32 1. Higher order Fourier analysis
Since e(−αn) fluctuates by at most ε on Pj, we can apply the triangle
inequality and conclude that
En∈[N](1A(n) − δ )e(−αn) ≤
1
N
J
j=1 n∈Pj
(1A(n) − δ ) + O(ε).
If ε is suﬃciently small depending on δ, we conclude that
(1.11)
J
j=1

n∈Pj
(1A(n) − δ )
δ2N.
On the other hand, as δ is the mean of 1A on [N], we have
n∈[N]
(1A(n) − δ ) = 0
and thus
J
j=1 n∈Pj
(1A(n) − δ ) = O(ε).
Adding this to (1.11) and noting that x + x = 2 max(x, 0) for real x, we
conclude (for ε small enough) that
J
j=1
max(
n∈Pj
(1A(n) − δ ), 0)
δ2N
and hence by the pigeonhole principle we can find j such that
max(
n∈Pj
(1A(n) − δ ), 0)
δ2Pj
or, in other words,
A ∩ Pj/Pj ≥ δ +
cδ2
for some absolute constant c 0, and Proposition 1.2.3 follows.
It is possible to rewrite the above argument in the ultralimit setting,
though it only makes the argument slightly shorter as a consequence. We
sketch this alternate formulation below.
Exercise 1.2.2. Let δ∗ be as above.
(i) Show that if N is an unbounded limit natural number, and A ⊂ [N]
is a limit subset whose density st(A/N) is strictly greater than δ∗,
then A contains a (limit) arithmetic progression n, n + r, n + 2r of
length three (with r = 0).
(ii) Show that there exists an unbounded limit natural number N and
a limit subset A ⊂ [N] of density st(A/N) = δ∗, which does not
contain any arithmetic progressions of length three.