text:
MS-ESS1-4 Construct a scientific explanation based on evidence from rock strata for how the geologic time scale is used to organize Earth's 4.6-billion-year-old history.
MS-ESS2-3 Analyze and interpret data on the distribution of fossils and rocks, continental shapes, and seafloor structures to provide evidence of the past plate motions.
HS-ESS1-5 Evaluate evidence of the past and current movements of continental and oceanic crust and the theory of plate tectonics to explain the ages of crustal rocks.
HS-ESS2-3 Develop a model based on evidence of Earth's interior to describe the cycling of matter by thermal convection.
The content and activities in this topic will work towards building an understanding of how the surface of the earth has changed over time by the process of plate tectonics.
Earth's Tectonic Plates
The earth's crust is broken into separate pieces called tectonic plates (Fig. 7.14). Recall that the crust is the solid, rocky, outer shell of the planet. It is composed of two distinctly different types of material: the less-dense continental crust and the more-dense oceanic crust. Both types of crust rest atop solid, upper mantle material. The upper mantle, in turn, floats on a denser layer of lower mantle that is much like thick molten tar.
Each tectonic plate is free-floating and can move independently. Earthquakes and volcanoes are the direct result of the movement of tectonic plates at fault lines. The term fault is used to describe the boundary between tectonic plates. Most of the earthquakes and volcanoes around the Pacific ocean basin—a pattern known as the "ring of fire"—are due to the movement of tectonic plates in this region. Other observable results of short-term plate movement include the gradual widening of the Great Rift lakes in eastern Africa and the rising of the Himalayan Mountain range. The motion of plates can be described in four general patterns:
Collision: when two continental plates are shoved together
Subduction: when one plate plunges beneath another (Fig. 7.15)
Spreading: when two plates are pushed apart (Fig. 7.15)
Transform faulting: when two plates slide past each other (Fig. 7.15)
The rise of the Himalayan Mountain range is due to an ongoing collision of the Indian plate with the Eurasian plate. Earthquakes in California are due to transform fault motion.
Geologists have hypothesized that the movement of tectonic plates is related to convection currents in the earth's mantle. Convection currents describe the rising, spread, and sinking of gas, liquid, or molten material caused by the application of heat. An example of convection current is shown in Fig. 7.16. Inside a beaker, hot water rises at the point where heat is applied. The hot water moves to the surface, then spreads out and cools. Cooler water sinks to the bottom.
Earth's solid crust acts as a heat insulator for the hot interior of the planet. Magma is the molten rock below the crust, in the mantle. Tremendous heat and pressure within the earth cause the hot magma to flow in convection currents. These currents cause the movement of the tectonic plates that make up the earth's crust.
Voice of the Sea
Simulate tectonic plate spreading by modeling convection currents that occur in the mantle.
Examine a map of the earth's tectonic plates. Based on evidence that has been found at plate boundaries, make some hypotheses about the movement of those plates.
Compare-Contrast-Connect
The earth has changed in many ways since it first formed 4.5 billion years ago. The locations of Earth's major landmasses today are very different from their locations in the past (Fig. 7.18). They have gradually moved over the course of hundreds of millions of years—alternately combining into supercontinents and pulling apart in a process known as continental drift. The supercontinent of Pangaea formed as the landmasses gradually combined roughly between 300 and 100 mya. The planet's landmasses eventually moved to their current positions and will continue to move into the future.
Plate tectonics is the scientific theory explaining the movement of the earth's crust. It is widely accepted by scientists today. Recall that both continental landmasses and the ocean floor are part of the earth's crust, and that the crust is broken into individual pieces called tectonic plates (Fig. 7.14). The movement of these tectonic plates is likely caused by convection currents in the molten rock in Earth's mantle below the crust. Earthquakes and volcanoes are the short-term results of this tectonic movement. The long-term result of plate tectonics is the movement of entire continents over millions of years (Fig. 7.18). The presence of the same type of fossils on continents that are now widely separated is evidence that continents have moved over geological history.
Evaluate and interpret several lines of evidence for continental drift over geological time scales.
Evidence for the Movement of Continents
The shapes of the continents provide clues about the past movement of the continents. The edges of the continents on the map seem to fit together like a jigsaw puzzle. For example, on the west coast of Africa, there is an indentation into which the bulge along the east coast of South America fits. The shapes of the continental shelves—the submerged landmass around continents—shows that the fit between continents is even more striking (Fig. 7.19).
Some fossils provide evidence that continents were once located nearer to one another than they are today. Fossils of a marine reptile called Mesosaurus (Fig. 7.20 A) and a land reptile called Cynognathus (Fig. 7.20 B) have been found in South America and South Africa. Another example is the fossil plant called Glossopteris, which is found in India, Australia, and Antarctica (Fig. 7.20 C). The presence of identical fossils in continents that are now widely separated is one of the main pieces of evidence that led to the initial idea that the continents had moved over geological history.
Evidence for continental drift is also found in the types of rocks on continents. There are belts of rock in Africa and South America that match when the ends of the continents are joined. Mountains of comparable age and structure are found in the northeastern part of North America (Appalachian Mountains) and across the British Isles into Norway (Caledonian Mountains). These landmasses can be reassembled so that the mountains form a continuous chain.
Paleoclimatologists (paleo = ancient; climate = long term temperature and weather patterns) study evidence of prehistoric climates. Evidence from glacial striations in rocks, the deep grooves in the land left by the movement of glaciers, shows that 300 mya there were large sheets of ice covering parts of South America, Africa, India, and Australia. These striations indicate that the direction of glacial movement in Africa was toward the Atlantic ocean basin and in South America was from the Atlantic ocean basin. This evidence suggests that South America and Africa were once connected, and that glaciers moved across Africa and South America. There is no glacial evidence for continental movement in North America, because there was no ice covering the continent 300 million years ago. North America may have been nearer the equator where warm temperatures prevented ice sheet formation.
Seafloor Spreading at Mid-Ocean Ridges
Convection currents drive the movement of Earth's rigid tectonic plates in the planet's fluid molten mantle. In places where convection currents rise up towards the crust's surface, tectonic plates move away from each other in a process known as seafloor spreading (Fig. 7.21). Hot magma rises to the crust's surface, cracks develop in the ocean floor, and the magma pushes up and out to form mid-ocean ridges. Mid-ocean ridges or spreading centers are fault lines where two tectonic plates are moving away from each other.
Mid-ocean ridges are the largest continuous geological features on Earth. They are tens of thousands of kilometers long, running through and connecting most of the ocean basins. Oceanographic data reveal that seafloor spreading is slowly widening the Atlantic ocean basin, the Red Sea, and the Gulf of California (Fig. 7.22).
The gradual process of seafloor spreading slowly pushes tectonic plates apart while generating new rock from cooled magma. Ocean floor rocks close to a mid-ocean ridge are not only younger than distant rocks, they also display consistent bands of magnetism based on their age (Fig. 7.22.1). Every few hundred thousand years the earth's magnetic field reverses, in a process known as geomagnetic reversal. Some bands of rock were produced during a time when the polarity of the earth's magnetic field was the reverse of its current polarity. Geomagnetic reversal allows scientists to study the movement of ocean floors over time.
Paleomagnetism is the study of magnetism in ancient rocks. As molten rock cools and solidifies, particles within the rocks align themselves with the earth's magnetic field. In other words, the particles will point in the direction of the magnetic field present as the rock was cooling. If the plate containing the rock drifts or rotates, then the particles in the rock will no longer be aligned with the earth's magnetic field. Scientists can compare the directional magnetism of rock particles to the direction of the magnetic field in the rock's current location and estimate where the plate was when the rock formed (Fig. 7.22.1).
Seafloor spreading gradually pushes tectonic plates apart at mid-ocean ridges. When this happens, the opposite edge of these plates push against other tectonic plates. Subduction occurs when two tectonic plates meet and one moves underneath the other (Fig. 7.23). Oceanic crust is primarily composed of basalt, which makes it slightly denser than continental crust, which is composed primarily of granite. Because it is denser, when oceanic crust and continental crust meet, the oceanic crust slides below the continental crust. This collision of oceanic crust on one plate with the continental crust of a second plate can result in the formation of volcanoes (Fig. 7.23). As the oceanic crust enters the mantle, pressure breaks the crustal rock, heat from friction melts it, and a pool of magma develops. This thick magma, called andesite lava, consists of a mixture of basalt from the oceanic crust and granite from the continental crust. Forced by tremendous pressure, it eventually flows along weaker crustal channels toward the surface. The magma periodically breaks through the crust to form great, violently explosive composite volcanoes—steep-sided, cone-shaped mountains like those in the Andes at the margin of the South American Plate (Fig. 7.23).
Continental collision occurs when two plates carrying continents collide. Because continental crusts are composed of the same low-density material, one does not sink under the other. During collision, the crust moves upward, and the crustal material folds, buckles, and breaks (Fig. 7.24 A). Many of the world's largest mountain ranges, like the Rocky Mountains and the Himalayan Mountains, were formed by the collision of continents resulting in the upward movement of the earth's crust (Fig. 7.24 B). The Himalayan Mountains were formed by the collision between Indian and Eurasian tectonic plates.
Ocean trenches are steep depressions in the seafloor formed at subduction zones where one plate moves downward beneath another (Fig. 7.24 C). These trenches are deep (up to 10.8 km), narrow (about 100 km), and long (from 800 to 5,900 km), with very steep sides. The deepest ocean trench is the Mariana Trench just east of Guam. It is located at the subduction zone where the Pacific plate plunges underneath the edge of the Filipino plate. Subduction zones are also sites of deepwater earthquakes.
Transform faults are found where two tectonic plates move past each other. As the plates slide past one another, there is friction, and great tension can build up before slippage occurs, eventually causing shallow earthquakes. People living near the San Andreas Fault, a transfom fault in California, regularly experience such quakes.
Recall that some volcanoes form near plate boundaries, particularly near subduction zones where oceanic crust moves underneath continental crust (Fig. 7.24). However, some volcanoes form over hot spots in the middle of tectonic plates far away from subduction zones (Fig. 7.25). A hot spot is a place where magma rises up from the earth's mantle toward the surface crust. When magma erupts and flows at the surface, it is called lava. The basalt lava commonly found at hot spots flows like hot, thick syrup and gradually forms shield volcanoes. A shield volcano is shaped like a dome with gently sloping sides. These volcanoes are much less explosive than the composite volcanoes formed at subduction zones.
Some shield volcanoes, such as the islands in the Hawaiian archipelago, began forming on the ocean floor over a hot spot. Each shield volcano grows slowly with repeated eruptions until it reaches the surface of the water to form an island (Fig. 7.25). The highest peak on the island of Hawai'i reaches 4.2 km above sea level. However, the base of this volcanic island lies almost 7 km below the water surface, making Hawai'i's peaks some of the tallest mountains on Earth—much higher than Mount Everest. Almost all of the mid-Pacific and mid-Atlantic ocean basin islands formed in a similar fashion over volcanic hot spots. Over millions of years as the tectonic plate moves, a volcano that was over the hot spot moves away, ceases to erupt, and becomes extinct (Fig. 7.25). Erosion and subsidence (sinking of the earth's crust) eventually causes older islands to sink below sea level. Islands can erode through natural processes such as wind and water flow. Reefs continue to grow around the eroded land mass and form fringing reefs, as seen on Kauaʻi in the main Hawaiian Islands (Fig. 7.26).
Eventually all that remains of the island is a ring of coral reef. An atoll is a ring-shaped coral reef or group of coral islets that has grown around the rim of an extinct submerged volcano forming a central lagoon (Fig. 7.27). Atoll formation is dependent on erosion of land and growth of coral reefs around the island. Coral reef atolls can only occur in tropical regions that are optimal for coral growth. The main Hawaiian Islands will all likely become coral atolls millions of years into the future. The older Northwestern Hawaiian Islands, many of which are now atolls, were formed by the same volcanic hot spot as the younger main Hawaiian Islands.
Question Set
Further Investigations


text:
This Is Anfield • Liverpool FC News • LFC Women •
Liverpool players lead PFA awards nominations
Three Liverpool players have been nominated for this year's PFA Player of the Year award, with two named in the Young Player of the Year award category.
Daniel Sturridge, captain Steven Gerrard and Luis Suarez looks dejected after Everton's third goal during the 221st Merseyside Derby Premiership match at Goodison Park. (Pic by David Rawcliffe/Propaganda)
Steven Gerrard, Daniel Sturridge and Luis Suarez are nominated alongside Chelsea's Eden Hazard, Manchester City's Yaya Toure and Southampton's Adam Lallana in the senior category.
Sturridge is also on the shortlist for the Young Player of the Year award, along with Raheem Sterling and Hazard.
Everton's Ross Barkley, Arsenal's Aaron Ramsey and Southampton's Luke Shaw make up the rest of the young nominees.
On announcing the two shortlists, PFA Chief Executive Gordon Taylor is quoted by the Daily Express as saying, "These Awards celebrate outstanding achievement in football and looking at this year's nominations it is clear that this has been an incredibly competitive year.
"Each of the six men nominated for the Players' Award could expect to be recognised by their fellow professionals, as they have each had a superb season individually and played significant roles in the success of their clubs this year.
"The quality of the Young Player nominations is a cause for real celebration. There are some extremely talented players coming through at the highest level of our domestic game and four of the six nominees already play for England, with Aaron Ramsey a stalwart of the Welsh national side."
In the PFA awards for women's football, Liverpool Ladies have three nominees amongst the shortlist.
Lucy Bronze, Natasha Dowie and Nicole Rosler join Arsenal Ladies' Siobhan Chamberlain and Jordan Nobbs, as well as Seattle Reign's Kim Little.
In the Women's Young Player of the Year award, Bronze is also nominated alongside teammate Martha Harris.
Arsenal's Danielle Carter and Nobbs, Manchester City Women's Isobel Christiansen and Everton Ladies' Alex Greenwood make up the rest of the Women's Young Player of the Year award nominees.
The winners will be announced at a ceremony at the Grosvenor House Hotel in London on Sunday, April 27th — the same day Liverpool face Chelsea at Anfield in the Premier League.
Who are your picks to win the awards? Let us know in the comments section below.
Tags Daniel Sturridge, Luis Suarez, PFA Player of the Year, Raheem Sterling, Steven Gerrard

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text:
\section{Introduction and results}

\subsection{Notation}

Let $d\geq 1$ and ${\mathcal P}({\rr^d})$ stand for the set of all probability measures on ${\rr^d}$.
For $\mu \in {\mathcal P}({\rr^d})$, we consider an i.i.d. sequence $(X_k)_{k\geq 1}$
of $\mu$-distributed random variables and, for $N \geq 1$, the empirical measure $$\mu_N:=\frac1N
\sum_{k=1}^N \delta_{X_k}.$$
As is well-known, by Glivenko-Cantelli's theorem, $\mu_N$ tends weakly to $\mu$ as $N\to \infty$
(for example in probability, see Van der Vaart-Wellner \cite{vdvw} for details and various modes
of convergence).  The aim of the paper is to quantify this convergence,
when the error is measured in some Wasserstein distance.
Let us set, for $p\geq 1$ and $\mu,\nu$ in ${\mathcal P}({\rr^d})$,
$$
{\mathcal T}_p(\mu,\nu) = \inf \left\{\left(\int_{{\rr^d}\times{\rr^d}} |x-y|^p \xi(dx,dy)  \right)
\; : \; \xi \in {\mathcal H}(\mu,\nu) \right\},
$$
where ${\mathcal H}(\mu,\nu)$ is the set of all probability measures on ${\rr^d}\times{\rr^d}$ with marginals
$\mu$ and $\nu$. See Villani \cite{v} for a detailed study of ${\mathcal T}_p$. The
Wasserstein distance ${\mathcal W}_p$ on ${\mathcal P}({\rr^d})$ is defined by ${\mathcal W}_p(\mu,\nu)={\mathcal T}_p(\mu,\nu)$
if $p\in (0,1]$ and ${\mathcal W}_p(\mu,\nu)=({\mathcal T}_p(\mu,\nu))^{1/p}$ if $p>1$.

