For an additive or multiplicative group G and subsets A ; B G we define A C B D fg 2 G W There exist a 2 A; b 2 B such that g D a C bgI and

Andrew Granville and József Solymosi

Abstract This is a survey on sum-product formulae and methods. We state old and new results. Our main objective is to introduce the basic techniques used to bound the size of the product and sumsets of finite subsets of a field.

Keywords Additive Combinatorics • Sum-Product problems Mathematics Subject Classification (2010): 11B75, 05B10

1 Introduction 1.1 A Few Definitions

For an additive or multiplicative group G and subsets A ; B G we define A C B D fg 2 G W There exist a 2 A; b 2 B such that g D a C bgI and

A B D fg 2 G W There exist a 2 A; b 2 B such that g D a bg:

We let r

.n/ WD #fa 2 A; b 2 B W n D a C bg; r

.n/ WD #fa 2 A; b 2 B W n D abg, and note that 0 r

.n/ minfjAj; jBjg since r

.n/ D jA \ .n B/j jAj and r

.n/ D jB \ .n A/j jBj. We write A.t/ O D P

a2A

e.at/ where e.u/ D e

.

A. Granville

Département de mathématiques et de statistiques, Université de Montréal, CP 6128 Succursale Centre-Ville, Montréal, QC, Canada H3C 3J7

e-mail:andrew@dms.umontreal.ca J. Solymosi ()

Department of Mathematics, University of British Columbia, 1984 Mathematics Road, Vancouver, BC, Canada V6T 1Z2

e-mail:solymosi@math.ubc.ca

© Springer International Publishing Switzerland 2016

A. Beveridge et al. (eds.),Recent Trends in Combinatorics, The IMA Volumes in Mathematics and its Applications 159, DOI 10.1007/978-3-319-24298-9_18

419

1.2 Multiplication Tables

We learnt to multiply by memorizing the multiplication tables; that is, we wrote down a table with the rows and columns indexed by the integers between 1 and N and the entries in the table were the row entry times the column entry.

Paul Erd˝os presumably learnt his multiplication tables rather more rapidly than the other students, and was left wondering: How many distinct integers are there in the N-by-N multiplication table? Note that if we take A D f1; 2; : : : ; Ng, then we are asking how big is A A? Or, more specifically, since the numbers in the N-by-N multiplication table are all N

, what proportion of the integers up to N

actually appear in the table? That is,

Does jA Aj=N

tend to a limit as N ! 1 ?

Erd˝os showed that the answer is, yes, and that the limit is 0. His proof comes straight from “ The Book ”.

Erd˝os’s proof is based on the celebrated result of Hardy and Ramanujan that “almost all” positive integers n N have log log N (not necessarily distinct) prime factors (here “almost all” means for all but o.N/ values of n N): Hardy and Ramanujan’s result implies that “almost all” products ab with a; b N have 2 log log N prime factors, whereas “almost all” integers N

have log log.N

/ log log N prime factors! The result follows from comparing these two statements.

1.3 The Motivating Conjectures

In fact one can show that jA Aj is large whenever A is an arithmetic progression or, more generally, when A is a generalized arithmetic progression of not-too-large dimension.

This led Erd˝os and Szemerédi to the conjecture that for any > 0, there exists c

> 0 such that for any finite set of integers A;

jA C Aj C jA Aj c

jAj

: (1)

1A.G.: In my primary school we took N D 12 which was the basic multiple needed for understanding U.K. currency at that time.

2Erd˝os claimed that the Supreme Being kept a book of all the best proofs, and only occasionally would allow any mortal to glimpse at “The Book.”

3Ageneralized arithmetic progression is the image of a lattice, that is:

CWD fa₀Ca₁n₁Ca₂n₂C CaknkW 0njNj1for1jkg;

whereN₁;N₂; : : : ;N_k are integers 2. This generalized arithmetic progression is said to have dimensionkandvolumeN₁N₂: : :N_k; and isproper if its elements are distinct.

Even more, one can conjecture that if jAj D jBj D jCj then

jA C Bj C jA Cj c

jAj

: (2) The above conjectures might hold for complex numbers even. Perhaps the most general version is

Either jA C Bj . jAjjBj /

or jA Cj . jAjjCj /

with no restrictions on the sizes of A; B, and C. The thinking in these conjectures is that if A C B is small then A must be “structured,” more precisely that it must look like a largish subset of a generalized arithmetic progression, and similarly if AC is small then log A must look like a largish subset of a generalized arithmetic progression, and that these two structures are incompatible.

2 Sum-Product for Real Numbers 2.1 Results Via Discrete Geometry

The second author proved (2) for D 8=11 [29] (see Theorem 1 below). We now prove (2) for D 3=4 . We begin by stating the

Szemerédi–Trotter Theorem. We are given a set C of m curves in R

such that

• Each pair of curves meet in

points;

• Any pair of points lie on

curves.

For any given set P of n points, there are m C 4

n C 4

.mn/

pairs .; / with point 2 P lying on curve 2 C .

