text:
\begin{document}
\title[Constants of the Kahane--Salem--Zygmund inequality]{Constants of the Kahane--Salem--Zygmund inequality asymptotically bounded by
$1$}
\author[D. Pellegrino]{Daniel Pellegrino}
\address{Departamento de Matem\'{a}tica \\
Universidade Federal da Para\'{\i}ba \\
58.051-900 - Jo\~{a}o Pessoa, Brazil.}
\email{daniel.pellegrino@academico.ufpb.br}
\author[A. Raposo Jr.]{Anselmo Raposo Jr.}
\address{Departamento de Matem\'{a}tica \\
Universidade Federal do Maranh\~{a}o \\
65085-580 - S\~{a}o Lu\'{\i}s, Brazil.}
\email{anselmo.junior@ufma.br}
\thanks{D. Pellegrino is supported by CNPq and Grant 2019/0014 Paraiba State Research
Foundation (FAPESQ) }
\subjclass[2010]{47A07, 15A60, 15A51 }
\keywords{Multilinear forms; Hadamard matrices; sequence spaces; Berlekamp's switching game}

\begin{abstract}
The Kahane--Salem--Zygmund inequality for multilinear forms in $\ell_{\infty}$
spaces claims that, for all positive integers $m,n_{1},...,n_{m}$, there
exists an $m$-linear form $A\colon\ell_{\infty}^{n_{1}}\times\cdots\times
\ell_{\infty}^{n_{m}}\longrightarrow\mathbb{K}$ \ ($\mathbb{K}=\mathbb{R}$ or
$\mathbb{C}$) of the type
\[
A(z^{(1)},...,z^{(m)})=\sum_{j_{1}=1}^{n_{1}}\cdots\sum_{j_{m}=1}^{n_{m}}\pm
z_{j_{1}}^{\left(  1\right)  }\cdots z_{j_{m}}^{\left(  m\right)  }\text{,}
\]
satisfying
\[
\Vert A\Vert\leq C_{m}\max\left\{  n_{1}^{1/2},\ldots,n_{m}^{1/2}\right\}
{\textstyle\prod\limits_{j=1}^{m}}n_{j}^{1/2}\text{,}
\]
for
\[
C_{m}\leq\kappa\sqrt{m\log m}\sqrt{m!}
\]
and a certain $\kappa>0.$ Our main result shows that given any $\epsilon>0$
and any positive integer $m,$ there exists a positive integer $N$ such that
\[
C_{m}<1+\epsilon\text{,}
\]
when we consider $n_{1},...,n_{m}>N$. In addition, while the original proof of
the Kahane--Salem--Zygmund relies in highly non-deterministic arguments, our
approach is constructive. We also provide the same asymptotic bound (which is
shown to be optimal in some cases) for the constant of a related
non-deterministic inequality proved by G. Bennett in 1977. Applications to
Berlekamp's switching game are given.

\end{abstract}
\maketitle
\tableofcontents

\section{Introduction}

All along this paper we shall denote $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$
and we shall represent $\mathbb{K}^{n}$ endowed with the $\ell_{p}$-norm by
$\ell_{p}^{n}$; the set of all positive integers shall be denoted by
$\mathbb{N}$. The investigation of bilinear forms and, more generally,
multilinear forms $A\colon\ell_{p_{1}}^{n}\times\cdots\times\ell_{p_{m}}
^{n}\longrightarrow\mathbb{K}$ with coefficients $\pm1$ has important
applications in several fields of mathematics and has been explored by several
authors in different contexts since the beginning of the mid last century.

The ideas that guide this topic rest on the search of unimodular multilinear
forms with relatively small norms, i.e., on the existence of multilinear forms
with coefficients $1$ and $-1$ (sometimes the coefficients are allowed to be
complex numbers with modulus $1$) and smallest possible norms. In 1931,
Bohnenblust and Hille \cite{bh} constructed an $m$-linear form $A_{n}
\colon\ell_{\infty}^{n}\times\cdots\times\ell_{\infty}^{n}\longrightarrow
\mathbb{C}$ with complex coefficients with modulus $1$ satisfying
\[
\left\Vert A_{n}\right\Vert \leq n^{(m+1)/2}
\]
and they have shown that the exponent $\left(  m+1\right)  /2$ is optimal,
i.e., it cannot be replaced by a smaller one (even if we multiply the
right-hand-side of the inequality by a positive constant). However, in many
instances, deterministic arguments are not available and probabilistic methods
come into play. In the 1970's and 1980's several authors have followed this
vein (see \cite{be2,carl,kah,man,var}) and nowadays these probabilistic
approaches are explored in different lines of research (see
\cite{alon,bayart,boasfootball,boas,bk,TBJS,dgm,defantlivro,defmas,mastylo,PRS}
and the references therein). In general, the probabilistic arguments are
enough to provide optimal exponents but the constants involved are not precise.

These classes of inequalities providing unimodular multilinear forms (and
polynomials) with small norms by means of probabilistic methods are usually
called Kahane--Salem--Zygmund inequalities (KSZ inequalities for short).
Recently, important far-reaching generalizations of these inequalities have
been developed by Masty\l o--Szwedek (\cite{mastylo}) and Defant--Masty\l o
(\cite{defmas}).

In this paper we shall be interested in the multilinear versions of the
KSZ\ inequality (see \cite[Lemma 6.1]{jfa} and \cite{alb,PRS}, which are
obtained following the approach of Boas in \cite[Theorem 4]{boas}; see also
\cite[Theorems 1.1 and 1.2]{man}). For the case of bilinear forms, a KSZ-type
inequality was independently proved by Bennett \cite{be1} in 1977 (see also
\cite{be2}), in a quite complete fashion. Applications of the KSZ inequality
can be found, for instance, in \cite{israel,bk,carl,vel,var}. The KSZ
inequality for multilinear forms asserts that for all positive integers $m,n$
and $p_{1},\ldots,p_{m}\in\lbrack1,\infty]$, there exists an $m$-linear form
$A_{n}\colon\ell_{p_{1}}^{n}\times\cdots\times\ell_{p_{m}}^{n}\longrightarrow
\mathbb{K}$ of the type
\[
A_{n}(z^{(1)},\ldots,z^{(m)})=

{\textstyle\sum\limits_{i_{1}=1}^{n}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n}}

\pm z_{i_{1}}^{(1)}\cdots z_{i_{m}}^{(m)}\text{,}
\]
such that
\begin{equation}
\Vert A_{n}\Vert\leq C_{m}n^{\frac{1}{\min\{\max\{2,p_{1}^{\ast}\},\ldots
,\max\{2,p_{m}^{\ast}\}\}}+\sum_{k=1}^{m}\max\left\{  \frac{1}{2}-\frac
{1}{p_{k}},0\right\}  }\text{,} \label{I1}
\end{equation}
with
\begin{equation}
C_{m}=\sqrt{32m\log\left(  6m\right)  }\sqrt{m!}\text{.} \label{I2}
\end{equation}
Above and henceforth, as usual, we consider $1/\infty=0$, the conjugate of $p$
is denoted by $p^{\ast}$, i.e., $p^{\ast}=p/\left(  p-1\right)  $.

For the case of bilinear forms Bennett's approach \cite[Proposition 3.2]{be1}
is more general, allowing different dimensions at the domain of the bilinear
forms. More precisely, Bennett's inequality affirms that there is a constant
$C\geq1$ such that for all $p_{1},p_{2}\in\lbrack1,\infty]$ and all positive
integers $n_{1},n_{2}$, there exists a bilinear form $A_{n_{1},n_{2}}
\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}}^{n_{2}}\longrightarrow\mathbb{R}$
with coefficients $\pm1$ satisfying
\begin{equation}
\left\Vert A_{n_{1},n_{2}}\right\Vert \leq C\max\left\{  n_{2}^{1/p_{2}^{\ast
}}n_{1}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n_{1}
^{1/p_{1}^{\ast}}n_{2}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}
}\right\}  \text{.} \label{I3}
\end{equation}
The KSZ inequality (following Bennett's style) was recently extended to
$m$-linear forms (\cite{alb,PRS}) as follows (see also \cite{defmas} for a
recent far-reaching generalization): let $m,n_{1},\ldots,n_{m}$ be positive
integers and $p_{1},\ldots,p_{m}\in\left[  1,\infty\right]  $. There exist a
constant $C_{m}$ and an $m$-linear form $A_{n_{1},...,n_{m}}\colon\ell_{p_{1}
}^{n_{1}}\times\cdots\times\ell_{p_{m}}^{n_{m}}\longrightarrow\mathbb{K}$ of
the type
\[
A_{n_{1},...,n_{m}}\left(  z^{(1)},\ldots,z^{(m)}\right)  =

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\pm z_{i_{1}}^{(1)}\cdots z_{i_{m}}^{(m)}\text{,}
\]
such that
\begin{equation}
\Vert A_{n_{1},...,n_{m}}\Vert\leq C_{m}\left(

{\textstyle\sum\limits_{k=1}^{m}}

n_{k}\right)  ^{\frac{1}{\min\{\max\{2,p_{1}^{\ast}\},\ldots,\max
\{2,p_{m}^{\ast}\}\}}}

{\textstyle\prod\limits_{k=1}^{m}}

n_{k}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{k}},0\right\}  }\text{.}
\label{I4}
\end{equation}

It is obvious that (\ref{I4}) is equivalent to write
\begin{equation}
\Vert A_{n_{1},...,n_{m}}\Vert\leq D_{m}\max_{k=1,\ldots,m}\left\{
n_{k}^{\frac{1}{\min\{\max\{2,p_{1}^{\ast}\},\ldots,\max\{2,p_{m}^{\ast}\}\}}
}\right\}

{\textstyle\prod\limits_{k=1}^{m}}

n_{k}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{k}},0\right\}  } \label{I5}
\end{equation}
for some constant $D_{m}\geq C_{m}$. A straightforward computation shows that
when $m=2$ and $p_{1},p_{2}\in\lbrack2,\infty]$, the inequality (\ref{I5})
recovers (\ref{I3}). The proof of the above inequalities follows the lines of
the proof of (\ref{I1}) and the constants $C_{m},D_{m}$ are essentially the
constant that appears in (\ref{I2}).

The exponent of $n$ in (\ref{I1}) is optimal (see \cite{PRS}; in fact this
result was previously proved by Mantero--Tonge \cite[Theorem 1.2]{man}) and,
when $p_{1},\ldots,p_{m}\in\lbrack2,\infty]$, all the exponents of
$n_{1},\ldots,n_{m}$ in all the expressions above are also sharp (see
\cite{alb}). On the other hand, the exact values of the constants
$C,C_{m},D_{m}$ are unknown; the approaches of Bennett, Boas and
Mantero--Tonge (\cite{be1,boas,man}) are non-constructive and rely on highly
non deterministic methods, giving no precise hint of the exact value of the
constants involved.

In this paper, among other results, we show that, provided that $p_{1}
=\cdots=p_{m}=\infty$, we have (\ref{I1}), (\ref{I3}) and (\ref{I5}) with
constants \textquotedblleft asymptotically dominated\textquotedblright\ by 1.
More precisely, our first main result provides the aforementioned asymptotic
bounds for (\ref{I3}) and (\ref{I5}) when $p_{1}=\cdots=p_{m}=\infty$ and
reads as follows:

\begin{theorem}
\label{090}Let a positive integer $m$ and $\epsilon>0$ be given. There exists
a positive integer $N$ such that, for all $n_{1},...,n_{m}>N$, there exists an
$m$-linear form $A_{n_{1},\ldots,n_{m}}\colon\ell_{\infty}^{n_{1}}\times
\cdots\times\ell_{\infty}^{n_{m}}\longrightarrow\mathbb{K}$ of the type
\[
A_{n_{1},\ldots,n_{m}}(z^{(1)},...,z^{(m)})=

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\pm z_{i_{1}}^{(1)}\cdots z_{i_{m}}^{(m)}\text{,}
\]
such that
\[
\Vert A_{n_{1},\ldots,n_{m}}\Vert\leq\left(  1+\epsilon\right)  \max\left\{
n_{1}^{1/2},\ldots,n_{m}^{1/2}\right\}

{\textstyle\prod\limits_{j=1}^{m}}

n_{j}^{1/2}.
\]

\end{theorem}

As an immediate consequence we have the same asymptotic bound for (\ref{I1}):

\begin{corollary}
Let a positive integer $m$ and $\epsilon>0$ be given. There exists a positive
integer $N$ such that, for all $n>N$, there exists an $m$-linear form
$A_{n}\colon\ell_{\infty}^{n}\times\cdots\times\ell_{\infty}^{n}
\longrightarrow\mathbb{K}$ of the type
\[
A_{n}(z^{(1)},...,z^{(m)})=

{\textstyle\sum\limits_{i_{1}=1}^{n}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n}}

\pm z_{i_{1}}^{(1)}\cdots z_{i_{m}}^{(m)}\text{,}
\]
such that
\[
\Vert A_{n}\Vert\leq\left(  1+\epsilon\right)  n^{\frac{m+1}{2}}.
\]

\end{corollary}

For the particular case of bilinear forms we prove a more general result
(which, in particular, shows that the constants of Bennett's inequality are
\textquotedblleft uniformly\textquotedblright\ asymptotically bounded by $1$).
For instance, among other results encompassing different dimensions
$n_{1},n_{2}$ (see Figure 2), we show that if $p_{1},p_{2}\in\lbrack1,\infty]$
and $\epsilon>0$, there is a positive integer $N$ (depending just on
$\epsilon$) such that, whenever $n>N$, there is a bilinear form $A\colon
\ell_{p_{1}}^{n}\times\ell_{p_{2}}^{n}\longrightarrow\mathbb{K}$ of the type
\[
A(z^{(1)},z^{(2)})=

{\textstyle\sum\limits_{i=1}^{n}}

{\textstyle\sum\limits_{j=1}^{n}}

\pm z_{i}^{(1)}z_{j}^{(2)}
\]
such that
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  \max\left\{  n^{1/p_{2}^{\ast}
}n^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n^{1/p_{1}^{\ast}
}n^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}  }\right\}
\]
and the constant $1$ is optimal whenever $p_{1},p_{2}\in\lbrack1,2]$.

