A Geometric Proof of the Circular Maximal Theorem

Ageometricproofofthecircularmaximaltheorem

W. Schlag1

1 Introduction

A well–known result of Bourgain [1] asserts that the circular maximal function

f(x) = sup f(x ty) dσ(y)(1) M 1 | − | 12. Simple examples show that this fails for p = 2. In this note we derive Bourgain’s result by geometric and combinatorial methods. In particular, we do not use the Fourier transform in any way. Our proof is based on a combinatorial argument from [6], which in turn uses Marstrand’s three circle lemma [7], and a lemma involving two circles that seems to originate in [10]. The three circle lemma was used in [7] to prove the following result, which is a simple consequence of Bourgain’s theorem: Suppose a planar set E has the property: for every point in the plane, E contains some circle with that point as center. Then E has positive measure. Thus we show here that Marstrand’s lemma, in combination with other ideas, does indeed allow one to establish the stronger maximal function estimate. Furthermore, we demonstrate in section 4 how to obtain the entire known range of Lp Lq estimates for → the circular maximal function (which is optimal possibly up to endpoints), see [9] and [11], by using the methods from sections 2 and 3. One — perhaps signiﬁcant — distinction from the techniques developed in [1], [8], [11], and [12], which involve the Fourier transform, is the fact that the methods presented here do not seem to yield estimates for the global maximal function f(x) = sup f(x ty) dσ(y). M 0

1 For the strong maximal function defined in terms of rectangles in Rn with sides parallel to the coordinate axes it was shown in [4] that weak Lp bounds are equivalent to certain geometric properties of collections of rectangles. [4] is related to this paper in so far as inequalities for maximal functions are proved by working directly with the associated geometric families. Moreover, in [3] page 37, A. Córdoba posed the problem of finding a geometric proof of Stein’s spherical maximal theorem [13] and he suggested that it might be possible to settle the two–dimensional case by studying families of annuli in the plane. Here it is shown that it is indeed possible to prove the correct bounds on the circular maximal function by a careful analysis of collections of annuli. This paper is organized as follows. Proposition 1.1 illustrates how maximal function estimates can be reduced to counting problems involving large families of thin annuli in the plane. We do not use the full equivalence as stated in Proposition 1.1, but only the fact that multiplicity estimates imply suitable Lp bounds. However, it might be of interest to know that the multiplicity bounds are indeed natural. Given a large collection of annuli with δ–separated centers, section 2 establishes estimates on the total number of annuli that can intersect a typical one. It turns out that these inequalities are essential in the analysis of the circular maximal function. In sections 3 and 4 they are used in combination with the three circle lemma to prove Bourgain’s theorem and the Lp Lq → bounds, respectively. In those sections the reader will find heuristic arguments which explain the underlying observations for the main results, i.e., Theorems 3.1 and 4.1. It is perhaps worth mentioning that we do not use the method of cell decomposition that was recently applied in [14] to prove a sharp maximal function estimate. The motivation for this paper was to adapt the method from [6], which was developed there for one–parameter families of circles, to the two–parameter setting of (1). It seems that the improvement over the method in [6] that was achieved in [14] using cells and the work required to pass from the one–parameter to the two–parameter case are of a different nature.

Definition 1.1 Let δ>0 be an arbitrary but fixed small number. By we shall always C 2 mean a family of circles with δ–separated centers lying in some fixed compact set of diameter < 1 . For our purposes we may assume that the circles in are in general position, 2 C in particular, all radii are distinct. Let

C = C(x, r)= y R2 : x y = r { ∈ | − | } Cδ = Cδ(x, r)= y R2 : r δ< x y

C δ µδ = χC . C ∈C will denote the following auxiliary maximal function: Mδ 1 δf(x) = sup δ f(y) dy. M 1

Proposition 1.1 The following are equivalent:

i. For every p>2 there exists a constant c(p) depending only on p so that

f p 2 c(p) f p 2 (3) Mδ L (R ) ≤ L (R )

for all f Lp(R2) and all δ>0. ∈ ii. Given δ>0 and , a family of circles with δ–separated centers, and a small number C 1 ρ>0, there exists with >c− for some constant c depending only A⊂C |A| ρ |C| ρ on ρ and so that

