Brownian windings, Stochastic Green’s formula and inhomogeneous magnetic impurities

Isao Sauzedde University of Warwick [email protected]

Abstract.

We give a general Green formula for the planar Brownian motion, which we apply to study the Aharonov–Bohm effect induced by Poisson distributed magnetic impurities on a Brownian electron in the presence of an inhomogeneous magnetic field.

2020 Mathematics Subject Classification:

Primary 60J65; 60K37 Secondary 60G17

1. Introduction

1.1. Stochastic Green’s formula

For a smooth loop $X=(X^{1},X^{2}):[0,T]\to\mathbb{R}^{2}$ and a point $z$ outside the range of $X$ , let ${\mathbf{n}}_{X}(z)\in\mathbb{Z}$ be the winding index of $X$ around $z$ . For any smooth differential $1$ -form $\eta=\eta_{1}{\mathop{}\!\mathrm{d}}x^{1}+\eta_{2}{\mathop{}\!\mathrm{d}}x^{2}$ , the Green formula states that

\int_{X}\eta=\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X}{\mathop{}\!\mathrm{d}}\eta,

where ${\mathop{}\!\mathrm{d}}\eta$ is the exterior derivative of $\eta$ . In other words, for two smooth functions $\eta_{1},\eta_{2}:\mathbb{R}^{2}\to\mathbb{R}$ ,

\int_{0}^{T}\eta_{1}(X_{t}){\mathop{}\!\mathrm{d}}X^{1}_{t}+\int_{0}^{T}\eta_{2}(X_{t}){\mathop{}\!\mathrm{d}}X^{2}_{t}=\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X}(z)(\partial_{1}\eta_{2}(z)-\partial_{2}\eta_{1}(z)){\mathop{}\!\mathrm{d}}z.

When the smooth loop is replaced with a Brownian one, such a formula cannot be written down directly. For its left-hand side, we do have a genuine candidate provided by the Stratonovich integrale of $\eta$ along $X$ . However, the index function ${\mathbf{n}}_{X}$ fails from being integrable on the vicinity of $X$ [12], and we need to use some kind of regularization in order to define the right-hand side. In such a framework, the Green formula is thus a convergence result rather than an equality.

In [13], Wendelin Werner proposed two alternative regularizations, for which he was able to prove that the Green formula holds with a convergence in probability.

In [11], I proposed two more regularizations, for which I proved that the Green formula holds with an almost sure limit, but only in the case $\partial_{1}\eta_{2}-\partial_{2}\eta_{1}=1$ .

The first goal of this paper is to extend such a formula to any differential $1$ -form $\eta$ with regularity $\mathcal{C}^{1+\epsilon}$ .

For an integer $x$ and a positive integer $k$ , let $[x]_{k}$ be equal to either $x\mathbbm{1}_{|x|\leq k}$ or $\max(\min(x,k),-k)$ (the following theorem holds for both choice).

Theorem 1.

Let $X:[0,T]\to\mathbb{R}^{2}$ be a Brownian motion, and let ${\mathbf{n}}_{X}$ be the winding function associated with the loop obtained by concatenation of $X$ with the straight line segment $[X_{T},X_{0}]$ between its endpoints. Then, almost surely, for all $\epsilon>0$ and all $f\in\mathcal{C}_{b}^{\epsilon}(\mathbb{R}^{2})$ ,

\int_{\mathbb{R}^{2}}[{\mathbf{n}}_{X}(z)]_{k}f(z){\mathop{}\!\mathrm{d}}z

converges as $k\to\infty$ .

Furthermore, if $\eta=\eta_{1}{\mathop{}\!\mathrm{d}}x^{1}+\eta_{2}{\mathop{}\!\mathrm{d}}x^{2}$ with $\eta_{1},\eta_{2}\in\mathcal{C}^{1+\epsilon}(\mathbb{R}^{2})$ is such that $f=\partial_{1}\eta_{2}-\partial_{2}\eta_{1}$ , almost surely,

\lim_{k\to\infty}\int_{\mathbb{R}^{2}}[{\mathbf{n}}_{X}(z)]_{k}f(z){\mathop{}\!\mathrm{d}}z=\int_{0}^{T}\eta\circ{\mathop{}\!\mathrm{d}}X+\int_{[X_{T},X_{0}]}\eta,

where the stochastic integral in the right hand side is to be understood in the sense of Stratonovich.

Corollary 2.

For all $x$ and $y$ in $\mathbb{R}^{2}$ , the same result holds if the planar Brownian motion is replaced with a planar Brownian loop or a planar Brownian bridge between distinct points.

We will denote this limit as $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z$ , since we want to think of it as to the integral of $n_{X}$ with respect to the measure $f(z){\mathop{}\!\mathrm{d}}z$ .

1.2. Magnetic impurities

In the theory of weak localization in 2 dimensional crystals, for which we refer to [2], one studies quasiclassical electrons moving inside a metal with magnetic impurities, in the presence of a magnetic fields which induces an Aharonov–Bohm effect on the electrons. In some regime of the parameters, the electron is usually modeled by a planar Brownian trajectory. In particular, for the computation of the weak-localization correction to the Drude conductivity, the electron is modeled by a Brownian loop (see e.g. [7]). The impurities are modeled by a Poisson process of points $\mathcal{P}$ with intensity $\rho{\mathop{}\!\mathrm{d}}z$ in the plane, and the Aharonov–Bohm effect is described by a phase shift $\exp(i\alpha\sum_{z\in\mathcal{P}}{\mathbf{n}}_{X}(z))$ .

In [4], the authors study the limit $\rho\to+\infty$ with $\kappa=\alpha\rho$ constant, and derive a formula for the phase shift averaged over both $\mathcal{P}$ and $X$ .

For an integrable function $f\in L^{1}(\mathbb{R}^{2})$ , $\frac{1}{\rho}\sum_{z\in\mathcal{P}}f(z)$ is a Monte–Carlo estimation for $\int_{\mathbb{R}^{2}}f(z){\mathop{}\!\mathrm{d}}z$ , and therefore

e^{i\kappa\int_{\mathbb{R}^{2}}f(z){\mathop{}\!\mathrm{d}}z}=\lim_{\rho\to\infty}\mathbb{E}^{\mathcal{P}}\big{[}e^{i\frac{\kappa}{\rho}\sum_{z\in\mathcal{P}}f(z)}\big{]}.

However, as it is noticed in [5], for a Brownian loop $X$ ,

\mathbb{E}^{X}\big{[}e^{i\kappa\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-3.40277pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-2.275pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.60416pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.31248pt}}\!\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z}\big{]}\neq\lim_{\rho\to\infty}\mathbb{E}^{X,\mathcal{P}}\big{[}e^{i\frac{\kappa}{\rho}\sum_{z\in\mathcal{P}}{\mathbf{n}}_{X}(z)}\big{]},

which is due to the lack of integrability of the function ${\mathbf{n}}_{X}$ .

As we proved in [11], the Monte–Carlo method fails in this situation: it is true that $X$ -almost surely, $\frac{1}{\rho}\sum_{z\in\mathcal{P}}{\mathbf{n}}_{X}(z)$ converges in distribution (with respect to $\mathcal{P}$ ) as $\rho\to\infty$ , but the limit is not deterministic –or should we say, not measurable with respect to $X$ . It is instead equal to the sum of $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z$ with a centered Cauchy distribution independent from $X$ . From this result, one can rigorously prove the formula obtained first in [5] for

\lim_{\rho\to\infty}\mathbb{E}^{X,\mathcal{P}}[e^{i\frac{\kappa}{\rho}\sum_{z\in\mathcal{P}}{\mathbf{n}}_{X}(z)}].

However, for the scales at play, the magnetic field which induces the Aharonov-Bohm effect cannot be considered as homogeneous in general [8]. Our second goal in this paper is to derive an asymptotic formula for the functional of $X$ given by

\lim_{\rho\to\infty}\mathbb{E}^{\mathcal{P}}[e^{i\frac{1}{\rho}\sum_{z\in\mathcal{P}}f(z){\mathbf{n}}_{X}(z)}],

for a non homogeneous magnetic field $f$ and a non homogeneous density of impurities.

Theorem 3.

Let $f,g\in\mathcal{C}^{\epsilon}_{b}(\mathbb{R}^{2})$ , with $g\geq 0$ . For $\rho>0$ , let $\mathcal{P}$ be Poisson process on $\mathbb{R}^{2}$ with intensity $\rho g(z){\mathop{}\!\mathrm{d}}z$ , and let $X$ be either a Brownian motion or a Brownian bridge with duration $1$ , independent from $\mathcal{P}$ . Then, $X$ -almost surely,

\lim_{\rho\to\infty}\mathbb{E}^{\mathcal{P}}[e^{i\frac{1}{\rho}\sum_{z\in\mathcal{P}}f(z){\mathbf{n}}_{X}(z)}]=\exp\Big{(}i\alpha\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z)g(z){\mathop{}\!\mathrm{d}}z-\frac{|\alpha|}{2}\int_{0}^{1}|f(X_{t})|g(X_{t}){\mathop{}\!\mathrm{d}}t\Big{)}

where $\mathbb{E}^{\mathcal{P}}$ is the expectation over $\mathcal{P}$ (conditional on $X$ ).

Although this formula is suited to the problem of magnetic impurities, the following alternative formulation might be more appealing to the reader.

Corollary 4.

Let $g\in\mathcal{C}^{\epsilon}_{b}(\mathbb{R}^{2})$ , with $g\geq 0$ . For $\rho>0$ , let $\mathcal{P}$ be Poisson process on $\mathbb{R}^{2}$ with intensity $\rho g(z){\mathop{}\!\mathrm{d}}z$ , and $X$ be either a Brownian motion or a Brownian bridge with duration $1$ , independent from $\mathcal{P}$ . Let also $\Gamma:[0,1]\to\mathbb{R}$ be a standard Cauchy process. Then, for all $(f_{1},\dots,f_{n})\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ , $X$ -almost surely, the $n$ -uple

\Big{(}\frac{1}{\rho}\sum_{z\in\mathcal{P}}f_{1}(z){\mathbf{n}}_{X}(z),\dots,\frac{1}{\rho}\sum_{z\in\mathcal{P}}f_{n}(z){\mathbf{n}}_{X}(z)\Big{)}

converges in distribution toward $(\xi(f_{1}),\dots,\xi(f_{n}))$ where

\xi(f)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z)g(z){\mathop{}\!\mathrm{d}}z+\frac{1}{2}\int_{0}^{1}f(X_{t})g(X_{t}){\mathop{}\!\mathrm{d}}\Gamma_{t}.

Remark 5.

Given $f,g\in\mathcal{C}^{\epsilon}_{b}(\mathbb{R}^{2})$ , there always exists a differential $1$ -form $\eta$ with regularity $\mathcal{C}^{1+\epsilon}$ such that $\partial_{1}\eta_{2}-\partial_{2}\eta_{1}=fg$ , so that $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z)g(z){\mathop{}\!\mathrm{d}}z$ can always be written as a stochastic integral.

Since all the results hold $X$ -almost surely, the assumptions that the functions are bounded can easily be lifted, but some of the intermediate results come with a quantitative version which depends upon the $L^{\infty}$ norms.

This paper is built in the continuity of two former papers from the same author, [11] and [9]. It is not necessary to read them to understand the present paper, but we will use some results from those papers, as well as from [10].

2. Notations

2.1. Differential forms and integrals

For $\alpha\in(0,1)$ , we define $\mathcal{C}^{\alpha}(\mathbb{R}^{2})$ as the set of functions $f:\mathbb{R}^{2}\to\mathbb{R}$ such that the semi-norm

|f|_{\mathcal{C}^{\alpha}}\coloneqq\sup_{\begin{subarray}{c}x,y\in\mathbb{R}^{2}\\ x\neq y\end{subarray}}\frac{f(x)-f(y)}{|x-y|}

is finite. We also define $\mathcal{C}_{b}^{\alpha}(\mathbb{R}^{2})=\mathcal{C}^{\alpha}(\mathbb{R}^{2})\cap L^{2}(\mathbb{R}^{2})$ , which we endow with the norm

\|f\|_{\mathcal{C}^{\alpha}_{b}}=\|f\|_{\infty}+|f|_{\mathcal{C}^{\alpha}}.

For a differential $1$ -form $\eta=\eta_{1}{\mathop{}\!\mathrm{d}}x^{1}+\eta_{2}{\mathop{}\!\mathrm{d}}x^{2}$ and $\alpha\in[0,1)$ , we write $\eta\in\mathcal{C}^{1+\alpha}(T^{*}\mathbb{R}^{2})$ if $\partial_{i}\eta_{j}\in\mathcal{C}^{\alpha}(\mathbb{R}^{2})$ for all $i,j\in\{1,2\}$ .

