L’INSTITUTFOURIER DE ANNALES

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A L

E S

E D L ’IN IT ST T U

F O U R

ANNALES

DE

L’INSTITUT FOURIER

Thomas W. KÖRNER

Fourier Transforms of Measures and Algebraic Relations on Their Supports Tome 59, n^o4 (2009), p. 1291-1319.

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FOURIER TRANSFORMS OF MEASURES AND ALGEBRAIC RELATIONS ON THEIR SUPPORTS

by Thomas W. KÖRNER

Abstract. — We investigate the relation between the rate of decrease of a Fourier transform and the possible algebraic relations on its support.

Résumé. — Si la transformée de Fourier d’une mesure décroît rapidement alors le support ne satisfait que très peu des relations algèbriques.

1. Non-technical introduction

This paper is fairly technical but deals with natural questions.

We work on the circle T = R/Z. A well known result tells us that, if a setE has positive Lebesgue measure, then E+E contains an interval.

It follows that there existx, y ∈E and integers mand nsatisfying some non-trivial equation

mx+ny= 0.

In other words, if a set has positive Lebesgue measure, it must be rich in short algebraic relations.

A closely related argument shows that any Borel measureµwhose Fourier transformµ(r)ˆ tends fairly rapidly to zero must have a support which is rich in fairly short algebraic relations. More specifically, ifµ(r) =ˆ O(|r|^−−q⁻¹), then we can findxj ∈suppµ and integersmj satisfying some non-trivial equation

q

X

j=1

m_jx_j= 0.

Keywords:Convolution, Fourier series.

Math. classification:42A16.

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In an earlier paper, I used a fairly simple probabilistic argument to construct a Borel measureµsuch thatµ(r) =ˆ O(|r|⁻²⁻¹^q⁻¹), but there do not existx_j ∈suppµand integersm_j satisfying some non-trivial equation

q

X

j=1

mjxj= 0.

There is a large gap between the results of the two paragraphs and both seem ‘natural’. However, in Theorem 2.4, I show that that, by using more complicated probabilistic arguments, we can construct a Borel measureµ such thatµ(r) =ˆ O(|r|⁻²⁻¹^q⁻¹), but there do not exist xj ∈suppµ and integersm_j satisfying some non-trivial equation

q+1

X

j=1

mjxj= 0.

If is small, the set







q+1

X

j=1

x_j : x_j ∈suppµ







has positive Lebesgue measure and this suggests that the new result is close to best possible or, at least, that it will be quite hard to improve.

On the other hand, if we deal with sets, I show (in Theorem 2.6) how to construct a closed setE such that theq-fold sum E+E+· · ·+E has positive Lebesgue measure but there do not existxj∈suppµand integers mj satisfying some non-trivial equation

2q−1

X

j=1

mjxj = 0.

2. Technical introduction

As stated earlier, we work on the circle T=R/Z. All measures will be Borel measures andmwill denote the Lebesgue measure. Ifµis a measure, we write

µ_[q] =µ∗µ∗ · · · ∗µ for theq-fold convolution ofµwith itself and

E_[q]=E+E+· · ·+E=







q

X

j=1

x_j : x_j∈E





 .

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As usualkfk1=R

T|f(t)|dt.

This paper, like its predecessor [4], centres round the following two simple observations.

Lemma 2.1. — Suppose thatµis a non-zero measure with supportE.

(i) IfP∞

r=−∞|ˆµ(r)|^q converges, then there exists a non-trivial intervalI such that everyx∈I can be written

x=x₁+x₂+· · ·+x_q withxj ∈E.

(ii) IfP∞

r=−∞|µ(r)|ˆ ^2qconverges, then there exists a setAof strictly pos- itive Lebesgue measure such that everyx∈Acan be written

x=x1+x2+· · ·+xq

withxj ∈E.

Proof. — (i) Observe that

|ˆµ_[q](r)|=|ˆµ(r)|^q

so dµ_[q] =f dm where f has an absolutely convergent Fourier series and so is continuous. The support off contains a non-trivial interval I and

suppf =suppµ[q] ⊆ {x1+x2+· · ·+xq : xj∈E}.

(ii) Observe thatdµ[q]=f dmwheref ∈L²(m)and argue much as in (i).

As I remarked earlier, part (i) of the next lemma is extremely well known, but, although part (ii) is a simple consequence, I do not know if it has been observed before.

Lemma 2.2. — (i) If E has strictly positive Lebesgue measure, then E+E contains a non-trivial interval.

(ii) If E has strictly positive Lebesgue measure, then we can find a non- trivial intervalI such that, wheneverx∈I, the equation

x1+x2=x

has uncountably many distinct solutions withx1, x2∈E.

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Proof. —

(i) IfE has strictly positive Lebesgue measure then we can find a closed set E^∗ ⊆ E with E^∗ having strictly positive measure. Thus, with- out loss of generality, we may assume that E is closed. We now know that the indicator function IE is a nontrivial L²(m) function soP∞

j=−∞|ˆIE(j)|² converges and we may apply Lemma 2.1 (i).

(ii) Suppose that the result is false. Then each intervalIcontains a point y such that the equation

x1+x2=y

has only countably many distinct solutions withx1, x2∈E. Thus we can find a countable dense sequencey_j and associated countable sets Ej such that, if

x1+x2=yj

withx₁, x₂∈E thenx₁, x₂∈E_j. Now observe thatE\S∞

j=1E_j is a set of strictly positive Lebesgue measure disobeying the conclusions of (i) which is impossible

Since every non-trivial interval contains a rational, Lemma 2.1 (i) implies the following result.

Lemma 2.3. — Suppose thatµ is a non-zero measure onT and q is a positive integer such that we can find anα >1/q and anA >0with

|µ(r)|ˆ 6A|r|^−α

for allr6= 0. Then we can find distinct pointsx₁, x₂, . . . , x_q∈suppµand mj∈Z, not all zero, such that

q

X

j=1

m_jx_j= 0.

In this paper we show how to prove the following result in the other direction.

