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The additive energy of a finite set $S\subseteq G$ in an abelian group $G$ counts additive quadruples, with order and repeated members allowed:At a fixed sum $s$, independently choosing two of its $r_{S+S}(s)$ ordered pairs gives $r_{S+S}(s{)}^2$ additive quadruples. Thus additive energy weights heavily represented sums more strongly than the cardinality of the sumset does.

For $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$, the sumset representation functions giveThe arithmetic progression has additional coincidences such as $0+3=1+2$. The set $B$ still has coincidences forced by commutativity, including $0+1=1+0$, so its additive energy is not merely the sixteen identical-pair comparisons.

The same additive energy can be counted using the difference set. Define $r_{S-S}(t)=|\{(a,b)\in S\times S:a-b=t\}|$. The identity $a+b=c+d$ is equivalent to $a-c=d-b$, and the coordinate rearrangement $(a,b,c,d)\mapsto (a,c,d,b)$ is a bijection. ConsequentlyFor $A$, the difference counts from $-3$ through $3$ are again $1,2,3,4,3,2,1$. For $B$, the zero difference has four representations and each of the twelve nonzero differences has one, giving $4^2+12\cdot 1^2=28$. The sumset representation functions for sums and differences can look different while their squared totals agree.

For every nonempty finite set $S\subseteq G$ in an abelian group $G$,The lower bound follows from the Cauchy-Schwarz inequality and the total sumset representation function count:For the upper bound, any three members $a,b,c$ determine at most one fourth member $d=a+b-c$ of an additive quadruple. See Yufei Zhao's MIT lecture 25, pages 1-3, for representation counts and the energy bounds.

In the two integer examples, the lower bounds are $256/7$ and $256/10$, and the upper bound is $64$. For the subgroup $H=\{0,2,4\}$ of $\mathbb{Z}/6\mathbb{Z}$, each element of $H$ has three sum representations, so $E(H)=3\cdot 3^2=27$ and both bounds are equalities.

For nonempty $S$, dividing additive energy by $|S{|}^3$ gives a relative measure between $1/K(S)$ and one. Small doubling constant therefore forces relatively large additive energy. A converse requires care: the Balog-Szemerédi-Gowers theorem extracts a large subset with small doubling constant, rather than asserting that all of $S$ has small doubling constant.

For fixed $c>0$ and prime $p$, if $A\subseteq {\mathbb{F}}_p^N$ has at least $c|A{|}^3$ additive quadruples, then $2A-2A$ contains a vector subspace of size at least $c^{\prime }(c,p)|A|$. Combine the Balog-Szemerédi-Gowers theorem, the Freiman-Ruzsa theorem over a finite field, and the Finite-field Bogolyubov lemma.

If a finite set $A$ has at least $\eta |A{|}^3$ additive quadruples, then it contains $A^{\prime }$ with $|A^{\prime }|\ge {\eta }^C|A|$ and $|A^{\prime }+A^{\prime }|\le {\eta }^{-C}|A^{\prime }|$ for an absolute constant $C$.

For a finite nonempty set $A$, its additive energy iswhere $r_{A+A}(s)$ counts ordered representations. Since ${\sum }_s{r}_{A+A}(s)=|A{|}^2$, Cauchy-Schwarz givesThus $|A+A|\le K|A|$ implies $E(A,A)\ge |A{|}^3/K$: a small doubling constant forces many additive quadruples.

A polynomial form of the Balog-Szemerédi-Gowers theorem gives the converse after passing to a subset. There are absolute constants $c,C>0$ such that if $E(A,A)\ge \eta |A{|}^3$, $0<\eta \le 1$, then some $A^{\prime }\subseteq A$ satisfiesThe exponent and constants here need not be optimal.

Precisely, a graph form of the Balog-Szemerédi-Gowers theorem states the following. If $A,B$ are finite sets of size $n$ in an abelian group, $\Gamma \subseteq A\times B$ has at least $\delta n^2$ edges, and its restricted sumsethas size at most $Kn$, then there are $A^{\prime }\subseteq A$, $B^{\prime }\subseteq B$ with $|A^{\prime }|,|B^{\prime }|\ge c{\delta }^{C}n$ andThese are uniform absolute constants, valid for $0<\delta \le 1$ and $K\ge 1$.

