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
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 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 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])
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.
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