Given a curve $X:[0,T]\to\mathbb{R}^{2}$ , we write

\int_{X}\eta\coloneqq\int_{0}^{T}\eta_{1}(X_{t}){\mathop{}\!\mathrm{d}}X^{1}_{t}+\int_{0}^{T}\eta_{2}(X_{t}){\mathop{}\!\mathrm{d}}X^{2}_{t},

where these integrals are to be understood either as classical integrals or as Stratonovich integrals, depending on the regularity of $X$ . No Itô integral will be involved in this paper, and all the stochastic integrals are to be understood in the sense of Stratonovich.

For $\eta\in\mathcal{C}^{1+\alpha}(T^{*}\mathbb{R}^{2})$ , we identify the $2$ -form ${\mathop{}\!\mathrm{d}}\alpha=(\partial_{1}\eta_{2}-\partial_{2}\eta_{1}){\mathop{}\!\mathrm{d}}x^{1}\wedge{\mathop{}\!\mathrm{d}}x^{2}$ with the signed measure $(\partial_{1}\eta_{2}-\partial_{2}\eta_{1}){\mathop{}\!\mathrm{d}}x$ , where ${\mathop{}\!\mathrm{d}}x$ is the Lebesgue measure on $\mathbb{R}^{2}$ .

For a bounded set $\mathcal{D}\subset\mathbb{R}^{2}$ and $f\in L^{1}_{loc}(\mathbb{R}^{2})$ , we use the unconventional notation

f(\mathcal{D})=\int_{\mathcal{D}}f(z){\mathop{}\!\mathrm{d}}z,

and $|\mathcal{D}|$ for the Lebesgue measure of $\mathcal{D}$ .

2.2. Winding

Given a curve $X$ on $\mathbb{R}^{2}$ , that is a continuous function from $[0,T]$ to $\mathbb{R}^{2}$ for some $T>0$ , we write $\bar{X}$ for the concatenation of $X$ with a straight line segment from $X_{T}$ to $X_{0}$ . Although the parameterisation of this line segment does not matter in the following, we will assume it is parameterized by $[T,T+1]$ at constant speed, unless $X$ is a loop (that is, a curve with $X_{T}=X_{0}$ ), in which case we set $\bar{X}=X$ .

Given a curve $X$ and a point $z$ outside the range of $\bar{X}$ , we write ${\mathbf{n}}_{X}(z)$ for the winding number of $\bar{X}$ around $z$ .

For a relative integer $k$ , we define

\mathcal{A}_{k}^{X}=\{z\in\mathbb{R}^{2}\setminus\operatorname{Range}(\bar{X}):{\mathbf{n}}_{X}(z)=k\}.

For $n>0$ , we also define

\mathcal{D}_{n}^{X}=\{z\in\mathbb{R}^{2}\setminus\operatorname{Range}(\bar{X}):{\mathbf{n}}_{X}(z)\geq n\}=\bigsqcup_{n\leq k<+\infty}\mathcal{A}_{k}^{X},

and

\mathcal{D}_{-n}^{X}=\{z\in\mathbb{R}^{2}\setminus\operatorname{Range}(\bar{X}):{\mathbf{n}}_{X}(z)\leq-n\}=\bigsqcup_{-\infty<k\leq-n}\mathcal{A}_{k}^{X}.

We also write $A_{k}^{X}$ (resp. $D_{k}^{X}$ ) for the Lebesgue measure of $\mathcal{A}_{k}^{X}$ (resp. $\mathcal{D}_{k}^{X}$ ), and we drop the superscript $X$ when there is no doubt about the curve we are talking about.

For a real number $z$ and a positive integer $n$ , we set

[x]_{n}=\left\{\begin{array}[]{cc}-n&\mbox{if }x\leq-n,\\ x&\mbox{if }-n\leq x\leq n,\\ n&\mbox{if }x\geq n.\\ \end{array}\right.

Once we have shown that the limit

\lim_{k\to\infty}\int_{\mathbb{R}^{2}}f(z)[n_{X}(z)]_{k}{\mathop{}\!\mathrm{d}}z

almost surely exists for all $f\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ , we will write $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{\mathbb{R}^{2}}f(z)n_{X}(z){\mathop{}\!\mathrm{d}}z$ for this limit.

For a locally finite set of points $\mathcal{P}$ , we define ${\mathbf{n}}_{X}(\mathcal{P})$ as the sum $\sum_{z\in\mathcal{P}}{\mathbf{n}}_{X}(z)$ . If we are also given a function $f$ on $\mathbb{R}^{2}$ , we define ${\mathbf{n}}_{X}(\mathcal{P},f)$ as the weighted sum

{\mathbf{n}}_{X}(\mathcal{P},f)=\sum_{z\in\mathcal{P}}{\mathbf{n}}_{X}(z)f(z).

2.3. Cauchy variables

The Cauchy distribution $\mathcal{C}(p,\sigma)$ with position parameter $p$ and scale parameter $\sigma>0$ is the probability distribution on $\mathbb{R}$ which has a density with respect to the Lebesgue measure given at $x$ by

\frac{1}{\pi\sigma}\frac{\sigma^{2}}{\sigma^{2}+(x-p)^{2}}.

A Cauchy random variable with position parameter $p$ and scale parameter $\sigma$ is a random variable distributed according to $\mathcal{C}(p,\sigma)$ . In ordre to unify some results, we will also write $\mathcal{C}(p,0)$ for a random variable which is actually deterministic and equal to $p$ .

Following [6, Definition 5.2]¹¹1As opposed to [6], we include the trivial case $\sigma=0$ in our definition. , we will say that a random variable $Z$ on $\mathbb{R}$ lies in the strong domain of attraction of a Cauchy distribution if there exists $\sigma\geq 0,\delta>0$ such that

\mathbb{P}(Z\geq x)\underset{x\to+\infty}{=}\frac{\sigma}{\pi x}+o(x^{-(1+\delta)}),\qquad\mathbb{P}(Z\leq-x)\underset{x\to+\infty}{=}\frac{\sigma}{\pi x}+o(x^{-(1+\delta)}).

It then follows from Lemma 5.1 and Theorem 1.2 in [6] that $Z$ follows a central limit theorem: if $(Z_{i})_{i\in\mathbb{N}}$ are i.i.d. copies of $Z$ , then there exists a unique $p$ such that

\frac{1}{N}\sum_{i=1}^{N}Z_{i}\Longrightarrow Y\sim\mathcal{C}(p,\sigma).

Notice that the same assumptions with $\delta=0$ are not sufficient for such a central limit theorem to hold.

The parameters $p$ and $\sigma$ such that $Y\sim\mathcal{C}(p,\sigma)$ are uniquely defined. We call them respectively the position parameter $p_{Z}$ of $Z$ , and the scale parameter $\sigma_{Z}$ of $Z$ .²²2When $Z$ is a Cauchy random variable, it belongs to the strong domain of attraction of a Cauchy distribution. There is thus two definitions of its position parameter, and two definitions of its scale parameter. Of course, the two definitions of its position parameter agree, and the two definitions of its scale parameter agree as well.

3. Former results

We will use the following results from [11], [9] and [10].

Lemma 3.1 (Lemma 5.2 in [11] ).

Assume $Z$ belongs to the strong attraction domain of a Cauchy distribution. Then, its position parameter $p_{Z}$ is equal to

\lim_{n\to\infty}\mathbb{E}[[Z]_{n}].

When $Y$ and $Z$ lie in the strong attraction domain of Cauchy distributions, or even when they are Cauchy random variables, but they are not independent, $Y+Z$ does not necessarily belong to the strong attraction domain of a Cauchy distribution. What might be even more surprising is that, even if $Y$ , $Z$ , and $Y+Z$ are Cauchy random variables, $p_{Y+Z}$ can differ from $p_{Y}+p_{Z}$ (see e.g. [3] for an explicit counter-example). Yet, the following lemma offers conditions weaker then independence under which additivity is restored.

Lemma 3.2 ( Lemma 5.3 in [11] ).

Let $n\geq 1$ and $Z_{1},\dots,Z_{n}$ be random variables which each lie in the strong attraction domain of a Cauchy distribution. Assume that there exists $\delta>0$ such that, for all $i,j\in\{1,\dots,n\}$ , $i\neq j$ ,

\mathbb{P}(|Z_{i}|\geq x\mbox{ and }|Z_{j}|\geq x)\underset{x\to+\infty}{=}o(x^{-(1+\delta)}).

Then, $Z=\sum_{i=1}^{n}Z_{i}$ lies in the strong attraction domain of a Cauchy distribution, and $p_{Z}=\sum_{i=1}^{n}p_{Z_{i}}$ .

The following lemma should be compared with the definition of the strong domain of attraction, where the random variable $Z$ is given by ${\mathbf{n}}_{X}(P)$ , with $X$ fixed and $P$ a random point distributed according to $\frac{1}{Z}\mathbbm{1}_{K}(z)f(z){\mathop{}\!\mathrm{d}}z$ (when $f\geq 0$ ), where $K$ is a convex set containing $\operatorname{Range}(X)$ .

Lemma 3.3 (Lemma 5 in [9]).

Let $X:[0,1]\to\mathbb{R}^{2}$ be a planar Brownian motion. For all $\beta<\frac{1}{2}$ , there exists $\delta>0$ such that almost surely, there exists $C$ such that for all bounded continuous function $f\in\mathcal{C}_{b}({\mathbb{R}}^{2})$ , for all $n\geq 1$ ,

\Big{|}2\pi nf(\mathcal{D}_{n})-\int_{0}^{1}f(X_{u}){\mathop{}\!\mathrm{d}}u\Big{|}\leq C(\omega_{f}(2\|X\|_{\mathcal{C}^{\beta}}n^{-\delta})+\|f\|_{\infty}n^{-\delta}),

where $\omega_{f}$ is the continuity modulus of $f$ , i.e. $\omega_{f}(r)=\sup_{x,y:|x-y|\leq r}|f(x)-f(y)|$ .

From symmetry of the Brownian motion, Lemma 3.3 also holds with $\mathcal{D}_{n}$ replaced with $\mathcal{D}_{-n}$ .

We will also need some $L^{p}$ control.

Lemma 3.4 ( Theorem 6.2 in [11] ).

For all $\delta<\frac{1}{2}$ and $p\geq 2$ , there exists a constant $C$ such that for all $N\geq 1$ ,

\mathbb{E}\big{[}\big{|}D_{N}-\tfrac{1}{2\pi N}\big{|}^{p}\big{]}^{\frac{1}{p}}\leq CN^{-1-\delta}.

Finally, the following lemma will be used to check the condition inside Lemma 3.2.

Lemma 3.5 (Theorem 1 in [10]).

Let $X,X^{\prime}:[0,1]\to\mathbb{R}^{2}$ be two independent Brownian motions, starting from equal or different points in the plane. Then, $n^{2}|\mathcal{D}^{X}_{n}\cap\mathcal{D}^{X^{\prime}}_{n}|$ almost surely converges as $n\to\infty$ .

A few more results will be used, but will be easier to formulate later.

4. Stokes formula

In this section, $X:[0,1]\to\mathbb{R}^{2}$ is a standard Brownian motion under $\mathbb{P}$ .

4.1. Existence of a limit

We will first prove the first part of Theorem 1:

Lemma 4.1.

Let $\epsilon>0$ . $\mathbb{P}$ -almost surely, for all $f\in\mathcal{C}_{b}^{\epsilon}(\mathbb{R}^{2})$ , the limits

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(x)f(x){\mathop{}\!\mathrm{d}}x\coloneqq\lim_{N\to\infty}\int_{\mathbb{R}^{2}}[{\mathbf{n}}_{X}(z)]_{N}f(z){\mathop{}\!\mathrm{d}}z\qquad\mbox{and}\qquad\lim_{N\to\infty}\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X}(z)\mathbbm{1}_{|{\mathbf{n}}_{X}(z)|\leq N}\ f(z){\mathop{}\!\mathrm{d}}z

exist and are equal. Almost surely, the application $f\mapsto{\mathbf{n}}_{X}(f)$ from $\mathcal{C}^{\epsilon}_{b}(\mathbb{R}^{2})$ to $\mathbb{R}$ is continuous.

Proof.

We fix $\beta\in\big{(}0,\tfrac{1}{2}\big{)}$ . Let $\delta>0$ be such that Lemma 3.3 holds, and let $\mathcal{E}$ be the full probability event on which $\|X\|_{\mathcal{C}^{\beta}}<\infty$ and Lemma 3.3 holds both for the sequence $\mathcal{D}_{n}$ and the sequence $\mathcal{D}_{-n}$ , with a corresponding random constant $C$ .

