text:
\section{Introduction}

Having a ring $R$ with two operations $+$ and $\cdot$ one can define the {\it sumset} of sets $A,B \subseteq R$ as
\[
	A+B = \{ a+b ~:~ a\in A,\, b\in B \}
\]
and, similarly,  the {\it product} set
\[
	AB = \{ a\cdot b ~:~ a\in A,\, b\in B \} \,.
\]
The {\it sum--product phenomenon} (see, e.g., \cite{TV}) predicts that additive and multiplicative structure cannot coexist up to some natural algebraic constrains.
This can be expressed in many different ways see, e.g., \cite{Bourgain_more}
and in our paper we consider just  one of them.
Let us formulate a particular case of the main result, which is contained in Theorem \ref{t:AP_in_G}
from Section \ref{sec:f=log}.

\begin{theorem}
		Let $S \subset \Z$ be a finite set, $l$ be an integer number, and let $\mathcal{P}^{(l)}$ be the set of primes in the segment $\{1,\dots, l\}$.
Then
the condition
\begin{equation}\label{cond:S_A_new_intr}
	\log |S| =o \left( \frac{l}{\log l} \right)
\end{equation}
implies
\begin{equation}\label{f:S_A_new_K_intr}
	|\{ (p,p',s,s') \in \mathcal{P}^{(l)} \times \mathcal{P}^{(l)} \times S \times S ~:~ ps = p's' \}| = o( |\mathcal{P}^{(l)}|^2 |S|) \,.
\end{equation}
In particular, if $|SS| \ll |S|$, then size of any arithmetic progression with the beginning at zero in $S$ does not exceed
	\begin{equation}\label{f:S_A_new_intr}
	|A| \ll \log |S| \cdot \log \log |S| \,.
	\end{equation}
\label{t:AP_in_G_intr}
\end{theorem}

The result above can be considered as an integer  analogue of Vinogradov's question about the least quadratic non--residue.
Namely,  having a  number $p$ one can take  the subgroup of squares $\mathcal{R} \subseteq \Z/p\Z$ with the product set $\mathcal{R}\mathcal{R}$ equals $\mathcal{R}$
and ask the question about the maximal length of the arithmetic progression with the beginning at zero, belonging to $\mathcal{R}$.
Size of this arithmetic progression is usually  denoted  as $n_p$ and it is known \cite{Paley2} that there are infinitely many primes such that
$$
	n_p \gg \log p \cdot \log \log p \,.
$$
On the other hand, GRH implies \cite{Ankeny} that $n_p = O(\log^2 p)$ (the best  unconditional bound belongs to Burgess \cite{Burgess} who proved
$n_p \ll p^{\frac{1}{4\sqrt{e}}+o(1)}$).
Thus in the integer case our Theorem \ref{t:AP_in_G_intr} gives upper bound \eqref{f:S_A_new_intr} of a comparable quality.

\bigskip

Our another  result is Theorem \ref{t:AS} from Section \ref{sec:f=log}.

\begin{theorem}
	Let $A,S \subset \Z$ be finite sets and $0\le \a <1/6$ be any number.
	Suppose that $|A+A| \le K|A|$ with
	\begin{equation}\label{cond:AS_1_intr}
	K \ll \exp(\log^{\a} |A|)
	\end{equation}
	and
	\begin{equation}\label{cond:AS_2_intr}
	|S| \le \exp\left(\frac{\log^{2-6\a}|A|}{\log \log |A|} \right) \,.
	\end{equation}
	Then for an absolute constant $C>0$ and a certain $a\in A$ one has
	\begin{equation}\label{f:AS_intr}
	|(A-a)S| \gg |S| \cdot \exp(C \log^{1-3\a} |A|) \,.
	\end{equation}
	In particular,
	$
	|(A-A)S| \gg |S| \cdot \exp(C \log^{1-3\a} |A|) \,.
	$
	\label{t:AS_intr}
\end{theorem}

The result above
can be considered as the first step towards the main conjecture from  \cite{BR-NZ} where authors do not assume that the additional condition \eqref{cond:AS_1_intr} takes place (also, see papers \cite{HR-NRI}, \cite{HR-NRII} in this direction).

The method of the proofs of Theorems \ref{t:AP_in_G_intr}, \ref{t:AS_intr} uses so--called GCD sums (see, e.g., \cite{Aisleitner}, \cite{BS}, \cite{Bloom}, \cite{Radziwll}), which are connected with a series of questions of the Uniform Distribution,  as well as Number Theory  in particular, with large values of the zeta function.
In our paper we follow beautiful exposition of random zeta functions approach from \cite{Radziwll}.
Thus our method extensively uses the integer arithmetic.
It is interesting to obtain some analogues of Theorems \ref{t:AP_in_G_intr}, \ref{t:AS_intr} for subsets of $\mathbb{R}$ or $\mathbb{C}$.

If one takes the function  $f(x) = \log x$, then
Theorem \ref{t:AS_intr}
can be considered as a repulsion result concerning the logarithmic function.
Namely, estimate \eqref{f:AS_intr} says that $|f(A-a) + \log S|$ must be significantly larger than $|S|$ for rather big sets $S$ as in \eqref{cond:AS_2_intr}.
The first results in the direction were obtained in \cite{HR-NR} for
general $k$--convex functions (that is having strictly monotone the first $k$ derivatives).
Recall \cite[Theorem 1.4]{HR-NR}.

\begin{theorem}
	Let $A$ be a finite set of real numbers contained in an interval $I$ and let $f$ be a function which is $k$--convex on $I$ for some $k \geq 1$.
	Suppose that $|A|>10 k$.
	Then if $|A+A-A| \leq K|A|,$  then we have
	$$
	\left|2^{k} f(A)-(2^{k}-1) f(A)\right| \geq \frac{|A|^{k+1}}{(C K)^{2^{k+1}-k-2}(\log |A|)^{2^{k+2}-k-4}}
	$$
	for some absolute constant $C>0$.
	\label{t:k-convex_stat_intr}
\end{theorem}

In this direction we obtain a result on common energy of an arbitrary set $S$ and the image of a $k$--convex function (the required definitions can be found in Section \ref{sec:def}).
Of course general Theorem \ref{t:E_f_intr}  below gives weaker bounds than Theorem \ref{t:AP_in_G_intr} in the particular case $f(x)=\log x$.

\begin{comment}

\begin{theorem}
	Let $f$ be function which is $k$--convex on a set $I$ for some $k \geq 1$.
Suppose that $|I+I-I|\le |I|^{1+\epsilon}$.
Then for all $l\le 2^k$, $\epsilon \le \frac{\log l}{l}$ one has
\begin{equation}\label{f:T_l_intr}
\T_{2^l} (f(I)) \ll |I|^{2^{l+1} - c \log l}  \,.
\end{equation}
for a certain absolute constant $c>0$.
\label{t:T_l_intr}
\end{theorem}

The result above implies

\end{comment}

\begin{theorem}
	Let $f$ be a function which is $k$--convex on a set $I$ for some $k \geq 1$.
	Suppose that $|I+I-I|\le |I|^{1+\epsilon}$.

	Then for any finite set $S\subset \mathbb{R}$ with  $|I| \ge |S|^\eps$, $\eps \gg 1/k$, $\epsilon \le \exp(-1/(c\eps))$  there is $\delta (\eps)>0$ such that
	\begin{equation}\label{f:E_f_intr}
	\E^{+} (f(I),S) \ll |I|^2 |S|^{1-\delta(\eps)} \,.
	\end{equation}
	In particular, $|f(I) + S| \gg |S|^{1+\delta(\eps)}$.
	\label{t:E_f_intr}
\end{theorem}

Using the Pl\"unnecke inequality (see estimate  \eqref{f:Plunnecke} below) one can show that to have growth as in \eqref{f:AS_intr} under the assumptions as in \eqref{cond:AS_1_intr}
Theorem \ref{t:k-convex_stat_intr} requires the condition
\begin{equation}\label{f:S_A_old}
	|S| \le \exp(O (\log |A| \cdot \log \log |A|))
\end{equation}
and our restriction \eqref{cond:AS_2_intr} is wider.
Theorem \ref{t:AP_in_G}, as well as  Proposition  \ref{p:E(P_z,w)} below require much weaker restrictions on $|S|$ but provide a smaller growth.

Finally, recall the main result from \cite{sh_as}, which can be considered as  a quantitative version of some results from \cite{Bourgain_more}.

\begin{theorem}
	Let $p$ be a primes number, $A, B, C \subseteq \mathbb{F}_{p}$ be arbitrary sets, and $k \geq 1$ be such that $|A||B|^{1+\frac{(k+1)}{2(k+4)} 2^{-k}} \leq p$ and
$$
|B|^{\frac{k}{8}+\frac{1}{2(k+4)}} \geq|A| \cdot C_{*}^{(k+4) / 4} \log ^{k}(|A||B|) \,,
$$
where $C_{*}>0$ is an absolute constant. Then
$$
\max \{|A B|,|A+C|\} \geq 2^{-3}|A| \cdot \min \left\{|C|,|B|^{\frac{1}{2(k+4)} 2^{-k}}\right\} \,,
$$
and for any $\alpha \neq 0$
$$
\max \{|A B|,|(A+\alpha) C|\} \geq 2^{-3}|A| \cdot \min \left\{|C|,|B|^{\frac{1}{2(k+4)} 2^{-k}}\right\} \,.
$$
\label{t:sh_as}
\end{theorem}

The result above takes place in $\mathbb{R}$ as well.
In this case we do not need any conditions containing the characteristic $p$.
The main difference between Theorems \ref{t:AS_intr}, \ref{t:E_f_intr} and Theorem \ref{t:sh_as} is that $A$ is large and $B$ is small in Theorem \ref{t:sh_as} but the opposite situation takes place in Theorem \ref{t:E_f_intr} (here $|A|=|I|=|f(I)|$) and similar in Theorem \ref{t:AS_intr}.

\bigskip

The author thanks Christoph  Aistleitner and Sergei Konyagin for useful discussions.