δ 1 ρ 1 δ C : µA >cλ− − δ− <λC (4) |{ δ ρ }| | | for all C and all 0 <λ 1. ∈A ≤ 3 Proof: Assume the second statement. For this implication we follow [6]. In view of Marcinkiewicz’s interpolation theorem it suﬃces to prove a restricted weak–type estimate for every p>2. Fix such a p and let δ>0, E R2 compact, and 0 <λ 1. Pick a ⊂ ≤ δ–net x in { j} 2 x R : δχE(x) >λ { ∈ M } and let = C(x ,r ) where r (1, 2) is chosen so that C { j j } j ∈

Cδ(x ,r ) E >λCδ(x ,r ) | j j ∩ | | j j | for all j. Applying (ii) with ρ = p 2 yields with property (4). Hence − A⊂C

δ 1 ρ 1 λ δ C E : µA c (λ/2)− − δ− C |{ ∩ δ ≤ ρ }| ≥ 2 | | for all C and thus ∈A

1 p 1 λδ µδA E cρ(λ/2) − δ− . 1 ρ 1 |A| E : µ cρ(λ/2) δ ≤| | { δA≤ − − − }

1 In view of >c− this implies |A| ρ |C|

1 2 1 2 p λ(δ ) p cρ E p . |C| | |

By our choice of x we ﬁnally conclude that { j}

2 1 1 λ x R : δχE(x) >λ p c(p) E p , |{ ∈ M }| ≤ | | as desired. To deduce the second statement from the ﬁrst, we shall use an argument that seems to originate in [9], cf. Lemma 2.1. Fix ρ>0 small. We claim that there exists cρ so that for half the circles C ∈C

δ 1 ρ 1 δ C : µC >cλ− − δ− <λC |{ δ ρ }| | | 4 for all 0 <λ 1. Assume that this fails with a choice of c to be speciﬁed below. Then ≤ ρ at least half the circles C satisfy ∈C

δ 1 ρ 1 1 δ C : µC >cλ− − δ− > λ C (5) |{ δ ρ }| 2 | |

j jρ for some dyadic λ =2− (0, 1] depending on C.Leta = ∞ 2− . We claim that ∈ ρ j=0 1 1¯ρ ¯ there is satisfying a− λ with some dyadic λ (0, 1] and so that (5) B⊂C |B| ≥ 2 ρ |C| ∈ holds with λ = λ¯ for all C . This is a simple application of the pigeon hole principle. ∈B Indeed, suppose our claim failed. Then

j card( C : C satisﬁes (5) for some λ =2− (0, 1] ) { ∈C ∈ } 1 1 ρ 1 < a− λ = , 2 ρ |C| 2|C| λ=2 j 1 − ≤ contrary to our assumption. Now let 1 ρ 1 E = µC >cλ¯− − δ− . { δ ρ } We distinguish two cases. Let p =1+√1+ρ.

p p+1 1 p+ρ 2 Case 1: E < (a c(p) 2 )− λ¯ δ | | ρ |C| Applying (3) to f = χE yields

1 1 1+ ρ 2 1 1 1 1 1 p − p p p p 2 p p 2− − aρ λ¯ δ λ¯(δ ) c(p) E , |C| ≤ 2 |B| ≤ | | which is a contradiction.

p p+1 1 p+ρ 2 Case 2: E (a c(p) 2 )− λ¯ δ | |≥ ρ |C| In this case we use duality. Note that the dual inequality to (3) is

1+ 2 p 1 a χ δ p 2 c(p)δ− p ( a ) p j C (yj ,ρj )L (R ) ≤ | j| j j for all families C(y ,ρ ) with δ–separated centers. We apply this to our family with { j j } C aj = 1. Then 1 ρ 1 1 1+ 2 1 c λ¯− − δ− E p µC p 2 c(p)δ− p p . ρ | | ≤ δ L (R ) ≤ |C| 5 This contradicts our assumption on E for large c , since p was chosen so that (1+ρ)p = | | ρ p + ρ. We prove Bourgain’s theorem, i.e., statement (i), by showing directly that a set A⊂C as in the second statement exists, cf. Theorem 3.1.