The paper is organized as follows. The proof of Theorem \ref{090} is presented
in Section 2, and the case of bilinear forms is considered in Section 3. In
Section 4 we present some remarks, including an improvement of the constants
of (\ref{I5}) and in the final section we apply our results to the
Gale--Berlekamp switching game.

\section{The proof of Theorem \ref{090}}

We recall that a Hadamard matrix of order $n$ is a square matrix
$\mathbf{H=}\left[  h_{ij}\right]  _{n\times n}$, with $h_{ij}\in\left\{
-1,1\right\}  $ for all $i,j$, such that
\[
\mathbf{HH}^{\top}=n\mathbf{I}_{n}\text{,}
\]
where $\mathbf{I}_{n}$ is the identity matrix of order $n$ and $\mathbf{H}
^{\top}$ is the transpose of $\mathbf{H}$. Thus, if $u_{1},\ldots,u_{n}$ are
the rows of $\mathbf{H}$, then the inner product of the rows is
\[
\left\langle u_{i},u_{j}\right\rangle =n\delta_{ij}
\]
for all $i,j$, where $\delta_{ij}$ denotes the Kronecker delta. In particular,
the rows of a Hadamard matrix are pairwise orthogonal. For more details we
refer to the book \cite{KJH}. We begin with the following lemma that adapts a
technique borrowed from \cite{bh} (see also \cite{PRS}):

\begin{lemma}
\label{LemaM1}Let $n_{1}\leq n_{2}\leq\cdots\leq n_{m}$ be positive integers
such that for each $k=2,\ldots,m$, there exists a Hadamard matrix
$\mathbf{H}_{n_{k}}=\left[  h_{ij}^{\left(  k\right)  }\right]  _{n_{k}\times
n_{k}}$. Then, the $m$-linear form
\[
A_{0}\colon\ell_{\infty}^{n_{1}}\times\cdots\times\ell_{\infty}^{n_{m}
}\longrightarrow\mathbb{C}
\]
defined by
\[
A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  =

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{i_{2}i_{3}}^{(3)}\cdots h_{i_{m-1}i_{m}}^{(m)}x_{i_{1}
}^{(1)}\cdots x_{i_{m}}^{(m)}
\]
has norm
\[
\left\Vert A_{0}\right\Vert \leq n_{m}^{1/2}

{\textstyle\prod\limits_{j=1}^{m}}

n_{j}^{1/2}\text{.}
\]

\end{lemma}

\begin{proof}
For $k=2,\ldots,m$, let $u_{i}^{\left(  k\right)  }$, $i=1,\ldots,n_{k}$, be
the rows of $\mathbf{H}_{n_{k}}$; hence
\begin{equation}
\left\langle u_{i}^{\left(  k\right)  },u_{j}^{\left(  k\right)
}\right\rangle =n_{k}\delta_{ij}\text{.} \label{M1}
\end{equation}
Let us consider the square matrices of order $n_{m}$ defined by
\[
\left[  h_{ij}^{\left(  k\right)  }\right]  _{n_{m}\times n_{m}}:=\left[
\begin{array}
[c]{cc}
\mathbf{H}_{n_{k}} & \mathbf{0}_{n_{k}\times\left(  n_{m}-n_{k}\right)  }\\
\mathbf{0}_{\left(  n_{m}-n_{k}\right)  \times n_{k}} & \mathbf{0}_{\left(
n_{m}-n_{k}\right)  \times\left(  n_{m}-n_{k}\right)  }
\end{array}
\right]
\]
for each $k=2,\ldots,m$. Define
\[
A_{0}\colon\ell_{\infty}^{n_{1}}\times\cdots\times\ell_{\infty}^{n_{m}
}\longrightarrow\mathbb{C}
\]
by
\[
A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  =

{\textstyle\sum\limits_{i_{1},\dots,i_{m}=1}^{n_{1},\ldots,n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{i_{2}i_{3}}^{(3)}\cdots h_{i_{m-1}i_{m}}^{(m)}x_{i_{1}
}^{(1)}\cdots x_{i_{m}}^{(m)}\text{,}
\]
and note that
\[
h_{i_{1}i_{2}}^{(2)}h_{i_{2}i_{3}}^{(3)}\cdots h_{i_{m-1}i_{m}}^{(m)}
\in\left\{  -1,1\right\}
\]
whenever $i_{k}\in\{1,\ldots,n_{k}\}$ for all $k=1,\ldots,m$. For each
$k=1,\ldots,m$, given $x^{(k)}$ in the closed unit ball $B_{\ell_{\infty
}^{n_{k}}}$, let us denote
\[
y^{(k)}=\left(  x_{1}^{(k)},\ldots,x_{n_{k}}^{(k)},0,\ldots,0\right)  \in
B_{\ell_{\infty}^{n_{m}}}\text{.}
\]
Then, by the H\"{o}lder inequality,
\begin{align*}
&  \left\vert A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  \right\vert
=\left\vert

{\textstyle\sum\limits_{i_{1},\dots,i_{m}=1}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{i_{2}i_{3}}^{(3)}\cdots h_{i_{m-1}i_{m}}^{(m)}y_{i_{1}
}^{(1)}\cdots y_{i_{m}}^{(m)}\right\vert \\
&  \leq

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\left\vert

{\textstyle\sum\limits_{i_{1},\dots,i_{m-1}=1}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{i_{2}i_{3}}^{(3)}\cdots h_{i_{m-1}i_{m}}^{(m)}y_{i_{1}
}^{(1)}\cdots y_{i_{m-1}}^{(m-1)}\right\vert \left\vert y_{i_{m}}^{\left(
m\right)  }\right\vert \\
&  \leq\left(

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

|y_{i_{m}}^{(m)}|^{2}\right)  ^{1/2}\cdot\left(

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\left\vert

{\textstyle\sum\limits_{i_{1},\dots,i_{m-1}=1}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{i_{2}i_{3}}^{(3)}\cdots h_{i_{m-1}i_{m}}^{(m)}y_{i_{1}
}^{(1)}\cdots y_{i_{m-1}}^{(m-1)}\right\vert ^{2}\right)  ^{1/2}\\
&  \leq n_{m}^{1/2}\left(

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-1}=1\\j_{1},\dots
,j_{m-1}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-2}i_{m-1}}
^{(m-1)}h_{j_{m-2}j_{m-1}}^{(m-1)}h_{i_{m-1}i_{m}}^{(m)}h_{j_{m-1}i_{m}}
^{(m)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}}\cdots y_{i_{m-1}}
^{(m-1)}\overline{y_{j_{m-1}}^{(m-1)}}\right)  ^{1/2}\text{.}
\end{align*}
Thus,
\begin{align*}
&  \left\vert A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  \right\vert \\
&  \leq n_{m}^{1/2}\left(

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-1}=1\\j_{1},\dots
,j_{m-1}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-2}i_{m-1}}
^{(m-1)}h_{j_{m-2}j_{m-1}}^{(m-1)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-1}}^{(m-1)}\overline{y_{j_{m-1}}^{(m-1)}}

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

h_{i_{m-1}i_{m}}^{(m)}h_{j_{m-1}i_{m}}^{(m)}\right)  ^{1/2}\\
&  =n_{m}^{1/2}\left(

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-1}=1\\j_{1},\dots
,j_{m-1}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-2}i_{m-1}}
^{(m-1)}h_{j_{m-2}j_{m-1}}^{(m-1)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-1}}^{(m-1)}\overline{y_{j_{m-1}}^{(m-1)}}\left\langle
u_{i_{m-1}}^{\left(  m\right)  },u_{j_{m-1}}^{\left(  m\right)  }\right\rangle
\right)  ^{1/2}
\end{align*}
and, by (\ref{M1}), we have
\begin{align*}
&  \left\vert A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  \right\vert \\
&  \leq n_{m}^{1/2}\left(

{\textstyle\sum\limits_{i_{m-1}=1}^{n_{m}}}

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-2}=1\\j_{1},\dots
,j_{m-2}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-2}i_{m-1}}
^{(m-1)}h_{j_{m-2}i_{m-1}}^{(m-1)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-2}}^{(m-2)}\overline{y_{j_{m-2}}^{(m-2)}}\left\vert
y_{i_{m-1}}^{(m-1)}\right\vert ^{2}n_{m}\right)  ^{1/2}\\
&  \leq n_{m}\left(

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-2}=1\\j_{1},\dots
,j_{m-2}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-3}i_{m-2}}
^{(m-2)}h_{j_{m-3}j_{m-2}}^{(m-2)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-2}}^{(m-2)}\overline{y_{j_{m-2}}^{(m-2)}}

{\textstyle\sum\limits_{i_{m-1}=1}^{n_{m}}}

h_{i_{m-2}i_{m-1}}^{(m-1)}h_{j_{m-2}i_{m-1}}^{(m-1)}\right)  ^{1/2}\\
&  =n_{m}\left(

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-2}=1\\j_{1},\dots
,j_{m-2}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-3}i_{m-2}}
^{(m-2)}h_{j_{m-3}j_{m-2}}^{(m-2)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-2}}^{(m-2)}\overline{y_{j_{m-2}}^{(m-2)}}\left\langle
u_{i_{m-2}}^{\left(  m-1\right)  },u_{j_{m-2}}^{\left(  m-1\right)
}\right\rangle \right)  ^{1/2}\text{.}
\end{align*}
Since
\begin{align*}
&  \left(

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-2}=1\\j_{1},\dots
,j_{m-2}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-3}i_{m-2}}
^{(m-2)}h_{j_{m-3}j_{m-2}}^{(m-2)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-2}}^{(m-2)}\overline{y_{j_{m-2}}^{(m-2)}}\left\langle
u_{i_{m-2}}^{\left(  m-1\right)  },u_{j_{m-2}}^{\left(  m-1\right)
}\right\rangle \right)  ^{1/2}\\
&  =n_{m-1}^{1/2}\left(

{\textstyle\sum\limits_{i_{m-2}=1}^{n_{m}}}

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-3}=1\\j_{1},\dots
,j_{m-3}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-3}i_{m-2}}
^{(m-2)}h_{j_{m-3}i_{m-2}}^{(m-2)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-3}}^{(m-3)}\overline{y_{j_{m-3}}^{(m-3)}}y_{i_{m-2}}
^{(m-2)}\overline{y_{i_{m-2}}^{(m-2)}}\right)  ^{1/2}\\
&  =n_{m-1}^{1/2}\left(

{\textstyle\sum\limits_{i_{m-2}=1}^{n_{m}}}

|y_{i_{m-2}}^{(m-2)}|^{2}

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-3}=1\\j_{1},\dots
,j_{m-3}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-3}i_{m-2}}
^{(m-2)}h_{j_{m-3}i_{m-2}}^{(m-2)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-3}}^{(m-3)}\overline{y_{j_{m-3}}^{(m-3)}}\right)
^{1/2}\text{,}
\end{align*}
we have
\begin{align*}
&  \left\vert A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  \right\vert \\
&  \leq n_{m}n_{m-1}^{1/2}\left(

{\textstyle\sum\limits_{i_{m-2}=1}^{n_{m}}}

{\textstyle\sum\limits_{\substack{i_{1},\dots,i_{m-3}=1\\j_{1},\dots
,j_{m-3}=1}}^{n_{m}}}

h_{i_{1}i_{2}}^{(2)}h_{j_{1}j_{2}}^{(2)}\cdots h_{i_{m-3}i_{m-2}}
^{(m-2)}h_{j_{m-3}i_{m-2}}^{(m-2)}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}\cdots y_{i_{m-3}}^{(m-3)}\overline{y_{j_{m-3}}^{(m-3)}}\right)  ^{1/2}
\end{align*}
and, repeating this procedure, we finally obtain
\begin{align*}
\left\vert A_{0}\left(  x^{(1)},\dots,x^{(m)}\right)  \right\vert  &  \leq
n_{m}n_{m-1}^{1/2}\cdots n_{2}^{1/2}\left(

{\textstyle\sum\limits_{i_{1}=1}^{n_{m}}}

\left\vert y_{i_{1}}^{(1)}\right\vert ^{2}\right)  ^{1/2}\\
&  \leq n_{m}^{1/2}

{\textstyle\prod\limits_{j=1}^{m}}

n_{j}^{1/2}\text{.}
\end{align*}

\end{proof}

Now we are able to prove Theorem \ref{090}.

\begin{proof}
[Proof of Theorem \ref{090}]It sufficies to consider $\mathbb{K}=\mathbb{C}$;
the case of real scalars is a straightforward consequence. We shall divide the
proof into three steps. Let $n_{1},\ldots,n_{m}$ be positive integers.