On $\mathcal{E}$ , for all $\epsilon>0$ , with $C^{\prime}=4\pi C,C^{\prime\prime}=C^{\prime}(1+|X|_{\mathcal{C}^{\beta}}^{\epsilon})$ , for all $f\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ ,

	$\displaystyle\Big{\|}f(\mathcal{D}_{n})-f(\mathcal{D}_{-n})\Big{\|}$	$\displaystyle\leq C^{\prime}n^{-1}(\omega_{f}(2\|X\|_{\mathcal{C}^{\beta}}n^{-\delta})+\\|f\\|_{\infty}n^{-\delta})$
		$\displaystyle\leq C^{\prime\prime}n^{-1}(\|f\|_{\mathcal{C}^{\epsilon}}n^{-\delta\epsilon}+\\|f\\|_{\infty}n^{-\delta}).$		(1)

Thus, on $\mathcal{E}$ , the sum

\sum_{n\geq 1}(f(\mathcal{D}_{n})-f(\mathcal{D}_{-n}))

is absolutely convergent. By applying an Abel summation, we obtain

\sum_{n=1}^{N}(f(\mathcal{D}_{n})-f(\mathcal{D}_{-n}))=\int_{\mathbb{R}^{2}}[{\mathbf{n}}_{X}(z)]_{N}f(z){\mathop{}\!\mathrm{d}}z,

so that the right-hand side is convergent on the event $\mathcal{E}$ .

Besides,

\Big{|}\int_{{\mathbb{R}}^{2}}[{\mathbf{n}}_{X}(z)]_{N}f(z){\mathop{}\!\mathrm{d}}z-\int_{{\mathbb{R}}^{2}}{\mathbf{n}}_{X}(z)\mathbbm{1}_{|{\mathbf{n}}_{X}(z)|\leq N}f(z){\mathop{}\!\mathrm{d}}z\Big{|}=N|f(\mathcal{D}_{N+1})-f(\mathcal{D}_{-N-1})|,

which, on the almost sure event $\mathcal{E}$ , converges toward $0$ as $N$ goes to infinity (by (1)).

The only thing that remains to be shown is the almost sure continuity of the application $f\mapsto\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(x)f(x){\mathop{}\!\mathrm{d}}x$ . Since it is clearly linear, it suffices to show that it is almost surely a bounded operator. By (1),

\sum_{n=1}^{N}|f(\mathcal{D}_{n})-f(\mathcal{D}_{-n})|

is bounded by $C^{(3)}\|f\|_{\mathcal{C}^{\epsilon}_{b}}$ for a random constant $\mathcal{C}^{(3)}$ which depends on $\epsilon,\ \beta$ and $\delta$ , but not on $f$ nor $N$ . Thus, $|\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(x)f(x){\mathop{}\!\mathrm{d}}x|\leq C^{(3)}\|f\|_{\mathcal{C}^{\epsilon}_{b}}$ , which concludes the proof. ∎

4.2. Strategy for the Stokes’ formula

In order to conclude the proof of Theorem 1, we now need to identify $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(x)f(x){\mathop{}\!\mathrm{d}}x$ with the Stratonovich integral $\int_{X}\eta+\int_{[X_{1},X_{0}]}\eta$ , when $f=\partial_{1}\eta_{2}-\partial_{2}\eta_{1}$ .

To this end, we decompose the trajectory $X$ into several pieces. First, we denote by $X^{(n)}$ the dyadic piecewise-linear approximation of $X$ with $2^{n}$ pieces: for $\lambda\in[0,1],\ i\in\{0,\dots,2^{n}-1\}$ , and $t=(i+\lambda)2^{-n}$ ,

X^{(n)}_{t}=X_{i2^{-n}}+\lambda(X_{(i+1)2^{-n}}-X_{i2^{-n}}).

For $i\in\{0,\dots,2^{n}-1\}$ , we also set $X^{i}$ , the restriction of $X$ to the interval $[i2^{-n},(i+1)2^{-n}]$ . Finally, set $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(x)f(x){\mathop{}\!\mathrm{d}}x$ the almost sure limit

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(z)f(z){\mathop{}\!\mathrm{d}}z=\lim_{N\to\infty}\int_{{\mathbb{R}}^{2}}[{\mathbf{n}}_{X^{i}}(z)]_{N}f(z){\mathop{}\!\mathrm{d}}z.

By Lemma 4.1, scale invariance, and translation invariance of the Brownian motion, almost surely, $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(x)f(x){\mathop{}\!\mathrm{d}}x$ is well -defined for all $n\geq 0$ , for all $i\in\{0,\dots,2^{n}-1\}$ , for all $f\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ .³³3Since we use the translation invariance, the function $f$ is replaced with the random function $z\mapsto f(z+X_{i2^{-n}})$ . This is not an issue, because Lemma 4.1 holds almost surely for all $f$ , and not the other way around.

Let us first sketch the strategy of our proof. First, notice that for all $z\in\mathbb{R}^{2}$ which does not belong to $\operatorname{Range}(X)$ nor to $\operatorname{Range}(X^{(n)})$ ,

{\mathbf{n}}_{X}(z)=\sum_{i=0}^{2^{n}-1}{\mathbf{n}}_{X_{i}}(z)+{\mathbf{n}}_{X^{(n)}}(z),

which essentially comes from the additivity of the winding index, with respect to the concatenation of loops. Thus, it is reasonable to expect that

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z=\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(z)f(z){\mathop{}\!\mathrm{d}}z+\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X^{(n)}}(z)f(z){\mathop{}\!\mathrm{d}}z.

By applying the standard Stokes’ formula on the last integral, we get

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z=\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(z)f(z){\mathop{}\!\mathrm{d}}z+\int_{X^{(n)}}\eta+\int_{[X_{1},X_{0}]}\eta.

As $n$ goes to infinity, we will see that the contribution from the small pieces (i.e. the sum over $i$ ) vanishes, whilst the integral along $X^{(n)}$ converges toward the Stratonovich integral $\int_{X}\eta$ , which gives the expected formula.

We will decompose the actual proof into the three following lemma, which we will prove in the three following subsections. Let $f\in\mathcal{C}^{\epsilon}_{b}(\mathbb{R}^{2})$ , and $\eta\in\mathcal{C}^{1+\epsilon}(T^{*}\mathbb{R}^{2})$ such that $f=\partial_{1}\eta_{2}-\partial_{2}\eta_{1}$ .

Lemma 4.2.

For all $n$ , almost surely,

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z=\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(z)f(z){\mathop{}\!\mathrm{d}}z+\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X^{(n)}}(z)f(z){\mathop{}\!\mathrm{d}}z.

(2)

Lemma 4.3.

As $n$ goes to infinity,

\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X_{i}}(z)f(z){\mathop{}\!\mathrm{d}}z

converges almost surely toward zero.

Lemma 4.4.

As $n$ goes to infinity, $\int_{X^{(n)}}\eta$ converges almost surely toward $\int\eta\circ{\mathop{}\!\mathrm{d}}X$ .

Of course, the conclusion that almost surely,

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z=\int_{X}\eta+\int_{[X_{1},X_{0}]}\eta,

and therefore that Theorem 1 holds, follows directly from these three lemma.

4.3. Additivity

Intuitively, the equality in Lemma 4.2 follows from integration of the almost-everywhere equality

{\mathbf{n}}_{X}(z)=\sum_{i=0}^{2^{n}-1}{\mathbf{n}}_{X^{i}}(z)+{\mathbf{n}}_{X^{(n)}}(z),

applied together with the Stokes formula for $X^{(n)}$ . However, neither ${\mathbf{n}}_{X}$ nor ${\mathbf{n}}_{X^{i}}$ are integrable, we have to deal with the cut-offs that allow to define $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z$ and the $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)f(z){\mathop{}\!\mathrm{d}}z$ : in general, for a finite $k$ ,

[{\mathbf{n}}_{X}(z)]_{k}\neq\sum_{i=0}^{2^{n}-1}[{\mathbf{n}}_{X^{i}}(z)]_{k}+[{\mathbf{n}}_{X^{(n)}}(z)]_{k}.

Proof of Lemma 4.2.

From linearity with respect to $f$ , we can and we do assume $f\geq 0$ . In the event that that the restriction of $f$ to $B(0,\|X\|_{\infty})$ is identically vanishing, the result is trivial, and we thus assume that

Z\coloneqq\int_{B(0,\|X\|_{\infty})}f(z){\mathop{}\!\mathrm{d}}z

is strictly positive.

Let $P$ be a random point in $\mathbb{R}^{2}$ those distribution conditional on $X$ admits a density with respect to the Lebesgue measure, given by

\frac{f(z)\mathbbm{1}_{B(0,\|X\|_{\infty})}(z)}{Z}.

Then, $X$ -almost surely, $P$ -almost surely,

{\mathbf{n}}_{X}(P)=\sum_{i=0}^{2^{n}-1}{\mathbf{n}}_{X_{i}}(P)+{\mathbf{n}}_{X^{(n)}}(P).

Notice that, for $N\geq 0$ , for $\tilde{X}$ equal to either $X$ , or to one of the $X^{i}$ , or to $X^{(n)}$ , it holds that

\mathbb{P}({\mathbf{n}}_{\tilde{X}}(P)\geq N|X)=\frac{1}{Z}f(\mathcal{D}^{\tilde{X}}_{N}),\qquad\mathbb{P}({\mathbf{n}}_{\tilde{X}}(P)\leq-N|X)=\frac{1}{Z}f(\mathcal{D}^{\tilde{X}}_{-N}).

Thus, Lemma 3.3 ensures that $X$ -almost surely, the random variable ${\mathbf{n}}_{\tilde{X}}(P)$ belong to the strong attraction domain of a Cauchy distribution for either $\tilde{X}=X$ or $\tilde{X}=X^{i}$ . As for $\tilde{X}=X^{(n)}$ , $|{\mathbf{n}}_{\tilde{X}}|$ is bounded by $2^{n}$ and therefore ${\mathbf{n}}_{\tilde{X}}(P)$ also belong to the strong attraction domain of a (degenerate, $\sigma=0$ ) Cauchy distribution.

Let us check that, $X$ -almost surely, we can apply Lemma 3.2 to the set of variables

(Z_{0},\dots,Z_{2^{n}-1},Z_{2^{n}})=({\mathbf{n}}_{X^{0}}(P),\dots,{\mathbf{n}}_{X^{2^{n}-1}}(P),{\mathbf{n}}_{X^{(n)}}(P)).

First, for $i\in\{0,\dots,2^{n}-1\}$ , for $x\geq 2^{n}$ ,

\mathbb{P}(|{\mathbf{n}}_{X^{i}}(P)|\geq x\mbox{ and }|{\mathbf{n}}_{X^{(n)}}(P)|\geq x)=0=o(x^{-(1+\delta)}).

Besides, for $i,j\in\{0,\dots,2^{n}-1\}$ , $i\neq j$ ,

	$\displaystyle\mathbb{P}(\|{\mathbf{n}}_{X^{i}}(P)\|\geq N\mbox{ and }\|{\mathbf{n}}_{X^{j}}(P)\|\geq N)$	$\displaystyle=\frac{1}{Z}f\big{(}\big{(}\mathcal{D}_{N}^{X^{i}}\cup\mathcal{D}_{-N}^{X^{i}}\big{)}\cap\big{(}\mathcal{D}_{N}^{X^{j}}\cup\mathcal{D}_{-N}^{X^{j}}\big{)}\big{)}$
		$\displaystyle\leq\frac{C\\|f\\|_{\infty}}{Z}\|N\|^{2},$

for a random constant $C$ . The last equality follows from Lemma 3.5, applied to the independent Brownian motions

\hat{X}_{i}:t\mapsto X_{(i+1-t)2^{-n}}-X_{(i+1)2^{-n}},\qquad\hat{X}_{j}:t\mapsto X_{(j+t)2^{-n}}-X_{(i+1)2^{-n}}.

Notice that the constant $C=C(n,i,j)$ depends upon $i$ and $j$ , but we can replace it with $C(n)=\max_{i,j}C(n,i,j)$ so that it only depends on $n$ . Furthermore, since there is only countably many couples $(i,j)$ , the previous inequality holds almost surely for all $(i,j)$ simultaneously.

Thus, we can indeed apply Lemma 3.2 to deduce that the, $X$ -almost surely, the position parameters add up:

p_{{\mathbf{n}}_{X}(P)}=\sum_{i=0}^{2^{n}-1}p_{{\mathbf{n}}_{X^{i}}(P)}+p_{{\mathbf{n}}_{X^{(n)}}(P)}.

(3)

Furthermore, since $|{\mathbf{n}}_{X^{(n)}}(P)|$ is bounded, $p_{{\mathbf{n}}_{X^{(n)}}(P)}$ is quickly checked to be equal $\mathbb{E}^{P}[{\mathbf{n}}_{X^{(n)}}(P)|X]$ , that is

p_{{\mathbf{n}}_{X^{(n)}}(P)}=\frac{1}{Z}\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X^{(n)}}(z)f(z){\mathop{}\!\mathrm{d}}z.

