text:
\section{Introduction and main results}

Let $G$ be an $r$-uniform hypergraph ($r$\emph{-graph} for short) with vertex
set $V$ and edge set $E$. Assume that $V=\left[  n\right]  :=\left\{
1,\ldots,n\right\}  $ and let $\mathbf{x}:=\left(  x_{1},\ldots,x_{n}\right)
\in\mathbb{R}^{n}$. Write $P_{G}\left(  \mathbf{x}\right)  $ for the
polynomial form of $G$
\[
P_{G}\left(  \mathbf{x}\right)  :
{\textstyle\sum\limits_{\left\{  i_{1},\ldots,i_{r}\right\}  \in E}}
x_{i_{1}}\cdots x_{i_{r}},
\]
and le
\[
\mu\left(  G\right)  :=\max_{\Delta^{n-1}}P_{G}\left(  \mathbf{x}\right)  ,
\]
where $\Delta^{n-1}\subset\mathbb{R}^{n}$ is the standard simplex:
\[
\Delta^{n-1}=\left\{  \mu\left(  G\right)  :\left(  x_{1},\ldots,x_{n}\right)
:x_{1}\geq0,\ldots,x_{n}\geq0\text{ and }x_{1}+\cdots+x_{n}=1\right\}  .
\]

We call $\mu\left(  G\right)  $ the \emph{MS-index }of $G$ in honor of Motzkin
and Straus, who introduced and studied $\mu\left(  G\right)  $ for $2$-graphs
in \cite{MoSt65}\footnote{The MS-index is commonly called the
\emph{Lagrangian}, a misty term that denies credit to Motzkin and Straus,
while Lagrange has nothing to do with the concept. Besides, the term
\emph{Lagrangian} has seven or so other meanings already in use elsewhere.}.
Let us note that the MS-index has a long-standing history in extremal
hypergraph theory (see, e.g., \cite{FrFu89} and \cite{Kee11} for more detailed discussion.)

Now, let
\[
\mu_{r}\left(  m\right)  :=\max\left\{  \mu\left(  G\right)  :G\text{ is an
}r\text{-graph with }m\text{ edges}\right\}  .
\]

The problem of finding $\mu_{r}\left(  m\right)  $ was first raised in 1989,
by Frankl and F\"{u}redi \cite{FrFu89}, who conjectured the exact value of
$\mu_{r}\left(  m\right)  $. During the years, their conjecture proved to be
rather hard: notwithstanding that it has been confirmed for most values of $m$
(see \cite{Tal02},\cite{TPWP16},\cite{Tyo17}), its toughest and most delicate
cases are still open.

However, even if completely solved, Frankl and F\"{u}redi's conjecture does
not provide an easy-to-use, closed-form expression for $\mu_{r}\left(
m\right)  $. In this regard, the following conjecture might be of interest:

\begin{conjecture}
\label{con1}Let $r\geq3$ and $G$ be an $r$-graph with $m$ edges. If $t\geq
r-1$ is the unique real number such that $m=\binom{t}{r}$, then $\mu\left(
G\right)  \leq mt^{-r},$ with equality if and only if $t$ is an integer.
\end{conjecture}

Note that the value of $\mu_{r}\left(  m\right)  $ conjectured by Frankl and
F\"{u}redi is quite close to $mt^{-r}$, and moreover, both values coincide if
$t$ is an integer. Tyomkyn \cite{Tyo17} called the latter case the\emph{
principal case\ }of the Frankl-F\"{u}redi's conjecture, and solved it for any
$r\geq4$ and $m$ sufficiently large; prior to that, Talbot \cite{Tal02} had
solved the principal case for $r=3$ and any $m$. Let us note that Talbot and
Tyomkyn contribute much more than the mentioned results, but neither of these
works imply a complete solution to Conjecture \ref{con1} for any $r$.

In this paper, we confirm Conjecture \ref{con1} whenever $3\leq r\leq5,$
thereby completely resolving the principal case of the Frankl-F\"{u}redi's
conjecture for these values of $r$. In addition, we show that Conjecture
\ref{con1} holds whenever $t\geq4\left(  r-1\right)  \left(  r-2\right)  $,
thereby giving an alternative proof of Tyomkyn's result and providing an
explicit bound\footnote{This bound is chosen for simplicity, and can be cut at
least by half. In contrast, Tyomkyn's proof of the principal case of the
Frankl-F\"{u}redi's conjecture provides no explicit bounds.}.

Our proofs are based on some seemingly novel bounds on elementary symmetric
functions, which, somewhat surprisingly, are just analytic results with no
relation to hypergraphs whatsoever. Theorems \ref{tub}, \ref{tlb}, and
\ref{tlub} below present the gist of this approach.\medskip

Given a vector $\mathbf{x}:=\left(  x_{1},\ldots,x_{n}\right)  $, write
$S_{k}\left(  \mathbf{x}\right)  $ for the $k$th elementary symmetric function
of $x_{1},\ldots,x_{n}$. Set $q\left(  \mathbf{x}\right)  :=x_{1}^{2
+\cdots+x_{n}^{2},$ and for nonnegative $\mathbf{x}$ set $\sigma\left(
\mathbf{x}\right)  :=x_{1}^{x_{1}}\cdots x_{n}^{x_{n}}$, with the caveat that
$0^{0}=1$.

Let us recall the most important case of the celebrated Maclaurin inequality
(\cite{M1729}, see also \cite{HLP88}, Theorem 52), that can be stated as:
\medskip

\emph{If }$k\geq2$\emph{ and }$\mathbf{x}\in\Delta^{n-1}$\emph{, then}
\begin{equation}
S_{k}\left(  \mathbf{x}\right)  <n^{-k}\binom{n}{k}, \label{MI
\end{equation}
\emph{unless the entries of }$\mathbf{x}$\emph{ are equal.\medskip}

Since $\mathbf{x}\in\Delta^{n-1}$ implies that $\sigma\left(  \mathbf{x
\right)  \geq1/n$, the following theorem strengthens inequality (\ref{MI})
under a mild restriction on the maximum entry of $\mathbf{x}$ (denoted by
$\left\vert \mathbf{x}\right\vert _{\max}$ hereafter):

\begin{theorem}
\label{tub}Let $k\geq3,$ $\mathbf{x}\in\Delta^{n-1}$, and $\sigma
=\sigma\left(  \mathbf{x}\right)  .$ If $\left\vert \mathbf{x}\right\vert
_{\max}\leq1/4\left(  k-2\right)  $, then
\[
S_{k}\left(  \mathbf{x}\right)  <\sigma^{k}\binom{1/\sigma}{k},
\]
unless the nonzero entries of $\mathbf{x}$ are equal.
\end{theorem}

It turns out that the bound in Theorem \ref{tub} is not an isolated exception,
but one of many interrelated similar bounds that avoid the parameter $n$
altogether (see the closing remarks of Section \ref{sym}.) In particular,
Theorem \ref{tub} can be matched with a very similar lower bound:

\begin{theorem}
\label{tlb}Let $k\geq3,$ $\mathbf{x}\in\Delta^{n-1}$, and $q=q\left(
\mathbf{x}\right)  $. If $\left\vert \mathbf{x}\right\vert _{\max}<1/\left(
k-1\right)  $, then
\[
S_{k}\left(  \mathbf{x}\right)  >q^{k}\binom{1/q}{k},
\]
unless the nonzero entries of $\mathbf{x}$ are equal.
\end{theorem}

The following theorem, crucial for tackling Conjecture \ref{con1},
incorporates an additional twist in order to weaken the constraint on
$\left\vert \mathbf{x}\right\vert _{\max}$:

\begin{theorem}
\label{tlub}Let $3\leq k\leq5,$ $\mathbf{x}:=\left(  x_{1},\ldots
,x_{n}\right)  \in\Delta^{n-1}$, and $\sigma=\sigma\left(  \mathbf{x}\right)
.$ If $x_{n}\leq\cdots\leq x_{1}\leq1/k$, the
\begin{equation}
\frac{\partial S_{k}\left(  \mathbf{x}\right)  }{\partial x_{1}}<k\sigma
^{k}\binom{1/\sigma}{k}, \label{inr3
\end{equation}
unless the nonzero entries of $\mathbf{x}$ are equal.
\end{theorem}

The rest of the paper is structured as follows: in Section \ref{sym}, we give
some results about symmetric functions and prove Theorems \ref{tub},
\ref{tlb}, and \ref{tlub}; at the end of the section we discuss possible
extensions of these theorems. In Section \ref{pfc}, we prove an upper bound on
$\mu\left(  G\right)  $ and then prove Conjecture \ref{con1} if $3\leq r\leq5$
or $t\geq4\left(  r-1\right)  \left(  r-2\right)  $.

\section{\label{sym}Some bounds on elementary symmetric functions}

Let $\mathbf{x}:=\left(  x_{1},\ldots,x_{n}\right)  \in\Delta^{n-1}$, and
assume hereafter that $x_{1}\geq\cdots\geq x_{n}.$ Set
\[
p\left(  \mathbf{x}\right)  :=x_{1}^{3}+\cdots+x_{n}^{3}\text{ \ \ and
\ \ }t\left(  \mathbf{x}\right)  :=x_{1}^{4}+\cdots+x_{n}^{4}.
\]
Whenever $\mathbf{x}$ is understood, we shorten $S_{k}\left(  \mathbf{x
\right)  ,$ $q\left(  \mathbf{x}\right)  ,$ $p\left(  \mathbf{x}\right)  ,$
$t\left(  \mathbf{x}\right)  ,$ $\sigma\left(  \mathbf{x}\right)  $ to
$S_{k},$ $q,$ $p,$ $t,$ $\sigma$.

We start with a few basic inequalities about $\sigma,$ $q,$ $p,$ $t,$ and
$\left\vert \mathbf{x}\right\vert _{\max}$.

\begin{proposition}
\label{pmi}Let $\mathbf{x}\in\Delta^{n-1}$, and let $n^{\prime}$ be the number
of nonzero entries of $\mathbf{x}$. The
\[
1/n^{\prime}<\sigma<q<p^{1/2}<t^{1/3}<\left\vert \mathbf{x}\right\vert _{\max
},
\]
unless the nonzero entries of $\mathbf{x}$ are equal, in which case,
equalities hold throughout.
\end{proposition}

\begin{proof}
Without loss of generality, assume that $\mathbf{x}$ is positive, and that the
entries of $\mathbf{x}$ are not all equal. First, the function $x^{c}$ is
strictly convex for $x>0$ and $c>1$; hence, Jensen's inequality implies tha
\
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}x_{i}<\left(
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}x_{i}^{2}\right)  ^{1/2}<\left(
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}x_{i}^{3}\right)  ^{1/3},
\]
yielding $q<p^{1/2}<t^{1/3}$. Likewise, since the function $\log
x$\footnote{Here and elsewhere $\log$ stands for \textquotedblleft logarithm
base $e$\textquotedblright.} is strictly concave for $x>0$, we see that
\[
\log\sigma
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}\log x_{i}<\log\left(
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}x_{i}\right)  =\log q,
\]
yielding $\sigma<q$. Further, since the function $x\log x$ is strictly convex
for $x>0$, we see tha
\[
\log\sigma=x_{1}\log x_{1}+\cdots+x_{n}\log x_{n}>n\left(  \frac{1}{n
\log\frac{1}{n}\right)  =\log1/n,
\]
yielding $\sigma>1/n$. Finally,
\[
t=x_{1}^{4}+\cdots+x_{n}^{4}<\left\vert \mathbf{x}\right\vert _{\max
^{3}\left(  x_{1}+\cdots+x_{n}\right)  =\left\vert \mathbf{x}\right\vert
_{\max}^{3},
\]
completing the proof of Proposition \ref{pmi}.
\end{proof}

