It is obtained by the inversion of an operator which is a
pertubation of a Sturm-Liouville operator. The method that we use is based on
a change of variables introduced in [BFG15] and in [Shc14].

###### Contents

  1. 1 Introduction and main result
  2. 2 Regularity of the equilibrium measure and Hilbert transform
  3. 3 Concentration inequality, proof of Theorem 1.5
  4. 4 Localization of the edge of a configuration
  5. 5 Laplace transform for smooth test functions, proof of Theorem 1.3
  6. 6 Inversion of $\mathcal{L}$
  7. 7 Regularity of the inverse of $\mathcal{L}$ and completion of the proof of Theorem 1.3
  8. A Appendix: proof of Theorem 6.2

##  1 Introduction and main result

The $\beta$-ensemble of dimension $N\geqslant 1$ with parameter $\beta>0$ and
potential $V$ is the probability measure on ${\mathbb{R}}^{N}$ given by

$d\mathbf{P}^{\beta,V}_{N}(x_{1},\ldots,x_{N})=\frac{1}{Z_{N}(V,\beta)}\prod_{i<j}|x_{i}-x_{j}|^{\beta}e^{-\sum_{i=1}^{N}V(x_{i})}dx_{1}\ldots
dx_{N}\,.$ (1)

The potential $V$ has to be chosen so that the partition function

$Z_{N}(V,\beta)=\int_{{\mathbb{R}}^{N}}\prod_{i<j}|x_{i}-x_{j}|^{\beta}e^{-\sum_{i=1}^{N}V(x_{i})}dx_{1}\ldots
dx_{N}$

is finite. This is the case for example if for some
$\beta^{\prime}>\max(1,\beta)$,

$\liminf_{|x|\to\infty}\frac{V(x)}{N\beta^{\prime}\ln|x|}>1\,,$ (2)

see [AGZ10, equation (2.6.2)]. The parameter $\beta$, which is allowed to
depend on $N$, is the so-called inverse temperature.

Under the special choice of $V_{G}(x)=\dfrac{x^{2}}{2}$, the measure (1) can
be seen as the joint law of the (unordered) eigenvalues of certain matrix
models:

  * •

For $\beta=1$ (resp. $\beta=2$), it is the law of the eigenvalues of the
Gaussian Orthogonal Ensemble (resp. Gaussian Unitary Ensemble),
[AGZ10][Theorem 2.5.2].

  * •

For general $\beta>0$, potentially depending on $N$, it is the law of the
spectrum of certain tri-diagonal random matrices as shown by Dumitriu and
Edelman in [DE02].

We consider here the high temperature regime where $\beta$ scales as $1/N$,
and write $\beta=\frac{2P}{N}$ for some $P>0$. The corresponding measure is
therefore

$d\mathbb{P}^{V,P}_{N}(x_{1},\ldots,x_{N})=\frac{1}{Z_{N}^{V,P}}\prod_{i<j}|x_{i}-x_{j}|^{\frac{2P}{N}}e^{-\sum_{i=1}^{N}V(x_{i})}dx_{1}\ldots
dx_{N}\,,$ (3)

with partition function

$Z_{N}^{V,P}=\int_{{\mathbb{R}}^{N}}\prod_{i<j}|x_{i}-x_{j}|^{\frac{2P}{N}}e^{-\sum_{i=1}^{N}V(x_{i})}dx_{1}\ldots
dx_{N}\,.$ (4)

It was shown in [GZ19] that under $\mathbb{P}^{V,P}_{N}$, the sequence of
empirical measures

$\hat{\mu}_{N}=\frac{1}{N}\sum_{i=1}^{N}\delta_{x_{i}}$

satisfies a large deviation principle at speed $N$ with strictly convex, good
rate function. As a consequence, $\hat{\mu}_{N}$ converges almost surely in
distribution towards a deterministic measure $\mu^{V}_{P}$ as $N$ goes to
infinity, meaning that almost surely, for every bounded continuous
$f:{\mathbb{R}}\to{\mathbb{R}}$,

$\int_{\mathbb{R}}fd\hat{\mu}_{N}\underset{N\rightarrow\infty}{\longrightarrow}\int_{\mathbb{R}}fd\mu^{V}_{P}\,.$

The limiting measure $\mu^{V}_{P}$ can be seen to have a density
$\rho^{V}_{P}$ which satisfies for almost every $x\in{\mathbb{R}}$

$V(x)-2P\int_{\mathbb{R}}\ln|x-y|\rho^{V}_{P}(y)dy+\ln\rho^{V}_{P}(x)=\lambda^{V}_{P}\,,$
(5)

where $\lambda^{V}_{P}$ is constant (see [GM22, Lemma 3.2] for example).

The $\beta$-ensemble in the regime $\beta
N\underset{N\rightarrow\infty}{\longrightarrow}2P>0$ has drawn a lot of
attention from the random matrix and statistical physics communities lately.
This regime was first considered by [CL97] with the study of Dyson Brownian
motion with vanishing repulsive coefficient scaled like $\dfrac{1}{N}$. Gases
of vortices were also studied with temperature proportional to $N$ in [BG99].
The limiting density was then described in the case of the quadratic potential
in [ABG12], as a crossover between the Wigner semicircle law (fixed $\beta>0$
case) and the Gaussian density (case $\beta=0$). The fluctuations of the
eigenvalues in the bulk and at the edge of a configuration were studied for
example in [BGP15],[NT18],[NT20],[Pak18], [Lam21]. These fluctuations were
shown to be described by Poisson statistics in this regime. Recently, Spohn
uncovered in [Spo20] a link between the study of the Classical Toda chain and
the $\beta$-ensemble in the high temperature regime, showing that the limiting
density of states of the classical Toda chain, distributed according to the
generalized Gibbs ensemble with polynomial potential, can be computed by means
of the limiting empirical measure of the $\beta$-ensemble at high temperature.
In [Maz22], the author established this relation using the matrix
representation of the $\beta$-ensemble and a moment method, and in [GM22] the
authors proved a large deviation principle for the empirical measure of the
Toda chain, establishing the previous result for potentials with polynomial
growth. See also [Spo22],[GM23],[MM22] for a similar link between the
Ablowitz-Ladik lattice and the circular $\beta$-ensemble at high temperature.
This relation can be further pushed to compute the limiting currents of the
Toda chain through the central limit theorem for the empirical measure in the
$\beta$ ensemble. The computation of these currents is a crucial step to the
derivation of a hydrodynamic equation for the Toda chain, and to the analysis
of the correlations of the locally conserved quantities at equilibrium through
linearized hydodynamics, see [Spo21].

The Central Limit Theorem for the fluctuations of the linear statistics of
$\beta$-ensembles was first established by [Joh98] for $\beta=2$ polynomial
potential, then generalized and further developed in the regime where $\beta$
is fixed in [Shc13], [BG13a], [BG13b],[BLS18],[LLW19]. Also an optimal local
law was found in this regime in [BMP22]. The CLT was obtained in the high-
temperature regime $\beta N\to 2P>0$ by Nakano and Trinh in [NT18, Theorem
4.9] for quadratic $V$, relying on the tridiagonal representation for the
$\beta$-ensemble with quadratic potential in [DE02]. In [HL21], the authors
prove the CLT in the case of the circular $\beta$-ensemble at high temperature
with general potential, using a normal approximation method involving the
spectral analysis of an operator associated to the limiting covariance
structure. Their method allowed them to derive a Berry-Esseen bound, i.e. a
speed of convergence of the fluctuations towards a Gaussian variable.

In this paper, we adapt part of the arguments of [HL21] to our setup. More
precisely, we show that for a class of regular, convex potentials $V$
satisfying a growth condition of the type

$\lim_{|x|\to\infty}\frac{V^{\prime\prime}(x)}{V^{\prime}(x)^{2}}=0\,,$

denoting $\nu_{N}=\hat{\mu}_{N}-\mu^{V}_{P}$ and considering test functions
$f$ belonging to the range of a certain integro-differential operator, the
scaled fluctuations of $\hat{\mu}_{N}$, defined by

$\sqrt{N}\nu_{N}(f):=\sqrt{N}\left(\int_{\mathbb{R}}fd\mu_{N}-\int_{\mathbb{R}}fd\mu^{V}_{P}\right)\,,$

converge in law towards centered Gaussian law with variance depending on $f$.

When considering the fixed temperature regime, i.e. $\beta$ fixed, one has to
renormalize the $x_{i}$’s by $\sqrt{N}$. It is shown in [AGZ10][Theorem 2.6.1]
that the measure

$\frac{1}{N}\sum_{i=1}^{N}\delta_{x_{i}/\sqrt{N}}$

satisfies a large deviation principle, and the limiting measure is
characterized in [AGZ10][Lemma 2.6.2] by an equation similar to (5). In fact,
the term $\ln\rho^{V}_{P}$ in the left-hand side of (5) is the only difference
in the equation characterizing the limiting measure in the fixed $\beta$ case.
We point out the very similar characterization of the equilibrium measure
corresponding to the minimization problem arising in [BGK16]. There again, the
limiting measure is compactly supported. The term $\ln\rho^{V}_{P}$ is of
prime importance because its presence implies that the support of
$\rho_{P}^{V}$ is the whole real line. It leads to technicalities to deal with
the behavior at infinity of most of the associated objects, namely dealing
with weighted Lebesgue spaces $L^{2}(\mu_{P}^{V})$ and the corresponding
Sobolev spaces $H^{k}(\mu_{P}^{V})$.

Our strategy is based on a change of variables in the partition function
$Z^{V,P}_{N}$ (4), used for the $\beta$-ensemble at fixed temperature
introduced in [BFG15] and [Shc14], and used in [Gui19] and in [BGK16] to
derive the loop equations and in [BLS18] to derive a CLT in the
$\beta$-ensemble with $\beta$ fixed. The outline of the argument goes as
follows: Take $\phi:{\mathbb{R}}\to{\mathbb{R}}$ smooth, vanishing fast enough
at infinity, and do the change of variables in $Z_{N}^{V,P}$,
$x_{i}=y_{i}+\frac{t}{\sqrt{N}}\phi(y_{i})$, $1\leqslant i\leqslant N$, to get

$Z_{N}^{V,P}=\int_{{\mathbb{R}}^{N}}\prod_{i<j}\left|y_{i}-y_{j}+\frac{t}{\sqrt{N}}(\phi(y_{i})-\phi(y_{j}))\right|^{2P/N}e^{-\sum_{i=1}^{N}V\left(y_{i}+\frac{t}{\sqrt{N}}\phi(y_{i})\right)}\prod_{i=1}^{N}\left(1+\frac{t}{\sqrt{N}}\phi^{\prime}(y_{i})\right)d^{N}\mathbf{y}\,.$

Expanding the different terms in this integral, one gets

$\displaystyle
Z_{N}^{V,P}=\int_{{\mathbb{R}}^{N}}\prod_{i<j}|y_{i}-y_{j}|^{\frac{2P}{N}}e^{-\sum_{i=1}^{N}V(y_{i})}e^{\frac{t}{\sqrt{N}}\left[\frac{2P}{N}\sum_{i<j}\frac{\phi(y_{i})-\phi(y_{j})}{y_{i}-y_{j}}+\sum_{i=1}^{N}\left(\phi^{\prime}(y_{i})-V^{\prime}(y_{i})\phi(y_{i})\right)\right]}e^{-\frac{t^{2}}{2}\sigma^{2}_{N}(\phi)}d^{N}\mathbf{y}\,,$

where the term $\sigma_{N}^{2}(\phi)$ converges towards a limiting variance
$\sigma^{2}(\phi)$ depending on $\phi$, $P$ and $V$. After dividing both parts
of the equation by $Z_{N}^{V,P}$, and because of equation (5) characterizing
$\mu^{V}_{P}$, one can deduce from the last equation the convergence of the
Laplace transform

$\mathbb{E}\left[e^{t\sqrt{N}\left(\nu_{N}(\Xi\phi)+\text{error
term}\right)}\right]\underset{N\rightarrow\infty}{\longrightarrow}\exp\Big{(}\frac{t^{2}}{2}\sigma^{2}(\phi)\Big{)}\,,$
(6)

where $\Xi$ is a linear operator acting on test functions and defined by

$(\Xi\phi)(x)=2P\int_{\mathbb{R}}\frac{\phi(x)-\phi(y)}{x-y}d\mu^{V}_{P}(y)+\phi^{\prime}(x)-V^{\prime}(x)\phi(x)\,.$
(7)

Once the error term is taken care of, (6) shows the central limit theorem for
test functions of the form $\Xi\phi$. Following [HL21], the operator
$\mathcal{L}$ given by

$\mathcal{L}\phi=\Xi\phi^{\prime}$ (8)

can be analyzed using Hilbert space techniques. In particular, the operator
$\mathcal{L}$, seen as an unbounded operator of the Hilbert space

$\mathsf{H}=\left\\{u\in L^{2}(\mu^{V}_{P})\ \Big{|}\ u^{\prime}\in
L^{2}(\mu^{V}_{P}),\int_{\mathbb{R}}u\rho^{V}_{P}dx=0\right\\},\qquad\Braket{u,v}_{\mathsf{H}}=\Braket{u^{\prime},v^{\prime}}_{L^{2}(\rho^{V}_{P})}\,,$

can be decomposed as

$-\mathcal{L}=\mathcal{A}+2P\mathcal{W}\,,$

where $\mathcal{A}$ is a positive Sturm-Liouville operator and $\mathcal{W}$
is positive and self-adjoint. Such a writing allows us to show that
$-\mathcal{L}$ is invertible, see Theorem 6.7.

We now state the assumptions we make on the potential $V$. Recall that a
probability measure $\mu$ supported on ${\mathbb{R}}$ satisfies the Poincaré
inequality if there exists $C>0$ such that for all
$f\in\mathcal{C}^{1}({\mathbb{R}})$ with compact support:

$\mathrm{Var}_{\mu}(f):=\int_{\mathbb{R}}\left(f-\int_{\mathbb{R}}fd\mu\right)^{2}d\mu\leqslant
C\int f^{\prime 2}d\mu\,.$ (9)

######  Assumptions 1.1.

The potential $V$ satisfies:

  * i)

$V\in\mathcal{C}^{3}({\mathbb{R}})$,
$V(x)\underset{|x|\rightarrow+\infty}{\longrightarrow}+\infty$,
$|V^{\prime}(x)|\underset{|x|\rightarrow+\infty}{\longrightarrow}+\infty$ and
is such that $\mu_{P}^{V}$ satisfies the Poincaré inequality (9).

  * ii)

For all polynomial $Q\in{\mathbb{R}}[X]$ and $\alpha>0$,
$Q\bigg{(}V^{\prime}(x)\bigg{)}e^{-V(x)}=\underset{|x|\to\infty}{o}(x^{-\alpha})$
.

  * iii)

Furthermore, for any sequence $x_{N}$ such that $|x_{N}|$ goes to infinity,
and for all real $a<b$, we have, as $N$ goes to infinity,

$\frac{1}{V^{\prime}(x_{N})^{2}}\sup_{a\leqslant x\leqslant
b}|V^{\prime\prime}(x_{N}+x)|\underset{N\rightarrow\infty}{\longrightarrow}0\,.$

  * iv)

The function $\dfrac{1}{V^{\prime 2}}$ is integrable at infinity.
$\dfrac{V^{\prime\prime}(x)}{V^{\prime}(x)}=\underset{|x|\rightarrow\infty}{O}(1)$
and $\dfrac{V^{(3)}(x)}{V^{\prime}(x)}=\underset{|x|\rightarrow\infty}{O}(1)$.

Taking $V=V_{\mathrm{conv}}+\phi$ with
$V_{\mathrm{conv}},\,\phi\in\mathcal{C}^{3}({\mathbb{R}})$ such that
$\phi^{(k)}$ is bounded for $k=0,\dots,3$, $V_{\mathrm{conv}}$ is convex with
$|V_{\mathrm{conv}}^{\prime}|\to+\infty$ at infinity, satisfying hypotheses
ii), iii) and iv) such that there exists $\varepsilon>0$ such that
$V_{\mathrm{conv}}-2Pf_{\varepsilon}$ is convex (see Lemma 2.4), then $V$
satisfies Assumptions 1.1.

Because i) implies that $V$ goes to infinity faster than linearly, we will see
that it ensures exponential decay at infinity of $\rho^{V}_{P}$. Recalling the
sufficient condition for $\mathbb{P}^{V,P}_{N}$ of equation (2) to be defined,
this first assumption implies that there exists $\alpha>0$ such that
$\liminf_{|x|\to\infty}\frac{V(x)}{|x|}>\alpha$. This guarantees in particular
that the $\beta$-ensemble (3) is well-defined for all $N\geqslant 1$ and
$P\geqslant 0$. We will use the fact that $\mu_{P}^{V}$ satisfies the Poincaré
inequality to ensure that $\mathsf{H}$ endowed with
$\braket{\cdot,\cdot}_{\mathsf{H}}$ is a Hilbert space.

The second assumption ensures that any power of $V^{\prime}$ and
$V^{\prime\prime}$ is in $L^{2}(\mu_{P}^{V})$ and $\rho_{P}^{V}$, which
behaves like $e^{-V}$ up to a sub-exponential factor, belongs to the Sobolev
space $H^{2}({\mathbb{R}})\subset\mathcal{C}^{1}({\mathbb{R}})$. Indeed, for
$k\leqslant 2$, using iv), $\rho_{P}^{V}\,{}^{(k)}$ behaves at infinity like
$(V^{\prime})^{k}\rho_{P}^{V}$ as shown in Lemma 2.2 which is in
$L^{2}({\mathbb{R}})$ by assumption ii).

Assumption iii) will be used to localize the minimum/maximum point of a
typical configuration $(x_{1},\ldots,x_{N})$ following the law
$\mathbb{P}^{V,P}_{N}$: this will be done in Corollary 4.2, which comes as a
consequence of [Lam21][Theorem 3.4]. More precisely, Corollary 4.2 establishes
that for sequences $(\alpha_{N}^{+})_{N},(\alpha_{N}^{-})_{N}$ going to
infinity, the random variables

$\alpha_{N}^{+}\left(\max_{1\leqslant j\leqslant
N}x_{j}-E_{N}^{+}\right)\qquad\text{and
}\qquad\alpha_{N}^{-}\left(\max_{1\leqslant j\leqslant
N}x_{j}-E_{N}^{-}\right)$

converge in distribution. For large $N$, the scalars $E_{N}^{+}$ and
$E_{N}^{-}$ can thus be seen as the edges of a typical configuration.
Furthermore,

$V(E_{N}^{\pm})\sim\ln N\,.$ (10)

We refer to Section 4 for detailed statements. The final step in the proof of
Theorem 1.3 consists in lifting the result of Proposition 5.1 from compactly
supported functions to more general functions.

We use Assumption iv) to control integral remainders in the proof of Theorem
7.1, ensuring that $\mathcal{L}^{-1}$ is regular enough i.e. that for
sufficiently smooth functions $f$,
$\Big{(}\mathcal{L}^{-1}f\Big{)}^{\prime}\in H^{2}({\mathbb{R}})$.

We will need another technical assumption to ensure that Taylor remainders
arising in the proof of Theorem 5.2 are negligible.

######  Assumption 1.2.

With the notations of Theorem 4.1, we have

$\sup_{d(x,I_{N})\leqslant 1}\left|V^{(3)}(x)\right|=o(N^{1/2})\,,$

where $I_{N}=\left[E_{N}^{-}-2;E_{N}^{+}+2\right]$.

Again taking $V=V_{\mathrm{conv}}+\phi$ with
$V_{\mathrm{conv}},\,\phi\in\mathcal{C}^{3}({\mathbb{R}})$ such that
$\phi^{(k)}$ is bounded for $k=0,\dots,3$, $V_{\mathrm{conv}}$ is convex with
$|V_{\mathrm{conv}}^{\prime}|\to+\infty$ at infinity, satisfying hypotheses
ii), iii), iv) and Assumption 1.2, such that there exists $\varepsilon>0$ such
that $V_{\mathrm{conv}}-2Pf_{\varepsilon}$ is convex (see Lemma 2.4), then $V$
satisfies Assumptions 1.1.

This last hypothesis is satisfied whenever $V_{\mathrm{conv}}$ is sufficiently
convex in a compact centered at 0 and is made so that $V_{\mathrm{conv}}$
compensate the small lack of concavity in the bulk of a function behaving like
$-\ln|x|$ at infinity (note that assuming that
$V^{\prime\prime}\geqslant\alpha$ for some $\alpha>0$ is sufficient). This is
the reason why we introduce the function $f_{\varepsilon}$, which are
functions with required growth at infinity and with second derivative as small
as desired. The main point that needs to be checked is that the measure
$\mu^{V}_{P}$ satisfies the Poincaré inequality, this will be done in
Proposition 2.6.

The type of potential $V_{\mathrm{conv}}$ that one can consider is typically
the convex polynomials or $\cosh(\alpha x)$. On the other hand a scaled
potentials like $e^{x^{2}}$ which have a faster growing derivative at infinity
and so doesn’t satisfy assumptions iii) and iv).

We are now able to state the main result, ie the central limit theorem for
functions belonging to the image of the operator $\mathcal{L}$ introduced in
(8).

######  Theorem 1.3.

Assume that $V$ satisfies Assumptions 1.1 and Assumption 1.2. Then for $\phi$
verifying the following conditions:

  * •

$\phi\in\mathcal{C}^{1}({\mathbb{R}})$

  * •

there exists $\varepsilon>0$,
$\phi(x)=\underset{|x|\rightarrow\infty}{O}(x^{-\frac{1}{2}-\varepsilon})$ and
$\phi^{\prime}(x)=\underset{|x|\rightarrow\infty}{O}(x^{\frac{1}{2}-\varepsilon})$
at infinity

  * •

$\displaystyle\int_{\mathbb{R}}\phi(x)d\mu_{P}^{V}(x)=0$

we have the convergence in law

$\sqrt{N}\nu_{N}(\phi)\to\mathcal{N}\Bigg{(}0,(\sigma^{V}_{P})^{2}(\phi)\Bigg{)}$
(11)

where the limiting variance $(\sigma^{V}_{P})^{2}(\phi)$ is given by

$(\sigma^{V}_{P})^{2}(\phi)=\braket{\phi,\mathcal{L}^{-1}\phi}_{\mathsf{H}}=\int_{\mathbb{R}}\Bigg{(}\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime\prime}(x)^{2}+V^{\prime\prime}(x)\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime}(x)^{2}\Bigg{)}d\mu^{V}_{P}(x)\\\
+P\iint_{{\mathbb{R}}^{2}}\Bigg{(}\frac{\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime}(x)-\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime}(y)}{x-y}\Bigg{)}^{2}d\mu^{V}_{P}(x)d\mu^{V}_{P}(y)\,.$
(12)

######  Remark 1.4.

Since $\nu_{N}(\phi+c)=\nu_{N}(\phi)$ for all constant $c\in{\mathbb{R}}$, the
assumption $\displaystyle\int_{\mathbb{R}}\phi(x)d\mu^{V}_{P}=0$ can be
dropped by replacing $\phi$ by
$\phi-\displaystyle\int_{\mathbb{R}}\phi(x)d\mu^{V}_{P}$ in the expression of
the limiting variance.

