Roth’s theorem
on 3-arithmetic progressions
CIMPA Research school
Shillong 2013
Anne de Roton
Institut Elie Cartan
Université de Lorraine
France
2
1. Introduction.
In 1953, K. Roth [6] proved the following theorem.
Theorem 1.1 (Roth (1953)). Let A be a set of natural integers with positive density. Then
A contains non trivial arithmetic progressions of length 3 (3-arithmetic progressions ).
Actually, Roth proved a stronger version of this theorem. He proved that there exist a
positive integer N0 and a positive constant c such that for any N ≥ N0 , any subset A of
[1, N ] with density larger than c/ log log N does contain a non trivial 3-arithmetic progression .
We shall explain the proof of this quantitative result and explain how this result have been
sharpened and generalized in various ways.
Definition 1.1. We call a three arithmetic progression {n, n + r, n + 2r} trivial if r = 0.
Of course any non empty set does contain a trivial 3-arithmetic progression. Thus we
have to exclude these trivial progressions to get a meaningful result.
The main ideas of the proof of Roth’s theorem are the following.
(1) Given a subset P of integers less than N of the form
P = {a + nb, n ∈ {0, 1, · · · , N1 − 1}} with a, b ∈ N
(2)
(3)
(4)
(5)
(P is called an arithmetic progression of length N1 ), any non trivial 3-arithmetic
progression in {n ∈ {0, 1, · · · , N1 −1} : a+nb ∈ A} is also a non trivial 3-arithmetic
progression in A. Thus having no non trivial 3-arithmetic progression in A implies
that we do not have any in any P ∩ A with P an arithmetic progression.
One would expect, for a ”random” set A ⊂ [1, N ] of cardinality δN , around δ 3 N 2
3-arithmetic progressions , whereas we only have δN trivial 3-arithmetic progressions
in A.
A non ”random” set has to be structured in some sense. More precisely, Roth
proves that a set with much less 3-arithmetic progressions than expected has to
concentrate on some large arithmetic progression P in A. To prove this part, Roth
uses the discrete Fourier transform.
If P is an arithmetic progression in A of length N1 on which A concentrates, the
subset A ∩ P is isomorphic to some set A1 ⊂ [1, N1 ] of cardinality δ1 N1 with δ1 >
δ. Furthermore, if A does non contain any non trivial 3-arithmetic progression ,
neither does A1 .
An iteration of the argument finishes the proof. If A is a subset of [1, N ] of cardinality δN with no nontrivial 3-arithmetic progressions and if δ and N are large
enough, then Roth constructs a sequence of subsets Ak ⊂ [1, Nk ] of integers with
cardinality δk Nk such that the sequence (Nk )k does not decrease too fast and the
sequence (δk )k increases much enough so that δk > 1 before Nk reaches 0. We thus
get an absurdity.
Remark 1.2. The first point is essential in the argument. Actually, all the argument
would work for any equation of the form ax + by = cz with a + b = c. These equations
3
are called invariant equations (invariant by translation and multiplication) according to
Rusza’s terminology [7] and [8].
2. Tools and notations
Given a positive integer N , we write ZN to denote Z/N Z. Given a subset of integers A,
we write AN for the set A ∩ [1, N ] and if A is finite, we write |A| for the number of elements
in A. Subsets of integers less than N will often be identified with the corresponding subsets
of ZN .
Let A ⊂ ZN such that |A| = δN where |A| denotes the number of elements in A and δ is
a positive real number. We denote by 1A the indicator function of A. Furthermore let fA
be the balanced function
(
1 − δ if n ∈ A,
fA (n) =
−δ
if n 6∈ A.
This name is motivated by the fact that
X
fA (n) = (1 − δ) |A| − δ(N − |A|) = (1 − δ)δN − δ(1 − δ)N = 0.
n≤N
Definition 2.1. Given f : ZN → C a function, we define the Fourier transform fˆ : ZN → C
of f by
1 X
kn
ˆ
f (k) =
f (n)e −
N n∈Z
N
N
2iπθ
where e(θ) denotes e .
We also define, for p ≥ 1 the Lp norm of f by
kf kp =
1 X
|f (n)|p
N n∈Z
!1/p
.
N
c
c
c
Note that fc
A (0) = 0 since 1A (0) = |A|/N whereas fA (k) = 1A (k) for k 6= 0.
The Fourier inversion formula gives
X
kn
b
f (n) =
f (k)e
.
