text:
\section{Introduction} \label{sec1}

\noindent The problem of Littlewood to which the title refers is the following question. If $A \subseteq \mathbb{Z}$ is a set of $n$ integers, what is the smallest possible value of the $L^1$-norm of the exponential sum over $A$. That is to say, what is
\[ I(n) := \min_{A \subseteq \mathbb{Z}, |A| = n} \int^{1}_0 \left| \sum_{n \in A} e^{2\pi i n \theta}\right| \, d\theta \; ?\]
The problem of determining the exact minimum, known as the strong Littlewood conjecture,
is still unresolved (the conjecture is that the minimum occurs when $A$ is
an arithmetic progression). However the bound $I(n) \gg \log n$, which is tight up to a constant factor, was obtained independently by the second-named author \cite{konyagin1} and by McGehee, Pigno and Smith \cite{MPS} some 20 years ago.\vspace{11pt}

\noindent There are various natural variants of the problem that one might consider. In this paper we will be interested in the discrete setting of functions defined on the group $\mathbb{Z}/p\mathbb{Z}$.
If $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ is a function, then instead of discussing
exponential sums we speak of the Fourier transform of $f$. This is defined for $r \in \mathbb{Z}/p\mathbb{Z}$ by \[ \widehat{f}(r) := \mathbb{E}_{x \in \mathbb{Z}/p\mathbb{Z}}f(x) e(rx/p).\]
Here $e(\theta) = e^{2\pi i \theta}$, as is customary, and $\mathbb{E}$ denotes the averaging operator, which in this instance is the same as $p^{-1}\sum_{x \in \mathbb{Z}/p\mathbb{Z}}$. If $A \subseteq \mathbb{Z}/p\mathbb{Z}$ is a set then we write $1_A$ for its characteristic function. We will often be concerned with the Fourier transform $\widehat{1}_A$.

\begin{question}\label{q1} Let $p$ be a prime, and let $A \subseteq \mathbb{Z}/p\mathbb{Z}$ be a set of size $\lfloor p/2\rfloor$. What is $S(p)$, the minimum possible value of
\[ \Vert \widehat{1}_A \Vert_1 := \sum_{r \in \mathbb{Z}/p\mathbb{Z}} |\widehat{1}_A(r)|?\]
\end{question}
\noindent There is no serious reason for restricting attention to sets of size exactly $\lfloor p/2\rfloor$, though clearly if sets of any size are allowed then $S(p)$ can equal 1. An obvious lower bound for $S(p)$ is $p^{-1}\lfloor p/2\rfloor = 1/2 + O(p
^{-1})$, since this is the contribution to $\Vert \widehat{1_A} \Vert_1$ from the term $r = 0$. It is already non-trivial to obtain a bound of the form $S(p) \rightarrow \infty$. We will obtain a very weak bound of this sort in \S \ref{sec2}, followed in \S \ref{sec4} by the following stronger result.

\begin{theorem}\label{mainthm1} We have $S(p) \gg (\log p/\log \log p)^{1/3}$.\end{theorem}

\noindent The best upper bound we have comes by considering the set $A = \{1,\dots,\lfloor p/2\rfloor\}$, which illustrates that $S(p) \ll \log p$. By analogy with the strong Littlewood conjecture, one might guess that this represents the truth.\vspace{11pt}

\noindent Theorem \ref{mainthm1} will be obtained as a straightforward consequence of the next result, which is the main theorem in our paper.

\begin{theorem}\label{mainthm2} Let $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ be a function with $\mathbb{E} f = 0$ and $\Vert \widehat{f} \Vert_1 \leq 1$.
Then \[ \min_{x \in \mathbb{Z}/p\mathbb{Z}} |f(x)| = O\big((\log \log p/\log p)^{1/3}\big).\]
\end{theorem}
\noindent The condition $\Vert \widehat{f} \Vert_1 \leq 1$ may be thought of as expressing a (rather strong) kind of ``continuity'' or ``smoothness'' of $f$. Indeed one knows from harmonic analysis on the circle that if $f : \mathbb{R}/\mathbb{Z} \rightarrow \mathbb{R}$ is twice continuously differentiable then $\widehat{f}(m) \ll |m|^{-2}$ for $m \in \mathbb{Z} \setminus \{0\}$, and hence $\widehat{f}$ lies in $l^1(\mathbb{Z})$. \vspace{11pt}

\noindent The conclusion of Theorem \ref{mainthm2} may be thought of as a type of intermediate value theorem.\vspace{11pt}

\noindent Taking $f(x) = cp/(p+1)\log p$ for $x = 0,1,\dots,\lfloor p/2\rfloor$ and $f(x) = -cp/(p-1)\log p$ for $x = -1,\dots,-\lfloor p/2\rfloor$, where $c$ is a suitably small constant, one sees that the bound in Theorem \ref{mainthm2} could not be improved
beyond $O(1/\log p)$. It may be that this is best possible.\vspace{11pt}

\noindent As a corollary of Theorem \ref{mainthm2}, we are able to obtain some information on a question of Gowers, the consideration of which was in fact the starting point of our investigations. Gowers asked whether, if $A \subseteq \mathbb{Z}/p\mathbb{Z}$ is a set of size $\lfloor p/2\rfloor$, there is some $x \in \mathbb{Z}/p\mathbb{Z}$ such that
\[ \left| |A \cap (A + x)| - p/4 \right| = o(p).\]
Note that $p/4$ is roughly the expected value of $|A \cap (A + x)|$. One interesting feature of this question, and of Theorem \ref{mainthm2}, is that similar statements do not hold in abelian groups $G$ more general than $\mathbb{Z}/p\mathbb{Z}$; for example if $G = \mathbb{F}_2^m$ and if $A$ is a hyperplane then clearly $|A \cap (A + x)|$ is always either $0$ or $|G|/2$. We answer a generalisation of Gowers' question in the affirmative, obtaining as a consequence of Theorem \ref{mainthm2} the following quantitative bound.

\begin{theorem}\label{mainthm3} Let $A,B \subseteq \mathbb{Z}/p\mathbb{Z}$ be subsets
of $\mathbb{Z}/p\mathbb{Z}$ with $|A| = \alpha p$, $|B| = \beta p$. Then there is some $x \in \mathbb{Z}/p\mathbb{Z}$ such that
\[ \big| |A \cap (B + x)| - \alpha \beta p \big| = O\left(p(\log \log p /\log p)^{1/3}\right).\]
\end{theorem}
\noindent If $f,g : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ are two functions then we define their convolution $f \ast g$ by
\[ f \ast g(x) := \mathbb{E}_y f(y) g(x - y).\] Note that
\[ |A \cap (B + x)| = p 1_A \ast 1_B^{\circ}(x),\] where we have written $f^{\circ}(x) = f(-x)$.
It is thus easy to see that Theorem \ref{mainthm3} is a special case of the following result:
\begin{theorem}\label{mainthm3.1} Let $f,g : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ be
two functions with $\Vert f \Vert_2, \Vert g \Vert_2\leq 1$. Then there is some $x \in \mathbb{Z}/p\mathbb{Z}$ such that
\[ \left| f \ast g(x)- \mathbb{E} f \mathbb{E} g\right|
 = O(\log \log p /\log p)^{1/3}.\]
\end{theorem}
\noindent For Gowers' original problem it is not clear how much one can hope for,
since there are no obvious examples of sets $A$ for which $|A \cap (A + x)|$
is always quite different to $p/4$.\vspace{11pt}

\noindent In \S \ref{sec6} and in \S \ref{sec7} we prove the following results.
\begin{theorem}\label{mainthm3.2}
For any prime $p$ there is a function $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$
such that $\mathbb{E} f=0$, $\Vert f \Vert_2 \leq 1$, and
\[ |f \ast f^{\circ}(x)| \geq c/\log p\] for all $x \in \mathbb{Z}/p\mathbb{Z}$.
\end{theorem}

\begin{theorem}\label{mainthm3.3}
For any prime $p\geq3$ there is a set $A \subseteq \mathbb{Z}/p\mathbb{Z}$,
$|A| = \lfloor p/2 \rfloor$, such that
\[ |A \cap (A + x) - p/4| \geq cp/\log p\log\log p\]
for all $x \in \mathbb{Z}/p\mathbb{Z}$.
\end{theorem}

\noindent It would be of interest to close the gap between Theorem \ref{mainthm3} and \ref{mainthm3.3}, and also to obtain the correct bound in Theorem \ref{mainthm2}.

\section{The continuity argument, and a simple lower bound}\label{sec2}

\noindent The goal of this section is to motivate our main argument, and in so doing to prove a weak version of Theorem \ref{mainthm2}.
\begin{theorem}
Let $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ be a function with $\mathbb{E} f = 0$ and $\Vert \widehat{f} \Vert_1 \leq 1$.
Then \[ \min_{x \in \mathbb{Z}/p\mathbb{Z}} |f(x)| = o(1).\]
\end{theorem}
\noindent By the Fourier inversion formula we have
\[ f(x) = \sum_{r \in \mathbb{Z}/p\mathbb{Z}} \widehat{f}(r) e(-rx/p).\] Since $\Vert \widehat{f} \Vert_1$ is so small, one might hope to have some rather direct control on this expression. Let us imagine, for a moment, that we know something even stronger about $\widehat{f
}$, namely that for some fairly small set $R \subseteq \mathbb{Z}/p\mathbb{Z}$ we have
\[ \sum_{r \notin R} |\widehat{f}(r)| \leq \epsilon/3.\]
By a well-known pigeonhole argument of Dirichlet, to be recalled in
\S \ref{sec3}, we may find some $t \neq 0$ such that the fractional parts
$\{tr/p\}$, $r \in R$, are all small (we define our fractional parts to lie between $-1/2$ and $1/2$). This implies that
$|e(tr/p) - 1|$ is always at most $\epsilon/3$, say. Thus we have
\begin{eqnarray*}
|f(x) - f(x+t)| & = & \sum_{r \in \mathbb{Z}/p\mathbb{Z}} \widehat{f}(r) \big( e(rx/p) - e(r(x+t)/p)\big) \\ & \leq & \sum_{r \in R} |\widehat{f}(r)| |1 - e(rt/p)| + 2 \sum_{r \notin R} |\widehat{f}(r)| \\ & \leq & \epsilon /3 + 2\epsilon/3 = \epsilon
\end{eqnarray*}
for all $x$. Since $\mathbb{E} f(x) = 0$, there must be some value of $x$ for which $f(x)$
and $f(x + t)$ have opposing signs. For this value of $x$ it is clear that $|f(x)| \leq \epsilon$.\vspace{11pt}

\noindent\textit{Remark.} What we have done here is identify a length, $t$, along which $f$ is somewhat continuous in the sense that $|f(x + t) - f(x)|$ is always small.\vspace{11pt}

\noindent We now proceed to show how the above argument may be modified to give a weak version of Theorem \ref{mainthm2}. It need not be the case that most of the $L^1$-norm of $\widehat{f}$ is concentrated on a few modes, but by using a certain pigeonholing argument one may select a set $R$ of modes so that the contribution of $\sum_{r \notin R} \widehat{f}(r) e(-rx/p)$ is small for many $x$. \vspace{11pt}

\noindent Let $\epsilon > 0$, and define a sequence $(\delta_j)_{j=0}^{\infty}$
by $\delta_0 = 1$ and $\delta_{l+1} = \epsilon^{2 + 1/\delta_l}2^{-6/\delta_l - 4}$
for $l \geq 0$. Take $J = \lceil 3/\epsilon\rceil$. Then there exists $l \in \{0,\dots,J\}$ such that
\[ \sum_{r : \delta_{l+1} \leq |\widehat{f}(r)| < \delta_l } |\widehat{f}(r)| \leq \epsilon/3.\]
For this value of $l$, define
\[ f^{(1)}(x) =  \sum_{r : |\widehat{f}(r)| \geq \delta_l } \widehat{f}(r) e(-rx/p),\]
\[ f^{(2)}(x) =  \sum_{r : \delta_{l+1} \leq |\widehat{f}(r)| < \delta_l } \widehat{f}(r) e(-rx/p)\] and
\[ f^{(3)}(x) =  \sum_{r : |\widehat{f}(r)| < \delta_{l+1}} \widehat{f}(r) e(-rx/p).\]
By construction we have the estimate
\begin{equation}\label{eq2.1} \Vert f^{(2)} \Vert_{\infty} \leq \epsilon/3,\end{equation}
and also the bound
\begin{equation}\label{eq2.2} \Vert f^{(3)} \Vert_2^2 = \sum_{r : |\widehat{f}(r)| < \delta_{l+1}} |\widehat{f}(r)|^2 \leq \delta_{l+1}.\end{equation}
Write $R$ for the set of frequencies occurring in $f^{(1)}$, that is to say
$R := \{ r : |\widehat{f}(r)| \geq \delta_l \}$, and set $k := |R|$. By the assumption on $\Vert \widehat{f} \Vert_1$ we have $k \leq 1/\delta_l$. Consider the Bohr set
\[ B(R,\eta) := \{x \in \mathbb{Z}/p\mathbb{Z} : |\{ x r /p \}|
\leq \eta \; \mbox{for all $r \in R$}\}.\]
By Dirichlet's argument we have the bound
\begin{equation}\label{eq2.3} |B(R,\eta)| \geq p\eta^k \geq p\eta^{1/\delta_l}.\end{equation}
For details, and more properties of Bohr sets, see \S \ref{sec3} (particularly Lemma \ref{lem4.00} (i) and its proof). \vspace{11pt}

\noindent Let $B := B(R,\epsilon/48)$. Suppose that $B\setminus\{0\}$ is nonempty,
and select an element $t$ from it. By the same argument we had earlier, we have
\[
|f^{(1)}(x) - f^{(1)}(x+t)| \leq  \sum_{r \in R} |\widehat{f^{(1)}}(r)| |1 - e(rt/p)|  \leq \epsilon/6.\]
Once again, since $\mathbb{E} f^{(1)}(x) = 0$, there is $x_0$ such that $f^{(1)}(x_0)$ and $f^{(1)}(x_0 + t)$ have opposing signs, which implies that $|f^{(1)}(x_0)| \leq \epsilon/6$. For any $t \in B$ we then have $|f^{(1)}(x_0 + t)| \leq \epsilon/3$. \vspace{11
pt}