As a tool to deal with the error term of equation (6), we establish a
concentration inequality for the empirical measure. This inequality is stated
in terms of the following distance over the set of probability distributions
$\mathcal{P}({\mathbb{R}})$.

For $\mu,\mu^{\prime}\in\mathcal{P}({\mathbb{R}})$ we define the distance

$d(\mu,\mu^{\prime})=\sup_{\begin{subarray}{c}\|f\|_{\text{Lip}}\leqslant 1\\\
\|f\|_{1/2}\leqslant 1\end{subarray}}\left\\{\left|\int fd\mu-\int
fd\mu^{\prime}\right|\right\\}\,,$ (13)

where $\|f\|_{\text{Lip}}$ denotes the Lipschitz constant of $f$, and
$\|f\|_{1/2}^{2}=\displaystyle\int_{\mathbb{R}}|t|\left|\mathcal{F}[f](t)\right|^{2}dt$,
where $\mathcal{F}$ denotes the Fourier transform on $L^{2}({\mathbb{R}})$
which takes the following expression
$\mathcal{F}[f](t)=\displaystyle\int_{\mathbb{R}}f(x)e^{-\mathrm{i}tx}dx$ for
$f$ in $L^{1}({\mathbb{R}})\cap L^{2}({\mathbb{R}})$.

We then have

######  Theorem 1.5.

There exists $K\in{\mathbb{R}}$ (depending on $P$ and on $V$), such that for
any $N\geqslant 1$ and $r>0$,

$\mathbb{P}^{V,P}_{N}\left(d(\hat{\mu}_{N},\mu^{V}_{P})>r\right)\leqslant
e^{-Nr^{2}\frac{P\pi^{2}}{2}+5P\ln N+K}\,.$ (14)

This result is the analog of [HL21, Theorem 1.4].

The paper is organized as follows. In Section 2 we discuss the regularity of
the equilibrium density $\rho^{V}_{P}$ under Assumption 1.1. In Section 3 we
prove Theorem 1.5. Section 4 is dedicated to the localization of the edge of a
typical configuration, mentioned in the discussion preceding the statement of
Assumption 1.2. We next prove in Section 5 the convergence of the Laplace
transform of $\sqrt{N}\nu_{N}(\mathcal{L}\phi)$ for general functions $\phi$
which establishes Theorem 1.3 for functions of the form $\mathcal{L\phi}$.
Section 6 is dedicated to the diagonalization and inversion of $\mathcal{L}$
given by (8). In Section 7, we show regularity properties of
$\mathcal{L}^{-1}$ to establish Theorem 1.3. We detail in Appendix A elements
of proof for the spectral theory of Schrödinger operators, used in Section 6.

Acknowledgments The authors wish to thank Alice Guionnet and Karol Kozlowski
for their helpful suggestions. We also thank Arnaud Debussche for pointing out
the link with Schrödinger operators theory and Gautier Lambert for pointing
out [Lam21]. We would also like to thank Jeanne Boursier, Corentin Le Bihan
and Jules Pitcho for their intuition about the regularity of the inverse
operator. We would like to thank Jean-Christophe Mourrat for telling us about
a more general framework for Poincaré inequalities.

##  2 Regularity of the equilibrium measure and Hilbert transform

In this section, we discuss the regularity properties of the equilibrium
density $\rho^{V}_{P}$, namely its decay at infinity and its smoothness, and
give formulas for its two first derivatives.

The Hilbert transform, whose definition we recall, plays a central role in the
analysis of the equilibrium measure. It is first defined on the Schwartz class
through $\forall\phi\in\mathcal{S}({\mathbb{R}}),\,\forall
x\in{\mathbb{R}},\;$

$\mathcal{H}[\phi](x):=\fint_{\mathbb{R}}\dfrac{\phi(t)}{t-x}dt=\lim_{\varepsilon\downarrow
0}\int_{|t-x|>\varepsilon}\dfrac{\phi(t)}{t-x}dt=\int_{0}^{+\infty}\dfrac{\phi(x+t)-\phi(x-t)}{t}dt,$
(15)

where $\displaystyle\fint$ denotes the Cauchy principal value integral, and
then extended to $L^{2}({\mathbb{R}})$ thanks to property ii) of Lemma 2.1:
$\|f\|_{L^{2}(dx)}=\dfrac{1}{\pi}\|\mathcal{H}[f]\|_{L^{2}(dx)}$. The last
expression in (15) is a definition where the integral converges in the
classical sense. We also recall the definition of the logarithmic potential
$U^{f}$ of a density of probability $f:{\mathbb{R}}\to{\mathbb{R}}$, given for
$x\in{\mathbb{R}}$ by

$U^{f}(x)=-\int_{\mathbb{R}}\ln|x-y|f(y)dy\,.$ (16)

Because we assume $f\in L^{1}({\mathbb{R}})$ to be nonnegative, $U^{f}$ takes
values in $[-\infty,+\infty)$. If $f$ integrates the function $\ln$, i.e
$\int_{\mathbb{R}}\ln|x|f(x)dx<+\infty$, then $U^{f}$ takes real values.
Additionally, one can check that the logarithmic potential and the Hilbert
transform of $f$ are linked through the distributional identity
$\big{(}U^{f}\big{)}^{\prime}=\mathcal{H}[f]$.

We recall in the next lemma some properties of the Hilbert transform that we
will use in the rest of the paper.

######  Lemma 2.1 (Properties of the Hilbert transform).

  * i)

Fourier transform: For all $\phi\in L^{2}({\mathbb{R}})$,
$\mathcal{F}\Big{[}\mathcal{H}[\phi]\Big{]}(\omega)=\mathrm{i}\pi\text{sgn}(\omega)\mathcal{F}[\phi](\omega)$
for all $\omega\in{\mathbb{R}}$.

  * ii)

As a consequence, $\dfrac{1}{\pi}\mathcal{H}$ is an isometry of
$L^{2}({\mathbb{R}})$, and $\mathcal{H}$ satisfies on $L^{2}({\mathbb{R}})$
the identity $\mathcal{H}^{2}=-\pi^{2}I$.

  * iii)

Derivative: For any $f\in H^{1}({\mathbb{R}})$, $\mathcal{H}[f]$ is also
$H^{1}({\mathbb{R}})$ and $\mathcal{H}[f]^{\prime}=\mathcal{H}[f^{\prime}]$.

  * iv)

For all $p>1$, the Hilbert transform can be extended as a bounded operator
$\mathcal{H}:L^{p}({\mathbb{R}})\to L^{p}({\mathbb{R}})$.

  * v)

Skew-self adjointness: For any $f,g\in L^{2}({\mathbb{R}})$,
$\Braket{\mathcal{H}[f],g}_{L^{2}({\mathbb{R}})}=-\langle
f,\mathcal{H}[g]\rangle_{L^{2}({\mathbb{R}})}$.

###### Proof.

We refer to [Kin09] for the proofs of these properties. ∎

As a consequence of [GZ19], $\hat{\mu}_{N}$ converges almost surely under
$\mathbb{P}^{V,P}_{N}$ towards the unique minimizer of the energy-functional
$\mathcal{E}_{P}^{V}$, defined for $\mu\in\mathcal{P}({\mathbb{R}})$ by

$\mathcal{E}_{P}^{V}(\mu)=\begin{cases}\displaystyle\int_{\mathbb{R}}\Big{[}V+\ln\Big{(}\dfrac{d\mu}{dx}\Big{)}\Big{]}d\mu-P\iint_{{\mathbb{R}}^{2}}\ln\big{|}x-y\big{|}d\mu(x)d\mu(y)\text{
if }\mu\ll dx\\\ +\infty\text{ otherwise }\end{cases}\,.$ (17)

(Here we wrote $\mu\ll dx$ for "$\mu$ is absolutely continuous with respect to
Lebesgue measure")

Consequently, following [GM22, Lemma 3.2], the density $\rho^{V}_{P}$ of
$\mu^{V}_{P}$ satisfies equation (5), which we rewrite here for convenience.

$V(x)-2P\int_{\mathbb{R}}\ln|x-y|\rho^{V}_{P}(y)dy+\ln\rho^{V}_{P}(x)=\lambda^{V}_{P}\,,$
(18)

where $\lambda^{V}_{P}$ is a constant (depending on $V$ and $P$). Using this
equation, we will show in the next lemma that $\rho^{V}_{P}$ decays
exponentially and is twice continuously differentiable via the representation:

$\forall
x\in{\mathbb{R}},\;\rho^{V}_{P}(x)=\exp\Big{(}-V(x)-2PU^{\rho_{P}^{V}}(x)-\lambda_{P}^{V}\Big{)}$

In the Gaussian potential case ie $V_{G}(x)=\dfrac{x^{2}}{2}$, an explicit
formula has been found [ABG12]:

$\rho_{P}^{V_{G}}(x)=\dfrac{\Gamma(P)}{P\sqrt{2\pi}}\dfrac{\exp\Big{(}-\dfrac{x^{2}}{2}\Big{)}}{\displaystyle\int_{0}^{+\infty}t^{P-1}e^{-\frac{t^{2}}{2}+ixt}dt}$

It has been established in [BGP15] that
$\sqrt{P+1}\rho_{P}^{V_{G}}(\sqrt{P+1}x)$ converges to the Gaussian
distribution when $P$ goes to zero and the semi-circle law when $P$ goes to
infinity. So in the Gaussian case, $\mu_{P}$ can be seen as an interpolation
between the Gaussian distribution and the semi-circular one. We now drop the
superscript of $\rho^{V}_{P}$ and $\mu_{P}^{V}$ and denote it $\rho_{P}$ and
$\mu_{P}$ for convenience. In the next lemma, we prove that $\rho_{P}$ has the
same regularity as $V$.

######  Lemma 2.2.

Under Assumption 1.1,

  * •

The support of $\mu_{P}$ is ${\mathbb{R}}$ and there exists a constant
$C^{V}_{P}$ such that for all $x\in{\mathbb{R}}$,

$\rho_{P}(x)\leqslant C^{V}_{P}(1+|x|)^{2P}e^{-V(x)}\,.$

  * •

The density $\rho_{P}$ is in $\mathcal{C}^{3}({\mathbb{R}})$ and we have

$\rho_{P}^{\prime}=-\Big{(}V^{\prime}+2P\mathcal{H}[\rho_{P}]\Big{)}\rho_{P}$
(19)

and

$\rho_{P}^{\prime\prime}=\Big{(}-2P\mathcal{H}[\rho_{P}]^{\prime}-V^{\prime\prime}+V^{\prime
2}+4P^{2}\mathcal{H}[\rho_{P}]^{2}+4PV^{\prime}\mathcal{H}[\rho_{P}]\Big{)}\rho_{P}\,.$
(20)

###### Proof.

For the first point, [GM22, Lemma 3.2] establishes that the support of
$\mu_{P}$ is the whole real axis, and that under the first condition of
Assumptions 1.1, we have the bound, valid for all $x\in{\mathbb{R}}$

$\rho_{P}(x)\leqslant\frac{K^{V}_{P}}{(1+|x|)^{2}}\,,$ (21)

with $K^{V}_{P}$ a positive constant. Using (18) and the fact that

$\ln|x-y|\leqslant\ln\big{(}1+|x|\big{)}+\ln\big{(}1+|y|\big{)}\,,$

we see that for all $x\in{\mathbb{R}}$,

$\rho_{P}(x)\leqslant C^{V}_{P}\exp\Big{(}-V(x)+2P\ln(1+|x|)\Big{)}\,,$ (22)

with

$C^{V}_{P}=\exp\Big{(}2P\int_{\mathbb{R}}\ln(1+|y|)\rho_{P}(y)dy+\lambda_{P}^{V}\Big{)}$

which is indeed finite by (21).

For the second point, we use that
$\big{(}U^{\rho_{P}}\big{)}^{\prime}=\mathcal{H}[\rho_{P}]$ weakly and
equation (18) to conclude on the distributional identity

$\rho_{P}^{\prime}=\Big{(}-V^{\prime}-2P\mathcal{H}[\rho_{P}]\Big{)}\rho_{P}\,.$

By the second point of Assumption 1.1,
$V^{\prime}(x)e^{-V(x)+2P\ln(1+|x|)}=o(x^{-1})$ as $|x|\rightarrow\infty$,
thus by (22), $V^{\prime}\rho_{P}\in L^{2}({\mathbb{R}})$. Also since
$\rho_{P}$ is $L^{2}({\mathbb{R}})$ and bounded, we deduce, by using that
$\mathcal{H}\big{[}L^{2}({\mathbb{R}})\big{]}=L^{2}({\mathbb{R}})$, that
$\mathcal{H}[\rho_{P}]\rho_{P}\in L^{2}({\mathbb{R}})$. Adding up these terms
we get $\rho_{P}\in H^{1}({\mathbb{R}})$. Because
$\mathcal{H}[\rho_{P}]^{\prime}=\mathcal{H}[\rho_{P}^{\prime}]$ in a weak
sense by Lemma 2.1, $\mathcal{H}[\rho_{P}]\in H^{1}({\mathbb{R}})$. By the
classical fact that $H^{1}({\mathbb{R}})$ is contained in the set of
$1/2$-Hölder functions $\mathcal{C}^{1/2}({\mathbb{R}})$, we have
$\mathcal{H}[\rho_{P}]\in\mathcal{C}^{1/2}({\mathbb{R}})$ and so
$U^{\rho_{P}}\in\mathcal{C}^{1,1/2}({\mathbb{R}})$, the set of functions in
$\mathcal{C}^{1}({\mathbb{R}})$ with derivative of class $1/2$-Hölder.
Using the fact that $V$ is continuously differentiable, the previous equation
for the weak derivative of $\rho_{P}$ then ensures that
$\rho_{P}\in\mathcal{C}^{1}({\mathbb{R}})$ and equation (19) holds in the
strong sense.

Differentiating (in a weak sense) equation (19) we obtain

$\rho_{P}^{\prime\prime}=\Big{(}-2P\mathcal{H}[\rho_{P}]^{\prime}-V^{\prime\prime}+V^{\prime
2}+4P^{2}\mathcal{H}[\rho_{P}]^{2}+4PV^{\prime}\mathcal{H}[\rho_{P}]\Big{)}\rho_{P}\,.$

The three first terms belong to $L^{2}({\mathbb{R}})$ for the same reasons as
before. Since $\rho_{P}\in H^{1}({\mathbb{R}})$n by Lemma 2.1iii) so is
$\mathcal{H}[\rho_{P}]\in H^{1}({\mathbb{R}})$, it is then bounded over
${\mathbb{R}}$ hence the two last term are in $L^{2}({\mathbb{R}})$ when
multiplied by $\rho_{P}$. Finally, we can conclude that $\rho_{P}\in
H^{2}({\mathbb{R}})$ and so that $\mathcal{H}[\rho_{P}]\in
H^{2}({\mathbb{R}})$ with
$\mathcal{H}[\rho_{P}]^{\prime\prime}=\mathcal{H}[\rho_{P}^{\prime\prime}]$
(in a weak sense). As before, we conclude that
$\rho_{P}\in\mathcal{C}^{2}({\mathbb{R}})$ and that equation (20) holds in a
strong sense. By the exact same method, we can show that
$\rho_{P}\in\mathcal{C}^{3}({\mathbb{R}})$. ∎

We next show that the Hilbert transform of $\rho_{P}$ is continuous and decays
at infinity.

######  Lemma 2.3.

Let $u\in L^{2}({\mathbb{R}})$ such that $\int_{\mathbb{R}}u(t)dt$ exists and
$f:t\mapsto tu(t)\in H^{1}({\mathbb{R}})$ then

$\mathcal{H}[u](x)\underset{|x|\rightarrow\infty}{\sim}\displaystyle\dfrac{-\int_{\mathbb{R}}u(t)dt}{x}.$

Moreover if $\displaystyle\int_{\mathbb{R}}u(t)dt=0$,
$\int_{\mathbb{R}}f(t)dt$ exists and $g:t\mapsto t^{2}u(t)\in
H^{1}({\mathbb{R}})$, then

$\mathcal{H}[u](x)\underset{|x|\rightarrow\infty}{\sim}\displaystyle\dfrac{-\int_{\mathbb{R}}tu(t)dt}{x^{2}}.$

As a consequence, we obtain that
$\mathcal{H}[\rho_{P}](x)\underset{|x|\rightarrow\infty}{\sim}\\-x^{-1}$ and
the logarithmic potential $U^{\rho_{P}}$ is Lipschitz bounded, with bounded
derivative $\mathcal{H}[\rho_{P}]$.

###### Proof.

Let $u\in L^{2}({\mathbb{R}})$, such that $\int_{\mathbb{R}}u(t)dt$ exists and
$f:t\mapsto tu(t)\in H^{1}({\mathbb{R}})$. Then

$x\mathcal{H}[u](x)+\int_{\mathbb{R}}u(t)dt=\int_{\mathbb{R}}\Big{[}\dfrac{xu(x+t)-xu(x-t)}{2t}+\dfrac{u(x+t)}{2}+\dfrac{u(x-t)}{2}\Big{]}dt=\mathcal{H}[f](x).$

Since $f\in H^{1}({\mathbb{R}})$, so is $\mathcal{H}[f]$, proving that it goes
to zero at infinity. Hence

$\mathcal{H}[u](x)\underset{|x|\rightarrow\infty}{\sim}\displaystyle\dfrac{-\int_{\mathbb{R}}u(t)dt}{x}$

Moreover if $\displaystyle\int_{\mathbb{R}}u(t)dt=0$,
$\int_{\mathbb{R}}f(t)dt$ exists and $g:t\mapsto t^{2}u(t)\in
H^{1}({\mathbb{R}})$, then by the same argument:

$x^{2}\mathcal{H}[u](x)=x\mathcal{H}[f](x)=\mathcal{H}[g](x)-\int_{\mathbb{R}}f(t)dt$

where $g(t)=t^{2}u(t)$. We deduce that
$\mathcal{H}[u](x)\underset{|x|\rightarrow\infty}{\sim}\displaystyle\dfrac{-\int_{\mathbb{R}}tu(t)dt}{x^{2}}$
since $\mathcal{H}[g]$ goes to zero at infinity. ∎

######  Lemma 2.4 (Asymptotic of the logarithmic potential).

We have the following asymptotic expansion at infinity
$U^{\rho_{P}}=\ln|x|+\underset{|x|\rightarrow\infty}{O}(1)$.

###### Proof.

Since
$\mathcal{H}[\rho_{P}](x)=\\-x^{-1}+\underset{|x|\rightarrow\infty}{O}(x^{-2})$,
and recalling that $U^{\rho^{V}_{P}}$ (defined by (16)) satisfies
$(U^{\rho^{V}_{P}})^{\prime}=\mathcal{H}[\rho^{V}_{P}]$, we deduce the result
by integrating $t\mapsto\mathcal{H}[\rho^{V}_{P}](t)-1/t$ in a neighborhood of
infinity. ∎

We conclude this section by stating the Poincaré inequality for the measure
$\mu_{P}$ under the assumption that $V$ is a bounded perturbation of a
strictly convex potential $V_{\mathrm{conv}}$.

######  Lemma 2.5.

Let $\varepsilon>0$, there exists a function
$f\in\mathcal{C}^{2}({\mathbb{R}})$ such that
$f_{\varepsilon}(x)-\ln|x|=\underset{|x|\rightarrow\infty}{O}(1)$, and
$\|f_{\varepsilon}^{\prime\prime}\|_{\infty}\leqslant\varepsilon$.

######  Proposition 2.6.

Assume that $V=V_{\mathrm{conv}}+\phi$, where
$V_{\mathrm{conv}}\in\mathcal{C}^{3}({\mathbb{R}})$ with $V_{\mathrm{conv}}$
convex, such that there exists $\varepsilon>0$ such that $\phi$ is bounded and
$V_{\mathrm{conv}}-2Pf_{\varepsilon}$ is convex ($f_{\varepsilon}$ being given
by Lemma 2.5). Then, the measure $\mu_{P}$ satisfies the Poincaré inequality:
there exists a constant $C>0$ such that for all
$f\in\mathcal{C}^{1}_{c}({\mathbb{R}})$,

$\operatorname{Var}_{\mu_{P}}(f)\leqslant
C\int_{\mathbb{R}}|f^{\prime}|^{2}d\mu_{P}\,.$ (23)

###### Proof.

We use the fact that if $\mu_{1},\mu_{2}$ are two absolutely continuous
probability measures supported on ${\mathbb{R}}$ such that
$\dfrac{1}{C}\leqslant\dfrac{d\mu_{1}}{d\mu_{2}}\leqslant C$ for some $C>0$
and $\mu_{1}$ satisfies Poincaré inequality with constant $C_{1}$ then so does
$\mu_{2}$ for some other constant. Indeed, in that case let
$f\in\mathcal{C}_{c}^{1}({\mathbb{R}})$, we have

$\displaystyle\operatorname{Var}_{\mu_{2}}(f)=\inf_{a}\int_{\mathbb{R}}\left(f-a\right)^{2}d\mu_{2}$
$\displaystyle\leqslant C\operatorname{Var}_{\mu_{1}}(f)\leqslant
C^{2}C_{1}\int_{\mathbb{R}}f^{\prime}d\mu_{2}.$

Here we take $d\mu_{2}(x):=\rho_{P}^{V}(x)dx$ and we want to compare it to a
measure $\mu_{1}$ supported on ${\mathbb{R}}$ defined by
$d\mu_{1}(x)=\dfrac{1}{Z}\exp\big{(}-W(x)\big{)}dx$ for some convex function
$W$. The measure $\mu_{1}$ then clearly verifies the Poincaré inequality. This
fact comes as a direct consequence of [BBCG08][Corollary 1.9], which states
that if a probability measure $\mu$ has a log-concave density on
${\mathbb{R}}$, then it satisfies (23). With the definition
$W:=V_{\mathrm{conv}}-2Pf_{\varepsilon}$ with $\varepsilon>0$ such that
$V_{\mathrm{conv}}-2Pf_{\varepsilon}$ is convex, $W-V-2PU^{\rho_{P}}$ is
bounded on ${\mathbb{R}}$. It is then not hard to see that
$\dfrac{1}{C}\leqslant\dfrac{d\mu_{1}}{d\mu_{P}}\leqslant C$ for some $C>0$
which allows to conclude that $\mu_{P}$ satisfies the Poincaré inequality.