\poubelle{
\vskip.2cm

It is a major challenge in quantization theory to develop a codebook of size $N$ with the smallest
possible distance (or averaged distance) to a $\mu$-distributed point, for example in signal theory
(see Dereich \cite{der} and Dereich-Scheutzow-Schottstedt \cite{dss}), in statistics
(see e.g. Biau-Devroye-Lugosi \cite{bdl} and Lalo\"e \cite{l}) or in finance
(see Pag\`es-Wilbertz \cite{pw}). A very close topic is the matching problem
(quantify the distance between two independent samples) as in the famous paper \cite{akt}
by Ajtai-Koml\'os-Tusn\`ady, see also the recent results of Barthe-Bordenave \cite{bb}
with applications to the salesperson problem. We refer to these nice papers for an
extensive introduction on this vast topic.
}

\vskip.2cm

The present paper studies the rate of convergence
to zero of ${\mathcal T}_p(\mu_N,\mu)$. This can be done in an asymptotic way, finding e.g.
a sequence $\alpha(N)\to 0$ such that
$\lim_N \alpha(N)^{-1}{\mathcal T}_p(\mu_N,\mu) <\infty$ a.s. or $\lim_N\alpha(N)^{-1}
\mathbb{E}({\mathcal T}_p(\mu_N,\mu))<\infty$. Here we will rather derive some non-asymptotic moment estimates
such as
$$
\mathbb{E}({\mathcal T}_p(\mu_N,\mu))\le \alpha(N) \quad \hbox{for all $N\geq 1$}
$$
as well as some non-asymptotic concentration estimates (also often called deviation inequalities)
$$
\Pr({\mathcal T}_p(\mu_N,\mu)\geq x)\le\alpha(N,x) \quad \hbox{for all $N\geq 1$, all $x>0$.}
$$
They are naturally related to moment (or exponential moment) conditions on the law $\mu$
and we hope to derive an interesting interplay between the dimension
$d\geq 1$, the cost parameter $p>0$ and these moment conditions.
Let us introduce precisely these moment conditions.
For $q>0$, $\alpha>0$, $\gamma>0$ and $\mu \in {\mathcal P}({\rr^d})$, we define
$$
M_q(\mu):= {\int_{\rd}} |x|^q \mu(dx) \quad \hbox{and} \quad {\mathcal E}_{\alpha,\gamma}(\mu):= {\int_{\rd}}
e^{\gamma|x|^\alpha} \mu(dx).
$$
We now present our main estimates, the
comparison with the existing results and methods will be developped after
this presentation. Let us however mention at once that our paper relies on some recent ideas
of Dereich-Scheutzow-Schottstedt \cite{dss}.

\subsection{Moment estimates}

We first give some $L^p$ bounds.

\begin{thm}\label{mainmom}
Let $\mu \in {\mathcal P}({\rr^d})$ and let $p>0$. Assume that $M_q(\mu)<\infty$ for some $q>p$.
There exists a constant $C$ depending only on $p,d,q$ such that, for all $N\geq 1$,
$$
\mathbb{E}\left({\mathcal T}_p(\mu_N,\mu)\right) \leq
C M_q^{p/q}(\mu)\left\{\begin{array}{ll}
N^{-1/2} +N^{-(q-p)/q}& \!\!\!\hbox{if $p>d/2$ and $q\ne 2p$},  \\[+3pt]
N^{-1/2} \log(1+N)+N^{-(q-p)/q} &\!\!\! \hbox{if $p=d/2$ and $q\ne 2p$}, \\[+3pt]
N^{-p/d}+N^{-(q-p)/q} &\!\!\!\hbox{if $p\in (0,d/2)$ and $q\ne d/(d-p)$}.
\end{array}\right.
$$
\end{thm}

Observe that when $\mu$ has sufficiently many moments (namely if
$q>2p$ when $p\geq d/2$ and $q> dp/(d-p)$ when $p\in (0,d/2)$),
the term $N^{-(q-p)/q}$ is small and can be removed.
We could easily treat, for example, the case $p>d/2$ and $q=2p$ but this would lead to some
logarithmic terms and the paper is technical enough.

\vskip.2cm

This generalizes \cite{dss}, in which only the case
$p\in [1,d/2)$ (whence $d\geq 3$) and $q>dp/(d-p)$ was treated.  The argument is also slightly
simplified.

\vskip.2cm

To show that Theorem \ref{mainmom} is really sharp, let us give examples where lower bounds can
be derived quite precisely.

\vskip.2cm

(a) If $a\ne b \in {\rr^d}$ and $\mu=(\delta_a+\delta_b)/2$, one easily checks (see e.g.
\cite[Remark 1]{dss}) that $\mathbb{E}({\mathcal T}_p(\mu_N,\mu)) \geq c N^{-1/2}$ for all $p\geq 1$.
Indeed, we have $\mu_N=Z_N\delta_a+(1-Z_N)\delta_b$ with $Z_N=N^{-1}\sum_1^N {{\bf 1}}_{\{X_i=a\}}$,
so that ${\mathcal T}_p(\mu_N,\mu)=|a-b|^p|Z_N-1/2|$, of which the expectation is of order $N^{-1/2}$.

\vskip.2cm

(b) Such a lower bound in $N^{-1/2}$ can easily be extended to any $\mu$ (possibly very smooth)
of which the support is of the form $A\cup B$ with $d(A,B)>0$
(simply note that ${\mathcal T}_p(\mu_N,\mu)\geq
d^p(A,B)|Z_N-\mu(A)|$, where $Z_N=N^{-1}\sum_1^N {{\bf 1}}_{\{X_i \in A\}}$).

\vskip.2cm

(c) If $\mu$ is the uniform distribution on $[-1,1]^d$, it is well-known and not difficult to prove
that for $p>0$, $\mathbb{E}({\mathcal T}_p(\mu_N,\mu)) \geq c N^{-p/d}$.
Indeed, consider a partition of $[-1,1]^d$ into (roughly) $N$ cubes with length $N^{-1/d}$.
A quick comptation shows that with probability greater than some $c>0$ (uniformly in $N$),
half of these cubes will not be charged by $\mu_N$. But on this event,
we clearly have ${\mathcal T}_p(\mu_N,\mu)\geq a N^{-1/d}$ for some $a>0$, because each time a cube is
not charged by $\mu_N$, a (fixed) proportion
of the mass of $\mu$ (in this cube) is at distance at least
$N^{-1/d}/2$ of the support of $\mu_N$. One easily concludes.

\vskip.2cm

(d) When $p=d/2=1$, it has been shown by Ajtai-Koml\'os-Tusn\'ady \cite{akt}
that for $\mu$ the uniform measure on $[-1,1]^d$,
${\mathcal T}_1(\mu_N,\mu) \simeq c (\log N / N)^{1/2}$ with high probability, implying that
$\mathbb{E}({\mathcal T}_1(\mu_N,\mu)) \geq c (\log N / N)^{1/2}$.

\vskip.2cm

(e) Let $\mu(dx)= c |x|^{-q-d}{{\bf 1}}_{\{|x|\geq 1\}}dx$ for some $q>0$. Then $M_r(\mu)<\infty$
for all $r\in(0,q)$ and for all $p\geq 1$, $\mathbb{E}({\mathcal T}_p(\mu_N,\mu)) \geq c N^{-(q-p)/q}$.
Indeed, $\P(\mu_N(\{|x|\geq N^{1/q}\})=0)= (\mu(\{|x|< N^{1/q}\}))^N
=(1-c/N)^N \geq c>0$ and $\mu(\{|x|\geq 2 N^{1/q}\})\geq c/N$.
One easily gets convinced that
${\mathcal T}_p(\mu_N,\mu)\geq N^{p/q}{{\bf 1}}_{\{\mu_N(\{|x|\geq N^{1/q}\})=0\}} \mu(\{|x|\geq 2 N^{1/q}\})$,
from which the claim follows.

\vskip.2cm

As far as {\it general laws} are concerned, Theorem \ref{mainmom} is really sharp:
the only possible improvements
are the following. The first one, quite interesting,
would be to replace $\log (1+N)$ by something like $\sqrt{\log (1+N)}$ when $p=d/2$
(see point (d) above). It is however not clear it is feasible in full generality. The second
one, which should be a mere (and not very interesting) refinement,
would be to sharpen the bound in $N^{-(q-p)/q}$ when $M_q(\mu)<\infty$:
point (e) only shows that there is $\mu$ with $M_q(\mu)<\infty$ for which we have a lowerbound in
$N^{-(q-p)/q - {\varepsilon}}$ for all ${\varepsilon}>0$.

\vskip.2cm

However, some improvements are possible when restricting the class of laws $\mu$.
First, when $\mu$ is the uniform distribution in $[-1,1]^d$, the results of Talagrand \cite{t,t2}
strongly suggest that when $d\geq 3$, $\mathbb{E}({\mathcal T}_p(\mu_N,\mu)) \simeq N^{-p/d}$ for all $p>0$,
and this is much better than $N^{-1/2}$ when $p$ is large.
Such a result would of course immediately extend to any distribution $\mu=\lambda \circ F^{-1}$,
for $\lambda$ the uniform distribution in $[-1,1]^d$ and
$F:[-1,1]^d\mapsto {\rr^d}$ Lipschitz continuous.
In any case, a smoothness assumption for $\mu$ cannot be sufficient, see point (b) above.

\vskip.2cm

Second, for irregular laws, the convergence can be much faster that $N^{-p/d}$ when
$p<d/2$, see point (a) above where, in an extreme case, we get $N^{-1/2}$ for all values of $p>0$.
It is shown by Dereich-Scheutzow-Schottstedt \cite{dss} (see also Barthe-Bordenave \cite{bb})
that indeed, for a singular law, $\lim_N N^{-p/d}\mathbb{E}({\mathcal T}_p(\mu_N,\mu))=0$.

\subsection{Concentration inequalities}

We next state some concentration inequalities.

\begin{thm}\label{maindev}
Let $\mu \in {\mathcal P}({\rr^d})$ and let $p>0$.
Assume one of the three following conditions:
\begin{align}\label{asexfort}
&\exists\; \alpha>p, \; \exists\;\gamma>0,\;  {\mathcal E}_{\alpha,\gamma}(\mu)<\infty,\\
\label{asexfaible}
\hbox{or} \quad &\exists\; \alpha\in(0,p), \;  \exists\;\gamma>0,\;  {\mathcal E}_{\alpha,\gamma}(\mu)<\infty,\\
\label{asmom}
\hbox{or} \quad &\exists\; q>2p , \;  M_q(\mu)<\infty.
\end{align}
Then for all $N\geq 1$, all $x\in (0,\infty)$,
$$
\P({\mathcal T}_p(\mu^N,\mu)\geq x) \leq a(N,x){{\bf 1}}_{\{x\leq 1\}}+b(N,x),
$$
where
$$
a(N,x)=C \left\{\begin{array}{ll}
\exp(-cNx^2) & \hbox{if $p>d/2$}, \\[+3pt]
\exp(-cN(x/\log(2+1/x))^2) & \hbox{if $p=d/2$}, \\[+3pt]
\exp(-cN x^{d/p}) & \hbox{if $p\in[1,d/2)$}\end{array}\right.
$$
and
$$
b(N,x)=C\left\{\begin{array}{ll}
\exp(-cNx^{\alpha/p}){{\bf 1}}_{\{x> 1\}} & \hbox{under \eqref{asexfort}}, \\[+3pt]
\exp(-c(Nx)^{(\alpha-{\varepsilon})/p}){{\bf 1}}_{\{x\leq 1\}}+\exp(-c (Nx)^{\alpha/p}){{\bf 1}}_{\{x>1\}}
& \hbox{$\forall \;{\varepsilon}\in(0,\alpha)$ under \eqref{asexfaible}},\\[+3pt]
N (Nx)^{-(q-{\varepsilon})/p}& \hbox{$\forall\;{\varepsilon}\in(0,q)$ under \eqref{asmom}}.\end{array}\right.
$$
The positive constants $C$ and $c$ depend only on $p,d$ and either
on $\alpha,\gamma,{\mathcal E}_{\alpha,\gamma}(\mu)$ (under \eqref{asexfort}) or
on $\alpha,\gamma,{\mathcal E}_{\alpha,\gamma}(\mu),{\varepsilon}$ (under \eqref{asexfaible})
or on $q,M_q(\mu),{\varepsilon}$ (under \eqref{asmom}).
\end{thm}

We could also treat the  {\it critical} case where ${\mathcal E}_{\alpha,\gamma}(\mu)<\infty$ with $\alpha=p$,
but the result we could obtain is slightly more intricate and not very satisfying for small
value of $x$ (even if good for large ones).

\begin{rk} When assuming \eqref{asexfaible} with $\alpha \in (0,p)$, we actually also prove that
$$
b(N,x)\leq C\exp(-c N x^2 (\log(1+N))^{-\delta})+C\exp(-c (Nx)^{\alpha/p}),
$$
with $\delta=2p/\alpha-1$, see Step 5 of the proof of Lemma \ref{znpex} below.
This allows us to extend the inequality
$b(N,x)\leq C\exp(-c (Nx)^{\alpha/p})$ to all values of $x\geq x_N$, for some (rather small)
$x_N$ depending on $N,\alpha,p$. But for very small values of $x>0$,
this formula is less interesting than that of Theorem \ref{maindev}.
Despite much effort, we have not been able
to get rid of the logarithmic term.
\end{rk}

We believe that these estimates are quite satisfying.
To get convinced, first observe that the {\it scales} seem to be the good ones.
Recall that $\mathbb{E}({\mathcal T}_p(\mu_N,\mu))=\int_0^\infty \P( {\mathcal T}_p(\mu^N,\mu)\geq x)dx$.

(a)  One easily checks that $\int_0^\infty a(N,x)dx \leq CN^{-p/d}$ if $p<d/2$,
$CN^{-1/2}\log (1+N)$ if $p=d/2$, and $CN^{-1/2}$ if $p>d/2$, as in Theorem \ref{mainmom}.

(b) When integrating $b(N,x)$ (or rather $b(N,x)\land 1$), we find $N^{-(q-{\varepsilon}-p)/(q-{\varepsilon})}$
under \eqref{asmom} and something smaller under \eqref{asexfort} or \eqref{asexfaible}.
Since we can take $q-{\varepsilon}>2p$,
this is less than $N^{-1/2}$ (and thus also less than $N^{-p/d}$ if $p<d/2$
and than $N^{-1/2}\log (1+N)$ if $p=d/2$).

\vskip.2cm

The {\it rates of decrease} are also satisfying in most cases.
Recall that in deviation estimates, we never get something better than
$\exp(-Ng(x))$ for some function $g$. Hence $a(N,x)$ is probably optimal.
Next, for $\bar Y_N$ the empirical mean of a family of centered i.i.d. random variables,
it is well-known that the {\it good} deviation inequalities are the following.

(a) If $\mathbb{E}[\exp(a|Y_1|^\beta)]<\infty$ with $\beta\geq 1$, then
$\Pr[|\bar Y_N|\geq x ] \leq Ce^{-cNx^2}{{\bf 1}}_{\{x\leq 1\}} + Ce^{-cNx^{\beta}}{{\bf 1}}_{\{x> 1\}}$,
see for example Djellout-Guillin-Wu \cite{dgw}, Gozlan \cite{goz} or Ledoux \cite{led},
using transportation cost inequalities.

(b)  If $\mathbb{E}[\exp(a|Y_1|^\beta)]<\infty$ with $\beta<1$, then
$\Pr[|\bar Y_N|\geq x ] \leq Ce^{-cNx^2} + Ce^{-c(Nx)^{\beta}}$,
see Merlev\`ede-Peligrad-Rio \cite[Formula (1.4)]{mpr} which is based on
results by Borovkov \cite{boro}.

(c) If $\mathbb{E}[|Y_1|^r]<\infty$ for some $r>2$, then
$\Pr[|\bar Y_N|\geq x ] \leq Ce^{-cNx^2} + CN (Nx)^{-r}$, see Fuk-Nagaev \cite{fn},
using usual truncation arguments.

\vskip.2cm

Our result is in perfect adequation with these facts
(up to some arbitratry small loss due to ${\varepsilon}$ under \eqref{asexfaible} and \eqref{asmom})
since ${\mathcal T}_p(\mu_N,\mu)$
should behave very roughly as the mean of the $|X_i|^p$'s, which e.g. has an exponential
moment with power $\beta:=\alpha/p$ under \eqref{asexfort} and \eqref{asexfaible}.

\subsection{Comments}

The control of the distance between the empirical measure of an i.i.d. sample and its true
distribution is of course a long standing problem central both in probability, statistics and
informatics with a wide number of applications: quantization (see Delattre-Graf-Luschgy-Pag\`es
\cite{dghp} and Pag\`es-Wilbertz \cite{pw} for recent results),
optimal matching (see Ajtai-Koml\'os-Tusn\'ady \cite{akt}, Dobri\'c-Yukich \cite{dy}, Talagrand
\cite{t2}, Barthe-Bordenave \cite{bb}), density estimation, clustering
(see Biau-Devroye-Lugosi \cite{bdl} and Lalo\"e \cite{l}),
MCMC methods (see \cite{roro} for bounds on ergodic averages), particle systems and approximations of partial differential
equations (see Bolley-Guillin-Villani \cite{bgv} and Fournier-Mischler \cite{fm}).
We refer to these papers for an extensive introduction on this vast topic.

