TITLE: Does $19,199,1999,\dotsc$ contain infinitely many prime numbers?
QUESTION [12 upvotes]: Are there infinitely many primes of the form $F_n =2\times10^n-1$? That is, does this sequence,
$$19,199,1999,\dotsc$$
contain infinitely many prime numbers?
I think about Dirichlet's theorem on arithmetic progressions, but the problem is difficult to start.
Edit: $F_n$ is prime for $n=1, 2, 3, 5, 7, 26, 27,\dotsc, 55347$ (A002957) and $n=1059002$ (Kamada's tables). $F_{n}$ for $n=6m+4$ is divisible by $7$.

REPLY [7 votes]: The current status of this conjecture is that it is not proven either way. It is similar to the mersenne primes conjecture, as well as the twin prime conjecture. At this point, the only infinitude proof concerning prime numbers of a specific kind gives an upper bound on the limit inferior of the difference between consecutive primes, which is somewhat vague and doesn't involve formulas as specific as yours. Ulam's spiral shows that there are many similar formulas, perhaps infinitely many, that often produce primes. Whether or not they do so infinitely often remains an open question.
It seems at this point that in order to prove such conjectures, we will need new tools and methodologies, as analytical number theory doesn't cut it and none of the other branches of mathematics have stepped in to pick up the slack. However, it is entirely possible that answers to these questions will begin to pop up within the next decade or even sooner. Once one such proof is assembled, others are sure to follow.


TITLE: Find all positive integers s.t. $10^m-8^n=2m^2$
QUESTION [1 upvotes]: Find all pairs of positive integers $(m,n)$ such that $10^m-8^n=2m^2$

REPLY [1 votes]: Hint:
$${10}^m - 8^n = {2^m}{5^m} - {2^{3n}}$$
so that
$$m^2 = {2^{m-1}}{5^m} - {2^{3n-1}}.$$
In particular:
$${2^{m-1}}{5^m} - {2^{3n-1}} \geq 1,$$
and
$${2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is a square, which means $m \equiv 1 \pmod 2$.
Suppose that $m > 1$.  Then the RHS of
$$m^2 = {2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is even while the LHS is odd (since $5^m - {2^{3n-m}}$ is always odd, unless $m = 3n$ [I'll deal with this case later]).  A contradiction.  Therefore, $m = 1$.
Substituting $m = 1$:
$$10 - {8^n} = 2$$
which gives $n = 1$.
Update:  If $m = 3n$, then
$$m^2 = {3^2}{n^2} = 2^{3n - 1}\left(5^{3n} - 1\right).$$
Since $\gcd(3,2) = 1$, then it follows that
$$2^{3n - 1} \mid n^2$$
which implies that $2^{3n - 1} \leq n^2$, contradicting $n^2 < 2^{3n - 1}$.
Therefore, in general, $m = 3n$ cannot occur.
Hence the only solution to the original problem is $(m, n) = (1, 1)$.


TITLE: Application of the Girsanov theorem
QUESTION [2 upvotes]: Let $B_t$ denote the standard brownian motion. In a homework exercise we are asked to use the Girsanov theorem to compute
\begin{equation}
\mathbb{E}\bigg((B_t-t)^2\exp\bigg(\int_0^t e^{-s}\, dB_s\bigg) \bigg) \tag{1}
\end{equation}
After reading about the Girsanov theorem I fail to understand how to apply it in this particular situation. My attempt so far is to first note that
$$
\mathbb{E}\bigg(\frac{1}{2}\int_0^t e^{-2s}\,ds\bigg) = \mathbb{E}\bigg(\frac{1}{4}\big(1-e^{-2t}\big) \bigg) < \infty , \ \forall t\geq 0
$$
hence the Novikov condition is satisfied. In the theory I've read, this means that if we set
$$
L_t = \exp\bigg(-\int_0^te^{-s}\,dB_s - \frac{1}{2}\int_0^te^{-2s}\,ds \bigg)
$$
and the denote by $Y_t = (B_t - t)^2$ and further if the expectation in equation (1) is with respect to some probability measure $\mathbb{P}$ there exists a probability measure $\mathbb Q$ such that
\begin{equation}
\mathbb E_{\mathbb Q}\bigg( Y_t \bigg) = \mathbb E_{\mathbb P}\bigg( Y_t L_t \bigg)
\end{equation}
Here is where I'm stuck, first, in equation (1) there is no minus sign in the exponential, nor is the quadratic term present. Where do I go from here?

REPLY [4 votes]: Set
$$q_T := \exp \left( \int_0^T e^{-s} dB_s - \frac{1}{2} \int_0^T e^{-2s} \, ds \right).$$
By Girsanov's theorem, the process
$$W_t := B_t - \int_0^t e^{-s} \, ds, \qquad t \leq T,$$
is a Brownian motion with respect to the probabillity measure $\mathbb{Q}:=\mathbb{Q}_T := q_T \, \mathbb{P}$. We have
\begin{align*} &\mathbb{E} \left[ (B_T-T)^2 \exp \left( \int_0^T e^{-s} \, dB_s \right) \right] \\&= \exp \left( \frac{1}{2} \int_0^T e^{-2s} \, ds \right) \mathbb{E} \left[ \left( W_T + \int_0^T e^{-s} \, ds-T \right)^2 q_T \right] \\ &= \exp\left( \frac{1-e^{-2T}}{4} \right) \mathbb{E}_{\mathbb{Q}}  \left[ \left( W_T +(1-e^{-T}-T)\right)^2 \right].\tag{1}\end{align*}
For brevity set $f(T) := 1-e^{-T}-T$, then
$$(W_T+f(T))^2 = W_T^2  + 2W_T f(T) +f(T)^2$$
and taking expectation (w.r.t to $\mathbb{Q}$) we get
$$\mathbb{E}_{\mathbb{Q}} ((W_T+f(T))^2) = \underbrace{\mathbb{E}_{\mathbb{Q}}(W_T^2)}_{=T} + 2 f(T) \underbrace{\mathbb{E}_{\mathbb{Q}}(W_T)}_{=0} + f(T)^2 = T+f(T)^2.$$
Plugging this into $(1)$ gives
\begin{align*} \mathbb{E} \left[ (B_T-T)^2 \exp \left( \int_0^T e^{-s} \, dB_s \right) \right] &= \exp \left( \frac{1-e^{-2T}}{4} \right) (T+f(T)^2) \\ &= \exp \left( \frac{1-e^{-2T}}{4} \right) (T+(1-e^{-T}-T)^2). \end{align*}


TITLE: Combinatorial proof of the sum of a "vertical" sequence of binomial coefficients
QUESTION [2 upvotes]: $ \textbf{Question:} $ Prove that for all $ n, \; k \in \mathbb{N} $
$$ \displaystyle \sum_{j = k}^{n} \binom{j}{k} = \binom{n + 1}{k + 1} $$
I have successfully proved this formula using induction. The book also provides an alternative way using a combinatorial proof as following. We know that $ \displaystyle \binom{n + 1}{k + 1} $ counts the class of all $ (k + 1) $-elements subsets of the $ (n + 1) $-elements set. But this class of subsets may be partitioned into subclasses corresponding to $ j = k, k + 1, \dots, n $ as follows.

The subclass of subsets with largest element equal to $ k + 1 $ is counted by $ \displaystyle \binom{k}{k}, $ the subclass of subsets with largest element equal to $ k + 2 $ is counted by $ \displaystyle \binom{k + 1}{k}, \dots, $ and the subclass of subsets with largest element equal to $ n + 1 $ is counted by $ \displaystyle \binom{n}{k}. $ The identity in questionfollows by the extended addition rule.

Can someone help me understand the part in the pink box? Each subset in the big class has $ k + 1 $ elements from the binomial coefficient $ \displaystyle \binom{n + 1}{k + 1}, $ but I don't understand what they mean when they say "subsets with largest element equal to $ k + 1. $" If the element in each subset is something like "people," how can there be "largest element?"

REPLY [1 votes]: The formulation  largest element $k+1$ in the proof indicates that the specific $(n+1)$-element set
\begin{align*}
[n+1]:=\{1,2,3,\ldots,n+1\}
\end{align*}
is considered. Considering $[n+1]$ is consistent with the complete text in the proof and quite usual in the context of a combinatorial proof.

When looking at the series of the LHS of
  \begin{align*}
\sum_{j=k}^n\binom{j}{k}=\binom{n+1}{k+1}\tag{1}
\end{align*}
  the key observation is that $(k+1)$-element subsets of $[n+1]$ have a largest element, which is at least $k+1$ and at most $n+1$. We use this fact to partition the subsets with respect to their largest element.
If we consider the $(k+1)$-element subsets with largest element equal to $j+1$ with $k+1\leq j+1\leq n+1$, we keep the element $j+1$ fix in the $(k+1)$-element subset and have $\binom{j}{k}$ possibilities to select the remaining $k$ elements from $[j]$.
Summing up this way the number of $(k+1)$-element subsets results in the LHS of (1).


text:
\begin{document}

\title[Nonparametric estimation for L\'evy processes]{Nonparametric adaptive estimation for pure jump L\'evy processes.}
\author[F. Comte]{F. Comte}

\author[V. Genon-Catalot]{V. Genon-Catalot$^{1}$}\thanks{$^1$ Universit\'e Paris V,  MAP5, UMR CNRS
8145. email: fabienne.comte@univ-paris5.fr and genon@math-info.univ-paris5.fr}

\begin{abstract} This paper is concerned with nonparametric estimation of the L\'evy density of a pure jump L\'evy process. The sample path is observed at $n$ discrete instants with fixed sampling interval. We construct a collection of estimators obtained by deconvolution methods and deduced from appropriate estimators of the characteristic function and its first derivative. We obtain a bound for the ${\mathbb L}^2$-risk, under general assumptions on the model. Then we propose a penalty function that allows to build an adaptive estimator. The risk bound for the adaptive estimator is obtained under additional assumptions on the L\'evy density. Examples of models fitting in our framework are described and rates of convergence of the estimator are discussed.
{\bf \today}
\end{abstract}
\maketitle

\noindent {\bf {\sc Keywords.}} Adaptive Estimation; Deconvolution; L\'evy process;
Nonparametric Projection Estimator.

\medskip\

\section{Introduction}
In recent years, the use of L\'evy processes for modelling purposes has become very popular in many areas and especially in the field of finance (see {\em e.g.} Eberlein and Keller (1995), Barndorff-Nielsen and Shephard (2001), Cont and Tankov (2004); see also Bertoin (1996) or Sato (1999) for a comprehensive study for these processes). The distribution of a L\'evy process is usually specified by its characteristic triple (drift, Gaussian component and L\'evy measure) rather than by the distribution of its independent increments. Indeed, the exact distribution of these increments is most often intractable or even has no closed form formula. For this reason, the standard parametric approach by likelihood methods is a difficult task and many authors have rather considered nonparametric methods. For L\'evy processes, estimating the L\'evy measure is of crucial importance since this measure specifies the jumps behavior. Nonparametric estimation of the L\'evy measure has been the subject of several recent contributions. The statistical approaches depend on the way observations are performed. For instance, Basawa and Brockwell (1982) consider non decreasing L\'evy processes and observations of jumps with size larger than some positive $\varepsilon$, or discrete observations with fixed sampling interval. They build nonparametric estimators of a distribution function linked with the L\'evy measure. More recently, Figueroa-L\'opez and Houdr\'e (2006) consider a continuous-time observation of a general L\'evy process and study penalized projection estimators of the L\'evy density based on integrals of functions with respect to the random Poisson measure associated with the jumps of the process. However,  their approach remains theoretical since these Poisson integrals are hardly accessible.

In this paper, we consider nonparametric estimation of the L\'evy measure for real-valued L\'evy processes of pure jump type, {\em i.e.} without drift and Gaussian component. We rely on the common assumption that the L\'evy measure admits a density $n(x)$ on ${\mathbb R}$ and assume that the process is discretely observed with fixed sampling interval $\Delta$.  Let $(L_t)$ denote the underlying L\'evy process and  $(Z_k^{\Delta}= L_{k\Delta}- L_{(k-1)\Delta}, k=1, \ldots, n)$ be the observed random variables which are independent and identically distributed. Under our assumption, the characteristic function of $L_{\Delta}= Z_1^{\Delta}$ is given by the following simple formula:
\begin{equation} \label{fc}
\psi_{\Delta}(u)= {\mathbb E}(\exp{i u Z_1^{\Delta}})=\exp{(\Delta \int_{{\mathbb R}}(e^{iux} -1) n(x) dx)}
\end{equation}
where  the unknown function  is the  L\'evy density  $n(x)$.  It  is therefore
natural to investigate the  nonparametric estimation of $n(x)$ using empirical
estimators  of  the charasteristic  functions  and  its  derivatives and  then
recover the L\'evy density by  Fourier inversion. This approach is illustrated
by Watteel and  Kulperger (2003) and Neumann and  Reiss (2000). However, these
authors   consider  general   L\'evy  processes,   with  drift   and  Gaussian
component. Hence, at least two  derivatives of the characteristic function are
necessary to reach the L\'evy density.  Moreover, the way Fourier inversion is
done in concrete is not detailed in these papers. In our case, under the assumption that $\int_{{\mathbb R}} |x| n(x) dx < \infty$, we get the simple relation:
\begin{equation}\label{basic}
g^*(u)= \int e^{iux}g(x)dx=-i\frac{\psi'_{\Delta}(u)}{\Delta\psi_{\Delta}(u)},
\end{equation}
with $g(x)=x n(x)$. This equation indicates that we can estimate $g^*(u)$ by using empirical counterparts of $\psi_{\Delta}(u)$ and $\psi'_{\Delta}(u)$ only. Then, the problem of recovering an estimator of $g$ looks like a classical deconvolution problem. We have at hand the methods used for estimating unknown densities of random variables observed with additive independent noise. This requires the additional assumption that $g$ belongs to ${\mathbb L}^{2}({\mathbb R})$.  However, the problem of deconvolution set by equation (\ref{basic}) is not standard and looks more like deconvolution in presence of unknown errors densities. This is due  to the fact that both the numerator and the denominator are unknown and have to be estimated from the same data. This is why our estimator of
$\psi_{\Delta}(u)$ is not a simple empirical counterpart. Instead, we use a truncated version analogous to the one used in Neumann (1997) and Neumann and Reiss (2000).

Below, we show how to adapt the deconvolution method described in Comte {\it et al.}~(2006). We consider an adequate sequence $(S_m, m=1, \ldots, m_n)$ of subspaces of ${\mathbb L}^{2}({\mathbb R})$ and build a collection of projection estimators $({\hat g}_m)$. Then using a penalization device, we select through a data-driven procedure the best estimator in the collection. We study the ${\mathbb L}^{2}$-risk of the resulting estimator under the asymptotic framework that $n$ tends to infinity. Although the sampling interval $\Delta$ is fixed, we keep it as much as possible in all formulae since the distributions of the observed random variables highly depend on $\Delta$.