To apply this theorem to additive energy, let $n=|A|$ and retain the sums with $r_{A+A}(s)\ge \eta n/2$. The discarded sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$ to the additive energy. Since $r(s)\le n$, the retained pairs number at least $\eta n^2/2$. There are at most $2n/\eta$ retained sums. They therefore define a bipartite graph of density at least $\eta /2$ with restricted sumset size at most $2n/\eta$. The graph theorem supplies large subsets $A^{\prime },B^{\prime }$ with polynomially bounded $|A^{\prime }+B^{\prime }|$.

For completeness, turn this cross-sumset bound into a doubling constant bound. The Ruzsa triangle inequality gives $|A^{\prime }-A^{\prime }|\le |A^{\prime }+B^{\prime }{|}^2/|B^{\prime }|$. If $D=|A^{\prime }-A^{\prime }|/|A^{\prime }|$, the Plünnecke inequality, applied to the pair $A^{\prime },-A^{\prime }$, gives $|A^{\prime }+A^{\prime }|\le D^2|A^{\prime }|$. The minimal-growth proof below in Question 5 applies to arbitrary finite pairs and proves this particular expansion estimate as well. All losses are fixed powers of ${\eta }^{-1}$, proving the displayed converse after increasing $C$.

The requested dense bipartite common-neighbour lemma is a dependent random choice assertion. Call an ordered pair in $V^2$ bad when it has fewer than $ϵn$ common neighbours. Choose $w\in W$ uniformly, and put $S=N(w)\subseteq V$. ThenFor each bad pair, the probability that both endpoints lie in $S$ is its number of common neighbours divided by $n$, which is less than $ϵ$. Thus the expected number $b(S)$ of bad ordered pairs in $S^2$ is at most $ϵm^2$. ConsequentlyChoose a neighbourhood $V^{\prime }=S$ attaining at least this expectation. It has $|V^{\prime }|\ge \delta m/\sqrt{2}\ge \delta m/2$, andHence at least the proportion $1-2ϵ/{\delta }^2$ of its ordered pairs have at least $ϵn$ two-edge connections. As usual these connections are counted by common neighbours, including a return walk for a diagonal ordered pair; assume $\delta >0$, as the formula requires.

The relevance to the graph theorem is that many two-edge connections turn a sparse collection of permitted sums into additive control on a large vertex subset. On a connection $a\to b\to a^{\prime }$, $a-a^{\prime }=(a+b)-(a^{\prime }+b)$ expresses a difference using two elements of the restricted sumset. The abundance of connections, followed by standard cleaning and further path counting, bounds the number of differences without throwing away most vertices. This is the role of dependent random choice in obtaining polynomial losses in the graph form of the Balog-Szemerédi-Gowers theorem.

We will prove a polynomial progression in a high-energy fourfold difference set. The route is to extract a large subset with a small difference set, construct a dense Freiman s-isomorphism model, and use the Bogolyubov lemma and the pigeonhole principle. To respect the proof requirement, every ingredient beyond the elementary rules for Freiman homomorphisms is established below.

Assume $n\ge 1$. Since three entries of an additive quadruple determine the fourth, $E(A)\le n^3$, so a nonvacuous hypothesis has $0<\theta \le 1$. Let $r(t)=|\{(a,b)\in A^2:a+b=t\}|$. Then $E(A)={\sum }_{t}r(t{)}^2$ and ${\sum }_{t}r(t)=n^2$. Define the popular sum set $S=\{t:r(t)\ge \theta n/2\}$. The contribution to additive energy from its complement is at most $\theta n^3/2$. Since $r(t)\le n$, the bipartite graph on two copies of $A$ whose edges have sums in $S$ has at least $\theta n^2/2$ edges. Moreover $|S|\le 2n/\theta$.

