36 1. Higher order Fourier analysis

Proof. By splitting f into real and imaginary parts, we may assume, with-

out loss of generality, that f is real. Rotating e(−αn), we may find a real

number θ such that

En∈[N]f(n)Re e(−αn + θ) ≥ δ.

We then express

Re e(−αn + θ) = 1 −

1

−1

1Et (n) dt

where

Et := {n ∈ [N] : Re e(−αn + θ) ≤ t}.

By Minkowski’s inequality, we thus have either

|En∈[N]f(n)| ≥ δ/2

or

1

−1

|En∈[N]f(n)1Et (n)| dt ≥ δ/2.

In the former case we are done (setting E = [N]), so suppose that the latter

holds. If all the Et were uniformly Fourier-measurable, we would now be

done in this case also by the pigeonhole principle. This is not quite true;

however, it turns out that most Et are uniformly measurable, and this will

be enough. More precisely, let ε 0 be a small parameter to be chosen

later, and say that t is good if one has

|Et+r\Et−r| ≤

2ε−1rN

for all r 0. Let Ω ⊂ [−1, 1] be the set of all bad t. Observe that for each

bad t, we have Mμ(t) ≥

ε−1,

where μ is the probability measure

μ(S) :=

1

N

|{n ∈ [N] : Re e(−αn + θ) ∈ S}|

and M is the Hardy-Littlewood maximal function

Mμ(t) := sup

r0

1

2r

μ([t − r, t + r]).

Applying the Hardy-Littlewood maximal inequality

|{t ∈ R : Mμ(t) ≥ λ}|

1

λ

μ ,

(see e.g. [Ta2011, §1.6] for a proof) we conclude that |Ω| ε. In particular,

if ε is small enough compared with δ, we have

[−1,1]\Ω

|En∈[N]f(n)1Et (n)| dt δ

and so by the pigeonhole principle, there exists a good t such that

|En∈[N]f(n)1Et (n)| δ.