Table Of ContentMORE DIFFERENCES THAN MULTIPLE SUMS
IMRE Z. RUZSA
Abstract. We compare the size of the difference set A A to that of the set kA of
k-fold sums. We show the existence of sets such that kA−< A Aαk with αk <1.
| | | − |
6
1
0 1. Introduction
2
The aim of this paper is to compare the size of the difference set A A and the size
n
−
a of
J
kA = A+...+A, k times
6
1 (we shall write k A = ka : a A for a set of multiples).
· { ∈ }
Much has been written about the most natural case k = 2. Freiman and Pigaev [1]
]
T proved that 2A 3/4 A A 2A 4/3. These are still the best exponents known,
N though there|is n|o re≤aso|n t−o ex|p≤ect|tha|t either of them is exact. For other aspects and
h. generalizations see the papers [9],[11],[5],[13] and the books [7],[14],[12].
t We will show the existence of sets of integers such that kA < A A αk with α < 1,
a k
m and of subsets of Z , the set of residues modulo q for all s|uffi|cien|tly−lar|ge q such that
q
[ A−A = Zq, |kA| < qαk.
1
As far as I know, the only paper to deal with this problem is Haight’s [3], who proved
v
6 the existence, for arbitrary prescribed positive integers k and l, of a q and a set A Z
q
4 such that A A = Z and kA avoids l consecutive residues, and used this to show⊂the
41 existence of a−set B oqf reals such that B B = R but kB is of measure 0 for all k.
−
0 Clearly if kA < qαk, then there will be gaps of size > q1−αk, so the above result
. | |
1 implies Haight’s. Ontheother hand, itisnot difficult todeduce our result fromHaight’s
0
either, so the two are essentially equivalent. I also acknowledge that, while the details
6
1 will be rather different, the main idea is taken from Haight’s paper.
v: Haight’s work remained rather unnoticed. A reason is that it was well ahead of its
i time, before additive combinatorics became a fashionable subject; it is not an easy
X
reading either.
r
a In Section 6 we shall consider the opposite question about the maximal possible size
of kA compared to A A.
−
2. The main results
We shall consider three ways of comparing sums and differences. For positive integers
k and q, q > 1 write
F (q) = min kA : A Z ,A A = Z ,
k q q
{| | ⊂ − }
G (q) = min kA : A Z,A A a+1,...,a+q for some a ,
k
| | ⊂ − ⊃ { }
1991 Mathematics Subje(cid:8)ct Classification. 11B13, 11B34. (cid:9)
Author was supported by ERC–AdG Grant No.321104 and Hungarian National Foundation for
Scientific Research (OTKA), Grants No.109789 , and NK104183.
1
2 IMREZ. RUZSA
H (q) = min kA : A Z, A A q .
k
{| | ⊂ | − | ≥ }
Put
logG (q)
k
α = inf .
k
q≥2 logq
Theorem 2.1.
logF (q) logG (q) logH (q)
k k k
lim = lim = lim
logq logq logq
logH (q)
k
= inf = α .
k
q≥2 logq
One possible quantity is missing from the list.
Problem 2.2. Is
logF (q)
k
inf = α ?
k
q≥2 logq
Theorem 2.3.
1 2−k α < 1
k
− ≤
for all k.
The exact value is not know except the obvious α = 1/2. The bound α 3/4 is
1 2
≥
Freiman and Pigaev’s [1]. The upper bound from the construction below will be of type
1-1/tower.
3. Properties of F,G,H
We list some properties of these functions that together will imply Theorem 2.1.
Lemma 3.1. (Monotonicity.) If q < q′, then
G (q) G (q′),
k k
≤
H (q) H (q′).
k k
≤
(Obvious, but important.)
Problem 3.2. Is F monotonically increasing?
k
Conjecture 3.3. No. Probably it depends on the multiplicative structure of q, not just
its size.
Lemma 3.4. (Submultiplicativity.) Let q = q q . We have
1 2
(3.1) F (q) F (q )F (q ) if gcd(q ,q ) = 1,
k k 1 k 2 1 2
≤
(3.2) G (q) G (q )G (q ) always,
k k 1 k 2
≤
(3.3) H (q) H (q )H (q ) always.
k k 1 k 2
≤
Proof. Let A ,A be sets that give the value of our function for q and q , resp.
1 2 1 2
Z Z Z
To see (3.1) notice that is isomorphic to the direct product , and the set
q q1 × q2
A = A A gives the bound for F(q).
