2 Fourier-analytic techniques

In this chapter we will assume that G is finite abelian.

G comes equipped with a group Ĝ of characters, i.e. homomorphisms γ:G→ℂ. In fact, Ĝ is isomorphic to G.

See Representation Theory notes for more information about characters and proofs of this as well as some of the facts below.

Example 2.1.

Notation. Given B⊆G nonempty, and any function g:B→ℂ, let

𝔼x∈Bg(x)=1|B|∑x∈Bg(x).

Lemma 2.2. Assuming that:

  • γ∈G^

Then
𝔼x∈Gγ(x)={1if γ=10otherwise,

and for all x∈G,

∑γ∈G^γ(x)={|G^|if x=00otherwise.

Proof. The first equality in eqch case is trivial. Suppose γ≠1. Then there exists y∈G with γ(y)≠1. Then

γ(y)𝔼z∈Gγ(z)=𝔼z∈Gγ(y+z)=𝔼z′∈Gγ(z′)

So 𝔼z∈Gγ(z)=0.

For the second part, note that given x≠0, there must by γ∈G^ such that γ(x)≠1, for otherwise G^ would act trivially on ⟨x⟩, hence would also be the dual group for G∕⟨x⟩, a contradiction. □

Definition 2.3 (Fourier transform). Given f:G→ℂ, define its Fourier transform f^:G^→ℂ by

f^(γ)=𝔼x∈Gf(x)γ(x)¯.

It is easy to verify the inversion formula: for all x∈G,

f(x)=∑γ∈G^f^(γ)γ(x).

Indeed,

∑γ∈G^f^(γ)γ(x)=∑γ∈G^𝔼y∈Gf(y)γ(y)¯γ(x)=𝔼y∈Gf(y)∑γ∈G^γ(x−y)⏟=|G| iff x=y=f(x)by Lemma 2.2

Given A⊆G, the indicator or characteristic function of A, 𝟙A:G→{0,1} is defined as usual.

Note that

𝟙A^(1)=𝔼x∈G𝟙A(x)1(x)=|A||G|.

The density of A in G (often denoted by α).

Definition (Characteristic measure). Given non-empty A⊆G, the characteristic measure μA:G→[0,|G|] is defined by μA(x)=α−1𝟙A(x).

Note that 𝔼x∈GμA(x)=1=μA^(1).

Definition (Balanced function). The balanced function fA:G→[−1,1] is given by fA(x)=𝟙A(x)−α. Note that 𝔼x∈GfA(x)=0=fA^(1).

Example 2.4. Let V≤𝔽pn be a subspace. Then for t∈𝔽pn^, we have

𝟙V^(t)=𝔼x∈𝔽pn𝟙V(x)e(−x⋅tp)=|V|pn𝟙V⊥(t)

where V⊥={t∈𝔽pn^:x⋅t=0 ∀⁡x∈V} is the annihilator of V. In other words, 𝟙V^(t)=μV⊥(t).

Example 2.5. Let R⊆G be such that each x∈G lies in R independently with probability 12. Then with high probability

sup⁡γ≠1|𝟙R^(γ)|=O( log ⁡⁡|G||G|).

This follows from Chernoff’s inequality: Given ℂ-valued independent random variables X1,X2,…,Xn with mean 0, then for all 𝜃>0, we have

ℙ(|∑i=1nXi|≥𝜃∑i=1n∥Xi∥L∞(ℙ)2)≤4exp⁡(−𝜃24).

Example 2.6. Let Q={x∈𝔽pn:x⋅x=0}⊆𝔽pn with p>2. Then

|Q|pn=1p+O(p−n2)

and sup⁡t≠0|𝟙Q^(t)|=O(p−n2).

Given f,g:G→ℂ, we write

⟨f,g⟩=𝔼x∈Gf(x)g(x)¯and⟨f^,g^⟩=∑γ∈G^f^(γ)g^(γ)¯.

Consequently,

∥f∥L2(G)2=𝔼x∈G|f(x)|2and∥f^∥l2(G^)2=∑γ∈G^|f^(γ)|2.