Finally, from Lemma 3.1, we deduce that $X$ -almost surely,

\displaystyle p_{n_{X}}(P)=\lim_{N\to\infty}\mathbb{E}^{P}\big{[}[{\mathbf{n}}_{X}(P)]_{N}\big{|}X\big{]}=\lim_{N\to\infty}\frac{1}{Z}\int_{\mathbb{R}^{2}}[{\mathbf{n}}_{X}(z)]_{N}f(z){\mathop{}\!\mathrm{d}}z=\frac{1}{Z}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z,

and similarly

p_{n_{X^{i}}}(P)=\frac{1}{Z}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)f(z){\mathop{}\!\mathrm{d}}z.

Thus, Equation 3 turns into

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z=\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)f(z){\mathop{}\!\mathrm{d}}z+\int_{\mathbb{R}^{2}}{\mathbf{n}}_{X^{(n)}}(z)f(z){\mathop{}\!\mathrm{d}}z,

as announced. ∎

4.4. Contribution from the small loops

We now prove that almost surely,

\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)f(z){\mathop{}\!\mathrm{d}}z\underset{n\to\infty}{\longrightarrow}0.

We will first need the following result, which should be compared with Lemma 3.3.

Lemma 4.5.

Let $\epsilon>0$ and $p\geq 1$ . There exists a constant $C$ and $\delta>0$ such that for all $f\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ and all $N\geq 1$ ,

\mathbb{E}\big{[}\big{|}f(\mathcal{D}^{X}_{N})-\frac{1}{2\pi N}\int_{0}^{1}f(X_{t}){\mathop{}\!\mathrm{d}}t\big{|}^{p}\big{]}^{\frac{1}{p}}\leq CN^{-1-\delta}\|f\|_{\mathcal{C}^{\epsilon}}.

Proof.

The proof is largely inspired from [9].

Let $T\geq 1$ , which we will later take to be a function of $N$ . For $i\in\{0,\dots,T-1\}$ , let $X^{i}$ be the restriction of $X$ to the interval $[iT^{-1},(i+1)T^{-1}]$ . Let $X^{pl}$ be the piecewise linear approximation of $X$ with $T$ pieces,

X^{pl}_{(i+\lambda)T^{-1}}=X_{iT^{-1}}+\lambda(X_{(i+1)T^{-1}}-X_{iT^{-1}}),\qquad i\in\{0,\dots,T-1\},\ \lambda\in[0,1].

For $i,j\in\{0,\dots,T-1\}$ , let

\mathcal{D}^{i}_{N}=\mathcal{D}^{X^{i}}_{N},\qquad\mathcal{D}^{i,j}_{N}=\big{(}\mathcal{D}^{X^{i}}_{N}\cup\mathcal{D}^{X^{i}}_{-N}\big{)}\cap\big{(}\mathcal{D}^{X^{j}}_{N}\cup\mathcal{D}^{X^{j}}_{-N}\big{)}.

For $z$ outside $\operatorname{Range}(X)\cup\operatorname{Range}(X^{pl})$ , we have

{\mathbf{n}}_{X}(z)=\sum_{i=0}^{T-1}n_{X^{i}}(z)+{\mathbf{n}}_{X^{pl}}(z),\quad|{\mathbf{n}}_{X^{pl}}(z)|\leq T.

It follows⁴⁴4See Section 3.2 in [11] for more details. that, for all $T,M,N\geq 1$ such that $T(M+1)<N$ ,

\mathcal{D}^{X}_{N}\subseteq\bigcup_{i=0}^{T-1}\mathcal{D}^{i}_{N-T-M(T-1)}\cup\bigcup_{\begin{subarray}{c}i,j=0\\ i\neq j\end{subarray}}^{T-1}\mathcal{D}^{i,j}_{M}\cup\operatorname{Range}(X)\cup\operatorname{Range}(X^{pl}),

and therefore

f(\mathcal{D}^{X}_{N})\leq\sum_{i=0}^{T-1}f(\mathcal{D}^{i}_{N-T-M(T-1)})+\sum_{\begin{subarray}{c}i,j=0\\ i\neq j\end{subarray}}^{T-1}f(\mathcal{D}^{i,j}_{M}).

We set $t\in(0,\tfrac{1}{3})$ , $m\in(\tfrac{1+t}{2},1-t)$ , $\alpha<\frac{1}{2}$ , $T=\lfloor N^{t}\rfloor$ , $M=\lfloor N^{m}\rfloor$ , and we assume that $N$ is large enough for the inequality $T(M+1)<N$ to hold. We also set $N^{\prime}=N-T-M(T-1)$ to ease notations.

Using the fact that $\mathcal{D}^{i}_{N^{\prime}}$ is contained inside the convex hull of $X^{i}$ , hence in the ball centered at $X_{\frac{i}{T}}$ with radius $\|X\|_{\mathcal{C}^{\alpha}}T^{-\alpha}$ , we deduce that $f$ is bounded above by $f(X_{\frac{i}{T}})+|f|_{\mathcal{C}^{\epsilon}}\|X\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}$ on $\mathcal{D}^{i}_{N^{\prime}}$ . Thus,

	$\displaystyle f(\mathcal{D}_{N})$	$\displaystyle\leq\sum_{i=0}^{T-1}f(X(\tfrac{i}{T}))\|\mathcal{D}^{i}_{N^{\prime}}\|+\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}\sum_{i=0}^{T-1}\|\mathcal{D}^{i}_{N^{\prime}}\|+\\|f\\|_{\infty}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|.$
		$\displaystyle\leq\frac{1}{2\pi NT}\sum_{i=0}^{T-1}f(X(\tfrac{i}{T}))+\\|f\\|_{\infty}\sum_{i=0}^{T-1}\big{\|}\tfrac{1}{2\pi NT}-\|\mathcal{D}^{i}_{N^{\prime}}\|\big{\|}+\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}\sum_{i=0}^{T-1}\|\mathcal{D}^{i}_{N^{\prime}}\|$
		$\displaystyle\hskip 310.13486pt+\\|f\\|_{\infty}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|$
		$\displaystyle\leq\frac{1}{2\pi N}\int_{0}^{1}f(X_{t}){\mathop{}\!\mathrm{d}}t+\frac{\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|_{\mathcal{C}^{\alpha}}^{\epsilon}T^{-\epsilon\alpha}}{2\pi N}+\\|f\\|_{\infty}\sum_{i=0}^{T-1}\big{\|}\tfrac{1}{2\pi NT}-\|\mathcal{D}^{i}_{N^{\prime}}\|\big{\|}$
		$\displaystyle\hskip 170.71652pt+\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}\sum_{i=0}^{T-1}\|\mathcal{D}^{i}_{N^{\prime}}\|+\\|f\\|_{\infty}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|.$

Writing $(f)^{p}_{+}$ for the positive part of $f$ , to the power $p$ , and using the triangle inequality in $L^{p}(\mathbb{P})$ , we obtain

	$\displaystyle\mathbb{E}\Big{[}\Big{(}f(\mathcal{D}_{N})-$	$\displaystyle\frac{1}{2\pi N}\int_{0}^{1}f(X_{t}){\mathop{}\!\mathrm{d}}t\Big{)}_{+}^{p}\Big{]}^{\frac{1}{p}}\leq\frac{\|f\|_{\mathcal{C}^{\epsilon}}T^{-\epsilon\alpha}}{2\pi N}\mathbb{E}[\\|X\\|_{\mathcal{C}^{\alpha}}^{\epsilon p}]^{\frac{1}{p}}+\\|f\\|_{\infty}\mathbb{E}\Big{[}\Big{\|}\frac{1}{2\pi N}-\|\mathcal{D}_{N^{\prime}}\|\Big{\|}^{p}\Big{]}^{\frac{1}{p}}$
		$\displaystyle+\|f\|_{\mathcal{C}^{\epsilon}}T^{-\epsilon\alpha}\mathbb{E}[\|\mathcal{D}_{N^{\prime}}\|^{2p}]^{\frac{1}{2p}}\mathbb{E}[\\|X\\|_{\mathcal{C}^{\alpha}}^{2p\epsilon}]^{\frac{1}{2p}}+\\|f\\|_{\infty}\mathbb{E}\Big{[}\Big{(}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|\Big{)}^{p}\Big{]}^{\frac{1}{p}}.$

We now use the asymptotic equivalence $N^{\prime}\sim_{N\to\infty}N$ and $\frac{1}{N}-\frac{1}{N^{\prime}}\sim_{N\to\infty}N^{t+m-2}$ , as well as Lemma 3.4, and the following estimations ([11, Lemma 2.4]): for all $p\geq 1$ , there exists a constant $C$ such that for all $N\geq 1$ ,

\mathbb{E}\Big{[}\Big{(}\sum_{i\neq j}|\mathcal{D}^{i,j}_{M}|\Big{)}^{p}\Big{]}^{\frac{1}{p}}\leq C\log(N+1)^{3+\frac{2}{p}}M^{-2}T^{1-\frac{1}{p}}.

We end up with

	$\displaystyle\mathbb{E}\Big{[}\Big{(}f(\mathcal{D}_{N})-\frac{1}{2\pi N}\int_{0}^{1}f(X_{t}){\mathop{}\!\mathrm{d}}t\Big{)}_{+}^{p}\Big{]}^{\frac{1}{p}}$	$\displaystyle\leq C\big{(}\|f\|_{\mathcal{C}^{\epsilon}}N^{-1-t\epsilon\alpha}+\\|f\\|_{\infty}N^{m+t-2}+\\|f\\|_{\infty}N^{-1-\delta}$
		$\displaystyle+\|f\|_{\mathcal{C}^{\epsilon}}N^{-1-t\epsilon\alpha}+\\|f\\|_{\infty}\log(N+1)^{3+\frac{2}{p}}N^{-2m+t-\frac{t}{p}}\big{)},$

for an arbitrary but fixed $\delta\in(0,\frac{1}{2})$ . The conditions on $t$ and $m$ ensures that all the exponents of $N$ are smaller than $-1$ , so that there exists $\delta^{\prime}$ and $C$ such that

\mathbb{E}\Big{[}\Big{(}f(\mathcal{D}_{N})-\frac{1}{2\pi N}\int_{0}^{1}f(X_{t}){\mathop{}\!\mathrm{d}}t\Big{)}_{+}^{p}\Big{]}^{\frac{1}{p}}\leq C\|f\|_{\mathcal{C}_{b}^{\alpha}}N^{-1-\delta^{\prime}}.

The negative part is treated in a similar way, and the lemma follows. ∎

Corollary 4.6.

Let $\epsilon>0$ and $p\geq 1$ . There exists a constant $C$ such that for all $f\in\mathcal{C}_{b}^{\epsilon}(\mathbb{R}^{2})$ , $\mathbb{E}[(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z)^{p}]^{\frac{1}{p}}\leq C\|f\|_{\mathcal{C}_{b}^{\epsilon}}$ .

Proof.

Let $C$ and $\delta$ be the constants of Lemma 4.5. Then, for all $f\in\mathcal{C}_{b}^{\epsilon}$ and $n$ ,

\mathbb{E}[|f(\mathcal{D}_{n})-f(\mathcal{D}_{-n})|^{p}]^{\frac{1}{p}}\leq 2Cn^{-1-\delta}\|f\|_{\mathcal{C}_{b}^{\epsilon}}.

By triangle inequality in $L^{p}$ ,

\mathbb{E}\Big{[}\Big{|}\sum_{n=1}^{\infty}(f(\mathcal{D}_{n})-f(\mathcal{D}_{-N}))\Big{|}^{p}\Big{]}^{\frac{1}{p}}\leq 2C\|f\|_{\mathcal{C}_{b}^{\epsilon}}\sum_{n=1}^{\infty}N^{-1-\delta}\leq C^{\prime}\|f\|_{\mathcal{C}_{b}^{\epsilon}},

as expected. ∎

With this estimation in hand, we can now prove Lemma 4.3.

Proof of Lemma 4.3.

For $i\in\{0,\dots,2^{n}-1\}$ , we define $\bar{f}^{i}:{\mathbb{R}}^{2}\to{\mathbb{R}}$ the constant function whose unique value is equal to $f(X_{i2^{-n}})$ , and $\tilde{f}^{i}=f-\bar{f}^{i}$ . Since for all $i$ , $f\mapsto\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X_{i}}{\mathbf{n}}_{X}(z)f(z){\mathop{}\!\mathrm{d}}z$ is linear, it suffices to show that both

\sum_{i=1}^{2^{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z)\bar{f}^{i}(z){\mathop{}\!\mathrm{d}}z=\sum_{i=1}^{2^{n}}f(X_{i2^{-n}})\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\qquad\mbox{ and }\qquad\sum_{i=1}^{2^{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z

almost surely converge toward $0$ as $n\to\infty$ .