\section{Definitions and preliminaries}
\label{sec:def}

	Let $\Gr$ be an abelian group.
Put
$\E^{+}(A,B)$ for the {\it common additive energy} of two sets $A,B \subseteq \Gr$
(see, e.g., \cite{TV}), that is,
$$
\E^{+} (A,B) = |\{ (a_1,a_2,b_1,b_2) \in A\times A \times B \times B ~:~ a_1+b_1 = a_2+b_2 \}| \,.
$$
If $A=B$, then  we simply write $\E^{+} (A)$ instead of $\E^{+} (A,A)$
and the quantity $\E^{+} (A)$ is called the {\it additive energy} in this case.
Sometimes we write $\E^{+}(f_1,f_2,f_3,f_4)$ for the additive energy of four real functions, namely,
$$
\E^{+}(f_1,f_2,f_3,f_4) = \sum_{x,y,z} f_1 (x) f_2 (y) f_3 (x+z)  f_4 (y+z) \,.
$$
Thus $\E^{+}(f_1,f_2,f_3,f_4)$ pertains to additive quadruples, weighed by the values of $f_1,f_2,f_3,f_4$.
It can be shown using the H\"older inequality (see, e.g., \cite{TV}) that
\begin{equation}\label{f:E_Ho}
\E^{+}(f_1,f_2,f_3,f_4) \le (\E^{+} (f_1) \E^{+} (f_1) \E^{+} (f_1) \E^{+} (f_1))^{1/4} \,.
\end{equation}
More generally,
we deal with
a higher energy
\begin{equation}\label{def:T_k_ab}
\T^{+}_k (A) := |\{ (a_1,\dots,a_k,a'_1,\dots,a'_k) \in A^{2k} ~:~ a_1 + \dots + a_k = a'_1 + \dots + a'_k \}|
\end{equation}
and similar $\T^{+}_k (f)$  for a general function $f$.
Sometimes we  use representation function notations like $r_{A+B} (x)$ or $r_{A+A-B}$, which counts the number of ways $x \in \Gr$ can be expressed as a sum $a+b$ or as a sum $a+a'-b$ with $a,a'\in A$, $b\in B$, respectively.
For example, $|A| = r_{A-A}(0)$ and  $\E^{+} (A) = r_{A+A-A-A}(0)=\sum_x r^2_{A+A} (x) = \sum_x r^2_{A-A} (x)$.
In the same way define the {\it common multiplicative energy} of two sets $A,B$
$$
\E^{\times} (A,B) =  |\{ (a_1,a_2,b_1,b_2) \in A\times A \times B \times B ~:~ a_1 b_1 = a_2 b_2 \}| \,,
$$
further $\T^\times_k (A)$, $\T^\times_k (f)$ and so on.

\bigskip

If $\Gr$ is an abelian group, then the Pl\"unnecke--Ruzsa inequality (see, e.g., \cite{TV}) takes place
\begin{equation}\label{f:Plunnecke}
	|nA-mA| \le \left( \frac{|A+A|}{|A|} \right)^{n+m} \cdot |A| \,,
\end{equation}
and
\begin{equation}\label{f:Plunnecke+}
|nA| \le \left( \frac{|A+A|}{|A|} \right)^{n} \cdot |A| \,.
\end{equation}

Now recall
our current  knowledge about the Polynomial Freiman--Ruzsa Conjecture, see \cite{Sanders_2A-2A}, \cite{Sanders_3log} and \cite{TV}.
We need a simple consequence of  \cite[Proposition 2.5, Theorem  2.7]{Sanders_3log}.
Recall that if $P_1,\dots, P_d \subset \Z$ are arithmetic progressions, then $Q:=P_1+\dots+P_d$ is a {\it generalized arithmetic progression} (GAP) of {\it dimension} $d$. A generalized arithmetic progression, $Q,$ is called to be {\it proper} if $|Q| = \prod_{j=1}^d |P_j|$. For properties of generalized arithmetic progressions consult, e.g., \cite{TV}.

\begin{theorem}
	Let $A \subset \Z$ be a finite set, $|A+A| \le K|A|$ and $\kappa>3$ be any constant.
	Then there is a proper GAP  $H$ of size at most $|A|\exp (O(\log^\kappa K))$ and dimension $O(\log^\kappa K)$ such that for a set of shifts $X$, $|X| \le \exp (O(\log^\kappa K))$ one has $A\subseteq H+X$.
\label{t:Sanders_3log}
\end{theorem}

\begin{comment}

\begin{corollary}
	Let $A \subseteq \Z$ be a set, $|A+A| \le K|A|$ and $\kappa>3$ be any constant.
	Suppose that $D\subseteq A-A$ is any set.
	Then there is an arithmetic progression $P =\{0,s,2s,\dots,ls\}$, $l\gg $ such that
\[
	|D\cap P| \gg |P|
\]
\end{corollary}
\begin{proof}
	Put $S=A-A$.
	By \eqref{f:Plunnecke}, we have $|S+S| \le K^4 |A| \le K^4 |S|$.
	Using Theorem \ref{t:Sanders_3log} applied to $S$
	we find a proper GAP  $H=P_1+\dots+P_d$ of size at most $|A|\exp (-O(\log^\kappa K))$ and dimension $d=O(\log^\kappa K)$ such that for a set of shifts $X$, $|X| \le \exp (O(\log^\kappa K))$ one has $S\subseteq H+X$.
Here $\kappa >3$ is any number.
	From the proof of Theorem \ref{t:Sanders_3log} we can assume that not just $H$ but even $H+H$  is also a proper GAP.
	It will coast us the factor $C^d$ in our final bound and this does not change any estimates.
	Without loss of generality we can assume that for $P=P_1$ one has $|P| \ge |H|^{1/d}$.
	Let $P' = P_2+\dots+P_d$, hence $H=P+P'$.
	By Katz--Koester trick \cite{kk} we know that for all $s\in S$ one has $r_{S-S} (s) \ge |A|$.
	Hence
$$
	|A| |D| \le \sum_{s\in D} r_{S-S} (s)
	  \le
	  \sum_{s\in D} |(H+X) \cap (H+X +s)|
	  \le
	  \sum_{p_1,p_2\in P'} \sum_{s\in D}
$$
and hence there is $x\in X$ and $j\in [2^d]$ such that $|S\cap (H_j+x)| \ge |S|/(2^d|X|)$.
Fix $H_j$ and with some abuse of notation we write $H$ for this $H_j$.
Thus we can assume that not only $H$ but even  $H+H$ is a proper generalized  arithmetic progression and it will cost us the factor $2^d$.

Without loss of generality we can assume that for $P_1$ one has $|P_1| \ge |H|^{1/d}$.
Also, there is $x\in X$ such that $|S\cap (H+x)| \ge |S|/(2^d |X|)$ and hence
\[
|S| \cdot \exp(-O(\log^\kappa K)) \le |S|/(2^d |X|) \le |S\cap (H+x)| \le \sum_{y\in P_2+\dots+P_d} |S\cap (P_1+y+x)| \,.
\]
This completes the proof.
$\hfill\Box$
\end{proof}

\end{comment}

All logarithms are to base $2.$ The signs $\ll$ and $\gg$ are the usual Vinogradov symbols.
For a positive integer $n,$
let
$[n]=\{1,\ldots,n\}.$

\section{The proof of the main result}
\label{sec:f=log}

Now we obtain Theorem \ref{t:AS_intr} from the Introduction.
Following the method from \cite{Radziwll} we recall some required definitions.

For each prime $p \in \mathcal{P}$ take a random variable  $X_p$, which is uniformly distributed on $S^{1}$ and let all $X_p$ be independent.
For every $n \in \mathbb{N}$, $n=p_1^{\o_1} \dots p_s^{\o_s}$, where $p_j \in \mathcal{P}$, $j\in [s]$ are different primes
put
$X_n:=\prod_{j=1}^s X_{p_j}^{\o_j}$.
Then define the random  zeta function by the formula  (let $\a$ be a real number, $\a >\frac{1}{2}$, say)
\begin{equation}\label{def:random_zeta}
	\zeta_{X}(\alpha):=\sum_{n \in \mathbb{N}} \frac{X_n}{n^{\alpha}} = \prod_{p \in \mathcal{P}} \left(1-\frac{X_p}{p^\a} \right)^{-1} \,.
\end{equation}
Using the product formula \eqref{def:random_zeta} one can compute the moments of the random zeta function \eqref{def:random_zeta}, see \cite{Radziwll} (or just similar  calculations in our Lemma \ref{l:moments_new} below).

\begin{lemma}
	Let $l$ be a positive integer.
	Then
$$
\log \mathbb{E}\left|\zeta_{X}(\alpha)\right|^{2 l} \ll\left\{\begin{array}{ll}
l \log \log l, & \alpha=1 \,, l\ge 3 \\
C (\alpha) l^{1 / \alpha}(\log l)^{-1}, & 1 / 2<\alpha<1 \,, l\ge 3 \\
l^{2} \log \left( \frac{1}{2\a-1}\right), & 1 / 2<\alpha \,, l\ge 1 \,,
\end{array}\right.
$$
where
$C (\alpha) = \frac{\a}{1-\a} + \frac{\a}{2\a-1}$.
\label{l:Radziwill_moments}
\end{lemma}

Also, for any function $g: \mathbb{Z} \to \mathbb{C}$ consider the following random analogue of its "multiplicative"\, Fourier transform
\begin{equation}\label{f:Fourier_random}
	\FF{g} (X) = \sum_{n \in \mathbb{N}} g(n) X_n \,.
\end{equation}
Clearly, we have an analogue of the Parseval identity
\begin{equation}\label{f:Parseval_random}
	\mathbb{E} |\FF{g} (X)|^2 = \| g\|^2_2 \,,
\end{equation}
and, moreover, for $k\ge 1$ one has
\begin{equation}\label{f:Tk_random}
	\mathbb{E} |\FF{g} (X)|^{2k} = \T^\times_k (g) \,.
\end{equation}
Further one can compute
$$
	 \mathbb{E} |\FF{g} (X) \zeta_X (\a)|^2 = \sum_{n_1,n_2, m_1,m_2 ~:~ n_1 m_1 = n_2 m_2} \frac{g(m_1) \ov{g} (m_2)}{(n_1 n_2)^\a}
	 =
$$
\begin{equation}\label{f:gcd}
	 =
	 \zeta (2\a) \sum_{m_1,m_2} g(m_1) \ov{g} (m_2) \cdot \frac{\mathrm{gcd} (m_1,m_2)^{2\a}}{(m_1 m_2)^\a}
\end{equation}
and hence GCD sum \eqref{f:gcd} can be interpreted as the multiplicative energy (see the definition of Fourier transform \eqref{f:Fourier_random}) of our weight $g$ with the random zeta function $\zeta_X (\a)$.
It is easy to see (consult estimate \eqref{tmp:25.06_1} below) that it can be converted further to the ordinary multiplicative energy of the function $g$ and the interval $[N]$.