N
k∈Z
N
Furthermore we note Parseval’s identities:
X
1 X
fˆ(k)ĝ(k) =
f (n)g(n) and
N
n∈Z
k∈Z
N
N
2
X 1 X
|f (n)|2 ,
fb(k) =
N
r∈Z
n∈Z
N
and Hölder’s inequality:
1 X
1
1
f (n)g(n) ≤ kf kp kgkp0 with + 0 = 1.
N n∈Z
p p
N
N
4
We shall also use a function counting weighted arithmetic progressions. Given f1 , f2 , f3 :
ZN → C, we define
X
Λ3 (f1 , f2 , f3 ) =
f1 (n)f2 (n + r)f3 (n + 2r).
n,r∈ZN
In particular, if 1A is the indicator function of a set A ⊂ ZN , then N 2 Λ3 (1A , 1A , 1A ) counts
the number of 3-arithmetic progressions in A (as a subset of ZN ).
One key formula in Roth’s argument is the following:
X
(2.1)
Λ3 (f1 , f2 , f3 ) =
fb1 (k)fb2 (−2k)fb3 (k).
k∈ZN
To prove this formula, we use the definition of the Fourier transform and get
3 X
X
X
1
k(n1 − 2n2 + n3 )
b
b
b
f1 (k)f2 (−2k)f3 (k) =
f1 (n1 )f2 (n2 )f3 (n3 )e −
N
N
k∈ZN n1 ,n2 ,n3 ∈ZN
k∈ZN
2
X
k(n1 − 2n2 + n3 )
1
1 X
e −
=
f1 (n1 )f2 (n2 )f3 (n3 )
N
N k∈Z
N
n1 ,n2 ,n3 ∈ZN
N
2
X
1
f1 (n1 )f2 (n2 )f3 (n3 )
=
N
n ,n ,n ∈Z
1
2
3
N
n1 +n3 =2n2
where in the last equality we used
nk
1 X
e
=
N k∈Z
N
(
1
0
if n = 0,
otherwise.
N
Finally, a change of variables gives
2
X
1
f1 (n1 )f2 (n2 )f3 (n3 ) = Λ3 (n1 , n2 , n3 ).
N
n ,n ,n ∈Z
1
2
3
N
n1 +n3 =2n2
The formula (2.1) will be the link between the lack of 3-arithmetic progressions in A and
the existence of a large Fourier coefficient of fA .
Notation 2.1. Given a real number x, we shall write kxk for the distance between x and
the nearest integer.
3. Proof of Roth’s theorem
Let N be a sufficient large prime integer and A be a subset of [1, N ].
The proof of Roth’s theorem will use the following steps:
(1) If there are not so many 3-arithmetic progressions in A, then there exists k such that
c fA (k) is big.
5
(2) If fc
(k)
is big then A concentrates on a large arithmetic progression.
A
(3) By iteration we will eventually reach an arithmetic progression P with relative density
of A in P strictly larger than 1, a contradiction, if A is large enough.
We may consider A as a subset of ZN but Λ3 (1A , 1A , 1A ) will count the number of
3-arithmetic progressions in A modulo N and this way, we may add some arithmetic
progressions. For instance, (3, 0, 4) is a 3-arithmetic progression in Z7 but not in N.
To prevent us from counting 3-arithmetic progressions in ZN which are not genuine 3
arithmetic progressions in N, we have two choices :
• either we choose f1 = 1A , f2 = f3 = 1B with B = {x ∈ A : N/3 < x < 2N/3},
which is close to the original argument of Roth;
• or we choose f2 = 1A and f1 = f3 = 1B with B the set of even or odd numbers
in A whichever is larger, which is the argument used in Gowers’ proof of Roth’s
theorem.
In the first case, if x ∈ A and y, z ∈ B, then 2y − N < N/3 < x + z and 2y + N >
7N/3 > 5N/3 > x + z thus a 3-arithmetic progression modulo N in A × B × B is a genuine
3-arithmetic progression in A. Furthermore, in this case, we have either that |B| ≥ |A|/4
or one of the sets A ∩ [1, N/3] and A ∩ [2N/3, N ] has cardinality at least 3|A|/8 which
means that the relative density of A on this set is at least 89 |A|/N , thus that we found an
arithmetic progression of length N1 = N/3 on which the relative density of A is 9/8 times
the density of A in [1, N ].
