Roth s theorem on 3-arithmetic progressions. CIMPA Research school Shillong 2013

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1 Roth s theorem on 3-arithmetic progressions CIPA Research school Shillong 2013 Anne de Roton Institut Elie Cartan Université de Lorraine France

2 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 0 and a positive constant c such that for any 0, any subset A of [1, ] with density larger than c/ log log 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 of the form P = {a + nb, n {0, 1,, 1 1}} with a, b P is called an arithmetic progression of length 1 ), any non trivial 3-arithmetic progression in {n {0, 1,, 1 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. 2) One would expect, for a random set A [1, ] of cardinality δ, around δ arithmetic progressions, whereas we only have δ trivial 3-arithmetic progressions in A. 3) A non random set has to be structured in some sense. ore 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. 4) If P is an arithmetic progression in A of length 1 on which A concentrates, the subset A P is isomorphic to some set A 1 [1, 1 ] of cardinality δ 1 1 with δ 1 > δ. Furthermore, if A does non contain any non trivial 3-arithmetic progression, neither does A 1. 5) An iteration of the argument finishes the proof. If A is a subset of [1, ] of cardinality δ with no nontrivial 3-arithmetic progressions and if δ and are large enough, then Roth constructs a sequence of subsets A k [1, k ] of integers with cardinality δ k k such that the sequence k ) k does not decrease too fast and the sequence δ k ) k increases much enough so that δ k > 1 before k 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, we write Z to denote Z/Z. Given a subset of integers A, we write A for the set A [1, ] and if A is finite, we write A for the number of elements in A. Subsets of integers less than will often be identified with the corresponding subsets of Z. Let A Z such that A = δ where A denotes the number of elements in A and δ is a positive real number. We denote by 1 A the indicator function of A. Furthermore let f A be the balanced function { 1 δ if n A, f A n) = δ if n A. This name is motivated by the fact that f A n) = 1 δ) A δ A ) = 1 δ)δ δ1 δ) = 0. n Definition 2.1. Given f : Z C a function, we define the Fourier transform ˆf : Z C of f by ˆfk) = 1 fn)e kn ) n Z where eθ) denotes e 2iπθ. We also define, for p 1 the L p norm of f by ) 1/p 1 f p = fn) p. n Z ote that f A 0) = 0 since 1 A 0) = A / whereas f A k) = 1 A k) for k 0. The Fourier inversion formula gives fn) = ) kn fk)e. k Z Furthermore we note Parseval s identities: 1 ˆfk)ĝk) = fn)gn) and k Z n Z fk) 2 = 1 fn) 2, n Z r Z and Hölder s inequality: 1 fn)gn) f p g p with 1 p + 1 = 1. p n Z 3

4 4 We shall also use a function counting weighted arithmetic progressions. Given f 1, f 2, f 3 : Z C, we define Λ 3 f 1, f 2, f 3 ) = f 1 n)f 2 n + r)f 3 n + 2r). n,r Z In particular, if 1 A is the indicator function of a set A Z, then 2 Λ 3 1 A, 1 A, 1 A ) counts the number of 3-arithmetic progressions in A as a subset of Z ). One key formula in Roth s argument is the following: 2.1) Λ 3 f 1, f 2, f 3 ) = f1 k) f 2 2k) f 3 k). k Z To prove this formula, we use the definition of the Fourier transform and get ) 3 f1 k) f 2 2k) f 1 3 k) = f 1 n 1 )f 2 n 2 )f 3 n 3 )e k Z k Z n 1,n 2,n 3 Z ) 2 1 = f 1 n 1 )f 2 n 2 )f 3 n 3 ) 1 e n 1,n 2,n 3 Z k Z ) 2 1 = where in the last equality we used 1 k Z e f 1 n 1 )f 2 n 2 )f 3 n 3 ) n 1,n 2,n 3 Z n 1 +n 3 =2n 2 ) { nk 1 if n = 0, = 0 otherwise. Finally, a change of variables gives ) 2 1 f 1 n 1 )f 2 n 2 )f 3 n 3 ) = Λ 3 n 1, n 2, n 3 ). n 1,n 2,n 3 Z n 1 +n 3 =2n 2 kn 1 2n 2 + n 3 ) ) kn 1 2n 2 + n 3 ) 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 f A. otation 2.1. Given a real number x, we shall write x for the distance between x and the nearest integer. 3. Proof of Roth s theorem Let be a sufficient large prime integer and A be a subset of [1, ]. 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 f A k) is big. )