1 2
×
To see (3.2) take the set A = A +q A .
1 1 2
·
To see (3.3) take the set A = A +m A with an integer m chosen sufficiently large
1 2
· (cid:3)
to avoid unwanted coincidences.
MORE DIFFERENCES THAN MULTIPLE SUMS 3
Problem 3.5. Does F (q) F (q )F (q ) hold for not coprime integers?
k k 1 k 2
≤
Monotonicity and submultiplicavity imply that
logG (q) logG (q) logH (q) logH (q)
k k k k
lim = inf = α , lim = inf .
k
logq q≥2 logq logq q≥2 logq
To prove the other equalities in Theorem 2.1 we show that these functions have the
same order of magnitude.
Lemma 3.6. For all q we have
(3.4) F (q) G (q),
k k
≤
(3.5) H (q) G (q),
k k
≤
(3.6) G (q) G (2q +1) 2kF (q).
k k k
≤ ≤
Proof. Inequalities (3.4) and (3.5) are evident.
To show (3.6), let A Z be a set such that A A = Z , kA = F (q). Define
q q k
⊂ − | |
A′ = n : q < n q, n mod q A .
{ − ≤ ∈ }
We claim that A′ A′ contains 2q+1 consecutive integers, namely those in the interval
−
[ q,q]. Indeed, if q m q, then there are x,y A′, 1 x,y q such that
− − ≤ ≤ ∈ ≤ ≤
x y m (mod q).
− ≡
Consequently one of x y, (x q) y, x (y q) will be equal to m, and all are
− − − − −
elements of A′ A′. To estimate kA′ observe that
− | |
kA′ [ k(q 1),kq].
⊂ − −
This interval can be covered by 2k intervals of length q, and in each our set has at mos
kA elements, hence kA′ 2k kA . (cid:3)
| | | | ≤ | |
These results partially show Theorem 2.1, except for the quantities involving H . For
k
H we shall give the following estimate.
k
Lemma 3.7.
(3.7) F (q) c (logq)k/2H (q).
k k k
≤
The proof of this lemma is relegated to Section 5. To prove our main result, Theorem
2.3, we shall work with F and G; the results about H are included because it is perhaps
the most natural quantity to consider.
Problem 3.8. Is F (q) c H (q)? Is H (q) F (q)?
k k k k k
≤ ≤
4. The construction
In this section we prove that α < 1. We start by proving the following, seemingly
k
weaker result.
Lemma 4.1. For every positive integer k and positive ε there is a positive integer q
and a set A Z such that A A = Z , kA < εq.
q q
⊂ − | |
4 IMREZ. RUZSA
Proof. We shall describe our set in the form
A = ϕ(x),x+ϕ(x) : x Z
q
{ ∈ }
via a function ϕ : Z Z . This guarantees A A = Z .
q q q
→ −
The set kA is the collection of all elements of the form
(4.1) u(x)ϕ(x)+v(x) x+ϕ(x) ,
xX∈Zq(cid:16) (cid:0) (cid:1)(cid:17)
where u,v are nonnegative integer-valued functions on Z , satisfying
q
u(x)+v(x) = k.
xX∈Zq(cid:0) (cid:1)
We define the level of such a pair (u,v) of functions as
l(u,v) = # x : u(x)+v(x) > 0 .
{ }
Clearly 1 l(u,v) k.
≤ ≤
For a function ϕ and 1 m k, let S (ϕ) denote the set of elements that have
m
≤ ≤
a representation of the form (4.1) with l(u,v) m (in particular, S (ϕ) = kA). The
k
≤
construction will proceed recursively. First we show how to find a modulus and a
function such that S (ϕ) < δq. Next we show that, given two numbers δ,δ′ such that
1
| |
0 < δ < δ′, a modulus and a function such that S (ϕ) < δq, we can find a modulus q′
m
| |
and a corresponding function ϕ′ such that S (ϕ′) < δ′q′.
m+1
| |
For the first step we will take a product of k + 1 different primes, q = p ...p and
0 k
identify Z with the direct product Z ... Z . We shall write elements of Z as
vectors, xq= (x ,...,x ), x Z . Ap0 ×pair (×u,v)pkof level 1 is supported by a sinqgle
0 k i ∈ pi
element x; necessarily v(x) = k u(x). Hence elements of S (ϕ) are of the form
1
−
u(x)ϕ(x)+ k u(x) x+ϕ(x) = kϕ(x)+ k u(x) x.