\bigskip

\noindent\textbf{First Step. }(See \cite[p. 295]{JSpencer}) Given $\delta>0$,
there exists a positive integer $N$ such that, for all $n_{k}>N$ there is a
Hadamard order $t_{k}$ such that
\begin{equation}
n_{k}\leq t_{k}<\left(  1+\delta\right)  n_{k}\text{, \ \ \ \ }k=1,\ldots
,m\text{.} \label{M2}
\end{equation}
As commented in \cite[p. 295]{JSpencer}, it suffices to observe that there are
Hadamard matrices of order $4^{i}12^{j}$ for all $i,j$. Since the sequence
defined recursively by
\[
x_{1}=1\text{ \ \ \ \ and \ \ \ \ }x_{n+1}=\min\left\{  \left\{  4^{i}
12^{j}:\left(  i,j\right)  \in\left\{  0,1,2,\ldots\right\}  \times\left\{
0,1,2,\ldots\right\}  \right\}  \backslash\left\{  x_{1},\ldots,x_{n}\right\}
\right\}
\]
is such that $\lim\limits_{n\rightarrow\infty}\dfrac{x_{n+1}}{x_{n}}=1$, a
straightforward calculation assures (\ref{M2}).

\begin{figure}[H]
\centering\includegraphics[scale=0.8]{grafico.pdf}\caption{Graphic of
$\frac{x_{n+1}}{x_{n}}$}
\end{figure}

\bigskip

\noindent\textbf{Second Step}. Notice that we can assume, without loss of
generality, $t_{1}\leq\cdots\leq t_{m}$. It follows from Lemma \ref{LemaM1}
that, for the integers $t_{1},\ldots,t_{m}$ of the first step, there exists an
$m$-linear form $A_{0}\colon\ell_{\infty}^{t_{1}}\times\cdots\times
\ell_{\infty}^{t_{m}}\longrightarrow\mathbb{C}$ with coefficients $\pm1$ such
that
\[
\left\Vert A_{0}\right\Vert \leq t_{m}^{1/2}

{\textstyle\prod\limits_{k=1}^{m}}

t_{k}^{1/2}\text{.}
\]

\bigskip

\noindent\textbf{Third Step. }If $n_{k}=t_{k}$ for all $k$, we just consider
$A_{n_{1},\ldots,n_{m}}=A_{0}$. Otherwise, we define
\[
A_{n_{1},\ldots,n_{m}}\colon\ell_{\infty}^{n_{1}}\times\cdots\times
\ell_{\infty}^{n_{m}}\longrightarrow\mathbb{C}
\]
by
\[
A_{n_{1},\ldots,n_{m}}\left(  \left(  z_{i_{1}}^{\left(  1\right)  }\right)
_{i_{1}=1}^{n_{1}},\ldots,\left(  z_{i_{m}}^{\left(  m\right)  }\right)
_{i_{m}=1}^{n_{m}}\right)  =A_{0}\left(  \left(  z_{1}^{\left(  1\right)
},\ldots,z_{n_{1}}^{\left(  1\right)  },0,\ldots,0\right)  ,\ldots,\left(
z_{1}^{\left(  m\right)  },\ldots,z_{n_{m}}^{\left(  m\right)  }
,0,\ldots,0\right)  \right)
\]
Then, given $x^{\left(  k\right)  }\in B_{\ell_{\infty}^{n_{k}}}$,
$k=1,\ldots,m$, we have
\begin{align*}
&  \left\vert A_{n_{1},\ldots,n_{m}}\left(  \left(  x_{i_{1}}^{\left(
1\right)  }\right)  _{i_{1}=1}^{n_{1}},\ldots,\left(  x_{i_{m}}^{\left(
m\right)  }\right)  _{i_{m}=1}^{n_{m}}\right)  \right\vert \\
&  =\left\vert A_{0}\left(  \left(  x_{1}^{\left(  1\right)  },\ldots
,x_{n_{1}}^{\left(  1\right)  },0,\ldots,0\right)  ,\ldots,\left(
x_{1}^{\left(  m\right)  },\ldots,x_{n_{m}}^{\left(  m\right)  }
,0,\ldots,0\right)  \right)  \right\vert \\
&  \leq\left\Vert A_{0}\right\Vert \leq t_{m}^{1/2}

{\textstyle\prod\limits_{k=1}^{m}}

t_{k}^{1/2}\\
&  \leq\left(  1+\delta\right)  ^{\frac{m+1}{2}}\max\left\{  n_{1}
^{1/2},\ldots,n_{m}^{1/2}\right\}

{\textstyle\prod\limits_{j=1}^{m}}

n_{j}^{1/2}
\end{align*}
and this ends the proof.
\end{proof}

\begin{remark}
A famous conjecture of Hadamard proposes that there exists a Hadamard matrix
of order $4k$ for every positive integer $k$. If this is true or, at least, if
the gap between two consecutive orders of Hadamard matrices is bounded by a
universal constant, a straightforward modification of our proof shows that
there exists an $m$-linear form $A_{n}\colon\ell_{\infty}^{n}\times
\cdots\times\ell_{\infty}^{n}\longrightarrow\mathbb{K}$ of the type
\[
A_{n}(z^{(1)},\ldots,z^{(m)})=

{\textstyle\sum\limits_{j_{1}=1}^{n}}

\cdots

{\textstyle\sum\limits_{j_{m}=1}^{n}}

\pm z_{j_{1}}^{(1)}\cdots z_{j_{m}}^{(m)}\text{,}
\]
such that
\[
\left\Vert A_{n}\right\Vert \leq\left(  1+O\left(  n^{-1/2}\right)  \right)
n^{\frac{m+1}{2}}\text{.}
\]

\end{remark}

\begin{corollary}
Let a positive integer $m$ and $\epsilon>0$ be given. There exists a positive
integer $N$ such that, for all $n>N$, there exists an $m$-linear form
$A_{n}\colon\ell_{\infty}^{n}\times\cdots\times\ell_{\infty}^{n}
\longrightarrow\mathbb{K}$ of the type
\[
A_{n}(z^{(1)},\ldots,z^{(m)})=

{\textstyle\sum\limits_{j_{1}=1}^{n}}

\cdots

{\textstyle\sum\limits_{j_{m}=1}^{n}}

\pm z_{j_{1}}^{(1)}\cdots z_{j_{m}}^{(m)}\text{,}
\]
satisfying
\[
\Vert A_{n}\Vert\leq\left(  1+\epsilon\right)  n^{\frac{m+1}{2}}\text{.}
\]

\end{corollary}

\section{Constants of Bennett's inequality}

In this section we shall show that, in most of the cases (see Figure 2), the
constants of Bennett's inequality (\ref{I3}) are \textquotedblleft
uniformly\textquotedblright\ asymptotically bounded by $1$ and, besides, the
constant $1$ will be shown to be sharp in some cases (see Corollary
\ref{op71}). We begin with a multipurpose simple lemma that shall be also used
in the next section. For the sake of simplicity, we use the notation
$\mathbb{N}_{k}$ to represent the set $\{k,k+1,k+2,\ldots\}.$

\begin{lemma}
\label{LemaB1}Let $k,m\in\mathbb{N}$ and let $f\colon\mathbb{N}_{k}
\longrightarrow\lbrack0,\infty)$ be an increasing function. If for all
$n\in\mathbb{N}_{k}$ there exists an $m$-linear form $A_{0}\colon\ell_{2}
^{n}\times\cdots\times\ell_{2}^{n}\longrightarrow\mathbb{K}$ with coefficients
$\pm1$ such that
\[
\left\Vert A_{0}\right\Vert \leq f\left(  n\right)  \text{,}
\]
then, for all $p_{1},\ldots,p_{m}\in\left[  2,\infty\right]  $ and all
$n_{1},\ldots,n_{m}\in\mathbb{N}_{k}$ there is an $m$-linear form $A\colon
\ell_{p_{1}}^{n_{1}}\times\cdots\times\ell_{p_{m}}^{n_{m}}\longrightarrow
\mathbb{K}$ with coefficients $\pm1$ such that
\[
\left\Vert A\right\Vert \leq\max\left\{  f\left(  n_{1}\right)  ,\ldots
,f\left(  n_{m}\right)  \right\}  n_{1}^{\frac{1}{2}-\frac{1}{p_{1}}}\cdots
n_{m}^{\frac{1}{2}-\frac{1}{p_{m}}}\text{.}
\]

\end{lemma}

\begin{proof}
Let $n=\max\left\{  n_{1},\ldots,n_{m}\right\}  $. From our hypothesis, there
is an $m$-linear form $A_{0}\colon\ell_{2}^{n}\times\cdots\times\ell_{2}
^{n}\longrightarrow\mathbb{K}$ of the type
\[
A_{0}\left(  x^{\left(  1\right)  },\ldots,x^{\left(  m\right)  }\right)  =

{\textstyle\sum\limits_{i_{1}=1}^{n}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n}}

\pm x_{i_{1}}^{\left(  1\right)  }\cdots x_{i_{m}}^{\left(  m\right)  }
\]
such that
\[
\left\Vert A_{0}\right\Vert \leq f\left(  n\right)  \text{.}
\]
Let us consider the $m$-linear form $A_{1}\colon\ell_{2}^{n_{1}}\times
\cdots\times\ell_{2}^{n_{m}}\longrightarrow\mathbb{K}$ defined by
\[
A_{1}\left(  x^{\left(  1\right)  },\ldots,x^{\left(  m\right)  }\right)
=A_{0}\left(  \left(  x_{1}^{\left(  1\right)  },\ldots,x_{n_{1}}^{\left(
1\right)  },0,\ldots,0\right)  ,\ldots,\left(  x_{1}^{\left(  m\right)
},\ldots,x_{n_{m}}^{\left(  m\right)  },0,\ldots,0\right)  \right)  \text{.}
\]
It is plain that
\begin{equation}
\left\Vert A_{1}\right\Vert \leq\left\Vert A_{0}\right\Vert \leq f\left(
n\right)  =\max\left\{  f\left(  n_{1}\right)  ,\ldots,f\left(  n_{m}\right)
\right\}  \text{.} \label{B1}
\end{equation}
Now, let us define $A\colon\ell_{p_{1}}^{n_{1}}\times\cdots\times\ell_{p_{m}
}^{n_{m}}\longrightarrow\mathbb{K}$ by
\[
A\left(  x^{\left(  1\right)  },\ldots,x^{\left(  m\right)  }\right)
=A_{1}\left(  x^{\left(  1\right)  },\ldots,x^{\left(  m\right)  }\right)
\text{.}
\]
In this case,
\begin{align}
\left\Vert A\right\Vert  &  =\sup_{\left\Vert x^{\left(  k\right)
}\right\Vert _{p_{k}}\leq1}\left\vert

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\pm x_{i_{1}}^{\left(  1\right)  }\cdots x_{i_{m}}^{\left(  m\right)
}\right\vert \nonumber\\
&  =n_{1}^{\frac{1}{2}-\frac{1}{p_{1}}}\cdots n_{m}^{\frac{1}{2}-\frac
{1}{p_{m}}}\sup_{\left\Vert x^{\left(  k\right)  }\right\Vert _{p_{k}}\leq
1}\left\vert

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\pm\frac{x_{i_{1}}^{\left(  1\right)  }}{n_{1}^{\frac{1}{2}-\frac{1}{p_{1}}}
}\cdots\frac{x_{i_{m}}^{\left(  m\right)  }}{n_{m}^{\frac{1}{2}-\frac{1}
{p_{m}}}}\right\vert \text{.} \label{B2}
\end{align}
For all $k=1,\ldots,m$, let $z^{\left(  k\right)  }=n_{k}^{\frac{1}{p_{k}
}-\frac{1}{2}}x^{\left(  k\right)  }$. By the H\"{o}lder inequality, since
$\left\Vert x^{\left(  k\right)  }\right\Vert _{p_{k}}\leq1$, we have
\[
\left\Vert z^{\left(  k\right)  }\right\Vert _{2}\leq1\text{.}
\]
It follows from (\ref{B1}) and (\ref{B2}) that
\begin{align*}
\left\Vert A\right\Vert  &  \leq n_{1}^{\frac{1}{2}-\frac{1}{p_{1}}}\cdots
n_{m}^{\frac{1}{2}-\frac{1}{p_{m}}}\sup_{\left\Vert z^{\left(  k\right)
}\right\Vert _{2}\leq1}\left\vert

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\pm z_{i_{1}}^{\left(  1\right)  }\cdots z_{i_{m}}^{\left(  m\right)
}\right\vert \\
&  =\left\Vert A_{1}\right\Vert n_{1}^{\frac{1}{2}-\frac{1}{p_{1}}}\cdots
n_{m}^{\frac{1}{2}-\frac{1}{p_{m}}}\\
&  \leq\max\left\{  f\left(  n_{1}\right)  ,\ldots,f\left(  n_{m}\right)
\right\}  n_{1}^{\frac{1}{2}-\frac{1}{p_{1}}}\cdots n_{m}^{\frac{1}{2}
-\frac{1}{p_{m}}}\text{.}
\end{align*}

\end{proof}

Now we prove another lemma which plays a fundamental role in this section.

\begin{lemma}
Let $\epsilon>0$. There exists a positive integer $N$ $\mathrm{(}$depending
just on $\epsilon\mathrm{)}$ such that, whenever $n>N$, there is a bilinear
form $A\colon\ell_{2}^{n}\times\ell_{2}^{n}\longrightarrow\mathbb{K}$ of the
type
\[
A\left(  z^{\left(  1\right)  },z^{\left(  2\right)  }\right)  =

{\textstyle\sum\limits_{i=1}^{n}}

{\textstyle\sum\limits_{j=1}^{n}}

\pm z_{i}^{\left(  1\right)  }z_{j}^{\left(  2\right)  }\text{,}
\]
such that
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  n^{1/2}\text{.}
\]