\noindent Set $X := x_0 + B$. Combining the above observations with \eqref{eq2.1} we have, if $x \in X$, the inequality
\[ |f(x)| \leq 2\epsilon/3 + |f^{(3)}(x)|.\]
To conclude the argument it remains to show that there is some $x \in X$ for which $|f^{(3)}(x)| \leq \epsilon/3$. To this end, note that if this were not the case then we should have
\begin{equation}\label{eq2.4} \Vert f^{(3)} \Vert_2^2 = \mathbb{E}_{x \in \mathbb{Z}/p\mathbb{Z}} f^{(3)}(x)^2  \geq \frac{|B|}{p}\mathbb{E}_{x \in X}f^{(3)}(x)^2 > |B|\epsilon^2/9 \geq \epsilon^{2 + 1/\delta_l}2^{-6/\delta_l - 4}p,\end{equation}
which contradicts \eqref{eq2.2}.\vspace{11pt}

\noindent The only condition that must be satisfied to make this argument work
is that $B\setminus\{0\}$ must be non-empty. By dint of \eqref{eq2.3} this will be so if $p(\epsilon/48)^{1/\delta_J} > 1$. Since $\delta_J$ is subject to a lower bound
depending only on $\epsilon$, this can certainly be satisfied with
$\epsilon = o(1)$, and in fact one can take $3/\epsilon = \log_{*}(p) + O(1)$. This is a very slowly growing function, of course: the reason for this is that $1/\delta_J$ looks like a tower of exponentials of height about $J$, this being forced by the need to have $\delta_{j+1}$ exponentially smaller than $\delta_j$.\vspace{11pt}

\noindent Before moving on, let us highlight the weakness in the above argument which we will be able to tighten up in subsequent sections, leading to a superior bound. It lies in the fact that \eqref{eq2.4} represents not just a lower bound for $\Vert f^{(3)} \Vert_2^2$, but in fact for $\mathbb{E}_{x \in X} |f^{(3)}(x)|^2$. In order to exploit this properly one needs to modify \eqref{eq2.2} appropriately. Writing $\beta(x) := 1_B(x)/|B|$ for the \textit{normalised Bohr cutoff} associated to
 $B$, our interest lies in finding a Fourier expression for
\[ \mathbb{E}_{x \in \mathbb{Z}/p\mathbb{Z}} |f^{(3)}(x)|^2 \beta(x + x_0) .\] This is not hard; indeed we have
\begin{eqnarray}\nonumber
\mathbb{E}_x |f^{(3)}(x)|^2 \beta(x + x_0) & = & \mathbb{E}_x \sum_{r,r'} \widehat{f^{(3)}}(r)\overline{\widehat{f^{(3)}}(r')} e((r' - r)x/p) \beta(x + x_0) \\ & = & \sum_{r,r'}
\widehat{f^{(3)}}(r)\overline{\widehat{f^{(3)}}(r')} \widehat{\beta}(r - r') e((r - r')x_0/p), \label{eq2.6}
\end{eqnarray}
an expression we will find very useful later on. In exploiting it we will require a few facts concerning Bohr sets and their Fourier transforms. We develop these in the next section.

\section{Properties of Bohr sets}
\label{sec3}
\noindent The material of this section is somewhat technical, but is rapidly becoming part of the foundation of additive combinatorics. It has its origins in the work of Bourgain \cite{Bou} on 3-term arithmetic progressions. See \cite{tao-exposition} for
a recent exposition of this work. For examples of subsequent work in a similar vein, see either \cite{green1} or \cite{tao1}, and for a more leisurely discussion of Bohr sets and the need for the results of this section, see \cite{green-finite-fields}. We
 work in the context of an arbitrary finite abelian group $G$, since this is most natural, though our interest is in the case $G = \mathbb{Z}/p\mathbb{Z}$. In fact, a lot of the material in this paper is valid in this more general setting, the key exception being the ``discrete intermediate value theorem'' of \S \ref{sec1}. The failure of this, of course, means that the key theorems as stated in \S \ref{sec1} are only valid when $G = \mathbb{Z}/p\mathbb{Z}$.\vspace{11pt}

\noindent If $f : G \rightarrow \mathbb{R}$ is a function and if $\gamma \in G^*$ is a character then we define the Fourier transform $\widehat{f}$ by
\[ \widehat{f}(\gamma) := \mathbb{E}_{x \in G} f(x) \gamma(x).\]
It is easy to tie this in with the definition we had in the case $G = \mathbb{Z}/p\mathbb{Z}$; in that group, all characters are of the form $x \mapsto e(rx/p)$ for some $r \in \mathbb{Z}/p\mathbb{Z}$.\vspace{11pt}

\noindent Now if $\Gamma \subseteq G^{\ast}$ is a set of $d$ characters, and if $\epsilon > 0$, then we define the Bohr set $B = B(\Gamma,\epsilon)$ by
\[ B(\Gamma,\epsilon) := \{ x \in G : |\textstyle \frac{1}{2\pi}\displaystyle \arg\gamma(x)| \leq \epsilon \;\; \mbox{for all $\gamma \in \Gamma$}\}.\]We also define $\beta = \beta_{\Gamma,\epsilon}$ (the \emph{normalised Bohr cutoff} associated to $B$) by $\beta(x) := 1_{B}(x)/|B|$.\vspace{11pt}

\noindent The next lemma details some well-known facts concerning the size of Bohr sets.

\begin{lemma}\label{lem4.00} Let $\Gamma \subseteq G^*$ be a set of $d$ characters, and let $\delta > 0$. Then we have the bounds
\begin{enumerate}
\item[\textup{(i)}] $|B(\Gamma,\delta)| \geq \delta^d |G|$;
\item[\textup{(ii)}] $|B(\Gamma,2 \delta)| \leq 5^d|B(\Gamma,\delta)|$.
\end{enumerate}
\end{lemma}
\noindent\textit{Proof. } For any $\eta$, write $S_{\eta}$ for the set of all
$y = (y_1,\dots,y_d) \in \mathbb{R}^d/\mathbb{Z}^d$ for which
$\Vert y_j\Vert_{\infty} \leq \eta/2$ for all $j = 1,\dots,d$ (to define the $\Vert \cdot \Vert_{\infty}$ norm on $\mathbb{R}/\mathbb{Z}$ we identify it with $(-\frac{1}{2},\frac{1}{2}]$).
Let $\Gamma=\{(\gamma_1,\dots,\gamma_d)\}$. Now if $x \in G$ write \[ v(x) \; = \; \textstyle \frac{1
}{2\pi} \displaystyle(\arg \gamma_1(x),\dots,\arg \gamma_d(x)) \; \in \;\mathbb{R}^d/\mathbb{Z}^d.\]
If $v(x)$ and $v(x')$ both lie in some translate $a + S_{\delta}$ then $x - x' \in B(\Gamma,\delta)$,
and so for fixed $x' \in G\cap v^{-1}((a + S_{\delta}))$ the map $x \mapsto x - x'$ defines an injection from $G \cap v^{-1}(a + S_{\delta})$ to $B(\Gamma,\delta)$. Hence
\begin{equation}\label{eq476} |v(G) \cap (a + S_{\delta})| \; \leq \; |B(\Gamma,\delta)|.\end{equation}
\textit{Proof of} (i). By a simple averaging there is some translate $a + S_{\delta}$ such that
\[ |v(G) \cap (a + S_{\delta})| \; \geq \; |S_{\delta}||v(G)| \; = \; |S_{\delta}| \; = \; \delta^d|G|.\] The result is now immediate from \eqref{eq476}.\vspace{11pt}

\noindent\textit{Proof of} (ii). From \eqref{eq476} one has
\[ \left|v(B(\Gamma,2 \delta)) \cap (a + S_{\delta})\right| \; \leq \;\left|v(G) \cap (a + S_{\delta})\right| \; \leq \; |B(\Gamma,\delta)|.\]
Now $v(B(\Gamma,2\delta)) \cap (a + S_{\delta})$ is empty unless $a \in S_{5\delta}$, and so
\begin{equation}\tag*{\usebox{\proofbox}}
|B(\Gamma,2\delta)| \; = \; \frac{1}{|S_{\delta}|} \int_{\mathbb{R}^d/\mathbb{Z}^d} \left|v(B(\Gamma,2 \delta)) \cap (a + S_{\delta})\right| da \; \leq \; |B(\Gamma,\delta)| \cdot \frac{|S_{5\delta}|}{|S_{\delta}|} \; = \; 5^d \cdot |B(\Gamma,\delta)|.
\end{equation}

\noindent Suppose now that $x \in B(\Gamma,\epsilon)$ and that $x' \in B(\Gamma,\epsilon')$, where $\epsilon' \ll \epsilon$. Then $x + x' \in B(\Gamma,\epsilon + \epsilon')$. Assuming that $B(\Gamma,\epsilon + \epsilon')$ is not much bigger than $B(\Gamma
,\epsilon)$, this means that $B(\Gamma,\epsilon) + B(\Gamma,\epsilon') \approx B(\Gamma,\epsilon)$. After passing to the associated normalised cutoff functions $\beta$ and $\beta'$, we might anticipate that $\beta \ast \beta' \approx \beta$. When true this property, which says that Bohr cutoffs are roughly invariant under convolution by narrower Bohr cutoffs, is extremely useful.\vspace{11pt}

\noindent Unfortunately the property does not always hold, since the size of $B(\Gamma,t)$ need not vary very smoothly with $t$. For example, suppose that $G = \mathbb{F}_5^n$ and that $\delta_1 < 1/5 < \delta_2$. Then if $|\Gamma|$ consists of $d$ linearly independent characters we have $|B(\Gamma, \delta_1)| = 5^{-d}|G|$ whilst $|B(\Gamma,\delta_2)| = (3/5)^d |G|$. \vspace{11pt}

\noindent Bourgain showed, and we shall repeat his argument, that for a fixed $\Gamma$ there are a plentiful supply of $\epsilon$ such that the function $|B(\Gamma,t)|$ does vary quite regularly for $t \approx \epsilon$, in the following sense.

\begin{definition}[Regular value]
Let $\Gamma$ be a set of $d$ characters. We say that $B(\Gamma,\epsilon)$ is \textup{regular value}, or that $\epsilon$ is a \textup{regular value} for $\Gamma$, if
\[ 1 - 100d |\kappa| \leq \frac{|B(\Gamma,(1 + \kappa)\epsilon)|}{|B(\Gamma,\epsilon)|}  \leq  1 + 100d|\kappa| \] whenever $|\kappa| \leq 1/100d$.
\end{definition}

\begin{proposition}\label{reg-prop} Let $\Gamma$, $|\Gamma| = d$, be a fixed set of characters. Let $\delta \in (0,1)$. Then there is $\epsilon \in [\delta,2\delta)$ which is regular for $\Gamma$.
\end{proposition}
\noindent\textit{Proof. }
Set $f(\alpha) = |B(\Gamma,2\alpha\delta)|$. Then
$f(\alpha)$ is a non-decreasing function on $[\frac{1}{2},1]$ and, by Lemma \ref{lem4.00},
\begin{equation} \label{fineq} f(1)  \leq  5^d f(1/2).\end{equation}  We wish to show that there is $\alpha \in [1/2,1]$ such that $1 - 100d |\kappa|  \leq  f((1 + \kappa)\alpha)/f(\alpha)  \leq  1 + 100 d |\kappa|$ for all $|\kappa| \leq 1/100d$. Suppose
 then that this is false. Observe that $\frac{1}{1 - x} \geq 1 + x$ when $x \geq 0$; hence for every $\alpha \in [1/2,1]$ there is $t_{\alpha} \in [0,1/100d]$ such that
\begin{equation}
\label{eq2} \left| \frac{f \left( (1 + t_{\alpha}) \alpha \right)}{f \left( (1 - t_{\alpha}) \alpha \right)} \right|  \geq  1 + 100 d t_{\alpha}
 \geq  e^{50 d t_{\alpha}}, \end{equation}
the last step following because $1 + x \geq e^{x/2}$ for $x \leq 1$.\vspace{11pt}

\noindent The next lemma is related to the Vitali covering lemma, but is so simple that we provide an independent proof. For an anecdotal discussion see \cite{croft}.
\begin{lemma}
\label{lemma6}
Suppose a finite collection of closed intervals $I_1,\ldots,I_k$ covers $[0,1]$. Then we can pick a subcollection $I_{i_1},\ldots,I_{i_m}$ whose members are disjoint except possibly at their endpoints, with total measure at least $1/2$.
\end{lemma}
\noindent\textit{Proof. } Without loss of generality suppose that the collection $I_1,\ldots,I_k$ is minimal in that if any $I_j$ is removed, the intervals no longer cover $[0,1]$. It is then easy to see that no point $x$ lies in three of the $I_j$, because there are two in
tervals $I_r$ and $I_s$ containing $x$ such that any other $I_t$ containing $x$ lies in $I_r \cup I_s$. But it is then easy to describe the form of the intervals exactly. Suppose that the $I_j = [a_j,b_j]$ with $a_1 \leq a_2 \leq \ldots \leq a_k$. Then \[
 a_1 \leq a_2 \leq b_1 \leq a_3 \leq b_2 \leq a_4 \leq \ldots \leq b_{k-1} \leq b_k. \]
It follows that the two collections $I_1 \cup I_3 \cup \ldots$ and $I_2 \cup I_4 \cup \ldots$ contain disjoint intervals. The result is now obvious. \hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent To apply Lemma \ref{lemma6}, recall \eqref{eq2}. By compactness we may take a finite set \[ \{\alpha_1,\ldots,\alpha_k\} \subseteq [ \textstyle\frac{1}{2} + \frac{1}{100d},1 - \frac{1}{100d} ] \] such that the intervals
$\left[ \left(1 - t_{\alpha_i}\right)\alpha_i,\left(1 + t_{\alpha_i}\right)\alpha_i \right]$ cover $\left[ \frac{1}{2} + \frac{1}{100d},1  -  \frac{1}{100d}  \right]$. Since  $t_{\alpha_i}  \leq  1/100d$, all of these intervals are contained in
$[\frac{1}{2},1]$. By Lemma \ref{lemma6},
 we can pick a disjoint subcollection of measure at least $\frac{1}{4}\left(1 - 1/100d\right) > \frac{1}{5}$. Letting these intervals correspond to $\{ \alpha_{1},\ldots,\alpha_{l} \}$, one has
$2 \sum_{i = 1}^l \alpha_i t_{\alpha_i}  >  1/5$,
and so $\sum_{i = 1}^l t_{\alpha_i}  >  1/10$.
Using this in \eqref{eq2} gives
\[ \prod_{i = 1}^l \left| \frac{f \left( (1 + t_{\alpha_i}) \alpha_i \right)}{f \left( (1 - t_{\alpha_i}) \alpha_i \right)} \right| \; \geq \; e^{50d \sum_{i} t_{\alpha_i}} \; > \; e^{5d}.\]
However the left hand side is at most $f(1)/f(1/2)$, and hence by \eqref{fineq} we have
\[ 5^d \; > \; e^{5d}.\]
This is a contradiction, and Proposition \ref{reg-prop} is established. \hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent\textit{Remark.} Actually, Proposition \ref{reg-prop} easily follows from
weak type estimates for the Hardy--Littlewood maximal function. Let
$\nu$ be a bounded non-decreasing function $\mathbb{R} \to \mathbb{R}$
with an associated Radon---Stieltjes measure
\[\nu[a,b)=\nu(b)-\nu(a)\quad(a<b).\]
Note that
\[\nu(\mathbb{R})=\lim_{b\to\infty}\nu[-b,b).\]
The Hardy--Littlewood maximal function is defined as
\[M_\nu(t)=\sup_{a<t<b}\frac{\nu(b)-\nu(a)}{b-a}.\]
The weak type estimates for $M_n(u)$ claim that for any $\alpha>0$
\begin{equation}\label{wti}
\mu\{t:\,M_\nu(t)>\alpha\}\le\frac{2\nu(\mathbb{R})}{\alpha}
\end{equation}
where $\mu$ is the Lebesgue measure. For an absolutely continuous $\nu$
this inequality is classical; a general case was established in \cite{luk}.
One can prove \ref{reg-prop}
applying (\ref{wti}) to the function $\nu$ such that $\nu(x)=\log f(e^x)$
for $x\in[-\log 2, 0]$, $\nu(x)=\nu(-\log 2)$ for $x<-\log 2$, and
$\nu(x)=\nu(0)$ for $x>0$. \vspace{11pt}