∎

######  Remark 2.7.

We will apply later inequality (23) to more general functions than
$\mathcal{C}^{1}_{c}({\mathbb{R}})$, namely functions of the weighted Sobolev
space $H^{1}(\rho_{P})$, defined in Section 6; which can be seen as the
completion of $\mathcal{C}^{\infty}_{c}({\mathbb{R}})$ with respect to the
norm $\|u\|_{L^{2}(\rho_{P})}+\|u^{\prime}\|_{L^{2}(\rho_{P})}$.

##  3 Concentration inequality, proof of Theorem 1.5

We prove in this section the concentration Theorem 1.5. Its proof is a direct
adaptation of Theorem 1.4 of [HL21], which shows the analogous estimate in the
circular setup. It is inspired by [MMS14] and based on a comparison between a
configuration $\mathbf{x}_{N}=(x_{1},\ldots,x_{N})$ sampled with
$\mathbb{P}_{N}^{V,P}$ and a regularized version
$\mathbf{y}_{N}=(y_{1},\ldots,y_{N})$, which we describe here.

######  Definition 3.1.

$y_{1}:=x_{1}$, and for $0\leqslant k\leqslant N-1$,
$y_{k+1}:=y_{k}+\max\\{x_{k+1}-x_{k},N^{-3}\\}.$

Note that the configuration $\mathbf{y}_{N}$ given by the previous definition
satisfies $y_{k+1}-y_{k}\geqslant N^{-3}$, and $\mathbf{y}_{N}$ is close to
$\mathbf{x}_{N}$ in the sense that

$\sum_{k=1}^{N}|x_{k}-y_{k}|\leqslant\dfrac{1}{2N}\,.$ (24)

Indeed, by construction we have
$|x_{k}-y_{k}|=y_{k}-x_{k}\leqslant(k-1)N^{-3}$, and we get the bound by
summing these inequalities.
The key point of the proof of Theorem 1.5 is comparing the empirical measure
$\hat{\mu}_{N}=\frac{1}{N}\sum_{i=1}^{N}\delta_{x_{i}},$ where
$\mathbf{x}_{N}$ follows $\mathbb{P}_{N}^{V,P}$, to the regularized measure

$\widetilde{\mu}_{N}:=\lambda_{N^{-5}}\ast\frac{1}{N}\sum_{i=1}^{N}\delta_{y_{i}},$
(25)

ie the convolution of $\lambda_{N^{-5}}$ and the empirical measure, where
$\lambda_{N^{-5}}$ is the uniform measure on $[0,N^{-5}]$. The interest of
introducing the measure $\widetilde{\mu}_{N}$ is that it is close to
$\hat{\mu}_{N}$, while having a finite energy
$\mathcal{E}_{P}^{V}(\widetilde{\mu}_{N})$, given by (17). Finally, notice
that the empirical measure doesn’t change when reordering
$x_{1},\ldots,x_{N}$, and thus we do not lose in generality for our purposes
in assuming that $x_{1}\leqslant\ldots\leqslant x_{N}$ in definition 3.1.

We now introduce a distance on $\mathcal{P}({\mathbb{R}})$ which is well-
suited to our context.

######  Definition 3.2.

For $\mu,\mu^{\prime}\in\mathcal{P}({\mathbb{R}})$ we define the distance
(possibly infinite) $D(\mu,\mu^{\prime})$ by

$\displaystyle D(\mu,\mu^{\prime})$
$\displaystyle:=\left(-\int\ln|x-y|d(\mu-\mu^{\prime})(x)d(\mu-\mu^{\prime})(y)\right)^{1/2}$
(26)
$\displaystyle=\left(\int_{0}^{+\infty}\frac{1}{t}\big{|}\mathcal{F}[\mu-\mu^{\prime}](t)\big{|}^{2}dt\right)^{1/2}.$
(27)

where the Fourier transform of a signed measure $\nu$ is defined by
$\mathcal{F}[\nu](x):=\displaystyle\int_{\mathbb{R}}e^{-\mathrm{i}tx}d(\mu-\mu^{\prime})(x)$

Let $f:{\mathbb{R}}\to{\mathbb{R}}$ with finite $1/2$ norm
$\|f\|_{1/2}:=\left(\int_{\mathbb{R}}|t|\left|\mathcal{F}[f](t)\right|^{2}dt\right)^{1/2}$.
By Plancherel theorem and Hölder inequality, for any
$\mu,\mu^{\prime}\in\mathcal{P}({\mathbb{R}})$, setting
$\nu=\mu-\mu^{\prime}$,

$\left|\int_{\mathbb{R}}fd\mu-\int_{\mathbb{R}}fd\mu^{\prime}\right|^{2}=\left|\dfrac{1}{2\pi}\int_{\mathbb{R}}|t|^{1/2}\mathcal{F}[f](t)\frac{\overline{\mathcal{F}[\nu](t)}}{|t|^{1/2}}dt\right|^{2}\leqslant\dfrac{1}{2\pi^{2}}\|f\|_{1/2}^{2}D^{2}(\mu,\mu^{\prime}).$

Therefore the metric $d$ defined in (13) is dominated by $D$:

$d(\mu,\mu^{\prime})\leqslant\dfrac{1}{\sqrt{2}\pi}D(\mu,\mu^{\prime}).$ (28)

The following lemma shows how the distance $D$ is related to the energy-
functional $\mathcal{E}^{V}_{P}$ defined in (17), we will write
$\mathcal{E}_{P}$ for simplicity.

######  Lemma 3.3.

We have for any absolutely continuous $\mu\in\mathcal{P}({\mathbb{R}})$ with
finite energy $\mathcal{E}^{V}_{P}(\mu)$,

$\mathcal{E}_{P}(\mu)-\mathcal{E}_{P}(\mu_{P})=PD^{2}(\mu,\mu_{P})+\int\ln\left(\frac{d\mu}{d\mu_{P}}\right)d\mu\,.$
(29)

###### Proof of Lemma 3.3.

Subtracting $\mathcal{E}_{P}(\mu)-\mathcal{E}_{P}(\mu_{P})$ we find

$\mathcal{E}_{P}(\mu)-\mathcal{E}_{P}(\mu_{P})=\int
Vd(\mu-\mu_{P})+\int\ln\frac{d\mu}{dx}d\mu-\int\ln\rho_{P}d\mu_{P}-P\iint\ln|x-y|d\mu(x)d\mu(y)\\\
+P\iint\ln|x-y|d\mu_{P}(x)d\mu_{P}(y)\,.$ (30)

Now, if $\nu$ is a signed measure of mass zero, integrating (18) we get

$\int
V(x)d\nu(x)-2P\iint\ln|x-y|d\nu(x)d\mu_{P}(y)+\int\ln(\rho_{P})(x)d\nu(x)=0\,.$

We take $\nu=\mu-\mu_{P}$, and get

$\int
V(x)d(\mu-\mu_{P})(x)=2P\iint\ln|x-y|d\mu(x)d\mu_{P}(y)-2P\iint\ln|x-y|d\mu_{P}(x)d\mu_{P}(y)\\\
-\int\ln(\rho_{P})(x)d\mu(x)+\int\ln(\rho_{P})(x)d\mu_{P}(x)\,.$

Plugging this last identity in (30), we find

$\mathcal{E}_{P}(\mu)-\mathcal{E}_{P}(\mu_{P})=-P\iint\ln|x-y|d\nu(x)d\nu(y)+\int\ln\left(\frac{d\mu}{d\mu_{P}}\right)(x)d\mu(x)$

which establishes the result. ∎

###### Proof of Theorem 1.5.

We first give a lower bound for the partition function $Z_{N}^{V,P}$ (4) of
$\mathbb{P}_{N}^{V,P}$. We rewrite it as

$Z^{V,P}_{N}=\int_{{\mathbb{R}}^{N}}\exp\Bigg{(}\dfrac{2P}{N}\sum_{i<j}\ln|x_{i}-x_{j}|-\sum_{i=1}^{N}\Big{[}V(x_{i})+\ln\rho_{P}(x_{i})\Big{]}\Bigg{)}d\rho_{P}(x_{1})\dots
d\rho_{P}(x_{N})\,,$

and apply Jensen inequality to obtain:

$\displaystyle\ln Z_{N}^{V,P}$
$\displaystyle\geqslant\int_{{\mathbb{R}}^{N}}\Bigg{(}\dfrac{2P}{N}\sum_{i<j}\ln|x_{i}-x_{j}|-\sum_{i=1}^{N}\Big{[}V(x_{i})+\ln\rho_{P}(x_{i})\Big{]}\Bigg{)}d\rho_{P}(x_{1})\dots
d\rho_{P}(x_{N})$ $\displaystyle\geqslant
P(N-1)\displaystyle\iint\ln|x-y|d\rho_{P}(x)d\rho_{P}(y)-N\int_{\mathbb{R}}\Big{[}V+\ln\rho_{P}\Big{]}d\rho_{P}$
$\displaystyle\geqslant-N\mathcal{E}_{P}^{V}\big{[}\mu_{P}\big{]}-P\iint\ln|x-y|d\rho_{P}(x)d\rho_{P}(y).$

Using this estimate and the fact that for $1\leqslant i,j\leqslant N$ we have
$|x_{i}-x_{j}|\leqslant|y_{i}-y_{j}|$, with
$\mathbf{y}_{N}=(y_{1},\ldots,y_{N})$ of definition 3.1, we deduce the bound
on the density of probability

$\frac{d\mathbb{P}^{V,P}_{N}}{d\mathbf{x}}(x_{1},\ldots,x_{N})\leqslant
e^{N\mathcal{E}_{P}(\mu_{P})+P\iint\ln|x-y|d\mu_{P}(x)d\mu_{P}(y)+\frac{P}{N}\sum_{i\neq
j}\ln|y_{i}-y_{j}|-\sum_{i=1}^{N}V(x_{i})}\,.$ (31)

Recalling (25), we now show the following estimate

$\sum_{i\neq j}\ln|y_{i}-y_{j}|\leqslant
2+N^{2}\iint\ln|x-y|d\widetilde{\mu}_{N}(x)d\widetilde{\mu}_{N}(y)+5N\ln
N+\frac{3}{2}N\,.$ (32)

Let $i\neq j$ and $u,v\in[0,N^{-5}]$. Since for $x\neq 0$ and
$|h|\leqslant\frac{|x|}{2}$, we have
$\big{|}\ln|x+h|-\ln|x|\big{|}\leqslant\frac{2|h|}{|x|}$, we deduce

$\big{|}\ln|y_{i}-y_{j}+u-v|-\ln|y_{i}-y_{j}|\big{|}\leqslant\frac{2|u-v|}{|y_{i}-y_{j}|}\leqslant\frac{2N^{-5}}{N^{-3}}=\dfrac{2}{N^{2}}.$

Thus, summing over $i\neq j$ and integrating with respect to $u$ and $v$, we
get

$\displaystyle\sum_{i\neq j}\ln|y_{i}-y_{j}|$ $\displaystyle\leqslant
2+\sum_{i\neq
j}\iint\ln|y_{i}-y_{j}+u-v|d\lambda_{N^{-5}}(u)d\lambda_{N^{-5}}(v)$
$\displaystyle=2+N^{2}\iint\ln|x-y|d\widetilde{\mu}_{N}(x)d\widetilde{\mu}_{N}(y)-N\iint\ln|u-v|d\lambda_{N^{-5}}(u)d\lambda_{N^{-5}}(v)\,.$

The last integral is equal to $-\frac{3}{2}-5\ln N$, so we deduce (32). We now
combine (31) and (32). Recall (17) and set

$c_{N}=P\left(\displaystyle\iint\ln|x-y|d\mu_{P}(x)d\mu_{P}(y)+3/2+2/N\right)\,.$

Then we get

$\displaystyle\frac{d\mathbb{P}^{V,P}_{N}}{d\mathbf{x}}(x_{1},\ldots,x_{N})$
$\displaystyle\leqslant e^{c_{N}+5P\ln
N}\exp{\left[N\left\\{\mathcal{E}_{P}(\mu_{P})-\mathcal{E}_{P}(\widetilde{\mu}_{N})+\int\left(V+\ln\frac{d\widetilde{\mu}_{N}}{dx}\right)d\widetilde{\mu}_{N}\right\\}-\sum_{i=1}^{N}V(x_{i})\right]}$
$\displaystyle=e^{c_{N}+5P\ln
N}\exp{\left[-NPD^{2}(\widetilde{\mu}_{N},\mu_{P})+N\int\left(V+\ln\rho_{P}\right)d\widetilde{\mu}_{N}-\sum_{i=1}^{N}V(x_{i})\right]}\,$

where we used equation (29) in the last equality. Using again equation (18) we
then see that the density
$\dfrac{d\mathbb{P}^{V,P}_{N}}{d\mathbf{x}}(x_{1},\ldots,x_{N})$ is bounded by

$e^{c_{N}+5P\ln
N}\exp{\left[-NPD^{2}(\widetilde{\mu}_{N},\mu_{P})+2PN\iint\ln|x-y|d(\widetilde{\mu}_{N}-\hat{\mu}_{N})(x)d\mu_{P}(y)\right]}\prod_{i=1}^{N}\rho_{P}(x_{i})\,.$

Recalling (16), we used that
$\displaystyle\iint\ln|x-y|d(\widetilde{\mu}_{N}-\hat{\mu}_{N})(x)d\mu_{P}(y)=-\int
U^{\rho_{P}}d(\widetilde{\mu}_{N}-\hat{\mu}_{N})$. As a consequence of the
bound on the density
$\dfrac{d\mathbb{P}^{V,P}_{N}}{d\mathbf{x}}(x_{1},\ldots,x_{N})$ we
established, we have for all $r>0$

$\mathbb{P}^{V,P}_{N}\left(D^{2}(\widetilde{\mu}_{N},\mu_{P})>r\right)\leqslant
e^{-NPr+c_{N}+5P\ln N}\int_{{\mathbb{R}}^{N}}\exp\left\\{-2PN\int
U^{\rho_{P}}d(\widetilde{\mu}_{N}-\hat{\mu}_{N})\right\\}\prod_{i=1}^{N}\rho_{P}(x_{i})dx_{i}\,.$
(33)

Next, we show that $-N\int U^{\rho_{P}}d(\widetilde{\mu}_{N}-\hat{\mu}_{N})$
is bounded. By Lemma 2.3, $U^{\rho_{P}}$ is differentiable with bounded
derivative $\mathcal{H}[\rho_{P}]$ on ${\mathbb{R}}$. As a consequence,

$\displaystyle\left|N\int
U^{\rho_{P}}d(\widetilde{\mu}_{N}-\hat{\mu}_{N})\right|$
$\displaystyle\leqslant\sum_{i=1}^{N}\int\left|U^{\rho_{P}}(y_{i}+u)-U^{\rho_{P}}(x_{i})\right|d\lambda_{N^{-5}}(u)$
$\displaystyle\leqslant\|\mathcal{H}[\rho_{P}]\|_{\infty}\left(\sum_{i=1}^{N}|y_{i}-x_{i}|+N\int
ud\lambda_{N^{-5}}(u)\right)$
$\displaystyle\leqslant\|\mathcal{H}[\rho_{P}]\|_{\infty}\Big{(}\dfrac{1}{2N}+N^{-4}/2\Big{)},$

where we used (24) in the last inequality. Therefore, we deduce from (33)

$\mathbb{P}^{V,P}_{N}\left(D^{2}(\widetilde{\mu}_{N},\mu_{P})>r\right)\leqslant
e^{-NPr+c_{N}+5P\ln
N+\frac{2P}{N}\|\mathcal{H}[\rho_{P}]\|_{\infty}}=e^{-NPr+5P\ln N+K_{N}}$ (34)

with
$K_{N}:=c_{N}+\dfrac{2P}{N}\|\mathcal{H}\left[\rho_{P}\right]\|_{\infty}$.
Since $(c_{N})_{N}$ is bounded, so is $(K_{N})_{N}$.
Finally, let $f$ be a Lipschitz bounded function with
$\|f\|_{\text{Lip}}\leqslant 1$, then, we have (as we did for $U^{\rho_{P}}$)

$\left|\int fd\hat{\mu}_{N}-\int fd\widetilde{\mu}_{N}\right|\leqslant
N^{-2}\,.$

Thus by (28)

$d(\hat{\mu}_{N},\mu_{P})\leqslant
d(\hat{\mu}_{N},\widetilde{\mu}_{N})+d(\widetilde{\mu}_{N},\mu_{P})\leqslant
N^{-2}+\frac{1}{\sqrt{2}\pi}D(\widetilde{\mu}_{N},\mu_{P})\,,$

and for any $N$ such that $r-N^{-2}\geqslant r/2$ (in particular $r-N^{-2}>0$)
we get

$\displaystyle\mathbb{P}^{V,P}_{N}\left(d(\hat{\mu}_{N},\mu_{P})>r\right)\leqslant\mathbb{P}^{V,P}_{N}\left(\frac{1}{2\pi^{2}}D^{2}(\widetilde{\mu}_{N},\mu_{P})>(r-N^{-2})^{2}\right)$
$\displaystyle\leqslant\mathbb{P}^{V,P}_{N}\left(\frac{1}{2\pi^{2}}D^{2}(\widetilde{\mu}_{N},\mu_{P})>r^{2}/4\right)\,,$

and the last term is bounded by $e^{-Nr^{2}\frac{P\pi^{2}}{2}+5P\ln N+K}$ for
some $K$ large enough, which concludes the proof. ∎

As a consequence of Theorem 1.5, we are able to control the quantities

$\zeta_{N}(\phi):=\iint_{{\mathbb{R}}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d(\hat{\mu}_{N}-\mu_{P})(x)d(\hat{\mu}_{N}-\mu_{P})(y)$
(35)

for a certain class of test functions $\phi$.

######  Corollary 3.4.

There exists $C,K>0$ such that for all
$\phi\in\mathcal{C}^{2}({\mathbb{R}})\cap H^{2}({\mathbb{R}})$ with bounded
second derivative, we have for $\varepsilon>0$ and $N$ large enough,

$\mathbb{P}_{N}^{V,P}\left(\sqrt{N}|\zeta_{N}(\phi)|\leqslant
N^{-1/2+\varepsilon}\right)\geqslant
1-\exp\left\\{-\frac{PN^{\varepsilon}}{2C\|\psi\|_{H^{2}({\mathbb{R}})}}+5P\ln
N+K\right\\}\,$

with
$N_{2}(\phi)=\|\phi^{\prime}\|_{L^{2}(dx)}+\|\phi^{\prime\prime}\|_{L^{2}(dx)}$.

###### Proof.

We follow the proof given in [Gui19][Cor. 4.16] and adapt it to our setting.
Let us denote by $\widetilde{\zeta_{N}}(\phi)$ the quantity

$\iint_{{\mathbb{R}}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d(\widetilde{\mu}_{N}-\mu_{P})(x)d(\widetilde{\mu}_{N}-\mu_{P})(y)\,.$

We have the almost sure inequality, by a Taylor estimate

$|\zeta_{N}(\phi)-\widetilde{\zeta_{N}}(\phi)|\leqslant
2N^{-2}\|\phi^{\prime\prime}\|_{\infty}\,.$ (36)

Thus, for any $\delta>0$,

$\displaystyle\mathbb{P}_{N}^{V,P}\left(|\zeta_{N}(\phi)|>\delta\right)$
$\displaystyle\leqslant\mathbb{P}_{N}^{V,P}\left(|\zeta_{N}(\phi)-\widetilde{\zeta_{N}}(\phi)|>\delta/2\right)+\mathbb{P}_{N}^{V,P}\left(|\widetilde{\zeta_{N}}(\phi)|>\delta/2\right)$
$\displaystyle\leqslant\mathbb{P}_{N}^{V,P}\left(2N^{-2}\|\phi^{\prime\prime}\|_{\infty}>\delta/2\right)+\mathbb{P}_{N}^{V,P}\left(|\widetilde{\zeta_{N}}(\phi)|>\delta/2\right)\,,$

where the first term of the right-hand side is either $0$ or $1$. With
$\delta=N^{-1+\varepsilon}$, $\varepsilon>0$, it is zero for $N$ large enough.
For such a choice of $\delta$, and for $N$ large enough,

$\mathbb{P}_{N}^{V,P}\left(|\zeta_{N}(\phi)|>N^{-1+\varepsilon}\right)\leqslant\mathbb{P}_{N}^{V,P}\left(|\widetilde{\zeta_{N}}(\phi)|>\frac{1}{2}N^{-1+\varepsilon}\right)\,.$

We next show that, for some $C>0$ independent of $\phi$, we have

$|\widetilde{\zeta_{N}}(\phi)|\leqslant
CD^{2}(\widetilde{\mu}_{N},\mu_{P})\|\phi\|_{H^{2}({\mathbb{R}})}\,.$ (37)

We begin by showing this inequality for $\psi\in\mathcal{S}({\mathbb{R}})$. By
using the inverse Fourier transform we have

$\displaystyle\widetilde{\zeta}_{N}$
$\displaystyle(\psi)=\dfrac{1}{2\pi}\iint\dfrac{\int
dt\mathcal{F}[\psi](t)e^{\mathrm{i}tx}-\int
dt\mathcal{F}[\psi](t)e^{\mathrm{i}ty}}{x-y}d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(x)d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(y)$
$\displaystyle=\dfrac{1}{2\pi}\int dt\mathrm{i}t\mathcal{F}[\psi](t)\iint
e^{\mathrm{i}ty}\dfrac{e^{\mathrm{i}t(x-y)}-1}{\mathrm{i}t(x-y)}d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(x)d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(y)$
$\displaystyle=\dfrac{1}{2\pi}\int dt\mathrm{i}t\mathcal{F}[\psi](t)\iint
e^{\mathrm{i}ty}\int_{0}^{1}d\alpha e^{\mathrm{i}\alpha
t(x-y)}d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(x)d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(y)$
$\displaystyle=\dfrac{1}{2\pi}\int
dt\mathrm{i}t\mathcal{F}[\psi](t)\int_{0}^{1}d\alpha\int e^{\mathrm{i}\alpha
tx}d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(x)\int
e^{\mathrm{i}(1-\alpha)ty}d\big{(}\widetilde{\mu}_{N}-\mu_{P}\big{)}(y)$

We then apply in order the triangular inequality, Cauchy-Schwarz inequality, a
change of variable and the fact that
$\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right]\right|^{2}$ is an
even function.