Further, note that $\partial S_{k}/\partial x_{i}$ is just the sum of all
products in $S_{k-1}$ that do not contain $x_{i}$; thus for every $i\in\left[
n\right]  $, we hav
\begin{equation}
\frac{\partial S_{k}}{\partial x_{i}}=S_{k-1}-x_{i}\frac{\partial S_{k-1
}{\partial x_{i}}. \label{MR
\end{equation}
In addition, a short argument shows that
\[
\frac{\partial S_{k}}{\partial x_{i}}-\frac{\partial S_{k}}{\partial x_{j
}=\left(  x_{j}-x_{i}\right)  \frac{\partial S_{k}}{\partial x_{i}\partial
x_{j}}.
\]
Hence, the assumption $x_{1}\geq\cdots\geq x_{n}$ implies that $\partial
S_{k}/\partial x_{1}\leq\cdots\leq\partial S_{k}/\partial x_{n}$. Moreover, we
see that $x_{i}=x_{j}$ if and only if $\partial S_{k}/\partial x_{i}=\partial
S_{k}/\partial x_{j}.$\medskip

Our proofs crucially rely on the weighted Chebyshev inequality (see, e.g.,
\cite{Bul03}, p. 161):\medskip

\textbf{Chebyshev's inequality. }\emph{Let }$\mathbf{w}:=\left(  w_{1
,\ldots,w_{n}\right)  \in\Delta^{n-1}$\emph{ be positive, and let }$a_{1
\leq\cdots\leq a_{n\text{ }}$\emph{. If }$b_{1}\geq\cdots\geq b_{n},$\emph{
then
\
{\textstyle\sum\limits_{i=1}^{n}}
w_{i}a_{i}b_{i}\le
{\textstyle\sum\limits_{i=1}^{n}}
w_{i}a_{i
{\textstyle\sum\limits_{i=1}^{n}}
w_{i}b_{i}.
\]
\emph{If }$b_{1}\leq\cdots\leq b_{n}$\emph{, then the opposite inequality
holds. In both cases equality holds if and only if }$a_{1}=\cdots=a_{n}$\emph{
or }$b_{1}=\cdots=b_{n}$\emph{.}

\subsection{Two recurrence inequalities}

The proof of Theorems \ref{tub} and \ref{tlb} reside on two recurrence
inequalities, stated in Propositions \ref{glb} and \ref{gub} below.

\begin{proposition}
\label{glb}If $k\geq2$ and $\mathbf{x}\in\Delta^{n-1}$, the
\begin{equation}
kS_{k}\geq\left(  1-\left(  k-1\right)  q\right)  S_{k-1}. \label{ina
\end{equation}
Equality holds if and only if the nonzero entries of $\mathbf{x}$ are equal.
\end{proposition}

\begin{proof}
Without loss of generality, we assume that $\mathbf{x}$ is positive, for
dropping out its zero entries does not alter $S_{2},\ldots,S_{n},$ and $q$.

First, multiplying equation (\ref{MR}) by $x_{i}$ and summing the results, we
ge
\[
kS_{k}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}\frac{\partial S_{k}}{\partial x_{i}}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}S_{k-1}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{2}\frac{\partial S_{k-1}}{\partial x_{i}}=S_{k-1}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{2}\frac{\partial S_{k-1}}{\partial x_{i}}.
\]
Now, let $w_{i}=a_{i}=x_{i}$ and $b_{i}=\partial S_{k-1}/\partial x_{i}$ for
all $i\in\left[  n\right]  ,$ and note that Chebyshev's inequality implies
that
\begin{equation
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}x_{i}\frac{\partial S_{k-1}}{\partial x_{i}}\le
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{2
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}\frac{\partial S_{k-1}}{\partial x_{i}}=\left(  k-1\right)  qS_{k-1},
\label{ina1
\end{equation}
so inequality (\ref{ina}) follows.

The sufficiency of the condition for equality in (\ref{ina}) is clear, so we
only prove its necessity. If equality holds in (\ref{ina}), then equality
holds in (\ref{ina1}). Hence, the condition for equality in Chebyshev's
inequality implies that
\[
x_{1}=\cdots=x_{n}\text{ \ \ or \ \ }\frac{\partial S_{k-1}}{\partial x_{1
}=\cdots=\frac{\partial S_{k-1}}{\partial x_{n}}.
\]
As noted above, in either case $x_{1}=\cdots=x_{n}$, completing the proof of
Proposition \ref{glb}.
\end{proof}

\medskip

\begin{proposition}
\label{gub}If $k\geq3$ and $\mathbf{x}\in\Delta^{n-1}$, the
\begin{equation}
kS_{k}\leq S_{k-1}-\left(  q-\left(  k-2\right)  p\right)  S_{k-2}.
\label{inb
\end{equation}
Equality holds if and only if the nonzero entries of $\mathbf{x}$ are equal.
\end{proposition}

\begin{proof}
Without loss of generality we assume that $\mathbf{x}$ is positive. As in the
proof of Proposition \ref{glb}, we see that
\[
kS_{k}=S_{k-1}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{2}\frac{\partial S_{k-1}}{\partial x_{i}}=S_{k-1}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{2}\left(  S_{k-2}-x_{i}\frac{\partial S_{k-2}}{\partial x_{i}}\right)
=S_{k-1}-qS_{k-2}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{3}\frac{\partial S_{k-2}}{\partial x_{i}}.
\]
Now, let $w_{i}=x_{i},$ $a_{i}=x_{i}^{2}$ and $b_{i}=\partial S_{k-2}/\partial
x_{i}$ for all $i\in\left[  n\right]  ,$ and note that Chebyshev's inequality
implies that
\begin{equation
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{3}\frac{\partial S_{k-2}}{\partial x_{i}}
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}x_{i}^{2}\frac{\partial S_{k-2}}{\partial x_{i}}\le
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{3
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}\frac{\partial S_{k-2}}{\partial x_{i}}=\left(  k-2\right)  pS_{k-2}.
\label{inb1
\end{equation}
so inequality (\ref{inb}) follows.

The sufficiency of the condition for equality in (\ref{inb}) is clear, so we
only prove its necessity. If equality holds in (\ref{inb}), then equality
holds in (\ref{inb1}). Hence, the condition for equality in Chebyshev's
inequality implies that
\[
x_{1}^{2}=\cdots=x_{n}^{2}\text{ \ \ or \ \ }\frac{\partial S_{k-2}}{\partial
x_{1}}=\cdots=\frac{\partial S_{k-2}}{\partial x_{n}}.
\]
In either case $x_{1}=\cdots=x_{n}$, completing the proof of Proposition
\ref{gub}.
\end{proof}

\subsection{Proofs of Theorems \ref{tub} and \ref{tlb}}

\begin{proof}
[\textbf{Proof of Theorem \ref{tub}.}]Set for short $x=x_{1}.$ Our proof
hinges on two claims:\medskip

\textbf{Claim 1. }\emph{If }$x\leq1/4\left(  k-2\right)  ,$ then for
$i=3,\ldots,k,$ we hav
\[
iS_{i}\leq S_{i-1}-\sigma\left(  1-\left(  i-2\right)  \sigma\right)
S_{i-2}.
\]

\emph{Proof.} Referring to Proposition \ref{gub}, it is enough to show that
\[
q-\left(  i-2\right)  p\geq\sigma\left(  1-\left(  i-2\right)  \sigma\right)
.
\]
To this end, le
\[
f\left(  z\right)  :=e^{z}-\left(  i-2\right)  e^{2z},
\]
and note that $f\left(  z\right)  $ is convex whenever $e^{z}\leq1/4\left(
i-2\right)  .$ Hence, in view of $x\leq1/4\left(  k-2\right)  $
\[
q-\left(  i-2\right)  p
{\textstyle\sum\limits_{j=1}^{n}}
x_{j}f\left(  \log x_{j}\right)  \leq f\left(
{\textstyle\sum\limits_{j=1}^{n}}
x_{j}\log x_{j}\right)  =\sigma-\left(  i-2\right)  \sigma^{2},
\]
proving Claim 1.\medskip

\textbf{Claim 2. }\emph{If }$x\leq1/4\left(  k-2\right)  ,$ then for
$i=3,\ldots,k,$ we have
\begin{equation}
iS_{i}\leq\left(  1-\left(  i-1\right)  \sigma\right)  S_{i-1}. \label{reci
\end{equation}

\emph{Proof. }We use induction on $i$. If $i=3,$ then Claim 1 yields
\[
3S_{3}=S_{2}-\left(  \sigma-\sigma^{2}\right)  =S_{2}-\sigma\left(
1-q\right)  =\left(  1-2\sigma\right)  S_{2}\text{;
\]
hence, the statement holds for $i=3.$ If $i>3$, then the induction assumption
implies that
\[
\frac{\left(  i-1\right)  S_{i-1}}{1-\left(  i-2\right)  \sigma}\geq\left(
i-1\right)  S_{i-1},
\]
because $1-\left(  i-2\right)  \sigma\geq1-\left(  k-2\right)  x>0.$ Now,
Claim 1 yields
\begin{align*}
iS_{k}  &  \leq S_{i-1}-\sigma\left(  1-\left(  i-2\right)  \sigma\right)
S_{i-2}\leq S_{i-1}-\sigma\left(  1-\left(  i-2\right)  \sigma\right)
\frac{\left(  i-1\right)  S_{i-1}}{\left(  1-\left(  i-2\right)
\sigma\right)  }\\
&  =\left(  1-\left(  i-1\right)  \sigma\right)  S_{i-1},
\end{align*}
completing the induction step and the proof of Claim 2.\medskip

To finish the proof of Theorem \ref{tub}, note that $1-\left(  i-1\right)
\sigma\geq1-\left(  k-1\right)  x>0.$ Thus we can multiply inequalities
(\ref{reci}) for $i=3,\ldots,k,$ gettin
\[
\frac{k!}{2}S_{3}\cdots S_{k}\leq\left(  1-2\sigma\right)  \cdots\left(
1-\left(  k-1\right)  \sigma\right)  S_{2}\cdots S_{k-1}.
\]
Now, using the fact $2S_{2}=1-q\leq1-\sigma$, we see that
\[
S_{k}\leq\frac{1}{k!}\left(  1-\sigma\right)  \left(  1-2\sigma\right)
\cdots\left(  1-\left(  k-1\right)  \sigma\right)  =\sigma^{k}\binom{1/\sigma
}{k},
\]
as desired. The condition for equality in this inequality follows from the
conditions for equality in Proposition \ref{gub}.Theorem \ref{tub} is proved.
\end{proof}

\emph{Remark. }The proof of Theorem \ref{tub} shows that its conclusion can be
strengthened t
\[
S_{k}\leq\frac{1}{k!}\left(  1-q\right)  \left(  1-2\sigma\right)
\cdots\left(  1-\left(  k-1\right)  \sigma\right)  .
\]
\medskip

\begin{proof}
[\textbf{Proof of Theorem \ref{tlb}.}]Proposition \ref{glb} implies that
\begin{equation}
S_{i}\geq\frac{1}{i}\left(  1-\left(  i-1\right)  q\right)  S_{i-1}
\label{lin
\end{equation}
for every $i=2,\ldots,k$. Since
\[
1-\left(  i-1\right)  q\geq1-\left(  k-1\right)  \left\vert \mathbf{x
\right\vert _{\max}>1-\frac{k-1}{k-1}=0,
\]
we can multiply inequalities (\ref{lin}) for $i=2,\ldots,k,$ obtainin
\[
S_{2}\cdots S_{k}\geq\frac{1}{k!}S_{1}\cdots S_{k-1}\left(  1-q\right)
\cdots\left(  1-\left(  k-1\right)  q\right)  =q^{k}\binom{1/q}{k},
\]
as desired.

The condition for equality in this inequality follows from the conditions for
equality in Proposition \ref{glb}. Theorem \ref{tlb} is proved.
\end{proof}

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

The proof of Theorem \ref{tlub} is the most involved one in this paper,
especially the case $k=5$. We give separate proofs for $k=3,4,5$, because a
compound one would be a harder read.

