text:
\begin{document}

\title[Existence of Bernstein-Sato polynomials]{Existence of Bernstein-Sato polynomials by using the analytic Gr\"obner fan}

\author{Rouchdi BAHLOUL}
\email{rouchdi@math.kobe-u.ac.jp}
\address{Department of Mathematics\\Faculty of Science\\Kobe University\\1-1, Rokkodai,\\Nada-ku\\Kobe 657-8501\\Japan}

\keywords{$V$-filtration, analytic Gr\"obner fan, Bernstein-Sato polynomial}

\begin{abstract}
In 1987, C.~Sabbah proved the existence of Bernstein-Sato polynomials
associated with several analytic functions. The purpose of this
article is to give a more elementary and constructive proof of the
result of C.~Sabbah based on the notion of the analytic Gr\"obner fan
of a $\D$-module.

\end{abstract}

\maketitle

This paper is a translation of \cite{comp}.

\section*{Introduction and statement of the main results}\label{sec:intro}

Fix two integers $n \ge 1$ and $p\ge 1$ and $v \in \N^p \smallsetminus \{0\}$. Let $x=(x_1,\ldots,x_n)$ and $s=(s_1,\ldots,s_p)$ be two systems of variables. Consider $f_1,\ldots, f_p \in \C\{x\}=\C\{x_1,\ldots, x_n\}$ and denote by $\Dn$ the ring of differential operators with coefficients in $\C\{x\}$.
For $b(s) \in \C[s]=\C[s_1,\ldots,s_p]$, consider the following identity:
\[(\star) \qquad b(s) f^s \in \D_n[s] f^{s+v},\]
where $f^{s+v}=f_1^{s_1+v_1} \cdots f_p^{s_p+v_p}$. A polynomial
$b(s)$ satisfying such an identity is called a Bernstein-Sato polynomial (associated with $f=(f_1,\ldots,f_p)$). The set of these polynomials form an ideal called the Bernstein-Sato ideal and denoted by $\B^v(f)$. Let us mention that usually, $v$ is taken as $(1,\ldots, 1)$ or $(0, \ldots, 0, 1, 0, \ldots,0)$ where $1$ is in the $j$-th position, $j \in \{1,\ldots, p\}$.\\
Let us give some historical recalls. In the case where $p=1$ and $f$ is a polynomial, I.N. Bernstein \cite{bernstein} showed that the ideal $\B^v(f)$ is not zero (in this case, in $(\star)$, $\Dn$ is replaced with the Weyl algebra $\An(\C)$, i.e. the ring of differential operators with polynomial coefficients). Again for $p=1$ but in the analytic case, the fact that $\B^v(f)$ is not zero was proved by J.~E.~Bj\"ork \cite{bjork} with similar methods as that of \cite{bernstein}. In the same case, let us cite M.~Kashiwara \cite{kashiwara} who published another proof and showed moreover that the unitary generator of the Bernstein-Sato ideal has rational roots.
Now, for $p\ge 2$ the proof in the polynomial case is an easy generalization of that by I.~N.~Bernstein, which can be found in \cite{lichtin}. In the analytic case, the proof of the existence of a non zero Bernstein-Sato polynomial was given by C.~Sabbah (\cite{sabbah1} and \cite{sabbah2}). Let us cite the contribution of A.~Gyoja \cite{gyoja} who showed moreover that $\B^v(f)$ contains a rational non zero element.

The goal of the present paper is a development of the proof by C.~Sabbah. More precisely, we can decompose the proof of C.~Sabbah into two main steps: the first one uses arguments similar to that used by M.~Kashiwara in the case $p=1$, the second one essentially consists in a finiteness result that reduces the problem to the first step. The second step of the proof by C.~Sabbah is based on an ``adapted fan''. The existence of such a fan is done in (\cite{sabbah1} theor. A.1.1). Unfortunately, there is a gap in the proof of theorem A.1.1. (see also the comments after th.~S1 in the present paper). In this paper, we propose a more elementary and constructive statement and proof of the second step, by avoiding the sensitive notion of ``adapted fan''.

In order to motivate the reading of this article, we recall (without all the details) the proof by C.~Sabbah. In this recall, we will emphasize the result (th. S1) for which we will give a more constructive statement. Let us mention that most of the notions or notations introduced below and useful for the sequel will be detailed in the next sections.

We denote by $\Dnp$ the ring of differential operators with coefficients in $\C\{x,t\}=\C\{x_1,\ldots,x_n, t_1,\ldots,t_p\}$. Following the method of B.~Malgrange \cite{malgrange}, we make $\Dnp$ act on $\C\{x\}[\frac{1}{f_1\cdots f_p}, s] f^s$. We denote by $I$ the (left\footnote{Throughout the text, ideal shall mean left ideal}) ideal annihilator of $f^s$ in $\Dnp$ and $M$ the quotient $M=\Dnp/I$.

For $j=1,\ldots,p$, denote by $V_j(\Dnp)$ the Kashiwara-Malgrange $V$-filtration associated with the variable $t_j$ et set $V=(V_1,\ldots, V_p)$ the (multi)filtration of $\Dnp$ indexed by $\Z^p$: for $w\in \Z^p$,
\[V_w(\Dnp)=\bigcap_{j=1}^p \{V_j\}_{w_j}(\Dnp).\]
It induces a filtration $V(M)$ on $M$ where for $w \in \Z^p$, $V_w(M)$ is the image of $V_w(\Dnp)$ by the projection $\Dnp \to M=\Dnp/I$.

For $j=1,\ldots,p$, we identify the filtration $V_j$ with the linear form on $\N^{2n+2p}$ given by $V_j(\alpha,\mu, \beta,\nu)=\nu_j -\mu_j$ (where $\alpha,\beta \in \N^n$, $\mu,\nu \in \N^p$ and $\alpha, \mu, \beta, \nu$ corresponds respectively to $x, t, \ddx, \ddt$).
Denote by $\UV=\sum_{j=1}^p \R_{\ge 0} V_j$. We identify $\UV$ with $(\R_{\ge 0})^p$. Each $L$ in $\UV \cap \N^p$ (i.e. $L$ with integral coefficients) gives rise to a natural filtration $V^L$ on $\Dnp$ and $M$ indexed by $\Z^p$ given by:
\[V^L_k(M) = \sum_{\{w\in \Z^p;\, L(w)\le k\} } V_w(M),\]
where $L(w)=l_1 w_1+\cdots+l_p w_p$ if $L=(l_1,\ldots,l_p) \in \N^p$.\\
\\
Now, let us recall the two main steps of the proof by C.~Sabbah.
\begin{description}
\item[Step 1]
\begin{theo*}(\cite{sabbah1} th. 3.1.1, see also \cite{gyoja} 2.9 and
  2.10)
For any $L \in \UV \cap \N^p$, there exists a non zero polynomial $b\in \C[\lambda]$ in one variable such that for any $k\in \Z$,
\[b\big(L(-\dt{1} t_1, \ldots, -\dt{p}t_p)-k\big) V^L_k(M) \subset
V^L_{k-1}(M).\]
\end{theo*}
The proof of this theorem uses arguments similar to that used by M.~Kashiwara \cite{kashiwara} in the case $p=1$.

Denote by $b_L$ the monic polynomial with minimal degree satisfying the previous identity. The contribution of A.~Gyoja \cite{gyoja} consists in the fact that $b_L$ has roots in $\Q_{<0}$.

\item[Step 2]

We are going to introduce two other filtratrions on $M$.
\begin{enumerate}
\item
Let $\sigma$ be a convex rational cone in $\R_{\ge 0}^p$. We denote by $\L(\sigma)$ the set of the primitive elements of the $1$-skeleton of $\sigma$ (i.e. $L\in \L(\sigma)$ if and only if the line generated by $L$ is in the $1$-skeleton of $\sigma$ and the coefficients of $L$ are integral and without a common factor $\ne 1$). For any $w\in Z^p$, we set:
\[{}^\sigma V_w(M)= \sum_{\{w'\in \Z^p \, | \forall L \in \L(\sigma)
  \, L(w')\le L(w)\}} V_{w'}(M).\]
(See figure \ref{fig:Vsigma} where $p=2$ and $\sigma$ is generated by $L_1$ and $L_2$: $m\in {}^\sigma V_w(M) \iff m$ is represented by an operator $P\in \Dnp$ which has a Newton diagram included in the cross-ruling.)

\begin{figure}[h!]
\centering

\includegraphics[angle=0, width=11cm]{Vsigma.eps}
\caption{${}^\sigma V_w(M)$}\label{fig:Vsigma}

\end{figure}

For any $w\in \Z^p$, it is easy to see that we have the following inclusions:
\[{}^\sigma V_w(M) \subseteq \bigcap_{L\in \sigma} V^L_{L(w)}(M)
\subseteq \bigcap_{L\in \L(\sigma)} V^L_{L(w)}(M),\]
the first one being trivial and the second one resulting from the fact that $\L(\sigma)$ is included in the closure of $\sigma$.

\begin{theo*}[S1](\cite{sabbah1} th. A.1.1 and prop. 2.2.1)
There exists a fan $\Sigma$ (called fan adapted to $V(M)$) made of convex rational polyhedral cones such that for any cone $\sigma\in \Sigma$ and any $w\in \Z^p$, we have:
\[{}^\sigma V_w(M) = \bigcap_{L\in \L(\sigma)} V^L_{L(w)}(M).\]
\end{theo*}
This is the theorem for which we will give a more elementary statement and proof. Let us say a word about the proof. In the appendix of \cite{sabbah1}, in collaboration with F.J.~Castro Jim\'enez, C.~Sabbah states the existence of a fan $\Sigma$ adapted to the filtration $V(M)$ (th.~A.1.1) then he shows, in prop. 2.2.1, that for any cone in such a fan, we have the previous equality. The exists a gap in the proof of theorem A.1.1. Indeed, the proof is based on a division with parameters which gives rise to formal power series in the derivation variables $\dxi$.
\item
Let $\Vb(M)$ be the filtration indexed by $\Z^p$ defined by:
\[\Vb_w(M)=\bigcap_{L\in \UV} V^L_{L(w)}(M).\]
A consequence of theorem~S1 is

\begin{cor*}[S2]\
\begin{enumerate}
\item
For any $w\in \Z^p$,
\[\Vb_w(M)= \bigcap_{L\in \L(\Sigma)} V^L_{L(w)}(M),\]
where $\L(\sigma)$ denotes the set of the primitive elements of the $1$-skeleton of $\Sigma$.
\item
There exists $\kappa \in \N^p$ such that for any $w\in \Z^p$,
\[V_w(M) \subset \Vb_w(M) \subset V_{w+\kappa}(M).\]
\end{enumerate}
\end{cor*}

Let us make some comments about the proof of this result. Assertion (a) of S2 trivially follows from S1, indeed:
\begin{eqnarray*}
\Vb_w(M) & = & \bigcap_{\sigma \in \Sigma} \big(\bigcap_{L\in \sigma}
V^L_{L(w)}(M) \big)\\
 & = & \bigcap_{\sigma \in \Sigma} \big( \bigcap_{L\in \L(\sigma)}
V^L_{L(w)}(M)\big) \text{ by S1}\\
 & = & \bigcap_{L\in \L(\Sigma)} V^L_{L(w)}(M).
\end{eqnarray*}
Concerning assertion (b), C.~Sabbah proves that if a fan $\Sigma$ satisfies the satement of theorem S1 then $\Vb(M)$ is a good $V(\Dnp)$ filtration, thus by a usual result (on the comparison of good filtrations), we obtain the wanted inclusions (remark that the first inclusion is trivial and it is the second one that we are interested in).
\end{enumerate}
Let us end the recall of the proof. Denote by $\delta$ the class of $1$ in the quotient $M=\Dnp/I$. Put
\[b(s)= \prod_{L\in \L(\Sigma)} \Big(\prod_{-L(v+\kappa)<k\le 0}
b_L\big(L(s)-k\big)\Big).\]
By assertions (a) and (b) of S2, we obtain:
\[b(-\ddt t) \delta \in \Vb_{-v-\kappa}(M) \subset V_{-v}(M),\]
which means that $b(s) \in \B^v(f)$.

\end{description}

Let us now state the main results of the present paper.

Let $I$ be an ideal in $\Dm$ (ring of differential operators with analytic coefficients in $m$ variables). In \cite{acg}, A.~Assi, F.J.~Castro Jim\'enez and M.~Granger introduced  the set $\U$ of the linear forms $L$ for which the naturally associated filtration on $\Dm$ is compatible with the non commutative structure of $\Dm$ (see paragraph 1.2). They studied the behaviour of the graded ring $\gr^L(I)$ when $L$ moves in $\U$. Consider the relation on $\U$ such that $L$ is in relation with $L'$ if the graded rings $\gr^L(h(I))$ and $\gr^{L'}(h(I))$ are equal (here, $h(I)$ denotes the homogenization of $I$, we shall recall it later). This relation, which is an equivalence relation, gives rise to a partition of $\U$ made of convex rational polyhedral cones. This partition is called the (analytic) Gr\"obner fan associated with $h(I)$ et denoted by $\E(h(I))$.

Now, put $m=n+p$ and resume the previous notations. As we will see, we can naturally include $\UV$ into $\U$. Denote by $\EV=\EV(h(I))$ the fan on $\UV$ obtained by restriction of $\E(h(I))$. Here are the results we aim to prove.

\begin{theoi}\label{theo}
For any cone $\sigma$ in $\EV$,
\[{}^\sigma V_w(M) = \bigcap_{L\in \L(\sigma)} V^L_{L(w)}(M).\]
\end{theoi}
The counterpart of corollary S2 follows by replacing the $1$-skeleton of $\Sigma$ with that of $\EV$. Indeed, the proof of corollary S2 works for any fan from the moment that it satisfies the statement of theorem S1.

\begin{theoi}\label{theo2}
For $p=2$, there exists $\kappa \in \N^2$ that we can compute from the Gr\"obner bases associated with each cone of $\EV$ (this computation shall be detailed in section 3) such that for any $w\in \Z^2$,
\[\Vb_{w+\kappa}(M) \subset V_w(M).\]
\end{theoi}
For $p\ge 3$, a generalization of this result seems to give rise to technical difficulties hard to solve.

Finally, here are the main contributions of the present paper.
\begin{itemize}
\item
A constructive statement and proof of the key theorem S1 which provides a more elementary and constructive approach of step 2 in the proof by C.~Sabbah and which avoids the notion of adapted fan.
\item
For $p=2$, a completely constructive proof of step 2 in the proof of C.~Sabbah.
\end{itemize}
I point out that this article is a part of my thesis \cite{bahloul} in which one can found another proof of the counterpart (i.e. with $\L(\EV)$ instead of $\L(\Sigma)$) of statement (a) of S2, without using theorem \ref{theo}.\\
\\
In a first section, we shall make some recalls concerning the division theorem in $\Dm$ as in \cite{acg}, the notions of standard basis and Gr\"obner fan. In section 2, we shall give the proof of theorem \ref{theo} and in section 3 the proof of theorem \ref{theo2}.

