Roth’s Theorem: Tom Sanders Reaches the Logarithmic Barrier

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I missed Tom by a few minutes at Mittag-Leffler Institute a year and a half ago

Suppose that R_n is a subset of \{1,2,\dots, n \} of maximum cardinality not containing an arithmetic progression of length 3. Let g(n)=n/|R_n|.

Roth proved that g(n) \ge log logn. Szemeredi and Heath-Brown improved it to g(n) \ge log^cn for some 0″ src=”″ alt=”c>0″ /> (Szemeredi’s argument gave c=1/4.) Jean Bourgain improved the bound in 1999 to c=1/2 and in 2008 to c=2/3 (up to lower order terms).

Erdös and Turan who posed the problem in 1936 described a set not containing an arithmetic progression of size n^c.  Salem and Spencer improved this bound to g(n) \le e^{logn/ loglogn}. Behrend’s upper bound from 1946 is of the form g(n) \le e^{C\sqrt {\log n}}. A small improvement was achieved recently by Elkin and is discussed here.  (Look also at the remarks following that post.)

In an earlier post we asked: How does g(n) behave? Since we do not really know, will it help talking about it? Can we somehow look beyond the horizon and try to guess what the truth is? (I still don’t know if softly discussing this or other mathematical problems is a fruitful idea, but it can be enjoyable.)

We even had a poll collecting people’s predictions about g(n).  Somewhat surprisingly 18.18% of answerers predicted that g(n) behaves like (\log n)^c for some c<1. Be the answer as it may be, reaching  the logarithmic barrier was considered extremely difficult.

A couple of months ago Tom Sanders was able to refine Bourgain’s argument and proved that g(n) \ge (\log n)^{3/4}. Very recently Tom have managed to reach the logarithmic barrier and to prove that

g(n) \ge (\log n)/(\log \log n)^{5}.

Quoting from his paper: “There are two main new ingredients in the present work: the first is a way of transforming sumsets introduced by Nets Katz and Paul Koester in 2008, and the second is a result on the L_p-invariance of convolutions due to Ernie Croot and Olof Sisask (2010).”

This is a truly remarkable result.

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