\bigskip

We follow the method from \cite{Radziwll}, \cite{Bloom}, \cite{Aisleitner} to give the proof of Lemma \ref{l:Radziwill} below.
Generally speaking, our bound \eqref{f:Radziwill} is close to the optimal one, see \cite{BS}.

\begin{lemma}
	Let $w:\Z \setminus\{0\} \to \mathbb{R}^{+}$ be  a non--negative function and $N$ be a positive integer.
	Then for any positive integer $s$ one has
\begin{equation}\label{f:Radziwill-}
\E^\times ([N],w) \ll N \| w\|_2^2 \exp\left( C\sqrt{s^{-1}\log \log N \cdot \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2)} + 2\log \log N \right)
	\ll
\end{equation}
\begin{equation}\label{f:Radziwill}
	\ll  N \| w\|_2^2 \exp\left( C \sqrt{\log \log N \cdot \log (\|w\|_1 \|w\|^{-1}_2)} + 2\log \log N \right) \,,
\end{equation}
	where $C>0$ is an absolute constant.
\label{l:Radziwill}
\end{lemma}
\begin{proof}
	Let $L=\log N$ and $\a \in (1/2,1]$.
	Using the Dirichlet principle, as well as estimate \eqref{f:E_Ho}, we find a positive number $U\le N$ such that
\[
	\E^\times ([N],w) \ll L^2  \sum_{U<n_1,n_2 \le 2U,\, m_1,m_2 ~:~ n_1 m_1 = n_2 m_2} w(m_1) w (m_2)
		\ll
\]
\[
		\ll
		L^2 U^{2\a} \sum_{U<n_1,n_2 \le 2U,\, m_1,m_2 ~:~ n_1 m_1 = n_2 m_2} \frac{w(m_1) w (m_2)}{(n_1 n_2)^\a} \,.
\]
	In terms of the random zeta function \eqref{def:random_zeta}, we see that the last sum is, clearly,
	does not exceed
$$
	\sum_{n_1,n_2, m_1,m_2 ~:~ n_1 m_1 = n_2 m_2} \frac{w(m_1) w (m_2)}{(n_1 n_2)^\a}  = \mathbb{E} |\FF{w} (X) \zeta_X (\a)|^2 \,.
$$
	Thus
\begin{equation}\label{tmp:25.06_1}
	\E^\times ([N],w) \ll 	L^2 U^{2\a} \mathbb{E} |\FF{w} (X) \zeta_X (\a)|^2
\end{equation}
	and our task is to estimate the last expectation.
	Let $l\ge 3$ be an integer parameter, which  we will choose later.
	Also, let $\T_{s+1} = \T^\times_{s+1} (w)$.
	Thanks to   identities  \eqref{f:Parseval_random}, \eqref{f:Tk_random}, Lemma \ref{l:Radziwill_moments}  and the H\"older inequality, we have
\begin{equation}\label{f:min_alpha--}
	\mathbb{E} |\FF{w} (X) \zeta_X (\a)|^2 \le  \mathbb{E}^{1-1/l} |\FF{w} (X)|^{2+2/(l-1)} \cdot \mathbb{E}^{1/l} |\zeta_X (\a)|^{2l}
	\le
\end{equation}
\begin{equation}\label{f:min_alpha-}
	\le
	\left(\mathbb{E} |\FF{w} (X)|^2 \right)^{\frac{s(l-1)-1}{sl}} \left(\mathbb{E} |\FF{w} (X)|^{2s+2} \right)^{\frac{1}{sl}} \cdot \mathbb{E}^{1/l} |\zeta_X (\a)|^{2l}
=
	\| w\|_2^2 \T^{\frac{1}{sl}}_{s+1}  \|w\|_2^{-\frac{2s+2}{sl}}  \cdot \mathbb{E}^{1/l} |\zeta_X (\a)|^{2l}
\end{equation}
\begin{equation}\label{f:min_alpha}
	\ll \| w\|_2^2  \exp\left( \frac{1}{ls} \log (\T_{s+1} \|w\|^{-2(s+1)}_2) + \min\left\{ \frac{C(\alpha) l^{1/\a}}{l \log l}, O\left( l\log\frac{1}{2\a-1}\right) \right\} \right)  \,.
\end{equation}
	Put
	$X= s^{-1} \log (\T_{s+1} \|w\|^{-2(s+1)}_2)  \ge 0$.
	First of all, take the second term in the minimum in \eqref{f:min_alpha}.
	In this case we see that the optimal choice of $l$ is $l\sim X^{1/2} \log^{-1/2} (1/(2\a-1))$.
	Hence
\[
	\mathbb{E} |\FF{w} (X) \zeta_X (\a)|^2 \ll \| w\|_2^2  \exp\left( O\left( X^{1/2} \log^{1/2} \frac{1}{2\a-1} \right) \right)  \,.
\]
	Now we take $\a = \frac{1}{2} + \frac{1}{\log N}$ (one can check that this choice of $\a$ allows us to choose  $l\ge 3$) and using $U\le N$, we get in view of \eqref{tmp:25.06_1} that
\[
	\E^\times ([N],w) \ll L^2 N \| w\|_2^2 \exp\left( O\left(\sqrt{s^{-1}\log \log N \cdot \log (\T_{s+1} \|w\|^{-2(s+1)}_2)} \right) \right) \,.
\]
	To obtain \eqref{f:Radziwill} just notice that $\T_{s+1} \le \|w\|_1^{2s} \| w\|_2^2$.
\begin{comment}
	Now we choose our parameter $\a$ in \eqref{f:min_alpha} to be close to one and  we use the first  term in the minimum in \eqref{f:min_alpha}.

	Concretely,
	let $\a = 1 - \frac{\log (1/\eps)}{\log N}$, hence $C(\a) \ll \frac{\log N}{\log (1/\eps)}$ and $l=X^\a$.
	Then bound \eqref{f:min_alpha}

	implies
	(we ignore small savings which gives us $\log l$ in \eqref{f:min_alpha} for simplicity)
\[
	\E^\times ([N],w) \ll \eps L N^2 \exp \left(C \exp \left(\frac{\log (1/\eps) \log X)}{\log N} + \log N^{1/\log (1/\eps)}\right) \right) \,,
\]
	where $C>0$ is an absolute constant.
\end{comment}
	This completes the proof.
$\hfill\Box$
\end{proof}

\bigskip

Using lemma above we obtain in particular, Theorem \ref{t:AS} from the Introduction.

\begin{theorem}
	Let $A,S \subset \Z$ be finite sets and $0\le \a <1/6$ be any number.
	Suppose that $|A+A| \le K|A|$ with
\begin{equation}\label{cond:AS_1}
	K \ll \exp(\log^{\a} |A|)
\end{equation}
	and
\begin{equation}\label{cond:AS_2}
	|S| \le \exp\left(\frac{\log^{2-6\a}|A|}{\log \log |A|} \right) \,.
\end{equation}
	Then there are at least $\exp(-O(\log^{1-6\a} |A|))$ elements $a\in A$ such that
\begin{equation}\label{f:AS}
	|(A-a)S| \gg |S| \cdot \exp(O(\log^{1-3\a} |A|)) \,.
\end{equation}
	In addition, if $|S+S| \le K_* |S|$, then \eqref{f:AS} takes place provided
\begin{equation}\label{cond:AS_2+}
	K_* \log |S| \le \exp\left(\frac{\log^{2-6\a}|A|}{\log \log |A|} \right) \,.
\end{equation}
\label{t:AS}
\end{theorem}
\begin{proof}
	Using Theorem \ref{t:Sanders_3log} we find a proper GAP  $H$ of size at most $|A|\exp (O(\log^\kappa K))$ and dimension $d=O(\log^\kappa K)$ such that for a set of shifts $X$, $|X| \le \exp (O(\log^\kappa K))$ one has $A\subseteq H+X$.
	Here $\kappa >3$ is any number.

	We have $H=P_1+ \dots + P_d$, where the sum is direct and all $P_j$ are arithmetic progressions.
	Without loss of generality we can assume that for $P=P_1$ one has $|P| \ge |H|^{1/d}$.
	Also, there is $x\in X$ such that $|A\cap (H+x)| \ge |A|/|X|$ and hence
\[
	|A| \cdot \exp(-O(\log^\kappa K)) \le |A|/|X| \le |A\cap (H+x)| \le \sum_{y\in P_2+\dots+P_d} |A\cap (P+y+x)| \,.
\]
	Thus there exists  $y\in P_2+\dots+P_d +x$ such that
\[
	|P| \cdot \exp(-O(\log^\kappa K)) \le |P| |A|/|H| \cdot \exp(-O(\log^\kappa K))
		=
\]
\begin{equation}\label{tmp:25.09_2}
		=
		|A| \cdot \exp(-O(\log^\kappa K)) (|P_2| \dots |P_d|)^{-1} \le  |A\cap (P+y)| \,.
\end{equation}
	For any $a\in A\cap (P+y)$, we have $D_*:=A\cap (P+y) - a \subseteq (A-A) \cap (P-P)$.
	Applying Lemma \ref{l:Radziwill}, the lower bound $|P| \ge |H|^{1/d}$
	and using the Holder inequality several times, as well as estimate \eqref{tmp:25.09_2}, we obtain
\[
	|D_* S| \ge \frac{|D_*|^2 |S|^2}{\E^\times (P-P,S)} \gg  \frac{|A\cap (P+y)|^2 |S|^2}{\E^\times (P,S)}
		\gg |S| |P| \cdot \exp(-O(\log^\kappa K + \sqrt{\log \log |P| \cdot \log |S|}))
\]
\[
	\gg |S| \cdot \exp\left(O\left(\frac{\log |A|}{\log^\kappa K} - \log^\kappa K - \sqrt{\log \log |A| \cdot \log |S|} \right) \right) \,.
\]
	Thanks to our conditions \eqref{cond:AS_1}, \eqref{cond:AS_2}, we obtain
\begin{equation}\label{tmp:30.09_1}
	|D_* S| \gg |S| \cdot \exp\left(O\left(\frac{\log |A|}{\log^\kappa K} - \sqrt{\log \log |A| \cdot \log |S|}\right)\right)
	\gg
	|S| \cdot\exp\left(O\left(\frac{\log |A|}{\log^\kappa K} \right)\right)
	\gg
\end{equation}
\begin{equation}\label{tmp:30.09_2}
	\gg |S| \cdot \exp(O(\log^{1-3\a} |A|))

\end{equation}
as required.