From symmetry, for all $i$ , $\mathbb{E}\Big{[}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X_{i}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z|(X_{s})_{s\leq\tfrac{i}{2^{n}}}\Big{]}=0$ . It follows that, for $i<j$ ,

\mathbb{E}\Big{[}f(X_{i2^{-n}})f(X_{j2^{-n}})\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X_{i}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X_{j}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\Big{]}=0.

Besides, from a simple scaling argument,

\mathbb{E}\Big{[}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}=2^{-2n}\mathbb{E}\Big{[}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}.

Notice $\mathbb{E}[(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z)^{2}]<\infty$ , which follows for example from the previous corollary.

We deduce that

	$\displaystyle\mathbb{E}\Big{[}\Big{(}\sum_{i=1}^{2^{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z)\bar{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}$	$\displaystyle=\sum_{i=1}^{2^{n}}\mathbb{E}\Big{[}f(X_{i2^{-n}})^{2}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}$
		$\displaystyle\leq 2^{-n}\\|f\\|^{2}_{\infty}\mathbb{E}\Big{[}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}.$

This $L^{2}$ convergence rate is sufficient to conclude to the almost sure convergence: for all $\epsilon^{\prime}>0$ ,

	$\displaystyle\mathbb{P}\Big{(}\exists n\geq n_{0}:\Big{\|}\sum_{i=1}^{2^{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z)\bar{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\geq\epsilon^{\prime}\Big{)}$	$\displaystyle\leq\frac{1}{\epsilon^{\prime 2}}\mathbb{E}\Big{[}\sup_{n\geq n_{0}}\Big{(}\sum_{i=1}^{2^{n}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X^{i}}{\mathbf{n}}_{X}(z)\bar{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}$
		$\displaystyle\leq\frac{2^{1-n_{0}}}{\epsilon^{\prime 2}}\\|f\\|^{2}_{\infty}\mathbb{E}\Big{[}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int_{X}{\mathbf{n}}_{X}(z){\mathop{}\!\mathrm{d}}z\Big{)}^{2}\Big{]}$
		$\displaystyle\underset{n_{0}\to\infty}{\longrightarrow}0.$

In order to deal with the sum involving $\tilde{f}^{i}$ , one must be a bit careful about the way we use the translation invariance and scale invariance of the Brownian motion. We set $\alpha<\frac{1}{2}$ and we define the event

\mathcal{R}=\{\|X\|_{\mathcal{C}^{\alpha}}\leq R\},

for a fixed $R\geq 1$ . Let $\hat{f}^{i}$ be the (random) function defined by

\hat{f}^{i}(X_{i2^{-n}}+z)=\left\{\begin{array}[]{ll}\tilde{f}^{i}(X_{i2^{-n}}+z)&\mbox{if }|z|\leq R2^{-\alpha n},\\ \tilde{f}^{i}(X_{i2^{-n}}+\frac{R2^{-\alpha n}}{|z|}z)&\mbox{otherwise.}\end{array}\right.

In particular, $\hat{f}^{i}$ satisfies the following properties:

$\scriptstyle\diamond$

$\hat{f}^{i}=\tilde{f}^{i}$ on $B=B(X_{i2^{-n}},R2^{-\alpha n})$ , so that, in the event $\mathcal{R}$ , $\hat{f}^{i}(\mathcal{D}^{i}_{n})=\tilde{f}^{i}(\mathcal{D}^{i}_{n})$ ,
$\scriptstyle\diamond$

$|\hat{f}^{i}|_{\mathcal{C}^{\epsilon}}\leq|f|_{\mathcal{C}^{\epsilon}}$ , and $\|\hat{f}^{i}\|_{\infty}\leq R^{\epsilon}2^{-\epsilon\alpha n}|f|_{\mathcal{C}^{\epsilon}}$ ,
$\scriptstyle\diamond$

As a random variable, $\hat{f}^{i}$ is measurable with respect to $\sigma(X_{i2^{-n}})$ .

Set also $\check{f}^{i}(z)=\hat{f}^{i}(X_{i2^{-n}}+2^{-\frac{n}{2}}z)$ , $\check{X}^{i}:s\in[0,1]\mapsto 2^{\frac{n}{2}}(X_{(i+s)2^{-n}}-X_{i2^{-n}})$ , which is a standard planar Brownian motion started from $0$ , independent from $X_{i2^{-n}}$ . Notice that $\|\check{f}^{i}\|_{\infty}=\|\hat{f}^{i}\|_{\infty}\leq R^{\epsilon}2^{-\epsilon\alpha n}|f|_{\mathcal{C}^{\epsilon}}$ and $|\check{f}^{i}|_{\mathcal{C}^{\epsilon}}=2^{-\frac{\epsilon n}{2}}|\hat{f}^{i}|_{\mathcal{C}^{\epsilon}}\leq 2^{-\frac{\epsilon n}{2}}|f|_{\mathcal{C}^{\epsilon}}$ , so that

\|\check{f}^{i}\|_{\mathcal{C}_{b}^{\epsilon}}\leq 2^{1-\epsilon\alpha n}|f|_{\mathcal{C}^{\epsilon}}.

On the event $\mathcal{R}$ , we have

\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z=2^{-n}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{\check{X}^{i}}(w)\check{f}^{i}(2^{-\frac{n}{2}}w){\mathop{}\!\mathrm{d}}w.

Using Corollary 4.6 with $p=1$ , we deduce

	$\displaystyle\mathbb{E}\Big{[}\mathbbm{1}_{\mathcal{R}}\Big{\|}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\Big{]}$	$\displaystyle=2^{-n}\mathbb{E}\left[\mathbb{E}\left[\big{\|}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{\check{X}^{i}}(w)\check{f}^{i}(2^{-\frac{n}{2}}w){\mathop{}\!\mathrm{d}}w\big{\|}\middle\|X_{i2^{-n}}\right]\right]$
		$\displaystyle\leq 2^{-n}\mathbb{E}\big{[}C\\|\check{f}^{i}\\|_{\mathcal{C}^{\epsilon}_{b}}\big{]}$
		$\displaystyle\leq C2^{1-n-\epsilon\alpha n}\|f\|_{\mathcal{C}^{\epsilon}}.$

Thus,

	$\displaystyle\mathbb{P}\Big{(}\mathcal{R}\mbox{ and }\exists n\geq n_{0}:\Big{\|}\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\geq\epsilon^{\prime}\Big{)}$	$\displaystyle\leq\frac{1}{\epsilon^{\prime}}\sum_{n=n_{0}}^{\infty}\sum_{i=0}^{2^{n}-1}\mathbb{E}\Big{[}\mathbbm{1}_{\mathcal{R}}\Big{\|}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\Big{]}$
		$\displaystyle\leq C_{\epsilon,\epsilon^{\prime},\alpha,R}2^{-\epsilon\alpha n_{0}}\|f\|_{\mathcal{C}^{\epsilon}}$
		$\displaystyle\underset{n_{0}\to\infty}{\longrightarrow}0.$

Since this holds for all $R$ , we deduce that $\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z$ almost surely converges toward $0$ as $n\to\infty$ , which concludes the proof. ∎

4.5. Stratonovich integral as a limit of integrals along piecewise-linear paths

It only remains to prove lemma 4.4 which for $\eta\in\mathcal{C}^{1+\epsilon}(T^{*}\mathbb{R}^{2})$ identifies the limit

\lim_{n\to\infty}\int_{X^{(n)}}\eta

with the Stratonovich integral of $\eta$ along $X$ , which is fairly classical. It is for example a direct consequence of the following lemma.

Lemma 4.7.

For a given dissection $\Delta=(t_{0}=0,t_{1},\dots,t_{n}=1)$ , and $X:[0,1]\to\mathbb{R}^{2}$ a Brownian motion, let $X_{\Delta}$ be the piecewise-linear approximation of $X$ associated with $\Delta$ : for $\lambda\in[0,1]$ and $t=\lambda t_{i}+(1-\lambda)t_{i+1}$ ,

X_{\Delta}(t)=\lambda X_{t_{i}}+(1-\lambda)X_{t_{i+1}}.

For $f\in\mathcal{C}^{1}(\mathbb{R}^{2})$ , let

	$\displaystyle I^{1}_{\Delta}(f)$	$\displaystyle=\sum_{[t_{i},t_{i+1}]\in\Delta}f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}(X^{1}(t_{i+1})-X^{1}(t_{i})),$
	$\displaystyle I^{2}_{\Delta}(f)$	$\displaystyle=\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{f(X_{t_{i+1}})+f(X_{t_{i}})}{2}(X^{1}(t_{i+1})-X^{1}(t_{i})),$
	$\displaystyle I^{3}_{\Delta}(f)$	$\displaystyle=\int_{0}^{1}f(X_{\Delta}(t)){\mathop{}\!\mathrm{d}}X_{\Delta}(t).$

Then, almost surely, for all $f\in\mathcal{C}^{1+\epsilon}(\mathbb{R}^{2})$ , as $|\Delta|\to 0$ ,

I^{2}_{\Delta}(f)-I^{1}_{\Delta}(f)\to 0\quad\mbox{and}\quad I^{3}_{\Delta}(f)-I^{1}_{\Delta}(f)\to 0.

Proof.

Let $\alpha\in\big{(}\tfrac{1}{2+\epsilon},\tfrac{1}{2}\big{)}$ . On the almost sure event $\|X\|_{\mathcal{C}^{\alpha}}<\infty$ , we have

	$\displaystyle\Big{\|}\sum_{[t_{i},t_{i+1}]\in\Delta}\Big{(}\frac{f(X_{t_{i+1}})+f(X_{t_{i}})}{2}-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}\Big{)}(X^{1}_{t_{i+1}}-X^{1}_{t_{i}})\Big{\|}$
	$\displaystyle\leq\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{1}{2}\Big{\|}f(X_{t_{i+1}})-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}+f(X_{t_{i}})-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}\Big{\|}\Big{\|}X^{1}_{t_{i+1}}-X^{1}_{t_{i}}\Big{\|}$
	$\displaystyle\leq\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{1}{4}\Big{\|}\underbrace{\nabla_{X_{t_{i+1}}-X_{t_{i}}}f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}+\nabla_{X_{t_{i}}-X_{t_{i+1}}}f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}}_{=0}\Big{\|}\Big{\|}X^{1}_{t_{i+1}}-X^{1}_{t_{i}}\Big{\|}$
	$\displaystyle\hskip 28.45274pt+\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{2}{2^{2+\epsilon}}\\|\nabla f\\|_{\mathcal{C}^{\epsilon}}\|X_{t_{i+1}}+X_{t_{i}}\|^{2+\epsilon}$
	$\displaystyle\leq 2^{-1-\epsilon}\\|f\\|_{\mathcal{C}^{1+\epsilon}}\\|X\\|_{\mathcal{C}^{\alpha}}^{2+\epsilon}\sum_{[t_{i},t_{i+1}]\in\Delta}\|t_{i+1}-t_{i}\|^{\alpha(2+\epsilon)}\underset{\|\Delta\|\to 0}{\longrightarrow}0.$

The second convergence is proved in a similar way:

	$\displaystyle\Big{\|}\sum_{[t_{i},t_{i+1}]\in\Delta}\Big{(}\int_{t_{i}}^{t_{i+1}}f(X_{\Delta}(s)){\mathop{}\!\mathrm{d}}X_{\Delta}(s)-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}(X^{1}_{t_{i+1}}-X^{1}_{t_{i}})\Big{)}\Big{\|}$
	$\displaystyle\leq\sum_{[t_{i},t_{i+1}]\in\Delta}\|X^{1}_{t_{i+1}}-X^{1}_{t_{i}}\|$
	$\displaystyle\hskip 28.45274pt\Big{\|}\int_{\frac{1}{2}}^{1}\Big{(}f(\lambda X_{t_{i}}+(1-\lambda)X_{t_{i+1}})+f((1-\lambda)X_{t_{i}}+\lambda X_{t_{i+1}})-2f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}\Big{)}{\mathop{}\!\mathrm{d}}\lambda\Big{\|}$
	$\displaystyle\leq 2^{-1-\epsilon}\\|f\\|_{\mathcal{C}^{1+\epsilon}}\\|X\\|_{\mathcal{C}^{\alpha}}^{2+\epsilon}\sum_{[t_{i},t_{i+1}]\in\Delta}\|t_{i+1}-t_{i}\|^{\alpha(2+\epsilon)}\underset{\|\Delta\|\to 0}{\longrightarrow}0.$

∎

This concludes the proof of Lemma 4.4, and therefore the proof of Theorem 1 as well. Before we conclude this section, we will shortly prove Corollary 2.

Proof of Corollary 2.

To keep the proof simple, we treat the case when $X:[0,1]\to\mathbb{R}^{2}$ is a Brownian loop started from $0$ . To deal with the case when $X$ is a Brownian bridge from $x$ to $y\neq x$ , one must also take into account the winding function of the triangle between $x$ , $y$ , and $X_{\frac{1}{2}}$ , but this is done in a straightforward way.

