text:
\begin{document}

\footer{{\bf Key words and phrases}: decoupling principle, symmetric tensor
products, random polynomials, multiple random series, multiple stochastic
integrals, random multilinear forms, random chaos, tail
inequalities, polarization, symmetrization, Banach space,
rearrangement invariant, Orlicz space, Lorentz space, Rademacher sequence,
U-statistics}
\footer{{\bf AMS (1980) Classification} {\em  Primary}: 60B11, 46M05 {\em
Secondary}: 60H07, 46E30, 60E15, 62H05,  62G30}
\vspace{10pt}

\begin{center}
{\large\bf CONTRACTION AND DECOUPLING INEQUALITIES FOR
MULTILINEAR FORMS AND U-STATISTICS}
\vspace{5pt}

{\sc V.H.~de la Pe\~na}$^{\dag}$
\footer
{$^{\dag}$ Research supported in part by N.S.F.\ Grant DMS 91-08006 and a grant
from the CNRS of France.}

{\small\it Department of Statistics, Columbia University,

New York, NY 10027.}
\vspace{5pt}

{\sc S.J.~Montgomery-Smith}$^{\ddag}$
\footer
{$^{\ddag}$ Research supported in part by N.S.F.\ Grants DMS 9001796 and DMS
9201357.}

{\small\it
Department of Mathematics, University of Missouri,

Columbia, MO 65211.}
\vspace{5pt}

{\sc Jerzy Szulga}

{\small\it Department of Mathematics, Auburn University,

Auburn, AL 36849
}
\end{center}
\vspace{7pt}

\parbox{6in}{\small We prove decoupling inequalities for
random polynomials in independent random variables with coefficients in vector
space. We use various means of comparison, including rearrangement invariant
norms (e.g., Orlicz and Lorentz norms), tail distributions, tightness,
hypercontractivity, etc. }

\renewcommand{\baselinestretch}{0.9}
{\small \tableofcontents }
\eject

\renewcommand{\baselinestretch}{1.2}
\vspace{7pt}

\section{Introduction}
\subsection{Background and scope of the paper}

Decoupling principles stem from the theory of martingale transforms (cf.
\cite{Bur:mf}). For homogeneous random forms of rank $k\ge 2$, decoupling
principles were introduced in \cite{McCT:realdec,Kwa:dec,McCT:bandec} (in some
special cases, they were known to Pisier, cf. \cite{McCT:dyadic}), and
subsequently became essential tools in multiple integration (cf.
\cite{KalS,Szu:Umli,KwaW:double,RosW:clock,RST}.
One of the most appealing interpretations of such an principle is the
reducibility of a study of multiple random series (respectively, of multiple
stochastic integrals) to a consecutive treatment of single random series
(respectively, of single stochastic integrals which allows one to treat a
multiple integral as an It\^o-type iterate integral).  The concept of a random
chaos goes back to N.\ Wiener \cite{Wie:chaos} (see also \cite{WieW}), who
elaborated what we call here a real-valued coupled Gaussian chaos. Decoupling
inequalities may be viewed as embedding-projection procedures, since  a
decoupled random chaos is nothing but a lacunary random chaos.  In comparison
to the classical $L^2$-theory of multiple summation or integration, decoupling
principles make up the lack of $L^2$-isometries.

Since the first
publication of the aforementioned decoupling principle, the theory has branched
into several directions. For example, comparison of tangent processes (cf.
\cite{Hit:tan,Pen:Wald,Pen:bound},) is akin to the classical decoupling
principle.  Further
contributions can be found, e.g.,  in
\cite{Kwa:dec,DeA:dec,Hit:tan,Zin:comp,NolP,Pen}.  In some of the
aforementioned papers (e.g., \cite{McCT:realdec,McCT:bandec,Kwa:dec,DeA:dec})
the symmetry of random variables is essential for the fulfillment of the
decoupling principle, while other papers (e.g.,
\cite{Hit:tan,Zin:comp,KalS}) point out the role of positivity.
Norms of $L^p$-spaces, or more general, of Orlicz spaces (basically, subject to
growth restriction) have become typical means of comparison of two classes of
vector random variables.

In this paper, we prove decoupling inequalities for random variables that are
not necessarily symmetric. Theorems~\ref{AB} and~\ref{F} in Section \ref{rob},
and Theorem~\ref{ABtail} in Section 3, are our main results.
The decoupling principle by means of probability tails,
Theorem \ref{ABtail}, immediately
ensures the parity of tightness of two types of chaoses (that Gaussian
decoupled and coupled chaoses are simultaneously tight was proved in
\cite{Kwa:dec}).

A number of decoupling results are obtained for arbitrary rearrangement
invariant norms and Orlicz functionals.
In particular, we provide one extended example regarding certain
Lorentz norms (important in the approximation theory).
Another application is the decoupling principle for U-statistics (a
result as in Theorem \ref{F} was  proven in \cite{Pen}).

The utilized techniques are based on ideas, borrowed from
\cite{Kwa:dec}, while some are taken from \cite{KwaW:book}.
Proofs are straightforward and point out the algebraic nature of decoupling
that is fruitfully merged with a widely understood context of convexity.   A
rule of thumb is that, in the field of random diagonal-free polynomials, a
``definable'' is ``decouplable''.  The obtained robust
constants are tightly estimated, and are sharper than constants known
before.

In the last section, we show tail probability decoupling results for
polynomials of symmetric random variables.  This section makes use of
techniques from \cite{AsmM}.

\subsection{Notation}
Random variables in this paper are defined on
a separable probability space $(\Omega, \ca A,\P)$ that is rich enough to carry
independent sequences.  A sequence of real random variables is
denoted by $\bm\xi = (\xi_1,\xi_2,\dots)$,
and a matrix of real random variables is denoted by
$\BX=[\bm\xi_1,\ldots,\bm\xi_n]$, where $\bm\xi_j = (\xi_{j1},\xi_{j2},\ldots)$.
We will make particular
use of one particular sequence, the Rademacher
sequence
$\beps=(\eps_i)$, where $\eps_i$ are independent random variables taking
values $\pm 1$ with probability $1/2$.

Let
$\BE=(\BE,\|\cdot\|)$ denote a real Banach space.
We will be considering $\BE$\ valued random variables, that is,
strongly measurable mappings from $\Omega$
into $\BE$.

\def\ik{{i_1,\dots,i_k}}

Let $k$ be a positive integer and $f = (f_\ik)$  be an array of vectors
from $\BE$\ taking only finitely many non-zero values.
Throughout the paper, all such arrays are assumed to
vanish on diagonals (We will say ``{\em diagonal-free}''), i.e.,
$f_\ik=0$, if
at least two indices $i_j,i_{j'}$ are equal.

The main object of our interest will be the following
$k$-homogeneous random polynomial
\[
Q(f;\BX) \df Q(f;\bm\xi_1,\ldots,\bm\xi_k)\df
\sum_{\ik}
f_\ik \xi_{1 i_1} \cdots \xi_{k i_k}.
\]
We will be desiring to compare this random polynomial with the `undecoupled'
version, that is
\begin{equation}\label{undecoupled}
Q(f; \bm\xi^k) \df
Q(f;\bm\xi,\ldots,\bm\xi)\df
\sum_{\ik}
f_\ik \xi_{i_1} \cdots \xi_{i_k}.
\end{equation}
The first term in the above definition will be introduced as a notational
convenience.  We will be quite free to stretch the use of this notation.
So for
example, we might write
\[
Q(f; \bm\xi^r , \bm\eta^{k-r})
=
Q(f; \bm\xi,\ldots,\bm\xi , \bm\eta,\ldots,\bm\eta)
=
\sum_{\ik}
f_\ik \xi_{i_1} \cdots \xi_{i_r} \eta_{i_{r+1}} \cdots \eta_{i_k}.
\]
We hope
to convince the reader of the value of this
notation, as it enables us to write many of the proofs in a more
compact form, and may ultimately lead to a clearer thinking on the
subject.  For the unconvinced reader, we hope that we have explained
the notation sufficiently that he will be able to rewrite all the
proofs and statements below in a more familiar form.

Many of the inequalities that we introduce require the array $f$\ to satisfy
certain symmetry conditions, and so for this reason we introduce the
symmetrized version of $f$:
\[\wh{f}_\ik
\df\rec{k!}\sum_\sigma f_{i_{\sigma_1},\ldots,i_{\sigma_k}}
\]
where the sum is taken over all permutations of the set
$[1,k]=\set{1,\ldots,k}$, and let $\wh{Q}(f;\cdot)=Q(\wh{f},\cdot)$. Note that
for the undecoupled random polynomial symmetry makes no change:
$\wh{Q}(f;\bm\xi^k)={Q}(f;\bm\xi^k)$.

In the sequel, we will occasionally refer to tetrahedral arrays, i.e.
$f$ such that $f_\ik = 0$, if indices fail to satisfy  $i_1<\ldots<i_k$.

We will make frequent use of
the following identity, which is known as the Mazur-Orlicz polarization
formula \cite{MazO}:
\begin{equation}\label{MO}
\wh{Q}(f;\bm\xi_1,\ldots,\bm\xi_k)=
\rec{k!}\Sum_{\ddl=(\dl_1,\ldots,\dl_k)\in \set{0,1}^k}
(-1)^{k-|\ddl|}Q(f;(\dl_1\bm\xi_1+\ldots+\dl_k\bm\xi_k)^k),
\end{equation}
where $|\bm\delta|=\sum_i \delta_i$. Switching to a
Rademacher sequence $\beps$, we can write
\begin{equation}\label{MORad}
\wh{Q}(f,\bm\xi_1,\ldots,\bm\xi_k)=
\Rec{k!}
\E\,
\eps_1\cdots \eps_k \,Q(f,(\sum_{i=1}^k\eps_i \bm\xi_i)^{k}),
\end{equation}
where the expectation is only over the Rademacher sequence $\bm\eps$.

\paragraph{Rearrangement invariant spaces}
By $(\L,\|\cdot\|_\L)$  we denote a rearrangement invariant Banach space of
integrable random variables (so that the norm of a random variable depends
only on its probability distribution), $\L\subset L^1(\P)$, defined on a
separable probability space $(\Omega, \ca A, \P)$ that is rich enough to carry
independent sequences.
For more information on  rearrangement invariant spaces, we refer the reader
to, e.g., \cite{LinT}.
The basic examples of rearrangement invariant spaces are $\L = L_p$,
for $1 \le p \le \infty$, (nothing more is needed in many parts of this paper),
Orlicz spaces and Lorentz spaces.  We will sometimes use the abbreviation `r.i.'
for `rearrangement invariant.'

The important property of rearrangement invariant spaces that we shall use
is the following:
\begin{equation}\label{L}
\mbox{\em Conditional expectations are contractions acting on $\L$}.
\end{equation}
The reader unfamiliar with r.i.\ spaces should note that this is true of
$L_p$.

We denote by $\L(\BE)$ the Banach space of $\BE$-valued random variables (i.e.,
strongly measurable mappings from $\Omega$ into $\BE$) whose norms belong to
$\L$, and let $\|\theta\|_{\L(\BE)}=\|\,\|\theta\|_{\BE}\,\|_{\L}$.
Thus if $\L = L_p$, then $\|\theta\|_{\L(\BE)} = \|\theta\|_{L_p(\BE)}
= (\E \|\theta\|_\BE ^p)^{1/p}$.
In
the sequel we sometimes omit the subscript indicating the space, if it
causes no ambiguity.

\section{Decoupling for r.i.\ norms}
\label{rob}
\paragraph{Interchangeability}
In the sequel, we will use several times the following elementary feature of
interchangeable random sequences $ \bm\xi_1,\ldots,\bm\xi_r$ (i.e., such that
each permutation has the same distribution).
Suppose that each $\bm\xi_k$\ is itself a sequence of
independent random variables.
Denote by $\ca G_r$ the
$\sigma$-field spanned by  $\sum_{j=1}^r
\bm\xi_j$.
Let $f$ be diagonal-free. Then if $j_1,\dots$, $j_k \le r$, then
\begin{equation}\label{inter}
\E[Q(f,\,\bm\xi_{j_1},\ldots,\bm\xi_{j_k})\,|\,\ca G_r]=
r^{-k}Q(f,(\bm\xi_1+\ldots+\bm\xi_r)^{k}).
\end{equation}
We should point out that the last term represents the random polynomial
$$ Q(f,(\bm\xi_1+\ldots+\bm\xi_r)^{k}) =
   \sum_\ik f_\ik (\xi_{1 i_1} + \dots + \xi_{r i_1}) \cdots
   (\xi_{1 i_k} + \dots + \xi_{r i_k}) .$$
Equation \refrm{inter} follows because
\[
\E  (f_\ik \xi_{j_1 i_1} \cdots \xi_{j_k i_k} | G_r)
= f_\ik \E(\xi_{j_1 i_1} | G_r) \cdots\E(\xi_{j_k i_k} | G_r)
\]
because $f$\ is diagonal free, and hence $\xi_{j_1 i_1},\dots$, $\xi_{j_k i_k}$
are independent if $f_\ik \ne 0$, and also because for $j \le r$
\[
\E(\xi_{j i} | G_r) = r^{-1}(\xi_{1 i} + \dots + \xi_{r i}) .
\]
We also point out the following easy consequence of the triangle inequality
for $\L$.
\begin{equation}\label{(0)}
\lnorm \wh{Q}(f,\BX) \rnorm_{\L(\BE)} \le \lnorm Q( f,\BX)\rnorm_{\L(\BE)}.
\end{equation}

Now we are ready to present our first decoupling inequality.  This result
allows us to decouple random polynomials in the rearrangement invariant
norm.
\begin{theorem}\label{AB}

Let $f = (f_\ik)$ be a diagonal free array of vectors from $\BE$.
Let $\bm\xi, \bm\xi_1,\ldots,\bm\xi_k$
be sequences of integrable independent real random variables.