\vskip.2cm

If many distances can be used to consider the problem, the Wasserstein distance is quite
natural, in particular in quantization or for particle approximations of P.D.E.'s.
However the depth of the problem was discovered only recently by Ajtai-Koml\'os-Tusn\'ady \cite{akt}, who considered
the uniform measure on the square, investigated thoroughly by Talagrand \cite{t2}.
As a review of the litterature is somewhat impossible, et us just say that the methods
involved were focused on two methods inherited by the definitions of the Wasserstein distance: the construction of a coupling or by duality to control a particular
empirical process.

\vskip.2cm

Concerning moment estimates (as in Theorem \ref{mainmom}),
some results can be found in
Horowitz-Karandikar \cite{hk}, Rachev-R\"uschendorf \cite{rr} and Mischler-Mouhot \cite{mm}.
But theses results are far from optimal, even when assuming that $\mu$ is compactly supported.
Very recently, strickingly clever  alternatives were considered by Boissard-Le Gouic \cite{blg}
and by Dereich-Scheutzow-Schottstedt \cite{dss}. Unfortunately, the construction of
Boissard-Le Gouic, based on iterative trees, was a little too complicated to yield sharp rates.
On the contrary, the method of \cite{dss}, exposed in details in the next section,
is extremely simple, robust, and leads to the almost optimal results exposed here.
Some sharp moment estimates were already obtained in \cite{dss} for a limited range of parameters.

\vskip.2cm

Concerning concentration estimates, only few results are available. Let us mention the work
of Bolley-Guillin-Villani \cite{bgv} and very recently by Boissard \cite{b}, on which we considerably improve. Our assumptions
are often much weaker (the reference measure $\mu$ was often assumed to satisfy some functional
inequalities, which may be difficult to verify and usually include more "structure" than mere integrability conditions) and
$\Pr[{\mathcal T}_p (\mu_N,\mu)\geq x]$ was estimated only for rather large values of $x$. In particular,
when integrating the concentration estimates of \cite{bgv},
one does never find the {\it good} moment estimates, meaning that the scales are not the good ones.

\vskip.2cm

Moreover, the approach of \cite{dss} is robust enough so that we can also give some good moment
bounds for the Wasserstein distance between the empirical measure of a Markov
chain and its invariant distribution (under some conditions). This could be useful for MCMC methods
because our results are non asymptotic.
We can also study very easily some $\rho$-mixing sequences (see Doukhan \cite{douk}),
for which only very few results exist, see Biau-Devroye-Lugosi \cite{blg}.
Finally, we show on an example how to use Theorem \ref{mainmom} to study some particle systems.
For all these problems, we might also obtain some concentration inequalities, but this would
need further refinements which are out of the scope of the present paper, somewhat already
technical enough, and left for further works.

\subsection{Plan of the paper}

In the next section, we state some general upper bounds of ${\mathcal T}_p(\mu,\nu)$,
for any $\mu,\nu \in {\mathcal P}({\rr^d})$, essentially taken from \cite{dss}.
Section \ref{prmm} is devoted to the proof of Theorem \ref{mainmom}.
Theorem \ref{maindev} is proved in three steps: in Section \ref{cp} we study
the case where $\mu$ is compactly supported and where $N$ is replaced by a Poisson$(N)$-distributed
random variable,
which yields some  pleasant independance properties. We show how to remove the randomization
in Section \ref{dp}, concluding the case where
$\mu$ is compactly supported. The non compact case is studied in Section \ref{nc}.
The final Section 7 is devoted to dependent random variables: $\rho$-mixing sequences,
Markov chains and a particular particle system.

\section{Coupling}

The following notion of distance, essentially taken from \cite{dss},
is the main ingredient of the paper.

\begin{nota}\label{dfdp}
(a) For $\ell\geq 0$, we denote by ${\mathcal P}_\ell$ the natural partition of $(-1,1]^d$ into $2^{d\ell}$
translations of $(-2^{-\ell},2^{-\ell}]^d$.
For two probability measures $\mu,\nu$ on $(-1,1]^d$ and for $p>0$, we introduce
$$
{\mathcal D}_p(\mu,\nu):= \frac{2^p-1}{2} \sum_{\ell\geq 1} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell} |\mu(F)-\nu(F)|,
$$
which obviously defines a distance on ${\mathcal P}((-1,1]^d)$, always bounded by $1$.

(b) We introduce $B_0:=(-1,1]^d$ and, for
$n\geq 1$, $B_n:=(-2^n,2^n]^d \setminus (-2^{n-1},2^{n-1}]^d$.
For $\mu \in {\mathcal P}({\rr^d})$ and $n\geq 0$, we denote by ${\mathcal R}_{B_n} \mu$ the probability measure
on $(-1,1]^d$ defined as the image of $\mu|_{B_n}/\mu(B_n)$ by the map $x \mapsto x/2^n$.
For two probability measures $\mu,\nu$ on ${\rr^d}$ and for $p>0$, we introduce
$$
{\mathcal D}_p(\mu,\nu):= \sum_{n\geq 0} 2^{pn} \big( |\mu(B_n)-\nu(B_n)|
+ (\mu(B_n)\land \nu(B_n)) {\mathcal D}_p({\mathcal R}_{B_n}\mu,{\mathcal R}_{B_n}\nu) \big).
$$
A little study, using that ${\mathcal D}_p \leq 1$ on ${\mathcal P}((-1,1]^d)$,
shows that this defines a distance on ${\mathcal P}({\rr^d})$.
\end{nota}

Having a look at ${\mathcal D}_p$ in the compact case, one sees that in some sense,
it measures distance of the two probability measures simultaneously at all the scales.
The optimization procedure can be made for all scales and outperforms the
approach based on a fixed diameter covering of the state space
(which is more or less the approach of Horowitz-Karandikar \cite{hk}). Moreover one
sees that the principal control is on $|\pi(F)-\mu(F)|$ which is a quite simple quantity. The
next results are slightly modified versions of estimates found in \cite{dss},
see \cite[Lemma 2]{dss} for the compact case and \cite[proof of Theorem 3]{dss}
for the non compact case. It contains the crucial remark that ${\mathcal D}_p$ is an upper bound
(up to constant) of the Wasserstein distance.

\begin{lem}\label{fonda}
Let $d\geq 1$ and  $p>0$.
For all pairs of probability measures $\mu,\nu$ on ${\rr^d}$, ${\mathcal T}_p(\mu,\nu)\leq \kappa_{p,d}
{\mathcal D}_p(\mu,\nu)$, with $\kappa_{p,d}:= 2^{p(1+d/2)}(2^p+1)/(2^p-1)$.
\end{lem}

\begin{proof}
We separate the proof into two steps.

\vskip.2cm

{\bf Step 1.} We first assume that $\mu$ and $\nu$ are supported in $(-1,1]^d$.
We infer from \cite[Lemma 2]{dss}, in which the conditions
$p\geq 1$ and $d\geq 3$ are clearly not used, that, since the diameter of $(-1,1]^d$ is $2^{1+d/2}$,
$$
{\mathcal T}_p(\mu,\nu) \leq 2^{p(1+d/2)-1} \sum_{\ell\geq 0} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell} \mu(F)
\sum_{C\; child\; of\; F}\left|\frac{\mu(C)}{\mu(F)}-\frac{\nu(C)}{\nu(F)}\right|,
$$
where ``$C$ child of $F$'' means that $C\in{\mathcal P}_{\ell+1}$ and $C \subset F$. Consequently,
\begin{align*}
{\mathcal T}_p(\mu,\nu) \leq& 2^{p(1+d/2)-1} \sum_{\ell\geq 0} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell}
\sum_{C\; child\; of\; F} \left(\frac{\nu(C)}{\nu(F)}|\mu(F)-\nu(F)|+|\mu(C)-\nu(C)|\right)\\
\leq& 2^{p(1+d/2)-1} \sum_{\ell\geq 0} 2^{-p\ell} \left( \sum_{F \in {\mathcal P}_\ell} |\mu(F)-\nu(F)|
+ \sum_{C \in {\mathcal P}_{\ell+1}} |\mu(C)-\nu(C)|\right)\\
\leq & 2^{p(1+d/2)-1}(1+2^p) \sum_{\ell\geq 1} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell} |\mu(F)-\nu(F)|,
\end{align*}
which is nothing but $\kappa_{p,d}{\mathcal D}_p(\mu,\nu)$. We used that
$\sum_{F \in {\mathcal P}_0} |\mu(F)-\nu(F)|=0$.

\vskip.2cm

{\it In \cite{dss}, Dereich-Scheutzow-Schottstedt use directly the formula with
the children to study the rate of convergence of empirical measures.
This leads to some (small) technical complications, and does not seem to improve the estimates.}

\vskip.2cm

{\it Step 2.} We next consider the general case.
We consider, for each $n\geq 1$, the optimal coupling $\pi_n(dx,dy)$
between ${\mathcal R}_{B_n}\mu$ and ${\mathcal R}_{B_n}\nu$ for ${\mathcal T}_p$. We define $\xi_n(dx,dy)$
as the image of $\pi_n$ by the map $(x,y)\mapsto (2^nx,2^ny)$, which clearly belongs
to ${\mathcal H}(\mu\vert_{B_n}/\mu(B_n),\nu\vert_{B_n}/\nu(B_n))$
and satisfies $\int\!\!\int |x-y|^p \xi_n(dx,dy) = 2^{np} \int\!\!\int |x-y|^p \pi_n(dx,dy)
=2^{np} {\mathcal T}_p({\mathcal R}_{B_n}\mu,{\mathcal R}_{B_n}\nu)$.

\vskip.2cm

Next, we introduce $q:=\frac12 \sum_{n\geq 0} |\nu(B_n)-\mu(B_n)|$
and we define
\begin{align*}
\xi(dx,dy)=&\sum_{n\geq 0} (\mu(B_n)\land\nu(B_n))\xi_n(dx,dy) + \frac{\alpha(dx)\beta(dy)}q,
\end{align*}
where
$$
\alpha(dx):= \sum_{n\geq 0}(\mu(B_n)-\nu(B_n))_+ \frac{\mu\vert_{B_n}(dx)}{\mu(B_n)} \hbox{ and }
\beta(dy):= \sum_{n\geq 0}(\nu(B_n)-\mu(B_n))_+ \frac{\nu\vert_{B_n}(dy)}{\nu(B_n)}.
$$
Using that
$$
q=\sum_{n\geq 0}(\nu(B_n)-\mu(B_n))_+=\sum_{n\geq 0}(\mu(B_n)-\nu(B_n))_+
=1-\sum_{n\geq 0} (\nu(B_n)\land\mu(B_n)),
$$
it is easily checked that $\xi \in {\mathcal H}(\mu,\nu)$.
Furthermore, we have, setting $c_p=1$ if $p\in(0,1]$ and $c_p=2^{p-1}$ if $p>1$,
\begin{align*}
\int\!\!\int |x-y|^p \frac{\alpha(dx)\beta(dy)}q
\leq& \frac 1 {q} \int\!\!\int c_p(|x|^p +|y|^p)
\alpha(dx)\beta(dy)\\
= &c_p\int |x|^p \alpha(dx) +  c_p \int |y|^p \beta(dy)\\
\leq & c_p \sum_{n\geq 0} 2^{pn} [(\mu(B_n)-\nu(B_n))_+ +(\nu(B_n)-\mu(B_n))_+ ]\\
=& c_p \sum_{n\geq 0} 2^{pn} |\mu(B_n)-\nu(B_n)|.
\end{align*}
Recalling that $\int\!\!\int |x-y|^p \xi_n(dx,dy) \leq 2^{np} {\mathcal T}_p({\mathcal R}_{B_n}\mu,{\mathcal R}_{B_n}\nu)$,
we deduce that
\begin{align*}
{\mathcal T}_p(\mu,\nu)\leq& \int\!\!\int |x-y|^p \xi(dx,dy)\\
\leq & \sum_{n\geq 0} 2^{np} \left(c_p|\mu(B_n)-\nu(B_n)|+
(\mu(B_n)\land\nu(B_n)){\mathcal T}_p({\mathcal R}_{B_n}\mu,{\mathcal R}_{B_n}\nu)\right).
\end{align*}
We conclude using Step 1 and that $c_p\leq \kappa_{p,d}$.
\end{proof}

When proving the concentration inequalities, which is very technical, it will be good to
break the proof into several steps to separate the difficulties and we will
first treat the compact case.
On the contrary, when dealing with moment estimates, the following formula will be
easier to work with.

\begin{lem}\label{ridic}
Let $p>0$ and $d\geq 1$. There is a constant $C$, depending only on $p,d$, such that
for all $\mu,\nu \in {\mathcal P}({\rr^d})$,
$$
{\mathcal D}_p(\mu,\nu) \leq C \sum_{n\geq 0} 2^{pn} \sum_{\ell\geq 0} 2^{-p\ell} \sum_{F\in{\mathcal P}_\ell} |\mu(2^nF\cap B_n)
-\nu(2^n F \cap B_n)|
$$
with the notation $2^n F = \{2^n x\;:\; x\in F\}$.
\end{lem}

\begin{proof}
For all $n\geq 1$, we have
$|\mu(B_n)-\nu(B_n)|=\sum_{F\in{\mathcal P}_0}|\mu(2^nF \cap B_n)-\nu(2^n F \cap B_n)|$ and
$(\mu(B_n)\land \nu(B_n)) {\mathcal D}_p({\mathcal R}_{B_n}\mu, {\mathcal R}_{B_n}\nu)$ is smaller than
\begin{align*}
&(\mu(B_n)\land \nu(B_n)) {\mathcal D}_p({\mathcal R}_{B_n}\mu, {\mathcal R}_{B_n}\nu)\\
\leq&\mu(B_n)\sum_{\ell\geq 1} 2^{-p\ell}\sum_{F\in{\mathcal P}_\ell} \left|\frac{\mu(2^n F \cap B_n)}{\mu(B_n)}
- \frac{\nu(2^n F \cap B_n)}{\nu(B_n)}\right|\\
\leq & \sum_{\ell\geq 1}2^{-p\ell} \sum_{F\in{\mathcal P}_\ell} \left|\mu(2^n F \cap B_n)-\nu(2^n F \cap B_n)\right|
+\left|1-\frac{\mu(B_n)}{\nu(B_n)} \right|\sum_{\ell\geq 1}2^{-p\ell} \sum_{F\in{\mathcal P}_\ell} \nu(2^n F \cap B_n).
\end{align*}
This last term is smaller than $2^{-p} |\mu(B_n)-\nu(B_n)| /(1-2^{-p})$ and this ends the proof.
\end{proof}

\section{Moment estimates}\label{prmm}

The aim of this section is to give the

\begin{preuve} {\it of Theorem \ref{mainmom}.} We thus assume that $\mu\in {\mathcal P}({\rr^d})$ and
that $M_q(\mu)<\infty$ for some $q>p$. By a scaling argument, we may assume that $M_q(\mu)=1$.
This implies that $\mu(B_n)\leq 2^{-q(n-1)}$ for all $n\geq 0$.
By Lemma \ref{fonda}, we have ${\mathcal T}_p(\mu_N,\mu)\leq \kappa_{p,d} {\mathcal D}_p(\mu_N,\mu)$, so that
it suffices to study $\mathbb{E}({\mathcal D}_p(\mu_N,\mu))$.