2 The two circle lemma

The following simple geometric lemma is well–known, see [1], [6], [7], and [14]. We refer the reader to [9], Lemma 4.2 for a proof of the statement below. Let C = C(x, r) and C = C(¯x, r¯). The notation

∆(C, C)=max( x x¯ r r¯ ,δ),d(C, C)= x x¯ + r r¯ || − |−| − || | − | | − |

will be used throughout. Note that x x¯ r r¯ = 0 if and only if the two circles are || − |−| − || internally tangent. If ∆(C, C)= we say that C and C are –tangent. This means that the shortest distance between the intersection points of C, C with the line joining their centers is equal to .

Lemma 2.1 Suppose x, y R2, 0 < x y 1 , and r, s (1, 2), 0 <δ<1. Then there ∈ | − |≤ 2 ∈ is an absolute constant A0 so that

i. Cδ(x, r) Cδ(y, s) is contained in a δ–neighborhood of an arc on C(x, r) of length ∩ ∆ x y A0 x y centered at the point x r sgn(r s) x−y . ≤ | − | − − | − | ii. the area of intersection satisﬁes

2 δ δ δ C (x, r) C (y, s) A0 . ∩ ≤ ∆ x y | − | The second part of Lemma 2.1 shows that the angle of intersection of C, C is propor- tional to ∆(C, C) x x¯ . This should indicate that it is important to know the size of | − | 6 ∆and the distance of the centers of intersecting circles. For this reason we introduce the set δ C = C : Cδ C = , δ ∆(C, C) 2, t x x¯ 2t , Ct { ∈C ∩ ∅ − ≤ ≤ ≤| − |≤ } where , t [δ, 1]. We shall make frequent use of the following simple observations. Firstly, ∈ δ C C iﬀ C C . Secondly, let y Cδ C , i.e. , x y r <δand x¯ y r¯ <δ. ∈Ct ∈Ct ∈ ∩ || − |− | || − |− | Then

r r¯ x y r + x¯ y r¯ + x y x¯ y | − |≤||− |− | || − |− | || − |−| − || < 2δ + x x¯ 2(δ + t) 4t. (6) | − |≤ ≤ In particular, d(C, C) 6t and 4t if C = . ≤ ≤ Ct ∅ In the following paragraph we give a heuristic discussion of the results in this section. Using the Fourier transform one obtains the well–known estimate, see [1] and [2],

1 f 2 2 log δ 2 f 2 2 . Mδ L (R ) | | L (R ) By the arguments in Proposition 1.1 this is equivalent to

δ b 1 1 C : µC > log δ λ− δ− <λδ |{ δ | | }| for some constant b, most circles C , and all λ (0, 1]. These estimates can improved. ∈C ∈ In fact, in Corollary 3.6 of [9] it was shown that there exists an absolute constant C0 so that

1 1 f 2 C t 2 log δ 2 f 2 2 Mδ L (B(x0,t)) ≤ 0 | | L (R ) 2 for all x0 R , 0

C δ t b 1 1 C : µC log δ λ− δ− t <λδ (7) |{ δ | | }| for at least half the circles in , all λ,, t [δ, 1], and a suitable constant b. This in turn C ∈ implies that

3 1 2 2 t C log δ b+1 (8) |Ct | δ δ | | 7 for at least half the circles C and all , t [δ, 1]. Indeed, let ∈C ∈ δ C j j+1 S = C : µCt [2 , 2 ) . j { δ ∈ } b j On the one hand, in view of (7), S log δ 2− t. On the other hand, the area estimate | j| | | in Lemma 2.1 implies, roughly speaking, that S splits into S /(δ2/√t) many rect- j | j| δ angles of size δ δ each of which is intersected by no more than 2j+1 many C C . × √t ∈Ct Thus 3 b j 1 2 C δ log δ 2− t j+1 b 2 t card( C t : C Sj = ) | | 2 log δ . { ∈C ∩ ∅} δ2/√t ∼| | δ δ Since clearly δ C C C : C S = , Ct ⊂ { ∈Ct ∩ j ∅} 1 2j δ 2 ≤ ≤ − (8) follows. It seems reasonable to conjecture that (7) and (8) should hold without the logarithmic factors. This would be optimal, as can be seen from the family of circles C with δ–separated centers in B(0,t) which are –tangent to the unit circle. Indeed, if C = C(x, r) with t/2 < x 0, see Proposition 2.1.

η ρ In section 3 it will turn out that this loss of − can be compensated for by a factor λ− for some small ρ>0. In view of Proposition 1.1 this is exactly what one can aﬀord to loose. The following statement is the main ingredient for the two circle lemma, Lemma 2.5.