In Section 2, we give assumptions and some preliminary properties. Section 3 contains examples of models included in our framework. Section 4 describes the statistical strategy. We present the projection spaces and define the collection of estimators. Proposition 4.1 gives the upper bound for the risk of a projection estimator on a fixed projection space. This proposition guides the choice of the penalty function and allows to discuss the rates of convergence of the projection estimators. Afterwards,  we  introduce a theoretical penalty (depending on the unknown characteristic function $\psi_{\Delta}$) and study the risk bound of a false estimator (actually not an estimator) (Theorem 4.1). Then, we replace the theoretical penalty by an estimated counterpart and give the upper bound of the risk of the resulting penalized estimator (Theorem 4.2). Section 6 gives some conclusions and open problems. Proofs are gathered in Section 6. In the Appendix, a fondamental result used in our proofs is recalled.

\section{Framework and assumptions.}

Recall that we consider the discrete time observation with sample step $\Delta$ of a L\'evy process $L_t$ with L\'evy density $n$ and characteristic function given by (\ref{fc}).
We assume that $(L_t)$ is a pure jump process with finite variation on compacts.

When the L\'evy measure $n(x)dx$ is concentrated on $(0, +\infty)$, then
$(L_t)$ has increasing paths and is called a subordinator.
We focus on the estimation of the real valued function
\begin{equation}
g(x)= x n(x),
\end{equation}
and introduce the following assumptions on the function $g$:
\begin{itemize}
\item[(H1)$\;\;\;\;\;$]  $\int_{\R} |x| n(x) dx < \infty.$
\item [(H2$(p)$)] For $p$ integer,  $\int_{\R}
  |x|^{p-1} |g(x)| dx <\infty$.
\item [(H3)$\;\;\;\;\;$] The function $g$ belongs to ${\mathbb L}_{2}(\R)$.
\end{itemize}
Note that (H1) is stronger than the usual assumption $\int (|x|\wedge 1) n(x)dx <+\infty$, and is also a moment assumption for $L_t$.
Under the usual assumption, (H2$(p)$) for $p\geq 1$ implies (H1) and (H2$(k)$) for $k\leq p$.

Our estimation procedure is based
on the random variables \begin{equation}\label{kiezi} Z_i^{\Delta}= L_{i\Delta}-
L_{(i-1)\Delta} , i= 1, \ldots, n,\end{equation}  which are independent,
identically distributed, with common characteristic function
$\psi_{\Delta}(u)$.

The moments of $Z_1^{\Delta}$ are linked with the function $g$. More precisely,
we have:
\begin{prop} Let $p \ge 1$ integer. Under (H2)($p$),  $\E|Z_1^{\Delta}|^{p} <\infty$. Moreover, setting, for $k=1, \dots p$, $M_k= \int_{\R} x^{k-1}g(x)dx$, we have ${\mathbb E}(Z_1^{\Delta})=\Delta M_1$, ${\mathbb E}[(Z_1^{\Delta})^2]=\Delta M_2+\Delta^2 M_1$, and more generally, $\E[(Z_{1}^{\Delta})^{l}]= \Delta\; M_l + o(\Delta)$ for all $l=1,\ldots, p$.
\end{prop}

\noindent {\bf Proof.} By the assumption,  the exponent of the exponential in (\ref{fc}) is $p$
times differentiable and, by derivating $\psi_{\Delta}$, we get the result.
$\Box$\\

Assumption (H1) yields the relation (\ref{basic}), which is the basis of our
estimation procedure.

We need a precise control of $\psi_{\Delta}$. For this, we introduce the assumption that, for $m_n$ an integer to be defined later, the following holds:
\begin{itemize} \item[(H4)$\;\;\;\;\;$]  $\forall x\in {\mathbb R}, \mbox{ we have }  c_\psi(1+x^2)^{-\Delta\beta/2} \leq |\psi_{\Delta}(x)|\leq C_\psi(1+x^2)^{-\Delta \beta/2},$
\end{itemize}  for some given constants $c_{\psi}, C_{\psi}$ and $\beta\geq 0$. Note that an assumption of this type is also considered in Neumann and Reiss~(2007).

For the adaptive version of our estimator, we need additional assumptions for $g$:
\begin{itemize} \item[(H5)$\;\;\;\;\;$]  There exists some positive $a$  such that $\int |g^*(x)|^2(1+x^2)^adx < +\infty$,
\end{itemize} and

\begin{itemize} \item[(H6)$\;\;\;\;\;$] $\int x^2g^2(x)dx<+\infty.$
\end{itemize}

We must set independent assumptions for $\psi_{\Delta}$ and $g$, since there may be no relation at all between these two functions (see the examples). Note that, in Assumption (H5), which is a classical regularity assumption, the knowledge of $a$ is not required.

\section{Examples.}\label{example}
\subsection{Compound Poisson processes.}
Let $L_t= \sum_{i=1}^{N_t} Y_i$, where $(N_t)$ is a Poisson process
with constant intensity $c$ and $(Y_i)$ is a sequence of i.i.d. random
variables with density $f$ independent of the process $(N_t)$. Then,
$(L_t)$ is a compound Poisson process with characteristic function
\begin{equation}
\psi_{t}(u)= \exp{c t \int_{\R}(e^{iux} -1) f(x) dx}.
\end{equation}
Its L\'evy density is $n(x)= c f(x)$. Assumptions (H1)-(H2)$(p)$ are
equivalent to $\E(|Y_1|^{p})<\infty$. Assumption (H3) is equivalent to
$\int_{\R} x^2 f^{2}(x) dx < \infty$, which holds for instance if $\sup_x f(x)<+\infty$ and ${\mathbb E}(Y_1^2)<+\infty$. We can compute the distribution
of $Z_1^{\Delta}=L_{\Delta}$ as follows:
\begin{equation}
  P_{Z_1^{\Delta}}(dz) = e^{-c \Delta} ( \delta_{0}(dz) + \sum_{n \ge 1} f^{*n}(z) \frac{(c \Delta)^{n}}{n!} dz ).
\end{equation}
We have the following bound:
\begin{equation}
1\geq |\psi_{\Delta}(u)| \ge e^{-2 c \Delta}.
\end{equation}
On this example, it appears clearly that we can not link the regularity assumption on $g$ and (H4) which holds with $\beta=0$.

\subsection{The L\'evy gamma process.}
Let $\alpha>0,\beta>0$. The L\'evy gamma process $(L_t)$ with parameters $(\beta,\alpha)$ is a subordinator such that, for all $t>0$, $L_t$ has distribution Gamma with parameters $(\beta t,\alpha)$, {\em i.e.} has density:
\begin{equation}
\frac{\alpha^{\beta t}}{\Gamma(\beta t)} x^{\beta t-1} e^{-\alpha x} 1_{x \ge 0}.
\end{equation}
The characteristic function of $Z_1^{\Delta}$ is equal to:
\begin{equation}
\psi_{\Delta}(u)= \left(\frac{\alpha}{\alpha-i u}\right)^{\beta \Delta}.
\end{equation}
The L\'evy density is $n(x)= \beta x^{-1} e^{-\alpha x}\1_{\{x>0\}}$ so that
$g(x)= \beta e^{-\alpha x}\1_{\{x>0\}}$ satisfies our assumptions. We have:
\begin{equation}
\frac{\psi_{\Delta}^{'}(u)}{\psi_{\Delta}(u)}= i \Delta \frac{\beta}{\alpha-i u}, \quad \quad
|\psi_{\Delta}(u)|= \frac{\alpha^{\beta \Delta}}{(\alpha^2+u ^2)^{\beta\Delta/2}}
\end{equation}

\subsection{Another class of subordinators}\label{other}
Consider the L\'evy process $(L_t)$ with L\'evy density
$$
n(x)= c x^{\delta-1/2} x^{-1} e^{-\beta x} 1_{x>0},
$$
where $(\delta,\beta,c)$ are positive parameters.
If $\delta> 1/2$, $\int_0^{+\infty}n(x)dx<+\infty$, and we recover compound Poisson processes.
If $0<\delta\leq 1/2$, $\int_0^{+\infty}n(x)dx=+\infty$ and $g(x)=xn(x)$ belongs to ${\mathbb L}^2({\mathbb R})\cap
{\mathbb L}^1({\mathbb R})$.
The case $\delta=0$, which corresponds to the L\'evy inverse Gaussian process does not fit in our framework.
For $0<\delta< 1/2$, we find
$$g^*(x)=c\frac{\Gamma(\delta+1/2)}{(\beta-ix)^{\delta+1/2}},$$ and $$|\psi_{\Delta}(x)|=\exp\left( -c\frac{\Delta\Gamma(\delta+1/2)}{1/2-\delta} [(\beta^2+x^2)^{-(\delta-1/2)/2}-\beta^{-(\delta-1/2)}]\right).$$
It is important to mention that $\psi_{\Delta}$ above does not satisfy assumption (H4) since \begin{equation}\label{equiv} |\psi_{\Delta}(x)|\sim_{x\rightarrow +\infty} K(\beta,\delta)\exp(-c\Delta\frac{\Gamma(\delta+1/2)}
{1/2-\delta}x^{-\delta+1/2})\end{equation}
where $K(\beta,\delta)=\exp\left( c\frac{\Delta\Gamma(\delta+1/2)}{1/2-\delta}\beta^{-(\delta-1/2)}\right)$.
Thus, it has an exponential rate of decrease.

\subsection{The bilateral Gamma process.}
This process has been recently introduced by K\"{u}chler and
Tappe~(2008).  Consider $X,Y$ two independent random variables,
$X$ with distribution $\Gamma(\beta,\alpha )$ and $Y$ with distribution
$\Gamma(\beta',\alpha')$. Then, $Z=X-Y$ has distribution bilateral gamma
with parameters $(\beta,\alpha,\beta',\alpha')$, that we denote by
$\Gamma(\beta,\alpha;\beta',\alpha')$. The characteristic function of $Z$ is equal
to:
\begin{equation}
\psi(u)= \left(\frac{\alpha}{\alpha-i u}\right)^{\beta} \left(\frac{\alpha'}{\alpha'+i
u}\right)^{\beta'}= \exp{( \int_{\R}(e^{iux} -1) n(x) dx)},
\end{equation}
with
$$
n(x)= x^{-1} g(x),
$$
and, for $x \in \R$,
$$
g(x)= \beta e^{-\alpha x} 1_{(0, +\infty)(x)} - \beta' e^{-\alpha'|x|} 1_{(-\infty,0)}(x).
$$
The bilateral Gamma process $(L_t)$ has characteristic function
$\psi_t(u)= \psi(u)^{t}$.

The method can be generalized and we may consider L\'evy processes on ${\mathbb R}$ obtained by bilateralisation of two subordinators.

\subsection{Subordinated Processes.}
 Let $(W_t)$  be a Brownian  motion, and let  $(Z_t)$ be an  increasing L\'evy
 process  (subordinator), independent  of  $(W_t)$. Assume  that the  observed
 process is $$L_t=W_{Z_t}.$$ We have $$\psi_{\Delta}(u)={\mathbb E}(e^{iuL_{\Delta}}) = {\mathbb E}(e^{-\frac{u^2}2 Z_{\Delta}}).$$ As $Z_t$ is
positive, we consider, for $\lambda \ge 0$,
$$\vartheta_{\Delta}(\lambda)   =  {\mathbb   E}(e^{-\lambda   Z_{\Delta}})  =
\exp\left( -\Delta  \int_0^{+\infty}(1-e^{-\lambda x})n_Z(x)dx\right),$$ where
$n_{Z}$  denotes  the  L\'evy density  of  $(Z_t)$.  Now  let us  assume  that
$g_Z(x)=xn_Z(x)$ is integrable over $(0,+\infty)$. We have:
\begin{eqnarray*} \log(\vartheta_{\Delta}(\lambda))&=& -\Delta\int_0^{+\infty}\frac{1-e^{-\lambda x}}x xn_Z(x)dx=
-\Delta\int_0^{+\infty}(\int_0^{\lambda}e^{-sx}ds) xn_Z(x)dx\\ &=& -\Delta \int_0^{\lambda}\left(
\int_0^{+\infty}e^{-sx}xn_Z(x)dx\right)ds. \end{eqnarray*}

Hence,
$$\psi_{\Delta}(u)=\exp\left(  -\Delta \int_0^{u^2/2}  \left( \int_0^{+\infty}
    e^{-sx}g_Z(x)dx\right) ds\right).$$
Moreover, it is possible to relate the L\'evy density $n_{L}$ of $(L_t)$ with the
L\'evy density $n_{Z}$ of $(Z_t)$ as follows. Consider $f$ a non negative
function on $\mathbb R$, with $f(0)=0$. Given the whole path $(Z_t)$, the jumps $\delta
L_s=W_{Z_s}-W_{Z_{s-}}$   are   centered   Gaussian  with   variance   $\delta
Z_s$. Hence,
\begin{eqnarray*}
{\mathbb    E}(\sum_{s\le    t}f(\delta L_s))&=   &\sum_{s\le    t}{\mathbb
  E}(\int_{\mathbb R} f(u)
\exp{(-u^{2}/2 \delta Z_s)}\frac{du}{\sqrt{2 \pi \delta Z_s}})\\
&=& t \int_{\mathbb R} f(u)du \left( \int_0^{+\infty}\exp{(-u^{2}/2 x)}\frac{n_{Z}(x)dx}{\sqrt{2 \pi x}}) \right).
\end{eqnarray*}
This gives $n_{L}(u)= \int_0^{+\infty}\exp{(-u^{2}/2 x)}\frac{n_{Z}(x)dx}{\sqrt{2 \pi x}}$. By the same tools, we see that
$$
{\mathbb  E}(\sum_{s\le  t}|\delta  L_s|)= \sqrt{2/\pi}{\mathbb  E}(\sum_{s\le
  t}\sqrt{\delta Z_s})= t\int_0^{+\infty}\sqrt{x}n_Z(x) dx.
$$
Therefore,  if the above  integral is  finite, the  process $(L_t)$  has finite
variation on compact sets and it holds that $\int_{\mathbb R} |u|n_L(u) du <\infty$.