\section{Recalls and preparatory results}\label{sec:rappels}

In the following paragraphs, we shall recall without giving proofs some notions and results which are helpful for the sequel.

\subsection{Homogenization}

As \cite{castro-narvaez} in the algebraic case, let us introduce the homogenized ring $\Dmz$ with which the authors of \cite{acg} introduced the analytic Gr\"obner fan.

In this paragraph and until 1.3.2, $x=(x_1,\ldots,x_m)$ and $\Dm$ denotes the ring of differential operators with coefficients in $\C\{x\}$. We define $\Dmz$ as the $\C\{x\}$-algebra generated by $\dx{1}, \ldots, \dx{m}, z$ where the only non trivial commutation relation are:
\[[\dx{i}, c(x)]=\dxsur{c(x)}{i} z \text{ for } i=1,\ldots,n \text{
 and } c(x) \in \C\{x\}.\]
In $\Dm$, we denote by $\deg(P)$ the total degree of $P$ in the $\dxi$. Consider the associated filtration. We can extend it to $\Dmz$ by considering the total degree in the $\dxi$ and $z$. This filtration provides to $\Dmz$ a structure of graded algebra:
\[\Dmz=\bigoplus_{d\in \N} \Dmz_d\quad  \text{ with }
\quad \Dmz_d=\!\!\!\!\bigoplus_{k+|\beta|=d}
\C\{x\}\ddx^\beta z^k,\]
where $\beta\in \N^m$, $\ddx^\beta=\dx{1}^{\beta_1} \cdots
\dx{m}^{\beta_m}$ and $|\beta|=\beta_1+\cdots+\beta_m$. Remark that the filtration of $\Dm$ given by $\deg$ gives rise to a Rees algebra which is isomorphic to $\Dmz$.\\
We say that an operator $P\in \Dmz$ is homogeneous (of degree $d$) if $P \in \Dmz_d$.\\
For $P\in \Dm$, we define its homogenization $h(P)\in \Dmz$ as follows.\\
Write $P=\sum_{\beta}c_\beta(x) \ddx^\beta$ and set $h(P)=\sum_\beta c_\beta(x)\ddx^\beta z^{d-|\beta|}$ where $d=\deg(P)$, thus $h(P)$ is homogeneous of degree $\deg(P)$.\\
Now, for an ideal $I$ in $\Dm$, we define $h(I)$ as the ideal of $\Dmz$ generated by the set of $h(P)$ for $P\in I$.

\subsection{Filtrations, divisions and standard bases}

We are going to recall the notions of a filtration on $\Dm$ and $\Dmz$ and the division theorem in $\Dmz$ as in \cite{acg}. We shall also recall the notion of standard basis and minimal reduced standard basis.

Let $\U$ be the set of the linear forms $L:\R^{2m} \to \R$,
$L(\alpha,\beta)=\sum_1^n e_i \alpha_i +\sum_1^n f_i \beta_i$ where for any $i=1,\ldots,n$, $e_i\le 0$ and $e_i+f_i\ge 0$. We extend $\U$ to $\R^{2m+1}$ by seting $L(\alpha,\beta,k)=L(\alpha,\beta)$.\\
For $P\in \Dmz$ (resp. $P\in \Dm$), write $P=
\sum_{\alpha,\beta,k} a_{\alpha, \beta, k} x^\alpha \ddx^\beta z^k$
(resp. with $a_{\alpha, \beta, k}=0$ for $k >0$ if $P\in \Dm$). We define the Newton diagram of $P$, denoted by $\DN(P)$ in $\N^{2m+1}$ (resp. in $\N^{2m}$), as the set of $(\alpha,\beta,k) \in \N^{2m+1}$ for which $a_{\alpha,\beta,k} \ne
0$.\\
Given $L\in \U$ and $P\in \Dmz$ (or $P\in \Dm$), we define $L$-order of $P$ as $\ord^L(P)=\max L(\DN(P))$. This order induces a filtration $F^L$ on $\Dmz$ or $\Dm$ indexed by $L(\N^{2m+1})$ given by:
\[F^L_k(\Dmz)=\{P\in \Dmz; \, \ord^L(P) \le k\}\]
and an associated graded ring $\dps\gr^L(\Dmz)= \bigoplus_{k\in
L(\N^{2m+1})} F^L_k(\Dmz)/F^L_{<k}(\Dmz)$.\\
For $P\in \Dmz$, we denote by $\sigma^L(P)$ the principal symbol of
$P$, i.e. the class of $P$ in the quotient $F^L_k(\Dmz)/F^L_{<k}(\Dmz)$ where
$k=\ord^L(P)$. If $J$ is an ideal in $\Dmz$, a filtration $F^L(J)$ is induced and gives rise to a graded ideal $\gr^L(J)$ of $\gr^L(\Dmz)$, which is generated by the set of the $\sigma^L(P)$ for $P\in J$.\\
\\
For a form $L\in \U$, we define two orders: $<_L$ on $\N^{2m}$ and $<_L^h$ on $\N^{2m+1}$:
\[(\alpha,\beta) <_L (\alpha',\beta') \iff
\begin{cases}
L(\alpha,\beta) < L(\alpha',\beta') \\
\text{or } \big(  \text{ equality and } |\beta|<|\beta'| \big) \\
\text{or } \big( \text{ equality and } (\alpha,\beta) >_0
(\alpha',\beta') \big),
\end{cases}\]
where $<_0$ is a well order compatible with sums, which is fixed for the sequel;

\[(\alpha,\beta,k) <_L^h (\alpha',\beta',k') \iff
\begin{cases}
k+|\beta|<k'+|\beta'| \\
\text{or } \big( \text{ equality and } (\alpha,\beta) <_L
(\alpha',\beta') \big).
\end{cases}\]

For $P\in \Dmz$, we define the leading exponent $\exp_{<_L^h}(P)= \max _{<_L^h}(\DN(P))$ and the leading monomial $\mp_{<_L^h}(P)=(x,\ddx,z)^{\exp_{<_L^h}(P)}$. We do the same for $P\in \Dm$ and with $<_L$ instead of $<_L^h$. Note that $\exp_\prec(PQ)=\exp_\prec(P) +\exp_\prec(Q)$ if $\prec$ is compatible with sums, which is the case for $<_L$ and $<_L^h$. Let us recall the division theorem in $\Dmz$ given in \cite{acg}.

Let $L\in\U$. Let $Q_1, \ldots, Q_r$ be a family of operators in $\Dm$. Denote by $(\Delta_1,\ldots, \Delta_r,\bar{\Delta})$ the partition of $\N^{2m+1}$ defined from the $\exp_{<_L^h}(Q_j)$:
\begin{itemize}
\item $\Delta_1=\exp_{<_L^h}(Q_1)+\N^{2m+1}$
\item $\Delta_j=(\exp_{<_L^h}(Q_j)+\N^{2m+1}) \smallsetminus
  (\bigcup_{i=1}^{i=j-1} \Delta_i)$ for $j=2,\ldots,r$
\item $\bar{\Delta}=\N^{2m+1} \smallsetminus (\bigcup_{j=1}^{j=r}
  \Delta_j)$
\end{itemize}
\begin{theo} (\cite{acg} Th. 7)\label{theo:div}
For any $P\in \Dmz$, there exists a unique $(q_1,\ldots,q_r,\\ R) \in
(\Dmz)^{r+1}$ such that:
\begin{enumerate}
\item $P=q_1 Q_1 + \cdots + q_r Q_r+R$
\item for any $j=1,\ldots,r$, if $q_j \ne 0$ then $\DN(q_j)+\exp_{<_L^h}(Q_j) \subset \Delta_j$
\item if $R \ne 0$ then $\DN(R) \subset \bar{\Delta}$.
\end{enumerate}
We call $R$ the remainder of the division of $P$ by the $Q_j$ w.r.t. $<_L^h$.
\end{theo}

\begin{cor}\
\begin{itemize}
\item
$\exp_{<_L^h}(P)=\max_{<_L^h}\{\exp_{<_L^h}(q_jQ_j),\, j=1,\ldots,r;
\, \exp_{<_L^h}(R)\}$.
\item
$\ord^L(P)=\max\{\ord^L(q_jQ_j),\, j=1,\ldots,r;\, \ord^L(R)\}$.
\end{itemize}
\end{cor}
Let $J$ be an ideal in $\Dmz$ and $Q_1,\ldots,Q_r \in J$. We say that $Q_1,\ldots,Q_r$ form a $<_L^h$-standard bases of $J$ if for any $P$ in $J$, the remainder of the division of $P$ by the $Q_j$ is zero. Consider the set of exponents of $J$: $\Exp_{<_L^h}(J)=\{\exp_{<_L^h}(P), \, P\in J\smallsetminus 0\}$. Given $Q_1,\ldots, Q_r \in J$, these two claims are equivalent (by the division theorem):
\begin{itemize}
\item The $Q_j$ form a $<_L^h$-standard basis of $J$.
\item $\Exp_{<_L^h}(J)=\bigcup_{j=1}^r (\exp_{<_L^h}(Q_j)+\N^{2m+1})$.
\end{itemize}
The existence of a standard basis holds by Dickson lemma which asserts that if a subset $E$ of $\N^q$ satisfies $E=E+\N^q$ (which is the case for $\Exp_{<_L^h}(J)$) then there exists $F\subset E$ finite such that $E=\cup_{e\in F} (e+\N^q)$.
It is easy to see that a standard basis is not unique in general, that is why there exists the notion of minimal reduced standard basis:
\begin{defin*}
Let $Q_1,\ldots,Q_r$ be a $<_L^h$-standard basis of $J\subset \Dmz$ and let $e_j=\exp_{<_L^h}(Q_j)$ for $j=1,\ldots,r$.
\begin{itemize}
\item
We says that it is minimal if for any finite subset $F$ of $\N^{2m+1}$, the following implication holds
\[\Exp_{<_L^h}(J)=\bigcup_{e\in F}(e+\N^{2m+1}) \Rightarrow
\{e_1,\ldots,e_r\} \subseteq F.\]
\item
We say that it is reduced if the $Q_j$ are unitary (i.e. the coefficient corresponding to the leading monomial is $1$) and if for any $j$,
\[(\DN(Q_j) \smallsetminus e_j) \subset (\N^{2m+1} \smallsetminus
\Exp_{<_L^h}(J)).\]
\end{itemize}
There exists a unique $<_L^h$-minimal reduced standard basis of an ideal in $\Dmz$.
\end{defin*}
Remark that if the ideal $J$ is homogemeous then the minimal reduced standard basis will be made of homogeneous elements. Let us end this paragraph by a result that we will apply in the next section.

\begin{lem}\label{lem:utile}
Let $<_1$ and $<_2$ be two orders on $\N^{2m+1}$ which allow one to make divisions in $\Dmz$ (for example $<_i=<_{L_i}^h$ with $L_1$ and $L_2$ being two forms). Let $\{Q_1,\ldots,Q_r\}$ be the minimal reduced standard basis of an ideal $J \subset \Dmz$ w.r.t. $<_1$. Suppose that for any $j$, $\exp_{<_1}(Q_j)$ and
$\exp_{<_2}(Q_j)$ are equal. Then $\{Q_1,\ldots,Q_r\}$ is also the minimal reduced standard basis of $J$ w.r.t. $<_2$.
\end{lem}
We omit the proof. Let us just say that what is important in a division is the set of the $\exp_<(Q_j)$ and not the order itself. Thus a division w.r.t. $<_1$ or to $<_2$ will give the same quotients and the same remainder.

\subsection{Gr\"obner fan}

In this paragraph, we shall recall the main result of \cite{acg} which describes the analytic Gr\"obner fan. We shall also introduce the $V$-Gr\"obner fan $\EV(h(I))$ which will be the main object of sections $2$ and $3$.

\subsubsection{}

Let $I$ be an ideal in $\Dm$. Consider $h(I)\subset \Dmz$ its homogenization. For $L$ and $L'$ in $\U$, we define the relation:
\[L \sim L' \iff \gr^L(\Dmz) = \gr^{L'}(\Dmz) \text{ and }
\gr^L(h(I)) = \gr^{L'}(h(I)).\]
This an equivalence relation on $\U$.

\begin{theo}\label{theo:event_Grob}\cite{acg}
The partition of $\U$ given by this relation is made of convex polyhedral rational cones. This partition denoted $\E(h(I))$ is called the (analytic) Gr\"obner fan of $h(I)$.\\
Moreover for any cone $\sigma \in \E(h(I))$, there exists
$Q_1,\ldots,Q_r \in h(I)$ homogeneous such that:
\begin{itemize}
\item
for any $L,L'$ in $\sigma$, $\sigma^L(Q_j)=\sigma^{L'}(Q_j)$ and $\exp_{<_L^h}(Q_j)=\exp_{<_{L'}^h}(Q_j)$ for any $j$.
\item
for any $L\in \sigma$, the set $\{Q_1,\ldots,Q_r\}$ is the minimal reduced standard basis of $h(I)$ w.r.t. $<_L^h$.
\end{itemize}
\end{theo}
By the second claim, we can see that on a cone $\sigma$, the set $\Exp_{<_L^h}(h(I))$ of exponents of $h(I)$ w.r.t. $<_L^h$ is constant when $L$ runs over $\sigma$.\\

\subsubsection{}

From now on, we shall work in $\Dnp$ and $\Dnpz$, that is $m=n+p$. For $j=1,\ldots,p$, we denote by $V_j \in \U$ the linear form given by: $V_j(\alpha,\mu,\beta,\nu)=\nu_j -\mu_j$ where $\alpha,\beta \in \N^n$ and $\mu,\nu\in \N^p$. This form gives rise to a filtration that we also denote by $V_j$ and which is nothing but the Kashiwara-Malgrange $V$-filtration associated with the variable $t_j$ (recall that in $\Dnp$, the variables are $x=(x_1,\ldots,x_n)$ and $t=(t_1,\ldots, t_p)$). We denote by $V$ the multifiltration $V=(V_1,\ldots,V_p)$.\\
We set $\UV\subset \U$ to be the subset of the linear forms $L$ of the form:
\[L=l_1 V_1+\cdots+l_p V_p,\]
with $(l_1,\ldots,l_p)\in (\R_{\ge 0})^p$, thus we shall identify $\UV$ and $(\R_{\ge 0})^p$.