In all three cases we assume that $\mathbf{x}$ is positive, and set $x=x_{1}$.
The proofs of the conditions for equality are straightforward and are
omitted.\medskip

\begin{proof}
[\textbf{Proof for }$k=3$.]Using equation (\ref{MR}) and Proposition
\ref{pmi}, we find that
\[
\frac{\partial S_{3}}{\partial x_{1}}=S_{2}-x\frac{\partial S_{2}}{\partial
x_{1}}=\frac{1-q}{2}-x\left(  1-x\right)  \leq\frac{1-\sigma-2x\left(
1-x\right)  }{2}.
\]
Since $x\left(  1-x\right)  $ is increasing for $x\leq1/2,$ it follows that
$\sigma\left(  1-\sigma\right)  \leq x\left(  1-x\right)  ,$ and so
\[
\frac{\partial S_{3}}{\partial x_{1}}\leq\frac{1-\sigma-2x\left(  1-x\right)
}{2}\leq\frac{1-\sigma-2\sigma\left(  1-\sigma\right)  }{2}=3\sigma^{3
\binom{1/\sigma}{3}.
\]
Theorem \ref{tlub} is proved for $k=3.$
\end{proof}

\medskip

\begin{proof}
[\textbf{Proof for }$k=4$.]Equation (\ref{MR}) implies tha
\begin{align*}
\frac{\partial S_{4}}{\partial x_{1}}  &  =S_{3}-x\frac{\partial S_{3
}{\partial x_{1}}=S_{3}-x\left(  S_{2}-x\frac{\partial S_{2}}{\partial x_{1
}\right)  =\frac{1-3q+2p}{6}-x\left(  \frac{1-q}{2}-x+x^{2}\right) \\
&  =\frac{1}{6}\left(  1-3q+2p-3x\left(  1-q\right)  +6x^{2}-6x^{3}\right)
\end{align*}

To establish (\ref{inr3}), we prove a chain of inequalities, consecutively
eliminating the parameters $p,$ $q,$ and $x$ from the right side of the above equation.

First, note that the function
\[
f\left(  z\right)  =e^{z}-e^{2z
\]
is convex whenever $e^{z}\leq1/4.$ Hence
\[
q-p
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}f\left(  \log x_{i}\right)  \geq f\left(
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}\log x_{i}\right)  =\sigma-\sigma^{2},
\]
yielding in tur
\begin{align*}
6\frac{\partial S_{4}}{\partial x_{1}}  &  \leq1-q-2\sigma+2\sigma
^{2}-3x\left(  1-q\right)  +6x^{2}-6x^{3}\\
&  =1-q\left(  1-3x\right)  -2\sigma+2\sigma^{2}-3x+6x^{2}-6x^{3}.
\end{align*}
Since $1-3x>0$ and $q\leq\sigma,$ we ge
\[
6\frac{\partial S_{4}}{\partial x_{1}}\leq1-3x\sigma-3\sigma+2\sigma
^{2}-3x+6x^{2}-6x^{3}.
\]
To finish the proof, we have to show that
\[
1-3x\sigma-3\sigma+2\sigma^{2}-3x+6x^{2}-6x^{3}\leq1-6\sigma+11\sigma
^{2}-6\sigma^{3}=24\sigma^{4}\binom{1/\sigma}{4},
\]
which, after some algebra, turns out to be equivalent to
\[
\left(  x-\sigma\right)  \left(  3x-2x^{2}+2\sigma-2x\sigma-2\sigma
^{2}\right)  \leq\left(  x-\sigma\right)  .
\]
Using the fact $x\leq\sigma,$ if suffice to prove that
\[
3x-2x^{2}+2\sigma-2x\sigma-2\sigma^{2}<1.
\]
However, $2\sigma-2x\sigma-2\sigma^{2}$ increases in $\sigma$ because
$\sigma\leq x\leq1/4<\left(  1-x\right)  /2.$ Hence
\[
3x-2x^{2}+2\sigma-2x\sigma-2\sigma^{2}\leq5x-6x^{2}\leq\frac{5}{4}-\frac
{6}{16}<1.
\]
Theorem \ref{tlub} is proved for $k=4.$
\end{proof}

\medskip

\begin{proof}
[\textbf{Proof for} $k=5$.]Equation (\ref{MR}) implies tha
\begin{align}
\frac{\partial S_{5}}{\partial x_{1}}  &  =S_{4}-x\frac{\partial S_{4
}{\partial x_{1}}=S_{4}-xS_{3}+x^{2}\frac{\partial S_{3}}{\partial x_{1
}=S_{4}-xS_{3}+x^{2}S_{2}-x^{3}\frac{\partial S_{2}}{\partial x_{1
}\nonumber\\
&  =\frac{1-6q+3q^{2}+8p-6t}{24}-x\frac{1-3q+2p}{6}+x^{2}\frac{1-q}{2
-x^{3}\left(  1-x\right) \nonumber\\
&  =\frac{1}{24}\left(  1-6q+3q^{2}+8\left(  1-x\right)  p-6t+12xq-12x^{2
q-4x+12x^{2}-24x^{3}+24x^{4}\right)  . \label{me5
\end{align}

To establish (\ref{inr3}), we prove a chain of inequalities, consecutively
eliminating the parameters $t,p,q,$ and $x$ from the right side of equation
(\ref{me5})\footnote{The reader may find that the tradeoffs between the stages
of the calculations result in weird numbers, but we were not able to spare
much room for elegance.}.

For a start, the following claim is used to eliminate $p$ and $t$:\medskip

\textbf{Claim 1 }\emph{If }$x<1/5,$ \emph{then
\begin{equation}
-\frac{106-160x}{25}q+8\left(  1-x\right)  p-6t\leq-\frac{106-160x}{25
\sigma+8\left(  1-x\right)  \sigma^{2}-6\sigma^{3}. \label{insig
\end{equation}

\emph{Proof.} Claim 1 follows from the fact that the function
\[
f\left(  z\right)  :=-\frac{106-160x}{25}e^{z}+8\left(  1-x\right)
e^{2z}-6e^{3z
\]
is concave whenever $e^{z}\leq1/5$. To verify this fact, we show that
\begin{equation}
f^{\prime\prime}\left(  z\right)  =e^{z}\left(  -\frac{106-160x}{25}+32\left(
1-x\right)  e^{z}-54e^{2z}\right)  \leq0. \label{concf
\end{equation}
Indeed, the expression $32\left(  1-x\right)  e^{z}-54e^{2z}$ is quadratic in
$e^{z}$, and thus it increases in $z$ whenever $e^{z}\leq8\left(  1-x\right)
/27$. On the other hand,
\[
e^{z}\leq1/5\leq8\left(  1-x\right)  /27,
\]
implying that
\[
32\left(  1-x\right)  e^{z}-54e^{2z}\leq\frac{32\left(  1-x\right)  }{5
-\frac{54}{25}=\frac{106-160x}{25}.
\]
The latter inequality clearly entails (\ref{concf}); therefore, $f\left(
z\right)  $ is concave.

Now, the concavity of $f\left(  z\right)  $ implies tha
\begin{align*}
-\frac{106-160x}{25}q+8\left(  1-x\right)  p-6t  &  =\sum x_{i}f\left(  \log
x_{i}\right)  \leq f\left(  x_{i}\log x_{i}\right) \\
&  =-\frac{106-160x}{25}\sigma+8\left(  1-x\right)  \sigma^{2}-6\sigma^{3},
\end{align*}
completing the proof of Claim 1.\medskip

To use Claim 1 we add and subtract the term $\frac{106-160x}{25}q$ in the
right side of (\ref{me5}), and note that
\[
-6q+\frac{106-160x}{25}q+12xq-12x^{2}q=-\frac{44-130x+300x^{2}}{25}q+\frac
{2}{5}xq.
\]
Thus, summarizing the current progress, Claim 1 implies that
\begin{align}
24\frac{\partial S_{5}}{\partial x_{1}}  &  \leq1-\frac{44-130x+300x^{2}
{25}q+3q^{2}+\frac{2}{5}xq-4x+12x^{2}-24x^{3}+24x^{4}\label{in1}\\
&  -\frac{106-160x}{25}\sigma+8\left(  1-x\right)  \sigma^{2}-6\sigma
^{3}.\nonumber
\end{align}

Our next goal is to eliminate $q$ in the right side of (\ref{in1}). To this
end, define the function
\[
g\left(  z\right)  :=-\frac{44-130x+300x^{2}}{25}z+3z^{2
\]
\medskip

\textbf{Claim 2} \emph{If }$x\leq1/5,$ \emph{then }$g\left(  q\right)  \leq
g\left(  \sigma\right)  .\medskip$

\emph{Proof. }Claim 2 follows from the fact that $g\left(  z\right)  $
decreases in $z$ whenever $z\leq x$. To prove this fact note that $g\left(
z\right)  $ is quadratic in $z$, and so it decreases whenever
\[
z\leq\frac{44-130x+300x^{2}}{150}.
\]
However, the stipulation $x\leq1/5$ entails that
\[
x\leq\frac{44-130x+300x^{2}}{150}.
\]
Thus, $g\left(  z\right)  $ decreases in $z$ whenever $z\leq x$. In
particular, the inequalities $\sigma\leq q\leq x$ imply that $g\left(
q\right)  \leq g\left(  \sigma\right)  $, completing the proof of Claim
2.\medskip

Applying Claim 2, we replace $q$ by $\sigma$ in the right side of (\ref{in1}),
and obtain
\begin{align*}
24\frac{\partial S_{5}}{\partial x_{1}}  &  \leq1-\frac{44-130x+300x^{2}
{25}\sigma+3\sigma^{2}-\frac{106-160x}{25}\sigma+8\left(  1-x\right)
\sigma^{2}-6\sigma^{3}+\frac{2}{5}xq\\
&  -4x+12x^{2}-24x^{3}+24x^{4}\\
&  =1-\frac{150-290x+300x^{2}}{25}\sigma+11\sigma^{2}-8x\sigma^{2}-6\sigma
^{3}-4x+\frac{62}{5}x^{2}-24x^{3}+24x^{4}\\
&  =1-6\sigma+11\sigma^{2}-6\sigma^{3}+\frac{58}{5}x\sigma-12x^{2
\sigma-8x\sigma^{2}-4x+\frac{62}{5}x^{2}-24x^{3}+24x^{4}.
\end{align*}
In the above derivation we also use the inequality $\frac{2}{5}xq\leq\frac
{2}{5}x^{2}$.

Therefore, to finish the proof of (\ref{inr3}), it remains to show that
\begin{align*}
&  1-6\sigma+11\sigma^{2}-6\sigma^{3}+\frac{58}{5}x\sigma-12x^{2
\sigma-8x\sigma^{2}-4x+\frac{62}{5}x^{2}-24x^{3}+24x^{4}\\
&  \leq1-10\sigma+35\sigma^{2}-50\sigma^{3}+24\sigma^{4}=120\sigma^{5
\binom{1/\sigma}{5},
\end{align*}
which is equivalent t
\[
\frac{58}{5}x\sigma-12x^{2}\sigma-8x\sigma^{2}-4x+\frac{62}{5}x^{2
-24x^{3}+24x^{4}\leq-4\sigma+24\sigma^{2}-44\sigma^{3}+24\sigma^{4}.
\]
After rearranging and factoring $\left(  x-\sigma\right)  $ out, we get
\begin{align*}
4\left(  x-\sigma\right)   &  \geq\frac{58}{5}\left(  x-\sigma\right)
\sigma+\frac{62}{5}\left(  x-\sigma\right)  \left(  x+\sigma\right) \\
&  -12\left(  x-\sigma\right)  \left(  x+\sigma\right)  \sigma-8\left(
x-\sigma\right)  \sigma^{2}-24\left(  x-\sigma\right)  \left(  x^{2
+x\sigma+\sigma^{2}\right) \\
&  +24\left(  x-\sigma\right)  \left(  x^{3}+x^{2}\sigma+x\sigma^{2
+\sigma^{3}\right)  .
\end{align*}
Since $x\geq\sigma$, it suffices to show tha
\begin{align}
2  &  >\frac{29}{5}\sigma+\frac{31}{5}\left(  x+\sigma\right)  -6\left(
x+\sigma\right)  \sigma-4\sigma^{2}-12\left(  x^{2}+x\sigma+\sigma^{2}\right)
+12\left(  x^{3}+x^{2}\sigma+x\sigma^{2}+\sigma^{3}\right) \nonumber\\
&  =\frac{31}{5}x-12x^{2}+12x^{3}+12\sigma^{3}-\left(  22-12x\right)
\sigma^{2}+\left(  12-18x+12x^{2}\right)  \sigma. \label{exp2
\end{align}
To this end, set
\[
h\left(  z\right)  :=12z^{3}-\left(  22-12x\right)  z^{2}+\left(
12-18x+12x^{2}\right)  z,
\]
and note that
\[
h^{\prime}\left(  z\right)  =36z^{2}-2\left(  22-12x\right)  z+\left(
12-18x+12x^{2}\right)  .
\]
Since $h^{\prime}\left(  z\right)  $ is quadratic in $z$ and $36>0,$ we see
that $h\left(  z\right)  $ is increasing if $z\leq z_{\min}$, where $z_{\min}$
is the smaller root of the equation $h^{\prime}\left(  z\right)  =0.$ However,
the stipulation $x\leq1/5$ easily implies that
\[
z_{\min}=\frac{22-12x-\sqrt{\left(  22-12x\right)  ^{2}-36\left(
12-18x+12x^{2}\right)  }}{36}>x,
\]
and therefore $h\left(  z\right)  $ is increasing in $z$ if $z\leq x$.

