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 H¨ older’s inequality (and the non-negativity of fstr + fsml),

En∈[N](fstr +

fsml)(n)3

≥

δ3.