From now on, for $L\in \UV$, we shall denote by $V^L$ the filtration associated with $L$ (following the notations of \cite{sabbah1}). Remark that for $L\in \UV$, we can identify $\gr^L(\Dnpz)$ to a subring of $\Dnp$. Indeed, after reordering the variables, assume that $L=l_1 V_1+\cdots +l_e V_e$ with $0\le e \le p$ and none of the $l_j$ is zero (by convention, if $e=0$ then $L=0$ and $\gr^L(\Dnpz)= \Dnpz$). In this case, we have
\[\gr^L(\Dnpz)= \C\{x_{e+1}, \ldots, x_p\}[x_1, \ldots, x_e][\dx{1}, \ldots, \dx{n}]\langle z\rangle,\]
in which the commutation relations coincides with those of $\Dnpz$. Thus we will consider that all the calculations are made in $\Dnpz$. Now, consider the restriction of $\E(h(I))$ to $\UV$. We obtain the $V$-Gr\"obner fan of $h(I)$, denoted $\EV(h(I))$. Similarly to \ref{theo:event_Grob}, we obtain:

\begin{cor}\label{cor:V_event_Grob}
For any cone $\sigma$ of $\EV(h(I))$, there exists $Q_1,\ldots,Q_r \in h(I)$ homogeneous such that:
\begin{itemize}
\item for any $L,L' \in \sigma$, $\sigma^L(Q_j)=\sigma^{L'}(Q_j)$
  and $\exp_{<_L^h}(Q_j)=\exp_{<_{L'}^h}(Q_j)$ for any $j$.
\item for any $L\in \sigma$, the set $\{Q_1,\ldots,Q_r\}$ is the minimal reduced standard basis of $h(I)$ w.r.t. $<_L^h$.
\end{itemize}
\end{cor}
We see that for any cone $\sigma$, there exists a set of $Q_j$ which, for any $L$ in $\sigma$, is the $<_L^h$-minimal reduced Gr\"obner basis of $h(I)$. We shall call this set {\bf the standard basis of $h(I)$ associated with $\sigma$}.

Let us end this paragraph with some remarks and notations.
A cone $\sigma$ of $\EV(h(I))$ is not necessarily open (for example, it may be ``semi open''). Thus here is the definition of $\L(\sigma)$:\\
We first consider the closure $\bar{\sigma}$ of $\sigma$.
There exists $L_1, \ldots,L_q \in \UV$ that we assume to be primitive (with $q\ge 1$ that may be greater than $p$) such that
\[\bar{\sigma}=\{L=r_1 L_1+\cdots r_q L_q; \, r_i \ge 0\}.\]
Assume $q$ to be minimal then $\L(\sigma)$ is the set $\{L_1,\ldots,L_q\}$. The set $\L(\EV(h(I)))$ is then nothing but the union of the $\L(\sigma)$ with $\sigma \in \EV(h(I))$ (remark that it is made of integral elements since the cones $\sigma$ are rational).

We define the interior of $\sigma$ as the set of the (strictly) positive combinations of the $L_i$. We denote by $\rangle L_1,\ldots, L_q \langle$ the open cone generated by the $L_i$. Moreover for $L_1$ and $L_2$ in $\UV$, we denote by $\rangle L_1,L_2 \rangle=\{r_1L_1 +r_2L_2; \, r_1>0, \, r_2\ge 0\}$ the semi open cone that contains $L_2$ and not $L_1$.

In the sequel, we shall write $\EV$ instead of $\EV(h(I))$.

\section{Proof of theorem \ref{theo}}\label{sec:demo1}

Let $\sigma$ be a cone of $\EV$. We have seen in Corollary \ref{cor:V_event_Grob} that there exists a family $Q_1,\ldots,Q_r$ in $h(I)$ which is the minimal reduced standard basis of $h(I)$ w.r.t. $<_L^h$, for any $L\in \sigma$. Now what
happens for a linear form $L$ in the closure of $\sigma$ but not in $\sigma$ (which may hold for an $L$ in $\L(\sigma)$)? Of course the $Q_j$ are not necessarily a standard basis of $h(I)$ w.r.t. $<_L^h$. However, it is possible, and this is the purpose of the following proposition, to construct an order that we will denote by $\lhd_L^\sigma$ (because it will depend on $L$ and $\sigma$) for which the $Q_j$ shall be the minimal reduced standard basis of $h(I)$. Here is the definition of the order in question.\\
First we fix a linear form $L_\sigma$ in the interior of $\sigma$, then for $(\alpha,\mu,\beta,\nu,k)$ and $(\alpha',\mu',\beta',\nu',k')$ in $\N^{n+p+n+p+1}$ we set:
\[(\alpha,\mu,\beta,\nu,k) \lhd_L^\sigma (\alpha',\mu',\beta',\nu',k)
\iff\]
\[\begin{cases}
k+|\beta|+|\nu|<k'+|\beta'|+|\nu'| \\
\text{or } \big( = \text{ and } L(\alpha,\mu,\beta,\nu)<
L(\alpha',\mu',\beta',\nu') \big) \\
\text{or } \big( = \text{ and } = \text{ and } (\alpha,\mu,\beta,\nu)
<_{L_\sigma} (\alpha',\mu',\beta',\nu') \big).
\end{cases}\]

\begin{prop}
Let $L \in \UV$ be in the closure of $\sigma \in \EV$ and let $Q_1,\ldots,Q_r$ be the standard basis of $h(I)$ associated with $\sigma$. Then for any $j=1,\ldots,r$ and for any $L'$ in $\rangle L, L_\sigma \rangle$,
\[\exp_{<_{L'}^h}(Q_j)=\exp_{\lhd_L^\sigma}(Q_j).\]
\end{prop}
As a {\bf consequence} of this proposition, we obtain that {\bf the $Q_j$ form the minimal reduced standard basis of $h(I)$ w.r.t. $\lhd_L^\sigma$}. Indeed, it suffices to apply lemma \ref{lem:utile}

\begin{proof}
For the reader's convenience, figure \ref{fig:Qj_sans_pentes} illustrates the situation of the proposition (with $p=2$, but for higher dimension, we could draw a similar figure by intersecting $\UV$ with the plan generated by $L$ and $L_\sigma$).

\begin{figure}[h!]
\centering

\includegraphics[angle=0, width=11cm]{diag_Qj.eps}
\caption{Newton diagram of a $Q_j$ associated with $\sigma=\rangle
  L_2, L_1 \rangle$}\label{fig:Qj_sans_pentes}

\end{figure}

\begin{eqnarray*}
\exp_{\lhd_L^\sigma}(Q_j) & = & \exp_{\lhd_L^\sigma}(\sigma^L(Q_j)) \text{ by definition of } \lhd_L^\sigma \text{ and by homogeneity of } Q_j\\
 & = & \exp_{<_{L_\sigma}^h}(\sigma^L(Q_j)) \text{ by definition of
   } \lhd_L^\sigma \text{ et } <_{L_\sigma}^h\\
 & = & \exp_{<_{L_\sigma}^h}(Q_j) \text{ by the following claim}\\
 & = & \exp_{<_{L'}^h}(Q_j) \text{ by corollary \ref{cor:V_event_Grob}}.
\end{eqnarray*}
To end the proof, it remains to proof the following equality.
\begin{aff*}
\[\exp_{<_{L_\sigma}^h}(\sigma^L(Q_j)) = \exp_{<_{L_\sigma}^h}(Q_j).\]
\end{aff*}
Since $\DN(\sigma^L(Q_j)) \subseteq \DN(Q_j)$, we have $\dps \exp_{<_{L_\sigma}^h}(\sigma^L(Q_j)) \le_{L_\sigma}^h \exp_{<_{L_\sigma}^h}(Q_j)$. Then, the reverse inequality suffices to prove the claim.
First let us prove the following:
\[(\star) \qquad \ord^L(Q_j)= L(\exp_{<_{L_\sigma}^h}(Q_j)).\]
For any $L''$ in $\rangle L, L_\sigma \rangle$, we have
\[\ord^{L''}(Q_j)= L''(\exp_{<_{L''}^h}(Q_j))=L''(\exp_{<_{L_\sigma}^h}(Q_j))\]
by definition of $<_{L''}^h$ and by corollary \ref{cor:V_event_Grob}. Thus for any fixed $m$ in $\DN(Q_j)$,
\[L''(\exp_{<_{L_\sigma}^h}(Q_j)) \ge L''(m).\]
Write $L''=(1-\varepsilon) L + \varepsilon L_\sigma$ (with $1 \ge \varepsilon >0$) and make $\varepsilon$ converge to $0$. By continuity, we obtain: $\dps L(\exp_{<_{L_\sigma}^h}(Q_j)) \ge L(m)$, the later holding for any $m$ in $\DN(Q_j)$. Thus the equality $(\star)$ is proved.
As a consequence and by definition of $\sigma^L(Q_j)$, we have
\[\exp_{<_{L_\sigma}^h}(Q_j) \in \DN(\sigma^L(Q_j)).\]
The wanted inequality then follows from the definition of $\exp_{<_{L_\sigma}^h}(\sigma^L(Q_j))$. The claim is proven.
\end{proof}

Recall that $I$ is an ideal in $\Dnp$, $M=\Dnp/I$ and $\delta$ is the class of $1$ in $M$.
Let $\sigma$ be a cone in $\EV$. Denote by $L_1,\ldots,L_q$ the elements of $\mathcal{L}(\sigma)$.
\begin{lem}\label{lem:cle}
Let $i_0 \in \{1,\ldots,q\}$, $m \in V^{L_{i_0}}_{\lambda_{i_0}}(M)$
with $\lambda_{i_0} \in \Q$ and $P \in \Dnp$ such that $P \delta=m$ and
$\ord^{L_{i_0}}(P)>\lambda_{i_0}$ then there exists $P' \in \Dnp$ such that
\begin{itemize}
\item $P-P' \in I$ i.e. $P' \delta=m$,
\item $\ord^{L_{i_0}}(P')<\ord^{L_{i_0}}(P)$,
\item $\ord^{L_i}(P') \le \ord^{L_i}(P)$ for $i\in \{1,\ldots,q\}
  \smallsetminus \{i_0\}$.
\end{itemize}
\end{lem}
In other terms, it is possible to decrease the order w.r.t. one $L_i$ without increasing the order w.r.t. the others $L_i$. Thanks to this lemma we can give a

\begin{proof}[Proof of theorem \ref{theo}]
Let $m \in \bigcap_{L \in \mathcal{L}(\sigma)}
V^L_{L(v)}(M)$ then for $i=1,\ldots,q$, there exists $P_i \in
\Dnp$ such that $P_i \delta=m$ and $\ord^{L_i}(P_i) \le L_i(v)$.
Set $\tilde{P}_1=P_1$. By applying the lemma with $i_0=1$ a finite number of times (the first time with $P=\tilde{P}_1$ and $\lambda_{i_0}=\ord^{L_{i_0}}(\tilde{P}_1)$), we construct $\tilde{P}_2$ such that $\tilde{P}_2\delta=m$ and
$\ord^{L_i}(\tilde{P}_2) \le L_i(v)$ for $i=1,2$. We start again the process with $i_0=3$ and $P=\tilde{P}_2$, etc. After a finite number of steps, we obtain $\tilde{P}_q \in \Dnp$ such that $\tilde{P}_q \delta=m$ and for any $i=1,\ldots,q$, $\ord^{L_i}(\tilde{P}_q) \le L_i(v)$. Finally we can conclude that $m \in
{}^{\sigma}V_v(M)$.
\end{proof}

To end this section, it remains to prove the previous lemma.
\begin{proof}[Proof of lemma \ref{lem:cle}]
To simplify the notations, we do the prove with $i_0=1$ and we set $\lambda=\lambda_{i_0}$ and $\lhd_{L_i}= \lhd_{L_i}^\sigma$. Denote by $Q_1,\ldots,Q_r$ the standard basis of $h(I)$ associated with the cone $\sigma$.

By hypothesis, there exists $P_1 \in \Dnp$ with $P_1 \delta=m$ and $\ord^{L_1}(P_1)\le \lambda$. There exists $l_0,l,l_1 \in \N$ such that $z^{l_0}h(P-P_1)=z^lh(P)-z^{l_1}h(P_1)$. We then set $H=z^lh(P)$, $H_1=z^{l_1}h(P_1)$ and $H_0=H-H_1$ and since $P-P_1$ is in $I$, we have $H_0 \in h(I)$.\\

Consider the division of $H_0$ by the $Q_j$ w.r.t. $\lhd_{L_1}$:
\[H_0=\sum_{j=1}^r q_j Q_j \text{ with }
\mathcal{N}(q_j)+\exp_{\lhd_{L_1}}(Q_j) \subset \Delta_j \text{ for any } j\]
where the $\Delta_j \subset \N^{2n+2p+1}$ form the partition of $\Exp_{\lhd_{L_1}}(h(I))$ associated with the leading exponents of the $Q_j$ (see the division theorem \ref{theo:div}).\\
Since for any $i,j$, the exponents $\exp_{\lhd_{L_1}}(Q_j)$ and $\exp_{\lhd_{L_i}}(Q_j)$ are equal, the previous division is also a division w.r.t. the orders $\lhd_{L_2}, \ldots, \lhd_{L_q}$. As a consequence, for any $i=1,\ldots,q$ and $j=1,\ldots,r$,
\[\ord^{L_i}(H_0) \ge \ord^{L_i}(q_j Q_j).\]
Denote by $J$ the set of the $j\in \{1,\ldots,r\}$ for which $\ord^{L_1}(H_0)=\ord^{L_1}(q_j Q_j)$, then we have
\[\sigma^{L_1}(H_0)=\sigma^{L_1}(H)= \sum_{j\in J} \sigma^{L_1}(q_j)
\sigma^{L_1}(Q_j).\]
Let us consider and denote $\dps W=\sum_{j\in J}
\sigma^{L_1}(q_j)Q_j$. It is an element of $h(I)$.\\
Put $H'=H-W$. We are going to prove the following claims
\begin{enumerate}
\item[(i)] $\ord^{L_1}(H') < \ord^{L_1}(H),$
\item[(ii)] $\ord^{L_i}(H') \le \ord^{L_i}(H)$ for $i=2,\ldots,q$.
\end{enumerate}

\begin{enumerate}
\item[(i)]
Clearly $\sigma^{L_1}(H)=\sigma^{L_1}(W)$. Therefore,
\[H'=(H-\sigma^{L_1}(H)) - (W -\sigma^{L_1}(W)).\]
We can easily see that the two terms in brackets have an $L_1$-order strictly less than that of $H$.

