Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 16 / 2

Assume $n>0$; the empty case is immediate. Write $r(s)=\#\{(a,b)\in A^2:a-b=s\}$. The additive energy in the hypothesis is ${\sum }_{s}r(s{)}^2$: rearranging $x+y=z+w$ gives $x-z=w-y$. Since $r(s)\le n$, the energy is at most $n^3$, so the substantive range is $0<\alpha \le 1$.

Let $S=\{s:r(s)\ge \alpha n/2\}$. The contribution of the other differences to the energy is at most $(\alpha n/2){\sum }_{s}r(s)=\alpha n^3/2$. ConsequentlyMake a bipartite graph with two copies of $A$, joining a left vertex $a$ to a right vertex $b$ when $a-b\in S$. Put $\rho =\alpha /2$. This graph has at least $\rho n^2$ edges, and $|S|\le n/\rho$.

We prove the needed dense-graph selection rather than quote the Balog-Szemerédi-Gowers theorem. A pair of left vertices is bad if it has fewer than $\tau n$ common right neighbours, where $\tau ={\rho }^2/32$. Choose a right vertex $y$ uniformly, let $U=N(y)$, and let $b(U)$ be the number of ordered bad pairs in $U$. Averaging the degrees gives $\mathbb{E}|U|\ge \rho n$. Every bad pair lies in fewer than $\tau n$ of these neighbourhoods, so $\mathbb{E}b(U)\le \tau n^2$. Therefore some $y$ satisfiesFor this neighbourhood, $|U|\ge \rho n/2$ and $b(U)\le \rho n|U|/16\le |U{|}^2/8$. Delete vertices having more than $|U|/4$ bad partners in $U$. At most half are deleted, so the remaining set $B$ has $|B|\ge |U|/2\ge \rho n/4$.

For any $b,b^{\prime }\in B$, at least $|U|/2$ vertices $v\in U$ are good partners of both. Each such $v$ has at least $\tau n$ common neighbours with $b$ and at least $\tau n$ with $b^{\prime }$. Independently choosing those neighbours gives at leastfour-edge walks $b,u,v,w,b^{\prime }$. Repeated vertices cause no difficulty. This proves the four-step path lemma for a dense bipartite graph in the required form.

The labels of such a walk are $s_1=b-u$, $s_2=v-u$, $s_3=v-w$, $s_4=b^{\prime }-w$, all in $S$, andFor a fixed pair $(b,b^{\prime })$, its label tuple uniquely recovers $u=b-s_1$, $v=u+s_2$, and $w=v-s_3$, so distinct walks have distinct label tuples. Choose one pair representing each element of the difference set $B-B$. Tuples belonging to different differences are disjoint because their alternating sum determines that difference. HenceWe have obtained explicit constants for the small-difference-set form of the Balog-Szemerédi-Gowers theorem:

We first prove the needed Petridis minimal-growth lemma. For finite nonempty sets $A,D$, choose a nonempty $X\subseteq A$ minimizing $K_0=|X+D|/|X|$. Thus $|Y+D|\ge K_0|Y|$ for every $Y\subseteq X$, including the empty set. For any finite $T=\{t_1,\dots ,t_r\}$, defineThe sets $X_i+t_i$ partition $X+T$. If $x\notin X_i$, then $x+t_i\in X+t_j$ for some $j<i$, and therefore $x+D+t_i\subseteq X+D+t_j$. The newly contributed portion of $X+D+t_i$ consequently has size at mostSum over $i$ to obtain $|X+D+T|\le K_0|X+T|$. Apply this successively with $T=(r-1)D$ to get $|X+rD|\le K_0^r|X|$.

Use $D=-A$. The hypothesis implies $K_0\le |A-A|/|A|\le C$, so $|X-2A|\le C^2|X|$. We also prove the Ruzsa triangle inequality: for finite nonempty $U,V,W$,For each $s\in U-W$, fix $u_s,w_s$ with $s=u_s-w_s$. The map $(s,v)\mapsto (u_s-v,v-w_s)$ is injective: summing its coordinates recovers $s$, and then its first coordinate recovers $v$. This proves the inequality.

Apply it with $U=W=2A$ and $V=X$. Both right-hand factors have cardinality $|X-2A|$, givingThus $\boxed{K=C^4\text{ suffices}}$. This is the required case of the Plünnecke-Ruzsa inequality, proved here using the minimal-growth argument and the explicit injection. If $A$ is empty the conclusion holds directly.

Let $\eta =|A|/N\ge \theta$, $f=1_A$, and use normalized averages on ${\mathbb{Z}}_N$. Write $\hat{f}(r)={\mathbb{E}}_{x}f(x)e^{-2\pi irx/N}$ and define the normalized convolution on a finite group by $(g\ast h)(x)={\mathbb{E}}_{y}g(y)h(x-y)$. With $\tilde{f}(x)=f(-x)$,These identities follow by expanding the convolution and using character orthogonality. Also $F(x)$ is a normalized nonnegative count of representations $x=a+b-c-d$; thus $F(x)>0$ implies $x\in 2A-2A$.

Take $R=\{r\ne 0:|\hat{f}(r)|\ge {\eta }^{3/2}/\sqrt{2}\}$. The Parseval identity on a finite group givesFor $x$ in the Bohr set in phase-distance convention $\mathcal{B}=\{x:\Vert rx/N{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/6\text{ for every }r\in R\}$, all large-frequency terms have real part at least half their magnitude. Since $F$ is real,Thus $\mathcal{B}\subseteq 2A-2A$. This proves the cyclic Bogolyubov lemma needed here, rather than citing it.

We now find an arithmetic progression in a cyclic Bohr set. Put $m=|R|$. If $m=0$, the whole group lies in $2A-2A$ and the result is immediate. Otherwise set $Q=\lfloor N^{1/(m+1)}\rfloor$. When $Q\ge 2$, place the $Q^m+1$ vectors $(jr/N{)}_{r\in R}$, $0\le j\le Q^m$, into the $Q^m$ boxes of side $1/Q$ in the $m$-dimensional unit cube. Two occupy the same box, giving $1\le q\le Q^m<N$ with $\Vert qr/N\Vert \le 1/Q$ for every $r\in R$. The arithmetic progressionlies in $\mathcal{B}$. Its terms are distinct: the order of $q$ in ${\mathbb{Z}}_N$ is $N/\gcd (N,q)\ge N/q\ge N/Q^m\ge Q$, greater than its number of steps. Its length is at least $Q/6\ge N^{1/(m+1)}/12$.

Let $D=\lceil 2{\theta }^{-2}\rceil$. Since $m\le D$, the exponent $1/(m+1)$ is at least $1/(D+1)$. If $Q=1$, then $N^{1/(m+1)}<2$ and the singleton $\{0\}$ already exceeds the required lower bound. This also covers $N=1$. Therefore one may takeThe arithmetic progression here is a set of distinct elements in the cyclic group; this part does not impose the additional integer nonwrapping condition used in Question 3.