\end{lemma}

\begin{proof}
It suffices to consider $\mathbb{K}=\mathbb{C}$. Let $\left[  h_{ij}\right]
_{n\times n}$ be a Hadamard matrix of order $n$. It is simple to show that the
bilinear form $A_{0}\colon\ell_{2}^{n}\times\ell_{2}^{n}\longrightarrow
\mathbb{C}$ given by
\begin{equation}
A_{0}(z^{(1)},z^{(2)})=

{\textstyle\sum\limits_{i=1}^{n}}

{\textstyle\sum\limits_{j=1}^{n}}

h_{ij}z_{i}^{(1)}z_{j}^{(2)}\text{,} \label{B3}
\end{equation}
has norm
\[
\Vert A_{0}\Vert\leq n^{1/2}\text{.}
\]
In fact, if $x^{\left(  1\right)  },x^{\left(  2\right)  }\in B_{\ell_{2}^{n}
}$, by the Cauchy-Schwarz inequality, we have
\begin{align*}
\left\vert A_{0}\left(  x^{\left(  1\right)  },x^{\left(  2\right)  }\right)
\right\vert  &  \leq

{\textstyle\sum\limits_{j=1}^{n}}

\left\vert

{\textstyle\sum\limits_{i=1}^{n}}

h_{ij}x_{i}^{(1)}\right\vert \left\vert x_{j}^{\left(  2\right)  }\right\vert
\\
&  \leq\left(

{\textstyle\sum\limits_{j=1}^{n}}

\left\vert x_{j}^{(2)}\right\vert ^{2}\right)  ^{1/2}\cdot\left(

{\textstyle\sum\limits_{j=1}^{n}}

\left\vert

{\textstyle\sum\limits_{i=1}^{n}}

h_{ij}x_{i}^{(1)}\right\vert ^{2}\right)  ^{1/2}\\
&  \leq\left(

{\textstyle\sum\limits_{j=1}^{n}}

\,

{\textstyle\sum\limits_{i,k=1}^{n}}

h_{ij}h_{kj}x_{i}^{(1)}\overline{x_{k}^{(1)}}\right)  ^{1/2}\\
&  =\left(

{\textstyle\sum\limits_{i,k=1}^{n}}

x_{i}^{(1)}\overline{x_{k}^{(1)}}n\delta_{ik}\right)  ^{1/2}\\
&  \leq n^{1/2}\text{.}
\end{align*}
We shall show that for other values of $n$, we have the same inequality, just
with an extra multiplicative \textquotedblleft asymptotic
factor\textquotedblright\ $\left(  1+\epsilon\right)  $.

As in the first step of the proof of Theorem \ref{090}, given $\delta>0$,
there exists a positive integer $N$ such that, whenever $n>N$, there is a
Hadamard matrix of order $t$ satisfying
\begin{equation}
n\leq t\leq n\left(  1+\delta\right)  \text{.} \label{B4}
\end{equation}
Now, let us consider a Hadamard matrix $\left[  h_{ij}\right]  _{t\times t}$
of order $t$. Let $A_{0}\colon\ell_{2}^{t}\times\ell_{2}^{t}\longrightarrow
\mathbb{C}$ be as in (\ref{B3}). Thus,
\[
\left\Vert A_{0}\right\Vert \leq t^{1/2}\text{.}
\]

If $n=t$, we just consider $A=A_{0}$. If $n<t$, we define
\[
A\colon\ell_{2}^{n}\times\ell_{2}^{n}\longrightarrow\mathbb{C}\text{,}
\]
by
\[
A\left(  \left(  z_{i}^{\left(  1\right)  }\right)  _{i=1}^{n},\left(
z_{j}^{\left(  2\right)  }\right)  _{j=1}^{n}\right)  =A_{0}\left(  \left(
z_{1}^{\left(  1\right)  },\ldots,z_{n}^{\left(  1\right)  },0,\ldots
,0\right)  ,\left(  z_{1}^{\left(  2\right)  },\ldots,z_{n}^{\left(  2\right)
},0,\ldots,0\right)  \right)  \text{.}
\]
Then, given $x^{\left(  1\right)  },x^{\left(  2\right)  }\in B_{\ell_{2}^{n}
}$, by (\ref{B4}),
\begin{align*}
\left\vert A\left(  \left(  x_{i}^{\left(  1\right)  }\right)  _{i=1}
^{n},\left(  x_{j}^{\left(  2\right)  }\right)  _{j=1}^{n}\right)
\right\vert  &  =\left\vert A_{0}\left(  \left(  x_{1}^{\left(  1\right)
},\ldots,x_{n}^{\left(  1\right)  },0,\ldots,0\right)  ,\left(  x_{1}^{\left(
2\right)  },\ldots,x_{n}^{\left(  2\right)  },0,\ldots,0\right)  \right)
\right\vert \\
&  \leq\left\Vert A_{0}\right\Vert \leq t^{1/2}\leq\left(  1+\delta\right)
^{1/2}n^{1/2}
\end{align*}
and the proof is complete.
\end{proof}

Combining the above result with Lemma \ref{LemaB1} we obtain the main result
of this section:

\begin{theorem}
\label{TeoB3}Let $\epsilon>0$ and $p_{1},p_{2}\in\lbrack2,\infty]$. There
exists a positive integer $N$ $\mathrm{(}$depending just on $\epsilon
\mathrm{)}$ such that, whenever $\max\left\{  n_{1},n_{2}\right\}  >N$, there
is a bilinear form $A\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}}^{n_{2}
}\longrightarrow\mathbb{K}$ of the type
\[
A(z^{(1)},z^{(2)})=

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\pm z_{i}^{(1)}z_{j}^{(2)}\text{,}
\]
such that
\begin{equation}
\Vert A\Vert\leq\left(  1+\epsilon\right)  \max\left\{  n_{2}^{1/p_{2}^{\ast}
}n_{1}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n_{1}
^{1/p_{1}^{\ast}}n_{2}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}
}\right\}  \text{.} \label{B5}
\end{equation}

\end{theorem}

Now we prove three corollaries that essentially cover the remaining cases (see
Figure \ref{Fig2}):

\begin{corollary}
\label{CorB4}Let $\epsilon>0$ and $1\leq p_{1}<2\leq p_{2}$ $\mathrm{(}1\leq
p_{2}<2\leq p_{1}\mathrm{)}$. There exists a positive integer $N$ $\mathrm{(}
$depending just on $\epsilon\mathrm{)}$ such that, whenever $n_{2}>N$ and
$n_{1}\leq n_{2}$ $\mathrm{(}n_{1}\geq N$ and $n_{2}\leq n_{1}\mathrm{)}$,
there is a bilinear form $A\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}}
^{n_{2}}\longrightarrow\mathbb{K}$ of the type
\[
A(x,y)=

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\pm x_{i}y_{j}\text{,}
\]
such that
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  \max\left\{  n_{2}^{1/p_{2}^{\ast}
}n_{1}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n_{1}
^{1/p_{1}^{\ast}}n_{2}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}
}\right\}  \text{.}
\]

\end{corollary}

\begin{proof}
Let us assume $1\leq p_{1}<2\leq p_{2}$ and $n_{1}\leq n_{2}$. Notice that in
this case
\[
\max\left\{  n_{2}^{1/p_{2}^{\ast}}n_{1}^{\max\left\{  \frac{1}{2}-\frac
{1}{p_{1}},0\right\}  },n_{1}^{1/p_{1}^{\ast}}n_{2}^{\max\left\{  \frac{1}
{2}-\frac{1}{p_{2}},0\right\}  }\right\}  =n_{2}^{1/p_{2}^{\ast}}\text{.}
\]
By Theorem \ref{TeoB3}, given $\epsilon>0$, there exists $N$ (depending just
on $\epsilon$) such that, if $n_{2}>N$, there is a bilinear form $A_{1}
\colon\ell_{2}^{n_{1}}\times\ell_{p_{2}}^{n_{2}}\longrightarrow\mathbb{K}$ of
the type
\[
A_{1}(z^{\left(  1\right)  },z^{\left(  2\right)  })=

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\pm z_{i}^{\left(  1\right)  }z_{j}^{\left(  2\right)  }\text{,}
\]
satisfying
\[
\Vert A_{1}\Vert\leq\left(  1+\epsilon\right)  n_{2}^{1/p_{2}^{\ast}}\text{.}
\]
Now, define $A\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}}^{n_{2}
}\longrightarrow\mathbb{K}$ by $A\left(  x,y\right)  =A_{1}\left(  x,y\right)
$ and note that
\[
\left\Vert A\right\Vert \leq\Vert A_{1}\Vert\leq\left(  1+\epsilon\right)
n_{2}^{1/p_{2}^{\ast}}\text{.}
\]
The case $1\leq p_{2}<2\leq p_{1}$ and $n_{2}\leq n_{1}$ is analogous.
\end{proof}

\begin{corollary}
\label{CorB5}Let $\epsilon>0$ and $p\in\lbrack1,\infty]$. There exists a
positive integer $N$ $\mathrm{(}$depending just on $\epsilon\mathrm{)}$ such
that, whenever $\max\left\{  n_{1},n_{2}\right\}  >N$, there is a bilinear
form $A\colon\ell_{p}^{n_{1}}\times\ell_{p}^{n_{2}}\longrightarrow\mathbb{K}$
of the type
\[
A(z^{(1)},z^{(2)})=

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\pm z_{i}^{(1)}z_{j}^{(2)}\text{,}
\]
such that
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  \max\left\{  n_{2}^{1/p^{\ast}
}n_{1}^{\max\left\{  \frac{1}{2}-\frac{1}{p},0\right\}  },n_{1}^{1/p^{\ast}
}n_{2}^{\max\left\{  \frac{1}{2}-\frac{1}{p},0\right\}  }\right\}  \text{.}
\]

\end{corollary}

\begin{proof}
It is suffices to consider $p\in\left[  1,2\right)  $, because the case
$p\in\left[  2,\infty\right]  $ is solved in Theorem \ref{TeoB3}. If
$n_{1},n_{2}\in\mathbb{N}$ and $\varepsilon_{ij}\in\left\{  -1,1\right\}  $
for each $\left(  i,j\right)  \in\left\{  1,\ldots,n_{1}\right\}
\times\left\{  1,\ldots,n_{2}\right\}  $, it is obvious that all bilinear
forms $A\colon\ell_{1}^{n_{1}}\times\ell_{1}^{n_{2}}\longrightarrow\mathbb{C}$
defined by
\begin{equation}
A\left(  z^{\left(  1\right)  },z^{\left(  2\right)  }\right)  =

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\varepsilon_{ij}z_{i}^{\left(  1\right)  }z_{j}^{\left(  2\right)  }
\label{B6}
\end{equation}
have norm precisely $1$. If $1\leq p<2$, by the Riesz-Thorin Theorem applied
to (\ref{B6}) and (\ref{B5}) with $p_{1}=p_{2}=2$, we can find a bilinear form
$A\colon\ell_{p}^{n_{1}}\times\ell_{p}^{n_{2}}\longrightarrow\mathbb{C}$
satisfying
\begin{align*}
\Vert A\Vert &  \leq\left(  1+\epsilon\right)  ^{2\left(  1-\frac{1}
{p}\right)  }\max\left\{  n_{1}^{1-\frac{1}{p}},n_{2}^{1-\frac{1}{p}}\right\}
\\
&  \leq\left(  1+\epsilon\right)  \max\left\{  n_{1}^{1-\frac{1}{p}}
,n_{2}^{1-\frac{1}{p}}\right\} \\
&  =\left(  1+\epsilon\right)  \max\left\{  n_{2}^{1/p^{\ast}}n_{1}
^{\max\left\{  \frac{1}{2}-\frac{1}{p},0\right\}  },n_{1}^{1/p^{\ast}}
n_{2}^{\max\left\{  \frac{1}{2}-\frac{1}{p},0\right\}  }\right\}  \text{.}
\end{align*}

\end{proof}

\begin{corollary}
\label{CorB6}Let $\epsilon>0$ and $1\leq p_{1}\leq p_{2}\leq2$ $\mathrm{(}
1\leq p_{2}\leq p_{1}\leq2\mathrm{)}$. There exists a positive integer
$N$,$\mathrm{\ }$depending just on $\epsilon,$ such that, whenever $n_{2}>N$
$\mathrm{(}n_{1}>N\mathrm{)}$ and $n_{1}\leq n_{2}$ $\mathrm{(}n_{2}\leq
n_{1}\mathrm{)}$ there is a bilinear form $A\colon\ell_{p_{1}}^{n_{1}}
\times\ell_{p_{2}}^{n_{2}}\longrightarrow\mathbb{K}$ of the type
\[
A(z^{(1)},z^{(2)})=

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\pm z_{i}^{(1)}z_{j}^{(2)}\text{,}
\]
such that
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  \max\left\{  n_{2}^{1/p_{2}^{\ast}
}n_{1}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n_{1}
^{1/p_{1}^{\ast}}n_{2}^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}
}\right\}  \text{.}
\]

\end{corollary}

\begin{proof}
If $1\leq p_{1}\leq p_{2}\leq2$ and $n_{1}\leq n_{2}$, then
\[
\max\left\{  n_{2}^{1/p_{2}^{\ast}}n_{1}^{\max\left\{  \frac{1}{2}-\frac
{1}{p_{1}},0\right\}  },n_{1}^{1/p_{1}^{\ast}}n_{2}^{\max\left\{  \frac{1}
{2}-\frac{1}{p_{2}},0\right\}  }\right\}  =n_{2}^{1/p_{2}^{\ast}}\text{.}
\]
By Corollary \ref{CorB5}, given $\epsilon>0$, there exists $N$ (depending just
on $\epsilon$) such that, if $n_{2}>N$, there is a bilinear form $A_{1}
\colon\ell_{p_{2}}^{n_{1}}\times\ell_{p_{2}}^{n_{2}}\longrightarrow\mathbb{K}$
of the type
\[
A_{1}(z^{\left(  1\right)  },z^{\left(  2\right)  })=