\item[(ii)]
Fix $i$ between $2$ and $q$.\\
By cor. \ref{cor:V_event_Grob}, for any $j$, we have
\begin{equation}\label{ident1}
\exp_{\lhd_{L_i}}(\sigma^{L_1}(Q_j))= \exp_{\lhd_{L_i}}(Q_j).
\end{equation}
On the other hand, by construction of the $q_j$, for any $j$: $\DN(q_j)+ \exp_{\lhd_{L_i}}(Q_j)\subset \Delta_j$, then for any $j$, the following holds
\begin{equation}\label{ident2}
\DN(\sigma^{L_1}(q_j))+\exp_{\lhd_{L_i}}(\sigma^{L_1}(Q_j))\subset
\Delta_j.
\end{equation}
Now, we have $\sigma^{L_1}(W)=\sum_{j\in J} \sigma^{L_1}(q_j)
\sigma^{L_1}(Q_j)$. By relation (\ref{ident2}), we can say that this writing is the result of the division of $\sigma^{L_1}(W)$ by $\{\sigma^{L_1}(Q_j), \, j\in J\}$ w.r.t. $\lhd_{L_i}$, therefore:
\begin{equation}\label{ident3}
\exp_{\lhd_{L_i}}(\sigma^{L_1}(W)) =\max_{j\in J}
\{\exp_{\lhd_{L_i}} (\sigma^{L_1}(q_j) \sigma^{L_1}(Q_j)) \}.
\end{equation}
In a similar way, we can prove
\begin{equation}\label{ident4}
\exp_{\lhd_{L_i}}(W) =\max_{j\in J}
\{\exp_{\lhd_{L_i}} (\sigma^{L_1}(q_j) Q_j) \}.
\end{equation}
As a consequence, thanks to (\ref{ident3}), (\ref{ident4}), (\ref{ident1}), we obtain the equality
$\exp_{\lhd_{L_i}}(\sigma^{L_1}(W)) =\exp_{\lhd_{L_i}}(W)$, which implies in particular $\ord^{L_i}(W)=\ord^{L_i}(\sigma^{L_1}(W))$.
Hence the following relations hold:
\begin{eqnarray*}
\ord^{L_i}(W) & = & \ord^{L_i}(\sigma^{L_1}(W)) \\
 & = & \ord^{L_i}(\sigma^{L_1}(H)) \text{ because }
 \sigma^{L_1}(W)=\sigma^{L_1}(H) \\
 & \le & \ord^{L_i}(H).
\end{eqnarray*}
Therefore, $\ord^{L_i}(H') \le \ord^{L_i}(H)$.
The two claims are proven.
\end{enumerate}
Now let us specialize $z=1$ (which a morphism between the algebras $\Dnpz$ and $\Dnp$) and set $P'=H'_{|z=1}=P-W_{|z=1}$. Since $W\in h(I)$, we have $W_{|z=1} \in I$ and $P'\delta=m$. After specialization, the claims (i) and (ii) become: $\ord^{L_i}(P') \le \ord^{L_i}(P)$ for any $i=1,\ldots,q$ with a strict inequality for $i=1$. The lemma is proven.
\end{proof}

\section{Proof of theorem \ref{theo2}}\label{sec:demo2}

In this section, we will give the proof of th. \ref{theo2}. We shall firstly state more precisely the theorem in question. We shall see that the proof consists essentially in a refined analysis of the previous lemma \ref{lem:cle}. Indeed, we shall do what we could call a control over the order w.r.t. to the form $V_1$.\\
Recall that in this section $p$ equals $2$.

\begin{nota*}
\
\begin{itemize}
\item
Let $L_1,L_2$ be two non zero forms in $\UV$. Write
$L_i=a_iV_1+b_iV_2$ with $a_i,b_i \ge 0$. We say that $L_1$ is lower
(resp. strictly lower) than $L_2$ if $b_1/a_1
\le b_2/a_2$ (resp. $b_1/a_1 < b_2/a_2$). We denote this notion by
$L_1 \le L_2$ (resp. $L_1 < L_2$). By convention,
$b/0=+\infty$, i.e. any form $L$ is lower than $V_2$.
\item
Let $L_1 \ne L_2$ be in $\UV$ and $H\in \Dnpz$. We say that $H$ is
$L_1$-homogeneous if $H=\sigma^{L_1}(H)$. We say that $H$ is
$(L_1,L_2)$-homogeneous if $H=\sigma^{L_1}(\sigma^{L_2}(H))$.
\item
Let $L$ be a form in $\UV$. We define $\lhd_L$ as the order on
$\N^{n+2+n+2+1}$ given by:
\[(\alpha,\mu,\beta,\nu,k) \lhd_L (\alpha',\mu',\beta',\nu',k) \iff\]
\[\begin{cases}
k+|\beta+\nu|<k'+|\beta'+\nu'| \\
\text{or } \big( = \text{ and } L(\alpha,\mu,\beta,\nu)<
L(\alpha',\mu',\beta',\nu') \big) \\
\text{or } \big( = \text{ and } = \text{ and } (\alpha,\mu,\beta,\nu)
<_{V_1} (\alpha',\mu',\beta',\nu') \big).
\end{cases}\]
Remark that with the notation of the previous sections, if we set $\sigma=\rangle V_1, L \langle$ (with $L\ne V_1$) then $\lhd_L=\lhd_L^\sigma$, and if $L=V_1$ then $\lhd_L=<_{V_1}^h$.
\end{itemize}
\end{nota*}

Let $\sigma$ be a cone of $\EV$ of (maximal) dimension $2$ and let $\{L_1,L_2\}=\mathcal{L}(\sigma)$ with $L_1 <L_2$. Let $Q_1,\ldots,Q_r$ be the standard basis of $h(I)$ associated with $\sigma$. We define $\kappa_\sigma^1 \in \N$ by:
\[\kappa_\sigma^1=\max\{ \ord^{V_1}(Q_j)-\ord^{V_1}(\sigma^{L_2}(Q_j))
, \, j=1,\dots,r\}.\]
With the previous notations, we have (see figure \ref{fig:kappa1})
\[\ord^{V_1}(\sigma^{L_2}(Q_j))=\ord^{V_1}(\exp_{\lhd_{L_2}}(Q_j)).\]

\begin{figure}[h!]
\centering

\includegraphics[angle=0, width=9cm]{kappa1.eps}
\caption{$\ord^{V_1}(Q_j)-
  \ord^{V_1}(\exp_{\lhd_{L_2}}(Q_j))$}\label{fig:kappa1}

\end{figure}

Now, we define $\kappa^1 \in \N$ as the maximum of all the $\kappa_\sigma^1$ for $\sigma \in \EV$ of maximal dimension. Here a restatement of theorem \ref{theo2}.

\begin{theobis*}
For any $w \in \Z^2$:
\[\Vb_w(M) \subset V_{w+(\kappa^1,0)}(M).\]
\end{theobis*}

\subsection{Control of the $V_1$-order}

Let $\sigma \in \EV$ be a cone of maximal dimension et let $L_1,L_2$ be its primitive generators. Let $m\in \Vb_w(M)$ with $w$ in $\Z^2$, in particular $m\in (V^{L_1}_{L_1(w)}(\Dnd)\delta) \cap (V^{L_2}_{L_2(w)}(\Dnd)\delta) $. Suppose given $P\in \Dnp$ such that $P\delta=m$ and $\ord^{L_1}(P)\le L_1(w)$ and such that $\ord^{L_2}(P) > L_2(w)$. Then we have shown in lemma \ref{lem:cle} how we can construct, in a finite number of steps, an operator $P_\sigma$ such that $\ord^{L_1}(P_\sigma) \le \ord^{L_1}(P)$ (i.e. the $L_1$-order has not increased) and $\ord^{L_2}(P_\sigma) \le L_2(w)$ (i.e. the $L_2$-order has decreased as much as possible). We can wonder what happens concerning the $V_1$-order of $P_\sigma$ compared with that of $P$. We are going to show that this order can be greater but in a controled way. That is the purpose of the next lemma.

\begin{lem}\label{lem:controleV1}
Let $\sigma$ be a cone of maximal dimension in $\EV(h(I))$ and $L_1
\ne L_2$ its primitive generators (which can be out of $\sigma$). Suppose $V_1 \le L_1 < L_2 \le V_2$.\\
Let $w\in \Z^2$ and $m \in V^{L_2}_{L_2(w)}(M)$. Let $P\in \Dnd$ be
such that $P\delta=m$ and $\ord^{L_1}(P)\le L_1(w)$ then we can construct $P_\sigma \in \Dnd$ from $P$ such that:
\begin{description}
\item[(i)] $P_\sigma -P \in I$
\item[(ii)] $P_\sigma \in {}^{\sigma}V_w(\Dnd)$, in particular:
  $\ord^{L_2}(P_\sigma)\le L_2(w)$
\item[(iii)] $\ord^{V_1}(P_\sigma) \le \max\{ \ord^{V_1}(P) \, ,\,
  w_1+\kappa_\sigma^1\}$.
\end{description}
\end{lem}
Statement $(iii)$ justifies the title of this paragraph.
\begin{proof}
If $\ord^{L_2}(P)\le L_2(w)$, it suffices to set $P_\sigma=P$.
Let us then assume that $\ord^{L_2}(P)>L_2(w)$ which implies
$\ord^{V_1}(\sigma^{L_2}(P)) \le w_1$.

By hypothesis, there exists $P_2 \in \Dnd$ satisfying $P_2\delta=m$ and
$\ord^{L_2}(P_2)\le L_2(w)$. We define $H_0=z^{l_0}h(P-P_2)
=z^lh(P)-z^{l_2}h(P_2)$ (there exists integers $l_0$, $l$, and $l_2$
for which such an equality holds), $H=z^lh(P)$ and $H_2=z^{l_2}h(P_2)$.
Let us restart the proof of lemma \ref{lem:cle} with the difference that we work with the form $L_2$ instead of $L_1$. Thus we consider the division of $H_0$ by the standard basis $Q_1,\ldots,Q_r$ w.r.t. the order $\lhd_{L_2}$:\\
$H_0=\sum_{j=1}^r q_j Q_j$ with $\ord^{L_2}(H_0) \ge
\ord^{L_2}(q_jQ_j)$. We denote by $J$ the set of $j$ in $\{1,\ldots,r\}$ for which the later is an equality. Then we set

$W=\sum_{j \in J} \sigma^{L_2}(q_j) Q_j$ et $H'=H-W$.\\
Now we are interesting in comparing $\ord^{V_1}(H)$ and $\ord^{V_1}(H')$.

\begin{affs*}
\
\begin{description}
\item[(a)] $\ord^{V_1}(W) \le w_1+\kappa_\sigma^1$
\item[(b)] $\ord^{V_1}(W)-\ord^{V_1}(\sigma^{L_2}(W)) \le
  \kappa^1_\sigma$
\end{description}
\end{affs*}
Let us prove these claims.\\
{\bf (a):} We have $\ord^{V_1}(\sigma^{L_2}(W))=
\ord^{V_1}(\sigma^{L_2}(H))=\ord^{V_1}(\sigma^{L_2}(P))$ and the later is upper bounded by $w_1$ then if {\bf (b)} is true then so it is for {\bf (a)}.\\
{\bf (b):} As in the previous paragraph, we can show that the
division of $W$ by $\{Q_j,\, j\in J\}$ w.r.t. $\lhd_{L_2}$ is: $W=\sum_{j \in J} \sigma^{L_2}(q_j) Q_j$. Therefore, there exists $j_1\in J$ such that $\exp_{\lhd_{L_2}}(W)=\exp_{\lhd_{L_2}}(m_1 Q_{j_1})$ where $m_1=\mp_{\lhd_{L_2}}(q_{j_1})$ is a monomial of $\sigma^{V_1}(\sigma^{L_2}(q_{j_1}))$. In particular, this implies
\begin{equation}\label{ident5}
\ord^{V_1}(\sigma^{L_2}(W))= \ord^{V_1}(m_1 \sigma^{L_2}(Q_{j_1})).
\end{equation}
Moreover,
\[\ord^{V_1}(W) \le \max\{ \ord^{V_1}(\sigma^{L_2}(q_j) Q_j);\, j \in
J \}.\]
Then let $j_2 \in J$ be such that $\ord^{V_1}(\sigma^{L_2}(q_{j_2})
Q_{j_2})= \max\{ \ord^{V_1}(\sigma^{L_2}(q_j) Q_j),\,  j\in J\}$. By setting $m_2=\mp_{\lhd_{L_2}}(q_{j_2})$, we obtain
\begin{equation}\label{ident6}
\ord^{V_1}(W)\le \ord^{V_1}(m_2 Q_{j_2}).
\end{equation}
Remark that we may have $j_1=j_2$. However, in any case (see figure \ref{fig:affirmations}), we have:

\begin{figure}[h!]
\centering

\includegraphics[angle=0, width=11cm]{affirmations.eps}
\caption{Illustration of the claims}\label{fig:affirmations}

\end{figure}

\begin{aff*}
\
\begin{description}
\item[(c)] $\ord^{V_1}(m_2 \sigma^{L_2}(Q_{j_2})) \le \ord^{V_1}(m_1
  \sigma^{L_2}(Q_{j_1}))$.
\end{description}
\end{aff*}
By using this last claim and identities (\ref{ident5}) and (\ref{ident6}), we obtain:
\begin{eqnarray*}
\ord^{V_1}(W) & = & \ord^{V_1}(W) - \ord^{V_1}(m_2 Q_{j_2})\\
 & & + \ord^{V_1}(m_2 Q_{j_2}) -\ord^{V_1}(m_2
 \sigma^{L_2}(Q_{j_2}))\\
 & & +\ord^{V_1}(m_2 \sigma^{L_2}(Q_{j_2}))-
 \ord^{V_1}(m_1 \sigma^{L_2}(Q_{j_1}))\\
 & & +\ord^{V_1}(\sigma^{L_2}(W))\\
 & \le & \ord^{V_1}(m_2 Q_{j_2})
 -\ord^{V_1}(m_2 \sigma^{L_2}(Q_{j_2})) +
 \ord^{V_1}(\sigma^{L_2}(W))\\
 & \le & \kappa^1_\sigma+ \ord^{V_1}(\sigma^{L_2}(W))
\end{eqnarray*}
This proves claim {\bf (b)}. It remains to prove claim {\bf (c)}.\\

The division of $\sigma^{L_2}(W)$ by $\{\sigma^{L_2}(Q_j),\, j\in J\}$
w.r.t. $\lhd_{L_2}$ is written as:
$\sigma^{L_2}(W)= \sum_{j\in J} \sigma^{L_2}(q_j) \sigma^{L_2}(Q_j)$. Therefore,
\[\exp_{\lhd_{L_2}}\big(\sigma^{L_2}(q_{j_2}) \sigma^{L_2}(Q_{j_2})
\big) \unlhd_{L_2} \exp_{\lhd_{L_2}}(\sigma^{L_2}(W)).\]
But
\[\exp_{\lhd_{L_2}}(\sigma^{L_2}(W))= \exp_{\lhd_{L_2}}(W)=
\exp_{\lhd_{L_2}}(m_1\sigma^{L_2}(Q_{j_1}))\]
and
\[\exp_{\lhd_{L_2}}\big(\sigma^{L_2}(q_{j_2}) \sigma^{L_2}(Q_{j_2})
\big)= \exp_{\lhd_{L_2}}(m_2\sigma^{L_2}(Q_{j_2}))\]
then
\[\exp_{\lhd_{L_2}}(m_2\sigma^{L_2}(Q_{j_2})) \unlhd_{L_2}
\exp_{\lhd_{L_2}}(m_1\sigma^{L_2}(Q_{j_1})).\]
But $\ord^{L_2}(m_2\sigma^{L_2}(Q_{j_2}))=
\ord^{L_2}(m_1\sigma^{L_2}(Q_{j_1}))$ then
\[\ord^{V_1}(m_2\sigma^{L_2}(Q_{j_2})) \le
\ord^{V_1}(m_1\sigma^{L_2}(Q_{j_1})).\]
Claim {\bf (c)} is proved.\\
Now let us see how claim {\bf (a)} allows one to prove the third claim of the lemma. We started with $H$ and we have constructed $H'=H-W$.
By {\bf (a)}, we have $\ord^{V_1}(H') \le
\max (\ord^{V_1}(H), w_1+\kappa_\sigma^1)$. We continue the same process with $H'$ instead of $H$, etc. The last element $H_\sigma$ that we obtain satisfies:
$H_\sigma-H \in h(I)$ et $\ord^{V_1}(H_\sigma) \le \max
(\ord^{V_1}(H), w_1+\kappa_\sigma^1)$.\\
We finally set $P_\sigma={H_\sigma}_{|z=1}$, and we have $P_\sigma-P \in
I$ et $\ord^{V_1}(P_\sigma) \le \max (\ord^{V_1}(P),
w_1+\kappa_\sigma^1)$. The lemma is proven.
\end{proof}