Theorem 2.4. — If q is an integer with q > 1 and ψ : N → R is a sequence of strictly positive numbers such thatψ(r)→ ∞asr→ ∞, then there exists a probability measureµsuch that

|ˆµ(r)|6|r|^−1/(2q) log(1 +|r|)1/2

ψ(|r|)

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for allr6= 0, but, given distinct pointsx₁, x₂, . . . , x_q+1∈suppµ, the only solution to the equation

q+1

X

j=1

m_jx_j= 0

withm_j∈Zis the trivial solutionm₁=m₂=· · ·=m_q+1= 0.

In [4] we proved a similar result with the equation Pq+1

j=1m_jx_j = 0re- placed by Pq

j=1m_jx_j = 0. Earlier I explained why the new result might be substantially more difficult to prove than the old. Observe that if, for example,ψ(r) = log(1 +|r|)^1/2

, then, by Lemma 2.1,







q+1

X

j=1

xj : xj ∈suppµ





 must have strictly positive Lebesgue measure.

The key lemma in our proof is the following.

Lemma 2.5. — Let q be an integer with q > 1 and ψ : N → R be a sequence of strictly positive numbers such thatψ(r)→ ∞asr→ ∞.

Supposem∈Z^q+1\ {0}andN is a positive integer such that

N >12



q+ 1 +

N

X

j=1

|mj|



.

Then, given closed intervals I_j = [(n_j −¹₂)/N,(n_j +¹₂)/N], with n_j an integer, such that

nj

N −nk

N > 6

N for16j < k6q+ 1

and > 0, we can find an infinitely differentiable function f with the following properties.

(i) f(t)>0for allt∈T. (ii) fˆ(0) = 1.

(iii) |fˆ(r)|6|r|^−1/(2q) log(1 +|r|)1/2

ψ(|r|)for allr6= 0.

(iv) Ifxj ∈suppf∩Ij then

q+1

X

j=1

mjxj6= 0.

(v) Ifx∈Twe can find ay∈suppf with|x−y|< .

If we deal with sets rather than measures we have the following result which excludes a natural conjecture.

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Theorem 2.6. — Ifqis an integer withq>1, then we can find a closed setE with the following properties.

(i) E_[q] has strictly positive Lebesgue measure.

(ii) The equation

2q−1

X

j=1

mjxj= 0

has no non-trivial solution with mj ∈ Z and the xj distinct points ofE.

Theµwe construct in the proof of Theorem 2.4 also has the property de- scribed in the next lemma, which furnishes a complement to Lemma 2.1 (ii).

Lemma 2.7. — Ifqis an integer withq>1andψ:N→Ris a sequence of positive numbers such that ψ(r) → ∞ as r → ∞, then there exists a probability measureµsuch that

|ˆµ(r)|6|r|^−1/(2q) log(1 +|r|)1/2

ψ(|r|) for allr6= 0, but the set







q

X

j=1

m_jx_j : x_j ∈suppµ, m_j∈Z





 has Lebesgue measure zero.

However, the method of [4] can be easily adapted to give a much simpler proof of this result.

Since the proof of Theorem 2.6 is substantially simpler than that of Theorem 2.4 we shall devote the next two sections to its proof. We give the fairly routine proof of Theorem 2.4 from Lemma 2.5 in section 5 and devote the rest of the paper to the proof of Lemma 2.5.

Like many others of my papers, this one owes a great deal to two re- markable papers [2] and [3] of Kaufman.

3. Sums and algebraic relations

We shall prove Theorem 2.6 by a Baire category argument. We use the Hausdorff metricd_F defined in the next lemma.

Definition 3.1. — Consider the space Fof non-empty closed subsets ofT. We set

d_F(E, F) = sup

e∈E

inf

f∈F|e−f|+ sup

f∈F

inf

e∈E|e−f|.

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It is well known that(F, d_F)is a complete metric space. (See, for example, Chapter II §21 VII and Chapter III §33 IV of [5].)

We need he following remarks.

Lemma 3.2. — (i) If E, F,GandH are closed then d_F(E+F, G+H)6d_F(E, G) +d_F(F, H).

(ii) SupposeEn,Fn, Eand F are closed sets with d_F(E_n, E), d_F(F_n, F)→0.

Thend_F(En+Fn, E+F)→0asn→ ∞.

(iii) SupposeEn andE are closed sets withd_F(En, E)→0. Then m(E)>lim sup

n→∞

m(E_n).

Proof. —

(i) Observe that, ife∈E,f ∈F,g∈Gandh∈H,

|(e+f)−(g+h)|6|e−g|+|f−h|

so, ife∈E,f ∈F,

e∈E, f∈Finf |(e+f)−(g+h)|6 inf

e∈E|e−g|+ inf

f∈F|f−h|

whence sup

g∈G, h∈H

inf

e∈E, f∈F|(e+f)−(g+h)|6sup

g∈G

e∈Einf |e−g|+ sup

h∈H

inf

f∈F|f−h|.

(ii) This follows directly from (i).

(iii) Given >0, we can find anη >0such that m E+ (−η, η)

< m(E) +. Whennis sufficiently large,

E_n⊆E+ (−η, η) so

m(En)< m(E) +.

Thuslim sup_n→∞m(E_n)6m(E) +for alland the result follows.

Definition 3.3. — Ifq>1, we defineH=Hq to be the subspace ofF consisting of those closed sets for whichm(E_[q])>1/2 with the inherited metricd_H=d_F|H×H.

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Lemma 3.2 tells us thatH is a closed subspace ofFand so(H, d_H)is a complete metric space. Since(T, m)∈H, the spaceH is non-empty.

We can thus deduce Theorem 2.6 from the Baire category version.

Theorem 3.4. — The collection ofE∈Hsuch that the equation

2q−1

X

j=1

mjxj= 0

has a non-trivial solution withm_j∈Zand thex_j distinct points ofEis of first category.

Since Z^2q−1 is countable, Theorem 3.4 follows in turn from the simpler result.

