2    Fourier-analytic techniques
In this chapter we will assume that 
is finite abelian.
 comes equipped with
a group Ĝ of characters,
i.e. homomorphisms .
In fact, Ĝ is
isomorphic to .
See  Representation Theory notes for more information about characters and proofs of this as well as some of
the facts below.
Example 2.1.  
    
- 
(i) 
If ,
then for any Ĝ,
we have an associated character ,
where .
- 
(ii) 
If ,
                                                                                    
                                                                                    
then any 
can be associated to a character .
 
 
Notation.  
Given 
nonempty, and any function ,
let 
 
 
Lemma 2.2.  
  Assuming that:
                                                                                    
                                                                                    
Then |  | 
and for all ,
|  | 
  
 
 
                                                                                    
                                                                                    
Proof. 
The first equality in eqch case is trivial. Suppose
. Then there
exists 
with .
Then
So .
For the second part, note that given ,
there must by  such
that , for otherwise
 would act trivially on
, hence would also be
the dual group for ,
a contradiction. □
 
                                                                                    
                                                                                    
Definition 2.3 (Fourier transform).  
Given ,
define its Fourier transform 
by 
 
 
 
It is easy to verify the inversion formula: for all ,
Indeed,
                                                                                    
                                                                                    
Given , the indicator or
characteristic function of ,
𝟙 is
defined as usual.
Note that 
| 𝟙𝟙 | 
The density of 
in  (often
denoted by ).
Definition (Characteristic measure).  
 Given non-empty ,
the characteristic measure 
is defined by 𝟙.
                                                                                    
                                                                                    
Note that .
 
 
Definition (Balanced function).  
 The balanced function
 is given by
𝟙. Note
that .
 
 
Example 2.4.  
 Let 
be a subspace. Then for ,
we have
𝟙𝟙𝟙
                                                                                    
                                                                                    
where  is the
annihilator of .
In other words, 𝟙.
 
 
Example 2.5.  
 Let 
be such that each 
lies in  independently
with probability .
Then with high probability 
| 𝟙 | 
This  follows  from  Chernoff’s inequality:  Given
-valued independent
random variables 
with mean ,
then for all ,
we have 
|  | 
 
 
Example 2.6.  
Let 
with .
Then 
and 𝟙.
 
 
Given ,
we write 
|  | 
Consequently, 
|  | 
  
Lemma 2.7.  
Assuming that:
Then 
                                                                                    
                                                                                    
-   
(i) 
(Parseval’s identity)
- 
(ii) 
(Plancherel’s identity)
  
 
 
Proof. 
Exercise (hopefully easy). □
 
Definition 2.8 (Spectrum).  
Let 
and . Define
the -large
spectrum of 
to be 
|  | 
 
 
 
Example 2.9.  
By Example 2.4, if 𝟙
with ,
then ,
| 𝟙𝟙 | 
 
 
                                                                                    
                                                                                    
Lemma 2.10.  
  Assuming that:
Then |  | 
  
 
 
Proof. 
By Parseval’s identity,
                                                                                    
                                                                                    
 
In particular, if 𝟙
for ,
then 
so 𝟙.
                                                                                    
                                                                                    
Definition 2.11 (Convolution).  
Given ,
we define their convolution 
by 
|  | 
 
 
 
Example 2.12.  
Given ,
| 𝟙𝟙𝟙𝟙𝟙𝟙 | 
In particular, 𝟙𝟙.
 
 
Lemma 2.13.  
  Assuming that:
Then |  | 
  
                                                                                    
                                                                                    
 
 
Proof. 
 
Example 2.14.  
|  | 
In particular, 
for any .
 
 
Theorem 2.15 (Bogolyubov’s lemma).  
Assuming that:
Then there exists 
of codimension 
such that 
.
  
 
 
Proof. 
Observe 
| 𝟙𝟙𝟙𝟙 | 
so wish to find 
such that 
for all . Let
𝟙 with
 and let
. By
Lemma 2.10, .
Fix .
                                                                                    
                                                                                    
𝟙𝟙𝟙𝟙
Note 
since 
for all 
and
𝟙𝟙𝟙𝟙𝟙
hence  (in
fact, )
for all 
and .
□
 
                                                                                    
                                                                                    
Example 2.16.  
The set 
(where  counts the
number of 1s in ) has
density , but there is no
coset  of any subspace
of codimension 
such that .
 
 
Lemma 2.17.  
  Assuming that:
Then there exists 
of codimension 
and 
such that 
  
 
 
Proof. 
Let  be
such that 𝟙,
and let .
Write  for
 for the
 distinct
cosets 
of .
Then
                                                                                    
                                                                                    
𝟙𝟙𝟙
By triangle inequality, .
But note that 
so , hence
there exists 
such that .
Then .
□
 
Notation.  
Given ,
write 
|  | 
 
                                                                                    
                                                                                    
 
Notation.  
Given ,
write 
to be distinguished from .
 
