Subsets of Fpn without three term arithmetic progressions have several large Fourier coefficients

Suppose that f : Fpn → [0, 1] satisfies Σaf(m) = θF ∈ [F , F ], where F = |Fpn | = p. In this paper we will show the following: Let fj denote the size of the jth largest Fourier coefficient of f . If fj < θ j1/2+δF, for some integer j satisfying J0(δ, p) < j < F , then S = support(f) contains a non-trivial three-term arithmetic progression. Thus, the result is asserting that if the Fourier transform decays rapidly enough (though not all that rapdily – in particular, not quite exponentially fast), then S is forced to have a three-term arithmetic progression. This result is similar in spirit to that appearing in [1]; however, in that paper the focus was on the “small” Fourier coefficients, whereas here the focus is on the “large” Fourier coefficients (furthermore, the proof in the present paper requires much more sophisticated arguments than those of that other paper). Here is a partial description of how this result was proved: First, to get our proof started, we reduced to the case where θ, δ and j satisfied certain nice constraints. For example, Meshulam’s theorem [2] was used to reduce to the case θ < 1/p; then, we reduced to the case δ < 2/3 by using a “dimension collapsing and cutting argument” from [1] which says that for special “smooth” functions f , the underlying set S must always be rich in three-term arithmetic progressions; and finally, this same argument (dimension collapsing) was used to reduce to the case where j < n. The rest of the proof used a type of RothMeshulam [2] iteration. Unfortunately, because our main theorem is