Székely provided a gorgeous proof of this result, straight from The Book , via geometric and random graph theory [27]. From this Elekes elegantly deduced the following in [7]:

Theorem 1. If A; B; C R, then jA C Bj C jA Cj 1

2 .jAj 1/

.jBjjCj/

: (3) Proof. If jA C BjjA Cj .

jBjjCj /

, then at least one of jA C Bj and jA Cj is

jBjjCj, and they are both jAj, so that their product is

jAj

jBjjCj. Hence

jA C Bj C jA Cj 1

2 jAj

. jBjjCj /

;

which implies the result. Hence we may assume that jA C BjjA Cj <

. jBjjCj /

.

Let P be the set of points .A C B/ .A C/; and C the set of lines y D c.x b/

where b 2 B and c 2 C. In the Szemerédi–Trotter Theorem we have

D

D 1 with

m D jBjjCj and n N WD jA C Bj jA Cj ;

since the set of points is [

.a C B; aC/. For fixed b 2 B and c 2 C, all of the points f .a C b; ac/ W a 2 Ag in P lie on the line y D c.x b/, so that

#f .; / W 2 P on 2 C g jAjm : Substituting this into the Szemerédi–Trotter Theorem we obtain

. jAj 1/ m 4 n C 4. mn /

4 N C 4. mN /

:

We assumed that N < m

=2

, so that N < . mN /

=2

< .2

1/. mN /

, and hence . jAj 1/ m

4.2 N /

. This implies that N > . jAj 1/

. jBjjCj /

=16 . u t Corollary 2.1. If A R, then

jA C Aj C jA Aj 1 2 jAj

:

Next we give an argument that improves this. It is still not the best result currently known in the direction of (1) but it only uses the Szemerédi–Trotter Theorem which has several advantages. The most important advantage is that incidence bounds between points and lines on a plane over any field K provide sum-product bounds in K : Even better, the point set where the incidence bounds are needed have a special Cartesian product structure. For example, on the complex plane, C

; it is quite easy to give a Szemerédi-Trotter type bound (with the same exponents) for lines and points of a Cartesian product like . A C B / . A C / above. The incidence bound for this special case appeared in [28]. Another example is Vinh’s work [32] who used a Szemerédi-Trotter type bound to obtain a different proof of Garaev’s sum-product estimate in finite fields (see Theorem 4 below).

Theorem 2. [29] If A ; B ; C ; D R with 0 62 C, then jA=CjjA C BjjC C Dj .jAjjCj/

.jBjjDj/

min

1; jA=Cj jBjjDj

:

Part of this follows from

Theorem 3. If A; B; C; D R with 0 62 C and jA=Cj jBjjDj, then jA=Cj

.jA C BjjC C Dj/

> 1

2 10

.jAjjCj/

.jBjjDj/; (4) and

jACj

. jA C BjjC C Dj /

1 10

.jAjjCj/

.jBjjDj/

log

.4 minfjAj ; jCjg / : We note some consequences:

Corollary 2.2. If A; C R with 0 62 C, then jA = Cj

jA C Cj

> 1

2 10

. jAjjCj /

(5) and

jACj

jA C Cj

1 10

. jAjjCj /

log

.4 minfjAj ; jCjg / : Hence

jAAj

jA C Aj

jAj

log

.4 jAj / I

in particular if jA C Aj jAj, then jAAj

jAj

= log.4jAj/.

Remark. If A D f1; : : : ; Ng, then jAAj N

=.log N/

.log log N/

for some ı D 1

D 0:08607 : : :. Hence some power of log in the denominator in this last result is unavoidable.

Proof of Theorem 3. Let V.k/ denote the set of m 2 C=A for which 2

r

.m/ <

2

, for k D 0; 1; 2; : : : Consider C D C

the set of lines y D mxCe with m 2 V.k/

which contain at least one point .x; y/ 2 B D. Each .x; y/ 2 B D lies on exactly jV.k/j lines of C

(as may be seen by taking e D b dm for each m 2 V .k/), so that

jV .k/jjBjjDj jC

j C 4jBjjDj C 4.jC

jjBjjDj/

by the Szemerédi–Trotter Theorem. Hence either jV.k/j 14 or

jV .k/jjBjjDj j C

j C 4. j C

jjBjjDj /

which implies that

j C

j minf 1

4 jV.k/ jjBjjDj ; 1

27 jV.k/ j

. jBjjDj /

g :

Now jV.k/j jC=Aj jBjjDj so that jC

j

jV.k/j

.jBjjDj/

.

Now consider the set of points .A C B/ .C C D/. If y D mx C e is a line in C

containing the point .b; d/, then it also contains the points .a C b; c C d/

whenever c=a D m with a 2 A; c 2 C. Hence each such line contains at least 2

points from .A C B/ .C C D/ and the Szemerédi–Trotter Theorem then yields 2

j C

j j C

j C 4 jA C BjjC C Dj C 4. j C

jjA C BjjC C Dj /

. Hence

.2

1/ j C

j maxf 80 jA C BjjC C Dj ; 4:2 . j C

jjA C BjjC C Dj /

g which implies that

jC

j 80 max

jA C BjjC C Dj

.2

1/ ; . jA C BjjC C Dj /

.2

1/

D 80 . jA C BjjC C Dj /

.2

1/

; where this last inequality follows since r

.m/ minfjAj ; jCjg which implies that

.2

1/

< r

.m/

jAjjCj jA C BjjC C Dj : Combining the deductions at the end of the last two paragraphs gives

jV.k/ j 6

10

.jA C BjjC C Dj/

.2

1/

.jBjjDj/

: (6)

Therefore X

rC=A.m/2^K

r

.m/ X

kK

X

m2V.k/

2

335 X

kK

2

.2

1/

. jA C BjjC C Dj /

. jBjjDj /

670 2

.2

1/

. jA C BjjC C Dj /

. jBjjDj /

;

and this is

jAjjCj provided 2

671. jA C BjjC C Dj /

=. jAjjCj /. jBjjDj /

. So select the smallest K for which this holds, so that

1

2 jAjjCj X

rC=A.m/<2^K

r

. m / < 2

jC = Aj < 1342 jC = Aj . jA C BjjC C Dj /

. jAjjCj /. jBjjDj /

;

and the first result follows.