{\textstyle\sum\limits_{i=1}^{n_{1}}}

{\textstyle\sum\limits_{j=1}^{n_{2}}}

\pm z_{i}^{\left(  1\right)  }z_{j}^{\left(  2\right)  }\text{,}
\]
satisfying
\[
\Vert A_{1}\Vert\leq\left(  1+\epsilon\right)  n_{2}^{1/p_{2}^{\ast}}\text{.}
\]
Now, define $A\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}}^{n_{2}
}\longrightarrow\mathbb{K}$ by $A\left(  x,y\right)  =A_{1}\left(  x,y\right)
$ and note that
\[
\left\Vert A\right\Vert \leq\Vert A_{1}\Vert\leq\left(  1+\epsilon\right)
n_{2}^{1/p_{2}^{\ast}}\text{.}
\]

\end{proof}

\begin{remark}
It is worth mentioning that there are still a few open cases concerning the
asymptotic behavior of the constants of Bennett's inequality. Our approach
does not cover bilinear forms $A\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}
}^{n_{2}}\longrightarrow\mathbb{K}$ with $1\leq p_{1}\leq p_{2}\leq2$
$\mathrm{(}1\leq p_{2}\leq p_{1}\leq2\mathrm{)}$ and $n_{1}>n_{2}$
$\mathrm{(}n_{2}>n_{1}\mathrm{)}$.
\end{remark}

\begin{figure}[H]
\centering\begin{minipage}{6,9cm}
\begin{tikzpicture}[x=2cm,y=2cm]
\draw[yellow, fill=yellow,opacity=.2] (1,1)--(1,4)--(2,4)--(2,2)--(1,1);
\draw[red, fill=red,opacity=.2] (1,1)--(4,1)--(4,2)--(2,2)--(1,1);
\draw[green, fill=green,opacity=.2] (2,2)--(4,2)--(4,4)--(2,4)--(2,2);
\draw[line width=1pt,color=green!80!black] (2,4)--(2,2)--(4,2);
\draw[line width=1pt,color=red] (1,1)-- (2,2);
\draw[->] (1,1)-- (1,4);
\draw[->] (1,1)-- (4,1);
\draw(1,1) node[below] {$1$};
\draw(2,1) node[below] {$2$};
\draw(4,1) node[below] {$p_{1}$};
\draw(1,1) node[left] {$1$};
\draw(1,2) node[left] {$2$};
\draw(1,4) node[left] {$p_{2}$};
\end{tikzpicture}
\end{minipage}
\begin{minipage}[bl]{5,8cm}
\begin{enumerate}
\item[{\color{green!80!black}$\blacksquare$}] Theorem \ref{TeoB3}
for all $n_{1}, n_{2}$
\item[{\color{green!20}$\blacksquare$}] Theorem \ref{TeoB3} for all $n_{1}
, n_{2}$
\item[{\color{red!20}$\blacksquare$}] Corollaries \ref{CorB4} and \ref{CorB6}
for $n_{2} \leq n_{1}$
\item[{\color{yellow!20}$\blacksquare$}] Corollaries \ref{CorB4}
and \ref{CorB6}
for $n_{1} \leq n_{2}$
\item[{\color{red}$\blacksquare$}] Corollary \ref{CorB5} for all $n_{1}
, n_{2}$
\end{enumerate}
\end{minipage}
\caption
{The uniform asymptotic behavior of the constants of Bennett's inequality.}
\label{Fig2}
\end{figure}

As a final corollary, we show that the asymptotic constant $1$ is optimal when
$n_{1}=n_{2}$ and $p_{1},p_{2}\in\lbrack1,2]$:

\begin{corollary}
\label{op71}Let $p_{1},p_{2}\in\lbrack1,\infty]$ and $\epsilon>0.$ There is a
positive integer $N$ $\mathrm{(}$depending just on $\epsilon\mathrm{)}$ such
that, whenever $n>N$, there is a bilinear form $A\colon\ell_{p_{1}}^{n}
\times\ell_{p_{2}}^{n}\longrightarrow\mathbb{K}$ of the type
\begin{equation}
A(z^{(1)},z^{(2)})=

{\textstyle\sum\limits_{i=1}^{n}}

{\textstyle\sum\limits_{j=1}^{n}}

\pm z_{i}^{(1)}z_{j}^{(2)}\text{,} \label{mengao}
\end{equation}
such that
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  \max\left\{  n^{1/p_{2}^{\ast}
}n^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n^{1/p_{1}^{\ast}
}n^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}  }\right\}
\]
and the constant $1$ is optimal whenever $p_{1},p_{2}\in\lbrack1,2]$.
\end{corollary}

\begin{proof}
All that remains to be proved is the optimality of the asymptotic constant
$1$. Let us show that $1$ cannot be replaced by a smaller constant when
$p_{1},p_{2}\in\lbrack1,2]$. Let $p=\max\{p_{1},p_{2}\}$. It is plain that for
all $A\colon\ell_{p_{1}}^{n}\times\ell_{p_{2}}^{n}\longrightarrow\mathbb{K}$
of the type (\ref{mengao}) we have
\[
n^{1/p^{\ast}}=\max_{i=1,...,n}\left(

{\textstyle\sum\limits_{j=1}^{n}}

\left\vert A\left(  e_{i},e_{j}\right)  \right\vert ^{p^{\ast}}\right)
^{\frac{1}{p^{\ast}}}\leq\left\Vert A\right\Vert .
\]
Therefore
\[
\left\Vert A\right\Vert \geq n^{1/p^{\ast}}=\max\left\{  n^{1/p_{2}^{\ast}
}n^{\max\left\{  \frac{1}{2}-\frac{1}{p_{1}},0\right\}  },n^{1/p_{1}^{\ast}
}n^{\max\left\{  \frac{1}{2}-\frac{1}{p_{2}},0\right\}  }\right\}  \text{,}
\]
and the proof is done.
\end{proof}

We conclude this section by remarking that, following the lines of the
statement of Theorem \ref{TeoB3}, we can re-state Theorem \ref{090} in a
slightly relaxed form as follows:

\begin{theorem}
Let a positive integer $m$ and $\epsilon>0$ be given. There is an $m$-linear
form $A_{n_{1},\ldots,n_{m}}\colon\ell_{\infty}^{n_{1}}\times\cdots\times
\ell_{\infty}^{n_{m}}\longrightarrow\mathbb{K}$ of the type
\[
A_{n_{1},\ldots,n_{m}}(z^{(1)},\ldots,z^{(m)})=

{\textstyle\sum\limits_{j_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{j_{m}=1}^{n_{m}}}

\pm z_{j_{1}}^{(1)}\cdots z_{j_{m}}^{(m)}\text{,}
\]
and a positive integer $N$ such that
\[
\Vert A_{n_{1},\ldots,n_{m}}\Vert\leq\left(  1+\epsilon\right)  \max\left\{
n_{1}^{1/2},\ldots,n_{m}^{1/2}\right\}

{\textstyle\prod\limits_{j=1}^{m}}

n_{j}^{1/2}\text{,}
\]
whenever
\[
\max_{i=1,\ldots,m}\left\{  \min\left\{  n_{j}:j\neq i\right\}  \right\}
>N\text{.}
\]

\end{theorem}

\section{Remarks}

We begin this section by remarking that well known probabilistic techniques
used in the proof of the classical Kahane--Salem--Zygmund inequalities,
combined with Lemma \ref{LemaB1}, improve the estimates obtained by
Albuquerque and Rezende in \cite{alb} for the remaining cases not covered in
the previous sections. In fact, combining the proof of Mantero and Tonge
\cite[Theorem 1.1]{man} for $p_{1}=\cdots=p_{m}=2$ with Lemma \ref{LemaB1} we
can easily improve the constants of (\ref{I4}) and (\ref{I5}). More precisely,
following the proof of \cite[Theorem 1.1]{man} (noting that the constant $K$
in the proof of \cite[Theorem 1.1]{man} can be chosen as the degree of
multilinearity (now denoted by $m$) and combining with Lemma \ref{LemaB1}, we
can replace the constants of (\ref{I4}) and (\ref{I5}) by
\[
C_{m}\leq2^{m+1}\sqrt{\left(  2m+1\right)  \log\left(  1+4m\right)  }\text{.}
\]
In the case of real scalars we can avoid the factor $2^{m}$ and get
\[
C_{m}\leq2\sqrt{\left(  2m+1\right)  \log\left(  1+4m\right)  }\text{.}
\]

However, it is still an open problem if it is possible to obtain constants
asymptotically dominated by $1$ in these cases.

We also observe that a simple adaptation of the arguments from Lemma
\ref{LemaB1} shows that the constants of the family of inequalities (\ref{I5})
are monotone in the following sense: if $C_{m,p}$ is the optimal constant of
(\ref{I5}) for $p_{1}=\cdots=p_{m}=p$, then
\[
q\geq p\geq2\Rightarrow C_{m,q}\leq C_{m,p}\text{.}
\]
We finish this section by remarking that the apparent gap between the results
from Theorems \ref{090} and \ref{TeoB3} is somewhat justified by the following
theorem which follows the exact same lines of the proof of Theorem \ref{090}:

\begin{theorem}
Let $m$ be a positive integer, $p_{1},p_{m}\in\left[  2,\infty\right]  $,
$p_{2}=\cdots=p_{m-1}=\infty$ and $\epsilon>0$. There exists a positive
integer $N$ such that, for any choice of positive integers $n_{1},\ldots
,n_{m}>N$, with $n_{m}=\max\left\{  n_{1},\ldots,n_{m}\right\}  $ and
$n_{1}=\min\left\{  n_{1},\ldots,n_{m}\right\}  $, there exists an $m$-linear
form $A\colon\ell_{p_{1}}^{n_{1}}\times\ell_{p_{2}}^{n_{2}}\times\cdots
\times\ell_{p_{m-1}}^{n_{m-1}}\times\ell_{p_{m}}^{n_{m}}\longrightarrow
\mathbb{K}$ of the type
\[
A(z^{(1)},\ldots,z^{(m)})=

{\textstyle\sum\limits_{i_{1}=1}^{n_{1}}}

\cdots

{\textstyle\sum\limits_{i_{m}=1}^{n_{m}}}

\pm z_{i_{1}}^{(1)}\cdots z_{i_{m}}^{(m)}\text{,}
\]
satisfying
\[
\Vert A\Vert\leq\left(  1+\epsilon\right)  n_{m}^{1/2}

{\textstyle\prod\limits_{k=1}^{m}}

n_{k}^{\frac{1}{2}-\frac{1}{p_{k}}}\text{.}
\]

\end{theorem}

\section{Application: Berlekamp's switching game}

Berlekamp's switching game (sometimes called Gale-Berlekamp switching game or
unbalancing lights problem, see \cite[Section 2.5]{alon} and \cite[Chapter
6]{spencer}) consists of an $n\times n$ square matrix of light bulbs set up at
an initial configuration $\Theta_{n}$. The board has $n$ row and $n$ column
switches, which invert the on-off state of each bulb (on to off and off to on)
in the corresponding row or column. Let $i(\Theta_{n})$ denote the smallest
final number of on-lights achievable by row and column switches starting from
$\Theta_{n}$. The goal is to find the value $R_{n}$ of $i(\Theta_{n}^{(0)})$
when $\Theta_{n}^{(0)}$ is (one of) the worst initial patterns, i.e.,
$i(\Theta_{n}^{(0)})\geq i(\Theta_{n})$ for all $\Theta_{n}$. Thus
\[
R_{n}:=\max\{i(\Theta_{n}):\Theta_{n}\text{ is an initial configuration of
}n\times n\text{ lights}\}\text{.}
\]
Sometimes the problem is posed as to find the maximum of the difference
between the state of the light bulbs (starting from one of the worst initial
patterns, as before), which we shall henceforth denote by $G_{n}$. It is
simple to check that
\[
R_{n}=\frac{1}{2}\left(  n^{2}-G_{n}\right)  \text{.}
\]
The determination of the exact value of $R_{n}$ seems to be conceivable only
for small values of $n$ due to involving combinatorial arguments. The exact
value of $R_{n}$ for $n$ up to $12$ was obtained by Carlson and Stolarski
(\cite{car}; see also \cite{TBJS,slo}). For bigger values of $n$, optimal
constructive approaches seem impracticable and no algorithm to construct such
a \textquotedblleft bad\textquotedblright\ configuration $\Theta_{n}$ seems to
be known. Thus, for bigger values of $n$, probabilistic (non-deterministic)
methods are used to provide estimates for $R_{n}$ and $G_{n}$. The natural
approach to modeling Berlekamp's switching game is by associating $+1$ to the
on-lights and $-1$ to the off-lights from the array of lights $\left(
a_{ij}\right)  _{i,j=1}^{n}$ and observing that
\[
G_{n}=\min\left\{  \max_{\left(  x_{i}\right)  _{i=1}^{n},\left(
y_{j}\right)  _{j=1}^{n}\in\{-1,1\}^{n}}\left\vert

{\textstyle\sum\limits_{i,j=1}^{n}}

a_{ij}x_{i}y_{j}\right\vert :a_{ij}=-1\text{ or }+1\right\}  \text{,}
\]
where $x_{i}$ and $y_{j}$ denote the switches of row $i$ and of column $j$, respectively.