With $(Z_t)$  a L\'evy-Gamma process, $g_Z(x)=\beta e^{-\alpha x}\1_{x>0}$. Then $\int_0^{+\infty}
e^{-sx}\beta e^{-\alpha x}dx= \beta/(\alpha+s)$, and $$\psi_{\Delta}(u)=\left(\frac \alpha{\alpha
+\frac{u^2}2}\right)^{\Delta\beta}.$$
This  model is  the Variance  Gamma stochastic  volatility model  described by
Madan and Seneta~(1990). As noted in K\"{u}chler and
Tappe~(2008), the  Variance Gamma distributions are special  cases of bilateral
Gamma distributions. The condition
$\int_0^{+\infty}\sqrt{x}n_Z(x) dx<\infty$ holds. We can compute, for instance
using the
norming constant for an inverse Gaussian density,
$$
n_L(u)=   \int_0^{+\infty}\exp{(-\frac{1}{2}(\frac{u^{2}}{x}    +   2   \alpha
  x)}\frac{\beta x^{-3/2}dx}{\sqrt{2 \pi }}= \beta (2\alpha)^{1/4})|u|^{-1} \exp{(-(2\alpha)^{1/2}|u|)}
$$

\section{Statistical strategy}

\subsection{Notations}
Subsequently we denote by
$u^*$ the Fourier transform of the function $u$ defined as
$u^*(y)=\int e^{iyx}u(x)dx,$ and by $\|u\|$, $<u,v>$, $u*v$ the quantities
$$\|u\|^2=\int |u(x)|^2 dx,$$
$$ <u,v>= \int u(x)\overline{v}(x)dx\mbox{ with }
z\overline{z}=|z|^2 \mbox{ and } u\star v(x)=\int u(y)\bar v(x-y)dy.$$
Moreover, we recall that for any integrable and square-integrable
functions $u, u_1, u_2$,
\begin{equation}\label{Fou1}
(u^*)^*(x)=2\pi u(-x) \mbox{ and }  \langle u_1, u_2\rangle = (2\pi)^{-1} \langle u_1^*,
u_2^*\rangle.
\end{equation}

\subsection{The projection spaces}
As we use projection estimators, we describe now the projection spaces.
Let us define $$\varphi(x)=\frac{\sin(\pi x)}{\pi x} \; \mbox{ and } \;
\varphi_{m,j}(x) = \sqrt{m} \varphi(mx-j),$$ where $m$ is an integer, that can be taken equal to
$2^{\ell}$. It is well known (see Meyer (1990), p.22) that
$\{\varphi_{m,j}\}_{j \in \mathbb{Z}}$ is an orthonormal basis of
the space of square integrable functions having Fourier transforms
with compact support included into $[-\pi m, \pi m]$. Indeed an elementary computation yields
\begin{equation}\label{phistar}
\varphi^*_{m,j}(x)= \frac{e^{ixj/m}}{\sqrt{m}} \1_{[-\pi m, \pi m]}(x).
\end{equation}
 We denote by
$S_m$ such a space:
\begin{eqnarray*}
S_m&=& {\rm Span}\{\varphi_{_{m,j}}, \; j\in \mathbb{Z}\}\;
=\{h\in \mathbb{L}^2(\mathbb{R}),  \mbox{supp}(h^*)
\subset[-m\pi,m\pi ]\}.
\end{eqnarray*}
We denote by $(S_m)_{m\in \mathcal M_n}$ the collection of linear spaces, where
$${\mathcal M}_n=\{1, \dots, m_n\}$$ and  $m_n\leq n$ is the maximal admissible value of $m$, subject to constraints to be precised later.

In practice, we should consider the truncated spaces $S_m^{(n)} =
{\rm Span}\{\varphi_{_{m,j}}, \; \; j\in \mathbb{Z}, \;\; |j|\leq
K_n \},$ where $K_n$ is an integer depending on $n$, and the
associated estimators. Under  assumption (H6), it is possible and does not change the main part of the study (see Comte {\it et al.}~(2006)). For the  sake of simplicity, we consider here sums over ${\mathbb Z}$.

\subsection{Estimation strategy}

We want to estimate $g$ such that \begin{equation}\label{fondam}
g^*(x)=-i\frac{\psi'_{\Delta}(x)}{\Delta\psi_{\Delta}(x)}=\frac{\theta_{\Delta}(x)}{\Delta\psi_{\Delta}(x)},\end{equation}
with $$ \psi_{\Delta}(x)={\mathbb E}(e^{ixZ_1^{\Delta}}),\;\; \theta_{\Delta}(x)=-i\psi'_{\Delta}(x)={\mathbb E}(Z_1^{\Delta}e^{ixZ_1^{\Delta}}).$$

The orthogonal projection $g_m$ of $g$ on
$S_m$ is given by \begin{equation}\label{projg} g_m=\sum_{j\in {\mathbb Z}}
a_{m,j}(g) \varphi_{m,j} \mbox{ with } a_{m,j}(g) =
\int_{\mathbb R} \varphi_{m,j}(x)g(x)dx=\langle \varphi_{m,j},g\rangle.
\end{equation}

We have at hand the empirical versions of $\psi_{\Delta}$ and
$\theta_{\Delta}$: $$\hat\psi_{\Delta}(x)=\frac 1n \sum_{k =1}^n e^{ixZ_{k}^{\Delta}}, \;\;
 \hat \theta_{\Delta}(x)=\frac 1n \sum_{k=1}^n Z_k^{\Delta}e^{ixZ_k^{\Delta}}.$$

Following Neumann~(1997) and Neumann and Reiss~(2007), we truncate $1/\hat\psi_{\Delta}$ and set \begin{equation}\label{psitilde} \frac 1{\tilde
\psi_{\Delta}(x)} = \frac 1{\hat\psi_{\Delta}(x)} \1_{|\hat\psi_{\Delta}(x)|>\kappa_\psi
n^{-1/2}}.\end{equation}

Now, for $t$ belonging to a space $S_m$ of the collection $(S_m)_{m\in {\mathcal M}_n}$, let us define
\begin{equation}\label{critere} \gamma_n(t)=\frac 1n\sum_{k=1}^n\left(\|t\|^2- \frac 1{\pi\Delta}
Z_k^{\Delta}\int e^{ixZ_k^{\Delta}}\frac{t^*(-x)}{\tilde \psi_{\Delta}(x)}dx \right),
\end{equation}
Consider $\gamma_n(t)$ as an approximation of the theoretical contrast
$$
\gamma_n^{th}(t)=\frac 1n\sum_{k=1}^n\left(\|t\|^2- \frac
1{\pi\Delta} Z_k^{\Delta} \int
e^{ixZ_k^{\Delta}}\frac{t^*(-x)}{\psi_{\Delta}(x)}dx \right),$$
The following sequence of equalities, relying on (\ref{Fou1}), explains the choice of the contrast:
\begin{eqnarray*} {\mathbb E}(\frac 1{2\pi\Delta} Z_k^{\Delta}\int e^{ixZ_k^{\Delta}}\frac{t^*(-x)}{\psi_{\Delta}(x)}dx) &=&
\frac 1{2\pi\Delta} \int \theta_{\Delta}(x)\frac{t^*(-x)}{\psi_{\Delta}(x)} dx
 = \frac 1{2\pi}\langle t^*, g^*\rangle =
 \langle
t,g \rangle.\end{eqnarray*} Therefore, we find that
$\mathbb{E}(\gamma_n^{th}(t))= \|t\|^2-2\langle g, t\rangle = \|t-g\|^2 -\|g \|^2$
is minimal when $t=g$. Thus, we define the estimator belonging to $S_m$ by
\begin{equation}\label{truncsanssel} \hat g_m = {\rm Argmin}_{t\in S_m} \gamma_n(t)\end{equation}
This estimator can also be written \begin{equation}\label{explicit} \hat g_m
 = \sum_{j\in {\mathbb Z}} \hat a_{m,j} \varphi_{m,j}, \mbox{ with } \hat
a_{m,j}= \frac 1{2\pi n\Delta}\sum_{k=1}^n
Z_{k}^{\Delta}\int e^{ixZ_k^{\Delta}} \frac{\varphi_{m,j}^*(-x)}{\tilde \psi_{\Delta}(x)}dx,\end{equation}
or $$\hat a_{m,j}= \frac 1{2 \pi \Delta}\int \hat\theta_{\Delta}(x)\frac{\varphi_{m,j}^*(-x)}{\tilde \psi_{\Delta}(x)}dx.$$

\subsection{Risk bound of the collection of estimators}
First, we recall a key Lemma, borrowed from Neumann~(1997) (see his Lemma 2.1):
\begin{lem}\label{neum}
 It holds that, for any $p\geq 1$,
$${\mathbb E}\left(\left|\frac 1{\tilde\psi_{\Delta}(x)}-\frac 1{\psi_{\Delta}(x)}\right|^{2p}\right) \leq C \left( \frac 1{|\psi_{\Delta}(x)|^{2p}} \wedge
\frac{n^{-p}}{|\psi_{\Delta}(x)|^{4p}}\right),$$ where $1/\tilde\psi_{\Delta}$ is defined by (\ref{psitilde}).
\end{lem}
\noindent Neumann's result is for $p=1$ but the extension to any $p$ is straighforward. See also Neumann and Reiss~(2007).
This lemma allows to prove the following risk bound.
\begin{prop}\label{first}
Under Assumptions {\rm (H1)-(H2)}$(4)${\rm -(H3)}, then for all $m$:
\begin{equation}\label{prelimi} {\mathbb E}(\|g-\hat g_m\|^2) \leq \|g-g_m\|^2 + K\frac{{\mathbb E}^{1/2}[(Z_1^{\Delta})^4] \int_{-\pi m}^{\pi m} dx /|\psi_{\Delta}(x)|^2}{n\Delta^2}.
\end{equation}
where $K$ is a constant.
\end{prop}
\noindent It is worth stressing that (H4) is not required for the above result. Therefore, it holds even for exponential decay of $\psi_{\Delta}$. \\

\noindent {\bf Proof of Proposition \ref{first}}. First with Pythagoras Theorem, we have
\begin{equation}\label{decen2} \|g-\hat g_m\|^2=\|g-g_m\|^2 + \|\hat g_m-g_m\|^2.\end{equation}
Let $$a_{m,j}(g)= \frac 1{2\pi \Delta}\int \theta_{\Delta}(x)\frac{\varphi_{m,j}^*(-x)}{\psi_{\Delta}(x)}dx.$$ Then, using Parseval's formula and (\ref{phistar}), we obtain
$$\|\hat g_m-g_m\|^2 = \sum_{j\in {\mathbb Z}} |\hat a_{m,j}-a_{m,j}(g)|^2= \frac 1{2\pi\Delta^2}\int_{-\pi m}^{ \pi m} \left|\frac{\hat\theta_{\Delta}(x)}{\tilde \psi_{\Delta}(x)} - \frac{\theta_{\Delta}(x)}{\psi_{\Delta}(x)}\right|^2 dx.$$
It follows that
\begin{eqnarray} \nonumber {\mathbb E}(\|\hat g_m-g_m\|^2)&\leq & \frac c{\Delta^2} \left\{ \int_{-\pi m}^{\pi m} {\mathbb E} \left|\hat \theta_{\Delta}(x)\left(\frac 1{\tilde \psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right)\right|^2 dx \right. \\  \nonumber && \left. \hspace{1cm} + \int_{-\pi m}^{\pi m}
\frac{{\mathbb E}|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^2}{|\psi_{\Delta}(x)|^2} dx\right\}\\  \label{repere}
&\leq &
\frac c{\Delta^2} \left\{ \int_{-\pi m}^{\pi m} \left( {\mathbb E}(|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^2
\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^2\right) dx \right. \\  \nonumber &&\hspace{-0.5cm}  \left. + \int_{-\pi m}^{\pi m} \left(\Delta^2 |g^*(x)\psi_{\Delta}(x)|^2  {\mathbb E}\left(\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^2\right) +\frac 1n \frac{{\mathbb E}[(Z_1^{\Delta})^2]}{|\psi_{\Delta}(x)|^2} \right)dx\right\}
\end{eqnarray}

The Schwarz Inequality yields
$$ {\mathbb E}\left(|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^2
\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^2\right)\leq
 {\mathbb E}^{1/2}(|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^4){\mathbb E}^{1/2}\left(
\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^4\right).$$
Then, with the Rosenthal inequality ${\mathbb E}(|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^4)\leq c{\mathbb E}[(Z_1^{\Delta})^4]/n^2$ and by using Lemma \ref{neum},
$${\mathbb E}\left(\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^4\right)\leq \frac{C}{|\psi_{\Delta}(x)|^4}$$ so that
\begin{eqnarray*} &&\int_{-\pi m}^{\pi m} \left( {\mathbb E}^{1/2}(|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^4) {\mathbb E}^{1/2}
\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^4\right) dx\leq  \frac{c{\mathbb E}^{1/2}[(Z_1^{\Delta})^4]}{n}\int_{-\pi m}^{\pi m} \frac{dx }{|\psi_{\Delta}(x)|^2}.\end{eqnarray*}
For the second term, we use Lemma \ref{neum}, to get
$${\mathbb E}\left(\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^2\right)\leq \frac{Cn^{-1}}{|\psi_{\Delta}(x)|^4}.$$
We obtain
\begin{equation}\label{fin} {\mathbb E}(\|\hat g_m-g_m\|^2)\leq
 \frac c{n\Delta^2} ({\mathbb E}^{1/2}[(Z_1^{\Delta})^4]+\Delta^2\|g\|_1^2+ {\mathbb E}[(Z_1^{\Delta})^2])\int_{-\pi m}^{\pi m} \frac{dx}{|\psi_{\Delta}(x)|^2} dx,
\end{equation}
where $\|g\|_1=\int |g(x)|dx$.
Therefore, gathering (\ref{decen2}) and (\ref{fin}) implies the result. $\Box$

\begin{rmk}
In papers concerned with deconvolution in presence of unknown error densities, the error characteristic function is estimated using a preliminary and independent set of data. This solution is possible here: we may split the sample and use the first half to obtain a preliminary and independent estimator of $\psi_{\Delta}$, and then estimate $g$ from the second half. This would simplify the above proof, but not the study of the adaptive case.
\end{rmk}

\subsection{Discussion about the rates}
Let us study some examples and use (\ref{prelimi}) to get a relevant choice of $m$.
We have $\|g-g_m\|^2=\int_{|x|\geq \pi m} |g^*(x)|^2dx$.
Suppose that $g$ belongs to the Sobolev class $${\mathcal S}(a, L)= \{f, \int
|f^*(x)|^2(x^2+1)^{a}dx\leq L\}.$$
Then, the bias term satisfies $$\|g-g_m\|^2=O(m^{-2a}).$$
Under (H4), the bound of the variance term satisfies $$\frac{\int_{-\pi m}^{\pi m} dx/|\psi_{\Delta}(x)|^2}{n\Delta}= O\left(\frac{m^{2\beta\Delta+1}}{n\Delta}\right).$$
The optimal choice for $m$ is $O((n\Delta)^{1/(2\beta\Delta + 2a+1})$ and the resulting rate for the risk is $(n\Delta)^{-2a/(2\beta\Delta + 2a+1})$. It is worth noting that the sampling interval $\Delta$ explicitely appears in the exponent of the rate. Therefore, for positive $\beta$, the rate is worse for large $\Delta$ that for small $\Delta$.