Let us also note that, by the Cauchy–Schwarz inequality

.jAjjCj/²Dˇˇ ˇˇˇ

X

n

rAC.n/ˇˇ ˇˇˇ

2

jACjX

n

rAC.n/²D jACjX

m

r_C=A.m/²4jACjX

k

V.k/2^2k:

By (6), and the fact that 2

r

.m/ minfjAj; jCjg we deduce the second result.

u

t

Proof of Theorem 2 when jC=Aj > jBjjDj. We may assume that jA C BjjC C Dj . jAjjCj /

=.118. jBjjDj /

/;

else we obtain the result by multiplying through by jA=Cj

> . jBjjDj /

. In the proof of Theorem 3 we note that if V.k/ > jBjjDj then

V.k/ 4 j C

j

jBjjDj 320 . jA C BjjC C Dj /

.2

1/

jBjjDj ; so we obtain P

rC=A.m/2^K

r

.m/

jAjjCj provided 2

max

. jA C BjjC C Dj /

. jAjjCj /. jBjjDj /

; jA C BjjC C Dj . jAjjBjjCjjDj /

jA C BjjC C Dj . jAjjBjjCjjDj /

; the last equality following from our assumption, and the result follows as in the

proof of Theorem 3. u t

2.2 Some Easier Ideas

The multiplicative energy of two finite sets A; B is defined as E .A; B/ D #fa

; a

2 A; b

; b

2 B W a

b

D a

b

g D X

a;b2B

jaA \ bAj:

By the Cauchy–Schwarz inequality we have E .A; B/

E .A; A/E .B; B/. We also can write

E .A; B/ D X

m

r

.m/

D X

n

r

.n/

D X

n

r

.n/r

.n/;

and hence, by the Cauchy–Schwarz inequality .jAjjBj/

D X

m

r

.m/

!

jABj X

m

r

.m/

D jABjE .A; B/: (7)

Similarly .jAjjBj/

jA=BjE .A; B/. Finally, if A

\ A

D ;, then r

.n/ D r

.n/ C r

.n/ and so, by the Cauchy–Schwarz inequality,

E .A

[ A

; B/ D X

m

r

.m/

2 X

m

.r

.n/

C r

.n/

/ D 2.E .A

; B/ C E .A

; B//: (8)

Proposition 2.3. (Solymosi, [30]) If A and B are finite sets of real numbers, not containing f 0 g, then

E .A; B/ 12 jA C AjjB C Bj log.3 minfjAj ; jBjg /; (9) and hence, by (7),

jA C AjjB C Bj minfjA=Bj ; jABjg . jAjjBj /

=.12 log.3 minfjAj ; jBjg /:

Remarks. Note that there are examples with 0 2 A [ B where this bound cannot hold. For example, if 0 2 B, then E .A; B/ #fa; a

2 A W a0 D 0 D a

0g D jAj

whereas the bound in Proposition 2.3 is smaller than jAj

if A and B are both arithmetic progressions with jBj jAj= log jAj.

Note also that this bound is, more-or-less, best possible in any example with jA C Aj jAj and jB C Bj jBj since, trivially, E .A; B/ jAjjBj.

Proof. We begin by proving this result when A and B are both finite sets of positive real numbers. Let m WD minfjAj; jBjg. If m D 1, then E .A; B/ D maxfjAj; jBjg and the result is easy, so we may assume m 2.

Let R

.`/ D f.a; b/ 2 A B W b D `ag which has size r

.`/, and note that r

.`/ m. Let L

WD f` W 2

r

.`/ < 2

g and K D Œlog m= log 2C1. Then

K1

X

kD0

X

`2Lk

r

.`/

D X

`

r

.`/

D E .A; B/;

and

K1

X

kD0 jLkjD1

X

`2Lk

r

.`/

K1

X

kD0

2

< 2

3 16m

3 : Hence there exists k with jL

j 2 for which

X

`2Lk

r

.`/

1 K

E .A; B/ 16m

3

:

Let L

D f `

< `

< : : : < `

g with r 2 ; we claim that the elements of

r1

[

iD1

.R

.`

/ C R

.`

// A B C A B

are distinct. For if i < j with a

C a

D a

C a

, then

`

a

C `

a

< `

.a

C a

/ `

.a

C a

/ < `

a

C `

a

I

and if .a C a

; `

a C `

a

/ D .x; y/, then a and a

are determined, and so unique.