\begin{lemma}\label{lem6.7} Let $\Gamma \subseteq G^*$ be a set of $d$ characters, and let $\delta,\epsilon > 0$. Suppose that $\eta$ is a regular value for $\Gamma$, and let $\beta = \beta_{\Gamma,\eta}$ be the normalised Bohr cutoff associated to
$B(\Gamma,\eta)$. Suppose that $y \in B(\Gamma,\eta')$, where $\eta' \leq \epsilon\eta/200d$. Then
\[ \mathbb{E}_{x \in G} |\beta(x + y) - \beta(x)|  \leq \epsilon.\]
\end{lemma}
\noindent\textit{Proof. } Set $\delta := \epsilon/200d$. If $\beta(x+y) - \beta(x) \neq 0$ then
$x \in B(\Gamma, \eta(1 + \delta)) \setminus B(\Gamma,\eta(1 - \delta))$. Thus, since $B = B(\Gamma,\eta)$ is regular, we have
\[ \sum_{x} |1_B(x+y) - 1_B(x)| \leq 200d\delta |B|,\]
which immediately implies the result.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent The following corollary will be very useful. It tells us that if $\widehat{\beta}(\gamma)$ is moderately large, then $\gamma(y) \approx 1$ for $y$ in a smaller Bohr set $B(\Gamma,\eta')$.

\begin{lemma}\label{lem3.6}
Let $B(\Gamma,\eta)$ be a regular Bohr set with normalised cutoff $\beta = \beta_{\Gamma,\eta}$. Suppose that $\kappa_1,\kappa_2 > 0$. Suppose that $\gamma$ is a character for which  $|\widehat{\beta}(\gamma)| \geq \kappa_1$, and that $y \in B(\Gamma,\eta
')$, where $\eta' \leq \kappa_1 \kappa_2 \eta/200d$. Then $|1 - \gamma(y)| \leq \kappa_2$.
\end{lemma}
\noindent\textit{Proof. } We have
\[ \kappa_1 |1 - \gamma(y)| \leq |\widehat{\beta}(\gamma)||1 - \overline{\gamma(y)}| = \big|\mathbb{E}_{x \in G} (\beta(x+y) - \beta(x))\gamma(x)\big|,\]
which is at most $\kappa_1 \kappa_2$ by Lemma \ref{lem6.7}. The result follows. \hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent\textit{Remark.} By employing appropriate smoothing devices, one may obtain a reasonably good quantitative description of the set of $\gamma$ for which $|\widehat{\beta_{\Gamma,\epsilon}}(\gamma)| \geq \delta$ when $B(\Gamma,\epsilon)$ is regular. If $
\Gamma = \{\gamma_1,\dots,\gamma_d\}$ these characters will all be of the form $\gamma_1^{a_1}\dots \gamma_d^{a_d}$, where the $a_i$ are integers bounded in absolute value by some $F(\delta)$. We will not need to use such information in this paper.

\section{Proofs of the main theorems}\label{sec4}
\noindent Our first task in this section will be to supply a proof of Theorem \ref{mainthm2}. We begin with some motivating remarks which, it will turn out, take us some distance into the proof itself. Let $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ be a function with
$\Vert \widehat{f} \Vert_1 \leq 1$, as in the statement of that theorem. Suppose to begin with that we have defined a splitting $f = f^{(1)} + f^{(2)}$ where
\[ f^{(1)}(x) := \sum_{\gamma \in \Gamma} \widehat{f}(\gamma) \overline{\gamma(x)}\]
for some reasonably small set $\Gamma$. If $f^{(1)}(x_0)$ is small then $f^{(1)}(x + x_0)$ will also be small provided $x \in B(\Gamma,\eta)$, for suitably small $\eta$. Our first observation is that much the same would be true if
$\Gamma = \{\gamma_1,\dots,\gamma_d\}$ were replaced by some set
\[ \{ \gamma_1^{a_1}\dots \gamma_d^{a_d} : |a_i| \leq M\},\] consisting of small combinations of elements of $\Gamma$. By remarks made at the end of \S \ref{sec3}, the characteristic function of this set rather resembles the function
$\widehat{\beta_{\Gamma,\eta}}$, for suitable $\eta$. Once this is noticed, one might think to redefine
\[ f^{(1)} := \sum_{\gamma} \widehat{\beta_{\Gamma,\eta}}(\gamma) \widehat{f}(\gamma) \overline{\gamma(x)} = f \ast \beta_{\Gamma,\eta},\]
which looks extremely natural. To begin one takes $\Gamma = \emptyset$; we add frequencies according to an iterative procedure.
\begin{lemma}\label{lem4.1}
Suppose that $B(\Gamma,\eta)$ is regular, that $|f^{(1)}(x_0)| \leq \epsilon$,
and that $x \in B(\Gamma,\eta')$ for some $\eta' \leq \epsilon^2 \eta/200d$.
Then $|f^{(1)}(x + x_0)| \leq 3\epsilon$.
\end{lemma}
\noindent\textit{Proof. } We have
\[
|f^{(1)}(x + x_0) - f^{(1)}(x_0)|  =   \big|\sum_\gamma \widehat{f}(\gamma)\widehat{\beta_{\Gamma,\eta}}(\gamma)\gamma(x_0)(\gamma(x) - 1)\big| \leq  \sup_{\gamma} \big|\widehat{\beta_{\Gamma,\eta}}(\gamma)\big| \big|1 - \gamma(x)\big|.\]
Now if $\big|\widehat{\beta_{\Gamma,\eta}}(\gamma)\big| \leq \epsilon$ then clearly
\begin{equation}\label{eq3.87} \big|\widehat{\beta_{\Gamma,\eta}}(\gamma)\big| \big|1 - \gamma(x)\big| \leq 2\epsilon.\end{equation}
If, however, $\big|\widehat{\beta_{\Gamma,\eta}}(\gamma)\big| \geq \epsilon$ and if $x \in B(\Gamma,\eta')$ then Lemma \ref
{lem3.6} tells us that $|1 - \gamma(x)| \leq \epsilon$, and so \eqref{eq3.87} holds in this case too.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent Now define
\begin{equation}\label{eq4.4a} f^{(2)} := f - f^{(1)}.\end{equation} Suppose that $f^{(1)}(x_0)$ is small. By Lemma \ref{lem4.1} we know that $f^{(1)}(x + x_0)$ is small for all $x \in B(R,\eta')$, provided $\eta'$ is not too large. Our hope is that $f^{(1)}(x + x_0)$ will also be small for some such $x$. To this end we look at
\begin{equation}\label{eq4.3} E := \mathbb{E}_{x \in G}  |f^{(2)}(x + x_0)|^2 \beta_{\Gamma,\eta'}(x) \leq \sum_{\gamma,\gamma'} |\widehat{f^{(2)}}(\gamma)||\widehat{f^{(2)}}(\gamma')||\widehat{\beta_{\Gamma,\eta}}(\gamma - \gamma')|,
\end{equation}
the inequality being an immediate consequence of \eqref{eq2.6}. Write
\[ U := \{ \chi : |\widehat{\beta_{\Gamma,\eta'}}(\chi)| \geq \epsilon^2\}.\] Observing that
\[ \sum_{\gamma} |\widehat{f^{(2)}}(\gamma)| = \sum_{\gamma} |\widehat{f}(\gamma)||1 - \widehat{\beta_{\Gamma,\eta}}(\gamma)| \leq \Vert \widehat{f} \Vert_1\sup_{\gamma} | 1 - \widehat{\beta_{\Gamma,\eta}}(\gamma)| \leq 2,\]
it is immediate from \eqref{eq4.3} that
\begin{equation}\label{eq4.4} E \leq 4\epsilon^2  + \sum_{\substack{\gamma,\gamma' \\ \gamma - \gamma' \in U}} |\widehat{f^{(2)}}(\gamma)||\widehat{f^{(2)}}(\gamma')||\widehat{\beta_{\Gamma,\eta'}}(\gamma - \gamma')| \leq 4\epsilon^2  + 2\sup_{\gamma}
\sum_{\gamma' : \gamma' - \gamma \in U} |\widehat{f^{(2)}}(\gamma')|.\end{equation}
Now if
\begin{equation}\label{eq4.5b} \sup_{\gamma} \sum_{\gamma' : \gamma' - \gamma \in U} |\widehat{f^{(2)}}(\gamma')| \leq \epsilon^2\end{equation}
then $E \leq 6\epsilon^2$, and there is indeed $x \in B(\Gamma,\eta')$ such that
$|f^{(2)}(x+x_0)| \leq \epsilon\sqrt{6} \leq 3 \epsilon$. For this $x$, we have
$|f(x+x_0)| \leq |f^{(1)}(x+x_0)| + |f^{(2)}(x+x_0)| \leq 6\epsilon$. \vspace{11pt}

\noindent If, on the other hand, there is an $\gamma$ such that
\begin{equation}\label{eq4.5a} \sum_{\gamma' : \gamma' - \gamma \in U} |\widehat{f^{(2)}}(\gamma')| \geq \epsilon^2\end{equation} then we must find an alternative argument. It is tempting to somehow ``add'' the offending characters $u + \gamma$, $u \in U$
, into the definition of $f^{(1)}$. Thus we might consider a new set $\widetilde{\Gamma} := \Gamma \cup \{\gamma\}$ and some $\widetilde{\eta} \ll \eta$ such that $\widehat{\beta_{\widetilde{\Gamma},\widetilde{\eta}}}(s) \approx 1$ for $s = u + \gamma$.
\vspace{11pt}

\noindent Once this is done, one may iterate the whole procedure starting with \eqref{eq4.4a}. If things are done carefully, the obstacle \eqref{eq4.5a} cannot be encountered more than $\epsilon^{-2}$ times since each new instance corresponds to at least
$\epsilon^2$ of the total $L^1$-mass of $\widehat{f}$, which we are assuming is at most $1$. Care must be taken to ensure that these portions of $L^1$-mass are all disjoint, but we will see later on that this can be arranged.\vspace{11pt}

\noindent Working out the details of the above argument, one obtains a bound $\min_x |f(x)| = O((\log p)^{-1/4 + \varepsilon})$ in Theorem \ref{mainthm2}. To obtain the superior exponent $\frac{1}{3}$, we exploit the fact that it is not necessary for $f^{(2)}$ itself to satisfy \eqref{eq4.5b}; we may perturb $f^{(2)}$ by an arbitrary function $f^{(3)}$ for which $\Vert \widehat{f^{(3)}} \Vert_1 \leq \epsilon$, since then
\begin{equation}\label{eq4.12} |f^{(3)}(x)| = \big| \sum_{\gamma} \widehat{f^{(3)}}(\gamma) \overline{\gamma(x)} \big| \leq \Vert \widehat{f^{(3)}} \Vert_1 \leq \epsilon\end{equation} for an arbitrary $x$.
Thus in effect we wish to write \begin{equation}\label{eq4.11} f = f^{(1)} + f^{(2)} + f^{(3)},\end{equation} where $f^{(2)}$ satisfies \eqref{eq4.5b} and \begin{equation}\label{eq4.8}\Vert \widehat{f^{(3)}} \Vert_1 \leq \epsilon.\end{equation} This, it transpires, can be achieved by a slight variant of the above iterative scheme in which the number of iterations is reduced to just $\epsilon^{-1}$. \vspace{11pt}

