30 1. Higher order Fourier analysis
so the phase α1n + α2(n + r) + α3(n + 2r) is trivial whenever (α1,α2,α3)
is of the form (α, −2α, α), and so the expectation in (1.9) is equal to 1.
Conversely, if (α1,α2,α3) is not of this form, then the phase is non-trivial,
and from Fourier analysis we conclude that the expectation in (1.9) vanishes.
We conclude that the left-hand side of (1.8) can be expressed as
f g h
Now using Plancherel’s theorem we have
f = f
(using normalised counting measure). Using this and H¨ older’s inequality
(and the fact that N is odd), we obtain the bounds
(1.10) |Λ(f, g, h)| ≤ f
and similarly for permutations of f, g, h on the right-hand side.
We could apply this directly to Λ(1A, 1A, 1A), but this is not useful, since
we seek a lower bound on this quantity rather than an upper bound. To
get such a lower bound, we split 1A = δ 1[N] + f, where f := 1A − δ 1[N] is
the mean zero portion of 1A, and use trilinearity to split Λ(1A, 1A, 1A) into
a main term Λ(δ 1[N],δ 1[N],δ 1[N]), plus seven other error terms involving
1A = δ 1[N] and f, with each error term involving at least one copy of f.
The main term can be computed explicitly as
Λ(δ 1[N],δ 1[N],δ 1[N])
Comparing this with (1.8), we conclude that one of the error terms must
also. For sake of concreteness, let us say that
|Λ(f, δ 1[N],f)|
the other cases are similar and are left to the reader.
From the triangle inequality we see that f, δ 1[N] have an
and so from (1.10) one has
|Λ(f, δ 1[N],f)| δ sup
and so we conclude that
for some ξ ∈ Z/N Z. Similarly for other error terms, though sometimes
one will need a permutation of (1.10) instead of (1.10) itself. The claim