Lemma 2.7. Assuming that:

  • f,g:G→ℂ

Then
  • (i) ∥f∥L2(G)2=∥f^∥l2(G^)2 (Parseval’s identity)
  • (ii) ⟨f,g⟩=⟨f^,g^⟩ (Plancherel’s identity)

Proof. Exercise (hopefully easy). □

Definition 2.8 (Spectrum). Let 1≥ρ>0 and f:G→ℂ. Define the ρ-large spectrum of f to be

Spec⁡ρ(f)={γ∈G^:|f^(γ)|≥ρ∥f∥1}.

Example 2.9. By Example 2.4, if f=𝟙V with V≤𝔽pn, then ∀⁡ρ>0,

Spec⁡ρ(𝟙V)={t∈𝔽pn^:|𝟙V^(t)|≥ρ|V|pn}=V⊥.

Lemma 2.10. Assuming that:

  • ρ>0

Then
|Spec⁡ρ(f)|≤ρ−2∥f∥22∥f∥12.

Proof. By Parseval’s identity,

∥f∥22=∥f^∥22=∑γ∈G^|f^(γ)|2≥∑γ∈Spec⁡ρ(f)|f^(γ)|2≥|Spec⁡ρ(f)|(ρ∥f∥1)2□

In particular, if f=𝟙A for A⊆G, then

∥f∥1=α=|A||G|=∥f∥22,

so |Spec⁡ρ(𝟙A)|≤ρ−2α−1.

Definition 2.11 (Convolution). Given f,g:G→ℂ, we define their convolution f∗g:G→ℂ by

f∗g(x)=𝔼y∈Gf(y)g(x−y)∀⁡x∈G.

Example 2.12. Given A,B⊆G,

𝟙A∗𝟙B(x)=𝔼y∈G𝟙A(y)𝟙B(x−y)=𝔼y∈G𝟙A(y)𝟙x−B(y)=|A∩(x−B)||G|=1|G|rA+B(x).

In particular, supp⁡(𝟙A∗𝟙B)=A+B.

Lemma 2.13. Assuming that:

  • f,g:G→ℂ

Then
f∗g^(γ)=f^(γ)g^(γ)∀⁡γ∈G^.

Proof.

f∗g^(γ)=𝔼x∈Gf∗g(x)γ(x)¯=𝔼x∈G𝔼[∈y]Gf(y)g(x−y⏟u)γ(x)¯=𝔼u∈G𝔼[∈y]Gf(y)g(u)γ(u+y)¯=f^(γ)g^(γ)□

Example 2.14.

𝔼x+y=z+wf(x)f(y)f(z)f(w)¯=∥f^∥l4(G^)4.

In particular,

∥𝟙A^∥l4(G^)4=E(A)|G|3

for any A⊆G.

Theorem 2.15 (Bogolyubov’s lemma). Assuming that:

  • A⊆𝔽pn be a set of density α

Then there exists V≤𝔽pn of codimension ≤2α−2 such that V⊆A+A−A−A.

Proof. Observe

2A−2A=supp⁡(𝟙A∗𝟙A∗𝟙−A∗𝟙−A⏟=:g),

so wish to find V≤𝔽pn such that g(x)>0 for all x∈V. Let S=Spec⁡ρ(𝟙A) with ρ=α2 and let V=⟨S⟩⊥. By Lemma 2.10, codim⁡(V)≤|S|≤ρ−2α−1. Fix x∈V.

g(x)=∑t∈𝔽pn^g^(t)e(x⋅t∕p)=∑t∈𝔽pn^|𝟙A^(t)|4e(x⋅t∕p)by Lemma 2.13=α4+∑t≠0|𝟙A^(t)|4e(x⋅t∕p)=α4+∑t∈S∖{0}|𝟙A^(t)|4e(x⋅t∕p)⏟(1)+∑t∉S|𝟙A^(t)|4e(x⋅t∕p)⏟(2)

Note (1)≥(ρα)4 since x⋅t=0 for all t∈S and

|(2)|≤sup⁡t∉S|𝟙A^(t)|2∑t∉S|𝟙A^|2≤sup⁡t∈S|𝟙A^(t)|2∑t∉S|𝟙A^|2≤(ρα)2∥𝟙A∥22by Parseval’s identity=ρ2α3

hence g(x)>0 (in fact, ≥α42) for all x∈V and codim⁡(V)≤2α−2. □

Example 2.16. The set A={x∈𝔽2n:|x|≥n2+n2} (where |x| counts the number of 1s in x) has density ≥18, but there is no coset C of any subspace of codimension n such that C⊆A+A(=A−A).