It will be understood that Cj = Cj(xj,rj)forj =0, 1,....

Lemma 2.2 Suppose C C1 , t 8, and that β 100. Then 2 ∈Cβτ ≥ ≥ 2 C1 C2 t t t . |C ∩C | δ2 √βτ

8 Proof: Let F (x, r)=(x x r r , x x r r ) be deﬁned on | − 1|−| − 1| | − 2|−| − 2| 2 Ω= (x, r) R [1, 2] : t x xj 2t, j =1, 2 . { ∈ × ≤| − |≤ }

x xi Suppose (x, r) Ωand let ei = x−x and σj = sgn(r rj). Then ∈ | − i| −

e1 σ1 DF(x, r)= −  e σ  2 − 2 and thus JF(x, r) (e σ ,e σ )=α.   ∼ 1 1 2 2 By we mean the angle [0,π] and JF2 denotes the sum of the squares of all ∈ 2 2 subdeterminants of DF. Suppose (x, r) Ωand F (x, r) < 4. Then there exist r × ∈ | | j so that r r < 4 and | j − j|

x x = r r for j =1, 2. | − j| | − j|

Moreover, x x t 8 and r r x x < 4 imply that sgn(r r )=σ . | − j|≥ ≥ || − j|−| − j|| − j j Thus

x x 2 = x x 2 + x x 2 2(x x) (x x) | 1 − 2| | − 1| | − 2| − 1 − · 2 − 2 2 = r r + r r 2σ σ r r r r + | − 1| | − 2| − 1 2| − 1|| − 2| +2σ σ x x x x (1 cos α) 1 2| − 1|| − 2| − 2 = r r +2σ σ x x x x (1 cos α), | 1 − 2| 1 2| − 1|| − 2| − and consequently, in view of the deﬁnition of ∆(C1,C2) and (6)

t2α2 βτ 50τ βτ. − We conclude that √βτ 1 JF on Ω F − (B(0, 4)). t ∩ Changing variables, or more precisely, using the coarea formula, see Theorem 3.2.11 in [5], we obtain

1 1 (F − (y) Ω) dy = JF(x, r) dxdr B(0,4) H ∩ Ω F 1(B(0,4)) ∩ − 2 √βτ 1 t Ω F − (B(0, 4)) . t | ∩ | 9 1 To bound the Hausdorﬀmeasure, note that diam(F − (y) Ω) t provided y t and ∩ | | 1 hence the length of the (algebraic) curve F − (y) Ωwill also be bounded by t. This clearly implies that 1 2 proj 2 (Ω F − (B(0, 2))) t , | R ∩ | √βτ as desired. Lemma 2.2 is suﬃcient for our purposes. However, we show below how to estimate C1 C2 in those cases where Lemma 2.2 does not apply, cf. Lemma 2.5. This can be |Ct ∩Ct | done using Lemma 2.3, which we shall use repeatedly in what follows. It is a quantitative version of the following simple observation: if Cj = C(xj, 3/4) are internally tangent to

C(0, 1) for j =1, 2 with the points of tangency being far apart, then C1 and C2 intersect each other transversely. This fact is of course well–known and has been used in [1] and [14]. See [12] for a harder version in the context of variable coeﬃcients.

C0 Lemma 2.3 Let C0,C1,C2 be three circles so that Cj for j =1, 2. Assume ∈Cβj τj

α = (sgn(r r )(x x ), sgn(r r )(x x )) 1 − 0 1 − 0 2 − 0 2 − 0 (β1 + β2)(τ1 + τ2) A0 (9) ≥ τ1τ2 for some suﬃciently large constant A0. Then

∆(C ,C )d(C ,C ) α2τ τ 1 2 1 2 ∼ 1 2 and thus, in particular, ∆(C ,C ) 2(β + β ). 1 2 ≥ 1 2

Proof: Let σ = sgn(r r ). Then j j − 0

x x 2 = x x 2 + x x 2 2(x x ) (x x ) | 1 − 2| | 1 − 0| | 2 − 0| − 1 − 0 · 2 − 0 r r 2 = r r 2 + r r 2 2(r r )(r r ) | 1 − 2| | 1 − 0| | 2 − 0| − 1 − 0 2 − 0 x x 2 r r 2 = x x 2 r r 2 + x x 2 r r 2 + | 1 − 2| −| 1 − 2| | 1 − 0| −| 1 − 0| | 2 − 0| −| 2 − 0| 10 +2σ σ ( r r r r x x x x )+ 1 2 | 1 − 0|| 2 − 0|−| 1 − 0|| 2 − 0| +2σ σ x x x x (1 cos α). 1 2| 1 − 0|| 2 − 0| −