Lemma 3.5. — Let m∈ Z^2q−1\ {0} and N >1. Let E(m, N) be the collection of ofE∈H such that the equation

2q−1

X

j=1

mjxj= 0

has a solution withx_j∈E [16j62q−1]and

|xj−x_k|>N⁻¹ for16j < k62q−1.

ThenE(m, N)is closed and has dense complement.

We split the proof of Theorem 3.4 into two parts, the easy Lemma 3.6 and the harder Lemma 3.7.

Lemma 3.6. — Supposem∈Z^2q−1\ {0}andN >1. ThenE(m, N)is open inH.

Proof. — We show that the complement ofE(m, N,I)is closed. Suppose that E_n ∈/ E(m, N,I) and d(E_n, E) → 0 as n → ∞ Then we can find x(n)∈E^2q−1_n such that|xj(n)−xi(n)|>N⁻¹ for alli6=j and

2q−1

X

j=1

mjxj(n) = 0.

The Bolzano–Weierstrass theorem tells us that, by extracting a subsequence if necessary, we may suppose thatx_j(n)→x_j as n→ ∞for each16j6 2q−1. Now|xj−xi|>N⁻¹ for alli6=j and

2q−1

X

j=1

mjxj = 0.

ThusEn ∈/ E(m, N).

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Thus our proof of Theorem 3.4 reduces to proving the following result.

Lemma 3.7. — Let m ∈ Z^2q−1\ {0}, let δ > 0 and let > 0. Then, given E ∈ H we can find an F ∈ H with d_H(E, F) < such that the equation

2q−1

X

j=1

mjxj= 0

has no solution withx_j ∈E [16j62q−1]and

|xj−xk|>δfor16j < k62q−1.

4. Completion of the proof of the theorem on sums

The main step in the construction for Lemma 3.7 is the following.

Lemma 4.1. — SupposeE1,E2, . . .Eq are closed subsets ofTsuch that m(E₁+E₂+· · ·+E_q)>1/2.

Then, given1>0 andm∈Z^2q−1\ {0}, we can find F1,F2, . . .Fq closed subsets ofTand₂>0with the following properties.

(i) m(F1+F2+· · ·+Fq)>1/2.

(ii) The Hausdorff distanced_F(E_j, F_j)< ₁for16j6q.

(iii) Ifx1 ∈Sq

s=1Fs xr ∈Sq

s=2Fs for2 6r62q−1,|xr−yr|62 for 16r62q−1 and|yj−yk|>δfor16j < k62q−1, we have

2q−1

X

j=1

mjyj6= 0.

Proof. — Choose0 < γ < ₁/4. We observe that the collection of open sets

E₁+ (γ,−γ)

+E₂+· · ·+E_q form an open cover of the compact set

E1+E2+· · ·+E_q−1+Eq. We can thus find a finite collection of points

e(r)∈E2×E3× · · · ×Eq [16r6N such that

N

[

r=1





q

X

j=2

e_j(r) + E₁+ (γ,−γ)



⊇E₁+E₂+· · ·+E_q−1+E_q.

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We now choose finite subsets F˜_j ofE_j [26j 6q]such that e(r)∈

q

Y

j=2

F˜j

for[16r6N]andd_F(Ej,F˜j)< γ. Automatically E1+ (γ,−γ)

+ ˜F2+· · ·+ ˜Fq ⊇E1+E2+· · ·+Eq−1+Eq. By perturbing each of the points in theF˜_jin turn by an amount less than γwe can find disjoint finite setsFj [26j 6q]such thatd_F(Ej, Fj)<2γ,

(iii)⁰ Ifyr∈Sq

s=2Fsfor16r62q−1and theyr are distinct, we have

2q−1

X

j=1

mjyj6= 0.

and

E₁+ (2γ,−2γ)

+F₂+· · ·+F_q ⊇E₁+E₂+· · ·+E_q−1+E_q. A simple argument shows that

m

E1+ [3γ,−3γ]

+F2+· · ·+Fq

>1/2.

SinceF₂,F₃, . . . ,F_q are finite, it follows that there is a finite setX such that if

y /∈X andyr∈

q

[

s=2

Fs for26r62q−1, with they_rdistinct then

m₁y+

2q−1

X

j=2

m_jy_j6= 0.

Now set

F1= E1+ [3γ,−3γ]

\ X+ (−η, η)

withη >0. Provided we takeη small enough, we haved_F(E1, F1)< 1and m(F₁+F₂+· · ·+F_q)>1/2.

Further, combining (iii)⁰ with the definition ofX we see that (iii)⁰⁰ Ify₁ ∈Sq

s=1F_s y_r ∈ Sq

s=2F_s for2 6r62q−1 and y₁, y₂, . . . , y_2q−1 are distinct then

2q−1

X

j=1

m_jy_j6= 0.

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Take K to be the collection of x ∈ T^2q−1 such that x₁ ∈ Sq s=1F_s, xr∈Sq

s=2Fsfor26r62q−1and|xj−xk|>δ/2for16j < k62q−1.

If we set

L=







x∈T^2q−1 :

2q−1

X

j=1

m_jx_j= 0





 ,

thenK andLare disjoint compact subsets of T^2q−1. A standard theorem now tells us that there exists an⁰₂>0such that, ifx∈Kand|yj−xj|6⁰₂ for16j62q−1, we have

2q−1

X

j=1

m_jy_j6= 0.

If we take 2 = min(⁰₂, δ)/4, then condition (iii) holds and the required

result follows.

We now show how to prove Lemma 3.7 from Lemma 4.1.

We first observe that, by repeated application of Lemma 4.1, with the various2q−1tuples obtained by permuting the entries ofmwe obtain the following version.

Lemma 4.2. — The result of Lemma 4.1 holds with condition (iii) of the conclusion replaced by the following.

(iii)⁰ Suppose16p62q−1. Ifxp ∈Sq

s=1Fs, xr ∈Sq

s=2Fsfor r6=p,

|xr−yr|62for16r62q−1, and|yj−yk|>δfor16j < k62q−1, we have

2q−1

X

j=1

mjyj6= 0.

Next observe that, by repeated application of Lemma 4.2 we can obtain the following version.