\noindent $\bullet$ Let us  consider the example of the compound process. In this case $\beta=0$, the upper bound of the mean integrated squared error is of order $O((n\Delta)^{-2a/(2a +1)})$, if $g$ belongs to the Sobolev
class ${\mathcal S}(a, L)$.
Note that if $g$ is analytic i.e. belongs to a class
$${\mathcal A}(\gamma, Q)=\{ f, \int (e^{\gamma x}+e^{-\gamma x})^2|f^*(x)|^2 dx\leq Q\},$$ then the
risk is of order $O(\ln(n\Delta)/(n\Delta))$ (choose $m=O(\ln(n\Delta))$).\\

\noindent $\bullet$ For the Levy Gamma process, we have a more precise result since we have $$|\psi_{\Delta}(u)|=\frac{\alpha^{\beta \Delta}}{(\alpha^2+u^2)^{{\beta\Delta}/2}}, \;\; g^*(x)=\frac \beta{\alpha-ix}.$$
Therefore $\int_{|x|\geq \pi m} |g^*(x)|^2dx=O(m^{-1})$ and $\int_{[-\pi m, \pi m]} dx/|\psi_{\Delta}(x)|^2 = O(m^{2\beta\Delta+1})$. The resulting rate is of order $(n\Delta)^{-1/(2\beta\Delta+2)}$ for a choice of $m$ of order $O((n\Delta)^{1/(2\beta\Delta+2)})$.\\

\noindent $\bullet$ For the Bilateral Gamma process with $(\beta,\alpha)=(\beta',\alpha')$, we have $$\psi_{\Delta}(u)=\frac{\alpha^{\beta\Delta}}{(\alpha^2+u^2)^{\beta\Delta}}, \;\; g^*(x)=\frac \beta{\alpha^2+x^2}.$$
Therefore $\int_{|x|\geq \pi m} |g^*(x)|^2dx=O(m^{-3})$ and $\int_{[-\pi m, \pi m]} dx/|\psi_{\Delta}(x)|^2 = O(m^{4\beta\Delta+1})$. The resulting rate is of order $(n\Delta)^{-3/(4\beta\Delta+4)}$ for a choice of $m$ of order $O((n\Delta)^{1/(4\beta\Delta+4)})$.\\

These examples illustrate that the relevant choice of $m$ depends on the unknown function, in particular on its smoothness. The model selection procedure proposes a data driven criterion to select $m$.\\

\noindent $\bullet$ Consider now the process described in Section \ref{other}.
In that case, it follows from (\ref{equiv}) that  $\int_{[-\pi m, \pi m]} dx/|\psi_{\Delta}(x)|^2 =O(m^{\delta+1/2}\exp(\kappa m^{1/2-\delta}))$  and $\int_{|x|\geq \pi m} |g^*(x)|^2dx=O(m^{-2\delta})$. In this case, choosing $\kappa m^{1/2-\delta}=\ln(n\Delta)/2$ gives the rate $[\ln(n\Delta)]^{-2\delta}$ which is thus very slow, but known to be optimal in the usual deconvolution setting (see Fan~(1991)). This case is not considered in the following for the adaptative strategy since it does not satisfy (H4).

\subsection{Study of the adaptive estimator}

We have to select an adequate value of $m$. For this, we start by defining the term
\begin{equation}\label{phipsi}
\Phi_{\psi}(m)=\int_{-\pi m}^{\pi m} \frac{dx}{|\psi_{\Delta}(x)|^2},
\end{equation}
and the following theoretical penalty
\begin{equation}\label{penalite} {\rm pen}(m)= \kappa (1+ {\mathbb E}[(Z_1^{\Delta})^2]/\Delta ) \frac{\Phi_{\psi}(m)}{n\Delta}. \end{equation}
We set
$$\hat m=\arg\min_{m\in {\mathcal M}_n} \left\{ \gamma_n(\hat
g_m) + {\rm pen}(m)\right\},$$ and study first the ``risk" of $\hat g_{\hat m}$.

\noindent Moreover we need the following assumption on the collection of models ${\mathcal M}_n=\{1, \dots, m_n\}$, $m_n\leq n$:
\begin{itemize}
\item[(H7)$\;\;\;\;\;$] $\exists \varepsilon, 0<\varepsilon <1, \;\; m_n^{2\beta \Delta}\leq Cn^{1-\varepsilon}$,
\end{itemize}
where $C$ is a fixed constant and $\beta$ is defined by (H4).

For instance, Assumption (H7) is fulfilled if:
\begin{enumerate} \item pen$(m_n)\leq C$. In such a case, we have
$m_n\leq C (n\Delta)^{1/(2\beta\Delta+1)}$.
\item $\Delta$ is small enough to ensure $2\beta\Delta<1$. In such a case we can take ${\mathcal M}_n=\{1, \dots, n\}$.
\end{enumerate}

\begin{rmk} Assumption (H7) raises a problem since it depends on the unknown $\beta$ and concrete implementation requires the knowledge of $m_n$. It is worth stressing that the analogous difficulty arises in deconvolution with unknown error density (see Comte and Lacour~(2008)). In the compound Poisson model, $\beta=0$ and nothing is needed. Otherwise one should at least know if $\psi_{\Delta}$ is in a class of polynomial decay. The estimator $\hat\psi_{\Delta}$ may be used to that purpose and to provide an estimator of $\beta$ (see e.g. Diggle et
Hall~(1993)).
\end{rmk}

Let us define $$\theta_{\Delta}^{(1)}(x)= {\mathbb E}(Z_1^{\Delta}\1_{|Z_1^{\Delta}|\leq k_n\sqrt{\Delta}} e^{ixZ_1^{\Delta}}),\;\;\;\; \theta_{\Delta}^{(2)}(x)= {\mathbb E}(Z_1^{\Delta}\1_{|Z_1^{\Delta}|> k_n\sqrt{\Delta}} e^{ixZ_1^{\Delta}})$$
so that $\theta_{\Delta}=\theta_{\Delta}^{(1)}+\theta_{\Delta}^{(2)}$ and analogously
$\hat \theta_{\Delta}=\hat \theta_{\Delta}^{(1)}+\hat\theta_{\Delta}^{(2)}$.
For any two functions $t, s$ in $S_m$, the contrast $\gamma_n$ satisfies:
\begin{eqnarray} \nonumber\gamma_n(t)-\gamma_n(s)&=&\|t-g\|^2- \|s-g\|^2 -2\nu_n^{(1)}(t-s) -2\nu_n^{(2)}(t-s)
\\ \label{dec} &&  - 2\sum_{i=1}^4 R_n^{(i)}(t-s),
\end{eqnarray}  with
\begin{eqnarray*} \nu_n^{(1)}(t)&=& \frac 1{2\pi \Delta} \int t^*(-x)  \frac{\hat \theta_{\Delta}^{(1)}(x)-\theta_{\Delta}^{(1)}(x)}{\psi_{\Delta}(x)} dx,\\
\nu_n^{(2)}(t)& =&\frac 1{2\pi \Delta}\int t^*(-x) \frac{\theta_{\Delta}(x)}{[\psi_{\Delta}(x)]^2} (\psi_{\Delta}(x)-\hat \psi_{\Delta}(x))dx,\\
R_n^{(1)}(t)&= &\frac 1{2\pi \Delta}  \int t^*(-x) (\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)) \left( \frac 1{\tilde \psi_{\Delta}(x)}-\frac 1{\psi_{\Delta}(x)}\right) dx\\
R_n^{(2)}(t) &=& \frac 1{2\pi \Delta}\int t^*(-x) \frac{\theta_{\Delta}(x)}{\psi_{\Delta}(x)} (\psi_{\Delta}(x)-\hat \psi_{\Delta}(x))\left( \frac 1{\tilde \psi_{\Delta}(x)}-\frac 1{\psi_{\Delta}(x)}\right) dx,\\
R_n^{(3)}(t)&=&\frac 1{2\pi \Delta} \int t^*(-x)  \frac{\hat \theta_{\Delta}^{(2)}(x)-\theta_{\Delta}^{(2)}(x)}{\psi_{\Delta}(x)} dx,\\
R_n^{(4)}(t)&=&-\frac 1{2\pi \Delta}\int t^*(-x) \frac{\theta_{\Delta}(x)}{\psi_{\Delta}(x)} \1_{|\hat \psi_{\Delta}(x)|\leq \kappa_{\psi}/\sqrt{n}} dx.
\end{eqnarray*}

Using this decomposition and Talagrand's inequality, we can prove

\begin{thm}\label{thmpenth}
Assume that assumptions {\rm (H1)-(H2)}$(8)${\rm -(H3)-(H7)} hold.

Then
$${\mathbb E}(\|\hat g_{\hat m}-g\|^2) \leq C \inf_{m\in {\mathcal M}_n} \left( \|g-g_m\|^2 +
{\rm pen}(m)\right) + K\frac{\ln^2(n)}{n\Delta},$$
where $K$ is a constant.
\end{thm}

\begin{rmk}
Assumption (H6) is satisfied for the Levy-Gamma process. For the compound Poisson process, it is equivalent to  $\int x^4f^2(x)dx<+\infty$, where $f$ denotes the density of $Y_i$ (see Section \ref{example}).

\end{rmk}

To get an estimator, we replace the theoretical penalty by: $$\widehat{{\rm pen}}(m)=\kappa'
\left(1+ \frac 1{n\Delta^2} \sum_{i=1}^n
(Z_i^{\Delta})^2\right) \frac{\int_{-\pi m}^{\pi m}dx /|\tilde
\psi_{\Delta}(x)|^2 dx }{n}.$$
In that case we can prove:
\begin{thm}\label{thmpenhat}
Assume that assumptions {\rm (H1)-(H2)}$(8)${\rm-(H3)-(H7)} hold and let $\tilde g =\hat g_{\widehat{\widehat m}}$
be the estimator defined with $\widehat{\widehat m}=\arg\min_{m\in {\mathcal M}_n} (\gamma_n(\hat g_m)+\widehat{{\rm pen}}(m))$.
Then
$${\mathbb E}(\|\tilde g-g\|^2) \leq C \inf_{m\in {\mathcal M}_n} \left( \|g-g_m\|^2 +
{\rm pen}(m)\right) + K'_{\Delta} \frac{\ln^2(n)}{n}$$
where $K'_{\Delta}$ is a constant depending on $\Delta$ (and on fixed quantities but not on $n$).
\end{thm}
Theorem \ref{thmpenhat} shows that the adaptive estimator automatically achieves the best rate that can be hoped. If $g$ belongs to the Sobolev ball ${\mathcal S}(a,L)$, and under (H4), the rate is automatically of order $O((n\Delta)^{-2a/(2\beta \Delta+2a+1)})$. See Section 4.5.

\begin{rmk} \begin{enumerate} \item  It is possible to extend our study of the adaptive estimator to the case $\psi_{\Delta}$ having exponential decay. Note that the faster $|\psi_{\Delta}|$ decays, the more difficult it will be to estimate $g$.
\item Few results on rates of convergence are available in the literature for this problem. The results of Neumann and Reiss~(2007) are difficult to compare with ours since the point of view is different.
\end{enumerate} \end{rmk}

\section{Proofs}
\subsection{Proof of Theorem \protect \ref{thmpenth}. }

Writing that $\gamma_n(\hat g_{\hat m})+ {\rm pen}(\hat m) \leq
\gamma_n(g_m)+ {\rm pen}(m)$ in view of (\ref{dec}) implies that
\begin{eqnarray*}
\|\hat g_{\hat m}-g\|^2 &\leq & \|g_m-g\|^2 + 2\nu_n^{(1)}(\hat g_{\hat m}-g_m)
 +2\nu_n^{(2)}(\hat g_{\hat m}-g_m) + 2\sum_{i=1}^4 R_n^{(i)}(\hat g_{\hat m}-g_m)\\
  && + {\rm pen}(m)- {\rm pen}(\hat m).
\end{eqnarray*}

Let us take expectations of both sides and bound each r.h.s. term.
\begin{eqnarray*}
|{\mathbb E}(\nu_n^{(1)}(\hat g_{\hat m}-g_m))| &\leq & \frac
1{16} {\mathbb E}(\|g_m-\hat g_{\hat m}\|^2) + 16 {\mathbb
E}\left[\sup_{t\in S_m+ S_{\hat m}, \|t\|=1}
|\nu_n^{(1)}(t)|^2\right] \\ &\leq & \frac 18 {\mathbb E}(\|g-\hat
g_{\hat m}\|^2) +\frac 18 \|g- g_m\|^2\\ && + 16 {\mathbb
E}\left(\sup_{t\in S_{m\vee\hat m}, \|t\|=1}
|\nu_n^{(1)}(t)|^2-p_1(m,\hat m)\right)_+ +16 {\mathbb E}(p_1(m,\hat
m)).
\end{eqnarray*}

