40 1. Higher order Fourier analysis
with the stated properties. This splits the left-hand side of (1.14) into 27
terms; but we can eliminate several of these terms:
Exercise 1.2.11. Show that all of the terms in (1.14) which involve at
least one copy of fpsd are of size O(1/F (M)). (Hint: Modify the proof of
Proposition 1.2.4.)
From this exercise we see that
(1.15) Λ(1A, 1A, 1A) = Λ(fstr + fsml,fstr + fsml,fstr + fsml) + O(1/F (M)).
Now we need to deal with fstr + fsml. A key point is the almost periodicity
of fstr + fsml:
Lemma 1.2.13 (Almost periodicity). For
δ,M
N values of r [−εN, εN],
one has
En∈[N]|(fstr + fsml)(n + r) (fstr + fsml)(n)| ε
(where we extend fstr,fsml by zero outside of [N]).
Proof. As fstr is Fourier-measurable, we can approximate it to an error of
O(ε) in
L1[N]
norm by a function
(1.16) g =
J
j=1
cje(αjn)
of Fourier complexity J OM,ε(1). From the smallness of fsml, we then
have
En∈[N]|(fstr + fsml)(n + r) (fstr + fsml)(n)|
En∈[N]|g(n + r) g(n)| + O(ε)
(where we extend g using (1.16) rather than by zero, with the error being
O(ε) when |r| εN). We can use (1.16) and the triangle inequality to
bound
En∈[N]|g(n + r) g(n)|
J
j=1
|e(αjr) 1|.
Using multiple recurrence, we can find
J,ε
N values of r [−εN, εN] such
that αjr
R/Z
ε/J for all 1 j J. The claim follows.
Now we can finish the proof of Roth’s theorem. As fstr + fsml has the
same mean as f, we have
En∈[N](fstr + fsml)(n) δ
and hence by older’s inequality (and the non-negativity of fstr + fsml),
En∈[N](fstr +
fsml)(n)3

δ3.
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