A new proof of Roth's theorem on arithmetic progressions

Abstract

We present a proof of Roth's theorem that follows a slightly different structure to the usual proofs, in that there is not much iteration. Although our proof works using a type of density increment argument (which is typical of most proofs of Roth's theorem), we do not pass to a progression related to the large Fourier coefficients of our set (as most other proofs of Roth do). Furthermore, in our proof, the density increment is achieved through an application of a quantitative version of Varnavides's theorem, which is perhaps unexpected.