Lemma 4.3. — SupposeE₁,E₂, . . .E_q are closed subsets ofTsuch that m(E₁+E₂+· · ·+E_q)>1/2.

Then, given >0,δ >0and m∈Z^2q−1\ {0}, we can find F1, F2, . . .Fq

closed subsets ofTandη >0 with the following properties.

(i) m(F1+F2+· · ·+Fq)>1/2.

(ii) The Hausdorff distanced_F(Ej, Fj)< for16j6q.

(iii) Suppose 1 6 p1 6 2q−1 and 1 6 q1 6 q. If xp1 ∈ Sq

s=1Fs xr ∈ S

s6=q1Fsforr6=p1and|xj−xk|>δfor16j < k62q−1, we have

2q−1

X

j=1

m_jx_j 6= 0.

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We now disentangle condition (iii) of Lemma 4.3.

Lemma 4.4. — Condition (iii) of Lemma 4.3 can be rewritten as follows.

(iii)⁰⁰ Ifxj ∈Sq

s=1Fsfor 16j 62q−1 and|xj−xk|>δ for1 6j <

k62q−1, we have

2q−1

X

j=1

mjxj 6= 0.

Proof. — By a simple counting argument there exists a16q16qsuch

thatFq1 contains at most one of thexj.

We can now prove Lemma 3.7.

Proof of Lemma 3.7 from Lemma 4.3. — SupposeE∈H. If we set E1=E2=· · ·=Eq =E

then, automatically,E1,E2, . . .Eq are closed subsets ofTsuch that m(E₁+E₂+· · ·+E_q)>1/2

and so, by Lemma 4.3 (supplemented by the observation of Lemma 4.4), we can findF₁,F₂, . . .F_q closed subsets ofTwith the following properties.

(i) m(F₁+F₂+· · ·+F_q)>1/2.

(ii) The Hausdorff distanced_F(Ej, Fj)< /qfor16j6q.

(iii) Ifx_j ∈Sq

s=1F_s for16j62q−1and |x_j−x_k|>δfor16j < k6 2q−1, we have

2q−1

X

j=1

mjxj 6= 0.

If we now set F = Sq

s=1F_s, then simple estimates (not the best possible) give

d_F(E, F)6

q

X

s=1

d_F(E, Fs) =

q

X

s=1

d_F(Es, Fs)<

and the remaining conclusions can be read off.

The following observation may be worth making.

Lemma 4.5. — IfE is as in Theorem 2.6 then the set







q−1

X

j=1

mjxj : mj∈Z, xj∈E





 has Lebesgue measure zero.

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Proof. — If not, we can find anm∈Z^q−1 such that F =







q−1

X

j=1

m_jx_j : x_j ∈E







has positive Lebesgue measure. But then F +F contains a non-trivial

interval which is impossible.

5. Proof of the main theorem from the main lemma Although we have simply demanded that ψ(r)→ ∞asr→ ∞in Theo- rem 2.4, we can demand rather better behaviour.

Lemma 5.1. — Ifψ˜:N→Ris a sequence of strictly positive numbers such that ψ(r)˜ → ∞ as r → ∞ and δ > 0 we can find an increasing sequence of strictly positive numbersψ(r)→ ∞ such that

(i) min( ˜ψ(r), δ)>ψ(r)for allr∈N,

(ii) 2ψ(n)>ψ(r)>ψ(n)for all2n>r>n>1.

Proof. — Immediate.

Throughout the rest of this paper q is a fixed integer with q > 1 and ψ : N → R is a fixed sequence of strictly positive numbers obeying the conditions of Lemma 5.1 with

δ= max

r>1 |r|^−1/(2q) log(1 +|r|)1/2−1

. We write

φ(r) =|r|^−1/(2q) log(1 +|r|)1/2

ψ(|r|).

Observe that0< φ(r)61for allr>1and there is a constantK>1such that

Kφ(n)>φ(r)>K⁻¹φ(n) for all2n>r>n>1.

Once again we use a Baire category argument but our metric space is a little more complicated than the Hausdorff metric space(F, D_F).

Definition 5.2. — We take G to be the set of ordered pairs (E, µ) where E is a non-empty closed set, and µ is a probability measure such that

(i) E⊇suppµ.

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(ii) |ˆµ(r)|φ(|r|)⁻¹→0 asr→ ∞.

If(E, µ),(F, τ)∈Gwe define d_G (E, µ),(F, τ)

=d_F(E, F) + sup

r∈Z

|ˆµ(r)−τˆ(r)|φ(|r|)⁻¹.

It is easy to check that(G, d_G)is a complete metric space. Since(T, m)∈ G, the space is non-empty.

Theorem 2.4 thus follows from its Baire category version.

Theorem 5.3. — The set E of (E, µ) such that there exist distinct points

x₁, x₂, . . . , x_q+1∈E, and integersm_j, not all zero such that

q+1

X

j=1

mjxj= 0

is of first category in(G, d_G).

It may be worth remarking that we shall use Baire category, not because it gives an apparently more general theorem, but because it makes the book keeping aspects of the proof rather easier. It should also be said that, even if the arguments of this section appear complicated, they are not deep.

In order to attack Theorem 5.3, we introduce some temporary definitions reflecting the conditions of Lemma 2.5. Ifm∈Z^q+1\ {0}we write

N0(m) = 12



q+ 1 +

q+1

X

j=1

|mj|



.

IfN >24(q+ 1) we writeJ(N)for the collection of ordered(q+ 1)tuples I= (I₁, I₂, . . . , I_q+1)

whereIj = [(nj+¹₂)/N,(nj−¹₂)/N], withnj an integer and

n_j N −n_k

N > 6

N for16j < k6q+ 1.

If m∈Z^q \ {0}, N >N₀(m)and I∈J(N)we writeE(m, N,I)for the set of(E, µ)with the property that ifxj∈E∩Ij then

q

X

j=1

mjxj6= 0.

By the definition of first category, it suffices to prove the following simpler result.