Lemma 2.17. Assuming that:

  • A⊆𝔽pn of density α

  • ρ>0

  • sup⁡t≠0|𝟙A^(t)|≥ρα

Then there exists V≤𝔽pn of codimension 1 and x∈𝔽pn such that
|A∩(x+V)|≥α(1+ρ2)|V|.

PIC

Proof. Let t≠0 be such that |𝟙A^(t)|≥ρα, and let V=⟨t⟩⊥. Write vj+V for j∈[p]={1,2,…,p} for the p distinct cosets vj+V={x∈𝔽pn:x⋅t=j} of V. Then

𝟙A^(t)=fA^(t)=𝔼x∈𝔽pn(𝟙A(x)−α)e(−x⋅t∕p)=𝔼j∈[p]𝔼x∈vj+V(𝟙A(x)−α)e(−j∕p)=𝔼j∈[p](|A∩(vj+V)||vj+V|−α⏟=aj)e(−j∕p)

By triangle inequality, 𝔼j∈[p]|aj|≥ρα. But note that 𝔼j∈[p]aj=0 so 𝔼j∈[p]aj+|aj|≥ρα, hence there exists j∈[p] such that aj+|aj|≥ρα. Then aj≥ρα2. □

Notation. Given f,g,h:G→ℂ, write

T3(f,g,h)=𝔼x,d∈Gf(x)g(x+d)h(x+2d).

Notation. Given A⊆G, write

2⋅A={2a:a∈A},

to be distinguished from 2A=A+A={a+a′:a,a′∈A}.

Lemma 2.18. Assuming that:

  • p≥3 prime

  • A⊆𝔽pn of density α>0

  • sup⁡t≠0|𝟙A^(t)|≤𝜀

Then the number of 3-term arithmetic progressions in A differs from α3(pn)2 by at most 𝜀(pn)2.

Proof. The number of 3-term arithmetic progressions in A is (pn)2 times

T3(𝟙A,𝟙A,𝟙A)=𝔼x,d∈𝔽pn𝟙A(x)𝟙(x+d)𝟙A(x+2d)=𝔼x,y∈𝔽pn𝟙A(x)𝟙A(y)𝟙A(2y−x)=𝔼y∈G𝟙A(y)𝔼x∈G𝟙A(x)𝟙A(2y−x)=𝔼y∈G𝟙A(y)𝟙A∗𝟙A(2y)=⟨𝟙2⋅A,𝟙A∗𝟙A⟩

By Plancherel’s identity and Lemma 2.13, we have

=⟨𝟙2⋅A^,𝟙A^2⟩=∑t𝟙2⋅A^(t)𝟙A^(t)2¯=α3+∑t≠0𝟙2⋅A^(t)𝟙A^(t)2¯

but

|∑t≠0𝟙2⋅A^(t)𝟙A^(t)2|≤sup⁡t≠0|𝟙A^(t)|∑t≠0|𝟙2⋅A^(t)||𝟙A^(t)|≤CSsup⁡t≠0|𝟙A^(t)|(∑t|𝟙2⋅A^(t)|2∑t|𝟙A^(t)|2)12≤𝜀∥𝟙2⋅A^∥2∥𝟙A^∥2=𝜀⋅α

by Parseval’s identity. □

Theorem 2.19 (Meshulam’s Theorem). Assuming that:

  • A⊆𝔽pn a set containing no non-trivial 3 term arithmetic progressions

Then |A|=O(pnlog⁡pn).

Proof. By assumption,

T3(𝟙A,𝟙A,𝟙A)=|A|(pn)2=αpn.

But as in (the proof of) Lemma 2.18,

|T3(𝟙A,𝟙A,𝟙A)−α3|≤sup⁡t≠0|𝟙A^(t)|⋅α,

so provided pn≥2α−2, i.e. T3(𝟙A,𝟙A,𝟙A)≤α32 we have sup⁡t≠0|𝟙A^(t)|≥α22.