Now, as A B C A B D .A C A/ .B C B/, we deduce that jA C AjjB C Bj D jA B C A Bj

r1

X

iD1

r

.`

/r

.`

/

.r 1/2

r

2 2

1 8

X

r

.`

/

1 8K

E . A ; B / 16m

3

: From this we deduce that

E .A; B/ 8 log 2m

log 2 jA C AjjB C Bj C 16

3 minfjAj; jBjg

;

and then (9) follows for m 2 after a little calculation, using the fact that jC C Cj 2 jCj 1 .

Now if A only has positive real numbers and 0 62 B, then write B D B

[ B

where B

D fb 2 B W ˙b > 0 g. Now E . A ; B / D E . A ; B / by definition so, by (8) and then the case in which we have already proved (9), we have

E .A; B/ 2E .A; B

/ C 2E .A; B /

12jA C Aj.jB

C B

j C jB C B j/ log.3 minfjAj; jBjg/

which implies (9), since B

C B

and B C B are evidently disjoint subsets of B C B (as their elements are of different signs) (Fig. 1).

b+d b d

c a a+c

l_i+1 l_i

A x B

Fig. 1 The sum of.a;b/ 2 `i and .c;d/ 2 `iC1 is a unique point of the Cartesian product .ACA/.BCB/.

Finally, if 0 62 A[B, then (9) follows similarly from this last result by partitioning

A as A

[ A . u t

Remark. We can deduce bounds on E .A; B/, when 0 2 A [ B, from Proposi- tion 2.3, using the following

If 0 62 A but 0 2 B D B

[ f 0 g then, by definition, E .A; B/ D E .A; B

/ C jAj

and B C B D .B

C B

/ [ B.

If 0 2 A D A

[ f 0 g and 0 2 B D B

[ f 0 g, then E .A; B/ D E .A

; B

/ C . jAj C jBj 1/

with A C A D .A

C A

/ [ A and B C B D .B

C B

/ [ B.

Corollary 2.4. If A is any finite set of real numbers, then

E .A; B/ 12 jA C Aj

log.3 jAj /; (10) and hence, by (7),

jA C Aj

minfjA=Aj ; jAAjg jAj

=.12 log.3 jAj /:

Proof. If 0 62 A, then our bound follows from setting B D A in (9). If 0 2 A, then we use the information in the previous remark, together with (9), to obtain

E .A; A/ D E .A

; A

/ C .2 jAj 1/

12 jA

C A

j

log.3 jAj / C .2 jAj 1/

D 12. jA C Aj jAj /

log .3 jAj / C .2 jAj 1/

D 12 jACAj

log .3 jAj / C 12 log .3 jAj /. jAj

2 jAjjACAj / C .2 jAj 1/

12 jA C Aj

log .3 jAj / C 12 log .3 jAj /.2 jAj 3 jAj

/ C .2 jAj 1/

as jA C Aj 2 jAj 1 , which yields (10). u t

A similar bound for complex numbers was obtained by Konyagin and Rudnev in [22]. Very recently Konyagin and Shkredov announced an improvement on the sum-product bound. They proved in [23] that

jA C Aj C jA Aj jAj

where

> c > 0 is an absolute constant.

2.3 Small Product Sets

From Corollary 2.2 it follows that if the sumset is very small then the product set

is almost quadratic. The opposite statement is surprisingly hard to prove. It was

Chang’s observation [6] that one can use a powerful tool, the Subspace Theorem,

to obtain such bound. For the history and more details about the Subspace Theorem

we refer to the excellent survey paper of Yuri Bilu [3].

An important variant of the Subspace Theorem was proved by Evertse, Schlick- ewei, and Schmidt [9]. We present the version with the best known bound due to Amoroso and Viada [1].

Theorem 2.5. Let K be a field of characteristic 0 , a subgroup of K

of rank r, and a

; a

; : : : ; a

2 K

. Then the number of solutions of the equation

a

z

C a

z

C C a

z

D 1 (11) with z

2 and no subsum on the left-hand side vanishing is at most

A.n; r/ .8n/

:

We are going to use the following result of Freiman (Lemma 1.14 in [11]).

Proposition 1. Let A C. If jAAj CjAj, then A is a subset of a multiplicative subgroup of C

of rank at most r, where r is a constant depending on C.

Theorem 2.6. Let A C with jAj D n. Suppose jAAj Cn: Then there is a constant C

depending only on C such that

jA C Aj n

2 C C

n:

Proof. We consider solutions of x

C x

D x

C x

with x

2 A. A solution of this equation corresponds to two pairs of elements from A that give the same element in A C A. Let us suppose that x

C x

¤ 0 (there are at most jAj D n solutions of the equation x

C x

D 0 with x

; x

2 A).

First we consider the solutions with x

D 0. Then by rearranging we get x

x

C x

x

D 1: (12)

By Proposition 1 and Theorem 2.5 there are at most s

. C / solutions of y

C y

D 1 with no subsum vanishing. Each of these gives at most n solutions of (12) since there are n choices for x

. There are only two solutions of y

C y

D 1 with a vanishing subsum, namely y

D 0 or y

D 0 , and each of these gives n solutions of (12). So we have a total of . s

. C / C 2/ n solutions of (12).