Thus, in view of $\sigma\leq x\leq1/5,$ we find that
\begin{align*}
12\sigma-18x\sigma-22\sigma^{2}+12x^{2}\sigma+12x\sigma^{2}+12\sigma^{3}  &
=h\left(  \sigma\right) \\
&  \leq h\left(  x\right)  =12x-40x^{2}+36x^{3}.
\end{align*}
Finally, for the right side of (\ref{exp2}) we obtain
\[
\frac{31}{5}x-12x^{2}+12x^{3}+12\sigma^{3}-\left(  22-12x\right)  \sigma
^{2}+\left(  12-18x+12x^{2}\right)  \sigma\leq\frac{91}{5}x-52x^{2
+48x^{3}<2.
\]
Theorem \ref{tlub} is proved for $k=5$.
\end{proof}

\subsection{Closing remarks}

The restriction on $\left\vert \mathbf{x}\right\vert _{\max}$ in Theorem
\ref{tub} can be somewhat relaxed. For example, for $r=3$ it is enough to
require that $\left\vert \mathbf{x}\right\vert _{\max}\leq3/8,$ while for
$r=4$ it is enough to have $\left\vert \mathbf{x}\right\vert _{\max}\leq
11/48$. It is challenging to find the weakest possible restriction on
$\left\vert \mathbf{x}\right\vert _{\max}$ for the conclusion of Theorem
\ref{tub} to hold.

It is unlikely that Theorem \ref{tlub} remains valid as is for sufficiently
large $r$; even the case $r=6$ is a challenge. Thus, it is interesting what
alterations are necessary to prove Conjecture \ref{con1} for $r>5$. Here is a
possibility for some progress:

Given $\mathbf{x}\in\Delta^{n-1}$ and real $t\geq-1$, defin
\[
\varphi_{t}\left(  \mathbf{x}\right)  :=\left\{
\begin{array}
[c]{cc
\left(
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}^{1+t}\right)  ^{1/t}\text{,} & \text{if }x\neq0\text{;}\\
\sigma\left(  \mathbf{x}\right)  , & \text{if }x=0.
\end{array}
\right.
\]
Note that if $\mathbf{x}$ is fixed and $t\in\left[  -1,\infty\right)  $, the
function $\varphi_{t}\left(  \mathbf{x}\right)  $ is continuous and
nondecreasing in $t.$ In addition, assuming that $0^{0}=1,$ we see that
\[
\varphi_{1}\left(  \mathbf{x}\right)  =q\left(  \mathbf{x}\right)  \text{;
\ \ \ }\varphi_{-1}\left(  \mathbf{x}\right)  =1/n\text{; \ \ \ }\lim
{}_{t\rightarrow\infty}\varphi_{t}\left(  \mathbf{x}\right)  =\left\vert
x\right\vert _{\max}.
\]

It seems possible to extend Theorems \ref{tub}, \ref{tlb}, and \ref{tlub}
using $\varphi_{t}\left(  \mathbf{x}\right)  $ instead of $q$ and $\sigma$.

\section{\label{pfc}Proof of Conjecture \ref{con1} for $3\leq r\leq5$ or
$t\geq4\left(  r-1\right)  \left(  r-2\right)  $}

Let $G$ be an $r$-graph of order $n$. A vector $\mathbf{x}\in\Delta^{n-1}$
such that $P_{G}\left(  \mathbf{x}\right)  =\mu\left(  G\right)  $ is called
an \emph{eigenvector} to $\mu\left(  G\right)  .$

Let $\mathbf{x}:=\left(  x_{1},\ldots,x_{n}\right)  \in\Delta^{n-1}$ be an
eigenvector to $\mu\left(  G\right)  $. Using Lagrange
multipliers\footnote{This argument is known from the times of Motzkin and
Straus, so we skip the details.}, one finds that
\begin{equation}
r\mu\left(  G\right)  =\frac{\partial P_{G}\left(  \mathbf{x}\right)
}{\partial x_{i}}
{\textstyle\sum\limits_{\left\{  i_{1},\ldots,i_{r-1},i\right\}  \in E}}
x_{i_{1}}\cdots x_{i_{r-1}}, \label{eeq
\end{equation}
for every $i\in\left[  n\right]  $ such that $x_{i}>0.$

We start with a simple lemma, valid for any $r\geq2$:

\begin{lemma}
\label{tsin}Let $G$ be an $r$-graph of order $n$, with $m$ edges. If
$\mathbf{x}\in\Delta^{n-1}$ is an eigenvector to $\mu\left(  G\right)  $, the
\begin{equation}
\mu\left(  G\right)  \leq m\sigma^{r}. \label{sin
\end{equation}

\end{lemma}

\begin{proof}
Clearly the lemma holds if $m=0$, so suppose that $m>0$. Likewise, without
loss of generality, suppose that $\mathbf{x}$ is positive. Let $\mu=\mu\left(
G\right)  $ and note that equations (\ref{eeq}) imply that
\begin{align*}
\mu\log\sigma^{r}  &  =r\m
{\textstyle\sum\limits_{i=1}^{n}}
x_{i}\log x_{i}
{\textstyle\sum\limits_{i=1}^{n}}
\frac{\partial P\left(  \mathbf{x}\right)  }{\partial x_{i}}x_{i}\log x_{i}
{\textstyle\sum\limits_{\left\{  i_{1},\ldots,i_{r}\right\}  \in E}}
x_{i_{1}}\cdots x_{i_{r}}\left(  \log x_{i_{1}}+\cdots+\log x_{i_{r}}\right)
\\
&
{\textstyle\sum\limits_{\left\{  i_{1},\ldots,i_{r}\right\}  \in E}}
x_{i_{1}}\cdots x_{i_{r}}\log x_{i_{1}}\cdots x_{i_{r}}.
\end{align*}
Since the function $x\log x$ is convex, we see that
\
{\textstyle\sum\limits_{\left\{  i_{1},\ldots,i_{r}\right\}  \in E}}
x_{i_{1}}\cdots x_{i_{r}}\log x_{i_{1}}\cdots x_{i_{r}}\geq m\left(  \frac
{\mu}{m}\right)  \log\frac{\mu}{m}=\mu\log\frac{\mu}{m}.
\]
Hence
\[
\mu\log\sigma^{r}\geq\mu\log\frac{\mu}{m},
\]
implying that $\mu\leq m\sigma^{r}$, as desired.
\end{proof}

\emph{Remark. }Note that $\mu\left(  G\right)  \leq mq^{r}$ is a weaker, yet
more usable consequence of bound (\ref{sin}).\medskip

With Theorem \ref{tlub} and Lemma \ref{tsin} in hand, it is not hard to prove
Conjecture \ref{con1} for $3\leq r\leq5$:$\medskip$

\begin{proof}
[\textbf{Proof of Conjecture \ref{con1} for }$3\leq r\leq5.$]Let $G$ be an
$r$-graph of order $n$ with $m$ edges, and let $\mu\left(  G\right)  =\mu
_{r}\left(  m\right)  .$ Suppose that $t>r-1$ is a real number satisfying
$m=\binom{t}{r}$. To prove Conjecture \ref{con1}, we have to show that
\begin{equation}
\mu\left(  G\right)  \leq mt^{-r}=\frac{1}{r!}\left(  1-\frac{1}{t}\right)
\cdots\left(  1-\frac{r-1}{t}\right)  .\label{exf
\end{equation}

Let $\mathbf{x}:=\left(  x_{1},\ldots,x_{n}\right)  \in\Delta^{n-1}$ be an
eigenvector to $\mu\left(  G\right)  $ and suppose that $x_{1}\geq\cdots\geq
x_{n}$. If $x_{n}=0$, replace $G$ with the subgraph $H\subset G$ induced by
the vertices with nonzero entries in $\mathbf{x}$. Clearly $\mu\left(
H\right)  =\mu\left(  G\right)  $, and $H$ has at most $m$ edges. Since the
right side of (\ref{exf}) decreases with $t$, it is enough to prove
(\ref{exf}) for $H$. Thus, without loss of generality, we assume that
$\mathbf{x}$ is positive.

Set for short $\mu=\mu\left(  G\right)  $, and assume for a contradiction that
(\ref{exf}) fails, that is, $mt^{-r}<\mu$. Now, Lemma \ref{tsin} implies that
$mt^{-r}<\mu\leq m\sigma^{r}$, and so $\sigma>1/t$.

On the other hand, equation (\ref{eeq}) yields
\[
r\mu x_{1}=\frac{\partial P_{G}\left(  \mathbf{x}\right)  }{\partial x_{1
}x_{1}\leq P_{G}\left(  \mathbf{x}\right)  =\mu\text{;
\]
therefore, $x_{1}\leq1/r$. With this provision, and supposing that $3\leq
r\leq5$, Theorem \ref{tlub} gives
\begin{equation}
r\mu=\frac{\partial P_{G}\left(  \mathbf{x}\right)  }{\partial x_{1}}\leq
\frac{\partial S_{r}\left(  \mathbf{x}\right)  }{\partial x_{1}}\leq
r\sigma^{r}\binom{1/\sigma}{r}. \label{ine
\end{equation}
Hence, in view of $\sigma>1/t$, we find that
\[
\mu\leq\frac{1}{r!}\left(  1-\sigma\right)  \cdots\left(  1-\left(
r-1\right)  \sigma\right)  <\frac{1}{r!}\left(  1-\frac{1}{t}\right)
\cdots\left(  1-\frac{r-1}{t}\right)  =mt^{-r}.
\]
This contradiction shows that $\mu\left(  G\right)  \leq mt^{-r}$.

It remains to prove the conditions for equality in Conjecture \ref{con1}.
Suppose that $\mu\left(  G\right)  =mt^{-r}$; thus, equalities hold throughout
in (\ref{ine}), and by Theorem \ref{tlub} the entries of $\mathbf{x}$ are
equal to $1/n$. Therefore, we find that
\[
\frac{1}{r!}\left(  1-\frac{1}{t}\right)  \cdots\left(  1-\frac{r-1
{t}\right)  =mt^{-r}=\mu\left(  G\right)  \leq\binom{n}{r}n^{-r}=\frac{1
{r!}\left(  1-\frac{1}{n}\right)  \cdots\left(  1-\frac{r-1}{n}\right)  ,
\]
yielding in turn $n\geq t$. Now, $mt^{-r}=\mu\left(  G\right)  \leq mn^{-r}$
implies that $t=n$.