\noindent We now turn to the proof proper of Theorem \ref{mainthm2}, beginning with a lemma summarising the above discussion.
\begin{lemma}\label{lem4.14}
Let $B = B(\Gamma,\eta)$ and $B' = B(\Gamma,\eta')$ be regular Bohr sets with normalised Bohr cutoffs $\beta= \beta_{\Gamma,\eta}$ and $\beta' = \beta_{\Gamma,\eta'}$. Suppose that $\eta' \leq \epsilon^2 \eta/200d$. Set \[ U := \{ \chi : |\widehat{\beta'}
(\chi)| \geq \epsilon^2\}.\]
Suppose that $f = f^{(1)} + f^{(2)} + f^{(3)}$, where $f^{(1)} = f \ast \beta$, that
\[ \sup_{\gamma} \sum_{\gamma' : \gamma' - \gamma \in U} |\widehat{f^{(2)}}(\gamma')| \leq \epsilon^2 \] and that
$\Vert \widehat{f^{(3)}} \Vert_1 \leq \epsilon$. Suppose that $\min_x |f^{(1)}(x)| \leq 2\epsilon$.
Then $\min |f(x)| \leq 8\epsilon$.
\end{lemma}
\noindent\textit{Proof. } Suppose that $|f^{(1)}(x_0)| \leq 2\epsilon$. Then Lemma \ref{lem4.1}
tells us that for any $x \in B(R,\eta')$ we have $|f^{(1)}(x + x_0)| \leq 4\epsilon$. Now by the discussion leading to \eqref{eq4.5b} we know that there is $x \in B(R,\eta')$ for
which $|f^{(2)}(x + x_0)| \leq 3\epsilon$, and finally from \eqref{eq4.12} we have $|f^{(3)}(x + x_0)| \leq \epsilon$ for any $x$ whatsoever. The result follows. \hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent The next lemma asserts that there is a decomposition of the type discussed in Lemma \ref{lem4.14}.
\begin{lemma}\label{lem4.33}
Let $\epsilon > 0$. Then there is a set $\Gamma$, $|\Gamma| \leq 2/\epsilon^2$, of characters, an $\eta \geq (\epsilon^9/2^{21})^{2/\epsilon}$ and an $\eta' \geq 2^{-10}\epsilon^3 \eta$ such that the hypotheses of Lemma \ref{lem4.14} are satisfied, with the possible exception of the inequality $\min_x |f^{(1)}(x)| \leq \epsilon$.
\end{lemma}
\noindent\textit{Proof. } We will have a kind of double iterative scheme, the \texttt{STEP}s of which will be indexed by pairs $(j,i)$. The two coordinates of $\mbox{\texttt{STEP}}$ will be denoted $\mbox{\texttt{STEP}}_1$ and $\mbox{\texttt{STEP}}_2$. The algorithm will proceed by first incrementing $\mbox{\texttt{STEP}}_2$, that is to
 say according to the following scheme:
\[ (0,0) \rightarrow (0,1) \rightarrow \dots \rightarrow (0,l_0) \rightarrow (1,0) \rightarrow (1,1) \rightarrow \dots \rightarrow (1,l_1) \rightarrow (2,0) \rightarrow \dots\]
Whilst $\mbox{\texttt{STEP}}_1 = j$ there will be a set $\Gamma_{j}$ of characters and regular
Bohr sets $B(\Gamma_{j},\eta_{j})$, $B(\Gamma_{j},\eta'_{j})$ with normalised Bohr cutoffs $\beta_{j} := \beta_{\Gamma_{j},\eta_{j}}$ and $\beta'_j := \beta_{\Gamma_{j},\eta'_{j
}}$. The parameters $\eta_{j}$ and $\eta'_j$ will be related by the inequalities
\begin{equation}\label{eq4.20} \epsilon^2 \eta_{j}/400d \leq \eta'_j \leq \epsilon^2 \eta_{j}/200d,
\end{equation}
in order that we may apply Lemma \ref{lem4.14}. In fact $\eta'_{j}$ can simply
be an arbitrary parameter in this range for which $B(\Gamma_j,\eta'_{j})$ is regular. There will also be disjoint sets $S_{j,i}$ of characters, $i = 1,2,\dots$. When $\mbox{\texttt{STEP}} = (j,
i)$ we will have occasion to discuss the union
\[ \Omega_{j} := \bigcup_{j' < j} \bigcup_{i\leq l_{j'}}S_{j',i}\]
of all characters which were defined whilst $\mbox{\texttt{STEP}}_1 < j$. Things will have been arranged so that $B(\Gamma_{j},\eta_{j}) \subseteq B(\Omega_{j},\epsilon)$; the set $\Omega_{j}$ should be thought of as containing small combinations of elements of $\Gamma_{j}$, and in fact the proof could be set up in such a way that this is made explicit (see the remarks at the end of \S \ref{sec3}). In proceeding from $\mbox{\texttt{STEP}} = (j,0)$ to $\mbox{\texttt{STEP}} = (j,i)$ we will have accrued a collection $\{\gamma_{j,1},\dots,\gamma_{j,i}\}$
of characters; these will be used to define $\Gamma_{j+1}$.\vspace{11pt}

\noindent To initialize the iteration set $\mbox{\texttt{STEP}} = (0,0)$, and define $\Gamma_0 = \emptyset$ and $\eta_0 = 1$.  Now suppose that $\mbox{\texttt{STEP}} = (j,i)$. Define
\[ f^{(1)}_j := f \ast \beta_{j},\] and set $g_j := f - f^{(1)}_j$. \vspace{11pt}

\noindent Define $U_{j} := \{ s : |\widehat{\beta'_j}(s)| \geq \epsilon^2\}$.
We will have reached $\mbox{\texttt{STEP}} = (j,i)$ by incrementing $\mbox{\texttt{STEP}}_2$, starting from $\mbox{\texttt{STEP}} = (j,0)$. During this process we will have defined disjoint sets $S_{j,1},S_{j,2},\dots,S_{j,i}$ of frequencies, and also a collection $\{\gamma_{j,1},\dots,\gamma_{j,i}\}$.
 The construction will be such that
\begin{equation}\label{eq4.16}
S_{j,i} \subseteq \gamma_{j,i} + U_{j}
\end{equation}
for all $i$.
We begin by asking whether or not
\begin{equation}\label{eq4.7} \sum_{\gamma' \in S_{j,1} \cup S_{j,2} \cup \dots \cup S_{j,i}} |\widehat{g_j}(\gamma')| \geq \epsilon.\end{equation}
If so, it is time to increment $\mbox{\texttt{STEP}}_1$ and to define $\Gamma_{j+1}$ and $\eta_{j+1}$. We set $\mbox{\texttt{STEP}} = (j+1,0)$, $\Gamma_{j+1} := \Gamma_j \cup \{\gamma_{j,1},\dots,\gamma_{j,i}\}$ and choose $\eta_{j+1}$,
\[  \epsilon^3\eta'_j/800|\Gamma_j| \leq \eta_{j+1} \leq \epsilon^3\eta'_j/400|\Gamma_j|,\] in such a way that the Bohr set $B(\Gamma_{j+1},\eta_{j+1})$ is regular. Clearly $\eta_{j+1} \leq \min(\eta_{j},\epsilon/2)$. Note that
\[ \Omega_{j+1} = \Omega_{j} \cup S_{j,1} \cup S_{j,2} \cup \dots \cup S_{j,i}.\]
\noindent\textit{Claim 1.} We have the inclusion \[B(\Gamma_{j+1},\eta_{j+1}) \subseteq B(\Omega_{j+1},\epsilon).\]
\noindent\textit{Proof. } Clearly \[ B(\Gamma_{j+1},\eta_{j+1}) \subseteq B(\Gamma_j,\eta_{j}) \subseteq B(\Omega_{j},\epsilon) \subseteq B(\Omega_{j+1},\epsilon),\] and so it suffices to check that $B(\Gamma_{j+1},\eta_{j+1}) \subseteq B(S_{j,i},\epsilon)$. For this, in view of \eqref{eq4.16}, we only need confirm that \[ B(\Gamma_{j+1},\eta_{j+1}) \subseteq B(\gamma_{j,i},\epsilon/2) \cap B(U_{j},\epsilon/2)\] for each $i$. The inclusion $B(\Gamma_{j+1},\eta_{j+1}) \subseteq B(\gamma_{j,i},\epsilon/2)$ is immediate from t
he fact that $\eta_{j+1} \leq \epsilon/2$. To see that $B(\Gamma_{j+1},\eta_{j+1}) \subseteq B(U_{j},\epsilon/2)$, we use Lemma \ref{lem3.6}. Indeed if $\gamma \in U_{j}$ and if $y \in B(\Gamma_j,\eta_{j+1})$ then by that lemma we have $|1 - \gamma(y)| \leq \epsilon/2$, which is exactly what we need.\vspace{11pt}

\noindent Now if \eqref{eq4.7} does not hold then we carry on incrementing $\mbox{\texttt{STEP}}_2$.  Split $g_{j}$ as $f^{(2)}_{j,i} + f^{(3)}_{j,i}$, where
\[ f^{(3)}_{j,i}(x) := \sum_{\gamma \in \Omega_{j} \cup S_{j,1} \cup \dots \cup S_{j,i}} \widehat{g_{j}}(\gamma)\overline{\gamma(x)}.\]
If \[ \sup_{\gamma} \sum_{\gamma' \in \gamma + U_{j}} |\widehat{f^{(2)}_{j,i}}(\gamma')| \leq \epsilon^2\] then \textbf{STOP}. Otherwise, there is a $\gamma_{j,i+1}$ with \begin{equation}\label{eq4.81} \sum_{\gamma' \in \gamma_{j,i+1} + U_{j}} |\widehat{f
^{(2)}_{j,i}}(\gamma')| \geq \epsilon^2.\end{equation} Set $\mbox{\texttt{STEP}} = (j,i+1)$, $S_{j,i+1} := (\gamma_{j,i+1} + U_{j}) \setminus (\Omega_{j} \cup S_{j,1} \cup \dots \cup S_{j,i})$. An important point to note is that
\begin{equation}\label{eq4.9} \sum_{\gamma' \in S_{j,i+1}} |\widehat{f^{(2)}_{j,i}}(\gamma')| \geq \epsilon^2:\end{equation} although this is ostensibly a stronger statement than \eqref{eq4.81}, this is merely an illusion since by construction the support
 of $\widehat{f_{j,i}^{(2)}}$ is disjoint from $\Omega_{j} \cup S_{j,1} \cup \dots \cup S_{j,i}$.\vspace{11pt}

\noindent\textit{Claim 2.} The algorithm terminates, and in fact we have the bound $|\Gamma_l| \leq 2\epsilon^{-2}$, independently of $l$.\\
\noindent\textit{Proof. } To each element $\gamma_{j,i}$, $j \leq l$ contained in $\Gamma_l$ is associated, by \eqref{eq4.9}, a set $S_{j,i}$ with the property that
\[ \sum_{\gamma \in S_{j,i}} |\widehat{f^{(2)}_{j,i-1}}(\gamma)| \geq \epsilon^2.\] This implies that
\[ 2\sum_{\gamma \in S_{j,i}} |\widehat{f}(\gamma)| \geq \sum_{\gamma \in S_{j,i}} |\widehat{f}(\gamma)||1 - \beta_{j}(\gamma)| = \sum_{\gamma \in S_{j,i}} |\widehat{f^{(2)}_{j,i-1}}(\gamma)| \geq \epsilon^2.\] Recalling that the sets $S_{j,i}$ are disjoint and that $\Vert \widehat{f} \Vert_1 \leq 1$, the claim follows.\vspace{11pt}

\noindent Suppose that the algorithm terminates when $\mbox{\texttt{STEP}}_1 = l$. The whole purpose of having an iteration somewhat more complicated than the one we outlined at the start of the section is that we can beat the crude bound $l \leq 2\epsilon^{-2}$ which follows from Claim 2. \vspace{11pt}

\noindent\textit{Claim 3.} We have $l \leq 2/\epsilon$.\\
\noindent\textit{Proof. } Set $S_j = \bigcup_i S_{j,i}$. For any $j$ we have \[ 2\sum_{\gamma \in S_{j}} |\widehat{f}(\gamma)| \geq \sum_{\gamma \in S_{j}} |\widehat{f}(\gamma)||1 - \widehat{\beta_{j}}(\gamma)| = \sum_{\gamma \in S_{j}} |\widehat{g_{j}}(\gamma)| \geq
\epsilon.\]
Since the sets $S_{j}$ are disjoint and $\Vert \widehat{f} \Vert_1 \leq 1$, we do indeed have $l \leq 2/\epsilon$.\vspace{11pt}

\noindent To conclude the proof of Lemma \ref{lem4.33}, suppose that $\mbox{\texttt{STEP}} = (l,m)$ at termination. By construction the decomposition
\[ f = f^{(1)}_{l,m} + f^{(2)}_{l,m} + f^{(3)}_{l,m}\] satisfies all of the properties required. It remains to check the bounds claimed. This is a simple matter; using the crude estimate $\eta_{j+1} \geq \epsilon^5\eta'_{j}/2^{11} \geq \epsilon^9\eta_{j}/
2^{21}$, we do indeed have $\eta_l \geq (\epsilon^9/2^{21})^{2/\epsilon}$.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent In order to apply Lemma \ref{lem4.33}, we must be able to exhibit a value of $x$ for which $|f^{(1)}(x)|$ is small. This is where we invoke a kind of ``discrete intermediate value theorem'' of the type used in \S \ref{sec1}. This argument, as the reader will appreciate, is specific to the case $G = \mathbb{Z}/p\mathbb{Z}$.

\begin{lemma}\label{lem4.77}
Let $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ be a function with $\mathbb{E}_x f(x) = 0$. Let $B = B(\Gamma,\eta)$ be a regular Bohr set with normalised cutoff $\beta$, and suppose that $f^{(1)} = f \ast \beta$. Suppose that $(\epsilon^2 \eta/200d)^d p > 1$. Then
$\min |f^{(1)}(x)| \leq 2\epsilon$.
\end{lemma}
\noindent\textit{Proof. } We invoke Lemma \ref{lem4.1}, or rather the proof of it, which tells us that if $\eta' = \epsilon^2 \eta/200 d$ and if $t \in B(\Gamma,\eta')$ then
\begin{equation}\label{eq4.21} |f^{(1)}(x + t) - f^{(1)}(x)| \leq 2\epsilon\end{equation}
for any $x$. The conditions of the lemma, together with the simple bound of Lemma \ref{lem4.00} (i), imply that $|B(\Gamma,\eta')| > 1$, so there is some $t \in B(\Gamma,\eta')$, $t \neq 0$. The elements $\{0,t,2t,3t,\dots, (p-1)t\}$ are in fact just the
elements of $\mathbb{Z}/p\mathbb{Z}$, listed once each. Since $\mathbb{E}_x f^{(1)}(x) = 0$, there must be a value of $j$ for which
$f^{(1)}(jt)$ and $f^{(1)}((j+1)t)$ have opposing signs. For this value of $j$,
\eqref{eq4.21} guarantees that $|f^{(1)}(jt)| \leq 2\epsilon$.\hfill{\usebox{\proofbox}}
\vspace{11pt}