This implies that (recall that d(C ,C ) 6τ ,see(6)) j 0 ≤ j

∆(C ,C )d(C ,C ) τ τ α2 + O(β τ + β τ + τ β + τ β ) 1 2 1 2 ∼ 1 2 1 1 2 2 1 2 2 1 = τ τ α2 + O((β + β )(τ + τ )) τ τ α2 1 2 1 2 1 2 ∼ 1 2 where we have used (9) in the last step. Since d(C ,C ) d(C ,C )+d(C ,C ) τ + τ 1 2 ≤ 1 0 0 2 1 2 the lemma follows.

A0 will denote the constants in Lemmas 2.1 and 2.3. Using Lemma 2.3 we can deal with the case β 100 that was left open in Lemma 2.2. The intuition behind Lemma 2.4 ≤ is as follows. If C and C are tangent, then any circle C C1 C2 has to intersect the 1 2 ∈Ct ∩Ct arc of minimal length on C that contains C C. 1 1 ∩ 2

Lemma 2.4 Suppose C C1 . Then 2 ∈Cβτ 2 C1 C2 t + β t t 2 . |C ∩C | δ τ Proof: We may assume that τ 4t. Indeed, let C = C(x, r) C1 C2 . Then ≤ ∈Ct ∩Ct τ x x x x + x x 4t.Let ≤| 1 − 2|≤| 1 − | | − 2|≤ β + (β + )(t + τ) γ0 = . τ ∼ tτ Suppose C C1 satisﬁes min( (¯x x ,x x ), (¯x x ,x x )) >Aγ . ∈Ct − 1 2 − 1 − 1 1 − 2 0 0 We apply Lemma 2.3 with C0 = C1, C1 = C2, C2 = C, β1 = β, β2 = , τ1 = τ, and τ = t to wit ∆(C,C ) 2( + β) > 2. In particular C C2 . We conclude that any 2 2 ≥ ∈ Ct C C1 C2 has to satisfy min( (¯x x ,x x ), (¯x x ,x x )) A γ . ∈Ct ∩Ct − 1 2 − 1 − 1 1 − 2 ≤ 0 0 In particular, the centers of all circles in C1 C2 are contained in a 4t 2tA γ rect- Ct ∩Ct × 0 0 2 angle centered at x and thus C1 C2 t γ , as claimed. 1 |Ct ∩Ct | δ2 0 The two circle lemma now follows easily from Lemmas 2.2 and 2.4.

11 Lemma 2.5 Suppose C C1 . Then 2 ∈Cβτ 2 C1 C2 t t t min( , ). (10) |C ∩C | δ2 τ √βτ Proof: As before we may assume that τ 4t. Moreover, we may also assume that ≤ 2 8 t. Indeed, since C1 C2 t either ≤ |Ct ∩Ct | δ2 1 1 or τ ≤ 10 √βτ ≤ 100 without loss of generality. Hence, if β 100 we apply Lemma 2.2 to conclude ≥ 2 C1 C2 t t t . |C ∩C | δ2 √βτ

If on the other hand β 100, then (10) follows from Lemma 2.4. ≤ The following lemma is a simple technical statement that we shall use repeatedly. It is based on the observation that two circles C ,C C will have to intersect provided 1 2 ∈Ct dist(Cδ Cδ,Cδ Cδ) is suﬃciently large (recall that all radii are [1, 2] and that the ∩ 1 ∩ 2 ∈ centers are no more than a distance 1/2 from each other).