Let $\L$ be a r.i. space of random variables, containing
$\xi_{1}\cdots \xi_{k}$ (hence, norms of all finitely
supported polynomials spanned by
 $\bm\xi_1,\ldots,\bm\xi_k$).
\begin{itemize}

\item[{\rm (A)}]
Assume that $\bm\xi,\bm\xi_1,\bm\xi_2,\ldots$ are independent and
identically distributed.

Then
\[
\lnorm   Q(f,\bm\xi^{k})\rnorm _{\L(\BE)}
\le A\lnorm  Q(f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm  _{\L(\BE)}
\]
where $A=A_k\sim (2k)^k$, or, if $\E\bm\xi=0,\, A_k=k^k$.

\item[{\rm (B)}]
Assume that $\bm\xi,\bm\xi_1,\ldots,\bm\xi_k$ are interchangeable.

Then
\[
\lnorm  \wh{Q}(f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm_{\L(\BE)} \le
B\lnorm {Q}(f,\bm\xi^{k})\rnorm_{\L(\BE)} ,
\]
where $B=B_k\sim k^k/k!$.
\end{itemize}
\end{theorem}

\Proof

During this proof, we will suppress the subscript $\L(\BE)$ on the norms.

{\bf (A)}:

\underline{ Step 1$^o$: Centering procedure}:
\vspace{5pt}

\def\ir{{i_1,\dots,i_r}}

Denote $\ov{\xi}=\xi-\E[\xi]$,  $\bm m=(m_1,m_2,\ldots)$,
where $m_i=\E[\xi_i]$.
For $1\le r\le k$, if $f = (f_\ir)$, then we have
\begin{equation}\label{centdec}
\lnorm  Q(f,\ov{\bm\xi}_1,\ldots,\ov{\bm\xi}_r)\rnorm
\le 2^r\,\lnorm Q(f,\bm\xi_1,\ldots,\bm\xi_r)\rnorm .
\end{equation}

Indeed, by interchangeability
\[
\begin{array}{c}
\displaystyle
\lnorm  Q(f,\ov{\bm\xi}_1,\ldots,\ov{\bm\xi}_r)\rnorm =
\lnorm
Q(f,\bm\xi_1-\bm m,\ldots,\bm\xi_r-\bm m)\rnorm \\
\displaystyle
=\lnorm  \Sum_{(\delta_1,\dots,\delta_r)\in\set{0,1}^r}
\sum_\ir f_\ir \xi_{1 i_1}^{\delta_1} \cdots \xi_{r i_r}^{\delta_r}
m_{1 i_1}^{1-\delta_1} \cdots m_{r i_r}^{1-\delta_r}\rnorm \\
\displaystyle
\le \sum_{j=0}^r {r \choose j} \lnorm \sum_\ir f_\ir
\xi_{1 i_1} \cdots \xi_{j i_j} m_{j+1, i_{j+1}} \cdots m_{r i_r} \rnorm \\
\displaystyle
= {\Sum_{j=0}^r}{r \choose j}
\lnorm  Q(f, \bm\xi_1,\ldots,\bm\xi_j, \bm m^{(r-j)})\rnorm .\\
\end{array}
\]

The latter expression is equal to
\[
{\Sum_{j=0}^r}{r \choose j}
\lnorm  Q( f, \bm\xi_1,\ldots,\bm\xi_j,
\E[\bm\xi_{j+1}|\BX_j],\ldots,
\E[\bm\xi_r|\BX_j])\rnorm
={\Sum_{j=0}^r}{r \choose j}
\lnorm  \E[Q( f, \bm\xi_1,\ldots,\bm\xi_j,
\bm\xi_{j+1},\ldots,
\bm\xi_r) | \BX_j ]\rnorm .
\]
where $\BX_j$ is the matrix $[\bm\xi_1,\ldots,\bm\xi_j]$.
Using the contractivity property \refrm{L}, we
estimate the above term from above by
\[
{\Sum_{j=0}^r}{r \choose j}
\lnorm  Q(f,\bm\xi_1,\ldots,\bm\xi_r)\rnorm
= 2^r \lnorm  Q(f,\bm\xi_1,\ldots,\bm\xi_r)\rnorm.
\]
\vspace{5pt}

\underline{Step 2$^o$: proving (A)}:
Arguing similarly to Step 1$^o$ above, we note that
\begin{eqnarray*}
\lnorm  Q(f, \bm\xi^k)\rnorm =
\lnorm   Q(f, (\ov{\bm\xi}+\bm m)^k)\rnorm
\le \Sum_{r=0}^k {k \choose r} \lnorm
Q( f, \ov{\bm\xi}^r, \bm m^{k-r})\rnorm := Q_0.
\end{eqnarray*}

Now, using \refrm{L}, and noting that
$\E(\ov{\bm\xi}_1 + \cdots + \ov{\bm\xi}_r | \ov{\bm\xi}_1) = \ov{\bm\xi}_1$,
which is identically
distributed to $\ov{\bm\xi}$, it follows that

\begin{eqnarray*}
\displaystyle
Q_0 \le  \Sum_{r=0}^k {k \choose r} \lnorm
Q(f, (\ov{\bm\xi}_1+\cdots +\ov{\bm\xi}_r)^r,
\bm m^{k-r})\rnorm
\end{eqnarray*}
Then, in virtue of \refrm{inter}, the latter expression is equal to
\[
\Sum_{r=0}^k {k \choose r} \lnorm   \,r^r\,
\E[\,Q(f,\ov{\bm\xi_1},\ldots, \ov{\bm\xi_r},
\bm m^{k-r})\,|\,\ca G_r]\rnorm
=:Q_1
\]

Using \refrm{L}  and  applying the centering procedure
\refrm{centdec}, the above term gets the following upper bounds:

\begin{eqnarray*}
Q_1\le
\Sum_{r=0}^k {k \choose r} \lnorm
 \,r^r\,
Q( f,\ov{\bm\xi_1},\ldots, \ov{\bm\xi_r},
\bm m^{k-r})\rnorm \\
\displaystyle
\le \Sum_{r=0}^k {k \choose r}  (2r)^r
\lnorm  \, \,
Q(f,\bm\xi_1,\ldots, \bm\xi_r, \bm m^{k-r})\rnorm
=: Q_2
\end{eqnarray*}

Then, by interchangeability, independence of columns,  and \refrm{L} again, we
keep estimating, as follows
\begin{eqnarray*}
Q_2= \Sum_{r=0}^k {k \choose r}   (2r)^r\,
\lnorm
\E[\,Q(f,\bm\xi_1,\ldots, \bm\xi_r,
\bm\xi_{r+1},\ldots,\bm\xi_k)\,|\,\BX_r\,]\rnorm \\
\le \Sum_{i=0}^k {k \choose r} (2r)^r\,
\lnorm
Q( f,\bm\xi_1,\ldots, \bm\xi_r,
 \,\bm\xi_{r+1},\ldots,\bm\xi_k)\rnorm \\
=A_k \lnorm  \,
Q( f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm .\\
\end{eqnarray*}
Clearly, $A_k\le (2k+1)^k$ (notice that $A_k\ge c (2k)^k$). If $\E\xi=0$, the
use of centering procedure, and the triangle inequality is superfluous,
hence the constant decreases to $k^k$.
\vspace{5pt}

(B): By the Mazur-Orlicz polarization formula \refrm{MO}, and \refrm{inter},
we obtain the following bounds
\[
\begin{array}{rl}
\lnorm  \wh{Q}(f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm
=&\lnorm  \Rec{k!} Q(f, \Sum_{\ddl}
(-1)^{k-|\ddl|}
(\dl_1\bm\xi_1+\ldots+\dl_k\bm\xi_k)^k)\rnorm \\

\le &\Sum_{\ddl}\lnorm  {\Rec{k!}}
Q( f, (\dl_1\bm\xi_1+\ldots+\dl_k\bm\xi_k)^k )\rnorm \\

=&\displaystyle
\Sum_{r=0}^k{k \choose r}
\lnorm  \Rec{k!}Q(f, (\bm\xi_1+\ldots+\bm\xi_r)^k)\rnorm \\

\le &\displaystyle
\Sum_{r=0}^k{k \choose r}
\lnorm  \Frac{r^k}{k!} Q(f,\bm\xi_1^k)\rnorm \\
= &
 B_k\lnorm Q(f,\bm\xi_1^k)\rnorm. \\
\end{array}
\]
That $B_k\sim k^k/k!$ is easy to verify.
The proof is completed.
\QED

\subsection{Extended multilinear forms and U-Statistics}

In this section, we show how to extend Theorem \ref{AB} to the so called
U-Statistics.  Let $k$\ be a positive integer.  Let $F = (F_\ik)$\
be an array of strongly Borel measurable functions $F_\ik:\R^k \to \BE$
such that
\[
\begin{array}{rl}
\mbox{\rm \hspace{-10pt}  (F0)\hspace{10pt} }&
           \mbox{\sf $F_\ik = 0$ if some $i_j$\
and $i_{j'}$ are identical for $j \ne j'$}\\
\mbox{\rm \hspace{-10pt}  (F1)\hspace{10pt} }&
           \mbox{\sf $F_\ik=0 $ for all but finitely many $(\ik)$;}\\
\end{array}
\]
Then we are going to consider U-Statistics, that is,
expressions of the form
\[
F(\bm\xi_1,\dots,\bm\xi_k)\df\sum_{\ik} F_\ik(\xi_{1i_1},\ldots,\xi_{ki_k}),
\]
where $\bm\xi_1,\dots,\bm\xi_k$ are real valued random variables.  (Here, $\R$\
could be replaced with any other measure space, but there is no loss of
generality to take it as $\R$.)  We are going to exercise the same notational
devices as for the random polynomials, so that the undecoupled U-Statistic is
written
\[
F(\bm\xi^k)\df\sum_{\ik} F_\ik(\xi_{i_1},\ldots,\xi_{i_k}) .
\]
As before, in order to prove the results, we require certain symmetry
properties to hold for $F$.  So we defined the symmetrized version of
$F$ as follows:
\[
\hat F(x_1,\dots,x_k)
\df {1\over k!} \sum_\sigma
F_{i_{\sigma_1},\dots,i_{\sigma_k}}(x_{i_{\sigma_1}},\ldots,xi_{i_{\sigma_k}}),
\]
where the sum runs over all permutations of $[1,k]$, and we set
\[
\hat F(\bm\xi_1,\dots,\bm\xi_k)
\df \sum_{\ik}
\hat F_\ik(\xi_{1i_1},\ldots,\xi_{ki_k}) .
\]
Decoupling results were proved by \cite{Pen} for Orlicz modulars (and
so by Note~\ref{riorl} below, one can obtain
results for all rearrangement invariant
spaces).  We will prove similar decoupling results, weakening some of
the hypotheses.

More interestingly, we are going to prove the decoupling results for
U-statistics as a corollary of Theorem~\ref{AB} which decouples random
polynomials.  The technique is to approximate the U-statistic as
a sum of random polynomials.  That is, let $D$\ be an integer,
and for $1 \le d \le D$, let $f^d = (f^d_\ik)$ be a diagonal free array
of vectors in $\BE$\ taking only finitely many non-zero values, and let
$(\bm\xi^d_1:1\le d \le D),\dots$, $(\bm\xi^d_k:1\le d \le D)$ be
sequences of independent random
variables.  Then we set
\begin{equation}\label{newtensor}
R(f;\bm\xi_1,\dots,\bm\xi_k) \df
\sum_{d=1}^D Q(f^d;\bm\xi^d_1,\dots,\bm\xi^d_k) .
\end{equation}

Then the remarkable thing is that the proof of Theorem~\ref{AB} works for
$R(f;\bm\xi_1,\dots,\bm\xi_k)$\ exactly as it does for
$Q(f;\bm\xi_1,\dots,\bm\xi_k)$, that is, we have the following result.

\begin{theorem}\label{ABnew}
Theorem \ref{AB} is valid, for the
multilinear form \refrm{newtensor}.
\end{theorem}

A version of the following result for Orlicz modulars
$\E\phi(\cdot)$, where $\phi$ was a moderately increasing function,
was proved in \cite{McCT:bandec}. In that paper, terms of the underlying sums
were sign-randomized, i.e., each $F(\bi, \cdot)$ was multiplied by  Walsh
functions $\eps_{i_1}\cdots \eps_{i_k}$. More precisely, the decoupling was
proved for
\begin{equation}\label{Fbeps}
(F\circ\beps)(\bm\xi_1,\dots,\bm\xi_k)\df
\sum_{\bi} \eps_{i_1}\cdots\eps_{i_k}F_\ik(\xi_{1i_1},\ldots,\xi_{ki_k}),
\end{equation}
where $\beps$ is independent of $\BX$.  That the
presence of Walsh functions is not necessary in the context of Orlicz modulars,
was shown in \cite{Pen}. We observe that the following result, generalizing
theorems in the mentioned papers, is implicit in the main decoupling principle.
Moreover,  constants remain the same. For the sake of completeness we give the
full proof.