\noindent\textit{Proof of Theorem \ref{mainthm2}.} Lemma \ref{lem4.33} tells us that there is a set $\Gamma$, $d := |\Gamma| \leq 2/\epsilon^2$, and an $\eta \geq (\epsilon^9/2^{21})^{2/\epsilon}$ for which the hypotheses of Lemma \ref{lem4.14} are satisfied.
If in addition $\min_x |f^{(1)}(x)| \leq 2\epsilon$, then the conclusion of that lemma
implies that $\min_x |f(x)| \leq 8\epsilon$. By Lemma \ref{lem4.77}, this will be the case if $(\epsilon^2 \eta/200d)^d p > 1$, which will be satisfied if $\epsilon
=C(\log p/\log \log p)^{1/3}$ for suitable $C$.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent It is a simple matter to deduce Theorem \ref{mainthm1} from Theorem \ref{mainthm2}. Recall that $S(p)$ is the smallest possible value of $\Vert \widehat{1_A} \Vert_1$, where $A \subseteq \mathbb{Z}/p\mathbb{Z}$ is a set with cardinality $\lfloor p/2 \rfloor$. \vspace{11pt}

\noindent\textit{Proof of Theorem \ref{mainthm1}.} Set $f(x) := (21_A(x) - 1)/\Vert (21_A - 1)^{\wedge} \Vert_1$, so that $\Vert \widehat{f} \Vert_1 = 1$, and apply Theorem \ref{mainthm2}. One obtains
\[ \Vert (21_A - 1)^{\wedge} \Vert_1^{-1} = \min_x |f(x)| = O(\log \log p/\log p)^{1/3}.\]
Now
\[ (21_A - 1)^{\wedge}(r) = \left\{ \begin{array}{ll} 2\widehat{1_A}(r) & \mbox{if $r \neq 0$} \\ O(1/p) & \mbox{if $r = 0$},   \end{array}  \right.\]
which means that
\[ \Vert (21_A - 1)^{\wedge} \Vert_1 \geq 2 \Vert \widehat{1_A} \Vert_1 + O(1).\] The result follows immediately.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent To conclude this section we give the proof of Theorem \ref{mainthm3.1} and hence, by the remarks in \S \ref{sec1}, answer the question of Gowers. \vspace{11pt}

\noindent\textit{Proof of Theorem \ref{mainthm3.1}.} Consider the functions
\[ f_0(x) := f(x) - \mathbb{E} f,\]
\[ g_0(x) := g(x) - \mathbb{E} g\]
and
\[ h(x) := f \ast g - \mathbb{E} f \mathbb{E} g.\]
It is easy to see that $\mathbb{E} f_0 = \mathbb{E} g_0 =0$, that $\Vert f_0 \Vert_2, \Vert g_0 \Vert_2 \leq 1$, and that
\[ h = f_0 \ast g_0.\]
In particular we have, by the Cauchy-Schwarz inequality and Parseval's identity, that
\[ \Vert \widehat{h} \Vert_1 \leq \Vert f_0 \Vert_2\Vert g_0 \Vert_2 \leq 1\]
(this is a special case of Young's inequality).
The result follows immediately from Theorem \ref{mainthm2}.\hfill{\usebox{\proofbox}}

\section{The proof of Theorem \ref{mainthm3.2}}\label{sec6}
\begin{lemma}\label{lem5.10} For any prime $p$ there is a function
$F : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ such that $\mathbb{E} F=0$,
$\Vert \widehat{F} \Vert_1\leq 1$, $\widehat{F}(r) = \widehat{F}(-r) \geq 0$
for all $r \in \mathbb{Z}/p\mathbb{Z}$, and $\min_x |F(x)|\gg(\log p)^{-1}$.
\end{lemma}
\noindent\textit{Proof. } We may, of course, assume that $p$ is sufficiently large. Consider to begin with the very simple function
\[ g_1(x) := \left\{ \begin{array}{ll} 1 & \mbox{if $|x/p| < 1/4$} \\ -1 & \mbox{otherwise}.  \end{array}\right.\]
We have $\mathbb{E} g_1 = O(1/p)$, whilst the Fourier transform $\widehat{g}_1(r)$ satisfies $\widehat{g}_1(r) = \widehat{g}_1(-r)$ and
\begin{equation}\label{eq5.1a} |\widehat{g}_1(r)| \leq \frac{1}{p|\sin(\pi r/p)|} \leq \frac{1}{2|r|} \end{equation}
for $|r| < p/2$. For suitable $C$, the function $F_1(x) := g_1(x)/C\log p$ satisfies all the requirements of the lemma except for the condition that $\widehat{F}_1(r)$ be non-negative (and the condition that $\mathbb{E} F_1 = 0$, though this can be achieved by a
very small perturbation of $F_1$).\vspace{11pt}

\noindent To construct a function with non-negative Fourier transform, we begin by considering an auxilliary function on $\mathbb{R}/\mathbb{Z}$. Define
\[ f_1(\theta) := -2 \log |2 \sin \pi \theta|\] for $\theta \neq 0$, and $f_1(0) = 0$.
Note that $f \in L^p$ for all $p \in [1,\infty)$. We claim that the Fourier transform of $f_1$ is given by
\begin{equation}\label{eq5.1b}
\widehat{f}_1(m) = \left\{ \begin{array}{ll} 1/|m| & \mbox{if $m \in \mathbb{Z} \setminus \{0\}$} \\ 0 & \mbox{otherwise}.  \end{array}\right.\end{equation}
Heuristically, this can be seen by observing the Taylor expansion
\[ -\log(1 - z) = z + \frac{z^2}{2} + \frac{z^3}{3} + \dots,\]
setting $z = e^{2\pi i \theta}$ and $z = e^{-2\pi i \theta}$, and adding. To justify the assertion rigourously, one could proceed as follows. Using Cauchy's formula for derivatives, one confirms that
\begin{equation}\label{eq5.1c} \frac{1}{2\pi i} \int_{\Gamma(\epsilon)} \frac{\log (1 - z)}{z^{n+1}} \, dz = -\frac{1}{n}\end{equation} for $n \geq 1$, where $\Gamma(\epsilon)$ is the contour consisting of the circle $|z| = 1$ indented to the left near $z
 = 1$ using a semicircle of radius $\epsilon$. The contribution from the semicircle of radius $\epsilon$ is $O(\epsilon \log(1/\epsilon))$, which tends to 0 as $\epsilon \rightarrow 0$. The remainder of the integral is
\[ \int_{\theta \in \mathbb{R}/\mathbb{Z} : |\theta| \geq \epsilon} \log(1 - e^{2\pi i \theta}) e^{-2\pi i n \theta} \, d\theta.\] This differs from the integral over all $\theta \in \mathbb{R}/\mathbb{Z}$ by an error of
\[ O \big( \int^{\epsilon}_0 |\log \theta | \, d\theta\big),\] which is $O(\epsilon \log(1/\epsilon))$. Letting $\epsilon \rightarrow 0$ in \eqref{eq5.1c}, then, we see that
\begin{equation}\label{eq5.1d} \int_{\theta \in \mathbb{R}/\mathbb{Z}} \log(1 - e^{2\pi i \theta}) e^{-2\pi i n\theta}\, d\theta = - \frac{1}{n}\end{equation} for $n \geq 1$. By a similar argument, this integral vanishes when $n \leq 0$. Furthermore by an almost
identical argument one may confirm that
\[ \int_{\mathbb{R}/\mathbb{Z}} \log(1 - e^{-2\pi i \theta}) e^{-2\pi i n\theta} = \left\{ \begin{array}{ll} 1/n & \mbox{if $n \leq -1$} \\ 0 & \mbox{otherwise}.\end{array}\right.\]
Adding this to \eqref{eq5.1d} establishes the claim \eqref{eq5.1b}.\vspace{11pt}

\noindent Now $f_1$ is useless to us as it stands, since $\widehat{f}_1$ does not lie in $l^1(\mathbb{Z})$. We must also transfer $f_1$ to a function defined on $\mathbb{Z}/p\mathbb{Z}$. We modify $f_1$ by defining
\begin{equation}\label{eq5.1h} f_2 := f_1 \ast \chi \ast \chi,\end{equation}
where
\[ \chi(\theta) = \frac{Cp}{2} 1_{|\theta| \leq 1/Cp}(\theta)\]
for some large constant $C$ to be specified later. Note that
\begin{equation}\label{eq5.1e} \widehat{f}_2(m) = \widehat{f}_1(m) \big(\frac{Cp\sin 2\pi m/Cp}{2\pi m}\big)^2,\end{equation} where the bracketed expression is to be interpreted as 1 if $m = 0$.
Consider the function $g_2 : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ defined by
\begin{equation}\label{eq5.1o} g_2(x) = \sum_{m \in \mathbb{Z}} \widehat{f}_2(m) e^{2\pi i m x/p}.\end{equation}
Since $\widehat{f}_2 \in l^1(\mathbb{Z})$, the series converges uniformly and
\[ g_2(x) = f_2(x/p)\] provided that $x \neq 0$. \vspace{11pt}

\noindent Now we have
\[ |f'_1(\theta)| = |2\pi \cot \pi \theta| \leq \frac{2}{|\theta|}\] for all $0 < |\theta| \leq 1/2$.
From the definition \eqref{eq5.1h}, it follows that
\begin{equation}\label{eq5.1i} |g_2(x) - f_1(x/p)| \leq \frac{2}{Cp} \sup_{|\theta - x/p| \leq 2/pC} |f'_1(\theta)| \leq 8/C\end{equation} provided that $x \neq 0$ and $C \geq 4$.\vspace{11pt}

\noindent All we need to know about $g_2(0)$ is that it is large, to which end the bound
\begin{equation}\label{eq5.1j} g_2(0) \geq g_2(1)\end{equation} will be quite sufficient.\vspace{11pt}

\noindent The Fourier transform $\widehat{g}_2(r)$ on $\mathbb{Z}/p\mathbb{Z}$ is given by
\begin{equation}\label{eq5.1f} \widehat{g}_2(r) =  \sum_{\substack{m \in \mathbb{Z} \\ m \equiv r \mdsub{p}}} \widehat{f}_2(m) \end{equation} (this is an instance of the Poisson summation formula, but in this case it follows from the inversion formula
\[ g_2(x) = \sum_r \widehat{g}_2(r) e^{2\pi i rx/p}\]
upon comparing with \eqref{eq5.1o} and recalling the uniqueness of Fourier expansion). In particular $\widehat{g}_2(r) = \widehat{g}_2(-r)$. Now $\widehat{f}_2(m)$ is real and non-negative, and hence so is $\widehat{g}_2(r)$ and we have
\[ \widehat{g}_2(r) \geq \widehat{f}_2(\overline{r}),\]
where $\overline{r}$ is the unique integer with $|\overline{r}| < p/2$ and $\overline{r} \equiv r \md{p}$. Using the inequality $\sin \theta/\theta \geq 1 - \theta^2/6$ and \eqref{eq5.1b}, we thus have
\begin{equation}\label{eq5.1ff} \widehat{g}_2(r) \geq \widehat{f}_1(\overline{r}) \big( 1 - \frac{2\pi^2 |\overline{r}|^2}{3C^2 p^2}\big)^2 \geq \frac{1}{|\overline{r}|}\big(1 - \frac{4}{C^2}\big)\end{equation} for $r \neq 0$.\vspace{11pt}

\noindent In the other direction, we have the estimate
\begin{equation}\label{eq5.1g} \Vert \widehat{g}_2 \Vert_1 = \sum_m \widehat{f}_2(m) = \sum_{\substack{m \in \mathbb{Z} \\ m \neq 0}}  \frac{1}{|m|}  \big(  \frac{Cp  \sin  2\pi  m/Cp}{2\pi  m}\big)^2  \leq  2\sum_{m=1}^{\infty}  \min(\frac{1}{m},  \frac{C^2
p^2}{4\pi^2m^3}) \ll \log p. \end{equation}
\noindent Now define
\[ g_3 := g_1 + g_2.\]
From \eqref{eq5.1a} and \eqref{eq5.1ff}, we see that $\widehat{g}_3(r) = \widehat{g}_3(-r)$, and that $\widehat{g}_3(r) \geq 0$ for all $r \neq 0$, provided that $C^2 \geq 8$. From \eqref{eq5.1a} and \eqref{eq5.1g} we see that
\begin{equation}\label{eq5.1m} \Vert \widehat{g}_3 \Vert_1 \ll \log p.\end{equation} We claim that if $C \geq 1000$ then \[ |g_3(x)| \geq 1/4 \qquad \mbox{for all $x$}.\]
To see this, observe from \eqref{eq5.1i} that if $1/4 < |x/p| \leq 1/2$ then
\[ g_3(x) = -1 + g_2(x) \leq -1 + f_1(x/p) + 8/C \leq -1 - \log 2 + 8\pi/C \leq -1.\]
If $0 < |x/p| < 1/4$ then we have
\[ g_3(x) = 1 + g_2(x) \geq 1 + f_1(x/p) - 8/C \geq 1 - \log 2 - 8\pi/C \geq 1/4.\]
Finally from \eqref{eq5.1j} we have
\[ g_3(0) = 1 + g_2(0) \geq 1 + g_2(1/p) \geq 1 + f_1(1/p) - 8/C \geq 1/4.\]
This proves the claim.\vspace{11pt}

\noindent The function $F$ (whose construction is our goal) will essentially be a rescaled version of $g_3$, but it is required to satisfy $\mathbb{E} F = 0$. To this end we note that
\[ \mathbb{E} g_1 = O(1/p),\]
whilst, from \eqref{eq5.1f}, we have
\[ \mathbb{E} g_2 = \sum_{m \equiv 0 \mdsub{p}} \widehat{f}_2(m) \leq 2\sum_{\substack{m =1 \\ m\equiv 0 \mdsub{p}}}^{\infty} \min(\frac{1}{m}, \frac{C^2p^2}{4\pi^2 m^3}) = O(1/p).\]
Thus $\mathbb{E} g_3 = O(1/p)$, and so if we define
\[ g_4 := g_3 - \mathbb{E} g_3\]
then we still have
\[ |g_4(x)| \geq 1/4 + O(1/p) \geq 1/5\]
for $p$ sufficiently large. Of course, since $\widehat{g}_4(r) = \widehat{g}_3(r)$ for $r \neq 0$, we also have $\widehat{g}_4(r) = \widehat{g}_4(-r) \geq 0$.\vspace{11pt}