The same kind of bounds are obtained for $\nu_n^{(2)}$ and the residuals leading to
\begin{eqnarray}\nonumber
\frac 28 {\mathbb E}( \|\hat g_{\hat m}-g\|^2) &\leq & \frac{14}8 \|g-g_m\|^2 +
16 \sum_{m'\in {\mathcal M}_n} {\mathbb E}\left(\sup_{t\in S_{m\vee m'}, \|t\|=1} |\nu_n^{(1)}(t)|^2-p_1(m,m')\right)_+ \\ \nonumber && +
16 {\mathbb E}\left(\sup_{t\in S_{m\vee\hat m}, \|t\|=1} |\nu_n^{(2)}(t)|^2-p_2(m,\hat m)\right)_+
\\  \nonumber   && + 16\sum_{i=1}^2 {\mathbb E}\left(\sup_{t\in S_{m\vee \hat m}, \|t\|=1}|R_n^{(i)}(t)|^2-p_1(m,\hat m)\right)
\\  \nonumber  && + 16\sum_{i=3}^4 {\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}|R_n^{(i)}(t)|^2\right) \\
&& \label{etape1} +  {\rm pen}(m) + {\mathbb E}(48p_1(m,\hat m)+
16p_2(m,\hat m) - {\rm pen}(\hat m)).
\end{eqnarray}
Next, definition of pen$(.)$ comes from the following constraint:
\begin{equation}\label{penpm} 48 p_1(m,m')+16p_2(m,m')\leq {\rm pen}(m')+ {\rm
pen}(m).\end{equation} This leads to
$${\rm pen}(m) + {\mathbb E}(48p_1(m,\hat m)+ 16p_2(m,\hat m) - {\rm pen}(\hat m)) \leq 2{\rm pen}(m).$$
First, we apply Talagrand's Inequality recalled in Lemma \ref{Concent} to
prove the following result:
\begin{prop}\label{nun1} Under the assumptions of Theorem \ref{thmpenth}, define $$p_1(m,m')=(4{\mathbb E}[(Z_1^{\Delta})^2] \int_{-\pi (m\vee m')}^{\pi(m\vee m')} |\psi_{\Delta}(x)|^{-2}dx)/(\pi n \Delta^2),$$ then
\begin{equation}\label{nunu} \sum_{m'\in {\mathcal M}_n} {\mathbb E}\left(\sup_{t\in S_{m\vee m'}, \|t\|=1} |\nu_n^{(1)}(t)|^2-p_1(m,m')\right)_+\leq \frac cn.\end{equation}
\end{prop}
\noindent Next we prove:
\begin{prop}\label{nun2} Under the assumptions of Theorem \ref{thmpenth}, define $p_2(m,m')=0$ if $-a+\beta\Delta\leq 0$ and $p_2(m,m')=(\int_{-\pi (m\vee m')}^{\pi(m\vee m')} |\psi_{\Delta}(x)|^{-2}dx)/n$ otherwise. Then
\begin{equation}\label{nunu2} {\mathbb E}\left(\sup_{t\in S_{m\vee \hat m}, \|t\|=1} |\nu_n^{(2)}(t)|^2-p_2(m,\hat m)\right)_+\leq \frac cn.\end{equation}
\end{prop}
For the residual terms, two type of results can be obtained.
\begin{prop}\label{rn12}
Under the assumptions of Theorem \ref{thmpenth}, for $i=1,2$,
$${\mathbb E}\left(\sup_{t\in S_{m\vee \hat m}, \|t\|=1}[R_n^{(i)}(t)]^2-p_1(m,\hat m)\right)\leq \frac{C}{n\Delta}.$$
\end{prop}
and
\begin{prop}\label{rn34}
Under the assumptions of theorem \ref{thmpenth}, for $i=3,4$
$${\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}[R_n^{(i)}(t)]^2\right)\leq c\frac{\ln^2(n)}{n\Delta}.$$
\end{prop}

Then the choice ${\rm pen}(m)$ given by (\ref{penalite}) gives,
following (\ref{etape1}) and (\ref{penpm}),
$$
\frac 14 {\mathbb E}( \|\hat g_{\hat m}-g\|^2) \leq  \frac{7}4 \|g-g_m\|^2 + 2{\rm pen}(m)
+ C\frac{\ln^2(n)}{n\Delta},$$
which is the result. $\Box$

\subsection{Proof of Proposition \ref{nun1}.}
Let $$\omega_t(z)=\frac{z\1_{\{|z|\leq k_n\sqrt{\Delta}\}}}{2\pi \Delta} \int e^{izx}\frac{t^*(-x)}{\psi_{\Delta}(x)}dx$$
and notice that $$\nu_n^{(1)}(t)= \frac 1{n} \sum_{k=1}^n \left[\omega_t(Z_k^{\Delta})- {\mathbb E}(\omega_t(Z_k^{\Delta}))\right].$$
To apply Lemma \ref{Concent}, we compute $M_1, H_1$ and $v_1$ defined therein.
First, we have
\begin{eqnarray*}
{\mathbb E}\left(\sup_{t\in S_{m}, \|t\|=1} |\nu_n^{(1)}(t)|^2\right) &\leq &
 {\mathbb E} \left(\sum_{j\in {\mathbb Z}} |\nu_n^{(1)}(\varphi_{m,j})|^2\right)\\ & = &   {\mathbb E} \left(\frac 1{2\pi \Delta^2} \int_{-\pi m}^{\pi m} \left|\frac{\hat\theta_{\Delta}^{(1)}(x)-
\theta_{\Delta}^{(1)}(x)}{\psi_{\Delta}(x)}\right|^2 dx\right)
\\ &\leq &  \frac{{\mathbb E}[(Z_1^{\Delta})^2]}{2\pi n\Delta^2} \Phi_{\psi}(m),

\end{eqnarray*}
where $\Phi_{\psi}(m)$ is defined in (\ref{phipsi}).
We can take, for $m^\star=m\vee m'$, $$H_1^2=
\frac{{\mathbb E}[(Z_1^{\Delta})^2]}{2\pi n\Delta^2} \Phi_{\psi}(m^\star).$$

Then it is easy to see that if $\|t\|=1$ and $t\in S_{m^{\star}}$, then
$$|\omega_t(z)|\leq \frac{k_n}{2\pi \sqrt{\Delta}}\int \left|\frac{t^*(-x)}{\psi_{\Delta}(x)} \right|dx \leq
\frac{k_n}{2\pi \sqrt{\Delta}}\sqrt{\Phi_{\psi}(m^{\star})}:=M_1.$$
Lastly, for $t\in S_m, \|t\|=1$, $t=\sum_{j\in {\mathbb Z}} t_{m,j}\varphi_{m,j}$
\begin{eqnarray*}
&&{\rm Var}(\omega_t ( Z_1^{\Delta} )) \leq \frac 1{(2\pi)^2\Delta^2}\iint {\mathbb E}\left( e^{i(u-v)Z_1^{\Delta}} (Z_1^{\Delta}\1_{|Z_1^{\Delta}|\leq k_n\sqrt{\Delta}})^2\right) \frac{t^*(-u)t^*(v)}{\psi_{\Delta}(u)\psi_{\Delta}(-v)} dudv\\ &=&
\frac 1{(2\pi \Delta)^2}\sum_{j,k} t_{m,j}t_{m,k}\iint {\mathbb E}\left( e^{i(u-v)Z_1^{\Delta}} (Z_1^{\Delta}\1_{|Z_1^{\Delta}|\leq k_n\sqrt{\Delta}})^2\right) \frac{\varphi_{m,j}^*(-u)\varphi_{m,k}^*(v)}{\psi_{\Delta}(u)\psi_{\Delta}(-v)} dudv.
\end{eqnarray*}
Denoting by
\begin{equation}\label{hdelta} h_{\Delta}^*(u)={\mathbb E}[e^{iuZ_1^{\Delta}} (Z_1^{\Delta}\1_{|Z_1^{\Delta}|\leq k_n\sqrt{\Delta}})^2], \end{equation}
we obtain:
\begin{eqnarray*}
{\rm Var}(\omega_t ( Z_1^{\Delta} )) \leq
&\leq &
\frac 1{(2\pi \Delta)^2}\left(\sum_{j,k} \left|\iint h_{\Delta}^*(u-v) \frac{\varphi_{m,j}^*(-u)\varphi_{m,k}^*(v)}{\psi_{\Delta}(u)\psi_{\Delta}(-v)} dudv\right|^2\right)^{1/2}\\
&=&
\frac 1{2\pi \Delta^2}\left(\iint_{[-\pi m,\pi m]^2}  \left| \frac{h_{\Delta}^*(u-v)}{\psi_{\Delta}(u)\psi_{\Delta}(-v)}\right|^2 dudv\right)^{1/2}
\end{eqnarray*}
where the last equality follows from the Parseval equality.  Next with the Schwarz inequality and the Fubini theorem, we obtain
\begin{eqnarray*}
{\rm Var}(\omega_t ( Z_1^{\Delta} )) &\leq & \frac 1{2\pi\Delta^2}\left(\iint_{[-\pi m,\pi m]^2}   \frac{|h_{\Delta}^*(u-v)|^2}{|\psi_{\Delta}(u)|^4} dudv\right)^{1/2}
\\  &=& \frac 1{2\pi \Delta^2}\left(\int_{[-\pi m,\pi m]}  \left| \frac{du}{\psi_{\Delta}(u)}\right|^4 du\int |h_{\Delta}^*(z)|^2dz \right)^{1/2}
\\ &\leq & \frac{\sqrt{\int_{-\pi m}^{\pi m} dx/|\psi_{\Delta}(x)|^4}}{2\pi \Delta} \frac{\|h_{\Delta}^*\|}{\Delta}.\end{eqnarray*}
Now we use the following Lemma:
\begin{lem}\label{bornehstar} Under the assumptions of Theorem \ref{thmpenth},
$$\|h^*_{\Delta}\|/\Delta \leq 2\sqrt{\pi} \left(\int x^2g^2(x)dx +
{\mathbb E}[(Z_1^{\Delta})^2]\|g\|^2\right)^{1/2}:=\xi.$$
\end{lem}
Thus, under (H5), $\xi$ is finite.
We set $$v_1=  \frac{\xi\sqrt{\int_{-\pi m^\star}^{\pi m^\star} dx/|\psi_{\Delta}(x)|^4}}{2\pi\Delta} .$$
Therefore, setting $\epsilon^2=1/2$, $$p_1(m,m')= 4{\mathbb E}[(Z_1^{\Delta})^2 /\Delta]\frac{\Phi_{\psi}(m^\star)}{2\pi n\Delta}(=2(1+2\epsilon^2)H_1^2).$$
Using (H4) and the fact that ${\mathbb E}[(Z_1^{\Delta})^2/\Delta]$ is bounded, we find
\begin{eqnarray*}{\mathbb E}\left(\sup_{t\in S_{m^{\star}}, \|t\|=1} |\nu_n^{(1)}(t)|^2-p_1(m,m')\right)_+ &\leq &
C\left(\frac{(m^\star)^{2\beta\Delta+1/2}}{n\Delta} e^{-K\sqrt{m^\star}} \right. \\ && \hspace{2cm} \left.
+ \frac{k_n^2 \Phi_{\psi}(m^\star)}{n^2\Delta} e^{-K'\sqrt{n}/k_n}\right).
\end{eqnarray*}
Here $K=K(c_\psi, C_\psi)$. Moreover, we take
\begin{equation}\label{choixkn} k_n=K'\sqrt{n}/((2\beta\Delta+3)\ln(n))
\end{equation}  and we obtain
$$\sum_{m'\in {\mathcal M}_n}{\mathbb E}\left(\sup_{t\in S_{m^\star}, \|t\|=1} [\nu_n^{(1)}(t)]^2-p_1(m,m')\right)_+ \leq \frac{K"}{n\Delta}.$$

\subsection{Proof of Proposition \ref{nun2}.}
The study of $\nu_n^{(2)}$ is slightly different.
\begin{eqnarray*}
{\mathbb E}\left(\sup_{t\in S_{m}, \|t\|=1} |\nu_n^{(2)}(t)|^2\right) &\leq &
\frac{1}{2 \pi n\Delta^2}\int_{-\pi m}^{\pi m}\frac{|\theta_{\Delta}(x)|^2}{|\psi_{\Delta}(x)|^4}dx
= \frac{1}{2 \pi n}\int_{-\pi m}^{\pi m}\frac{|g^*(x)|^2}{|\psi_{\Delta}(x)|^2}dx.
\end{eqnarray*}
With assumptions (H4) and (H5), we can see that if $-a+\beta\Delta\leq 0$, then
$$\int_{-\pi m}^{\pi m}\frac{|g^*(x)|^2}{|\psi_{\Delta}(x)|^2}dx\leq \int_{-\pi m}^{\pi m} |g^*(x)|^2(1+x^2)^a
\frac{(1+x^2)^{-a+\beta\Delta}}{c_{\psi}^2}dx\leq \frac 1{c_\psi^2} \int |g^*(x)|^2(1+x^2)^a dx\leq \frac{L}{c_\psi^2}.$$
In that case, we simply take $p_2(m,m')=0$ and write
$${\mathbb E}\left(\sup_{t\in S_{m\vee \hat m}, \|t\|=1} [\nu_n^{(2)}]^2(t)\right) \leq {\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1} [\nu_n^{(2)}]^2(t)\right) \leq \frac{L}{n c_\psi^2}.$$
Now we study the case $-a+\beta\Delta>0$ and find the constants $H=H_2, v=v_2, \epsilon=\epsilon_2$ to apply Lemma  \ref{Concent}. Consider $$\tilde\omega_t(z)=(1/2\pi\Delta)\int e^{izu}t^*(-u)\{\theta_{\Delta}(u)/[\psi_{\Delta}(u)]^2\} du.$$ As
$$\int_{-\pi m}^{\pi m}\frac{|g^*(x)|^2}{|\psi_{\Delta}(x)|^2}dx\leq \frac{L}{c_\psi^2} m^{-2a+2\beta\Delta},$$
we take $$H_2^2= \frac{L}{2\pi c_\psi^2} \frac{(m^{\star})^{-2a+2\beta\Delta}}n.$$
Next, we have $$M_2=\sqrt{n}H_2$$ and we use the rough bound $v_2=nH_2^2$. Moreover, we take $\epsilon^2_2=(-2a+2\beta\Delta+2)\ln(m^{\star})/K_1$.
There exists $m_0$, such that for $m^\star\geq m_0$, $$2(1+2\epsilon_2^2)H_2^2\leq \Phi_{\psi}(m^\star)/n.$$
We set $p_2(m,m')=\Phi_{\psi}(m^\star)/n.$ Introducing
$$W_n(m,m')=\left[\sup_{t\in S_{m\vee m'}, \|t\|=1} |\nu_n^{(2)}|^2(t)-p_2(m,m')\right]_+,$$
we find that
\begin{eqnarray*}
\sum_{m'\in {\mathcal M}_n} {\mathbb E}(W_n(m,m')) &=& \sum_{m'| m^\star\leq m_0} {\mathbb E}(W_n(m,m'))
+\sum_{m'| m^\star> m_0} {\mathbb E}(W_n(m,m'))\\
&\leq & \sum_{m'| m^\star \leq m_0} [{\mathbb E}(\sup_{t\in S_{m^\star}, \|t\|=1} |\nu_n^{(2)}(t)|^2-2(1+2\epsilon_2^2) H_2^2]_+)\\ && + \sum_{m'| m^\star \leq m_0} |p_2(m,m')-2(1+2\epsilon_2^2)H_2^2| \\ & & +
 \sum_{m'| m^\star> m_0} {\mathbb E}([\sup_{t\in S_{m^\star}, \|t\|=1} |\nu_n^{(2)}(t)|^2-2(1+2\epsilon_2^2) H_2^2
]_+).\end{eqnarray*}
Therefore
\begin{eqnarray*}
\sum_{m'\in {\mathcal M}_n} {\mathbb E}(W_n(m,m')) & \leq &
2\sum_{m'\in {\mathcal M}_n}{\mathbb E}([\sup_{t\in S_{m^\star}, \|t\|=1} |\nu_n^{(2)}(t)|^2-2(1+2\epsilon_2^2) H_2^2]_+)
\\&&  + \sum_{m'| m^\star \leq m_0} |p_2(m,m')-2(1+2\epsilon_2^2)H_2^2| \\
&\leq & 2\sum_{m'\in {\mathcal M}_n}{\mathbb E}([\sup_{t\in S_{m^\star}, \|t\|=1} |\nu_n^{(2)}(t)|^2-2(1+2\epsilon_2^2) H_2^2]_+) + \frac{C(m_0)}n.\end{eqnarray*}
Talagrand's Inequality again can be then applied and gives that
$$\sum_{m'\in {\mathcal M}_n}{\mathbb E}([\sup_{t\in S_{m^\star}, \|t\|=1} |\nu_n^{(2)}(t)|^2-2(1+2\epsilon_2^2) H_2^2]_+) \leq \frac Cn.$$
The result for $\nu_n^{(2)}$ in this case follows then by saying as for $\nu_n^{(1)}$ that
$${\mathbb E}\left(W_n(m,\hat m)\right) \leq
\sum_{m'\in {\mathcal M}_n}{\mathbb E}(W_n(m,m')).$$