In the proof we will use the fact that
any inequality involving norms of functions of discrete
r.v.'s, that converge to some limits, is preserved for these
limits.
Fix $\bm i$, say $\bm i=(1,\ldots,k)$. Consider $X=F(\xi_1,\cdots,\xi_k)$.
We may assume that the probability space $(\Omega,\ca F,\P)$  and $\ca F$ is
spanned by $\xi_1,\xi_2,\ldots$. Also, we may assume that it is separable,
i.e., $\ca F_n=\sigma\{\bigcup_n \ca F_n\}$, where $\ca F_n$ are finite
$\sigma$-fields.  Put $\xi_i^n=\E[\xi|\ca F_n]$. Thus
$\E[X|\ca F_n]\to \E[X|\ca F]=X$ a.s. and in $\L$.

\begin{theorem}\label{F}
Let $F:\N^k\times\R^k\to\BE$  satisfy {\rm (F0) -- (F1)}, and also
and additional condition:
\[
\begin{array}{rl}
\mbox{\rm \hspace{-10pt}  (F2)\hspace{10pt} }&
           \mbox{\em $F(\bi;\xi_{i_1},\ldots,\xi_{i_k})\in \L$
                                      for every $\bi\in\N^k$.}\\
\end{array}
\]
Let
$\bm\xi_1,\bm\xi_2,\ldots$ be sequences of independent
random variables.
\begin{itemize}
\item[(A$'$)]
Let $\|\cdot\|$ be a r.i. norm. Let $\bm\xi_1,\bm\xi_2,\ldots $ be
independent and identically
distributed. Then
\[
\left\|\,F(\bm\xi^k)\,\right\|\le A
\left\|\,\,F(\BX)\,\right\|,
\]
where $A$ is the constant from Theorem \ref{AB}.(A).
\item[(B$'$)]
Let $\bm\xi_1,\ldots,\bm\xi_k$ be interchangeable (in particular, i.i.d.)
and $\|\cdot\|$ be a r.i.\ norm.  Then
\[
\left\lnorm \,\wh{F}(\BX)\,\right \rnorm
\le B_k \left\lnorm \,F(\bm\xi^k)\,\right\rnorm.
\]
where $B$ is the constant from Theorem \ref{AB}.(B).
\end{itemize}
\end{theorem}
\Proof
By Note~\ref{riorl},
we may assume that the rearrangement space $\L$\ is separable.
In that case, we may assume without loss of generality that $\bm\xi_1,
\bm\xi_2,\ldots$\ are real discrete random variable.

Thus
we may assume that the random variables are defined on a product probability
space
\[
(\prod _{ij}\Omega_{ij}, \ca
(F^{\otimes N})^{\otk},(\P^{\otimes \N})^{\otk}),
\]
where $\Omega_{ij}$ are
equal, $\Omega_i=\prod_j\Omega_{ij}$, and the superscript $\otimes$
indicates the product $\sigma$-field and the product probability,
respectively. So, let
\[
\xi_{i}=\sum_m x_{im}\I{A_{im}},
\]
where $A_{i1},A_{i2},\ldots\subset\Omega_i$ are bases of rectangular sets that
form a disjoint finite partition of $\Omega$, and let $(A_{sim})$,
$s=1,\ldots,k$, be independent copies of $(A_{im})$.  Put
$I_{sim}=\I{A_{sim}}$, hence
\[
\xi_{si}=\sum_m x_{im} I_{sim},\qquad s=1,\ldots,k.
\]
Then
\[
\sum_{\ik}F_{ik}(\xi_{1i_1},\ldots,\xi_{ki_k})=
\sum_{m_1,\dots,m_k} \sum_{\ik}
F_\ik(x_{i_1m_1},\ldots, x_{i_km_k})
I_{1i_1m_1}\cdots I_{ki_km_k}.
\]
Now, we can apply Theorem \ref{ABnew}, and the proof is complete.
\QED

\subsection{An example in a certain Lorentz space}\label{lorentzeg}

 Motivated by the results in \cite{Pen:Wald} where
the problem as to when expectation results imply tail probability results is
treated,
we obtained the following asymptotic tail probability comparison.

\begin{proposition}\label{limsup}
Let $\xi$ or $\eta$ be the norm
of $\wh{F}(\bm\xi^k)$ or $\wh{F}(\BX)$, and let $W:[0,\infty) \to [0,\infty)$\
be an increasing function
such that there exists constants $p>1$\ and $c>0$\ such that
$W(st) \le c s^p W(t)$\ for all $0<s<1$\ and all $t>0$.  Then there is
a constant $C$, depending only on $p$\ and $c$, such that
\[
\limsup_{t\to \infty} W(t) \P(\xi \ge t) \le
\limsup_{t\to\infty} W(t) \P(\eta \ge C t ).
\]
\end{proposition}

This result is a consequence of Theorem~\ref{F},
and follows by arguments from the theory of Lorentz-Zygmund spaces.
If $\xi$ is a random variable, let $F(t)=\P(|\xi|\ge t)$, and define
the decreasing
rearrangement of $\xi$\ to be the
function $\xi^*(t)\df \sup\{s:F(s) > t \}$\ (i.e. the
right-continuous inverse of $F$).
Obviously, $|\xi|$ and $\xi^*$ are equidistributed.
When $\xi$ is integrable,  an average operator is often considered
\[
\xi^{**}(t)=\frac{1}{t}\int_0^t \xi^*(u)\, du,
\]
which corresponds to  a rearrangement invariant norm for every $t>0$.
Therefore, our decoupling inequalities for U-statistics hold for
$\Phi(X)=(\|X\|)^{**}$. That is, denoting by $\xi$ or $\eta$ the norm
of $\wh{F}(\bm\xi^k)$ or $\wh{F}(\BX)$, we have
\[
\xi^{**}(t) \le C \eta^{**}(t)
\]
for some constant $C>0$.

Now consider the Lorentz-Zygmund space  defined by the quasi-norm
$|||f|||=\sup_x w(x) \xi^*(x)$, where $w:[0,1]\to
[0,\infty)$\ is an increasing function.
Note that $|||f||| \le 1$\ if and only if
$ \sup_t W(t) \P(|f| \ge t) \le 1$, where $w(t)=\Frac{1}{W^{-1}(1/t)}$.
If $W$\ satisfies the relation given in Proposition~\ref{limsup}, then
for some constant $c$, the function $w$\ satisfies
the relation $w(x)\le c a^{-1/p} w(xa)$\ for $a \le 1$.  Then it is possible
to show that $|||f||| \le |||f^{**}||| \le C |||f|||$.  Indeed, the first
inequality is obvious, and for the second:
\[
w(x) f^{**}(x) = w(x) \int_0^1 f^*(xa) \, da
\le c \int_0^1 a^{-1/p} w(xa) f^*(xa) \, da
\le \frac {cp}{p-1} ||| f ||| .
\]
Thus, to show Proposition~\ref{limsup},
let $F(t)=\P(\xi\ge t)$\ and $G(t)=\P(\eta\ge t)$.
Then
\[
\sup_t W(t) F(t)\le
\sup_t W(t) G(Ct) .
\]
If we now set $w(x)=0$, for $x\ge x_0$, the same argument applies,
and letting $x_0 \to \infty$, we obtain
\[
\limsup_{t\to \infty} W(t) F(t)\le \limsup_{t\to\infty} W(t) G( C t ).
\]

\section{A discourse on probability tails}
\subsection{$L^p$-estimates imply tail estimates}
\subsubsection{Auxiliary results}
The following result can be found in \cite{AsmM}.

\begin{lemma}\label{Lemma 2.1}
Let $\{X;~X_i\}$ be a sequence of positive i.i.d.\ random variables.
Then, for all positive integers $n$,
all $\alpha > 0 $
and all $0\le \theta \le n$
\begin{eqnarray*}
\P(X \ge \alpha) \ge {\theta \over n} \quad
&\Rightarrow \quad \P(\sup_{1\le i\le n} X_i  \ge \alpha)
\ge {\theta\over 1+\theta},&  \\
\P(X \ge \alpha) \le {\theta \over n} \quad
&\Rightarrow \quad
\P(\sup_{1\le i\le n} X_i \ge \alpha )  \le \theta. &
\end{eqnarray*}
\end{lemma}

\Proof
To show the first inequality, observe first that for $\theta > 0$,
$$ (1 - {\theta \over n} )^n \le {1 \over (1+\theta ) }.$$
Hence, by independence assumption,
\[
\begin{array}{rl}
 \P(\sup_j X_j \ge \alpha )
      &= 1 - \P(\sup_j X_j < \alpha ) \\
      &= 1- \displaystyle\prod_{j=1}^n \P(X_j < \alpha ) r\\
      &\displaystyle\ge 1- (1-{\theta \over n} )^n
\ge 1-{1\over (1+\theta )} = {\theta \over (1+ \theta )}. \\
\end{array}
\]
The second inequality is easy: from the imposed condition, one gets
$$
\P(\sup_{1\le j\le n}X_j \ge \alpha ) \le
\sum_{i=1}^n \P( X_i  \ge \alpha ) \le \theta .
$$
The proof is completed.
\QED

The following result can be found in \cite[Chap. 4]{LedT}.

\begin{lemma}\label{Lemma 2.2}
Consider a positive random variable $Z $ such that
$ \|Z\|_q \le C\|Z\|_p$\ for $q>p>0$. Then,
\[
\P(Z>t) \le (2C^p)^ {q\over p-q} \qquad
\Rightarrow \qquad \|Z\|_p \le 2^{1/p} t
\quad\hbox{and}\quad
\|Z\|_q \le 2^{1/p} Ct.
\]
\end{lemma}
Putting together Lemmas \ref{Lemma 2.1} and \ref{Lemma 2.2} we get the
following.

\begin{lemma}\label{Lemma 2.3}
Let $\{X;~X_i\}$ be a sequence of positive i.i.d.\ random variables.
Assume that there exists a constant $c$ such that for $0<p<q<\infty$
$$
\|\sup_{1\le i \le n}\| X_i\|\|_q \le c\|\sup_{1\le i \le n}\|X_i\|\|_p.
$$
Then, letting
$\theta = (2c^p)^{q\over p-q} $,
\[
\P(X\ge t) \le
{\theta \over n} \quad\Rightarrow\quad
\|\sup_{ 1\le i\le n} X \|_p \le 2^{1\over p}t.
\]
\end{lemma}

For later reference, we also include the next lemma.
\begin{lemma}\label{Lemma 2.4}
Let $\{X;~X_i\}$ be a sequence of positive i.i.d.\ random variables.
Then,
\[
\|\sup_{1\le i \le n} X_i \|_p
\le t
\quad\Rightarrow\quad \P(X\ge 2^{1\over p} t) \le {1 \over n}.
\]
\end{lemma}

\Proof
Use Chebychev's inequality and Lemma \ref{Lemma 2.1} with $\theta = 1$.
\QED

\subsubsection{Main result}
Now we are ready to prove an  extension of a result from \cite{AsmM}
that  deals
with strict tail probability comparisons for
pairs of random variables.

\begin{theorem}\label{2.1}
Let $(X,\,X_i)$ and $(Y,\, Y_i)$ be sequences of positive
i.i.d.\ random variables.
For some $0<p<q$\ and all positive integers $n$ assume that

\begin{equation}\label{(0.1)}
\|\sup_{1\le i \le n} X_i\|_q \le c_1\|\sup_{1\le i \le n} X_i \|_p,
\end{equation}
and
\[
\|\sup_{1\le i \le n} Y_i\|_p \le c_2\|\sup_{1\le i \le n} X_i\|_p.
\]
Then there exists $c_3$, depending only on $p$, $q$, $c_1$\ and $c_2$\
such that for all $t \ge 0$
\[
\P(Y\ge c_3t) \le c_3 \P(X\ge t).
\]
\end{theorem}

\Proof
Given an arbitrary $\alpha = \alpha_1 > 0$ with
\begin{equation}\label{(2.6)}
\P(Y\ge \alpha_1) > 0,
\end{equation}
choose $\mu $ to be the
smallest positive integer satisfying
\begin{equation}\label{(2.7)}
{1\over 2\mu} \le \P(Y\ge \alpha_1)\le {1\over \mu}.
\end{equation}
>From Lemma \ref{Lemma 2.1} it follows that
$$
\P(\sup_{1\le j \le \mu}Y_j\ge \alpha_1) \ge {1\over 3}.
$$
Hence, by Chebychev's inequality,
$$
{1\over \alpha_1^p }
\|\sup_{1\le j \le \mu}Y_j\|_p^p \ge {1\over 3},
$$
which, by assumptions, yields
$$
{c_{2}^p\over \alpha_1^p }
\|\sup_{1\le j \le \mu}X_j\|_p^p \ge {1\over 3},
$$
and, consequently, for any $\alpha_2 > 0$,

$$
{1\over \alpha_2^p }\|\sup_{1\le j \le \mu}X_j\|_p^p \ge
{\alpha_1^p\over 3\alpha_2^p c_{2}^p}.
$$
In particular, if $\alpha_2^p = {\alpha_1^p\over 6 c_{2}^p}$,
we get from the latter inequality that
$$
{1\over \alpha_2^p }
\|\sup_{1\le j \le \mu}X_j\|_p^p \ge 2.
$$
Now, Lemma 2.3 implies that
$$
\P(X\ge {\alpha_1\over (6^{1\over p}
c_{2} )}) = \P(X\ge \alpha_2)\ge
 (2 c_1^p)^{q\over p-q} {1\over \mu} .
$$
Finally, \refrm{(2.7)} gives,
\begin{equation}\label{(2.14)}
\P(X\ge {\alpha_1\over (6^{1\over p} c_{2})}) \ge
 (2 c_1^p)^{q\over p-q} \P(Y \ge \alpha_1 ).
\end{equation}
Note that \refrm{(2.14)} holds for all $\alpha_1$ for which \refrm{(2.6)}
holds. For any other $\alpha_1 > 0$, \refrm{(2.14)} holds trivially.
\QED