Finally, if $t$ is an integer, then taking $G$ to be the complete $r$-graph of
order $t$ and $\mathbf{x}\in\Delta^{t-1}$ to be the vector with all entries
equal to $1/t$, we see that $\mu_{r}\left(  m\right)  =mt^{-r}$, completing
the proof of Conjecture \ref{con1} for $3\leq r\leq5$.
\end{proof}

Our proof of Conjecture \ref{con1} for $t\geq4\left(  r-1\right)  \left(
r-2\right)  $ is similar, so we omit a few details.\medskip

\begin{proof}
[\textbf{Proof of Conjecture \ref{con1} for }$t\geq4\left(  r-1\right)
\left(  r-2\right)  .$]Let $G$ be an $r$-graph of order $n$ with $m$ edges,
and let $\mu\left(  G\right)  =\mu_{r}\left(  m\right)  .$ Suppose that
$t\geq4\left(  r-1\right)  \left(  r-2\right)  $ is a real number satisfying
$m=\binom{t}{r}$. To prove Conjecture \ref{con1}, we show that $\mu\left(
G\right)  \leq mt^{-r}$. To this end, select an eigenvector $\mathbf{x
:=\left(  x_{1},\ldots,x_{n}\right)  $ to $\mu\left(  G\right)  $ with
$x_{1}\geq\cdots\geq x_{n}>0$.

First note that if $x_{1}>\left(  r-1\right)  /t,$ then
\[
\frac{\left(  1-x_{1}\right)  ^{r-1}}{\left(  r-1\right)  !}<\frac{1}{\left(
r-1\right)  !}\left(  1-\frac{r-1}{t}\right)  ^{r-1}<\frac{1}{\left(
r-1\right)  !}\left(  1-\frac{1}{t}\right)  \cdots\left(  1-\frac{r-1
{t}\right)  =rmt^{-r},
\]
and therefor
\[
r\mu\left(  G\right)  =\frac{\partial P_{G}\left(  \mathbf{x}\right)
}{\partial x_{1}}\leq\frac{\partial S_{r}\left(  \mathbf{x}\right)  }{\partial
x_{1}}<\frac{\left(  1-x_{1}\right)  ^{r-1}}{\left(  r-1\right)  !}<rmt^{-r},
\]
proving that $\mu\left(  G\right)  <mt^{-r}$.

On the other hand, if $x_{1}\leq\left(  r-1\right)  /t,$ then the premise
$t\geq4\left(  r-1\right)  \left(  r-2\right)  $ implies that $x_{1
\leq1/4\left(  r-2\right)  $; therefore, Theorem \ref{tub} yields
\[
\mu\left(  G\right)  =P_{G}\left(  \mathbf{x}\right)  \leq S_{r}\left(
\mathbf{x}\right)  \leq\sigma\binom{1/\sigma}{r}.
\]
This inequality and Lemma \ref{tsin} imply that $\mu\left(  G\right)  \leq
mt^{-r}$.

The proof of the condition for equality in $\mu\left(  G\right)  \leq mt^{-r}$
is omitted.
\end{proof}

\bigskip

\textbf{Acknowledgement. }A preliminary version of this work has been reported
at the International workshop on spectral hypergraph theory held in November,
2017 at Anhui University, Hefei, P.R. China. I am grateful to the organizers,
and particularly to Prof. Yi-Zheng Fan, for wonderful experience.

\bigskip

redpajama_set_name: RedPajamaArXiv


text:
TOP STORIES / Culture
First contemporary African art museum opens in Cape Town
The Zeitz Museum of Contemporary Art Africa (MOCAA), the first ever museum of modern African art, has opened to much fanfare in Cape Town, South Africa. Art lovers are over the moon – but there has also been controversy.
Zeitz MOCAA - a spectacular museum for contemporary African art
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Another art installation that hangs from the ceiling, "Divider" by Lungiswa Gqunta, is made of 130 beer bottles suspended by plastic strings. The young South African artist, whose work explores resistance against apartheid and colonialism, uses everyday items that, when combined, can also become weapons.
Off the street
An entire room is dedicated to Cyrus Kabiru's self-portraits. The series entitled "C-Stunners" show the Kenyan artist wearing various masks and glasses created from objects he found in the streets of Nairobi.
The image shows detail of a sculpture by South Afcrica's Athi-Patra Ruga titled "Proposed Model for Tseko Simon Nkoli Memorial," a work that veers between contemporary art, fashion and performance. Several large-format paintings by South African painter Penny Siopis are also visible in the background.
This unnamed metal sculpture by South African Kyle Morland is set up next to the museum restaurant. The young sculptor creates works of art from aluminum, steel and recycled materials to challenge preconceived hierarchies inherent in conversations around material culture.
Skulpturengarten mit Ausblick
Zeitz MOCAA's top floor also has a space for outdoor sculptures. Visitors have a prime view of downtown Cape Town and Table Mountain through the convex faceted windows. At a price of 15 euros for an annual ticket, or free admission on some holidays for less frequent visits, the museum is bound to draw crowds from far and near.
Author: Julia Jaki (db)
Some people say they look like honeycomb cells, others feel reminded of an insect's compound eyes. The blue-tinted convex windows of the Zeitz Museum of Contemporary Art Africa (MOCAA) have already become the building's trademark. There's no getting around superlatives when writing about Cape Town's newest museum: with 100 exhibition rooms, the Zeitz MOCAA is Africa's largest museum, and more importantly, it's the first museum dedicated to contemproary African art on the continent.
Its significance should not be underestimated, says Mark Coetzee, the museum director and head curator. "It finally acknowledges the huge contribution that artists from our continent have made in the dialogue of visual practice around the world and locally," he told DW.
Building MOCAA in Cape Town, Africa's "whitest" city, created quite a stir in the African art world during the four-year planning phase. But former Puma CEO and founder of the musuem, Jochen Zeitz, argues the choice was a stroke of luck. "Of course cities like Nairobi were also in the running, but the operators of the Waterfront had this empty building, and we were looking for a home for the collection," Zeitz told DW.
A spectacular space
Three white men in Africa
In 2013, Zeitz and the operators of Victoria & Alfred Waterfront decided to create the museum. The deal: the Waterfront would foot the bill for the 32 million euro ($38 million) restructuring of the old grain silos, a relatively affordable sum for such a large project, while businessman, philanthropist and art patron Zeitz pledged his considerable collection of contemporary African art as a permanent loan. Zeitz has long been collecting contemporary African art by the young avant-garde – as opposed to traditional masks or cult objects – and especially works that question the art world's eurocentrism.
Read more:Defining Black art in the age of Black Power
Four yeas later, the metamorphosis from a 1921 grain silo is complete after London architect Thomas Heatherwick created a vast industrial-style exhibition space.
The storage silo's structures were cut open in the middle of the building, making them look like huge organ pipes that dominate the soaring atrium. A dragon-like sculpture by coveted South African artist Nicholas Hlobo hovers in the boundless space. The 2002 Zeitz Collection, with artworks by Zanele Muholi, Athi-Patra Ruga, Kudzanai Chiurai and El-Anatsui, to name just a few, in on display on nine floors.
Thomas Heatherwick and Jochen Zeitz
The fact that three white men – Zeitz, Heatherwick and Coetzee –were responsible for the creation of a museum for African art was criticized, though usually behind closed doors. Coetzee dismisses any criticism, saying the MOCAA will eventually be run by up to 150 people, with 21 curators already on staff. "Many people are running this museum, who are making the decisions about content, about narrative, about the storytelling and I think that is something that is very important to remember," he said.
Zeitz says he only created the platform: it is now up to the African artists to pick up the megaphone. "We live in a globalized world and we must overcome black/white cliches," Zeitz said. "We are only here to create opportunities."
Reaction to criticism
Art needs well-heeled investors, says artist Atang Tshikare. Busily preparing for an exhibition in Amsterdam in his studio in Cape Town's Woodstock neighborhood, he told DW the Zeitz MOCAA is good for the city because it brings "a lot of different artists to a central place that is accessible to people from around the world."
The Victoria & Albert Waterfront is a popular shopping destination for locals and tourists
The museum is also focused on art education in addition to exhibitions, which will help it counter potential criticism.
In this regard, the Zeitz MOCAA aims to draw locals and tourists alike by making admission free on Wednesday mornings and many holidays – while children under the age of 18 also get in free. An annual ticket costs the equivalent of 15 euros, not much more than a single ticket.
Whether the museum can live up to its claim of offering a representative cross section of contemporary African art remains to be seen – especially after the exhibitions move away from the core Zeitz Collection and the next generation of curators call the shots.
Defining Black art in the age of Black Power
What was the role of the African-American artist in the 1960s and 70s? A new exhibition at London's Tate Modern explores the many answers to this complex question. (12.07.2017)
Robin Rhode: Painting in black and white, but only in art
Renowned South African street art performer Robin Rhode employs his black-and-white paintings to lay down the marker against racial segregation. (10.03.2017)
Cape Town shines as tourist destination
South Africa's top tourist hub Cape Town expects a 13 percent jump in tourist arrivals this year to 1.77 million, officials said on Thursday. (27.07.2017)
Cape Town is now home to the largest museum on the continent dedicated to contemporary African art. The converted grain silo from the 1920s is an artwok in itself. (22.09.2017)
African art: a unique investment opportunity
Author Julia Jaki (db)
Related Subjects South Africa
Keywords South Africa, art
Permalink https://p.dw.com/p/2kTpM
South Africa: the state of the arts 25 years after apartheid 08.05.2019
South Africa's young democracy is viewed throughout the world as an exemplary pluralistic rainbow nation. But as the country approaches its most important elections in 25 years, does art imitate life?
Welcome to the latest edition of Eco Africa 28.06.2019
On this week's special edition of Eco Africa, we meet eco heroes including a photographer who launched a campaign to ban plastic bags in Kenya and a man who's been planting trees in Burkina Faso to stop desertification.
What a Drag! The Queen of Cape Town 04.06.2019
A look at South Africa's stand-out queer scene: Odidi Mfenyana adopted the on-stage drag persona "Odidiva the African Queen" decades ago. He is now a larger than life celebrity in Cape Town's fashion, art, and gay community.
'Oh Boy' director Jan-Ole Gerster finally follows up with 'Lara'
Following his hit Berlin-set debut "Oh Boy," director Jan-Ole Gerster presents his new film "Lara," an unsettling portrait of a desperate woman, at the Munich Film Festival. He spoke to DW about the pressure to succeed.
Hard times for democracy, says historian Timothy Snyder
Is liberal democracy at its end? How should we deal with the rise of populism fueled by fake news on the internet? Yale historian and world-renowned author Timothy Snyder discusses these burning issues with DW.
Cosmic sound - Brian Eno
British musician Brian Eno has composed music for films and space. An asteroid has just been named for him. He's reissuing his legendary album "Apollo" to coincide with the 50th anniversary of the moon landing. A talk about the sound of the universe:
A festival of stars
The Starmus Festival has seen greats such as Hans Zimmer, Brian May, Neil Armstrong and Stephen Hawking. This global festival of science communication and art is taking place in Zurich for the fifth time this year. The star of 2019 is the moon.
YouTube in schools: A digital revolution in the classroom
Teachers and parents might not have noticed, but students don't only use YouTube for fun. According to a new study, about half of them watch videos to learn things as well. How does this affect the educational program?