Put $\delta =\theta /2$ and $\tau ={\delta }^2/16$. Here is a dependent random choice argument producing the small difference set. Independently choose two right vertices, allowing repetition, and let $X$ be their common left neighbourhood. The Cauchy-Schwarz inequality gives $\mathbb{E}|X|\ge {\delta }^{2}n$. Call an ordered pair of left vertices bad if its common right neighbourhood has fewer than $\tau n$ elements. The expected number $b(X)$ of bad ordered pairs in $X$ is at most ${\tau }^2{n}^2$, because a fixed bad pair survives both random choices with probability at most ${\tau }^2$. ThusChoose a realization attaining at least this expectation. Then $|X|\ge 3{\delta }^{2}n/4$, and $b(X)\le {\delta }^{2}n|X|/64\le |X{|}^2/48$. Remove from $X$ every vertex having more than $|X|/4$ bad partners in $X$. At most $|X|/12$ vertices are removed. The remaining set $B$ has $|B|\ge {\delta }^{2}n/2$, and any $a,b\in B$ have at least $|X|/2$ vertices $z\in X$ for which both $(a,z)$ and $(z,b)$ are good.

For such a $z$, there are at least $\tau n$ right vertices $y$ adjacent to $a,z$, and at least $\tau n$ right vertices $y^{\prime }$ adjacent to $z,b$. These paths givea representation with four members of $S$. For fixed $a,b$, the tuple of four sums determines $y,z,y^{\prime }$ uniquely, so there are at least $|X|{\tau }^2{n}^2/2$ distinct tuples. Select one ordered pair $a,b$ for each member of $B-B$. Tuples belonging to different differences are disjoint. Counting all possible tuples in $S^4$ provesIn particular, with $m=|B|$, we have $m\ge {\theta }^{2}n/8$ and $|B-B|\le Km$ for $K=2^{25}{\theta }^{-12}$. This is the required small-difference set form of the Balog-Szemerédi-Gowers theorem, proved by counting paths rather than cited.

We next control higher iterated sumsets. Choose a nonempty $X\subseteq B$ minimizing $\rho =|X-B|/|X|$, so $\rho \le K$. We prove the Petridis minimal-growth lemma in the form $|X-B+C|\le \rho |X+C|$ for finite $C$. Order $C=\{c_1,\dots ,c_t\}$, and let $X_i$ consist of those $x\in X$ for which $x+c_i$ is new in the union of the translates $X+c_j$. Then ${\sum }_i|X_i|=|X+C|$. Every member of $(X\setminus X_i)-B+c_i$ already occurs in an earlier translate of $X-B$. By minimality, $|(X\setminus X_i)-B|\ge \rho |X\setminus X_i|$, also when the complement is empty. Thus the new contribution from $X-B+c_i$ is at most $\rho |X_i|$. Sum over $i$ to prove the lemma. Iterating with $C=-(j-1)B$ gives $|X-jB|\le K^j|X|$.

For completeness, the Ruzsa triangle inequality in the needed form isFor each $d\in U-W$, fix $d=u_d-w_d$. The map $(d,v)\mapsto (u_d-v,v-w_d)$ is injective: adding the coordinates recovers $d$, and then $v$. Take $U=kB$, $W=\ell B$, $V=X$ and apply the preceding bounds. We obtain the Plünnecke-Ruzsa inequalityIn particular $T=8B-8B$ has at most $K^{16}m$ elements.

We now prove a slightly wasteful version of the Ruzsa modeling lemma, avoiding any assumption about the diameter of $B$. Set $q=8|T|$. Choose a prime $p$ larger than all $|t|$, $t\in T$, and much larger than $q$. Arbitrarily large primes exist by the elementary argument that a prime divisor of one plus the product of all primes up to a given bound exceeds that bound. For a uniformly chosen nonzero multiplier $r$ modulo $p$, each nonzero $t\in T$ has $rt$ uniform among the nonzero residues. Call a multiplier forbidden if some such $rt$ agrees modulo $p$ with $jq$, where $|jq|<p$. There are at most $2p/q+1\le 3p/q$ such residues. For $p\ge 4$, a union bound therefore makes the probability of being forbidden less thanChoose a nonforbidden $r$. Represent $rB$ by integers in $[0,p-1]$ and partition that interval into sixteen consecutive pieces of diameter less than $p/16$. One piece contains the images of a subset $B^{\prime }\subseteq B$ with $|B^{\prime }|\ge m/16$. Let $\varphi (b)$ be the integer representative of $rb$ in that piece, reduced modulo $q$.