\subsection{Proof of Proposition \ref{rn12}.}

First define $\Omega(x)=\Omega_1(x)\cap\Omega_2(x)$ with
\begin{eqnarray*} \Omega_1(x)&=&\left\{|\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|\leq 8{\mathbb E}^{1/2}[(Z_1^{\Delta})^2] (\log^{1/2}(n) n^{-1/2}\right\}, \\ \Omega_2(x)&=&\left\{ \left|\frac 1{\tilde \psi_{\Delta}(x)}-\frac 1{\psi_{\Delta}(x)}\right|\leq 1/(\log^{1/2}(n)n^{\omega}|\psi_{\Delta}(x)|^2)\right\}.
\end{eqnarray*}
Then split: $R_n^{(1)}(t)=R_n^{(1,1)}(t)+R_n^{(1,2)}(t)$ where
$$R_n^{(1,1)}(t)=\frac 1{2\pi \Delta} \int t^*(-x)(\hat\theta_{\Delta}-\theta_{\Delta})(x)\left(\frac 1{\tilde \psi_{\Delta}(x)}-\frac 1{\psi_{\Delta}(x)}\right)\1_{\Omega(x)}dx$$ and $R_n^{(1,2)}(t)$ the integral on the complement of $\Omega(x)$.
$$
{\mathbb E}\left(\sup_{t\in S_{m\vee \hat m}, \|t\|=1}|R_n^{(1)}(t)|^2\right) \leq
2{\mathbb E}\left(\sup_{t\in S_{m\vee \hat m}, \|t\|=1}|R_n^{(1,1)}(t)|^2\right)+
2 {\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}|R_n^{(1,2)}(t)|^2\right)
$$

\begin{eqnarray*}
&& {\mathbb E}\left(\sup_{t\in S_{m\vee \hat m},
\|t\|=1}|R_n^{(1,1)}(t)|^2\right)\\ & \leq & \frac 1{2\pi \Delta^2}{\mathbb E}\left(\int_{-\pi (m\vee \hat m)}^{\pi (m\vee \hat m)}  |\hat\theta_{\Delta}(x)-\theta_{\Delta}(x)|^2\left|\frac 1{\tilde\psi_{\Delta}(x)}-\frac 1{\psi_{\Delta}(x)}\right|^2\1_{\Omega(x)} dx \right) \\
 &\leq  &
\frac{8({\mathbb E}[(Z_1^{\Delta})^2]/\Delta)}{2\pi n \Delta}{\mathbb E}\left(\int_{-\pi (m\vee \hat
m)}^{\pi (m\vee \hat m)}  n^{-2\omega}
\frac{dx}{|\psi_{\Delta}(x)|^4} \right) \leq \frac{4{\mathbb E}[(Z_1^{\Delta})^2]}{\pi \Delta}{\mathbb E}(\frac{\Phi_{\psi}(m\vee \hat m)}{n\Delta}) \leq {\mathbb E}(p_1(m,\hat m)),
\end{eqnarray*}
under the condition $-2\omega + (1-\varepsilon)\leq 0$. Therefore we choose $\omega=(1-\varepsilon)/2$. Note that if $\beta=0$ the decomposition is useless and the residual is straightforwardly negligible.

On the other hand, Lemma (\ref{neum}) yields:  $${\mathbb E}^{1/4}\left[\left|\frac 1{\tilde\psi_{\Delta}(x)} - \frac 1{\psi_{\Delta}(x)}\right|^8
\right]\leq \frac{C_{\Delta}}{n|\psi_{\Delta}(x)|^4}.$$
Now, we find
\begin{eqnarray*}
&& {\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}|R_n^{(1,2)}(t)|^2\right) \\ &\leq&
 \frac 1{2\pi \Delta^2}\int_{-\pi m_n}^{\pi m_n}  {\mathbb P}^{1/2}(\Omega(x)^c){\mathbb E}^{1/4}[(\hat\theta_{\Delta}(x)-\theta_{\Delta}(x))^8] {\mathbb E}^{1/4}\left[\left|\frac 1{\tilde\psi_{\Delta}(x)} - \frac 1{\psi_{\Delta}(x)}\right|^8\right] dx\\ &\leq & \frac{C{\mathbb E}^{1/4}[(Z_1^{\Delta})^8]}{2\pi n^2}\int_{-\pi m_n}^{\pi m_n}\frac{{\mathbb P}^{1/2}(\Omega(x)^c)}{|\psi_{\Delta}(x)|^4}dx
 \\ &\leq & \frac{C{\mathbb E}^{1/4}[(Z_1^{\Delta})^8]n^{2(1-\varepsilon)+1-b}}{n^2}\leq \frac{C'_{\Delta}}{n} \mbox{ if } {\mathbb P}(\Omega(x)^c)\leq n^{-2b} \mbox{ and } 2(1-\varepsilon)-b\leq 0.
 \end{eqnarray*}
We take $b=2(1-\varepsilon)$.
In fact,
$${\mathbb P}(\Omega(x)^c)\leq {\mathbb P}(\Omega_1(x)^c) + {\mathbb P}(\Omega_2(x)^c).$$  We use
the Markov Inequality to bound ${\mathbb P}(\Omega_2(x)^c)$:
\begin{eqnarray*}
{\mathbb P}(\Omega_2(x)^c) &\leq & \log^p(n) n^{2p\omega} |\psi_{\Delta}(x)|^{4p}{\mathbb E}
\left(\left|\frac 1{\tilde\psi_{\Delta}(x)} -\frac 1{\psi_{\Delta}(x)}\right|^{2p}\right)
\\ &\leq & \log^p(n) n^{2p\omega-p}.
\end{eqnarray*}
The choice of $p$ is thus constrained by $2p\omega-p=-p(1-2\omega)< -4(1-\varepsilon)$ that is $p>4(1-\varepsilon)/\varepsilon$, e.g. $p=5(1-\varepsilon)/\varepsilon$.

We use the decomposition of $\theta_{\Delta}(x)=\theta_{\Delta}^{(1)}(x)+\theta_{\Delta}^{(2)}(x)$ with
$$k_n\sqrt{\Delta} = \frac{\sqrt{n{\mathbb E}[(Z_1^{\Delta})^2]}}{8\sqrt{\log(n)}}.$$ We use the Bernstein Inequality to bound ${\mathbb P}(\Omega_1(x)^c)$. If $X_1, \dots, X_n$ are i.i.d. variables with variance less than $v^2$ and such that $|X_i|\leq c$, then for $S_n=\sum_{i=1}^n X_i$, we have:
$${\mathbb P}(|S_n-{\mathbb E}(S_n)|\geq n\epsilon)\leq 2 \exp\left(-\frac{n\epsilon^2/2}{v^2+c\epsilon}\right).$$ This yields
\begin{eqnarray*}
{\mathbb P}(\Omega_1(x)^c) &\leq & {\mathbb P}\left(|\hat\theta^{(1)}_{\Delta}(x)-\theta_{\Delta}^{(1)}(x)|\geq
4\sqrt{{\mathbb E}[(Z_1^{\Delta})^2]\log(n)/n}\right) \\ && + {\mathbb P}\left(|\hat\theta^{(2)}_{\Delta}(x)-\theta_{\Delta}^{(2)}(x)|\geq 4\sqrt{{\mathbb E}[(Z_1^{\Delta})^2]\log(n)/n}\right)\\ &\leq & n^{-16/3} + \frac{n}{16{\mathbb E}[(Z_1^{\Delta})^2]\log(n)} {\mathbb E}(|\hat\theta^{(2)}_{\Delta}(x)-\theta^{(2)}_{\Delta}(x)|^2)
\\ &\leq & n^{-16/3} + \frac{{\mathbb E}[(Z_1^{\Delta})^2\1_{|Z_1^{\Delta}|\geq k_n\sqrt{\Delta}}]}
{16{\mathbb E}[(Z_1^{\Delta})^2]\log(n)}  \\ &\leq &
n^{-16/3} + \frac{8^4{\mathbb E}[(Z_1^{\Delta})^6]\log^2(n)}{16{\mathbb E}^3[(Z_1^{\Delta})^2]n^2} \\
&\leq & n^{-16/3} + \frac{c}{n^2\Delta^2}.
\end{eqnarray*}
This gives the result of Proposition \ref{rn12} for $R_n^{(1)}$.
The study of $R_n^{(2)}$ follows the same line and is omitted.

\subsection{Proof of Proposition \ref{rn34}.}
First we study $R_n^{(3)}$.
\begin{eqnarray*}
{\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}|R_n^{(3)}(t)|^2\right) & \leq &
\frac 1{4\pi^2\Delta^2} {\mathbb E}\left[\sup_{t\in S_{m_n}, \|t\|=1}\left|\int (\hat\theta_{\Delta}^{(2)}(x)-\theta_{\Delta}^{(2)}(x)) \frac{t^*(-x)}{\psi_{\Delta}(x)}dx\right|^2\right] \\
&\leq &
\frac 1{2\pi\Delta^2}\int_{-\pi m_n}^{\pi m_n}  {\mathbb E}[|\hat\theta_{\Delta}^{(2)}(x)-\theta_{\Delta}^{(2)}(x)|^2]  \frac{dx}{|\psi_{\Delta}(x)|^2} \\ &=&
\frac 1{2\pi\Delta^2}\int_{-\pi m_n}^{\pi m_n}  \frac{{\rm Var}(Z_1^{\Delta}\1_{|Z_1^{\Delta}|\geq k_n\sqrt{\Delta} })}n \frac{dx}{|\psi_{\Delta}(x)|^2} \\  &\leq &
\frac{{\mathbb E}[(Z_1^{\Delta})^8]\Phi_{\psi}(m_n)}{2\pi n k_n^6 \Delta^4}
\\ & \leq &
\frac{K {\mathbb E}[(Z_1^{\Delta})^8]\ln^6(n)}{n^{2+\varepsilon} \Delta^4},
\end{eqnarray*}
using the choice of $k_n$ given by (\ref{choixkn}). \\
Next,
\begin{eqnarray*}
{\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}|R_n^{(4)}(t)|^2\right)& \leq & \frac 1{2\pi\Delta} \int_{-\pi m_n}^{\pi m_n} |g^*(x)|^2 {\mathbb P}(|\hat\psi_{\Delta}(x)|\leq \kappa_{\psi} / \sqrt{n}) dx
\leq \frac c{n\Delta}.
\end{eqnarray*}
If $|\psi_{\Delta}(x)|\geq 2\kappa_{\psi}/\sqrt{n}$, then
\begin{eqnarray*} {\mathbb P}(|\hat \psi_{\Delta}(u)|\leq \kappa_{\psi}n^{-1/2})&\leq &
{\mathbb P}(|\hat \psi_{\Delta}(u)-\psi_{\Delta}(u)|\leq |\psi_{\Delta}(u)|-\kappa_{\psi}n^{-1/2})\\
&\leq & {\mathbb P}(|\hat \psi_{\Delta}(u)-\psi_{\Delta}(u)|\geq \frac 12 |\psi_{\Delta}(u)|)\\
&\leq & \exp(-cn|\psi_{\Delta}(u)|^2)
\end{eqnarray*} for some $c>0$, where the last inequality follows from Bernstein's Inequality.

Now, it follows from (H4) that $|\psi_{\Delta}(u)|\geq c_{\psi}(1+u^2)^{-\Delta \beta/2}$. Therefore, for $|u|\leq \pi m_n$ with $m_n^{2\beta\Delta} \leq Cn^{1-\varepsilon}$ by (H7), $$|\psi_{\Delta}(u)|\geq c'm_n^{-\beta\Delta}\geq 2\kappa_{\psi}n^{-1/2}.$$

Moreover, with the previous remarks, $\exp(-cn|\psi_{\Delta}(u)|^2)\leq \exp(-cn^{\varepsilon})$ and thus
$$\int_{-\pi m_n}^{\pi m_n} |g^*(x)|^2 {\mathbb P}(|\hat\psi_{\Delta}(x)|\leq \kappa / \sqrt{n}) dx\leq \|g^*\|^2\exp(-cn^{\varepsilon}).$$
Therefore
$$ {\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1}|R_n^{(4)}(t)|^2\right)\leq \frac c{n\Delta}.$$

\subsection{Proof of Lemma \ref{bornehstar}}
Let us denote by $P_{\Delta}$ the distribution of $Z_1^{\Delta}$ and define
$\mu_{\Delta}(dz)= \Delta^{-1}z P_{\Delta}(dz)$. Let us set
$\mu(dx)=g(x)dx$. Equation (\ref{fondam}) states that
$$
\mu_{\Delta}^{*}= \mu^{*} P_{\Delta}^{*}.
$$
Hence, $\mu_{\Delta}= \mu \star P_{\Delta}$. Therefore,
$\mu_{\Delta}$ has a density given by
$$
 \int g(z-y) P_{\Delta}(dy)= \E g(z- Z_1^{\Delta}).
$$
Moreover, we have, for any compactly supported function  $t$:
\begin{equation*}
\frac{1}{\Delta} \E(Z_1^{\Delta} t(Z_1^{\Delta}))= \int t(z) \E g(z- Z_1^{\Delta}) dz= \int
\E(t(x+Z_1^{\Delta}) g(x) dx.
\end{equation*}
Hence, we apply first Parseval formula:
\begin{eqnarray*} \|h_{\Delta}^*\|^2&=& \int |h_{\Delta}^*(x)|^2dx=2\pi \int
h_{\Delta}^2(x)dx =2\pi \Delta \int z^2\1_{|z|\leq
k_n\sqrt{\Delta}}{\mathbb E}^2(g(z-Z_1^{\Delta})) dz \\ &\leq &
2\pi \Delta {\mathbb E}\left(\int z^2\1_{|z|\leq
k_n\sqrt{\Delta}}g^2(z-Z_1^{\Delta})dz \right)
\\& \leq & 2\pi \Delta {\mathbb E}\left( \int (x+Z_1^{\Delta})^2 g^2(x)dx \right) \leq
4\pi\Delta {\mathbb E}\left(\int (x^2+(Z_1^{\Delta})^2)g^2(z)dz \right) \\ &\leq & 4\pi
\Delta\left(  \int x^2g^2(x)+{\mathbb
E}[(Z_1^{\Delta})^2]\|g\|^2\right).
\end{eqnarray*}
This ends the proof. $\Box$