− −
We will achieve that whene(cid:0)ver u(x) =(cid:1)(cid:0)j, the j’t(cid:1)h coordinate(cid:0)of this su(cid:1)m will vanish. To
this end we put
1 k j k 1
ϕ(x ,...,x ) = x , − x ,..., − x ,..., − x ,0 .
0 k 0 1 j k−1
− k k k
(cid:18) (cid:19)
Here division in the j’th coordinate is meant modulo p , and in order that this make
j
sense we assume p > k for all j.
j
The number of elements where the j’th coordinate vanishes is exactly q/p , conse-
j
quently we have
1
S (ϕ) q < δq
1
| | ≤ p
j
X
if we select primes so that p > (k +1)/δ.
j
For the inductive step, assume that for some 1 m < k we are given two numbers
≤
δ,δ′ such that 0 < δ < δ′, a modulus q and a function such that S (ϕ) < δq. We shall
m
| |
construct a modulus q′ and a corresponding function ϕ′ such that S (ϕ′) < δ′q′.
m+1
Let t be the number of pairs (u,v) of level m+1 on Z . Our ne|w number| will be of
q
the form
q′ = qp p ...p ,
1 2 t
MORE DIFFERENCES THAN MULTIPLE SUMS 5
with distinct primes pj, not dividing q. We identify Zq′ with the direct product Zq
xZp1 ×Z...fo×rZip>t.0W. eThsheaflulnwcrtiitoeneϕle′mweinlltsalosof Zbqe′ daesfivneecdtocros,orxdi=na(txe0w,ixs1e,..a.s,xt), x0 ∈ Z×q,
i ∈ pi
ϕ′(x) = ϕ (x),...,ϕ (x) .
0 t
We put ϕ (x) = ϕ(x ).
0 0 (cid:0) (cid:1)
Given a pair (u′,v′) on Z′, we define its shadow on Z by the formula
q q
u(x) = u′(x,x ...,x ), v(x) = v′(x,x ...,x ).
1 t 1 t
x1X,...,xt x1X,...,xt
Clearly the level of (u,v) does not exceed the level of (u′,v′).
Elements of S (ϕ′) are of the form
m+1
(4.2) u′(x)ϕ′(x)+v′(x) x+ϕ′(x) ,
xX∈Z′q(cid:16) (cid:0) (cid:1)(cid:17)
with pairs (u′,v′) of level at most m+1. The 0’th coordinate of this sum is exactly
u(x)ϕ(x)+v(x) x+ϕ(x) ,
xX∈Zq(cid:16) (cid:0) (cid:1)(cid:17)
where (u,v) is the shadow of (u′,v′). In particular, if the level of (u,v) is at most m,
then the 0’th coordinate is an element of S (ϕ).
m
Now we consider the case when the level of (u,v), as well as of (u′,v′), is m+1. Let
(u ,v ),...,(u ,v ) be a list of all pairs (u,v) of level m+1. We shall achieve that the
1 1 t t
j’th coordinate vanish whenever the shadow of (u′,v′) is (u ,v ).
j j
Observe thatthelevel ofapair(u′,v′)andthatofitsshadowcanbeequal onlyifthere
isnocoincidenceamongthe0’thcoordinateofthoseelements forwhichu′(x)+v′(x) > 0;
the sum defining the shadow has always at most one nonzero term. Consequently for
all x = (x ,x ...,x ) either u′(x),v′(x) = (0,0) or u′(x),v′(x) = u (x ),v (x ) .
0 1 t j 0 j 0
Thus all nonzero terms in the sum (4.2) are of the form
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
u (x )ϕ′(x)+v (x ) x+ϕ′(x) .
j 0 j 0
The j’th coordinate of this summand is
(cid:0) (cid:1)
u (x )ϕ (x)+v (x ) x +ϕ (x) .
j 0 j j 0 j j
This will vanish if we define
(cid:0) (cid:1)
vj(x0) if u (x )+v (x ) > 0,
ϕ (x) = −uj(x0)+vj(x0) j 0 j 0
j
(0 if uj(x0)+vj(x0) = 0,
the division being understood modulo p .
j
This construction ensures that either the 0’th coordinate is in S (ϕ) or another
m
coordinate vanishes. Hence
S (ϕ′) S (ϕ) t 1 t 1
| m+1 | | m | + < δ + < δ′,
q′ ≤ q p p
j j
j=1 j=1
X X
if we choose primes satisfying p > t/(δ′ δ).
j
−
To prove the Lemma we start with δ = ε/(k+1) and proceed by finding moduli and
functions with S (ϕ) /q < (m+1)ε/(k+1). After k steps we have the desired bound
m
for the size of S| (ϕ) =|kA. (cid:3)
k
6 IMREZ. RUZSA
Remark 1. FortheinitialstepIknowseveral constructions, someofwhichyieldsmaller
values of q; I chose this one because it anticipates the inductive step.