In the second case by parity, we also have that a 3-arithmetic progression modulo N is a
genuine 3-arithmetic progression provided N is odd. In this case we have |B| ≥ |A|/2.
If A is a subset of N of upper density strictly larger than 2δ which does not contain any
3-arithmetic progresion, we choose for B the set of odd or even integers in A whichever
is larger as in the second case and, for simplicity, f1 = f2 = f3 = 1BN =: f . We have for
infinitely many N :
1 X
f (n) ≥ δ, #{3-arithmetic progressions in AN } ≥ N 2 Λ3 (f, f, f )
N n∈Z
N
and
Λ3 (f, f, f ) =
X
k∈ZN
|BN |
fb(k)2 fb(−2k) =
.
N2
Since fb(0)3 = (|BN |/N )3 is much greater than |BN |/N 2 , it means that the function f must
have large non-zero Fourier coefficients.
Remark 3.1. Note that N 2 fb(0)3 measures the expected number of 3-arithmetic progressions in B since we have N 2 triples (x, y, z) in 3-arithmetic progression modulo N and the
3
probability that (x, y, z) ∈ BN
is (|BN |/N )3 .
6
3.1. Few 3APs implies big Fourier coefficient. Let’s make the statement ”the function
f must have large non-zero Fourier coefficients” more precise. We have the following
proposition.
Proposition 3.2. Let N be P
a large odd integer, α be a positive real number and f : ZN → C
1
be a function satisfying N n∈ZN f (n) = α and Λ3 (f, f, f ) < α3 /2. Then there exist
k ∈ ZN \ {0} such that |fˆ(k)| ≥ α3 /56.
P
Proof. Suppose f satisfies N1 n∈ZN f (n) ≥ α and Λ3 (f, f, f ) < α3 /2.
Write f = f1 + f2 , with f1 = α1ZN . Then
X
α3
3
Λ3 (f1 , f1 , f1 ) ≥ α whereas Λ3 (f, f, f ) =
.
Λ3 (fi , fj , f` ) <
2
3
(i,j,`)∈{1,2}
We must therefore have for at least one triple (i, j, `) ∈ {1, 2}3 \ {(1, 1, 1)} satisfying
α3
.
14
P
Since Λ3 (fi , fj , f` ) = k∈ZN fbi (k)fbj (−2k)fb` (k) and at least one among i, j, ` must be equal
to 2, say ` = 2, we have by Cauchy inequality,
α3
sup |fb2 (k)|kfbi k2 kfbj k2 ≥
.
14
k
Λ3 (fi , fj , f` ) ≥
Since kf1 k2 ≤ N 1/2 and, using the fact that f is a bounded function, kf2 k2 ≤ kf k2 +kf1 k2 ≤
2N 1/2 , we obtain
α3
sup |fb2 (k)| ≥
.
56
k
Since fb2 (k) = fˆ(k) when k 6= 0, we have the announced result.
3.2. Big Fourier coefficient implies density increment.
Lemma 3.3. Let x be a real number. Then |e(x) − 1| ≤ 2πkxk where kxk denotes the
distance between x and the nearest integer.
Proof. The function (x 7→ e(x)) is 1-periodic so this is enough to prove the result for
x ∈ [0, 1]. If x ∈ [0, 1/2], we use
|e(x) − 1| = e2iπx − e2iπ0 ≤ 2π|x − 0| = 2πkxk
whereas for x ∈ [1/2, 1], we use
|e(x) − 1| = e2iπx − e2iπ ≤ 2π|x − 1| = 2πkxk.
Proposition 3.4. Let σPbe a positive real number, N be a large integer and g : ZN → R
be a function satisfying n∈ZN g(n) = 0 and kgk∞ ≤ 1. Assume that there exists k ∈ ZN
such that |b
g (k)| ≥ σ. Then there exists an arithmetic progression P of length at least
such that g has mean value at least σ/8 on P .
√
σ N
,
4π
7
Proof. Note that gb(0) = 0. Assume that for some k 6= 0, |b
g (k)| ≥ σN with σ > 0. Then
N
1 X
nk g(n)e
≥ σ.
N n=1
N √
By Dirichlet’s theorem there exist an integer h ∈ [1, N ] such that
hk ≤ √1 .