	To obtain \eqref{cond:AS_2+}  just repeat the previous calculations and use Lemma \ref{l:Radziwill} with the parameter $s=1$.
	By Solymosi's result \cite{soly} we know that $\E^\times (S) \ll |S+S|^2 \log |S| \ll K^2_* |S|^2 \log |S|$
	and we arrive to an analogue of  \eqref{tmp:30.09_1}, \eqref{tmp:30.09_2}
\[
	|D_* S| \gg |S| \cdot \exp\left(O\left(\frac{\log |A|}{\log^\kappa K} - \sqrt{\log \log |A| \cdot \log (K_*^2 \log |S|)}\right)\right) \,.
\]
	This completes the proof.
$\hfill\Box$
\end{proof}

\bigskip

Now consider another zeta function, which allows to make calculations above  better and even simpler.
Let $\a >0$ be a real number and $z$ be a positive integer.
Then
\begin{equation}\label{def:random_zeta2}
	\mathcal{Z}_{X}(\alpha):=\prod_{z \le p < 2z} \left(1+\frac{X_p}{p^\a} \right) \,.
\end{equation}
Denote by $\mathcal{P}_z$ the set of all primes in $[z,2z)$ and let $g$ be any non--negative function.
Since the support of
$\mathcal{Z}_{X}(\alpha)$
coincides with  all possible products of primes from $\mathcal{P}_z$ and $1$, we see that
the function $\mathcal{Z}_{X}(\alpha)$ can be used to calculate the common energy of the set $\mathcal{P}_z$ with any function $g$, namely,
\begin{equation}\label{f:E(g,Z)}
	\E^\times (g,\mathcal{P}_z) < 4^\a z^{2\a} \cdot \mathbb{E} |\FF{g} (X) \mathcal{Z}_{X}(\alpha)|^2 \,.
\end{equation}
Thus to
compute
$\E^\times (g,\mathcal{P}_z)$ we need to estimate all moments of the function  $\mathcal{Z}_{X}(\alpha)$ similar to Lemma  \ref{l:Radziwill_moments}.

\begin{lemma}
	Let $\a>0$ be any real number, $l$ be a positive integer and $l\le z^\a$.
	Then
\[
	\log \mathbb{E} |\mathcal{Z}_{X}(\alpha)|^{2l} \ll \frac{l^2 z^{1-2\a}}{\log z} \,.
\]
\label{l:moments_new}
\end{lemma}
\begin{proof}
	In view of the fact that all the variables $X_p$, $p\in \mathcal{P}_z$ are independent,  we have
\[
	\mathbb{E} |\mathcal{Z}_{X}(\alpha)|^{2l} = \prod_{z \le p < 2z} \mathbb{E} \left(1+\frac{X_p}{p^\a} \right)^l \left(1+\frac{\ov{X}_p}{p^\a} \right)^{l}
	:=
	\prod_{z \le p < 2z} E_{l} (p) \,,
\]
	and
\[
	E_{l} (p) = \frac{1}{2\pi} \int_{0}^{2\pi} \left(1+\frac{e^{i\theta}}{p^\a} \right)^l \left(1+\frac{e^{-i\theta}}{p^\a} \right)^{l}\, d\theta
	=
	\sum_{n=0}^{l} \binom{l}{n}^2 \frac{1}{p^{2\a n}} \,.
\]
	Using the condition $l\le z^\a$, we obtain $\log E_{l} (p) \le 2l^2/p^{2\a}$.
	Hence
\[
	\log \mathbb{E} |\mathcal{Z}_{X}(\alpha)|^{2l} \le 2l^2 \sum_{z \le p < 2z} p^{-2\a} \ll \frac{l^2 z^{1-2\a}}{\log z} \,.
\]
	This completes the proof.
$\hfill\Box$
\end{proof}

\bigskip

Now we formulate an analogue of Lemma \ref{l:Radziwill} allowing to calculate the  common energy of the set $\mathcal{P}_z$ with a general weight $w$.

\begin{proposition}
	Let $w:\Z \setminus\{0\} \to \mathbb{R}^{+}$ be  a non--negative function and $s,z$ be positive integers.
	Suppose that
\begin{equation}\label{cond:l}
	\log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2) \le \frac{s z}{\log z} \,.
\end{equation}
Then for any $\a>0$ the following holds
\begin{equation}\label{f:Radziwill-_new}
	\E^\times (\mathcal{P}_z,w) \ll z^{2\a} \| w\|_2^2 \exp\left( C z^{1/2-\a}   \sqrt{s^{-1} \log^{-1} z \cdot \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2)} \right) \,.
\end{equation}
	In particular, for any $\eps>0$ one has
\begin{equation}\label{f:Radziwill-_new_eps}
\E^\times (\mathcal{P}_z,w) \ll (\eps z)^{2} \| w\|_2^2 \exp\left( C \eps^{-1} z^{-1/2}   \sqrt{s^{-1} \log^{-1} z \cdot \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2) }
 \right) \,.
\end{equation}
\label{p:E(P_z,w)}
\end{proposition}
\begin{proof}
	Let

	$X=s^{-1}  \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2)$.
	Choose  $l=(X\log z /z)^{1/2} z^\a$.
	Thanks to our assumption \eqref{cond:l}, we have $l \le z^\a$.
	Using Lemma \ref{l:moments_new} as in lines \eqref{f:min_alpha--}---\eqref{f:min_alpha}, combining with  bound \eqref{f:E(g,Z)}, we get
\[
\E^\times (\mathcal{P}_z,w)  \ll z^{2\a} \| w\|_2^2  \exp\left( \frac{1}{ls} \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2) + \frac{l z^{1-2\a}}{\log z}  \right)
\ll
\]
\begin{equation}\label{f:min_alpha_new}
\ll
	z^{2\a} \| w\|_2^2  \exp\left( C z^{1/2-\a}   \sqrt{s^{-1} \log^{-1} z \cdot \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2)} \right) \,.
\end{equation}
	Taking $\a = 1 - \frac{\log (1/\eps)}{\log z}$, we obtain
\begin{equation}\label{f:min_alpha_new'}
	\E^\times (\mathcal{P}_z,w)  \ll  (\eps z)^{2} \|w\|_2^2 \exp\left( C \eps^{-1} z^{-1/2}   \sqrt{s^{-1} \log^{-1} z \cdot \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2) } \right)

\end{equation}
\begin{comment}
	Finally, to obtain \eqref{f:Radziwill-_new_K} we use a simple covering (see, e.g., \cite[Excercise 1.1.8]{TV})  to a set $X$, $|X| \ll \log^2 z$ such that
	$[z,2z] \subseteq X\mathcal{P}_z$.
	Hence taking $\a = 1 - \frac{\log (|X|/\eps)}{\log z}$, we derive as in estimates \eqref{f:min_alpha_new}, \eqref{f:min_alpha_new'}
\[
	\E^\times ([z,2z],w) \ll |X|^2 \E^\times (\mathcal{P}_z,w)
		\ll
\]
\[
		\ll
			(\eps z)^{2} \|w\|_2^2 \exp\left( C |X| \eps^{-1} z^{-1/2}   \sqrt{s^{-1} \log^{-1} z \cdot \log (\T^\times_{s+1} (w) \|w\|^{-2(s+1)}_2) } \right)
\]
\end{comment}
	as required.
$\hfill\Box$
\end{proof}

\bigskip

Now we derive an upper bound for the common energy of the set of the primes in a segment and an arbitrary set.
It shows that in a sense the primes  "repulse"\, the other sets.

\begin{theorem}
	Let $S \subset \Z$ be a set, $l$ be an integer number, and $\mathcal{P}^{(l)} := [l] \cap \mathcal{P}$.
	Then for any $d\neq 0$ the condition
	\begin{equation}\label{cond:S_A_new}
	\log |S| \ll \frac{\eps^{} l}{\log l}
	\end{equation}
	implies
	\begin{equation}\label{f:S_A_new_K}
		\E^{\times} (d \cdot \mathcal{P}^{(l)},S) \le \eps |\mathcal{P}_l|^2 |S| \,.
	\end{equation}
\label{t:AP_in_G}
\end{theorem}
\begin{proof}
\begin{comment}
	We use the arguments of the proof of Theorem \ref{t:AS}, combining with estimate \eqref{f:Radziwill-_new_eps} of Proposition \ref{p:E(P_z,w)} instead of bounds \eqref{f:Radziwill-}--\eqref{f:Radziwill} of Lemma \ref{l:Radziwill}.
	Let $A = \{ 0,d,\dots, dl \}$.
	Put $z=[l/2]$ and consider the set $\mathcal{P}_z \subseteq [l]$.
	Take $s=1$ and $\eps^2 = c/(M\log^2 |A|)$, where $c>0$ is a sufficiently small constant.
	We have $|\mathcal{P}_z| \gg |A|/\log |A|$.
	If \eqref{f:S_A_new} does not hold, then applying Proposition \ref{p:E(P_z,w)} and the Cauchy--Schwarz inequality,  we derive
	$$
	M|S| \ge |SS| \ge |\mathcal{P}_z S| \ge \frac{|\mathcal{P}_z|^2 |S|^2}{\E^\times (\mathcal{P}_z, S)}
	\gg
	$$
	$$
	\gg c^{-1} M |S| \exp\left(-Cc^{-1} M^{1/2} \log |A| \cdot \sqrt{\frac{\log |S|}{|A|\log |A|}}  \right)  \gg c^{-1} M |S|
	$$
	and this is a contradiction because the constant $c$ can be taken sufficiently small.
	Clearly, we can assume that assumption \eqref{cond:l} of Proposition \ref{p:E(P_z,w)} takes place because otherwise with result follows immediately.
\end{comment}

	Take any $z\le [l/2]$.
	By estimate \eqref{f:Radziwill-_new} of Proposition \ref{p:E(P_z,w)}, we get for any $\eps_* >0$
\[
	\E^\times (\mathcal{P}_z,S) \ll (\eps_* z)^{2}  |S| \exp\left( C \eps^{-1}_* z^{-1/2} \sqrt{ \log^{-1} z \cdot \log |S| } \right) \,.
\]
	Summing over $z$ of the form $2^j\le l/2$,

	 we obtain
\[
	\E^{\times} (d \cdot \mathcal{P}^{(l)},S) \ll \eps^2_* l^2 |S|  \exp\left( C \eps^{-1}_* l^{-1/2}   \sqrt{ \log^{-1} l \cdot \log |S| } \right) \,.
\]
	Now put $\eps^2_*  = \frac{\eps}{\log^2 l}$.
	Then thanks to our assumption \eqref{cond:S_A_new}, we have \eqref{f:S_A_new_K}.
	This completes the proof.
	$\hfill\Box$
\end{proof}

\bigskip

Of course in Theorem \ref{t:AP_in_G} one can consider more general arithmetic  progressions as well but in this case one should control the beginning and the step of such progression, simultaneously.