\subsubsection{Contraction for multipliers}

Condition \refrm{(0.1)} yields an example of a class of random variables
with the so called Marcinkiewicz-Paley-Zygmund property (MPZ in short). The
concept  was studied in \cite{KraS:hyp}, and can be traced back to  \cite{PalZ}
and \cite{MarZ}. A family $\ca Z\in L^q_+$ of random variables is said to be in
the class $MPZ(q)$ (in short: {\em have MPZ}), if one of the following
equivalent conditions is satisfied:

\begin{eqnarray}
&\exists~ p<q \mbox{ (equivalently, $\forall~ q\le p$)~ }~
m_{q,p}\df\sup_{Z\in \ca Z} \Frac{\|Z\|_q}{\|Z\|_p}<\infty
\label{MPZ1}&\\
\rule{6mm}{0pt} \nonumber\\
&\exists ~ \delta>0~~
\inf_{Z\in \ca Z} \P(Z>\delta\|Z\|_q)>\delta&
\label{MPZ2}
\end{eqnarray}

That is, \refrm{(0.1)} involves $\ca Z=\set{\sup_{1\le i\le n} \|X_i\|:n\in
\N}$. Also, in \cite{KraS:hyp} it was shown that the space of diagonal-free
random polynomials of finite degree, spanned by symmetric random variables
with, so called, semi-regular distribution,
has MPZ. A random variable $\xi$ is
said to have the {\em semi-regular distribution}, if its tail
$G(t)=\P(|\xi|>t)$
satisfies the relation,
\[
V(a)=\limsup_{t\to\infty} G(at)/G(t)<1
\]
for some (or all) $a>1$ (by convention, $0/0=0$). For example, any bounded
random variable has semi-regular distribution. In particular, (the norm of) any
normed space-valued Rademacher polynomial of degree $d$ has MPZ with the
constant $m_{qp}=[2 (q-1)/(p-1)]^{d}$ (\cite[Corollary 2.7]{KraS:hyp}).

Now, the essence of Theorem \ref{2.1} is that  the continuity of a certain
operator, once is fulfilled by means of $L^p$-norms, will be also fulfilled by
means of probability tails. We will illustrate this concept by the following
result.
\begin{theorem}\label{contrRad}
Let $f_\ik$ be a,
finitely supported, diagonal-free
array, taking values in a Banach space
$\BE$.  Let $\bm\xi = (\xi_1,\dots)$\ be a sequence
of symmetric independent random variables.
\begin{enumerate}
\item[{\rm (i)}] ({\bf Contraction Inequality})
There is a constant $c>0$ such that
\[
\P(\|Q(f, \bm (s\bm \xi)^k)\|>c t)\le
c \P(\|Q(f, \bm \xi)\|>t),\qquad t>0,
\]
where $\bm s\bm\xi=(s_i\xi_i)$ and $\|\bm s\|_\infty=\sup_i |s_i|\le 1$.
\item[{\rm (ii)}] ({\bf Maximal Inequality})
There is a constant $C>0$\ such that
\[
\P(\sup_{m_1,\dots,m_k }\|T_{m_1,\dots,m_k}Q(f, \bm \xi^k )\|>C t) \le C
\P(\| Q( f, \bm \xi^k)\|>t),\qquad t>0,
\]
where
\[
T_{m_1,\dots,m_k} Q(f;\BX) = \sum_{i_1\le m_1,\dots, i_k\le m_k}
f_{\ik} \xi_{1i_1}\cdots\xi_{ki_k}.
\]
\end{enumerate}
\end{theorem}
\Proof
Let us first prove~(i).
In virtue of symmetry assumption and Fubini's theorem, it suffices to
give the proof for the
case when $\BX$\ is a matrix of Rademacher random variables.
Let $Q_1,\dots,Q_n$\ be independent copies of $Q(f,\bm)$.
Then the vector $(Q_1,\dots,Q_n)$\ is a Rademacher homogeneous
polynomial of degree $k$\ taking values in $\ell_n^\infty(\BE)$.
Similarly, let $R_1,\dots,R_n$\ be independent copies of $Q(f, \bm s\bm \xi)$.
>From the contraction principle for $L^p$-norms, which may be found in
\cite[Remark 2.9]{KraS:hyp} (essentially, it is due
to \cite{Kwa:dec}), it follows that for all $p \ge 1$
\[
\| \|(R_1,\dots,R_n)\|_{\ell_n^\infty(\BE)} \|_p
\le c
\| \|(Q_1,\dots,Q_n)\|_{\ell_n^\infty(\BE)} \|_p .
\]
>From the observation that $\|(Q_1,\dots,Q_n)\|_{\ell_n^\infty(\BE)}
= \sup_{1 \le i \le n} \|Q_i\|_{\BE}$, and using the fact that
Rademacher polynomials are MPZ, and also citing Theorem~\ref{contrRad}
above, the result
follows.

The proof of part~(ii) is the same, using the corresponding result
for $L^p$-norms of polynomials for the Rademacher random variables,
which follows easily from
\cite{McCT:realdec} and L\'evy's inequality.

\begin{remark}\label{contrRAD:rem}\rm
In fact, Theorem~\ref{contrRad} is also valid for the sign randomized
U-Statistics as in equation~\refrm{Fbeps}.  The proof is identical.
\end{remark}

\begin{theorem}\label{compRad}[ Comparison Inequality]
Let $(f_{i_1,\dots,i_k})$ be a diagonal-free, finitely supported, diagonal free
array.  Let
$\bm\xi = (\xi_i)$\ and $\bm\eta = (\eta_i)$\ be sequences
of symmetric independent random variables.
such
that, for some constant $A>0$,
\[
\P(|\xi_i|> t)\le A\P(|\eta_i|>t),\qquad t>0, ~i\in \N.
\]
Then, for some constant $K=K(c,d,A)$,
\[
\P(\|Q(f, \bm \xi^k)\|> t)\le
K \P(K \|Q(f, \bm \eta^k)\|>t),\qquad t>0.
\]
\end{theorem}
\Proof
We have
\[
Q(f, \bm \xi^k)\D Q(f, (\beps|\bm \xi|)^k),
\]
and
\[
Q(f, \bm \eta^k)\D Q(f, (\beps|\bm \eta|)^k),
\]
where $\beps$ is a Rademacher sequence independent of $\bm\xi$ and $\bm\eta$.

If $A=1$, then we may replace each $|\xi_i|$ and $|\eta_i|$ by their decreasing
rearrangements $|\xi_i|^*$ and $|\eta_i|^*$, respectively. The assumption yields
$|\xi_i|^*\le |\eta_i|^*$ a.s. Hence, by Theorem~\ref{contrRad}(i),
the inequality follows.

Let $A>1$. Then there exist a sequence $\bm\alpha=(\alpha_i)$ of i.i.d. random
variables, independent of $\bm\xi$, such that
$\P(\alpha_i=1)=1/K,\,\P(\alpha_i=0)=1-1/K$, so that
$\P(\alpha_i|\xi_i|>t)=\P(|\xi_i|>t)/K$. Therefore, by the first part of the
proof,
\[
c\P(\|Q(f, (\beps\bm\alpha|\bm \xi|)^k)\|> t)\le
\P(K \|Q(f, (\beps|\bm \eta|)^k)\|>t),\qquad t>0.
\]
Conditioning on $\bm\xi$, it remains to prove that,
for every polynomial $Q$, and every diagonal free array $f_\ik$,
\begin{equation}\label{compeps}
\P(\|Q(f, \beps^k)\|> m t)\le
m\P(\|Q(f, (\beps\bm\alpha)^k)\|> t)
\end{equation}
for some constant $m=m_k$.
Let $\bm\beta=\beps\bm\alpha$.  Then it is clear that $\bm\beta$\ is
semiregular, as defined earlier, and hence homogeneous random
polynomials of degree $k$\
over $\beta$\ have MPZ.  Furthermore, the comparison inequality is
true for $L^p$\ for $p\ge 1$\ (see, for example,
\cite[Theorem~2.13]{KraS:hyp})  Hence arguing as in the proof of
Theorem~\ref{contrRad}, we obtain \refrm{compeps}.
\QED

\subsection{Decoupling for tails}
In order
to prove any tail inequality of the type $\P(\xi>t)\le K\P(\eta>t)$, where
$\xi,\eta$ are real random variables, it is enough to prove it for an
arbitrarily chosen conditional probability
\[
\P \left[\,\xi>t\,|\,\ca G \,\right]\le
K\P \left[\,\eta>t\,|\,\ca G\,\right].
\]
This observation was used in proving the inequality
\cite[(6.9.5)]{KwaW:book}.
Denote by $\ca G$ the
$\sigma$-field spanned by all random variables of the form $\sum_{j=1}^i
h(\bm\xi_j)$ (in other words, by the random point measure
\[
 \sum_{j=1}^i\delta_{\bm\xi_j}
\]
on $(\R^\N)^k$ (cf. \cite[p.\ 182]{KwaW:book}). Then
$(\bm\xi_1,\ldots,\bm\xi_k)$ is concentrated on a finite
permutation invariant subset of  $(\R^\N)^k$. Now, \refrm{inter}
can be rewritten, as follows (recall the notation, preceding \refrm{inter}).
\begin{equation}\label{newinter}
\E[Q(f,\bm\xi_{j_1},\ldots,\bm\xi_{j_k})\,|\,\ca G]=
k^{-k}Q(f,(\bm\xi_1+\ldots+\bm\xi_k)^{k}),
\end{equation}

\begin{theorem}
\label{ABtail}
Let $f,\bm\xi,\bm\xi_1,\ldots,\bm\xi_k$ be as in
Theorem \ref{AB} (but \underline{we do not assume integrability}).
\begin{itemize}
\item[{\rm (A$''$)}]
Let $\bm\xi,\bm\xi_1,\bm\xi_2,\ldots$ be
independent and symmetric. Then, there exists a constant $A''$,
depending only on $k$, such that, for all $t\ge 0$,
\[
\P(\lnorm   Q( f,\bm\xi^{k})\rnorm\|\ge A'' t)\le
A''\P(\lnorm  Q( f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm\ge t).
\]
\item[{\rm (B$''$)}]
Let $\bm\xi,\bm\xi_1,\ldots,\bm\xi_k$ be interchangeable.
Then, there exists some constant $B''$,
depending only on $k$, such that, for all $t\ge 0$,
\[
\P(\lnorm  \wh{Q}(f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm\ge B''t)
\le
B''\P(\lnorm Q(f,\bm\xi^{k})\rnorm\ge t).
\]
\end{itemize}
\end{theorem}
\Proof

{\bm (A$''$)}: By symmetry, using Theorem \ref{contrRad}.(ii), with
$\eta=\xi_1+\ldots+
\xi_k$ and $A=k$, we obtain that
\[
\P(Q(f, \bm\xi^k)\rnorm \ge t K) \le
 K\P[ k^k Q( f, (\bm\xi_1+\ldots+\bm\xi_k)^k)\rnorm \ge t K)
\]
By \refrm{newinter}, and inequality \cite[(6.9.5)]{KwaW:book},
the latter quantity
can be estimated from below by
\[
c_k\,K\,
\P(k^{2k} k^k Q(f,  \bm\xi_1,\ldots,\bm\xi_k)\rnorm \ge t),
\]
which completes the proof of (A$''$).

