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 nontrivial,
and from Fourier analysis we conclude that the expectation in (1.9) vanishes.
We conclude that the lefthand side of (1.8) can be expressed as
α∈
1
N
Z/Z
ˆ(α)ˆ(−2α)ˆ(α).
f g h
Now using Plancherel’s theorem we have
α∈
1
N
Z/Z

ˆ(α)2
f = f
2
L2(Z/N
Z)
(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
L2(Z/N Z)
g
L2(Z/N Z)
sup
ξ∈Z/N Z
ˆ(ξ)
h
and similarly for permutations of f, g, h on the righthand 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])
δ3.
Comparing this with (1.8), we conclude that one of the error terms must
have magnitude
δ3
also. For sake of concreteness, let us say that
Λ(f, δ 1[N],f)
δ3;
the other cases are similar and are left to the reader.
From the triangle inequality we see that f, δ 1[N] have an
L2(Z/N
Z)
norm of
O(δ1/2),
and so from (1.10) one has
Λ(f, δ 1[N],f) δ sup
ξ∈Z/N Z

ˆ(ξ),
f
and so we conclude that

ˆ(ξ)
f
δ2
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
follows.