For x

¤ 0 we get

x

x

C x

x

x

x

D 1: (13)

Again by Proposition 1 and Theorem 2.5, the number of solutions of this with no

vanishing subsum is at most s

. C / n. If we have a vanishing subsum, then x

D x

which is a case we excluded earlier or x

D x

and then x

D x

; or x

D x

and then

x

D x

: So we get at most 2 n

solutions of (13) with a vanishing subsum (these are

the x

C x

D x

C x

identities).

So, in total, we have at most 2 n

C s . C / n solutions of x

C x

D x

C x

with x

2 A. Suppose jA C Aj D k and A C A D f ˛

; : : : ; ˛

g. We may assume that

˛

D 0 . Recall that we ignore sums a

C a

D 0 . Let

P

D f . a ; b / 2 A A W a C b D ˛

g ; 2 i k : Then

X

jP

j n

n D n . n 1/:

Also, a solution of x

C x

D x

C x

corresponds to picking two values from P

where x

C x

D ˛

. Thus 2n

C s.C/n

X

jP

j

1 k 1

X

jP

j

!

n

. n 1/

k 1 by the Cauchy-Schwarz inequality. The bound for k D jA C Aj follows.

2.4 Upper Bounds in the Sum-Product Inequality

One obvious way to obtain upper bounds is to select A to be a largish subset of f 1; : : : ; xg with lots of multiplicative structure. For example, we could let A be the set of integers x all of whose prime factors are y, so that jAj D .x; y/; jAAj .x

; y/, and jA C Aj 2x. Roughly .x; y/ D x..e C o.1//=u log u/

when x D y

; so that jAAj = jAj

D .1=2 C o.1//

and jA C Aj = jAj

D .u log u=.e C o.1///

=x.

We select u so that these are roughly equal, that is, u D

1 C

, and thus y .log x/

. Therefore jAj D x

2

. Hence we have an infinite family of examples in which, if jAj D N then

maxfjA C Aj; jAAjg N

o.1/

log logN

:

We can obtain this result without using any “machinery”: Let A be the set of N WD

u

integers composed of no more than u not necessarily distinct prime factors y. Then A Œ1; y

so that jA C Aj 2 y

, whereas jAAj D

2u

. We select u D Œey

=2 log y so that, by Stirling’s formula and the prime number theorem,

N D

e . y = log y / u

e

D

2y

e

D .u.log u/

/

;

and therefore u log N= log log N. Now, by similar calculations we find that jAAj and jA C Aj D jAj

=2

e

D N

o.1/

log logN

:

3 Sum-Product Inequalities over Finite Fields

The Szemerédi–Trotter Theorem does not hold over F

which renders all of the above results moot in this setting. However such results in finite fields are the most applicable, so we will now pursue this. The first thing to note is that we must modify (1) when the set A is large, hence Garaev conjectured that if A F

then

jA C Aj C jA Aj minfjAj

= p p; p

pjAjg ı

jAj

: (14)

3.1 Upper Bounds in F

We begin by showing that the lower bound in Garaev’s conjecture cannot, in general, be increased:

Proposition 3.1. For any given integers I ; J ; N with 1 N I ; J p, and N dIJ = pe, there exist A B ; C F

, with jAj D N ; jBj D J ; jCj D I such that

jA C Bj < 2 jBj and jA Cj < 2 jCj :

In particular, for any given N ; 1 N p, there exists A F

with jAj D N such that

maxfjA C Aj; jA Ajg minfjAj

; 2 p

pjAj C 1g:

Remark. If we have maxfjA C Aj; jA Ajg jAj

for all sets A F

of size N, then the second part of Proposition 3.1 implies that N p

.

Proof. Let C WD fg

; : : : ; g

g where g is a primitive root mod p, and A

WD C \ B

for each x 2 F

where B

WD x C f1; : : : ; Jg. Now X

x

jA

j D X

iD1

X

#fx 2 F

W x D g

jg D IJ ;

so that there exists x with jA

j IJ=p. Let A be any subset of A

of size N, and

B D B

. Therefore A C A A C B B C B D f 2x C 2; : : : ; 2x C 2Jg, A A

C C fg

; : : : ; g

g so that jA C Aj jA C Bj jB C Bj < 2J and jA Aj

jA Cj jC Cj < 2I, which completes the proof of the first part. Now, taking I D J D d p

pNe we find that jA C Aj ; jA Aj 2 d p

pNe 1, which implies the

second part. u t

3.2 A Little Cauchying

Let us make note of a couple of inequalities, for characteristic functions of sets: By Cauchy we obtain

0

@ X

j

ˇˇ ˇˇ A j

p ˇˇ ˇˇ ˇˇ

ˇˇ B O j

p ˇˇ ˇˇ

1 A

2

X

j

ˇˇ ˇˇ A j

p ˇˇ ˇˇ

X

j

ˇˇ ˇˇ B j

p

ˇˇ ˇˇ

D pjAj pjBj:

(15) By Cauchy we obtain

ˇˇ ˇˇ ˇ

X

a2A

X

b2B

e kab

p ˇˇ ˇˇ ˇ

2

X

a2A

1 X

a2A

ˇˇ ˇˇ ˇ

X

b2B

e kab

p ˇˇ ˇˇ ˇ

2

jAj X

a2Fp

ˇˇ ˇˇ ˇ

X

b2B

e kab

p ˇˇ ˇˇ ˇ

2

D jAj pjBj ; (16)

by Parseval.