So by Lemma 2.17 with ρ=α2, there exists V≤𝔽pn of codimension 1 and x∈𝔽pn such that |A∩(x+V)|≥(α+α24)|V|.

We iterate this observation: let A0=A, V0=𝔽pn, α0=|A0||V0|. At the i-th step, we are given a set Ai−1⊆Vi−1 of density αi−1 with no non-trivial 3 term arithmetic progressions. Provided that pdim⁡(Vi−1)≥2αi−1−2, there exists Vi≤Vi−1 of codimension 1, xi∈Vi−1 such that

|(A−xi)∩Vi|≥(αi−1+(αi−1)24)|Vi|.

Set Ai=(A−xi)∩Vi⊆Vi, has density ≥αi−1+(αi−1)24, and is free of non-trivial 3 term arithmetic progressions.

Through this iteration, the density increases from α to 2α in at most α(α24)=4⋅α−1 steps.

2α to 4α in at most 2α((2α)24)=2α−1 steps and so on.

So reaches 1 in at most

4α−1(1+12+14+18+⋯)≤8α−1

steps. The argument must end with dim⁡(Vi)≥n−8α−1, at which point we must have had pdim⁡(Vi)<2αi−12≤2α−2, or else we could have continued.

But we may assume that α≥2p−n4 (or α−2<2pn2) whence pn−8α−1≤pn2, or n2≤2α−1. □

At the time of writing, the largest known subset of 𝔽3n containing no non-trivial 3 term arithmetic progressions has size (2.2202)n.

We will prove an upper bound of the form (2.756)n.

Theorem 2.20 (Roth’s theorem). Assuming that:

  • A⊆[N]={1,…,N}

  • A contains no non-trivial 3 term arithmetic progressions

Then |A|=O(Nlog⁡ log ⁡N).

Example 2.21 (Behrend’s example). There exists A⊆[N] of size at least |A|≥exp⁡(−c log ⁡N)N containing no non-trivial 3 term arithmetic progressions.

Lemma 2.22. Assuming that:

  • A⊆[N] of density α>0

  • N>50α−2

  • A contains no non-trivial 3 term arithmetic progressions

  • p a prime in [N3,2N3]

  • let A′=A∩[p]⊆ℤ∕pℤ

Then one of the following holds:
  • (i) sup⁡t≠0|𝟙A′^(t)|≥α210 (where the Fourier coefficient is computed in ℤ∕pℤ)
  • (ii) There exists an interval J⊆[N] of length ≥N3 such that |A∩J|≥α(1+α400)|J|

Proof. We may assume that |A′|=|A∩[p]|≥α(1−α200)p since otherwise

|A∩[p+1,N]|≥αN−(α(1−α200)p)=α(N−p)+α2200p≥(α+α2400)(N−p)

so we would be in Case (ii) with J=[p+1,N]. Let A″=A′∩[p3,2p3]. Note that all 3 term arithmetic progressions of the form (x,x+d,x+2d)∈A′×A″×A″ are in fact arithmetic progressions in [N].

If |A′∩[p3]| or |A′∩[2p3,p]| were at least 25|A′|, we would again be in case (ii). So we may assume that |A″|≥|A′|5.

Now as in Lemma 2.18 and Theorem 2.19,

α″p=|A″|p2T3(𝟙A′,𝟙A″,𝟙A″)=α′(α″)2+∑t𝟙A′^(t)𝟙A ″^(t)¯𝟙2⋅A ″^(t)

where α′=|A′|p and α″=|A″|p. So as before,

α′α″2≤sup⁡t≠0|𝟙A′(t)|⋅α ″⁡,

provided that α″p≤12α′(α″)2, i.e. 2p≤α′α″. (Check this is satisfied).