From linearity, it suffices to prove the result when restricted to functions $f\geq 0$ . Furthermore, since the result is trivial in the event $f_{|{B(0,\|X\|_{\infty})}}=0$ , we assume $\int_{B(0,\|X\|_{\infty})}f(z){\mathop{}\!\mathrm{d}}z>0$ .

Let $X^{1}$ be the restriction of $X$ to $[0,\tfrac{1}{2}]$ , $X^{2}$ its restriction to $[\tfrac{1}{2},1]$ , and $\hat{X}_{2}:t\in[0,\tfrac{1}{2}]\mapsto X_{1-t}$ . Then, the distribution of $X^{1}$ (resp. $\hat{X}^{2}$ ) admits a density with respect to the density of a standard planar Brownian motion defined on $[0,\tfrac{1}{2}]$ . Using scale invariance, we can apply Theorem 1 to both $X^{1}$ and $\hat{X}^{2}$ . We deduce that, for $i\in\{1,2\}$ , for all $\epsilon>0$ , almost surely, for all $f\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ ,

\int_{\mathbb{R}^{2}}[{\mathbf{n}}_{X^{i}}(z)]_{k}f(z){\mathop{}\!\mathrm{d}}z

converges as $k\to\infty$ , and the limits are almost surely equal to respectively $\int_{X^{1}}\eta+\int_{[X_{\frac{1}{2}},0]}\eta$ and $\int_{X^{2}}\eta-\int_{[X_{\frac{1}{2}},0]}\eta$ , where $\eta$ is such that $\partial_{1}\eta_{2}-\partial_{2}\eta_{1}=f$ .

Now we need to show that almost surely, for all $f\in\mathcal{C}^{\epsilon}(\mathbb{R}^{2})$ , $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{1}}(z)f(z){\mathop{}\!\mathrm{d}}z$ and $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{2}}(z)f(z){\mathop{}\!\mathrm{d}}z$ add up properly, for which we proceed as in Lemma 4.2, introducing again a random point $P$ . Going through the same arguments as in the proof of Lemma 4.2, we see that it suffices to show that, $X$ -almost surely,

|\mathcal{D}^{X^{1}}_{\pm N}\cap\mathcal{D}^{X^{2}}_{\pm N}|=o(N^{-1-\delta}),

(4)

for the four possible couple of signs in front of $N$ , and for some $\delta>0$ .

To prove (4), we further decompose $X^{1}$ and $X^{2}$ by setting $X^{11}$ (resp. $X^{12}$ , $X^{21}$ , $X^{22}$ ) the restriction of $X$ to the interval $[0,\tfrac{1}{4}]$ (resp. $[\tfrac{1}{4},\tfrac{1}{2}]$ , $[\tfrac{1}{2},\tfrac{3}{4}]$ , $[\tfrac{3}{4},1]$ ). Then, $\mathcal{D}^{X^{i}}_{\pm N}\subseteq\mathcal{D}^{X^{i1}}_{\pm N^{\prime}}\cup\mathcal{D}^{X^{i2}}_{\pm N^{\prime}}$ where $N^{\prime}=\lfloor N/2\rfloor$ .

We show that almost surely, $|\mathcal{D}^{X^{11}}_{N^{\prime}}\cap\mathcal{D}^{X^{21}}_{N^{\prime}}|=O(N^{-2})$ , the 15 other intersections are treated either similarly. Conditionally on $X_{\frac{1}{2}}$ , $X^{11}$ and $X^{21}$ are independen. Furthermore, both their distribution, conditional on $X_{\frac{1}{2}}$ , have a density with respect to the distribution of a standard Brownian motion with duration $\frac{1}{4}$ , started respectively from $0$ and $X_{\frac{1}{2}}$ . Thus, it suffices to show that for all $y$ , $|\mathcal{D}^{X^{11}}_{N^{\prime}}\cap\mathcal{D}^{X^{21}}_{N^{\prime}}|=O(N^{-2})$ when $X^{11}$ and $X^{21}$ are independent Brownian motions started respectively from $0$ and $y$ . This follows directly from 3.5, with a scaling of $\frac{1}{2}$ . ∎

5. Magnetic impurities

In this section, we fix a function $g\in\mathcal{C}^{\epsilon}_{b}(\mathbb{R}^{2})$ . For all $\lambda>0$ , we define $\mathcal{P}_{\lambda}$ a Poisson process on $\mathbb{R}^{2}$ with intensity $\lambda g(z){\mathop{}\!\mathrm{d}}z$ , independent from $X$ , and $\Gamma:[0,T]\to\mathbb{R}$ a standard Cauchy process, independent from $X$ . We write $\mathbb{E}^{\mathcal{P}}$ the expectation with respect to $\mathcal{P}_{\lambda}$ , $\mathbb{E}^{X}$ the one with respect to $X$ , $\mathbb{E}^{\Gamma}$ the expectation with respect to $\Gamma$ and $\mathbb{E}=\mathbb{E}^{X}\otimes\mathbb{E}^{\mathcal{P}}\otimes\mathbb{E}^{\Gamma}$ the expectation on the product space (although none of the variables we consider depend on both $\mathcal{P}$ and $\Gamma$ , so truly $\mathbb{E}=\mathbb{E}^{X}\otimes\mathbb{E}^{\mathcal{P}}$ or $\mathbb{E}=\mathbb{E}^{X}\otimes\mathbb{E}^{\Gamma}$ , whichever is relevant).

For a function $f\in\mathcal{C}_{b}^{\epsilon}(\mathbb{R}^{2})$ , we define

\xi_{\lambda}(f)=\frac{1}{\lambda}\sum_{z\in\mathcal{P}_{\lambda}}f(z){\mathbf{n}}_{X}(z),

as well as

\xi(f)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)\ f\cdot g(z){\mathop{}\!\mathrm{d}}z+\frac{1}{2}\int_{0}^{1}f\cdot g(X_{t}){\mathop{}\!\mathrm{d}}\Gamma_{t}.

Notice that $\Gamma$ almost surely has a finite $p$ -variation for all $p>1$ (see [1, Theorem 4.1]). Since $X$ -almost surely, $(f\cdot g)\circ X\in\mathcal{C}^{\frac{\epsilon}{4}}([0,1])$ , the integral $\int_{0}^{1}fg(X_{t}){\mathop{}\!\mathrm{d}}\Gamma_{t}$ is well-defined as a Young integral.

The main result from this section is the following

Lemma 5.1.

Let $f,g\in\mathcal{C}^{\epsilon}_{b}({\mathbb{R}}^{2})$ be continuous and bounded functions. Assume that $g$ takes non-negative values. Let

G_{\beta,f,g}\coloneqq\sum_{k\neq 0}\int_{\mathcal{A}_{k}}(e^{ik\beta f(z)}-1)g(z){\mathop{}\!\mathrm{d}}z.

Then, $X$ -almost surely, as $\beta\to 0$ ,

G_{\beta,f,g}\underset{\beta\to 0}{=}i\beta\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)fg(z){\mathop{}\!\mathrm{d}}z-\frac{|\beta|}{2}\int_{0}^{1}|f(X_{t})|g(X_{t}){\mathop{}\!\mathrm{d}}t+o(\beta).

(5)

Before we dive into the proof of this lemma, we first explain with it implies both Theorem 3 and Corollary 4.

Lemma 5.1 implies Theorem 3 and Corollary 4.

Since the function $\min(|{\mathbf{n}}_{X}\cdot f|,1)$ is integrable against the intensity measure $\lambda g{\mathop{}\!\mathrm{d}}z$ of $\mathcal{P}_{\lambda}$ , we can use Campbell’s theorem, which gives

\mathbb{E}^{\mathcal{P}}[e^{i\alpha\xi_{\lambda}(f)}]=\exp\Big{(}\sum_{k\neq 0}\int_{\mathcal{A}_{k}}(e^{ik\frac{\alpha}{\lambda}f(z)}-1)\lambda g(z){\mathop{}\!\mathrm{d}}z\Big{)}=\exp(\lambda G_{\beta,f,g}),

where $\beta=\frac{\alpha}{\lambda}$ .

Besides, conditional on $X$ , $\int_{0}^{1}f(X_{t})g(X_{t}){\mathop{}\!\mathrm{d}}\Gamma(t)$ is a centered Cauchy random variable with scale parameter $\int_{0}^{1}|f(X_{t})|g(X_{t}){\mathop{}\!\mathrm{d}}t$ , whilst $\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)fg(z){\mathop{}\!\mathrm{d}}z$ is deterministic. It follows that

\mathbb{E}^{\Gamma}[e^{i\alpha\xi(f)}]=e^{i\alpha\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-3.40277pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-2.275pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.60416pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.31248pt}}\!\int{\mathbf{n}}_{X}(z)fg(z){\mathop{}\!\mathrm{d}}z-\frac{|\alpha|}{2}\int_{0}^{1}|f(X_{t})|g(X_{t}){\mathop{}\!\mathrm{d}}t},

Thus, Lemma 5.1 implies Theorem 3.

Furthermore, since both $\xi_{\lambda}(f)$ and $\xi(f)$ are linear in $f$ , one can use the Cramér-Wold device to deduce Corollary 4 from its special case $n=1$ . By Lévy’s continuity theorem, this specific case is equivalent to the statement that $X$ -almost surely, for all $\alpha\in\mathbb{R}$ ,

\mathbb{E}^{\mathcal{P}}[e^{i\alpha\xi_{\lambda}(f)}]\underset{\lambda\to\infty}{\longrightarrow}\mathbb{E}^{\Gamma}[e^{i\alpha\xi(f)}].

From our previous computation, this amount to show that $X$ almost surely, for all $\alpha\in\mathbb{R}$ ,

\exp(\lambda G_{\beta,f,g})\underset{\lambda\to\infty}{\longrightarrow}\exp\Big{(}i\alpha\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)fg(z){\mathop{}\!\mathrm{d}}z-\frac{|\alpha|}{2}\int_{0}^{1}|f(X_{t})|g(X_{t}){\mathop{}\!\mathrm{d}}t\Big{)},

which follows again from Lemma 5.1. ∎

Proof of Lemma 5.1.

From symmetry, we can assume $\beta>0$ . Performing an Abel summation, we obtain

	$\displaystyle G_{\beta,f,g}$	$\displaystyle=\sum_{k=1}^{\infty}\Big{(}\int_{\mathcal{D}_{k}}e^{i\beta kf}(1-e^{-i\beta f})g{\mathop{}\!\mathrm{d}}z+\int_{\mathcal{D}_{-k}}e^{-i\beta kf}(1-e^{i\beta f})g{\mathop{}\!\mathrm{d}}z\Big{)}$
		$\displaystyle=\sum_{k=1}^{\infty}(\phi_{k,\beta}(\mathcal{D}_{k})+\phi_{-k,\beta}(\mathcal{D}_{-k})),$

where

\phi_{k,\beta}=e^{i\beta kf}(1-e^{-\operatorname{sgn}(k)i\beta f})g.

The two terms in (5) comes from two different parts in this last sum: the term $i\beta\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z)g(z){\mathop{}\!\mathrm{d}}z$ comes from the bulk of the sum, that is the part with $k$ of the order of $1$ . The second term comes from the tail of the sum, or more precisely from the part of the sum when $k$ is of the order of $\beta^{-1}$ . We will split the sum into several parts. For $n,N\in\mathbb{N}\cup\{\infty\}$ with $n<N$ , we set

G_{\beta,f,g}^{n,N}=\sum_{k=n+1}^{N}(\phi_{k,\beta}(\mathcal{D}_{k})+\phi_{-k,\beta}(\mathcal{D}_{-k})).

For $N_{1}=N_{1}(\beta)$ and $N_{2}=N_{2}(\beta)$ which will be set later on, we decompose $G_{\beta,f,g}$ into three parts,

G_{\beta,f,g}=\underbrace{G_{\beta,f,g}^{0,N_{1}}}_{\text{bulk}}+\underbrace{G_{\beta,f,g}^{N_{1},N_{2}}}_{\text{tail}}+\underbrace{G_{\beta,f,g}^{N_{2},\infty}}_{\text{end}}.

As $\beta\to 0$ , both $N_{1}$ and $\beta N_{2}$ will slowly diverge toward $\infty$ . In particular, $N_{1}(\beta)<<\beta^{-1}<<N_{2}(\beta)$ . The reason why we need to treat the end part in a separate way is that its convergence toward $0$ is not absolute, in the sense that the

\sum_{k=N_{2}+1}^{\infty}|\phi_{k,\beta}(\mathcal{D}_{k})+\phi_{-k,\beta}(\mathcal{D}_{-k})|

does not converge toward zero as $\beta\to 0$ , and one must be a bit careful when dealing with this term. The general term (without the absolute values) slowly oscillates between positive and negative values, and we must take advantage of compensations.