\subsection{End of the proof}

\begin{proof}[Proof of theorem 2 bis]
Denote by $L_0=V_1<\cdots <L_q=V_2$ the primitive elements of the $1$-skeleton of $\EV$. For each $i=1,\ldots,q$, denote by
$\sigma_i \in \EV$ the (maximal) cone containing the open cone generated by $L_{i-1}$ and $L_i$.\\
Let $m \in \Vb_w(M)$.\\
Let us prove by an induction on $i$ that for any $i=0,\ldots,q$, there exists $T_i \in \Dnd$ satisfying:
\begin{itemize}
\item $T_i \delta=m$
\item $T_i \in V^{L_i}_{L_i(w)}(M)$
\item $\ord^{V_1}(T_i) \le w_1+\kappa^1$.
\end{itemize}
For $i=0$: $m\in \Vb_w$ then in particular $m \in V^{V_1}_{V_1(w)}$
(note that $V_1(w)=w_1$) then there exists $T_0$ such that $T_0 \delta=m$
and $\ord^{V_1}(T_0)\le w_1 \le w_1+\kappa^1$.\\
Assume the statement to be true at rank $i-1$.\\
Let us apply lemma \ref{lem:controleV1} with $\sigma=\sigma_i$ and $P=T_{i-1}$. Then set $T_i=P_\sigma$ (notations of the lemma). Thanks to the lemma in question, $T_i$ satisfies:
\begin{itemize}
\item $T_i \delta=m$
\item $T_i \in V^{L_i}_{L_i(w)}(M)$
\item $\ord^{V_1}(T_i) \le \max(\ord^{V_1}(T_{i-1}),w_1+\kappa^1)
  = w_1+\kappa^1$
\end{itemize}
Thus, the claim is true for any $i$. The particular case $i=q$ gives us: $m= T_q \delta$, $\ord^{V_2}(T_q)\le w_2$ and $\ord^{V_1}(T_q)
\le w_1+\kappa^1$, which means $m \in V_{w+(\kappa^1,0)}(M)$.
\end{proof}

\begin{rem}
The way we constructed $\kappa$ shows that it is not unique in general. Indeed, by working with $V_2$ instead of $V_1$, we could construct some $\kappa$ of the form $(0,\kappa^2)$.
\end{rem}


TITLE: Does this sequence of polynomials have a limit?
QUESTION [7 upvotes]: Consider the sequence of polynomials $p_n$ defined as follows: $p_n$ is the unique polynomial of degree $2n+1$ satisfying
$$p_n(0) = 0$$
$$p_n(1) = 1$$
$$p_n^{(k)}(0) = p_n^{(k)}(1) = 0 \text{ for $k=1$ to $n$}$$
where $p_n^{(k)}$ is the $k$th derivative of $p_n$. For example,
$$p_0(x)=x$$
$$p_1(x)=-2 x^3+3 x^2$$
$$p_2(x)=6 x^5-15 x^4+10 x^3$$
$$p_3(x)=-20 x^7+70 x^6-84 x^5+35 x^4$$
$$p_4(x)=70 x^9-315 x^8+540 x^7-420 x^6+126 x^5$$
$$p_5(x)=-252 x^{11}+1386 x^{10}-3080 x^9+3465 x^8-1980 x^7+462 x^6$$
To give some intuition, here is an animated plot of $p_n$ for $n=0$ to $50$.

How can I determine, with proof, $\lim_{n \to \infty} p_n(x)$ (if it exists)? I've never worked with functions implicitly defined in this way before, and I have no idea where to begin.

REPLY [0 votes]: @Michael has already found a very elegant answer, but I would like to contribute an alternate approach that I stumbled across. It turns out that the polynomials $p_n$ are precisely the degree $2n+1$ Bernstein polynomial approximants of $u(x-1/2)$, where $u$ is the unit step function (and $u(0) = 1/2$). These are well-known to converge to the original function.


text:
\begin{document}
\maketitle

\begin{center}
\parbox{0.8\hsize}{ \parindent=0.5 cm
In this paper, with the notion of independent identically
distributed random variables under sub-linear expectations initiated
by Peng, we derive three kinds of strong laws of large numbers for
capacities. Moreover, these theorems are natural and fairly neat
extensions of the classical Kolmogorov's strong law of large numbers
to the case where probability measures are no longer additive.
Finally, an important feature of these strong laws of large numbers
is to provide a frequentist perspective on capacities.}
\end{center}
\textbf{Keywords:} capacity, strong law of large number,
independently and identically distributed(IID), sub-linear
expectation.

\textbf{AMS 1991 subject classifications. 60H10, 60G48.}

\section{ \textbf {Introduction}}

  The classical strong laws of large numbers (strong LLN) as
fundamental limit theorems in probability theory play an important
role in the development of probability theory and its applications.
The key in the proofs of these limit theorems is the additivity of
the probabilities and the expectations. However, such additivity
assumption is not reasonable in many areas of applications because
many uncertain phenomena can not be well modeled using additive
probabilities or additive expectations. More specifically, motivated
by some problems in mathematical economics, statistics, quantum
mechanics and finance, a number of papers have used non-additive
probabilities (called capacities) and nonlinear expectations (for
example Choquet integral/expectation, $g$-expectation) to describe
and interpret the phenomena (see for example, Chen and Epstein [1],
Feynman [4], Gilboa [5], Huber [16], Peng [8,9], Schmeidler [18],
Wakker [19], Walley and Fine [20], Wasserman and Kadane [21]).
Recently, motivated by the risk measures, super-hedge pricing and
modelling uncertainty in finance, Peng [11,12,13,14,15] initiated
the notion of independent and identically distributed (IID) random
variables under sub-linear expectations. Under this framework, he
proved the weak law of large numbers and  the central limit theorems
(CLT). However, Peng's techniques can not be extended to prove the
strong laws of large numbers. In this paper, we develop new
approaches to solving this problem. We obtain three strong laws of
large numbers for capacities in this framework. All of them are
natural and fairly neat extensions of the classical Kolmogorov's
strong law of large numbers, but the proofs here are different from
the original proofs of the classical strong law of large numbers.

Now we describe the problem in more detail. For a given set $\CP$ of
multiple prior probability measures on $(\Omega,\CF)$, we define a
pair $(\CV, v)$ of capacities by
$$\CV(A):=\sup_{P\in \CP}P(A),\;\quad  v(A):=\inf_{P\in\CP}P(A), \quad \forall A\in\CF.$$
The corresponding Choquet integrals/expecations $(C_{\CV},C_v)$
are defined by
$$
C_V[X]:=\int_0^\infty V(X\ge t)dt+\int_{-\infty}^0[V(X\ge t)-1]dt
$$
where $V$ is replaced by $v$ and $\CV$ respectively.

The pair of so-called maximum-minimum expectations $(\CE,\Ce)$ are
defined by
$$\CE[\xi]:=\sup_{P\in\CP}E_P[\xi], \quad \Ce[\xi]:=\inf_{P\in\CP}
E_{P}[\xi].$$ Here and in the sequel, $E_P$ denotes  the classical
expectation under probability $P.$

In general, the relation between Choquet integral and maximum
-minimum expectations is as follows: For any random variable $X,$
$$
\CE[X]\leq C_\CV[X],\quad\quad C_v[X]\leq \Ce[X].
$$
Note that under some very special assumptions on $\CP$ and $\CV$,
both inequalities could become equalities (see for example Gilboa
[5], Huber [16], Schmeidler [18]).

Given a sequence $\{X_i\}_{i=1}^\infty$ of IID random variables for
capacities, the earlier papers related to strong laws of large
numbers are Dow and Werlang [2] and Walley and Fine [20]. However,
the more general results for strong laws of large numbers for
capacities were given by Marinacci [6,7] and Epstein and Schneider
[3]. They show that, on full set, any cluster point of empirical
averages lies between the lower Choquet integral $C_v[X_1]$ and the
upper Choquet integral $C_\CV[X_1]$ with probability one under
capacity $v$. That is
$$v\left(\omega\in\Omega:
C_v[X_1]\leq\liminf\limits_{n\rightarrow\infty}\frac
1n\sum\limits_{i=1}^n X_i(\omega)
\leq\limsup\limits_{n\rightarrow\infty}\frac1n\sum\limits_{i=1}^nX_i(\omega)\leq
C_{\CV}[X_1]\right)=1.$$ Marinacci [6,7] obtains his result under
the assumptions that $\CV$ is a totally monotone capacity on a
Polish space $\Omega$, random variables $\{X_n\}_{n\ge 1}$ are
bounded or continuous. Epstein and Schneider [3] also show the same
result under the assumptions that $\CV$ is rectangular and  the set
$\CP$ is finite.

Since the gap  between the Choquet integrals $C_v[X]$ and $C_V[X]
$ is bigger than that of the maximum-minimum expectations  $
\Ce[X]$ and $ \CE[X]$ for all $X$, it is of interest to ask
whether we can obtain a more precise result if the Choquet
integrals/expctations in the above equality are replaced by
maximum-minimum expectations. That is
$$
v\left(\omega\in \Omega:
\Ce[X_1]\leq\liminf\limits_{n\to\infty}\frac1n\sum\limits_{i=1}^n
X_i(\omega) \leq\limsup\limits_{n\to\infty}\frac
1n\sum\limits_{i=1}^nX_i(\omega)\leq \CE[X_1]\right)=1.
$$
The first result in this paper is to show that the above equality is
still true in Peng's framework even under some weaker assumptions.
Furthermore, motivated by this result, we establish two new laws of
large numbers. The first is to show that there exist two cluster
points of empirical averages which reach the minimum expectation
$\Ce[X_1]$ and the maximum expectation $\CE[X_1]$ respectively under
capacity $\CV.$ That is
$$
\CV\left(\omega\in \Omega: \limsup
\limits_{n\to\infty}\frac1n\sum\limits_{i=1}^nX_i(\omega)=
\CE[X_1]\right)=1;$$
$$
\CV\left(\omega\in
\Omega:\liminf\limits_{n\to\infty}\frac1n\sum\limits_{i=1}^n
X_i(\omega)= \Ce[X_1]\right)=1.$$ The second  is to prove that the
cluster set of empirical averages coincides with the interval
between minimum expectation $\Ce[X_1]$ and maximum expectation
$\CE[X_1].$ That is, if $C(\{x_n\})$ is the cluster set of
$\{x_n\},$ then
$$
\CV\left( \omega\in\Omega: C\left(\left\{\frac1n\sum\limits_{i=1}^n
X_i(\omega)\right\}\right)=\left[\Ce[X_1],\;\;
\CE[X_1]\,\right]\right)=1.
$$
Obviously, if capacity $\CV$ or $v$ in the above results is a
probability measure, all of our main results are natural and fairly
neat extension of the classical Kolmogorov's strong law of large
numbers. Moreover, an important feature of our strong laws of large
numbers is to provide a frequentist perspective on capacities.

Finally, our results also imply that $[\Ce[X_1], \CE[X_1] ]$ is the
smallest interval of all intervals in which the limit point of
empirical averages lies with probability (capacity $v$) one.

\section{Notation and Lemmas}

In this section, we will recall briefly the notion of IID random
variables and sub-linear expectations initiated by Peng.

Let $(\CW,\CF)$ be a measurable space, and $\mathcal {H}$ be a set
of random variables on $(\CW,\CF).$

\begin{definition}
A functional $\mathbb{E}$ on $\mathcal {H}
\mapsto(-\infty,+\infty)$ is called a sub-linear expectation, if
it satisfies the following properties: for all $X,Y\in\mathcal
{H},$
\begin{description}
\item{(a)} Monotonicity: $X\geq Y$ implies
$\mathbb{E}[X]\geq\mathbb{E}[Y]$.
 \item{(b)} Constant preserving:
$\mathbb{E}[c]=c,\forall c\in\mathbb{R}$.
\item{(c)} Sub-additivity:
$\mathbb{E}[X+Y]\leq\mathbb{E}[X]+\mathbb{E}[Y]$.
\item{(d)} Positive homogeneity: $\mathbb{E}[\lambda
X]=\lambda\mathbb{E}[X],\forall\lambda\geq0$.
\end{description}
\end{definition}

Given a sub-linear expectation $\CE$, let us denote the conjugate
expectation $\Ce$ of  sub-linear $\CE$  by
$$
\Ce[X]:=-\CE[-X], \quad  \forall X\in \mathcal {H}.
$$
Obviously, for all $X\in\mathcal {H},$ $\Ce[X]\leq \CE[X].$

Furthermore, let us denote a pair $(\CV,v)$ of capacities by
$$
\CV(A):=\CE[I_A], \quad  v(A):= \Ce[I_A],\quad \forall A\in\CF.
$$
It is easy to check that
$$
\CV(A)+v(A^c)=1,\quad \forall A\in\CF,
$$
where $A^c$ is the complement set of $A.$
\begin{definition}\label{definition1}
A set function $V$: $\CF\rightarrow[0,1]$ is called  a continuous
capacity if it satisfies
\begin{description}
\item {(1)} $V(\phi)=0,V(\Omega)=1$. \item {(2)} $V(A)\leq V(B),$
whenever $A\subset B$ and $A,B\in\CF$. \item {(3)}
$V(A_{n})\uparrow V(A)$,if $A_{n}\uparrow A$. \item {(4)}
$V(A_{n})\downarrow V(A)$, if $A_{n}\downarrow A$, where $A_{n},
A\in \CF$.
\end{description}
\end{definition}

\begin{assumption} Throughout this paper, we assume that $(\CV,v)$ is a pair of continuous capacities
generated by sub-linear expectation $\CE$ and its conjugate
expectation $\Ce. $
\end{assumption}

 The following  example shows that the maximum expectation generated by multiple prior probability measures
is a sub-linear expectation and the associated capacities are
continuous (see for example, Huber and Strassen [17]).

\begin{example} Let $\CP$ be a weakly compact set of probability measures defined on $(\CW,\CF)$. For
any random variable $\xi$,  we denote upper expectation by
$$
\CE[\xi]=\sup_{Q\in\CP}E_Q[\xi].$$ Then $\CE[\cdot]$ is a
sub-linear expectation. Moreover, the capacities $V$ and $v$ given
by
$$
\CV(A)=\sup_{Q\in\CP}Q(A), \quad v(A)=\inf_{Q\in\CP}Q(A), \quad
\forall A\in\CF$$ are continuous.
\end{example}

The following is the notion of IID random variables under
sub-linear expectations  introduced by Peng.

\begin{definition} {\bf(Peng [11])}

\begin{description}
\item{\bf Independence:} Suppose that $Y_1,Y_2,\cdots,Y_n$ is a
sequence of random variables such that $Y_i \in\mathcal {H}.$ Random
variable $Y_n$ is said to be independent of
$X:=(Y_1,\cdots,Y_{n-1})$ under $\mathbb{E}$, if for each measurable
function $\varphi$ on $R^n$ with $\varphi(X,Y_n)\in \mathcal {H}$
and $\varphi(x,Y_n)\in\mathcal {H}$ for each $x\in
\mathbb{R}^{n-1},$ we have
$$\mathbb{E}[\varphi(X,Y_n)]=\mathbb{E}[\overline{\varphi}(X)],$$
where $\overline{\varphi}(x):=\mathbb{E}[\varphi(x,Y_n)]$ and
$\overline{\varphi}(X)\in\mathcal {H}$.