$\displaystyle|\widetilde{\zeta}_{N}(\psi)|$
$\displaystyle\leqslant\dfrac{1}{2\pi}\int_{\mathbb{R}}dt\left|t\mathcal{F}[\psi](t)\right|\int_{0}^{1}d\alpha\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right](\alpha
t)\right|.\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right]\big{(}(1-\alpha)t\big{)}\right|$
$\displaystyle\leqslant\dfrac{1}{2\pi}\int_{\mathbb{R}}dt\left|t\mathcal{F}[\psi](t)\right|\Big{(}\int_{0}^{1}d\alpha\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right](\alpha
t)\right|^{2}\Big{)}^{\frac{1}{2}}\Big{(}\int_{0}^{1}d\alpha\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right]\big{(}(1-\alpha)t\big{)}\right|^{2}\Big{)}^{\frac{1}{2}}$
$\displaystyle\leqslant\dfrac{1}{2\pi}\int_{\mathbb{R}}dt\left|t\mathcal{F}[\psi](t)\right|\int_{0}^{1}d\alpha\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right](\alpha
t)\right|^{2}$
$\displaystyle\leqslant\dfrac{1}{2\pi}\int_{0}^{+\infty}dt\left|t\mathcal{F}[\psi](t)\right|\int_{0}^{1}\dfrac{td\alpha}{t\alpha}\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right](\alpha
t)\right|^{2}+\dfrac{1}{2\pi}\int_{-\infty}^{0}dt\left|t\mathcal{F}[\phi](t)\right|\int_{0}^{1}\dfrac{-td\alpha}{-t\alpha}\left|\mathcal{F}\left[\widetilde{\mu}_{N}-\mu_{P}\right](\alpha
t)\right|^{2}$
$\displaystyle\leqslant\dfrac{1}{2\pi}\int_{\mathbb{R}}dt\left|t\mathcal{F}[\psi](t)\right|D^{2}(\widetilde{\mu}_{N},\mu_{P})$
$\displaystyle\leqslant\dfrac{1}{2\pi}\Big{(}\int_{\mathbb{R}}dt\left|t\mathcal{F}[\psi](t)\right|^{2}(1+t^{2})\Big{)}^{\frac{1}{2}}\Big{(}\int_{\mathbb{R}}\dfrac{dt}{1+t^{2}}\Big{)}^{\frac{1}{2}}D^{2}(\widetilde{\mu}_{N},\mu_{P})$
$\displaystyle\leqslant\dfrac{1}{2\sqrt{\pi}}D^{2}(\widetilde{\mu}_{N},\mu_{P})N_{2}(\psi)$
$\displaystyle\leqslant\dfrac{1}{2\sqrt{\pi}}D^{2}(\widetilde{\mu}_{N},\mu_{P})\|\psi\|_{H^{2}({\mathbb{R}})}$

By density of $\mathcal{S}({\mathbb{R}})$ in $H^{2}({\mathbb{R}})$, and since
$\widetilde{\zeta}_{N}:\Big{(}H^{2}({\mathbb{R}}),\|\cdot\|_{H^{2}({\mathbb{R}})}\Big{)}\rightarrow{\mathbb{R}}$
is continuous, the inequality still holds for $\phi$. Thus, using equation
(34),

$\mathbb{P}_{N}^{V,P}\left(|\widetilde{\zeta_{N}}(\phi)|>\frac{1}{2}N^{-1+\varepsilon}\right)\leqslant\mathbb{P}_{N}^{V,P}\left(D^{2}(\widetilde{\mu}_{N},\mu_{P})>\frac{N^{-1+\varepsilon}}{2C\|\phi\|_{H^{2}({\mathbb{R}})}}\right)\leqslant\exp\left\\{-P\frac{N^{\varepsilon}}{2C\|\phi\|_{H^{2}({\mathbb{R}})}}+5P\ln
N+K\right\\}\,,$

which concludes the proof. ∎

##  4 Localization of the edge of a configuration

In [Lam21][Theorem 1.8, Theorem 3.4], Lambert was able to control the edge
(i.e the minimum and the maximum) of a typical configuration
$(x_{1},\ldots,x_{N})$ distributed according to $\mathbb{P}^{V,P}_{N}$, by
showing that the random measure

$\Xi_{N}:=\sum_{j=1}^{N}\delta_{\varphi_{N}^{-1}(x_{j})}$

converges in distribution towards a Poisson point process for a function
$\varphi_{N}$ which takes the form

$\varphi_{N}(x):=E_{N}+\alpha_{N}^{-1}x\,.$

Before being more precise on the construction of $(E_{N})_{N}$ and
$(\alpha_{N})_{N}$, we explain, following [Lam21], how one can use this
convergence to localize the edge of a typical configuration
$(x_{1},\ldots,x_{N})$. Let us assume for a moment that $\Xi_{N}$ converges
towards a Poisson point process with intensity $\theta(x)=e^{-x}$, with
$E_{N}\to+\infty$. In particular, the random variable

$\Xi_{N}(t,+\infty)$

converges in distribution towards a Poisson random variable with mean
$\int_{t}^{+\infty}e^{-x}dx$. Combined with the equalities

$\displaystyle\mathbb{P}^{V,P}_{N}\bigg{(}\Xi_{N}(t,+\infty)=0\bigg{)}$
$\displaystyle=\mathbb{P}^{V,P}_{N}\bigg{(}\forall\ 1\leqslant j\leqslant N,\
\varphi_{N}^{-1}(x_{j})=\alpha_{N}(x_{j}-E_{N})\leqslant t\bigg{)}$
$\displaystyle=\mathbb{P}^{V,P}_{N}\bigg{(}\alpha_{N}\left(\max_{1\leqslant
j\leqslant N}x_{j}-E_{N}\right)\leqslant t\bigg{)}\,,$

we deduce that for all $t\in{\mathbb{R}}$

$\mathbb{P}^{V,P}_{N}\left(\alpha_{N}\left(\max_{1\leqslant j\leqslant
N}x_{j}-E_{N}\right)\leqslant
t\right)\underset{N\rightarrow\infty}{\longrightarrow}\exp(-e^{-t})\,.$

Therefore, the random variable

$\alpha_{N}\left(\max_{1\leqslant j\leqslant N}x_{j}-E_{N}\right)$

converges in distribution to the Gumbel law, showing that the maximum of a
configuration is of order $E_{N}$. Furthermore, as will be clear from the
construction of $\alpha_{N}$ and $E_{N}$, $\alpha_{N}$ is positive, and goes
to infinity as $N$ goes to infinity.

Replacing in the previous analysis $\theta(x)=e^{x}$ and $E_{N}\to-\infty$, we
would have deduced in the same fashion that

$\alpha_{N}\left(\min_{1\leqslant j\leqslant N}x_{j}-E_{N}\right)$

converges in law.

With the above notations, we can apply [Lam21][Theorem 3.4] to our context.

######  Theorem 4.1.

Let $v=\pm$. There exists $(E_{N}^{v})_{N},\,(\alpha_{N}^{v})_{N}$ sequences
of real numbers with $|E_{N}^{v}|\to+\infty$, $\alpha_{N}^{v}>0$ for large
enough $N$, satisfying $V^{\prime}(E_{N}^{v})=\alpha_{N}^{v}v$, such that:

  * a)

$\dfrac{Ne^{-V(E_{N}^{v})+2P\ln|E_{N}^{v}|+\lambda_{V}^{P}}}{\alpha_{N}^{v}}\underset{N\rightarrow\infty}{\longrightarrow}1$
(recall $\lambda^{V}_{P}$ is defined through equation (5)),

  * b)

$\frac{\ln(\alpha_{N}^{v})}{N}\underset{N\rightarrow\infty}{\longrightarrow}0$
and
$\alpha_{N}^{v}|E_{N}^{v}|\underset{N\rightarrow\infty}{\longrightarrow}+\infty$
,

  * c)

For all compact $K\subset{\mathbb{R}}$,

$(\alpha_{N}^{v})^{-2}\sup_{x\in
K}\left|V^{\prime\prime}(\varphi_{N}(x))\right|\underset{N\rightarrow\infty}{\longrightarrow}0\,.$

As a consequence, the random measure $\Xi_{N}$ converges in distribution as
$N\to\infty$ to a Poisson point process with intensity $\theta(x)=e^{-vx}$.

###### Proof.

We prove it in the case $v=+$, the case where $v=-$ being similar. We show
that there exists a sequence $(E_{N}^{+})_{N}$ going to $+\infty$ satisfying
$f(E_{N}^{+})=-\ln N$, where we defined the function $f$ by

$f(x)=-V(x)+2P\ln|x|+\lambda_{P}^{V}-\ln|V^{\prime}(x)|\,.$

Recalling Assumption 1.1 i), $|V^{\prime}|$ goes to infinity at infinity, thus
$\alpha_{N}^{+}=V^{\prime}(E_{N}^{+})\to+\infty$ (in the case $v=-1$ we would
have looked for a sequence $(E_{N}^{-})_{N}$ going to $-\infty$ and
$\alpha_{N}^{-}=-V^{\prime}(E_{N}^{-})$).
As a consequence of Assumptions 1.1,ii), one shows that $\ln|V^{\prime}|$ is
negligible with respect to $V$ at infinity. Therefore, because
$\dfrac{\ln|x|}{V(x)}\underset{|x|\rightarrow\infty}{\longrightarrow}0$,

$f(x)=-V(x)+\underset{x\rightarrow+\infty}{o}(V(x))\,.$ (38)

Because $f(x)\underset{x\rightarrow+\infty}{\longrightarrow}-\infty$ there
exists $(E_{N}^{+})_{N}$ going to infinity such that for all $N\geqslant 1$,
$f(E_{N}^{+})=-\ln N$. Setting $x=E_{N}^{+}$ in (38), we obtain that
$-V(E_{N}^{+})\sim f(E_{N}^{+})=-\ln N$. Property c) follows from Assumptions
1.1, point ii), along with the fact that $\alpha_{N}^{-1}$ stays bounded.

It remains to show that
$\dfrac{\ln(\alpha_{N}^{+})}{N}=\dfrac{\ln|V^{\prime}(E_{N}^{+})|}{N}\underset{N\rightarrow\infty}{\longrightarrow}0$.
By construction, we have

$\dfrac{\ln|V^{\prime}(E_{N}^{+})|}{N}=\dfrac{\ln\Big{(}Ne^{-V(E_{N}^{+})+2P\ln
N+\lambda_{P}}\Big{)}}{N}=-\dfrac{V(E_{N}^{+})}{N}+o(1)\,.$

Using that $V(E_{N}^{+})\sim\ln N$, we can conclude that
$\ln|V^{\prime}(E_{N}^{+})|=o(N)$ which concludes the proof. ∎

By the discussion preceding Theorem 4.1, we deduce

######  Corollary 4.2 (Edge of a configuration).

Let $E_{N}^{\pm}$, $\alpha_{N}^{\pm}:=|V^{\prime}(E_{N}^{\pm})|$ be the
sequences of Theorem 4.1 associated with $v=\pm 1$. Then, both random
variables

$\alpha_{N}^{+}\left(\max_{1\leqslant j\leqslant N}x_{j}-E_{N}^{+}\right)$

and

$\alpha_{N}^{-}\left(\min_{1\leqslant j\leqslant N}x_{j}-E_{N}^{-}\right)$

converge to a Gumbel law, whose distribution function is given for $t\geqslant
0$ by $\mathcal{G}([0,t])=\exp(e^{-t})$. Furthermore, $V(E_{N}^{\pm})\sim\ln
N$ and
$\alpha_{N}^{\pm}\underset{N\rightarrow\infty}{\longrightarrow}\pm\infty$.

######  Remark 4.3.

Note that[Lam21][Theorem 3.4] applies for $V$ of class $\mathcal{C}^{2}$
outside of a compact set, allowing to take $V(x)=|x|^{a}$ for $a>1$. In this
case, we find $E_{N}^{\pm}\sim\pm(\ln N)^{1/a}$. If $V(x)=\cosh(x)$, we find
$E_{N}^{+}\sim-E_{N}^{-}\sim\arg\cosh(\ln N)\sim\ln\ln N$.

The next lemma will be convenient in the proof of Theorem 5.2 when dealing
with error terms.

######  Lemma 4.4.

With the notations of Corollary 4.2, we have

$\mu_{P}([E_{N}^{-},E_{N}^{+}]^{c})=o(N^{-1/2})\,.$

###### Proof.

Let $0<\delta<1$, to be specified later. We have

$\displaystyle\int_{E_{N}^{+}}^{+\infty}\rho_{P}dx=\int_{E_{N}^{+}}^{+\infty}(\rho_{P})^{\delta}(\rho_{P})^{1-\delta}dx$
$\displaystyle\leqslant\int_{\mathbb{R}}(\rho_{P})^{\delta}dx\sup_{[E_{N}^{+},+\infty[}(\rho_{P})^{1-\delta}\,.$

By the first inequality of Lemma 2.2, the integral is finite. Also from the
same inequality, we have for some constant $C^{\prime}$ and $x$ big enough
$\rho_{P}(x)\leqslant C^{\prime}e^{-\frac{3}{4}V(x)}$. Because $V$ is
increasing in a neighborhood of $+\infty$, we get for $N$ large enough

$\sup_{[E_{N}^{+},+\infty[}(\rho_{P})^{1-\delta}\leqslant C^{\prime
1-\delta}e^{-(1-\delta)\frac{3}{4}V(E_{N}^{+})}\,.$

Taking $\delta>0$ such that $\frac{1}{2}-(1-\delta)\frac{3}{4}=:-\gamma<0$ and
using that $V(E_{N}^{+})=\ln N+o(\ln N)$ (established in the proof of Theorem
4.1),

$\displaystyle\sqrt{N}\int_{E_{N}^{+}}^{+\infty}\rho_{P}dx\leqslant
Ke^{-\gamma\ln N+(1-\delta)\frac{3}{4}o(\ln N)}\,,$

and the right-hand side goes to zero as $N$ goes to infinity. We deal with the
integral $\int_{-\infty}^{E_{N}^{-}}\rho_{P}dx$ in the same way.

∎

######  Remark 4.5.

We could improve the proof to show that
$\mu_{P}([E_{N}^{-},E_{N}^{+}]^{c})\sim\dfrac{1}{N}$ but showing that it is
$o(N^{\frac{1}{2}})$ is sufficient for what we need and requires less
carefulness.

##  5 Laplace transform for smooth test functions, proof of Theorem 1.3

Section 3 allows us to justify in Proposition 5.1 the heuristics we gave in
equation (6) for $\phi$ having compact support. We will then extend in Theorem
5.2 this result to a more general set of functions, by an approximation by
compactly supported functions, using Corollary 4.2.

######  Proposition 5.1.

For $\phi\in\mathcal{C}^{1}({\mathbb{R}},{\mathbb{R}})$ with compact support,
we have for any real $t$, as $N$ goes to infinity,

$\mathbb{E}^{V,P}_{N}\left[e^{t\sqrt{N}\nu_{N}(\Xi\phi)}\right]\to\exp\left\\{\frac{t^{2}}{2}q_{P}(\phi)\right\\},$
(39)

where $\Xi\phi$ is given by equation (7), and $q_{P}(\phi)$ is given by

$q_{P}(\phi):=\int_{\mathbb{R}}\bigg{(}\phi^{\prime}(x)^{2}+V^{\prime\prime}(x)\phi(x)^{2}\bigg{)}d\mu_{P}(x)+P\iint_{{\mathbb{R}}^{2}}\Big{(}\frac{\phi(x)-\phi(y)}{x-y}\Big{)}^{2}d\mu_{P}(x)d\mu_{P}(y)\,.$
(40)

###### Proof.

Let $\phi\in\mathcal{C}_{c}^{1}({\mathbb{R}},{\mathbb{R}})$, and let
$t\in{\mathbb{R}}$. We perform in equation (4) the change of variables
$x_{i}=y_{i}+\frac{t}{\sqrt{N}}\phi(y_{i})$, $1\leqslant i\leqslant N$, which
is a diffeomorphism for $N$ big enough. We thus have

$Z_{N}^{V,P}=\int\prod_{1\leqslant i<j\leqslant
N}\left|y_{i}-y_{j}+\frac{t}{\sqrt{N}}\big{(}\phi(y_{i})-\phi(y_{j})\big{)}\right|^{2P/N}.e^{-\sum_{i=1}^{N}V\left(y_{i}+\frac{t}{\sqrt{N}}\phi(y_{i})\right)}.\prod_{i=1}^{N}\left(1+\frac{t}{\sqrt{N}}\phi^{\prime}(y_{i})\right)d^{N}\mathbf{y},$
(41)

and we develop separately the different terms of this integral. The first term
can be written as:

$\prod_{i<j}\left|y_{i}-y_{j}\right|^{2P/N}\prod_{i<j}\left|1+\frac{t}{\sqrt{N}}\frac{\phi(y_{i})-\phi(y_{j})}{y_{i}-y_{j}}\right|^{2P/N},$

The second product above, setting
$\Delta\phi_{i,j}:=\frac{\phi(y_{i})-\phi(y_{j})}{y_{i}-y_{j}}$ and using
Taylor-Lagrange theorem, equals

$\exp\bigg{(}\frac{2P}{N}\sum_{i<j}\ln\left|1+\frac{t}{\sqrt{N}}\frac{\phi(y_{i})-\phi(y_{j})}{y_{i}-y_{j}}\right|\bigg{)}=\exp\bigg{(}\frac{2P}{N}\sum_{i<j}\left(\frac{t}{\sqrt{N}}\Delta\phi_{i,j}-\frac{t^{2}}{2N}(\Delta\phi_{i,j})^{2}+R_{N,1}(i,j)\right)\bigg{)},$

where we noticed that $1+\frac{t}{\sqrt{N}}\Delta\phi_{i,j}\geqslant
1-\frac{t}{\sqrt{N}}\|\phi^{\prime}\|_{\infty}>0$ if $N$ is big enough, and
where

$|R_{N,1}(i,j)|\leqslant\frac{|t|^{3}}{3N^{3/2}}\|\phi^{\prime}\|_{\infty}^{3}.$

Again by Taylor-Lagrange theorem, the second term in (41) equals

$\exp\bigg{(}-\sum_{i=1}^{N}\left(V(y_{i})+\frac{t}{\sqrt{N}}V^{\prime}(y_{i})\phi(y_{i})+\frac{t^{2}}{2N}V^{\prime\prime}(y_{i})\phi(y_{i})^{2}+R_{N,2}(i)\right)\bigg{)}$

where
$R_{N,2}(i)=\frac{t^{3}}{6N^{3/2}}V^{(3)}\left(y_{i}+\frac{t\theta_{i}}{\sqrt{N}}\phi(y_{i})\right)\phi(y_{i})^{3}$
for some $\theta_{i}\in[0,1]$, thus for $N$ large enough

$|R_{N,2}(i)|\leqslant\frac{|t|^{3}}{6N^{3/2}}\|\phi\|_{\infty}^{3}\sup_{d(x,\text{supp
}\phi)\leqslant 1}|V^{(3)}(x)|.$

The last term reads

$\prod_{i=1}^{N}\left(1+\dfrac{t}{\sqrt{N}}\phi^{\prime}(y_{i})\right)=\exp\bigg{(}\sum_{i=1}^{N}\left(\frac{t}{\sqrt{N}}\phi^{\prime}(y_{i})-\frac{t^{2}}{2N}\phi^{\prime}(y_{i})^{2}+R_{N,3}(i)\right)\bigg{)},$

with
$|R_{N,3}(i)|\leqslant\frac{t^{3}}{3N^{3/2}}\|\phi^{\prime}\|_{\infty}^{3}$.
Dividing both sides of equation (41) by $Z^{V,P}_{N}$ we get

$\mathbb{E}_{N}^{V,P}\bigg{[}\exp\left\\{t\sqrt{N}\bigg{(}P\iint_{{\mathbb{R}}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d\hat{\mu}_{N}(x)d\hat{\mu}_{N}(y)+\int_{\mathbb{R}}(\phi^{\prime}-V^{\prime}\phi)d\hat{\mu}_{N}\bigg{)}\right\\}\times\exp\left\\{K_{N}(t,\phi)\right\\}\\\
\times\exp\left\\{\frac{t^{2}}{2}\left(-P\iint_{{\mathbb{R}}^{2}}\left(\frac{\phi(x)-\phi(y)}{x-y}\right)^{2}d\hat{\mu}_{N}(x)d\hat{\mu}_{N}(y)-\int_{\mathbb{R}}(V^{\prime\prime}\phi^{2}+\phi^{\prime
2})d\hat{\mu}_{N}\right)\right\\}\bigg{]}=1,$

with $|K_{N}(t,\phi)|\leqslant\frac{c(t,\phi)}{\sqrt{N}}$ where
$c(t,\phi)\geqslant 0$ is independent of $N$. This bound shows that taking the
limit $N\to\infty$ we can get rid of $K_{N}$:

$\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\bigg{[}\exp\left\\{t\sqrt{N}\bigg{(}P\iint_{{\mathbb{R}}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d\hat{\mu}_{N}(x)d\hat{\mu}_{N}(y)+\int_{\mathbb{R}}(\phi^{\prime}-V^{\prime}\phi)d\hat{\mu}_{N}\bigg{)}\right\\}\\\
\times\exp\left\\{\frac{t^{2}}{2}\left(-P\iint_{{\mathbb{R}}^{2}}\left(\frac{\phi(x)-\phi(y)}{x-y}\right)^{2}d\hat{\mu}_{N}(x)d\hat{\mu}_{N}(y)-\int_{\mathbb{R}}(V^{\prime\prime}\phi^{2}+\phi^{\prime
2})d\hat{\mu}_{N}\right)\right\\}\bigg{]}=1.$

Using Fubini’s theorem (the function $(x,y)\mapsto\frac{\phi(x)-\phi(y)}{x-y}$
being bounded continuous on ${\mathbb{R}}^{2}$), the first line in the
expectation value can be rewritten as $e^{t\sqrt{N}\Lambda_{N}}$ with

$\Lambda_{N}:=2P\iint_{{\mathbb{R}}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d\mu_{P}(x)d(\hat{\mu}_{N}-\mu_{P})(y)+\int_{\mathbb{R}}(\phi^{\prime}-V^{\prime}\phi)d(\hat{\mu}_{N}-\mu_{P})+P\zeta_{N}(\phi)$
(42)

where we used equation (5) and $\zeta_{N}(\phi)$ is given by (35). Let
$F:\mathcal{P}({\mathbb{R}})\to{\mathbb{R}}$ be defined by

$F(\mu)=-P\iint_{{\mathbb{R}}^{2}}\left(\frac{\phi(x)-\phi(y)}{x-y}\right)^{2}d\mu(x)d\mu(y)-\int_{\mathbb{R}}(V^{\prime\prime}\phi^{2}+\phi^{\prime
2})d\mu\,.$ (43)

It is continuous for the topology of weak convergence since all the functions
in the integrals are bounded continuous. So far we have established that

$\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\Lambda_{N}+\frac{t^{2}}{2}F(\hat{\mu}_{N})}\right]=1,$

with $\Lambda_{N}$ given by (42). We now replace in the latter equation the
term $F(\hat{\mu}_{N})$ by its limiting expression, $F(\mu_{P})$. Fix a metric
that is compatible with the weak convergence of probability measures on
${\mathbb{R}}$. For example,

