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\title{\bf Energies and structure of additive sets}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address

\author{Shkredov Ilya \thanks{Supported by Russian Scientific Foundation grant RSF 14--11--00433.}\\
\small Steklov Mathematical Institute\\[-0.8ex]
\small ul. Gubkina, 8\\[-0.8ex]
\small 119991, Moscow, Russia\\
%\and
\small IITP RAS\\[-0.8ex]
\small Bolshoy Karetny per., 19\\[-0.8ex]
\small 127994, Moscow, Russia\\
\small\tt ilya.shkredov@gmail.com
}



\date{\dateline{??, 2014}{??, 2014}\\
\small Mathematics Subject Classifications: 11B13, 11B30}

\begin{document}

\maketitle



\begin{abstract}
    In the paper
    we prove that any sumset or difference set has large $\E_3$ energy.
    Also, we give a full description of  families of  sets having
%    the form $H\dotplus \L$, where $H$ has small $H+H$ and $\L$ has random
    %structure
%    behaviour as well as the family of disjoint union of sets $H_j$ having small $H_j+H_j$ in terms of some energies.
    critical relations between some kind of energies such as $\E_k$, $\T_k$ and Gowers
    %$U^3$--
    norms.
    In particular, we give
%    criterions
    criteria
    for a set to be a \\
$\bullet~$ set of the form $H\dotplus \L$, where $H+H$ is small and $\L$ has "random structure", \\
$\bullet~$ set equals a disjoint union of sets $H_j$, each $H_j$ has small doubling,\\
$\bullet~$ set having large subset $A'$ with $2A'$ is equal to a set with small doubling
and $|A'+A'| \approx |A|^4 / \E(A)$.
%, $n\in \N$.


  \bigskip\noindent \textbf{Keywords:} additive combinatorics; additive energy; sumsets; Gowers norms
\end{abstract}





\section{Introduction}
\label{sec:introduction}



Let $\Gr = (\Gr,+)$ be an abelian group.
For two sets $A,B\subseteq \Gr$ define the {\it sumset} as
$$A+B := \{ x\in \Gr ~:~ x=a+b \,,a\in A\,,b\in B \}$$
and, similarly, the {\it difference} set
$$A-B := \{ x\in \Gr ~:~ x=a-b \,,a\in A\,,b\in B \} \,.$$
Also denote the {\it additive energy} of a set $A$ by
$$
    \E(A) = \E_2 (A) = |\{ a_1-a'_1 = a_2-a'_2 ~:~ a_1,a'_1,a_2,a'_2 \in A \}| \,,
$$
and
%the
{\it $\E_k (A)$ energy} as
\begin{equation}\label{f:start_E_k}
    \E_k (A) = |\{ a_1-a'_1 = a_2-a'_2 = \dots = a_k-a'_k ~:~ a_1,a'_1,\dots, a_k,a'_k \in A \}| \,.
\end{equation}
The special case $k=1$
%of (\ref{f:start_E_k})
gives us $\E_1 (A) = |A|^2$
because there is no any restriction in the set from (\ref{f:start_E_k}).
So, the cardinality of a set can be considered as a degenerate sort of energy.
Note that a trivial upper bound for $\E_k (A)$ is $|A|^{k+1}$.
Now recall a well--known Balog--Szemer\'{e}di--Gowers Theorem \cite{TV}.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set, and $K\ge 1$ be a real number.
    Suppose that $\E(A) \ge |A|^3 / K$.
    Then there is $A' \subseteq A$ such that $|A'| \gg |A|/K^C$ and
\begin{equation}\label{f:BSzG_introduction}
     |A'-A'| \ll K^C |A'| \,,
\end{equation}
    where $C>0$ is an absolute constant.
\label{t:BSzG_introduction}
\end{theorem}



So, Balog--Szemer\'{e}di--Gowers Theorem can be considered as a result
about
the structure
of sets $A$ having the extremal (in terms of its cardinality or $\E_1(A)$ in other words) value of $\E(A)$.
Namely,
any of such a set has a subset $A'$ with the extremal value
%(in terms of its cardinality)
of cardinality of its difference set.
These sets $A'$ are called sets with {\it small doubling}.
On the other hand, it is easy to obtain, using the Cauchy--Schwarz inequality,
that any $A$ having subset $A'$
%, satisfying
such that
(\ref{f:BSzG_introduction}) holds, automatically has polynomially large energy
$\E(A) \gg_K |A|^3$ (see e.g. \cite{TV}).
Moreover, the structure of sets with small doubling is known more or less thanks to a well--known
Freiman's theorem (see \cite{TV} or \cite{Sanders_2A-2A}).
Thus,
%on the one hand
Theorem \ref{t:BSzG_introduction}
%gives
%even more structure
finds subsets of $A$
with rather rigid structure
and,
% on the other hand,
actually,
it is a criterium for a set $A$ to be a set with large (in terms of $\E_1(A)$) additive energy:
$\E(A) \sim_K |A|^3 \sim_K (\E_1 (A))^{3/2}$.


In the paper we consider another extremal relations between different energies and describe the structure
of sets having these critical relations.
Such kind of theorems have plenty of applications.
It is obvious for Balog--Szemer\'{e}di--Gowers theorem, see
e.g. \cite{TV}, \cite{BK_AP3}, \cite{Gow_4}, \cite{Gow_m}, \cite{KSh}, \cite{M_R-N_S} and so on;
for recent
%results
applications using
%on
critical relations between energies  $\E_2 (A)$ and $\E_3 (A)$,
see e.g. \cite{s_ineq}, \cite{s_mixed} and others.
Before formulate our main results let us recall a beautiful theorem of Bateman--Katz
\cite{BK_AP3}, \cite{BK_struct}
which is another example of theorems are called "structural"\, by us.



\begin{theorem}
Let $A \subseteq \Gr$ be a symmetric set, $\tau_0$, $\sigma_0$ be nonnegative real numbers
and $A$ has the property that for any $A_*\subseteq A$, $|A_*| \gg |A|$ the following holds
$\E(A_*) \gg \E (A) = |A|^{2+\tau_0}$.
Suppose that $\T_4 (A) \ll |A|^{4+3\tau_0+\sigma_0}$.
Then there exists a function $f_{\tau_0} : (0,1) \to (0,\infty)$ with $f_{\tau_0} (\eta) \to 0$ as $\eta \to 0$
and a number $0\le \a \le \frac{1-\tau_0}{2}$ such that there are sets $X_j,H_j\subseteq \Gr$, $B_j\subseteq A$,
$j\in [|A|^{\a-f_{\tau_0} (\sigma_0)}]$ with
\begin{equation}\label{f:BK_1}
    |H_j| \ll |A|^{\tau_0+\a+ f_{\tau_0} (\sigma_0)} \,,\quad \quad |X_j| \ll |A|^{1-\tau_0-2\a+ f_{\tau_0} (\sigma_0)} \,,
\end{equation}
\begin{equation}\label{f:BK_2}
    |H_j-H_j| \ll |H_j|^{1+f_{\tau_0} (\sigma_0)} \,,
\end{equation}
\begin{equation}\label{f:BK_3}
    |(X_j+H_j) \cap B_j| \gg |A|^{1-\a-f_{\tau_0} (\sigma_0)} \,,
\end{equation}
and $B_i \cap B_j = \emptyset$ for all $i\neq j$.
\label{t:BK_structural}
\end{theorem}


Here $\T_4 (A)$ is the number of solutions of the equation $a_1+a_2+a_3+a_4=a'_1+a'_2+a'_3+a'_4$,
$a_1,a_2,a_3,a_4,a'_1,a'_2,a'_3,a'_4 \in A$
and this characteristic is another sort of energy.
One can check that any set satisfying (\ref{f:BK_1})---(\ref{f:BK_3}), $\E(A) = |A|^{2+\tau_0}$
and all another conditions of the theorem is an example of a set having $\T_4 (A) \ll |A|^{4+3\tau_0+\sigma_0}$.
Note that if $\E (A) = |A|^{2+\tau_0}$ then by the H\"{o}lder inequality one has $\T_4 (A) \ge |A|^{4+3\tau_0}$.
Thus, Theorem \ref{t:BK_structural} gives us a full description of sets having critical relations between
a pair of two energies:
$\E(A)$ and $\T_4 (A)$.


There are two opposite extremal cases in Theorem \ref{t:BK_structural} : $\a = 0$ and $\a = \frac{1-\tau_0}{2}$.
For simplicity consider the situation when $\Gr = \f_2^n$.
In the case $\a = 0$ by Bateman--Katz result  our set $A$,  roughly speaking, is close to a set of the form
$H \dotplus \Lambda$, where $\dotplus$ means the direct sum,
$H\subseteq \mathbf{F}_2^n$ is a subspace, and $\Lambda \subseteq \mathbf{F}_2^n$ is a dissociated set (basis),
$|\L| \sim |A| /|H| \sim |A|^{1-\tau_0}$.
These sets are interesting in its own right being counterexamples in many problems of additive combinatorics.
The reason for this
is trivial.
Indeed, the sets have rather mixed properties: on the one hand they contain translations $H+\la$, $\la \in \L$
of really structured set $H$ but on the other hand they have also some random properties, for example, its Fourier
coefficients (see the definition in section \ref{sec:definitions}) are small.
Our first result says that a set $A$ is close to a set of the form $H \dotplus \Lambda$ iff there is
the critical relation between $\E_3 (A)$ and $\E(A)$, that is $\E_3(A) \gg |A|\E(A)$.
For more rigorous
% more precisely,
formulation
see Theorem \ref{t:H+L_description}.


In the situation $\a = \frac{1-\tau_0}{2}$, $\Gr = \f_2^n$ our set $A$ looks like a union
of (additively) disjoint subspaces $H_1,\dots,H_k$ (see example (iii) from \cite{Sanders_survey2})
with $k = |A|^{\frac{1+\tau_0}{2}}$.
Such sets
%are
can be
called {\it self--dual} sets, see \cite{s_mixed}.
In our second result we show that, roughly speaking,
any such a set has critical relation between $\E_3 (A)$
(more precisely, the product
%between
$\E(A) \cdot \E_4(A)$)
and so--called
Gowers $U^3$--norm of the set $A$ (see the definition in section \ref{sec:Gowers}) and vice versa.
Theorem \ref{t:self-dual} contains the
%precise
exact
formulation.



These two structural results on sets having critical relations between a pair of its energies are the hearth of our paper.
In the opposite of Theorem \ref{t:BK_structural}
almost
all bounds of the paper are polynomial, excluding, of course,
the dependence on the number $k$ of the considered energies
$\E_k$, $\T_k$ or $U^k$ if its appear.
%But
Moreover
the first
%one
structural theorem
hints us a partial  answer to the following
%further
important  question.
%Now
Consider the difference set $D=A-A$ or the sumset $S=A+A$ of an arbitrary set $A$.
What can we say nontrivial about the energies of $D$, $S$ in terms of the energies of $A$?
In view of
%the first of the main examples
the first example
%of section \ref{sec:structural}
above,
that is $\Gr = \f_2^n$, $A=H\dotplus \L$,
$|\L| = K$, $\E(A) \sim |A|^3 / K$
we cannot hope to obtain a nontrivial bound for additive energy of $D$ or $S$ because in the case
$D=S=H \dotplus (\L+\L)$, and
hence it has a similar structure to $A$ with $\L$ replacing by $\L+\L$.
On the other hand, we know that the sets of the form $H\dotplus \L$ have large $\E_3$ energy.
Thus, one can hope to
%prove
obtain
a good lower bound for $\E_3 (D)$ and $\E_3 (S)$.
It turns out to be the case and we prove it in section \ref{sec:sumsets1}.
Roughly speaking, our result asserts that if $|D|=K|A|$, $\E(A) \ll |A|^3 /K$ then
\begin{equation}\label{f:start_E3D}
    \E_3 (D) \gg K^{7/4} |A|^4 \,,
\end{equation}
and a similar inequality for $A+A$.
%provided some regularity assumptions on $A$ hold.



The paper is organized as follows.
We start  with definitions and notations used in the paper.
In the next section we give several characterisations of sets of the form $A=H\dotplus \L$, where
%roughly,
$H$ is a set with small doubling, $\L$ is a "dissociated"\, set.
Also we consider a "dual"\, question on sets having critical relations between $\T_4$ and $\E$ energies,
that is the situation when $\T_4 (A)$ is large in terms on $\E(A)$.
It was proved that, roughly speaking, $A$ contains a large subset $A'$ such that the sequence $A', 2A', 3A', \dots$
is stabilized at the second step, namely, $A'+A'$ is a set with small doubling
and, besides, $|A'+A'| \approx |A|^4 / \E (A)$
in the
%situation iff
only case when
$\T_4 (A) \gg |A|^2 \E(A)$, see Theorem \ref{t:T_3_and_E_critical}.


Section \ref{sec:sumsets1} contains the proof of inequality (\ref{f:start_E3D})
and we make some preliminaries to this in section \ref{sec:sumsets}.
For example, we obtain in the section an interesting characterisation of sumsets $S=A+A$ or difference sets $D=A-A$ with
extremal cardinalities of intersections
$$|A| \le |D\cap (D+x_1) \cap \dots \cap (D+x_s)| \le |A|^{1+o(1)} \,,$$
and, similarly, for $S$,
see Theorem \ref{t:dichotomy_DS}.
It turns out that for such sets $D$, $S$ the set $A$
should have either very small $O(|A|^{k+o(1)})$
energy $\E_k (A)$
or very large $\gg |A|^{3-o(1)}$ additive energy.
In other words either $A$ has "random behaviour"\,
or, in contrary, is very structured.
%is close to a random set or to a set with small doubling.
Clearly, both situations are realized: the first one in the situation when
$A$ is a fair random set (and hence $A\pm A$ has almost no structure) and the second one
if $A$ is a set with small doubling, say.


In section \ref{sec:Gowers} we consider  some simple properties of Gowers norms of
{\it the characteristic function of a set $A$} and
prove a preliminary result on the connection of $\E(A)$ with $\E(A\cap (A+s))$, $s\in A-A$,
see Theorem \ref{t:E(A_s)}.
%the following result
It gives a partial counterexample to a famous construction of Gowers \cite{Gow_4}, \cite{Gow_m} of uniform sets
with non-uniform intersections $\E(A\cap (A+s))$
(see the definitions in \cite{Gow_4}, \cite{Gow_m} or \cite{TV}).
We show that although all sets $A\cap (A+s)$ can be non-uniform but there is always $s\neq 0$
such that $\E(A\cap (A+s)) \ll |A\cap (A+s)|^{3-c}$, $c>0$, provided by some weak conditions take place.
This question was asked to the author by T. Schoen.


In the next section we develop the investigation from the previous one and
characterize all sets with critical relation between Gowers $U^3$--norm and the energies  $\E,\E_4$ or $\E_3$.
Also we consider some questions on finding in $A$ a family of disjoint sets $A\cap (A+s)$ or its large disjoint subsets.


A lot of results of the paper such as Bateman--Katz theorem  are
%require
proved under
some regular conditions
on $A$.
For example, the
%condition
assumption
from Theorem \ref{t:BK_structural} require that
for  all $A_*\subseteq A$, $|A_*| \gg |A|$ the following holds $\E(A_*) \gg \E (A)$.
We call the conditions as connectedness of our set $A$
(see the definitions from sections \ref{sec:preliminaries}, \ref{sec:Gowers})
and prove in the appendix that any set contains some large connected subset.
Basically, we
%just
generalize the method from \cite{s_doubling}.


Thus, we have characterized two extremal
%cases
situations
of Theorem \ref{t:BK_structural} in terms of energies.
Is there some similar characterisation for other cases of the result?
Do exist criteria in terms of
%some
energies for another families of sets?
Finally,
are there
further
%another
characteristics of sumsets/difference sets which separate it from arbitrary sets?








\section{Definitions}
\label{sec:definitions}



Let $\Gr$ be an abelian group.
If $\Gr$ is finite then denote by $N$ the cardinality of $\Gr$.
It is well--known~\cite{Rudin_book} that the dual group $\FF{\Gr}$ is isomorphic to $\Gr$ in the case.
Let $f$ be a function from $\Gr$ to $\mathbb{C}.$  We denote the Fourier transform of $f$ by~$\FF{f},$
\begin{equation}\label{F:Fourier}
  \FF{f}(\xi) =  \sum_{x \in \Gr} f(x) e( -\xi \cdot x) \,,
\end{equation}
where $e(x) = e^{2\pi i x}$
and $\xi$ is a homomorphism from $\FF{\Gr}$ to $\R/\Z$ acting as $\xi : x \to \xi \cdot x$.
We rely on the following basic identities
\begin{equation}\label{F_Par}
    \sum_{x\in \Gr} |f(x)|^2
        =
            \frac{1}{N} \sum_{\xi \in \FF{\Gr}} \big|\widehat{f} (\xi)\big|^2 \,,
\end{equation}
\begin{equation}\label{svertka}
    \sum_{y\in \Gr} \Big|\sum_{x\in \Gr} f(x) g(y-x) \Big|^2
        = \frac{1}{N} \sum_{\xi \in \FF{\Gr}} \big|\widehat{f} (\xi)\big|^2 \big|\widehat{g} (\xi)\big|^2 \,,
\end{equation}
and
\begin{equation}\label{f:inverse}
    f(x) = \frac{1}{N} \sum_{\xi \in \FF{\Gr}} \FF{f}(\xi) e(\xi \cdot x) \,.
\end{equation}
If
$$
    (f*g) (x) := \sum_{y\in \Gr} f(y) g(x-y) \quad \mbox{ and } \quad
        %(f\circ g) (x) := \sum_{y\in \Gr} \ov{f(y)} g(y+x)
        (f\circ g) (x) := \sum_{y\in \Gr} f(y) g(y+x)
$$
 then
\begin{equation}\label{f:F_svertka}
    \FF{f*g} = \FF{f} \FF{g} \quad \mbox{ and } \quad \FF{f \circ g} = \FF{f^c} \FF{g} = \ov{\FF{\ov{f}}} \FF{g} \,,
    %(\F{fg}) (x) = \frac{1}{N} (\F{f} * \F{g}) (x) \,.
\end{equation}
where for a function $f:\Gr \to \mathbb{C}$ we put $f^c (x):= f(-x)$.
 Clearly,  $(f*g) (x) = (g*f) (x)$ and $(f\c g)(x) = (g \c f) (-x)$, $x\in \Gr$.
 The $k$--fold convolution, $k\in \N$  we denote by $*_k$,
 so $*_k := *(*_{k-1})$.
% It is unimportant but write for definiteness
% $$(f \c_k f) (x) := \sum_{y_1,\dots,y_k} f(y_1) \dots f(y_k) f(x+y_1+\dots+y_k)\,.$$



We use in the paper  the same letter to denote a set
$S\subseteq \Gr$ and its characteristic function $S:\Gr\rightarrow \{0,1\}.$
Clearly, $S$ is the characteristic function of  a set iff
\begin{equation}\label{f:char_char}
    \FF{S} (x) = N^{-1} (\ov{\FF{S}} \c \FF{S}) (x) \,.
\end{equation}
Write $\E(A,B)$ for the {\it additive energy} of two sets $A,B \subseteq \Gr$
(see e.g. \cite{TV}), that is
$$
    \E(A,B) = |\{ a_1+b_1 = a_2+b_2 ~:~ a_1,a_2 \in A,\, b_1,b_2 \in B \}| \,.
$$
If $A=B$ we simply write $\E(A)$ instead of $\E(A,A).$
Clearly,
\begin{equation}\label{f:energy_convolution}
    \E(A,B) = \sum_x (A*B) (x)^2 = \sum_x (A \circ B) (x)^2 = \sum_x (A \circ A) (x) (B \circ B) (x)
    \,.
\end{equation}
and by (\ref{svertka}),
%we have
%the following holds
\begin{equation}\label{f:energy_Fourier}
    \E(A,B) = \frac{1}{N} \sum_{\xi} |\FF{A} (\xi)|^2 |\FF{B} (\xi)|^2 \,.
\end{equation}
Let
$$
   \T_k (A) := | \{ a_1 + \dots + a_k = a'_1 + \dots + a'_k  ~:~ a_1, \dots, a_k, a'_1,\dots,a'_k \in A \} |
    =
        \frac{1}{N} \sum_{\xi} |\FF{A} (\xi)|^{2k}
        %\,.
$$
and more generally
$$
    \T_k (A_1,\dots,A_k) :=
        | \{ a_1 + \dots + a_k = a'_1 + \dots + a'_k  ~:~ a_1,a'_1 \in A_1, \dots, a_k, a'_k \in A_k \} | \,.
$$
Let also
$$
    \sigma_k (A) := (A*_k A)(0)=| \{ a_1 + \dots + a_k = 0 ~:~ a_1, \dots, a_k \in A \} | \,.
$$
Notice that for a symmetric set $A$ that is $A=-A$ one has $\sigma_2
(A) = |A|$ and $\sigma_{2k} (A) = \T_k (A)$.
Having a set $P\subseteq A-A$ we write $\sigma_P (A) := \sum_{x\in P} (A\c A) (x)$.
%and moreover $\sigma_P (A,B) := \sum_{x\in P} (A\c B) (x)$.




 For a sequence $s=(s_1,\dots, s_{k-1})$ put
$A^B_s = B \cap (A-s_1)\dots \cap (A-s_{k-1}).$
If $B=A$ then write $A_s$ for $A^A_s$.
Let
\begin{equation}\label{f:E_k_preliminalies}
    \E_k(A)=\sum_{x\in \Gr} (A\c A)(x)^k = \sum_{s_1,\dots,s_{k-1} \in \Gr} |A_s|^2
\end{equation}
and
\begin{equation}\label{f:E_k_preliminalies_B}
\E_k(A,B)=\sum_{x\in \Gr} (A\c A)(x) (B\c B)(x)^{k-1} = \sum_{s_1,\dots,s_{k-1} \in \Gr} |B^A_s|^2
\end{equation}
be the higher energies of $A$ and $B$.
The second formulas in (\ref{f:E_k_preliminalies}), (\ref{f:E_k_preliminalies_B})
can be considered as the definitions of $\E_k(A)$, $\E_k(A,B)$ for non integer $k$, $k\ge 1$.
Similarly, we write $\E_k(f,g)$ for any complex functions $f$, $g$
and more generally
$$
    \E_k (f_1,\dots,f_{k}) = \sum_x (f_1 \c f_1) (x) \dots (f_k \c f_k) (x) \,.
$$
Putting $\E_1 (A) = |A|^2$.
%As above for a
For a set $P\subseteq \Gr$ write $\E^P_k (A) := \sum_{s\in P} |A_s|^k$, $\E^P (A) := \E^P_2 (A)$.
%and for a set $\mathcal{P} \subseteq \Gr^{k-1}$ put
%$$
%    \E^{\mathcal{P}}_k (A) := \sum_{(s_1,\dots,s_{k-1}) \in \mathcal{P}} |A_s|^2 \,.
%$$
We
%write
put
$\E^*_k (A)$ for $\E^*_k (A) = \sum_{s\neq 0} |A_s|^k$.



Clearly,
\begin{eqnarray}\label{f:energy-B^k-Delta}
\E_{k+1}(A,
B)&=&\sum_x(A\c A)(x)(B\c B)(x)^{k}\nonumber \\
&=&\sum_{x_1,\dots, x_{k-1}}\Big (\sum_y A(y)B(y+x_1)\dots
B(y+x_{k})\Big )^2 =\E(\Delta_k (A),B^{k}) \,,
 \end{eqnarray}
where
$$
    \Delta (A) = \Delta_k (A) := \{ (a,a, \dots, a)\in A^k \}\,.
$$
We also put $\Delta(x) = \Delta (\{ x \})$, $x\in \Gr$.
%Note that $B\subseteq A$ iff $\D (B) \subseteq \D (A)$.


%We shall write $\sum_x$ and $\sum_{\xi}$ instead of $\sum_{x\in \Gr}$ and $\sum_{\xi \in \FF{\Gr}}$ for simplicity.

\bigskip

Quantities $\E_k (A,B)$ can be written in terms of generalized convolutions.


\begin{definition}
   Let $k\ge 2$ be a positive number, and $f_0,\dots,f_{k-1} : \Gr \to \C$ be functions.
%Write $F$ for the vector $(f_0,\dots,f_{k-1})$ and $x$ for vector $(x_1,\dots,x_{k-1})$.
Denote by
$${\mathcal C}_k (f_0,\dots,f_{k-1}) (x_1,\dots, x_{k-1})$$
the function
$$
    %\Cf_k(F) (x) =
    \Cf_k (f_0,\dots,f_{k-1}) (x_1,\dots, x_{k-1}) = \sum_z f_0 (z) f_1 (z+x_1) \dots f_{k-1} (z+x_{k-1}) \,.
$$
Thus, $\Cf_2 (f_1,f_2) (x) = (f_1 \circ f_2) (x)$.
%Put $C_1 (f) = \sum_z f(z)$.
If $f_1=\dots=f_k=f$ then write
$\Cf_k (f) (x_1,\dots, x_{k-1})$ for $\Cf_k (f_1,\dots,f_{k}) (x_1,\dots, x_{k-1})$.
\end{definition}


In particular, $(\Delta_k (B) \c A^k) (x_1,\dots,x_k) = \Cf_{k+1} (B,A,\dots,A) (x_1,\dots,x_k)$, $k\ge 1$.