For a given $k\neq 0$ , as $\beta\to 0$ , uniformly in $z$ ,

\phi_{k,\beta}(z)=\operatorname{sgn}(k)i\beta f(z)g(z)+O(\beta^{2}),

and it follows that

\phi_{k,\beta}(\mathcal{D}_{k})+\phi_{-k,\beta}(\mathcal{D}_{-k})=i\beta((fg)(\mathcal{D}_{k})-(fg)(\mathcal{D}_{-k}))+O(\beta^{2}).

For $k\geq 1$ , let $C_{k}$ be such that for all $\beta\in(0,1)$ ,

|\phi_{k,\beta}(\mathcal{D}_{k})+\phi_{-k,\beta}(\mathcal{D}_{-k})-i\beta((fg)(\mathcal{D}_{k})-(fg)(\mathcal{D}_{-k}))|\leq C_{k}\beta^{2},

and set $N_{1}(\beta)=\min(\lfloor\beta^{-\frac{1}{3}}\rfloor,\sup\{N:\forall k\leq N,C_{k}\leq\beta^{-\frac{1}{3}}\})$ .

Then,

\Big{|}G_{\beta,f,g}^{0,N}-i\beta\sum_{k=1}^{N_{1}}((fg)(\mathcal{D}_{k})-(fg)(\mathcal{D}_{-k}))\Big{|}\leq\sum_{k=1}^{N_{1}}C_{k}\beta^{2}\leq\beta^{\frac{4}{3}}=o(\beta).

Besides, $N_{1}\underset{\beta\to 0}{\longrightarrow}+\infty$ , and Theorem 1 implies that

\sum_{k=1}^{N_{1}}((fg)(\mathcal{D}_{k})-(fg)(\mathcal{D}_{-k}))\underset{\beta\to 0}{\longrightarrow}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z)g(z){\mathop{}\!\mathrm{d}}z.

Therefore,

G_{\beta,f,g}^{0,N}=i\beta\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X}(z)f(z)g(z){\mathop{}\!\mathrm{d}}z+o(\beta).

(6)

We now look at the tail part of $G_{\beta,f,g}$ . Let $\delta>0$ and $C$ (random) be such that for all $N\neq 0$ and $\phi\in\mathcal{C}^{\epsilon}_{b}$ ,

\Big{|}\phi(\mathcal{D}_{N})-\frac{1}{2\pi|N|}\int_{0}^{1}\phi(X_{u}){\mathop{}\!\mathrm{d}}u\Big{|}\leq C\|\phi\|_{\mathcal{C}_{b}^{\epsilon}}N^{-1-\delta}.

Recall that the existence of such a couple $(\delta,C)$ is provided by Lemma 3.3. Let $N_{2}=N_{2}(\beta)$ be any integer-valued function such that $\beta N_{2}\underset{\beta\to 0}{\longrightarrow}+\infty$ and $\beta N_{2}^{1-\delta}\underset{\beta\to 0}{\longrightarrow}0$ .

For all $\phi,\psi\in\mathcal{C}^{\epsilon}_{b}$ , $|\phi\psi|_{\mathcal{C}^{\epsilon}}\leq|\phi|_{\mathcal{C}^{\epsilon}}\|\psi\|_{\infty}+\|\phi\|_{\infty}|\psi|_{\mathcal{C}^{\epsilon}}$ . We deduce that for all $k$ and $\beta$ ,

	$\displaystyle\\|\phi_{k,\beta}\\|_{\infty}$	$\displaystyle\leq\\|e^{i\beta kf}\\|_{\infty}\\|1-e^{i\beta f}\\|_{\infty}\\|g\\|_{\infty}\leq\beta\\|f\\|_{\infty}\\|g\\|_{\infty},$
	$\displaystyle\|\phi_{k,\beta}\|_{\mathcal{C}^{\epsilon}}$	$\displaystyle\leq\|e^{i\beta kf}\|_{\mathcal{C}^{\epsilon}}\\|1-e^{i\beta f}\\|_{\infty}\\|g\\|_{\infty}+\\|e^{i\beta kf}\\|_{\infty}\|1-e^{i\beta f}\|_{\mathcal{C}^{\epsilon}}\\|g\\|_{\infty}+\\|e^{i\beta kf}\\|_{\infty}\\|1-e^{i\beta f}\\|_{\infty}\|g\|_{\mathcal{C}^{\epsilon}}$
		$\displaystyle\leq k\beta^{2}\|f\|_{\mathcal{C}^{\epsilon}}\\|f\\|_{\infty}\\|g\\|_{\infty}+\beta\|f\|_{\mathcal{C}^{\epsilon}}\\|g\\|_{\infty}+\beta\\|f\\|_{\infty}\|g\|_{\mathcal{C}^{\epsilon}},$

so that

\|\phi_{k,\beta}\|_{\mathcal{C}^{\epsilon}_{b}}\leq\beta(1+k\beta)(1+\|f\|_{\mathcal{C}^{\epsilon}_{b}})\|f\|_{\mathcal{C}^{\epsilon}_{b}}\|g\|_{\mathcal{C}^{\epsilon}_{b}}.

We deduce that, for all $k>0$ ,

\Big{|}\phi_{k,\beta}(\mathcal{D}_{k})+\phi_{-k,\beta}(\mathcal{D}_{-k})-\frac{1}{2\pi k}\int_{0}^{1}(\phi_{k,\beta}(X_{u})+\phi_{-k,\beta}(X_{u})){\mathop{}\!\mathrm{d}}u\Big{|}\leq 2C(1+\|f\|_{\mathcal{C}^{\epsilon}})\|f\|_{\mathcal{C}^{\epsilon}}\|g\|_{\mathcal{C}^{\epsilon}}\beta(1+k\beta)k^{-1-\delta},

and there exists constants $C^{\prime}=C^{\prime}(f,g),\ C^{\prime\prime}=C^{\prime\prime}(f,g)$ such that for all $N_{2}\geq N_{1}$ ,

	$\displaystyle\Big{\|}G_{\beta,f,g}^{N_{1},N_{2}}-$	$\displaystyle\frac{1}{2\pi}\sum_{k=N_{1}+1}^{N_{2}}\frac{1}{k}\int_{0}^{1}(\phi_{k,\beta}(X_{u})+\phi_{-k,\beta}(X_{u})){\mathop{}\!\mathrm{d}}u\Big{\|}$
		$\displaystyle\leq C^{\prime}\sum_{k=N_{1}+1}^{N_{2}}\beta(1+k\beta)k^{-1-\delta}\leq C^{\prime\prime}\beta(N_{1}^{-\delta}+\beta N_{2}^{1-\delta})=o(\beta).$

The remaining part of the analysis is standard calculus. Set

\psi_{k,\beta}=e^{i\beta kf}\operatorname{sgn}(k)i\beta fg.

Then, for $\beta\leq\|f\|_{\infty}$ ,

	$\displaystyle\Big{\|}\sum_{k=N_{1}+1}^{N_{2}}\frac{\phi_{k,\beta}-\psi_{k,\beta}}{k}\Big{\|}$	$\displaystyle=\|g\|\Big{\|}\sum_{k=N_{1}+1}^{N_{2}}\frac{1}{k}e^{i\beta kf}(1-e^{-\operatorname{sgn}(k)i\beta f}-\operatorname{sgn}(k)i\beta f)\Big{\|}$
		$\displaystyle\leq\|g\|\sum_{k=N_{1}+1}^{N_{2}}\frac{1}{k}\frac{\beta^{2}f^{2}}{2}\leq C_{f,g}\|\log(\beta)\|\beta^{2}=o(\beta).$

It follows that

	$\displaystyle G_{\beta,f,g}^{N_{1},N_{2}}$	$\displaystyle=\frac{1}{2\pi}\sum_{k=N_{1}+1}^{N_{2}}\frac{1}{k}\int_{0}^{1}(\psi_{k,\beta}(X_{u})+\psi_{-k,\beta}(X_{u})){\mathop{}\!\mathrm{d}}u+o(\beta)$
		$\displaystyle=-\frac{\beta}{\pi}\sum_{k=N_{1}+1}^{N_{2}}\int_{0}^{1}f(X_{u})g(X_{u})\frac{\sin(k\beta f(X_{u}))}{k}{\mathop{}\!\mathrm{d}}u+o(\beta)$
		$\displaystyle=-\frac{\beta}{\pi}\sum_{k=1}^{N_{2}}\int_{0}^{1}f(X_{u})g(X_{u})\frac{\sin(k\beta f(X_{u}))}{k}{\mathop{}\!\mathrm{d}}u+o(\beta).$

The last line follows from the fact that

\Big{|}\frac{\beta}{\pi}\sum_{k=1}^{N_{1}}\int_{0}^{1}f(X_{u})g(X_{u})\frac{\sin(k\beta f(X_{u}))}{k}{\mathop{}\!\mathrm{d}}u\Big{|}\leq\|f\|_{\infty}^{2}\|g\|_{\infty}\beta^{2}N_{1}=o(\beta).

For $s\leq 0$ , let

\Phi(s)=\left\{\begin{array}[]{ll}\int_{0}^{1}f(X_{u})g(X_{u})\frac{\sin(sf(X_{u}))}{s}{\mathop{}\!\mathrm{d}}u&\mbox{for }s\neq 0\\ \int_{0}^{1}f(X_{u})^{2}g(X_{u}){\mathop{}\!\mathrm{d}}u&\mbox{for }s=0,\end{array}\right.

so that $\Phi$ is continuous on $[0,\infty)$ and

G_{\beta,f,g}^{N_{1},N_{2}}=-\frac{\beta^{2}}{\pi}\sum_{k=1}^{N_{2}}\Phi(\beta k)+o(\beta).

(7)

For all $R>0$ ,

\Big{|}\beta\sum_{k=1}^{\lfloor R\beta^{-1}\rfloor}\Phi(\beta k)-\int_{0}^{R}\Phi(s){\mathop{}\!\mathrm{d}}s\Big{|}\leq\beta\|f\|_{\infty}^{2}\|g\|_{\infty}+\omega_{\Phi,[0,R]}(\beta),

where $\omega_{\Phi,[0,R]}(\beta)=\sup_{s,t\in[0,R]}|\Phi(s)-\Phi(t)|$ is the continuity modulus of $\Phi$ .

Since $\beta+\omega_{\Phi,[0,R]}(\beta)\to 0$ for all $R>0$ , there exists a function $R_{\beta}$ such that $R_{\beta}\to\infty$ as $\beta\to 0$ and $\beta+\omega_{\Phi,[0,R_{\beta}]}(\beta)\to 0$ . We fix such a function, and set $N_{2}=\beta^{-\frac{2}{2-\delta}}\wedge(\beta^{-1}R_{\beta})$ . This way, we do have $\beta N_{2}\underset{\beta\to 0}{\longrightarrow}+\infty$ and $\beta N_{2}^{1-\delta}\underset{\beta\to 0}{\longrightarrow}0$ .

We obtain

\Big{|}\beta\sum_{k=1}^{N_{2}}\Phi(\beta k)-\int_{0}^{\beta^{-1}N_{2}}\Phi(s){\mathop{}\!\mathrm{d}}s\Big{|}=o(1).

(8)

To estimate this last integral, there is two things we must be careful about. First, because of the $\operatorname{sinc}$ function in the definition of $\Phi$ , the function $\Phi$ is not integrable on $[0,+\infty)$ so we cannot naively replace the bound $\beta^{-1}N_{2}$ with its limit. Secondly, when manipulating the integral, we must be extra careful at the vicinity of $f(X_{u})=0$ .

Recall that for $x\neq 0$ , $\lim_{C\to\infty}\int_{0}^{C}\frac{\sin(sx)}{s}{\mathop{}\!\mathrm{d}}s=\operatorname{sgn}(x)\frac{\pi}{2}$ . Performing an integration by part, we deduce that for all $x$ and $C>0$ ,

	$\displaystyle\Big{\|}\int_{0}^{C}\frac{\sin(sx)}{s}{\mathop{}\!\mathrm{d}}s-\operatorname{sgn}(x)\frac{\pi}{2}\Big{\|}$	$\displaystyle=\Big{\|}\lim_{C^{\prime}\to\infty}\int_{C}^{C^{\prime}}\frac{\sin(sx)}{s}{\mathop{}\!\mathrm{d}}s\Big{\|}$
		$\displaystyle=\Big{\|}\frac{\cos(Cx)}{Cx}-\lim_{C^{\prime}\to\infty}\int_{C}^{C^{\prime}}\frac{\cos(sx)}{s^{2}x}{\mathop{}\!\mathrm{d}}s\Big{\|}$
		$\displaystyle\leq\frac{2}{C\|x\|}.$

It follows that

$\displaystyle\Big{\|}\int_{0}^{\beta^{-1}N_{2}}\Phi(s){\mathop{}\!\mathrm{d}}s$	$\displaystyle-\frac{\pi}{2}\int_{0}^{1}\|f(X_{u})\|g(X_{u}){\mathop{}\!\mathrm{d}}u\Big{\|}$
	$\displaystyle=\Big{\|}\int_{0}^{1}f(X_{u})g(X_{u})\Big{(}\int_{0}^{\beta^{-1}N_{2}}\frac{\sin(sf(X_{u}))}{s}{\mathop{}\!\mathrm{d}}s-\operatorname{sgn}(f(X_{u}))\frac{\pi}{2}\Big{)}{\mathop{}\!\mathrm{d}}u\Big{\|}$
	$\displaystyle\leq\int_{0}^{1}\|f(X_{u})\|g(X_{u})\frac{2}{{\beta^{-1}N_{2}}\|f(X_{u})\|}{\mathop{}\!\mathrm{d}}u$
	$\displaystyle=O(\beta N_{2}^{-1})=o(1).$	(9)

Combining (7), (8) and (9), we obtain

G^{N_{1},N_{2}}_{\beta,f,g}=-\frac{\beta}{2}\int_{0}^{1}|f(X_{u})|g(X_{u}){\mathop{}\!\mathrm{d}}u+o(\beta).