$d_{\text{Lip}}(\mu,\nu)=\sup\left|\int fd\mu-\int fd\nu\right|\,,$ (44)

where the supremum runs over $f:{\mathbb{R}}\to{\mathbb{R}}$ bounded and
Lipschitz with $\|f\|_{\infty}\leqslant 1$ and Lipschitz constant
$|f|_{\text{Lip}}\leqslant 1$. By the large deviations principle for
$(\hat{\mu}_{N})_{N}$ under the probability (3) established by [GZ19, Theorem
1.1], for all $\delta>0$ the event
$\\{d_{\text{Lip}}(\hat{\mu}_{N},\mu_{P})>\delta\\}$ has (for $N$ big enough)
probability smaller than $e^{-Nc_{\delta}}$ where $c_{\delta}>0$. Hence,

$\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\Lambda_{N}+\frac{t^{2}}{2}F(\hat{\mu}_{N})}\right]=\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[\mathbf{1}_{\\{d_{\text{Lip}}(\hat{\mu}_{N},\mu_{P})\leqslant\delta\\}}e^{t\sqrt{N}\Lambda_{N}+\frac{t^{2}}{2}F(\hat{\mu}_{N})}\right].$

By continuity of $F$ there is some $\varepsilon(\delta)$ which goes to $0$ as
$\delta\to 0$ such that, for $d_{\text{Lip}}(\nu,\mu_{P})\leqslant\delta$, we
have $|F(\nu)-F(\mu_{P})|\leqslant\varepsilon(\delta)$. Taking the
(decreasing) limit as $\delta$ goes to zero we deduce

$\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\Lambda_{N}+\frac{t^{2}}{2}F(\hat{\mu}_{N})}\right]=\lim_{\delta\to
0}\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[\mathbf{1}_{\\{d_{\text{Lip}}(\hat{\mu}_{N},\mu_{P})\leqslant\delta\\}}e^{t\sqrt{N}\Lambda_{N}}\right]e^{\frac{t^{2}}{2}F(\mu_{P})}.$

But the same large deviations argument shows that

$\lim_{\delta\to
0}\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[\mathbf{1}_{\\{d_{\text{Lip}}(\hat{\mu}_{N},\mu_{P})\leqslant\delta\\}}e^{t\sqrt{N}\Lambda_{N}}\right]=\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\Lambda_{N}}\right].$

Thus, we have shown that

$\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\left(2P\iint_{{\mathbb{R}}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d\mu_{P}(x)d(\hat{\mu}_{N}-\mu_{P})(y)+\int_{\mathbb{R}}(\phi^{\prime}-V^{\prime}\phi)d(\hat{\mu}_{N}-\mu_{P})+P\zeta_{N}(\phi)\right)}\right]=e^{-\frac{t^{2}}{2}F(\mu_{P})}\,,$
(45)

Which establishes that
$\sqrt{N}\Lambda_{N}=\sqrt{N}\Big{(}\nu_{N}(\Xi\phi)+P\zeta_{N}(\phi)\Big{)}$
converges in law towards a centered Gaussian random variable with announced
variance. We finally get rid of the remaining term $\zeta_{N}(\phi)$, using
Corollary 3.4: taking $\varepsilon=1/4$ for example, we see in particular that
$\sqrt{N}\zeta_{N}(\phi)$ converges in probability towards zero. The
conclusion follows from Slutsky’s lemma. ∎

We now extend the result of Proposition 5.1 to a more general set of
functions. With the notations of Proposition 5.1, we have

######  Theorem 5.2.

Let $\phi\in H^{2}({\mathbb{R}})\cap\mathcal{C}^{2}({\mathbb{R}})$ such that
$\phi^{\prime\prime}$ is bounded. Additionally, suppose that
$V^{(3)}\phi^{2}$, $V^{\prime\prime}\phi\phi^{\prime}$,
$V^{\prime\prime}\phi^{2}$ and $V^{\prime}\phi$ are bounded. Then, recalling
(40) we have the convergence in distribution as $N$ goes to infinity

$\sqrt{N}\nu_{N}(\Xi\phi)\to\mathcal{N}(0,q_{P}(\phi))\,.$

###### Proof.

For $N\geqslant 1$, let $E_{N}^{-},E_{N}^{+}$ be given by Corollary 4.2. Let
$\chi_{N}:{\mathbb{R}}\to[0,1]$ be $\mathcal{C}^{2}$ with compact support such
that

$\chi_{N}(x)=1\text{ for }x\in[E_{N}^{-}-1,E_{N}^{+}+1]\text{ and
}\chi_{N}(x)=0\text{ for }x\in[E_{N}^{-}-2,E_{N}^{+}+2]^{c}$

and such that, denoting $\phi_{N}=\phi\chi_{N}$,
$\sup_{N}\|\phi_{N}^{(k)}\|_{\infty}+\|\phi_{N}^{(k)}\|_{L^{2}({\mathbb{R}})}<+\infty$
for $k=0,1,2$ (we assumed $\phi\in H^{2}({\mathbb{R}})$, in particular $\phi$
and $\phi^{\prime}$ is bounded and such a $\chi_{N}$ exists). The point of
cutting $\phi$ outside the set $[E_{N}^{-}-1,E_{N}^{+}+1]$ is that with high
probability, the empirical measure $\hat{\mu}_{N}$ doesn’t see the difference
between $\phi$ and $\phi_{N}$.

The support of $\phi_{N}$ is then contained in $[E_{N}^{-}-2,E_{N}^{+}+2]$,
and we now argue that the proof of Proposition 5.1 can be adapted so that

$\sqrt{N}\nu_{N}(\Xi\phi_{N})\to\mathcal{N}(0,q_{P}(\phi))\,.$ (46)

Similarly as in Proposition 5.1, we perform in $Z_{N}^{V,P}$ the change of
variables $x_{i}=y_{i}+\frac{t}{\sqrt{N}}\phi_{N}(y_{i})$, $1\leqslant
i\leqslant N$, which is the same as before, but with $\phi$ replaced by
$\phi_{N}$. First, with $I_{N}:=[E_{N}^{-}-2,E_{N}^{+}+2]$, the error term

$K_{N}(t,\phi_{N})\leqslant
2\frac{t^{3}}{3N^{1/2}}\|\phi_{N}^{\prime}\|_{\infty}^{3}+\frac{t^{3}}{6N^{1/2}}\|\phi_{N}\|_{\infty}\sup_{d(x,I_{N})\leqslant
1}|V^{(3)}(x)|$

of the proof of Proposition 5.1 is still going to zero, because of our choice
of $\chi_{N}$ and Assumption 1.2. As previously, we then have

$\lim_{N\to\infty}\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\Lambda_{N}(\phi_{N})+\frac{t^{2}}{2}F_{N}(\hat{\mu}_{N})}\right]=1$
(47)

with

$\Lambda_{N}(\phi_{N}):=2P\iint_{{\mathbb{R}}^{2}}\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}d\mu_{P}(x)d(\hat{\mu}_{N}-\mu_{P})(y)+\int_{\mathbb{R}}(\phi_{N}^{\prime}-V^{\prime}\phi_{N})d(\hat{\mu}_{N}-\mu_{P})+P\zeta_{N}(\phi_{N})\,,$

where $\zeta_{N}$ is given by (35), and

$F_{N}(\hat{\mu}_{N})=-P\iint_{{\mathbb{R}}^{2}}\left(\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}\right)^{2}d\hat{\mu}_{N}(x)d\hat{\mu}_{N}(y)-\int_{\mathbb{R}}(V^{\prime\prime}\phi_{N}^{2}+\phi_{N}^{\prime
2})d\hat{\mu}_{N}\,.$

Taking again the distance $d_{\text{Lip}}$ defined in (44), one can check that
for $\mu$, $\nu$ probability measures over ${\mathbb{R}}$,

$\left|F_{N}(\mu)-F_{N}(\nu)\right|\leqslant C_{N}d_{\text{Lip}}(\mu,\nu)\,,$

where $C_{N}$ is a term depending on the norms
$\|\phi_{N}^{\prime}\|_{\infty},\|\phi_{N}^{\prime\prime}\|_{\infty}$,
$\|V^{\prime\prime}\phi_{N}^{2}\|_{\infty}$ and
$\|(V^{\prime\prime}\phi_{N}^{2})^{\prime}\|_{\infty}$. The choice of
$\chi_{N}$ and the fact that $\phi$ is chosen so that $V^{(3)}\phi^{2}$ and
$V^{\prime\prime}\phi\phi^{\prime}$ are bounded guarantee that
$\|(V^{\prime\prime}\phi_{N}^{2})^{\prime}\|_{\infty}$ is bounded in $N$. The
other norms are easily bounded by hypothesis. Therefore $C_{N}$ can be seen to
be uniformly bounded in $N$, and we find some $C\geqslant 0$ independent of
$N$ such that

$\left|F_{N}(\mu)-F_{N}(\nu)\right|\leqslant Cd_{\text{Lip}}(\mu,\nu)\,.$

As in proposition 5.1, we use the large deviation principle for
$(\hat{\mu}_{N})$ to deduce

$\lim_{N\to+\infty}\mathbb{E}^{V,P}_{N}\left[e^{t\sqrt{N}\Lambda_{N}(\phi_{N})+\frac{t^{2}}{2}F_{N}(\hat{\mu}_{N})}\right]=\lim_{N\to+\infty}\mathbb{E}^{V,P}_{N}\left[e^{t\sqrt{N}\Lambda_{N}(\phi_{N})}\right]e^{\frac{t^{2}}{2}F_{N}(\mu_{P})}\,.$

By dominated convergence, $F_{N}(\mu_{P})$ converges to $F(\mu_{P})$, the
function $F$ being given by (43). This shows the convergence as $N$ goes to
infinity

$\lim_{N\to+\infty}\mathbb{E}^{V,P}_{N}\left[e^{t\sqrt{N}\Lambda_{N}(\phi_{N})}\right]=e^{-\frac{t^{2}}{2}F(\mu_{P})}\,,$

and $\sqrt{N}\Big{(}\nu_{N}(\Xi\phi_{N})+P\zeta_{N}(\phi_{N})\Big{)}$
converges towards a centered Gaussian variable with variance
$-F(\mu_{P})=q_{P}(\phi)$. Because
$\sup_{N}\|\phi_{N}\|_{H^{2}({\mathbb{R}})}$ is finite, we can apply again
Corollary 3.4 to deduce the convergence in law (46). We now have the
ingredients to conclude, by showing that the characteristic function

$\mathbb{E}^{V,P}_{N}\left[e^{\mathrm{i}t\sqrt{N}\nu_{N}(\Xi\phi)}\right]=\mathbb{E}^{V,P}_{N}\left[e^{\mathrm{i}t\sqrt{N}\int\Xi\phi
d\hat{\mu}_{N}}\right]e^{-\mathrm{i}t\sqrt{N}\int\Xi\phi d\mu_{P}}$

converges to the characteristic of a Gaussian variable with appropriate
variance. By Corollary 4.2, the probability under $\mathbb{P}^{V,P}_{N}$ of
the event
$\mathcal{E}_{N}=\bigg{\\{}x_{1},\ldots,x_{N}\in[E_{N}^{-}-1,E_{N}^{+}+1]\bigg{\\}}$
converges to $1$. Along with the convergence (46), we deduce

$\displaystyle
e^{-\frac{t^{2}}{2}q_{P}(\phi)}=\lim_{N}\mathbb{E}_{N}^{V,P}\left[e^{\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\hat{\mu}_{N}}\right]e^{-\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\mu_{P}}=\lim_{N}\mathbb{E}_{N}^{V,P}\left[\mathbf{1}_{\mathcal{E}_{N}}e^{\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\hat{\mu}_{N}}\right]e^{-\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\mu_{P}}\,,$

Where we used

$\left|\mathbb{E}^{V,P}_{N}\left[\mathbf{1}_{\mathcal{E}_{N}^{c}}e^{\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\hat{\mu}_{N}}\right]e^{-\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\mu_{P}}\right|\leqslant\mathbb{P}^{V,P}_{N}(\mathcal{E}_{N}^{c})\xrightarrow[N\to+\infty]{}0\,.$

Using that $\phi_{N}=\phi$ on $J_{N}={[E_{N}^{-}-1,E_{N}^{+}+1]}$,

$\displaystyle\int\Xi\phi_{N}d\mu_{P}$
$\displaystyle=2P\iint\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}d\mu_{P}(x)d\mu_{P}(y)+\int(\phi_{N}^{\prime}-V^{\prime}\phi_{N})d\mu_{P}$
$\displaystyle=2P\iint_{J_{N}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d\mu_{P}(x)d\mu_{P}(y)+2P\iint_{(J_{N}^{2})^{c}}\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}d\mu_{P}(x)d\mu_{P}(y)$
$\displaystyle+\int_{J_{N}}(\phi^{\prime}-V^{\prime}\phi)d\mu_{P}+\int_{J_{N}^{c}}(\phi\chi_{N}^{\prime}+\phi^{\prime}\chi_{N}-V^{\prime}\phi\chi_{N})d\mu_{P}\,.$

By boundedness of $(\|\phi_{N}^{\prime}\|_{\infty})_{N}$, the second term is
bounded by

$C_{P}\iint_{(J_{N}^{2})^{c}}d\mu_{P}d\mu_{P}\leqslant
2C_{P}\mu_{P}(J_{N}^{c})=o(N^{-1/2})\,,$

where we used the union bound and Lemma 4.4. By the same estimate and the fact
that $\chi_{N}$ can be chosen so that $(\|\chi_{N}^{\prime}\|_{\infty})_{N}$
is bounded, and because $\phi^{\prime}$, $V^{\prime}\phi$ are bounded, the
last term is also $o(N^{-1/2})$. By the previous arguments, we also conclude
that

$2P\iint_{(J_{N}^{2})^{c}}\frac{\phi(x)-\phi(y)}{x-y}d\mu_{P}(x)d\mu_{P}(y)+\int_{J_{N}^{c}}(\phi^{\prime}-V^{\prime}\phi)d\mu_{P}=o(N^{-1/2})\,,$

thus

$\displaystyle\int\Xi\phi_{N}d\mu_{P}=\int\Xi\phi d\mu_{P}+o(N^{-1/2})\,,$

and so far we have

$e^{-\frac{t^{2}}{2}q_{P}(\phi)}=\lim_{N}\mathbb{E}_{N}^{V,P}\left[\mathbf{1}_{\mathcal{E}_{N}}e^{\mathrm{i}t\sqrt{N}\int\Xi\phi_{N}d\hat{\mu}_{N}}\right]e^{-\mathrm{i}t\sqrt{N}\int\Xi\phi
d\mu_{P}}\,.$

Finally, on $\mathcal{E}_{N}$, using $\phi_{N}=\phi$ and that $\hat{\mu}_{N}$
is supported in $J_{N}$,

$\displaystyle\int\Xi\phi_{N}d\hat{\mu}_{N}$
$\displaystyle=2P\iint_{J_{N}^{2}}\frac{\phi(x)-\phi(y)}{x-y}d\mu_{P}(x)d\hat{\mu}_{N}(y)+2P\iint_{(J_{N}^{2})^{c}}\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}d\mu_{P}(x)d\hat{\mu}_{N}(y)+\int_{J_{N}}(\phi^{\prime}-V^{\prime}\phi)d\hat{\mu}_{N}$
$\displaystyle=2P\iint\frac{\phi(x)-\phi(y)}{x-y}d\mu_{P}(x)d\hat{\mu}_{N}(y)+\int(\phi^{\prime}-V^{\prime}\phi)d\hat{\mu}_{N}+o(N^{-1/2})\,,$

Where in the second line we used, using Lemma 4.4 again, that

$\iint_{(J_{N}^{2})^{c}}\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}d\mu_{P}(x)d\hat{\mu}_{N}(y)=\iint_{J_{N}\times
J_{N}^{c}}\frac{\phi_{N}(x)-\phi_{N}(y)}{x-y}d\mu_{P}(x)d\hat{\mu}_{N}(y)=o(N^{-1/2})\,,$

and the same estimate holds for $\phi_{N}$ replaced by $\phi$. Therefore,

$e^{-\frac{t^{2}}{2}q_{P}}(\phi)=\lim_{N}\mathbb{E}_{N}^{V,P}\left[\mathbf{1}_{\mathcal{E}_{N}}e^{\mathrm{i}t\sqrt{N}\int\Xi\phi
d\hat{\mu}_{N}}\right]e^{-\mathrm{i}t\sqrt{N}\int\Xi\phi d\mu_{P}}\,.$

This establishes that

$\lim_{N}\mathbb{E}^{V,P}_{N}\left[e^{\mathrm{i}t\sqrt{N}\int\Xi\phi
d\hat{\nu}_{N}}\right]=e^{-\frac{t^{2}}{2}q_{P}(\phi)}\,,$

which concludes the proof. ∎

######  Remark 5.3.

Taking $\phi$ such that $\phi^{\prime}$ satisfies the conditions of Theorem
5.2, we then have

$\mathbb{E}_{N}^{V,P}\left[e^{t\sqrt{N}\nu_{N}(\mathcal{L}\phi)}\right]\underset{N\rightarrow\infty}{\longrightarrow}\exp\left\\{\frac{t^{2}}{2}q_{P}(\phi^{\prime})\right\\},$
(48)

where the operator $\mathcal{L}$ is defined as
$\mathcal{L}\phi:=\Xi\phi^{\prime}$, ie

$\mathcal{L}\phi=2P\int_{\mathbb{R}}\frac{\phi^{\prime}(x)-\phi^{\prime}(y)}{x-y}d\mu_{P}(y)+\phi^{\prime\prime}(x)-V^{\prime}(x)\phi^{\prime}(x)\,.$
(49)

Note that
$q_{P}^{V}(\phi^{\prime})=\big{(}\sigma_{P}^{V}\big{)}^{2}(\mathcal{L}\phi)$
where $\sigma_{P}^{V}$ is defined in (12). By Theorem 7.1, the class of
functions in $\mathcal{L}^{-1}(\mathcal{T})$ where

$\mathcal{T}:=\left\\{f\in\mathcal{C}^{1}({\mathbb{R}}),\exists\varepsilon>0,\,f(x)=\underset{|x|\rightarrow\infty}{O}\left(x^{-\frac{1}{2}-\varepsilon}\right),\;f^{\prime}(x)=\underset{|x|\rightarrow\infty}{O}\left(x^{-\frac{1}{2}-\varepsilon}\right),\int_{\mathbb{R}}f\rho_{P}=0\right\\}$

satisfies (48). This proves Theorem 1.3.

We now prove a more compact formula for the variance such as the one appearing
in [HL21].

######  Lemma 5.4.

The following equality holds for all $\phi\in\mathcal{T}$

$\braket{\mathcal{L}^{-1}\phi,\phi}_{\mathsf{H}}=\big{(}\sigma_{P}^{V}\big{)}^{2}(\phi):=\int_{\mathbb{R}}\Bigg{(}\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime\prime}(x)^{2}+V^{\prime\prime}(x)\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime}(x)^{2}\Bigg{)}d\mu_{P}(x)\\\
+P\iint_{{\mathbb{R}}^{2}}\Bigg{(}\frac{\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime}(x)-\big{(}\mathcal{L}^{-1}\phi\big{)}^{\prime}(y)}{x-y}\Bigg{)}^{2}d\mu_{P}(x)d\mu_{P}(y)$
(50)

###### Proof.

It suffices to show that
$\big{(}\sigma_{P}^{V}\big{)}^{2}(\mathcal{L}\phi)=\braket{\mathcal{L}\phi,\phi}_{\mathsf{H}}$
for all $\phi$, such that $\phi^{\prime}\in H^{2}({\mathbb{R}})$.

$\braket{\mathcal{L}\phi,\phi}_{\mathsf{H}}=-\int_{\mathbb{R}}\Big{(}\dfrac{(\phi^{\prime}\rho_{P})^{\prime}}{\rho_{P}}\Big{)}^{\prime}\phi^{\prime}\rho_{P}-2P\int_{\mathbb{R}}\mathcal{H}[\phi^{\prime}\rho_{P}]^{\prime}\phi^{\prime}\rho_{P}$

Proceeding to integration by parts in the first integral leads to

$-\int_{\mathbb{R}}\Big{(}\dfrac{(\phi^{\prime}\rho_{P})^{\prime}}{\rho_{P}}\Big{)}^{\prime}\phi^{\prime}\rho_{P}=\int_{\mathbb{R}}\Big{(}\dfrac{(\phi^{\prime}\rho_{P})^{\prime}}{\rho_{P}}\Big{)}^{2}\rho_{P}=\int_{\mathbb{R}}\phi^{\prime\prime
2}\rho_{P}+2\phi^{\prime}\phi^{\prime\prime}\rho_{P}^{\prime}+\phi^{\prime
2}\big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\big{)}^{2}\rho_{P}\\\
=\int_{\mathbb{R}}\phi^{\prime\prime 2}\rho_{P}-\phi^{\prime
2}\dfrac{\rho_{P}^{\prime\prime}}{\rho_{P}}\rho_{P}+\phi^{\prime
2}\big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\big{)}^{2}\rho_{P}$

Since

$\dfrac{\rho_{P}^{\prime\prime}}{\rho_{P}}=\Big{(}-V^{\prime\prime}-2P\mathcal{H}[\rho_{P}]^{\prime}+V^{\prime
2}+4P^{2}\mathcal{H}[\rho_{P}]^{2}+4PV^{\prime}\mathcal{H}[\rho_{P}]\Big{)}=-V^{\prime\prime}-2P\mathcal{H}[\rho_{P}]^{\prime}+\Big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\Big{)}^{2}$

we obtain

$\braket{\mathcal{L}\phi,\phi}_{\mathsf{H}}=\int_{\mathbb{R}}\phi^{\prime\prime
2}\rho_{P}+V^{\prime\prime}\phi^{2}\rho_{P}-2P\int_{\mathbb{R}}\mathcal{H}[\phi^{\prime}\rho_{P}]^{\prime}\phi^{\prime}\rho_{P}+2P\int_{\mathbb{R}}\mathcal{H}[\rho_{P}]^{\prime}\phi^{\prime
2}\rho_{P}$

To conclude, we just have to show that

$\iint_{{\mathbb{R}}^{2}}\Big{(}\dfrac{\phi^{\prime}(x)-\phi^{\prime}(y)}{x-y}\Big{)}^{2}d\mu_{P}(x)d\mu_{P}(y)=2\int_{\mathbb{R}}\mathcal{H}[\rho_{P}]^{\prime}\phi^{\prime
2}\rho_{P}-\mathcal{H}[\phi^{\prime}\rho_{P}]^{\prime}\phi^{\prime}\rho_{P}$

First

$\int_{\mathbb{R}}\mathcal{H}[\phi^{\prime}\rho_{P}]^{\prime}\phi^{\prime}\rho_{P}=\int_{\mathbb{R}}\int_{\mathbb{R}}\dfrac{\phi^{\prime}(x)\phi^{\prime}(y)}{(y-x)^{2}}d\mu_{P}(x)d\mu_{P}(y)$

Secondly

$\int_{\mathbb{R}}\mathcal{H}[\rho_{P}]^{\prime}\phi^{\prime
2}\rho_{P}=\dfrac{1}{2}\iint_{{\mathbb{R}}^{2}}\dfrac{\phi^{\prime
2}(x)+\phi^{\prime 2}(y)}{(y-x)^{2}}d\mu_{P}(x)d\mu_{P}(y)$

which allows to conclude that
$\big{(}\sigma_{P}^{V}\big{)}^{2}(\mathcal{L}\phi)=\braket{\mathcal{L}\phi,\phi}_{\mathsf{H}}$.
∎

##  6 Inversion of $\mathcal{L}$

This section is dedicated to the definition of $\mathcal{L}$ given by (8) and
its domain and then we focus on its inversion. We rely heavily on results of
Appendix A: the diagonalization of the operator $\mathcal{A}$ by the use of
the theory of Schrödinger operators.