{\bf (B$''$)}: By the
Mazur-Orlicz polarization formula \refrm{MO}, and \refrm{inter},
we obtain the following estimates
\[
\begin{array}{rl}
\P(\lnorm
\wh{Q}(f,\bm\xi_1,\ldots,\bm\xi_k)\rnorm \ge t)
=&\P(\lnorm  \Rec{k!}Q( f,\Sum_{\ddl}
(-1)^{k-|\ddl|}
(\dl_1\bm\xi_1+\ldots+\dl_k\bm\xi_k)^k)\rnorm \ge t)\\

\le &\Sum_{\ddl}
\P(2^k\lnorm  {\Rec{k!}} Q(f, (\dl_1\bm\xi_1+\ldots+\dl_k\bm\xi_k)^k)
\rnorm\ge t) \\

=&\displaystyle
\Sum_{i=0}^k{k \choose i}
\P(2^k\lnorm  \Rec{k!}Q(f, (\bm\xi_1+\ldots+\bm\xi_i)^k) \rnorm\ge t). \\
\end{array}
\]
By  \refrm{newinter}, with $j_1=\ldots=j_k$,  and the inequality
\cite[(6.9.5)]{KwaW:book}, we estimate the above expression from
above by
\[
\begin{array}{rl}
 &\displaystyle
c_k^{-1} \Sum_{i=0}^k{k \choose i}
\lnorm\P(\Frac{(2i)^k}{k!} Q( f,\bm\xi_1^k)\rnorm\|\ge t) \\
\le  &
c_k^{-1}2^k \P(\Frac{(2k)^k}{k!}\lnorm Q(f,\bm\xi_1^k) \rnorm\ge t), \\
\end{array}
\]
which completes the proof.
\QED

\begin{remark}\label{ABtail:rem}\rm
While the symmetry
assumption is irrelevant in condition (B$''$) (or in (B), before), the
symmetrization procedure used in the proof of (A) fails.
The reason is, that we use the conditioning on $\ca G$, which
destroys the independence, which is essential in applications
of \refrm{centdec}.
\end{remark}

\section{Notes}
{~~~~~~~~}
\begin{enumerate}
\item
The inverse estimate in \refrm{centdec} is not true, in general, even if $k=1$.
For example, let $\xi_1,\ldots,\xi_n$ be Bernoulli random variables with
$p=\P(\xi_1=1)=1/2$, and $f(i)=1$. Then $\E|f\ov{\bm\xi}|^2= n/4$ and
$\E|f\bm\xi|^2= (n+n^2)/4$.
\item
The symmetry of functions $f$ is essential in
Theorem \ref{AB}.(B) and its analogs, as was pointed
out in \cite{McCT:bandec}. The Bourgain's counterexample, given there, involves
$\BE=\ell^2\otimes\ell^2$ endowed with the projective norm $\|\bm
a\|=\inf\set{\sum_{i,j}\|a^1_i\|\cdot\|a^2_j\|:\bm a= \sum_{i,j} a^1_i\otimes
a^2_j}$, Rademacher chaoses, and tetrahedral functions $f$.
However, the inequalities (B) of both
Theorems \ref{AB} and \ref{F} hold for tetrahedral Rademacher
chaoses induced by $\bm\xi$ and $\BX$ (with  independent columns),
whenever $\BE$ is (a) a Banach lattices with no subspace isomorphic
to $c_0$, or (b) a UMD-space.
\item
The full analog of
Theorem \ref{AB} is valid in locally convex spaces.
\item
The decoupling results from Section~2 can be carried over to
linear spaces over the
field of complex numbers.  To obtain similar results for Section~3 is
more difficult.  One approach is to show that if $\beps$\ denotes a sequence
of independent Rademacher random variables, and if $\bm \sigma$\ denotes a
sequence of independent Steinhaus random variables (that
is, $\sigma_i$\ is uniformly distributed over the complex unit circle), then
$
\| Q(f; \beps^k) \| \approx \| Q(f; \bm \sigma^k)\|
$.
We omit the details of the development.
\item
In the case when the tail decoupling holds, i.e., in  Theorems \ref{ABtail},
\ref{F}.(A$'''$) and (B$'''$), we obtain the comparison of tightness. That
is, for a family of functions $\set{f:f\in F}$, we have that, if one type
of chaos $\set{Q_d(f):f\in F}$ is tight, so is the other,
$\set{\wt{Q_d}(f):f\in F}$, subject to restrictions listed in the above
theorems. That remark also applies to functions $f$ taking values in a
locally convex space.
\item
In the context discussed above, we immediately obtain the comparison of
generalized Orlicz modulars, i.e., functionals of the form
$\Phi(\cdot)=\E\phi(\|\cdot\|)$, where $\phi$ is a nondecreasing
function on the positive half-line, $\phi(0)=0$.
\item
Multiple stochastic integrals of deterministic multivariate functions
(cf., e.g., \cite{KalS}) can be seen as limits of multilinear random
forms. Therefore if $\bm\xi,\bm\xi_1,\bm\xi_k$ are stochastic
processes with independent increments, and the symbols
$\ll f\, \bm\xi_1\otimes\cdots\otimes\bm\xi_d\rr$ and
$\ll f\,\bm\xi^{\otimes k}\rr$ are understood as such integrals,
then all decoupling inequalities carry over word-for-word.
\item\label{riorl}
Our decoupling inequalities involve a certain means of
domination. Essentially, we show that the domination
by means of $L^p$-norms yields the same for probability tails.
The passing from one to another type of domination may be of an
intrinsic interest. Recall the definition of $f^{**}$\ mentioned in
Section~\ref{lorentzeg}. Let us note the following
result, which can be applied in a wider context than ours.

Suppose that
$\xi$ and $\eta$\ are two given non-negative random variables, and define
quantities $c_1,\dots,c_5$\ below.
\begin{enumerate}
\item[{\rm (i)}]  Let $c_1$\ be the smallest constant such that
for every Orlicz function,
$\|\xi\|_{\phi}\le c_1 \|\eta\|_{\phi}$;
\item[{\rm (ii)}] Let $c_2$\ be the smallest constant such that
for all $t>0$, if $\phi_t(x)=(x-1)_+/t$, then
$\|\xi\|_{\phi_t}\le c_2 \|\eta\|_{\phi_t}$;
\item[{\rm (iii)}] Let $c_3$\ be the smallest constant such that
$\xi^{**} \le c_3 \eta^{**}$;
\item[{\rm (iv)}] Let $c_4$\ be the smallest constant such that
for every r.i. norm, $\|\xi\|\le c_4 \|\eta\|$;
\item[{\rm (v)}] Let $c_5$\ be the smallest constant such that
for every separable r.i. norm, $\|\xi\|\le c_5 \|\eta\|$.
\end{enumerate}
Then $c_1 = c_2 \le c_3 = c_4 = c_5 \le 2 c_1$.  Indeed, inequalities
$c_2 \le c_1 \le
c_4$\ and $c_3 \le c_5 \le c_4$\ are obvious.  That $c_1 \le c_2$\ follows
immediately from the formula
\[ \phi(x) = \int_0^\infty \phi_t(x) \, d(\phi'(t)) .\]
That $c_4 \le c_3$\
was proved in \cite[Proposition 2.a.8]{LinT}.
That $c_3 \le 2c_2$\ follows from the formula
$\|\xi\|_{\phi_t} \le \xi^{**}(t) \le 2\|\xi\|_{\phi_t}$.
To show the left hand side,
suppose that $\xi^{**}(t) \le 1$.
Then
\[ \int_0^t \xi^*(s) \, ds \le t .\]
Thus we have that $\xi^*(t) \le 1$, and hence
\[
\E\phi_t(\xi) = {1\over t}\int_0^1 (\xi^*(s)-1)_+ \, ds =
   {1\over t}\int_0^t (\xi^*(s)-1)_+ \, ds \le {1\over t}\int_0^t \xi^*(s)
   \, ds \le 1 .
\]
To show the right hand side, suppose that $\|\xi\|_{\phi_t}\le 1$.
Thus
\[ \int_0^{t_0} (\xi^*(s) - 1) \, ds \le t ,\]
where $t_0 = \P(\xi > 1)$.  If $t_0 \ge t$, then it follows that
\[ \int_0^{t} (\xi^*(s) - 1) \, ds \le t ,\]
from whence it follows that
\[ \int_0^{t} \xi^*(s) \, ds \le 2t .\]
If $t_0 < t$, then
\[ \int_0^t \xi^*(s) \, ds = \int_0^{t_0} (\xi^*(s) - 1) \, ds
   + \int_0^{t_0} \, ds + \int_{t_0}^t \xi^*(s) \, ds
   \le 2t ,\]
because $\xi^*(s) \le 1$\ if $s > t_0$.
\item
A decoupling principle for multivalued functions (proved in \cite{Pen})
also follows from our
basic decoupling inequalities. Suppose that
$F(\cdot,\bm\xi)$ is a countably multivalued function, i.e., a countable family
of functions $\ca F_{\bi}$ is associated with each $\bi$.  In equivalent terms,
one may think of a decision function $\tau:D\times \N^k\to
\prod_{\bi\in\N^k}\ca F_{\bi}$ ($D$ is countable). Then the statements of
Theorem \ref{F} hold uniformly with respect to $\tau$, that is, the norm
$\|F(\cdot)\|$ is replaced by $\sup_\tau\sup_d\|\tau(d,\cdot)(\cdot)\|$.
The theorem follows for a finite collection of decision functions
$\set{\tau_1,\ldots,\tau_n}$, since this means the replacement of
the underlying Banach space $\BE$ by another Banach space $\ell_n^\infty(\BE)$.
In the full statement we need the banach lattice $\L(\ell^\infty)$ to
satisfy the property ``{\em $\sup_n\|x_n\|=\|\sup _n x_n\|$, for an
increasing sequence of
nonnegative vectors}''.  In view of the preceding note, we may choose
a family of Orlicz spaces, and the required property holds.

Other sequential functionals on $\R^D$, e.g.
$\ell^p$, Orlicz $\ell^\psi$, etc., yield numerous variations of Theorem
\ref{F}.
\end{enumerate}

\begin{remark}\rm
 This paper represents the combination of the papers
\cite{PenM} and \cite{Szu:dechom}
\end{remark}


TITLE: Show that algebraic integers in $\mathbb{Q}[\zeta_3]$ are exactly $\mathbb{Z}[\zeta_3]$.
QUESTION [3 upvotes]: If $\zeta_3$ is a primitive cube root of unity, then I'm trying to show that $\mathbb{Z}[\zeta_3]$ is the ring of algebraic integers in $\mathbb{Q}[\zeta_3]$. I have shown that $\mathbb{Z}[\zeta_3]$ is contained in the ring of algebraic integers, but I am unsure of how to prove the converse. I'm aware that there are general proofs for the case where $\zeta$ is a primitive $p$-th root of unity where $p$ is any prime, but I'm wondering if there is a more elementary proof that will work for $p = 3$.

REPLY [3 votes]: One immediate way is to show that $ \mathbf Z[\zeta_3] $ is an Euclidean domain by giving an explicit algorithm for performing Euclidean division there. This, of course, immediately implies that it is integrally closed in its field of fractions, which is $ \mathbf Q(\zeta_3) $; and clearly it is integral over $ \mathbf Z $, so it is the full ring of integers. Another way is to use the well-known result about the classification of quadratic rings of integers: the ring of integers of $ \mathbf Q(\sqrt{d}) $, for squarefree $ d $ is $ \mathbf Z[(1 + \sqrt{d})/2] $ for $ d \equiv 1 \pmod{4} $, and $ \mathbf Z[\sqrt{d}] $ otherwise. Now, recall that $ \zeta_3 = (-1 + \sqrt{-3})/2 $, and you are done.


TITLE: $f:\mathbb{R}^m\to\mathbb{R}, f(tx) = |t|f(x)$, prove $f(x) = 0$ for all $x$
QUESTION [0 upvotes]: Let $f:\mathbb{R}^m\to\mathbb{R}$ such that $f(tx) = |t|f(x)$ for
  $x\in\mathbb{R}^m$ and $t\in\mathbb{R}$. If $f$ is differentiable at
  the origin, then $f(x) = 0$ for all $x$.

By the differencialibity criterion, $f$ is differentiable if there exists $r(v)$ such that
$$f(a+v) = f(a) + \frac{\partial f}{\partial x_1}(a) v_1 + \cdots + \frac{\partial f}{\partial x_m}(a)v_m + r(v)$$
such that $\lim_{v\to 0}\frac{r(v)}{|v|} = 0$
Let's look at it in the point $a=0$, so:
$$f(0+v) = f(0) + \frac{\partial f}{\partial x_1}(0) v_1 + \cdots + \frac{\partial f}{\partial x_m}(0)v_m + r(v)$$
By the function relation in the question, $f(0) = f(0.x) = |0|f(x) = 0$, so:
$$f(0+v) = 0 + \frac{\partial f}{\partial x_1}(0) v_1 + \cdots + \frac{\partial f}{\partial x_m}(0)v_m + r(v)$$
By the chain rule:
$$f(tx) = |t|f(x) \implies tf'(tx) = |t|f'(x) \implies f'(0) = f'(0\cdot x) \implies ?$$
I don't know how to get $f'(0)$ from above. Also, is the above derivation right? I'm confused because $x\in \mathbb{R}^m$.
So how can I prove that $f(x) = 0$ for all $x$?

REPLY [3 votes]: A simple continuity argument shows $f(0) = 0$.
Assume that $f$ is differentiable at $0$. The directional derivative of $f$ at $0$ in the direction of a unit vector $\nu$ is given by $$\lim_{t \to 0} \frac{f(t\nu) - f(0)}{t} = \lim_{t \to 0} \frac{|t|}{t} f(\nu)$$ which exists if and only if $f(\nu) = 0$. Since $f$ is differentiable at $0$ the limit must exist, so we conclude $f(\nu) = 0$ for each unit vector $\nu$.
Finally if $x \not= 0$, then $\nu = x/|x|$ is a unit vector and $$f(x) = f(|x|x/|x|) = |x| f(x/|x|) = 0.$$

REPLY [0 votes]: If $v\in\mathbb R$, the function $\ell_v:\mathbb R \to \mathbb R^m$ defined by $\ell_v(t) = tv$ is differentiable. Becaause $f$ is differentiable, then $g:\mathbb R \to \mathbb R$ given by $g(t) = f(tv) = |t|f(v)$ is also differentiable (it is just $f\circ \ell_v$). Remember: $f(v)$ is a constant and $t\mapsto |t|$ is not differentiable at zero on the real line. So the only way for $g$ to be differentiable is if $f(v) =0$.