3.3 Lower Bounds in F

Theorem 4. (Garaev) If A; B; C F

with 0 62 C, then jA C Bj jA Cj jAj

4 min

jAjjBjjCj p ; 2 p

: Remark. Taking I D J D Πp

pN in Proposition 3.1, we obtain examples with jA C Bj jA Cj 4 pjAj. Therefore Theorem 4 is best possible, up to a factor of 8, when jAjjBjjCj 2 p

. In particular for jAj D jBj D jCj 2 p

.

Theorem 4 and its proof remain valid, with suitable modifications, in F

F

(changing both occurrences of p to q D p

in the lower bound). If we select a set D F

such that jD C Dj ; jDDj minfjDj

; pg, then taking A D B D C D D F

we have jA C Bj ; jACj p minfjDj

; pg D minfjAj

= p ; p

g so that jA C AjjAAj

minfjAj

= q ; q

g. Therefore Theorem 4 is best possible up to a constant factor when

q

jAj q

, in this setting.

First by letting C ! 1=C above, and then by taking A D B D C we deduce:

Corollary 3.2. If A; B; C F

with 0 62 C, then jA C Bj jA=Cj jAj

4 min

jAjjBjjCj p ; 2p

:

If A F

with 0 62 A, then

jA C Aj jAAj; jA C Aj jA=Aj jAj 4 min

jAj

p ; 2p

:

If A is a multiplicative subgroup of F

, then

jA C Aj min jAj

4p ; p=2

:

Proof of Theorem 4. For any a 2 A; b 2 B; c 2 C we have a distinct solution to

u=c C b D v (17)

with u 2 AC ; c 2 C ; b 2 B ; v 2 V D A C B, where u D ac and v D a C b. Hence jAjjBjjCj is no more than the total number of solutions of (3.4), which equals

X

u2AC

X

c2C

X

b2B

X

v2ACB

1 p

p1

X

jD0

e

j.u=c C b v/

p

D 1 p

p1

X

jD0

B O j

p

V O j

p X

u2AC

X

c2C

e ju = c

p

jBjjCjjA C BjjACj

p C 1

p max

k¤0

ˇˇ ˇˇ ˇ

X

u2AC

X

c2C

e ku=c

p ˇˇ ˇˇ ˇ

X

j¤0

ˇˇ ˇˇ B O j

p ˇˇ ˇˇ ˇˇ

ˇˇ V O j

p ˇˇ ˇˇ

jBjjCjjA C BjjACj

p C p

pjBjjCjjA C BjjACj

by (15) with A replaced by V , and by (16) with A replaced by AC and B replaced by

1=C, and the result follows. u t

Theorem 5. Suppose that A; B; C; D F

, and C does not contain 0.

If jA C BjjC C DjjA=Cj

jBjjDj p

, then

jA C BjjC C Dj . jAjjCj /

jBjjDj

4p

:

If jA C BjjC C DjjA=Cj

jBjjDj > p

, then

jA C BjjC C DjjA=Cj p 2 jAjjCj:

Corollary 3.3. If 0 62 A F

and jA C AjjA=AjjAj p

, then jA C Aj jAj

=2p.

If jA C Aj jAj and jAj p

, then jAAj ; jA=Aj p=2 (by Corollary 3.2).

If jA C Aj jAj and .2p/

< jAj p

, then jA=Aj > p=2

.

Remark. The first part is stronger than Garaev’s jAAjjA C Aj jAj

=2p in this range.

Proof of Theorem 5. Let us look at solutions to u b D m .v d / with b 2 B ; d 2 D ; u 2 U D A C B ; v 2 V D C C D ; m 2 M D A = C. For each . a ; b ; c ; d / 2 A B C D we have the (distinct) solution . b ; d ; u ; v; m / D . b ; d ; aC b ; cC d ; a = c / , so there are at least jAjjBjjCjjDj solutions. On the other hand we can give an exact count via the exponential sum

X

b;d;u;v;m

1 p

p1

X

jD0

e

j.u b m.v d//

p

D jBjjDjjUjjVjjMj C Error

p ;

where

jErrorj max

i¤0

ˇˇ ˇˇ ˇ

X

m2M

D O im

p

V O im

p ˇˇ ˇˇ ˇ

ˇˇ ˇˇ ˇˇ

p1

X

jD1

U O j

p

B O j

p ˇˇ ˇˇ

ˇˇ : By Cauchy–Schwarz this gives

jErrorj

X

m2M

ˇˇ ˇˇ D O im

p

ˇˇ ˇˇ

X

m2M

ˇˇ ˇˇ V O im

p ˇˇ ˇˇ

p1

X

jD1

ˇˇ ˇˇ U O j

p ˇˇ ˇˇ

p1

X

jD1

ˇˇ ˇˇ B O j

p ˇˇ ˇˇ

which is pjDjpjVjpjUjpjBj by (15). Hence we have proved jAjjBjjCjjDj jBjjDjjUjjV jjMj

p C p p

jBjjDjjUjjVj ;

and the result follows. u t

Theorem 6 ([17]). Suppose that A; B; C; D F

, and A; C do not contain 0.

