116 1. Higher order Fourier analysis
where B is the Bohr set
B := {n Z/NZ : |e(nξ/N) 1| ε for all ξ S}.
Clearly, if f is non-negative, then g is also. Now we look at upper bounds
on g. Clearly,
g(n) Em1,m2∈Bν(n + m1 m2)
so by Fourier expansion
g
L∞(Z/NZ)

ξ∈Z/NZ
|Em∈Be(ξB)|2|ˆ(ξ)|.
ν
Let us make the Fourier-pseudorandomness assumption
(1.55) sup
ξ=0
|ˆ(ξ)| ν = oN→∞(1).
Evaluating the ξ = 0 term on the RHS separately, we conclude that
g
L∞(Z/NZ)
1 + oN→∞(
ξ∈Z/NZ
|Em∈Be(ξB)|2).
By Plancherel’s theorem we have
ξ∈Z/NZ
|Em∈Be(ξB)|2
= |B|/N.
From the Kronecker approximation theorem we have
|B|/N
(ε/10)|S|
(say). Finally, if we assume (1.54) we have |S|
ε−q.
Putting this all
together we obtain the pointwise bound
g 1 + oN→∞;q,ε(1).
Finally, we see how g approximates f. From Fourier analysis one has
ˆ(ξ) g =
ˆ(ξ)|Em∈Be(ξB)|2
f
and so
f g
u2(Z/NZ)
= sup
ξ∈Z/NZ
|
ˆ(ξ)|(1
f
|Em∈Be(ξB)|2).
The frequencies ξ that lie outside ξ give a contribution of at most ε by the
definition of S, so now we look at the terms where ξ S. From the definition
of B and the triangle inequality we have
|Em∈Be(ξB) 1| ε
in such cases, while from the measure nature of ν we have
|
ˆ(ξ)|
f En∈Z/NZν(n) = 1 + oN→∞(1).
Putting this all together, we obtain
f g
u2(Z/NZ)
ε + oN→∞(1).
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