Berlekamp's switching game has several natural variants (see, for instance,
\cite{BS,BRUA,PST} and the references therein); we also refer to \cite{kalai}
for a recent related result. We shall be interested in its extension to higher
dimensions (see \cite{ara}). Let $m\geq2$ be an integer and let an
$n\times\cdots\times n$ array $\left(  a_{j_{1}\cdots j_{m}}\right)  $ of
lights be given, each either on ($a_{j_{1}\cdots j_{m}}=1$) or off
($a_{j_{1}\cdots j_{m}}=-1$). Let us also suppose that for each $k=1,\ldots,m$
and each $j_{k}=1,\ldots,n$ there is a switch $x_{j_{k}}^{(k)}$ so that if the
switch is pulled ($x_{j_{k}}^{(k)}=-1$) all of the corresponding lights
$a_{j_{1}\ldots j_{m}}$ (with $j_{k}$ fixed) are switched: on to off or off to
on. The goal is to maximize the difference between the number of lights that
are on and the number of lights that are off.

\begin{figure}[ptb]
\centering\includegraphics[scale=0.56]{Figure.pdf}\caption{Three-dimensional
Berlekamp's switching game for $n=5$.}
\end{figure}

As in the two-dimensional case, maximizing the difference between the number
of on-lights and off-lights is equivalent to estimating
\[
\max_{(x_{j_{1}}^{(1)})_{j_{1}=1}^{n},\ldots,(x_{j_{m}}^{(m)})_{j_{m}=1}
^{n}\in\{-1,1\}^{n}}\left\vert

{\textstyle\sum\limits_{j_{1},\ldots,j_{m}=1}^{n}}

a_{j_{1}\ldots j_{m}}x_{j_{1}}^{(1)}\cdots x_{j_{m}}^{(m)}\right\vert
\]
and the extreme problem consists of estimating
\[
S_{m,n}=\min\left\{  \max_{(x_{j_{1}}^{(1)})_{j_{1}=1}^{n},\ldots,(x_{j_{m}
}^{(m)})_{j_{m}=1}^{n}\in\{-1,1\}^{n}}\left\vert

{\textstyle\sum\limits_{j_{1},\ldots,j_{m}=1}^{n}}

a_{j_{1}\ldots j_{m}}x_{j_{1}}^{\left(  1\right)  }\cdots x_{j_{m}}^{\left(
m\right)  }\right\vert :a_{j_{1}\ldots j_{m}}=1\text{ or }-1\right\}  \text{,}
\]
It is folklore (a consequence of the Krein--Milman Theorem) that
\begin{equation}
S_{m,n}=\min\left\Vert A_{m,n}\colon\ell_{\infty}^{n}\times\cdots\times
\ell_{\infty}^{n}\longrightarrow\mathbb{R}\right\Vert \text{,} \label{uzz}
\end{equation}
with
\[
A_{m,n}\left(  x^{(1)},\ldots,x^{(m)}\right)  =

{\textstyle\sum\limits_{j_{1},\ldots,j_{m}=1}^{n}}

a_{j_{1}\ldots j_{m}}x_{j_{1}}^{(1)}\cdots x_{j_{m}}^{(m)}\text{.}
\]

As an illustrative example of how the estimates of the constants of the KSZ
inequality are associated to this framework, note that, combining (\ref{uzz})
and Lemma \ref{M1}, we have $G_{16}\leq64$ and hence $R_{16}\geq96,$ improving
the estimate $R_{16}\geq94$ given by \cite[Theorem 8]{carl}.

Let us choose, as usual in this setting, the notation $o(1)$ to represent
that, in the variable $n$, we have $\lim\limits_{n\rightarrow\infty}o(1)=0$.
The results of Section 2 show that
\[
S_{m,n}\leq\left(  1+o(1)\right)  n^{\frac{m+1}{2}}\text{,}
\]
and this result improves the best known constants to this problem, which were
obtained by the KSZ inequality, i.e.,
\[
S_{m,n}\leq\left(  \sqrt{32m\log\left(  6m\right)  }\sqrt{m!}\right)
n^{\frac{m+1}{2}}\text{.}
\]

Combining our estimates with those in \cite[Theorem 3.2]{ara} and
\cite[Theorem 2.5.1]{alon}, we obtain
\[
2^{-\frac{1}{2}\psi\left(  m\right)  -\frac{1}{2}\gamma}

{\textstyle\prod\limits_{k=2}^{m}}

\left(  \dfrac{\Gamma\left(  \frac{3k-2}{2k}\right)  }{\Gamma\left(  \frac
{3}{2}\right)  }\right)  ^{\frac{k}{2k-2}}+o\left(  1\right)  \leq
\dfrac{S_{m,n}}{n^{\frac{m+1}{2}}}\leq1+o\left(  1\right)  \text{,}
\]
where $\psi$ is the digamma function and $\gamma$ is the Euler-Mascheroni
constant. For instance, for $m=2,3$ we have
\begin{align*}
0.797+o\left(  1\right)   &  \leq\dfrac{S_{2,n}}{n^{3/2}}\leq1+o\left(
1\right)  \text{,}\\
0.694+o\left(  1\right)   &  \leq\dfrac{S_{3,n}}{n^{2}}\leq1+o\left(
1\right)  \text{.}
\end{align*}

\bigskip

\medskip\noindent\textbf{Acknowledgment.} The authors thank Fernando Vieira
Costa Jr and Janiely Silva for their support and generosity.

config: arxiv


TITLE: What will happen when Russia shuts down the GPS stations?
QUESTION [2 upvotes]: Russian officials are seriously talking about shutting down US GPS ground stations within their borders and Deputy Prime Minister of Russia tweeted on the subject.
What will happen to the GPS? My assumption would be that its accuracy will get worse, at least in some parts of the globe. If it is correct, then where, and by how much exactly?

REPLY [1 votes]: There are two types of GPS ground stations for the USA system.  Main stations about 5 around the world provide corrections to timing data due to relativistic effects, clock drift, ephemeris, or ionospheric delays that are timing constants rebroadcast by the satellites to receivers for corrections.  The types of stations in Russia are for augmented gps called Global-DGPS .  DGPS improves local GPS accuracy from about 3m to 10 cm.  However this lack of support in Russia's areas only effects civilian use of the US system in those areas.  The Chinese, EU, and Russian systems won't be effected.

"Employing a large ground network of real-time reference receivers,
  innovative network architecture, and award-winning real-time data
  processing software, the GDGPS System provides decimeter (10 cm)
  positioning accuracy and sub-nanosecond time transfer accuracy
  anywhere in the world, on the ground, in the air, and in space,
  independent of local infrastructure." link


TITLE: Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_* \leq 1$
QUESTION [1 upvotes]: Given $\mathrm A \in \mathbb R^{m \times n}$, $$\begin{array}{ll} \text{maximize} & \langle \mathrm A , \mathrm X \rangle\\ \text{subject to} & \| \mathrm X \|_* \leq 1\end{array}$$

Though I know something about the spectral norm, I know almost nothing about the nuclear norm,  dual norms, convex analysis, etc.  Since I am utterly unqualified to answer this on my own, I post this question.

Related:

Show that the dual norm of the spectral norm is the nuclear norm

Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_2 \leq 1$

REPLY [3 votes]: I assume that $\| X \|_*$ is the nuclear norm of $X$. Then the function $f$ defined by
$$
f(A) = \sup_{\| X \|_* \leq 1} \langle A, X \rangle
$$
is by definition the dual of the nuclear norm. But a standard result is that the dual of the nuclear norm is the spectral norm. Thus
$$
f(A) = \sigma_\max(A),
$$ the largest singular value of .


TITLE: Generating function for $15, 66,189,420,795,\dots$
QUESTION [0 upvotes]: I am to derive the formula for $15,66,189,420, 795,\dots$
I have the common pattern as $3m(2n+1)$ where $n$ and $m$ are integers. For example,
$66=3\cdot2\cdot(2\cdot5+1)$
Will this end up in me deriving the correct formula or is there another pattern? Also could someone help me with the next step.

REPLY [2 votes]: You seem to be on the right track. The numbers are
$$3\cdot 1\cdot 5$$
$$3\cdot 2\cdot 11$$
$$3\cdot 3\cdot 21$$
$$3\cdot 4\cdot 35$$
$$3\cdot 5\cdot 53$$
The differences in the last column are $6,10,14,18$ , forming an arithmetic progression. So, we have $$a_n=3n\cdot (2n^2+3)=6n^3+9n$$


TITLE: Order of Inverse Operations
QUESTION [1 upvotes]: so this is a very simple question but I am having a tough time with it.
So it's finals week and I'm studying up for an Algebra 2 final. The only part I am having trouble with is finding the inverse of a function. It doesn't confuse me that I have to basically undo everything that is happening to x (or any variable) but it confuses me what order I have to undo everything in.
Say I have the expression $f(x)= \sqrt{x-3}$. I would first substitute $f(x)$ with y and then switch the x and y so I'd be finding x. After that I'd square x and then add 3 to x to make the equation $y = x^2 + 3$.
But what if I had the expression $f(x)= \sqrt{x-3} + 5$? Would I subtract 5 and then square the x and 5 or would I square x and then add 3 and subtract 5?
Thanks.

REPLY [1 votes]: Use the order of operations to track what happens to $x$. In your example, $f(x) = \sqrt{x-3} + 5$, we have (starting with $x$):

subtract $3$
take the (positive) square root
add $5$.

To undo these and solve for $x$, we would undo each, in reverse order. (You noted this much in your question).

subtract $5$
square
add $3$


TITLE: Grobner Basis and Surfaces
QUESTION [5 upvotes]: Grobner basis are really good at describing polynomial systems of equations with 0-dimensional zero sets. In a sense, Grobner basis yields a better/simplified description of such systems because the Grobner polynomials (in appropriate monomial order) let one explicitly compute the zeros.
However, suppose the ideal $I\subseteq k[x_1,\ldots,x_n]$ has a positive dimensional zero set, such as a curve or surface. What can Grobner basis tell you about these zero sets? I'm particularly interested in the curves and surfaces case (as I am intersecting the flat 3-torus with various varieties).

REPLY [2 votes]: Chapter $15$ of Eisenbud's Commutative Algebra: with a View Toward Algebraic Geometry contains many applications of Gröbner bases, and many of these are not special to $0$-dimensional ideals.  An initial list can be found in $\S15.1$ (p. $318$), while a more detailed treatment occurs in $\S15.10$ (p. $355$).  I will list a couple here.  Let $k$ be a field and $S = k[x_1, \ldots, x_r]$ be a polynomial ring.
Ideal membership: The most fundamental problem that Gröbner bases solve is that of ideal membership: given an ideal $I \trianglelefteq S$ and $f \in S$, how can we tell if $f \in I$?  One can show that the remainder of $f$ upon division by a Groebner basis is unique, so we have that $f \in I$ iff its remainder is $0$, just as in the single variable case.
The geometric significance is that if $S/I$ is the coordinate ring of an irreducible affine algebraic set (with $I = \sqrt{I}$), a polynomial $f \in S$ induces the zero function on $V$ iff $f \in I$.  So this allows us to determine when two polynomials $f,g \in S$ induce the same regular function when restricted to $V$: $f = g$ on $V$ iff $f-g \in I$.
Hilbert functions: Given a projective variety $V$, let $M$ be its homogeneous coordinate ring, which is a graded module over $S$.  Let $M_s$ denote the $s^\text{th}$ graded piece of $M$.  Then $H_M(s) := \dim_k(M_s)$ is the Hilbert function of $M$ (or of $V$).  Gröbner bases provide a simple way of computing the Hilbert function that only requires considering monomial ideals.  (Here $\DeclareMathOperator{\init}{in} \init(N)$ is the initial submodule with respect to some monomial ordering.)
Theorem 15.26. Let $M$ be a finitely generated graded $S$-module, given by generators and relations as $M = F/N$, where $F$ is a free module with a homogeneous basis and $N$ is a submodule generated by homogeneous elements.  The Hilbert function of $M$ is the same as the Hilbert function of $F/\init(N)$.
For sufficiently large integers the Hilbert function agrees with a polynomial with rational coefficients called the Hilbert polynomial.  The Hilbert polynomial carries several pieces of information about a projective variety, such as its dimension, degree, and arithmetic genus.  Some other information that can be extracted from the Hilbert polynomial is mentioned in this MO thread.
Primary decomposition: From a Gröbner basis, one can compute the primary decomposition of an ideal $I \trianglelefteq S$, i.e., one can write $I = Q_1 \cap \cdots \cap Q_m$ where the $Q_i$ are primary.  In geometric terms, this allows one to decompose an affine algebraic set into irreducible varieties since
$$
\mathbb{V}(I) = \mathbb{V}(Q_1 \cap \cdots \cap Q_m) = \mathbb{V}(Q_1) \cup \cdots \cup \mathbb{V}(Q_m) \, .
$$
(Technically, primary decomposition carries even more information than the classical geometric picture, since $\mathbb{V}(I) = \mathbb{V}(\sqrt{I})$ if $k$ is algebraically closed.)
Eisenbud includes quite a few other applications of Gröbner bases, too, such as computation of syzygies, free resolutions, and important functors like $\text{Hom}, \text{Ext}$, and $\text{Tor}$.