If jA C BjjC C DjjBjjDj p

, then

jACj

jA C BjjC C Dj .jAjjCj/

jBjjDj=p:

If jA C BjjC C DjjBjjDj p

, then

jACjjA C BjjC C Dj pjAjjCj :

Remark. We claim that these bounds can be obtained trivially if jAjjCj . jBjjDj /

p

: The second case cannot hold since

jAjjCj . jBjjDj /

D jAjjBjjCjjDjjBjjDj jA C BjjC C DjjBjjDj p

; but then jACj

jA C BjjC C Dj jAjjCj jAj

jBj

jCj

jDj

D . jAjjCj /

jBjjDj = p . p

=. jAjjCj . jBjjDj /

/

. jAjjCj /

jBjjDj = p.

Proof. There exists m 2 AC such that r

. m / jAjjCj = jACj. Now in the set f.u; v/ 2 .A C B/ .C C D/; .b; d/ 2 B D W .u b/.v d/ D mg we evidently have the distinct points ..a C b; c C d/; .b; d// for every b 2 B; d 2 D, and a 2 A; c 2 C with ac D m; a total of jBjjDjr

.m/ jAjjBjjCjjDj=jACj points.

To get an exact count, write U D A C B; V D C C D, to obtain X

b;d;u;v

X

r;s rsDm

1 p

X

i

e

i.u b r/

p

1 p

X

j

e

j.v d s/

p

D 1 p

X

i;j

U O i

p

B O i

p

V O j

p

D O j

p X

r;s rsDm

e

.ir C js/

p

:

The i D j D 0 term yields

jUjjBjjVjjDj, since there are exactly p 1 solutions to rs D m. If j D 0 but i ¤ 0, then our final sum equals 1, so that the sum over i ¤ 0 is

jVjjDj . jUjjBj pjU \ Bj /. Similarly with i D 0. Finally if i ¤ 0 and j ¤ 0, then the final term is 2 p

p in absolute value by a well-known result on Kloosterman sums, and the total contribution is therefore

2

p

X

i¤0

ˇˇ ˇˇ U O i

p

B O i

p ˇˇ ˇˇ X

j¤0

ˇˇ ˇˇ V O j

p

D O j

p ˇˇ ˇˇ ;

and by Cauchying the square of this is 4

p

X

i

ˇˇ ˇˇ U O i

p ˇˇ ˇˇ

X

i

ˇˇ ˇˇ B O i

p

ˇˇ ˇˇ

X

j

ˇˇ ˇˇ V O j

p ˇˇ ˇˇ

X

j

ˇˇ ˇˇ D O j

p

ˇˇ ˇˇ

D 4pjUjjBjjDjjVj :

Putting this altogether we obtain jAjjBjjCjjDj

jACj p C 1

p

jUjjBjjVjjDj C 2 p

pjUjjBjjDjjVj;

which implies the result. u t

Corollary 3.4. Suppose that 0 62 A F

. If jA C AjjAj p

, then jAAjjA C Aj jAj

= p

p:

If jA C AjjAj p

, then

jAAjjA C Aj

pjAj

: This is only non-trivial if jAj p

.

4 Ruzsa–Plunnecke Type Inequalities

We begin with a key result of Ruzsa:

Proposition 4.1. If X ; A

; : : : ; A

F

, then there exists a non-empty Y X such that

jY C A

C : : : C A

j

jYj

Y

jX C A

j jXj : Corollary 4.2. If A; B; C F

, then

jA ˙ Bj jA C CjjB C Cj jCj :

Proof. We can define an injective map W .A B/ C ! .A C C/ .B C C/ as follows, so that the inequality jA Bj jA C CjjB C Cj=jCj holds: If 2 A B, fix a

2 A; b

2 B such that D a

b

and then define .; c/ D .a

C c; b

C c/.

The map is injective since if u D a

C c and v D b

C c then D u v and then c D u a

.

For the other case take k D 2; A

D A; A

D B; X D C in Proposition 4.1 to obtain that there exists non-empty Y C such that

jA C Bj jY C A C Bj jA C Cj

jCj jB C Cj

jCj jY j jA C Cj

jCj jB C Cj jCj jCj:

u t Corollary 4.3. If X; A

; : : : ; A

F

, then there exists Z X such that jZj

jXj and

jZ C A

C : : : C A

j

jZj 2

Y

jX C A

j

jXj :

Proof. By Proposition 4.1 we know that there exists a set Z X for which the inequality holds, so let Z

be the largest subset of X for which this inequality is satisfied and suppose that jZ

j

jXj. Apply Corollary 4.2 with X

D X nZ

in place of X. Noting that jX

j > jXj =2 , and each jX

C A

j jX C A

j we deduce that there exists a non-empty Y X

such that jY C A

C : : : CA

j < 2

jYj Q

iD1

. jX C A

j = jXj /:

Now let Z D Z

[ Y so that jZ C A

C : : : C A

j jZ

C A

C : : : C A

j C jY C A

C : : : C A

j, which is 2

Q

iD1

. jX C A

j = jXj / times jYj C jZ

j D jZj, and thus our inequality is satisfied by Z which is larger than Z

, contradicting the hypothesis. u t Corollary 4.4. For any a ; b 2 F

and A ; B F

we have

jaA ˙ bBj jA C AjjB C Bj jaA \ bBj ; and

jaA ˙ bBj jA C AjjB C Bj max

r

.n/ :