\noindent At last we may define $F$ by
\[ F(x) := g_4(x)/\Vert \widehat{g}_4 \Vert_1.\] The asserted properties of $F$ are immediate from the definition, the facts we have assembled about $g_4$, and \eqref{eq5.1m}.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent We conclude this section by proving Theorem \ref{mainthm3.2}, which asserted the existence of a function $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ with $\mathbb{E} f = 0$, $\Vert f \Vert_2 \leq 1$, and $|f \ast f^{\circ}(x)| \geq c/\log p$ for all $x$.\vspace{11pt}

\noindent\textit{Proof of Theorem \ref{mainthm3.2}.} Let $F$ be the function
constructed in Lemma \ref{lem5.10}. Take the function $f$
with $\widehat{f}=\sqrt{\widehat{F}}$. Then
\[\mathbb{E} f = \widehat{f}(0) = \sqrt{\widehat{F}(0)}
= \sqrt{\mathbb{E} F} = 0,\]
\[\mathbb{E} f^2 = \sum_r \widehat{f}(r)^2 = \Vert \widehat{F} \Vert_1\leq 1,\]
and
\[ f \ast f^{\circ} = F.\]
The theorem follows. \hfill{\usebox{\proofbox}}

\section{An example for Gowers' question}\label{sec7}

\noindent In this section we prove Theorem \ref{mainthm3.3}, which asserted that there is a set $A \subseteq \mathbb{Z}/p\mathbb{Z}$ with $|A| = \lfloor p/2\rfloor$ and
\[ |A \cap (A + x) - p/4| \geq c/\log p\log\log p\] for all $x$. Noting that
\[ |A \cap (A + x) - p/4| = p f \ast f^{\circ}(x),\]
where $f := 1_A - 1/2$, we see that this is a matter of finding a function $f$ in Theorem \ref{mainthm3.2} which somehow ``resembles a set''. The function $f$ constructed at the end of \S \ref{sec6} need not have this property. Note, however, that we could have exercised considerable freedom in our choice of $f$: any function such that $\widehat{f}(r) = \xi_r \sqrt{\widehat{F}(r)}$, where $\xi_r \in \{-1,1\}$, would have done. By choosing the signs $\xi_r$ at random we may force $f$ to behave well in $L^{\infty}$, and hence to resemble a set. The rest of the section is devoted to the details of such an argument.\vspace{11pt}

\noindent Suppose, then, that $\xi = \{\xi_r\in\{\pm1\},\, 0\leq r < p/2\}$ is a sequence of independent random signs and that $f=f_{\xi}$ is the function whose Fourier transform satisfies $\widehat{f}(r) = \xi_r \sqrt{\widehat{F}(r)}$, $0 \leq r < p/2$, and $\widehat{f}(r) = \widehat{f}(-r)$.\vspace{11pt}

\noindent Let us reiterate the properties enjoyed by $f_{\xi}$: It satisfies $\mathbb{E} f_{\xi} = 0$, $\Vert f_{\xi} \Vert_2 \leq 1$, and $|f_{\xi} \ast f_{\xi}^{\circ}(x)| \gg 1/\log p$ for all $x$.

\begin{lemma}\label{lem6.1} There is an absolute constant $c$ and a choice of signs $\xi$ such that
\begin{equation}\label{largedev}
\mathbb{E}_{x \in \mathbb{Z}/p\mathbb{Z}} \exp\left(cf_{\xi}(x)^2\right) \ll 1.
\end{equation}
\end{lemma}

\noindent\textit{Proof.} Let $(a_r)_{0 \leq r < p/2}$ be any sequence of real numbers. It is well-known (a simple instance of Khintchine's inequality, see for example \cite{kah}) that
\[ \mathbb{E}_{\xi}| \sum_r a_r \xi_r |^{2k} \leq \big(C_0 k \sum_r a_r^2\big)^k \]
for some absolute $C_0$. It follows that for any $c \geq 0$ we have
\begin{equation}\label{eq3.394} \mathbb{E}_{\xi} \exp \big( c \big(\sum_r a_r \xi_r \big)^2\big) \leq \sum_{k \geq 0}\frac{1}{k!} \big(cC_0 k \sum_r a_r^2 \big)^k.\end{equation}
Now for any $x \in \mathbb{Z}/p\mathbb{Z}$ we have
\[ f_\xi(x)=\sum_{0 < r < p/2}2\xi_r\sqrt{\widehat{F}(r)}\cos(2\pi xr/p).\]
For each fixed $x$ this has the form
\[ \sum_r a_r \xi_r\]
where
\[ a_r = 2\sqrt{\widehat{F}(r)} \cos (2\pi x r/p).\] Since $\Vert \widehat{F} \Vert_1 \leq 1$, we have
\[ \sum_r a_r^2 \leq 4.\]
In view of \eqref{eq3.394} this means that
\[ \mathbb{E}_{\xi} \exp(cf_{\xi}(x)^2) \leq \sum_{k \geq 0} \frac{1}{k!}(4cC_0k)^k.\]
By Stirling's formula this series converges provided that $4c C_0 < 1/e$, an inequality which can certainly be achieved by selecting $c$ sufficiently small.\vspace{11pt}

\noindent Averaging over $x$, we see that
\[ \mathbb{E}_x \mathbb{E}_{\xi} \exp(c f_{\xi}(x)^2) \ll 1.\] In particular there is at least one choice of the signs $\xi$ for which the claimed inequality is true.\hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent The next lemma is a technical fact used in the proof of Lemma \ref{lem6.3} below.

\begin{lemma}\label{lem6.2} Let $f_1, f_2 : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ be two functions. Then the following inequality holds:
\begin{equation}\label{stab}
\Vert f_1 \ast f_1^{\circ}- f_2 \ast f_2^{\circ}\Vert_{\infty} \leq \|f_1-f_2\|_2
\left(\|f_1\|_2+\|f_2\|_2\right).
\end{equation}
\end{lemma}

\noindent\textit{Proof.} By the Cauchy--Schwarz inequality we have
\[ |f*g(x)| \leq \|f\|_2\|g\|_2\] for any two functions $f,g$ and for any $x \in \mathbb{Z}/p\mathbb{Z}$.
Setting $g:=f_1-f_2$ we have
$$|(f_1 \ast f_1^{\circ})(x)-(f_2 \ast f_2^{\circ})(x)| \leq |f_2*g^{\circ}(x)| +
|g* f_1^{\circ}(x)| \leq \|f_2\|_2\|g\|_2 + \|f_1\|_2\|g\|_2,$$
as required. \hfill{\usebox{\proofbox}}\vspace{11pt}

\noindent The next lemma is a stronger version of Theorem \ref{mainthm3.2}.

\begin{lemma}\label{lem6.3} There is a function $f : \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{R}$ such that $\mathbb{E} f = 0$, $\Vert f \Vert_2 \ll 1$, $|f \ast f^{\circ}(x)| \gg 1/\log p$ for all $x$, and with the additional property that $\Vert f \Vert_{\infty} \ll \sqrt{\log \log
p}$.
\end{lemma}

\noindent\textit{Proof.} Take a function $f_{\xi}$ satisfying the conclusion of Lemma \ref{lem6.1}, that is to say for which the inequality
\begin{equation}\label{eq4.72} \mathbb{E}_{x} \exp(c f_{\xi}(x)^2) \ll 1\end{equation} is satisfied. Let $C$ be a large absolute constant, and consider the set
\[ A := \{x : |f_{\xi}(x)| \geq C\sqrt{\log \log p}\}.\]
It follows from \eqref{eq4.72} that, for a suitable choice of $C$, we have
\begin{equation}\label{exceptional-size} \mathbb{E} 1_A \ll (\log p)^{-10}.\end{equation}
Choose an arbitrary set $A'$ such that $A \subseteq A'$ and
\begin{equation}\label{exact-size} (\log p)^{-10} \ll \mathbb{E} 1_{A'} \ll (\log p)^{-10}.\end{equation}
Define $f$ by
\[ f(x) := \left\{ \begin{array}{ll} f_{\xi}(x) & \mbox{if $x \notin A'$}; \\ \kappa & \mbox{if $x \in A'$},\end{array}\right.\]
where $\kappa$ is selected so that $\mathbb{E} f = 0$. \vspace{11pt}

\noindent Observe that
\[ |\kappa| = |\mathbb{E}_{x \in A'} f_{\xi}(x)| \leq \mathbb{E}_{x \in A'} |f_{\xi}(x)|.\]
From \eqref{eq4.72} one may deduce without difficulty that
\[ \mathbb{E}_{x \in S} |f_{\xi}(x)| \ll \big( \log (1/\E1_S) \big)^{1/2}\]
for any nonempty set $S \subseteq \mathbb{Z}/p\mathbb{Z}$. Hence, in view of \eqref{exact-size}, we have the bound
\[ |\kappa| \ll \sqrt{\log \log p}.\]
This immediately implies that $\Vert f \Vert_{\infty} \ll \sqrt{\log \log p}$. \vspace{11pt}

\noindent To obtain the upper bound on $\Vert f \Vert_2$, it suffices to note that
\[ \Vert f - f_{\xi} \Vert_2 = \Vert (f_{\xi} - \kappa) 1_A \Vert_2 \ll \sqrt{\log p} (\mathbb{E} 1_A)^{1/2} \ll (\log p)^{-4}.\]
Furthermore, this inequality together with Lemma \ref{lem6.2} implies that
\[ \Vert f \ast f^{\circ} - f_{\xi} \ast f_{\xi}^{\circ}\Vert_{\infty} \ll (\log p)^{-4}.\]
In view of the fact that
\[ |f_{\xi} \ast f_{\xi}^{\circ}(x)| \gg 1/\log p\] for all $x  \in \mathbb{Z}/p\mathbb{Z}$,
we obtain the desired lower bound
\begin{equation}\tag*{\usebox{\proofbox}} |f \ast f^{\circ}(x)| \gg 1/\log p.\end{equation}

\noindent\textit{Remark.} A well-known conjecture of Koml\'os implies that if $(a_{ij})_{1 \leq i,j \leq n}$ are any scalars then there is a choice of signs $\xi_j$ so that
\[ \big| \sum_{j} a_{ij} \xi_j \big| \leq C \big( \sum_{j} a_{ij}^2 \big)^{1/2},\]
for some absolute constant $C$. This conjecture, if true, would allow us to replace the $\sqrt{\log \log p}$ appearing in Lemma \ref{lem6.3} by an absolute constant, and to remove all subsequent appearances of $\log \log p$. The best result towards Koml\'os' conjecture is due to Banaszczyk \cite{bana}; it may be that his ideas may be applied in our context to reduce the $\log \log p$ factors to something a little smaller, but it does not seem worth the effort of pursuing this line of inquiry.\vspace{11pt}

\noindent Using a simple linear transformation, Lemma \ref{lem6.3} has the following consequence.

\begin{lemma}\label{lem6.4} There exists a function $f: \mathbb{Z}/p\mathbb{Z} \rightarrow [0,1]$
such that $\mathbb{E} f=1/2$ and
\begin{equation}\label{lem6.4bd} |f \ast f^{\circ}(x) - 1/4| \gg 1/\log p\log\log p \end{equation} for all $x \in \mathbb{Z}/p\mathbb{Z}$.\hfill{\usebox{\proofbox}}
\end{lemma}

\noindent To complete the proof Theorem \ref{mainthm3.3} we use probabilistic arguments again, in a fairly standard manner. We choose a set $B$ by selecting each $x \in \mathbb{Z}/p\mathbb{Z}$ to lie in $B$ independently at random with probability $f(x)$, and then show that with high probability $|B|$ is close to $p/2$ and $1_B \ast 1_B^{\circ}(x)$ is close to $f \ast f^{\circ}(x)$ whenever $x \neq 0$. We then modify $B$ by adding or deleting a small number of elements to give a set $A$ with $|A| = \lfloor p/2 \rfloor$.\vspace{11pt}

\noindent For each $x \in \mathbb{Z}/p\mathbb{Z}$ we write $X_x$ for the random variable which is $1$ with probability $f(x)$ and $0$ with probability $1 - f(x)$. To prove the statements we need, it is natural to use one of the standard large-deviation type inequalities. Amongst inequalities of this sort, one particularly suited to our purpose is Hoeffding's inequality (see, for example, \cite{green-deviation-notes}).

\begin{lemma}[Hoeffding's inequality]
Suppose that $Y_1,\dots, Y_m$ are independent random variables such that $|Y_i| \leq 1$ for each $i$. Write $Y := m^{-1}(Y_1 + \dots + Y_m)$ and $\mu := \mathbb{E} Y$. Then for any $t \geq 0$ we have
\[ \mathbb{P}(|Y - \mu| \geq t) \leq 2 e^{-mt^2/2}.\]
\end{lemma}
\noindent It follows immediately from this, applied with $m = p$ and $Y_x = X_x$, that
\begin{equation}\label{b-size} \mathbb{P}(| |B| - p/2 | \geq p^{2/3}) \ll e^{-p^{1/3}/2}.\end{equation}
Fix $x \in \mathbb{Z}/p\mathbb{Z}$, $x \neq 0$, and look at the expression
\[ 1_B \ast 1_B^{\circ}(x) = p^{-1}\sum_y X_y X_{y - x}.\]
Writing
\[ Z_y := X_y X_{y - x},\]
we have
\[ p^{-1}\sum_y \mathbb{E} Z_y = f \ast f^{\circ}(x).\]
Hoeffding's inequality does not immediately apply to the family $(Z_y)_{y \in \mathbb{Z}/p\mathbb{Z}}$, since these random variables are not independent. However we may partition the family into three subfamilies $(Z_y)_{y \in S_i}$, $i = 1,2,3$, where $|S_i| \geq p/5$, and within each family the random variables $Z_y$ are jointly independent. This may be achieved by forming the graph on vertex set $\mathbb{Z}/p\mathbb{Z}$ in which verex $i$ is joined to vertex $j$ if $i - j = x$; this is a cycle of length $p$, and we partition its vertex set into 3 classes each having size at least $p/5$.\vspace{11pt}