Hence

sup⁡t≠0|𝟙A′^(t)|≥α′α ″⁡2≥12(α(1−α200))2⋅25≥α210.□

Lemma 2.23. Assuming that:

  • m∈ℕ

  • φ:[m]→ℤ∕pℤ be given by x↦tx for some t≠0

  • 𝜀>0

Then there exists a partition of [m] into progressions Pi of length li∈[𝜀m2,𝜀m] such that
diam⁡(φ(Pi))=max⁡x,y∈Pi|φ(x)−φ(y)|≤𝜀p

for all i.

Proof. Let u=⌊m⌋ and consider 0,t,2t,…,ut. By Pigeonhole, there exists 0≤v<w≤usuch that |wt−vt|=|(w−v)t|≤pu. Set s=w−v, so |st|≤pu. Divide [m] into residue classes modulo s, each of which has size at least ms≥m4. But each residue class can be divided into arithmetic progressions of the form a,a+s,…,a+ds with 𝜀u2<d≤𝜀u. The diameter of the image of each progression under φ is |dst|≤dpu≤𝜀upu=𝜀p. □

Lemma 2.24. Assuming that:

  • A⊆[N] of density α>0

  • p a prime in [N3,2N3]

  • let A′=A∩[p]⊆ℤ∕pℤ

  • |𝟙A′^(t)|≥α220 for some t≠0

Then there exists a progression P⊆[N] of length at least α2N500 such that |A∩P|≥α(1+α80)|P|.

Proof. Let 𝜀=α240π, and use Lemma 2.23 to partition [p] into progressions Pi of length

≥𝜀p2≥α240πN32≥α2N500

and diam⁡(φ(Pi))≤𝜀p. Fix one xi from each of the Pi. Then

α220≤|fA′^(t)|=|1p∑i∑x∈PifA′(x)e(−xt∕p)|=1p|∑i∑x∈PifA′(x)e(−xit∕p)+∑i∑x∈PifA′(x)(e(−xt∕p)−e(−xit∕p))|≤1p∑i|∑x∈PifA′(x)|+1p∑i∑x∈Pi|fA′(x)||e(−xt∕p)−e(−xit∕p)⏟≤2π𝜀since |t(x−xi)|≤𝜀p|

So

∑i|∑x∈PifA′(x)|≥α240p.

Since fA′ has mean zero,

∑i(|∑x∈PifA′(x)|+∑x∈PifA′(x))≥α240p,

hence there exists i such that

|∑x∈PifA′(x)|+∑x∈PifA′(x)≥α280|Pi|

and so

∑x∈PifA′(x)≥α2160|Pi|.□

Definition 2.25 (Bohr set). Let Γ⊆G^ and ρ>0. By the Bohr set B(Γ,ρ) we mean the set

B(Γ,ρ)={x∈G:|γ(x)−1|<ρ ∀⁡γ∈Γ}.

We call |Γ| the rank of B(Γ,ρ), and ρ its width or radius.

Example 2.26. When G=𝔽pn, then B(Γ,ρ)=⟨Γ⟩⊥ for all sufficiently small ρ.

Lemma 2.27. Assuming that:

  • Γ⊆G^ of size d

  • ρ>0

Then
|B(Γ,ρ)|≥(ρ8)d|G|.

Proposition 2.28 (Bogolyubov in a general finite abelian group). Assuming that:

  • A⊆G of density α>0

Then there exists Γ⊆G^ of size at most 2α−2 such that A+A−A−A⊇B(Γ,ρ).

Proof. Recall 𝟙A∗𝟙A∗𝟙−A∗𝟙−A(x)=∑⁡γ∈G^|𝟙A^(γ)|4γ(x).

Let Γ∈Spec⁡α2(𝟙A), and note that, for x∈B(Γ,12) and γ∈Γ, Re⁡(γ(x))>0. Hence, for x∈B(Γ,12),

Re⁡∑γ∈G^|𝟙A^(γ)|4γ(x)=Re⁡∑γ∈Γ|𝟙A^(γ)|4γ(x)⏟≥α4+Re⁡∑γ∉Γ|𝟙A^(γ)|4γ(x)

and

|Re⁡∑γ∉Γ|𝟙A^(γ)|4γ(x)|≤sup⁡γ∉Γ|𝟙A^(γ)|2∑γ∉Γ|𝟙A^(γ)|2≤(α2⋅α)2⋅α=α42.□