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Lemma 5.4. — If m ∈ Z^q+1\ {0}, N > N₀(m) and I ∈ J(N), then E(m, N,I)is open and dense in (G, d_G).

Proof of Theorem 5.3 from Lemma 5.4. — It suffices to show that E= \

m∈Z^q+1\{0}

\

N>N₀(m)

\

I∈J(N)

E(m, N,I), and, since

E⊆E(m, N,I), we need only show that

E⊇ \

m∈Z^q+1\{0}

\

N>N0(m)

\

I∈J(N)

E(m, N,I).

To this end, let

(E, µ)∈ \

m∈Z^q+1\{0}

\

N>N₀(m)

\

I∈J(N)

E(m, N,I).

Supposem˜ ∈Z^q \ {0} andx1, x2, . . .xq+1 are distinct points inE. If we chooseN˜ >N₀( ˜m)with

N˜ >48 1 + max

16i,j6q+1|xi−xj|⁻¹ , then we can find˜I∈J( ˜N)such that xj ∈I˜j. Since

(E, µ)∈E( ˜m,N ,˜ ˜I) we have

q+1

X

j=1

mjxj6= 0.

Thus(E, µ)∈Eand we are done.

We split the proof of Lemma 5.4 into two parts, the easy Lemma 5.5 and the harder Lemma 5.6 (this depends on Lemma 2.5 which we still have to prove).

Lemma 5.5. — If m ∈ Z^q+1\ {0}, N > N0(m) and I ∈ J(N), then E(m, N,I)is open in(G, d_G).

Proof. — Imitate the proof of Lemma 3.6.

Thus the proof of Lemma 5.4 reduces to the proof of the next lemma.

Lemma 5.6. — If m ∈ Z^q+1\ {0}, N > N0(m) and I ∈ J(N), then E(m, N,I)is dense in(G, d_G).

The proof of Lemma 5.6 from Lemma 5.6 will occupy the rest of this section. The next lemma merely serves to establish notation.

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Lemma 5.7. — Let K :R→R be an infinitely differentiable function with the following properties.

(i)⁰ K(x)>0 for allx∈R. (ii)⁰ R

RK(x)dx= 1.

(iii)⁰ K(x) = 0for|x|>1/4.

IfM is a positive integer and we defineK_M :T→Rby

KM(t) =

(M K(M t) if|t|61/(4M),

0 otherwise,

thenKM is an infinitely differentiable function having the following prop- erties.

(i) KM(t)>0for allt∈T. (ii) R

TK_M(t)dt= 1.

(iii) KM(t) = 0for|t|>1/(4M).

(iv) |KˆM(r)|61for allr.

(v) There exists a constantA, independent of M, such that |Kˆ_M(r)| 6 A(M/r)²for allr6= 0.

Proof. — This is entirely straightforward.

Lemma 5.8. — Given(E, µ)∈Eand >0, we can find(F, τ)∈Esuch thatdτ =gdmwithg infinitely differentiable.

Proof. — Observe that(K_M ∗µ, E+ [−M⁻¹, M⁻¹])∈E and d (KM∗µ, E+ [−M⁻¹, M⁻¹]),(E, µ)

→0

asM → ∞.

Proof of Lemma 5.6 from Lemma 2.5. — In view of Lemma 5.8, it is sufficient to show that, given (E, µ) ∈ E such that dµ = gdm with g infinitely differentiable and > 0, we can find (F, τ) ∈ E(m, N,I) with d (E, µ),(F, τ)

< .

Note that, since g is infinitely differentiable there exists a constant A such that

|ˆg(r)|6A|r|⁻³

for allr6= 0 and, since(E, µ)∈E, there exists a constantB such that

|ˆg(r)|6Bφ(|r|) for allr6= 0.

To this end, observe that givenη >0(to be fixed later), Lemma 2.5 tells us that, sinceηψ(|r|)→ ∞asr→ ∞, we can find an infinitely differentiable functionf with the following properties.

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(i) f(t)>0for allt∈T. (ii) fˆ(0) = 1.

(iii) |fˆ(r)|6η|r|^−1/(2q) log(1 +|r|)^1/2

ψ(|r|) =ηφ(|r|)for allr6= 0.

(iv) Ifx_j ∈suppf∩I_j then

q+1

X

j=1

m_jx_j6= 0.

(v) Ifx∈Twe can find ay∈suppf with|x−y|< η.

Note also that, sinceg is infinitely differentiable, there exists a constant Asuch that

|ˆg(r)|6A|r|⁻³

for allr6= 0. Finally we observe thatr⁻²φ(r)→0asr→ ∞so there exists aC >0such that

Cφ(r)> r⁻² for allr>1.

Set h(t) =g(t)f(t)and choose some 1 > δ >0 (to be fixed later). We seek to estimateˆh(r). Ifr6= 0

|ˆh(r)−ˆg(r)|=

X

m6=0

fˆ(r−m)ˆg(m)

6 X

m6=0

|fˆ(r−m)||ˆg(m)|

= X

06=|m|6|r|/2

|fˆ(r−m)||ˆg(m)|+ X

|m|>|r|/2

|fˆ(r−m)||ˆg(m)|

Using the remarks about the behaviour of φ at the beginning of this section, we have

|fˆ(r−m)|6ηφ(|r−m|)6ηKφ(r) whenever|m|6|r|/2 and

|f(rˆ −m)|6ηφ(|r−m|)6η whenever|r−m| 6= 0. Thus

|h(r)ˆ −g(r)|ˆ 6KAηφ(|r|) X

06=|m|6|r|/2

|m|⁻³+Aη X

|m|>|r|/2

|m|⁻³

6KAηX

m6=0

|m|⁻³+Aη X

|m|>|r|/2

|m|⁻³

6η(10KAφ(|r|) + 10Ar⁻²)610Aη(K+C)φ(|r|) 6δφ(|r|)

for allr6= 0 provided only that we chooseη small enough.