Let $P>0$ be fixed. We introduce the operators $\mathcal{A}$ and
$\mathcal{W}$, acting on sufficiently smooth functions of $L^{2}(\rho_{P})$,
by

$\mathcal{A}\phi=-\dfrac{\Big{(}\phi^{\prime}\rho_{P}\Big{)}^{\prime}}{\rho_{P}}=-\left(\phi^{\prime\prime}+\frac{\rho_{P}^{\prime}}{\rho_{P}}\phi^{\prime}\right)\quad\text{and}\quad\mathcal{W}\phi=-\mathcal{H}\big{[}\phi^{\prime}\rho_{P}\big{]}\,.$
(51)

One can show that the operator $\mathcal{A}$ corresponds to the operator
verifying:

$\braket{\phi,\psi}_{\mathsf{H}}=\int_{\mathbb{R}}\phi^{\prime}\psi^{\prime}d\mu_{P}=\int_{\mathbb{R}}\phi\mathcal{A}\psi
d\mu_{P}=\braket{\phi,\mathcal{A}\psi}_{L^{2}(\mu_{P})}$

We first show the following decomposition of $\mathcal{L}$.

######  Lemma 6.1.

For $\phi$ twice differentiable we have the following pointwise identity

$-\mathcal{L}\phi=\mathcal{A}\phi+2P\mathcal{W}\phi\,.$ (52)

###### Proof.

We write for $x\in{\mathbb{R}}$

$2P\int_{\mathbb{R}}\frac{\phi^{\prime}(x)-\phi^{\prime}(y)}{x-y}\rho_{P}(y)dy=-2P\phi^{\prime}(x)\mathcal{H}[\rho_{P}](x)+2P\mathcal{H}[\phi^{\prime}\rho_{P}](x)\,.$
(53)

Then,

$\mathcal{L}\phi=\phi^{\prime\prime}-V^{\prime}\phi^{\prime}-2P\phi^{\prime}\mathcal{H}[\rho_{P}]+2P\mathcal{H}\big{[}\phi^{\prime}\rho_{P}\big{]}\,.$

By (19) we have
$-V^{\prime}-2P\mathcal{H}[\rho_{P}]=\frac{\rho_{P}^{\prime}}{\rho_{P}}$,
which concludes the proof. ∎

In order to state the next theorem, whose proof we detail in the Appendix, we
introduce the following Sobolev-type spaces. Let

$H^{1}_{V^{\prime}}({\mathbb{R}}):=\Big{\\{}u\in
H^{1}({\mathbb{R}}),\,uV^{\prime}\in L^{2}({\mathbb{R}})\Big{\\}}\,.$

We now define

$\mathcal{D}(\mathcal{S})=\Big{\\{}u\in
H^{1}_{V^{\prime}}({\mathbb{R}}),-u^{\prime\prime}+(w_{P}+\alpha)u\in
L^{2}({\mathbb{R}})\Big{\\}}$

and

$\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A}):=\rho_{P}^{-1/2}\mathcal{D}(\mathcal{S})$

and its homogeneous counterpart

$\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A}):=\Big{\\{}u\in\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A}),\,\int_{\mathbb{R}}u\rho_{P}dx=0\Big{\\}}\,.$

Finally, we let $L^{2}_{0}(\rho_{P})$ be the subset of $L^{2}(\rho_{P})$ of
zero mean functions with respect to $\rho_{P}$.

We detail the proof of the following theorem in Appendix A which is based on
Schrödinger operators theory.

######  Theorem 6.2 (Diagonalization of $\mathcal{A}$ in
$L^{2}_{0}(\rho_{P})$).

There exists a sequence $0<\lambda_{1}<\lambda_{2}<\ldots$ going to infinity,
and a complete orthonormal set $(\phi_{n})_{n\geqslant 1}$ of
$L^{2}_{0}(\rho_{P})$ of associated eigenfunctions for $\mathcal{A}$, meaning
that

  * •

$\operatorname{span}\\{\phi_{n},\,n\geqslant 1\\}$ is dense in
$L^{2}_{0}(\rho_{P})$,

  * •

For all $i,j$, $\Braket{\phi_{i},\phi_{j}}_{L^{2}(\rho_{P})}=\delta_{i,j}$,

  * •

For all $n\geqslant 1$, $\mathcal{A}\phi_{n}=\lambda_{n}\phi_{n}$.

Furthermore, each $\phi_{n}$ is in
$\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$,
$\mathcal{A}:\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})\to
L^{2}_{0}(\rho_{P})$ is bijective, and we have the writing, for $u\in
L^{2}_{0}(\rho_{P})$

$\mathcal{A}^{-1}u=\sum_{n\geqslant
1}\lambda_{n}^{-1}\Braket{u,\phi_{n}}_{L^{2}(\rho_{P})}\phi_{n}\,.$

We see the operators $\mathcal{A}$ and $\mathcal{W}$ as unbounded operators on
the space

$\mathsf{H}=\Big{\\{}u\in
H^{1}(\rho_{P})\,|\,\int_{\mathbb{R}}u\rho_{P}dx=0\Big{\\}}$

endowed with the inner product
$\Braket{u,v}_{\mathsf{H}}=\Braket{u^{\prime},v^{\prime}}_{L^{2}(\rho_{P})}$.
This defines an inner product on $\mathsf{H}$ and makes it a complete space:
it can be seen that $H^{1}(\rho_{P})$ is the completion of
$\mathcal{C}_{c}^{\infty}({\mathbb{R}})$ with respect to the inner product
$\Braket{u,v}_{L^{2}(\rho_{P})}+\Braket{u^{\prime},v^{\prime}}_{L^{2}(\rho_{P})}$.
The space $\mathsf{H}$ is then the kernel of the bounded (with respect to
$\|\cdot\|_{\mathsf{H}}$) linear form,
$\langle\widetilde{1},\cdot\rangle_{L^{2}(\rho_{P})}$ on $H^{1}(\rho_{P})$,
and both inner products are equivalent on $\mathsf{H}$ because of the Poincaré
inequality, Proposition 2.6. The use of $\mathsf{H}$ is motivated by the fact
that both $\mathcal{A}$ and $\mathcal{W}$ are self-adjoint positive on this
space as we show in Lemma 6.4.

In the next proposition, we deduce from Theorem 6.2 the diagonalization of
$\mathcal{A}$ in $\mathsf{H}$.

######  Proposition 6.3 (Diagonalization of $\mathcal{A}$ in $\mathsf{H}$).

With the same eigenvalues $0<\lambda_{1}<\lambda_{2}<\ldots$ as in Theorem
6.2, there exists a complete orthonormal set $(\psi_{n})_{n\geqslant 1}$ of
$\mathsf{H}$ formed by eigenfunctions of $\mathcal{A}$.

###### Proof.

With $(\phi_{n})_{n\geqslant 1}$ of Theorem 6.2,

$\displaystyle\delta_{i,j}=\langle\phi_{i},\phi_{j}\rangle_{L^{2}(\rho_{P})}$
$\displaystyle=\frac{1}{\lambda_{j}}\langle\phi_{i},\mathcal{A}\phi_{j}\rangle_{L^{2}(\rho_{P})}$
$\displaystyle=\frac{1}{\lambda_{j}}\langle\phi_{i}^{\prime},\phi_{j}^{\prime}\rangle_{L^{2}(\rho_{P})}$
$\displaystyle=\frac{1}{\lambda_{j}}\langle\phi_{i},\phi_{j}\rangle_{\mathsf{H}}.$

With $\psi_{n}=\frac{1}{\sqrt{\lambda_{n}}}\phi_{n}$, $(\psi_{n})_{n\geqslant
1}$ is then orthonormal with respect to the inner product of $\mathsf{H}$. To
show that $\operatorname{span}\\{\psi_{n},n\geqslant 1\\}$ is dense in
$\mathsf{H}$, let $u\in\mathsf{H}$ be such that for all $j\geqslant 1$,
$\langle u,\phi_{j}\rangle_{\mathsf{H}}=0$. In the last series of equalities,
replace $\phi_{i}$ by $u$: we see that $u$ is orthogonal to each $\phi_{j}$ in
$L^{2}(\rho_{P})$, thus $u$ is a constant as shown in the proof of Lemma A.10,
and because $u\in\mathsf{H}$ it has zero mean against $\rho_{P}$. This shows
that $u=0$. ∎

We set for what follows
$\mathcal{D}(\mathcal{A}):=\left\\{u\in\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})\,|\,\mathcal{A}u\in\mathsf{H}\right\\}$
and
$\mathcal{D}(\mathcal{W}):=\left\\{u\in\mathsf{H}\,|\,\mathcal{W}u\in\mathsf{H}\right\\}$.

######  Lemma 6.4.

The following properties hold:

  * •

The operator $\mathcal{W}:\mathcal{D}(\mathcal{W})\to\mathsf{H}$ is positive:
for all $\phi\in\mathcal{D}(\mathcal{W})$,

$\langle\mathcal{W}\phi,\phi\rangle_{\mathsf{H}}=\dfrac{1}{2}\|\phi^{\prime}\rho_{P}\|_{1/2}^{2}\geqslant
0\,,$

with equality only for $\phi=0$, where the $1/2$-norm of $u$ is given by

$\|u\|_{1/2}^{2}=\int_{\mathbb{R}}|x|.\left|\mathcal{F}[u](x)\right|^{2}dx\,.$

  * •

Both $\mathcal{A}$ and $\mathcal{W}$ are self-adjoint for the inner product of
$\mathsf{H}$.

###### Proof.

To prove the first point, let $\phi\in\mathcal{D}(\mathcal{W})$. Then,

$2\pi\Braket{\mathcal{W}\phi,\phi}_{\mathsf{H}}=-2\pi\Braket{\mathcal{H}[\phi^{\prime}\rho_{P}]^{\prime},\phi^{\prime}\rho_{P}}_{L^{2}(dx)}=-\Braket{ix\mathcal{F}\big{[}\mathcal{H}[\phi^{\prime}\rho_{P}]\big{]},\mathcal{F}[\phi^{\prime}\rho_{P}]}_{L^{2}(dx)}\\\
=\pi\Braket{}{x}{\mathcal{F}[\phi^{\prime}\rho_{P}],\mathcal{F}[\phi^{\prime}\rho_{P}]}_{L^{2}(dx)}=\pi\|\phi^{\prime}\rho_{P}\|_{1/2}^{2}\geqslant
0\,,$

and because $\phi$ is in $\mathsf{H}$, this last quantity is zero if and only
if $\phi$ vanishes.

For the second point, let $u,v\in\mathcal{D}(\mathcal{W})$. Using Plancherel’s
isometry and i) of Lemma 2.1,

$\displaystyle\Braket{\mathcal{W}u,v}_{\mathsf{H}}=\Braket{(\mathcal{W}u)^{\prime},v^{\prime}\rho_{P}}_{L^{2}(dx)}=\frac{1}{2}\Braket{}{x}{\mathcal{F}[u^{\prime}\rho_{P}],\mathcal{F}[v^{\prime}\rho_{P}]}_{L^{2}(dx)}\,,$

and this last expression is symmetric in $(u,v)$. The proof of the self-
adjointness of $\mathcal{A}$ follows from integration by parts. ∎

######  Definition 6.5 (Quadratic form associated to $-\mathcal{L}$).

We define for all $u,v\in\mathsf{H}\cap\mathcal{C}_{c}^{\infty}({\mathbb{R}})$
the quadratic form associated to $-\mathcal{L}$ by

$q_{-\mathcal{L}}(u,v)=\braket{\mathcal{A}u,\mathcal{A}v}_{L^{2}(\rho_{P})}+2P\braket{\mathcal{F}[u^{\prime}\rho_{P}],\mathcal{F}[v^{\prime}\rho_{P}]}_{L^{2}(|x|dx)}$

Note that for all
$u,v\in\mathsf{H}\cap\mathcal{C}_{c}^{\infty}({\mathbb{R}})$,
$q_{-\mathcal{L}}(u,v)=\braket{-\mathcal{L}u,v}_{\mathsf{H}}$ and that
whenever $u\in\mathcal{D}(\mathcal{A})\cap\mathcal{D}(\mathcal{W})$,

$q_{-\mathcal{L}}(u,u)=\braket{\mathcal{A}u,u}_{\mathsf{H}}+2P\braket{\mathcal{W}u,u}_{\mathsf{H}}\geqslant\lambda_{1}(\mathcal{A})\|u\|_{\mathsf{H}}^{2}$
(54)

by Proposition 6.3 and Lemma 6.4. After extending the $q_{-\mathcal{L}}$ to
its form domain $Q(\mathcal{L})$ which is equal to
$\Big{\\{}u\in\mathsf{H},\mathcal{A}u\in
L^{2}(\rho_{P}),\,\mathcal{F}[u^{\prime}\rho_{P}]\in
L^{2}(|x|dx)\Big{\\}}=\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$. The
equality comes from the fact that
$\mathcal{A}^{-1}\Big{(}L^{2}_{0}(\rho_{P})\Big{)}=\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$,
that $\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})\subset\mathsf{H}$ and
that $\mathcal{F}[u^{\prime}\rho_{P}]\in L^{2}(x^{2}dx)$ whenever
$u\in\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$, indeed
$u^{\prime}\rho_{P}\in H^{1}({\mathbb{R}})$ because
$(u^{\prime}\rho_{P})^{\prime}=-\rho_{P}\mathcal{A}u\in L^{2}({\mathbb{R}})$.
We now define $\mathcal{D}(\mathcal{L})$ the domain of definition of
$-\mathcal{L}$ by:

$\mathcal{D}(\mathcal{L}):=\Big{\\{}u\in Q(\mathcal{L}),v\mapsto
q_{-\mathcal{L}}(u,v)\text{ can be extended to a continuous linear form on
}\mathsf{H}\Big{\\}}$

######  Proposition 6.6.

$\mathcal{D}(\mathcal{L})=\mathcal{D}(\mathcal{A})\cap\mathcal{D}(\mathcal{W})$.

###### Proof.

Let $u\in\mathcal{D}(\mathcal{L})$, by Riesz’s theorem there exists
$f_{u}\in\mathsf{H}$, such that
$q_{-\mathcal{L}}(u,v)=\braket{f_{u},v}_{\mathsf{H}}$ for all
$v\in\mathsf{H}$, we set $-\mathcal{L}u:=f_{u}$, it is called the Friedrichs
extension of $-\mathcal{L}$. Then for all
$v\in\mathsf{H}\cap\mathcal{C}_{c}^{\infty}({\mathbb{R}})$, by integration by
part we get:

$\braket{-\mathcal{L}u,v}_{\mathsf{H}}=q_{-\mathcal{L}}(u,v)=\braket{u,\mathcal{A}v}_{\mathsf{H}}+2P\braket{u,\mathcal{W}v}_{\mathsf{H}},$

hence we deduce the distributional identity
$-\mathcal{L}u=\mathcal{A}u+2P\mathcal{W}u$. Since
$u\in\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$, $\mathcal{W}u\in
H^{1}(\rho_{P})$ implying that $\mathcal{A}u\in\mathsf{H}$ and then that
$\mathcal{W}u\in\mathsf{H}$. The converse is trivially true. ∎

We are now ready to state the main theorem of this section, that is the
inversion of $\mathcal{L}$ on $\mathcal{D}(\mathcal{L})$.

######  Theorem 6.7 (Inversion of $\mathcal{L}$).

$-\mathcal{L}:\mathcal{D}(\mathcal{L})\longrightarrow\mathsf{H}$ is bijective.
Furthermore, $(-\mathcal{L})^{-1}$ is continuous from
$(\mathsf{H},\|.\|_{\mathsf{H}})$ to
$(\mathcal{D}(\mathcal{L}),q_{-\mathcal{L}})$.

###### Proof.

Let $f\in\mathsf{H}$, since $\braket{f,.}_{\mathsf{H}}$ is a linear form on
$Q(\mathcal{L})=\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$ which is, by
(54), continuous with respect to $q_{-\mathcal{L}}$, one can applies Riesz’s
theorem so there exists a unique
$u_{f}\in\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$, such that for all
$v\in\mathsf{H}$, $\braket{f,v}_{\mathsf{H}}=q_{-\mathcal{L}}(u_{f},v)$.
Since, $u_{f}$ is clearly in $\mathcal{D}(\mathcal{L})$ by definition of the
Friedrichs extension of $-\mathcal{L}$, we have $-\mathcal{L}u=f$. ∎

######  Remark 6.8.

We can diagonalize $\mathcal{L}$ by the same argument we used in Appendix A to
diagonalize $\mathcal{A}$ in $L^{2}_{0}(\rho_{P})$.

We now state a result that could allow one to extract more regularity for
$\mathcal{L}^{-1}$ by the help of an explicit form that uses Fredholm
determinant theory for Hilbert-Schmidt operators, the reader can refer to
[GGK12].

######  Definition 6.9 (Fredholm determinant).

Let $\mathcal{U}$ be a self-adjoint Hilbert-Schmidt operator, we denote the
Fredholm determinant by $\det(I+\mathcal{U})$.

######  Theorem 6.10 (Determinant formula for $\mathcal{L}^{-1}$).

For all $u\in\mathsf{H}$, such that
$x\mapsto\dfrac{1}{\rho_{P}(x)}\displaystyle\int_{x}^{+\infty}u(t)\rho_{P}(t)dt$
is integrable at $+\infty$, we have:

$\mathcal{L}^{-1}u=\mathcal{A}^{-1}u-\rho_{P}^{-1/2}\mathcal{R}\big{[}\rho_{P}^{1/2}\mathcal{A}^{-1}u\big{]}$
(55)

where $\mathcal{R}$ is the kernel operator defined for all $v\in
L^{2}({\mathbb{R}})$ by:

$\mathcal{R}[v](x)=\int_{\mathbb{R}}R(x,y)v(y)dy$

where

$R(x,y)=\displaystyle\dfrac{1}{\det(I+\mathcal{K})}\sum_{n\geqslant
0}\dfrac{1}{n!}\int_{{\mathbb{R}}^{n}}\det_{n+1}\begin{bmatrix}K(x,y)&K(x,\lambda_{b})\\\
K(\lambda_{a},y)&K(\lambda_{a},\lambda_{b})\end{bmatrix}_{a,b=1\dots
n}d\lambda_{1}\dots d\lambda_{n}$

where $\mathcal{K}$ is the kernel operator defined for all $w\in
L^{2}(\rho_{P})$ by:

$\mathcal{K}[w](x)=\int_{\mathbb{R}}K(x,y)w(y)dy$ (56)

with

$K(x,y)=-2P\rho_{P}(x)\rho_{P}(y)\ln\Big{|}1-\dfrac{y}{x}\Big{|}.$ (57)

###### Proof.

Let $f\in\mathsf{H}$, there exists a unique $u\in\mathcal{D}(\mathcal{A})$
such that $\mathcal{A}u=f$. Since
$(u^{\prime}\rho_{P})^{\prime}=\rho_{P}\mathcal{A}u\in L^{2}({\mathbb{R}})$,
hence $u^{\prime}\rho_{P}\in H^{1}({\mathbb{R}})$ so
$u^{\prime}(x)\rho_{P}(x)\underset{|x|\rightarrow+\infty}{\longrightarrow}0$.
By definition, $-\dfrac{(u^{\prime}\rho_{P})^{\prime}}{\rho_{P}}=f\text{ hence
}$

$(\mathcal{A}^{-1}f)^{\prime}(x)\rho_{P}(x)=u^{\prime}(x)\rho_{P}(x)=\int_{x}^{+\infty}f(t)\rho_{P}(t)dt.$
(58)

Using the fact that $\int_{\mathbb{R}}u(x)\rho_{P}(x)dx=0$, integrating again
we get:

$u(x)=-\int_{x}^{+\infty}\dfrac{ds}{\rho_{P}(s)}\int_{s}^{+\infty}f(t)\rho_{P}(t)dt+C$

where
$C=\displaystyle\int_{\mathbb{R}}\rho_{P}(x)dx\int_{x}^{+\infty}\dfrac{ds}{\rho_{P}(s)}\int_{s}^{+\infty}f(t)\rho_{P}(t)dt$.
Now let $g\in\mathsf{H}$, there exists a unique
$v\in\mathcal{D}(\mathcal{L})$, such that
$-\mathcal{L}v=\mathcal{A}v+2P\mathcal{W}v=g$ and then
$v+2P\mathcal{W}\mathcal{A}^{-1}v=\mathcal{A}^{-1}g$. using (58), we get:

$\mathcal{W}\mathcal{A}^{-1}v(x)=\fint_{\mathbb{R}}\dfrac{ds}{s-x}\int_{s}^{+\infty}dtv(t)\rho_{P}(t)$

By Sokhotski-Plejmel formula, we have:

$\fint_{\mathbb{R}}\dfrac{ds}{s-x}\int_{s}^{+\infty}dtv(t)\rho_{P}(t)=\underset{M\rightarrow+\infty}{\lim}\lim_{\varepsilon\rightarrow{0}}\int_{-M}^{M}\dfrac{ds}{2}\Big{\\{}\dfrac{1}{x-s+\mathrm{i}\varepsilon}+\dfrac{1}{x-s-\mathrm{i}\varepsilon}\Big{\\}}\int_{s}^{+\infty}dtv(t)\rho_{P}(t)$

We then proceed to an integration by part:

$\fint_{\mathbb{R}}\dfrac{ds}{s-x}\int_{s}^{+\infty}dtv(t)\rho_{P}(t)=\underset{M\rightarrow+\infty}{\lim}\lim_{\varepsilon\rightarrow{0}}\Big{[}-\dfrac{\ln\big{(}(x-s)^{2}+\varepsilon^{2}\big{)}}{2}\int_{s}^{+\infty}dtv(t)\rho_{P}(t)\Big{]}_{-M}^{M}\\\
-\int_{\mathbb{R}}ds\ln|x-s|v(s)\rho_{P}(s)$