A difference of eight sums of these representatives has absolute value less than $p/2$. If its original difference in $B$ is zero, it is a multiple of $p$, hence zero. Conversely, if its reduction modulo $q$ is zero, it equals $jq$ with $|jq|<p$; our choice of $r$ forces its original difference to be zero. This proves that $\varphi$ is a Freiman s-isomorphism of order eight. Injectivity follows by padding a one-term equality to eight terms with a fixed element. Consequently $D=\varphi (B^{\prime })\subseteq G=\mathbb{Z}/q\mathbb{Z}$ has subset densityNo primality of $q$ is needed.

Here is the cyclic Bogolyubov lemma with all its Fourier analysis on a finite abelian group details. Use $\hat{f}(r)={\mathbb{E}}_{x}f(x)e(-rx/q)$, $f=1_D$, and normalized convolution on a finite group. A finite geometric series proves orthogonality, and expanding the finite sums proves the Parseval identity on a finite group, Fourier inversion on a finite group, and $\widehat{f\ast g}=\hat{f}\hat{g}$. For $\tilde{f}(x)=f(-x)$, the nonnegative functionhas support exactly $2D-2D$ and Fourier coefficients on a finite abelian group $|\hat{f}(r){|}^4$. Put $\Gamma =\{r\ne 0:|\hat{f}(r)|\ge {\alpha }^{3/2}/\sqrt{8}\}$. The Parseval identity on a finite group gives $|\Gamma |\le 8{\alpha }^{-2}$. The total fourth moment of the frequencies outside $\Gamma \cup \{0\}$ is at most ${\alpha }^4/8$.

If $\Vert rx/q\Vert \le 1/16$ for every $r\in \Gamma$, then $\mathrm{Re}e(rx/q)\ge \cos (\pi /8)>3/4$. The zero frequency contributes ${\alpha }^4$ with no loss. Separating the large and small frequencies in Fourier inversion on a finite group yieldsThus $2D-2D$ contains the Bohr set expressed in distance coordinates by $\Vert rx/q\Vert \le 1/16$. This is consistent with the existing character-distance convention for a Bohr set: here it is enough that each character lies in the corresponding short arc around one.

We finally prove the needed nonwrapping progression in a cyclic Bohr set, including for composite $q$. Write $d=|\Gamma |$ and $Q=\lfloor q^{1/(d+1)}\rfloor$. Partition the $d$-dimensional unit cube into $Q^d$ boxes. The $Q^d+1$ points with coordinates $jr/q$ modulo one, $0\le j\le Q^d$, give by the pigeonhole principle an integer $1\le t\le Q^d$ with $\Vert rt/q\Vert \le 1/Q$ for all $r\in \Gamma$. ThenThese points are distinct modulo $q$, since their integer representatives lie between zero and $Q^{d+1}/16\le q/16$. Their number is at least $q^{1/(d+1)}/32$. For $d=0$ the same argument simply chooses $t=1$.

The Freiman s-isomorphism induces a bijection $\Phi :2B^{\prime }-2B^{\prime }\to 2D-2D$: define it on a representation by taking the same two positive and two negative images. Equality of representations uses an order-four relation, and injectivity uses the converse relation. If three consecutive members $y_{j-1},y_j,y_{j+1}$ of $P$ satisfy $y_{j-1}+y_{j+1}=2y_j$, their preimages satisfy the same equality, since its expansion is an equality of eight-term sums in $B^{\prime }$. Hence the inverse image of $P$ is an ordinary arithmetic progression in $2B^{\prime }-2B^{\prime }\subseteq 2A-2A$, with distinct terms.