Lemma 2.6 Let C ,C C such that sgn(r r ) = sgn(r r ), and (x x, x x) 1 2 ∈Ct − 1 − 2 1 − 2 − ≥ A . Then C C = . 0 t 1 ∩ 2 ∅ Proof : Consider the case r>r1,r2. We may assume that x = 0. By Lemma 2.1,

C C R , ∩ i ⊂ i

xi an –neighborhood of an arc on C centered at r x , i =1, 2, of length A0 t . By our | i| assumptions, R R = . In particular, p = x + r x1 R exterior(C ) and 1 2 1 1 1 x1 1 2 ∩ ∅ | | ∈ ⊂ p = x + r x2 R exterior(C ). Since r > 1 > x (recall that the centers lie in a 2 2 2 x2 2 1 i i | | ∈ ⊂ | | set of diameter < 1 ), 0 interior(C ) interior(C ) and thus the segment (0,p ) intersects 2 ∈ 1 ∩ 2 2 C1 in an interior point of C2,sayq1. Hence the arc p1q1 on C1 intersects C2. Finally, the

case r

12 In Lemma 2.7 we apply the two circle lemma in order to bound the total number of circles that can intersect a given one. This will be crucial in proving the multiplicity

t estimate (4). Lemma 2.7 states, roughly speaking, that an unwanted power of can be cut in half at the cost of eliminating half the circles.

Lemma 2.7 Suppose that for some 1 α,ρ 0 and some constant A ≥ ≥ 3 1 2 α C 2 t t ρ A t− (11) |Ct |≤ δ δ for all C , , t [δ, 1]. ∈C ∈ Fix 0 <ν 1 and assume ν + ρ 1+α. Then there exists , 1 and a ≤ ≤ A⊂C |A| ≥ 2 |C| constant bν so that

3 α 1 2 2 +ν C 1 2 t t ρ 3ν (b A) 2 t− 2 − 2 |Ct |≤ ν δ δ for all C , , t [δ, 1]. ∈A ∈

Proof: Assume false. Then for at least half the circles C in C 3 α 1 2 2 +ν C 1 2 t t ρ 3ν (b A) 2 t− 2 − 2 (12) |Ct | ν δ δ for some choice of dyadic , t [δ, 1] depending on C. This will lead to a contradiction ∈ if bν is suﬃciently large. The idea is as follows. Suppose (12) holds for a ﬁxed choice of , t [δ, 1] and for all C where 1 . Consider the set ∈ ∈B⊂C |B| ≥ 2 |C|

S = (C, C ,C ):C ,C,C C . 0 { 1 2 ∈B 1 2 ∈Ct }

Clearly,

C 2 card(S0) min t , (13) C ≥|B| ∈B |C |

which in turn can be estimated by (12). To bound card(S0) from above, we assume that the majority of (C, C ,C ) S satisfy C C1 . Using Lemma 2.3 we will see below 1 2 ∈ 0 2 ∈Ct 13 that this is the most significant case. Now we count by choosing first C1, then C2, and finally C:

card(S ) C1 C2 . (14) 0 ≤ |Ct ∩Ct | C1 C1 C2 ∈C ∈Ct The cardinality of the intersection is estimated by Lemma 2.5, whereas C1 is controlled |Ct | by our hypothesis (11). The reader will easily check that the bounds (13) and (14) obtained in this way agree if ν = 0. The terms involving ν are of a technical nature. They arise because we apply the pigeon hole principle to make the above argument rigorous. The details are as follows.

jν Let aν = j∞=0 2− . We claim that for some (ﬁxed) choice of , t (12) holds for at 2 1 ν least (8a )− many circles C. This follows from a standard pigeon hole argument. ν |C| Indeed, if our claim failed then

j k card( C : (12) holds for some =2− ,t=2− [δ, 1] ) { ∈C ∈ } ν 2 1 ν < (8a )− t ν t |C| =2 j 1 /4 t=2 k 1 − ≤ ≤− ≤ ∞ ∞ 2 1 lν kν 1 < (8a )− 2− 2− , ν |C| ≤ 2|C| k=0 l= 2 − contradicting our assumption. Now ﬁx , t as in the claim and let be the set of circles B for which (12) holds with those values. Thus

2 ν a− . (15) |B| ν |C| Deﬁne

S = (C, C ,C ):C ,C,C C , sgn(r r )=sgn(r r ) . (16) 0 { 1 2 ∈B 1 2 ∈Ct − 1 − 2 }

1 C 2 Clearly, S0 minC t . | |≥ 4 |B| ∈B |C | Case 1: The majority of (C, C ,C ) S satisfy 1 2 ∈ 0 (x x, x x) A . 1 − 2 − ≤ 0 t 14 Let S1 be the set of those triples. Then, in view of (12),