\vskip.2cm

For a Borel subset $A\subset {\rr^d}$, since
$N\mu_N(A)$ is Binomial$(N,\mu(A))$-distributed, we have
$$
\mathbb{E}(|\mu_N(A)-\mu(A)|) \leq \min \left\{2\mu(A),\sqrt{\mu(A)/N}\right\}.
$$
Using the Cauchy-Scharz inequality and that $\#({\mathcal P}_\ell)=2^{d\ell}$, we deduce that
for all $n\geq 0$, all $\ell \geq 0$,
$$
\sum_{F\in{\mathcal P}_\ell} \mathbb{E}(|\mu_N(2^nF \cap B_n)-\mu(2^nF \cap B_n)|)
\leq \min\left\{ 2\mu(B_n), 2^{d\ell/2} (\mu(B_n)/N)^{1/2}\right\}.
$$
Using finally Lemma \ref{ridic} and that $\mu(B_n)\leq 2^{-q(n-1)}$, we find
\begin{align}\label{topcool}
\mathbb{E}({\mathcal D}_p(\mu_N,\mu))\leq C \sum_{n\geq 0} 2^{pn} \sum_{\ell\geq 0} 2^{-p\ell} \min\left\{
2^{-qn}, 2^{d\ell/2} (2^{-qn}/N)^{1/2}\right\}.
\end{align}

{\bf Step 1.} Here we show that for all ${\varepsilon}\in(0,1)$, all $N\geq 1$,
\begin{align*}
\sum_{\ell\geq 0} 2^{-p\ell} \min\left\{ {\varepsilon}, 2^{d\ell/2} ({\varepsilon}/N)^{1/2}\right\}
\leq& C
\left\{\begin{array}{lll}
\min\{{\varepsilon},({\varepsilon}/N)^{1/2}\} & \hbox{if} & p>d/2, \\[+3pt]
\min\{{\varepsilon},({\varepsilon}/N)^{1/2}\log(2+{\varepsilon} N)\}& \hbox{if} & p=d/2, \\[+3pt]
\min\{{\varepsilon},{\varepsilon} ({\varepsilon} N)^{-p/d}\} & \hbox{if} & p\in(0,d/2).
\end{array}\right.
\end{align*}
First of all, the bound by $C{\varepsilon}$ is obvious in all cases (because $p>0$).
Next, the case $p>d/2$ is immediate.
If $p\leq d/2$, we introduce $\ell_{N,{\varepsilon}}:= \lfloor \log(2+{\varepsilon} N)/(d\log 2)\rfloor$,
for which $2^{d \ell_{N,{\varepsilon}}} \simeq 2+{\varepsilon} N$ and get an upper bound in
$$
({\varepsilon} /N)^{1/2} \sum_{\ell \leq \ell_{N,{\varepsilon}}} 2^{(d/2-p)\ell} + {\varepsilon} \sum_{\ell \geq \ell_{N,{\varepsilon}}} 2^{-p\ell}.
$$
If $p=d/2$, we find an upper bound in $$({\varepsilon} /N)^{1/2}\ell_{N,{\varepsilon}}
+ C {\varepsilon} 2^{-p\ell_{N,{\varepsilon}}} \leq C({\varepsilon} /N)^{1/2}\log(2+{\varepsilon} N) +C {\varepsilon} (1+{\varepsilon} N)^{-1/2}\leq
C({\varepsilon} /N)^{1/2}\log(2+{\varepsilon} N)$$ as desired. If $p\in(0,d/2)$, we get an upper bound in
$$
C({\varepsilon} /N)^{1/2} 2^{(d/2-p)\ell_{N,{\varepsilon}}}+ C {\varepsilon} 2^{-p\ell_{N,{\varepsilon}}} \leq C ({\varepsilon} /N)^{1/2} (2+{\varepsilon} N)^{1/2-p/d}
+ C {\varepsilon} (2+{\varepsilon} N)^{-p/d}.
$$
If ${\varepsilon} N \geq 1$, then $(2+{\varepsilon} N)^{1/2-p/d}\leq (3{\varepsilon} N)^{1/2-p/d}$ and the conclusion follows.
If now ${\varepsilon} N \in (0,1)$, the result is obvious because $\min\{{\varepsilon},{\varepsilon} ({\varepsilon} N)^{-p/d}\}={\varepsilon}$.

\vskip.2cm

{\bf Step 2:} $p>d/2$. By \eqref{topcool} and Step 1 (with ${\varepsilon}=2^{-qn}$), we find
\begin{align*}
\mathbb{E}({\mathcal D}_p(\mu_N,\mu))\leq C \sum_{n\geq 0} 2^{pn} \min\left\{2^{-qn},(2^{-qn}/N)^{1/2}\right\}
\leq C \left\{\begin{array}{lll}
N^{-1/2} & \hbox{ if } & q>2p, \\[+3pt]
N^{-(q-p)/q}    & \hbox{ if } & q\in (p,2p).
\end{array}\right.
\end{align*}
Indeed, this is obvious if $q>2p$, while the case $q\in (p,2p)$ requires to
separate the sum in two parts $n \leq n_N$ and $n>n_N$ with $n_N=\lfloor \log N /(q\log 2)\rfloor$.
This ends the proof when $p>d/2$.
\vskip.2cm

{\bf Step 3:} $p=d/2$. By \eqref{topcool} and Step 1 (with ${\varepsilon}=2^{-qn}$), we find
\begin{align*}
\mathbb{E}({\mathcal D}_p(\mu_N,\mu))\leq C \sum_{n\geq 0} 2^{pn} \min\left\{2^{-qn},(2^{-qn}/N)^{1/2}\log(2+2^{-qn} N)
\right\}.
\end{align*}

If $q>2p$, we immediately get a bound in
\begin{align*}
\mathbb{E}({\mathcal D}_p(\mu_N,\mu))\leq
C \sum_{n\geq 0} 2^{(p-q/2)n} N^{-1/2} \log(2+N) \leq C \log(2+N) N^{-1/2},
\end{align*}
which ends the proof (when $p=d/2$ and $q>2p$).

\vskip.2cm

If $q\in (p,2p)$, we easily obtain, using that $\log(2+x)\leq 2\log x$ for all $x\geq 2$,
an upper bound in
\begin{align*}
\mathbb{E}({\mathcal D}_p(\mu_N,\mu))\leq
& C \sum_{n\geq 0} {{\bf 1}}_{\{N < 2.2^{nq}\}} 2^{(p-q)n}
+ C \sum_{n\geq 0} {{\bf 1}}_{\{N \geq 2.2^{nq}\}}2^{(p-q/2)n} N^{-1/2} \log(N2^{-nq})\\
\leq& C N^{-(q-p)/q} + CN^{-1/2}  \sum_{n=0}^{n_N}  2^{(p-q/2)n} (\log N - nq \log 2)\\
=:&CN^{-(q-p)/q} +CN^{-1/2} K_N,
\end{align*}
where $n_N=\lfloor \log (N/2) / (q \log 2)\rfloor$. A tedious exact computation shows that
\begin{align*}
K_N=&\log N \frac{2^{(p-q/2)(n_N+1)}-1}{2^{(p-q/2)}-1} \\
&- q\log 2 \left[(n_N+1)\frac{2^{(p-q/2)(n_N+1)}-1}{2^{(p-q/2)}-1} + \frac{n_N+1}{2^{(p-q/2)}-1}
- \frac{2^{(p-q/2)(n_N+2)}-2^{(p-q/2)}}{(2^{(p-q/2)}-1)^2}   \right].
\end{align*}
Using that the contribution of the middle term of the second line is negative and
the inequality $\log N - (n_N+1)q\log 2 \leq \log 2$ (because
$(n_N+1)q\log 2 \geq \log (N/2)$), we find
$$
K_N \leq C 2^{(p-q/2)n_N}
\leq C N^{p/q-1/2}.
$$
We finally have checked that $\mathbb{E}({\mathcal D}_p(\mu_N,\mu))\leq CN^{-(q-p)/q}+CN^{-1/2}N^{p/q-1/2}
\leq C N^{-(q-p)/q} $, which ends the proof when  $p=d/2$.

\vskip.2cm

{\bf Step 4:} $p\in (0,d/2)$. We then have, by \eqref{topcool} and Step 1,
\begin{align*}
\mathbb{E}({\mathcal D}_p(\mu_N,\mu)) \leq& C\sum_{n\geq 0} 2^{pn} \min\left\{2^{-qn},
2^{-qn(1-p/d)} N^{-p/d}\right\}.
\end{align*}

If $q>dp/(d-p)$, which implies that
$q(1-p/d)>p$, we immediately get an upper bound by $C N^{-p/d}$, which
ends the proof when $p<d/2$ and $q>dp/(d-p)$

\vskip.2cm

If finally $q\in (p,dp/(d-p))$, we
separate the sum in two parts $n \leq n_N$ and $n>n_N$ with $n_N=\lfloor \log N /(q\log 2)\rfloor$
and we find a bound in $C N^{-(q-p)/q}$ as desired.
\end{preuve}

\section{Concentration inequalities in the compact poissonized case}\label{cp}

It is technically advantageous to first consider the case where the size of the
sampling is Poisson distributed, which implies some independence properties.
If we replace $N$ (large) by a Poisson$(N)$-distributed random variable,
this should not change much the problem, because a Poisson$(N)$-distributed random variable
is close to $N$ with high probability.

\begin{nota}\label{fg}
We introduce the functions $f$ and $g$ defined on $(0,\infty)$ by
$$
f(x)=(1+x)\log(1+x)-x \quad \hbox{and} \quad g(x)=(x \log x - x + 1) {{\bf 1}}_{\{x\geq 1\}}.
$$
Observe that $f$ is increasing, nonnegative, equivalent to $x^2$ at $0$ and to
$x\log x$ at infinity. The function $g$ is positive and increasing
on $(1,\infty)$.
\end{nota}

The goal of this section is to check the following.

\begin{prop}\label{devpoisscomp}
Assume that $\mu$ is supported in $(-1,1]^d$. Let
$\Pi_N$ be a Poisson measure on ${\rr^d}$ with intensity measure $N\mu$ and introduce
the associated empirical measure $\Psi_N=(\Pi_N({\rr^d}))^{-1}\Pi_N$. Let
$p\geq 1$ and $d\geq 1$.
There are some positive constants $C,c$ (depending only on $d,p$)
such that for all $N\geq 1$, all $x\in (0,\infty)$,
$$
\P\left(\Pi_N({\rr^d}){\mathcal D}_p(\Psi_N,\mu)\geq N x\right)
\leq C \left\{\begin{array}{ll}
\exp(-N f(c x))&\hbox{if $p>d/2$,}\\[+3pt]
\exp\left(- N f(c x/\log(2+1/x))\right) &\hbox{if $p=d/2$,}\\[+3pt]
\exp\left(- Nf(c x)\right) + \exp\left(-c N x^{d/p}\right)&\hbox{if $p\in[1,d/2)$.}
\end{array}\right.
$$
\end{prop}

We start with some easy and well-known concentration inequalities for the Poisson distribution.

\begin{lem}\label{devpoiss}
For $\lambda>0$ and $X$ a Poisson$(\lambda)$-distributed random variable, we have

(a) $E(\exp(\theta X)) = \exp(\lambda (e^\theta -1))$ for all $\theta \in {\mathbb{R}}$;

(b) $E(\exp(\theta|X-\lambda|)) \leq 2 \exp(\lambda(e^\theta-1-\theta) )$ for all $\theta>0$;

(c) $\P(X>\lambda x) \leq \exp(-\lambda g(x))$ for all $x>0$;

(d) $\P(|X-\lambda|>\lambda x) \leq 2 \exp(-\lambda f(x))$ for all $x>0$;

(e) $\P(X>\lambda x) \leq \lambda$ for all $x>0$.
\end{lem}

\begin{proof}
Point (a) is straightforward. For point (b), write
$E(\exp(\theta|X-\lambda|))\leq e^{\theta \lambda}\mathbb{E}(\exp(-\theta X))+  e^{-\theta \lambda}
\mathbb{E}(\exp(\theta X))$, use (a) and that $\lambda(e^{-\theta}-1+\theta)\leq \lambda(e^\theta-1-\theta)$.
For point (c), write $\P(X>\lambda x)\leq e^{- \theta \lambda x}\mathbb{E}[\exp(\theta X)]$,
use (a) and optimize in $\theta$. Use the same scheme to deduce (d) from (b).
Finally, for $x>0$, $\P(X>\lambda x)\leq\P(X>0)=1-e^{-\lambda}\leq \lambda$.
\end{proof}

We can now give the

\begin{preuve} {\it of Proposition \ref{devpoisscomp}.}
We fix $x>0$ for the whole proof.
Recalling Notation \ref{dfdp}-(a), we have
\begin{align*}
\Pi_N({\rr^d})&{\mathcal D}_p(\Psi_N,\mu) = C\sum_{\ell \geq 1} 2^{-p\ell} \sum_{F\in {\mathcal P}_\ell}
|\Pi_N(F) - \Pi_N({\rr^d})\mu(F)| \\
\leq &  C|\Pi_N({\rr^d})-N| + C\sum_{\ell \geq 1} 2^{-p\ell} \sum_{F\in {\mathcal P}_\ell} |\Pi_N(F) - N\mu(F)|\\
\leq & C|\Pi_N({\rr^d})-N| + C (N+\Pi_N({\rr^d}))2^{-p \ell_0} +
C\sum_{\ell=1}^{\ell_0} 2^{-p\ell} \sum_{F\in {\mathcal P}_\ell} |\Pi_N(F) - N\mu(F)|
\end{align*}
for any choice of $\ell_0\in{\mathbb{N}}$. We will choose $\ell_0$ later, depending on
the value of $x$.
For any nonnegative family $r_\ell$ such that $\sum_1^{\ell_0} r_\ell\leq 1$, we thus have
\begin{align*}
{\varepsilon}(N,x):=&\P\left(\Pi_N({\rr^d}){\mathcal D}_p(\Psi_N,\mu)\geq Nx\right) \\
\leq & \P\left(|\Pi_N({\rr^d})-N| \geq c Nx\right)
+ \P\left(\Pi_N({\rr^d})\geq N(c x 2^{p \ell_0} -1)\right)\\
& +\sum_{\ell=1}^{\ell_0} \P\left(\sum_{F\in {\mathcal P}_\ell} |\Pi_N(F) - N\mu(F)|
\geq c N x 2^{p\ell}r_\ell \right).
\end{align*}
By Lemma \ref{devpoiss}-(c)-(d), since $\Pi_N({\rr^d})$ is Poisson$(N)$-distributed,
$\P(\Pi_N({\rr^d})\geq N(c x 2^{p \ell_0} -1))\leq \exp(-N g(c x 2^{p \ell_0} -1))$ and
$\P(|\Pi_N({\rr^d})-N| \geq c Nx) \leq 2 \exp(-Nf(cx))$. Next, using that
the family $(\Pi_N(F))_{F\in{\mathcal P}_\ell}$ is independent, with $\Pi_N(F)$ Poisson$(N\mu(F))$-distributed,
we use Lemma \ref{devpoiss}-(a) and that $\#({\mathcal P}_\ell)=2^{\ell d}$ to obtain, for any $\theta>0$,
$$
\mathbb{E}\left(\exp\left(\theta\sum_{F\in {\mathcal P}_\ell} |\Pi_N(F) - N\mu(F)| \right)\right)
\leq \prod_{F\in {\mathcal P}_\ell} 2e^{N\mu(F)(e^\theta-\theta-1)}
\leq 2^{2^{d\ell}} e^{N(e^\theta-\theta-1)}.
$$
Hence
\begin{align*}
\P\left(\sum_{F\in {\mathcal P}_\ell} |\Pi_N(F) - N\mu(F)| \geq c N x2^{p\ell} r_\ell \right) \leq &
\exp\left(-c \theta N x2^{p\ell} r_\ell\right) 2^{2^{d\ell}} \exp\left(N(e^\theta-\theta-1)\right).
\end{align*}
Choosing $\theta= \log(1+c x2^{p\ell} r_\ell  )$, we find
\begin{align*}
\P\left(\sum_{F\in {\mathcal P}_\ell} |\Pi_N(F) - N\mu(F)| \geq c N x2^{p\ell} r_\ell \right) \leq
&2^{2^{d\ell}} \exp(-Nf(c x 2^{p\ell}r_\ell )).
\end{align*}
We have checked that
\begin{align*}
{\varepsilon}(N,x)\leq 2 \exp(-Nf(cx)) + \exp(-N g(c x 2^{p \ell_0} -1)) +
\sum_{\ell=1}^{\ell_0}2^{2^{d\ell}} \exp(-Nf(c x 2^{p\ell}r_\ell )).
\end{align*}
At this point, the value of $c>0$ is not allowed to vary anymore.
We introduce some other positive constants $a$ whose value may change from line to line.
\vskip.2cm

{\bf Case 1: $cx>2$.} Then we choose $\ell_0=1$ and $r_1=1$. We have
$cx2^{p\ell_0}-1 = 2^pcx -1 \geq (2^p-1)cx+1$ whence  $g(c x 2^{p \ell_0} -1)\geq g((2^p-1)cx+1)
=f((2^p-1)cx)$. We also have $\sum_{\ell=1}^{\ell_0}2^{2^{d\ell}} \exp(-Nf(c x 2^{p\ell}r_\ell ))
=2^{2^d}\exp(-Nf(2^pc x))$. We finally get
${\varepsilon}(N,x) \leq C \exp(-Nf(ax))$, which proves the statement (in the three cases, when $cx>2$).