\item{\bf Identical distribution:} Random variables $X$ and $Y$
are said to be identically distributed, denoted by
$X\overset{d}{=}Y$, if for each $\varphi$ such that $\varphi(X),
\; \varphi(Y)\in \mathcal {H}$,
$$\mathbb{E}[\varphi(X)]=\mathbb{E}[\varphi(Y)].$$

\item{\bf IID random variables:} A sequence of random variables
$\{X_i\}_{i=1}^\infty$ is said to be IID, if $X_i\overset{d}{=}X_1$
and $X_{i+1}$ is independent of $Y:=(X_1,\cdots,X_i)$ for each $i\ge
1.$
\end{description}
\end{definition}

The following lemma shows the relation between Peng's independence
and pairwise independence in Marinacci [6,7].

\begin{lemma} Suppose that $X,Y \in \mathcal {H}$ are two random variables.
$\CE$ is a sub-linear expectation and $(\CV,v)$ is the pair of
capacities generated by  $\CE.$ If  random variable $X$ is
independent of $Y$ under $\CE$, then $X$ is also pairwise
independent of $Y$ under capacities $\CV$ and $v,$ i.e. for all
subsets $D$ and $G\subset \CR,$
$$
V(X\in D, Y\in G)=V(X\in D)V(Y\in G)
$$
holds for both capacity $\CV$ and $v$.
\end{lemma}

{\bf \Proof.} If we choose $\varphi(x,y)=I_D(x)I_G(y)$ for $\CE, $
by the definition of Peng's independence, it is easy to obtain
$$
\CV(X\in D, Y\in G)=\CV(X\in D)\CV(Y\in G).
$$
Similarly,  if we choose $\varphi(x,y)=-I_D(x)I_G(y)$ for $\CE,$
it is easy to obtain
$$
v(X\in D, Y\in G)=v(X\in D)v(Y\in G).
$$
The proof is complete.

 Borel-Cantelli Lemma is still true for capacity under some
assumptions.

\begin{lemma}
Let $\{A_n,n\geq1\}$ be a sequence of events in $\CF$ and $(\CV,v)$
be a pair of capacities generated by sub-linear expectation $\CE$.
 \begin{description}
 \item{(1)} If $\sum\limits_{n=1}^\infty\CV(A_n)<\infty,$ then $\CV\left(\bigcap\limits_{n=1}^\infty\bigcup\limits_{i=n}^\infty
 A_i\right)=0.$
\item{(2)} Suppose that  $\{A_n,n\geq1\}$ are pairwise independent
with respect to $v$, i.e. $$v\left(\bigcap\limits_{i=1}^\infty
A_i^c\right)=\prod_{i=1}^\infty v(A_i^c).$$ If
$\sum\limits_{n=1}^\infty{\CV}(A_n)=\infty $, then
$$\CV\left(\bigcap\limits_{n=1}^\infty\bigcup\limits_{i=n}^\infty A_i\right)=1.$$
\end{description}
\end{lemma}

{\bf \Proof.}
$$\begin{array}{lcl}
 0&\leq& \CV(\bigcap\limits_{n=1}^\infty\bigcup\limits_{i=n}^\infty
 A_i)\\
 &\leq& \CV(\bigcup\limits_{i=n}^\infty
 A_i)\\
 &\leq &\sum\limits_{i=n}^\infty\CV(A_i) \to 0,\;  \hbox{ as}\; n\to
 \infty.
 \end{array}
 $$
 The proof of (1) is complete.

$$
\begin{array}{lcl}
0&\leq &1-\CV(\bigcap\limits_{n=1}^\infty\bigcup\limits_{i=n}^\infty A_i)\\
&=& 1-\lim\limits_{n\to \infty}\CV(\bigcup\limits_{i=n}^\infty A_i)\\
&=& \lim\limits_{n\to \infty}[1-\CV(\bigcup\limits_{i=n}^\infty A_i)]\\
& =&\lim\limits_{n\to \infty}v(\bigcap\limits_{i=n}^\infty A_i^c)\\
&=&\lim\limits_{n\to \infty}\prod\limits_{i=n}^\infty
v(A_i^c)\\
&=&\lim\limits_{n\to
\infty}\prod\limits_{i=n}^\infty(1-\CV(A_i))\\
&\leq&\lim\limits_{n\to \infty}\prod\limits_{i=n}^\infty{\rm
exp}(-\CV(A_i))\\
&=&\lim\limits_{n\to \infty}{\rm
exp}(-\sum\limits_{i=n}^\infty\CV(A_i))=0.
\end{array}.$$ We complete the proof of (2).

With the notion of IID under sub-linear expectation, Peng shows a
weak law of large numbers for sub-linear expectation (see Theorem
5.1 in Peng \cite{peng5}).

\begin{lemma}{\bf (Peng [14])} Let $\{X_i\}_{i=1}^\infty$ be a sequence of IID random
variables with finite means $\overline\mu:=\CE[X_1], \;\;
\underline \mu:=\Ce[X_1].$  Suppose $\CE[|X_1|^{1+\alpha}]
<\infty$ for some $\alpha>0.$  Then for any bounded, continuous
and linear growth function $\phi,$
$$
\CE\left[\phi \left(\frac 1n \sum_{i=1}^nX_i\right)\right]\to
\sup_{\underline \mu\leq x\leq\overline \mu}\phi(x), \;  \hbox{
as}\; n\to \infty.
$$
\end{lemma}

\section{The Main Results }
We now prove our main results, let us first prove the following
Lemma.

\begin{lemma} Given sub-linear expectation $\CE,$ let $\{X_i\}_{i=1}^\infty$ be  a sequence of
independent random variables such that $\sup\limits_{i\ge
1}\CE[|X_i|^{1+\alpha}]<\infty$ for some constant $\alpha>0.$
 Suppose that there exists a constant $c>0$
such that
$$|X_n-\CE[X_n]|\leq c\frac{n}{\log(1+n)}, \quad n=1,2,\cdots.$$
 Then there exists a sufficiently large number $m>1$ such that
$$\sup_{n\ge 1}\CE\left[\exp\left(\frac{m\log(1+n)}{n} S_n\right)\right]<\infty,$$
where $S_n:=\sum\limits_{i=1}^n[X_i-\CE[X_i]].$
\end{lemma}

{\bf \Proof.} It is easy to prove that $\sup\limits_{i\ge
1}\CE[|X_i|^{1+\alpha}]<\infty $ implies $\sup\limits_{i\ge
1}\CE[|X_i|]<\infty$ and $\sup\limits_{i\ge
1}\CE[|X_i-\CE[X_i]|^{1+\alpha}]<\infty.$

For given $\alpha >0,$  noting that
$\frac{{n}^\alpha}{[\log(1+n)]^{1+\alpha}}\to \infty$ as
$n\to\infty,$  thus, there exist a constant $n_0>0$ and a
sufficiently large number $m>1$ such that
$$
\frac{{n}^\alpha}{[\log(1+n)]^{1+\alpha}}\ge m^{1+\alpha},\;\; n\ge
n_0.
$$
So
\begin{equation}\label{m0}
\left(\frac{m\log(1+n)}{n}\right)^{1+\alpha}\leq \frac 1{n}, \quad
n\ge n_0.
\end{equation}

On the other hand, it can be directly verified that the following inequality holds:

\begin{equation}\label{m01}
 e^x\leq 1+x+{|x|^{1+\alpha}}e^{2|x|},\quad  x\in \CR,\; 1\ge\alpha >0.
 \end{equation}
Set $x=\frac{m\log(1+n)}{n}[X_k-\CE[X_k]]$ and put it to
(\ref{m01}).

It suffices
 to obtain that, for each $n\ge k,$ $k=1,2,\cdots,$

$$
\begin{array}{lcl}
\exp\left(\frac{m\log(1+n)}{n}[X_k-\CE[X_k]]\right)&\leq &
1+ \frac{m\log(1+n)}{n}[X_k-\CE[X_k]]\\
&+&\left(\frac{m\log(1+n)}{n}\right)^{1+\alpha}|X_k-\CE[X_k]|^{1+\alpha}
e^{ \left (\frac{2m\log(1+n)}{n}|X_k-\CE[X_k]|\right)}.
\end{array}
$$
By assumption, it is easy to obtain
$$ \frac{m\log(1+n)}{n}|X_k-\CE[X_k]|\leq
c m.$$ This with (\ref{m0}), we have
$$
\exp\left(\frac{m\log(1+n)}{n}[X_k-\CE[X_k]]\right)
 \leq 1+\frac{m\log(1+n)}{n}[X_k-\CE[X_k]]+ \frac{|X_k-\CE[X_k]|^{1+\alpha}}{n}e^{2c m}.
 $$
Taking expectation on both sides of the above equality and setting
$$L:=\sup_{k\ge 1}\CE[|X_k-\CE[X_k]|^{1+\alpha}],$$ we have
$$
\CE\left[\exp\left(\frac{m\log(1+n)}{n}[X_k-\CE[X_k]]\right)\right]\leq
1+\frac L{n}e^{2c m}
$$
By the independence of $\{X_i\}_{i=1}^n,$
$$\begin{array}{lcl}
 \CE\left[\exp\left(\frac{m\log(1+n)}{n} S_n\right)\right]&=&
\prod\limits_{k=1}^n\CE\left[\exp\left(\frac{m\log(1+n)}{n}
[X_k-\CE[X_k]]\right)\right]\\
&\leq &\left(1+\frac{L}{n}e^{2c m}\right)^n\to e^{Le^{2c\,
m}}<\infty, \; \hbox{as} \; n\to \infty.
\end{array}
$$
 The proof is complete.

\begin{theorem} Let $\{X_i\}_{i=1}^\infty$ be a sequence of IID random
variables for sublinear expectation $\CE$. Suppose
$\CE[|X_1|^{1+\alpha}]<\infty$ for some $\alpha\in (0, 1].$ Set
$\overline \mu:=\CE[X_1],$ $\underline \mu=\Ce[X_1]$  and $
S_n:=\sum\limits_{i=1}^n X_i.$ Then
\begin{description}
\item{(I)}
\begin{equation}\label{M}
\CV\left(  \{\liminf\limits_{n\to\infty} S_n/n <\underline
\mu\}\bigcup \{\limsup\limits_{n\to\infty} S_n/n
>\overline \mu\}\right)=0.
\end{equation}
Also
 \item{}
\begin{equation}\label{M2}
v\left( \underline \mu\leq \liminf\limits_{n\to\infty} S_n/n \leq
\limsup\limits_{n\to\infty} S_n/n \leq \overline \mu\right)=1.
\end{equation}
\item{(II)}\label{M1}
$$\CV\left( \limsup\limits_{n\to\infty}S_n/n= \overline
\mu\right)=1,\quad \CV\left( \liminf\limits_{n\to\infty} S_n/n=
\underline \mu\right)=1.
$$
 \item{(III)} Suppose that  $C(\{x_n\})$ is the cluster set of a sequence of
 $\{x_n\}$ in $\CR,$ then
 $$
 \CV\left (C(\{S_n/n\})=[\underline\mu, \overline
 \mu]\right)=1.
 $$
\end{description}
\end{theorem}

{\bf \Proof.} First, it is easy to show that argument (3) is
equivalent to the conjunction of
\begin{equation}\label{F1}
\CV\left(\limsup\limits_{n\to\infty} S_n/n >\overline \mu\right)=0,
\end{equation}
\begin{equation}\label{F2}
\CV\left(\liminf\limits_{n\to\infty} S_n/n <\underline \mu\right)=0.
\end{equation}
In fact, writing $A:=\{\limsup\limits_{n\to\infty} S_n/n> \overline
\mu\},$ $B:=\{\liminf\limits_{n\to\infty} S_n/n < \underline \mu\},$
the equivalence can be proved from the inequality $$ \max\{
V(A),V(B)\}\leq V(A\bigcup B)\leq V(A)+V(B).$$

We only need to prove (\ref{F1}), since the proof of (\ref{F2}) is
similar for the reason of symmetry equation.

We shall finish the proof of (\ref{F1}) by two steps.

{\bf  Step 1}  Assume that there exists a constant $c> 0$ such that
$|X_n-\overline \mu|\leq \frac{cn}{\log(1+n)}$ for $n\ge 1.$ Thus,
$\{X_i\}_{i=1}^\infty$ satisfies the assumptions of Lemma 4.