\noindent Hoeffding's inequality now applies, and we have
\[ \mathbb{P}\big( \big| |S_i|^{-1}\sum_{y \in S_i} Z_y - |S_i|^{-1}\sum_{y \in S_i} \mathbb{E} Z_y \big| \geq t \big) \leq 2 e^{-p t^2/10} \]for $i = 1,2,3$. It follows using the triangle inequality that
\[ \mathbb{P}\big( \big| p^{-1} \sum_{y \in \mathbb{Z}/p\mathbb{Z}} Z_y - p^{-1} \sum_{y \in \mathbb{Z}/p\mathbb{Z}} \mathbb{E} Z_y \big| \geq t \big) \leq 6 e^{-p t^2/10},\] that is to say
\[ \mathbb{P}\big( \big| (1_B \ast 1_B^{\circ})(x) - (f \ast f^{\circ})(x) \big| \geq t \big) \leq 6e^{-p t^2/10}.\]
Taking $t = p^{-1/3}$, we see from this and \eqref{b-size} that with probability close to 1 we have both
\begin{equation}\label{eq-19a} \big| |B| - p/2 \big| \leq p^{2/3}\end{equation}
and
\begin{equation}\label{eq-19b} \big| 1_B \ast 1_B^{\circ}(x) - f \ast f^{\circ}(x) \big| \leq p^{-1/3}\end{equation} for all $x \neq 0$. Fix a specific set $B$ verifying \eqref{eq-19a} and \eqref{eq-19b}.
Form a set $A$ by adding to or deleting from $B$ arbitrary elements so that $|A| = \lfloor p/2\rfloor$. In view of \eqref{eq-19a} we have
\[ \Vert 1_A - 1_B \Vert_2 \leq 2p^{-1/6},\] and hence by Lemma \ref{lem6.2} we infer that
\[ \Vert 1_A \ast 1_A^{\circ} - 1_B \ast 1_B^{\circ} \Vert_{\infty} \leq 4p^{-1/6}.\] It follows from this, \eqref{eq-19b} and \eqref{lem6.4bd} that
\[ | 1_A \ast 1_A^{\circ}(x) - 1/4 | \gg 1/\log p \log\log p\] for all $x \neq 0$. This inequality is manifestly true for $x = 0$ as well, and this completes the proof of Theorem \ref{mainthm3.3}.\hfill{\usebox{\proofbox}}

\section{A recent development}

\noindent We have recently learnt that Sanders \cite{sanders} has combined some of our methods with some new ideas, and has been able to improve the bound in Theorem \ref{mainthm2} to $(\log p)^{-1/2 + \epsilon}$.

redpajama_set_name: RedPajamaArXiv


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A whole new "Star Wars" world has been revealed, coming to Disney theme parks. At the D23 Expo 2017, Disney Parks & Resorts this evening gave ITM a preview of the massive, impressively detailed model of Star Wars Land, opening at Walt Disney World and Disneyland in 2019.
EXCLUSIVE: Don't miss our interview with Imagineer Scott Trowbridge, revealing new details about Star Wars Land.
The model, hand-crafted by Disney's talented Imagineers, reveals all the interplanetary nooks and crannies that will ultimately be explored by thousands of fans. From the Millennium Falcon ride to the battle of the First Order and the shops to the restaurants, Star Wars Land will no doubt be quite the immersive experience.
There, two signature attractions will await guests. One will let guests take control of the famed Millennium Falcon on a "customized" secret mission while the other puts guests squarely in the middle of a battle between the First Order and the Resistance on an abandoned Rebel base.
Stormtroopers were on hand to help debut the first look at the Star Wars Land exhibit.
The pavilion itself feels like Star Wars Land. There are maquettes, artwork, and other sculptures on display showing off various elements of the upcoming theme park expansion.
Check back to ITM tomorrow morning for an interview with the creative director of Star Wars Land, Imagineer Scott Trowbridge, followed by even more "Star Wars" excitement on Saturday as Walt Disney Studios, Lucasfilm, and Disney Parks & Resorts host two panels revealing even more from the highly-anticipated theme park expansion.
This is a developing story. Keep checking back throughout the evening for a closer look at the Star Wars Land model.
Video, photos, and reporting by Mike Celestino.
Why does it look like Episode 1 ?
The above game have a lot of features and this is advance level of game. This game is not play in all mobiles and we can see the requriements of the game while installing in the mobile. Any how, I want essay rush and at the same time this is really good for me because i'm big craze of playing games.
waiting that settle upon in good shape your needs.


An additional "medically fragile child" who had a confirmed case of adenovirus at the Wanaque Center for Nursing and Rehabilitation died Saturday night, the state's health department said.
"This latest death was an individual who had a confirmed case of adenovirus and had become ill before October 22", explains health officials.
On Friday, an eighth child was confirmed dead following the viral outbreak, but officials say they do not now have laboratory confirmation of adenovirus in this child that's killed other pediatric patients at the facility.
The new diagnosis of adenovirus brings the total number of cases at the rehabilitation centre to 26. A staff member also became ill, and there have been eight pediatric deaths.
The nine children who died in the outbreak that began September 26 range in age from toddler to young adult, with majority under age 18. Among the 25 children sickened, nine have died, the Health Department said.
The Department of Health's Communicable Disease service remains on site to monitor the outbreak.
Both the staff at the facility as well as state authorities are continuous in the efforts to contain the spread of the virus.
This 1981 electron microscope image made available by the Centers for Disease Control and Prevention shows a group of adenovirus virions. Based on an inspection conducted in August, CMS reported that "it was determined that the facility failed to provide a clean and homelike physical environment for their residents. The strain has been particularly associated with disease in communal living arrangements and can be more severe", the department said in a statement.


text:
klaus mikaelson real name
Klaus then falls asleep, and when he wakes up later, he finds that Cami is dead. Niklaus Mikaelson is the Original Werewolf-Vampire Hybrid. Aurora tells them that Rebekah is at the bottom of the ocean. Until then, my sacrifice will allow you to grow. Still pretending to be Alaric, he attacks her at the 60's decade dance by compelling several students. They both are shown to care very deeply for each other and have possible feelings for one another. Klaus is almost killed in the struggle as Mikael stakes him with the white oak stake. She terrorized the French Quarter and killed many of his best men which made him nervous of her power. Due to his long existence, Klaus has seen countless battles and is very experienced in battle. They fought against Lucien together with their family and defeated him, and by the end of the season, while Klaus is imprisoned, Hayley now searches for a way to reunite the family. Characters. Despite this, he shows no remorse for his actions in regards to others in the same aspect, such as with Marcel, knowing the feeling of being secluded for an extended period of time. But, Klaus reminded her, if he dies, Caroline and everyone she knows dies. Klaus made his return to Mystic Falls in 500 Years of Solitude. Klaus is upset that his friendship and family bond with Marcel had ended, as he understands Marcel's anger and vengeance, however, (silently) he cannot make excuses for his ex-best friend's behavior, the look on Klaus' face moments before Marcel stabbed him with Papa Tunde's blade; clearly indicates that he plans to make Marcel suffer for his treachery against him and his half-siblings. Hayley tells him that one day their daughter might go through the same thing as Kayla is going through now. Yours, Klaus" was sent to the Salvatore Boarding School. In The Ties That Bind, Klaus appears in one of Bonnie's dreams as she tries to open the locked coffin. Klaus and Elijah get spelled and their minds are sent to a Chambre de Chasse where they meet Aurora her brother Tristan. Rebekah and Klaus are later seen in a box at the Theater, and Rebekah heads towards the lobby to find Marcel. He talks to Cami and he begs her to drink human blood because he doesnt want to lose her. A few moments later Klaus and Mikael go to fight Dahlia but end up by losing the only weapon they believe that can kill her. however, unlike Esther, he never underestimated members of his family, or his enemies, he committed immoral acts; he killed Elijah's girlfriend Gia, and daggered him with Papa Tunde's Blade, compelled Marcel to kill Rebekah, (who is in the body of the late Eva Sinclair) if she tries to escape the compound, had Dahlia reactivate the Crescent Curse on the Crescent Wolf Pack and Hayley. He took him to Lafayette Cemetery No. Despite their warnings of Mikael's coming, he refused to flee. Klaus was emotionally fragile enough from a truly confusing evening that he nearly burst into tears. Caroline tries to defend her actions by saying that she couldn't let Bonnie die but then realizes what she has done. Elijah tells him that if he will not do anything, Elijah will do it in stead. Klaus listens to each story, keeping a smile on his face still. stating that his aim has been improving, when he misses a shot; however, Mikael intervenes and scolds Elijah for encouraging his brother and snatches the bow from Niklaus, stating that he wasn't man enough to hold a weapon. Klaus was there on his side, together with Rebekah and Elijah. As Klaus is left alone and the curtain about to lift, Mikael appears behind him, poking his back with the White Oak. He then comes in the house and talks to his family before the time comes and spends time with them in a final dinner. In Red Door, Klaus reunites with Kol. He tells her the only way for her to find peace is for her to move on and forget. He goes home with Freya when he finds out that Lucien helped Aurora escaping. Hayley fled into the woods. Klaus pursued him and, in front of all of their guests, yet again daggered his brother with Elijah's help. Klaus later arrives at the location of Hayley's disappearance and asks who took Hayley and ho killed her attackers she say she does not know. He told Klaus he'd find out what Jane-Anne had wanted for him. She is also angry at Klaus for refusing to call off his quest to kill Tyler and refuses to forgive him. In that moment, Klaus took the Governor's son and threw him off the second story, killing him immediately. Klaus hugs him and tells him that he understands, because after he fled New Orleans he thought he was dead and it took him years before he could even speak his name. Dahlia questions why he is going through all of this. He, Elijah and Rebekah walk around New Orleans and talk to each other as final goodbyes. Elijah tells him they worked together to build up the city, but it was originally Niklaus' vision. Finn wanted to know the big secret they were hiding. Klaus then go to Marcel's loft, and tells Vincent and Josh that Marcel is dead. Klaus initially believed her to be a fling of Tyler's whilst in the Appalachians, however it was revealed she came to Mystic Falls in a deliberate attempt to have Klaus murder his hybrids, in exchange for information on her family. Instead, Klaus offers to use himself to link to Dahlia, as he is an Original Hybrid. Rebekah argued that she'd acted with human decency and Klaus said no, the most human thing in that moment would have been if she could stomach standing by watching April die knowing she couldn't help her. In 1492, he finally got the chance to break his curse when he met Katerina Petrova, the final piece necessary to break the curse. Cami provokes her and she and Aurora fight, but Aurora quickly gets the upper hand, and pushes Cami into Lucien's uninvited penthouse, causing the young vampire to bleed, and Klaus tells Aurora to stop. In the battle to save the baby from the witches Klaus kills Abigail but it is Marcel who rescues the baby. Morgan was born in London, but lived in Swansea for 11 years. All these actions of his was partly as revenge/punishment for betraying him/for trying to take his daughter away from him, and mainly to gain Dahlia's trust to take her down. Rebekah later shows up to stall Marcel giving Klaus time to find Thomas and snapped his neck. He told her that she was also all that he had, and it was a beautiful symmetry. A few minutes later he finds out that the Strix captured Rebekah. In The Map of Moments, Klaus is shown in a flashback with Marcel and Kol, as he caught the latter stealing the Paragon Diamond. Even before Klaus became an Original Vampire, later the Original Hybrid, he was a fairly skilled swordsman, even though he rarely uses a sword, due to his supernatural powers, Klaus' swordsman skills have not diminished. Klaus calls Freya asking for a locator spell. Later he has to deal with Hayley who is angry with him because of his order to kill the werewolves. 1, where Sophie was waiting. Rebekah calls him again and informs Klaus of that Katie is the girlfriend of Thierry, making him the key element in the plan. She tells the crowd what Klaus has done to his siblings and that he ruins everything he touches. Klaus remains pinned to the wall as Hayley gives birth to his daughter. Aurora snaps her neck and asks Klaus who should she kill first. Klaus continually avoids Elena messages or visions to Bonnie and Stefan. Kol is Klaus' maternal younger half-brother. Kol tries desperately to reason with Marcel but he grabs Kol and bites his neck then throws him. He spotted Cami, admiring the same artist. He is drinking a lot and tells Elijah that he has been dreaming about their father. He then returns to Marcel and asks him what happened next. Klaus thanks Hayley for convincing him that it was right to help Kayla, telling her that they should decide things together as partners what is best for Hope. Klaus Mikaelson : I bowed down to you, brother, to make up for daggering you for the greater good of our plan to reclaim our home. He asks Klaus if he has forgotten that Genevieve tortured their sister and as a result she is now gone forever. She agrees and they end up having sex. Once the vervain is out of Marcel's system, Klaus compels him to go after Rebekah. Klaus gets a call from Diego who informs him that Papa Tunde is dead and they start to understand that their witch problems are only getting started. Klaus befriends him to find out his weaknesses so he could destroy him. Klaus is ranting about his disappointment in his siblings and he finds a mysterious drawing of Cami's; she mocks him by saying that it's a plot to destroy him and then a tattoo design. His relationship with Klaus is currently broken as Marcel lives in exile and tries to find a way to get New Orleans back under his rule. Even five years after his imprisonment, Klaus' personality has not changed much as Marcel had failed to break him emotionally. Klaus Baudelaire, character in the A Series of Unfortunate Events books; Klaus Heisler, character on American Dad! So how is … Klaus cares very deeply for Rebekah, but he seems to have a lot difficulty showing her that it is true. It was later revealed that when he had attacked Meredith's grandfather before the books started, he had in fact been trying to capture Meredith and her twin brother Cristian.
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klaus mikaelson real name 2021

redpajama_set_name: RedPajamaCommonCrawl


This inviting 3bd/3bth villa offers loads of development potential as it sits in a well established neighbourhood just across from the renowned Lime Grove Lifestyle Centre. Operating primarily as a holiday rental, this private residence or rental investment property has many advantages. It is located just a stroll away from many amenities including a supermarket, bank, 24hr medical centre, shopping plazas and the beach.
Entry via the main entrance leads into the living and dining room which have an exit out to the backyard where the new owner can install a swimming pool and deck. To the left of the living area is a bedroom with partial bathroom which can easily be converted into a fully ensuite bedroom.
There is a kitchen with a small lookout into the living area. This partition can certainly be removed for a more open plan style. To the rear of the kitchen is is a laundry area and an external entrance, which facilitates easily offloading groceries.
Along the hallway is a bedroom with an adjoining bathroom with shower. The master bedroom overlooks the backyard and has an external entrance. It would be ideal to build a deck adjoining this room allowing for a night cap or a cup of tea under the stars. The master bathroom is well laid out with a shower.
The backyard is extremely spacious and offers loads of possibilities when it comes to development. Plans for improvement could include a pool, deck and even a gazebo.