\vskip.2cm

{\bf Case 2: $cx\leq 2$.} We choose $\ell_0$ so that $(1+2/(cx))\leq 2^{p\ell_0}\leq 2^p (1+2/(cx))$,
i.e.
$$
\ell_0:=\lfloor \log(1+2/(cx))/(p\log 2) \rfloor+1.
$$
This implies that
$c x 2^{p \ell_0} \geq 2 + cx$. Hence $g(c x 2^{p \ell_0} -1)\geq g(1+cx)=f(cx)$.
Furthermore, we have $c x 2^{p \ell}r_\ell \leq c x 2^{p \ell_0}\leq 2^p( 2+cx)\leq 2^{p+2}$
for all $\ell \leq \ell_0$, whence $f(c x 2^{p\ell}r_\ell) \geq a x^2 2^{2p\ell} r_\ell^2$
(because $f(x)\geq a x^2$ for all $x\in [0,2^{p+2}]$).
We thus end up with (we use that $2^{2^{d\ell}}\leq \exp(2^{d\ell})$)
$$
{\varepsilon}(N,x) \leq 3\exp(-Nf(cx))+ \sum_{\ell=1}^{\ell_0}\exp(2^{d\ell} - N a  x^2 2^{2p\ell}r^2_\ell).
$$
Now the value of $a>0$ is not allowed to vary anymore, and we introduce $a'>0$,
whose value may change from line to line.

\vskip.2cm

{\it Case 1.1: $p>d/2$.}
We take $r_\ell:=(1-2^{-\eta})2^{- \eta\ell}$ for some $\eta>0$ such that
$2(p-\eta)>d$. If $Nx^2\geq 1$, we easily get
\begin{align*}
{\varepsilon}(N,x) \leq& 3\exp(-Nf(cx))+ \sum_{\ell=1}^{\ell_0}\exp(2^{d\ell} - N a' x^2 2^{2(p-\eta)\ell})\\
\leq&  3\exp(-Nf(cx)) + C \exp(-a'Nx^2) \\
\leq& C \exp(-Nf(a'x)).
\end{align*}
The last inequality uses that $y^2\geq f(y)$ for all $y>0$. If finally $Nx^2\leq 1$,
we obviously have
\begin{align*}
{\varepsilon}(N,x)\leq 1 \leq \exp(1-Nx^2)\leq C \exp(-Nx^2) \leq C \exp(-Nf(x)).
\end{align*}
We thus always have
${\varepsilon}(N,x)\leq C \exp(-Nf(a'x))$ as desired.

\vskip.2cm

{\it Case 2.2:  $p=d/2$.}
We choose $r_\ell:=1/\ell_0$. Thus, if $a N(x/\ell_0)^2 \geq 2$, we easily find
\begin{align*}
{\varepsilon}(N,x) \leq& 3\exp(-Nf(cx))+ \sum_{\ell=1}^{\ell_0}\exp(2^{d\ell}(1 - a N (x/\ell_0)^2)\\
\leq&  3\exp(-Nf(cx)) + C \exp(-a'N(x/\ell_0)^2)\\
\leq & 3\exp(-Nf(cx)) + C \exp(-Nf(a'x/\ell_0))\\
\leq & C \exp(-Nf(a'x/\ell_0))
\end{align*}
because $\ell_0 \geq 1$ and $f$ is increasing.
If now $a N(x/\ell_0)^2 < 2$, we just write
\begin{align*}
{\varepsilon}(N,x)\leq 1 \leq \exp(2-aN(x/\ell_0)^2)\leq C \exp(-aN(x/\ell_0)^2)
\leq C \exp(-Nf(ax/\ell_0)).
\end{align*}
We thus always have ${\varepsilon}(N,x)\leq C \exp(-Nf(a'x/\ell_0))$.
Using that $\ell_0 \leq C \log(2+1/x)$, we immediately conclude that
${\varepsilon}(N,x)\leq C \exp(-Nf(a'x/\log(2+1/x)))$ as desired.

\vskip.2cm

{\it Case 2.3: $p\in[1,d/2)$.} We choose $r_\ell:= \kappa 2^{(d/2-p)(\ell-\ell_0)}$ with
$\kappa=1/(1-2^{p-d/2})$.
For all $\ell\leq \ell_0$,
\begin{align*}
2^{dl}-a Nx^2 2^{2p\ell}r_\ell^2 =& -a \kappa^2 N x^{d/p} 2^{2p\ell}\left[2^{(d-2p)(\ell-\ell_0)} x^{2-d/p}
- 2^{(d-2p)\ell}/(N  a x^{d/p})\right]\\
\leq & -a\kappa^2Nx^{d/p}2^{2p\ell}\left[b 2^{(d-2p)\ell}- 2^{(d-2p)\ell}/(N a \kappa^2 x^{d/p})\right]
\end{align*}
where the constant $b>0$ is such that $2^{-(d-2p)\ell_0}\geq b x^{d/p-2}$ (the existence of
$b$ is easily checked).
Hence if $N a \kappa^2 x^{d/p}\geq 2/b$, we find
\begin{align*}
2^{dl}-a Nx^22^{2p\ell}r_\ell^2 \leq & -a b \kappa^2 Nx^{d/p}2^{d\ell} b/2
\end{align*}
and thus, still using that $N x^{d/p}\geq 2/(ab\kappa^2)$,
$$
\sum_{\ell=1}^{\ell_0}\exp(2^{d\ell} - N c^2 x^2 2^{2p\ell}r^2_\ell) \leq C\exp(-a' N x^{d/p}).
$$
Consequently, we have ${\varepsilon}(N,x) \leq 3\exp(-Nf(cx)) +C\exp(-a' N x^{d/p})$
if  $N a \kappa^2 x^{d/p}\geq 2/b$. As usual, the case where  $N a \kappa^2 x^{d/p}\leq 2/b$
is trivial, since then
\begin{align*}
{\varepsilon}(N,x)\leq 1 \leq \exp(2/b -N a \kappa^2 x^{d/p}) \leq C \exp(-a' N x^{d/p}).
\end{align*}
This ends the proof.
\end{preuve}

\section{Depoissonization in the compact case}\label{dp}

We next check the following compact version of Theorem \ref{maindev}.

\begin{prop}\label{devcomp}
Assume that $\mu$ is supported in $(-1,1]^d$. Let $p>0$ and $d\geq 1$ be fixed.
There are some positive constants $C$ and $c$
(depending only on $p,d$) such that for all $N\geq 1$, all
$x\in (0,\infty)$,
$$
\P\left[{\mathcal D}_p(\mu_N,\mu)\geq x\right] \leq {{\bf 1}}_{\{x\leq 1\}}C \left\{\begin{array}{ll}
\exp(-c N x^2) & \hbox{if $p>d/2$};\\[+3pt]
\exp\left(- c N (x/\log(2+1/x))^2\right)& \hbox{if $p=d/2$};\\[+3pt]
\exp\left(-c N x^{d/p}\right)& \hbox{if $p \in (0,d/2)$}.
\end{array}\right.
$$
\end{prop}

We will need the following easy remark.

\begin{lem}\label{pasdur}
For all $N\geq 1$,
for $X$ Poisson$(N)$-distributed, for all
$k\in \{0,\dots, \lfloor \sqrt N \rfloor\}$,
$$
\P[X=N+k] \geq \kappa_0 N^{-1/2} \hbox{ where } \kappa_0= e^{-2}/\sqrt 2.
$$
\end{lem}

\begin{proof}
By Perrin \cite{p}, we have $N! \leq e \sqrt N (N/e)^N$.
Thus
\begin{align*}
\P[X=N+k] = & e^{-N} \frac{N^{N+k}}{(N+k)!}
\geq e^{-N-1} \frac{N^{N+k}}{ \sqrt{N+k} ((N+k)/e)^{N+k}}
\geq \frac1{\sqrt{2N}} \left(\frac{N}{N+k} \right)^{N+k}e^{k-1}.
\end{align*}
Since $\log(1+x)\leq x$ on $(0,1)$, we have
$((N+k)/N)^{N+k}\leq \exp(k+k^2/N)\leq \exp(k+1)$, so that
$\P[X=N+k] \geq e^{-2}/\sqrt{2N}$.
\end{proof}

\begin{preuve} {\it of Proposition \ref{devcomp}.}
The probability indeed vanishes if $x>1$, since ${\mathcal D}_p$ is smaller than $1$
when restricted to probability measures on $(-1,1]^d$.

\vskip.2cm

{\bf Step 1.}
We introduce a Poisson measure $\Pi_N$ on
${\rr^d}$ with intensity measure $N\mu$ and the associated empirical measure
$\Psi_N=\Pi_N/\Pi_N({\rr^d})$. Conditionally on $\{\Pi_N({\rr^d})=n\}$, $\Psi_N$ has the same law as
$\mu_n$ (the empirical measure of $n$ i.i.d. random variables with law $\mu$). Consequently,
$$
\P\left[\Pi_N({\rr^d}){\mathcal D}_p(\Psi_N,\mu) \geq Nx\right]
=\sum_{n\geq 0} \P[\Pi_N({\rr^d})=n]\P\left[n{\mathcal D}_p(\mu_n,\mu)\geq N x\right].
$$
By Lemma \ref{pasdur} (since $\Pi_N({\rr^d})$ is Poisson$(N)$-distributed),
$$
\frac 1 {\sqrt N}\sum_{k=0}^{\lfloor \sqrt N \rfloor} \P\left[(N+k){\mathcal D}_p(\mu_{N+k},\mu)\geq N x\right]
\leq \kappa_0^{-1} \P\left[\Pi_N({\rr^d}){\mathcal D}_p(\Psi_N,\mu) \geq Nx\right],
$$
which of course implies that (for all $N\geq 1$, all $x>0$),
\begin{align*}
\frac 1 {\sqrt N}\sum_{k=0}^{\lfloor \sqrt N \rfloor} \P\left[{\mathcal D}_p(\mu_{N+k},\mu)\geq x\right]
\leq \kappa_0^{-1} \P\left[\Pi_N({\rr^d}){\mathcal D}_p(\Psi_N,\mu) \geq Nx\right].
\end{align*}

{\bf Step 2.} Here we prove that there is a constant $A>0$ such that for any $N\geq 1$, any
$k \in \{0,\dots, \lfloor \sqrt N \rfloor\}$, any $x > A N^{-1/2}$,
$$
\P\left[{\mathcal D}_p(\mu_{N},\mu)\geq x\right] \leq \P\left[{\mathcal D}_p(\mu_{N+k},\mu)\geq x/2\right].
$$
Build $\mu_n$ for all values of $n\geq 1$ with the same i.i.d. family of $\mu$-distributed
random variables $(X_k)_{k\geq 1}$. Then a.s.,
$$
|\mu_{N+k}-\mu_N|_{TV}\leq \left|\frac{k}{N(N+k)}\sum_1^N \delta_{X_j}\right|_{TV}
+ \left|\frac 1{N+k}\sum_{N+1}^{N+k}\delta_{X_j}\right|_{TV}\leq \frac{k}{N+k}\leq \frac{1}{\sqrt N}.
$$
This obviously implies (recall Notation \ref{dfdp}-(a))
that ${\mathcal D}_p(\mu_{N},\mu_{N+k}) \leq C N^{-1/2}$ a.s. (where $C$ depends only on $p$).
By the triangular inequality,
${\mathcal D}_p(\mu_N,\mu) \leq {\mathcal D}_p(\mu_{N+k},\mu) + C N^{-1/2}$, whence
$$
\P\left[{\mathcal D}_p(\mu_{N},\mu)\geq x\right] \leq \P\left[{\mathcal D}_p(\mu_{N+k},\mu)\geq
x - CN^{-1/2} \right]
\leq \P\left[{\mathcal D}_p(\mu_{N+k},\mu)\geq x/2\right]
$$
if $x- CN^{-1/2}\geq x/2$, i.e. $x\geq 2C N^{-1/2}$.

\vskip.2cm

{\bf Step 3.} Gathering Steps 1 and 2, we deduce that for all $N\geq 1$, all $x>AN^{-1/2}$,
$$
\P\left[{\mathcal D}_p(\mu_{N},\mu)\geq x\right] \leq \frac 1 {\sqrt N}
\sum_{k=0}^{\lfloor \sqrt N \rfloor} \P\left[{\mathcal D}_p(\mu_{N+k},\mu)\geq x/2\right]
\leq C  \P\left[\Pi_N({\rr^d}){\mathcal D}_p(\Psi_N,\mu) \geq Nx/2\right].
$$
We next apply Proposition \ref{devpoisscomp}. Observing that, for
$x \in (0,1]$,

(i) $\exp(-Nf(cx/2))\leq \exp(-cNx^2)$ (case $p>d/2$),

(ii) $\exp(-Nf(cx/2\log(2+2/x)))\leq \exp(-cN(x/\log(2+1/x)^2)$ (case $p=d/2$),

(iii) $\exp(-Nf(cx/2)) + \exp(c N (x/2)^{d/p})\leq \exp(-cNx^{d/p})$ (case $p\in (0,d/2)$)

\noindent
concludes the proof when $x>AN^{-1/2}$. But the other case is trivial, because for
$x \leq A N^{-1/2}$,
$$
\P[{\mathcal D}_p(\mu_{N},\mu)\geq x]\leq 1 \leq \exp(A^2-Nx^2)\leq C\exp(-Nx^2),
$$
which is also smaller than $C\exp(-N(x/\log(2+1/x))^2)$ and than $C\exp(-Nx^{d/p})$ (if $d>2p$).
\end{preuve}

\section{Concentration inequalities in the non compact case}\label{nc}

Here we conclude the proof of Theorem \ref{maindev}.
We will need some concentration estimates for the Binomial distribution.

\begin{lem}\label{bin}
Let $X$ be Binomial$(N,p)$-distributed. Recall that $f$ was defined in Notation \ref{fg}.

(a) $\P[|X-Np|\geq N p z]\leq ({{\bf 1}}_{\{p(1+z) \leq 1\}}+{{\bf 1}}_{\{z \leq 1\}}) \exp(-Npf(z))$
for all $z>0$.

(b) $\P[|X-Np|\geq N p z]\leq Np$ for all $z>1$.

(c) $\mathbb{E}(\exp(-\theta X))=(1-p+pe^{-\theta})^N \leq \exp(-N p (1-e^{-\theta}))$ for $\theta>0$.
\end{lem}

\begin{proof}
Point (c) is straightforward. Point (b) follows from the fact that for $z>1$,
$\P[|X-Np|\geq N p z]=\P[X\geq Np(1+z)]\leq \P[X\ne 0]=1-(1-p)^N \leq pN$. For point (a),
we use Bennett's inequality \cite{ben}, see Devroye-Lugosi
\cite[Exercise 2.2 page 11]{dl}, together with the obvious facts that
$\P[X-Np\geq N p z]=0$ if $p(1+z)>1$ and  $\P[X-Np\leq -N p z]=0$ if $z>1$.
The following elementary tedious computations also works:
write  $\P[|X-Np|\geq N p z] = \P(X\geq Np(1+z)) + \P(N-X\geq N(1-p+zp))
=:\Delta(p,z)+\Delta(1-p,zp/(1-p))$, observe that $N-X\sim$ Binomial$(N,1-p)$.
Use that $\Delta(p,z)\leq  {{\bf 1}}_{\{p(1+z)\leq 1\}}\exp(-\theta N p(1+z))(1-p+pe^\theta)^N$
and choose
$\theta=\log ((1-p)(1+z)/(1-p-pz))$, this gives $\Delta(p,z)\leq {{\bf 1}}_{\{p(1+z)\leq 1\}}
\exp(-N[p(1+z)\log (1+z)+(1-p-pz)\log((1-p-pz)/(1-p)) ] )$. A tedious study shows that
$\Delta(p,z)\leq {{\bf 1}}_{\{p(1+z)\leq 1\}} \exp(-Npf(z))$ and that
$\Delta(1-p,zp/(1-p))\leq {{\bf 1}}_{\{z \leq 1\}}\exp(-Npf(z))$.
\end{proof}

We next estimate the first term when computing
${\mathcal D}_p(\mu_N,\mu)$.

\begin{lem}\label{znpex}
Let $\mu \in {\mathcal P}({\rr^d})$ and $p>0$. Assume \eqref{asexfort}, \eqref{asexfaible} or \eqref{asmom}.
Recall Notation \ref{dfdp} and put $Z_N^p:=\sum_{n\geq 0}2^{pn}|\mu_N(B_n)-\mu(B_n)|$.
Let $x_0$ be fixed.
For all $x>0$,
\begin{align*}
\P[Z_N^p& \geq x ] \leq C \exp(-c N x^2){{\bf 1}}_{\{x\leq x_0\}}\\
&+ C\left\{\begin{array}{ll}
\exp(-cNx^{\alpha/p}){{\bf 1}}_{\{x> x_0\}} & \hbox{under \eqref{asexfort}}, \\[+3pt]
\exp(-c(Nx)^{(\alpha-{\varepsilon})/p}){{\bf 1}}_{\{x\leq x_0\}}+\exp(-c (Nx)^{\alpha/p}){{\bf 1}}_{\{x>x_0\}}
& \hbox{$\forall \;{\varepsilon}\in(0,\alpha)$ under \eqref{asexfaible}},\\[+3pt]
N (Nx)^{-(q-{\varepsilon})/p}& \hbox{$\forall\;{\varepsilon}\in(0,q)$ under \eqref{asmom}}.\end{array}\right.
\end{align*}
\end{lem}

\begin{proof}
Under \eqref{asexfort} or \eqref{asexfaible}, we assume that $\gamma=1$ without loss of generality
(by scaling), whence
${\mathcal E}_{\alpha,1}(\mu)<\infty$ and thus $\mu(B_n)\leq C e^{-2^{(n-1)\alpha}}$ for all $n\geq 0$.
Under \eqref{asmom}, we have $\mu(B_n)\leq C 2^{-qn}$ for all $n\geq 0$.
For $\eta>0$ to be chosen later
(observe that $\sum_{n\geq 0} (1-2^{-\eta})2^{-\eta n}=1$), putting $c:=1-2^{-\eta}$
and $z_n:= c x 2^{-(p+\eta)n}/\mu(B_n)$,
\begin{align*}
\P\left(Z_N^p\geq x \right)
\leq& \left(\sum_{n\geq 0} {{\bf 1}}_{\{z_n \leq 2\}}
\P\left[ |N\mu_N(B_n)-N\mu(B_n)| \geq N \mu(B_n) z_n\right]\right)\land 1 \\
&+ \left(\sum_{n\geq 0}  {{\bf 1}}_{\{z_n > 2\}}
\P\left[ |N\mu_N(B_n)-N\mu(B_n)| \geq N \mu(B_n) z_n\right]\right)\land 1 \\
=:&  \left(\sum_{n\geq 0} I_n(N,x)\right)\land 1 +  \left(\sum_{n\geq 0} J_n(N,x)\right)\land 1.
\end{align*}
From now on, the value of $c>0$ is not allowed to vary anymore.
We introduce another positive constant $a>0$ whose value may change from line to line.