To prove (\ref{F1}), we shall show that for any $\epsilon>0,$
\begin{equation}\label{R1}
\CV\left(\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\{S_k/k\ge
\overline \mu+\epsilon\}\right)=0.
\end{equation}
In fact, by Lemma 4, for $\epsilon >0,$ let us choose $m>
1/\epsilon$ such that
$$\sup_{n\ge 1}\CE\left[\exp\left( \frac{m\log(1+n)}{n}\sum_{k=1}^n(X_k-\CE[X_k])\right)\right]<\infty.$$
By Chebyshev's inequality,
$$\begin{array}{lcl}
\CV(S_n/n\ge \overline \mu+\epsilon) &=& \CV(\frac{S_n-n\overline
\mu}n\ge \epsilon)\\
&=&
\CV\left(\frac{m\log(1+n)}{n}\sum\limits_{k=1}^n(X_k-\overline\mu)\ge
\epsilon m\log(1+n)\right)\\
&\leq & e^{-\epsilon m\log(1+n)}\CE\left[\exp\left(
\frac{m\log(1+n)}{n}\sum\limits_{k=1}^n(X_k-\overline
\mu)\right)\right]\\
 &\leq& \frac 1{(1+n)^{\epsilon m}}\sup\limits_{n\ge 1}\CE\left[\exp\left(
\frac{m\log(1+n)}{n}\sum\limits_{k=1}^n(X_k-\overline
\mu)\right)\right].
\end{array}$$
Since $\epsilon m>1,$ $\sup\limits_{n\ge 1}\CE\left[\exp\left(
\frac{m\log(1+n)}{n}\sum\limits_{k=1}^n(X_k-\overline
\mu)\right)\right]<\infty.$ Following from the convergence of
$\sum\limits_{n=1}^\infty \frac 1{(1+n)^{\epsilon m}},$  we have
$$\sum_{n=1}^\infty \CV( S_n/n\ge \overline
\mu+\epsilon)<\infty.$$
 Using the first Borel-Cantelli Lemma, we have
$$
\CV\left(\limsup\limits_{n\to\infty} S_n/n \geq\overline
\mu+\epsilon\right)=0 \quad\forall \epsilon >0,
$$
which implies
$$
\CV\left(\limsup\limits_{n\to\infty} S_n/n > \overline \mu\right)=0,
$$
and
$$
v\left(\limsup\limits_{n\to\infty} S_n/n \leq \overline
\mu\right)=1.
$$
Similarly, considering the sequence $\{-X_i\}_{i=1}^\infty,$ by step
1, it suffices to obtain
$$
\CV\left(\limsup\limits_{n\to\infty}(-S_n)/n >\CE[-X_1]\right)=0.
$$
Hence,
$$
\CV\left(\liminf\limits_{n\to\infty}S_n/n <-\CE[-X_1]\right)=0.
$$
But $\underline\mu=-\CE[-X_1],$ hence
$$
\CV\left(\liminf\limits_{n\to\infty}S_n/n <\underline\mu\right)=0,
$$
and
$$
v\left(\liminf\limits_{n\to\infty}S_n/n \ge \underline\mu\right)=1
.$$

{\bf Step 2.} Write
$$\overline X_n:=(X_n-\overline
\mu)I_{\{|X_n-\overline\mu|\leq \frac
{cn}{\log(1+n)}\}}-\CE\left[(X_n-\overline
\mu)I_{\{|X_n-\overline\mu|\leq
\frac{c\;n}{\log(1+n)}\}}\right]+\overline \mu.$$

Immediately, $\CE[\overline X_n]=\overline \mu.$ Moreover, it is
easy to check that $\{\overline X_i\}_{i=1}^\infty$ satisfies the
assumptions in Lemma 4.

 Indeed, obviously for each $n\ge 1,$
$$|\overline X_n-\overline
\mu|\leq \frac{2c\;n}{\log(1+n)}.$$

On the other hand, for each $n\ge 1,$ it easy to check that
$$|\overline X_n-\overline \mu|\leq |X_n-\overline \mu|+\CE[|X_n-\overline
\mu|].$$  Then
$$\CE[|\overline X_n-\overline \mu|^{1+\alpha}]\leq
2^{1+\alpha}\left(\CE[|X_n-\overline{\mu}|^{1+\alpha}]+(\CE[|X_n-\overline{\mu}|])^{1+\alpha}\right)<\infty.
$$
Setting $\overline S_n:=\sum\limits_{i=1}^n\overline X_i,$
immediately,
\begin{equation}\label{R}
\frac{1}{n}S_n\leq \frac{1}{n}\overline S_n+ \frac 1n
\sum_{i=1}^n|X_i-\overline
\mu|I_{\{|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\}}+\frac1n
\sum_{i=1}^n\CE\left[|X_i-\overline
\mu|I_{\{|X_i-\overline\mu|>\frac{c\,i}{\log(1+i)}\}}\right].
\end{equation}
Since $\{X_i-\overline \mu\}_{i=1}^\infty$ have common
distributions, we have

$$\begin{array}{lcl}
\sum\limits_{i=1}^\infty\frac{\CE\left[|X_i-\overline
\mu|I_{\{|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\}}\right]}{i}
&\leq& \sum\limits_{i=1}^\infty\frac{[\log(1+i)]^\alpha}{c^\alpha\,
i^{1+\alpha }} \CE[|X_i-\overline \mu|^{1+\alpha}]\\
&=&\CE[|X_1-\overline \mu|^{1+\alpha}]\left(\frac 1
c\right)^\alpha\sum\limits_{i=1}^\infty\frac{[\log(1+i)]^\alpha}
{i^{1+ \alpha }}<\infty.\end{array}$$ By Kronecker Lemma,
\begin{equation}\label{E1}
\frac 1n \sum_{i=1}^n\CE\left[|X_i-\overline
\mu|I_{\{|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\}}\right]\to 0.
\end{equation}
Furthermore, write
$A_i:=\{|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\}$ for $i\ge 1.$
It suffices now to prove that
$$\CV\left(\bigcap_{n=1}^\infty\bigcup_{i=n}^\infty A_i\right)=0.$$
In fact, by Chebyshev's inequality,
$$\CV\left(|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\right)\leq
\left(\frac{\log(1+i)}{c\;i}\right)^{1+\alpha}\CE[|X_i-\overline\mu|^{1+\alpha}]
$$
Hence,
$$
\sum_{i=1}^\infty\CV\left(|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\right)<\infty
$$
and by the first Borel-Cantelli Lemma, we have
$\CV\left(\bigcap\limits_{n=1}^\infty\bigcup\limits_{i=n}^\infty
A_i\right)=0.$\\
This implies that $\omega\not\in
\bigcap\limits_{n=1}^\infty\bigcup\limits_{i=n}A_i,$ the sequence $$
\sum_{i=1}^n\frac{|X_i-\overline
\mu|I_{\{|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\}}}i$$ converges
almost surely with respect to $\CV$ as $n\rightarrow\infty$.
$$
$$
Applying Kronecker Lemma again,
\begin{equation}\label{E2}
\frac 1n \sum_{i=1}^n\left(|X_i-\overline
\mu|I_{\{|X_i-\overline\mu|>\frac{c\;i}{\log(1+i)}\}}\right)\to 0,
\hbox{ a.s} \quad \CV.
\end{equation}
Setting $\lim\sup$ on both side of (\ref{R}),  then by (\ref{E1})
and (\ref{E2}), we have
 $$\limsup\limits_{n\to\infty}
S_n/n\leq \limsup\limits_{n\to\infty} \overline S_n/n,\quad
\hbox{a.s.}\quad \CV.$$ Since $\{\overline X_n\}_{n=1}^\infty$
satisfies the assumption of Step 1, by Step 1,
$$\CV(\limsup\limits_{n\to\infty}\overline S_n/n> \overline
\mu)=0,$$ and also $$v(\limsup\limits_{n\to\infty}\overline
S_n/n\leq \overline \mu)=1.$$
Therefore, the proof of (I) is
complete.

To prove (II), if $\overline{\mu}=\underline{\mu}$, it is trivial.
Suppose $\overline{\mu}>\underline{\mu}$, we only need to prove
that there exists an increasing subsequence $\{n_k\}$ of $\{n\}$
such that for any $0<\epsilon< \overline \mu-\underline \mu,$
\begin{equation}\label{R2}
\CV\left(\bigcap_{m=1}^{\infty}\bigcup_{k=m}^{\infty}\{S_{n_k}/{n_k}\ge
\overline \mu-\epsilon\}\right)=1.
\end{equation}
 Then  we have
 $$
\CV\left(\limsup\limits_{k\to\infty} S_{n_k}/{n_k}\ge \overline
\mu \right)=1.$$ This with (I) suffices to  yield the desired
result (II).

Indeed, choosing $n_k=k^k$ for $k\ge 1 $ and setting $\overline
S_n:=\sum\limits_{i=1}^n(X_i-\overline \mu),$ we have
$$\begin{array}{lcl}
\CV\left(\frac{ S_{n_k}-S_{n_{k-1}} } {n_k-n_{k-1}}\ge \overline
\mu-\epsilon\right)
&=& \CV\left(\frac{ S_{n_k-n_{k-1}} }
{n_k-n_{k-1}}\ge \overline \mu-\epsilon\right)\\
&=&\CV\left(\frac{ S_{n_k-n_{k-1}}-(n_k-n_{k-1})\overline \mu }
{n_k-n_{k-1}}\ge -\epsilon\right) \\
&=& \CV\left(\frac{ \overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}}\ge
-\epsilon\right)\\
&\ge& \CE[\phi(\frac{ \overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}})]
\end{array}
$$
where $\phi(x)$ is defined by
$$
\phi(x)=\left\{
\begin{array}{l}
1-e^{-(x+\epsilon)},\quad x\ge -\epsilon;\\
0,      \quad\quad\quad\quad  \quad  x<-\epsilon.
\end{array}
\right.
$$
Considering the sequence of IID random variables $\{X_i-\overline
\mu\}_{i=1}^\infty.$
 Obviously $$\CE[X_i-\overline \mu]=0,\quad \Ce[X_i-\overline
\mu]=-(\overline \mu-\underline \mu).
$$
Applying Lemma 3, we have, $n_k-n_{k-1}\to\infty$ as $k\to \infty$
and
$$\CE[\phi(\frac{ \overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}})]\to
\sup_{-(\overline \mu-\underline \mu)\leq y\leq
0}\phi(y)=\phi(0)=1-e^{-\epsilon}>0.$$ Thus
 $$\sum_{k=1}^\infty \CV\left(\frac{ S_{n_k}-S_{n_{k-1}} } {n_k-n_{k-1}}\ge
\overline \mu-\epsilon\right)\ge \sum_{k=1}^\infty\CE[\phi(\frac{
\overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}})]=\infty.
$$
Note the fact that $\{
 S_{n_k}-S_{n_{k-1}}\}_{k\ge 1}$ is a sequence of independent random variables for $k\ge 1$.
$$
$$
Using the second Borel-Cantelli Lemma, we have
 $$
\limsup\limits_{k\to\infty}\frac{ S_{n_k}-S_{n_{k-1}} }
{n_k-n_{k-1}}\ge \overline \mu-\epsilon,\quad \hbox{a.s.} \; \CV.
$$
But
$$
\frac{S_{n_k}}{n_k}\ge \frac{ S_{n_k}-S_{n_{k-1}} }
{n_k-n_{k-1}}\cdot\frac{n_k-n_{k-1}}{n_k}-\frac{|S_{n_{k-1}}|}{n_{k-1}}\cdot\frac{{n_{k-1}}}{n_k}.$$
Note the fact that
$$
\frac{n_k-n_{k-1}}{n_k}\to 1, \quad \frac{{n_{k-1}}}{n_k}\to 0, \;
\hbox{as} \; k\to \infty
$$
and from
$$
\limsup\limits_{n\to\infty} S_n/n\leq \overline \mu, \quad
\limsup\limits_{n\to\infty} (-S_n)/n\leq -\underline \mu ,
$$
we have $$\limsup\limits_{n\to\infty} |S_n|/n\leq\max\{ |\overline
\mu|,|\underline \mu|\}, \hbox{a.s.}\; \CV.$$ Hence,

$$
\limsup\limits_{k\to\infty}\frac{S_{n_k}}{n_k}\ge
\limsup\limits_{k\to\infty}\frac{ S_{n_k}-S_{n_{k-1}}}
{n_k-n_{k-1}}\lim\limits_{k\to\infty}\frac{n_k-n_{k-1}}{n_k}-
\limsup\limits_{k\to\infty}\frac{|S_{n_{k-1}}|}{n_{k-1}}\lim\limits_{k\to\infty}\frac{{n_{k-1}}}{n_k}.
$$
We conclude that
$$
\limsup\limits_{k\to\infty}\frac{S_{n_k}}{n_k}\ge \overline
\mu-\epsilon, \hbox{ a.s. }\; \CV. $$ Since $\epsilon$ is
arbitrary, we have
$$\CV\left(\limsup\limits_{k\to\infty} S_{n_k}/{n_k}\ge
\overline \mu \right)=1.
$$
By (I), we know $\CV\left(\limsup\limits_{n\to\infty} S_{n}/{n}>
\overline \mu \right)=0$, thus
$$\begin{array}{lcl}
 \CV\left(\limsup\limits_{n\to\infty} S_{n}/{n}= \overline \mu \right)
&=&\CV\left(\limsup\limits_{n\to\infty} S_{n}/{n}= \overline \mu
\right)+\CV\left(\limsup\limits_{n\to\infty} S_{n}/{n}> \overline
\mu \right)\\
&\ge & \CV\left(\limsup\limits_{n\to\infty} S_{n}/{n}\ge \overline
\mu \right)=1.
\end{array}$$
Considering the sequence of $\{-X_n\}_{n=1}^\infty,$  we have
$$\CV\left(\limsup_{n\to\infty} (-S_{n})/{n}=\CE[-X_1]\right)=1.$$
Therefore,
$$\CV\left(\liminf_{n\to\infty} S_{n}/{n}=-\CE[-X_1]\right)=1.$$
But $\underline \mu=-\CE[-X_1],$ thus
$$\CV\left(\liminf_{n\to\infty} S_{n}/{n}=\underline \mu\right)=1.
$$
The proof of (II) is complete.