To remove the multiplicative constant from the length bound, let $R=\lceil 8{\alpha }_0^{-2}\rceil$ and $\beta =1/(R+1)$. The length just proved is at least $c_{\theta }{n}^{\beta }$, where $c_{\theta }={\theta }^{2\beta }/32>0$. Choose $N_0\ge 3$ so that $c_{\theta }{n}^{\beta }\ge n^{\beta /2}$ for $n\ge N_0$, and set $\gamma =\min \{\beta /2,\log 3/\log N_0\}$. For $2\le n<N_0$, two distinct members of $A$ give a nonzero $u\in A-A$, and $\{-u,0,u\}\subseteq 2A-2A$ has length three, at least $n^{\gamma }$. For $n=1$, the singleton $\{0\}$ suffices. In every nonempty case,with $\gamma >0$ depending only on $\theta$. If empty sets are permitted, the required length is zero and there is no substantive assertion.

The Balog-Szemerédi-Gowers theorem states that if a finite set $A$ has at least $\delta |A{|}^3$ additive quadruples, then it has a subset $A^{\prime }\subseteq A$ such thatwhere $C$ is an absolute constant.

Consider the graphThe hypothesis says exactly that $\Gamma$ has at least $\delta N^3$ additive quadruples. By the Balog-Szemerédi-Gowers theorem, it contains ${\Gamma }^{\prime }$ with $|{\Gamma }^{\prime }|{\gg }_{\delta }N$ and $|{\Gamma }^{\prime }+{\Gamma }^{\prime }|{\ll }_{\delta }|{\Gamma }^{\prime }|$.

Part b gives a Freiman s-isomorphism from ${\Gamma }^{\prime }$ to a set $D\subseteq \mathbb{Z}$, where it is enough to take any fixed $s\ge 2$. The set $D$ has bounded doubling, with a bound depending only on $\delta$. Part c therefore provides a proper generalized arithmetic progression $Q$ of rank $O_{\delta }(1)$ such that

Write $Q$ as a proper parameter box. Since $|Q|{\gg }_{\delta }N$, one of its $O_{\delta }(1)$ side lengths tends to infinity with $N$. Averaging over all lines parallel to that side gives a line on which $D\cap Q$ has density bounded below in terms of $\delta$. The Szemerédi theorem quoted in the question then gives, once $N$ is sufficiently large in terms of $k$ and $\delta$, a nonconstant $k$-term arithmetic progression in $D$.

The inverse Freiman isomorphism sends it to a $k$-term arithmetic progression in ${\Gamma }^{\prime }\subseteq \Gamma$, because each relation between three consecutive terms is an additive-quadruple relation. Write this progression asIts first-coordinate difference cannot be zero: the graph of a function has only one point above each $x$. Hence $d\ne 0$. On the nonconstant progressionwe haveTaking $a=e/d$ and $b=y_0-(e/d)x_0$, both in $\mathbb{Q}$, proves the claim.

For $s\in {\mathbb{F}}_p^n$, letThe number $T$ of ordered three-term arithmetic progressions in $A$ isBy the Cauchy-Schwarz inequality,The last sum is the additive energy of $A$, namely the number of additive quadruples. If $T\ge \eta |A{|}^2$, then

Part (iii) gives at least ${\eta }^2|A{|}^3$ additive quadruples. By the Balog-Szemerédi-Gowers theorem, there is $A^{\prime }\subseteq A$ withThe Freiman-Ruzsa theorem places $A^{\prime }$ in a coset progression $P$ whose rank is bounded in terms of $\eta$ and which satisfiesThus $A^{\prime }$ has density at least $C_1(\eta {)}^{-1}$ inside $P$.

The Szemerédi theorem in a bounded-rank coset progression says that, for fixed rank and density, every sufficiently large such progression has a nontrivial four-term progression in every subset of that density. If $|A|\ge C(\eta )$ for a sufficiently large $C(\eta )$, then $|P|\ge |A^{\prime }|\ge c(\eta )|A|$ crosses this threshold. It gives a nontrivial four-term progression in $A^{\prime }$, which is also contained in $A$. This proves the claim with a constant depending only on $\eta$.