TITLE: Combinatorics ball-counting problem.
QUESTION [2 upvotes]: Let there be two persons P1, P2. Now let there be $6n$ balls so that $2n$ balls are red, blue and green respectively. In how many ways can P1 and P2 get $3n$ balls?
And how about $2pn$ balls in $p$ colours?
Thanks :)

REPLY [2 votes]: Suppose that there are $2pn$ balls in $p$ colors, $2n$ balls of each color. For $k=1,\dots,p$ let $x_k$ be the number of balls of color $k$ received by $P_1$; then the question asks for the number of solutions to $$\sum_{k=1}^px_k=pn\tag{1}$$ in non-negative integers $x_1,\dots,x_p$, subject to the condition that $x_k\le 2n$ for $k=1,\dots,p$. It’s well-known that the number of solutions to $(1)$ in non-negative integers without any upper bound conditions is $$\binom{pn+p-1}{p-1}=\binom{pn+p-1}{pn}\;.\tag{2}$$ From this total we must subtract the solutions that exceed the $2n$ limit in some term. In general this is a somewhat messy inclusion-exclusion calculation of a kind discussed several times on this site; see, for instance, the answer to this question. For $p=3$, however, it’s not bad.
If $p=3$, $(1)$ becomes $$x_1+x_2+x_3=3n\;,\tag{3}$$ and at most one of the terms can exceed $2n$. There is an easy bijection between solutions to $(3)$ in non-negative integers with $x_1>2n$ and solutions to $y_1+y_2+y_3=3n-(2n+1)=n-1$ in non-negative integers, and there are $$\binom{(n-1)+3-1}{3-1}=\binom{n+1}2$$ of the latter. Similarly, there are $\binom{n+1}2$ solutions to $(3)$ with $x_2>2n$ and another $\binom{n+1}2$ with $x_3>2n$. Thus, the answer to the original question (with $p=3$) is
$$\begin{align*}
\binom{3n+2}2-3\binom{n+1}2&=\frac12\Big((3n+2)(3n+1)-3n(n+1)\Big)\\\\
&=3n^2+3n+1\;.
\end{align*}$$


TITLE: automorphisms of fat points
QUESTION [12 upvotes]: Let $k$ be an algebraically closed field. I am looking for an
easily quotable description of automorphism groups of $\mathrm{Spec} k[x]/(x^n)$. I could compute explicit matrix representations for several small $n$, but a more intrinsic description would be desirable.

REPLY [13 votes]: Smooth points are all alike; every fat point is happy in its own way.
                                                                                "Anna Karenina", Leo Tolstoy
Let $k$ be an algebraically closed field. Denote $A_n:=k[x]/x^{n+2}$, $G_n:=\text{Aut}A_n$, and let $N_n\subset G_n$ stand for the unipotent radical of $G_n$. Since there is no problem in finding a faithful matrix representation of $G_n$, I guess, Dima asks about a coordinate-free description of $G_n$.
We will mainly deal with $N_n$ because $G_n=N_n\rtimes T$ is a semidirect product, where $\text{G}_m\simeq T\subset G_n$ has no canonical choice (though all tori are conjugated in $G_n$).
Answer. We assume for simplicity $k$ of characteristic $0$. For sufficiently big $n$, $N_{n+1}$ is a central extension (with that very centre) of $N_n$. This extension is determined by a certain orbit of the group $\text{Out}N_n$ naturally acting on the projective plane
${\mathbb P}_kH^2(N_n,k)$. The orbits are: a dense one, two $1$-dimensional orbits $L'$ and $H'$ dense in the projective lines $L$ and $H$, and a couple of fixed points $f_0,f_1\in L$, where $f_0$ is the intersection $L\cap H$. Our choice determining the extension $0\to k\to N_{n+1}\to N_n\to1$ is $H'$. The action of $T$ can be easily defined on the way.
The interesting stuff appears at the end, when the point becomes really fat.
Boring stuff. Every element $g\in G_n$ is determined by the polynomial $f_g:=gx$, where $f_g(x)=f_0x+f_1x^2+\dots+f_nx^{n+1}$ with $f_i\in k$ and $f_0\ne0$. The composition in $G_n$ is expressed in these terms as
$f_{g_1g_2}(x)\equiv f_{g_2}\big(f_{g_1}(x)\big)\mod x^{n+2}$. Note that
$g\in N_n$ iff $f_0=1$ and $g\in T$ iff $f_1=\dots=f_n=0$. Sometimes, it is more convenient to use other coordinates: $f_g(x)=x\Big(h_g(x)+\displaystyle\sum_{i=0}^n(p_1x)^i\Big)$, where
$h_g(x)=\displaystyle\sum_{i=2}^np_ix^i$. So, $p_1=f_1$ and $p_i=f_i-f_1^i$ for $i\ge2$.
The Lie algebra $L_n$ of $G_n$ is formed by all the derivations from $A_n$ to $A_n$, i.e., by all the derivations from $k[x]$ to $k[x]$ that preserve $\text{Ideal}(x^{n+2})$. The formula $df(x)=f'(x)dx$ correctly defines a derivation $d\in L_n$ by its value $dx\in A_nx$. Taking $d_ix=x^{i+1}$ for $i=0,1,\dots,n$, we obtain a basis of $L_n$ subject to $[d_i,d_j]=(j-i)d_{i+j}$ for $i+j\le n$ and $[d_i,d_j]=0$ for $i+j>n$. The Lie algebra $M_n$ of $N_n$ is simply spanned by $d_1,\dots,d_n$.
0. $N_0=1$ and $G_0=\text{G}_m$.
1. $N_1=\text{G}_a$ and $G_1=N_1\rtimes\text{G}_m$ is the semidirect product with respect to $\text{G}_m=\text{Aut}\text{G}_a$.
2. $N_2=\text{G}_a\oplus\text{G}_a$ (the corresponding coordinates are, for instance, $p_1,p_2$) and $G_1=N_1\rtimes\text{G}_m$, where the action of $c_0\in\text{G}_m$ is given by the rule $c_0\cdot(p_1,p_2):=(c_0p_1,c_0^2p_2)$.
Preliminaries. The homomorphism $\pi:A_n\to A_{n-1}$ induces the surjective homomorphisms $\pi:G_n\to G_{n-1}$ and $\pi:N_n\to N_{n-1}$ whose kernel $\text{G}_a\simeq C_n\subset N_n$, formed by all $c\in N_n$ with $f_c=x(1+fx^n)$, $f\in k$, lies in the centre of $N_n$ and coincides with this centre unless $n=2$.
A few more well-known and easy facts. Under the assumption that there is a rational section of $\pi$ (which is valid in our case of $X_h=N_{n+1}$ and $X=N_n$), a central extension $0\to k\to X_h\overset\pi\to X\to1$ of algebraic groups is given by an (arbitrary) element $h\in H^2(X,k)$ of rational $2$-cohomologies of $X$ (the action of $X$ on $k=\text{G}_a$ is trivial). The
groups $\text{Aut}X$ and $\text{G}_m$ act on $X$ and on $k$, hence, on $H^2(X,k)$. Assuming that $k$ coincides with the centre of $X_{h_i}$ for both $i=1,2$, the groups $X_{h_1}$ and $X_{h_2}$ are isomorphic iff $h_1$ and $h_2$ are in a same orbit of $\text{Out}X\times\text{G}_m$ (the inner automorphisms of $X$ act trivially on $H^2(X,k)$).
In our case, the central extension is nontrivial. Denote by
$0\ne[h_n]\in H^2(N_n,k)$ an element providing the extension $\pi:N_{n+1}\to N_n$. By the above, the group $N_{n+1}$ is in fact given by the choice of the orbit of $[h_n]$ with respect to the action of $\text{Out}N_n$ on
${\mathbb P}_kH^2(N_n,k)$. Our task is therefore to indicate this orbit and to explain how $T\simeq\text{G}_m$ acts on $N_{n+1}$ (already knowing how does $T$ act on $N_n$).
The homomorphism $\pi$ induces a $k$-linear map
$\pi^*:H^2(N_{n-1},k)\to H^2(N_n,k)$ and, unless $n=2$, a homomorphism $\pi:\text{Out}N_n\to\text{Out}N_{n-1}$. Another elementary remark: an element
$\alpha\in\text{Out}N_{n-1}$ belongs to the image $\pi\text{Out}N_n$ (i.e., is liftable) iff $[h_{n-1}]\in{\mathbb P}_kH^2(N_{n-1},k)$ is a fixed point of $\alpha$.
Using the Chevalley-Eilengebrg complex of $M_n$, one can calculate $H^2(M_n,k)\simeq H^2(N_n,k)$ for $n\ge2$ (the action of $M_n$ on $k$ is trivial), getting the following result. For any $n\ge4$, we have the cocycle $a:=d_2^*\wedge d_3^*$; for any $n\ge6$, we have the cocycle
$b:=d_2^*\wedge d_5^*-3d_3^*\wedge d_4^*$; for any $n\ge2$, we have the cocycle
$h_n:=\displaystyle\sum_{1\le i<j\le n\atop i+j=n+1}(j-i)d_i^*\wedge d_j^*$. Now, $1$-dimensional $H^2(M_2,k)$ and $H^2(M_3,k)$ are spanned respectively by $[h_2]$ and $[h_3]$, $2$-dimensional $H^2(M_4,k)$ and $H^2(M_5,k)$ are spanned respectively by $[a],[h_4]$ and $[a],[h_5]$, and a $3$-dimensional $H^2(M_n,k)$ is spanned by $[a],[b],[h_n]$ when $n\ge6$. Note also that (when it makes sense)
$\pi^*:[a]\mapsto[a]$, $\pi^*:[b]\mapsto[b]$, and $\pi^*:[h_n]\mapsto0$.
First steps. I claim (without proof as it is bulky) that $\text{Out}N_n$ is $4$-dimensional.
Since $N_2$ is in fact a $2$-dimensional $k$-linear space, we conclude that $\text{Out}N_2=\text{GL}_2k$. In view of $\dim_kH^2(N_2,k)=1$, we have a unique possible extension, thus obtaining the famous Heisenberg group $N_3$. The group $\text{Out}N_2$ is entirely liftable, and $\text{Aut}N_3$ is formed by
$3\times 3$-matrices
$G:=\left[\begin{smallmatrix}g&0\\{*}&\det g\end{smallmatrix}\right]$, where $g\in\text{GL}_2k$. Taking $g=1$, we obtain the normal subgroup $N_2\subset\text{Aut}N_3$ of inner automorphisms of $N_3$, hence,
$\text{Out}N_3=\text{GL}_2k$ is $4$-dimensional, as claimed. The action of $T$ on $N_3$ will be defined later.
Since $\dim_kH^2(N_3,k)=1$, we again get a unique possible extension $N_4$, the entirely liftable $\text{Out}N_3$, and $\text{Out}N_4=\text{GL}_2k$.
When $\dim_kH^2(N_4,k)=2$, we are stuck.
Lemma. For $n\ge5$, there is a $1$-dimensional subgroup $\text{G}_a\simeq U_n\subset\text{Out}N_n$ such that
$\text{Out}N_n=U_n\cdot\pi\text{Out}N_{n+1}$, $U_n\cap\pi\text{Out}N_{n+1}=1$, $\pi^3U_n=1$, and $\pi^2U_n\ne1$.
In place of a proof, we only define how $u\in k$ acts on $f(x)=x\Big(1+\displaystyle\sum_{i=1}^nf_ix^i\Big)\in N_n$. (Let me skip some reasons allowing to get rid of an unpleasant straightforward verification.) We use the coordinates $p_1:=f_1$ and $p_i:=f_i-f_1^i$ for $i\ge2$. Consider the "quadratic" polynomial $q_f(x):=q_{f,0}+q_{f,1}x+q_{f,2}x^2$, where
$$q_{f,0}:=p_2,\qquad q_{f,1}:=(n-3)p_3-2(n-4)p_2p_1,$$
$$q_{f,2}:=\frac{(n-2)(n-3)}2p_4-(n-3)(n-4)p_3p_1-\frac{(3n-14)(n-1)}4p_2^2+\frac{(n-4)(n-9)}2p_2p_1^2.$$
The action of $u\in k$ on $f$ is given by the rule $u\cdot f:=f+ux^{n-1}q_f$.
Fat point fun.  Using Lemma and omitting the $\pi$'s, we can write $\text{Out}N_n=U_nU_{n+1}U_{n+2}T$. As $U_{n+1},U_{n+2},T$ are liftable, they fix the point $[h_n]\in H^2(N_n,k)$. As $U_n$ is not liftable, the orbit $(\text{Out}N_n)[h_n]=U_n[h_n]$ is $1$-dimensional. The action of $U_n,U_{n+1},U_{n+2}$ on $[a]$ and $[b]$ is trivial because it can be performed at the level of $H^2(N_{n-3},k)$, where $U_n,U_{n+1},U_{n+2}$ vanish. It is easy to see that $[a]$ and $[b]$ are fixed points of $T$ of different weights. Finally, we can get the picture of $\text{Out}N_n$-orbits in
${\mathbb P}_kH^2(N_n,k)$. There is a dense orbit which is the complement of two lines $L$ and $H$. The intersection $L\cap H$ is a fixed point $f_0$ (either $[a]$ or $[b]$; no idea which one). The line $L$ spanned by $[a]$ and $[b]$ contains therefore one more fixed point $f_1$. Our choice is the $1$-dimensional orbit $H'$ such that the compliment $H\setminus H'$ in its closure $H$ is a single point $f_0$.
I forgot to define the action of $T$ on $N_n$. For big $n$, it is easy: just take any torus $T\subset\text{Aut}N_n$. In order to get the action of $T$ for small $n$, simply apply a number of $\pi$'s.
Conclusion. When we killed $[h_n]$ by inserting it as a centre, a new $[h_{n+1}]$ was born, and the other cohomologies $[a]$ and $[b]$ survived. What if we kill $[a]$ or $[b]$, once or several times. How probable is that we obtain a group of automorphisms of some fat point? (The algebras $A_n$ provide possibly the simplest series of artinian local algebras. Of course, there is no hope to classify all local artinian algebras. Nevertheless, my question may have a tame nature.)