We are delighted to share that Blankstone Sington, an independent Investment Management and Stockbroking firm based in Liverpool, have successfully launched DION's WIN Client Portal to end investors. Blankstone Sington and Dion have been working together since 1994, and we already benefiting from Dion's full front to back office Wealth Management solutions. Blankstone Sington will roll out the Client Portal to all of their clients and intermediaries.
Liverpool based Investment Managers and Stockbrokers Blankstone Sington was founded by Michael Blankstone along with his business partner Reggie Sington in 1976. Independently owned, as a private client boutique we have taken our stockbroking heritage and evolved into a modern, forward thinking investment manager.
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He says people moving out of Michigan, and fewer people having kids, are to blame for the drop. Guthrie says the decline could pose a problem for schools in the state.
"Schools are presumably competing for a fewer number of students to fill their schools," he said.
Guthrie says the decrease could also impact per-pupil funding for schools.
"When you have declining numbers of 5 to 17 year olds you're going to be having declining enrollment," he said.
Guthrie said the school-aged population should begin to grow again as more millennials have children. He said he anticipates that shift will happen by 2024.
Michigan has shifted $4.5 billion from K-12 education funding to universities and community colleges since 2010. That's according to a new report from the Michigan League for Public Policy.
The funds have been appropriated from the School Aid Fund, established in 1955 as an amendment to the Michigan Constitution. The School Aid Fund is used exclusively to fund K-12 schools. Funds for post-secondary education, on the other hand, come from the General Fund.


text:
Altona is a settlement on the island of Saint Croix in the United States Virgin Islands. It is an eastern suburb of Christiansted.

References

Populated places in Saint Croix, U.S. Virgin Islands

redpajama_set_name: RedPajamaWikipedia


text:
When I was in prison, I was wrapped up
When I was in prison, I was wrapped up in all those deep books. That Tolstoy crap - people shouldn't read that stuff.
There are dreadful punishments enacted
There are dreadful punishments enacted against thieves; but it were much better to make such good provisions, by which every man might be put in a method how to live, and so to be preserved from the fatal necessity of stealing and dying for it.
It is more dangerous that even a guilty
It is more dangerous that even a guilty person should be punished without the forms of law than that he should escape.
Shyness is the prison of the heart.
Spanish Proverb
I have never been contained except I
I have never been contained except I made the prison.
You utter a vow, or forge a signature,
You utter a vow, or forge a signature, and you may find yourself bound for life to a monastery, a woman, or prison.
Bronislaw Malinowski
When the Nazis came for the communists,
When the Nazis came for the communists, I remained silent; I was not a communist. When they locked up the social democrats, I remained silent; I was not a social democrat. When they came for the trade unionists, I did not speak out; I was not a trade unionist. When they came for me, there was no one left to speak out.
Reverend Martin Niemoeller
To my mind, to kill in war is not a whit
To my mind, to kill in war is not a whit better than to commit ordinary murder.
Punishment, that is the justice for the
Punishment, that is the justice for the unjust.
Saint Aurelius Augustine (Augustine of Hippo)
In a civilized society, all crimes are
In a civilized society, all crimes are likely to be sins, but most sins are not and ought not to be treated as crimes.
Geoffrey Fisher
It is the spirit and not the form of law
It is the spirit and not the form of law that keeps justice alive.
Earl Warren
He had drawn many a thousand of these
He had drawn many a thousand of these rations in prisons and camps, and though he'd never had an opportunity to weight them on scales, and although, being a man of timid nature, he knew no way of standing up for his rights, he, like every other prisoner, had discovered long ago that honest weight was never to be found in the bread-cutting. There was short weight in every ration. The only point was how short. So every day you took a look to soothe your soul - today, maybe, they haven't snitched any.
Aleksander Solzhenitsyn
In prisons, those things withheld from
In prisons, those things withheld from and denied to the prisoner become precisely what he wants most of all.
Eldridge Cleaver
The public have more interest in the
The public have more interest in the punishment of an injury than he who receives it.
Marcus Porcius Cato
The penalty for laughing in a courtroom
The penalty for laughing in a courtroom is six months in jail; if it were not for this penalty, the jury would never hear the evidence.
Do not lay on the multitude the blame
Do not lay on the multitude the blame that is due to a few.
Ovid (Publius Ovidius Naso)
Nor stony tower, nor walls of beaten
Nor stony tower, nor walls of beaten brass, Nor airless dungeon, nor strong links of iron, Can be retentive to the strength of spirit; But life, being weary of these worldly bars, Never lacks power to dismiss itself.
We who live in prison, and in whose
We who live in prison, and in whose lives there is no event but sorrow, have to measure time by throbs of pain, and the record of bitter moments.
A Sunday school is a prison in which
A Sunday school is a prison in which children do penance for the evil conscience of their parents.
Whatever you think of de Sade, he was a
Whatever you think of de Sade, he was a complex figure and we should not look for easy answers with him. He was, strangely perhaps, against the death penalty, and he was never put in prison for murders or anything like that.


Trails in Lugones (España). 1871 results.
Below you'll see a map with Hiking trails related with search Lugones (España). You can browse the map and zoom in to see the trails detail, or click the trail pushpin to go to a detail map with the trail of your interest. Or you can see the trails list below, with a snippet of the trail description and a link to the trail's detail map.
Longitud del recorrido: 16.004 km.
Longitud del recorrido: 9.204 km.
Longitud del recorrido: 21.688 km.
Longitud del recorrido: 3.227 km.
Longitud del recorrido: 27.911 km.
Longitud del recorrido: 12.071 km.
Longitud del recorrido: 17.174 km.


text:
Cops Say They Only Use Facial Recognition for Leads, But It's Often the Sole Basis for Arrests
A massive report highlights fundamental problems with facial recognition and how police lie about its uses.
Thomas Germain
Photo: Ian Waldie / Staff (Getty Images)
In 2018, a man was arrested for stealing a pair of socks from a T.J. Maxx store in New York City, even though he was signed in at a hospital for the birth of his child when the crime was committed. The police never checked the man's alibi, and the case rested on a single piece of evidence. Apparently, a police officer had run security footage through a facial recognition database and found a "possible match." The officer texted a photo of the man to a witness who said "that's the guy." After six months in jail, the man pled guilty, though he maintains his innocence.
The Best Security Camera for Most People
Unfortunately, this story represents a growing trend, not a tragic exception. The anecdote opens a landmark report on facial recognition from the Georgetown Law Center on Privacy & Technology, which found that law enforcement agencies are often using the technology as the sole basis for putting people in jail despite claiming they don't and despite a growing body of evidence that the technology has serious problems with accuracy and bias.
The report, titled "A Forensic Without the Science: Face Recognition in U.S. Criminal Investigations," focuses on the technical shortcomings of facial recognition software, which are many and multiform, according to the authors.
"Police have used face recognition for more than 20 years based on the assumption that it is a reliable identification tool," said Clare Garvie, author of the report, and training resource counsel with the National Association of Criminal Defense Lawyers. "Not only has that assumption never been tested, there is every reason to posit that face recognition doesn't produce reliable leads and in fact may put people at risk of misidentification and wrongful arrest."
Studies show that facial recognition is less accurate when tasked with identifying the faces of women and people of color—particularly people with darker complexions. Those problems are so severe that some experts have argued it should be banned altogether.
Even under the best conditions, the people using the technology introduce their own problems. As the report puts it, "algorithms and humans may compound the other's mistakes." Users make false assumptions about the tech's accuracy. And when a facial recognition scan turns out a lineup of possible suspects, the people interpreting those results are biased by facial features such as demographics and expressions.
"Over its past 20 years of use, face recognition algorithms have improved, but the people running the searches have not," Garvie said. "Police departments are still not required to implement policies and training programs to guard against cognitive bias, misuse, and mistake."
Law enforcement officials and companies who make facial recognition software acknowledge those issues, and often say the technology should only be used to generate leads. But investigators ignore that caution, Garvie and company found. The report highlights multiple cases where police used facial recognition evidence as probable cause to make arrests.
A key problem is there are no clear guidelines about how facial recognition technology should be used, in part because the technology is new, but also because we don't fully understand facial recognition's flaws. If and when those guidelines are developed, they will likely come in response to the prosecution of innocent people, they wrote.
Due to a lack of transparency, we don't know how many people have been put in jail thanks to facial recognition, but according to the report, the number is growing. Law enforcement often fails to disclose when facial recognition is used in an investigation, which a problem for defendants in legal cases, because they can't dispute evidence derived from flawed technology if they don't know about it. The report argues that the results of facial recognition are so unreliable that they should be considered Brady material—a legal term for evidence that prosecutors are required to disclose because it helps defendants.
According to Garvie, failing to disclose the details about facial recognition in a criminal case violates people's legal rights. "This is a due process violation, one that has been happening since 2001," she said.
We've known about facial recognition's problems for a long time, but law enforcement and large swaths of the public continue to put blind trust in the technology.
"There is still a tendency to place undue faith in an artificial neutral, mathematics-based approach to solving hard problems or eliminating human error from decision-making," Garvie said. "This is the wrong approach."


Jane Grech, creative director of Jane Grech Dance Centre. Photo by Marina Birch.
Dance students aren't the only ones who have to focus for this new school year ahead. Dance studio owners, directors and teachers, too, have to plan ahead for 2017 with thoughts on studio retention, fresh class and routine ideas, and students' happiness.
Dance Informa turns to two sources – Jane Grech, creative director of Jane Grech Dance Centre in Adelaide and author at Dance Studio Success; and Karen Malek, president of Australian Teachers of Dancing and managing director of Transit Dance. With their advice, you can ensure 2017 will be the best year yet for your studio!
Summer break shouldn't leave you exhausted and unmotivated; rather, it should leave you feeling refreshed and raring to go for the year ahead! You've probably spent much of the break reflecting and relaxing with family, but hopefully you've also allotted time to listening to new music and thinking about exciting class and routine ideas.
Karen Malek. Photo by Belinda's Photography.
What to focus on in 2017.
Of course, each dance studio may have different areas of focus, depending on the studio's goals, but there are some things that every studio owner and teacher can focus on this year.
And be careful that your energy doesn't only go toward your studio's most talented dancers. Even those who may not win at every competition are an important asset; their happiness keeps them coming back, year after year.
Studio owners: Make it the best year yet.
Grech suggests choosing one word for which to focus your efforts and attention this year. This word, she says, will serve as a guide to keep you on track and motivate you toward achieving your goals – personal and business.
Teachers: Make it the best year yet.
Dance teachers, too, can have goals of their own, for their classes, their students and their personal growth as a leader.
"Keep attending ongoing professional development, as this will refresh you and give you new ideas to stay one step ahead," Malek recommends.
Success is relative and can differ from studio to studio, teacher to teacher. But define how your goals can help you reach what is successful for your studio, this year. Perhaps it is to place at a regional competition, or to master student retention. Maybe success to you is to get more parents involved this year, or to renovate the studio's bathroom.
Karen Malek leading a master class at the Southern School of Dance in Hong Kong. Photo by Rebecca McMullin.
If you enter the 2017 school year with a focus word, goals, refreshed excitement and/or gratitude, you're already one step closer to success. That energy toward bettering your studio and its teachers and dancers shows that you care about trying to have a great year, and it's contagious, too! Define your goals, and you will reap the benefits.
Cheers to 2017 and to making it your best year yet!


- There is no upgrade path for "NGX R65 CoreXL LA". Perform a clean installation of any version R70 and above.
00426948 It is not possible to upgrade Power-1 appliances that run with "R65 CoreXL" image. It is necessary to "Restore to Factory Defaults" and choose the non-CoreXL R65 image. Only then, the upgrade can be performed. After upgrade, CoreXL can be enabled via "cpconfig" menu.
- 3rd party clusters are not supported.
Note: IPSO OS with Check Point R70 version and above fully supports CoreXL.
Connections will be dropped during any ClusterXL upgrade method (including "Full Connectivity Upgrade"; refer to sk107042) if the number of CoreXL FW instances is different on cluster members. Before starting any upgrade, configure the same number of CoreXL FW instances on all cluster members. In addition, refer to sk42096 - Cluster member is stuck in 'Ready' state.
CoreXL is supported on StandAlone machine only from R70 and above.
If any of the above features/settings is enabled/configured in SmartDashboard, then CoreXL acceleration will be automatically disabled on the Gateway (while CoreXL is still enabled). In order to preserve consistent configuration, before enabling one of the unsupported features, deactivate CoreXL via "cpconfig" menu and reboot the Gateway (in cluster setup, CoreXL should be deactivated on all members).
In Security Gateways R77.30 and lower, when CoreXL is enabled, VPN traffic inspection occurs only in global CoreXL FW instance #0 (fw_worker_0). By design, global CoreXL FW instance #0 (fw_worker_0) always runs on the CPU core with highest ID (as allowed by the current CoreXL license).
Note: Starting in R80.10, VPN Multi-Core feature allows CoreXL to inspect VPN traffic on all CoreXL FW instances. Refer to sk118097.
- In Security Gateways R77.30 and lower, when CoreXL is enabled, VoIP control connections are processed only in global CoreXL FW instance #0 (fw_worker_0). By design, global CoreXL FW instance #0 (fw_worker_0) always runs on the CPU core with highest ID (as allowed by the current CoreXL license).
- Maxumal number of CoreXL FW instances is limited.
Refer to sk98737 - ATRG: CoreXL - section "Architecture".


Annually, the Australian Financial Review compiles a list of 'Australia's Top 100 Restaurants'. The top 100 is decided by the best Chef's and restauranteurs in Australia from a shortlist of 500 restaurants created by a team of critics and editors.
The prestigious awards program is designed to support and celebrate the rising standards of excellence and professionalism in Australian dining.
In our first year being included in the list, we were honoured to be featured 55th amongst the best culinary dining experiences the country has to offer.
We would like to thank the judges for recognising Oakridge Wines and congratulate the other finalists on their success.
To make a reservation at Oakridge Wines, call (03) 9738 9900.