redpajama_set_name: RedPajamaCommonCrawl


text:
The David Suzuki Foundation Reports That The Majestic Whooping Crane Faces An Uncertain Future
Whooping cranes follow an ultralight airplane on a 1,218-mile migration route from Necedah National Wildlife Refuge to Florida.
It's Spring! The cranes and Canadian geese who wintered in southern U.S. states are flying north to their homes in Canada. If you are lucky enough to live along the flyways, enjoy the magnificent sight of flocks flying overhead.
The whooping crane has long been the flagship of the conservation movement. The majestic bird—North America's tallest—flies more than 8,000 kilometers each year, breeding in northern Alberta and wintering in marshes along the Texas coast. Still, its future remains uncertain.
Today we think of species becoming endangered when sprawling big box stores and subdivisions displace remnant natural areas within and around our cities or when large-scale industrial resource extraction fragments wilderness areas.
Whooping cranes have a longer history. In the late 1800s, Canadians began farming the prairies, expelling cranes from their breeding habitat. At the same time, Americans drained west coast marshes, where whooping cranes overwintered. Hunting also contributed to their decline.
North America was once home to more than 10,000 cranes. By 1938, the population reached an all-time low of 14 known adults.
In the early 1940s, the conservation movement leapt into action.
Over seventy years later, after significant breeding-area protection, captive breeding, aircraft-led migration and relocation efforts, the whooping crane has made a slow, often tenuous comeback.
Today there are several captive-bred and non-migratory populations, but only one—shared by the U.S. and Canada—that is self-sustaining and wild. That group breeds every year in Wood Buffalo National Park, which straddles the Northwest Territories and Alberta, and winters in Texas, flying about 4,000 kilometers each way. Although the population has been steadily increasing—there were 74 nests counted in Wood Buffalo National Park in 2013 and a few breeding pairs outside the park—it still faces a number of perilous risks in Texas and Alberta—not un-coincidently, the oil kings of North America.
Despite the many years and millions of dollars dedicated to the recovery of the whooping crane, continued habitat degradation darkens its recovery horizon. According to COSEWIC, Canada's independent science assessment body of at-risk species, U.S. chemical spills are one of the biggest threats: "…the greatest concern is in the Gulf Intracoastal Waterway on the Texas coast. Numerous oil and gas wells and connecting pipelines are located in bay and upland sites near the cranes' winter habitat, and many barges carrying dangerous, toxic chemicals travel the Gulf Intracoastal Waterway daily through Whooping Crane winter habitat. A spill or leak of these substances could contaminate or kill the cranes' food supply or poison the cranes."
Although the birds have a national protected area in Canada to breed, they still need to survive along their migration and stopover route—right over the Alberta tar sands. One of the most serious threats to the population is juvenile birds dying during their migration from Wood Buffalo to Texas. They risk landing in toxic tailings ponds, flying into power lines and exposure to water and air contaminated by the bitumen extraction process.
The fact that there is only one self-sustaining wild population of crane heightens the risk of a single, catastrophic event.
Crane recovery efforts cannot be done in isolation. National parks that protect Canadian breeding areas are important, but insufficient if the cranes can't safely make their way back to Texas for their blue crab feast.
For the whooping crane—and for Canada's over 300 other at-risk species—we need to find ways to maintain healthy, functioning ecosystems even where development occurs.
Source: Our friends at the David Suzuki Foundation – "solutions are in our nature". <http://davidsuzuki.org/issues/wildlife-habitat/science/species-at-risk-act/majestic-whooping-crane-faces-uncertain-future/>


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text:
Digital Glass Makes an Impact
Posted July 21st, 2017 by Sean Brown & filed under Technology, Press Release, General.
Largest Rigid Rear Projection Screen in the Middle East Installed in Dubai Mall
As with all projection applications, there is more to a cutting edge display than just the projector. An effective display solution relies on several factors that are essential to maintain a setup that serves your requirements to the best standard possible. One such component is the projection screen, something which Pro Display and Eclipse Installations know all too well. One of Eclipse Installations' most recent projects required an innovative alternative to conventional large video walls in order to avoid the dissecting bezels that cuts content into parts. The Chalhoub Group were seeking a single screen solution for their luxury shoe store, Level Shoes in Dubai Mall to display high quality content, without the usual join lines that are so often seen on traditional video walls.
Pro Display rose to the challenge by supplying our unique optical screen technology for ultra-short throw rear projection applications – Digital Glass. Digital Glass is a rather special screen that displays incredibly high-quality images. Manufactured using a special process that coats our advanced proprietary optical projection layer directly onto ultra-clear glass or cast acrylic, Digital Glass is a cost effective large format Ultra HD screen that is specifically designed for Ultra Short Throw applications, allowing it to be used in areas where space is limited and at a premium. In this case, it served as a high-quality alternative to video walls alongside an Epson EB-L1405U projector, which is a large-venue projector that combines a laser light source and 3LCD technology for powerful, uncompromising images, meeting The Chalhoub Group's requirements.
Epson comments, "Eclipse Installations' concern was to find a trusted vendor that could provide a high-quality projector for this prestigious project. They decided to meet with Epson's team in Dubai to understand whether they could provide an apt solution and, for the Rear Projection surface, the UK based company Pro Display." Pro Display's reputation in AV innovation is one to behold. We pride ourselves on our commitment to the development of our technologies, operating for over 25 years in optical screen Research & Development. Digital Glass falls amongst one of our newest innovations, catching the interest of major projector manufacturers all over the world for its outstanding image quality and performance.
David Fell, the director of Eclipse Installations, comments on the collaboration, "clients come to us because they know we source the very best equipment possible for the solutions we design and, with that, that we will not compromise on performance or quality; choosing to work with Epson and Pro Display for a project like this was therefore a natural choice." He adds, "Quality and performance have always been our criteria when choosing our technology partners."
Pro Display are proud to have contributed to this amazing project, and look forward continuing to collaborate with Eclipse Installations and Epson in the future. Ben Hadfield, Pro Display's sales lead on this project explains "Pro Display have a great working relationship with partners. We have worked hard over the years to build a solid global reseller network which sees our products sold all over the world. We are delighted to be able to work alongside companies such as Epson and Eclipse Installations at each stage of a project's life, allowing us to be a part of some of the world's most exciting and pioneering AV installations, as well as putting us in a position to offer advice and support, ensuring the project is completed to the highest of standards."
Testing this Digital Glass screen produced amazing images
Getting ready to load the Digital Glass screen for dispatch
Pro Display offers a solid range of display focused products that will excite your customers. In addition to the product mentioned in this article, there is a wealth of information about our audio/visual options available to you on our website. If you have some ideas about how you would like to integrate our technology into your displays, please get in touch using the form below, or call us on +44 (0)1226 740 663.
Tags: Retail, Window Display, Digital Signage, Projection, Digital Glass
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text:
Q: How to select unique values from a column using np.unique in python I have the following dataframe called Trees.
FieldSP Common_name     FIA_code   DBH          Height     Year
BP      Bishop Pine     120        11.3435      45.76023   2014
BP      Bishop Pine     120        21.17852     126.86418  2014
CB      CA Bay/Laurel   981        21.19289     114.02973  2011
CB      CA Bay/Laurel   981        24.73787     138.85386  2011

The following lines of code are my attempts at getting the unique value for the Year column. There is only one value in this column that is the same for all rows in the dataframe.
Year = np.unique(Trees['Year'])

Year = Trees['Year'].iloc[0]

Year = np.unique(Trees[Trees.Year])

if Year == 2014:
    Trees.loc[(Trees['FieldSP'] == 'RW') & (Trees['DBH'] <= 39.37), 'FieldSP'] = 'Rwa'
    Trees.loc[(Trees['FieldSP'] == 'RW') & (Trees['DBH'] <= 39.37), 'FIA_code'] = '211a'
    Trees.loc[(Trees['FieldSP'] == 'RW') & (Trees['DBH'] > 39.37), 'FieldSP'] = 'Rwb'
    Trees.loc[(Trees['FieldSP'] == 'RW') & (Trees['DBH'] > 39.37), 'FIA_code'] = '211b'
elif Year == 2011:
    print('COP 2011 has not been added yet!')
    quit()
else:
    print('the COP year in the treelist is not one we know about. fix it')
quit()

This line is being run for multiple dataframes, all with different years (but the year is constant for each dataframe). So I don't want to have to hard code in the value for the year.
I keep getting the following error message when running my script. So it seems that np.unique is creating an empty array. And I don't know how to fix this. Is there a way to select the unique value from the Year column?
postfse.py:184: DeprecationWarning: The truth value of an empty array is ambiguous. Returning False, but in future this will result in an error. Use `array.size > 0` to check that an array is not empty.
  if Year == 2014:
postfse.py:189: DeprecationWarning: The truth value of an empty array is ambiguous. Returning False, but in future this will result in an error. Use `array.size > 0` to check that an array is not empty.
  elif Year == 2011:
the year in the treelist is not one we know about. fix it

A: Replace
Year = np.unique(Trees[Trees.Year])
with Year = np.unique(Trees['Year'])
That is if you have to use numpy. I'd just delete the line of code instead, considering I already have Year.
Your question may not be giving the the entire picture of the problem. Are you getting into a situation where a dataframe has no rows?

redpajama_set_name: RedPajamaStackExchange


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text:
Weekly Roundup...4 January 2022
Published on Jan 4 2022 1:14 PM in Retail tagged: Jumbo / Ahold Delhaize / Lidl Switzerland
Ahold Delhaize has announced that it has commenced the €1 billion share buyback programme.
Ahold Delhaize has commenced the €1 billion share buyback programme announced on 15 November 2021. The company expects to complete the programme before the end of 2022. The share buyback programme aims to reduce the capital of Ahold Delhaize by cancelling all or part of the common shares acquired through the programme. The company focuses on maintaining a balanced approach between funding growth in key channels and returning excess liquidity to shareholders to support the financial framework of its 'Leading Together' strategy.
Dutch retailer Jumbo has announced a strategic partnership with Gorillas in the Netherlands and Flanders, focusing on instant delivery of groceries ordered online in larger cities. The partnership will allow both companies to benefit from each other's expertise and strengthen their market positions. The agreement will see Jumbo supplying a large part of its range to Gorillas, including various a-brand, private-label, organic, and exclusive La Place products. The deal also has provision for delivering orders within minutes in urban areas in the future. Currently, Gorillas operates in eleven Dutch cities and two Belgian cities. Commenting on the partnership, Jumbo CEO Frits van Eerd, said, "Like Jumbo, Gorillas stands for a high level of service and distinguishes itself through a customer-oriented approach. Thanks to this collaboration, we can further expand our service and increase the choice for our online customers."
Lidl Switzerland has introduced Nutri-Score labelling for certain own-label products to help consumers make more informed purchases and opt for healthier food options. The voluntary labelling indicates the nutritional balance of a product's composition on a scale of A to E, and has been widely used in France since its introduction in 2017. In Switzerland, the Federal Office of Food Safety and Veterinary Affairs (USAV) came out in favour of the Nutri-Score in 2019. Now, Lidl Switzerland has rolled out the label for its 'Harvest Basket' and 'Trattoria Alfredo' private-label brands, with plans to expand it to other product areas.
Weekly Round Up... 8 June 2021
With Global Motor Neurone Disease (MND) day just around the corner the Irish Motor Neurone Disease Association (IMNDA) has announced its plans to host a virtual Drink Tea for MND event on 21 June. If...
Weekly Roundup... 13 April 2021
A leading trade union representative has described the planned store reorganisation strategy at Dutch retailer Jumbo as "ridiculous", according to reports, suggesting that job or wage cuts are likely...
Did you know?... 04 August, 2020
Did you know?... SuperValu has announced that its Signature Tastes ClemenGold box of mandarins has switched to sustainable packaging. The South African ClemenGold fruit are now boxed in recyclable car...
Weekly Round-Up... 10 September, 2019
Spar has recently teamed up with the Republic of Ireland senior international team to offer four lucky school children, and four of their closest friends, the opportunity to meet with and be trained b...


text:
Hancocks to move to new level after management buy-out
By Gaelle Walker2012-11-26T09:00:00+00:00
The 50-year-old family-owned cash and carry Hancocks has been sold to its management and H2 Equity Partners (H2) in a move which is hoped to "take the company to the next level".
Established in 1962 by Ray Hancock, the confectionery specialist wholesaler operates 18 cash and carry depots and has a turnover in excess of £100m. The current owners, Ray's sons Andrew and Adrian Hancock, will be stepping down from the company.
Former owner and chairman Andrew Hancock said the buy-out was "great news" for customers, employees and suppliers.
"From a sweet shop in Shepshed, our family has grown the business to a group that serves many thousands of businesses with their confectionery supplies," he said. "We now hand the baton on with pride, knowing that Hancocks can only go from strength to strength."
While current supplier arrangements will remain in place, plans to expand the national depot network, develop both the branded and own label ranges and grow online activities will be drawn up over the coming months.
Mark Watson, who will remain as chief operating officer, said he was "confident of a bright future" for the confectionery company. "We have some exciting plans and opportunities to take the business forward with H2 at the helm," he added.
Blakemore sales dip due to Covid-19 disruptions
The group saw a 5% dip in sales for the year to 24 April.
Parfetts expands Aintree depot to improve delivered offer
Parfetts has completed an extension of its Aintree depot as part of its growth strategy to increase its delivered offering.