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A similar but simpler argument shows that

|ˆh(0)−ˆg(0)|6δ

provided only that we choose η small enough. If we now set H(t) =

|ˆh(0)|⁻¹h(t)then, since

|Hˆ(r)−ˆg(r)|6|ˆh(r)−g(r)|ˆ +|1−ˆh(0)⁻¹|(|ˆg(r)|+|ˆh(r)−ˆg(r)|), it follows that, provided we pickδsmall enough,

sup

r

φ(r)⁻¹|Hˆ(r)−ˆg(r)|< /2.

Takingτ =HdmandF=E∩suppf we see that(F, τ)∈Eby construction. Providedηis small enough, condition (v) implies thatd_H(E, F)< /2 and so, using the conclusion of the previous paragraph,d (E, µ),(F, τ)

< . Condition (iv) shows that(F, τ)∈E(m, N,I)so we are done.

6. Preparations for the main lemma

Before we start the start the proof of Lemma 2.5 in earnest we need to do some cleaning up.

Lemma 6.1. — If y₁, y₂, . . .y_m are distinct points of T, > 0 and φ is as specified at the beginning of section 5, we can find an infinitely differentiable functionf with the following properties.

(iii) |fˆ(r)|6φ(|r|)for allr6= 0.

(iv) Ify_k ∈/ suppf for16k6m.

(v) Ifx∈Twe can find ay∈suppf with|x−y|< .

Proof. — (The reader may prefer to supply her own proof.) Choose K in Lemma 5.7 In such a way that

K(x) =kKk_∞ for|x|61/16 and set

LM(t) =kKk⁻¹_∞M⁻¹KM(t).

If we set

g(t) = 1−

m

X

j=1

L_M(t−y_k)

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then

|ˆg(r)|6

(_mkKk_∞

M for|r|6M

mAkKk∞

M M

r

2

for|r|>M Ifη >0 then, provided only thatM is large enough, we have

m(A+ 1)kKk_∞

M 6ηM^−3/4 and so

|ˆg(r)|6η|r|^−3/4

for all r 6= 0. If we now set f = kgk⁻¹₁ g then, provided that M is large enough, all the conditions of the lemma follow.

Lemma 6.1 gives a proof of Lemma 2.5 in the particular case when all them_j except one are zero. Lemma 2.5 is also trivial in the case when

0∈/

q

X

j=1

m_jI_j

since we can then takef = 1. Thus we need only prove the following version of Lemma 2.5

Lemma 6.2. — Letφbe as specified at the beginning of section 5 and let >0. Supposem∈Z^q+1,m1, m26= 1andN is a positive integer such that

N >12



q+ 1 +

N

X

j=1

|mj|



.

Suppose further that we are givenI_j = [(n_j−¹₂)/N,(n_j+¹₂)/N], withn_j an integer[16j6q+ 1], such that

nj

N −nk

N > 6

N for16j < k6q+ 1 and

0∈

q

X

j=1

mjIj,

Then we can find an infinitely differentiable functionf with the following properties.

(iii) |fˆ(r)|6φ(|r|)for allr6= 0.

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(iv) Ifx_j ∈suppf∩I_j then

q+1

X

j=1

mjxj6= 0.

(v) Ifx∈Twe can find ay∈suppf with|x−y|< . Lemma 6.2 follows in turn from the following result.

Lemma 6.3. — Suppose the hypotheses of Lemma 6.2 hold and in ad- dition we are given infinitely differentiablegj [16j6q+ 1]such that

(i)⁰ 1>g_j(t)>0 for allt∈T.

(ii)⁰ gj(t) = 0ift /∈[(nj−¹₂−)/N,(nj+¹₂+)/N].

(iii)⁰ g_j(t) = 1ift∈[(n_j−¹₂)/N,(n_j+¹₂)/N].

Then we can find infinitely differentiable functionsfj with the following properties.

(i)⁰⁰ fj(t)>0 for allt∈T. (ii)⁰⁰ fˆj(0) = ˆgj(0).

(iii)⁰⁰ |fˆj(r)−gˆj(r)|6(q+ 1)⁻¹φ(|r|)for allr6= 0.

(iv)⁰⁰ Ifx_j∈suppf_j then

q+1

X

j=1

mjxj6= 0.

(v) Ifx∈Ij we can find ay∈suppfj with|x−y|< .

Proof of Lemma 6.2 from Lemma 6.3. — Choose gj satisfying conditions (i)⁰, (ii)⁰ and (iii)⁰ and setg= 1−Pq+1

j=1gj. If we choosefj satisfying the conclusions of Lemma 6.3 and setf =g+Pq+1

j=1f_j, thenf satisfies the

conclusions of Lemma 6.2.

We can deduce Lemma 6.3 from a result on sums of point masses. Here and elsewhere we write|E| for the number of elements in a finite setE.

Lemma 6.4. — Suppose the hypotheses of Lemma 6.3 hold. Then we can findN0,N1,B >1 andγ >0with the following properties. Ifn>N1

we can find finite sets of pointsE_j [16j6q+ 1]such that writing µj =|Ej|⁻¹kgjk1

X

x∈Ej

δx

the following conditions hold.

(1) |ˆµ_j(r)−gˆ_j(r)|62⁻¹(q+ 1)⁻¹φ(|r|)for all|r|6N₀.

(2) |ˆµj(r)|+|ˆgj(r)|6(q+ 1)⁻¹φ(|r|)for allN06|r|6n^2(q+1).

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(3) If x_j∈E_j+ [−4γn^−q,4γn^−q]then

q+1

X

j=1

mjxj6= 0.

(4) If x∈I_j we can find ay∈E_j with|x−y|< .

Proof of Lemma 6.3 from Lemma 6.4. — letM =M(n)be the integer satisfying

(γn^−q)⁻¹+ 1>M >(γn^−q)⁻¹

and setfj=µj∗KM whereKM is defined as in Lemma 5.7. Automatically thef_j satisfy the conclusions of Lemma 6.3. with the possible exception of (iii)⁰⁰. We now show that (iii)⁰⁰holds, provided only thatnis large enough.