3 α+2ν 1 C 2 t t ρ 3ν S1 min t bνA t− − . (17) | |≥8|B| C |C | |B| δ δ ∈B On the other hand,

3 α t t ρ S A t− . (18) | 1| |B| δ δ Indeed,

C S1 card( C2 t : (x1 x, x2 x) A0 )(19) | |≤ C { ∈C − − ≤ t} C C1 ∈B ∈Ct 3 1 2 C 2 t max t A0 C |B| ∈B |C | δ δ 3 3 1 2 α 1 2 2 t t ρ 2 t A t− |B| δ δ δ δ

by (11). To bound the cardinality of the set in (19), simply observe that the centers x2 will lie in a rectangle of size A √t t centered at x. ∼ 0 × Clearly, (17) and (18) contradict each other for large bν. Case 2: The majority of (C, C ,C ) S satisfy 1 2 ∈ 0 (x x, x x) >A . (20) 1 − 2 − 0 t Let

S = (C, C ,C ):C ,C,C C , sgn(r r) = sgn(r r), (x x, 2 { 1 2 ∈B 1 2 ∈Ct 1 − 2 − 1 − x x) >A ,β δ ∆(C ,C ) 2β,τ x x 2τ . 2 − 0 t − ≤ 1 2 ≤ ≤| 1 − 2|≤ } 1 ν Here β,τ [δ, 1] are chosen by applying the pigeon hole principle with weights a− β and ∈ ν 1 ν aν− τ , respectively so that

2 ν 2 ν C 2 card(S2) aν− (βτ) card(S0) aν− (βτ) min t . (21) C |B| ∈B |C | 15 By (20) and the intersecting circles lemma, Lemma 2.6, any (C, C ,C ) S satisﬁes 1 2 ∈ 2 C C1 . Thus Lemma 2.5 and (11) imply that 2 ∈Cβτ

card(S ) C1 C2 2 ≤ |Ct ∩Ct | C1 C1 C2 ∈C ∈Cβτ 1 2 3 α 2 β τ 2 τ ρ t A τ − min( , ). (22) |C| δ δ β δ2 τ √βτ Since (C, C ,C ) S satisfy (20), Lemma 2.3 implies that 2β ∆(C ,C ) 4.More- 1 2 ∈ 2 ≥ 1 2 ≥ over, τ x x x x + x x 4t. Now it is easy to see that (21) and (22) are ≤| 1 − 2|≤| 1 − | | − 2|≤ incompatible. Indeed, ﬁrst note that (22) is the same as

2 α t τ τ ρ card(S ) A τ − . 2 |C| δ δ3 β On the other hand, (12), (15), and (21) imply

3 α+2ν 2 ν 2 ν t t ρ 3ν card(S ) a− a− (βτ) β A t− − 2 ν |C| ν ν δ δ 2 1+α ρ ν α+ν α 4 t τ t − − β τ ρ a− b A τ − . ν ν |C| δ δ3 τ β Since 1 + α ρ + ν, the upper and lower bound will contradict each other for large b . ≥ ν Proposition 2.1 is the main result of this section. Starting from the trivial bound

t2 C , (23) |Ct | δ2

we iterate Lemma 2.7 in order to get as close to the t –improvement as possible, see the discussion following Lemma 2.1 above.

Proposition 2.1 Let be as above and let η>0 be a small number. Then there exist a C 1 constant c , a subset so that c− , and η A⊂C |A| ≥ η |C|

3 1 2 C 2 t η c − (24) |At|≤ η δ δ 16 for all C and all , t [δ, 1]. In particular, every C satisﬁes ∈A ∈ ∈A

C δ t 2 1 1 η C : µA >c λ− δ− t− <λδ (25) |{ δ η }|

for all , t [δ, 1], 0 <λ 1. ∈ ≤

j Proof: Let α = 1 2 for j =0, 1, 2,... and ν = η/6. We claim that there exist , j 2 3 Aj ⊂C j j 2− so that for all C j (bν being the constant from Lemma 2.7) |A |≥ |C| ∈A 3 1 2 αj C 2 t t η/2 ( ) b t− | Aj t|≤ ν δ δ for all j satisfying α 2η/3. j ≥ This follows by induction using Lemma 2.7 with ρ =3ν.Forj = 0 simply observe

that 3 2 1 2 α0 C t 2 t t ρ b t− |Ct | δ2 ≤ ν δ δ for all C = . For the induction step note that ∈A0 C 1 2 αj 1 + ν αj 1 = αj 2 − ≤ 3 −