\subsection{Proof of Theorem \ref{thmpenhat}}

Let us define the sets $$\Omega_1=\left\{ \forall m\in {\mathcal M}_n, \int_{-\pi m}^{\pi m} \left|\frac 1{\tilde\psi_{\Delta}(x)} - \frac 1{\psi_{\Delta}(x)}\right|^2dx \leq k_1 \int_{-\pi m}^{\pi m} \frac {dx}{|\psi_{\Delta}(x)|^2}\right\}$$ and
$$ \Omega_2=\left\{ \left|\frac{\frac 1n \sum_{i=1}^n [Z_i^{\Delta }]^2}{{\mathbb E}[(Z_i^{\Delta})^2]}-1\right|\leq k_2\right\}.$$
Take $0<k_1<1/2$ and $0<k_2<1$. On $\Omega_1$, we have, $\forall m\in {\mathcal M}_n$, $$\int_{-\pi m}^{\pi m} \frac{dx}{|\tilde \psi_{\Delta}(x)|^2} \leq (2k_1+2)\int_{-\pi m}^{\pi m} \frac{dx}{|\psi_{\Delta}(x)|^2} \mbox{ and }
\int_{-\pi m}^{\pi m} \frac{dx}{|\psi_{\Delta}(x)|^2} \leq \frac 2{1-2k_1}\int_{-\pi m}^{\pi m} \frac{dx}{|\tilde \psi_{\Delta}(x)|^2}$$
and on $\Omega_2$, we find $$\frac 1n \sum_{i=1}^n [Z_i^{\Delta }]^2\leq (1+k_2){\mathbb E}[(Z_1^{\Delta})^2] \mbox{ and } {\mathbb E}[(Z_1^{\Delta})^2]\leq \frac 1{1-k_2}\frac 1n \sum_{i=1}^n [Z_i^{\Delta }]^2.$$
Il follows that, on $\Omega_1\cap\Omega_2:=\Omega_{1,2}$, we can choose $\kappa'$ large enough to ensure $$48p_1(m,\widehat{\widehat m})+16p_2(m,\widehat{\widehat m})+\widehat{{\rm pen}}(m)-\widehat{{\rm pen}}(\widehat{\widehat m}) \leq C(a,b){\rm pen}(m).$$
This allows to extend the result of Theorem \ref{thmpenth} as follows: $\forall m\in {\mathcal M}_n$, $${\mathbb E}\left( \|\tilde g-g\|^2\1_{\Omega_{1,2}}\right)\leq C\left( \|g-g_m\|^2 + {\rm pen}(m)\right) + \frac{K\ln^2(n)}{n\Delta}.$$
Next we need to prove that \begin{equation}\label{compl} {\mathbb E}\left( \|\tilde g-g\|^2\1_{\Omega_{1,2}^c}\right)\leq \frac{K'}n.\end{equation}
First, we prove that ${\mathbb P}(\Omega_{1,2}^c)\leq c/n^2$ by proving that ${\mathbb P}(\Omega_1^c)\leq c/n^2$ and
${\mathbb P}(\Omega_2^c)\leq c/n$.

\begin{eqnarray*}
{\mathbb P}((\Omega_1)^c) &\leq & \sum_{m\in {\mathcal M}_n} {\mathbb P}\left(
\int_{-\pi m}^{\pi m} \left|\frac 1{\tilde\psi_{\Delta}(x)} - \frac 1{\psi_{\Delta}(x)}\right|^2dx > k_1 \int_{-\pi m}^{\pi m} \frac {dx}{|\psi_{\Delta}(x)|^2}\right) \\ &\leq & \sum_{m\in {\mathcal M}_n} {\mathbb E}\left[\left(\frac{
\int_{-\pi m}^{\pi m} \left|\frac 1{\tilde\psi_{\Delta}(x)} - \frac 1{\psi_{\Delta}(x)}\right|^2dx}{k_1 \Phi_{\psi}(m)}\right)^p\right]\\ &\leq & \sum_{m\in {\mathcal M}_n} \frac{(2\pi m)^{p-1}}{(k_1\Phi_{\psi}(m))^p} {\mathbb E}\left( \int_{-\pi m}^{\pi m} \left| \frac 1{\tilde\psi_{\Delta}(x)} - \frac 1{\psi_{\Delta}(x)}\right|^{2p}dx\right)\\ &\leq &
\sum_{m\in {\mathcal M}_n} C_pm^{p-1}n^{-p}\frac{\int_{-\pi m}^{\pi m} dx/|\psi_{\Delta}(x)|^{4p}}{(\Phi_{\psi}(m))^p} \\ &\leq & \sum_{m\in {\mathcal M}_n} C'_p n^{-p} m^{(p-1) -p(2\beta\Delta+1) + 4p\beta\Delta+1}
= \sum_{m\in {\mathcal M}_n} C'_p m^{2p\beta\Delta}n^{-p}\\
&\leq & C" n^{1-p+p(1-\varepsilon)}\leq C"n^{1-p\varepsilon}.
\end{eqnarray*}
As $m^{2\beta\Delta+1}/(n\Delta)$ is bounded $m^{2p\beta\Delta}n^{-p}=O(n^{2p\beta\Delta/(2\beta\Delta+1) -p})=O(n^{-p/(2\beta\Delta+1)})$. Therefore, choosing $p=3/\varepsilon$ ensures that $n^{1-p\varepsilon)}=n^{-2}$ and ${\mathbb P}(\Omega_1^c)\leq C/n^2$.

On the other hand, $${\mathbb P}[\Omega_2^c]\leq \frac 1{k_2^p{\mathbb E}^p[(Z_1^{\Delta})^2]} {\mathbb E}\left(\left|\frac 1n\sum_{i=1}^n [(Z_i^{\Delta})^2-{\mathbb E}[(Z_1^{\Delta})^2]]\right|^p\right).$$
Here the choice $p=4$ gives ${\mathbb P}[\Omega_2^c]=O(1/n^2)$ with a simple variance inequality, provided that ${\mathbb E}[(Z_1^{\Delta})^8]<+\infty$.

\noindent Next, we write that $$\|g-\tilde g\|^2=\|g-g_{\widehat{\widehat m}}\|^2 + \|g_{\widehat{\widehat m}}-\hat g_{\widehat{\widehat m}}\|^2\leq \|g\|^2 + \sum_{j\in {\mathbb Z}} |\hat a_{\widehat{\widehat m},j}- a_{\widehat{\widehat m},j}(g)|^2$$ and
\begin{eqnarray*} \sum_{j\in {\mathbb Z}} |\hat a_{\widehat{\widehat m},j}- a_{\widehat{\widehat m},j}(g)|^2&=&\sum_{j\in {\mathbb Z}} |\nu_n^{(1)}(\varphi_{\widehat{\widehat m},j}) + \nu_n^{(2)}(\varphi_{\widehat{\widehat m},j}) + \sum_{k=1}^4 R_n^{(k)}(\varphi_{\widehat{\widehat m},j})|^2 \\
&\leq & C\sum_{j\in {\mathbb Z}}\{ |\nu_n^{(1)}(\varphi_{\widehat{\widehat m},j})|^2 + |\nu_n^{(2)}(\varphi_{\widehat{\widehat m},j})|^2 + \sum_{k=1}^4 |R_n^{(k)}(\varphi_{\widehat{\widehat m},j})|^2\}\\
&= & C\left\{ \sup_{t\in S_{\widehat{\widehat m}}, \|t\|=1} |\nu_n^{(1)}(t)|^2
+\sup_{t\in S_{\widehat{\widehat m}}, \|t\|=1} |\nu_n^{(2)}(t)|^2 \right.\\
&& \left. + \sum_{k=1}^4 \sup_{t\in S_{\widehat{\widehat m}}, \|t\|=1}
|R_n^{(k)}(t)|^2 \right\}
\end{eqnarray*}
It follows that, ${\mathbb E}(\|g\|^2\1_{\Omega_{1,2}^c})= \|g\|^2{\mathbb P}(\Omega_{1,2}^c)\leq c/n$, and for $k=3,4$,
$${\mathbb E}\left(\sup_{t\in S_{\widehat{\widehat m}}, \|t\|=1} |R_n^{(k)}(t)|^2\1_{\Omega_{1,2}^c}\right) \leq
{\mathbb E}\left(\sup_{t\in S_{m_n}, \|t\|=1} |R_n^{(k)}(t)|^2\right)\leq C/n$$
as it has been proved previously.
Lastly,
\begin{eqnarray*} {\mathbb E}\left(\sup_{t\in S_{\widehat{\widehat m}}, \|t\|=1} |\nu_n^{(1)}(t)|^2\1_{\Omega_{1,2}^c} \right)
&\leq &{\mathbb E}\left(\sup_{t\in S_{\widehat{\widehat m}}, \|t\|=1} \left\{|\nu_n^{(1)}(t)|^2-{\rm pen}(\widehat{\widehat m})\right\}\right)_+  \\
&& + {\mathbb E}\left({\rm pen}(\widehat{\widehat m}) \1_{\Omega_{1,2}^c} \right)\\ &\leq & c( \frac 1{n\Delta} + n{\mathbb P}(\Omega_{1,2}^c))\leq \frac{c'}n
\end{eqnarray*}
using the proof of Theorem \ref{thmpenth} and the fact that pen$(.)$ is less than $O(n)$. The same line can be followed for the other terms.

\section{Appendix}
\begin{lem}
\label{Concent} Let $Y_1, \dots, Y_n$ be independent random
variables, let $\nu_{n,Y}(f)=(1/n)\sum_{i=1}^n [f(Y_i)-{\mathbb E}(f(Y_i))]$ and
let ${\mathcal F}$ be a countable class of
uniformly bounded measurable functions. Then for $\epsilon^2>0$
\begin{eqnarray*}
\mathbb{E}\Big[\sup_{f\in {\mathcal
F}}|\nu_{n,Y}(f)|^2-2(1+2\epsilon^2)H^2\Big]_+ &\leq &  \frac
4{K_1}\left(\frac vn e^{-K_1\epsilon^2 \frac{nH^2}v} +
\frac{98M^2}{K_1n^2C^2(\epsilon^2)} e^{-\frac{2K_1
C(\epsilon^2)\epsilon}{7\sqrt{2}}\frac{nH}{M}}\right),
\end{eqnarray*}
with $C(\epsilon^2)=\sqrt{1+\epsilon^2}-1$, $K_1=1/6$, and
$$\sup_{f\in {\mathcal F}}\|f\|_{\infty}\leq M, \;\;\;\;
\mathbb{E}\Big[\sup_{f\in {\mathcal F}}|\nu_{n,Y}(f)|\Big]\leq
H, \; \sup_{f\in {\mathcal F}}\frac{1}{n}\sum_{k=1}^n{\rm
Var}(f(Y_k)) \leq v.$$\end{lem} This result follows from the
concentration inequality given in Klein and Rio~(2005) and
arguments in Birg\'e and Massart~(1998) (see the proof of their
Corollary 2 page 354). It can be extended to the case where ${\mathcal F}$ is a unit ball of a linear space.

{\small

config: arxiv


TITLE: Is it always possible to factorize a symmetric matrix $xx^T$ into $Avv^T$?
QUESTION [0 upvotes]: For a given matrix $xx^T$ and a given positive definite matrix $A$, can we always find a vector $v$ so that $xx^T = Avv^T$?

REPLY [0 votes]: Presumably, $x$ is a column vector.
If $x=0$, surely you can pick $v=0$.
If $x\ne0$, the equation $xx^T=Avv^T$ implies that both $Av$ and $v$ are nonzero vectors parallel to $x$. It follows that $x$ is an eigenvector of $A$. Conversely, if $Ax=\lambda x$, then $xx^T=Avv^T$ when $v=\frac{1}{\sqrt{\lambda}}x$.


TITLE: Prove that if g is injective, f is injective
QUESTION [0 upvotes]: $f \colon A \to \mathbb R$ be a function (where $A$ is some set) and define the function
$g \colon A \to \mathbb R$ as
$g(x) = 3 (f(x))^2 + 1.$
Prove if $g$ is injective then $f$ is injective
How do I prove the composite of the function is injective? I know for injection every $x$ value equals a y value $(x = y)$ for injection. Is this correct:
$3(f(y))^2 + 1= 3(f(x))^2 + 1 = f(y) = f(x)$

REPLY [1 votes]: You have that $g(x)$ is inyective. Then if you have $g(x)=g(y)$ then $x=y$. What does it mean on $f(x)$.
Suposse
$$f(x)=f(y) $$
then
$$g(x)=3f(x)^3+1=3f(y)^3+1=g(y)$$
which implies
$$x=y$$
Since $g$ is inyective.
in general if you have that $g=h\circ f$ is inyective then $f$ is inyective.

REPLY [0 votes]: Since g is injective, we have
$$\forall x,y \in A \,\,\, if  \,\,\ g(x)=g(y) \to x=y  \,\,\,\, (1)$$
Now assume $f(x)=f(y)$, we want to prove, in this case also $x$ equals to $y$. We have,
$$f(x)=f(y) \to (f(x))^2=(f(y))^2 \to 3(f(x))^2=3(f(y))^2 \\
\to 3(f(x))^2+1=3(f(y))^2+1 \to g(x)=g(y)$$
Therefore, using equation (1), we conclude $x=y$, so
$$f(x)=f(y) \to x=y$$
and therefore f is injective.


TITLE: Tensor Product: Identification
QUESTION [0 upvotes]: This is meant as note.
Given a measure space and a Hilbert space.
Then there's an identification:
$$\mathcal{L}^2(\mu)\hat{\otimes}\mathcal{H}\cong\mathcal{L}^2_\mathcal{H}(\mu):\quad \varphi\otimes\eta\leftrightarrow\eta\varphi$$
In particular, this yields:
$$\mathcal{L}^2(\Omega;\mu)\hat{\otimes}\mathcal{L}^2(\Omega';\mu')\cong\mathcal{L}^2(\Omega\times\Omega';\mu\otimes\mu')$$
How to prove this carefully?