\section{On general  $k$--convex functions}
\label{sec:k-convex}

In \cite[Theorem 1.3]{HR-NR} authors obtained the following growth result for sequences of the form $A = f([N])$, where $f$ is  an arbitrary $k$--convex function.

\begin{theorem}
	Let $k\ge 2$ be an integer and let $A$ be a $k$--convex sequence.
	Then
	\[
	|2^k A - (2^k-1) A| \gg \frac{|A|^{k+1}}{2^{k^2}} \,.
	\]
	\label{t:k-convex_intr}
\end{theorem}

Thus Theorem \ref{t:k-convex_stat_intr} from the Introduction can be considered as  a "statistical"\, version of Theorem \ref{t:k-convex_intr}.
Also, notice that the dependence on $k$ in Theorem \ref{t:k-convex_intr} is better.

In this Section we show how Theorem \ref{t:k-convex_stat_intr} implies an upper bound for the higher energy of any $k$--convex function.
Basically, we repeat the combination of the arguments from \cite[Theorem 13]{RS} and \cite[Theorem 23]{s_energy}.

\begin{theorem}
	Let $f$ be a function which is $k$--convex on a set $I$ for some $k \geq 1$.
	Suppose that $|I+I-I|\le |I|^{1+\epsilon}$.
	Then for all $l\le 2^k$, $\epsilon \le \frac{\log l}{l}$ one has
	\begin{equation}\label{f:T_l}
	\T^{+}_{2^l} (f(I)) \ll |I|^{2^{l+1} - c \log l}  \,.
	\end{equation}
	for a certain absolute constant $c>0$.
	\label{t:T_l}
\end{theorem}
\begin{proof}
	Put $A=f(I)$.
	Let $\T_{2^j} := \T^{+}_{2^j}(A)$ and $\T_1 = |A|^2$.
	Our task is to prove for any $j\in [l]$ that
	\begin{eqnarray}\label{f:ind}
	\T_{2^j} \le \frac{\T_{2^{j-1}} |A|^{2^j}}{Q} \,,
	\end{eqnarray}
	where $Q=|A|^{\frac{c \log j}{j}}$ because it clearly implies \eqref{f:T_l}.
	Suppose not.
	Put $L=O(k \log |A|)$.
	By the dyadic Dirichlet principle and the H\"older inequality in the form \eqref{f:E_Ho} there is a number $\D >0$ and a set $P = \{ x \in \Z ~:~ \D <  r_{2^{j-1}A} (x) \le 2 \D \}$ such that
	\begin{equation}\label{tmp:19.09_3}
	L^4 \D^4 \E^{+}(P) \ge \T_{2^j} \ge \frac{|A|^{2^j} \T_{2^{j-1}}}{Q} \ge \frac{(\D |P|)^2 \D^2 |P|}{Q} \,.
	\end{equation}
	Indeed, we can assume that \eqref{f:T_l} does not hold (otherwise there is nothing to prove) and

	thus  by our condition $j\le l\le 2^k$ one has
	\[
	|A|^{2^{j-1}-1} \ge \D \gg |A|^{2^{j-1}-c \log j}
	\]
	and hence we do indeed have the upper bound \eqref{tmp:19.09_3} with the quantity $L$.

	Further from \eqref{tmp:19.09_3}, we obtain
	$\D \ge L^{-4} \T_{2^j} |A|^{-3\cdot 2^{j-1}}$ and
	\[
	\E^{+}(P) \gg L^{-4} \frac{|P|^3}{Q} := \frac{|P|^3}{Q_1} \,.
	\]
	Also notice that $\D^4 \E^{+}(P) \le \D^2 |P| (\D |P|)^2 \le \T_{2^{j-1}} (\D |P|)^2$ and hence from \eqref{tmp:19.09_3}, we get
	\begin{equation}\label{tmp:19.09_4}
	\D |P| \ge \frac{|A|^{2^{j-1}}}{L^2 Q^{1/2}} \,.
	\end{equation}
	Similarly, $\D^4 \E^{+}(P) \le (\D^2 |P|) |A|^{2^j-2} |P|^2 \le \T_{2^{j-1}} |A|^{2^j-2} |P|^2$ and thus from \eqref{tmp:19.09_3}, we derive
	\begin{equation}\label{tmp:19.09_4'}
	|P| \ge \frac{|A|}{L^2 Q^{1/2}} \,.
	\end{equation}
	By the   Balog--Szemer\'edi--Gowers Theorem (see, e.g., \cite{TV}),

	we find $P_* \subseteq P$ such that $|P_*| \gg |P| Q_1^{-C_*}$, and  $|P_*+P_*| \ll Q^{C_*}_1 |P_*|$.
	Here $C_* > 1$ is an absolute constant, which may change from line to line.
	By the definition of the set $P$, we have
	\[
	\D |P_*| \le \sum_{x\in P_*} r_{2^{j-1}A} (x) = \sum_{x_1,\dots, x_{2^{j-1}-1}\in A} r_{P_*-A} (x_1 +\dots +x_{2^{j-1}-1}) \,.
	\]
	Hence there is a shift $x$ and a set $A_* \subseteq A \cap (P_*-x)$ such that
	\begin{equation}\label{tmp:16.09_1}
	|A_*| \ge \D |P_*| / |A|^{2^{j-1}-1} \gg |A| (LQ)^{-C_*} \,.
	\end{equation}
	Here we have used bound \eqref{tmp:19.09_4}.
	The set $A_*$ has the form $A_* = f(S)$, where $S\subseteq I$ is a set of the same size.
	Clearly,
	$$
	|S+S-S|\le |I+I-I| \le |I|^{1+\epsilon} = |A|^{1+\epsilon}/|A_*| \cdot |S| := K|S| \,.
	$$
	Applying Theorem \ref{t:k-convex_stat_intr} with a parameter $t = t(j) \le k$, which we will choose later, combining with  inequality \eqref{f:Plunnecke}, we obtain
	\[
	\frac{|A_*|^{t+1}}{(C K)^{2^{t+1}-t-2}(\log |A_*|)^{2^{t+2}-t-4}} \le |2^{t} A_* - (2^{t}-1)A_*| \le |2^{t} P_* - (2^{t}-1) P_*|
	\ll
	\]
	\begin{equation}\label{tmp:03.10_1}
	\ll
	Q^{(2^{t+1}-1)C_*}_1 |P_*| \,.
	\end{equation}
	Thanks to estimate \eqref{tmp:16.09_1}, we know that $K \ll (LQ)^{C_*} |A|^\epsilon$.
	By the assumption $\epsilon \le \frac{\log l}{l}$ and hence  $K \ll (LQ)^{C_*}$ (with another constant $C_*$ of course) by our choice of $Q$.
	Using this

	estimate,
	as well as both inequalities  from \eqref{tmp:16.09_1}, combining with \eqref{tmp:19.09_4} and the lower bound $|P_*| \gg |P| Q_1^{-C_*}$, we derive from \eqref{tmp:03.10_1}
	\[
	\D |P_*| \cdot |A|^{t+1-2^{j-1}} Q^{-C_* 2^t}_1  \le \left( \frac{\D |P_*|}{A|^{2^{j-1}-1}}  \right)^{t+1} \ll Q^{C_* 2^t}_1 |P_*| \,.
	\]
	Hence
	$$
	\D |A|^{t+1-2^{j-1}} \ll Q^{C_* 2^t}_1
	$$
	and in view of \eqref{tmp:19.09_3}, we get

	\[
	|A|^{2^{j-1}- (t+1)} Q^{C_* 2^t}_1 |A|^{3\cdot 2^{j-1}} \ge \T_{2^j} \ge |A|^{2^{j+1} - c \log j} \,.
	\]
	Now take the parameter $t$ as $t(j) = \log j$.
	It follows that for sufficiently large constant $C'$ we get  $Q\gg |A|^{\frac{\log j}{C' j}}$.
	This completes the proof.
	$\hfill\Box$
\end{proof}

\bigskip

Theorem \ref{t:T_l}	 can be used to obtain a series of lower bounds for various combinations of {\it different} sets see, e.g., \cite[Corollary 1.5]{HR-NR}.
We restrict ourself  by just one consequence.
Much more stronger results for subsets of $\Z$ were obtained in \cite{HR-NRI}, \cite{HR-NRII}.

\begin{corollary}
	Let $m$ be a positive integer, $A_1,\dots, A_{2^m} \subset \mathbb{R}$ be sets of the same size $|A_1|$, $|A_j A_j| \ll |A_j|$, $j\in [2^m]$.
	Then for any non--zero shifts $z_1,\dots,z_{2^m}$ one has
	\[
	|(A_1+z_1) \dots (A_{2^m}+z_{2^m})| \gg |A_1|^{c \log m} \,.
	\]
\end{corollary}
\begin{proof}
	For any $z\neq 0$ consider the function $f_z (x) = \log (z+e^x)$.
	Then $f_z$ is $k$--convex for any $k$.
	Also, for $I=\log A$, where $A$ is any of the sets $A_j$, $j\in [2^m]$ one has in view of \eqref{f:Plunnecke} that $|I+I-I|\ll |I|$.
	Applying Theorem \ref{t:T_l} for  $f=f_z$, and $l=m$, we see that $\T^\times_{2^m} (A+z) \ll |A|^{2^{m+1}- c \log m}$.
	Hence by the H\"older inequality
	\[
	|A_1|^{2^{m+1}} \le |(A_1+z_1) \dots (A_{2^m}+z_{2^m})| \cdot \sum_x r^2_{(A_1+z_1) \dots (A_{2^m}+z_{2^m})} (x)
	\le
	\]
	\[
	\le
		|(A_1+z_1) \dots (A_{2^m}+z_{2^m})| \cdot \left(\prod_{j=1}^{2^m} \T^\times_{2^m} (A_j+z_j) \right)^{1/2^m}
	\ll
	|(A_1+z_1) \dots (A_{2^m}+z_{2^m})| \cdot |A_1|^{2^{m+1}- c \log m}
	\]

	as required.
	$\hfill\Box$
\end{proof}

\bigskip

Now we obtain a new incidence result for one--parametric curves.