\vskip.2cm

{\bf Step 1: bound of $I_n$.}
Here we show that under \eqref{asmom} (which is of course implied
by \eqref{asexfort} or \eqref{asexfaible}), if $\eta \in (0,q/2-p)$, there is $A_0>0$ such that
$$
\sum_{n\geq 0} I_n(N,x) \leq C\exp(-a N x^2){{\bf 1}}_{\{x\leq A_0\}} \quad \hbox{if $N x^2 \geq 1$.}
$$
This will obviously imply that for all $N\geq 1$, all $x>0$,
$$
\left(\sum_{n\geq 0} I_n(N,x)\right)\land 1 \leq C \exp(-a N x^2){{\bf 1}}_{\{x\leq A_0\}}.
$$

First, $\sum_{n\geq 0} I_n(N,x)=0$ if $z_n>2$ for all $n\geq 0$. Recalling that
$\mu(B_n)\leq C2^{-qn}$, this is the case if
$x\geq (2C/c)\sup_{n\geq 0}2^{(p+\eta-q)n}=(2C/c):=A_0$.
Next, since $N\mu_N(B_n)\sim$ Binomial$(N,\mu(B_n))$, Lemma \ref{bin}-(a) leads us to
$$
I_n(N,x) \leq 2 {{\bf 1}}_{\{z_n\leq 2\}} \exp(-N\mu(B_n)f(z_n))
\leq 2 \exp(-N \mu(B_n)z_n^2/4)),
$$
because $f(x) \geq x^2/4$ for $x\in [0,2]$. Since finally
$\mu(B_n) z_n^2/4 \geq a x^2 2^{(q-2p-2\eta)n}$, we easily conclude,
since $q-2p-2\eta>0$ and since $Nx^2 \geq 1$, that
$$
\sum_{n\geq 0} I_n(N,x) \leq C \sum_{n\geq 0} \exp(-a N x^22^{(q-2p-2\eta)n}){{\bf 1}}_{\{x\leq A_0\}}
\leq C \exp(-a N x^2){{\bf 1}}_{\{x\leq A_0\}}.
$$
\vskip.2cm

{\bf Step 2: bound of $J_n$ under \eqref{asexfort} or \eqref{asexfaible} when $x\leq A$.}
Here we fix $A>0$ and prove that if $\eta>0$ is small enough, for all $x\in (0,A]$ such that
$Nx^2\geq 1$,
\begin{align*}
\sum_{n\geq 0} J_n(N,x) \leq & C  \left\{
\begin{array}{ll}
\exp(-a N x^2)  & \hbox{under \eqref{asexfort},}\\[+3pt]
\exp(-a N x^2)+\exp(- a (N x)^{(\alpha-{\varepsilon})/p}) & \hbox{$\forall\;{\varepsilon}\in(0,\alpha)$
under \eqref{asexfaible}.}
\end{array}\right.
\end{align*}
This will imply, as usual, that for all $N\geq 1$, all $x>0$,
\begin{align*}
\left(\sum_{n\geq 0} J_n(N,x)\right)\land 1 \leq & C  \left\{
\begin{array}{ll}
\exp(-a N x^2)& \hbox{under \eqref{asexfort},}\\[+3pt]
\exp(-a N x^2)+\exp(- a (N x)^{(\alpha-{\varepsilon})/p}) & \hbox{$\forall\;{\varepsilon}\in(0,\alpha)$
under \eqref{asexfaible}.}\\[+3pt]
\end{array}\right.
\end{align*}

By Lemma \ref{bin}-(a)-(b) (since $z_n> 2$ implies
${{\bf 1}}_{\{\mu(B_n)(1+z_n)\leq 1\}}+{{\bf 1}}_{\{z_n\leq 1\}} \leq {{\bf 1}}_{\{z_n\leq 1/\mu(B_n)\}}$),
\begin{align*}
J_n(N,x)\leq & {{\bf 1}}_{\{2<z_n\leq 1/\mu(B_n) \}}
\min\left\{\exp(-N\mu(B_n)f(z_n)),N\mu(B_n) \right\} \\
\leq & {{\bf 1}}_{\{z_n \mu(B_n)\leq 1\}}
\min \left\{\exp\left(-a N \mu(B_n) z_n \log [2\lor z_n]\right),  N\mu(B_n)\right\}
\end{align*}
because $f(y)\geq a y \log y \geq a y \log [2\lor y]$ for $y> 2$.
Since $\mu(B_n) \leq Ce^{-2^{(n-1)\alpha}}$, we get
$$
J_n(N,x)\leq C
\min \{\exp(-aN x 2^{-(p+\eta)n} \log [2\lor( a x2^{-(p+\eta)n}e^{2^{(n-1)\alpha}})]),Ne^{2^{-(n-1)\alpha}}\}.
$$
A straightforward computation shows that there is a
constant $K$ such that
for $n \geq n_1:=\lfloor K(1+\log \log (K/x))\rfloor$,
we have
$\log (a x2^{-(p+\eta)n}e^{2^{(n-1)\alpha}})\geq 2^{(n-1)\alpha}/2$.
Consequently,
\begin{align*}
\sum_{n\geq 0}J_n(N,x) \leq& C n_1 \exp(-aN x2^{-(p+\eta)n_1}) +
C \sum_{n> n_1} \min\left\{\exp(-aN x 2^{(\alpha-p-\eta)n}), e^{-2^{(n-1)\alpha}}   \right\}\\
=& C J^1(N,x)+C J^2(N,x).
\end{align*}
We first show that $J^1(N,x)\leq Ce^{-aNx^2}$ (here we actually could get something much better).
First, since
$n_1=\lfloor K+K\log \log (K/x)\rfloor$ and $x \in [0,A]$, we clearly have e.g.
$x2^{-(p+\eta)n_1} \geq a x^{3/2}$. Next, $Nx^2\geq 1$ implies that
$1/x \leq (Nx^{3/2})^2$. Thus
$$
J^1(N,x) \leq C(1+\log\log(C(Nx^{3/2})^2))\exp(-aNx^{3/2}) \leq
C\exp(-aNx^{3/2}) \leq \exp(-aNx^2).
$$
We now treat $J^2(N,x)$.

\vskip.2cm

{\it Step 2.1.} Under \eqref{asexfort}, we immediately get, if $\eta \in (0, \alpha-p)$
(recall that $x\in[0,A]$),
$$
J^2(N,x)\leq \sum_{n\geq 0} \exp(-aN x 2^{(\alpha-p-\eta)n})\leq C \exp(-aNx) \leq C\exp(-aNx^2),
$$
where we used that $x \leq A$ and $Nx^2\geq 1$ (whence $Nx\geq 1/A$).

\vskip.2cm

{\it Step 2.2.} Under \eqref{asexfaible}, we first write
\begin{align*}
J^2(N,x) \leq &\sum_{n\geq 0} \min\left\{\exp(-aN x 2^{(\alpha-p-\eta)n}),e^{-2^{(n-1)\alpha}}\right\}
\leq  n_2 \exp(-cNx 2^{(\alpha-p-\eta)n_2})+ Ne^{-2^{(n_2-1)\alpha}}.
\end{align*}
We choose $n_2:=\lfloor \log(Nx)/((p+\eta)\log 2)  \rfloor$, which yields us to
$2^{(n_2-1)\alpha}\geq (Nx)^{\alpha/(p+\eta)}/2^{2\alpha}$ and
$(Nx) 2^{(\alpha-p-\eta)n_2}\leq (Nx)^{\alpha/(p+\eta)}$.
Consequently (recall that $x\in(0,A]$),
\begin{align*}
J^2(N,x)\leq & C (1+\log(Nx)+N)  \exp(-a (Nx)^{\alpha/(p+\eta)})
\leq C(1+N)  \exp(-a (Nx)^{\alpha/(p+\eta)}).
\end{align*}
For any fixed ${\varepsilon}\in(0,\alpha)$, we choose $\eta>0$ small enough so that
$\alpha/(p+\eta)\geq (\alpha-{\varepsilon})/p$ and
we conclude that (recall that $Nx\geq 1/A$ because $Nx^2\geq 1$ and $x\leq A$)
\begin{align*}
J^2(N,x)\leq & C(1+N)  \exp(-a (Nx)^{(\alpha-{\varepsilon})/p}) \leq C  \exp(-a (Nx)^{(\alpha-{\varepsilon})/p}).
\end{align*}
The last inequality is easily checked, using that  $Nx^2\geq 1$ implies that $N\leq (Nx)^2$.

\vskip.2cm

{\bf Step 3: bound of $J_n$ under \eqref{asmom}.} Here we show that
for all ${\varepsilon}\in(0,q)$, if $\eta>0$ is small enough,
$$
\sum_{n\geq 0}J_n(N,x)\leq C N \left(\frac1{Nx}\right)^{(q-{\varepsilon})/p} \hbox{ if $Nx \geq 1$}.
$$
As usual, this will imply that for all $x>0$, all $N\geq 1$,
$$
\left(\sum_{n\geq 0}J_n(N,x)\right)\land 1
\leq C N \left(\frac1{Nx}\right)^{(q-{\varepsilon})/p}.
$$

\vskip.2cm

Exactly as in Step 2, we get from Lemma \ref{bin}-(a)-(b) that
\begin{align*}
J_n(N,x) \leq &
\min \left\{\exp\left(-a N \mu(B_n)z_n \log [2\lor z_n]\right),  N\mu(B_n)\right\}.
\end{align*}
Hence for $n_3$ to be chosen later, since $a N \mu(B_n)z_n= a Nx2^{-(p+\eta)n}$,
\begin{align*}
\sum_{n\geq 0}J_n(N,x)\leq& C\sum_{n=0}^{n_3} \exp(- a Nx2^{-(p+\eta)n}) + C N \sum_{n>n_3} 2^{-qn}\\
\leq & Cn_3 \exp(-a Nx2^{-(p+\eta)n_3}) + C N 2^{-qn_3}.
\end{align*}
We choose $n_3:= \lfloor (q-{\varepsilon})\log(Nx)/(pq\log 2) \rfloor$, which implies that
$2^{-qn_3} \leq 2^q (Nx)^{-(q-{\varepsilon})/p}$ and that $2^{-(p+\eta)n_3} \geq (Nx)^{-(q-{\varepsilon})(p+\eta)/(pq)}$. Hence
\begin{align*}
\sum_{n\geq 0}J_n(N,x)\leq& C \log(Nx) \exp(-a(Nx)^{1-(q-{\varepsilon})(p+\eta)/(pq)} )
+CN (Nx)^{-(q-{\varepsilon})/p}.
\end{align*}
If $\eta \in (0,p{\varepsilon}/(q-{\varepsilon}))$, then $1-(q-{\varepsilon})(p+\eta)/(pq)>0$, and thus
$$\log(Nx) \exp(-a(Nx)^{1-(q-{\varepsilon})(p+\eta)/(pq)} ) \leq C (Nx)^{-(q-{\varepsilon})/p}.$$
This ends the step.

\vskip.2cm

{\bf Step 4.} We next assume \eqref{asexfort}
and prove that for all $x\geq A_1:=2^{p}[M_p(\mu)+(2 \log {\mathcal E}_{\alpha,1}(\mu))^{p/\alpha}]$,
$$
\Pr[Z_N^{p} \geq x ] \leq C \exp(-a N x^{\alpha/p}).
$$

A simple computation shows that for any $\nu \in {\mathcal P}({\rr^d})$,
$\sum_{n\geq 0} 2^{pn}\nu(B_n) \leq 2^p M_p(\nu)$, whence
$Z_N^{p} \leq 2^p M_p(\mu)+ 2^p N^{-1}\sum_1^N |X_i|^p\leq
2^p M_p(\mu)+ 2^p [N^{-1}\sum_1^N |X_i|^\alpha]^{p/\alpha}$. Thus
$$
\Pr[Z_N^{p} \geq x ] \leq \Pr\left[N^{-1}\sum_1^N |X_i|^\alpha \geq [x2^{-p} - M_p(\mu)]^{\alpha/p}\right].
$$
Next, we note that for $y\geq 2 \log {\mathcal E}_{\alpha,1}(\mu)$,
$$
\Pr\left[N^{-1}\sum_1^N |X_i|^\alpha \geq y\right]
\leq \exp(-N y + N \log{\mathcal E}_{\alpha,1}(\mu) ) \leq \exp(-N y/2).
$$
The conclusion easily follows, since $x\geq A_1$ implies that
$y:=[x2^{-p} - M_p(\mu)]^{\alpha/p}\geq 2 \log {\mathcal E}_{\alpha,1}(\mu)$
and since $y \geq [x2^{-p-1}]^{\alpha/p}-[M_p(\mu)]^{\alpha/p}$.