To conclude that
$\mathcal{W}\mathcal{A}^{-1}v(x)=-\int_{\mathbb{R}}ds\ln|x-s|v(s)\rho_{P}(s)$,
we just need to show that

$\ln(x)\int_{x}^{+\infty}dtv(t)\rho_{P}(t)\underset{|x|\rightarrow\infty}{\longrightarrow}0$

which can be seen by Cauchy-Schwarz inequality:

$\Big{|}\ln(x)\int_{x}^{+\infty}dtv(t)\rho_{P}(t)\Big{|}\leqslant|\ln(x)|\|v\|_{L^{2}(\rho_{P})}.\rho_{P}(x)^{1/4}\Big{(}\int_{\mathbb{R}}\rho_{P}(t)^{1/2}dt\Big{)}^{1/2}.$

In this inequality, we used that $\rho_{P}$ is decreasing in a neighborhood of
$+\infty$, hence

$\ln|x|\int_{s}^{+\infty}dtv(t)\rho_{P}(t)\underset{x\rightarrow+\infty}{\longrightarrow}0$

the exact same argument allows us to conclude when $x$ goes to $-\infty$.
Using the fact that $\int_{\mathbb{R}}v(t)\rho_{P}(t)dt=0$, we obtain the
following equality:

$v-2P\int_{\mathbb{R}}ds\ln|x-s|v(s)\rho_{P}(s)=\mathcal{A}^{-1}g:=h.$

Now setting $\tilde{v}(t)=\rho_{P}^{1/2}(t)v(t)$ and
$\tilde{h}=\rho_{P}^{1/2}(t)h(t)$, we obtain
$\tilde{v}+\mathcal{K}[\tilde{v}]=\tilde{h}$ where $\mathcal{K}$ is defined in
(56). Since its kernel (defined in (57)) belongs to $L^{2}({\mathbb{R}}^{2})$,
$\mathcal{K}$ is Hilbert-Schmidt. Hence by Fredholm determinant theory:

$\tilde{v}=\tilde{h}-\mathcal{R}[\tilde{h}]$

or
$\mathcal{L}^{-1}g=\mathcal{A}^{-1}g-\rho_{P}^{-1/2}\mathcal{R}\big{[}\rho_{P}^{1/2}\mathcal{A}^{-1}g\big{]}$
as expected. ∎

##  7 Regularity of the inverse of $\mathcal{L}$ and completion of the proof
of Theorem 1.3

Since we have proven the central limit theorem for functions of the type
$\mathcal{L}\phi$ with $\phi$ regular enough and satisfying vanishing
asymptotic conditions at infinity, we exhibit a class of functions $f$ for
which $\mathcal{L}^{-1}f$ is regular enough to satisfy conditions of Theorem
5.2. We define $\mathcal{T}$ the subset of $\mathsf{H}$ defined by

$\mathcal{T}:=\left\\{f\in\mathcal{C}^{1}({\mathbb{R}}),\exists\varepsilon>0,\,f(x)=\underset{|x|\rightarrow\infty}{O}\left(x^{-\frac{1}{2}-\varepsilon}\right),\;f^{\prime}(x)=\underset{|x|\rightarrow\infty}{O}\left(x^{-\frac{1}{2}-\varepsilon}\right),\int_{\mathbb{R}}f\rho_{P}=0\right\\}$

######  Theorem 7.1.

For all $f\in\mathcal{T}$, there exists a unique
$u\in\mathcal{C}^{3}({\mathbb{R}})$ such that $u^{\prime}\in
H^{2}({\mathbb{R}})$ with $u^{(3)}$ bounded wich verifies:

  * •

$u^{\prime}(x)=\underset{|x|\rightarrow\infty}{O}\left(\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}\right)$

  * •

$u^{\prime\prime}(x)=\underset{|x|\rightarrow\infty}{O}\left(\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}\right)$

  * •

$u^{(3)}(x)=\underset{|x|\rightarrow\infty}{O}\left(\dfrac{1}{x^{\frac{1}{2}+\varepsilon}}\right)$

such that $f=\mathcal{L}u$.

###### Proof.

Let $f\in\mathcal{T}\subset\mathsf{H}$, then since $-\mathcal{L}$ is bijective
from $\mathcal{D}(\mathcal{L})\rightarrow\mathsf{H}$, there exists a unique
$u\in\mathcal{D}(\mathcal{L})$ such that $-\mathcal{L}u=f$ ie:

$-u^{\prime\prime}-\dfrac{\rho_{P}^{\prime}}{\rho_{P}}u^{\prime}-2P\mathcal{H}[u^{\prime}\rho_{P}]=f$
(59)

Hence we have

$-(u^{\prime}\rho_{P})^{\prime}=\rho_{P}\Big{(}f+2P\mathcal{H}[u^{\prime}\rho_{P}]\Big{)}.$
(60)

Since
$u\in\mathcal{D}(\mathcal{L})\subset\\{u\in\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A}),\mathcal{A}u\in\mathsf{H}\\}$,
the functions $u^{\prime}\rho_{P}$ and its distributional derivatives
$(u^{\prime}\rho_{P})^{\prime}=-\rho_{P}\mathcal{A}u$ and
$(u^{\prime}\rho_{P})^{\prime\prime}=-\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\rho_{P}^{1/2}.\Big{(}\rho_{P}^{1/2}\mathcal{A}u\Big{)}-\rho_{P}\Big{(}\mathcal{A}u\Big{)}^{\prime}$
are in $L^{2}({\mathbb{R}})$. In particular $u^{\prime}\rho_{P}$ goes to zero
at infinity, and $\mathcal{H}[u^{\prime}\rho_{P}]\in
H^{2}({\mathbb{R}})\subset\mathcal{C}^{1}({\mathbb{R}})$. So we can integrate
(60) on $[x,+\infty[$ , since by Lemma 2.3, the right-hand side behaves like a
$\underset{|x|\rightarrow\infty}{O}\left(\dfrac{\rho_{P}(x)}{|x|^{\frac{1}{2}+\varepsilon}}\right)$,
to get the following expression

$u^{\prime}(x)\rho_{P}(x)=\int_{x}^{+\infty}\dfrac{\rho_{P}(t)}{\rho_{P}^{\prime}(t)}(f+2P\mathcal{H}[u^{\prime}\rho_{P}]).\rho_{P}^{\prime}(t)dt$
(61)

From this expression, we can see that
$u^{\prime}\in\mathcal{C}^{2}({\mathbb{R}})$ so we just have to check the
integrability condition at infinity and the fact that $u^{(3)}$ is bounded.
After proceeding to an integration by parts, which is permitted by the
previous argument, we obtain:

$u^{\prime}(x)=-\dfrac{\rho_{P}(x)}{\rho_{P}^{\prime}(x)}\Big{(}f(x)+2P\mathcal{H}[u^{\prime}\rho_{P}](x)\Big{)}-\dfrac{1}{\rho_{P}(x)}\int_{x}^{+\infty}\Bigg{(}\dfrac{\rho_{P}(t)}{\rho_{P}^{\prime}(t)}(f+2P\mathcal{H}[u^{\prime}\rho_{P}])\Bigg{)}^{\prime}\rho_{P}(t)dt$
(62)

and we define
$R_{1}(x):=\displaystyle\dfrac{1}{\rho_{P}(x)}\int_{x}^{+\infty}\Bigg{(}\dfrac{\rho_{P}(t)}{\rho_{P}^{\prime}(t)}(f+2P\mathcal{H}[u^{\prime}\rho_{P}])\Bigg{)}^{\prime}\rho_{P}(t)dt$,
we will need to show that $R_{1}$ is a remainder of order
$\underset{x\rightarrow+\infty}{O}\left(\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\right)$
at infinity. In this case, we will have
$u^{\prime}(x)=\underset{x\rightarrow+\infty}{O}\left(\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}\right)$
which will be useful for the following. If we reinject (62) in (59), we find:

$u^{\prime\prime}=-(f+2P\mathcal{H}[u^{\prime}\rho_{P}])-\dfrac{\rho^{\prime}_{P}}{\rho_{P}}\Big{(}-\dfrac{\rho_{P}}{\rho_{P}^{\prime}}\Big{(}f+2P\mathcal{H}[u^{\prime}\rho_{P}]\Big{)}-R_{1}\Big{)}=\dfrac{\rho_{P}^{\prime}}{\rho_{P}}R_{1}$
(63)

Hence

$u^{\prime\prime}(x)=\dfrac{\rho_{P}^{\prime}}{\rho_{P}^{2}}(x)\int_{x}^{+\infty}\rho_{P}(t)dt\Bigg{\\{}\underbrace{\Big{(}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}\Big{)}^{\prime}(t)}_{=\underset{t\rightarrow+\infty}{O}\Big{(}\frac{V^{\prime\prime}(t)}{V^{\prime}(t)^{2}}\Big{)}}\underbrace{\big{[}f+2P\mathcal{H}[u^{\prime}\rho_{P}]\big{]}(t)}_{=\underset{t\rightarrow+\infty}{O}\Big{(}t^{-\frac{1}{2}-\varepsilon}\Big{)}}+\underbrace{\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(t)}_{=\underset{t\rightarrow+\infty}{O}\Big{(}\frac{1}{V^{\prime}(t)}\Big{)}}\underbrace{\big{[}f^{\prime}-2P\mathcal{H}[\rho_{P}\mathcal{A}u]\big{]}(t)}_{=\underset{t\rightarrow+\infty}{O}\Big{(}t^{-\frac{1}{2}-\varepsilon}\Big{)}}\Bigg{\\}}.$

The fact that
$\mathcal{H}[\rho_{P}\mathcal{A}u](t)=\underset{t\rightarrow+\infty}{O}(t^{-2})$
comes again from lemma 2.3. Finally we have that,

$u^{(3)}(x)=\Big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}^{2}}\Big{)}^{\prime}(x)\rho_{P}(x)R_{1}(x)-\Big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}^{2}}\Big{)}(x)\Bigg{(}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(f+2P\mathcal{H}[u^{\prime}\rho_{P}])\Bigg{)}^{\prime}(x)\rho_{P}(x)\\\
=\Big{(}\underbrace{\dfrac{\rho_{P}^{\prime\prime}}{\rho_{P}}-2\dfrac{\rho_{P}^{\prime
2}}{\rho_{P}^{2}}\Big{)}(x)}_{=\underset{x\rightarrow+\infty}{O}\Big{(}V^{\prime}(x)^{2}\Big{)}}R_{1}(x)-\underbrace{\Big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\Big{)}(x)\Bigg{(}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(f+2P\mathcal{H}[u^{\prime}\rho_{P}])\Bigg{)}^{\prime}(x)}_{=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{V^{\prime\prime}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}+x^{-\frac{1}{2}-\varepsilon}\Big{)}}$

The second term is
$\underset{x\rightarrow+\infty}{O}\Big{(}x^{-\frac{1}{2}-\varepsilon}\Big{)}$
by the assumption that
$\dfrac{V^{\prime\prime}}{V^{\prime}}(x)=\underset{|x|\rightarrow\infty}{O}(1)$.
Hence, we just have to check that
$R_{1}(x)=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}$
to establish that $u^{\prime}$, $u^{\prime\prime}$, $u^{(3)}$ are in
$L^{2}({\mathbb{R}})$. By a comparison argument, we control $R_{1}$ by
controlling

$I_{1}(x):=\int_{x}^{+\infty}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)}dt$

By integration by parts:

$I_{1}(x):=-\underbrace{\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(x)}_{=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}}-\int_{x}^{+\infty}\rho_{P}(t)\Big{(}\dfrac{1}{t^{\frac{1}{2}+\varepsilon}V^{\prime}}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}\Big{)}^{\prime}(t)dt\\\
=\underset{x\rightarrow+\infty}{O}\Big{(}\frac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}-\int_{x}^{+\infty}\rho_{P}(t)dt\Bigg{\\{}\dfrac{-\frac{1}{2}-\varepsilon}{t^{\frac{3}{2}+\varepsilon}V^{\prime}(t)}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(t)-\dfrac{V^{\prime\prime}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{2}}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}+\dfrac{1}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)}\Big{(}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}\Big{)}^{\prime}(t)\Bigg{\\}}$
(64)

By the same argument as before, the last integral is of the form
$\displaystyle\int_{x}^{+\infty}\underset{t\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{2}}\Big{)}dt$
so if
$I_{2}(x):=\displaystyle\int_{x}^{+\infty}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{2}}dt=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}$
then so is $I_{1}$. By integration by parts, we obtain:

$I_{2}(x)=\rho_{P}(x)\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(x)-\int_{x}^{+\infty}\rho_{P}(t)dt\Bigg{\\{}\Big{(}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}\Big{)}^{\prime}(t)\dfrac{1}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{2}}-\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(t)\Big{(}\dfrac{1}{t^{\frac{3}{2}+\varepsilon}V^{\prime}(t)^{2}}+\dfrac{2V^{\prime\prime}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{3}}\Big{)}\Bigg{\\}}\\\
=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}+\int_{x}^{+\infty}\underset{t\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{3}}\Big{)}dt$

The last integral is a
$\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}$
because, again, by integration by part:

$\int_{x}^{+\infty}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{3}}dt=\rho_{P}(x)\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{3}}\dfrac{\rho_{P}}{\rho_{P}^{\prime}}(x)-\int_{x}^{+\infty}\underset{t\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{4}}\Big{)}$

and finally

$\int_{x}^{+\infty}\dfrac{\rho_{P}(t)}{t^{\frac{1}{2}+\varepsilon}V^{\prime}(t)^{4}}dt\leqslant\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\int_{x}^{+\infty}\dfrac{dt}{V^{\prime}(t)^{2}}=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}$

In the final step, we used the fact that
$x\mapsto\dfrac{\rho_{P}(x)}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}$ is
decreasing in a neighborhood of $+\infty$ (which can be checked by
differentiating) and that $x\mapsto\dfrac{1}{V^{\prime}(x)^{2}}$ is integrable
at $\infty$ by assumption iv). Hence
$R_{1}(x)=\underset{x\rightarrow+\infty}{O}\Big{(}\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)^{2}}\Big{)}$
(the exact same result can be shown at $-\infty$), which leads to the fact

$u^{\prime}(x)=\underset{|x|\rightarrow+\infty}{O}\Big{(}\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}\Big{)},\hskip
28.45274ptu^{\prime\prime}(x)=\underset{|x|\rightarrow+\infty}{O}\Big{(}\dfrac{1}{x^{\frac{1}{2}+\varepsilon}V^{\prime}(x)}\Big{)}\hskip
11.38092pt\text{and}\hskip
11.38092ptu^{(3)}(x)=\underset{|x|\rightarrow+\infty}{O}\Big{(}\dfrac{1}{x^{\frac{1}{2}+\varepsilon}}\Big{)}$
(65)

which establishes that these functions are in $L^{2}$ in a neighborhood of
$\infty$. Since we already showed that
$u\in\mathcal{C}^{3}({\mathbb{R}})\subset H^{3}_{\text{loc}}({\mathbb{R}})$,
it establishes that $u\in
H^{3}({\mathbb{R}})\cap\mathcal{C}^{3}({\mathbb{R}})$ with $u^{(3)}$ bounded.
To complete the proof we just have to show that $(u^{\prime})^{2}V^{(3)}$,
$u^{\prime}u^{\prime\prime}V^{\prime\prime}$,
$(u^{\prime})^{2}V^{\prime\prime}$ and $u^{\prime}V^{\prime}$ are bounded
which is easily checked by (65) and Assumption 1.1 iv). ∎

######  Remark 7.2.

We choose here the functions that vanishes at infinity at worst like
$|x|^{-1/2-\varepsilon}$, but functions like
$x\mapsto|x|^{-1/2}\ln^{-1/2-\varepsilon}|x|$ or
$x\mapsto|x|^{-1/2}\ln^{-1/2}|x|\ln^{-1/2-\varepsilon}\ln|x|$ also work, the
proof being the same. The only hypotheses that we use is that $f\in
H^{1}({\mathbb{R}})\cap\mathcal{C}^{1}({\mathbb{R}})$, that
$f^{\prime}=\underset{|x|\rightarrow+\infty}{O}\left(f(x)\right)$ and that $f$
is decreasing (resp. increasing) in a neighborhood of + (resp.-) $\infty$.

##  Appendix A Appendix: proof of Theorem 6.2

In order to analyze $\mathcal{A}$, we let, for $u\in L^{2}({\mathbb{R}})$,

$\mathcal{S}u:=\rho_{P}^{1/2}\mathcal{A}\rho_{P}^{-1/2}u\,.$

Note that
$u\in\left(L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\right)\mapsto\rho_{P}^{-1/2}u\in\left(L^{2}(\rho_{P}),\|.\|_{L^{2}(\rho_{P})}\right)$
is an isometry. It turns out that it will be easier to study first the
operator $\mathcal{S}$ in order to get the spectral properties of
$\mathcal{A}$.

######  Proposition A.1.

The operator $\mathcal{S}$ is a Schrödinger operator: it admits the following
expression for all $u\in C_{c}^{2}({\mathbb{R}})$:
$\mathcal{S}u=-u^{\prime\prime}+w_{P}u$ with

$w_{P}=\frac{1}{2}\left(\frac{1}{2}V^{\prime
2}-V^{\prime\prime}+2PV^{\prime}\mathcal{H}[\rho_{P}]-2P\mathcal{H}[\rho_{P}^{\prime}]+2P^{2}\mathcal{H}[\rho_{P}]^{2}\right)=\dfrac{1}{2}\Big{[}(\ln\rho_{P})^{\prime\prime}+\dfrac{1}{2}(\ln\rho_{P})^{\prime
2}\Big{]}\,.$

Furthermore, $w_{P}$ is continuous and we have
$w_{P}(x)\underset{\infty}{\sim}\dfrac{V^{\prime}(x)^{2}}{4}\underset{|x|\rightarrow\infty}{\longrightarrow}+\infty$.

###### Proof.

We compute directly

$\displaystyle\dfrac{\Big{(}\rho_{P}\big{(}\rho_{P}^{-1/2}u\big{)}^{\prime}\Big{)}^{\prime}}{\rho_{P}}$
$\displaystyle=\big{(}\rho_{P}^{-1/2}u\big{)}^{\prime\prime}+\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\big{(}\rho_{P}^{-1/2}u\big{)}^{\prime}$
$\displaystyle=\big{(}\rho_{P}^{-1/2}u^{\prime}-\dfrac{1}{2}\rho_{P}^{-3/2}\rho_{P}^{\prime}u\big{)}^{\prime}+\rho_{P}^{\prime}\rho_{P}^{-3/2}u^{\prime}-\dfrac{1}{2}\rho_{P}^{-5/2}\big{(}\rho_{P}^{\prime}\big{)}^{2}u$
$\displaystyle=\rho_{P}^{-1/2}u^{\prime\prime}+\dfrac{1}{4}\rho_{P}^{-5/2}\big{(}\rho_{P}^{\prime}\big{)}^{2}u-\dfrac{1}{2}\rho_{P}^{-3/2}\rho_{P}^{\prime\prime}u$
$\displaystyle=\rho_{P}^{-1/2}\Big{[}u^{\prime\prime}+\dfrac{1}{4}\rho_{P}^{-2}\big{(}\rho_{P}^{\prime}\big{)}^{2}u-\dfrac{1}{2}\rho_{P}^{-1}\rho_{P}^{\prime\prime}u\Big{]}$
$\displaystyle=\rho_{P}^{-1/2}\Bigg{(}u^{\prime\prime}-\dfrac{1}{2}\Big{[}\big{(}\dfrac{\rho_{P}^{\prime\prime}}{\rho_{P}}\big{)}-\dfrac{1}{2}\big{(}\dfrac{\rho_{P}^{\prime}}{\rho_{P}}\big{)}^{2}\Big{]}u\Bigg{)}$
$\displaystyle=\rho_{P}^{-1/2}\Bigg{(}u^{\prime\prime}-\dfrac{1}{2}\Big{[}(\ln\rho_{P})^{\prime\prime}+\dfrac{1}{2}(\ln\rho_{P})^{\prime
2}\Big{]}u\Bigg{)}=\rho_{P}^{-1/2}\Bigg{(}u^{\prime\prime}-w_{P}u\Bigg{)}\,.$

Now, using Lemma 2.2, we have

$w_{P}=\frac{1}{2}\left(\frac{1}{2}V^{\prime
2}-V^{\prime\prime}+2PV^{\prime}\mathcal{H}[\rho_{P}]-2P\mathcal{H}[\rho_{P}^{\prime}]+2P^{2}\mathcal{H}[\rho_{P}]^{2}\right)\,.$

Notice that $\mathcal{H}[\rho_{P}^{\prime}]$ and $\mathcal{H}[\rho_{P}]$ are
bounded since they belong to $H^{1}({\mathbb{R}})$, as we showed in Lemma 2.2
that $\rho_{P}$ is $H^{2}({\mathbb{R}})$. Along with Assumption 1.1 iii) and
Lemma 2.3, we deduce $w_{P}(x)\underset{\infty}{\sim}\dfrac{1}{4}V^{\prime
2}(x)$. ∎

######  Remark A.2.

Note that the function $w_{P}$ need not be positive on ${\mathbb{R}}$. In
fact, neglecting the terms involving the Hilbert transforms of $\rho_{P}$ and
$\rho_{P}^{\prime}$, $w_{P}$ would only be positive outside of a compact set.
However, using the positivity of $\mathcal{A}$, which will be shown further in
the article, we can show that the operator $-u^{\prime\prime}+w_{P}u$ is
itself positive on $L^{2}({\mathbb{R}})$. It can also be checked that, by
integration by parts, $\mathcal{S}$ is self-adjoint on
$\mathcal{C}_{c}^{\infty}({\mathbb{R}})$ with the inner product of
$L^{2}({\mathbb{R}})$.

We now introduce an extension of $\mathcal{S}$ by defining its associated
bilinear form.

######  Definition A.3 (Quadratic form associated to $\mathcal{S}$).

Let $\alpha\geqslant 0$ such that $w_{P}+\alpha\geqslant 1$. We define the
quadratic form associated to $\mathcal{S}+\alpha I$, defined for all
$u\in\mathcal{C}_{c}^{\infty}({\mathbb{R}})$ by

$q_{\alpha}(u,u):=\int_{\mathbb{R}}u^{\prime
2}dx+\int_{\mathbb{R}}u^{2}(w_{P}+\alpha)dx$

This quadratic form can be extended to a larger domain denoted by
$Q(\mathcal{S}+\alpha I)$, called the form domain of the operator
$\mathcal{S}+\alpha I$. By the theory of Schrödinger operators, it is well-
known (see [Dav96][Theorem 8.2.1]) that such a domain is given by

$Q(\mathcal{S}+\alpha I)=\left\\{u\in
H^{1}({\mathbb{R}}),u(w_{P}+\alpha)^{1/2}\in
L^{2}({\mathbb{R}})\right\\}=\left\\{u\in H^{1}({\mathbb{R}}),uV^{\prime}\in
L^{2}({\mathbb{R}})\right\\}=:H^{1}_{V^{\prime}}({\mathbb{R}})\,.$

The space $H^{1}_{V^{\prime}}({\mathbb{R}})$ can be seen to be the completion
of $\mathcal{C}_{c}^{\infty}({\mathbb{R}})$ under the norm $q_{\alpha}$. Now
that the quadratic form associated to $\mathcal{S}+\alpha I$ has been extended
to its form domain, it is possible to go back to the operator and extend it by
its Friedrichs extension.