 
Lemma 2.18.  
  Assuming that:
     
- 
prime 
-      
of density  
-      
Then the number of 3-term arithmetic progressions in
 differs from
 by at
most .
 
 
 
Proof. 
The number of 3-term arithmetic progressions in
 is
times
𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙
                                                                                    
                                                                                    
By Plancherel’s identity and Lemma 2.13, we have
𝟙𝟙𝟙𝟙𝟙𝟙
but
𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙𝟙
by Parseval’s identity. □
 
                                                                                    
                                                                                    
Theorem 2.19 (Meshulam’s Theorem).  
  Assuming that:
Then .
 
 
 
Proof. 
By assumption, 
| 𝟙𝟙𝟙 | 
But as in (the proof of) Lemma 2.18, 
| 𝟙𝟙𝟙𝟙 | 
so provided ,
i.e. 𝟙𝟙𝟙
we have 𝟙.
So by Lemma 2.17 with ,
there exists 
of codimension 1 and 
such that .
We iterate this observation: let ,
,
. At the
-th step, we are
given a set 
of density 
with no non-trivial 3 term arithmetic progressions. Provided that
, there
exists  of
codimension ,
 such
that 
|  | 
                                                                                    
                                                                                    
Set ,
has density ,
and is free of non-trivial 3 term arithmetic progressions.
Through            this            iteration,            the            density            increases            from
to
in                                                             at                                                             most
steps.
to
in                                                             at                                                             most
steps and so on.
So reaches 
in at most 
steps. The argument must end with ,
at which point we must have had ,
or else we could have continued.
But we may assume that 
(or )
whence ,
or .
□
                                                                                    
                                                                                    
 
At the time of writing, the largest known subset of
containing no non-trivial 3 term arithmetic progressions has size
.
We will prove an upper bound of the form .
Theorem 2.20 (Roth’s theorem).  
  Assuming that:
Then .
 
 
 
                                                                                    
                                                                                    
Example 2.21 (Behrend’s example).  
 There exists
 of size
at least 
containing no non-trivial 3 term arithmetic progressions. 
 
 
Lemma 2.22.  
  Assuming that:
Then one of the following holds:
                                                                                    
                                                                                    
-   
(i) 
𝟙
(where the Fourier coefficient is computed in )
- 
(ii) 
There exists an interval 
of length 
such that 
  
 
 
Proof. 
We may assume that 
since otherwise
so we would be in Case (ii) with .
Let . Note that all 3 term arithmetic
progressions of the form  are in
                                                                                    
                                                                                    
fact arithmetic progressions in .
If  or
 were at least
, we would again be in case
(ii). So we may assume that .
Now as in Lemma 2.18 and Theorem 2.19,
𝟙𝟙𝟙𝟙𝟙𝟙
where 
and . So
as before, 
| 𝟙 | 
                                                                                    
                                                                                    
provided that ,
i.e. .
(Check this is satisfied).
Hence 
| 𝟙 | 
 
Lemma 2.23.  
  Assuming that:
Then there exists a partition of 
into progressions 
of length 
such that 
|  | 
for all .
  
 
 
Proof. 
Let 
and consider .
By Pigeonhole, there exists such
that .
Set ,
so .
Divide 
into residue classes modulo ,
each of which has size at least .
But each residue class can be divided into arithmetic progressions of the form 
with .
The diameter of the image of each progression under 
is .
                                                                                    
                                                                                    
□
 
Lemma 2.24.  
Assuming that:
     
- 
of density  
-      
a prime in  
-      
let  
-      
𝟙
for some  
 
Then there exists a progression 
of length at least 
such that .
 
 
 
                                                                                    
                                                                                    
Proof. 
Let , and use
Lemma 2.23 to partition 
into progressions 
of length 
and . Fix
one  from
each of the .
Then
                                                                                    
                                                                                    
So 
Since 
has mean zero, 
|  | 
                                                                                    
                                                                                    
hence there exists 
such that 
|  | 
and so 
 
Definition 2.25 (Bohr set).  
Let 
and . By the
                                                                                    
                                                                                    
Bohr set  we
mean the set 
|  | 
We call 
the rank of ,
and 
its width or radius.
 
 
 
Example 2.26.  
When ,
then  for all
sufficiently small .
 
 
                                                                                    
                                                                                    
Lemma 2.27.  
Assuming that:
 
 
 
                                                                                    
                                                                                    
Proposition 2.28 (Bogolyubov in a general finite abelian group).  
Assuming that:
Then there exists 
of size at most 
such that 
.
  
 
 
Proof. 
Recall 𝟙𝟙𝟙𝟙𝟙.
Let 𝟙, and note
that, for 
and ,
. Hence,
for ,
| 𝟙𝟙𝟙 | 
and 
| 𝟙𝟙𝟙 |