\vskip.2cm

{\bf Step 5.} Assume \eqref{asexfaible} and put $\delta:= 2p/\alpha -1$.
Here we show that for all $x>0$, $N\geq 1$,
$$
\Pr[Z_N^{p} \geq x ] \leq C \exp(-a(Nx)^{\alpha/p})+C \exp(-a Nx^2 (\log (1+N))^{-\delta}).
$$

{\it Step 5.1.} For $R>0$ (large) to be chosen later, we introduce the probability measure $\mu^R$
as the law of $X{{\bf 1}}_{\{|X|\leq R\}}$. We also denote by $\mu_N^R$ the corresponding empirical
measure (coupled with $\mu_N$ in that the $X_i$'s are used for $\mu_N$ and the
$X_i{{\bf 1}}_{\{|X_i|\leq R\}}$'s are chosen for $\mu_N^R$).
We set $Z_N^{p,R}:=\sum_{n\geq 0}2^{pn}|\mu_N^R(B_n)-\mu^R(B_n)|$ and first observe that
$|Z^p_N-Z^{p,R}_N| \leq 2^p N^{-1}\sum_1^N |X_i|^p{{\bf 1}}_{\{|X_i|>R\}} + 2^p \int_{\{|x|>R\}}|x|^p\mu(dx)$.
On the one hand, $\int_{\{|x|>R\}}|x|^p\mu(dx)\leq \exp(-R^\alpha/2) \int |x|^p e^{|x|^\alpha/2}
\mu(dx) \leq C \exp(-R^\alpha/2)$ by \eqref{asexfaible} (with $\gamma=1$).
On the other hand, since $\alpha \in (0,p]$,  $\sum_1^N |X_i|^p{{\bf 1}}_{\{|X_i|>R\}}\leq
(\sum_1^N |X_i|^\alpha{{\bf 1}}_{\{|X_i|>R\}})^{p/\alpha}$. Hence if
$x\geq A \exp(-R^\alpha/2)$, where $A:=2^{p+1}C$,
\begin{align*}
\Pr\left(|Z^p_N-Z_N^{p,R}|\geq x\right)\leq& \Pr\left(N^{-1}\sum_1^N |X_i|^p{{\bf 1}}_{\{|X_i|>R\}}
\geq x2^{-p-1}  \right)\\
\leq & \Pr\left(\sum_1^N |X_i|^\alpha{{\bf 1}}_{\{|X_i|>R\}}\geq  (Nx2^{-p-1})^{\alpha/p} \right)\\
\leq & \exp(-  (Nx2^{-p-1})^{\alpha/p}/2)\mathbb{E}\left[\exp\left(|X_1|^\alpha{{\bf 1}}_{\{|X_1|>R\}}/2\right)
\right]^N.
\end{align*}
Observing that $\mathbb{E}[\exp(|X_1|^\alpha{{\bf 1}}_{\{|X_1|>R\}}/2)]\leq 1+
\mathbb{E}[\exp(|X_1|^\alpha/2){{\bf 1}}_{\{|X_1|>R\}}] \leq 1 + C\exp(-R^\alpha/2)$ by \eqref{asexfaible}
and using that $\log(1+u)\leq u$, we deduce that for all $x\geq 2^{p+1} C \exp(-R^\alpha/2)$,
\begin{align*}
\Pr\left(|Z^p_N-Z_N^{p,R}|\geq x\right)
\leq & \exp\left(-  (Nx2^{-p-1})^{\alpha/p}/2+ CN \exp(-R^\alpha/2) \right).
\end{align*}
With the choice
\begin{equation}\label{dfR}
R:= (2\log (1+N))^{1/\alpha},
\end{equation}
we finally find
\begin{align*}
\Pr\left(|Z^p_N-Z_N^{p,R}|\geq x\right)
\leq & \exp\left(-  (Nx2^{-p-1})^{\alpha/p}/2 + C \right) \leq C \exp\left(-  a(Nx)^{\alpha/p}\right)
\end{align*}
provided $x\geq A \exp(-R^\alpha/2)$, i.e. $(N+1)x \geq A$. As usual, this immediately extends
to any value of $x>0$.

\vskip.2cm

{\it Step 5.2.} To study $Z_N^{p,R}$, we first observe that since $\mu^R(B_n)=0$ if $2^{n-1}\geq R$,
we have $2^{pn}\mu^R(B_n) \leq  (2R)^{p-\alpha/2}2^{\alpha n/2}\mu^R(B_n)$ for all $n\geq 0$.
Hence $Z_N^{p,R}\leq (2R)^{p-\alpha/2}Z_N^{\alpha/2,R}$. But $\mu^R$ satisfies ${\int_{\rd}} \exp(|x|^\alpha)
\mu^R(dx)<\infty$ uniformly in $R$, so that we may use Steps 1, 2 and 4 (with $p=\alpha/2<\alpha$)
to deduce that for all $x>0$, $\Pr(Z_N^{\alpha/2,R}\geq x) \leq C \exp(-a N x^2)$. Consequently,
$\Pr(Z_N^{p,R}\geq x) \leq C \exp(-a N (x/R^{p-\alpha/2})^2)$. Recalling \eqref{dfR}
and that $\delta:= 2p/\alpha -1$, we see that
that $\Pr(Z_N^{p,R}\geq x) \leq C \exp(-a Nx^2 (\log (1+N))^{-\delta})$. This ends the step.

\vskip.2cm

{\bf Conclusion.} Recall that $x_0>0$ is fixed.

\vskip.2cm

First assume \eqref{asexfort}. By Step 4,
$\Pr[Z_N^{p} \geq x ] \leq C\exp(-aNx^{\alpha/p})$ for all $x\geq A_1$. We deduce from
Steps 1 and 2 that for $x\in (0,A_1)$, $\Pr[Z_N^{p} \geq x ]\leq C\exp(-aNx^2)$.
We easily conclude that for all $x>0$, $\Pr[Z_N^{p} \geq x ]\leq C\exp(-aNx^2){{\bf 1}}_{\{x\leq x_0\}}
+  C\exp(-aNx^{\alpha/p}){{\bf 1}}_{\{x>x_0\}}$ as desired.

\vskip.2cm

Assume next \eqref{asexfaible}. By Step 5,
$\Pr[Z_N^{p} \geq x ] \leq C \exp(-a Nx^2 (\log (1+N))^{-\delta})+C\exp(-a(Nx)^{\alpha/p})$.
But if $x \geq x_0$, we clearly have $(Nx)^{\alpha/p} \leq C
Nx^2 (\log (1+N))^{-\delta}$ because $\alpha<p$,
so that $\Pr[Z_N^{p} \geq x ] \leq C\exp(-a(Nx)^{\alpha/p})$. If now $x \leq x_0$, we use
Steps 1 and 2 to write $\Pr[Z_N^{p} \geq x] \leq C\exp(-aNx^2)+C\exp(-a(Nx)^{(\alpha-{\varepsilon})/p})$.

\vskip.2cm

Assume finally \eqref{asmom}. By Steps 1 and 3,
$\Pr[Z_N^{p} \geq x ]\leq C\exp(-aNx^2) + C N (Nx)^{-(q-{\varepsilon})/q}$
for all $x>0$.
But if $x\geq x_0$, $\exp(-aNx^2)\leq \exp(-aNx)\leq C (Nx)^{-(q-{\varepsilon})/q}
\leq C N (Nx)^{-(q-{\varepsilon})/q}$. We conclude that for all $x>0$,
$\Pr[Z_N^{p} \geq x ]\leq C\exp(-aNx^2){{\bf 1}}_{\{x\leq x_0\}} + C N (Nx)^{-(q-{\varepsilon})/q}$ as desired.
\end{proof}

We can now give the

\begin{preuve} {\it of Theorem \ref{maindev}.}
Using Lemma \ref{fonda}, we write
\begin{align*}
{\mathcal T}_p(\mu_N,\mu)\leq& \kappa_{p,d} {\mathcal D}_p(\mu_N,\mu)\\
\leq& \kappa_{p,d} \sum_{n\geq 0} 2^{pn}|\mu_N(B_n)-\mu(B_n)|
+  \kappa_{p,d}\sum_{n\geq 0} 2^{pn} \mu(B_n){\mathcal D}_p({\mathcal R}_{B_n}\mu_N,{\mathcal R}_{B_n}\mu)\\
=:&\kappa_{p,d}( Z_N^p +  V_N^p).
\end{align*}
Hence
\begin{align*}
\Pr({\mathcal T}_p(\mu_N,\mu)\geq x) \leq \Pr(Z_N^p \geq x/(2\kappa_{p,d})) +\Pr(V_N^p \geq x/(2\kappa_{p,d})).
\end{align*}
By Lemma \ref{znpex} (choosing $x_0:=1/(2\kappa_{p,d})$), we easily find
$\Pr(Z_N^p \geq x/(2\kappa_{p,d}))
\leq Ce^{-cNx^2}{{\bf 1}}_{\{x\leq 1\}} +b(N,x) \leq a(N,x){{\bf 1}}_{\{x\leq 1\}}+b(N,x)$,
these quantities being defined in the statement of Theorem \ref{maindev}.
We now check that there is $A>0$ such that for all $x>0$,
\begin{align}\label{toprove}
\Pr[V_N^p \geq x/(2\kappa_{p,d}) ] \leq a(N,x){{\bf 1}}_{\{x \leq A\}}.
\end{align}
This will end the proof, since one easily checks that
$a(N,x){{\bf 1}}_{\{x \leq A\}}\leq a(N,x){{\bf 1}}_{\{x \leq 1\}}+b(N,x)$
(when allowing the values of the constants to change).

\vskip.2cm

Let us thus check \eqref{toprove}.
For $\eta>0$ to be chosen later, we set $c:=(1-2^{-\eta})/(2\kappa_{p,d})$
and $z_n:= c x 2^{-(p+\eta)n}/\mu(B_n)$. Observing that
$\sum_{n\geq 0} (1-2^{-\eta})2^{-\eta n}=1$), we write
\begin{align*}
\P\left(V_N^p \geq x/(2\kappa_{p,d}) \right)
\leq&\left( \sum_{n\geq 0}
\P\left[ {\mathcal D}_p({\mathcal R}_{B_n}\mu_N,{\mathcal R}_{B_n}\mu) \geq z_n\right]\right)\land 1
=:  \left(\sum_{n\geq 0} K_n(N,x)\right)\land 1.
\end{align*}
From now on, the value of $c>0$ is not allowed to vary anymore.
We introduce another positive constant $a>0$ whose value may change from line to line.
We only assume \eqref{asmom} (which is implied by \eqref{asexfort} or \eqref{asexfaible}).
We now show that if $\eta>0$ is small enough,
\begin{align}\label{toprove2}
\sum_{n\geq 0} K_n(N,x) \leq C \exp(-a N h(x)){{\bf 1}}_{\{x \leq A\}} \quad  \hbox{if $Nh(x)\geq 1$,}
\end{align}
where $h(x)=x^2$ if $p>d/2$, $h(x)=(x/\log(2+1/x))^2$ if $p=d/2$ and $h(x)=x^{d/p}$ if
$p<d/2$.
This will obviously imply as usual that for all $x>0$,
$$
\left(\sum_{n\geq 0} K_n(N,x)\right)\land 1 \leq C \exp(-a N h(x)){{\bf 1}}_{\{x\leq A\}}
$$
and thus conclude the proof of \eqref{toprove}. We thus only have to prove \eqref{toprove2}.

\vskip.2cm

Conditionally on $\mu_N(B_n)$, ${\mathcal R}_{B_n}\mu_N$ is the empirical measure of $N\mu_N(B_n)$ points
which are ${\mathcal R}_{B_n}\mu$-distributed. Since ${\mathcal R}_{B_n}\mu$ is supported in
$(-1,1]^d$, we may apply Proposition \ref{devcomp} and obtain
\begin{align*}
K_n(N,x)\leq& C\mathbb{E}\left[{{\bf 1}}_{\{z_n\leq 1\}}
\exp\left(- a N\mu_N(B_n) h(z_n)\right)\right]
\leq C {{\bf 1}}_{\{z_n\leq 1\}} \exp(-N\mu(B_n)(1- e^{-a h(z_n)}))
\end{align*}
by Lemma \ref{bin}-(c). But the condition $z_n\leq 1$ implies that $h(z_n)$
is bounded (by a constant depending only on $p$ and $d$), whence
$$
K_n(N,x)\leq C{{\bf 1}}_{\{z_n\leq 1\}} \exp(- a N\mu(B_n)h(z_n)).
$$
By \eqref{asmom}, we have $\mu(B_n)\leq C 2^{-qn}$. Hence
if $x>A:=C/c$, we have $z_n\geq (c/C)x2^{(q-p-\eta)n} >1$ for all $n\geq 1$ (if
$\eta\in (0,q-p)$) and thus
$\sum_{n\geq 0} K_n(N,x)=0$
as desired.

\vskip.2cm

Next, we see that $\theta \mapsto \theta h(x/\theta)$ is decreasing, whence for all $x\leq A$,
$$
K_n(N,x)\leq C \exp(- a N2^{-qn}h(c x 2^{(q-p-\eta)n}/C))\leq
C \exp(- a N2^{-qn}h(x 2^{(q-p-\eta)n})).
$$
We now treat separately the three cases.

\vskip.2cm

{\bf Step 1: case $p>d/2$.} Since $h(x)=x^2$, we have, if $\eta \in (0,q/2-p)$,
\begin{align*}
\sum_{n\geq 0} K_n(N,x)
\leq C \sum_{n\geq 0} \exp(-a N x^2 2^{n(q-2p-2\eta)})\leq C \exp(-a Nx^2)
\end{align*}
if $Nx^2\geq 1$.

{\bf Step 2: case $p=d/2$.} Since $h(x)=(x/\log(2+1/x))^2$, we have, if $\eta \in (0,q/2-p)$,
\begin{align*}
\sum_{n\geq 0} K_n(N,x) \leq& C \sum_{n\geq 0} \exp\left(-a N x^2 2^{(q-2p-2\eta)n}/
\log^2[2+1/(x2^{(q-p-\eta)n})]\right)\\
\leq  & C \sum_{n\geq 0} \exp(-a N h(x) 2^{n(q-2p-2\eta)})\\
\leq & C\exp(-a N h(x))
\end{align*}
if $N h(x) \geq 1$. The third inequality only uses that
$\log^2(2+1/(x2^{n(q-p-\eta)})) \leq \log^2(2+1/x)$.

\vskip.2cm

{\bf Step 3: case $p<d/2$.} Here $h(x)=x^{d/p}$.
Since $p<d/2$ and $q>2p$, it holds that $q(1-p/d)-p>0$.
We thus may take $\eta \in (0,q(1-p/d)-p)$ (so that $q(d/p-1)-d-d\eta/p>0$)
and we get
\begin{align*}
\sum_{n\geq 0} K_n(N,x) \leq& C \sum_{n\geq 0} \exp(-a N x^{d/p} 2^{n(q(d/p-1)-d-d\eta/p)})
\leq  C\exp(-a N x^{d/p})
\end{align*}
if  $N x^{d/p} \geq 1$.
\end{preuve}

\poubelle{
\section{Normalized convergence}\label{tcl}

Here we give the

\begin{preuve} {\it of Proposition \ref{maintcl}.}
Recall that $\mu$ is supported in $(-1,1]^d$.
Consider a centered Gaussian random measure $W$ on
$(-1,1]^d$, with covariance $\mathbb{E}(W(A)W(B))=\mu(A\cap B)$
for all $A,B \in {\mathcal B}((-1,1]^d)$. Define $G$ by $G(A)=W(A)-\mu(A)W((-1,1]^d)$.
Then it is well-known, see e.g. Dudley \cite[Pages 900]{d}
that for any finite family ${\mathcal A} \subset {\mathcal B}((-1,1]^d)$,
$(\sqrt N (\mu_N(A)-\mu(A)))_{A\in{\mathcal A}}$ goes in law to $(G(A))_{A\in{\mathcal A}}$.

Recall from Notation \ref{dfdp}-(a) that
\begin{align*}
Z_N:=N^{1/2}{\mathcal D}_p(\mu_N,\mu)= \frac{2^p-1}{2}\sum_{\ell \geq 1} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell}N^{1/2}
|\mu_N(F)-\mu(F)|
\end{align*}
and set
\begin{align*}
Z:= \frac{2^p-1}{2}\sum_{\ell \geq 1} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell} |G(F)|.
\end{align*}
We claim that $Z_N$ goes in law to $Z$ as $N \to \infty$. Indeed, setting
\begin{align*}
Z_N^K:=\frac{2^p-1}{2}\sum_{\ell=1}^K 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell}N^{1/2}
|\mu_N(F)-\mu(F)| \quad \hbox{and} \quad
Z^K:= \frac{2^p-1}{2}\sum_{\ell=1}^K2^{-p\ell} \sum_{F \in {\mathcal P}_\ell} |G(F)|,
\end{align*}
we know that $Z_N^K$ goes in law to $Z^K$ for each $K$. To conclude, it is enough to check that
\begin{align}\label{tch}
\lim_{K\to \infty} \left( \mathbb{E}(|Z-Z^K|)+\sup_{N\geq 1} \mathbb{E}(|Z_N-Z_N^K|)\right)=0.
\end{align}
Since $\mathbb{E}(|\mu_N(F)-\mu(F)|) \leq (\mu(F)/N)^{1/2}$ and since
$\sum_{F \in {\mathcal P}_\ell} (\mu(F))^{1/2} \leq 2^{\ell d/2}$ by the Cauchy-Schwarz inequality,
we easily check that
$$
\mathbb{E}\left(|Z_N-Z_N^K|\right)\leq C \sum_{\ell>K} 2^{-p\ell} \sum_{F \in {\mathcal P}_\ell} (\mu(F))^{1/2}
\leq C \sum_{\ell>K} 2^{-(p-d/2)\ell} \leq C 2^{-(p-d/2)K}.
$$
Since $\mathbb{E}(|G(F)|)\leq (\mu(F))^{1/2}$, we get $\mathbb{E}(|Z-Z^K|)\leq C2^{-(p-d/2)K}$ similarly.
Recalling that $p>d/2$, we deduce \eqref{tch}. The same computation
shows that $\mathbb{E}(Z)<\infty$, whence $Z<\infty$ a.s.
This ends the proof.
\end{preuve}
}

\section{The dependent case}

We finally study a few classes of dependent sequences of random variables. We only
give some moment estimates. Concentration inequalities might be obtained, but this should be
much more complicated.