text:
Makae's poor form persists
Tumelo Makae
LESOTHO'S top mountain bike (MTB) cyclists Tumelo Makae continued with his poor form after failing to rise to the occasion during the 2021 Otztaler Cross Country Olympic MTB Festival in Austria last weekend.
Makae failed to finish the race only completing three out of seven laps. The performance saw him 59th out of 70 riders in the men's elite category. This was Makae second official race this year.
His first race was a fortnight ago in the 2021 Verona MTB International Cup in Italy where he came 53rd out of 190 in the senior men's category. He also failed to finish the race managing just three of the eight laps.
Makae is based in Switzerland where he is under a Lesotho Tokyo 2020 Olympic Athletes Solidarity Scholarship funded by the Lesotho National Olympic Committee (LNOC).
His scholarship to train at World Cycling Training Centre (WCTC) commenced in 2017 and was expected to end in August last year. However, it was extended until the forthcoming Tokyo Olympics. This after the Olympics which were initially set for July and August last year were postponed to this year on the back of the Covid-19 pandemic.
Federation of Cycling Lesotho (CFL) public relations officer Malefane Morie said he was unsure whether or not Makae's form has been affected by the high level of competition.
"It is true that for the past two recent races he didn't perform well but I am not in a position to say whether it the completion was too tough for him or the problem is with the rider himself. I think his coach is better placed to comment on that.
"For me, I don't think the competition has been too tough for him based on the knowledge that we sent him to Switzerland because he is our top rider. He has been in Europe since 2017 and my understanding is that he should now be better equipped to compete at that level," Morie said.
Moshe announces CFL presidency bid
LNOC donates to athletes


text:
"The CDF is doing great harm to the Church"
It is now over five years since the Vatican first got their claws into me. Most of the time I believe I have moved on, that my life has been interesting for those years, that I have moved into a different world, got to know many interesting people. But once in a while a wave of anger comes over me. I know that anger, if it gets a grip on a person, is destructive, so I usually manage to let it go. The anger is never to do with me being censored or forbidden to exercise my ministry as a priest. It always has to do with the unjust procedures of the Vatican, and the fact that no Church authority, bishop or religious superior, is willing to highlight this publicly.
Every now and again I find it helpful to write a little about it. So here goes.
A Synod of Bishops has been recognised, since the early Church, as one of the main sources of wisdom and authority for the Church. In the last couple of centuries Synod have been neglected, as more and more emphasis was put on the authority of the pope. But the Vatican Council, in the middle of the last century, reinstated Synods of Bishops. Unfortunately, while they were held occasionally during the papacy of John Paul and Benedict, they were so tightly controlled that they became meaningless. As we know, Francis is trying hard to put them back into a central place in the life of the Church.
But there was a good one held a few years after the Council, in 1971. It produced an excellent document on Justice in the World. I quote below from this document, written forty six years ago:
"While the Church is bound to give witness to justice, she recognizes that anyone who ventures to speak to people about justice must first be just in their eyes. Hence we must undertake an examination of the modes of acting and of the possessions and life style found within the Church herself. Within the Church rights must be preserved. No one should be deprived of his ordinary rights because he is associated with the Church in one way or another.
We urge that women should have their own share of responsibility and participation in the community life of society and likewise of the Church. We propose that this matter be subjected to a serious study employing adequate means: for instance, a mixed commission of men and women, religious and lay people, of differing situations and competence.
The Church recognizes everyone's right to suitable freedom of expression and thought. This includes the right of everyone to be heard in a spirit of dialogue which preserves a legitimate diversity within the Church.
The form of judicial procedure should give the accused the right to know his accusers and also the right to a proper defense. To be complete, justice should include speed in its procedure.
In other words, as the last piece that I have highlighted shows, it is official Church teaching that everyone has a right to freedom of expression and thought, and a right to be heard. An accused person should know his accusers, and have the opportunity to defend him/herself. And justice should be done speedily. In all these ways, in my dealings with the Vatican, they have blatantly violated their own teaching.
Of course this 1971 teaching was not knew. The Fourth Lateran Council in 1215 said much the same thing:
He who is the object of an enquiry should be present at the process, and, unless absent through contumacy, should have the various headings of the enquiry explained to him, so as to allow him the possibility of defending himself. As well, he is to be informed not only of what the various witnesses have accused him of, but also of the names of those witnesses. (Fourth Lateran Council, 1215)
Like most others in my situation, I still do not know who my accuser was, even though I understand it was someone in a high position in the Irish Church.
My experience and contact during these recent years have taught me that I am by no means the only one who has suffered in this way at the hands of the Church, there are, and have been in the past, countless others.
The Marist priest and theologian,Sean Fagan, who died last July, was a person who suffered greatly at the hands of the Vatican. The quote below is from an address he gave to Catholics for a Changing Church in 2009. Because of the pressure he was under, he requested that it would not be published until after his death. I read it in the latest issue of their magazine RENEW.
"The Vatican ignores basic human rights in its procedures; the right to be heard, the right to know your accusers, the right not to have the same individuals as prosecutor and judge. Those accused are never addressed personally, but only through their superiors, who can command them to silence. The CDF is doing great harm to the Church."
The main impact of all of this for me is that I have very little respect for the Vatican establishment any more. I do know Francis is trying to change things,and I hope he will have some success, though clearly the opposition is strong. Unless there is real change in the Vatican, and particularly in the Congregation for the Doctrine of the Faith, it is hard to see the Catholic Church regaining any of the credibility it has lost in the last twenty of thirty years. Sean Fagan was right. The CDF is doing great harm to the Church
I am further disillusioned by the fact that bishops and religious superiors, including my own, are unwilling to stand up and name this blatant and damaging behaviour of the Vatican. What price staying silent?
The Papal Nuncio is Departing The Resurrection and the Consecration.


Ashley Reda earned her undergraduate degree from Louisiana State University. She holds a Bachelor of Science Degree in Kinesiology, and a Master of Science in Physician Assistant Studies from University of Alabama at Birmingham. Ashley is certified by the National Commission on the Certification of Physician Assistants. She is a member of the American Academy of Physician Assistants, Louisiana Academy of Physician Assistants, and the American Association of Surgical Physician Assistants.


text:
Business|Beyond 3-D Printers' Magic, Possible Legal Wrangling
https://nyti.ms/1e8sZay
Beyond 3-D Printers' Magic, Possible Legal Wrangling
By Phyllis Korkki
When reports first appeared that computers could produce three-dimensional objects — from toys to auto parts to household items — it sounded like a page from a science fiction novel.
But the era of 3-D printers is upon us. For a mere $1,299, plus shipping, you can even buy one from Staples to use at home.
There's still a gee-whiz aspect to the technology, but once that fades away, it's likely to set off something else: lawsuits. That warning comes from two law professors in a paper to appear early next year in The Georgetown Law Journal.
The 3-D printing "will do for physical objects what MP3 files did for music," wrote Deven R. Desai, associate professor at the Thomas Jefferson School of Law, and Gerard N. Magliocca, professor at the Robert H. McKinney School of Law at Indiana University.
Using computer modeling software, 3-D printers can reproduce objects using layers of materials like rubber, plastics, ceramics and metals. Some websites share software to build these objects; the attitude of many of the software makers is: "I designed this cool thing, and I want you to be able to print it," Professor Desai said in an interview.
Credit...Michael Waraksa
But just as people copy music files, it seems probable that they will do the same with objects — a tool, say, or a piece of furniture that may be covered by a patent. All patents are available to the public, and it would be possible for a knowledgeable person to pore over a patent file and create software that can reproduce the invention described, Professor Desai said. Also, 3-D scanners can scan some objects and translate them into computer models, to be modified or printed.
So what is a patent owner seeking to stop an infringement to do, given that tracking down people in their homes would be extremely difficult?
One option would be to go after the makers of the printing hardware, but that would be a misguided approach centered on a general-purpose technology with many legal uses, Professor Desai said. Patent holders could also sue the websites that host the software that enables the printers to manufacture the objects, but this, too, could stymie perfectly legal inventions and end up putting a stranglehold on innovation, he said.
Just as record companies were unable to stop music file-sharing, manufacturers will not be able to prevent the proliferation of 3-D printing, he said. While violation of patents is a concern, and there may be ways to sue some individual lawbreakers, the best way to handle this threat, he said, may well be to embrace the new technology and the new markets it opens.
People who use unauthorized music-sharing sites know that the files they download may be poor in quality or corrupt, or even contain viruses; that's why they are willing to pay for their music on sites like iTunes. Similarly, manufacturers can set themselves up as authorized dealers for 3-D software and material, Professor Desai said, so that "consumers would know they were getting a trusted product."
A main advantage of 3-D printing is that users can customize items to their personal needs — for example, by adjusting the sizes and shapes of parts. Manufacturers could customize their mass-market products for people using 3-D printers and promote them as having superior quality, Professor Desai said.
Is the government likely to take an aggressive approach toward 3-D printing violations? That's hard to know, but past efforts by the government to stop illegal taping of movies and television shows, along with illegal downloading of music, have not been very effective, and the same seems likely to be true of 3-D printing, Professor Desai said. The march of technology is just too insistent.