Quantities $\E_k (A)$ and $\T_k (A)$ are "dual"\, in some sense.
For example in \cite{s_mixed}, Note 6.6 (see also \cite{SS1}) it was proved that
$$
    \left( \frac{\E_{3/2} (A)}{|A|} \right)^{2k}
        \le
            \E_k (A) \T_k (A) \,,
$$
provided by $k$ is even.
Moreover, from (\ref{F:Fourier})---(\ref{f:inverse}), (\ref{f:char_char}) it follows that
%$\E_{2k} (\FF{A}, \dots, \FF{A}, \ov{\FF{A}}, \dots, \ov{\FF{A}})
$\t{\E}_{2k} (\FF{A}) := \sum_{x} ( \ov{\FF{A}} \c \FF{A})^{k} (x) (\FF{A} \c \ov{\FF{A}})^{k} (x)
= N^{2k+1} \T_k (A)$ and $\T_k (|\FF{A}|^2) = N^{2k-1} \E_{2k} (A)$.


\bigskip

%We conclude with few comments regarding the notation used in this paper.
For a positive integer $n,$ we set $[n]=\{1,\ldots,n\}.$
Let $x$ be a vector. By $\| x \|$ denote the number of components of $x$.
All logarithms are to base $2.$ Signs $\ll$ and $\gg$ are the usual Vinogradov's symbols
and if the bounds depend on some parameter $M$ {\it polynomially} then we write $\ll_M$, $\gg_M$.
If for two numbers $a$, $b$ the following holds $a \ll_M b$, $b \ll_M a$ then we write $a \sim_M b$.
In particular, $a \sim b$ means $a\ll b$ and $b\ll a$.

All polynomial bounds in the paper can be obtained in explicit way.
%Finally, with a slight abuse of notation we use the same letter to denote a set $S\subseteq \Gr$
%and its characteristic function $S:\Gr\rightarrow \{0,1\}.$


%%We conclude with few comments regarding the notation used in this paper.
%For a positive integer $n,$ we set $[n]=\{1,\ldots,n\}$.
%All logarithms used in the paper are to base $2.$
%By  $\ll$ and $\gg$ we denote the usual Vinogradov's symbols.
%If $p$ is a prime number then write $\F_p$ for $\Z/p\Z$ and $\F^*_p$ for $(\Z/p\Z) \setminus \{ 0 \}$.

\newpage


\section{Preliminaries}
\label{sec:preliminaries}




Let us begin with the famous Pl\"{u}nnecke--Ruzsa inequality (see  \cite{petridis} or \cite{TV}, e.g.).

\begin{lemma}
Let $A\subseteq \Gr$ be a set.
Then for all positive integers $n,m$ the following holds
\begin{equation}\label{f:Plunnecke}
    |nA-mA| \le K^{n+m} |A| \,.
\end{equation}
\label{l:Plunnecke}
\end{lemma}



%This
We need in several quantitative versions
%will require an application
of the Balog--Szemer\'{e}di--Gowers Theorem.
The first symmetric variant
%with the least
%The version below
is due to T. Schoen \cite{schoen_BSzG}.
% and provides the best bounds in the statement at the moment.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set, $K\geq{1}$ and $\E(A)\geq{\frac{|A|^{3}}{K}}$.
    Then there is $A'\subseteq A$ such that
    $$
        |A'| \gg{\frac{|A|}{K}}\,,
    $$
    and
    $$
        |A'-A'| \ll K^4 |A'| \,.
    $$
\label{BSG}
\end{theorem}


\bigskip


Also we need in
%two
a version
of Balog--Szemer\'{e}di--Gowers theorem in the asymmetric form, see \cite{TV}, Theorem 2.35.
%section 2.5.


%\begin{theorem}
%    Let $A,B\subseteq \Gr$ be two sets, $K\ge 1$ and $\E(A,B) \ge |A|^{3/2} |B|^{3/2} / K$.
%    Then there are $A'\subseteq A$, $B' \subseteq B$ such that
%    $$
%        |A'| \gg |A| / K\,, \quad \quad |B'| \gg |B| / K \,,
%    $$
%    and
%    $$
%        |A' + B'| \ll K^7 |A|^{1/2} |B|^{1/2} \,.
%    $$
%\label{t:BSzG_1.5}
%\end{theorem}


%To do this we need in an asymmetric version of Balog--Szemer\'{e}di--Gowers theorem, see \cite{TV}, Theorem 2.35.


\begin{theorem}
    Let $A,B\subseteq \Gr$ be two sets, $|B| \le |A|$, and $M\ge 1$ be a real number.
    Let also $L=|A|/|B|$ and $\eps \in (0,1]$ be a real parameter.
    Suppose that
\begin{equation}\label{cond:BSzG_as}
    \E (A,B) \ge \frac{|A| |B|^2}{M} \,.
\end{equation}
    Then there are two sets $H\subseteq \Gr$, $\L \subseteq \Gr$ and $z\in \Gr$ such that
\begin{equation}\label{f:BSzG_as_1}
    |(H+z) \cap B| \gg_\eps M^{-O_\eps (1)} L^{-\eps} |B| \,,
    \quad \quad
    |\L| \ll_{\eps} M^{O_\eps (1)} L^\eps \frac{|A|}{|H|} \,,
\end{equation}
\begin{equation}\label{f:BSzG_as_2}
    |H - H| \ll_{\eps} M^{O_\eps (1)} L^\eps \cdot |H| \,,
\end{equation}
     and
\begin{equation}\label{f:BSzG_as_3}
    |A\cap (H+\L)| \gg_{\eps} M^{-O_\eps (1)} L^{-\eps} |A| \,.
\end{equation}
\label{t:BSzG_as}
\end{theorem}


%First of all we need a lemma on the connection between $\E_3 (A)$
%and $\E (A,A_s)$, see e.g  \cite{SS1}.
The next lemma is a special case of  Lemma 2.8 from \cite{SV}.
In particular, it gives us a connection between $\E_3 (A)$
and $\E (A,A_s)$, see e.g  \cite{SS1}.




\begin{lemma}
    Let $A\subseteq \Gr$ be a set.
    Then for every $k,l\in \N$
$$
    \sum_{s,t:\atop \|s\|=k-1,\,\, \|t\|=l-1} \E(A_s,A_t)=\E_{k+l}(A)
    %\,,
    \,.
$$
%    where $\|x\|$ denote the number of components of vector $x$.
    In particular,
$$
    \E_3 (A) = \sum_s \E (A,A_s) \,.
$$
\label{l:E_3_A_s}
\end{lemma}


Now recall a lemma from \cite{SS3}, \cite{s_mixed}.

\begin{lemma}
\label{corpop}
    Let $A$ be a subset of an abelian group, $Q \subseteq A-A$.
    % and $\sum_{s\in P_*} |A_s| = \eta |A|^2$, $\eta \in (0,1]$.
    Then
    \begin{equation*}
        \sum_{s\in Q} |A\pm A_s| \geq \frac{\sigma^2_{Q} (A) |A|^2}{\E_3(A) }
        %\,.
         %\text{~~ and ~~}\, \sum_{s\in P'} |A\pm A_s| \gg |A|^6\E^{-1}_3(A).
    \end{equation*}
    and
\begin{equation}\label{f:corpop2}
    \E_3 (Q,A,A) \cdot \E_3 (A) \ge \frac{\E^2 (A) \sigma^4_{Q} (A)}{|A|^{6}}
        \,.
%            =
%        \eta^4 |A|^2 \E^2 (A) \,.
\end{equation}
\end{lemma}


\bigskip

Let also give a simple
%Recall
Corollary 18 from \cite{s_ineq}.

\begin{lemma}
    Let $A\subseteq \Gr$ be a set.
    Then
$$
    \sum_{s} \frac{|A_s|^2}{|A\pm A_s|} \le \frac{\E_3 (A)}{|A|^2}
    \,.
$$
\label{l:E_3_weight}
\end{lemma}





%A simple lemma was proved  in
We give a
%slight
small
generalization of  Proposition 11 from \cite{SS1},
%basically,
see also \cite{M_R-N_S}.


\begin{lemma}
    Let $A\subseteq \Gr$ be a set, $n,m\ge 1$ be positive integers.
    Then
    \begin{equation}\label{tmp:12.05.2014_1}
        |A^{n+m} - \D(A)| \ge |A|^m |A^{n} - \D(A)| \,,
    \end{equation}
    and
    \begin{equation}\label{tmp:12.05.2014_2}
        |A^{n+m} + \D(A)| \ge |A|^m \max\{ |A^{n} + \D(A)|, |A^{n} - \D(A)| \} \,.
    \end{equation}
    In particular,
    \begin{equation}\label{f:A^2_pm_p1}
        |A^2 - \D(A)| = \sum_{s\in A-A} |A-A_s| \ge |A| |A-A| \,,
    \end{equation}
    and
    \begin{equation}\label{f:A^2_pm_p2}
        |A^2 + \D(A)| = \sum_{s\in A-A} |A+A_s| \ge |A| \max\{ |A+A|,|A-A| \} \,.
    \end{equation}
\label{l:A^2_pm}
\end{lemma}
\begin{proof}
    In view of \cite{SS1}, Proposition 11 it remains to prove
    the second bound from (\ref{tmp:12.05.2014_2})
%    with minus
    in the case $m=1$ only, namely, that
    $|A^{n+1} + \D(A)| \ge |A| |A^{n} - \D(A)|$, $n\ge 1$.
    But $(a_1+a,\dots,a_{n}+a,a_{n+1}+a) \in A^{n+1} + \D(A)$ iff $a_{n+1}\in A_{s_1,\dots,s_n}$,
    where $s_j = a_j-a_{n+1}$, $(s_1,\dots,s_n) \in A^n - \D(A)$.
    Thus
$$
    |A^{n+1} + \D(A)| = \sum_{(s_1,\dots,s_n) \in A^n - \D(A)} |A+A_{s_1,\dots,s_n}|
        \ge
            |A| |A^{n} - \D(A)|
$$
    and the result follows.
%$\hfill\Box$
\end{proof}


\bigskip

We will use  the Katz--Koester trick \cite{kk}
\begin{equation}\label{f:KK_trick}
    A - A_s \subseteq (A-A)_{-s}\,, \quad \quad \quad A + A_s \subseteq (A+A)_s \,,
    %A+A\cap (s-A) \subseteq
\end{equation}
and its generalization (see e.g. \cite{SV})
\begin{equation}\label{f:KK_trick_new}
    A - A_{\v{x}} \subseteq (A-A)_{-\v{x}} \,, \quad \quad \quad A + A_{\v{x}} \subseteq (A+A)_{\v{x}}
    %\,.
\end{equation}
%where $\| x \|$
very often.

\bigskip


Finally, recall some results from \cite{s_mixed}.
We begin with an analog of a definition from \cite{s_doubling}.


\begin{definition}
    Let $\a > 1$ be a real number, $\beta,\gamma \in [0,1]$.
    A set $A\subseteq \Gr$ is called $(\a,\beta,\gamma)$--connected
    if for any $B \subseteq A$, $|B| \ge \beta|A|$
    %one has
    the following holds
    $$
        \E_\a (B) \ge \gamma \left( \frac{|B|}{|A|} \right)^{2\a} \E_\a (A) \,.
    $$
\label{def:conn}
\end{definition}

Thus, a set from Theorem \ref{t:BK_structural} is a $(2,\beta,\gamma)$--connected set
with $\beta,\gamma \gg 1$.
The H\"{o}lder inequality implies that if
%Note also that if
$\E_\a (A) \le \gamma^{-1} |A|^{2\a} |A-A|^{1-\a}$ then $A$ is $(\a,\beta,\gamma)$--connected for any $\beta$.
As was proved in  \cite{s_doubling} that for $\a=2$ {\it every} set $A$ always contains large connected subset.
For
%larger
integers $\a>2$, see the Appendix.


\bigskip


Our first lemma from \cite{s_mixed} (where some operators were used in the proof)
provides  a nontrivial lower bound for $\E_s (A)$, $s\in [1,2]$
in terms of $\E(A)$.


\begin{lemma}
    Let $A\subseteq \Gr$ be a set, and $\beta,\gamma \in [0,1]$.
    Suppose that $A$ is $(2,\beta,\gamma)$--connected with $\beta \le 1/2$.
    Then for any $s\in [1,2]$ the following holds
\begin{equation}\label{f:connected}
    \E_s (A) \ge 2^{-5} \gamma |A|^{1-s/2} \E^{s/2} (A) \,.
\end{equation}
\label{l:connected}
\end{lemma}


The second lemma from \cite{s_mixed} provides an upper bound for eigenvalues of some operators.
To avoid of using the operators notation we formulate the result in the following way.



\begin{lemma}
    Let $A\subseteq \Gr$ be a set.
    Then for an arbitrary function $f : A \to \C$ one has
\begin{equation}\label{f:eigen_A}
        \E(A,f) \le \E^{1/2}_3 (A) \| f\|_2^2 \,.
\end{equation}
    Further, there is a set $A'\subseteq A$, $|A'| \ge |A|/2$, namely,
\begin{equation}\label{f:A'_def}
    A' := \{ x ~:~ ((A*A) \c A) (x) \le 2 \E(A) |A|^{-1} \}
\end{equation}
    such that for any
    function $f : A' \to \C$ the following holds
\begin{equation}\label{f:eigen_A'}
    \E(A,f) \le \frac{2\E(A)}{|A|} \cdot \| f\|_2^2 \,.
\end{equation}
    Moreover for any even real function $g$ there is a set $A'\subseteq A$, $|A'| \ge |A|/2$
    %(is defined in formula (\ref{f:A'_def}))
    such that for any
    function $f : A' \to \C$
    %the following holds
    one has
\begin{equation}\label{f:eigen_A''}
    \sum_x g(x) (\ov{f} \c f)(x) = \sum_x g(x) (f\c \ov{f})(x)
        \le 2 |A|^{-1} \sum_x g(x) (A\c A) (x) \cdot \| f\|_2^2 \,.
\end{equation}
\label{l:eigen_A'}
\end{lemma}


Note that
%for the characteristic functions
if $f = \t{A}$,
%of sets from
$\t{A} \subseteq A$
then
bound
(\ref{f:eigen_A}) can be obtained using the Cauchy--Schwarz inequality.
%$\E^2 (A,A_s) \le $
Further, estimate  (\ref{f:eigen_A''}) is a generalization of (\ref{f:eigen_A'})
which was
%obtained
proved
in \cite{s_mixed}, see Lemma 44. Bound (\ref{f:eigen_A''}) can be obtained in a similar way.


\bigskip

%It remains to prove the only statement where the opera

We finish the section noting a generalization of formula (\ref{f:corpop2}) of Lemma \ref{corpop}.
That is just a part of  Lemma 4.2 from \cite{s_mixed}.



\begin{lemma}
    Let $A,B\subseteq \Gr$ be finite sets, $S\subseteq \Gr$ be a set such that
    $A+B \subseteq S$.
    Suppose that $\psi$ is a
    function on $\Gr$.
    Then
\begin{equation}\label{f:T_A,B}
    |B|^2 \cdot \left( \sum_{x} \psi(x) (A\c A)(x) \right)^2
        \le
            \E_3(B,A) \sum_{x} \psi^2 (x) (S\c S)(x) \,.
\end{equation}
\label{l:T_A,B}
\end{lemma}

\section{Structural results}
\label{sec:structural}




%We need in an auxiliary statement.

In this section we obtain several general structural results,
some of which have applications to sum--products phenomenon, for example.
These results are closely related to the Balog--Szemer\'{e}di--Gowers Theorem.
We adopt the convention of writing $\Gr$ as an additive group.
The proofs follow the arguments from \cite{SS1} and \cite{SS2}.


Now we formulate the first result of the section.


\begin{proposition}
    Let $A\subseteq \Gr$ be a finite set, and $M\ge 1$, $\eta \in (0,1]$ be real numbers.
    Let $\E(A) = |A|^3/K$.
    Suppose that for some set $P\subseteq A-A$ the following holds
\begin{equation}\label{cond:eta}
    \sum_{s\in P} (A\c A) (s) = \eta |A|^2 \,,
\end{equation}
    and
\begin{equation}\label{cond:M}
    \sum_{s\in P} |A \pm A_s| \le M K |A|^2 \,.
\end{equation}
    Then
    for any $\eps \in (0,1)$,
    there are two sets $H\subseteq \Gr$, $\L \subseteq \Gr$ and $z\in \Gr$ such that
\begin{equation}\label{f:E_3_and_E_critical_1'}
    |(H+z) \cap A| \gg_{M,\eta^{-1},K^\eps} \frac{\E(A)}{|A|^2} \,,
    \quad \quad
    |\L| \ll_{M,\eta^{-1},K^\eps} \frac{|A|}{|H|} \,,
\end{equation}
\begin{equation}\label{f:E_3_and_E_critical_2'}
    |H - H| \ll_{M,\eta^{-1},K^\eps} |H| \,,
\end{equation}
     and
\begin{equation}\label{f:E_3_and_E_critical_3'}
    |A\bigcap (H+\L)| \gg_{M,\eta^{-1},K^\eps} |A| \,.
\end{equation}
\label{p:A^2-D(A)}
\end{proposition}
\begin{proof}
Using Lemma \ref{corpop} with $Q=P$, we see that
\begin{equation}\label{tmp:08.05.2014_1}
    \E_3 (A) \ge \frac{\eta^2 |A|^4}{M K} \,.
\end{equation}
We have
\begin{equation*}
\sum_{s ~:~ |A_s|<\frac{|A|\eta^2}{2KM}} \E(A,A_s)
    \leq {\left(\frac{|A|\eta^2}{2KM}\right)\sum_s|A_s||A|}
=\frac{|A|^4\eta^2}{2KM} \,.
\end{equation*}
Applying Lemma \ref{l:E_3_A_s}, that is the formula $\E_3 (A) = \sum_s \E(A,A_s)$,
combining it with (\ref{tmp:08.05.2014_1}), we get
\begin{equation}\label{TMP:05.10.2013}
    \sum_{s ~:~ |A_s| \ge 2^{-1} \eta^2 M^{-1} K^{-1} |A| } \E(A,A_s) \ge \frac{\eta^2 |A|^4}{2M K} \,.
\end{equation}
Put
$$
    \mu := \max_{s ~:~ |A_s| \ge 2^{-1} \eta^2 M^{-1} K^{-1} |A| } \frac{\E(A,A_s)}{|A| |A_s|^2 } \,.
$$
Using (\ref{TMP:05.10.2013}), we have
$$
    \mu |A| \E (A) \ge \mu |A| \cdot \sum_{s ~:~ |A_s| \ge 2^{-1} \eta^2 M^{-1} K^{-1} |A| } |A_s|^2
        \ge
            \frac{\eta^2 |A|^4}{2M K} \,.
$$
Thus, $\mu \ge \frac{\eta^2}{2M}$.
Hence there is an $s$
%such that
with
$|A_s| \ge 2^{-1} \eta^2 M^{-1} K^{-1} |A|$
and
such that\\
${\E(A,A_s) \ge 2^{-1} M^{-1} \eta^2|A||A_s|^2}$.
Applying the asymmetric version of Balog--Szemer\'{e}di--Gowers Theorem \ref{t:BSzG_as},
we find two sets $\L,H$ such that
(\ref{f:E_3_and_E_critical_1'})---(\ref{f:E_3_and_E_critical_3'}) take place.
%$A\approx_{M,\eta^{-1},K^\eps} \L \dotplus H$.\footnote{Recall the definition \eqref{approxdefn} of this approximate %equality of sets.}
This completes the proof.
%$\hfill\Box$
\end{proof}



\bigskip
We write the fact that sets $A,H,\L$ satisfy (\ref{f:E_3_and_E_critical_1'})---(\ref{f:E_3_and_E_critical_3'})
with $\eta \gg 1$ as
\begin{equation}
    A \approx_{M,K^\eps} \L \dotplus H.
\label{approxdefn}
\end{equation}
Note that the degree of polynomial dependence in formula (\ref{approxdefn}) is a function on $\eps$.



\begin{example}
    Let $H\subseteq \mathbf{F}_2^n$ be a subspace and $\Lambda \subseteq \mathbf{F}_2^n$ be a dissociated set (basis).
    Put $A=H \dotplus \Lambda$, where $\dotplus$ means the direct sum, and $|\Lambda| = K$.
    Detailed discussion of the example can be found, e.g. in \cite{s_mixed}.
    If $s\in H$ then $A_s = A$ and hence $A+A_s = A+A$.
    If $s\in (A+A) \setminus H$ then $A_s$ is the disjoint union of two shifts of $H$
    and thus $|A+A_s| \le 2|A|$.
    Whence
    $$
        \sum_{s\in A+A} |A+A_s| \le |H| |A+A| + 2|A+A||A| \ll K|A|^2 \,,
    $$
    and $\E(A) \sim |A|^3 /K$.
    It means that condition (\ref{cond:M}) takes place in the case
    %and by the construction we have, obviously,
    $A=H \dotplus \Lambda$.
\end{example}



Taking $P=A-A$ and applying Proposition \ref{p:A^2-D(A)}
as well as formulas (\ref{f:A^2_pm_p1}), (\ref{f:A^2_pm_p2}) of Lemma \ref{l:A^2_pm},
we obtain the following consequence.


\begin{corollary}
    Let $A\subseteq \Gr$ be a set, $M\in{\mathbb{R}}$, $\eps \in (0,1)$ and $\E(A) = |A|^3/K$.
    Then either
    $$
        |A^2 \pm \D(A)| \ge M K |A|^2
    $$
    or
    $A \approx_{M,K^{\eps}} \L \dotplus H$.
\label{c:A^2-D(A)}
\end{corollary}



Note that for any set $A\subseteq \Gr$ with $\E(A) = |A|^3/K$
the inequality $|A^2 \pm \D(A)| \ge K |A|^2$ follows from Lemma \ref{corpop}
and a trivial
%inequality
estimate
$\E_3(A) \le |A| \E(A)$.
%On the other hand ???????
We will deal
%develop the idea
with the reverse condition $\E_3(A) \gg |A| \E(A)$
in Proposition \ref{p:E_3_and_E_critical}
and Theorem \ref{t:E_3_and_E_critical} below.



The next corollary shows that if  a set $A$ is not close
to a set of the form $\L \dotplus H$ then there is some imbalance
(in view of Pl\"{u}nnecke--Ruzsa inequality (\ref{f:Plunnecke}))
between doubling constant
and additive energy of $A$ or $A-A$.



\begin{corollary}
    Let $A\subseteq \Gr$ be a set, $M$, $\eps \in (0,1)$ be real numbers and
    $|(A-A) \pm (A-A)| \gg |A-A|^3 / |A|^2$.
    Then either
    $$
        \E(A) \gg \frac{M^{1/2} |A|^4}{|A-A|}
            \quad \mbox{ or } \quad \E(A-A) \gg \frac{M |A-A|^4}{|(A-A) \pm (A-A)|}
        %\quad \mbox{ or }
    $$
    or
    $A \approx_{M,K^{\eps}} \L \dotplus H$, where
    %$K = |A|^3 / \E(A)$.
    $K = |A-A|/|A|$.
\label{c:3A}
\end{corollary}
\begin{proof}
Put $D=A-A$.
% and $|D| = K|A|$.
Suppose that $\E(A) \ll \frac{M^{1/2} |A|^4}{|D|}$ because otherwise we are done.
In view of Corollary \ref{c:A^2-D(A)} one can assume that $\sum_{s} |A-A_s| \ge M^{1/2} |A| |D|$.
Thus, by the Katz--Koester trick (\ref{f:KK_trick})
%\cite{kk}
\begin{equation}\label{f:KK_trick'}
    A - A_s \subseteq (A-A)_{-s}\,, \quad \quad \quad A + A_s \subseteq (A+A)_s \,,
    %A+A\cap (s-A) \subseteq
\end{equation}
and the Cauchy--Schwarz inequality, we get
$$
    |D| \E(D) \ge \left( \sum_{s\in D} (D\c D) (s) \right)^2 \ge \left( \sum_{s\in D} |A-A_s| \right)^2
        \ge (M^{1/2} |A| |D|)^2 \,.
$$
Hence
$$
    \E(D) \ge M^{} |D|^{} |A|^2 \gg \frac{M^{} |D|^4}{|(A-A) \pm (A-A)|}
$$
where the assumption of the corollary has been used.
This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


The quantities $\E(A\pm A)$ (and hence $|A^2 \pm \D(A)|$ in view of Lemma \ref{l:A^2_pm},
see also Proposition \ref{pr:E_k(D)_simple} below)
appear in sum--products results (in multiplicative form).
For example, in \cite{M_R-N_S} the following
%result was obtained.
theorem was proved.


\begin{theorem}
    Let $A,B\subseteq \R$ be finite sets.
Then
\begin{equation*}\label{f:main_intr_2_new}
    |B+AA|^3 \gg \frac{|B| \E^{\times} (AA)}{\log |A|} \,.
\end{equation*}
\end{theorem}


Here $\E^\times (A) := |\{ a_1 a_2 = a_3 a_4 ~:~ a_1, a_2, a_3, a_4 \in A \}|$.
Thus, by the obtained results, we have, roughly, that  either $\E^\times (A)$, $\E^\times (AA)$
can be estimated better then
by Lemma \ref{l:A^2_pm}, see formulas (\ref{f:A^2_pm_p1}), (\ref{f:A^2_pm_p2})
or $A$ has the rigid structure $A \approx \L \cdot H$.
Usually, the last case is easy to deal with.
Similar methods were used in \cite{M_R-N_S}.