\subsection{$\rho$-mixing stationary sequences}

A stationary sequence of random variables $(X_n)_{n\ge 1}$ with common law $\mu$ is said to be
$\rho$-mixing, for some $\rho:{\mathbb{N}}\to{\mathbb{R}}^+$ with $\rho_n\to 0$, if for all $f,g \in L^2(\mu)$
and all $i,j\geq 1$
$$\mathbb{C}{\rm ov}\,(f(X_i),g(X_j))\le \rho_{|i-j|}\sqrt{\mathbb{V}{\rm ar}\,(f(X_i))\mathbb{V}{\rm ar}\,(g(X_j))}.$$
We refer for example to Rio \cite{rio}, Doukhan \cite{douk} or Bradley \cite{bradbook}.

\begin{thm}\label{mix}
Consider a stationary sequence of random variables $(X_n)_{n\ge 1}$ with common law $\mu$
and set $\mu_N:=N^{-1}\sum_1^N \delta_{X_i}$.
Assume that this sequence is $\rho$-mixing, for some $\rho:{\mathbb{N}}\to{\mathbb{R}}^+$ satisfying
$\sum_{n\geq 0} \rho_n<\infty$. Let $p>0$ and assume that $\mu \in M_q({\rr^d})$ for
some $p>q$.
There exists a constant $C$ depending only on $p,d,q, M_q(\mu),\rho$ such that, for all $N\geq 1$,
$$
\mathbb{E}\left({\mathcal T}_p(\mu_N,\mu)\right) \leq
C\left\{\begin{array}{ll}
N^{-1/2} +N^{-(q-p)/q}& \!\!\!\hbox{if $p>d/2$ and $q\ne 2p$},  \\[+3pt]
N^{-1/2} \log(1+N)+N^{-(q-p)/q} &\!\!\! \hbox{if $p=d/2$ and $q\ne 2p$}, \\[+3pt]
N^{-p/d}+N^{-(q-p)/q} &\!\!\!\hbox{if $p\in (0,d/2)$ and $q\ne d/(d-p)$}.
\end{array}\right.
$$
\end{thm}

This is very satisfying: we get the same estimate as in the independent case.
The case $\sum_{n\geq 0} \rho_n=\infty$ can also be treated (but then the upper bounds will be
less good and depend on the rate of decrease of $\rho$).
Actually, the $\rho$-mixing condition is slightly too strong (we only need
the covariance inequality when $f=g$ is an indicator function), but it is best adapted
notion of mixing we found in the litterature.

\begin{proof}
We first check that for any Borel subset $A \subset {\rr^d}$,
$$
\mathbb{E}[|\mu_N(A)-\mu(A)|] \leq \min\{2\mu(A),C\mu(A)N^{-1/2}\}.
$$
But this is immediate: $\mathbb{E}[\mu_N(A)]=\mu(A)$ (whence $\mathbb{E}[|\mu_N(A)-\mu(A)|]\leq 2\mu(A)$) and
\begin{align*}
\mathbb{V}{\rm ar}\, \mu_N(A)=& \frac{1}{N^2} \sum_{i,j\leq N} \mathbb{C}{\rm ov}\,({{\bf 1}}_A(X_i),{{\bf 1}}_A(X_i))\\
\leq& \frac{1}{N^2} \sum_{i,j\leq N} \rho_{|i-j|}\mathbb{V}{\rm ar}\,({{\bf 1}}_A(X_1))\\
\leq& \frac{\mu(A)(1-\mu(A))}{N^2}\sum_{i,j\leq N} \rho_{|i-j|}.
\end{align*}
This is smaller than $C \mu(A)/N$
as desired, since $\sum_{i,j\leq N} \rho_{|i-j|}\leq N \sum_{k\geq 0} \rho_k= C N$.
Once this is done, it suffices to copy (without any change) the proof of Theorem \ref{mainmom}.
\end{proof}

\subsection{Markov chains}

Here we consider a ${\rr^d}$-valued Markov chain $(X_n)_{n \geq 1}$ with transition kernel $P$
and initial distribution $\nu \in {\mathcal P}({\rr^d})$ and we set $\mu_N:=N^{-1}\sum_{1}^N \delta_{X_n}$.
We assume that it admits a unique invariant
probability measure $\pi$ and the following $L^2$-decay property
(usually related to a Poincar\'e inequality)
\begin{align}\label{decay}
\forall \;n\geq 1,\; \forall \;f\in L^2(\pi),
\quad \|P^nf-\pi(f)\|_{L^2(\pi)}\le C\rho_n \|f-\pi(f)\|_{L^2(\pi)}
\end{align}
for some sequence $(\rho_n)_{n\geq 1}$ decreasing to $0$.

\begin{thm}\label{markov} Let $p\geq 1$, $d\geq 1$ and $r> 2$ be fixed.
Assume that our Markov chain $(X_n)_{n \geq 0}$ satisfies
\eqref{decay} with a sequence $(\rho_n)_{n\geq 1}$ satisfying $\sum_{n\geq 1} \rho_n<\infty$.
Assume also that the initial distribution $\nu$ is absolutely continuous with respect to $\pi$
and satisfies $\|d\nu/d\pi\|_{L^r(\pi)}<\infty$. Assume finally that $M_q(\pi)<\infty$ for some
$q>p r/(r-1)$. Setting $q_r:=q(r-1)/r$ and $d_r=d(r+1)/r$, there is a constant $C$ such that for
all $N\geq 1$,
$$
\mathbb{E}_\nu\left({\mathcal T}_p(\mu_N,\pi)\right) \leq
C\left\{\begin{array}{ll}
N^{-1/2} +N^{-(q_r-p)/q_r}& \hbox{if $p>d_r/2r$ and $q_r\ne 2p$},  \\[+3pt]
N^{-1/2} \log(1+N)+N^{-(q_r-p)/q_r} & \hbox{if $p=d_r/2r$ and $q_r\ne 2p$}, \\[+3pt]
N^{-p/d}+N^{-(q_r-p)/q_r} &\hbox{if $p\in (0,d_r/2)$ and $q_r\ne d_r/(d_r-p)$}.
\end{array}\right.
$$
\end{thm}

Once again, we might adapt the proof to get a complete picture
corresponding to other decay than $L^2$-$L^2$ and to slower mixing rates $(\rho_n)_{n\geq 1}$.

\begin{proof}
We only have to show that for any $\ell \geq 0$, any $n\geq 0$,
\begin{align*}
\Delta_{n,\ell}^N:=&\sum_{F\in{\mathcal P}_\ell}\mathbb{E}_\nu\left( |\mu_N(2^n F \cap B_n)-\pi(2^n F \cap B_n)| \right)\\
\leq& C \min \left\{(\pi(B_n))^{(r-1)/r}, [2^{d_r \ell}(\pi(B_n))^{(r-1)/r}/N]^{1/2} \right\}.
\end{align*}
Since $M_q(\pi)<\infty$ (whence $\pi(B_n)\leq C 2^{-qn}$), we will deduce that
$$
\Delta_{n,\ell}^N
\leq C \min \left\{2^{-q_r n}, 2^{d_r \ell/2}(2^{-q_rn}/N)^{1/2} \right\}.
$$
Then the rest of the proof is exactly the same as that of Theorem \ref{mainmom},
replacing everywhere $q$ and $d$ by $q_r$ and $d_r$.

\vskip.2cm

We first check that $\Delta_{n,\ell}^N \leq  C (\pi(B_n))^{(r-1)/r}$. Using that
$\|d\nu/d\pi\|_{L^r(\pi)}<\infty$, we write
$$
\mathbb{E}_\nu(\mu_N(B_n))=\frac1N\sum_{i=1}^N\mathbb{E}_\pi\left[\frac{d\nu}{d\pi}(X_0)1_{\{X_i\in B_n\}}\right]
\leq  \|d\nu/d\pi\|_{L^r(\pi)}  \pi(B_n)^{(r-1)/r}.
$$

We next consider a Borel subset $A$ of ${\rr^d}$ and check that
$$
\mathbb{E}_\nu(|\mu_N(A)-\pi(A)|)\le C  (\pi(A))^{(r-1)/(2r)} N^{-1/2}.
$$
To do so, as is usual when working with Markov chains or covariance properties (see \cite{blg}),
we introduce $f=1_A-\pi(A)$ and write
\begin{align*}
\mathbb{E}_\nu(|\mu_N(A)-\pi(A)|)=\frac1N\mathbb{E}_\nu\left(\left|\sum_{i=1}^N f(X_i)\right|\right)
\le \frac1N \left( \sum_{i,j=1}^N\mathbb{E}_\nu(f(X_i)f(X_j))\right)^{1/2}.
\end{align*}
For $j\geq i$, it holds that
\begin{align*}
\mathbb{E}_\nu(f(X_i)f(X_j))=&\mathbb{E}_\nu[f(X_i)P^{j-i}f(X_i)]
=\mathbb{E}_\pi\left[\frac{d\nu}{d\pi}(X_0)f(X_i).P^{j-i}f(X_i)\right].
\end{align*}
Using the H\"older inequality (recall that $\|d\nu/d\pi\|_{L^r(\pi)}<\infty$ with $r>2$)
and \eqref{decay}, we get
\begin{align*}
\mathbb{E}_\nu(f(X_i)f(X_j))\leq& \|d\nu/d\pi\|_{L^r(\pi)} \|f\|_{L^{2r/(r-2)}(\pi)}\|P^{j-i}f\|_{L^2(\pi)}
\leq C \rho_{j-i} \|f\|_{L^{2r/(r-2)}(\pi)}\|f\|_{L^2(\pi)} .
\end{align*}
But for $s>1$, $\|f\|_{L^s(\pi)}\leq C_s(\pi(A)+(\pi(A))^s)^{1/s}\leq C_s (\pi(A))^{1/s}$,
we find $\mathbb{E}_\nu(f(X_i)f(X_j))\leq C \rho_{j-i}(\pi(A))^{(r-1)/r}$ and thus
\begin{align*}
\mathbb{E}_\nu(|\mu_N(F)-\pi(F)|)\le \frac CN \left(\sum_{i,j=1}^N \rho_{|i-j|} (\pi(F))^{(r-1)/2r} \right)^{1/2}
\leq C  (\pi(F))^{(r-1)/(2r)} N^{-1/2}
\end{align*}
as desired. We used that $\sum_{i,j=1}^N \rho_{|i-j|}\leq C N$.

\vskip.2cm

We can finally conclude that
\begin{align*}
\Delta_{n,\ell}^N \leq & CN^{-1/2} \sum_{F \in {\mathcal P}_\ell} (\pi(2^n F \cap B_n))^{(r-1)/(2r)}
\leq  CN^{-1/2} 2^{d_r \ell / 2} (\pi(B_n))^{(r-1)/(2r)}
\end{align*}
by the H\"older inequality (and because $\# {\mathcal P}_\ell=2^{d\ell}$), where $d_r=d (r+1)/r$
as in the statement.
\end{proof}

\subsection{Mc Kean-Vlasov particles systems}
Particle approximation of nonlinear equations has attracted a lot of attention in the past
thirty years. We will focus here on the following ${\rr^d}$-valued nonlinear S.D.E.
$$
dX_t=\sqrt{2}dB_t-\nabla V(X_t)dt-\nabla W*u_t(X_t)dt,\qquad X_0=x
$$
where $u_t= Law(X_t)$ and $(B_t)$ is and ${\rr^d}$-Brownian motion. This is a probabilistic
representation of the so-called Mc Kean-Vlasov equation, which has been studied
in particular by Carillo-Mac Cann-Villani \cite{cmcv}, Malrieu \cite{m} and Cattiaux-Guillin-Malrieu
\cite{cgm} to which we refer for further motivations and existence and uniqueness of
solutions. We will mainly consider here the case where $V$ and $W$ are convex (and if $V=0$ the
center of mass is fixed) and $W$ is even.
To fix the ideas, let us consider only two cases:

(a) $Hess\, V\ge\beta Id>0$, $Hess\, W\ge 0$.

(b) $V(x)=|x|^\alpha$ for $\alpha>2$, $Hess\, W\ge 0$.

\vskip.2cm

The particle system introduced to approximate the nonlinear equation is the following.
Let $(B^i_t)_{t\geq 0}$ be $N$ independent Brownian motion. For $i=1,\dots,N$, set $X^{i,N}_0=x$ and
$$dX^{i,N}_t=\sqrt{2}dB^i_t-\nabla V(X^{i,N}_t)dt-\frac1N\sum_j\nabla W(X^{i,N}_t-X^{j,N}_t)dt.$$
Usual propagation of chaos property is usually concerned with control of
$${\mathcal W}_2(Law(X^{1,N}_t),u_t)$$
uniformly (or not) in time. It is however very natural to consider rather a control of
$${\mathcal W}_2(\hat u^N_t,u_t)$$
where $\hat u^N_t=\frac 1N\sum_{i=1}^N\delta_{X^{i,N}_t}$, as in Bolley-Guillin-Villani \cite{bgv}.

\vskip.2cm

To do so, and inspired by the usual proof of propagation of chaos, let us consider nonlinear
independent particles
$$
dX^i_t=\sqrt{2}dB^i_t-\nabla V(X_t^i)dt-\nabla W*u_t(X_t^i)dt,\qquad X^i_0=x
$$
(driven by the same Brownian motions as the particle system)
and the corresponding empirical measure $u^N_t=\frac 1N\sum_{i=1}^N\delta_{X^{i}_t}$. We then have
$$
{\mathcal W}_2(\hat u^N_t,u_t)\le {\mathcal W}_2(\hat u^N_t,u^N_t)+{\mathcal W}_2(u^N_t,u_t).
$$
Then following \cite{m} in case (a) and \cite{cgm} in case (b), one easily gets (for
some time-independent constant $C$)
$$
\mathbb{E}({\mathcal W}_2^2(\hat u^N_t,u^N_t))\le \frac1N\mathbb{E}\left(\sum_{i=1}^N|X^{i,N}_t-X^i_t|^2\right)\le C\alpha(N)
$$
where $\alpha(n)=N^{-1}$ in case (a) and $\alpha(N)=N^{-1/(\alpha-1)}$ in case (b). It is not hard to
prove here that the nonlinear particles have infinitely many moments (uniformly in time)
so that combining Theorem \ref{mainmom} with the previous estimates gives
$$
\sup_{t\geq 0}\mathbb{E}({\mathcal W}_2(\hat u^N_t,u_t))\le C(\alpha(N)+\beta(N))
$$
where $\beta(N)=N^{-1/2}$ if $d=1$, $\beta(N)=N^{-1/2}\log(1+N)$ if $d=2$ and
$\beta(N)=N^{-1/d}$ if $d\geq 3$.

redpajama_set_name: RedPajamaArXiv


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Priority No 2 Aim: To sustain the rigorous and robust teaching of phonics, including early identification of pupils who need interventions to achieve the Year 1 phonics standard.
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text:
package com.nec.strudel.bench.auction.entity;

public class BidId {
    private int sellerId;
    private int itemNo;
    private int bidNo;

    public BidId(ItemId itemId, int bidNo) {
        this.sellerId = itemId.getSellerId();
        this.itemNo = itemId.getItemNo();
        this.bidNo = bidNo;
    }

    public BidId() {
    }

    public ItemId getItemId() {
        return new ItemId(sellerId, itemNo);
    }

    public int getSellerId() {
        return sellerId;
    }

    public void setSellerId(int sellerId) {
        this.sellerId = sellerId;
    }

    public int getItemNo() {
        return itemNo;
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    public void setItemNo(int itemNo) {
        this.itemNo = itemNo;
    }

    public void setBidNo(int bidNo) {
        this.bidNo = bidNo;
    }

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        return bidNo;
    }

    @Override
    public String toString() {
        return sellerId + "."
                + itemNo + "."
                + bidNo;
    }

    private static final int HASH_BASE = 31;

    @Override
    public int hashCode() {
        int code = sellerId * HASH_BASE + itemNo;
        return code * HASH_BASE + bidNo;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj) {
            return true;
        }
        if (obj instanceof BidId) {
            BidId id = (BidId) obj;
            return sellerId == id.sellerId
                    && itemNo == id.itemNo
                    && bidNo == id.bidNo;
        }
        return false;
    }
}

redpajama_set_name: RedPajamaGithub


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