5 2) If f A k) is big then A concentrates on a large arithmetic progression. 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 Z but Λ 3 1 A, 1 A, 1 A ) will count the number of 3-arithmetic progressions in A modulo and this way, we may add some arithmetic progressions. For instance, 3, 0, 4) is a 3-arithmetic progression in Z 7 but not in. To prevent us from counting 3-arithmetic progressions in Z which are not genuine 3 arithmetic progressions in, we have two choices : either we choose f 1 = 1 A, f 2 = f 3 = 1 B with B = {x A : /3 < x < 2/3}, which is close to the original argument of Roth; or we choose f 2 = 1 A and f 1 = f 3 = 1 B 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 < /3 < x + z and 2y + > 7/3 > 5/3 > x+z thus a 3-arithmetic progression modulo 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, /3] and A [2/3, ] has cardinality at least 3 A /8 which means that the relative density of A on this set is at least 9 A /, thus that we found an 8 arithmetic progression of length 1 = /3 on which the relative density of A is 9/8 times the density of A in [1, ]. In the second case by parity, we also have that a 3-arithmetic progression modulo is a genuine 3-arithmetic progression provided is odd. In this case we have B A /2. 5 If A is a subset of 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, f 1 = f 2 = f 3 = 1 B =: f. We have for infinitely many : and 1 n Z fn) δ, #{3-arithmetic progressions in A } 2 Λ 3 f, f, f) Λ 3 f, f, f) = 2 B fk) f 2k) =. 2 k Z Since f0) 3 = B /) 3 is much greater than B / 2, it means that the function f must have large non-zero Fourier coefficients. Remark 3.1. ote that 2 f0) 3 measures the expected number of 3-arithmetic progressions in B since we have 2 triples x, y, z) in 3-arithmetic progression modulo and the probability that x, y, z) B 3 is B /) 3.

6 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 be a large odd integer, α be a positive real number and f : Z C be a function satisfying 1 n Z fn) = α and Λ 3 f, f, f) < α 3 /2. Then there exist k Z \ {0} such that ˆfk) α 3 /56. Proof. Suppose f satisfies 1 n Z fn) α and Λ 3 f, f, f) < α 3 /2. Write f = f 1 + f 2, with f 1 = α1 Z. Then Λ 3 f 1, f 1, f 1 ) α 3 whereas Λ 3 f, f, f) = Λ 3 f i, f j, f l ) < α3 2. i,j,l) {1,2} 3 We must therefore have for at least one triple i, j, l) {1, 2} 3 \ {1, 1, 1)} satisfying Λ 3 f i, f j, f l ) α3 14. Since Λ 3 f i, f j, f l ) = k Z fi k) f j 2k) f l k) and at least one among i, j, l must be equal to 2, say l = 2, we have by Cauchy inequality, sup f 2 k) f i 2 f j 2 α3 k 14. Since f 1 2 1/2 and, using the fact that f is a bounded function, f 2 2 f 2 + f /2, we obtain sup f 2 k) α3 k 56. Since f 2 k) = ˆfk) when k 0, we have the announced result Big Fourier coefficient implies density increment. Lemma 3.3. Let x be a real number. Then ex) 1 2π x where x denotes the distance between x and the nearest integer. Proof. The function x ex)) is 1-periodic so this is enough to prove the result for x [0, 1]. If x [0, 1/2], we use whereas for x [1/2, 1], we use ex) 1 = e 2iπx e 2iπ0 2π x 0 = 2π x ex) 1 = e 2iπx e 2iπ 2π x 1 = 2π x. Proposition 3.4. Let σ be a positive real number, be a large integer and g : Z R be a function satisfying n Z gn) = 0 and g 1. Assume that there exists k Z such that ĝk) σ. Then there exists an arithmetic progression P of length at least σ, 4π such that g has mean value at least σ/8 on P.

7 7 Proof. ote that ĝ0) = 0. Assume that for some k 0, ĝk) σ with σ > 0. Then 1 ) nk gn)e σ. n=1 By Dirichlet s theorem there exist an integer h [1, ] such that hk 1. We set = σ 1 and P = {h, 2h,..., h}. Then P = σ 1/2 /4π) and 4π 1 ) nk σ ĝk) = gn)e n Z 1 1 = gn + x)e x P n Z 1 1 = gn + x)e n Z x P 1 1 n Z x P gn + x)e n Z x P Using Lemma 3.3, we get for x P n + x) k ) n + x) k ) gn + x) e n + x) k ) e ) nk. )) nk ) xk e 1 2π xk 2π hk where x = lh with l ) 2π σ ) O.

8 8 Combining this we have that 1 1 nk gn + x)e n Z x P ) ĝk) ĝk) 1 σ 2 + O n Z 1 σ ) 1 2 O. gn + x)e x P )) 1 ) nk e ) xk 1) ow, 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 1 ) nk gn + x)e = 1 1 ) nk gn + x)e θ n Z x P n Z x P = 1 1 ) ) nk gn + x) e θ + 1 n Z x P 1 1 ) ) ) nk = R gn + x) e θ + 1 n Z x P = 1 1 ) ) nk gn + x)r e θ + 1. n Z x P This last quantity is larger than σ/2 O1/ ), thus there exists n Z such that 1 ) ) nk gn + x)r e θ + 1 σ ) 1 2 O. x P Since R e nk θ) + 1 ) [0, 2] and does not depend on x, we have 1 gn + x) σ ) 1 4 O. x P Therefore, on the arithmetic progression {n, n + h,, n + h} of length σ, g 4π has mean value at least σ/8 provided is large enough Iteration. Let A be a subset of of upper density strictly larger than 2δ. We choose for A 0 = B the set of odd or even integers in A whichever is larger. For a given integer, we identify B with the corresponding subset of Z and, for simplicity, write f = 1 B. Then the number of 3-arithmetic progressions in A as a subset of [1, ] is larger than Λ 3 f, f, f). We have for infinitely many, 1 n Z fn) δ.