\bigskip


Now we obtain another structural result.
Using Lemma \ref{corpop} as well as a trivial estimate $\E_3 (A) \le |A| \E(A)$
one can derive Proposition \ref{p:A^2-D(A)} from Proposition \ref{p:E_3_and_E_critical}
below.



\begin{proposition}
    Let $A\subseteq \Gr$ be a set, and $M\ge 1$ be a real number.
    Suppose that
\begin{equation}\label{cond:E_3_and_E_critical}
    \E_3 (A) \ge \frac{|A| \E(A)}{M} \,.
\end{equation}
    Then there is $A' \subseteq A$ such that
\begin{equation}\label{f:E_3_and_E_critical_1}
    |A'| \gg \frac{\E(A)}{|A|^2 (M \log M)^5}
\end{equation}
    and
\begin{equation}\label{f:E_3_and_E_critical_2}
    |A' - A'| \ll M^{15} \log^{16} M \cdot |A'| \,.
\end{equation}
    Further, take any $\eps \in (0,1)$ and
    put $K:=\frac{|A|^3}{\E (A)} $.
    Then $A \approx_{M,K^\eps} \L \dotplus H$.
%    there are two sets $H\subseteq \Gr$, $\L \subseteq \Gr$ and $z\in \Gr$ such that
%\begin{equation}\label{f:E_3_and_E_critical_1'}
%    |(H+z) \cap A| \gg_{M,K^\eps} \frac{\E(A)}{|A|^2} \,,
%        \quad \quad |\L| \ll_{M,K^\eps} \frac{|A|}{|H|} \,,
%\end{equation}
%\begin{equation}\label{f:E_3_and_E_critical_2'}
%    |H - H| \ll_{M,K^\eps} |H| \,,
%\end{equation}
%     and
%\begin{equation}\label{f:E_3_and_E_critical_3'}
%    |A\bigcap (H+\L)| \gg_{M,K^\eps} |A| \,.
%\end{equation}
\label{p:E_3_and_E_critical}
\end{proposition}
\begin{proof}
First of all prove (\ref{f:E_3_and_E_critical_1}),
(\ref{f:E_3_and_E_critical_2}). Let
$$
    P_j = \{ x ~:~ 2^{j-1} |A| / (2^{2} M)
        < |A_x| \le 2^{j} |A| / (2^{2} M) \} \,, \quad j\in [L] \,,
$$
where $L=[\log(4M)]$. By the pigeonhole principle there is $j\in [L]$
such that
$$
    \frac{|A| \E(A)}{2M L} \le \frac{\E_3 (A)}{2L} \le \sum_{x\in P_j} |A_x|^3 \,.
$$
Put $P=P_j$ and $\D =  2^{j} |A| / (2^{2} M)$.
Thus
\begin{equation}\label{tmp:04.02.2013_2}
    \frac{8M \E(A)}{2^{2j} |A|L} \le \sum_x P(x) (A\c A) (x) = \sum_x A(x) (A\c P) (x) \,.
\end{equation}
Hence, by the Cauchy--Schwarz inequality
\begin{equation}\label{tmp:05.02.2013_1}
    \frac{2^{6} M^2 \E^2 (A)}{2^{4j} |A|^3 L^2} \le \E(A,P) \le (\E (A))^{1/2} (\E (P))^{1/2} \,.
\end{equation}
Note that
$$\E(A)\geq{\sum_{x\in{P}}|A_x|^2}\geq{\frac{|P||A|^22^{2j-2}}{2^4M^2}},$$
and therefore
%$|P| \le 2^3 2^{-j} |A|$.
\begin{equation}
|P| \le 2^{6} 2^{-2j} M^{2} \E(A) |A|^{-2}.
\label{Pbound}
\end{equation}
 It follows that
\begin{equation}\label{tmp:08.02.2013_4}
    \E(P) \ge \frac{2^{12} M^4 (\E (A))^3}{2^{8j} |A|^6 L^4} \ge \frac{|P|^3}{2^{6} M^2 2^{2j} L^4}
        \ge
            \frac{|P|^3}{2^{10} M^4 L^4} := \mu |P|^3 \,.
\end{equation}
By Theorem \ref{BSG} there is $P'\subseteq P$ such that
$|P'| \gg \mu |P|$ and $|P'-P'| \ll \mu^{-4} |P'|$.
Note that
$$\sum_{x\in{A}} (A\c P') (x)=\sum_{x\in{P'}} (A\c A) (x)\geq{\frac{|P'|2^{j-3}|A|}{M}},$$
and so there exists  $x\in A$ such that the set $A' := A\cap (P'+x)$ has the size at
least $|P'| 2^{j-3} M^{-1}$. We have
\begin{equation}\label{tmp:08.02.2013_5}
    |A'-A'| \le |P' - P'| \ll \mu^{-4} |P'| \ll \mu^{-4} 2^{-j} M |A'| \ll M^{15} L^{16} |A'| \,.
\end{equation}
Finally, from (\ref{tmp:04.02.2013_2}), say, one has
\begin{equation}\label{tmp:06.02.2013_1}
    |P| \gg \frac{M^2 \E(A)}{2^{3j} |A|^2 L} \gg \frac{\E(A)}{M |A|^2 L}
\end{equation}
and because
$$
    |A'| \ge |P'| 2^{j-3} M^{-1} \gg \mu |P| \cdot 2^{j-3} M^{-1}
$$
the result follows.


To obtain
(\ref{f:E_3_and_E_critical_1'})---(\ref{f:E_3_and_E_critical_3'}),
that is $A \approx_{M,K^\eps} \L \dotplus H$, $K = |A|^3 \E^{-1} (A)$,
note that by the first inequality of (\ref{tmp:05.02.2013_1}) and the bound
$|P| \le 2^{6} 2^{-2j} M^{2} \E(A) |A|^{-2}$, we have
$$
    \E(A,P) \ge \frac{2^{6} M^2 \E^2 (A)}{2^{4j} |A|^3 L^2}
        \ge
            \frac{|A| |P|^2}{2^6 L^2 M^2} \,.
$$
Also, by the definition of the number $K$, and inequality  \eqref{tmp:06.02.2013_1} the following holds
$$|A|/|P| \ll ML K \ll_M K.$$
Applying the asymmetric version of
Balog--Szemer\'{e}di--Gowers Theorem \ref{t:BSzG_as} with $A=A$, $B=P$, and
recalling (\ref{tmp:06.02.2013_1}), we obtain the required inequalities,
excepting the first inequality of (\ref{f:E_3_and_E_critical_1'}),
where it remains to
%should
 replace $P$ by $A$.
Put $H' = (H+z) \cap P$.
We have $|H'| \gg_{M,K^\eps} |P|$.
Thus, by the definition of the number $\D$ and estimate (\ref{tmp:06.02.2013_1}), we obtain
\begin{equation}\label{tmp:04.05.2014_1}
    \sum_{x\in A} (A \c H') (x) = \sum_{x\in H'} (A\c A) (x) \ge 2^{-1} \D |H'|
        \gg_{M,K^\eps} \D |P|
        \gg_{M,K^\eps} \frac{\E (A)}{|A|} \,.
\end{equation}
Hence there is $w\in A$ such that
\begin{equation}\label{tmp:04.05.2014_2}
    |(H+w) \cap A| \ge |(H'+w) \cap A| \gg_{M,K^\eps} \frac{\E (A)}{|A|^2} \,.
\end{equation}
This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


Assumption (\ref{cond:E_3_and_E_critical}) of the Proposition \ref{p:E_3_and_E_critical}
is a generalisation of the usual condition $\E (A) \ge \frac{|A|^3}{M}$
(because $\E(A) |A|^3 M^{-1} \le \E^2 (A) \le \E_3 (A) |A|^2$)
and $\E_3 (A) \ge \frac{|A|^4}{M}$
(because $\E_3 (A) \ge |A|^4 M^{-1} \ge |A| \E(A) M^{-1}$).
Further, one can check that the same consequences
(\ref{f:E_3_and_E_critical_1'})---(\ref{f:E_3_and_E_critical_3'}) hold
if we replace condition (\ref{cond:E_3_and_E_critical}) by
$\E_s (A) \ge |A| \E_{s-1} (A) / M$, $s\ge 3$.
%Moreover, the same arguments give the following.
Let us write the correspondent result.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set, $s\ge 3$ be a positive integer, and $M\ge 1$ be a real number.
    Suppose that
\begin{equation}\label{cond:E_3_and_E_criticalt}
    \E_s (A) \ge \frac{|A| \E_{s-1} (A)}{M} \,.
\end{equation}
    Take any $\eps \in (0,1)$ and put $K:=\frac{|A|^s}{\E_{s-1} (A)} $.
    Then $A \approx_{s,\, M,\, K^\eps} \L \dotplus H$, $|H| \gg_{s,\, M,\, K^\eps} |A|/K$.
\label{t:E_3_and_E_critical}
\end{theorem}
\begin{proof}
The arguments almost repeat the proof of Proposition \ref{p:E_3_and_E_critical}, so we
skip some details.
Using dyadic pigeonholing and the assumption, we find $P\subseteq A-A$, $\D < |A_x|\le 2\D$, $x\in P$
with
$$
    M^{-1} |A| \E_{s-1} (A) \le \E_{s} (A) \ll_{\log M} \sum_{x\in P} |A_x|^s \ll_{\log M} \D^{s-1} \sigma_P (A) \,.
$$
Thus, by the Cauchy--Schwarz inequality
$$
    M^{-2} |A| \E^2_{s-1} (A) \D^{-2(s-1)} \ll_{\log M} \E(A,P) \,.
$$
On the other hand $|P| \D^{s-1} \le \E_{s-1} (A)$ and hence
$$
    \E(A,P) \gg_{\log M} M^{-2} |A| |P|^2 \,.
$$
Note that by (\ref{cond:E_3_and_E_criticalt}) and our choice of the set $P$, we have
$$
    \frac{|A|}{|P|} \ll_{M} \frac{\D^{s}}{\E_{s-1} (A)} \le \frac{|A|^s}{\E_{s-1} (A)} \,,
$$
and, again,
$$
    |P| \sim_{\log M} \E_s (A) \D^{-s} \ge  M^{-1} |A| \E_{s-1} (A) \D^{-s} = \frac{|A|^{s+1}}{MK \D^{s}}
        \ge
            \frac{|A|}{MK} \,.
$$
After that apply the asymmetric version of Balog--Szemer\'{e}di--Gowers Theorem \ref{t:BSzG_as}
and an analog of the arguments from (\ref{tmp:04.05.2014_1})---(\ref{tmp:04.05.2014_2}).
This concludes
%the sketch of
the proof.
\end{proof}
%$\hfill\Box$



\bigskip

%Because of
%$$
%    \E_s (A) \le |A| \E_{s-1} (A) \le \dots \le |A|^k \E_{s-k} (A) \,,
%$$
%where $1\le k \le s-2$ it follows that
The more general assumption
%that
$\E_s (A) \ge |A|^k \E_{s-k} (A)  / M^k$
implies that for some $j \in [k]$ one has $\E_{s-j+1} (A) \ge |A| \E_{s-j}(A)  / M$.
Thus, we have considered
%the general
the common
case.
Note, finally, that estimates (\ref{f:E_3_and_E_critical_1}), (\ref{f:E_3_and_E_critical_2})
are the best possible.
Indeed, take $\Gr = \mathbf{F}^n_2$, $A=H\dotplus \L$, where $H\le \mathbf{F}^n_2$ is a linear subspace
and $\L$ is a dissociated set (basis).
%\begin{remark}
%\label{r:E_k_E_gen}
%\end{remark}


\bigskip

Now we can prove a criterium for sets having critical relation between $\E (A) $ and $\E_3 (A)$ energies.

\begin{theorem}
    Let $A\subseteq \Gr$ be a set, and $M\ge 1, \eps \in (0,1)$ be real numbers.
    Put $K:=\frac{|A|^3}{\E (A)}$.
    Then
\begin{equation*}\label{cond:E_3_and_E_critical_cor}
    \E_3 (A) \gg_{M,K^\eps} |A| \E(A)
\end{equation*}
    iff
$$
    A \approx_{M,K^\eps} \L \dotplus H \,.
$$
\label{t:H+L_description}
\end{theorem}
\begin{proof}
    In view of Proposition \ref{p:E_3_and_E_critical} it remains to prove that if
$A \approx_{M,K^\eps} \L \dotplus H$ then $\E_3 (A) \gg_{M,K^\eps} |A| \E(A)$.
    Put $A_1 = A\cap (H+\L)$.
    We have $|A_1| \gg_{M,K^\eps} |A|$.
    Then
$$
    |A|^2 |H|^2 \ll_{M,K^\eps} |A_1|^2 |H|^2 \le \E(A_1,H) |A_1-H| \le \E(A,H) |\L| |H-H| \ll_{M,K^\eps} \E(A,H) |A| \,.
$$
    Thus
$$
    (|A| |H|^2)^3 \ll_{M,K^\eps} \left( \sum_x (A\c A)(x) (H\c H) (x) \right)^3
        \le
            \E_3 (A) \E^2_{3/2} (H)
                \le
                    \E_3 (A) |H|^5 \,.
$$
In other words
$$
    \E_3 (A) \gg_{M,K^\eps} |A|^3 |H| \gg_{M,K^\eps} \E(A) |A| \,.
$$
To get the last estimate
%follows from
we have used
the fact $|H| \gg_{M,K^\eps} \E (A) |A|^{-2}$.
 This completes the proof.
%$\hfill\Box$
\end{proof}



\bigskip


Recall that
$$
   \T_k (A) :=| \{ a_1 + \dots + a_k = a'_1 + \dots + a'_k  ~:~ a_1, \dots, a_k, a'_1,\dots,a'_k \in A \} |
   \,.
%   =
%    \sum_x r^2_{A+\dots+A} (x) \,.
$$
We conclude the section proving
a ``dual"\, analogue of Proposition \ref{p:E_3_and_E_critical},
that is we replace the condition on $\E_3 (A)$ with a similar condition for $\T_4 (A)$
and moreover for $\T_s (A)$.
Again, the proof follows the arguments from \cite{SS1}.



\begin{theorem}
    Let $A\subseteq \Gr$ be a set, and $M\ge 1$ be a real number.
    Suppose that
\begin{equation}\label{cond:T_3_and_E_critical}
    \T_4 (A) \ge \frac{|A|^4 \E(A)}{M} \,.
\end{equation}
    Then there is $A' \subseteq A$ such that
\begin{equation}\label{f:T_3_and_E_critical_1}
    |A'| \gg \frac{|A|}{M^3 \log^{\frac{16}{3}} |A|} \,,
\end{equation}
    and
\begin{equation}\label{f:T_3_and_E_critical_2}
    |nA' - mA'| \ll (M^3 \log^4 |A|)^{4(n+m)} M |A'| \cdot \frac{|A|^3}{\E(A)}
\end{equation}
for every $n,m\in \N$.


Moreover, if
\begin{equation}\label{cond:T_3_and_E_critical_s}
    \T_{2s} (A) \ge \frac{|A|^{2s} \T_s (A)}{M} \,,
\end{equation}
$s\ge 2$ then formulas (\ref{f:T_3_and_E_critical_1}),
(\ref{f:T_3_and_E_critical_2}) take place.
%
Conversely, bounds  (\ref{f:T_3_and_E_critical_1}), (\ref{f:T_3_and_E_critical_2}) imply that
$\T_{2s} (A) \gg_{M,\,\log |A|,\,s} |A|^{2s} \T_s (A)$.
\label{t:T_3_and_E_critical}
\end{theorem}
\begin{proof}
Put
%$\E = \E(A)$,
$\T_s = \T_s (A)$, $\T_{2s} = \T_{2s} (A)$, $a=|A|$,
$L_s = [\log (16Ma^{2s-1}/\T_s)] \ll_s \log a$.
%, $L=L_2 = [\log (16Ma^3/\E)]$.
Let
$$
    P_j = \{ x ~:~ 2^{j-1} \T_s / (2^{4} M a^s)
        < (A *_{s-1} A) (x) \le 2^{j} \T_s / (2^{4} M a^s) \} \,, \quad j\in [L_s] \,.
$$
Put $f_j (x) = P_j (x) (A *_{s-1} A) (x)$.
Thus,
\begin{equation}\label{tmp:04.05.2014_3}
    (A *_{s-1} A) (x) = \sum_{j=1}^{L_s} f_j (x) + \Omega (x) (A *_{s-1} A)(x) \,,
\end{equation}
where
$\Omega = \{ x ~:~ (A *_{s-1} A) (x) \le 2^{-4} M^{-1} \T_s a^{-s} \}$.
Substituting formula (\ref{tmp:04.05.2014_3}) into the identity
$$
    \T_{2s} (A) = \sum_{x} ((A *_{s-1} A) \c (A *_{s-1} A))^2 (x)
$$
and using assumptions (\ref{cond:T_3_and_E_critical}), (\ref{cond:T_3_and_E_critical_s}),
combining with the definition of sets $P_j$, $\Omega$,
we have
$$
    2^{-1} \T_{2s} (A) \le \sum_{j_1,j_2,j_3,j_4 = 1}^{L_s} \sum_x ( f_{j_1} \c f_{j_2} ) (x) ( f_{j_3} \c f_{j_4} ) (x) \,.
$$
Applying the H\"{o}lder inequality, we get
$$
    2^{-1} L_s^{-3} \T_{2s} (A) \le \sum_{j=1}^{L_s} \sum_x ( f_{j} \c f_{j} ) (x) ( f_{j} \c f_{j} ) (x) \,.
$$
By the pigeonhole principle there is $j\in [L_s]$ such that
\begin{equation}\label{tmp:08.02.2013_6}
    \frac{a^{2s} \T_s}{2M L_s^4} \le \frac{\T_{2s}}{2L_s^{4}}
        \le \sum_x ( f_{j} \c f_{j} ) (x) ( f_{j} \c f_{j} ) (x) \,.
\end{equation}
Put $P=P_j$, $f=f_j$ and $\D =  2^{j} \T_s / (2^{4} M a^s)$.
Thus
\begin{equation}\label{tmp:08.02.2013_3}
    \frac{a^{2s} \T_s}{2M L_s^4 \D^4} \le \E (P) \,.
\end{equation}
%and by the Cauchy--Schwartz inequality
%\begin{equation}\label{tmp:08.02.2013_2}
%    \frac{a^4 \E^2}{2^2 M^2 L^6 \D^6 |P|} \le \E(P) \,.
%\end{equation}
Clearly, $|P| \le 4\T_s \D^{-2}$.
Using the last inequality, the definition of the number $\D$
and bound (\ref{tmp:08.02.2013_3}), we obtain
$$
    \E (P) \ge \frac{a^{2s} \T_s}{2M L_s^4 \D^4} \ge |P|^3 \frac{a^{2s} \D^2 }{2^7 M L_s^4 \T_s^2}
        \ge
            |P|^3 \frac{2^{2j}}{2^{15} M^3 L_s^4}
                \ge
                    \frac{|P|^3}{2^{15} M^3 L_s^4} := \mu |P|^3 \,.
$$
To estimate the size of $P$ we note by (\ref{tmp:08.02.2013_3}) that
%Note, finally, that bound (\ref{tmp:08.02.2013_3}) implies
\begin{equation}\label{tmp:08.02.2013_7}
    |P|^3 \ge \frac{a^{2s} \T_s}{2M L_s^4 \D^4} \,.
\end{equation}
%To estimate the size of $P$ we note by (\ref{tmp:08.02.2013_6}) that
%\begin{equation}\label{tmp:08.02.2013_7}
%    |P| \D \ge \frac{\E}{2M L^3 a} \,.
%\end{equation}
After that use
%the
arguments (\ref{tmp:08.02.2013_4})---(\ref{tmp:08.02.2013_5})
of the proof of Proposition \ref{p:E_3_and_E_critical}.
By Theorem \ref{BSG} there is $P'\subseteq P$ such that
$|P'| \gg \mu |P|$ and $|P'-P'| \ll \mu^{-4} |P'|$.
Applying Pl\"{u}nnecke--Ruzsa inequality (\ref{f:Plunnecke}), we obtain
\begin{equation}\label{tmp:08.02.2013_8}
    |nP'-mP'| \ll \mu^{-4(n+m)} |P'|
\end{equation}
for every $n,m\in \N$.
We have
$$
    \D |P'| \le \sum_x (A *_{s-1} A) (x) P'(x) = \sum_{x_1,\dots,x_{s-1}\in A} (A \c P') (x_1+\dots+x_{s-1}) \,.
$$
By (\ref{tmp:08.02.2013_7})
and the definition of the number $\D$
there is $x\in (s-1)A$ such that the set $A' := A\cap (P'-x)$ has the size at
least
$$
    |A'|
        %\ge |P'| |P|^{-1} \frac{\E}{4M L^3 a^2}
        \gg
            |P'| \D a^{-(s-1)}
                \ge
            \mu |P| \D a^{-(s-1)}
                \gg
                    \mu \left( \frac{\T_s}{M L_s^4 \D a^{s-3}} \right)^{1/3}
                        \gg
                            \frac{2^{5j/3} a}{M^3 L_s^{16/3}}
                                \gg
                                    \frac{a}{M^3 L_s^{16/3}} \,.
$$
We have by (\ref{tmp:08.02.2013_8}) that
%by Pl\"{u}nnecke--Ruzsa inequality that
$$
    |nA'-mA'| \le |nP' - mP'| \ll \mu^{-4(n+m)} |P'|
        \ll 
            \mu^{-4(n+m)} |A'| a^{s-1} \D^{-1} 
        \ll
$$
$$
        \ll 
            \mu^{-4(n+m)} M |A'| \cdot \frac{a^{2s-1}}{\T_s}
$$
for every $n,m\in \N$.


Conversely, applying bound (\ref{f:T_3_and_E_critical_2}) with $n=m=s$, combining with the Cauchy--Schwarz inequality,
we obtain
$$
    |A'|^{4s} \le \left( \sum_x ( (A' *_{s-1} A') \c (A' *_{s-1} A')  ) (x) \right)^2
        \le
            \T_{2s} (A') |sA'-sA'|
                \ll_s
$$
$$
                \ll_s
                    (M^{3} \log^{4} |A|)^{8s} M \cdot |A'| \frac{|A|^{2s-1}}{\T_s (A)} \T_{2s} (A) \,.
$$
Using (\ref{f:T_3_and_E_critical_1}), we get
$$
    \T_{2s} (A) \gg_s \T_s (A) |A'|^{4s-1} |A|^{-(2s-1)} (M^{3} \log^{4} |A|)^{-8s} M^{-1}
        \gg_s
$$
$$
        \gg_s
            \T_s (A) |A|^{2s} M^{-36s+2} \log^{-54s} |A| \,.
$$
In other words, $\T_{2s} (A) \gg_{M,\,\log |A|,\,s} |A|^{2s} \T_s (A)$.
This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


So, we have proved in Theorem \ref{t:T_3_and_E_critical} that, roughly speaking, $A'-A'$
is a set with small (in terms of the parameter $M$) doubling and vice versa.
Thus, $A'$ does not equal to a set with small doubling but $A'-A'$ does.
Results of such a sort were obtained in \cite{SS1}, \cite{s_ineq} and \cite{s_mixed}.
Note,
%finally,
that we need in multiple $|A|^3 \E^{-1} (A)$ in (\ref{f:T_3_and_E_critical_2}),
because by (\ref{f:T_3_and_E_critical_1}) and the Cauchy--Schwarz inequality,
we have the same lower bound for $|A'-A'|$.


It is easy to see that an analog of Theorem \ref{t:T_3_and_E_critical}
takes place if one replace (\ref{cond:T_3_and_E_critical})
onto condition
%$\T_{2s} (A) \ge |A|^{2s} \T_s (A) / M$, $s\ge 2$,
%further,
$\T_s (A) \ge |A|^{2(s-2)} \E(A) / M$,
where $s$ is an even number, $s\ge 4$,
and, further,  even  more general relations between $\T_k$ energies
can be reduced to the last case (and Theorem \ref{t:T_3_and_E_critical})
 via a trivial estimate
$ \T_s (A) \le |A|^2 \T_{s-1} (A)$.
%obtained
We do not need in such generalizations in the paper.