First observe that, provided only thatnis large enough, standard results on approximate identities tell us that

|fˆj(r)−µˆj(r)|62⁻¹(q+ 1)⁻¹φ(|r|) and so, using (1),

|ˆg_j(r)−fˆ_j(r)|6(q+ 1)⁻¹φ(|r|)

for all|r|6N0provided only thatnis large enough. Next we note that

|fˆj(r)|=|ˆµj(r)||KˆM(r)|6|µˆj(r)|

so, using (2),

|fˆj(r)|+|ˆgj(r)|6(q+ 1)⁻¹φ(|r|) whence

|ˆµj(r)−fˆj(r)|6(q+ 1)⁻¹φ(|r|) for allN06|r|6n^2(q+1).

Note that, since gj is infinitely differentiable, there exists a constant C such that

|ˆgj(r)|6C|r|⁻¹

for allr6= 0. Thus, provided only thatnis large enough,

|ˆgj(r)|6(q+ 1)⁻¹φ(|r|)/2

for alln^2q+ 16|r|. Using the equality|fˆj(r)|=|ˆµj(r)||KˆM(r)|, we observe that

|fˆj(r)|6(q+ 1)⁻¹φ(|r|)/2

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and so

|fˆ_j(r)−ˆg_j(r)|6(q+ 1)⁻¹φ(|r|)

for alln^2q+ 16|r|and so we are done.

7. Completion of the proof of the main lemma In this final section we obtain Lemma 6.4 by means of a probabilistic construction. All parts of following theorem are well known (see for example [1]) but it may be helpful to recall the proofs.

Theorem 7.1. — (i) IfY is a real valued random variable with|Y|6 1andEY = 0 then

Ee^λY 6e^λ².

(ii) IfY₁,Y₂. . . are independent real valued random variable with|Y_k|61 andEYk= 0then

Pr

n

X

k=1

Yk >y

!

6e^−y²^/4n.

(iii) If Z₁, Z₂ . . . are independent complex valued random variable with

|Zk|61andEZk = 0then

Pr

n

X

k=1

Z_k

>y

!

64e^−y²^/4n

(iv) Suppose U1, U2 . . . are independent identically distributed random variables taking values onT. If

Pr U1∈[a, b)

=µ [a, b) for some probability measureµthen

Pr





n⁻¹

n

X

j=1

e^irU^k−µ(r)ˆ

>y



64e^−y²^/(16n).

(v) Suppose06α61 andW1,W2, . . . are independent complex valued random variables with|W_k|61 and|EW_k|6αis as in (iv). Then

Pr





n⁻¹

n

X

j=1

W_j

>α+y



64e^−y²^/(16n).

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Proof. —

(i) The result is immediate if|λ|>1. If|λ|61, Ee^λY =

∞

X

r=0

EY^rλ^r r! = 1 +

∞

X

r=2

EY^rλ^r

r! 6= 1 +

∞

X

r=2

|λ|^r r! 6

∞

X

r=0

|λ|^2r r! =e^λ².

(ii) Observe that the random variablese^λY^k are independent so Ee^λ

Pn j=1Y_k

=E

n

Y

j=1

e^λY^k=

n

Y

j=1

Ee^λY^k 6e^nλ².

Thus by a Tchebychev estimate Pr

n

X

k=1

Y_k>y

!

6e^−λyEe^λ Pn

j=1Yk

=e^nλ²^−λy and settingλ=y/2nwe have the desired result.

(iii) Apply part (ii) to<Zk,−<Zk,=Zk and−=Zk. (iv) Observe that

Ee^irU^k= ˆµ(r),

so applying part (iii) withZk = e^irU^k−µ(r)ˆ

/2 gives the required result.

(v) Observe that Pr

n⁻¹

n

X

k=1

W_k

>|µ(r)|ˆ +y

!

6Pr

n⁻¹

n

X

k=1

W_k

>|EW₁|+y

!

6Pr

n⁻¹

n

X

k=1

(W_k−EW_k)

>y

!

64e^−y²^/(16n) as in part (iv).

We now state our probabilistic version of Lemma 6.4.

Lemma 7.2. — Suppose the hypotheses of Lemma 6.3 hold. Set M = Pq+1

j=1|mj|. Then we can findN0,N1,B>1and γ >0 with the following properties such that whenevern>N₁ the following is true.

SupposeXjkare independent random variables taking values onT[16 j 6q+ 1, 1 6k 6n] such thatXjk has probability densitykgjk⁻¹₁ gj. If 26j6q+ 1 take

Ej={Xjk : 16k6n}

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set

E˜₁={Xjk : 16k6n}

and

E1=







x∈E1 : 0∈/ [−4M γn^−q,4M γn^−q] +m1x+

q+1

X

j=2

mjEj





 .

If we take

µ_j =|Ej|⁻¹kgjk1

X

x∈Ej

δ_x

then, with probability at least1/2, following conditions hold.

(1) |ˆµj(r)−gˆj(r)|62⁻¹(q+ 1)⁻¹φ(|r|)for all|r|6N0.

(2) |ˆµ_j(r)|+|ˆg_j(r)|6(q+ 1)⁻¹φ(|r|)for allN₀6|r|6n^2(q+1). (3) If xj∈Ej+ [−4γn^−q,4γn^−q], then

q+1

X

j=1

m_jx_j6= 0.

(4) If x∈I_j we can find ay∈E_j with|x−y|< .

Since any event which has strictly positive probability must have an instance Lemma 7.2 follows from Lemma 6.4.

Most of Lemma 7.2 is easy to prove.

j=1|mj|. Then we can find N₀⁰, N₁⁰ and B⁰ > 1 such that whenever n>N₁⁰ andB>B⁰ the following is true.

SupposeX_jkare independent random variables taking values onT[16 j6q+ 1, 16k6n]such that Xjk has probability densitykgjk⁻¹₁ gj and supposeγ >0. If26j6q+ 1take

Ej={Xjk : 16k6n}

set

E˜1={Xjk : 16k6n}

and

E₁=







x∈E₁ : 0∈/ [−4M γn^−q,4M γn^−q] +m₁x+

q+1

X

j=2

m_jE_j





 .