TITLE: Covariant derivative of a $(1,1)$ tensor field
QUESTION [7 upvotes]: I am trying to understand covariant derivatives in GR.
So far, I understand that if $Z$ is a vector field, $\nabla Z$ is a $(1,1)$ tensor field, i.e. the map $\eta,X \mapsto \eta(\nabla_XZ)$ is linear in $\eta$ and $X$. The map $Z\mapsto\nabla_XZ$ is not a tensor field since, by the Leibniz rule, it is not linear in $Z$: $$\nabla_X(fZ)=(\nabla_Xf)Z+f(\nabla_XZ)$$
where $\nabla_Xf:=X(f)$.
My lecture notes then state that the action of $\nabla$ is extended to general $(r,s)$ tensor fields by the Leibniz rule, giving an $(r,s+1)$ tensor field as a result. The example given is $\nabla \eta$, with $\eta$ a $(0,1)$ tensor field:$$(\nabla\eta)(X,Y)\equiv\nabla_X\eta(Y):=\nabla_X(\eta(Y))-\eta(\nabla_XY)$$
which is easily seen to be linear in $X$, while the first terms from the Leibniz rule cancel to give linearity in $Y$ as well, so this is a $(0,2)$ tensor.
They then suggest to try and define $\nabla T$ for $T$ a $(1,1)$ tensor. My attempt at this is: $$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))+T(\nabla_X\eta,Y)-T(\eta,\nabla_XY)$$
Again, linearity in $X$ is easy and the Leibniz rule gives linearity in $Y$. However, this doesn't seem to be linear in $\eta$. First, I find that:
$$\begin{eqnarray}
\nabla_X(f\eta+\omega) (Y)&=& \nabla_X((f\eta+\omega)(Y)) - (f\eta+\omega)(\nabla_XY)\\
&=& (\nabla_Xf)\eta(Y)+f\nabla_X(\eta(Y))+\nabla_X(\omega(Y))-f\eta(\nabla_XY)-\omega(\nabla_XY)\\
&=& (\nabla_Xf)\eta(Y) + f \nabla_X\eta(Y)+\nabla_X\omega(Y)
\end{eqnarray}$$
hence $\nabla_X(f\eta+\omega) = (\nabla_Xf)\eta + f \nabla_X\eta+\nabla_X\omega$. Therefore:
$$\begin{eqnarray}
\nabla_XT(f\eta+\omega,Y)&=& \nabla_X(T(f\eta+\omega,Y))+T(\nabla_X(f\eta+\omega),Y)-T(f\eta+\omega,\nabla_XY)\\
&=& (\nabla_Xf)(T(\eta,Y))+f\nabla_X(T(\eta,Y))+\nabla_X(T(\omega,Y))+(\nabla_Xf)T(\eta,Y)\\&&+fT(\nabla_X\eta,Y)+T(\nabla_X\omega,Y)-fT(\eta,\nabla_XY)-T(\omega,\nabla_XY)\\
&=& 2(\nabla_Xf)(T(\eta,Y)) + f\nabla_XT(\eta,Y)+\nabla_XT(\omega,Y)
\end{eqnarray}$$
where the term $2(\nabla_Xf)(T(\eta,Y))$ shows that $\nabla_XT(\eta,Y)$ is not linear in $\eta$. However, if I defined it as:$$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))-T(\nabla_X\eta,Y)+T(\eta,\nabla_XY)$$
instead, then the same calculation shows that this is linear in $\eta$, but no longer linear in $Y$. So I am not sure which, if either, of them is the correct definition.
UPDATE: Thanks to Michael Albanese's answer I now understand that the correct definition is $$\nabla_XT(\eta,Y):=\nabla_X(T(\eta,Y))-T(\nabla_X\eta,Y)-T(\eta,\nabla_XY)$$
which is indeed linear in $\eta$, $X$ and $Y$. For the benefit of others, I'll mention that my confusion arose because I was trying to reverse engineer the $+$ and $-$ signs from the coordinate basis formula:
$$\begin{eqnarray}
T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s};\rho} =T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s},\rho}&+\Gamma^{\mu_1}_{\sigma\rho}T^{{\sigma}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s}}+\ldots+\Gamma^{\mu_r}_{\sigma\rho}T^{{\mu_1}\ldots{\sigma}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\nu_s}}\\ &-\Gamma^{\sigma}_{\nu_1\rho}T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\sigma}\ldots{\nu_s}}-\ldots-\Gamma^{\sigma}_{\nu_s\rho}T^{{\mu_1}\ldots{\mu_r}}_{\phantom{{\mu_1}\ldots{\mu_r}}{\nu_1}\ldots{\sigma}}
\end{eqnarray}$$
But I now see how they follow from the correct definition. In the case of a $(1,1)$ tensor we have $T^{\mu}_{\phantom{\mu}\nu\,;\rho}:=\nabla_{e_\rho}T(f^\mu,e_\nu)$, where I am using $f^\mu$ and $e_\nu$ for $\mathrm{d}x^\mu$ and $\frac{\partial}{\partial x^\nu}$. Hence we have:
$$\begin{eqnarray}
\nabla_{e_\rho}T(f^\mu,e_\nu) :=& \nabla_{e_\rho}(T(f^\mu,e_\nu))-T(\nabla_{e_\rho}f^\mu,e_\nu)-T(f^\mu,\nabla_{e_\rho}e_\nu)
\end{eqnarray}$$
The first and third terms are easily evaluated:
$$\nabla_{e_\rho}(T(f^\mu,e_\nu))=e_\rho(T(f^\mu,e_\nu))=e_\rho(T^{\mu}_{\phantom{\mu}\nu})=T^{\mu}_{\phantom{\mu}\nu,\rho}$$
$$T(f^\mu,\nabla_{e_\rho}e_\nu)=T(f^\mu,\Gamma^{\sigma}_{\nu\rho}e_\sigma)=\Gamma^{\sigma}_{\nu\rho}T(f^\mu,e_\sigma)=\Gamma^{\sigma}_{\nu\rho}T^\mu_{\phantom{\mu}\sigma}$$
For the second term, since $f^\mu$ is a $(0,1)$ tensor field, $\nabla f^\mu$ is a $(0,2)$ tensor field and hence $\nabla_{e_\rho}f^\mu$, for fixed $e_\rho$, is a $(0,1)$ tensor field. We evaluate:
$$\begin{eqnarray}
(\nabla_{e_\rho} f^\mu)_\lambda=\nabla_{e_\rho}f^\mu(e_\lambda) &:= &\nabla_{e_\rho}(f^\mu(e_\lambda))-f^\mu(\nabla_{e_\rho}e_\lambda)  \\
&= & e_\rho(\delta^\mu_\lambda)-f^\mu(\Gamma_{\lambda\rho}^\sigma e_\sigma)\\
&= & 0 -\Gamma_{\lambda\rho}^\sigma f^\mu( e_\sigma)\\
&= & - \Gamma_{\lambda\rho}^\sigma \delta^\mu_\sigma\\
&= & - \Gamma_{\lambda\rho}^\mu \\
&= & - \Gamma_{\sigma\rho}^\mu \delta^\sigma_\lambda\\
&= &  -\Gamma_{\sigma\rho}^\mu f^\sigma( e_\lambda)\\
\end{eqnarray}$$
Therefore $\nabla_{e_\rho}f^\mu =  -\Gamma_{\sigma\rho}^\mu f^\sigma$ and hence:
$$T(\nabla_{e_\rho}f^\mu,e_\nu)=T(-\Gamma_{\sigma\rho}^\mu f^\sigma,e_\nu)=-\Gamma_{\sigma\rho}^\mu T(f^\sigma,e_\nu)=-\Gamma_{\sigma\rho}^\mu T^\sigma_{\phantom{\sigma}\nu}$$
Thus, we have:
$$\nabla_\rho T^{\mu}_{\phantom{\mu}\nu}=T^{\mu}_{\phantom{\mu}\nu,\rho}+\Gamma_{\sigma\rho}^\mu T^\sigma_{\phantom{\sigma}\nu}-\Gamma^{\sigma}_{\nu\rho}T^\mu_{\phantom{\mu}\sigma}$$ as expected.

REPLY [6 votes]: In general, if $T$ is an $(r, s)$ tensor field, then $\nabla T$ is an $(r, s+1)$-tensor field given by
$$(\nabla T)(\eta_1, \dots, \eta_r, X, Y_1, \dots, Y_s) = (\nabla_XT)(\eta_1, \dots, \eta_r, Y_1, \dots, Y_s)$$
where the latter expression is intrinsically defined by the equation
\begin{align*}
\nabla_X(T(\eta_1, \dots, \eta_r, Y_1, \dots, Y_s)) =&\ (\nabla_XT)(\eta_1, \dots, \eta_r, Y_1, \dots, Y_s)\\
&+ \sum_{i=1}^rT(\eta_1, \dots, \eta_{i-1}, \nabla_X\eta_i, \eta_{i+1}, \dots, \eta_r, Y_1, \dots, Y_s)\\
&+ \sum_{j=1}^sT(\eta_1, \dots, \eta_r, Y_1, \dots, Y_{j-1}, \nabla_XY_j, Y_{j+1}, \dots, Y_s).
\end{align*}
In particular, if $T$ is a $(1, 1)$ tensor, then $(\nabla_XT)(\eta, Y)$ is intrisically defined by the equation
$$\nabla_X(T(\eta, Y)) = (\nabla_XT)(\eta, Y) + T(\nabla_X\eta, Y) + T(\eta, \nabla_XY)$$
and therefore the correct expression for $(\nabla_XT)(\eta, Y)$ is
$$(\nabla_XT)(\eta, Y) = \nabla_X(T(\eta, Y)) - T(\nabla_X\eta, Y) - T(\eta, \nabla_XY).$$


TITLE: Let $f$ be a polynomial with at least $k$ different roots, prove that $f'$ hast at least $k-1$ different roots
QUESTION [2 upvotes]: I don't know how to write the proof to this problem:

Let $f$ be a polynomial with at least $k$ different roots, prove that $f'$ has at least $k-1$ different roots.

It's featured among a few exercises made to apply Rolle's theorem, but this one is about a general result and asks for higher degree of formality I guess... I don't know how to write the proof although I have the following intuition:
$f(x)=x^k + g(x)$
$f'(x)=kx^{k-1} + g'(x)$
Any suggestions?

REPLY [2 votes]: Regards @Gioacchino . I may not give full detail. We can use induction. For $k=2$, then the function $f(x) = (x-r_{1})(x-r_{2})g(x) $ has $$f'(x) = (x-r_{2})g(x) + (x-r_{1})g(x) + (x-r_{1})(x-r_{2})g'(x) $$
The possibilities for $g(x)$ are $g(x) = H(x)$,  or  $g(x) = (x-r_{1})^{m}H(x)$, or $g(x) = (x-r_{2})^{m}H(x)$ or $g(x) =(x-r_{1})^{m} (x-r_{2})^{n} H(x)$. With these possibilities, you could check that $f'(x)$ will have at least $k-1 = 1$ root.
Now let the statement be true for $k = 2, ... , n$. We would like to prove for $k=n+1$ :
$$ f(x) = (x-r_{1})...(x-r_{n+1})g(x)  $$

$$ f'(x) = h(x) + (x-r_{1}) \left[ h'(x) \right]  $$

with

$$h(x) = (x-r_{2})...(x-r_{n+1})g(x)$$

Notice that $h(x)$ is a polynomial with at least $n$ different roots, so $h'(x)$ must have at least $n-1$ different roots. We can write $h'(x)$ of the form

$$ h'(x) = (x-r_{m}) G_{m}(x), \:\:\: m = 2,...,n+1  $$

From here it can be seen that $f'(x)$ has at least $k-1 = n$ different roots.
Regards, Arief


TITLE: Reducing double summation to geometric series
QUESTION [1 upvotes]: How can I reduce this summation to a geometric series?
$\displaystyle\sum\limits_{i=0}^n x^{25i}\cdot\displaystyle\sum\limits_{j=i}^n x^{5j}$
I'm a little confused since the second summation begins at $i$.

REPLY [0 votes]: You can compute the inner sum explicitly after noticing that
\begin{align*}
\sum_{j = i}^n x^{5j} &= x^{5i}\sum_{j = 0}^{n - i}x^{5j} \\
&= x^{5i} \sum_{j = 0}^{n - i} (x^5)^j
\end{align*}
Once you've done this, a similar reindexing on the outer series will allow you to compute the entire sum.


TITLE: Why is $\limsup_{n\rightarrow \infty} (1/n) \log(a_n) < 0 \Rightarrow \sum_{n=0}^{\infty} a_n < \infty$ true?
QUESTION [1 upvotes]: Let $(a_n)$ be a real values series.
Why is the following implication true?
$$\limsup_{n\rightarrow \infty} (1/n) \log(a_n) < 0   \Rightarrow   \sum_{n=0}^{\infty} a_n < \infty$$

REPLY [10 votes]: Let's assume $a_n > 0$ to avoid the problem of $\log 0$, and the necessity of taking the absolute value. If
$$\limsup_{n\to\infty} \frac{1}{n}\log a_n = \lambda < 0,$$
then there is an $n_0$ such that, for all $n \geqslant n_0$, you have
$$\begin{align}
\frac{1}{n}\log a_n &< \frac{\lambda}{2}\\
\Rightarrow (a_n)^{1/n} &< e^{\lambda/2} < 1\\
\Rightarrow a_n < e^{n\lambda/2},
\end{align}$$
and hence
$$\sum_{n=n_0}^N a_n < \sum_{n=n_0}^N e^{n\lambda/2} < \frac{e^{n_0\lambda/2}}{1-e^{\lambda/2}}$$
for all $N \geqslant n_0$. Thus the sequence of partial sums is bounded, hence $\sum\limits_{n=0}^\infty a_n < \infty$.