%\section{Energies of sumsets: preliminaries}
\section{Sumsets: preliminaries}
\label{sec:sumsets}


Let $A \subseteq \Gr$ be a set.
Before studying the energies of sumsets or difference sets we concentrate on a following related question,
which was asked to the author by Tomasz Schoen.
% asked the author about
Namely, what can be proved nontrivial concerning lower bounds for $|A\pm A_s|$, $s\neq 0$?
The connection with $A\pm A$ is obvious in view of Katz--Koester trick (\ref{f:KK_trick}).
We start  with a result in the direction.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set.
    %%$(2,\beta,\gamma)$--connected set with $\beta \le 1/2$.
    %Put $D=A-A$, $S=A+A$.
    %Take two sets $D,S$ such that $A-A \subseteq D$, $A+A \subseteq S$.
%    Moreover, without any assumptions on $A$,
    If $\E_3 (A) \ge 2|A|^3$ then
\begin{equation}\label{f:E_3(D)_0}
        \left( \max_{s\neq 0} |A\pm A_s| \right)^3 \gg \frac{|A|^{10}}{|A-A| \E^2 (A)} \,.
\end{equation}
    Now suppose that $A$ is $(3,\beta,\gamma)$--connected with $\beta \le 1/2$,
    and $\E_3 (A) \ge 2^4 \gamma^{-1} |A|^3$.
    Then
\begin{equation}\label{f:E_3(D)_0'}
    \left( \max_{s\neq 0} |A\pm A_s| \right)^2 \gg \gamma \frac{|A|^{5}}{\E (A)} \,.
\end{equation}
\label{t:E_3(D)-}
\end{theorem}
\begin{proof}
    Write $\E = \E(A)$, $\E_3 = \E_3 (A)$, and $a=|A|$.
    Let us begin with (\ref{f:E_3(D)_0}).
    Denote by $\o$ the maximum in (\ref{f:E_3(D)_0}).
    By the Cauchy--Schwarz inequality  and formula (\ref{f:eigen_A}) of Lemma \ref{l:eigen_A'}, we obtain
\begin{equation}\label{f:E_3(D)_first}
    a^2 |A_s|^2 \le \E(A,A_s) |A \pm A_s| \le \E^{1/2}_3 |A_s| |A \pm A_s| \,.
\end{equation}
    Multiplying the last inequality by $|A_s|$, summing over $s\neq 0$
    and using the assumption
    %$K\le 2^{-1} |A|^{1/2}$,
    $\E_3 (A) \ge 2|A|^3$,
    we get
\begin{equation}\label{tmp:22.01.2014_1}
    a^2 \E^{1/2}_3 \ll \o \E \,.
\end{equation}
    On the other hand, by Lemma \ref{corpop}, we have
\begin{equation}\label{tmp:22.01.2014_2}
    a^6 \ll \E_3 \sum_{s\in A-A,\, s\neq 0} |A \pm A_s| \le \E_3 |A-A| \o \,.
\end{equation}
    Combining (\ref{tmp:22.01.2014_1}) with (\ref{tmp:22.01.2014_2}), we obtain
$$
    \o^3 \gg \frac{a^{10}}{|A-A| \E^2}
$$
    as required.


    Now let us obtain (\ref{f:E_3(D)_0'}).
    Using Lemma \ref{l:eigen_A'}, we find $A'$, $|A'| \ge |A|/2$ such that estimate (\ref{f:eigen_A'}) takes place.
    As in (\ref{f:E_3(D)_first}), we get
\begin{equation}\label{f:E_3(D)_first'-}
    a^2 |A'_s|^2 \le \E(A,A'_s) |A \pm A'_s| \le \frac{2\E}{a} |A'_s| |A \pm A_s| \,.
\end{equation}
    By assumption $\E_3 (A) \ge 2^4 \gamma^{-1} |A|^3$.
    Using the connectedness, we obtain
\begin{equation}\label{tmp:09.04.2014_I''}
    \E_3 (A') \ge \gamma \frac{|A'|^6}{|A|^6} \E_3 (A) \ge 2 |A'|^3 \,.
\end{equation}
    Multiplying
    %the last inequality
    inequality (\ref{f:E_3(D)_first'-})
    by $|A'_s|$, summing over $s \neq 0$, we have in view of (\ref{tmp:09.04.2014_I''}) that
\begin{equation}\label{tmp:22.01.2014_3}
    a^2 \E_3 (A') \ll a^{-1} \E \E^* (A') \o \,.
    %\,.
\end{equation}
    Combining the last formula with first inequality from (\ref{tmp:09.04.2014_I''}), we get
\begin{equation}\label{tmp:09.04.2014_I'}
    a^2 \gamma \E_3 (A) \ll a^{-1} \E \E^* (A') \o \,.
\end{equation}
    On the other hand, summing the first estimate from  (\ref{f:E_3(D)_first'-}) over $s\neq 0$
    and applying Lemma \ref{l:E_3_A_s}, we see that
    % it follows that
\begin{equation}\label{tmp:09.04.2014_I'''}
    a^2 \E^* (A') \le \o \E_3 \,.
\end{equation}
    Combining (\ref{tmp:09.04.2014_I'}), (\ref{tmp:09.04.2014_I'''}), we obtain
$$
    \o^2 \gg \gamma \frac{a^5}{\E} \,.
$$
    This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip

From (\ref{f:E_3(D)_0}) it follows that if $|A-A| = K|A|$, $\E(A) \ll |A|^3 /K$, $\E_3 (A) \ge 2|A|^3$
then there is
$s\neq 0$ such that $|A-A_s| \gg K^{1/3} |A|$ as well as there exists
$s\neq 0$ with $|A+A_s| \gg K^{1/3} |A|$.
It improves a trivial lower bound $|A\pm A_s| \ge |A|$.
Using bound
%(\ref{f:E_3(D)_1}) or
(\ref{f:E_3(D)_0'})
%, say,
one can show that there is $s\neq 0$ such that $|A-A_s| \gg K^{1/2} |A|$
(here $\E(A) \ll |A|^3 /K$),
provided by some connectedness assumptions take place.
%(the similar follows from (\ref{tmp:22.01.2014_3}), (\ref{tmp:22.01.2014_4}) and $(2,\beta,\gamma)$--connectedness).


We need in lower bounds on $\E_k (A)$ in Theorem \ref{t:E_3(D)-} and Proposition \ref{p:f:E_4(D,A)} below
to be separated from a very natural simple example, which can be called a "random sumset"\, case.
Namely, take a random $A\subseteq \Gr$ and consider $A\pm A$.
This "random sumset"\, has almost no structure (provided by $A\pm A$ is not a whole group, of course)
and we cannot say something useful in the situation.
It does not contradict Theorem \ref{t:E_3(D)-} and Proposition \ref{p:f:E_4(D,A)} (see also the results of the next section)
because the energies $\E_k (A)$ are really small in the case.

\bigskip

%Let us give an application of Theorem \ref{t:E_3(D)}.
%Suppose that $\G$ is a multiplicative subgroup of the set of invertible elements of
%a finite field $\F_p^* = \F_p \setminus \{ 0\}$ with prime $p$.
%\bigskip


Now we give another proof  of estimate (\ref{f:E_3(D)_0'}) which can be derived from
inequality (\ref{f:E_4(D,A)}), case $k=2$ below.
Actually, it gives us even stronger inequality, namely, $|A^2 - \D (A_s)| \gg |A|^5 \E^{-1} (A)$
for some $s\neq 0$, or, more generally, (see formulas (\ref{tmp:06.05.2014_1}), (\ref{tmp:06.05.2014_2}) below)
\begin{equation}\label{f:max_s,k}
    \max_{s\neq 0} |A^k \pm \D (A_s)|
        \gg
            \max_{r\ge 1} \left\{ \frac{|A|^{2k+1} \E_{r+1} (A)}{\E_r (A) \E_{k+1} (A)} \right\}
    \,,
\end{equation}
provided by some connectedness assumptions take place.
In particular, taking $r=k$ and $r=1$ in the previous formula, we get
$$
    \max_{s\neq 0} |A^k \pm \D (A_s)|
        \gg
            \max \left\{ \frac{|A|^{2k+1}}{\E_{k} (A)} \,, \frac{|A|^{2k-1} \E(A)}{\E_{k+1} (A)} \right\} \,.
    %\,.
$$
% take place.

\begin{proposition}
    Let $A\subseteq \Gr$ be a set, $k\ge 2$ be a positive integer.
    %%$(2,\beta,\gamma)$--connected set with $\beta \le 1/2$.
    %Put $D=A-A$, $S=A+A$.
    Take two sets $D,S$ such that $A-A \subseteq D$, $A+A \subseteq S$.
    Then
\begin{equation}\label{f:E_4(D,A)}
    \gamma |A|^{2k+1} \ll_k \sum_{x\neq 0} (A\c A)^k (x) d_k (x) \le \sum_{x\neq 0} (A\c A)^k (x) (D\c D)^k (x) \,,
\end{equation}
\begin{equation}\label{f:E_4(D,A)'}
    \gamma |A|^{2k+1} \ll_k \sum_{x\neq 0} (A\c A)^k (x) s_k (x) \le \sum_{x\neq 0} (A\c A)^k (x) (S\c S)^k (x) \,,
\end{equation}
    provided by $A$ is $(k+1,\beta,\gamma)$--connected with $\beta \le 1/2$
    and
    $\E_{k+1} (A) \ge 2^{k+2} \gamma^{-1} |A|^{k+1}$.
    Here $d_k (x) = \sum_\a D_x (\a) (D \c D_x)^{k-1} (\a)$, $s_k (x) = \sum_\a S_x(\a) (S_x * D)^{k-1} (\a)$.
\label{p:f:E_4(D,A)}
\end{proposition}
\begin{proof}
    Using Lemma \ref{l:eigen_A'}, we find $A'$, $|A'| \ge |A|/2$ such that estimate (\ref{f:eigen_A''}) takes place
    with $g(z) = (A \c A)^k (z)$.
    It follows that
\begin{equation}\label{tmp:06.05.2014_1}
    |A|^{2k} |A'_x|^2 \le |A^k \pm \D(A'_x)| \E_{k+1} (A'_x, A)
        \le |A^k \pm \D(A_x)| \frac{2\E_{k+1} (A)}{|A|} |A_x| \,.
\end{equation}
%    where $\| x \| = k$.
    Multiplying the last inequality by $|A'_x|^{k-1}$ and
    summing
    %the last inequality
    over $x\neq 0$ (to obtain (\ref{f:max_s,k}) multiply by $|A'_x|^{r-1}$), we get
\begin{equation}\label{tmp:06.05.2014_2}
    \gamma |A|^{2k+1} \E_{k+1} (A) \ll |A|^{2k+1} \E^*_{k+1} (A')
        \ll \E_{k+1} (A) \sum_{x \neq 0} |A^k \pm \D(A_x)| |A_x|^k \,.
\end{equation}
    Here we have used  the fact
$$
    \E_{k+1} (A') \ge \gamma \frac{|A'|^{2(k+1)}}{|A|^{2(k+1)}} \E_{k+1} (A) \ge 2 |A'|^{k+1}
$$
    and the assumption $\E_{k+1} (A) \ge 2^{k+2} \gamma^{-1} |A|^{k+1}$.
    %we get
    %Because of
    Note that by Katz--Koester trick (\ref{f:KK_trick_new}), one has $|A^k - \D(A_x)| \le d_k (x)$, $|A^k + \D(A_x)| \le s_k (x)$
    (or just see the proof of Proposition \ref{pr:E_k(D)_simple} below).
    Finally,  $d_k (x) \le |D_x|^k$, $s_k (x) \le |S_x|^k$
    and  we obtain (\ref{f:E_4(D,A)}).
    This completes the proof.
%$\hfill\Box$
\end{proof}



\bigskip

Using Katz--Koester trick or just the Cauchy--Schwarz inequality
one can show that (\ref{f:E_4(D,A)}) trivially takes place in the case $k=1$
without any conditions on connectedness of $A$ or lower bounds for any sort of energy.
Under the assumptions of Proposition \ref{p:f:E_4(D,A)} from (\ref{f:E_4(D,A)}) it follows that
$$
    \gamma^2 |A|^{4k+2} \ll_k \E^*_{2k} (A) \E^*_{2k} (D)
$$
and similarly for $S$.
Some weaker results
but without any conditions on $A$
were obtained in \cite{SS1}.


%\bigskip



\bigskip

Results above give an interesting
%dichotomic
corollary on non--random sumsets/difference sets.
Put $D=A-A$, $S=A+A$.
Then for an arbitrary positive integer $k$ and any elements $a_1,\dots,a_k\in A$, we have
\begin{equation}\label{f:A,D_inclusion}
    A \subseteq (D + a_1) \bigcap (D + a_2) \dots \bigcap (D+a_k) \,, \quad
    A \subseteq (S - a_1) \bigcap (S - a_2) \dots \bigcap (S-a_k)
    \,.
\end{equation}
In particular, there is $x\in D$, $x\neq 0$ such that $|D_x|, |S_x| \ge |A|$
(also it follows  from Katz--Koester inclusion (\ref{f:KK_trick_new})).
By Corollary \ref{c:A^2-D(A)} (see also Lemma \ref{l:A^2_pm}) one can improve it to
%that
$|D_x|, |S_x| \ge K^\eps |A|$,
% is possible,
where $K=|A|^3 \E^{-1} (A)$.
%We have  obtained
%an improvement.
Theorem \ref{t:E_3(D)-} gives us
even stronger result.


\begin{corollary}
Let $A \subseteq \Gr$ be a set, $D=A-A$, $S=A+A$, $\E (A) = |A|^3/K$,
%$|D|=K|A|$, $|S|=L|A|$.
Suppose that $A$ is $(3,\beta,\gamma)$--connected, $\beta \le 1/2$,
and $\E_3 (A) \ge \gamma^{-1} 2^{4} |A|^3$.
Then there is $x\neq 0$ such that $|D_x|,|S_x| \gg \gamma^{1/2} K^{1/2} |A|$.
\label{c:Dx}
\end{corollary}


One can
%obtain
get
an analog of Corollary \ref{c:Dx} for multiple intersections (\ref{f:A,D_inclusion})
but another types of energies will require in the case.
Nevertheless, some weaker inequality of the form $|D_{\v{x}}| \ge K^{\eps} |A|$ can be obtained,
using Proposition \ref{p:E_3_and_E_critical} and
%Remark \ref{r:E_k_E_gen}.
Theorem \ref{t:E_3_and_E_critical}.
Here $K=|A|^{k+1} \E^{-1}_k (A)$, $k\ge 2$.
Interestingly, we do not even need in any connectedness in this weaker result.


\begin{proposition}
    Let $A\subseteq \Gr$ be a set,  $k\ge 2$ be a positive integer, $c\in (0,1]$ be a real number,
    $\E_{k} (A) \ge k^2 \E_{k-1} (A)$,
    and $K=|A|^{k+1} \E^{-1}_k (A) \le |A|^{1-c}$, $K_1 = |A|^3 \E^{-1} (A)$.
    Then for all sufficiently small $\eps = \eps (c) >0$
    there is $\v{x} = (x_1,\dots,x_{k-1})$ with distinct $x_j$, $j\in [k]$ such that
\begin{equation}\label{}
    |D_{\v{x}}|\,, |S_{\v{x}}| \gg_k |A| \cdot \min\{ K^{\eps}, c_{K^{\eps}} K_1 \} \,,
\end{equation}
    where the constant $c_{K^{\eps}}$ satisfies $c_{K^{\eps}} \gg_{K^{\eps}} 1$
    and, again, the degree of the polynomial dependence is a function on $c$.
\label{p:DxSx}
\end{proposition}
\begin{proof}
Suppose not.
Take any set $\mathcal{P} \subseteq A^{k-1} - \D(A)$.
Applying Lemma \ref{l:E_3_A_s}, one has
$$
    |A|^2 \left( \sum_{\v{x} \in \mathcal{P}} |A_{\v{x}}| \right)^2
        =
    \left( \sum_{\v{x} \in \mathcal{P}} \sum_z (A\c A_{\v{x}}) (z) \right)^2
        \le
            \E_{k+1} (A) \sum_{\v{x} \in \mathcal{P}} |A \pm A_{\v{x}}|
                \le
$$
\begin{equation}\label{tmp:09.04.2014_1}
                \le
                    \E_{k+1} (A) |\mathcal{P}| \cdot \max_{\v{x} \in \mathcal{P}} |A \pm A_{\v{x}}| \,.
                    %K^{\eps} |A| \,.
\end{equation}
Note that the assumption $\E_{k} (A) \ge k^2 \E_{k-1} (A)$ implies
\begin{equation}\label{tmp:09.04.2014_2}
    \E_k (A) = \sum_{\v{x}} |A_{\v{x}}|^2 \le \sum'_{\v{x}} |A_{\v{x}}|^2 + \binom{k-1}{2} \E_{k-1} (A)
        \le
            2 \sum'_{\v{x}} |A_{\v{x}}|^2 \,,
\end{equation}
where the sum $\sum'$ above is taken over $\v{x} = (x_1,\dots,x_{k-1})$ with distinct $x_j$.
Now take $\mathcal{P}$ such that  $\D < |A_{\v{x}}| \le 2 \D$ for $\v{x} = (x_1,\dots,x_{k-1}) \in \mathcal{P}$,
where all $x_j$, $j\in [k]$ are distinct
and
$$
    \sum_{\v{x} \in \mathcal{P}} |A_{\v{x}}|^2 \gg \frac{\E_{k} (A)}{\log K} \,.
$$
Of course, such $\mathcal{P}$ exists by the pigeonhole principle and bound (\ref{tmp:09.04.2014_2}).
Using the last inequality, and recalling (\ref{tmp:09.04.2014_1}), we obtain
$$
    \frac{|A|^2 \E_{k} (A)}{\log K}
        \ll
            |A|^2 |\mathcal{P}| \D^2
        \ll
            \E_{k+1} (A) K^{\eps} |A| \,.
$$
In other words, $\E_{k+1} (A) \gg_{K^\eps} |A| \E_k (A)$.
Put $M=\max\{ K^\eps, k\}$.
%\max\{ \log |A|, K^\eps \}$.
Applying Proposition \ref{p:E_3_and_E_critical} as well as
% Remark \ref{r:E_k_E_gen},
Theorem \ref{t:E_3_and_E_critical},
we see that
$A \approx_{M} \L \dotplus H$.
%%Without loosing of generality, one can suppose that $H\subseteq A$.
%In the former case
After that put $A' = A\cap (H+\L)$.
Then $|A'| \gg_{M} |A|$ and $|H| \gg_M |A| /K$.
Further, as in the proof of Theorem \ref{t:H+L_description}, we get
$$
    %\E (A',H)
    \sum_{\v{x},\,\|x\|=k-1} \Cf_{k} (A') (\v{x}) \Cf_{k} (H) (\v{x})
        =
            \sum_s (A' \c H)^k (s)
    \ge \frac{|H|^{k} |A'|^{k}}{|A'-H|^{k-1}} \gg_{M} \frac{|H|^k |A|^k}{|\L+H-H|^{k-1}}
%    \gg_M
$$
$$
    \gg_{M} \frac{|H|^k |A|^k}{(|\L||H-H|)^{k-1}}
        \gg_{M}
            |A| |H|^k
            %\,,
$$
and hence
\begin{equation}\label{tmp:09.04.2014_3}
    \sum_{\v{x},\,\|x\|=k-1} |A'_{\v{x}}| |H_{\v{x}}| \gg_{M} |A| |H|^k \,.
\end{equation}
In particular, there are at least $\gg_{M} |H|^{k-1}$ elements
$\v{x} \in H^{k-1}- \D(H)$
% \subseteq A^{k-1}- \D(A)$
such that
$|A'_{\v{x}}| \gg_{M} |A|$.
We can suppose that the summation in (\ref{tmp:09.04.2014_3}) is taken over
$\v{x} = (x_1,\dots,x_{k-1})$ with distinct $x_j$ because the rest is bounded by
$$
    \binom{k-1}{2} \sum_{x} (A\c H)^{k-1} (x) \le \binom{k-1}{2} |H|^{k-1} |A| \ll_{M} |A| |H|^k \,.
$$
The last estimate follows from the assumption $K \le |A|^{1-c}$.
Choosing any such $\v{x}$ and using the Cauchy--Schwarz inequality, we obtain
$$
    |A \pm A_{\v{x}}| \ge |A' \pm A'_{\v{x}}| \ge \frac{|A'|^2 |A'_{\v{x}}|^2}{\E(A',A'_{\v{x}})}
        \gg_{M}
            \frac{|A|^4}{\E(A)} = K_1 |A|
$$
and in view of Katz--Koester trick (\ref{f:KK_trick_new}),
we see that $|D_{\v{x}}|\,, |S_{\v{x}}|$ are huge for large $K_1$.
This concludes the proof.
%$\hfill\Box$
\end{proof}




\bigskip

%Let us derive a simple consequence of
Using Proposition \ref{p:DxSx} one can derive an interesting dichotomy
result.



\begin{theorem}
    Let $A\subseteq \Gr$ be a set, $D=A-A$, $S=A+A$, $k\ge 2$, and
    %$M\ge 1$ be a real number.
    $M\ge 1$, $\eps \in (0,1)$ be real numbers.
    Put $K=|A|^{k+1} \E^{-1}_k (A)$.
    %, $M=K^\eps$.
    Suppose that for any vector $\v{x} = (x_1,\dots,x_{k-1})$ with distinct $x_j$, $j\in [k]$
    the following holds
    \begin{equation*}\label{}
        |D_{\v{x}}| \le M|A| \quad \mbox{or, similarly, } \quad  |S_{\v{x}}| \le M|A| \,.
    \end{equation*}
    Then either $\E_k (A) \ll_{M,\, |A|^\eps,\, k} |A|^k$ or $\E(A) \gg_{M,\, |A|^\eps,\, k} |A|^3$.
    Again, the degree of the polynomial dependence is a function on $\eps$.
\label{t:dichotomy_DS}
\end{theorem}
\begin{proof}
    Put $K_1 = |A|^3 \E^{-1} (A)$.
    Suppose, in contrary, that $\E_k (A) \gg_{M,\, |A|^\eps,\, k} |A|^k$ and\\ $\E(A) \ll_{M,\, |A|^\eps,\, k} |A|^3$.
    Then $K \ll_{M,\, |A|^\eps,\, k} |A|^{}$ and $K_1 \gg_{M,\, |A|^\eps,\, k} 1$.
%    Suppose that $K=|A|^{k+1} \E^{-1}_k (A) \gg_{M,\, k} |A|^{}$.
%    Then $\E_k (A) \ll_{M,\, k} |A|^k$.
    Trivially, $\E_k (A) \le |A|^{k-2} \E(A)$
    and hence
    $$|A| \gg_{M,\, |A|^\eps,\, k} K \ge K_1 \gg_{M,\, |A|^\eps,\, k} 1 \,.$$
    Finally, by the  upper bound for the parameter $K$
    we see that
    the number $c$
    which is defined as $K = |A|^{1-c}$ can by taken depending  on $\eps$ only.
    Thus,
    everything follows from Proposition \ref{p:DxSx}, the only thing we need to consider is the situation when
    $\E_k (A) \le k^2 \E_{k-1} (A)$.
    But in the case
$$
    |A|^k \le \E_k (A) \le k^2 \E_{k-1} (A) \le k^2 |A|^{k-3} \E(A) \,.
$$
    In other words $\E(A) \gg_k |A|^3$ and this concludes the proof.
%$\hfill\Box$
\end{proof}


%A simple consequence of Corollary
%Suppose that $|D_{\v{x}}|\,, |S_{\v{x}}|$, $\| v \| = k-1$ are not much larger than $|A|$.
%Then using \ref{p:DxSx} is the following, we


\bigskip

Thus, if $|D_{\v{x}}|\,, |S_{\v{x}}|$ are not much larger than $|A|$
then either $A$ is close to what we called a "random sumset"\, or, on the contrary, is very structured.
Clearly, the both situations are realized.

%\input{sumsets2}

\section{Energies of sumsets}
\label{sec:sumsets1}



Let $A \subseteq \Gr$ be a set.
Throughout the section we put $D=A-A$ and $S=A+A$.
As was explained in the introduction
% that
 one can hope to prove a good lower bound for $\E_3 (D)$, $\E_3 (S)$.
It will be done in Theorem \ref{t:E_3(D)} below but before this we formulate a simple preliminary
lower bound for $\E^D_k (D)$, $\E^D_k (S)$.
Similar
%almost trivial
lower bounds for $\E^D_2 (D)$, $\E^D_2 (S)$
were given
%in formula (\ref{f:A^k-D(A)_for_E_k}) of Lemma \ref{l:A^k-D(A)_for_E_k} and also
in Corollary 5.6 of paper \cite{M_R-N_S}.
Further, it was proved in \cite{SS1} (see Remark 8) that $\sigma_{k+1} (D) \ge |A^k - \D(A)|$.
Now we obtain a similar lower bound for $\E^{D}_k (D)$.
Recall that by $\E^{D}_1 (D)$ we mean $\sigma_{3} (D) = \sigma_D (D)$, that is $\sum_{x\in D} (D \c D) (x)$.


\begin{proposition}
    Let $A\subseteq \Gr$ be a set.
    Put $D=A-A$, $S=A+A$.
%    Take two sets $D,S$ such that $A-A \subseteq D$, $A+A \subseteq S$.
    Then for all $k\ge 1$ one has
    \begin{equation}\label{f:E_k(D)_simple}
        \E^D_k (D) \ge |A^{k+1} - \D(A)|
                %\ge |A|^{k-1} |A^2 - \D(A)|
            \ge |A-A| |A|^k \,,
    \end{equation}
    and, similarly,
    $$
        \sum_{x} S(x) (S * D)^{k} (x) \ge |A^{k+1} + \D(A)| \ge |A|^k \max\{ |A-A|, |A+A| \} \,,
    $$
    \begin{equation}\label{f:E_k(D)_simple'}
        \E^D_k (S) \ge |A|^{k-1} |A^2 + \D(A)| \ge |A|^k \max\{ |A-A|, |A+A| \} \,.
    \end{equation}
\label{pr:E_k(D)_simple}
\end{proposition}
\begin{proof}
    The second estimates in (\ref{f:E_k(D)_simple}), (\ref{f:E_k(D)_simple'})
     follow from Lemma \ref{l:A^2_pm}.
    Further, it is easy to get (or see e.g. \cite{SS1}) that
$$
    |A^{k+1} - \D(A)| \le \sum_{x_1,\dots,x_{k+1}} D(x_1) \dots D(x_{k+1}) \prod_{i\neq j} D(x_i - x_j)
        \le
$$
$$
        \le
            \sum_{x_1,\dots,x_{k+1}} D(x_1) \dots D(x_{k+1}) D(x_1-x_2) \dots D(x_1-x_{k+1})
                =
                    \sum_{x} D(x) (D\c D)^{k} (x)
                        =
                            \E^D_{k} (D)
                            %\,.
$$
as required.
Similarly,
$$
%    |A|^{k-1} |A^2 + \D(A)|
%        \le
            |A^{k+1} + \D(A)| \le \sum_{x_1,\dots,x_{k+1}} S(x_1) \dots S(x_{k+1}) \prod_{i\neq j} D(x_i - x_j)
        \le
$$
$$
        \le
            \sum_{x_1,\dots,x_{k+1}} S(x_1) \dots S(x_{k+1}) D(x_1-x_2) \dots D(x_1-x_{k+1})
                =
                    \sum_{x} S(x) (S* D)^{k} (x)
                            \,.
$$
Finally, by Lemma \ref{l:A^2_pm}, we get
$$
    \E^D_k (S) = \sum_{s\in D} (S\c S)^k (s) \ge \sum_{s\in D} |A+A_s|^k
        \ge |A|^{k-1} \sum_{s\in D} |A+A_s|
            =
$$
$$
            =
                |A|^{k-1} |A^2 + \D(A)| \ge |A|^k \max\{ |A-A|, |A+A| \} \,.
$$
This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


Interestingly, that some sort of sumset, namely, $A^n \pm \D(A)$
gives a lower bound for an energy, although, usually, an energy
provides lower bounds for cardinality of sumsets via the Cauchy--Schwarz inequality.
The trick allows to obtain a series of results in \cite{SS1}---\cite{SS3}.
%
%
Although bounds (\ref{f:E_k(D)_simple}), (\ref{f:E_k(D)_simple'}) are very simple they can be tight in some cases.
For example, take $A$ to be a dissociated set
%($k=1,2$)
or, in contrary, a very structural set as a subspace.
% (any $k$).