To prove (III), obviously by (I), we have
$$
 \CV\left (C(\{S_n/n\})\subset[\underline\mu, \overline
 \mu]\right)=1.
$$
We still need to prove
$$
 \CV\left (C(\{S_n/n\})\supset[\underline\mu, \overline
 \mu]\right)=1.
$$
If $\overline{\mu}=\underline{\mu}$, it is trivial. Suppose
$\overline{\mu}>\underline{\mu},$  we only need to prove that, for
any rational number $b\in (\underline \mu, \overline \mu)$
$$
\CV\left(\liminf_{n\to\infty}|S_{n}/{n}-b|=0\right)=1.
$$
To do so, we only need to prove that there exists an increasing
subsequence $\{n_k\}$ of $\{n\}$ such that for any rational number
$b\in (\underline \mu, \overline \mu)$
\begin{equation}\label{R3}
\CV\left(\bigcap_{m=1}^{\infty}\bigcup_{k=m}^{\infty}\{|S_{n_k}/{n_k}-b|\leq\epsilon\}\right)=1.
\end{equation}
Indeed, for any $0<\epsilon \leq \min\{ \overline\mu -b,
b-\underline \mu\},$ let us choose $n_k=k^k$ for $k\ge 1.$

Setting $\overline S_n:=\sum\limits_{i=1}^n(X_i-b),$ then
$$\begin{array}{lcl}
\CV\left(|\frac{ S_{n_k}-S_{n_{k-1}} } {n_k-n_{k-1}}-b|\leq
\epsilon\right) &=& \CV\left(|\frac{ S_{n_k-n_{k-1}} }
{n_k-n_{k-1}}-b|\leq \epsilon\right)\\
&=&\CV\left(|\frac{ S_{n_k-n_{k-1}}-(n_k-n_{k-1}) b }
{n_k-n_{k-1}}|\leq \epsilon\right) \\
&=& \CV\left(|\frac{ \overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}}|\leq
\epsilon\right)\\
&\ge& \CE[\phi(\frac{ \overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}})]
\end{array},
$$
where $\phi(x)$ is defined by
$$
\phi(x)=\left\{
\begin{array}{l}
1-e^{|x|-\epsilon},\quad\quad |x|\leq \epsilon;\\
0,      \quad\quad\quad\quad  \quad\;\;  |x|>\epsilon.
\end{array}
\right.
$$
Considering the sequence of IID random variables
$\{X_i-b\}_{i=1}^\infty,$
 obviously $$\CE[X_i-b]=\overline\mu-b>0,\quad \Ce[X_i-
b]=\underline \mu-b<0.
$$
Applying Lemma 3, we have, as $k\to \infty$,
$$\CE[\phi(\frac{ \overline S_{n_k-n_{k-1}}} {n_k-n_{k-1}})]\to
\sup_{\underline \mu-b\leq y\leq \overline\mu-b
}\phi(y)=\phi(0)=1-e^{-\epsilon}>0.$$ Thus
 $$\sum_{k=1}^\infty \CV\left(|\frac{ S_{n_k}-S_{n_{k-1}} }
 {n_k-n_{k-1}}-b|\leq
\epsilon\right)\geq \sum_{k=1}^\infty\CE[\phi(\frac{ \overline
S_{n_k-n_{k-1}}} {n_k-n_{k-1}})]=\infty.
$$
Note the fact that the sequence of
$\{S_{n_k}-S_{n_{k-1}}\}_{k\geq1}$ is independent for all $k\ge 1$ .
Using the second Borel-Cantelli Lemma, we have
 $$
\liminf_{k\to\infty}|\frac{ S_{n_k}-S_{n_{k-1}} }
{n_k-n_{k-1}}-b|\leq \epsilon,\quad \hbox{a.s.} \; \CV.
$$
But
\begin{equation}\label{ED}
\left|\frac{S_{n_k}}{n_k}-b\right|\leq \left|\frac{
S_{n_k}-S_{n_{k-1}}
}{n_k-n_{k-1}}-b\right|\cdot\frac{n_k-n_{k-1}}{n_k}+\left[\frac{|S_{n_{k-1}}|}{n_{k-1}}+|b|\right]\frac{{n_{k-1}}}{n_k}.
\end{equation}
Noting the following fact,
$$
\frac{n_k-n_{k-1}}{n_k}\to 1, \quad \frac{{n_{k-1}}}{n_k}\to 0,\;
\hbox{as} \; k\to \infty
$$
and
$$
\limsup_{n\to\infty} S_n/n\leq \overline \mu, \quad
\limsup_{n\to\infty} (-S_n)/n\leq -\underline \mu,
$$
which imply
$$
\limsup_{n\to\infty} |S_n|/n\leq \max\{|\overline \mu|, |\underline
\mu|\}<\infty.
$$
Hence, from inequality(\ref{ED}), for any $\epsilon >0,$
$$
\liminf_{k\to\infty}|\frac{S_{n_k}}{n_k}-b|\leq \epsilon, \hbox{
a.s.}\; \CV.
$$ That is
$$\CV\left(\liminf_{n\to \infty} |S_{n}/{n}-b|\leq \epsilon\right)=1.$$
Since $\epsilon$ is arbitrary, we have

$$\CV\left(\liminf_{n\to \infty} |S_{n}/{n}-b|=0\right)=1.
$$
The proof of (III) is complete.

config: arxiv


TITLE: Inequality insde algorithm design (kleinberg), Disjoint Path
QUESTION [0 upvotes]: I am reading the "Algorithm Design" (Kleinberg et al). Inside this book, at the chapter 11.6 there is an inequality I have failed to resolve.
Considering $\beta=m^{(1/3)}$ and $|I|>=1$ (it is a set) divide the follower inequality by $\beta^2$:
$
\beta^2|I^*|<=2(\beta^3|I|+m)+\beta|I|)
$
The book say that this division results:
$
|I^*|<=(4m^{1/3}+1)|I|
$
I try several time to resolve this inequality but I have never succeeded in referring to the above formula.
For more details on this inequality you can see the Kleinberg book at p. 630.
Thanks,
Marco

REPLY [1 votes]: We have of course $\beta^3 = m$, so we can rewrite the original expression as
$\beta^2 |I^*| \leq 2(\beta^3|I| + \beta^3) + \beta |I|$.
Then dividing by $\beta^2$,
$|I^*| \leq 2 \beta |I| + 2 \beta + \frac{|I|}{\beta} = 2m^{1/3} |I| + 2m^{1/3} + \frac{|I|}{m^{1/3}}$.
The two short steps that are glossed over in the proof are to notice that since $|I| \geq 1$, $2m^{1/3} \leq 2m^{1/3}|I|$, and (since presumably $m \geq 1$) that $\frac{|I|}{m^{1/3}} \leq |I|$. Then we have
$2m^{1/3} |I| + 2m^{1/3} + \frac{|I|}{m^{1/3}} \leq 2m^{1/3} |I| + 2m^{1/3} |I| + |I| = (4m^{1/3} + 1)|I|$, as desired.


TITLE: Smallest proposition given a state $\psi$
QUESTION [1 upvotes]: Today, in my lecture the following was mentioned:

Given a state $| \psi \rangle$ in a Hilbert Space $H$, the smallest proposition which is true given this state is given by the projection operator $| \psi \rangle \langle \psi |$, which is the smallest projection operator that projects onto the one-dimensional subspace $| \psi \rangle$.

There are few things I don't understand here. What is meant by smallest projection operator? Also why is the outer product $| \psi \rangle \langle \psi |$ the smallest?

REPLY [0 votes]: This terminology is not familiar to me, but I can think of only one possibility :
For Hermitian operators, there is a notion of positivity (and therefore an order). In this case we would say that a hermitian projector $P$ is smaller than another $Q$ if :
$$\forall |\varphi\rangle, \langle \varphi|P|\varphi\rangle\leq \langle \varphi|Q|\varphi\rangle$$
We see that this is equivalent to the range of $P$ being included in that of $Q$. As any closed subspace containing $|\psi\rangle$ will contain the range of $P$, we see that $P$ is the smallest orthogonal projector with $P|\psi\rangle = |\psi\rangle$.


TITLE: Books in ANT and Galois Theory with medium difficulty
QUESTION [4 upvotes]: I'm interested in learning Algebraic Number Theory and Galois Theory, so I was suggested Ireland Rosen and Stewart for them respectively. But the problem with the  books is that the text is very very easy to understand and almost all the exercises are trivial, so I couldn't understand things properly from this.
Then I was suggested to work on Problems in Algebraic Number Theory by Murty, and I'm finding it to be very hard. Since it's supposed to be an "introductory first read" on ANT, does the fact I'm finding Murty hard suggest I lack the required mathematical maturity to solve the problems from it ?
Also what are some Galois Theory/Algebraic Number Theory somewhere between these two ? The exposition needn't be too hard to follow, but the exercises should at least take some time to solve.

REPLY [3 votes]: I might consider this set of notes written by Matt Baker for some algebraic number theory (minus Galois theory). The preface states

“Algebraic Number Theory is often presented in either a very elementary way which does not take full advantage of students’ backgrounds in Abstract Algebra, or in a very abstract and high-powered way which runs the danger of divorcing itself in the students’ minds from the concrete origins and applications of the subject. We have tried to steer a middle ground between these approaches.”

This seems like it might be a good fit for you. With regard to difficulty related to your knowledge of algebra, I would suggest you keep a copy of Dummit and Foote next to you while you read. If you don’t understand an argument, take a break and read a bit in D&F about the objects being used and then come back to whatever you end up reading.
(Moreover, D&F has some Galois theory in it. It can be a little long-winded at times depending on what you’re looking for, but it is there.)


TITLE: Infinite sums of reciprocal power: $\sum\frac1{n^{2}}$ over odd integers
QUESTION [5 upvotes]: The infinite series I need to solve is
$$\sum_{n=1,3,5...}^{\infty}\frac{1}{n^{2}}$$
and because the point of interest lies in the value of odd n,
the infinite series can be expressed as
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$
This came up in a quantum mechanics problem involving the expectation value of the Hamiltonian.
Is there a good idea to verify the solution is indeed $$\frac{\pi^{2}}{8}$$
or is this something with which I must refer to a math table?
Any good ideas would be helpful.

REPLY [0 votes]: To use $\zeta(2)=\frac{\pi^2}{6}$ (have a look at this historical question) is a fast way to go, but plenty of approaches are possible. An Euler-style proof follows.
Step 1. The function $f(x)=\cos(\pi x)$ is an entire function having simple roots at $\mathbb{Z}+\frac{1}{2}$.
Step 2. The Mittag-Leffler theorem then gives the Weierstrass product:
$$ \cos(\pi z) = \prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2}\right) \tag{2a}$$
where the LHS can be written as a Taylor series:
$$ \cos(\pi z) = 1-\frac{\pi^2 z^2}{2}+\frac{\pi^4 z^4}{24}-\ldots\tag{2b}$$
Step 3. By comparing the coefficient of $z^2$ in the RHS of $(2a)$ and in the RHS of $(2b)$ we get:
$$ -\sum_{n\geq 0}\frac{4}{(2n+1)^2} = -\frac{\pi^2}{2}\tag{3} $$
and your claim immediately follows.


TITLE: How should one interpret an interval like $[2,1]$?
QUESTION [9 upvotes]: Normally, when we're dealing with intervals $[a,b]$, it is at least implied that $a\leq b$. In this case, there are multiple equivalent ways to define what the interval actually is:

$[a,b]=\{x\in\mathbb{R}\colon a\leq x\leq b\}$
$[a,b]:=\operatorname{conv}(\{a,b\})$.

While the former definition is the one usually taught, the latter seems rather more elegant to me, though the convex-hull operator is itself defined by a more complicated set-builder notation. As both are equivalent, the distinction is unnecessary.
It gets interesting when we have $[a,b]$ with $a>b$. Usually, I would consider this simply wrong, but sometimes you have a general interval $[a,b]$ and then it can be tedious and contraproductive to always have to require $b\geq a$. What such an interval is comes out differently from both definitions:

Always the empty set. This is the usual interpretation according to Wikipedia, but it's not necessarily very useful.
The set with "correctly-ordered" borders $[b,a]$.

The latter result seems more useful to me, for instance we have always
$$\mu([a,b]) = \mu([b,a])$$
$$
  \int\limits_a^b\mathrm{d}x\ f(x) = \int_{[a,b]}\!\!\!\mathrm{d}x\ f(x)\cdot\operatorname{sgn}(b-a)
$$
and so on.
What reasons are there to not just always define intervals this way?

REPLY [3 votes]: Because of the usual left to right reading order in European languages, $[a,b]$ has a useful connotation of order. However, your suggestion certainly has merit.  If it were the current convention, there would be no strong arguments for change.
However, it is not the current convention.  In elementary mathematics, changing the convention would instantly make most books very slightly obsolete.  That might make the idea attractive to publishers, but surely not to anyone else.
History has left us with quite a number of unfortunate notations.  If one introduces a new notation, or a new interpretation of a notation, ordinarily one of the following happens: (i) The new notation is not widely adopted, and quickly dies or (ii) The new notation or interpretation is attractive to many, and is adopted by many.  In case (ii), we end up with two notations or interpretations.


TITLE: inequalities like x^{p/q}>3x^{r/t}
QUESTION [0 upvotes]: I don't remember how to resolve in the classical way an equation like
$$x^{4/5}>3x^{1/2}$$
I think to do:
$$x^{4/5}>3x^{1/2} \iff \frac{x^{4/5}}{x^{1/2}}>3 \iff x^{3/10}>3 $$
(obviously for $x^{1/2}\neq 0$)
 but I'm not sure is the correct way...

REPLY [1 votes]: If u multiply or divide the unknown in an inequality it might change depending on x, for a proper way x^4/5 - 3x½ >0 , then take common x½ and you get two inequalities and take the intersection of both the answers, if you do so u will see you might have missed a set of answer.


TITLE: Flux of a vector field.
QUESTION [2 upvotes]: In $\mathbb{R}^3 \setminus \{0\}$, it's given the vector field $$\vec{E}(\vec{r}) = g(r) \vec{r}$$
where $g$ is some function of class $\mathcal{C}^\infty$ defined in $[0, + \infty[$, $\vec{r} = (x,y,z)$ and $r = \sqrt{x^2 + y^2 + z^2}$. Also, $\mathrm{div} \hspace{.5pt}\vec{E}(\vec{r}) = r g'(r) + 3g(r)$. I have to compute the flux of $\vec{E}$ through the sphere $S_a^2$.
My 1st attempt: Denoting by $B_a$ the ball centered in the origin with radius $a$, and using the divergence theorem, I have: $$\int_{S_a^2} \vec{E} \cdot \vec{N} \hspace{.5pt}\mathrm{d}S = \int_{B_a} r g'(r) + 3g(r) \hspace{.5pt} \mathrm{d}V$$
Using spherical coordinates, I would calculate: $$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \left(r g'(r) + 3g(r) \hspace{.5pt}\right)r^2 \sin{\varphi} \mathrm{d}r  \mathrm{d}\varphi  \mathrm{d}\theta = 4\pi \int_{0}^{a} \left(r g'(r) + 3g(r) \hspace{.5pt}\right)r^2 \mathrm{d}r$$
My 2nd attempt: By definition. $$\int_{S_a^2} \vec{E} \cdot \vec{N} \hspace{.5pt}\mathrm{d}S = \int_{[0,2\pi] \times [0, \pi]} \vec{E}(X(\theta, \varphi))\cdot (-a^2 \sin \varphi \cos \varphi, -a^2 \sin^2 \varphi \sin \theta, -a^2 \sin^2 \varphi \cos \theta) \mathrm{d}\theta \mathrm{d}\varphi$$
where $X$ is a parametrization of the sphere using spherical coordinates. However, I'm having trouble computing $\vec{E}(X(\theta,\varphi))$. Does not make any sense to me. I would make $(x,y,z) = (r \sin \varphi \cos \theta, \cdots )$ or $(x,y,z) = (a \sin \varphi \cos \theta, \cdots )$? In the first case, $r$ is a variable but the integral is only on $\theta$ and $\varphi$.
There is no way to give an answer without it being in terms of integrals of $g$, is it? If someone know an easy way to solve this, it is also welcome. Thank you.

REPLY [2 votes]: It would be the latter $(x,y,z)=(a\sin\phi\cos\theta,...)$, here $r=a$ is fixed, because this is a surface integral $(dS)$, rather than a volume integral $(dV)$.
Now with regards to the first integral, I think were in luck:
$4\pi\int_0^a(rg'(r)+3g(r))r^2dr=4\pi\int_0^a r^3g'(r)+3r^2g(r)dr=4\pi\int_0^a \frac{d}{dr}(r^3g(r))dr=4\pi a^3g(a)$