(10)

We finally look at the end part of $G_{\beta,f,g}$ . Since the $\mathcal{C}^{\epsilon}$ norm of $\phi_{k,\beta}$ becomes arbitrarily large as $k$ goes to infinity, one cannot directly rely on Lemma 3.3. For a positive integer $j$ , we decompose $G^{j^{2}N_{2},(j+1)^{2}N_{2}}_{\beta,f,g}$ into

	$\displaystyle G^{j^{2}N_{2},(j+1)^{2}N_{2}}_{\beta,f,g}$	$\displaystyle=\underbrace{\sum_{k=j^{2}N_{2}+1}^{(j+1)^{2}N_{2}}(\phi_{k,\beta}(\mathcal{D}_{(j+1)^{2}N_{2}})-\phi_{-k,\beta}(\mathcal{D}_{-(j+1)^{2}N_{2}}))}_{H^{j}_{\beta,f,g}}$
		$\displaystyle\hskip 56.9055pt+\underbrace{\sum_{k=j^{2}N_{2}+1}^{(j+1)^{2}N_{2}}(\phi_{k,\beta}(\mathcal{D}_{k})-\phi_{k,\beta}(\mathcal{D}_{(j+1)^{2}N_{2}})-\phi_{k,\beta}(\mathcal{D}_{-k})+\phi_{-k,\beta}(\mathcal{D}_{-(j+1)^{2}N_{2}})}_{K^{j}_{\beta,f,g}}.$

We have

	$\displaystyle\Big{\|}\sum_{k=j^{2}N_{2}+1}^{(j+1)^{2}N_{2}}\phi_{k,\beta}(\mathcal{D}_{(j+1)^{2}N_{2}})\Big{\|}$	$\displaystyle=\Big{\|}\int_{\mathcal{D}_{(j+1)^{2}N_{2}}}\sum_{k=j^{2}N_{2}+1}^{(j+1)^{2}N_{2}}e^{-i\beta kf(z)}(1-e^{-i\beta f(z)})g(z){\mathop{}\!\mathrm{d}}z\Big{\|}$
		$\displaystyle=\Big{\|}\int_{\mathcal{D}_{(j+1)^{2}N_{2}}}e^{-i\beta(j^{2}N_{2}+1)f(z)}(1-e^{-i\beta((j+1)^{2}N_{2}-j^{2}N_{2})f(z)})g(z){\mathop{}\!\mathrm{d}}z\Big{\|}$
		$\displaystyle\leq\int_{\mathcal{D}_{(j+1)^{2}N_{2}}}2\|g(z)\|{\mathop{}\!\mathrm{d}}z$
		$\displaystyle\leq 2\\|g\\|_{\infty}D_{(j+1)^{2}N_{2}}.$

Using again Lemma 3.3 with $f=1$ , we deduce that almost surely, there exists $C$ such that for all $N$ , $D_{N}\leq\frac{C}{N}$ . It follows that

|H^{j}_{\beta,f,g}|\leq\frac{4C\|g\|_{\infty}}{(j+1)^{2}N_{2}},

which yields

\Big{|}\sum_{j=1}^{\infty}H^{j}_{\beta,f,g}\Big{|}\leq\frac{4C\|g\|_{\infty}}{N_{2}}\sum_{j=2}^{\infty}\frac{1}{j^{2}}=o(\beta).

As for $K^{j}_{\beta,f,g}$ , using the fact that the sequences $(\mathcal{D}_{k})_{k\geq 1}$ and $(\mathcal{D}_{-k})_{k\geq 1}$ are nested, we have

	$\displaystyle K^{j}_{\beta,f,g}$	$\displaystyle=\sum_{k=j^{2}N_{2}+1}^{(j+1)^{2}N_{2}}\|\phi_{x,\beta}\|(D_{k}-D_{(j+1)^{2}N_{2}}+D_{-k}-D_{-(j+1)^{2}N_{2}})$
		$\displaystyle\leq\sum_{k=j^{2}N_{2}+1}^{(j+1)^{2}N_{2}}\beta\\|f\\|_{\infty}\\|g\\|_{\infty}(D_{k}-D_{(j+1)^{2}N_{2}}+D_{-k}-D_{-(j+1)^{2}N_{2}}).$

Let $C,\delta>0$ such that for all $N\neq 0$ ,

\big{|}D_{N}-\frac{1}{2\pi|N|}\big{|}\leq CN^{-1-\delta}.

Then, for all $k\in\{j^{2}N_{2}+1,\dots,(j+1)^{2}N_{2}\}$ ,

0\leq D_{k}-D_{(j+1)^{2}N_{2}}\leq\frac{1}{2\pi k}-\frac{1}{2\pi(j+1)^{2}N_{2}}+2Ck^{-1-\delta}\leq C^{\prime}\big{(}\frac{1}{j^{3}N_{2}^{2}}+(j^{2}N_{2})^{-1-\delta}\big{)}.

We deduce

|K^{j}_{\beta,f,g}|\leq C^{\prime\prime}\|f\|_{\infty}\|g\|_{\infty}N_{2}^{-1}j^{-2},

and it follows that

\sum_{j=1}^{\infty}|K^{j}_{\beta,f,g}|=o(\beta).

Finally, we have

|G^{N_{2},\infty}_{\beta,f,g}|\leq\sum_{j=1}^{\infty}|G^{j^{2}N_{2},(j+1)^{2}N_{2}}_{\beta,f,g}|\leq\sum_{j=1}^{\infty}|K^{j}_{\beta,f,g}|+\sum_{j=1}^{\infty}|H^{j}_{\beta,f,g}|=o(\beta).

(11)

We conclude the proof by putting together (6), (10) and (11). ∎

6. Funding

I am pleased to acknowledge support from the ERC Advanced Grant 740900 (LogCorRM), and later from the EPSRC grant EP/W006227/1 .

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	$\displaystyle\Big{\|}f(\mathcal{D}_{n})-f(\mathcal{D}_{-n})\Big{\|}$	$\displaystyle\leq C^{\prime}n^{-1}(\omega_{f}(2\|X\|_{\mathcal{C}^{\beta}}n^{-\delta})+\\|f\\|_{\infty}n^{-\delta})$
		$\displaystyle\leq C^{\prime\prime}n^{-1}(\|f\|_{\mathcal{C}^{\epsilon}}n^{-\delta\epsilon}+\\|f\\|_{\infty}n^{-\delta}).$		(1)

	$\displaystyle f(\mathcal{D}_{N})$	$\displaystyle\leq\sum_{i=0}^{T-1}f(X(\tfrac{i}{T}))\|\mathcal{D}^{i}_{N^{\prime}}\|+\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}\sum_{i=0}^{T-1}\|\mathcal{D}^{i}_{N^{\prime}}\|+\\|f\\|_{\infty}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|.$
		$\displaystyle\leq\frac{1}{2\pi NT}\sum_{i=0}^{T-1}f(X(\tfrac{i}{T}))+\\|f\\|_{\infty}\sum_{i=0}^{T-1}\big{\|}\tfrac{1}{2\pi NT}-\|\mathcal{D}^{i}_{N^{\prime}}\|\big{\|}+\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}\sum_{i=0}^{T-1}\|\mathcal{D}^{i}_{N^{\prime}}\|$
		$\displaystyle\hskip 310.13486pt+\\|f\\|_{\infty}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|$
		$\displaystyle\leq\frac{1}{2\pi N}\int_{0}^{1}f(X_{t}){\mathop{}\!\mathrm{d}}t+\frac{\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|_{\mathcal{C}^{\alpha}}^{\epsilon}T^{-\epsilon\alpha}}{2\pi N}+\\|f\\|_{\infty}\sum_{i=0}^{T-1}\big{\|}\tfrac{1}{2\pi NT}-\|\mathcal{D}^{i}_{N^{\prime}}\|\big{\|}$
		$\displaystyle\hskip 170.71652pt+\|f\|_{\mathcal{C}^{\epsilon}}\\|X\\|^{\epsilon}_{\mathcal{C}^{\alpha}}T^{-\epsilon\alpha}\sum_{i=0}^{T-1}\|\mathcal{D}^{i}_{N^{\prime}}\|+\\|f\\|_{\infty}\sum_{i\neq j}\|\mathcal{D}^{i,j}_{M}\|.$

	$\displaystyle\mathbb{E}\Big{[}\mathbbm{1}_{\mathcal{R}}\Big{\|}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\Big{]}$	$\displaystyle=2^{-n}\mathbb{E}\left[\mathbb{E}\left[\big{\|}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{\check{X}^{i}}(w)\check{f}^{i}(2^{-\frac{n}{2}}w){\mathop{}\!\mathrm{d}}w\big{\|}\middle\|X_{i2^{-n}}\right]\right]$
		$\displaystyle\leq 2^{-n}\mathbb{E}\big{[}C\\|\check{f}^{i}\\|_{\mathcal{C}^{\epsilon}_{b}}\big{]}$
		$\displaystyle\leq C2^{1-n-\epsilon\alpha n}\|f\|_{\mathcal{C}^{\epsilon}}.$

	$\displaystyle\mathbb{P}\Big{(}\mathcal{R}\mbox{ and }\exists n\geq n_{0}:\Big{\|}\sum_{i=0}^{2^{n}-1}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\geq\epsilon^{\prime}\Big{)}$	$\displaystyle\leq\frac{1}{\epsilon^{\prime}}\sum_{n=n_{0}}^{\infty}\sum_{i=0}^{2^{n}-1}\mathbb{E}\Big{[}\mathbbm{1}_{\mathcal{R}}\Big{\|}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\!\int{\mathbf{n}}_{X^{i}}(z)\tilde{f}^{i}(z){\mathop{}\!\mathrm{d}}z\Big{\|}\Big{]}$
		$\displaystyle\leq C_{\epsilon,\epsilon^{\prime},\alpha,R}2^{-\epsilon\alpha n_{0}}\|f\|_{\mathcal{C}^{\epsilon}}$
		$\displaystyle\underset{n_{0}\to\infty}{\longrightarrow}0.$

	$\displaystyle\Big{\|}\sum_{[t_{i},t_{i+1}]\in\Delta}\Big{(}\frac{f(X_{t_{i+1}})+f(X_{t_{i}})}{2}-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}\Big{)}(X^{1}_{t_{i+1}}-X^{1}_{t_{i}})\Big{\|}$
	$\displaystyle\leq\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{1}{2}\Big{\|}f(X_{t_{i+1}})-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}+f(X_{t_{i}})-f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}\Big{\|}\Big{\|}X^{1}_{t_{i+1}}-X^{1}_{t_{i}}\Big{\|}$
	$\displaystyle\leq\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{1}{4}\Big{\|}\underbrace{\nabla_{X_{t_{i+1}}-X_{t_{i}}}f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}+\nabla_{X_{t_{i}}-X_{t_{i+1}}}f\Big{(}\frac{X_{t_{i+1}}+X_{t_{i}}}{2}\Big{)}}_{=0}\Big{\|}\Big{\|}X^{1}_{t_{i+1}}-X^{1}_{t_{i}}\Big{\|}$
	$\displaystyle\hskip 28.45274pt+\sum_{[t_{i},t_{i+1}]\in\Delta}\frac{2}{2^{2+\epsilon}}\\|\nabla f\\|_{\mathcal{C}^{\epsilon}}\|X_{t_{i+1}}+X_{t_{i}}\|^{2+\epsilon}$
	$\displaystyle\leq 2^{-1-\epsilon}\\|f\\|_{\mathcal{C}^{1+\epsilon}}\\|X\\|_{\mathcal{C}^{\alpha}}^{2+\epsilon}\sum_{[t_{i},t_{i+1}]\in\Delta}\|t_{i+1}-t_{i}\|^{\alpha(2+\epsilon)}\underset{\|\Delta\|\to 0}{\longrightarrow}0.$