TITLE: Is $(n-1)+\max\limits_{1\leqslant i\leqslant n} x_i\geqslant \left(1-\left(1-\frac{1}{n}\right)^n\right)\sum\limits_{i=1}^{n}x_i$?
QUESTION [1 upvotes]: Let $n>0$ and $x_i>0$ for all $i\in \{1,\ldots,n\}$ be integer numbers.
I would like to compare $$(n-1)+\max\limits_{1\leqslant i\leqslant n} x_i,$$ and $$\left(1-\left(1-\frac{1}{n}\right)^n\right)\sum\limits_{i=1}^{n}x_i.$$
I think, after trying some examples, that
$$(n-1)+\max\limits_{1\leqslant i\leqslant n} x_i\leqslant \left(1-\left(1-\frac{1}{n}\right)^n\right)\sum\limits_{i=1}^{n}x_i,$$ but I can't prove this. I can't show the opposite also as I was unable to come up with a counterexample.

REPLY [2 votes]: For $n = 1$, both quantities are exactly equal to $x_1$.
Now suppose $n \ge 2$.
If $x_i = 1$ for all $i \in \{1,\ldots,n\}$, then we have:
$$(n-1)+\displaystyle\max_{1 \le i \le n}x_i = (n-1)+1 = n$$
$$\left(1-(1-\tfrac{1}{n})^n\right)\displaystyle\sum_{i = 1}^{n}x_i = \left(1-(1-\tfrac{1}{n})^n\right)n < n$$
So in this case, $(n-1)+\displaystyle\max_{1 \le i \le n}x_i > \left(1-(1-\tfrac{1}{n})^n\right)\displaystyle\sum_{i = 1}^{n}x_i$.
On the other hand, if $x_i = 4$ for all $i \in \{1,\ldots,n\}$, then we have:
$$(n-1)+\displaystyle\max_{1 \le i \le n}x_i = (n-1)+4 = n+3 \le \dfrac{5}{2}n$$
$$\left(1-(1-\tfrac{1}{n})^n\right)\displaystyle\sum_{i = 1}^{n}x_i = \left(1-(1-\tfrac{1}{n})^n\right) \cdot 4n > 4\left(1-\tfrac{1}{e}\right)n > \dfrac{5}{2}n$$
So in this case, $(n-1)+\displaystyle\max_{1 \le i \le n}x_i < \left(1-(1-\tfrac{1}{n})^n\right)\displaystyle\sum_{i = 1}^{n}x_i$.
Therefore, no comparison can be made if $n \ge 2$.


TITLE: Topological spaces $(X,\tau)$ where $|\text{Cont}(X,X)| = |X|$
QUESTION [2 upvotes]: Let $(X,\tau)$ be a topological space. Let $\text{Cont}(X,X)$ denote the set of continuous functions $f:X\to X$.
What can be said about spaces $(X,\tau)$ where $|\text{Cont}(X,X)| = |X|$? For instance, is it impossible that they are zero-dimensional? (Note that trivially spaces with $|\text{Cont}(X,X)| = |X|$ must be infinite, because all constant functions as well as the identity are members of $\text{Cont}(X,X)$.)

REPLY [5 votes]: Suppose that $\lambda$ is an infinite cardinal with $\lambda^{\aleph_{0}}=\lambda$. Then Saharon Shelah has shown in his paper Existence of endo-rigid Boolean algebras that there is some Boolean algebra $B$ with $|B|=\lambda$ but where the only endomorphisms $f:B\rightarrow B$ are the trivial endomorphisms. When we transfer by Stone duality these facts over to topological spaces, for each infinite cardinal $\lambda$ with $\lambda^{\aleph_{0}}=\lambda$, there is a compact totally disconnected space $X$ with $w(X)=\lambda$ (here $w$ denotes the weight of the space) but where the only continuous maps $f:X\rightarrow X$ are the mappings where there is some partition $A_{1},...,A_{n}$ of $X$ into clopen sets where $f|_{A_{i}}$ is a constant function or where $f|_{A_{i}}$ is the identity function for each $i\in\{1,...,n\}$ (i.e. the only mappings are the trivial mappings). Take note that if $X$ is a compact space, then $w(X)\leq|X|$ and if $X$ is compact and totally disconnected, then $|\mathfrak{B}(X)|\leq|X|$ where $\mathfrak{B}(X)$ is the Boolean algebra of clopen subsets of $X$. Therefore if $X$ is a compact zero-dimensional space where the only mappings $f:X\rightarrow X$ are the trivial mappings, then $|C(X,X)|=|X|$ since there are at most $|X|$ ways to partition $X$ into finitely many clopen sets.


TITLE: Squaring across an inequality with an unknown number
QUESTION [3 upvotes]: This should be something relatively simple. I know there's a trick to this, I just can't remember it.
I have an equation
$$\frac{3x}{x-3}\geq 4.$$
I remember being shown at some point in my life that you could could multiply the entire equation by $(x-3)^2$
in order to get rid of the divisor. However,
$$3x(x-3)\geq 4(x-3)^2.$$
Really doesn't seem to go anywhere.
Anyone have any clue how to do this, that could perhaps show me/ point me in the right direction?

REPLY [1 votes]: I'll give a more mechanical way to solve the inequality that avoids the (neat!) trick of multiplying by $(x-3)^2$. But note that this is essentially the same as what Bill and Eivind showed. From
$$
\frac{3x}{x-3} \geq 4 ,
$$
we get
$$
\frac{3x}{x-3} - 4 \geq 0 \hspace{.2in} \implies \frac{3x-4x+12}{x-3} \geq 0.
$$
Simplifying this, we get
$$
\frac{12-x}{x-3} \geq 0.
$$
From this point, the solution is similar to Eivind's. This inequality implies one of the following two cases:

$12 - x \geq 0$ and $x-3 > 0$. That is, $3 < x \leq 12$.
$12 - x \leq 0$ and $x - 3 < 0$. That is, $x < 3$ and $x \geq 12$; clearly this is impossible.

Hence the valid solutions are $3 < x \leq 12$.


TITLE: Is it possible / allowed to use L'Hôpitals rule for products?
QUESTION [4 upvotes]: In our readings, we had L'Hôpitals rule and defined it like that:

$\lim_{x\rightarrow x_{0}}\frac{f'(x)}{g'(x)}$

Because we had it in our readings, we are allowed to use this to find limit of functions.
Now my question is, is it possible to use this rule for products? If yes, do you think I would be allowed to do it (since we have dicussed this rule in our reading...)?
Actually, a fraction is a product at the same time, isn't it?
Because we can also write:
$\lim_{x\rightarrow x_{0}}f'(x)*\frac{1}{g'(x)}$
and it would be called product, or am I totally wrong here?
How would you use L'Hôpitals rule for products? Possible at all?
I could imagine it has something to do with fraction and reciprocal.
But not sure about that.

REPLY [2 votes]: You know that you can use $\frac 0 0$ and $\frac \infty \infty$ in L'Hôpital's rule. Thus, if you have $0\cdot\infty$ where $f(k)\rightarrow0$ and $g(k)\rightarrow\infty$, you can rewrite $\lim_{k\rightarrow c}{f(k)\cdot g(k)}$ as $\lim_{k\rightarrow c}{\frac {f(k)}{\frac 1 {g(k)}}}$. Now the problem is no longer in the form $0\cdot \infty$ but $\frac 0 0$, and thus you can apply L'Hôpital's rule.


TITLE: Is this a theorem (is it correct?)?
QUESTION [2 upvotes]: My instruction notes have specified a theorem of matrix transpose that be there two compatible matrices $A$ and $B$ in respect of their sums and products, then:
$(AB)^T =A^TB^T$
So I set on to verify if indeed this is correct:
Let $A = \begin{pmatrix}2 & 2 \\ 6 & 4\end{pmatrix}  $
Let $B = \begin{pmatrix}1 & 0 \\ 7 & 3\end{pmatrix}  $
$A^T=\begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}  $
$B^T=\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}  $
$(AB) = \begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}$=$\begin{pmatrix}16 & 6 \\ 34 & 12\end{pmatrix}  $
hence $(AB)^T=\begin{pmatrix}16 & 34 \\ 6 & 12\end{pmatrix}$
and....
$A^TB^T=\begin{pmatrix}2 & 6 \\ 2 & 4\end{pmatrix}\begin{pmatrix}1 & 7 \\ 0 & 3\end{pmatrix}=\begin{pmatrix}2 & 32 \\ 2 & 26\end{pmatrix}$
$\therefore (AB)^T\neq A^TB^T $
Is the theorem wrong or am having a rough time understanding?

REPLY [1 votes]: The assertion is false; what is true is that
$$
(AB)^T=B^TA^T
$$
which can be easily proved. Actually, the product $A^TB^T$ need not be defined when the product $AB$ is defined, so the equality doesn't make sense unless both matrices are square. However, the equality doesn't generally hold in this case.
Let's see what happens if you also have, for two square matrices, $(AB)^T=A^TB^T$:
\begin{gather}
A^TB^T=B^TA^T\\
(A^TB^T)^T=(B^TA^T)^T\\
(B^T)^T(A^T)^T=(A^T)^T(B^T)^T\\
BA=AB
\end{gather}
Conversely, it's obvious that from $AB=BA$ you can deduce $(AB)^T=A^TB^T$.
So the equality you read in your instructions is wrong.


TITLE: Why is : $ \ \ V \times_U W = p^{-1} (V) \bigcap q^{-1} (W) $?
QUESTION [1 upvotes]: Let $ f: Y \to X $ and $ g : Z \to X $ be two morphisms of schemes.
Suppose we know that $ Y \times_X Z $ exists, and let $ p $ dénote its projection on $ Y $, and $ q $ its projection on $ Z $.
Let $U, V $ and $ W $ be open sets in $X, Y $ and $Z$ respectively, such that $f (V) \subset U $ and $g (W) \subset U $.
Question :
How to establish that $ V \times_U W $ exists and is canonically identified with the open set $ p^{-1} (V) \bigcap q^{-1} (W) $ of $ Y \times_X Z $, combining universal properties of fiber products and universal properties of open immersions ?
Thank you in advance.

REPLY [0 votes]: To show that $ V \times_U W $ exists, you have to find an $U$-scheme $Z$ satisfing the universal property of $ V \times_U W $. By the way, two $U$-schemes $Z,Z'$ satisfing the universal property of $ V \times_U W $ are canonically isomorphic in the category of $U$ schemes. Now, $p^{-1}(V)\cap q^{-1} (W)$ is canonically a $U$-scheme, and you can verify, as a really simple exercise, that it satisfies the universal property of $ V \times_U W $. Try it step by step, and post if blockers any.


TITLE: pullback on vector fields and 1-forrms
QUESTION [0 upvotes]: On a Riemannian manifold $M$ one can identify 1-forms and vector fields by
$$ \alpha(p) = \langle X(p),\cdot\rangle_p $$
Since we can perform a pullback on both 1-forms and vector fields I expected that for a diffeomorphism $\phi:M\to M$ we have
$$\phi^*\alpha(p)=\langle\phi^*X(p),\cdot\rangle_p$$
However, since
$$\phi^*\alpha(p)=\alpha(\phi(p)\textrm D \phi^{-1}(\phi(p))
= \langle X(\phi(p)),\textrm D\phi^{-1}(\phi(p))\cdot\rangle_{\phi(p)}$$
and
$$\langle\phi^*X(p),\cdot\rangle_p=\langle\textrm D\phi^{-1}(\phi(p))X(\phi(p)),\cdot\rangle_p$$
my guess seems to be wrong. Is there anything wrong with my calculation?

REPLY [1 votes]: Are you sure about your definition of the pullback of a one-form? My definition would be
$$
(\phi^*\alpha)_p(Z_p) = \alpha_{\phi(p)}(D \phi_p (Z_p)) = \langle X_{\phi(p)}, D \phi_p (Z_p) \rangle_{\phi(p)},
$$
your expressions with $D\phi^{-1}$ don't make much sense (note that $D\phi^{-1}_{\phi(p)}$ maps into $T_pM$, so you can't take the scalar product at $\phi(p)$ on the r.h.s.)  ...
Now, in order to make your identifications (which depend on the metric!) compatilbe with taking the pullback by a diffeomorphism $\phi$, you have to assume that $\phi$ is not only a diffeomorphism but also an isometry, i.e.
$$
\langle \cdot , \cdot \rangle_p = \langle D\phi(p) \cdot , D\phi(p) \cdot \rangle_{\phi(p)}
$$
Just plug in this into the calculations and you'll be fine :)


TITLE: Which point do all possible circles pass through?
QUESTION [1 upvotes]: $P$ is a variable point on side $BC$ of triangle $ABC$. $M, N$ are respectively on $AB, AC$ such that $PM||AC$,$PN||AB$. Prove that as $P$ varies on $BC$, then the circumcircle of $AMN$ passes through a fixed point.

I predict that the fixed point would be the "$A$ dumpty point", i.e. the intersection point of the circumcircle of $BOC$ and the $A$ symmedian. But I couldn't prove it. Can anyone help proving this, or otherwise, if the fixed point I mentioned is incorrect, then what is it, and can you prove it? Thanks for helping.

REPLY [1 votes]: Hint:
Notice that $PMAN$ is a paralelogram and that $BM:MA = AN:NC$, which means there is a spiral similarity which takes $B-M-A$ to $A-N-C$ (wherever $P$ is) and let $S$ be a center of this spiral similarity. What can you deduce for $S$?