If we take

µj =|Ej|⁻¹kgjk1

X

x∈Ej

δx

then, with probability at least3/4, the following conditions hold.

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(1)⁰ |ˆµ_j(r)−ˆg_j(r)|62⁻¹(q+1)⁻¹φ(|r|)for all|r|6N₀⁰ and26j6q+1.

(2)⁰ |ˆµj(r)|+|ˆgj(r)| 6 (q+ 1)⁻¹φ(|r|) for all N0 6 |r| 6n^2(q+1) and 26j6q+ 1.

(3) Ifxj∈Ej+ [−4γn^−q,4γn^−q], then

q+1

X

j=1

m_jx_j6= 0.

(4)⁰ If26j6q+ 1 andx∈Ij we can find ay∈Ej with|x−y|< . Proof. — Observe that (3) is always true by virtue of the definition of E1. The weak law of large numbers tells us that, provided only that n is large enough, condition (4)⁰ will hold with probability at least7/8.

Since gj is once continuously differentiable we can find aCj such that

|gˆ_j(r)|6C_j|r|⁻¹ for all|r|>0and so we can find anN₀⁰ such that

|gˆj(r)|64⁻¹(q+ 1)⁻¹φ(|r|) for all|r|6N₀⁰ and26j6q+ 1.

By Theorem 7.1

Pr |ˆµj(r)−gˆj(r)|>Bn^−1/2(logn)^1/2

= Pr

n⁻¹

n

X

k=1

e^irX^k−µ(r)ˆ

>Bn^−1/2(logn)^1/2

!

64e^−B^log^n/(16n). Thus, if we chooseB>64(q+ 1), we have

Pr |ˆµj(r)−gˆj(r)|>Bn^−1/2(logn)^1/2

64n^−4(q+1)

for allrand all26j6q+ 1. Thus provided only thatnis large enough, Pr |ˆµ_j(r)−gˆ_j(r)|>Bn^−1/2(logn)^1/2

64n^−4(q+1) will hold with probability at least7/8.

Using the results of the two previous paragraphs we see that conditions (1)⁰ and (2)⁰ will both hold (with probability at least 7/8) provided only

thatnis large enough. The result follows.

We now prove the harder part of Lemma 7.2.

j=1|m_j|. Then we can findN₀⁰⁰,N₁⁰⁰,B⁰⁰>1andγ >0 such that when- evern>N₁⁰⁰andB >B1the following is true.

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SupposeX_jkare independent random variables taking values onT[16 j 6q+ 1, 1 6k 6n] such thatXjk has probability densitykgjk⁻¹₁ gj. If 26j6q+ 1 take

Ej={Xjk : 16k6n}

set

E˜₁={Xjk : 16k6n}

and

E1=







x∈E˜1 : 0∈/ [−4M γn^−q,4M γn^−q] +m1x+

q+1

X

j=2

mjEj





 .

If we take

µ1=|E1|⁻¹kgjk1

X

x∈E1

δx

then, with probability at least3/4, following conditions hold.

(1)⁰⁰|µˆ1(r)−gˆ1(r)|62⁻¹(q+ 1)⁻¹φ(|r|)for all|r|6N₀⁰⁰.

(2)⁰⁰|µˆ₁(r)|+|ˆg₁(r)|6(q+ 1)⁻¹φ(|r|)for allN₀⁰⁰6|r|6n^2(q+1). (4)⁰⁰Ifx∈I1we can find ay∈E1 with|x−y|< .

Proof. — Let E_∗=E1\E˜1

=







x∈E˜1 : 0∈[−4M γn^−q,4M γn^−q] +m1x+

q+1

X

j=2

mjEj





 and

τ=n⁻¹kgk1

X

x∈E∗

δ_x.

The main step of the proof involves finding an upper bound forτˆ(r)which holds with high probability independent of the choice ofγ.

First observe that, if we set Wk =e^irX^1k when X1k ∈E_∗ andWk = 0 otherwise, then

ˆ

τ(r) =kgk₁

n

X

k=1

Z_k theWk satisfy the conditions of Theorem 7.1 (v).

SinceX2k has density functiong2/kg2k1 it follows that, to first order in δt

Pr(m2X2k∈[t, t+δt]) =G(t)δt

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whereGis differentiable density function with first and second derivatives bounded by someK1 depending only on m2andg2. Thus

Pr(m2X2k∈[t, t+δt] : for some 16k6n)}=nG(t)δt to first order inδtand

Pr (m2E2+m3E3+· · ·+mq+1Eq+1)∩[t, t+δt]6=∅

=nG∗H(t)δt for someH. We observe thatG∗H is differentiable density function with first and second derivatives bounded byK1. It follows that, iftis fixed Pr (t+m2E2+m3E3+· · ·+mq+1Eq+1)∩[−4M γn^−q,4M γn^−q]6=∅

=Fγ(t) whereF_γhas continuous first and second derivatives bounded bykFk⁻¹_γ K₁. Thus the density functionGγ ofX11 given that

(m₁X₁₁+m₂E₂+m₃E₃+· · ·+m_q+1E_q+1)∩[−4M γn^−q,4M γn^−q]6=∅ has a continuous derivative bounded byK2, whereK2is independent both ofγandn.

We now have

|EWk|= Pr(Wk 6= 0)

E(Wk|Wk6= 0)

=|Gˆγ(r)|6K2

|r|

forr6= 0. Using Theorem 7.1 (v), we see that that, if takeB>64(q+ 1) then provided we takenlarge enough, there is a probability at least31/32 that

(1) |ˆτ(r)|6K₂|r|⁻¹+Bn^−1/2(logn)^1/2 for all16|r|6n^2(q+1).

Next we observe that

P r(X1k ∈E_∗)6n^q×(8M γn^−q) = 8M γ

so the expected number of points inE^∗is no greater than 8M γ. Since yPr(Y >y)6EY ,

it follows that givenη >0(to be fixed later) we can chooseγso small that with probability at least31/32E^∗ contains at mostηnpoints and so

(2) kτk6ηkgk₁.

If we set

µ=n⁻¹kgk1

X

x∈E1

δx,