Proof of Theorem 2.3. By virtue of Theorem 2.1, to prove the upper bound it is suffi-
cient to find a single q such that G (q) < q; and by inequality (3.6), it suffices to find
k
a single q such that F (q) < q/k, which is the previous lemma with ε = 1/k.
k
To demonstrate the lower bound we show that for any finite set in any group we have
kA A A 1−2−k.
| | ≥ | − |
We use induction on k. The case k = 1 is evident. To go from k to k + 1 we use the
inequality [9] (see also [7],[14],[12])
X Y Z Y X Y Z
| || − | ≤ | − || − |
with Y = Z = A, X = kA. (cid:3)
−
5. From integers to residues
In this section we prove Lemma 3.7.
We start with an arbitrary set A of integers, and in several steps we turn it into a
set of residues modulo q. Our tool will be the following projection-like transformation,
which depends on a real parameter t:
π (n) = [q tn ].
t
{ }
(We suppress the parameter q, which will be fixed through the section.) The values of
π are integers in [1,q), and π has a quasi-additivity property:
t t
π (x+y) = π (x)+π (y)+r, r 0,1, q,1 q .
t t t
∈ { − − }
Lemma 5.1. Let S be a set of integers, S = q. There is a t (0,1) such that
| | ∈
π (S) > q/3.
t
| |
Proof. We select t [0,1) randomly with uniform distribution. Let z be the number of
∈
pairs m,n S such that π (m) = π (n). For a fixed pair (m,n) the probability that
t t
∈
π (m) = π (n) is 1 if m = n, and at most 2/q if m = n. To see the latter claim note
t t
6
that if π (m) = π (n), then
t t
tm tn < 1/q,
|{ }−{ }|
hence t(m n) < 2/q, which has probability 2/q. Hence the expectation of z is
k − k
2
q + q(q 1) < 3q.
≤ q −
Select any t for which z < 3q. For an integer j [0,q) let r be the number of integers
j
∈
n S such that π (n) = j. The inequality of arithmetic and square means yields
t
∈
( r )2 q2
z = r2 j = ,
j ≥ π (S) π (S)
|Pt | | t |
X
hence π (S) q2/z > q/3 as wanted. (cid:3)
t
| | ≥
MORE DIFFERENCES THAN MULTIPLE SUMS 7
Lemma 5.2. Let A Z be a nonempty set, A tq, 0 < t < 1, and let k be a positive
q
integer. There are se⊂ts B ,...,B Z such|tha|t≥
1 k q
⊂
A+B +...+B = Z
1 k q
and
1/k
logq
B m = .
i
| | ≤ t
& '
(cid:18) (cid:19)
Proof. Select B randomly, with equal probability from all q m-element subsets of
1 m
Z . The probability that an element of Z is not in A+B is
q q 1
(cid:0) (cid:1)
q A q
−| | < (1 t)m.
m m −
(cid:18) (cid:19) (cid:18) (cid:19)
.
Hence the expectation of Z (A+B ) is < (1 t)mq. Fix B so that
q 1 1
| \ | −
Z (A+B ) < (1 t)mq.
q 1
| \ | −
Now repeat the process with A+B in the place of A to find B , and so on. After k
1 2
steps the number of elements outside A+B +...+B will be
1 k
< (1 t)mkq < e−tmkq < 1.
−
(cid:3)
The case k = 1 of this lemma is a theorem of Lorentz [6] (see also [4]).
Proof of Lemma 3.7. Let A be a set of integers such that A A q and kA = H (q).
k
| − | ≥ | |
Put A = π (A) with a number t such that π (A A) > q/3. The quasi-additivity
1 t t
| − |
property implies that
π (A A) (A A )+ 0, 1,q,q 1 .
t 1 1
− ⊂ − { − − }
Let A Z be the image of A . The above inclusion shows that (A A )+ 0, 1
2 q 1 2 2
⊂ − { − }
contains the image of π (A A), hence
t
−
A A π (A A) /2 > q/6.