\bigskip


Now we formulate the main result of the section concerning lower bounds for some energies of sumsets/difference sets.
Again we need in lower bounds on $\E_k (A)$ in Theorem \ref{t:E_3(D)} below
to be separated from the "random sumset"\, case.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set.
    %%$(2,\beta,\gamma)$--connected set with $\beta \le 1/2$.
    %Put $D=A-A$, $S=A+A$.
    Take two sets $D,S$ such that $D=A-A$, $S=A+A$.
    Then
\begin{equation}\label{f:E_3(D)_1}
    \E^2_3 (D,A,A),\,~ \E^2_3 (S,A,A)
        \ge \frac{|A|^{13}}{|A-A|^2 \E (A)} \,,
        %\ge |A|^5 \E (A) \,,
\end{equation}
    and
\begin{equation}\label{f:E_3(D)_2}
    (\E^D_3 (D))^4 \ge \max \left\{ |D|^{12}, \frac{|A|^{45}}{\E^9 (A) |D|^2} \right\} \,,
        \quad \quad
    (\E^D_3 (S))^4 \ge \max \left\{ |S|^{12}, \frac{|A|^{45}}{\E^9 (A) |D|^2} \right\} \,.
\end{equation}
    Further, let $\beta,\gamma \in [0,1]$ be real numbers,  $\beta \le 1/2$.
    If $A$ is $(2,\beta,\gamma)$--connected then
\begin{equation}\label{f:E_3(D)_2.5}
    \E^2_3 (D,A,A),\,~ \E^2_3 (S,A,A)
        \gg
            \gamma |A|^5 \E (A) \,.
\end{equation}
    Suppose that $A$ is $(3/2,\beta,\gamma)$ and $(2,\beta,\gamma)$--connected, correspondingly.
    Then
\begin{equation}\label{f:E_3(D)_3}
    \E^D_3 (D),\, \E^D_3 (S)
        \gg
            \gamma \frac{|A|^{33/4} \E_{3/2} (A)}{\E^{9/4} (A) \log |A|} \,,
        \quad \quad \quad
    \E^D_3 (D),\, \E^D_3 (S)
                \gg
                    \gamma \frac{|A|^{17/2}}{\E^{3/2} (A) \log |A|} \,,
\end{equation}
correspondingly.
\label{t:E_3(D)}
\end{theorem}
\begin{proof}
    Write $\E = \E(A)$, $\E_3 = \E_3 (A)$, and $a=|A|$.
    Let us obtain bounds (\ref{f:E_3(D)_1}), (\ref{f:E_3(D)_2.5}).
    Using Lemma \ref{l:eigen_A'}, we find $A'$, $|A'| \ge |A|/2$ such that estimate (\ref{f:eigen_A'}) takes place.
    As in the proof of inequalities (\ref{f:E_3(D)_first}), (\ref{f:E_3(D)_first'-}), we get
\begin{equation}\label{f:E_3(D)_first'}
    a^2 |A'_s|^2 \le \E(A,A'_s) |A \pm A'_s| \le \frac{2\E}{a} |A'_s| |A \pm A_s| \,.
\end{equation}
    Multiplying the last inequality by $|A'_s|$, summing over $s$ and using Katz--Koester trick, we have
\begin{equation}\label{tmp:22.01.2014_3}
    a^2 \E_3 (A') \ll a^{-1} \E \cdot \E_3 (D,A)
    %\,.
\end{equation}
    and similar for $S$.
    On the other hand by the second part of Lemma \ref{corpop}, we obtain
\begin{equation}\label{tmp:22.01.2014_4}
    \left( \frac{a^4}{|A-A|} \right)^2 a^2 \ll \E^2 (A') a^2 \ll \E_3 (A') \cdot \E_3 (D,A)
    %\,.
\end{equation}
    and using the first part of the lemma, we have the same bound for the set $S$
\begin{equation}\label{tmp:22.01.2014_4+}
    \left( \frac{a^2}{|A-A|} \right)^2 a^6 \ll \sigma^2_{\t{D}} (A') (a')^2 \left( \frac{(a')^2}{2|D|} \right)^2
            \le \E_3 (A') \sum_{x \in \t{D}} |S_x| |A_x|^2
        \le \E_3 (A') \cdot \E_3 (S,A) \,,
\end{equation}
    where $\t{D} := \{ x\in D ~:~ |A'_x| \ge (a')^2 /2|D| \}$, $\sigma_{\t{D}} (A') \ge (a')^2 /2$
    (for details, see \cite{SS3}).
    Another way to prove the same is to use Lemma \ref{l:T_A,B} with $A=B=A'$, $\psi (x) = (A'\c A') (x)$.
    Combining (\ref{tmp:22.01.2014_3}) and (\ref{tmp:22.01.2014_4}), (\ref{tmp:22.01.2014_4+}), we get
$$
    \E^2_3 (D,A)\,,~ \E^2_3 (S,A)
        \gg \frac{|A|^{13}}{|A-A|^2 \E (A)}
        %\,.
        %\gg |A|^5 \E (A) \,.
$$
    Using the tensor trick (see \cite{TV} or \cite{s_mixed}), we have (\ref{f:E_3(D)_1}).
    If $A$ is $(2,\beta,\gamma)$--connected then
    $$
        \E(A') \gg \gamma \E(A)
    $$
    and combining the last inequality with (\ref{tmp:22.01.2014_3})
    and the second bound from (\ref{tmp:22.01.2014_4}),
    we get (\ref{f:E_3(D)_2.5})
    (to obtain lower bound for $\E^2_3 (S,A)$
    %we have used
    one should use
    Lemma \ref{l:T_A,B}).



    It remains to prove (\ref{f:E_3(D)_2}) and (\ref{f:E_3(D)_3}).
    Returning to (\ref{f:E_3(D)_first'})---(\ref{tmp:22.01.2014_3}), we obtain
    % it follows that
$$
    a^3 \E_3 (A') \ll \E \sum_s |A'_s|^2 |A' \pm A'_s|
$$
    or, by the H\"{o}lder inequality
\begin{equation}\label{tmp:22.01.2014_5}
    a^9 \E_3 (A') \ll \E^3 \E^D_3 (D) \,.
\end{equation}
    On the other hand, by Lemma \ref{corpop} for any $P\subseteq A'-A'$, we have
$$
    a^2 \sigma^2_P (A') \ll \E_3 (A') \sum_{s\in P} |A' \pm A'_s| \,.
$$
    Applying the H\"{o}lder inequality, we get
\begin{equation}\label{tmp:22.01.2014_5'}
    a^6 \sigma^6_P (A') \ll \E^3_3 (A') |P|^2 \E^D_3 (D)
    %\,.
\end{equation}
    and similarly for $S$.
    Now choose $P\subseteq A'-A'$ such that $P = \{ s\in A'-A' ~:~ \rho < |A'_s| \le 2 \rho \}$
    for some positive number $\rho$ and such that
$$
    \sum_{s\in P} |A'_s|^{3/2} \gg \frac{\E_{3/2} (A')}{\log a} \,.
$$
    Of course, the set $P$ exists by Dirichlet principle.
    Combining the last inequality with (\ref{tmp:22.01.2014_5'}), we obtain
\begin{equation}\label{tmp:22.01.2014_6}
     a^6 \E^4_{3/2} (A') \log^{-4} a \ll a^6 ( |P| \rho^{3/2} )^4 \ll \E^3_3 (A') \E^D_3 (D)
     \,.
\end{equation}
    Using estimates (\ref{tmp:22.01.2014_5}), (\ref{tmp:22.01.2014_6}), we have
\begin{equation}\label{tmp:22.01.2014_7}
    (\E^D_3 (D))^4 \E^9 \gg a^{33} \E^4_{3/2} (A') \log^{-4} a \,.
\end{equation}
    Thus
$$
    (\E^D_3 (D))^4 \,,~ (\E^D_3 (S))^4 \gg \frac{a^{45}}{\E^9 |D|^2 \log^4 a} \,.
$$
    Applying the tensor trick again, we get (\ref{f:E_3(D)_2}).
    To obtain (\ref{f:E_3(D)_3}) recall that $A$ is $(3/2,\beta,\gamma)$--connected set.
    Hence by (\ref{tmp:22.01.2014_7}), we obtain
$$
    (\E^D_3 (D))^4 \E^9 \gg \gamma^4 a^{33} \E^4_{3/2} (A) \log^{-4} a
$$
    and the first formula of (\ref{f:E_3(D)_3}) follows.
    Further, because $A$ is $(2,\beta,\gamma)$--connected set then
    using Lemma \ref{l:connected} for $A'$ as well as (\ref{tmp:22.01.2014_7}), we have
$$
    (\E^D_3 (D))^2 \E^3 \gg \gamma^2 a^{17} \log^{-2} a
$$
    and the last estimate coincide with the second inequality in (\ref{f:E_3(D)_3}).
    This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


From (\ref{f:E_3(D)_2}), the definition of the number $K$ as $|D|=K|A|$
and the assumption $\E(A) \ll |A|^3 /K$, we get
\begin{equation}\label{f:E_3_74}
    \E_3 (D) \gg K^{7/4} |A|^4 \,.
\end{equation}
An upper bound here is $K^{2} |A|^4$ and it follows from the main example of section \ref{sec:structural},
that is $\Gr = \f_2^n$, $A=H\dotplus \L$.
%Probably, it is
%May be it is the right order of the lower bound for $\E_3 (D)$.
Note also that the second inequality in formula (\ref{f:E_3(D)_3}) is weak but do not depends on the size of $A-A$
or on energy $\E_{3/2} (A)$.


\bigskip


%Note, finally, that
As
for dual quantities $\T_k (D)$, $\T_k (S)$,
our example $A=H\dotplus \L$ shows that
there are not nontrivial lower bounds for $\T_k (A\pm A)$, in general,
which are better than
simple consequences  of Katz--Koester
$$
    \T_k (D) \ge |A|^2 \T_{k-1} (D) \ge \dots \ge |A|^{2(k-2)} \E (D) \ge |A|^{2k-2} |D| \,,
$$
$$
    \T_k (S) \ge |A|^2 \T_{k-1} (S,\dots,S,D) \ge \dots \ge |A|^{2(k-2)} \E (D) \ge |A|^{2k-2} |D|
    %\,.
$$
%because of
(just use $(D\c D)(x) \ge |A| D(x)$ and $(S\c S) (x) \ge |A| D(x)$).
The reason is that the structure of $A\pm A$ is similar
to the structure of $A$ in the case, of course.
Nevertheless,  it was proved in \cite{SS1}, Lemma 3 that
\begin{equation}\label{f:Lev-}
    |A|^{4k} \le \E_{2k} (A) \T_k (A\pm A) \,,
\end{equation}
and also in \cite{s_mixed}, Note 6.6 that
\begin{equation}\label{f:Lev}
    \left( \frac{\sum_{x\in P} (A\c A) (x)}{|A|} \right)^{4k} \le \E_{2k} (A) \T_k (P) \,,
\end{equation}
where $P\subseteq D$ is any set.
So, if we know something on $\E_{2k} (A)$ then it gives us a new information about $\T_k (A\pm A)$.
Trivially, formula (\ref{f:Lev-}) implies that $\T_k (A\pm A) \ge |A|^{2k+2} / \E(A)$.
Again, the last inequality is sharp as our main example $A=H\dotplus \L$ shows.

\bigskip

Vsevolod F. Lev asked the author about an analog of (\ref{f:Lev}) for different sets $A$ and $B$.
Proposition \ref{p:Lev_question} below is our result in the direction.
The proof is in spirit of \cite{s_mixed}.
For simplicity we consider the case $k=2$ only.
The case of greater powers of two is considered similarly if one take $M^2,M^4,\dots$
or just see the proof of Theorem 6.3 from \cite{s_mixed} (the case of any $k$).
We do not insert the full proof because we avoid to use the operators from \cite{s_ineq}, \cite{s_mixed}
in the paper which is considered to be elementary.


\begin{proposition}
    Let $k\ge 2$ be a power of two, $A,B\subseteq \Gr$ be two sets, and $P\subseteq A-B$.
    Then
$$
    \left( \frac{\sum_{x\in P} (A \c B) (x)}{|A|^{1/2} |B|^{1/2}} \right)^{4k}
        \le
            \E_{2k} (A,\dots,A,B,\dots,B) \T_k (P) \,.
$$
\label{p:Lev_question}
\end{proposition}
\begin{proof}
    Let $k=2$.
    %Consider
    Define
    the matrix
$$
    M(x,y) = P(x-y) A(x) B(y)
$$
    and calculate its rectangular norm (e.g., see section \ref{sec:Gowers})
$$
    \la^4_1 (M) \le \sum_j \la^4_j (M)
        =
    \sum_{x,y,x',y'} M(x,y) M(x',y) M(x,y') M(x',y')
        =
$$
$$
        =
            \sum_{x,x'\in A}\, \sum_{y,y'\in B} P(x-y) P(x'-y) P(x-y') P(x'-y')
        =
$$
$$
    =
        \sum_{\a,\beta,\gamma} \Cf_4 (B,A,A,B) (\a,\beta,\gamma) P(\a) P(\beta) P(\a-\gamma) P(\beta-\gamma) \,,
$$
where $\la_j (M)$ the singular numbers of $M$.
Clearly,
$$
    \la_1 (M) \ge \frac{\sum_{x\in P} (A \c B) (x)}{|A|^{1/2} |B|^{1/2}} \,.
$$
Thus, by the Cauchy--Schwarz inequality, we get
$$
    \left( \frac{\sum_{x\in P} (A \c B) (x)}{|A|^{1/2} |B|^{1/2}} \right)^8
        \le
            \E_4 (A,A,B,B) \E (P)
$$
    as required.
    This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip






We end this section showing that there is a different way to prove our Theorem \ref{t:E_3(D)} using
slightly bigger sets $D_x$ or $S_x$ not $A\pm A_x$.
%
%
%Let us begin with a lemma,
The proof based on a lemma,
which demonstrates,
%The next lemma shows that,
in particular, that $A\pm A$ contains approximately $|A|^3 \E^{-1} (A)$
almost disjoint translates of $A$, roughly.
In the proof we use arguments from \cite{ALON}.



\begin{lemma}
    Let $A,B \subseteq \Gr$ be two sets.
    Then there are
    \begin{equation}\label{f:s_bound}
        s \ge 2^{-4} |A| |B|^2 \E^{-1} (A,B)
    \end{equation}
    disjoint sets $A_j \subseteq A+b_j$, $|A_j|\ge |A|/2$, $b_j \in B$, $j\in [s]$.

Moreover, for any set $S\subseteq A+B$ put $\sigma = \sum_{x\in S} (A * B) (x)$.
Suppose that $\sigma \ge 16 |B|$.
Then there are
\begin{equation}\label{f:s_bound+}
    s \ge 2^{-8} \sigma^3 |A|^{-2} |B|^{-1} \E^{-1} (A,B)
\end{equation}
disjoint sets $S_j \subseteq S \cap (A+b_j)$,
$|S_j|\ge 2^{-3} \sigma |B|^{-1}$, $b_j \in B$, $j\in [s]$.
\label{l:sumsets_disjoint}
\end{lemma}
\begin{proof}
Let us begin with (\ref{f:s_bound}).
Put $S=A+B$.
Our arguments is a sort of an algorithm.
At the first step of the algorithm take $A_1 = A+b$, where $b\in B$ is any element of $B$.
Suppose that we have constructed $k$ disjoint sets $A_1,\dots,A_k$.
If there is $b\in B$ such that $|(A+b) \setminus \bigsqcup_{j=1}^k A_j| \ge |A|/2$
then put $A_{k+1} = (A+b) \setminus \bigsqcup_{j=1}^k A_j$ and take $b_{k+1} = b$.
Suppose that our algorithm stops after $s$ steps.
If $s\ge |B|/2$ then we are done.
Put $U = \bigsqcup_{j=1}^s A_j$ and $B_* = B\setminus \{ b_1,\dots,b_s \}$.
Then $s|A|/2 < |U| \le s |A| $ and $|B_*| \ge |B|/2$.
We have
$$
    2^{-2} |A| |B| \le 2^{-1} |A| |B_*| \le \sum_x (A \c U) (x) B_* (x) \,.
$$
Using the Cauchy--Schwarz inequality, we obtain
$$
    2^{-4} |A|^2 |B|^2 \le \E(A,B) |U| \le \E(A,B) s |A|
    %\,.
$$
and the required lower bound for $s$ follows.

Let us prove the second part of the lemma.
Put $a=|A|$, $b=|B|$.
First of all note that
$$
    \sigma = \sum_{x\in S} (A * B) (x) = \sum_{x\in B} |S \cap (A+x)|
$$
and hence there is $x\in B$ such that $|S\cap (A+x)| \ge \sigma b^{-1}$.
Put $b_1 = x$, and let $S_1 \subseteq S\cap (A+b_1)$ be an arbitrary set of size $\lceil \eps \sigma b^{-1} \rceil$,
where $\eps = 1/8$.
After that using the arguments as above, we construct a family of disjoint sets
$S_j \subseteq S \cap (A+b_j)$, $|S_j|\ge \eps \sigma b^{-1}$, $b_j \in B$, $j\in [s]$.
If $s \ge \sigma/(4a)$ then we are done.
If not then
%considering
put $U=\bigsqcup_{j=1}^s S_j$ and $B_* = B\setminus \{ b_1,\dots,b_s \}$.
We have $B_* \subseteq B'_* \bigcup B''_*$, where
$$B'_* = \{ x\in B_* ~:~ (A \c U) (x) \ge \eps \sigma b^{-1} \} \,,$$
$$B''_* = \{ x\in B_* ~:~ |(A+x) \setminus S| \ge a-2\eps \sigma b^{-1} \} \,.$$
Further, because
$$
    (a-2\eps \sigma b^{-1}) |B''_*| \le \sum_{x\in B''_*} |(A+b) \setminus S|
        \le
            \sum_{x\in B} |(A+x) \setminus S|
                \le
                    ab - \sum_{x\in B} |(A+x) \bigcap S|
                        =
                            ab - \sigma
                            %\,.
$$
we see that
$
    |B''_*| \le (b-\sigma/a)(1+4\eps \sigma/(a b)).
$
Thus,
$$
    |B'_*| \ge b-s -(b-\sigma/a)(1+4\eps \sigma/(a b))
        \ge
            \sigma/a - 4\eps \sigma/a - s
                \ge
                    \sigma/4a \,.
$$
Finally, we obtain
$$
    \eps \sigma b^{-1} \cdot \sigma/4a \le \sum_x (A \c U) (x) B'_* (x) \le \sum_x (A \c U) (x) B (x)
$$
and hence, in view of the condition $\sigma \ge 16 b$ the following holds
$$
    (\eps \sigma b^{-1})^2 \cdot (\sigma/4a)^2
        \le
            |U| \E(A,B)
                \le
                    (2 \eps \sigma b^{-1}) s \E(A,B) \,.
$$
Whence, $s\ge 2^{-8} \sigma^3 a^{-2} b^{-1} \E^{-1} (A,B)$.
This concludes the proof.
%$\hfill\Box$
\end{proof}


\bigskip

Now let us show how to get (\ref{f:E_3(D)_2.5}), for example.
Applying Lemma \ref{l:sumsets_disjoint} with $A=A$, $B=-A$, we obtain
\begin{equation}\label{tmp:11.05.2014_1}
    \E_3 (D,A,A) \ge \sum_{j=1}^s \E_3 (A_j,A,A) \gg |A|^3 \E^{-1} (A) \E_3 (A)
\end{equation}
provided by $(3,\beta,\gamma)$--connectedness assumptions, $\beta,\gamma \gg 1$
(by the way bound (\ref{tmp:11.05.2014_1}) is tight as our main example $H\dotplus \Lambda$ shows).
%Finally,
On the other hand, we have by Lemma \ref{corpop} that
$$
    \E^2 (A) |A|^2 \le \E_3 (A) \E_3 (D,A,A) \,.
$$
Combining the last two bounds, we get (\ref{f:E_3(D)_2.5}).


\bigskip


Using similar arguments and the second part of Lemma \ref{l:sumsets_disjoint}, we obtain
the following consequence, which shows, in particular,
that the popular difference sets \cite{Gow_m}, \cite{TV}
have some structure in the sense that they have large
%sort of
energy
of some sort.

\begin{corollary}
    Let $A\subseteq \Gr$ be a set, $P\subseteq A-A$.
    Suppose that $A$ is a $(3,\beta,\gamma)$--connected set,
    $\beta \le 2^{-3} \sigma_P (A) |A|^{-1}$.
    Then
\begin{equation}\label{f:E_3(P,A,A)}
    \E_3 (P,A,A) \ge 2^{-9} \gamma^{1/2} \sigma^5_P (A) \E(A) |A|^{-9} \,.
\end{equation}
\end{corollary}
\begin{proof}
Let
%$P = A-A$ and
$\sigma = \sigma_P (A)$.
On the one hand, using Lemma \ref{l:sumsets_disjoint} with $A=A$, $B=-A$, we construct the family
of disjoint sets $P_j \subseteq P \cap (A-a_j)$, $a_j \in A$, $|P_j| \ge 2^{-3} \sigma |A|^{-1}$,
the number $s$ satisfies (\ref{f:s_bound+}).
Put $A_j = P_j + a_j \subseteq A$.
Thus, by connectedness of our set $A$, we have
$$
    \E_3 (P,A) \ge \sum_{j=1}^s \E_3 (P_j,A) = \sum_{j=1}^s \E_3 (A_j,A)
        \ge
            \gamma \left( \frac{|A_j|}{|A|} \right)^6 \E_3 (A)
                \ge
                2^{-18} \gamma \frac{\sigma^6}{|A|^{12}} \E_3 (A) \,.
$$
On the other hand, applying Lemma \ref{corpop}, we get
$$
    \sigma^4 \E^2 (A) |A|^{-6}
        \le
            \E_3 (A) \E_3 (P,A) \,.
$$
Combining the last two inequalities, we  obtain
bound (\ref{f:E_3(P,A,A)}).
%the result.
This concludes the proof.
%$\hfill\Box$
\end{proof}


\section{On Gowers norms}
\label{sec:Gowers}


    The notion of Gowers norms was introduced in papers \cite{Gow_4,Gow_m}.
    At the moment it is a very important tool of investigation in wide class
    of problems of additive combinatorics
    (see e.g. \cite{GT_great}---\cite{Gow_m}, \cite{Samorod_false}, \cite{Samorod})
    as well as in ergodic theory (see e.g. \cite{Austin}, \cite{Fu}, \cite{HK1}, \cite{HK2},
    \cite{Tao1}, \cite{Tao2}, \cite{TZ}, \cite{Z1}, \cite{Z2}).
    Recall the definitions.




  Let
  %$\Gr=(\Gr,+)$
  $G$
  be a finite
  set,
%  Abelian group with additive group operation $+$
  and $N=|G|$.
Let also $d$ be a positive integer, and
$$\{ 0,1 \}^d = \{ \omega = (\omega_1,\dots, \omega_d) ~:~ \omega_j \in \{0,1\},\, j=1,2,\dots,d \}$$
be the ordinary $d$---dimensional cube.
For $\omega \in \{0,1\}^d$ denote by $|\omega|$
the sum
$\omega_1 + \dots + \omega_d$.
Let also $\mathcal{C}$ be the operator of complex conjugation.
Let $\v{x} = (x_1,\dots,x_d), \v{x}'=(x'_1,\dots,x'_d)$ be two arbitrary vectors from $G^d$.
By $\v{x}^\o = (\v{x}^\o_1, \dots, \v{x}^\o_d)$ denote the vector
\begin{displaymath}
    \v{x}^\o_i = \left\{ \begin{array}{ll}
                        x_i & \mbox{ if } \o_i = 0, \\
                        x'_i & \mbox{ if } \o_i = 1. \\
                        \end{array} \right.
\end{displaymath}
Thus $\v{x}^\o$ depends on $\v{x}$ and $\v{x}'$.