TITLE: Error analysis / propagation : trigonometry
QUESTION [0 upvotes]: If the hypotenuse of a triangle is (1536 +- 3)m long, and the (non right-angle) angle measured from the ground is (22.2 +-0.1) degrees, what is the height of the triangle, and the error in this?
sin(theta) = opp/hypotenuse = h/x
h = x sin(theta)
dh = ?

REPLY [1 votes]: Assume the values are normally distributed without correlation (i.e. $\Delta(x+y) = \sqrt{\Delta x^2 + \Delta y^2}$), then
\begin{align}
\Delta h
&= \Delta ( x \sin \theta ) \\
&= \sqrt{ (\Delta x \cdot \sin \theta)^2 + (x \cdot \Delta (\sin \theta))^2 } \tag{product rule} \\
&= \sqrt{ (\Delta x \cdot \sin \theta)^2 + (x\cos \theta \cdot \Delta \theta)^2 }. \tag{$\tfrac d{d\phi}\sin \phi = \cos \phi$}
\end{align}


TITLE: Determining a number is perfect square
QUESTION [0 upvotes]: Let $88k + 41$ be an integer where $k \in \mathbb{Z}$. Is there any value for $k$ such that $88k+41$ is a perfect square? I tried to find that using C++ program but it wasn't successful. I don't know how to find such $k$ or prove that it's impossible.

REPLY [1 votes]: $\!\bmod 11\!:\ a^{\large 2}\! = 88k\!+\!41 \equiv -3\,\overset{(\ \ )^{\Large 5}}\Longrightarrow\, a^{\large 10} \equiv (-3)^{\large 5} \equiv (-3)(-2)^{\large 2} \equiv -1\,$ contra little Fermat


text:
{\bf Problem.} Find the largest real number $c$ such that \[x_1^2 + x_2^2 + \dots + x_{101}^2 \geq cM^2\]whenever $x_1,x_2,\ldots,x_{101}$ are real numbers such that $x_1+x_2+\cdots+x_{101}=0$ and $M$ is the median of $x_1,x_2,\ldots,x_{101}.$
{\bf Level.} Level 5
{\bf Type.} Intermediate Algebra
{\bf Solution.} Since the inequality is always true for $M = 0,$ it suffices to consider the case $M \neq 0.$

For a particular $c$ and for any tuple $(x_1, \dots, x_{101})$ satisfying the conditions, the tuple $(-x_1, \dots, -x_{101})$ satisfies the conditions as well, so we may assume that $M > 0.$ Finally, we may assume that $x_1 \le x_2 \le \dots \le x_{101},$ so that $M = x_{51}.$

We want to find the largest $c$ such that the inequality \[x_1^2 + x_2^2 + \dots + x_{101}^2 \ge cx_{51}^2\]always holds, where $x_1 \le x_2 \le \dots \le x_{101}$ and $x_1 + x_2 + \dots + x_{101} = 0.$ Therefore, fixing a value of $x_{51},$ we should write inequalities that minimize $x_1^2 + x_2^2 + \dots + x_{101}^2.$

To compare the terms on the left-hand side to $x_{51}^2,$ we deal with the terms $x_1^2 + x_2^2 + \dots + x_{50}^2$ and $x_{51}^2+x_{52}^2+\dots+x_{101}^2$ separately.

By Cauchy-Schwarz, \[(1 + 1 + \dots + 1)(x_1^2+x_2^2+\dots+x_{50}^2) \ge (x_1+x_2+\dots+x_{50})^2,\]so \[x_1^2 + x_2^2 + \dots + x_{50}^2 \ge \tfrac{1}{50}\left(x_1+x_2+\dots+x_{50}\right)^2.\]We have $x_1+x_2+\dots+x_{50} = -x_{51}-x_{52} -\dots - x_{101}\le -51x_{51} $ because $x_{51} \le x_{52} \le \dots \le x_{101}.$ Since $x_{51} > 0,$ both $x_1 + x_2 + \dots + x_{50}$ and $-51x_{51}$ are negative, so we can write \[\begin{aligned} x_1^2+x_2^2+\dots+x_{50}^2 &\ge \tfrac{1}{50} (x_1+x_2+\dots+x_{50})^2\\ & \ge\tfrac{1}{50} \left(-51x_{51}\right)^2 \\ &= \tfrac{51^2}{50} x_{51}^2. \end{aligned}\]On the other hand, since $0 < x_{51} \le x_{52} \le \dots \le x_{101},$ we simply have \[x_{51}^2 + x_{52}^2 + \dots + x_{101}^2 \ge 51x_{51}^2.\]Putting all this together gives \[(x_1^2 + x_2^2 + \dots + x_{50})^2 + (x_{51}^2 + x_{52}^2 + \dots + x_{101}^2) \ge \left(\tfrac{51^2}{50} + 51\right) x_{51}^2 = \tfrac{5151}{50} x_{51}^2.\]Equality holds when $x_1 = x_2 = \dots = x_{50} = -\tfrac{51}{50}$ and $x_{51} = x_{52} = \dots = x_{101} = 1,$ so the answer is $\boxed{\tfrac{5151}{50}}.$

set_name: MATH


TITLE: Breaking chocolate problem and Euler characteristic.
QUESTION [1 upvotes]: There is the following Problem:

Given an $m \times n$ chocolate bar, where you can only break it along the gridlines and only break one piece at the time. What is the minimum amout of steps needed to break it into $1\times1$ pieces?

Answer: You always need $m \times n -1$ Steps.

Now, my gut is telling me that this is related to the Euler characteristic. If that is the case, where can i find more information as to why/how that is the case?
P.S. This is not a homework problem, just a question about the more general principles and ways of seeing it.

REPLY [2 votes]: I like your thought, there is indeed a (very simple) Euler characteristic description, although I think it turns out to be a little disappointing because it's really not much different than the "square counting" solution that is linked in the comments.
But here goes anyway.
From the assumption given, each "break" is carried out on a single piece, breaking that piece along a "fault line" which follows the grid. Let's examine how the numbers of faces, edges, and vertices change from "before" the break to "after" the break.
The number of faces is the the same before and after the break (namely $m \times n$).
Before the break, assume that the fault line has $k$ edges, it therefore also has $k+1$ vertices. After the break, each of those $k$ edges and $k+1$ vertices along the fault line has doubled, producing $2k$ edges and $2k+2$ vertices.
The numbers of vertices and edges not along the fault line is the same before and after the break.
The net change in the Euler characteristic is therefore
\begin{align*}
\chi_{\text{after}} - \chi_{\text{before}} &= (\text{vertices after} - \text{vertices before}) - (\text{edges after} - \text{edges before}) \\
  &= ((2k+2)-(k+1)) - (2k - k) \\
  &= 1
\end{align*}
That is to say: after each break, the Euler characteristic has increased by $1$. Since the Euler characteristic started at $1$ and ended at $m \times n$, there are $m \times n - 1$ breaks.


TITLE: Experimental bounds on microcausality violation
QUESTION [13 upvotes]: In "The Great Soviet Encyclopedia", 3rd Edition from 1970-1979, (evidently an old book), some V. I. Grigor’ev has a well-written little note on microcausality. Towards the end he states an experimental bound on the vialotation of microcausality. According to that ancient encyclopedia article, if such violation occurs at all then it must be on scales smaller than $10^{-15} cm$.
I don't see a citation for that number in his text (I don't have the original book in my hand though, to be frank) and I am wondering which experiment Grigor’ev might have been thinking of.
And: if there are decent such experiments indeed, then probably there are more recent numbers on whatever that experiment actually measured.
My question is if anyone might be able to help me with finding whatever useful there might be to say about experimental bounds of the kind Grigor’ev seems to be referring to, or anything similar.
Maybe I should say that I am not trying to find such violations. Rather, I am in some discussion with some mathematicians interested in axiomatic QFT, and I'd find it entertaining to be able to accompany the statement of the microcausality axiom (or a similar axiom) with something substantial that has the word "experiment" in it.

REPLY [1 votes]: The text you refer to is vintage but not ancient as there are many who still remember their own work from those times.  As far as I know the experimental data you seek was not the result of a laboratory experiment specifically designed to test the limits of microcausality but were rigorous thought experiments using the physical data and mathematics available at the time.  As no violations could be found, predicted or inferred within the precision of data at the time, it was concluded any violation must be on scales smaller than $10^{−15}$ cm.  You need not, however, feel bound by these vintage results as we no doubt have better tools and ideas to address the question now.


text:
\section{Numerical Examples} \label{sec:examples}

In this section, we present numerical examples to verify the properties
of the proposed methods. Since our purpose is to validate
the proposed deep learning method, we employ synthetic data generated
from known dynamical systems with known time-dependent inputs. The
training data are generated by solving the known system with high
resolution numerical scheme, e.g., 4th-order Runge Kutta with
sufficiently small time steps. Our proposed learning method is then
applied to the training data set. Once the learned model is
constructed, we conduct system prediction using the model with new
initial conditions and new external inputs. The prediction results are
then compared with the reference solution obtained by solving the
exact system with the same new inputs. Also, to clearly examine the
numerical errors, we only present the tests where the training
data do not contain noises.

In all the examples, we generate the training
data set \eqref{data_set} with $K^{(i)}\equiv 2$, $\forall i$, i.e., each trajectory
only contains two data points. For each of the $i$-th entry in the
data set,
the first data entry is
randomly sampled from a domain $I_\x$ using uniform distribution.
The second data entry is produced by
solving the underlying reference dynamical system with a time step
$\delta^{(i)}\in I_{\Delta}=[0.05, 0.15]$ and subject to a
parameterized external input in the form of \eqref{local_param},
whose parameters \eqref{Gamma} are uniformly sampled from a domain
$I_{\ggamma}$. The sampling domains $I_\x$ and $I_{\ggamma}$ are
problem specific and listed separately for each example.

The DNNs in all the examples use activation function
$\sigma(x)=\tanh(x)$ and are trained by minimizing the mean squared
loss function in \eqref{eq:loss}. The network training is conducted
by using Adam algorithm \cite{kingma2014adam} with the open-source
Tensorflow library \cite{tensorflow2015}.
Upon satisfactory training, the learned models are used to conduct
system prediction, in the form of \eqref{eq:prediction}, with a
constant step size $\delta_n=0.1$.

\subsection{Linear Scalar Equation with Source}
	Let us first consider  the following scalar equation
	\label{exmp:scalar}
	\begin{equation}
		\frac{dx}{dt}=-\alpha(t)x+\beta(t),
              \end{equation}
              where the time-dependent inputs $\alpha(t)$ and $\beta(t)$ are locally
parameterized with polynomials of degree $2$, resulting the local
parameter set \eqref{Gamma} $\ggamma_n\in\R^{n_b}$ with $n_b=3+3=6$. We build a neural
network model consisting of $3$ hidden layers with $80$ nodes per
layer. The model is trained with
$20,000$ data trajectories randomly sampled, with uniform
distribution, in the state variable domain $I_{\x}=[-2,2]$ and the
local parameter domain $I_{\ggamma}=[-5,
5]^6$.

After the network model is trained, we use it to conduct system
prediction. In \figref{fig:ex1_scalar}, the prediction result with a
new initial condition $x_0=2$ and new external inputs
$\alpha(t)=\sin(4t)+1$ and
$\beta(t)=\cos(t^2/1000)$ is shown, for time up to $T=100$.
The reference solution is also shown for comparison. It can be seen
that
the network model produces accurate prediction for this relatively
long-term integration.

	\begin{figure}[!htb]

		\begin{center}
                  \includegraphics[width=\linewidth]{./Figures/Scalar_x_4_interpolate.pdf}
        \caption{DNN model prediction of \eqref{exmp:scalar} with
          external inputs $\alpha(t)=\sin(4t)+1$ and
          $\beta(t)=\cos(t^2/1000)$ and an initial condition $x_0 = 2$.
          Comparison of long-term neural network model prediction (labelled ``NN'')
          with the reference solution.}
          	\label{fig:ex1_scalar}
		\end{center}

	\end{figure}

For this relatively simple and low-dimensional system, its learning can be
effectively conducted by other standard approximation method, as
discussed in Remark \ref{remark1}. With the same quadratic polynomial
for local parameterization as in the DNN modeling, which results in $\ggamma_n\in
[-5,5]^6$, we employ tensor Legendre orthogonal polynomials in total
degree space, which is a standard multi-dimensional approximation
technique, for the approximation of the one-step evolution operator
in \eqref{evo_mod}. In \figref{fig:ex1_polynomial_approx}, the
prediction results by the polynomial learning model are shown, for a case with external inputs
$\alpha(t)=\sin(t/10)+1$ and $\beta(t)=\cos(t)$.
In \figref{fig:ex1_polynomial_approx}(a), the prediction result
obtained by 2nd-degree polynomial learning model is shown. We observe
good agreement with the reference solution. In
\figref{fig:ex1_polynomial_approx}(b), the numerical errors at $T=100$
are shown for the polynomial learning model with varying degrees. We observe
that the errors decay exponentially fast when the degree of polynomial
is increased. Such kind of exponential error convergence is expected
for approximation of smooth problems, such as this example.
        \begin{figure}[!htb]
	\centering
	\begin{subfigure}[b]{0.48\textwidth}
	\begin{center}
	\includegraphics[width=0.99\linewidth]{./Figures/example-1-compare-solutions}
	\caption{System prediction.}
	\end{center}
	\end{subfigure}
        \begin{subfigure}[b]{0.48\textwidth}
	\begin{center}
	\includegraphics[width=0.99\linewidth]{./Figures/example-1-increasing-degree}
	\caption{Errors vs. polynomial degree.}
	\end{center}
	\end{subfigure}
        \caption{Polynomial learning model for \eqref{exmp:scalar} with
          $\alpha(t)=\sin(t/10)+1$ and $\beta(t)=\cos(t)$. (a) Comparison
          of the model prediction with reference solution. (b)
          Relative error in prediction at $T=100$
          for increasing polynomial degree in the polynomial learning
          model. In all models piecewise quadratic polynomials are
          used for local parameterization.}
        \label{fig:ex1_polynomial_approx}
        \end{figure}