REPLY [0 votes]: Please, if you find a flaw let me know.
First by Pettis' measurability criterion one has:
$$l(\eta\varphi)=l(\eta)\varphi\in\mathcal{M}(\mu)\implies\eta\varphi\in\mathcal{M}(\mu)$$
or alternatively one can check directly:
$$\sigma_n\in\mathcal{S}(\mu):\quad\sigma_n\to\varphi\implies\eta\sigma_n\to\eta\varphi$$
So the identification is well-defined as:
$$\int\|\eta\varphi\|^2\mathrm{d}\mu=\|\eta\|^2\int|\varphi|^2\mathrm{d}\mu<\infty$$
Next by monotone convergence it is an embedding:
$$\sum_{\varepsilon\delta}\kappa_\varepsilon\lambda_\delta\delta\varepsilon=0\implies0=\|\sum_{\varepsilon\delta}\kappa_\varepsilon\lambda_\delta\delta\varepsilon\|^2=\sum_{\varepsilon\delta}|\kappa_\varepsilon|^2|\lambda_\delta|^2\|\delta\|^2\int|\varepsilon|^2\mathrm{d}\mu\implies\sum_{\varepsilon\delta}\kappa_\varepsilon\lambda_\delta\varepsilon\otimes\delta=0$$
Finally characteristic functions write as:
$$X=\eta\chi\hat{=}\chi\otimes\eta$$
But simple functions are dense:
$$F\in\mathcal{L}^2_\mathcal{H}(\mu):\quad\int\|F-S_n\|^2\mathrm{d}\mu\to0\quad(S_n\in\mathcal{S}_\mathcal{H}(\mu))$$
So the image is indeed all.


TITLE: Spivak's Calculus on Manifolds - Statement of Lemma 2-10 is incorrect?
QUESTION [3 upvotes]: In Spivak's Calculus on Manifolds, there is a Lemma 2-10 that is later used to prove the Inverse Function Theorem.

Lemma 2-10 : Let $A \subset \mathbb{R}^n$ be a rectangle and let $f : A \to \mathbb{R}^n$ be continuously differentiable. If there is a number $M$ such that
  $| D_j f^i (x) | \leq M$ for all $x$ in the interior of $A$, then
  $$
|f(x)-f(y)| \leq n^2 M |x-y|
$$
  for all $x,y \in A$.

To prove this lemma, we first write
$$
f^i(y)-f^i(x) = \sum_{j=1}^n (f^i(y^1,\dots,y^j,x^{j+1},\dots,x^n) - f^i(y^1,\dots,y^{j-1},x^j,\dots,x^n)).
$$
Applying the mean-value theorem, we get
$$
f^i(y^1,\dots,y^j,x^{j+1},\dots,x^n) - f^i(y^1,\dots,y^{j-1},x^j,\dots,x^n) = (y^j - x^j) \cdot D_j f^i(z_{ij})
$$
for some $z_{ij}$. Then, Spivak uses the hypothesis in the lemma to say that the right hand side has absolute value less than or equal to $M \cdot |y^j - x^j|$.
But strictly speaking we cannot use this hypothesis yet, because it is possible that both $x$ and $y$ are boundary points such that $z_{ij}$ is also a boundary point. For example, $A \subset \mathbb{R}^2$ could be $[0,1]\times[0,1]$ and $x$ and $y$ could be $(0,0)$ and $(1,0)$, respectively. Then
$$
f^i(1,0) - f^i(0,0) = (f^i(1,0) - f^i(0,0)) + (f^i(1,0)-f^i(1,0))\\
\Rightarrow f^i(1,0) - f^i(0,0) = (1-0) \cdot D_1 f^i (z_{i1})
$$
where $z_{i1}$ lies on the $x$-axis and so is a boundary point of $A$.
So, the lemma should be stated more accurately as follows:

Lemma 2-10 : Let $A \subset \mathbb{R}^n$ be a rectangle and let $f : A \to \mathbb{R}^n$ be continuously differentiable. If there is a number $M$ such that
  $| D_j f^i (x) | \leq M$ for all $x$ in $A$, then
  $$
|f(x)-f(y)| \leq n^2 M |x-y|
$$
  for all $x,y \in A$.

The same proof as earlier should work, provided $z_{ij}$ is appropriately chosen from $A \cup \text{boundary $A$}$.
Note: Spivak has defined a function $f : A \subset \mathbb{R}^n \to \mathbb{R}^m$ to be differentiable, where $A$ is any subset, if $f$ can be extended to a differentiable function on some open set containing $A$.
I would like to know whether my reasoning is accurate. Thank you in advance.
EDIT: @rogerl has clarified my original question in the comments below. I would like to add another question though: is it necessary that $f$ be continuously differentiable? Can't it just be differentiable?
To apply the Mean Value Theorem on an interval $[a,b]$ we only require that the function be continuous on $[a,b]$ and differentiable on $(a,b)$. The proof does not seem to require continuity of the derivative.
The proof continues as follows:
$$
|f^i (y) - f^i (x)| \leq \sum_{j=1}^n |y^j - x^j | \cdot M \leq nM |y - x|
$$
since each $|y^j - x^j| < |y-x|$. Finally
$$
|f(y) - f(x)| \leq \sum_{i=1}^n |f^i (y) - f^i (x)| \leq n^2 M |y - x|.
$$

REPLY [1 votes]: Note: this is just an elaboration of the comments above by @rogerl and @MarianoSuárez-Álvarez

Since we are taking $f$ to be continuously differentiable, we can use the continuity of the partial derivatives of the component functions, $D_jf^i$, to extend the condition $$|D_jf^i(x)| \leq M\quad \forall\ x \in \operatorname{interior} A$$ to all $x \in \operatorname{boundary} A$ as well. So, there is no problem in assuming that the above condition holds only for $x$ in the interior of the rectangle $A$.
Regarding the second question, if we relax the conditions on $f$ to just differentiability on $A$ instead of continuous differentiability, then we first prove the lemma for all $x,y \in \operatorname{interior} A$. Then, we use the continuity of $f$ to extend the condition $$|f(x)-f(y)| \leq n^2 M|x-y|\quad \forall\ x,y \in \operatorname{interior} A$$
to all $x,y \in \operatorname{boundary}
A$. So, the answer is, yes, it is possible to relax the conditions on $f$; however, Spivak is not interested in minimal smoothness hypotheses and he will usually assume all the functions are smooth unless otherwise specified.


TITLE: Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$
QUESTION [3 upvotes]: Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong

REPLY [0 votes]: We can solve this problem completely, without looking at anything larger than a quartic polynomial.
Let $f(x)=\sqrt{4+\sqrt{4-x}}$.  The equation we are trying to solve is $f(f(x))=x$.
Observe that $f(x)$ is strictly decreasing from $f(0)=\sqrt6\gt0$ to $f(4)=2\lt4$, which means the equation $f(x)=x$ has a unique solution, which is relatively easy to find:
$$\begin{align}
x=\sqrt{4+\sqrt{4-x}}
&\implies x^2=4+\sqrt{4-x}\\
&\implies (x^2-4)^2=4-x\\
&\implies x^4-8x^2+x-12=0\\
&\implies(x^2-x-3)(x^2+x-4)=0
\end{align}$$
There are two roots that satisfy $0\le x\le4$:
$$x={1+\sqrt{13}\over2}\quad\text{and}\quad x={\sqrt{17}-1\over2}$$
Only the first of these satisfies the equation $x=\sqrt{4+\sqrt{4-x}}$; the other one, $x=(\sqrt{17}-1)/2$, satisfies the equation $x=\sqrt{4-\sqrt{4-x}}$.
Since $x=(1+\sqrt{13})/2$ satisfies $f(x)=x$, it also satisfies $f(f(x))=x$, so we have found $a$ solution of our equation.  To see that there are no other solutions, we can appeal to calculus.
From $f(x)=\sqrt{4+\sqrt{4-x}}$ we have
$$f'(x)={-1\over4\sqrt{4+\sqrt{4-x}}\sqrt{4-x}}$$
from which it will turn out to be convenient to note that, for $x\le3$, $|f'(x)|\le|f'(3))|={1\over4\sqrt5}\lt1$
Meanwhile, since $f(x)$ is decreasing on $0\le x\le4$, the function $F(x)=f(f(x))$ is increasing, hence
$$2\lt\sqrt{4+\sqrt{4-\sqrt6}}=f(f(0))\le f(f(x))\le f(f(4))=\sqrt{4+\sqrt2}\lt3$$
(The crude outer bounds, $2$ and $3$, are just there to simplify some of the arithmetic we're about to do.)  Thus any solution to $x=f(f(x))$ satisfies $2\lt x\lt3$, which means $f(x)\lt f(2)=\sqrt{4+\sqrt2}\lt3$.
Now suppose the equation $f(f(x))=x$ had two solutions, say $x=a$ and $x=b$ with $a\lt b$.  Then by the Mean Value Theorem applied to $F(x)=f(f(x))$, $F'(c)=1$ for some $c\in(a,b)$.  But $2\lt a\lt c\lt b\lt 3$ implies $f(c)\lt3$, and we thus see that
$$|F'(c)|=|f'(f(c))f'(c)|\lt|f'(3)|^2\lt1$$
which is a contradiction.  We thus conclude that $x={1+\sqrt{13}\over2}$ is the only solution to the equation $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$


TITLE: Why do we still use perturbation theory, when we have advanced numerical methods and fast computers?
QUESTION [16 upvotes]: If my question sounds ignorant or even insulting, I apologise. I may be completely wrong, since I'm not a theoretical physicist.
So, I understand why perturbation theory was originally used in quantum mechanics and even earlier in other fields. There were just no better ways to solve some problems (at least approximately). And it's a great method for some problems with 'small' perturbations.
But now we have fast computers and advanced numerical methods. But not only do researchers still use PT for particular problems, much of theoretical work in quantum mechanics is still based on first-order (or second-order at best) PT. Take Fermi's golden rule for example.
But PT (first or second order, which are used) does not always give a good approximation, and I even heard that its convergence has not been proven in general.
So please, tell me, is there some work being done on moving beyond PT in the theoretical body of quantum mechanics, solid state physics and other fields? Or maybe, there is some advanced PT with high-order terms better suited to computer implementations?

REPLY [4 votes]: For the same reason we still use $\sqrt{1+x}\approx (1+\frac{1}{2}x)$: because in many circumstance one is not terribly interested by a long series of decimals but rather seeks to understand more qualitative aspects of the answer.
Perturbation theory for the eigenvalues/eigenvectors is a kind of sophisticated version of a series expansion, since one obtains eigenvalues and eigenvectors as a series in the powers of the perturbation.
Much in the way that, in a very large number of cases it’s enough to use $\sqrt{2}\approx 1.41$ or even $\sqrt{2}\approx 1.4$,  there’s is rarely something fascinating with perturbation theory beyond the second order (there are exceptions).    [As an aside, interesting problems often occur when perturbation theory is not applicable.]
If your Hilbert space is large - say $10^6$ for the purpose of argument although this is not particularly big - then the computer will spit out $10^6$ eigenvectors; each of these eigenvectors has $10^6$ components, most of which will be very small (else it’s not a perturbation).   If a particular basis state enters in the exact eigenvector with probability $10^{-10}$, it’s likely this state is not terribly important to understand the properties of the eigenvector: your computer will have slaved to produce $10^6$ complicated eigenvectors, for which most of the components are of little interest.  If you change the strength of the perturbation by just a bit, you have to redo the calculation to obtain a whole new set of eigenvectors, which will not differ much from the previous calculation.
With perturbation theory, on the other hand, the perturbative parameter is explicit so you can easily assess how things will change as you change this parameter.  Basically, you don’t need to redo the calculation if you want to know $\sqrt{9.1}\approx 3.017,\sqrt{9.2}\approx 3.033$ or $\sqrt{9.3}\approx 3.050$: you can see how things change using
$$
\sqrt{9+\epsilon}\approx 3\left(1+\frac{\epsilon}{18}\right)\, .
$$
If you need a numerically more accurate answer you can always include the term in $\epsilon^2$.  It’s only if you really need accuracy that you would use a calculator.


TITLE: How to find the number of $5$-digit numbers whose digits are nondecreasing
QUESTION [3 upvotes]: I was working on a problem:

"Find the number of those $5$-digit numbers whose digits are non-decreasing."

I was able to calculate the number of decreasing $5$-digit numbers, which I found to be $252$. I subtracted this from $90,000$ five-digit numbers to obtain an answer of $89,748$. However, I am unsure whether this is correct. Can anyone confirm?

REPLY [1 votes]: The flaw in your attempt is that not every number whose digits are not strictly decreasing has nondecreasing digits.  For instance, the number $32856$ is a five-digit number with digits that are not strictly decreasing.  However, it is also a number in which the digits are not nondecreasing.
A five-digit number in which the digits are nondecreasing is completely determined by how many times each digit appears.  For instance, if $2$ appears once, $3$ appears twice, $5$ appears once, and $7$ appears once, the number must be $23357$.
Notice that if $0$ did appear in a five-digit number with nondecreasing digits, it would have to be the leading digit, making the number less than five digits.  Therefore, $0$ cannot appear in a five-digit number with nondecreasing digits.
Since the leading digit of a five-digit number cannot be zero, the number of five-digit numbers with nondecreasing digits is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 5 \tag{1}$$
in the nonnegative integers, where $x_i$ is the number of times digit $i$ appears in the number.
A particular solution of equation 1 corresponds to the placement of $9 - 1 = 8$ addition signs in a row of five ones.  For instance,
$$+ 1 + 1 1 + + 1 + + 1 + +$$
corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 2$, $x_4 = 0$, $x_5 = 1$, $x_6 = 0$, $x_7 = 1$, $x_8 = 0$, $x_9 = 0$, and the number with nondecreasing digits $23357$.
The number of solutions of equation 1 is the number of ways we can place eight addition signs in a row of five ones, which is
$$\binom{5 + 9 - 1}{9 - 1} = \binom{13}{8}$$
since we must choose which eight of the thirteen positions required for five ones and eight addition signs will be filled with addition signs.


TITLE: Question about the conditions that satisfy the Alternating Series Test.
QUESTION [1 upvotes]: The series $\sum_{n=1}^{\infty}(-1)^{n+1}u_n$ converges if all three of the following conditions are satisfied:

The $u_n$'s are all positive.
The positive $u_n$'s are (eventually) nonincreasing: $u_n\ge u_{n+1}$ for all $n\ge N$.
$u_n \to 0$

This is the definition of the Alternating Series Test from my textbook. I understand the second and third conditions but I do not really understand the significance of the first condition. Since this series is an alternating series, the signs switch from positive to negative so how can all $u_n$'s be positive? Could someone explain the significance of the first condition?

REPLY [3 votes]: You're confusing the individual terms $a_n = (-1)^{n+1}u_n$ with the $u_n$ themselves. In fact, we only have that the series is alternating because the $u_n$ are all positive: that means that the $a_n$ change sign every time! As an example for why we need positivity, let's say we picked $u_n = (-1)^{n+1} \frac 1n$. Then our series becomes $$\sum_{n=1}^\infty (-1)^{2n+2} \frac 1n = \sum_{n=1}^\infty \frac 1n$$ By not restricting the $u_n$ to be positive, our sequence (the $a_n = \frac 1n$) no longer alternates signs.