\begin{theorem}
	Let $f$ be a function which is $k$--convex on a set $I$ for some $k \geq 1$.
	Suppose that $|I+I-I|\le |I|^{1+\epsilon}$ and $\epsilon \le \frac{\log k}{k}$.
	Then for any finite sets $B,C\subset \mathbb{R}$ with  $|I| \ge |B|^\eps$, $\eps \gg 1/k$ and $\epsilon \le \exp(-1/(c\eps))$
	there is $\delta (\eps) \ge \exp(-\exp (O(1/\eps)))>0$ such that
	\begin{equation}\label{f:inc_f}
	|\{ (i,b,c)\in I\times B \times C ~:~ f(i) + b = c \}| \ll \sqrt{|B||C|} |I| \cdot |B|^{-\d(\eps)} \,.
	\end{equation}
	\label{t:inc_f}
\end{theorem}
\begin{proof}
	Put $A=f(I)\cup (-f(I))$ and let $\sigma$ be cardinality of the set on the left--hand side of \eqref{f:inc_f}.
	Using the Cauchy--Schwarz inequality several times, we obtain for any $j$
	\[
	\sigma^{2^j} \le |C|^{2^{j-1}} |B|^{2^{j-1}-1} \sum_x r_{2^j A} (x) r_{B-B} (x) \,.
	\]
	Applying the Cauchy--Schwarz  inequality one more time, we get
	\[
	\sigma^{2^{j+1}} \le |C|^{2^{j}} |B|^{2^{j}-2} \E^{+}(B) \T_{2^{j}} (A) \,.
	\]
	Now suppose that $j\le 2^k$.
	Then by Theorem \ref{t:T_l} and the trivial bound $\E^{+}(B)\le |B|^3$, we obtain
	\[
	\sigma^{2^{j+1}} \ll |C|^{2^{j}} |B|^{2^{j}} \cdot |B|  |I|^{2^{j+1} - c\log j} \,.
	\]
	It gives us
	\[
	\sigma \ll \sqrt{|B||C|} |I| \cdot \left( \frac{|B|}{|I|^{c\log j}} \right)^{2^{-(j+1)}}
	\]
	By our assumption $|I| \ge |B|^\eps$ and hence taking $j \gg \exp(1/(c\eps))$, we derive
	\[
	\sigma \ll \sqrt{|B||C|} |I| \cdot |B|^{-2^{-(j+1)}}
	\]
	as required.
	Here $\delta(\eps) \sim \exp(-\exp (1/c\eps))$.
	This completes the proof.
	$\hfill\Box$
\end{proof}

\bigskip

The incidence result above implies Theorem \ref{t:E_f_intr} from the Introduction.

\begin{corollary}
	Let $f$ be a function which is $k$--convex on a set $I$ for some $k \geq 1$.
	Suppose that $|I+I-I|\le |I|^{1+\epsilon}$.

	Then for any finite set $B\subset \mathbb{R}$ with  $|I| \ge |B|^\eps$, $\eps \gg 1/k$, $\epsilon \le \exp(-1/(c\eps))$  there is $\delta (\eps)>0$ such that
	\begin{equation}\label{f:E_f}
	\E^{+} (f(I),B) \ll |I|^2 |B|^{1-\delta(\eps)} \,.
	\end{equation}
	In particular, $|f(I) + B| \gg |B|^{1+\delta(\eps)}$.
	\label{c:E_f}
\end{corollary}
\begin{proof}
	Let $\tau>0$ be  a real number and
	$$S_\tau = \{ s\in \mathbb{R} ~:~ |\{ (i,b) \in I \times B ~:~ f(i)+b = s\}| \ge \tau \} \,.$$
	Using Theorem \ref{t:T_l}, we have
	\[
	\tau |S_\tau| \le |\{ (i,b,s) \in I \times B \times S_\tau ~:~ f(i)+b = s\}| \ll \sqrt{|B| |S_\tau|} |B|^{-\delta(\eps)} \,.
	\]
	By summation we obtain \eqref{f:E_f} and  the bound $|f(I) + B| \gg |B|^{1+\delta(\eps)}$ follows from the Cauchy--Schwarz inequality.
	This completes the proof.
	$\hfill\Box$
\end{proof}

redpajama_set_name: RedPajamaArXiv


The MFB S280 industrial cabinet has been further enhanced with the option of lining the enclosure with Thermobreak insulation foam to reduce the amount of heat entering into the cabinet.
Siemon's cabinet solutions and accessories are now available in light grey.
The 007 series server racks and cabinets are available from Rack World Systems.
IoT and expansion at the edge of the network is changing how businesses manage their data. Edge data centres mean data can be managed where it's created, and they're helping organisations change and improve the way they work.
MFB Class B and Class C security enclosures have been tested for certification by the Australian Government's Security Construction and Equipment Committee (SCEC). Both versions of the security enclosure, in both floor-standing and wall-mounted styles, are now approved for supply.
Pentair Schroff Interscale C cases feature an interlocking tabbed construction that provides integrated EMC protection and secures with only two screws.
Emerson Network Power has announced the availability of the SmartCabinet, a pre-engineered integrated rack solution designed for deployment to edge of IT locations.
The Rittal AE compact wall-mounted enclosure is available in over 75 variants including powder-coated steel in light grey RAL7035, 304 and 316 stainless steel with 400 grain finish.
NextComputing has added the Vigor ED Portable Workstation for maximum performance and storage in tough environments to its range.
RackLink has announced its growing range of AC/DC rack distribution boards, designed with a 3 or 4RU form factor which can accommodate up to 24-pole DIN-mount hardware.
The APC NetShelter SV enclosure by Schneider Electric is an adaptable server rack designed to maintain good airflow for IT infrastructure.
The Siemon V800 cabinet system is designed to provide a robust enclosure for both high-density data centre environments as well as floor distributors.
The Altronics 19″ rack systems range is designed for audio, data and communications installations.
ERNTEC's SE frame enclosure system is suitable for 19/21/23″ and industrial enclosures for applications that require up to IP54 and EMI/RFI.
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In 1976, Omahaki feral sheep were reported to be found around the confluence of the Ngaruroro and the Taruarau Rivers, in the Eastern Ruahine State Forest, and on the Big Hill and Omahaki stations. There appeared to be several hundred sheep spread over an area of some 7000 hectares of scrubland, screes and bluffs.
In his report on feral sheep in New Zealand (see » About Feral Sheep for reference), scientist Tony Whitaker noted that the origin of this flock was uncertain but it was known to have existed there for over fifty years. It was probably the remnant of a much larger feral flock that had its beginning in the Merino sheep that had been grazed on the open tops of the Ruahines in the previous century. (In the early 1960s feral sheep were know to exist in six different Ruahine catchments but were now reduced to this one area in the northeast.) There were indications that the proportion of black sheep in the Omahaki flock was increasing over the years.
Some Omahaki sheep have been brought out of the area, but often they are run with the very similar » Mohaka sheep – which has led to a questioning of the purity of some individual animals.


JIMTalks is a refreshing look at how creation and brain science work together to help us understand some of the most challenging passages of Scripture. It offers insights into many aspects of everyday life and brings the Life Model challenge for maturity into family and church life. You want to share this with your friends!
Every Sunday Dr. Jim Wilder emerges with his best thought of the week and teaches a select group in Pasadena, CA.
Dr. Wilder unpacks familiar Bible passages in new ways to include lessons linking science, psychology and theology together.
Over forty years in brain science is applied to unlocking the Bible to give you a fresh perspective on interpretation and practical application.
Volume 31 includes an amazing sermon on prayer and the brain recorded at the Joy Starts Here book tour in Austin, TX. You will want everyone who has ever struggled with prayer to hear this message.
Dr. Jim Wilder is one of the thinkers behind the Life Model, from design to application. He delights in finding difficult passages in Scripture and making them obvious and practical.
Every Sunday Dr. Wilder emerges with his best thought of the week and teaches a select group in Pasadena, CA. His topics are based on familiar Bible passages but not the way you've heard them before.
You will be introduced to practical neurotheology (brain-science and the Bible). If you want to be in on the thinking that will develop into groundbreaking programs in the near future, get JIMTalks!


text:
« Florida Governor Shuts Down Schools as Hurricane Irma Looms | Main | As Hurricane Irma Lashes Florida, Educators and Schools Offer Refuge »
How a Florida School District Prepared for Multiple Hurricane Irma Scenarios
By Denisa R. Superville on September 9, 2017 4:28 PM
Michael Eugene spent the first part of his career in Cleveland advising superintendents when to close schools because of snowstorms. In Los Angeles, it was earthquakes.
In Florida, it's hurricanes. And in recent days, Eugene has been helping to prepare his district for what is forecast to be the most devastating hurricane to hit the state in a generation.
Eugene, the chief operations officer at the Orange County school district in central Florida, leads a group of chief officers—from facilities, instruction, and technology—who track hurricanes as they develop in the Atlantic Ocean or Gulf of Mexico and make the recommendation to the superintendent on whether—and when—the district should close. The team also makes the recommendation for when superintendent should reopen schools.
Over the last several days, as Hurricane Irma gained speed and strength and barreled its way through the Caribbean, Eugene's team jumped into high gear, taking on full-time storm watch and preparation.
They followed weather models four times a day, getting more "forensic" and technical models from the National Weather Service than the general public sees in the press, he said. They huddled at 4 p.m. each day to pore over hurricane maps to understand Irma's possible paths and impacts on the school district. They talked daily with the county's office of emergency management and joined conference calls with meteorologists who explained the details behind the various weather models.
Figuring out Irma's likely path and behavior has been very challenging. Early models showed the hurricane possibly hitting South Florida and veering west, while another showed Irma veering east.
"With the size of a storm like this, inside of it is a hurricane, but just outside of it is tropical storm force winds—and those are two very different impacts for us," he said.
Making Sure There's Fuel in Tanks; Money Queued Up for Payroll
In addition to the "policy group," more than 20 departments, who are part of an incident management team that includes emergency management and transportation, are involved in planning for a storm like this. This group puts together the plan on how to prepare for and carry out the preparedness and recovery.
But the team doesn't just make recommendations about how and when to close: They make sure that teachers can get paid if there is prolonged closure.
So, before Florida Gov. Rick Scott decided late Thursday to shut down all K-12 schools, colleges, and universities in the state, Eugene and his team were juggling a range of preparations:
Lining up emergency contracting and spending authority from Superintendent Barbara Jenkins, so they can move quickly to spend large sums of money in the wake of the storm.
Creating a special emergency fund for disaster-related spending.
Ensuring all fuel tanks were at 100 percent, so if there's a gas shortage or other supply chain problem after the storm, the district can function and its buses can run.
Making sure staff will get paid if schools close for an extended period.
Ensuring that food services, which supplies food to evacuees at shelters, has enough supplies for at least five days. Food services also set up a "nutrition access program" so that if schools are closed for an extended period, the district can set up food trucks at specific locations to give students access to meals.
Raising spending limits for some departments, such as maintenance staff, in case they need to go out and spend large sums of money.
"The preventative-based approach helps us prepare in advance so we are not making those decisions in the midst of a crisis," Eugene said in an interview Friday night. "That allows you to be more level-headed. It allows to you to be more strategic."
Each department has a specific readiness strategy on the steps it's taking to get ready and what it will need to do to get back up and running, he said.
"That kind of coordination improves our effectiveness. It improves our efficiency and our consistency," he said.
When some school districts in south Florida made the decision as early as Tuesday to close Thursday and Friday, Orange County did not.
Eugene and his team were still following weather models that showed that the storm was not likely to impact Orange County until Monday with tropical storm force winds—meaning that they could complete the school week and close on Monday.
Some parents were anxious about that decision, Superintendent Jenkins said.
But she knew that her staff was using data, hard facts, and what the weather experts were saying to inform their recommendations to her.
Readying for What Comes After the Storm
Now that the district has shut down for Hurricane Irma, the team's work continues. They were back in the district Saturday, to go over the most recent projections about the fast-moving and rapidly-changing weather system, and think about the earliest day they will be able to reopen. (They decided to stay closed on Tuesday.)
One of the first things they must calculate: Are there enough teachers—including possible substitutes—available? Looking at how many teachers in the district submitted requests for leave next week will be one key factor.
Other things the team will evaluate before deciding whether to recommend that Superintendent Jenkins reopen schools:
Are the buildings and buses in a good and safe condition?
Beyond having enough teachers, are there bus drivers available? Have cafeteria workers returned in order to run food services?
Are students able to come to school?
Eugene, who has been with Orange County since 2009, says all the departments have to work in tandem. If the internet goes down, the IT department may need facilities and finance to get up and running again.
"The way we think here in Orange County is that none of us is doing it on our own; we all do it as one family," he said.
Members of department readiness teams train during the year to test their preparedness, run through hypotheticals, and come up with tactical response plans, he said.
Eugene says he thinks all Florida districts that face the constant threats of hurricanes have some level of disaster preparedness and emergency response systems in place—the same way districts in the Midwest have structures to deal with snowstorms and California districts focus on earthquakes. The Miami-Dade school district runs disaster response scenarios with its staff.
Having dealt with all three, Eugene thinks hurricanes are more nerve-wracking.
"A hurricane places greater stress on staff because of the anticipation of its arrival and because one would think it's predictable—that you can see a hurricane coming three to five days away," he said.
"The unanticipated part of a hurricane is whether it turns, and that's a very common conversation, and it's a very important [reason] for why we focus on those models," he continued. "It's the potential for that storm to turn and you're not affected at all. Whereas, an earthquake is an earthquake. When it hits, you are not told about it, and so there isn't the stress of the anticipation. But it definitely hits with a level of certainty."
Photo caption: Chief officers with the Orange County, Fla., school system meeting this week to discuss preparations for Hurricane Irma. Standing left to right: Jesus Jara, deputy superintendent, Michael Eugene, chief operations officer, and Scott Howat, chief communications officer. (Photo courtesy of the Orange County school district.)