9 Given such an integer, we define α by α = 1 n Z fn). Then, according to Proposition 3.2, either we have Λ 3 f, f, f) α 3 /2 or there exist k Z \ {0} such that ˆfk) α 3 /56. In the first case, A must contain some non trivial 3-arithmetic progression otherwise we would have Λ 3 f, f, f) B / 2 = α/. In the second case, we write g = f α1 Z and apply Proposition 3.4 with σ = α 3 /56. Thus we find an arithmetic progression P = {n + lh, l 1 } in Z on which the mean value of g is larger than σ/8. We write 1 = P σ /4π) and A 1 = {l [1, 1 ] : n + lh P A}. We get a subset A 1 of Z 1 and with A 1 α + σ ) ) 1 δ + δ3 1 =: δ + c 2 δ 3 ) δ3 424π =: c 1δ 3. We iterate the argument on A 1. We get that either Λ 3 1 A1, 1 A1, 1 A1 ) A 1 2 δ + c 2 δ 3 ) 3 > δ + c δ 3 ) A 1 2 in which case we have some non trivial 3-arithmetic progressions in A 1, thus in A 0 = B and in A; or there exists a subset A 2 of Z 2 such that 2 c 1 δ + c 2 δ 3 ) 3 1 c 1 δ 3 ) 1+1/2 1/4 and A 2 2 δ + 2c 2 δ 3. 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 A k of Z k such that A k δ + kc 2 δ 3 and k c 1 δ 3) ) 2 k k = c 1 δ 3) 21 2 k ) 1 k After k = δ 3 /c 2 iterations, we get a relative density larger than 1 on a progression of length larger than minc 1 δ 3, 1) 2 1/2k which is larger than 1 if is large enough depending on δ, c 1 and c 2 ). This is a contradiction, thus the first case must occur and A does contain 3AP. The condition c 1 δ 3 ) 21 1/2k) 1/2k 1 with k 1/c 2 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 be a large odd integer, δ be a positive real number and f : Z R + be a function bounded by such that 1 n Z fn) δ. Then there exist some positive constant c = cδ, ) such that Λ 3 f, f, f) cδ, ). 2 k. 9

10 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 Quantitative improvements. Since 1953, significative quantitative improvements of Roth s theorem have been made. If r 3 ) denotes the maximum size of a subset of positive integers less than with no nontrivial 3-arithmetic progression, Roth proved that lim sup r 3 ) log log. uch later, Heath-Brown [4] 1987) and Szemerédi [10] 1990) improved independantly) this result by showing that r 3 ) Clog ) c for some small positive c and some large constant C. By considering Bohr sets where previous arguments had used arithmetic progressions, Bourgain obtained r 3 ) C log log /log in [1] 1999) and r 3 ) Clog log ) 2 log ) 2/3 in [2] 2008). Very recently, Sanders obtained the best known result so far by proving that log log )5 r 3 ) C log. 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 Roth s theorem in other sets. Roth s argument not only prove that a subset of integers less than with not too small upper density does contain some non trivial 3- arithmetic 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 with no non trivial 3-arithmetic progression. We shall talk about Salem and Spencer s, Behrend s and Elkin s constructions.

11 References [1] Bourgain, J. On triples in arithmetic progression. Geom. Funct. Anal. 9, no ): [2] Bourgain, J. Roth s theorem on progressions revisited. J. Anal. ath ): [3] Green, B. Roth s theorem in the primes. Ann. of ath. 2) ): [4] Heath-Brown, D. R. Integer sets containing no arithmetic progressions. J. London ath. Soc. 2) 35, no ): [5] Helfgott, H. A., de Roton A. Improving Roth s theorem in the primes, Int. ath. Res. otices ), pp [6] Roth, K. F. On certain sets of integers. J. London ath. Soc ): [7] Ruzsa, I. Z. Solving a linear equation in a set of integers. I. Acta Arith., 653), 259? ). [8] Ruzsa, I. Z. Solving a linear equation in a set of integers. II. Acta Arith., 724), 385? ). [9] T. Sanders, On Roth s theorem on progressions, Ann. of ath. 2), to appear, arxiv: [10] Szemerédi, E. Integer sets containing no arithmetic progressions. Acta ath. Hungar. 56, no ): [11] Varnavides, P. On certain sets of positive density. J. London ath. Soc ):