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.
Previous Page Next Page