redpajama_set_name: RedPajamaCommonCrawl


text:
OnePlus 7 Case Renders Contrast With Earlier Leak; Show A Headphone Jack?
By Ramish Zafar
With the launch of Samsung's Galaxy S10 lineup behind us, it's time to see what other flagships will grace the market this year. Surprisingly, it's the OnePlus 7 that has started to leak early on in the year. OnePlus launched the OnePlus 6t last year, and the device featured new features such as a virtual fingerprint sensor. However, with Samsung having one-upped everyone with the Galaxy S10's ultrasonic fingerprint sensor, other manufacturers are surely feeling the heat. Is OnePlus among them? We'll find out soon enough, so take a look below for more details.
OnePlus 7 Alleged Case Renders Show A Triple Rear Camera Setup And A Headphone Jack
OnePlus made a bold move with the OnePlus 6t; a move that wasn't popular with a section of the company's fan base. After holding out for a while, the company removed the 3.5mm earphone jack on its smartphone, and then justified the decision later by claiming that the move paved the way for wireless charging on the smartphone.
Related Galaxy Note 10 Will Have A 12MP Primary Lens With 1.4µm Pixels
The issue of the headphone jack is an interesting one, as we take a look at what is allegedly a third leak for the OnePlus 7. The first one was from prolific tipster Steve Hemmerstoffer, who started off the season by sharing renders of the OnePlus 7. These renders show a smartphone without a 3.5mm earphone jack and with minimal cutouts on the top. These, as you'll see below, are different from today's alleged leak.
Today's leak shows two different cutouts at the top of the case. We're certain that one of these is for pop-up front camera that's rumored to make it on the OnePlus 7. The second appears to be a 3.5mm earphone jack, and this is where things get interesting. The earlier render leak from Steve shows no such port, but today's images beg to differ. This leaves only a handful of possibilities on the table. One of these is that either the earlier one or today's leak is fake.
The other possibility concerns their source. While Steve's likely managed to get his hands on what OnePlus intends the OnePlus 7 to look like at the end of the smartphone's development cycle, the case renders depict what the device looks like right now. After all, case manufacturers need to be kept up to date with a smartphone's design so that they can tailor their products accordingly. Safe to say, one feature that is showing an increasing probability of making it on the OnePlus 7 is a pop-up front camera.
Thoughts? Let us know what you think in the comments section below and stay tuned. We'll keep you updated on the latest.
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Will strive to bring back 'President', 'PM' post in JK: Omar; Modi condemns remark.
National Conference leader Omar Abdullah Monday said his party will strive hard for restoring the autonomy of Jammu and Kashmir that could include having a 'Sadar-e-Riyasat' (President) and 'Wazir-e-Azam' (Prime Minister), a statement which evoked sharp condemnation from Prime Minister Narendra Modi.
Addressing a rally at Bandipore in North Kashmir, Omar, a former chief Minister, said his party will not allow any attacks on J and K's special status and will work to get back what was "infringed upon" including the coveted posts of Sadr-e-Riyasat' (president) and prime minister for the state.
Modi condemned Omar's remark and demanded to know from the Congress and the 'grand opposition alliance' parties if they supported the NC leader on the autonomy issue.Addressing a rally in Secunderabad in Telangana, Modi asked the 'Mahagathbandhan' leaders to make their stand clear on the issue.
Modi said there were media reports that Omar has stated there should be a separate prime minister for Kashmir."Two prime ministers for Hindustan? Do you agree with it? Congress has to answer and all the 'Mahagathbandhan' partners have to answer. What are the reasons and how dare he(Omar) say that", he said.
Modi said he also wants to ask Trinamool Chief and West Bengal Chief Minister Mamata Banerjee, former prime minister H D Deve Gowda, Andhra Pradesh Chief Minister N Chandrababu Naidu and NCP supremo Sharad Pa war whether they agreed with Omar's statement.
"I want to ask Bengal Didi who shouts a lot, do you agree with it?. Answer the people U-turn (Chandrababu) Babu with whom Farooq Abdullah campaigned recently in Andhra Pradesh. Do you think Naidu should get votes?" NCP's Sharad Pawar and Gowda, whose son is Karnataka Chief Minister H D Kumaraswamy. should also answer, he said.
"Would you like to go with them (Mahagathbandhan) or break away from them?", Modi asked.
At his rally, Omar said the accession of the state with the Union of India was accomplished in lieu of various Constitutional safeguards, which if tampered with, will raise serious questions on accession.
"We are not like any another state of India. Every other state got merged with the Union of India. We joined our hands with the Union on certain conditions unlike other states of India. Does any other state in India have its own flag and constitution? Our handshake with Indian Union was subject to certain conditions, he said.
Omar said after 70 years, forces inimical to the state's special status were trying to backtrack from the conditions. "The move will also initiate debate on the fundamentals of accession, there is no second thought about it. It takes two hands to clap," he said.
"…we will fight any attempt that is aimed to fiddle with our state's status. We won't allow any more attack on our special status. On the contrary, we will strive to get back what was infringed upon. We will work towards getting back the coveted posts of Sadr-e-Riyasat' and Prime Minister for our state," he said.
The National Conference has always advocated restoration of autonomy for the state and had passed a resolution for the same in July 2000 in the state assembly. It was, however, not agreed to by the then BJP-led NDA government at the Centre.
Rebuffing the statement of BJP president Amit Shah against Article 35-A, Omar also said no one knew about Article-35A until he raised it in the assembly.
"I inquired from the then chief minister about the efforts that were being made by the state government to defend it in the Supreme Court. Only then it came into the realm of debate. In 1947 certain assurances were given to us in the form of securing our identity and the assurances were for all times.
"Before Shah, Arun Jaitley also had made some uncouth remarks on Article 35-A. Let me put it across, that we won't allow that to happen. Article 370 and Article 35-A are the articles of faith for us, he said. The two articles relate to special status for the state.
Previous articleMore than 113 milion people suffer 'acute hunger': UN.
Next articleAny attempt to abrogate Article 370 would be saying goodbye to J&K: Sajad Lone.


In the year since scooter startup Lime launched, people have taken 6 million rides on its trademark dockless scooters and bikes, according the company's first annual report.
By the end of 2017, the number of US bikes from bike-share companies more than doubled to about 100,000, according to the National Association of Transportation Officials. Forty-four percent of these bikes, it said, were dockless, meaning they could be parked on the edge of a sidewalk or on a local bike rack and get unlocked with a smartphone app.
Lime announced that it landed $335 million in investments from Uber and Alphabet. Earlier this year, Uber appointed one of its staffers to head a New Modalities division, which specifically focuses on scooters and bikes.
When coupled with public transportation, bike-sharing costs about 80 percent less than owning a car, according to Lime, which has made it a popular transit option among people in low-income brackets. In Washington, DC, for example, Lime says it has seen a 20 percent increase of rides on dockless vehicles from people who earn less than $35,000 a year, while usage of docked bikes has remained flat.
Even though these e-vehicles might be easy on people's wallets, they've also been a nuisance in some cities, as they're often left in the middle of sidewalks and even in the San Francisco Bay. The involvement of more established transportation companies like Uber and Lyft could give these electric vehicle startups the structure they need as they arrive in more cities.
Previous articleApple is now the world's first $1 trillion company, following successful Q3 results from the company.