\subsection{Predator-prey Model with Control}

We now consider the following Lotka-Volterra Predator-Prey model with
a time-dependent input $u(t)$:
	\begin{equation}	\label{exmp:pred}
		\begin{split}
			\dfrac{d x_1}{dt}&= x_1- x_1 x_2+u(t),\\
			\dfrac{d x_2}{dt}&=- x_2+ x_1 x_2.
		\end{split}
	\end{equation}

The local parameterization for the external input is conducted using
quadratic polynomials, resulting in $\ggamma_n\in\R^3$. More
specifically, we set $I_{\ggamma}=[0, 5]^3$ and the state variable
space $I_{\x}=[0, 5]^2$.
The DNN learning model consists of $3$ hidden layers, each of which
with $80$ nodes. The network training is conducted using $20,000$ data
trajectories randomly sampled from $I_{\x}\times I_{\ggamma}$.
In \figref{fig:pred_1}, we plot its prediction result for a case with
$u(t)=\sin(t/3)+\cos(t)+2$, for time up to $T=100$, along with the
reference solution. It can be seen that the DNN model prediction
agrees very well with the reference solution. The numerical error
fluctuates at the level of $O(10^{-3})$, for this relatively long-term
prediction.

\begin{figure}[!htb]
	\centering
	\begin{subfigure}[b]{0.49\textwidth}
		\begin{center}
        \includegraphics[width=\linewidth]{./Figures/Pred_u1_2_projection.pdf}
        \caption{System prediction of $x_1$.}
	\label{fig:pred_1}
		\end{center}
	\end{subfigure}

	\begin{subfigure}[b]{0.49\textwidth}
		\begin{center}
			\includegraphics[width=1.0\linewidth]{./Figures/Predator_err_x1_2_interpolation.pdf}
			\caption{Error in prediction for $x_1$}
		\end{center}
	\end{subfigure}

        \caption{DNN learning model for \eqref{exmp:pred}.
          Comparison of its prediction result for $x_1$ with
          $u(t)=\sin(t/3)+\cos(t)+2$ against reference solution.
          Results for $x_2$ are very similar and not shown.}
\end{figure}

\subsection{Forced Oscillator}

We now consider a forced oscillator
\begin{equation}
  	\label{exmp:pend}
	\begin{split}
		\frac{dx_1}{dt}&=x_2,\\
		\frac{dx_2}{dt}&=-\nu(t)\,x_1-k\, x_2+f(t),
	\end{split}
\end{equation}
where the damping term $\nu(t)$ and the forcing $f(t)$ are
time-dependent processes.
Local parameterization for the inputs is conducted using quadratic
polynomials. More specifically, the training data are generated
randomly by sampling from  state variable space
$I_\x=[-3, 3]^2$ and local parameterization space
$I_{\ggamma}=[-3,3]^6$. Similar to other examples, the DNN
contains $3$ hidden layers with $80$ nodes in
each hidden layer. System prediction using the trained network model
is shown in \figref{fig:pend_3}, for rather arbitrarily chosen
external inputs $\nu(t)=\cos(t)$
and $f(t)=t/50$. Once again, we observe very good agreement with the
reference solution for relatively long-term simulation up to $T=100$.
\begin{figure}[!htb]
	\centering
	\begin{subfigure}[b]{0.48\textwidth}
		\begin{center}
			\includegraphics[width=1.0\linewidth]{./Figures/Pendulum_x1_3_interpolate.pdf}
			\caption{$x_1(t)$}
		\end{center}
	\end{subfigure}
	\begin{subfigure}[b]{0.48\textwidth}
		\centering
		\includegraphics[width=1.0\linewidth]{./Figures/Pendulum_x2_3_interpolate.pdf}
		\caption{$x_2(t)$}
	\end{subfigure}
	\caption{DNN model prediction of \eqref{exmp:pend} with inputs
          $\nu(t)=\cos(t)$ and $f(t)=t/50$.
          }
	\label{fig:pend_3}
\end{figure}

\subsection{PDE: Heat Equation with Source}

We now consider a partial differential equation (PDE). In particular,
the following heat equation with a source term,
	\begin{equation} \label{exmp:heat}
		\begin{split}
			&u_t = u_{xx}+q(t,x), \quad x\in [0, 1],\\
			&u(0, x)=u_0(x), \\
                        & u(t,0) = u(t,1) = 0,
		\end{split}
	\end{equation}
where $q(t, x)$ is the source term varying in both space and time. We
set the source term to be
	\begin{equation*}
		q(t, x)= \alpha(t) e^{-\frac{(x-\mu)^2}{\sigma^2}},
	\end{equation*}
	where $\alpha(t)$ is its time varying amplitude and parameter
        $\mu$ and $\sigma$ determine its the spatial profile.

The learning of \eqref{exmp:heat} is conducted in a discrete
space. Specifically, we employ $n=22$ equally distributed grid points in
the domain $[0,1]$,
$$
x_j = j/(n-1), \qquad j=1,\dots, n.
$$
Let
$$
\u(t) = \left[u(t, x_2), \cdots, u(t,x_{n-1})\right]^\dagger,
$$
we then seek to construct a DNN model to discover
the dynamical behavior of the solution vector $\u(t)$. Note that the
boundary values $u(x_1) = u(x_n)=0$ are fixed in the problem setting
and  to be included in the learning model.

Upon transferring the learning of the PDE \eqref{exmp:heat} into
learning of a finite dimensional dynamical system of $\u\in\R^d$, where
$d=n-2 = 20$, the DNN learning method discussed in this paper can be
readily applied. Training data are synthetic data generated by solving
the system \eqref{exmp:heat} numerically. In particular, we employ
second-order central difference scheme using the same grid points
$\{x_j\}$. The trajectory data are generated by randomly sample
$\u\in\R^{20}$ in a specific domain $I_\u = [0,2]^{20}$. Quadratic
polynomial interpolation is used in local parameterization of the time
dependent source term, resulting in 3-dimensional local representation
for the time dependent coefficient $\alpha(t)$. Random sampling in
domain $I_\alpha = [-2,2]^3$, $I_\mu = [0,3]$, $I_\sigma = [0.05,
0.5]$ is then used to generate the synthetic training data set, for
the parameters
$\alpha$, $\mu$, and $\sigma$, respectively.

The DNN network model thus consists of a total of $25$
inputs. Because of
curse-of-dimensionality, constructing accurate approximation in 25
dimensional space is computational expensive via traditional methods
such as polynomials, radial basis, etc. For DNN, however, 25 dimension
is considered low and accurate network model can be readily
trained. Here we employ a DNN with $3$ hidden layers, each of which
with $80$ nodes.
Upon successful training of the DNN model, we conduct system
prediction for a new source term (not in training data set), where
$\alpha(t)=t-\lfloor t \rfloor$ is a saw-tooth discontinuous function,
$\mu=1$, and $\sigma=0.5$.

The system prediction results are shown in \figref{fig:heat_3}, along
with the reference solution solved from the underlying PDE. We observe
excellent agreement between the DNN model prediction to the reference
solution. It is worth noting that the DNN model, once trained, can be readily used to
predict system behavior for other time dependent inputs.
	\begin{figure}[!htb]
		\centering
		\begin{subfigure}[b]{0.48\textwidth}
			\begin{center}
				\includegraphics[width=1.0\linewidth]{./Figures/Heat_u2_4_interpolate.pdf}
				\caption{Solution evolution at $x=0.5$}
			\end{center}
		\end{subfigure}
		\begin{subfigure}[b]{0.48\textwidth}
			\begin{center}
				\includegraphics[width=1.0\linewidth]{./Figures/Heat_snapshot_4.pdf}
				\caption{Solution profile at $t=2$}
			\end{center}
		\end{subfigure}
		\begin{subfigure}[b]{0.48\textwidth}
			\begin{center}
				\includegraphics[width=1.0\linewidth]{./Figures/Heat_contour_predict_30_4.pdf}
				\caption{Reference solution contours
                                  over time}
			\end{center}
		\end{subfigure}
		\begin{subfigure}[b]{0.48\textwidth}
			\begin{center}
				\includegraphics[width=1.0\linewidth]{./Figures/Heat_contour_exact_30_4.pdf}
				\caption{DNN prediction contours
                                  over time}
			\end{center}
		\end{subfigure}

		\caption{System prediction of \eqref{exmp:heat} with
                  $\alpha(t)=t- \lfloor t \rfloor$, $\mu=1$, and
                  $\sigma=0.5$. Comparison between the predictions by the
                  DNN model and the reference solution.}
		\label{fig:heat_3}
	\end{figure}

config: arxiv


TITLE: Points of continuity of Baire class one functions
QUESTION [10 upvotes]: This is an idle question motivated by two comments I made to a previous MO question (which I just searched for, unsuccessfully).  That question asked if the characteristic function of the rationals is a pointwise limit of continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$.
My first reply was that the answer is no: a function is said to be of Baire class one if it is a pointwise limit of continuous functions, and (so I remembered) every Baire class one function has a dense set of points of continuity, whereas of course $\bf{1}_{\mathbb{Q}}$ is discontinuous everywhere.
Then I checked up on this result (I am not an analyst, if that is not already clear) and realized that I had misremembered it slightly: the Baire Characterization Theorem says that a real function is of Baire class one iff for every nonempty closed subset $S \subset \mathbb{R}$, the restriction of $f$ to $S$ has at least one point of continuity.  (Note that in my comment I required that $S$ be perfect, i.e., without isolated points, because some sources state it this way.  This hypothesis is however superfluous because an isolated point of a topological space is a point of continuity for every real-valued function on that space.)  Taking $S = [a,b]$, we find that either (i) our BC1 function restricted to $S$ is continuous at $c \in (a,b)$ -- meaning that $f$ itself is continuous at $c$, since $S$ is a neighborhood of $c$ -- or (ii) $f|_S$ is continuous at $a$ -- meaning that $f$ itself is right-continuous at $a$, or finally (iii) $f|_S$ is continuous at $b$ -- so $f$ itself is left-continuous at $b$.  So we find that any BC1 function has a dense set of points at which it is either left-continuous or right-continuous (or both).  This is good enough to answer the question because $\bf{1}_{\mathbb{Q}}$ has no points of left- or right-continuity.
But later I wondered whether my original claim was actually incorrect:

Do there exist BC1 functions for which the set of points of continuity is not dense?  Or even empty?

I did of course try looking on the internet, but I didn't find a reference which gave a solid treatment of Baire classes of real functions: could someone supply one?  Moreover, I do see that by taking intervals of the form $(-\infty,b]$ and $[a,\infty)$ a BC1 function must have infinitely many points at which it is left-continuous and infinitely many points at which it is right-continuous.  And then I got stumped.  Probably this is in the category of things that were much better understood almost one hundred years ago than they are now...

REPLY [13 votes]: This is impossible. Baire proved that if a function defined on $\mathbb R$ is of Baire class 1, then it is continuous everywhere except, possibly, for a meagre set. And by another Baire's theorem a complement of a meagre set in $\mathbb R$ is dense.
An elementary exposition of this and related results can be found in the nice little book by Oxtoby.
Edit. Another good reference which covers Baire's theorems and provides some historical background is "The calculus gallery" by W. Dunham.


TITLE: Proving function has a specific value in its domain
QUESTION [3 upvotes]: If $a$ and $b$ $\in \Bbb{R}$, $a \neq b$, and $f(x)=(x-a)^2 (x-b)^2 + x$. Show that $f$ gets the value $\frac{a+b}{2}$ at some $x$ in its domain.
I'm trying to prove this by the intermediate-value theorem. Since $f$ is continuous in its domain, we can say that $f(x) = \frac{a+b}{2}$ exists if $$f(x_1)\leq f(x)=\frac{a+b}{2}\leq f(x_2)$$
is true for some $x_1 , x_2$ in the domain of $f$. One way to show this, is to differentiate $f$ with respect to $x$ and find $f_{min}$ and/or $f_{max}$, but since $f'$ is quite convoluted I don't think its necessarily the best idea. We can plug in $0$ and show that $f(0)=a^2 b^2 > f(x)$, i.e $\frac{a+b}{2}$ is not the largest value of $f$. Other than that I don't know how to proceed.

REPLY [4 votes]: Let $a<b$
Then $f(a)=a<b=f(b)$
Now you can use the intermediate value theorem since $a<\frac{a+b}{2}<b$

REPLY [2 votes]: $f(a)=a, f(b)=b$ apply the intermediate value theorem