N N
√
We set M = d σ4πN e ≥ 1 and P = {h, 2h, . . . , M h}. Then |P | = M ≥ σN 1/2 /(4π) and
1 X
nk g(n)e
σ ≤ |b
g (k)| = N
N n∈ZN
1 X 1 X
k
g(n + x)e (n + x)
=
M
N
N
x∈P
n∈ZN
1 X 1 X
k =
g(n + x)e (n + x)
N
M x∈P
N n∈ZN
1 X 1 X
nk
k
−e
g(n + x) e (n + x)
≤
N
M
N
N
x∈P
n∈ZN
1 X 1 X
nk +
g(n + x)e
.
N
M
N n∈ZN
x∈P
Using Lemma 3.3, we get for x ∈ P
xk
e
≤ 2π xk −
1
N N
hk ≤ 2πM N (where x = lh with l ≤ M )
2πM
≤ √
N
σ
1
≤ +O √
.
2
N
8
Combining this we have that
1 X 1 X
1 X 1 X
nk nk
xk
g(n + x)e
g (k)| − g(n + x)e
e
−1 ≥ |b
N
N
M x∈P
N M x∈P
N
N
n∈ZN
n∈ZN
σ
1
≥ |b
g (k)| −
+O √
2
N
σ
1
≥ −O √
.
2
N
Now, we take a real number θ such that the sum on the left hand side of this inequality
has argument θ/(2π). Using that g is a balanced function and a real function, we get
1 X 1 X
nk nk
1 X 1 X
g(n + x)e
g(n + x)e
−θ
=
N
N
M
N
M
N
x∈P
x∈P
n∈ZN
n∈ZN
X
X
1
nk
1
g(n + x) e
−θ +1
=
N n∈Z M x∈P
N
N
!
1 X 1 X
nk
=<
−θ +1
g(n + x) e
N n∈Z M x∈P
N
N
X
1
nk
1 X
=
g(n + x)< e
−θ +1 .
N n∈Z M x∈P
N
N
√
This last quantity is larger than σ/2 − O(1/ N ), thus there exists n ∈ ZN such that
1 X
nk
σ
1
g(n + x)< e
−θ +1 ≥ −O √
.
M x∈P
N
2
N
−
θ
+
1
∈ [0, 2] and does not depend on x, we have
Since < e nk
N
1 X
σ
1
g(n + x) ≥ − O √
.
M x∈P
4
N
Therefore, on the arithmetic progression {n, n + h, · · · , n + M h} of length M ≥
has mean value at least σ/8 provided N is large enough.
√
σ N
,
4π
g
3.3. Iteration. Let A be a subset of N of upper density strictly larger than 2δ. We choose
for A0 = B the set of odd or even integers in A whichever is larger. For a given integer N ,
we identify BN with the corresponding subset of ZN and, for simplicity, write f = 1BN .
Then the number of 3-arithmetic progressions in AN as a subset of [1, N ] is larger than
Λ3 (f, f, f ). We have for infinitely many N ,
1 X
f (n) ≥ δ.
N n∈Z
N
9
P
Given such an integer N , we define α by α = N1 n∈ZN f (n). Then, according to Proposition 3.2, either we have Λ3 (f, f, f ) ≥ α3 /2 or there exist k ∈ ZN \ {0} such that
fˆ(k) ≥ α3 /56.
In the first case, A must contain some non trivial 3-arithmetic progression otherwise we
would have Λ3 (f, f, f ) ≤ |BN |/N 2 = α/N . In the second case, we write g = f − α1ZN
and apply Proposition 3.4 with σ = α3 /56. Thus we find an arithmetic progression P =
{n + lh, l ≤ √
N1 } in ZN on which the mean value of g is larger than σ/8. We write
N1 = |P | ≥ σ N /(4π) and A1 = {l ∈ [1, N1 ] : n + lh ∈ P ∩ A}. We get a subset A1 of
ZN1 with
δ3
σ
N1 ≥ δ +
N1 =: (δ + c2 δ 3 )N1
|A1 | ≥ α +
8
8 × 56
and
√
√
δ3 N
N1 ≥
=: c1 δ 3 N .
424π
We iterate the argument on A1 . We get that either
Λ3 (1A1 , 1A1 , 1A1 ) ≥
|A1 |2 (δ + c2 δ 3 )3
3 |A1 |
>
(δ
+
c
δ
) 2
2
2N 2
N
in which case we have some non trivial 3-arithmetic progressions in A1 , thus in A0 = B
and in A; or there exists a subset A2 of ZN2 such that
N2 ≥ c1 (δ + c2 δ 3 )3
p
N1 ≥ (c1 δ 3 )1+1/2 N 1/4
|A2 |
≥ δ + 2c2 δ 3 .