text:
The World Turned Upside Down — Or, Why Americans Dont Bow to Kings
The red-coated army band played The World Turned Upside Down as they marched out of their entrenchments. They were about to surrender to the victorious American and French forces at Yorktown, Virginia. The British army commander, Lord Charles Cornwallis, had been trapped with his back to the York River for three weeks.
When the French fleet under Admiral Comte de Grasse defeated the British fleet in the Battle of the Capes, Cornwallis escape hatch was slammed shut. The surrender ceremony for Cornwallis army took place on this day, October 19, 1781. Gen. George Washingtons Yorktown victory effectively ended the six-year American Revolutionary War.
To be sure, we date our independence from the signing of the great Declaration. Our countrys birthday is rightly July 4, 1776. But without Washingtons military success against British and Hessian forces, the Fourth of July might have been only a stumper question in a British Commonwealth game of Trivial Pursuit.
The Declaration that Congress approved, as drafted by Thomas Jefferson, with some amendment, contained lofty ideals and immortal words, to be sure, but that Declaration required men brave enough to risk their lives in order to give it force. Gen. Washington, Gen. Lafayette, Gen. Rochambeau, and a combined force of Americans and French soldiers were just brave enough to give it force.
Giving it force is what happened on this day 229 years ago. From that time, America has been free of kings, free of the need to bow down before any hereditary monarchs, free at last.
No small part of why President Obamas bowing to various foreign despots---like the odious King Abdullah of Saudi Arabia---is so disturbing is that it demonstrates his lack of understanding of his own countrys history.
Britains Winston Churchill knew Americas history better than Barack Obama does. Thats why Churchill called us the Great Republic.
Does it really matter in the modern world whether or not a President knows his countrys history? Well, yes. First, Barack Obama like every new President hopes to be a great one. No President-elect ever tells an interviewer My model for my administration will be Franklin Pierce. Millard Fillmore has very few fans, even in his home state of New York.
The fact is: weve never had a great President who was not steeped in U.S. history. President Franklin D. Roosevelt rode with Prime Minister Winston Churchill to George Washingtons home at Mount Vernon on New Years Day, 1942. The U.S. had just joined Britain in what they agreed to call the Second World War. You would think FDR and Winston would have bigger things on their minds than a tourist attraction.
But Roosevelt knew that millions of Americans still harbored strong anti-British sentiments. With his acute sense of history, he knew that it would help with U.S. public opinion for the British Prime Minister to be seen laying a wreath on the tomb of George Washington. And so they did.
Second, the National Archives Building, down the street from FRCs Washington headquarters, tells us the past is prologue. In other words, American history did not begin on January 20, 2009. The National Archives Building contains the recordsin words carved in stoneof the Glory and Romance of our History.
President Obama has said Brits think their country is exceptional. So do the Greeks. So do the Belgians. Therefore, if America is exceptional, we are no more exceptional than any other country. Thats what Barack Obama thinks.
American history proves that concept wrong. President Reagan used to challenge skeptical foreign students. The United States after World War II had an army and navy of twelve million. We had the only atomic bomb in the world. Ours was the only country that had suffered virtually no destruction at home.
Yet the only territory we asked for were the cemeteries in France and Belgium where we buried our dead. Could you say that of the USSR? Or of any other country that had just forced its enemies to surrender unconditionally?
Or, consider President Lincoln visiting the Confederate White House on April 4, 1865. Jefferson Davis, the Southern leader, had fled just hours before. Lincoln walked through the smoldering streets of Richmond, Virginia. He understood that anyone could have shot him from any second story window. Yet, fearless, he went there. And just as fearlessly he said of the defeated rebels: Let `em up easy.
A crowd of freed slaves surrounded him, cheering and praying, offering praises to God, and singing a powerful hymn:
Oh, all ye people clap your hands,
And with triumphant voices sing;
No force the mighty power withstands
Of God, the universal King.
Lincoln was deeply moved, as was the tiny band of U.S. sailors who formed his honor guard. One elderly black man approached the President and bowed down before him.
The President gently reprimanded him: Dont kneel to me. That is not right. You must kneel to God only, and thank Him for the liberty you will hereafter enjoy. I am but Gods humble instrument; but you may rest assured that as long as I live no one shall put a shackle on your limbs, and you shall have all the rights which God has given to every other free citizen of this Republic.
Such a scene could have occurred in no other country on earth. It is one of the hundreds of reasons America is exceptional. Here, we kneel to God alone. Thats another reason to remember and be grateful for Yorktown Day.
Prosecutors Ask Court to Acquit Geert Wilders »
« The Social Conservative Review: The Insider's Guide to Pro-Family News--October 21, 2010


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text:
Theodore Roosevelt, Jr. (født 13. november 1887, død 12. juli 1944) var en amerikansk politiker og forretningsmand. Han var den ældste søn af Theodore Roosevelt.

Liv
Roosevelt var den tabende guvernørkandidat i New York (1924), guvernør over Puerto Rico (1929-32) og generalguvernør på Filippinerne (1932-33). Efter at hans fætter Franklin D. Roosevelt blev præsident, trak Theodore sig tilbage fra politik – han var republikaner og Franklin var demokrat – og blev forretningsmand, blandt andet hos American Express og Doubleday.

I 2. verdenskrig var han var den første amerikanske general på stranden på D-Dag. Han døde af et hjerteanfald kort derefter og blev posthumt tildelt den amerikanske Medal of Honor.

Andet
I filmen Den Længste Dag spilles brigadegeneral Roosevelt af Henry Fonda.

Eksterne henvisninger

Politikere fra USA
Filippinernes politik
Guvernører fra USA
Puerto Rico
Puerto Ricos guvernører
Amerikanere i 2. verdenskrig

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NEW YORK – Attorney General Eric T. Schneiderman and New York State Comptroller Thomas P. DiNapoli announced today the sentencing of Jimmie Buie, a Brooklyn resident, for stealing over $22,000 in pension benefits issued by the New York State and Local Employees Retirement System to his deceased mother, Sandra Buie, between May 2011 and December 2012. Jimmie Buie was convicted of Attempted Grand Larceny in the Third Degree and sentenced to 1 ½ to 3 years in state prison by the Honorable Elizabeth Foley in Kings County Supreme Court.
According to statements made in court, Jimmie Buie's grandmother, Thelma Hairston, was a New York State pensioner who retired in 1995. At the time of her retirement, Hairston selected to receive her pension as a Pop Up Joint Allowance Half, which provided her with a reduced pension benefit, but allowed her to pass on pension payments to her daughter, Sandra Buie, following Hairston's death. When Hairston died in 1996, Sandra Buie began to receive Hairston's share of monthly pension benefits.


text:
Touro Law Center Library
United States Constitutional Law
United States Constitutional Law: Home
This guide is intended to assist the user with basic constitutional law research, as well as to provide the researcher with further resources to consult. The most important sources are outlined including primary sources, treatises, hornbooks and more.
Treatises, Hornbooks, Encyclopedias & Dictionaries
Law Reviews & Journals
Authoritative Websites & Online Guides
Sample Searches
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Chat with a Touro Law librarian - and get the guidance, advice, and help you need to get your research off on the right foot.
Chat is available only during regular reference hours. Please check our hours here.
If you're researching after hours, please e-mail your questions to libraryreference@tourolaw.edu
The study of constitutional law involves the interpretation of the United States Constitution. The United States Supreme Court determines what is and what is not "constitutional." Therefore, the study of U.S. Supreme Court case law is an integral part of researching constitutional law. Many secondary sources are available that seek to explain, discuss and challenge the Constitution and how it has been interpreted over the last 200 plus years.
Many of the principles embodied in the Constitution involve individual rights and liberties, freedom of expression and religion, as well as due process and equal protection. Other issues involve states' rights, the relationship between state and federal governments, and the system of checks and balances among and between the three branches of the federal government.
When researching constitutional law, one should look to the primary sources, which consist of the Constitution itself, as well as the case law interpreting the Constitution. A useful starting point is to consult a secondary source such as a treatise, hornbook, encyclopedia or dictionary. These sources seek to explain and define the principles of constitutional law. Law reviews and journals, which also are valuable secondary sources that can aid in the understanding of constitutional principles, often discuss new or anticipated constitutional challenges and concerns. These sources are available at Gould Law Library.
About This Libguide
This guide is intended to assist the user with basic constitutional law research, as well as to provide the researcher with further resources to consult. The most important sources on constitutional law that are available at Gould Law Library and online are outlined. This list is not intended to be exhaustive. This guide concludes by taking the user, step-by-step, through two sample constitutional law searches utilizing these sources.
Laura Ross
Next: Primary Sources >>
Last Updated: Jul 27, 2017 2:54 PM
URL: https://guides.tourolaw.edu/UnitedStatesconstitutionallaw
Tags: amendments, bill of rights, constitutional law


text:
Island Opinion
Ross Gordon James
November 13, 1956 – July 7, 2019
Ross passed away peacefully at home surrounded by family and close friends on July 7, 2019.
Raised in Nanaimo, he moved to Vancouver where he began his career as a Customs Broker. After several years, he became a partner in Omnitrans where he continued his career until his death.
Ross always had an adventurous spirit and curiosity about life. He grew up sailing and continued that in his later life. He enjoyed go-carting and racing cars, was an avid reader, loved music and was a person people wanted to be around.
He will be lovingly missed by his wife, Leslie, children Sean, Carli, Lindsay (Jeff) and Michael (Sarah), eight grandchildren, three nephews, mother Muriel James and sister Lesley Barth (Gord). Predeceased by father Robert James.
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text:
Q: How to get SQLCMD to output errors and warnings only? How can you get SQLCMD, when executing a SQL script file, to just output any errors or warnings it encounters?
I essentially dont want information based messages to be output.

A: By default, SQLCMD sends all non-error messages and error messages to stdout. Hence, attempting to redirect output won't help.
The first thing you need to do in order to get only one or the other type of messages (error or non-error) is to tell SQLCMD to separate them by sending the error messages (severity level 11 or higher) to stderr. You do this by using the -r command-line switch, specifying 0 as the option for that switch (i.e. -r0). At this point there is no visible difference in terms of running SQLCMD and seeing messages of any type displayed on the screen.
The next part is to filter out the regular messages going to stdout.  This can be done by redirecting the stdout messages to somewhere via >, but where to? You could do > file.txt, but I doubt you want a file of the messages that you didn't want to begin with. Fortunately, DOS has the NUL keyword (that is not a typeo: it has one L, not two) that works like /dev/null in Unix. Meaning you can use the following to redirect output to nowhere: > NUL.
The following will execute the PRINT command and show no output as no errors are generated, and no file is created that would contain the output of the PRINT command:
CD %TEMP%

SQLCMD -E -Q "print 1;" -r0 > NUL

But the following displays an error message as those are not being redirected to NUL:
CD %TEMP%

SQLCMD -E -Q "print a;" -r0 > NUL

Returns:
Msg 128, Level 15, State 1, Server DALI, Line 1
The name "a" is not permitted in this context. Valid expressions are constants,
constant expressions, and (in some contexts) variables. Column names are not permitted.

A: The following will work:
sqlcmd -U user -P pass -S Server -Q "sp_who" -r0 1> test.log
The output from the query will be put into the log and nothing is printed on the screen.
More details at the MSDN article on sqlcmd.

A: Found this on SQLServerCentral
sqlcmd -E -i"install.sql" -r1 2> install-err.log 1> install.log

http://www.sqlservercentral.com/Forums/Topic536968-146-1.aspx

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