######  Theorem A.4 (Friedrichs extension of $\mathcal{S}+\alpha I$).

There exists an extension $(S+\alpha I)_{F}$ of the operator
$\mathcal{S}+\alpha I$, called the Friedrichs extension of $\mathcal{S}+\alpha
I$ defined on $\mathcal{D}\Big{(}(S+\alpha I)_{F}\Big{)}=\Big{\\{}u\in
H^{1}_{V^{\prime}}({\mathbb{R}}),-u^{\prime\prime}+(w_{P}+\alpha)u\in
L^{2}({\mathbb{R}})\Big{\\}}$ .

###### Proof.

We denote

$\mathcal{D}\Big{(}(S+\alpha I)_{F}\Big{)}\\\ :=\Big{\\{}v\in
H^{1}_{V^{\prime}}({\mathbb{R}}),u\in
H^{1}_{V^{\prime}}({\mathbb{R}})\longmapsto q_{\alpha}(u,v)\text{ can be
extended to a continuous linear form on }L^{2}({\mathbb{R}})\Big{\\}}$

If $v\in\mathcal{D}\Big{(}(S+\alpha I)_{F}\Big{)}$, by Riesz’s theorem there
exists a unique $f_{v}\in L^{2}({\mathbb{R}})$ such that
$q_{\alpha}(u,v)=\braket{u,f_{v}}_{L^{2}(dx)}$ holds for all $u\in
L^{2}({\mathbb{R}})$ and we can set $(S+\alpha I)_{F}v:=f_{v}$. Note that it
is indeed a way of extending $\mathcal{S}+\alpha I$ since for all
$u,v\in\mathcal{C}_{c}^{\infty}({\mathbb{R}})$,
$q_{\alpha}(u,v)=\braket{u,(\mathcal{S}+\alpha I)v}_{L^{2}(dx)}$.

We want to show that $\mathcal{D}\Big{(}(S+\alpha I)_{F}\Big{)}=\Big{\\{}u\in
H^{1}_{V^{\prime}}({\mathbb{R}}),-u^{\prime\prime}+(w_{P}+\alpha)u\in
L^{2}({\mathbb{R}})\Big{\\}}$. Let $f\in\mathcal{D}\Big{(}(S+\alpha
I)_{F}\Big{)}$ and $g:=(S+\alpha I)_{F}f\in L^{2}({\mathbb{R}})$. By
definition of $q_{\alpha}$, for all
$u\in\mathcal{C}_{c}^{\infty}({\mathbb{R}})$:

$\int_{\mathbb{R}}gudx=\int_{\mathbb{R}}f^{\prime}u^{\prime}dx+\int_{\mathbb{R}}(w_{P}+\alpha)fudx=-\int_{\mathbb{R}}fu^{\prime\prime}dx+\int_{\mathbb{R}}(w_{P}+\alpha)fudx$

Therefore in the sense of distributions, we get
$-f^{\prime\prime}+(w_{P}+\alpha)=g$ which is a function in
$L^{2}({\mathbb{R}})$, hence $f\in\Big{\\{}u\in
H^{1}_{V^{\prime}}({\mathbb{R}}),-u^{\prime\prime}+(w_{P}+\alpha)u\in
L^{2}({\mathbb{R}})\Big{\\}}$. Conversely, if $f\in
H^{1}_{V^{\prime}}({\mathbb{R}})$ such that
$-f^{\prime\prime}+(w_{P}+\alpha)f\in L^{2}({\mathbb{R}})$, it is possible to
extend $u\mapsto q_{\alpha}(f,u)$ to a continuous linear form on
$L^{2}({\mathbb{R}})$ by

$u\mapsto\int_{\mathbb{R}}u\Big{(}-f^{\prime\prime}+f(w_{P}+\alpha)\Big{)}dx$

which concludes the fact that $\mathcal{D}\Big{(}(S+\alpha
I)_{F}\Big{)}=\Big{\\{}u\in
H^{1}_{V^{\prime}}({\mathbb{R}}),-u^{\prime\prime}+(w_{P}+\alpha)u\in
L^{2}({\mathbb{R}})\Big{\\}}$. ∎

In the following, we will deal only with $(\mathcal{S}+\alpha
I)_{F}:\mathcal{D}\Big{(}(S+\alpha I)_{F}\Big{)}\longrightarrow
L^{2}({\mathbb{R}})$ and denote it $\mathcal{S}+\alpha
I:\mathcal{D}(\mathcal{S}+\alpha I)$.

######  Remark A.5.

Note that in the previous proof, the application of Riesz’s theorem doesn’t
allow to say that $(\mathcal{S}+\alpha I):v\in\Big{(}\mathcal{D}(S+\alpha
I),\|.\|_{q_{\alpha}}\Big{)}\mapsto
f_{v}\in\Big{(}L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\Big{)}$, where
$\|.\|_{q_{\alpha}}$ stands for the norm associated to the bilinear positive
definite form $q_{\alpha}$, is continuous. It can be seen by the fact that
$v\in\Big{(}\mathcal{D}(S+\alpha I),\|.\|_{q_{\alpha}}\Big{)}\mapsto
q(.,v)\in\Big{(}L^{2}({\mathbb{R}})^{\prime},\|.\|_{L^{2}(dx)^{\prime}}\Big{)}$,
where $L^{2}({\mathbb{R}})^{\prime}$ stands for the topological dual of
$L^{2}({\mathbb{R}})$ equipped with its usual norm, is not continuous. Indeed
the $\|.\|_{q_{\alpha}}$ norm doesn’t control the second derivative of $v$ and
hence doesn’t provide any module of continuity for the
$L^{2}({\mathbb{R}})$-extended linear form $q(.,v)$.

Also note that, even though it would be convenient that
$\mathcal{D}\Big{(}(S+\alpha
I)_{F}\Big{)}=L^{2}({\mathbb{R}},(w_{P}+\alpha)^{2}dx)\cap
H^{2}({\mathbb{R}})$ it is not true without more properties on $w_{P}$. Such a
result holds, for example when $w_{P}$ belongs to $B_{2}$, the class of
reverse Hölder weights, see [ABA07][Theorem 1.1].

######  Theorem A.6 (Inversion of $\mathcal{S}+\alpha I$).

For every $f\in L^{2}({\mathbb{R}})$, there exists a unique
$u\in\mathcal{D}\Big{(}(S+\alpha I)_{F}\Big{)}$ such that $(\mathcal{S}+\alpha
I)u=f$. Furthermore, the map $(S+\alpha I)^{-1}$ is continuous from
$\big{(}L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\big{)}$ to
$\big{(}\mathcal{D}(S+\alpha I),\|.\|_{q_{\alpha}}\big{)}$.

###### Proof.

Let $f\in L^{2}({\mathbb{R}})$, the map $u\longmapsto\braket{u,f}_{L^{2}(dx)}$
is continuous on
$\big{(}H_{V^{\prime}}^{1}({\mathbb{R}}),\|.\|_{q_{\alpha}}\big{)}$ which is a
Hilbert space. Therefore by Riesz’s theorem, there exists a unique $v_{f}\in
H^{1}_{V^{\prime}}({\mathbb{R}})$ such that for all $u\in
H^{1}_{V^{\prime}}({\mathbb{R}})$,
$\braket{f,u}_{L^{2}(dx)}=q_{\alpha}(v_{f},u)$ from which we deduce that, in
the sense of distributions, $f=-v_{f}^{\prime\prime}+(w_{P}+\alpha)v_{f}$
which implies that $v_{f}\in\mathcal{D}(S+\alpha I)$. Since
$v_{f}\in\mathcal{D}(S+\alpha I)$, we have then for all $u\in
L^{2}({\mathbb{R}})$,
$\braket{f,u}_{L^{2}(dx)}=q_{\alpha}(v_{f},u)=\Braket{(\mathcal{S}+\alpha)v_{f},u}_{L^{2}(dx)}$,
hence $(\mathcal{S}+\alpha I)v_{f}=f$. Finally, by Riesz’s theorem, $f\in
L^{2}({\mathbb{R}})\mapsto v_{f}\in H^{1}_{V^{\prime}}({\mathbb{R}})$ is
continuous hence so is $(\mathcal{S}+\alpha I)^{-1}$. ∎

######  Remark A.7.

It would be tempting to use Banach’s isomorphism theorem to say that since
$(\mathcal{S}+\alpha I)^{-1}$ is bijective and continuous, so must be
$\mathcal{S}+\alpha I$. But since $\big{(}\mathcal{D}(S+\alpha
I),\|.\|_{q_{\alpha}}\big{)}$ is not a Banach space (it’s not closed in
$H^{1}_{V^{\prime}}({\mathbb{R}})$) we can’t apply it.

We are now able to diagonalize the resolvent of $\mathcal{S}$.

######  Theorem A.8 (Diagonalization of $(\mathcal{S}+\alpha I)^{-1}$).

There exists a complete orthonormal set $(\psi_{n})_{n\geqslant 0}$ of
$L^{2}({\mathbb{R}})$ (meaning that

$\overline{\operatorname{span}\\{\psi_{n},\ n\geqslant
0\\}}^{\|.\|_{L^{2}(dx)}}=L^{2}({\mathbb{R}})$

and $\langle\psi_{i},\psi_{j}\rangle_{L^{2}(dx)}=\delta_{i,j}$), where each
$\psi_{n}\in\mathcal{D}(S+\alpha I)$ and
$\big{(}\mu_{n}(\alpha)\big{)}_{n\geqslant 0}\in[0,1]^{\mathbb{N}}$ with
$\mu_{n}(\alpha)\underset{N\rightarrow\infty}{\longrightarrow}0$ such that
$(\mathcal{S}+\alpha I)^{-1}\psi_{n}=\mu_{n}(\alpha)\psi_{n}$ for all
$n\geqslant 0$. We also have
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\big{(}\mathcal{S}+\alpha
I\big{)}^{-1}\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{L}\big{(}L^{2}(dx)\big{)}}\leqslant
1$.

###### Proof.

By Proposition A.1, $w_{P}+\alpha$ is continuous and goes to infinity at
infinity. By Rellich criterion [RS78][see Theorem XIII.65], the unit ball of
$\mathcal{D}(S+\alpha I)$, ie the set

$\Big{\\{}u\in\mathcal{D}(S+\alpha I),\,\int_{\mathbb{R}}u^{\prime
2}+\int_{\mathbb{R}}(w_{P}+\alpha)u^{2}\leqslant 1\Big{\\}}$

considered as a subset of $L^{2}({\mathbb{R}})$ is relatively compact in
$\big{(}L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\big{)}$. Hence, we can conclude
that the injection $\iota:\big{(}\mathcal{D}(S+\alpha
I),\|.\|_{q_{\alpha}}\big{)}\longrightarrow\big{(}L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\big{)}$
is a compact operator. Since $(S+\alpha
I)^{-1}:\big{(}L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\big{)}\longrightarrow\big{(}\mathcal{D}(S+\alpha
I),\|.\|_{q_{\alpha}}\big{)}$ is continuous then $(\mathcal{S}+\alpha I)^{-1}$
is compact from $\big{(}L^{2}({\mathbb{R}}),\|.\|_{L^{2}(dx)}\big{)}$ to
itself. The fact that $(\mathcal{S}+\alpha I)^{-1}$ is self-adjoint and
positive allows us to apply the spectral theorem to obtain
$\big{(}\mu_{n}(\alpha)\big{)}_{n\geqslant 0}$ positive eigenvalues verifying
$\mu_{n}(\alpha)\underset{N\rightarrow\infty}{\longrightarrow}0$ by
compactness and a Hilbertian basis $(\psi_{n})_{n\geqslant 0}\in
L^{2}({\mathbb{R}})^{\mathbb{N}}$, such that for all $n\geqslant 0$,
$(\mathcal{S}+\alpha I)^{-1}\psi_{n}=\mu_{n}(\alpha)\psi_{n}$. It is then easy
to see that for all $n$, $\psi_{n}\in\mathcal{D}(S+\alpha I)$ since they
belong to the range of $(\mathcal{S}+\alpha I)^{-1}$. Finally, since for all
$\phi\in L^{2}({\mathbb{R}})$, $\Braket{(\mathcal{S}+\alpha
I)\phi,\phi}_{L^{2}(dx)}\geqslant\|\phi\|_{L^{2}(dx)}^{2}$, the spectrum of
$(\mathcal{S}+\alpha I)^{-1}$ is contained in $[0,1]$. It allows us to
conclude that
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|(\mathcal{S}+\alpha
I)^{-1}\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{L^{2}(dx)}\leqslant
1$. ∎

Since for all $u\in H^{1}_{V^{\prime}}({\mathbb{R}})$, $(S+\alpha)u\in
L^{2}({\mathbb{R}})$ iff $\mathcal{S}u\in L^{2}({\mathbb{R}})$, if we define
$\mathcal{D}(\mathcal{S})$ in the same manner that we did before,
$\mathcal{D}(\mathcal{S})=\mathcal{D}(S+\alpha I)$. It is now straightforward
to see how to extend $\mathcal{A}=\rho_{P}^{-1/2}\mathcal{S}\rho_{P}^{1/2}$ on
$\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A}):=\rho_{P}^{-1/2}\mathcal{D}(\mathcal{S})$
equipped with the norm $\|.\|_{q_{\alpha},\rho_{P}}$ to
$\big{(}L^{2}(\rho_{P}),\|.\|_{L^{2}(\rho_{P})}\big{)}$. The norm
$\|.\|_{q_{\alpha},\rho_{P}}$ is defined for all
$u\in\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A})$ by

$\|u\|_{q_{\alpha},\rho_{P}}=\int_{\mathbb{R}}u^{\prime
2}\rho_{P}dx+\int_{\mathbb{R}}u^{2}(w_{P}+\alpha)\rho_{P}dx\,.$

It is easy to see that $(\mathcal{A}+\alpha I)^{-1}$ is continuous.

######  Remark A.9.

The kernel of $\mathcal{A}$ is generated by the function $\widetilde{1}$.
Indeed if $\phi\in\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A})$ is in the
kernel of $\mathcal{A}$ then

$0=-\dfrac{\big{(}\phi^{\prime}\rho_{P}\big{)}^{\prime}}{\rho_{P}}\Rightarrow\exists
c\in{\mathbb{R}},\,\phi^{\prime}=\dfrac{c}{\rho_{P}}$

But since $\phi^{\prime}$ is in $L^{2}(\rho_{P})$ then $c=0$ which implies
that $\phi$ is constant. We must restrict $\mathcal{A}$ to the orthogonal of
$Ker\mathcal{A}$ with respect to the inner product of $L^{2}(\rho_{P})$, ie

$\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A}):=\left\\{u\in\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A})\,|\,\int_{\mathbb{R}}u\rho_{P}=0\right\\}\,.$

Doing so makes $\mathcal{A}$ injective.

Before inverting $\mathcal{A}$, we need the following lemma:

######  Lemma A.10.

The following equality holds

$(\mathcal{A}+\alpha
I)\Big{(}\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})\Big{)}=L^{2}_{0}(\rho_{P}):=\Big{\\{}u\in
L^{2}(\rho_{P}),\,\int_{\mathbb{R}}u\rho_{P}dx=0\Big{\\}}$

###### Proof.

Let $\phi=\widetilde{c}$ for $c\in{\mathbb{R}}$, $(\mathcal{A}+\alpha
I)\phi=\widetilde{\alpha c}$ then $(\mathcal{A}+\alpha
I)({\mathbb{R}}.\widetilde{1})={\mathbb{R}}\widetilde{1}$. Hence since
$\mathcal{A}+\alpha I$ is self-adjoint with respect to the inner product of
$L^{2}(\rho_{P})$ and that ${\mathbb{R}}\widetilde{1}$ is stable by
$\mathcal{A}+\alpha I$, then $(\mathcal{A}+\alpha
I)\Big{(}({\mathbb{R}}.\widetilde{1})^{\perp}\cap\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A})\Big{)}\subset({\mathbb{R}}.\widetilde{1})^{\perp}$.
For the converse, let $u\in({\mathbb{R}}.\widetilde{1})^{\perp}$, since
$\mathcal{A}+\alpha I$ is bijective, there exists
$v\in\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A})$ such that
$u=(\mathcal{A}+\alpha I)v$. For all $w\in{\mathbb{R}}.\widetilde{1}$,

$0=\Braket{u,w}_{L^{2}(\rho_{P})}=\Braket{(\mathcal{A}+\alpha
I)v,w}_{L^{2}(\rho_{P})}=\Braket{v,(\mathcal{A}+\alpha I)w}_{L^{2}(\rho_{P})}$

Hence $v\in\big{[}(\mathcal{A}+\alpha
I)({\mathbb{R}}\widetilde{1})\big{]}^{\perp}={\mathbb{R}}\widetilde{1}^{\perp}$
and so $({\mathbb{R}}.\widetilde{1})^{\perp}\subset(\mathcal{A}+\alpha
I)\Big{(}({\mathbb{R}}.\widetilde{1})^{\perp}\Big{)}$. ∎

It is easy to see that $L_{0}^{2}(\rho_{P})$ is a closed subset of
$L^{2}(\rho_{P})$ as it is the kernel of the linear form $\phi\in
L^{2}(\rho_{P})\mapsto\Braket{\phi,\widetilde{1}}_{L^{2}(\rho_{P})}$, making
it a Hilbert space.

######  Proposition A.11 (Diagonalization and invertibility of $\mathcal{A}$).

There exists a complete orthonormal set of
$\left(L_{0}^{2}(\rho_{P}),\Braket{.,.}_{L^{2}(\rho_{P})}\right)$,
$(\phi_{n})_{n\in\mathbb{N}}\in\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})^{\mathbb{N}}$
such that $\mathcal{A}\phi_{n}=\lambda_{n}\phi_{n}$ (meaning that

$\overline{\operatorname{span}\\{\phi_{n},\ n\geqslant
0\\}}^{\|.\|_{L^{2}(\rho_{P})}}=L^{2}_{0}(\rho_{P})$

and $\langle\phi_{i},\phi_{j}\rangle_{L^{2}(\rho_{P})}=\delta_{i,j}$).
Furthermore,
$\mathcal{A}:\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})\longrightarrow
L^{2}_{0}(\rho_{P}):=\Big{\\{}u\in
L^{2}(\rho_{P}),\,\int_{\mathbb{R}}u\rho_{P}dx=0\Big{\\}}$ is bijective,
$\mathcal{A}^{-1}$ is continuous when considered as an operator of
$L^{2}_{0}(\rho_{P})$.

###### Proof.

Since $(\mathcal{S}+\alpha I)^{-1}$ considered as an operator of
$L^{2}({\mathbb{R}})$, is compact so is $(\mathcal{A}+\alpha I)^{-1}$ on
$L^{2}(\rho_{P})$ and since $\mathcal{A}$ is self-adjoint, by the spectral
theorem, $(\mathcal{A}+\alpha I)^{-1}$ is diagonalizable. With the notations
of Theorem A.8, $(\mathcal{A}+\alpha I)^{-1}$ has eigenvalues
$\big{(}\mu_{n}(\alpha)\big{)}_{n\geqslant 0}$ and corresponding
eigenfunctions
$\phi_{n}=\rho_{P}^{-1/2}\psi_{n}\in\mathcal{D}_{L^{2}({\mathbb{R}})}(\mathcal{A})$.
Hence for all $n\in\mathbb{N}$, $\mathcal{A}\phi_{n}=\lambda_{n}\phi_{n}$ with
$\lambda_{n}:=\big{(}\dfrac{1}{\mu_{n}(\alpha)}-\alpha\big{)}$. Now,

$\lambda_{n}\|\phi_{n}\|_{L^{2}(\rho_{P})}=\int_{\mathbb{R}}(\mathcal{A}\phi_{n})\phi_{n}\rho_{P}dx=-\int_{\mathbb{R}}(\rho_{P}\phi_{n}^{\prime})^{\prime}\phi_{n}=\int_{\mathbb{R}}\phi_{n}^{\prime
2}\rho_{P}\geqslant\,0\,.$

Furthermore, the kernel of $\mathcal{A}$ is ${\mathbb{R}}.\widetilde{1}$, thus
the spectrum of $\mathcal{A}$ restricted to
$\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$ is positive. But since
$(\mathcal{A}+\alpha I)^{-1}$ is a compact operator of $L^{2}(\rho_{P})$ and
that $(\mathcal{A}+\alpha I)$ maps ${\mathbb{R}}.\widetilde{1}^{\perp}$ to
${\mathbb{R}}.\widetilde{1}^{\perp}$ with respect to the inner product of
$L^{2}(\rho_{P})$ (see lemma A.10), then $\big{(}\mathcal{A}+\alpha
I\big{)}^{-1}$ is compact as an operator from $L^{2}_{0}(\rho_{P})$ to itself.
By Fredholm alternative, for every $\lambda\in{\mathbb{R}}$ $\lambda\neq 0$,
either $(\mathcal{A}+\alpha I)^{-1}-\lambda I$ is bijective either $\lambda\in
Sp\big{(}(\mathcal{A}+\alpha I)^{-1}\big{)}$. These conditions are equivalent
to: either $\mathcal{A}+(\alpha-\dfrac{1}{\lambda})I$ is bijective as an
operator from $\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})$ to
$L^{2}_{0}(\rho_{P})$, either $-\alpha+\dfrac{1}{\lambda}\in
Sp\big{(}\mathcal{A}\big{)}$. If we set $\lambda=\dfrac{1}{\alpha}$ then
either $\mathcal{A}$ is bijective either $0\in Sp(\mathcal{A})$, since the
latter is wrong then
$\mathcal{A}:\mathcal{D}_{L^{2}({\mathbb{R}}),0}(\mathcal{A})\rightarrow
L^{2}_{0}(\rho_{P})$ is bijective. The spectrum of $\mathcal{A}$ is
$\left(\dfrac{1}{\mu_{n}(\alpha)}-\alpha\right)_{n\geqslant
0}\subset(\lambda_{1},+\infty)\subset(0,+\infty)$, where $\lambda_{1}$ is the
smallest eigenvalue, hence we deduce that
${\left|\kern-1.07639pt\left|\kern-1.07639pt\left|\mathcal{A}^{-1}\right|\kern-1.07639pt\right|\kern-1.07639pt\right|}_{\mathcal{L}\left(L^{2}(\rho_{P})\right)}\leqslant\lambda_{1}^{-1}$.
∎

## References