%Let $X$ be a non--empty finite set, $Z : X \to \C$ be a function.
%Denote by $\mathbb{E} Z = \mathbb {E}_x Z$ the sum $\frac{1}{|X|} \sum_{x\in X} Z(x)$.



    Let $f : G^d \to \C$ be an arbitrary function.
    We will write $f(\v{x})$ for $f(x_1,\dots,x_d)$.


\begin{definition}
    {\it Gowers $U^d$--norm} (or $d$--uniformity norm) of the function $f$ is the following expression
\begin{equation}\label{f:G_norm_d'}
    \| f \|_{U^d}
            =
                \left( N^{-2d} \sum_{\v{x}\in G^d}\, \sum_{\v{x}' \in G^d}
                    \prod_{\o \in \{ 0,1 \}^d} \mathcal{C}^{|\o|} f (\v{x}^{\o}) \right)^{1/2^d} \,.
\end{equation}
\end{definition}


A sequence of $2^d$ points $\v{x}^\o \in G^d$, $\o \in \{ 0,1 \}^d$
is called {\it $d$--dimensional cube in $G^d$}
or just a $d$--dimensional cube.
Thus the summation in formula (\ref{f:G_norm_d'})
is taken over all cubes of $G^d$.
For example, $\{ (x,y), (x',y), (x,y'), (x',y') \}$, where $x,x',y,y'\in G$
is a two--dimensional cube in $G \m G$.
In the case Gowers norm is called rectangular norm.





For $d=1$ the expression above gives a semi--norm but for $d\ge 2$
Gowers norm is a norm.
In particular, the triangle inequality holds \cite{Gow_m}
\begin{equation}\label{e:triangle}
    \| f+g \|_{U^d} \le \| f \|_{U^d} + \| g \|_{U^d} \,.
\end{equation}
One can prove also (see \cite{Gow_m}) the following monotonicity relation.
Let $f_{x_d} (x_1, \dots, x_{d-1}) := f (x_1, \dots, x_{d})$. Then
\begin{equation}\label{e:Gowers_mon}
    N^{-1} \sum_{x_d \in G} \| f_{x_d} \|^{2^{d-1}}_{U^{d-1}} \le \| f \|^{2^{d-1}}_{U^d}
\end{equation}
for all $d\ge 2$.






If $\Gr = (\Gr,+)$ is a finite
Abelian group with additive group operation $+$, $N=|\Gr|$
then one can "project"\, the norm above onto the group
$\Gr$ and obtain the ordinary ("one--dimensional") Gowers norm.
In other words, we put the function $f(x_1,\dots,x_d)$ in formula
(\ref{f:G_norm_d'})
equals "one--dimensional"\, function  $f(x_1,\dots,x_d) := f({\rm pr} (x_1,\dots,x_d))$,
where ${\rm pr} (x_1,\dots,x_d) = x_1+\dots+x_d$.
%(Here we write
%$\omega \cdot \v{h} := \omega_1 h_1 + \dots +\omega_d h_d$
%for $\v{h}=(h_1,\dots,h_d) \in \Gr^d$).
Denoting the obtained norm as $U^d$,
%and
we
have
%obtain
an analog of (\ref{e:Gowers_mon}), see \cite{Gow_m}, \cite{TV}
\begin{equation}\label{e:Gowers_mon_1}
    \| f \|_{U^{d-1}} \le \| f \|_{U^d}
\end{equation}
for all $d\ge 2$.
It is convenient to write
\begin{equation}\label{f:G_norm_d'_UU-}
    \| f \|_{\U^d} =
    N^{-d+1}
                    \sum_{\v{x}\in \Gr^d}\, \sum_{\v{x}' \in \Gr^d}
                    \prod_{\o \in \{ 0,1 \}^d} \mathcal{C}^{|\o|} f ( {\rm pr} (\v{x}^{\o}))
    =
\end{equation}
\begin{equation}\label{f:G_norm_d'_UU}
    =
        \sum_x \sum_{h_1,\dots,h_{d}}\, \prod_{\o \in \{ 0,1 \}^{d}}
                \mathcal{C}^{|\o|} f(x + \o\cdot \v{h}) \,.
\end{equation}
In the case $f=A$, where $A\subseteq \Gr$ is a set, we have by formula (\ref{f:G_norm_d'_UU}) that
$$
    \| A \|_{\U^d} = \sum_{s_1,\dots,s_d} |A_{\pi(s_1,\dots,s_d)}| \,,
$$
where $\pi(s_1,\dots,s_d)$ is a vector with $2^d$ components, namely,
$$
    \pi(s_1,\dots,s_d) = \left( \sum_{j=1}^d s_j \eps_j \right)\,, \quad \quad \eps_j \in \{ 0,1 \}^d \,.
$$
Note also
\begin{equation}\label{f:Gowers_sq_A}
    \| A \|_{\U^{d+1}} = \sum_{s_1,\dots,s_d} |A_{\pi(s_1,\dots,s_d)}|^2 \,.
\end{equation}
Further, $\| A \|_{\U^1} = \E_1 (A) = |A|^2$ and $\| A \|_{\U^2} = \E (A)$.

\bigskip

In definitions (\ref{f:G_norm_d'}), (\ref{f:G_norm_d'_UU-})
%, (\ref{f:G_norm_d'_UU})
we have used the size of the set $G$/group $\Gr$.
The results of the paper are local, in the sense that they  do not use
cardinality of the
container group $\Gr$.
Thus it is natural to ask about the possibility to obtain an analog of (\ref{e:Gowers_mon_1}), say,
without any $N$ in the definition.
That is our simple result in the direction.


\begin{proposition}
    Let $A \subseteq \Gr$ be a set.
    Then for any integer $k\ge 2$ one has
\begin{equation}\label{f:Gowers_A}
    \| A \|_{\U^{k+1}} \ge \frac{\| A \|^{(3k-2)/(k-1)}_{\U^{k}}}{\| A \|^{2k/(k-1)}_{\U^{k-1}}} \,.
\end{equation}
In particular,
\begin{equation}\label{f:Gowers_A_0}
        \| A \|_{\U^3} \ge \frac{\E^4 (A)}{|A|^8} \,.
\end{equation}
\label{p:Gowers_A}
\end{proposition}
\begin{proof}
We have
$$
    \| A \|_{\U^k} = \sum_{s_1,\dots,s_k} |A_{\pi(s_1,\dots,s_k)}|
        =
            \sum_{s_1,\dots,s_{k-1}} \sum_{s_k} \sum_z A_{\pi(s_1,\dots,s_{k-1})} (z) A_{\pi(s_1,\dots,s_{k-1})} (z+s_k)
                =
$$
\begin{equation}\label{tmp:01.02.2014_1}
                =
                    \sum_{s_1,\dots,s_{k-1}} \sum_z A_{\pi(s_1,\dots,s_{k-1})} (z) \cdot |A_{\pi(s_1,\dots,s_{k-1})}| \,.
\end{equation}
%Hence
%\begin{equation}\label{tmp:01.02.2014_1+}
%    2^{-1}
%\end{equation}
Thus, if the summation in (\ref{tmp:01.02.2014_1}) is taken over the set
\begin{equation}\label{tmp:Q_def}
    Q_k := \{ (s_1,\dots,s_{k-1}) ~:~ |A_{\pi(s_1,\dots,s_{k-1})}| \ge \| A \|_{\U^k} (2k \| A \|_{\U^{k-1}} )^{-1} \}
    %\,.
\end{equation}
then it gives us $(1-1/2k)$ proportion of the norm $\| A \|_{\U^k}$.
Let us estimate the size of $Q_k$.
Clearly,
$$
    |Q_k| \| A \|_{\U^k} (2k \| A \|_{\U^{k-1}} )^{-1}
        \le
            \sum_{s_1,\dots,s_{k-1}} |A_{\pi(s_1,\dots,s_{k-1})}| = \| A \|_{\U^{k-1}}
            %\,.
$$
and whence $|Q_k| \le 2k \| A \|^2_{\U^{k-1}} \| A \|^{-1}_{\U^k}$.
Certainly, the same bound holds for
cardinality of any set of tuples $(s_{i_1},\dots,s_{i_{k-1}})$
defined similar to (\ref{tmp:Q_def}) and  having the size $k-1$.
%any tuple from $(s_1,\dots,s_k)$ having the size $k-1$.
Hence, by the standard  projection results, see e.g. \cite {Bol_Th}, we see that the summation
in (\ref{tmp:01.02.2014_1}) is taken over a set $\mathcal{S}$ of vectors $(s_1,\dots,s_k)$ of size at most
$
     ( 2k \| A \|^2_{\U^{k-1}} \| A \|^{-1}_{\U^k} )^{k/(k-1)}
$.
Returning to (\ref{tmp:01.02.2014_1}) and using the Cauchy--Schwarz inequality
as well as formula (\ref{f:Gowers_sq_A}), we obtain
$$
    2^{-2} \| A \|^2_{\U^k}
        \le
           \left( \sum_{(s_1,\dots,s_k) \in \mathcal{S}} |A_{\pi(s_1,\dots,s_k)}| \right)^2
            \le
                |\mathcal{S}| \sum_{s_1,\dots,s_k} |A_{\pi(s_1,\dots,s_k)}|
                    \le
$$
$$
                    \le
                        ( 2k \| A \|^2_{\U^{k-1}} \| A \|^{-1}_{\U^k} )^{k/(k-1)} \| A \|_{\U^{k+1}} \,.
$$
The last inequality implies that
$$
    \| A \|_{\U^{k+1}} \ge C_k \frac{\| A \|^{(3k-2)/(k-1)}_{\U^{k}}}{\| A \|^{2k/(k-1)}_{\U^{k-1}}} \,,
$$
where $0<C_k<1$ depends on $k$ only.
Using the tensor trick we obtain the result.
This completes the proof.
%$\hfill\Box$
\end{proof}



\begin{remark}
    Estimate (\ref{f:Gowers_A}) is sharp as an example of a sufficiently dense random subset of a group $\Gr$ shows.
    For higher Gowers norms one can obtain  by induction a similar sharp inequality
    $\| A\|_{\U^k} \ge \E(A)^{2^k-k-1} |A|^{-(3\cdot2^k-4k-4)}$.
    It demonstrates expected exponential (in terms of $\E(A)$) growth  of the norms.
\end{remark}


In the next section we will need in a statement, which is generalizes lower bound for $U^3$--norm (\ref{f:Gowers_A_0}).

\begin{lemma}
     Let $A,B\subseteq \Gr$ be two sets.
     Then
\begin{equation}\label{f:U^3(A,B)}
    \sum_{s_1,s_2} \left( \sum_{x} A(x) B(x+s_1) A(x+s_2) B(x+s_1+s_2) \right)^2
        \ge
            \frac{\E^4 (A,B)}{|A|^4 |B|^4} \,.
\end{equation}
\end{lemma}
\begin{proof}
We use the same arguments as in the proof of Proposition \ref{p:Gowers_A}.
One has
$$
    \E (A,B) = \sum_{s_1,s_2} \sum_{x} A(x) B(x+s_1) A(x+s_2) B(x+s_1+s_2)
        =
$$
$$
        =
            \sum_{x} \sum_{s_1} B^A_{s_1} (x) |B^A_{s_1}| = \sum_x \sum_{s_2} A_{s_2} (x) |B_{s_2}| \,.
$$
Because of $\sum_s |A_s| = |A|^2$, $\sum_s |B_s| = |B|^2$, we get
%Thus,
for the set $\mathcal{S}$ above that\\ $|\mathcal{S}| \ll (|A|^2 |B|^2 \E^{-1} (A,B))^2$.
Thus
$$
    \E^2 (A,B)
        \ll
            |\mathcal{S}| \sum_{s_1,s_2} \left( \sum_{x} A(x) B(x+s_1) A(x+s_2) B(x+s_1+s_2) \right)^2
$$
and the result follows.
%This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip

Using (\ref{f:Gowers_A_0}) and the Cauchy--Schwarz inequality, we have a consequence.

\begin{corollary}
    Let $A\subseteq \Gr$ be a set and $|A - A|\le K|A|$ or $|A + A|\le K|A|$.
    Then
    $$
        \| A \|_{\U^3} \ge \frac{|A|^4}{K^4} \,.
    $$
\label{c:U^3&doubling}
\end{corollary}


Inequality (\ref{f:Gowers_A_0}) gives us a relation between $\| A\|_{\U^3} = \sum_s \E (A_s)$ and $\E(A)$.
W.T. Gowers (see \cite{Gow_m}) constructed a set $A$ having a random behavior in terms of $\E(A)$
(more precisely, he constructed a uniform set, that is having small Fourier coefficients, see \cite{Gow_m})
such that for all $s$ the sets $A_s$ have
non--random (non--uniform) behavior in terms of $\E(A_s)$.
Nevertheless, it is natural to ask about the possibility to find an $s\neq 0$ with a weaker notion of randomness,
that is $\E (A_s) \ll |A_s|^{3-c}$, $c>0$.
This question was asked to the author by T. Schoen.
We give an affirmative answer on it.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set, $\E(A) = |A|^3 /K$.
Suppose that for all $s\neq 0$ the
following holds
\begin{equation}\label{cond:A_s_bounded}
    |A_s| \le \frac{M|A|}{K} \,,
\end{equation}
 where $M\ge 1$ is a real number.
 Let $K^4 \le M |A|$.
 %$K^2 \le M |A|^2$.
Then there is $s\neq 0$ such that $|A_s| \ge |A|/2K$ and
\begin{equation}\label{f:E(A_s)}
    \E (A_s) \ll \frac{M^{93/79}}{K^{1/198}} \cdot |A_s|^3 \,.
\end{equation}
\label{t:E(A_s)}
\end{theorem}
\begin{proof}
Let
$$
    P %= P_c
        := \{ s ~:~ |A_s| \ge |A|/2K \} \,.
$$
%where $c>0$ will be specified later.
Find the number $L$ satisfying $L := \max_{s\in P} |A_s|^3 \E^{-1} (A_s)$.
In other words for all $s \in P$, one has $\E(A_s) \ge |A_s|^3 / L$.
Our task is to find a lower bound for $L$.



Put
$$
    \Cf(x,y) = \Cf_3 (A) (x,y) := |A \cap (A-x) \cap (A-y)|
$$
and
$$
    \t{\Cf} (x,y) = \Cf_4 (A) (x,y,x-y) := |A \cap (A-x) \cap (A-y) \cap (A-x + y)|
    \,.
$$
%(we will choose the sign later, probably plus).
Clearly,
\begin{equation}\label{f:C_A_s}
    \t{\Cf} (x,y) \le \Cf(x,y) \le \min\{ |A_x|, |A_y| \} \,.
\end{equation}
for any $x,y$.
Put also
$$
    \mathcal{P} := \{ (x,y) ~:~ \t{\Cf} (x,y) \ge |A|/(4KL) \} \,.
$$
We will write $\mathcal{P}_x := \mathcal{P} \cap (\{x\} \m \Gr)$,
and $\mathcal{P}_y := \mathcal{P} \cap (\Gr \m \{y\})$.
%Having a vector $v \in \Gr^2$ we put
%$\mathcal{P}^v :=
Put also
$$
    \mathcal{P}^\la := \mathcal{P} \cap \{(x,y) ~:~ x - y = \lambda \} \,.
$$


Our first lemma says that the size of $\mathcal{P}$ and some characteristics of the set
can be
%calculated
estimated
in terms of $L$ and $M$.
%precisely more or less.


\begin{lemma}
%Let $c\le 1/2$.Then
We have
\begin{equation}\label{f:P_size1}
    \frac{|A|^2}{4LM^2} \le |\mathcal{P}| \le 4L|A|^2 \,.
\end{equation}
Further, for any nonzero $y$ and $\la$ the following holds
\begin{equation}\label{f:P_size2}
    |\mathcal{P}_y|\,, |\mathcal{P}^\la| \le \frac{4M^2 L|A|}{K} \,.
\end{equation}
\label{l:P_size}
\end{lemma}
\begin{proof}
%Because of $c\le 1/2$,
By the Cauchy--Schwarz inequality, we have
$$
    \sum_{s\in P} |A_s|^3 \ge 2^{-1} \E_3 (A) \ge 2^{-1} |A|^4 / K^2 \,.
$$
Hence
$$
    \frac{|A|^4}{2K^2 L}
        \le
    \frac{1}{L} \sum_{s\in P} |A_s|^3
        \le
            \sum_{s\in P} \E(A_s) \le \sum_s \E (A_s)
            =
$$
$$
            =
            \sum_s \sum_{z,x,y} A_s (z) A_s (z+x) A_s (z+y) A_s (z+x-y) = \sum_{x,y} \t{\Cf}^2 (x,y) \,.
$$
We can assume that $|A| \ge 4 K L^{1/2}$ because otherwise  the result is trivial
in view of the condition $K^4 \le M |A|$.
%$K^2 \le M |A|^2$.
%Because of
Since
$$
    \sum_{x,y} \t{\Cf} (x,y) = \E (A) = \frac{|A|^3}{K} \,,
$$
%we get
it follows
%that
by (\ref{f:C_A_s}) and the assumption  $|A| \ge 4 K L^{1/2}$ that
$$
    \frac{|A|^4}{4K^2 L} \le \sum_{0\neq (x,y) \in \mathcal{P}} \t{\Cf}^2 (x,y)
        \le
            \left( \frac{M|A|}{K} \right)^2 |\mathcal{P}| \,.
$$
In other words $\frac{|A|^2}{4LM^2} \le |\mathcal{P}|$.
On the other hand
$$
    \frac{|A|}{4KL} |\mathcal{P}| \le \sum_{(x,y) \in \mathcal{P}} \t{\Cf} (x,y) \le \sum_{x,y} \t{\Cf} (x,y)
        =
            \E(A) = \frac{|A|^3}{K}
$$
and we obtain the required upper bound for the size of $\mathcal{P}$.


Further, for any fixed $y \neq 0$ the following holds
$$
    |\mathcal{P}_y| \le \frac{4KL}{|A|} \sum_x \t{\Cf} (x,y) = \frac{4KL}{|A|} |A_{- y}| |A_y| \le \frac{4M^2 L|A|}{K} \,.
$$
Finally,
$$
    |\mathcal{P}^\la| = \sum_{(x,y) ~:~ x - y = \lambda} \mathcal{P} (x,y)
        \le
            \frac{4KL}{|A|} \sum_{(x,y) ~:~  x - y = \lambda}\, \sum_x \t{\Cf} (x,y)
                =
                    \frac{4KL}{|A|} |A_\la|^2
                    %|A_{\pm \la}|
                    \le \frac{4M^2 L|A|}{K}
$$
as required.
%$\hfill\Box$
\end{proof}


\bigskip


Now, we show that some norm of $\mathcal{P}$ is huge.
Actually, we use the function $\Cf$ not $\t{\Cf}$ in the proof.



\begin{lemma}
    One has
\begin{equation}\label{f:P_norm}
    \frac{|A|^{5}}{2^9 K L^4 M^6}
        \le
            \sum_{y,x',y'} |\sum_x \mathcal{P} (x,y+x) \mathcal{P} (x'+x,y'+x) |^2 \,.
\end{equation}
\label{l:P_norm}
\end{lemma}
\begin{proof}
As in the proof of Lemma \ref{l:P_size}, we get
$$
    \frac{|A|^3}{4KLM} \le \sum_{(x,y) \in \mathcal{P}} \Cf (x,y)
        =
            \sum_z A(z) \sum_{x,y} \mathcal{P} (x,y) A(z+x) A(z+y) \,.
$$
Using the Cauchy--Schwarz inequality, we obtain
$$
    \frac{|A|^5}{16 K^2 L^2 M^2}
        \le
            \sum_z A(z) \sum_{x,y}
                \sum_{x',y'} \mathcal{P} (x,y) A(z+x) A(z+y) \mathcal{P} (x',y') A(z+x') A(z+y')
    \le
$$
$$
    \le
        \sum_{x,y,x',y'} \mathcal{P} (x,y+x) \mathcal{P} (x'+x,y'+x) \Cf_4  (A) (y,x',y')
            \,.
$$
Applying the Cauchy--Schwarz inequality again, we have
$$
    \frac{|A|^{10}}{2^8 K^4 L^4 M^4}
        \le
            \sum_{y,x',y'} |\sum_x \mathcal{P} (x,y+x) \mathcal{P} (x'+x,y'+x) |^2
                \m
                    \E_4 (A)
                        \le
$$
$$
                        \le
                \sum_{y,x',y'} |\sum_x \mathcal{P} (x,y+x) \mathcal{P} (x'+x,y'+x) |^2
        \m
            \frac{2M^2 |A|^5}{K^3}
$$
because $K^3\le K^4 \le M |A| \le M^2 |A|$ and hence
\begin{equation}\label{f:E_4_tmp'}
    \E_4 (A)
        \le
            |A|^4 + \frac{M^2 |A|^2}{K^2} \E (A)
                \le
                    \frac{2M^2 |A|^5}{K^3} \,.
\end{equation}
%and we are done.
This concludes the proof of the lemma.
%$\hfill\Box$
\end{proof}


\bigskip


In terms of the sets $\mathcal{P}^\la$ we can rewrite expression (\ref{f:P_norm}) as
$$
    \sum_{y,x',y'} |\sum_x \mathcal{P} (x,y+x) \mathcal{P} (x'+x,y'+x) |^2
        =
            \sum_{y,x',y'} | \sum_x \mathcal{P}^{-y} (x) \mathcal{P}^{x'-y'} (x'+x) |^2
                =
$$
$$
                =
                    \sum_{y,x',y'} | \sum_x \mathcal{P}^{y} (x) \mathcal{P}^{y'} (x'+x) |^2
                        =
                            \sum_{\la,\mu} \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu}) \,.
$$
In view of estimate (\ref{f:P_norm}), a trivial inequality
$$
    \sum_{\la,\mu} \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu})
        \le
            \sum_{\la,\mu} |\mathcal{P}^{\la}|^2 |\mathcal{P}^{\mu}|
                \ll
                    |\mathcal{P}|^2 \cdot \max_{\la \neq 0} |\mathcal{P}^{\la}|
$$
and bounds for size of $\mathcal{P}^{\la}$ of Lemma \ref{l:P_size}
the sets $\mathcal{P}^\la$ should look, firstly, like some sets with small doubling
(more precisely as sets with  large additive energy) and,
secondly, the large proportion of such sets must correlate to each other.
A model example here is $\mathcal{P}^\la (x) = Q(x+\a(\la))$, where $\a$ is an arbitrary function
and $Q$ is an arithmetic progression of size approximately $\Theta (|A|/K)$.
Now we prove that the example is the only case, in some sense.


\bigskip


Put $l = \log (LM)$.
Note that $l\le \log (KM)$ because if $L\ge K$ then the result is trivial.
We can suppose that
\begin{equation}\label{cond:L_tmp}
    2^{17} M^6 K^3 L^7 \le |A|
\end{equation}
because otherwise the result follows
%trivially
immediately
in view of the condition $M |A| \ge K^4$.
Reducing zero terms, we have
$$
    \frac{|A|^{5}}{2^{10} K L^4 M^6}
        \le
    \sum_{\la,\mu \neq 0} \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu})
$$
because
%our choice of parameters,
Lemma \ref{l:P_size}, estimate (\ref{cond:L_tmp}), a bound
$$
    |\mathcal{P}^{0}| \le 4KL |A|^{-1} \sum_{x} |A_x| = 4KL |A|
$$
and a calculation
$$
    2(4KL|A|)^2 4L |A|^2 \le \frac{|A|^{5}}{2^{10} K L^4 M^6} \,.
$$
Below we will assume that any summation is taken over
%sets $\mathcal{P}^{\la}, \mathcal{P}^{\mu}$  with
nonzero indices
%elements of
$\la$, $\mu$.
%Whence
By Lemma \ref{l:P_size} and
a trivial estimate
$$
    \sum_{\la} |\mathcal{P}^{\la}| = |\mathcal{P}| \le 4 L |A|^2 \,,
        \quad \quad
    \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu})
        \le
            |\mathcal{P}^{\la}| |\mathcal{P}^{\mu}|^2
$$
the following holds
$$
    \frac{|A|^{5}}{2^{11} K L^4 M^6}
            \le
    \sum_{\la,\mu ~:~ |\mathcal{P}^{\la}|,\, |\mathcal{P}^{\mu}| \ge \D_*} \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu}) \,,
$$
where $\D_* = \frac{|A|}{2^{16} K L^6 M^6}$.
Using the pigeonhole principle and Lemma \ref{l:P_size}, we find a number $\D$
such that $\D_* \le \D \le \frac{4M^2 L|A|}{K}$ and
\begin{equation}\label{tmp:10.06.2013_1}
    \frac{|A|^{5}}{lK L^4 M^6}
        \ll
            \sum_{\mu ~:~ \D < |\mathcal{P}^{\mu}| \le 2\D} \,\,
                \sum_{\la :~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu}) \,.
\end{equation}
From  (\ref{tmp:10.06.2013_1}) and the Cauchy--Schwarz inequality one can see
%, we can suppose
that the summation in the
formula is taken over
$$
    \E(\mathcal{P}^{\mu})
        \gg
            \frac{|\mathcal{P}^{\mu}|^3}{l^2 L^{14} M^{16}}
            %\frac{|\mathcal{P}^{\mu}|^3}{l L^{9} M^{8}}
                := \eps |\mathcal{P}^{\mu}|^3 \,.
$$
By
%formula
inequality
(\ref{f:P_size1}) of Lemma \ref{l:P_size} there is $\mu$ with
\begin{equation}\label{tmp:11.03.2014_1}
    \zeta \frac{|A|^{3} \D}{K}
        :=
    \frac{|A|^{3} \D}{lK L^5 M^6}
        \ll
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, \mathcal{P}^{\mu}) \,.
\end{equation}
Put $Q=\mathcal{P}^{\mu}$.
We have $\E(Q) \ge \eps |Q|^3$.
Applying a trivial general
%inequality
bound
$$
    \E(A,B) \le |A| |B| \m \max_{x} |A \cap (B-x)| \,,
$$
we get by Lemma \ref{l:P_size}
$$
    \frac{|A|^{3} \D}{lK L^5 M^6}
        \ll
            \D \frac{4M^2 L|A|}{K} \m \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} \max_{x} |\mathcal{P}^{\la} \cap (Q-x)| \,.
$$
%Suppose that for
Given
an arbitrary $\la$ let the maximum in the last formula is attained at point $x:=\a(\la)$.
Thus, we have
$$
    \frac{|A|^{2}}{l L^6 M^8}
        \ll
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} |\mathcal{P}^\la \cap (Q-\a(\la))|
                =
                    \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                        \sum_{x} \mathcal{P} (x,x-\la) Q(x+\a(\la)) \,.
$$
%Thus,
Hence
we find a set $Q$ of the required form, that is having large additive energy and which is
correlates with sets $\mathcal{P}^{\la}$.
Now, we transform the obtained information into some knowledge about the original set $A$.