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Oedipus at Colonus
by Donald L. Wasson
Oedipus at Colonus was the third play of the Oedipus trilogy written by the great Greek tragedian Sophocles (c. 496 - c. 406 BCE). Although written in the years prior to his death, it would finally be presented by his son Iophon at a dramatic competition in 401 BCE. The play's sequel Antigone was actually written years earlier in 441 BCE. Oedipus at Colonus accounts for the final years of the fallen king, 20 years after his exile from Thebes. Blind, weak and dressed in rags, he accepted his fate and wandered from town to town as an outcast accompanied only by his young daughter Antigone. Arriving outside Athens at Colonus, he is befriended by the king of Athens, Theseus, who offers him protection. Oedipus speaks of a prophecy that says whatever city grants him sanctuary will be given special protection. Knowledge of this prophecy comes to the attention of Creon, his brother-in-law, and his son Polyneices who want to take advantage of the blinded king. Both had coldly refused him shelter in the past but now travel to Colonus to offer him sanctuary.
Along with Aeschylus and Euripides, Sophocles represents the greatest of the Greek playwrights. In the 4th and 5th centuries BCE, Greek tragedians performed their plays in outdoor theaters at various festivals and rituals in a series competitions. The purpose of these tragedies was to not only entertain but also to educate the Greek citizen, to explore a problem. Along with a chorus of singers to explain the action, there were actors often three (later four or more and always male) who wore masks and costumes. Although he was often considered a passionless observer of life, classicist Edith Hamilton in her book The Greek Way believed Sophocles was the embodiment of what we believe to be Greek. "He is direct, lucid, simple, reasonable. Excess - the word is not to be mentioned in his presence. Restraint is his as no other writer's" (199).
by Mark Cartwright (CC BY-NC-SA)
Sophocles was born into a wealthy family in the small Athenian suburb of Colonus, the site of his play. He was active in Athenian public life serving as treasurer, general and envoy. Later, because he was so respected, he became a member of a group of magistrates given the task of organizing finances and domestic affairs after the disastrous defeat of Athens at Syracuse (412-411 BCE). Although active in Athenian political circles, his plays rarely contain any references to current events or issues, making the dating of his plays difficult. He had two sons (one by his wife and one by his mistress); Iophon and Ariston both became playwrights.
The myth
The purpose of these tragedies was to not only entertain but also to educate the Greek citizen.
All three plays are based on an old myth which was quite familiar to the audience. In Oedipus the King, Oedipus fulfills a prophecy that foretold that he would murder his father and marry his mother. At Oedipus' birth, his father, hoping to avoid his own fate, had his son taken away. Providence, however, interceded, and the young boy was raised by the king and queen of Corinth. Unfortunately, Oedipus learned of the prophecy and unknowingly believing it meant his adoptive parents, he left Corinth. On route to Thebes, he got into an altercation and killed a man who happened to have been the Theban king, his real father. Upon arriving in the city, he saved the people by solving the riddle of the Sphinx and was rewarded by marrying the now-widowed queen (his mother). Years later when the truth was finally revealed, the queen committed suicide and Oedipus blinded himself and went into exile. Although Sophocles chose to exile Oedipus, various authors have viewed the outcome differently; both Homer and Euripides have him remain in Thebes although relinquishing the throne. Now, in Sophocles' final play the story continued as Oedipus realized that he was near death and decided to choose a final resting place.
Oedipus at Colonus has a rather small cast of central characters:
Oedipus
his daughter Antigone
his other daughter Ismene
Polyneices
a messenger
and, of course, the chorus.
To many, the Oedipus in Oedipus at Colonus is not the same character as in Oedipus the King. Paul Roche in his translation of the Oedipus plays believed in Oedipus at Colonus the former king finally accedes to his fate and thereby rises to a restored sense of dignity, becoming a vehicle of divine justice. To Michael Grant in his The Classical Greeks, the storyline of Oedipus has been modified so that the former king does not sin but has been sinned against. Editor Moses Hadas in his book Greek Drama said that at the time of the play's writing Athens was beginning its gradual decline, and Sophocles returned to the story of Oedipus as a farewell to both the city and the stage. Theseus is included in the play "to glorify Athenian traditions of justice and generosity…" (167).
The play begins at a grove in Colonus just outside Athens. As Antigone and Oedipus enter the grove, the former king turns to his daughter and says:
My daughter - daughter of the blind old man - where have we come to now, Antigone? What lands are these or holdings of what city? Who will be kind to Oedipus this evening and give alms to the wanderer? (145)
Recognizing the area as Athens, she describes the grove to her father and guides him to a large, flat rock. A lone stranger approaches from a distance and reprimands Oedipus for sitting on the rock, for it is considered to be holy, an area dedicated to the master horseman Colonus. He tells them that they must move and informs them that the land is governed by Theseus, son of Aegeus. Oedipus requests that the stranger go and seek Theseus, telling him that a little kindness will gain him much. After the stranger leaves, Oedipus turns to Antigone and says:
I have first bent my knees in this new land, therefore be mindful of me and of Apollo. For when he gave me oracles of evil, he also spoke of this, a resting place I should find home among the sacred Furies; that I might round out there my bitter life. (150)
He tells her that the place that accepts him will receive benefits but the ones that have driven him away will be cursed. After Antigone and Oedipus move away from the sacred rock, the chorus enters (they had been sent by the stranger) and, upon seeing the father and daughter, immediately asks the old king and daughter to come forward. They tell Oedipus to identify himself. Oedipus only replies that he is an exile and begs them not to ask anything more. After further questioning, he finally admits that he is the ruined Oedipus and adds that he is cursed. Their reaction is immediate: "Away with you? Out with you? Leave our country." (156) Oedipus pleads with them to give him shelter. Antigone speaks on her father's behalf asking them to have mercy on a beaten man. Understanding Oedipus' plea, they decide to leave the matter in the hands of Theseus.
Theatre Masks
As they wait for Theseus' arrival, Ismene, Oedipus' second daughter, appears, bringing news from Thebes. Her brother Polyneices has been betrayed, stripped of his power and exiled, fleeing to Argos where he hopes to raise an army and return to conquer Thebes. With the problems brewing in Thebes, Ismene wonders how long it will be before the gods have pity on him. She speaks of a prophecy from the Delphi oracle that says "… you shall be much solicited by our people before your death - and after - for their welfare" (163). She adds that the gods that once threw him out will now sustain him. Over time Oedipus has realized that he had acted hastily and now understands that his punishment has been far greater than his crime. Ismene warns him that Creon is on his way to take him back to settle him near Thebes. Hearing the conversation between Ismene and Oedipus, the chorus leader speaks:
Oedipus, you are surely worth our pity: you, and your children, too. And since you claim also to be a savior of our land, I'd like to give you counsel for good luck. (167)
Soon after Ismene leaves, Theseus arrives at the grove and immediately recognizes Oedipus. He wonders what the old blinded king wants. Oedipus speaks of his sons and the family curse and how the Delphi oracle compels them to summon him. Realizing his plight, Oedipus asks only to remain in Athens.
I was expelled from my own land by my own sons, and now, as a parricide, my return is not allowed. [...] Here I shall prevail over those who banished me. (174, 176)
Theseus promises that he will not betray Oedipus and no one will be allowed to take him against his will. Soon after Theseus leaves, Creon approaches with a number of soldiers. He advances towards Oedipus, Antigone, and the chorus. He turns to Oedipus and asks him to come home:
I see from your eyes that my arrival has been a cause of sudden fear to you. Do not be fearful and say nothing hostile. I have not come for any hostile action. [...] I grieve for your unhappiness, old man. I see you ravaged as you are, a stranger everywhere, never at rest with only a girl to serve you in your need. (180-181)
He says that poor Antigone has been leading a beggar's life, never knowing marriage; it is a disgrace. Of course, Oedipus does not believe him, calls him a rascal, and asks him to leave him and his daughter alone. However, Creon gives him a dire warning and informs him that he has already taken Ismene captive. His soldiers grab Antigone. The soldiers drag Antigone away, and as Creon grabs Oedipus, Theseus arrives with a small army of his own. He is told that Antigone and Ismene have been taken prisoner. Turning to Creon, he warns him:
You shall never leave this land of Attica until you produce those girls here in my presence. (190)
Creon responds saying he did not know Athens was a place for exiles. He claims that Oedipus had placed a curse on him and his people; it was an injury that deserved retaliation. Theseus tires of Creon's excuses, and shortly after Theseus and Creon leave, the Athenian king returns with both Antigone and Ismene. Oedipus is elated:
I have what is dearest to me in the world to die now would not be so terrible since you are near me. (197)
However, his happiness is short-lived, for he received word that a man claiming to be a relative has been asking about him in Athens. Theseus said the man was found praying at the altar of Poseidon. Oedipus realizes that it must be his son Polyneices. Antigone begs his father to listen Polyneices, and Oedipus agrees. Polyneices arrives and addresses his father. He tells them that he is a fugitive, too. He was banished from Thebes and had found asylum in Argos, where he hopes to raise an army and return to Thebes. He asks for forgiveness and begs his father to come with him. He will defeat his brother and bring Oedipus home to Thebes. Oedipus rejects his offer and informs him that he will not defeat his brother. Polyneices will die by his brother's hands, and his brother will die by his.
I abominate and disown you wretched scum. Go with this malediction I here pronounce for you that you will never master you native land by force of arms. (207)
As Polyneices departs, Oedipus and his daughters hear thunder. Oedipus realizes that the end is near and asks for Theseus in order to give him his blessing and fulfill his promise for Athens. The thunder is from Zeus, and it is a sign that it is time for him to die. Speaking to his daughters, he tells them that he will take only Theseus with him to the place where he will die, and adds that the Athenian king must not disclose it to anyone. Oedipus takes his daughters and Theseus to where the angel of death, Hermes, and Persephone lead him.
Greek Tragedy Mask
Shortly, a messenger arrives to speak to the chorus. Oedipus is dead. He tells of the old king's final moments. With his daughters weeping, he told them they no longer had to bear the burden of caring for him, but he reminded them that they were not to witness his death. Theseus was the only one permitted to stay and see his final resting place. Later, when Theseus returned to a solemn Antigone and Ismene, they asked if they could see his final resting spot, but he said no. Antigone regrettably accepts her father's last wishes and simply adds that she will return to Thebes with a hope to stop the war between her brothers.
Oedipus redeemed
As he approached his own death, Sophocles wrote the play as the long and unpopular war between Sparta and Athens was being fought and Spartan forces sat on the edge of the city. The play was a final homage to his city and beloved stage. In Oedipus at Colonus the blinded, fallen king dies. After 20 years as an outcast, Oedipus had arrived at Colonus with the hope that his odyssey was finally at an end. However, his death was not to be in vain. He was befriended by Theseus who was promised that Athens would receive special protection for having granted him sanctuary. Unfortunately, he learned that he was being pursued by his old nemesis Creon and his treacherous son Polyneices. In the end, he and his daughters were saved, and he finally found the rest that he had sought. In Oedipus at Colonus, he came to realize that his suffering far exceeded his crime. He had acted in haste. Now, as he approached his death, he understood that he was not the victimizer but a victim. He was able to die with a restored sense of dignity.
Grant, M. The Classical Greeks. (Scribner, 1989).
Hadas, M. Greek Drama. (Bantam Classics, 2013).
Hamilton, E. The Greek Way. (W. W. Norton & Company, 2017).
Hornblower, S. The Oxford Classical Dictionary. (Oxford University Press, 2012).
Martin, T.R. Ancient Greece. (Yale University Press, 2013).
Roche, P. The Oedipus Plays of Sophocles. (Mentor, 1963).
Sophocles. Sophocles I. (University of Chicago Press, 2013).
Donald L. Wasson
Donald has taught Ancient, Medieval and U.S. History at Lincoln College (Normal, Illinois)and has always been and will always be a student of history, ever since learning about Alexander the Great. He is eager to pass knowledge on to his students.
Ancient Greek Literature
Greek literature has influenced not only its Roman neighbors to...
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Oedipus the King (429-420 BCE), also known as Oedipus...
Sophocles of Kolōnos (c. 496 - c. 406 BCE) was one of the most...
Oedipus & the Sphinx of Thebes
A red-figure tondo showing Oedipus and the sphinx of Thebes. Vulci...
Sophocles I: Antigone, Oedipus the King, Oedipus at Colonus
published by University of Chicago Press (19 April 2013)
Theban Plays, The: "Oedipus the King", "Oedipus at Colonus", "Antigone"...
Everyman (20 January 1994)
THE THEBAN PLAYS:KING OEDIPUS,OEDIPUS AT COLONUS,ANTIGONE.Translated...
Penguin Books (20 January 2020)
Three Theban Plays:
OUP USA (20 January 1963)
The Theban Plays: "Oedipus the King", "Oedipus at Colonus", "Antigone"...
Translated by Ruth Fainlight, Translated by Robert J. Littman By (author) Sophocles
JOHNS HOPKINS UNIVERSITY PRESS (20 January 2020)
Wasson, D. L. (2018, March 08). Oedipus at Colonus. Ancient History Encyclopedia. Retrieved from https://www.ancient.eu/Oedipus_at_Colonus/
Wasson, Donald L. "Oedipus at Colonus." Ancient History Encyclopedia. Last modified March 08, 2018. https://www.ancient.eu/Oedipus_at_Colonus/.
Wasson, Donald L. "Oedipus at Colonus." Ancient History Encyclopedia. Ancient History Encyclopedia, 08 Mar 2018. Web. 20 Jan 2020.
Written by Donald L. Wasson, published on 08 March 2018 under the following license: CC BY-NC-SA. This license lets others remix, tweak, and build upon this content non-commercially, as long as they credit the author and license their new creations under the identical terms. Please note that content linked from this page may have different licensing terms.
c. 496 BCE - c. 406 BCE
Life of Greek tragedy poet Sophocles.
Sophocles' tragedy 'Oedipus the King' is first performed.
Sophocles' tragedy 'Oedipus at Colonus' is first performed.


text:
El Saneal es una ranchería del municipio de Navojoa ubicada en el sur del estado mexicano de Sonora, en la zona del valle del río Mayo. Según los datos del Censo de Población y Vivienda realizado en 2010 por el Instituto Nacional de Estadística y Geografía (INEGI), El Saneal tiene un total de 1,164 habitantes.

Geografía
El Saneal se sitúa en las coordenadas geográficas 26°54'38" de latitud norte y 109°23'08" de longitud oeste del meridiano de Greenwich, a una elevación de 62 metros sobre el nivel del mar.

Véase también

 Municipio de Navojoa
 Anexo:Localidades de Sonora

Referencias

Localidades del estado de Sonora

redpajama_set_name: RedPajamaWikipedia