Proof. In Corollary 4.2 replace A by aA, B by bB, and take C D .x C aA/ \ bB for some x 2 F

. We note that aA C C x C aA C aA which has the same size as A C A and, similarly, bB C C bB C bB which has the same size as B C B. The first result follows taking x D 0. Now jCj D r

.x/, so writing x D n and changing b to

b, we get our second result. u t

Corollary 4.5. For any a; b 2 F

and A; B F

we have

jaA ˙ bBj jA C Bj

jbA \ aBj ; and

jaA ˙ bBj jA C Bj

max

r

.n/ :

Proof. In Corollary 4.2 now replace A by aA, B by bB, and take C D . x C bA / \ aB for some x 2 F

. We note that aA C C aA C aB which has the same size as A C B and, similarly, bB C C x C bB C bA which also has the same size as A C B. The first result follows taking x D 0 . Now jCj D r

. x / , so changing b to b, we get

our second result u t

5 Lower Bounds on the Size of A C tB

Lemma 5.1. If A; B F

with jAjjBj > p, then

D F

[ f1g.

Proof. As jAjjBj > p and each of jAj ; jBj p hence jAj ; jBj > 1, that is, jAj, jBj 2. Hence 0 D .a a/=.b

b

/ and 1 D .a

a

/=.b b/. If t ¤ 0, then there are jAjjBj > p numbers a C tb so that two must be congruent mod p. Taking

their difference implies the result. u t

Remark. One might expect that if A; B F

with jAjjBj > p then AB C AB D F

. However if A D B D fm .mod p/ W .m=p/ D 1 g where p is a prime 3 .mod 4/, then evidently 0 62 AB C AB (and here jAj ; jBj D .p 1/=2). Hence the best we can hope for is that if jAjjBj > p then AB C AB C AB D F

, and perhaps AB C AB D F

. Glibichuk [14] proved that if jAjjBj 2p then 8AB D F

(so that if jAjjBj p then 8AB D F

, since then jA C AjjBj 2p so that 16AB 8.A C A/B D F

, unless A is an arithmetic progression, which can be handled).

Let T D

n f 0; 1g. We are interested in the size of A C tB when jAj ; jBj > 1.

Evidently jA C tBj jAj jBj with equality if and only if t 62 T [ f 0 g.

Let R.t/ D R

.t/ denote the number of solutions a; c 2 A; b; d 2 B to a C tb D c C td. We always have the the “diagonal solutions” where a D c and b D d to a C tb D c C td, so that R.t/ jAjjBj. Equality holds, that is, R.t/ D jAjjBj, if and only if t 62 T [ f 0 g. Hence

jA C tBj D jAj jBj ” t 62 T [ f 0 g ” R.t/ D jAjjBj : (18) There is a link between jA C tBj that holds no matter what, which is given by the Cauchy–Schwarz inequality: Let r

.n/ D #fa 2 A; b 2 B W n D a C tbg so that

. jAj jBj /

D X

n2ACtB

r

.n/

!

jA C tBj R

.t/; (19)

since R

.t/ D P

n

r

.n/

.

Proposition 5.2. For any finite sets A; B; S with 0 62 S, there exists t 2 S for which jA C tBj > 1

2 minfjSj ; jAj jBjg : If A ; B 2 F

, then we also have

jA C tBj > 1 2 min

p; jAj jBj jSj p

:

Proof. Note that jA C 0Bj D jAj and R.0/ D jAj jBj

. Since R.t/ jAjjBj for all t hence

X

t2S

.R.t/ jAjjBj/X

t6D0

.R.t/ jAjjBj/D#fa;c2A;b;d2BW b¤danda¤cg D.jAj² jAj/.jBj² jBj/:

(20)

Therefore there exists t 2 S with R.t/ . jAj

jAj /. jBj

jBj /

jSj C jAj jBj < 2jAjjBj max jAjjBj

jSj ; 1

;

whence, by (19),

jAj jBj 2 jA C tBj max jAjjBj

jSj ; 1

and so the first result follows.

When we are working mod p, we have

R . t / D X

n

r

. n /

D 1 p

p1

X

jD0

j O A . j = p / j

j O B . jt = p / j

. jAj jBj /

p ;

taking the j D 0 term, since every term is non-negative. If jAjjBj > p, then R . t / >

jAj jBj and so T D F

n f 0 g by (18) giving another proof of the lemma above.

Now, rearranging (20) we obtain X

t2S

R.t/ . jAj jBj /

p

X

t6D0

R.t/ . jAj jBj /

p

D pjAjjBj

1 jAj

p 1 jBj p

;

so there exists t 2 S with

jAj jBj 2jA C tBj max p

jSj ; jAj jBj p

by (19), and the result follows. u t

Corollary 5.3. Suppose that jAj ; jBj 2. If

D F

, then there exist a

; a

2 A; b

; b

2 B such that

j.a

a

/B C .b

b

/Aj > 1

2 min fp; jAj jBjg : If

¤ F

, then there exist a

; a

2 A; b

; b

2 B such that

j.a

a

C b

b

/B C .b

b

/Aj D jAj jBj:

In other words, the elements .a

a

C b

b

/b C .b

b

/a with a 2 A; b 2 B,

are distinct.