\bigskip



Using the definition of the set $\mathcal{P}$, we obtain
%and multiplying the last estimate by $\
$$
    \frac{|A|^{3}}{l L^7 M^8 K}
        \ll
            \sum_{x,\la} Q(x+\a(\la)) \sum_z A (z) A (z+x) A (z+x-\la) A (z+\la)
                =
$$
\begin{equation}\label{tmp:10.06.2013_2}
                =
                    \sum_{z,\la} (Q\circ A_{-\la}) (z-\a(\la)) A_\la (z) \,.
\end{equation}
We know that $\E(Q) \ge \eps |Q|^3$.
By Balog--Szemer\'{e}di--Gowers Theorem \ref{BSG} we find $Q'\subseteq Q$, $|Q'| \gg \eps |Q|$
such that  $|Q'-Q'| \ll {\eps}^{-4} |Q'|$.
We
%can suppose
will prove shortly
that the set $Q$ in (\ref{tmp:10.06.2013_2}) can be replaced by a set $\t{Q}$, namely
\begin{equation}\label{tmp:10.06.2013_3}
    c(\eps) \frac{|A|^{3}}{K}
    %= \frac{|A|^{3}}{l L^4 M^6 K}
        \ll
            \sum_{z,\la} (\t{Q} \circ A_{-\la}) (z-\a(\la)) A_\la (z) \,,
\end{equation}
where $c(\eps)>0$ is some constant depends on $L$ and $M$ only
and $\t{Q}$ has small doubling.
Indeed, starting with (\ref{tmp:11.03.2014_1}),
put $Q'_1 = Q'$, and define inductively disjoint sets $Q'_j \subseteq Q$,
$B_j = \bigsqcup_{i=1}^j Q'_j$, $\bar{B}_j = Q\setminus B_j$, $j\in [s]$, applying Theorem \ref{BSG} to $\bar{B}_j$.
Put also  $\E(\bar{B}_j) = \nu_j |Q|^3 \le 8 \nu_j \D^3$.
If at some stage $j$
\begin{equation}\label{tmp:11.03.2014_2}
    \zeta \frac{|A|^{3} \D}{2K}
        >
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, \bar{B}_j)
\end{equation}
then we stop the algorithm.
Let the procedure works exactly $s$ steps
and put $B= B_s$, $\bar{B} = Q\setminus B$.
%Suppose that $|\bar{B}| \le \kappa \D$, $\kappa = 2^{-8} L^{-2} M^{-2}$, $s\ll \eps^{-4} \log 1/\kappa$.
We claim that $s\ll \eps^{-1}$.
To prove this note that if (\ref{tmp:11.03.2014_2}) does not hold then
$$
    \zeta \frac{|A|^{3} \D}{K} \ll \E^{1/2} (\bar{B}_j) \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} |\mathcal{P}^{\la}|^{3/2}
        \ll
            \nu^{1/2}_j \D \frac{M^2L |A|}{K} L |A|^2 \,.
$$
In other words,
$\E (\bar{B}_j) \gg \eps |Q|^3$ and, hence,
$|Q'_j| \gg \eps |Q|$,   $|Q'_j-Q'_j| \ll {\eps}^{-4} |Q|$.
It means, in particular, that after $s\ll \eps^{-1}$ number of steps our algorithm
%either
stops indeed.
%or we find $\bar{B}$ such that $|\bar{B}| \le \kappa \D$.
%In the former situation at the last step, we get
At the last step, we get by the
%procedure
construction
that
$$
    \zeta \frac{|A|^{3} \D}{2K}
        \le
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} \E (\mathcal{P}^{\la},Q)
                - \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} \E (\mathcal{P}^{\la}, \bar{B}, \bar{B})
        \le
$$
$$
        \le
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} \E (\mathcal{P}^{\la}, B, B)
                    +
                2\sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                        \E (\mathcal{P}^{\la}, B, \bar{B}) ) \,.
$$
Let us prove
%that
the following estimate
$$
    \zeta \frac{|A|^{3} \D}{K}
        \ll
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, B) \,.
$$
If not then by the Cauchy--Schwarz inequality and the choice of $\bar{B}$,
%we have
we obtain
$$
    \left( \zeta \frac{|A|^{3} \D}{K} \right)^2
        \ll
            \left( \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*} \E (\mathcal{P}^{\la}, B, \bar{B}) \right)^2
                \le
                    \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, B) \cdot  \zeta \frac{|A|^{3} \D}{2K}
$$
and we %are done.
get a contradiction.
Hence
%, we have
the following holds
$$
    \zeta \frac{|A|^{3} \D}{K}
        \ll
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, B) \,.
$$
%Using
Applying
the H\"{o}lder inequality, we find a set $Q'_j$ such that
$$
    \zeta \frac{|A|^{3} \D}{s^2 K}
        \ll
            \sum_{\la ~:~ |\mathcal{P}^{\la}| \ge \D_*}
                    \E (\mathcal{P}^{\la}, Q'_j) \,.
$$
So, putting $Q':=Q'_j$ we get (\ref{tmp:10.06.2013_3}) with $c(\eps) \gg \frac{\zeta \eps^2}{lL^2 M^2}$.
Of course, the summation in the
%last
obtained
formula can be taken just over $\la$ with $|A_\la| \gg c(\eps) \frac{|A|}{K}$
and we will assume this.


\bigskip


Denote by $\Omega$ the set
$$
    \Omega := \{ (z,\la) ~:~ A_\la (z) =1 \,, \mbox{ and } (\t{Q} \circ A_{-\la}) (z-\a(\la)) \ge \frac{c(\eps)|A|}{2K} \} \,.
$$
From (\ref{tmp:10.06.2013_3}) and our assumption (\ref{cond:A_s_bounded})
%Lemma \ref{l:P_size},
we have $|\Omega| \gg c(\eps) M^{-1} |A|^2$.
On the other hand, considering $\Omega_\la := \{ z~:~ (z,\la) \in \Omega \}$ for any fixed $\la$, one has
$$
    |\Omega_{\la}| \frac{|A|}{K} c(\eps) \ll \sum_z A_\la (z) (\t{Q} \circ A_{-\la}) (z-\a(\la))
        \le
            |A_\la| |Q|
                \ll
                    |A_\la| \D
                \ll \frac{M^3 L |A|^2}{K^2} \,.
$$
Hence there are at least $\gg c^2 (\eps) K M^{-4} L^{-1} |A|$ sets $A_\la$
such that there exists some $z=z(\la)$ with $(\t{Q} \circ A_{\la}) (z-\a(\la)) \gg c(\eps) |A|/2K$.
Denote the set of these $\la$ by $T$.
For any such $A_\la$ there exists a shift of the set $\t{Q}$ such that
$|A_\la \cap (\t{Q} + w(\la))| \gg c(\eps) |A|/2K \gg_{\eps,M} |A_\la|$.
Put $A'_\la := A_\la \cap (\t{Q} + w(\la))$.
We have by Lemma \ref{l:Plunnecke} that for any $\la_1,\la_2 \in T$ the following holds
$$
    |A'_{\la_1}+A'_{\la_2}| \le |\t{Q}+\t{Q}| \ll \eps^{-8} \frac{M^2 L|A|}{K} \,.
$$
In particular,
$$
    \E(A'_{\la_1}, A'_{\la_2}) \gg \frac{c^4(\eps) \eps^8 |A|^3}{M^2 L K^3} \,.
$$
%$Q \approx A_\la$ and hence for any $\la_1,\la_2 \in T$ the following holds $A_{\la_1} \approx A_{\la_2}$.
%The sign $\approx$ means the closeness in Ruzsa's sense : the sets $A$, $B$
%of approximately the same size $A\approx B$ iff $|A-B| \ll |A|$, see \cite{Green_sum-prod}.
Finally, using (\ref{f:E_4_tmp'}) as well as Lemma \ref{l:E_3_A_s} with $k=l=2$, we obtain
$$
    \frac{|A|^5}{l^{64} M^{458} L^{395} K}
        \ll
    \frac{\zeta^8 \eps^{24} |A|^5}{l^8 M^{26} L^{19} K}
        \ll
    \frac{c^8(\eps) \eps^8 |A|^5}{M^{10} L^3 K}
        \ll
    |T|^2 \cdot \frac{c^4(\eps) \eps^8 |A|^3}{M^2 L K^3}
        \ll
            \sum_{\la_1, \la_2 \in T} \E(A'_{\la_1}, A'_{\la_2})
            \le
$$
$$
            \le
    \sum_{\la_1, \la_2 \in T} \E(A_{\la_1}, A_{\la_2}) \le \E_4 (A) \le \frac{2M^2 |A|^5}{K^3}
$$
with
the required lower bound for $L$.
%a contradiction.
This completes the proof of Theorem \ref{t:E(A_s)}.
%$\hfill\Box$
\end{proof}


\bigskip

We finish the section by analog of Definition \ref{def:conn},
which we will use in the
%next section.
the last part of the paper.


\begin{definition}
For $\beta,\gamma \in [0,1]$ a  set $A$ is called  $U^k (\beta,\gamma)$--connected
    if for any $B \subseteq A$, $|B| \ge \beta|A|$
    %one has
    the following holds
    $$
        \| B \|_{\U^k} \ge \gamma \left( \frac{|B|}{|A|} \right)^{2^k} \| A \|_{\U^k} \,.
    $$
\label{def:Uk-conn}
\end{definition}


Again, if, say, $\gamma^{-1} |A|^8 / |A\pm A|^4 \ge \|A \|_{\U^3}$ then by inequality (\ref{f:Gowers_A_0})
one can  see that $A$ is $U^3 (\beta,\gamma)$--connected for any $\beta$.
The existence of $U^k (\beta,\gamma)$--connected subsets in an arbitrary set
is discussed in the Appendix.


\section{Self--dual sets}
\label{sec:structural2}



Inequality (\ref{f:Gowers_A_0}) gives us a relation between $\| A\|_{\U^3}$ and $\E(A)$.
It attaints at a random subset $A$ of $\Gr$,
where by randomness we mean that each element of $A$ belongs to the set with probability $\E(A)/|A|^3$.
%Also
On the other hand,
it is easy to see that an upper bound takes place
\begin{equation}\label{f:U_3_E_3}
    \| A \|_{\U^3} = \sum_s \E(A_s) \le \sum_s |A_s|^3 = \E_3 (A) \,.
\end{equation}
A weaker estimate follows from (\ref{f:U_3_E_3}) combining with  the Cauchy--Schwarz inequality
\begin{equation}\label{f:U_3_E_3'}
    \| A \|^2_{\U^3} \le \E_4 (A) \E(A) \,.
\end{equation}
In the section we consider sets having critical relations between $\| A \|_{\U^3}$ and $\E_4 (A)$, $\E(A)$
that is the sets satisfying the reverse inequality to (\ref{f:U_3_E_3'})
(actually, we use just a slightly stronger estimate then reverse to (\ref{f:U_3_E_3})).
It turns out that they are exactly which we called self--dual sets.
%The following result gives the full description of self--dual sets.


\bigskip


%\begin{lemma}
%    Let $X_1,\dots,X_n$ are Bernoulli random variables, and $a>0$.
%    Then
%\begin{equation}\label{f:large_deviations}
%    \mathbf{P} \{ |X_1+\dots+X_n| > a \} < 2e^{-2a^2/n} \,.
%\end{equation}
%\label{l:large_deviations}
%\end{lemma}


Let us recall a result on large deviations.
The following variant
%of this inequality
can be found in \cite{GreenA+A}.



\begin{lemma}
    Let $X_1,\dots,X_n$ be independent random variables with $\mathbb{E} X_j = 0$
    and $\mathbb{E} |X_j|^2 = \sigma_j^2$.
    Let $\sigma^2 = \sigma_1^2 + \dots + \sigma_n^2$.
    Suppose that for all  $j\in [n]$, we have $|X_j| \le 1$.
    Let also $a$ be a real number such that  $\sigma^2 \ge 6 na$.
    Then
    $$
        \mathbb{P} \left( \left| \frac{X_1+\dots + X_n}{n} \right| \ge a \right) \le 4 e^{-n^2 a^2 / 8\sigma^2} \,.
    $$
\label{l:large_deviations}
\end{lemma}


We need in a combinatorial lemma.


\begin{lemma}
    Let $\D$, $\sigma$, $C>1$ are positive numbers, $t$ be a positive integer,
    and $M_1, \dots, M_t$ be sets, $\D \le |M_j| \le C\D$, $j\in [t]$,
    $\sigma \le 10^{-4} t^2 \D$,
    where
$$
    \sigma:= \sum_{i,j=1}^t |M_i \bigcap M_j| \,.
$$
    Then
%%with probability greater than  $1/2$
    there are at least $\frac{t^2 \D}{16(2C+1) \sigma}$ disjoint sets
    $\t{M}_l \subseteq M_{i_l}$
    such that
    $|\t{M}_l| \ge \frac{\D}{4(2C+1)}$.
\label{l:disjoint_probab}
\end{lemma}
\begin{proof}
    We will choose our sets $\t{M}_i$ deterministically from a randomly chosen family.
    The family will be taken with probability at least $1/4$.


    First of all, we note that
\begin{equation}\label{tmp:25.03.2014_1}
    10^{-4} t^2 \D \ge \sigma \ge \sum_{i=1}^t |M_i| \ge t \D \,.
\end{equation}
    Put $p=t\D 2^{-1} \sigma^{-1}$.
    In view of (\ref{tmp:25.03.2014_1}), we get $p\in (0,1/2]$.
    Let us form a new family of sets taking a set $M_i$ from $M_1,\dots,M_t$ uniformly and independently
    with probability $p$.
    Denote the obtained family as $M'_1,\dots, M'_s$.
    By Lemma \ref{l:large_deviations} and bound (\ref{tmp:25.03.2014_1}), we have after some calculations that
\begin{equation}\label{tmp:25.03.2014_2}
    2^{-1} pt \le s \le 2pt
\end{equation}
    with probability at least $3/4$.
    Further the expectation of $\sigma$ equals
$$
    \mathbb{E} \sum_{i,j=1}^t |M_i \bigcap M_j|
        =
            \sum_x \sum_{i=1}^t \mathbb{E} M_i (x) + \sum_x \sum_{i,j=1,\,i\neq j}^t \mathbb{E} M_i (x) M_j (x)
                =
$$
$$
                =
                    p \sum_{i=1}^t |M_i| + p^2 \sum_{i,j=1,\,i\neq j}^t |M_i \bigcap M_j|
                        \le
                            Cpt\D + p^2 \sigma
                                \le
                                    (2C+1)p^2 \sigma
$$
by our choice of $p$.
Hence, by Markov inequality, with probability at least $1/2$ one has
$$
    \sum_{i,j=1}^s |M'_i \bigcap M'_j| \le (4C+2)p^2 \sigma
$$
and by the Cauchy--Schwarz inequality, we get
$$
    |\bigcup_{i=1}^s M'_i| \ge \frac{(\sum_{i=1}^s |M'_i| )^2}{\sum_{i,j=1}^s |M'_i \cap M'_j|}
        \ge
            \frac{s^2 \D^2}{(4C+2)p^2 \sigma} \ge \frac{2s^2 \sigma}{(2C+1)t^2} := q \,.
$$
Now find disjoint subsets $\t{M}_i \subseteq M'_i$, $i\in [s]$,
taking it consequently and using greedy choice.
Thus, we have at most $s$ nonempty sets $\t{M}_i$
such that
$\bigsqcup_i  \t{M}_i = \bigcup_{i=1}^s M'_i$.
%Thus
Hence
\begin{equation}\label{tmp:25.03.2014_3}
    2 \sum_{i ~:~ |\t{M}_i| \ge q (2s)^{-1}} |\t{M}_i| \ge \sum_{i} |\t{M}_i|
        =
            |\bigcup_{i=1}^s M'_i|
                \ge q \,.
\end{equation}
%%Finally, we put $\t{M}_i = M''_i$.
By our choice of parameters and estimates (\ref{tmp:25.03.2014_2}) the following holds
$$
    |\t{M}_i| \ge \frac{q}{2s} = \frac{s\sigma}{(2C+1)t^2} \ge \frac{p\sigma}{(4C+2) t} = \frac{\D}{(8C+4)} \,.
$$
Similarly, the number $n$ of the sets $\t{M}_i$ can be estimated from (\ref{tmp:25.03.2014_3})
$$
    n \ge \frac{q}{2\D} = \frac{s^2 \sigma}{(2C+1)t^2 \D} \ge \frac{p^2 \sigma}{(8C+4) \D} = \frac{t^2 \D}{(32C+16) \sigma} \,.
$$
This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


Let us remark an interesting consequence of Lemma \ref{l:disjoint_probab}.

\bigskip




\begin{corollary}
    Let $A\subseteq \Gr$ be a
    $(2,\beta,\gamma)$--connected set, $\beta\le 0.5$ be a constant.
    %Suppose that for some real number $\D$ the set
    %$P=\{ x~:~ (A\c A) (x) \sim \}$
    Then there is a set
    $P \subseteq \{ x~:~ (A\c A) (x) \ge \D \}$
    satisfies
    $\E^P (A) \gg \E (A) \log^{-1} |A|$,
    %Then there are
    and there are
    $k \gg  \gamma |A| \D^{-1} \log^{-1} |A|$ disjoint sets
    $\t{A}_j \subseteq A_{s_j}$ with  $|\t{A}_j| \gg \D$.
\label{c:disjoint_P}
\end{corollary}
\begin{proof}
    Using Lemma \ref{l:eigen_A'}, we find $A'$, $|A'| \ge |A|/2$ such that estimate (\ref{f:eigen_A'}) takes place.
    We want to apply Lemma \ref{l:disjoint_probab} to the sets $A'_s \subseteq A_s$, $s\in P'$,
    where $P'=\{ x~:~ (A'\c A') (x) \sim \Delta \}$, $\E^{P'} (A') \gg \E (A') \log^{-1} |A|$.
    Of course, such set $P'$ exists by the pigeonhole principle.
    To apply Lemma \ref{l:disjoint_probab}, we need to calculate the quantity $\sigma$
    $$
        \sigma := \sum_{s,t\in P'} |A'_s \bigcap A'_t| = \sum_{x,y} \Cf_3 (A') (x,y) P'(x) P'(y) \,.
    $$
    By the last identity and estimate (\ref{f:eigen_A'}) (for details, see \cite{s_mixed}), we get
    $$
        \sigma \ll \frac{\E(A)}{|A|} \cdot |P'| \,.
    $$
    Applying Lemma \ref{l:disjoint_probab}, we find disjoint sets $\t{A}_j \subseteq A'_{s_j} \subseteq A_{s_j}$,
    $s_j \in P'$, $j\in [k]$ such that
    $$
        k \gg \frac{|P'|^2 \D |A|}{\E(A) |P'|} \gg \gamma \frac{|A|}{\D \log |A|} \,.
    $$
    In the last inequality we have used   $(2,\beta,\gamma)$--connectedness  of $A$.
    To complete the proof note that $P' \subseteq \{ x~:~ (A\c A) (x) \ge \D \}$.
%    This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip

Clearly, the bound on $k$ in Corollary \ref{c:disjoint_P} is the best possible up to logarithms.
Calculating $\E(A,A_j)/|A_j|$ and comparing its with $\E_3$ (see \cite{s_mixed}) one can obtain an alternative
proof of lower bounds for $|A\pm A_s|$ as of section \ref{sec:sumsets}.
Another result on a family
%with
of
disjoint $A_s$ is proved in Proposition \ref{p:disjoint_via_A-A_s} below.
%  Theorem \ref{t:E_3(D)-}.


\bigskip

Now we are able to
% prove
obtain
the main result of the section.


\begin{theorem}
    Let $A\subseteq \Gr$ be a set, and $M\ge 1$ be a real number.
    Put $l=\log |A|$.
    Suppose that $A$ is
    $U^3(\beta,\gamma)$ and
    $(2,\beta,\gamma)$--connected with $\beta \le 0.5$.
    Then inequality
\begin{equation}\label{cond:E_3_and_E_critical_cor}
%    \| A \|_{\U^3} \gg_{M} \E_3 (A)
    \| A \|^2_{\U^3} \gg_{M} \E_4 (A) \E(A)
\end{equation}
    takes place iff there is  a positive real $\D \sim_{M,\,l} \E_3(A) \E(A)^{-1}$ and a set
    $$
        P \subseteq \{ s\in A-A ~:~ \D < |A_s| \} \,,
    $$
    such that $|P| \gg_{M,\,l} |A|$, $P=-P$, further,
\begin{equation}\label{f:self-dual0}
    \E^P (A) \gg_{M,\,l} \E(A) \,,\quad \quad \E^P_3 (A) \gg_{M,\,l} \E_3 (A)
        \,,\quad \quad \E^P_4 (A) \gg_{M,\,l} \E_4 (A) \,.
\end{equation}
    and
    such that for any $s\in P$ there is $H^s \subseteq A_s$,
    $|H^s| \gg_{M,\,l} \D$,
    with
\begin{equation}\label{f:self-dual1}
    |H^s - H^s| \ll_{M,\,l} |H^s| \,,
\end{equation}
    and $\E(A,H^s) \ll_{M,\,l} |H^s|^3$.\\
    Moreover there are disjoint sets $H_j \subseteq A_{s_j}$, $|H_j| \gg_{M,\,l} \D$,
    $s_j \in P$,  $j\in [k]$
    such that all $H_j$ have small doubling property (\ref{f:self-dual1}),
    $\E(A,H_j) \ll_{M,\,l} |H_j|^3$ and  $k\gg_{M,\,l} |A| \D^{-1}$.
%    All signs $\ll$ depends on $\beta,\gamma$.
\label{t:self-dual}
\end{theorem}
\begin{proof}
    Put $a=|A|$, $\E = \E(A)$, $\E_3 = \E_3 (A)$, $\E_4 = \E_4 (A)$.
    %Without loosing of generality suppose that $0\in A$.
    %We begin
    Let us begin with the necessary condition.
    Using Lemma \ref{l:eigen_A'}, we find $A'$, $|A'| \ge |A|/2$ such that estimate (\ref{f:eigen_A'}) takes place.
%    We will use definition (\ref{f:A'_def}) of the set $A'$ at the end of the proof.
    Because of $A$ is
    $U^3(\beta,\gamma)$ and
    $(2,\beta,\gamma)$--connected with $\beta \le 0.5$,
    we have $\| A' \|_{\U^3} \sim \| A \|_{\U^3}$ and $\E(A') \sim \E(A)$.
    Combining assumption (\ref{cond:E_3_and_E_critical_cor})
    with the Cauchy--Schwarz inequality, we get
\begin{equation}\label{f:E_4_begin}
    \| A \|^2_{\U^3} \gg_{M} \E_4 (A) \E(A) \ge \E^2_3 (A)
    %\,.
\end{equation}
    In particular, by the last inequality and (\ref{f:U_3_E_3}), (\ref{f:U_3_E_3'}), we obtain
$$
    \E^2_3 (A') \ge \| A' \|^2_{\U^3} \gg \| A \|^2_{\U^3} \ge_M \E^2_3 (A) \,,
$$
$$
    \E_4 (A') \E(A') \ge \| A' \|^2_{\U^3} \gg \| A \|^2_{\U^3} \ge_M \E_4 (A) \E(A) \ge \E_4 (A) \E(A')
$$
    and, hence, $\E_3 (A') \sim_M \E_3 (A)$, $\E_4 (A') \sim_M \E_4 (A)$.
    With some abuse of the notation we will use the same letter $A$ for $A'$ below.
    By Lemma \ref{l:E_3_A_s}, we have
\begin{equation}\label{f:23.03.2014_1}
    \| A \|_{\U^3} = \sum_s \E (A_s) = \sum_s \sum_t (A_s \c A_s)^2 (t)
        \gg_{M} \sum_s |A_s|^3 = \sum_s \E(A,A_s) = \E_3 \,.
\end{equation}
    One can assume that the summation in the last formula is taken over $s$ such that
    $|A_s| \gg_M \E_3 \E^{-1}$
    and $\E (A_{s}) \gg_M |A_{s}|^3$, $|A_{s}|^3 \gg_M \E(A,A_s)$.
    Let us consider the condition $\E (A_{s}) \gg_M |A_{s}|^3$.
    By Balog--Szemer\'{e}di--Gowers Theorem we can find $H^s \subseteq A_{s}$ with
    $|H^s| \gg_M |A_{s}|$, and $|H^s - H^s| \ll_M |H^s|$.
    %Because of $0\in A$, we get $H^s \subseteq A-A$.
    Loosing a logarithm $l=\log a$ we can assume that the summation in (\ref{f:23.03.2014_1}) is taken over
    $|A_s|$, $\D < |A_s| \le 2 \D$,
    $\Delta \gg_{M,\,l} \E_3 \E^{-1}$ and $\E (A_{s}) \gg_{M,\,l} |A_{s}|^3$, $|A_{s}|^3 \gg_{M,\,l} \E(A,A_s)$.
    %, of course.
    By $P$ denote the set of such $s$.
    Thus, $|P| \D^3 \gg_{M,\,l} \E_3$ and
    %, clearly,
    it is easy to check that
    $P=-P$.
    %Clearly,
    Note also that $\E(A,H^s) \ll_{M,\,l} |H^s|^3$ for any $s\in P$.
    We have
    \begin{equation}\label{f:27.04.2014_1}
        \sum_{s,t\in P} (A_s \c A_s)^2 (t) \gg_M \E_3 \,.
    \end{equation}
    Returning to (\ref{f:E_4_begin}), we obtain
    $$
        (|P| \D^3)^2 \gg_{M,\,l} \max \{ |P| \D^4 \E, \E_4 |P| \D^2  \}
    $$
    and hence $|P| \D^2 \gg_{M,\,l} \E$, $|P| \D^4 \gg_{M,\,l} \E_4$.
    Thus, $\D \sim_{M,\,l} \E_3 \E^{-1}_2$ and
    $$
        \E_4 \sim_{M,\,l} \D \E_3 \sim_{M,\,l} \D^2 \E \sim_{M,\,l} \D^4 |P| \,.
    $$
    So, we know all energies $\E$, $\E_3$, $\E_4$ if we know $|P|$ and $\D$.
    Let us estimate
    %$|P|$.
    the size of the $P$.
    %On the one hand,
    Taking any $s\in P$, we get by Lemma \ref{l:eigen_A'} that
    $$
        \D^3 \ll_{M,\,l} \E(A_s) \le \E(A,A_s) \ll \E a^{-1} \D \ll_{M,\,l} |P| \D^3 a^{-1}
        %\,.
    $$
    %In other words,
    or
    $|P| \gg_M a$.
%    We will show further then $|P| \ll_M a$.
    So, we have proved (\ref{f:self-dual0})---(\ref{f:self-dual1}).