2 2 t
| − | ≥ | − |
Similarly kA is contained in the image of π (kA)+ 0,1,...,k 1 , hence
2 t
{ − }
kA k kA = kH (q).
2 k
| | ≤ | |
By Lemma 5.2, applied to the set A , there are sets B ,B Z such that (A A )+
2 1 2 q 2 2
B + B = Z and B < c√logq. Our set will be A = A⊂ + (B B ). T−his set
1 2 q i 3 2 1 2
satisfies A A = Z| a|nd ∪ −
3 3 q
−
kA kA k(B B ) < (clogq)k/2 kA k(clogq)k/2H (q).
3 2 1 2 2 k
| | ≤ | || ∪− | | | ≤
(cid:3)
8 IMREZ. RUZSA
6. The other side
In this section we consider the question about the maximal possible size of kA com-
pared to A A. Most results and proofs are completely analogous, and we shall not
−
give details.
For positive integers k and q, q > 1 write
f (q) = min A A : A Z ,kA = Z ,
k q q
{| − | ⊂ }
g (q) = min A A : A Z,kA a+1,...,a+q for some a ,
k
| − | ⊂ ⊃ { }
h (q) = min A A : A Z, kA q .
(cid:8) k (cid:9)
{| − | ⊂ | | ≥ }
Put
logg (q)
k
β = inf .
k
q≥2 logq
Theorem 6.1.
logf (q) logg (q) logh (q)
k k k
lim = lim = lim
logq logq logq
logh (q)
k
= inf = β .
k
q≥2 logq
Again, I cannot decide whether
logf (q)
k
inf = β .
k
q≥2 logq
The proof of this result proceeds through analogues of Lemmas 3.1, 3.4, 3.6, 3.7.
Lemma 6.2. (Monotonicity.) If q < q′, then
g (q) g (q′),
k k
≤
h (q) h (q′).
k k
≤
Problem 6.3. Is f monotonically increasing?
k
Conjecture 6.4. No. Probably it depends on the multiplicative structure of q, not just
its size.
Lemma 6.5. (Submultiplicativity.) Let q = q q . We have
1 2
(6.1) f (q) f (q )f (q ) if gcd(q ,q ) = 1,
k k 1 k 2 1 2
≤
(6.2) g (q) g (q )g (q ) always,
k k 1 k 2
≤
(6.3) h (q) h (q )h (q ) always.
k k 1 k 2
≤
Lemma 6.6. For all q we have
(6.4) f (q) g (q),
k k
≤
(6.5) h (q) g (q),
k k
≤
(6.6) g (q) g (2q +1) 4f (q).
k k k
≤ ≤
Lemma 6.7.
(6.7) f (q) c (logq)2/kh (q).
k k k
≤
MORE DIFFERENCES THAN MULTIPLE SUMS 9
Problem 6.8. Is f (q) c h (q)? Is h (q) f (q)?
k k k k k
≤ ≤
Theorem 6.9. (a):
2 1 2
β
k − k2 ≤ k ≤ k
for all k.
(b): kβ is increasing.
k
Proof. For the lower estimate we show that
kA < A A k2/(2k−1)
| | | − |
for every finite set in any commutative group. Write A = n, A A = tn. By a
| | | − |
Plu¨nnecke-type inequality (see e.g. [8],[10],[7],[14],[13]) we get
(6.8) kA tkn,
| | ≤
and obviously
(6.9) kA < nk.
| |
By multiplying the k’th power of (6.8) and (k 1)’th power of (6.9) and taking k2’th
−
root we get the desired bound.
For the upper estimate take a generic set without any coincidence among the k-fold
sums.
Claim (b) follows from the fact that kA 1/k is a decreasing function of k, see [2]. (cid:3)
| |
Claim (b) above leaves two possibilities: either always β < 2/k, or β = 2/k after a
k k
point.
Problem 6.10. Is always β < 2/k?
k
Conjecture 6.11. Yes.
As far as I know, the only known case is k = 2. I think the case k = 4 is particularly
interesting:
Problem 6.12. Is always 4A A A 2?
| | ≤ | − |
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10 IMREZ. RUZSA
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Alfr´ed R´enyi Institute of Mathematics, Budapest, Pf. 127, H-1364 Hungary
E-mail address: [email protected]
