Alex Iosevich. September 17, 2003

Transcription

1 ROTH S THEOREM ON ARITHMETIC PROGRESSIONS Alex Iosevich September 7, 003 The purpose of this paper is to provide a simple and self-contained exposition of the celebrated Roth s theorem on arithmetic progressions of length three. The original result is proved in [Roth53], while the proof given below is very similar to the exposition of Roth s original argument given in [GRS990]. Definition. We say that a subset of positive integers A has positive upper density if () lim N sup A [, N] N > 0. Roth s Theorem. If A is a subset of positive integers of positive upper density, then A contains a three term arithmetic progression. Basic setup. Let S(n) denote the largest number of integers in [, n] that can be chosen so that no three term arithmetic progression is formed. Let S(n) () c = lim n n. The existence of this limit exists follows from the fact that S : Z + R is a sub-additive function. The (easy) details are left to the reader. Let (3) ɛ = c 0 6, and let m be large enough so that (4) c S(n) n < c + ɛ for m + n. Typeset by AMS-TEX

2 Let N be sufficiently large in the sense that will become clear below. Let A {,..., N} with A Nc which contains no arithmetic progressions of length three. Let (5) A = {u, u,..., u r }. Let A even denote the set of even elements of A, (6) A even = {v, v,..., v s }; A even = {v, v,..., v s }. By assumption, (7) Nc r N(c + ɛ), and (8) N(c ɛ) s N(c + ɛ). Fourier transform. Let (9) Â(α) = r e(αu i ), i= and (0) Â even (α) = where s e(αv j ), j= () e(t) = e πit. As we shall see below, the key to the proof of Roth s theorem is to show that away from α = 0, Â(α) is much smaller than N, the number of terms in the sum. The idea behind this is that the enemy of an exponential sum is presence of arithmetic progression. These progressions play the same role in the discrete Fourier transforms as that of lack of curvature in the decay properties of the Fourier transform of a surface carried measure. The basic idea. Let denote the sum over α = () e(αu) = N if u = 0, and 0 otherwise. i N, i = 0,,..., N. Observe that

3 The key observation is that (3) Â(α)(Â even ) ( α) = N#{(i, j, k) : u i v j v k = 0; u i A, v j, v k A even}. Why is this important? If u i v j v k = 0, then the set (4) {v j, u i, v k } is an arithmetic progression of length three, except for the trivial cases when (5) v j = u i = v k. By assumption our A does not contain arithmetic progressions of length three, therefore the right hand side of (3) equals (6) Ns < 3cN, where the inequality follows by (8). We shall obtain a contradiction in the following manner. First, (7) Â(0)(Â even ) (0) = rs > c 3 N 3. On the other hand, (8) Â(0)(Â even ) (0) < Â(α)(Â even ) ( α) + α 0 Â(α)(Â even ) ( α) = I + II. We already know that (9) I 3cN. We will see that (0) II 8ɛcN 3. This will be accomplished as follows. We will show that when α 0, () Â(α) 6ɛN. This is the main estimate of the paper which we shall carry out shortly. With () in tow, we complete the proof in the following way. We observe that ( ) () II Â(α) (Â even ( α)) 6ɛN (Â even ( α)), max α 0 3

4 so (0) would follow if we could show that (3) (Â even ( α)) 3cN. (4) To establish (3) observe that the left hand side of (3) equals s s j= k= by (8) and (). Using (7), (8), (9), and (0) we see that e(α(v j v k )) = Ns 3cN (5) c 3 N 3 3cN + 8ɛcN 3, which is not possible if N is sufficiently large. Thus the proof of Roth s theorem is reduced to establishing (). Estimate (). We need the following basic fact about diophantine approximation which can be found in any book on elementary number theory and/or deduced easily from the pigeon-hole principle. For α arbitrary and M > 0, there exist integers p, q with (6) α = p q + β, q M and q β M. (7) We also need the following elementary estimate which we do not prove... (e(x) + e( x)) = cos(x) x. The first step in establishing the estimate () is smearing. More precisely, we show that (8) Â(α) e(α(u + iq)) m + A m M Nm M. A We shall then estimate the smeared part by showing that (9) e(α(u + iq)) m + 5Nɛ. A Combining (8) and (9) yields (). 4

5 (30) We first establish (8), which is a bit easier. From (7) we deduce that m + [e(α + iγ) e(α)] (mγ) Let M denote the largest integer smaller than N. Let α 0 and let p, q, β be as in (6) above. Then (3) e(α(u + iq)) = e(αu + i(βq)). It follows that [e(α(u + iq)) e(αu)] m + = e(αu) m +. e(α(u + iq)) (3) (mβq) m M, and (8) follows instantly. We now turn our attention to (9). Let (33) W s = {s + iq : i m} calculated modulo N. It follows that (34) m + A e(α(u + iq)) = e(αs) W s A m +. (35) We want to show that e(αs) W s A m + 5Nɛ which will establish (9). From (35), (34) and (8) it will then follow that (36) Â(α) m N + 5Nɛ 6Nɛ M for a sufficiently large N which will establish (). 5

6 We now prove (35). Let (37) E s = W s A m + c. Observe that for mq s N mq, W s forms an arithmetic progression of length m+ in {,,..., N}. It follows that (38) W s A (m + )(c + ɛ) by (7). It follows that for these values of s, E s ɛ. For the other mq values of s we use the trivial bound E s. Our next observation is that the average value of E s is positive. Indeed, since each a A appears in exactly m + sets W s, we have (39) W s A = A m + m + m + = A, so (40) E s N = A N c, which is a positive number. Now, using (40) and the discussion following (38), (4) E s 0 s :E s >0 E s (Nɛ + mq) 4ɛN + 4mM 5Nɛ if N is sufficiently large. For α 0, e(αs) = 0, so e(αs) W s A m + = e(αs)e s (4) E s 5Nɛ. This establishes (35) and proof of Roth s theorem is complete. 6

7 References [GRS990] R. Graham, B. Rothschild, and J. Spencer, Ramsey theory, John Wiley and Sons, Inc. (990). [Roth53] K. Roth, On certain sets of integers, J. London Math Soc. 8 (953),