N2
and
We iterate the argument until the first case occurs. If at some point, the first case occurs,
we have some non trivial 3APs in A. If, after k iterations we still have no occurance of the
first case, then we get that there exists a subset Ak of ZNk such that
|Ak |
≥ δ + kc2 δ 3
Nk
and Nk ≥ c1 δ 3
1
(1+ 12 +···+ k−1
)
2
1
N 2k = c1 δ 3
2(1−2−k )
1
N 2k .
After k = dδ −3 /c2 e iterations, we get a relative density larger than 1 on a progression of
k
length larger than min(c1 δ 3 , 1)2 N 1/2 which is larger than 1 if N is large enough (depending
on δ, c1 and c2 ). This is a contradiction, thus the first case must occur and A does contain
k
k
3AP . The condition (c1 δ 3 )2(1−1/2 ) N 1/2 ≥ 1 with k ≥ 1/c2 δ 3 and some precise study of
admissible values for the constants give a quantitative version of Roth’s theorem.
Remark 3.5. Following the same argument, we can actually prove the following more general result:
+
Theorem 3.6. Let N be a large odd integer,
P δ be a positive real number and f : ZN → R
1
be a function bounded by M such that N n∈ZN f (n) ≥ δ. Then there exist some positive
constant c = c(δ, M ) such that Λ3 (f, f, f ) ≥ c(δ, M ).
10
4. Improvements and generalizations of Roth’s theorem
According to the remaining time, we shall spend more or less time on the following
issues.
4.1. Quantitative improvements. Since 1953, significative quantitative improvements
of Roth’s theorem have been made. If r3 (N ) denotes the maximum size of a subset of
positive integers less than N with no nontrivial 3-arithmetic progression , Roth proved
that
N
lim sup r3 (N ) ≤
.
log log N
N →∞
Much later, Heath-Brown [4] (1987) and Szemerédi [10] (1990) improved (independantly)
this result by showing that
r3 (N ) ≤ CN (log N )−c
for some small positive c and some large constant C. By considering Bohr sets where
previous arguments had used arithmetic progressions, Bourgain obtained
p
r3 (N ) ≤ CN log log N /log N
in [1] (1999) and r3 (N ) ≤ CN (log log N )2 (log N )−2/3 in [2] (2008).
Very recently, Sanders obtained the best known result so far by proving that
r3 (N ) ≤ CN
(log log N )5
.
log N
We shall briefly explain some of the ideas which led to some of these improvements. We
shall mainly talk about Bohr sets which were considered by Bourgain and replaced large
arithmetic progressions in his argument.
4.2. Roth’s theorem in other sets. Roth’s argument not only prove that a subset of
integers less than N with not too small upper density does contain some non trivial 3arithmetic progressions. It also leads to the fact that such a set must contain many non
trivial 3-arithmetic progressions. We shall explain how Varnavides [11] proved such a result
and how this result is used to prove Roth’s theorem in the primes, for instance [3], [5].
We shall also briefly give the corresponding results in a continuous setting.
5. Large sets with no 3-arithmetic progressions
In this chapter, we shall briefly explain the constructions of large sets of integers less than
N with no non trivial 3-arithmetic progression . We shall talk about Salem and Spencer’s,
Behrend’s and Elkin’s constructions.
11
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
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Bourgain, J. “Roth’s theorem on progressions revisited.” J. Anal. Math. 104 (2008): 155-192.
Green, B. “Roth’s theorem in the primes.” Ann. of Math. (2) 161 (2005): 1609-1636.
Heath-Brown, D. R. “Integer sets containing no arithmetic progressions.” J. London Math. Soc. (2)
35, no. 3 (1987): 385-394.
Helfgott, H. A., de Roton A. ”Improving Roth’s theorem in the primes”, Int. Math. Res. Notices 4
(2011), pp. 767-783.
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T. Sanders, On Roth’s theorem on progressions, Ann. of Math. (2), to appear, arXiv:1011.0104.
Szemerédi, E. “Integer sets containing no arithmetic progressions.” Acta Math. Hungar. 56, no. 1–2
(1990): 155-158.
Varnavides, P. “On certain sets of positive density.” J. London Math. Soc. 34 (1959): 358-360.