    %Note also that
    Further
    $$
        \sum_{s,t\in P} |H^s \cap H^t| \le \sum_{s,t\in P} |A_s \cap A_t|  := \sigma \,.
    $$
    Applying Lemma \ref{l:disjoint_probab}, we find disjoint sets
    $H_j \subseteq A_{s_j}$, $|H_j| \gg_M |A_{s_j}|$, $|H_j-H_j| \le |H_j-H_j| \ll_{M,\,l} |H_j|$,
    $|H_j| \gg \D \sim_{M,\,l} \E_3(A) \E(A)^{-1}$,
    $j\in [k]$ and $k\gg |P|^2 \D \sigma^{-1}$.
    Arguing as in Corollary \ref{c:disjoint_P}, we get $\sigma \ll \E |P| |A|^{-1}$
    and hence $k \gg |P| \D |A| \E^{-1} \gg_{M,\,l} |A| \D^{-1}$.
    Of course the
    %obtained
    last
    bound on $k$ is the best possible up to constants depending on $M$, $l$.
    We have obtained the necessary condition.



    Let us prove the sufficient condition.
    Using the Cauchy--Schwarz inequality and formulas (\ref{f:self-dual0})---(\ref{f:self-dual1}), we have
$$
    \| A \|^2_{\U^3} \ge \left( \sum_{s\in P} \E(A_s) \right)^2 \ge
        \left( \sum_{s\in P} \E(H_s) \right)^2
            \gg_{M,\,l}
            \left( \sum_{s\in P} |H_s|^3 \right)^2
            \gg_{M,\,l}
%                \left( \sum_{s\in P} |A_s|^3 \right)^2
%                    =
$$
$$
            \gg_{M,\,l}
%                    =
%                ( \E^P_3 (A) )^2
%                    \ge
                       |P|^2 \D^6
                            \gg_{M,\,l}
                                \E(A) \E_4 (A)
$$
as required.
This completes the proof.
%$\hfill\Box$
\end{proof}


\bigskip

\begin{remark}
%    Actually, the calculations of the proof of Theorem \ref{t:self-dual} show much
%    the set $A$ has much stronger structure than formulas (\ref{f:self-dual1}), (\ref{f:self-dual2})
%    guarantee. For example, we know that $|P| \sim_M |A|$.
    In the statement of Theorem \ref{t:self-dual}
    %contains
    there is
    the set of popular differences $P$
    and the structure of $A$ is described in terms of the set $P$.
    Although, we have obtained a criterium it can be named as a weak structural result.
    Perhaps, a stronger version avoiding using of the set $P$ takes place.
    Namely,
    under the hypothesis  of
    Theorem \ref{t:self-dual} there are disjoint
    sets $H_j \subseteq A_{s_j}$, $|H_j| \gg_{M,\,l} \D$,
    $\D \sim_{M,\,l} \E_3(A) \E(A)^{-1}$,  $j\in [k]$
    such that (\ref{f:self-dual1})
    %takes place
    holds
    and
    %such that
    \begin{equation}\label{f:self-dual2_conj}
%    \sum_{j=1}^k |H_j| \E (H_j) \gg_M \E_3 (A) \,.
    \sum_{j=1}^k |H_j|^4 \gg_{M,\,l} \E_3 (A) \,,
    \quad \quad \sum_{j=1}^k |H_j|^3 \gg_{M,\,l} \E(A) \,,
    \quad \quad \sum_{j=1}^k |H_j|^5 \gg_{M,\,l} \E_4(A)\,.
\end{equation}
    It is easy to see that it is a sufficient condition.
    Indeed, because the sets $H_j \subseteq A$ are disjoint, we have
$$
    \| A \|_{\U^3} \ge \sum_{j=1}^k \| H_j \|_{\U^3} \,.
$$
Using the assumption $|H_j-H_j| \ll_{M,\,l} |H_j|$,
the first bound from (\ref{f:self-dual2_conj}), as well as Corollary \ref{c:U^3&doubling}, we obtain
$$
    \| A \|_{\U^3} \gg_{M,\,l} \sum_{j=1}^k |H_j|^4 \gg_{M,\,l} \E_3 (A)
$$
and, similarly, by the second and the third inequality of (\ref{f:self-dual2_conj}), we get
$$
    \| A \|^2_{\U^3} \gg_{M,\,l} \E (A) \E_4 (A)
$$
as required.
\end{remark}

\bigskip

\begin{example}
    Let $A\subseteq \Gr$ be a set having small Wiener norm, that is
    the following quantity $\| A \|_W := N^{-1} \sum_{\xi} |\FF{A} (\xi)| := M$
    is small.
    Then for any $B\subseteq A$, applying the Parseval identity,  one has
$$
    |B| = \sum_{x} B(x) A(x) = N^{-1} \sum_{\xi} \FF{B} (\xi) \ov{\FF{A} (\xi)} \,.
$$
    Using the H\"{o}lder inequality twice (see also \cite{KSh}), we get
$$
    |B|^4 \le \left( N^{-1} M \sum_{\xi} |\FF{B} (\xi)|^2 |\FF{A} (\xi)| \right)^2
        \le
            M^2 |B| \E(A,B)
$$
or, in other words,
\begin{equation}\label{f:E(A,B)_Wiener}
    \E(A,B) \ge \frac{|B|^3}{M^2} \,.
\end{equation}
    By
    the
    multiplicative property of Wiener norm, we have $\| A_s \|_W \le M^2$.
    %it follows that
    In particular, $\E(A_s) \ge \frac{|A_s|^3}{M^4}$.
    Hence $\| A \|_{\U^3} \ge M^{-4} \E_3 (A)$.
    Further, $\E(A) \ge M^{-2} |A|^3$, $\E_3 (A) \ge M^{-4} |A|^4$
    and hence
    $$
        \| A \|^2_{\U^3} \ge M^{-8} \E^2_3 (A) \ge M^{-16} \E(A) \E_4 (A) \,.
    $$
    Thus, an application of Theorem \ref{t:self-dual}
    %implies
    gives us
    that $A$ has very explicit structure
    (2--connectedness follows from (\ref{f:E(A,B)_Wiener}) and $U^3$--connectedness can be obtained
    via formula (\ref{f:U^3(A,B)}) in a similar way).
    Another structural result on sets from $\F_p$ with small Wiener norm was given in \cite{KSh}.
\end{example}


\bigskip

If Theorem \ref{t:E(A_s)} does not hold that is $\E(A_s) \gg |A_s|^3$ for all $s$ then, clearly,
$\| A \|_{\U^3} \gg \E_3 (A)$
and we can try to apply our structural Theorem \ref{t:self-dual}.
On the other hand, if $A$ is a self--dual set, that is a disjoint union of sets with small doubling
then for any $s\neq 0$ one has exactly $\E(A_s) \gg |A_s|^3$.
It does not contradict to Theorem \ref{t:E(A_s)} because  condition (\ref{cond:A_s_bounded}).


\bigskip

Roughly speaking, in the proof of Theorem \ref{t:self-dual} we found disjoint subsets of $A_s$,
containing huge amount of
%$\E_3$
the
energy (see also Corollary \ref{c:disjoint_P}).
One can ask about the possibility to find some number of disjoint $A_s$ (and not its subsets)
in general situation.
Our next statement answer the question affirmatively.


\begin{proposition}
    Let $A\subseteq \Gr$ be a set, $D\subseteq A-A$.
    Put
\begin{equation}\label{cond:A^2-D(A)}
    \sigma := \sum_{s \in D} |A-A_s| \,.
\end{equation}
    Then there are at least $l \ge |D|^2/(4\sigma)$ disjoint sets $A_{s_1}, \dots, A_{s_l}$.
    In particular, if
\begin{equation}\label{cond:A^2-D(A)'}
        |A^2 - \Delta (A)| \le \frac{|A-A|^2}{M}
\end{equation}
    then there are at least $l \ge M/4$ disjoint sets $A_{s_1}, \dots, A_{s_l}$.
\label{p:disjoint_via_A-A_s}
\end{proposition}
\begin{proof}
    Our arguments is a sort of an algorithm.
    By (\ref{cond:A^2-D(A)}) there is $s_1 \in D$ such that $|A-A_{s_1}| \le \sigma/|D|$.
    Put $D_1 = D\setminus (A-A_{s_1})$.
    If $|D_1| < |D|/2$ then terminate the algorithm.
    If not then by an obvious estimate
    $$
        \sum_{s\in D_1} |A-A_s| \le \sigma
    $$
    we find $s_2 \in D_1$ such that
    $$
        |A-A_{s_2}| \le \frac{\sigma}{|D_1|} \le \frac{2\sigma}{|D|} \,.
    $$
    Put $D_2 = D_1 \setminus (A-A_{s_2})$.
    If $|D_1| < |D|/2$ then terminate the algorithm.
    And so on.
    At the last step, we obtain the set $D_l = D\setminus \bigcup_{j=1}^l (A-A_{s_j})$, $|D_l|<|D|/2$.
    It follows that
    $$
        \frac{|D|}{2} \le |\bigcup_{j=1}^l (A-A_{s_j})| \le \sum_{j=1}^l |A-A_{s_j}| \le l \frac{2\sigma}{|D|} \,.
    $$
    Thus $l \ge |D|^2/(4\sigma)$.
%
%
    Finally, recall that
\begin{equation}\label{f:A-A_s}
    t\in A-A_s \quad \mbox{ iff } \quad A_t \cap A_s \neq \emptyset \,.
\end{equation}
    Thus all constructed sets  $A_{s_1}, \dots, A_{s_l}$ are disjoint.


    To get (\ref{cond:A^2-D(A)'}) put $D=A-A$ and recall that by Lemma \ref{l:A^2_pm}
    the following holds $|A^2 - \D(A)| = \sum_{s\in A-A} |A-A_s|$.
    This completes the proof.
%$\hfill\Box$
\end{proof}

\bigskip

One can ask is it true that not only $\E_3$ energy but $U^3$--norm of sumsets or difference sets is large?
It is easy to see that the answer is no, because of our basic example $A=H\dotplus \L$, $|\L| = K$.
In the case $\E(A) \sim |A|^3/K$, $\E_3 (A) \sim |A|^4 /K$ but $\| A\|_{\U^3} \sim |A|^4 /K^2$
and similar for $A\pm A$.






\section{Appendix}
\label{sec:appendix}


In the section we prove that any set contains a relatively large connected subset.
The case $k=2$ of Proposition \ref{p:connected_k} below was proved in \cite{s_doubling}
(with slightly worse constants)
and we begin with
%an asymmetric
a wide generalization.
% of the case.


\begin{definition}
    Let $X$, $Y$ be two nonempty sets, $|X|=|Y|$.
    A nonnegative symmetric function $q(x,y)$, $x\in X$, $y\in Y$
    is called {\bf weight} if the correspondent
    matrix $q(x,y)$ is nonnegatively defined.
\end{definition}
Having two sets $A$ and $B$ put $\E^q (A,B) := \sum_{x,y} q(x,y) A(x) B(y)$, $\E^q (A) := \E^q (A,A)$.
%We call $q(x,y)$ to be {\it weight} if $\E^q (A)
Clearly, $\E^q (A,B) \le  |A| |B| \| q \|_\infty$.
The main property of any weight is the following.

\begin{lemma}
%Further, because of the matrix $q(x,y)$ nonnegatively defined it is easy to see that
Let $q$ be  a weight.
Then for any sets $A,B$, one has
$$
    (\E^q (A,B))^2 \le \E^q (A) \E^q (B) \,.
$$
\label{l:weight_property}
\end{lemma}

\begin{example}
    Clearly, the function $q(x,y)=(B\c B)^k (x-y)$ for any set $B$
    and an arbitrary positive integer $k$
    is a weight.
    Further, by the construction of Gowers $U^d$--norms it follows that
\begin{equation}\label{f:q_Gowers}
    q_d (x_1,x'_1) = \sum_{x_2,\dots,x_d \in \Gr}\, \sum_{x'_2,\dots,x'_d \in \Gr}\,
                    \prod_{\o \in \{ 0,1 \}^d} f ({\rm pr} (\v{x}^{\o}))
\end{equation}
is also a weight for any nonnegative function $f$.
    In formula (\ref{f:q_Gowers}), we have $\v{x} = (x_1,\dots,x_d)$, $\v{x}' = (x'_1,\dots,x'_d)$,
    and
    ${\rm pr} (y_1,\dots,y_d) := y_1 + \dots + y_d$.
%    and the summation is taken over $x_2,\dots,x_d,x'_2,\dots,x'_d$.
    Another example of a weight is
 \begin{equation}\label{f:q_Gowers+}
    q^*_d (x,y)
        =
            \sum_{h_1,\dots,h_{d-1}}\, \prod_{\o \in \{ 0,1 \}^{d-1},\, \o \neq 0}
                f(x + \o\cdot \v{h}) f(y + \o\cdot \v{h}) \,,
 \end{equation}
    where $f$ is an arbitrary nonnegative function again and $\v{h} = (h_1,\dots,h_{d-1})$.
\label{ex:weights}
\end{example}


For two sets $S,T \subseteq \Gr$, $S\neq \emptyset$, $T\subseteq S$ put $\mu_{S} (T) = |T|/|S|$.
Now we prove a general lemma on connected sets and quantities $\E^q$, where $q$ is a weight.

\begin{lemma}
    Let $A,B \subseteq \Gr$ be two sets, $\beta_1,\beta_2,\rho \in (0,1]$ be real numbers,
    $\beta_1 \le \beta_2$, $\rho < \beta_1 / \beta_2$.
    Let $q$ be a weight.
    Suppose that $\E^q (A) \ge c |A|^2 \| q \|_\infty$, $c\in (0,1]$.
    Then there is $A'\subseteq A$ such that for any subset $\t{A} \subseteq A'$,
    $\beta_1 |A'| \le |\t{A}| \le \beta_2 |A'|$ one has
\begin{equation}\label{f:connected_AB+}
    \E^q (\t{A}) \ge \rho^2 \mu^2_{A'} (\t{A}) \cdot \E^q (A')
    %\,.
    \,,
\end{equation}
    and besides
\begin{equation}\label{f:connected_AB_E+}
     \E^q (A') > (1-\beta_2 \rho)^{2s} \E^q (A) \,,
\end{equation}
    where $s\le \log (1/c) ( 2 \log (\frac{1-\beta_2 \rho}{1-\beta_1} ))^{-1}$.
\label{l:connected_AB+}
\end{lemma}
\begin{proof}
    Put $b=\| q \|_\infty$.
    We use an inductive procedure
    in the proof.
%    At the first step
    Let us describe the first step of our algorithm.
    Suppose that (\ref{f:connected_AB+}) does not hold for some set $C \subseteq A$,
    $\beta_1 |A| \le |C| \le \beta_2 |A|$.
    Put $A^1 = A\setminus C$.
    Then $|A^1| \le (1-\beta_1) |A|$.
    %Applying
    Using
    %the Cauchy--Schwartz inequality
    Lemma \ref{l:weight_property}, we get
$$
    \E^q (A) = \E^q (C,A) + \E^q (A^1,A) < \rho \mu_A (C) \E^q (A) + \E^{1/2}_q (A^1) \E^{1/2}_q (A) \,.
$$
    Hence
$$
    \E^q (A^1) > \E^q (A) (1-\mu_A (C) \rho)^2 \ge \E^q (A) (1- \beta_2 \rho)^2 \,.
$$
    After that applying the same arguments to the set $A^1$, find a subset $C \subseteq A^1$ such that (\ref{f:connected_AB+}) does not hold (if it exists) and so on.
    We obtain a sequence of sets $A\supseteq A^1 \supseteq \dots \supseteq A^s$,
    and $|A^s| \le (1-\beta_1)^s |A|$.
    So, at the step $s$, we have
\begin{equation}\label{tmp:02.04.2014_1}
    c|A|^2 b (1-\beta_2 \rho)^{2s} \le \E^q (A) (1-\beta_2 \rho)^{2s} < \E^q (A^s) \le |A^s|^2 b
        \le
            (1-\beta_1)^{2s} |A|^2 b \,.
\end{equation}
    Thus, our algorithm must stop after at most $s\le \log (1/c) ( 2 \log (\frac{1-\beta_2 \rho}{1-\beta_1} ))^{-1}$ number of steps.
    Putting $A' = A^s$, we see that inequality (\ref{f:connected_AB+}) takes place for any $\t{A} \subseteq A'$ with
    $\beta_1 |A'| \le |\t{A}| \le \beta_2 |A'|$.
    Finally, by the second estimate in (\ref{tmp:02.04.2014_1}), we obtain (\ref{f:connected_AB_E+}).
This concludes the proof.
%$\hfill\Box$
\end{proof}


\bigskip


Let us formulate a useful particular case of Lemma \ref{l:connected_AB+}.


\begin{lemma}
    Let $A,B \subseteq \Gr$ be two sets, $\beta_1,\beta_2,\rho \in (0,1]$ be real numbers,
    $\beta_1 \le \beta_2$, $\rho < \beta_1 / \beta_2$.
    Suppose that $\E(A,B) \ge c|A|^2 |B|$, $c\in (0,1]$.
    Then there is $A'\subseteq A$ such that for any subset $\t{A} \subseteq A'$,
    $\beta_1 |A'| \le |\t{A}| \le \beta_2 |A'|$ one has
\begin{equation}\label{f:connected_AB}
    \E (\t{A},B) \ge \rho^2 \mu^2_{A'} (\t{A}) \cdot \E(A',B)
    %\,.
    \,,
\end{equation}
    and besides
\begin{equation}\label{f:connected_AB_E}
     \E(A',B) > (1-\beta_2 \rho)^{2s} \E(A,B) \,,
\end{equation}
    where $s\le \log (1/c) ( 2 \log (\frac{1-\beta_2 \rho}{1-\beta_1} ))^{-1}$.
\label{l:connected_AB}
\end{lemma}


Lemma \ref{l:connected_AB} implies the required statement, generalizing the result from \cite{s_doubling}.


\begin{proposition}
    Let $A \subseteq \Gr$ be a set,  $\beta \in (0,1)$ be real numbers, and $k\ge 2$ be an integer.
    Put $c=\E_k(A) |A|^{-(k+1)}$.
    Then there is $A'\subseteq A$ such that
    \begin{equation}\label{p:connected_k_energy}
        \E_k (A',A) > (1- 2^{-1} \beta)^{2s} \E_k (A) \,,
    \end{equation}
    where $s\le \log (1/c) ( 2 \log (\frac{2-\beta}{2-2\beta} ))^{-1}$,
    and $A'$ is $(k,\beta,\gamma)$--connected with
    \begin{equation}\label{p:connected_k_gamma}
        \gamma \ge 2^{-(2sk+2k-2s)} \beta^{2k} (2-\beta)^{2s(k-1)} \,.
    \end{equation}
    In particular,
    \begin{equation}\label{p:connected_k_card}
        |A'| \ge (1-2^{-1} \beta)^s c^{1/2} |A| \,.
    \end{equation}
\label{p:connected_k}
\end{proposition}
\begin{proof}
%Recall
Note
that $T\subseteq S$ iff $\D (T) \subseteq \D (S)$.
Applying Lemma \ref{l:connected_AB} with $A=\D(A)$, $B=A^{k-1}$, $\beta_1 = \beta$, $\beta_2 = 1$, $\rho = \beta_1/(2\beta_2) = \beta/2$,
and using formula (\ref{f:energy-B^k-Delta}), we find a set $A'\subseteq A$ such that
for any subset $\t{A} \subseteq A'$, $\beta |A'| \le |\t{A}|$ one has
\begin{equation}\label{tmp:connected_AB}
    \E_k (\t{A},A) \ge \rho^2 \mu^2_{A'} (\t{A}) \cdot \E_k (A',A)
    %\,.
    \,,
\end{equation}
and
\begin{equation}\label{tmp:connected_AB_E}
    \E_k (A',A) > (1- 2^{-1} \beta)^{2s} \E_k (A) \,,
\end{equation}
where $s\le \log (1/c) ( 2 \log (\frac{1-\rho}{1-\beta} ))^{-1}$.
We have obtained inequality (\ref{p:connected_k_energy}).
From (\ref{tmp:connected_AB}), (\ref{tmp:connected_AB_E}) and the H\"{o}lder inequality, we get
$$
    \E_k (\t{A}) \ge \rho^{2k} \mu^{2k}_{A'} (\t{A}) \E^k_k (A',A) \E^{-(k-1)}_k (A)
        \ge
            (2^{-1} \beta)^{2k} (1- 2^{-1} \beta)^{2s(k-1)} \mu^{2k}_{A'} (\t{A}) \E_k (A',A)
                \ge
%                \,.
$$
$$
                \ge
                    (2^{-1} \beta)^{2k} (1- 2^{-1} \beta)^{2s(k-1)} \mu^{2k}_{A'} (\t{A}) \E_k (A')
                        \,.
$$
Thus, the set $A'$ is $(k,\beta,\gamma)$--connected with $\gamma$ satisfying (\ref{p:connected_k_gamma}).
To obtain (\ref{p:connected_k_card}) just apply (\ref{tmp:connected_AB_E}) and a trivial upper bound
for $\E_k (A',A)$
$$
    |A'|^2 |A|^{k-1} \ge \E_k (A',A) > (1- 2^{-1} \beta)^{2s} \E_k (A) = (1- 2^{-1} \beta)^{2s} c|A|^{k+1}
$$
as required.
This completes the proof.
%$\hfill\Box$
\end{proof}


%\begin{remark}
    In view of Lemma \ref{l:connected_AB+} and Example \ref{ex:weights} one can obtain an analog of Proposition \ref{p:connected_k} for $U^k (\beta,\gamma)$--connected sets, see Definition \ref{def:Uk-conn}.
    We leave the details to an interested reader.
    %Further, in Theorem \ref{t:self-dual}, say, we in need in sets which are simultaneously connected in different %senses.   It is easy to see
%\end{remark}



\subsection*{Acknowledgements}
The author is grateful to Vsevolod F. Lev and Tomasz Schoen
%Sergey Konyagin and Misha Rudnev
for useful discussions.
%and, especially, Tomasz Schoen for very useful and fruitful explanations and  discussions.




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\end{document} 