3 since αj 1 = αj 6ν = η. Furthermore, ρ/2+3ν/2=ρ = η/2. Finally, we have − 2 ≥ 1+αj 1 5η/6=ν+ρ, which establishes our claim. Thus the ﬁrst part of the proposition − ≥ follows by letting = where j is maximal with α 2η/3. A Aj0 0 j0 ≥ Assume that the second part of the proposition is false. Then there is a circle C ∈A and numbers , t,λ so that

C δ t 2 1 1 η C : µA >c λ− δ− t− >λδ. |{ δ η }|

This implies, in view of Lemma 2.1, that

2 2 η C δ δ C δ At cη t− µδ = C C t , Cδ | ∩ | |A |√t C C ∈At

which contradicts (24) if cη is large.

17 3 The three circle lemma

The following lemma is essentially Marstrand’s three circle lemma, cf. Lemma 5.2 in [7]. It is a quantitative version of the following fact, known in incidence geometry as the circles of Apollonius: Given three circles which are not internally tangent at a single point, there are at most two other circles that are internally tangent to the three given ones.

Lemma 3.1 Fix circles C ,C ,C , some [δ, 1], and positive numbers λ ,λ ,λ . Let 1 2 3 ∈ 12 13 23 3 2 S = x R B(xj,): r (1, 2) with xi x ri r < { ∈ \ ∃ ∈ || − |−| − || j=1 for i =1, 2, 3 and e (x, r) e (x, r) λ for 1 i

xi x where we have set ei(x, r)= x −x sgn(ri r). Then | i− | −

2 1 S (λ λ λ )− . | | 12 13 23

For a proof of Lemma 3.1 the reader is referred to [7] or [9]. These references deal

with the case λ12 = λ13 = λ23, but the same arguments apply. [6] contains a version of Lemma 3.1 (with λ = λ = λ ) that gives further information on the set 12 13 23 ≥ t of circles under consideration and, moreover, applies to families of curves satisfying the cinematic curvature condition from [12]. The proof of Theorem 3.1 below follows a scheme that originates in [6]. The basic idea is to pass from the three circle lemma to bounds on the number of tangencies oc- curring in an arbitrarily large collection of circles by means of a simple result of extremal graph theory. Roughly speaking, this amounts to counting a suitable set of quadruples of circles in two different ways, as explained in the following paragraph. The Kolasa–Wolff argument splits naturally into two cases: Given a typical annulus Cδ, either the majority of intersections of Cδ with other annuli are concentrated on a small set of Cδ or they are spread out over an arc on Cδ of sufficient length. In view of the nondegeneracy condition

18 in the circles of Apollonius, Marstrand’s lemma can be applied only in the second case. However, Kolasa and Wolffobserved that for one–parameter families (they consider C circles with δ–separated radii), it suffices to consider a single annulus in the first case. Roughly speaking, if their multiplicity estimate failed on a sufficiently small set of a fixed

δ δ 1 annulus C , then the number of annuli intersecting C would have to exceed δ− , which is the trivial bound on card( ), see [6] section 4 or [14] section 3 for details. One can show C that this type of argument does not apply to the two–parameter families of circles that arise in the analysis of (1). More precisely, we will require stronger bounds than the ones given by (23). This was the original motivation for considering a two circle argument and it is also were Proposition 2.1 becomes important. Note that one cannot expect to use the two circle lemma in the one–parameter setting of [6] and [14]. Indeed, Lemma 2.5 estimates a neighborhood of a curve which might contain the centers of all circles in if C 1 <δ− . However, it turns out that the bounds (24) are also needed in the second case, |C| i.e., that part of the argument involving Lemma 3.1.

Theorem 3.1 Let be as above and let ρ>0 be a small number. Then there exists C 1 so that >c− for some constant c and so that A⊂C |A| ρ |C| ρ δ 1 ρ 1 C : µA >cλ− − δ− λδ (26) |{ δ ρ }| ≤ for all C and all 0 <λ 1. ∈A ≤ In view of Proposition 1.1 this result implies Bourgain’s theorem. The heuristics behind the proof of Theorem 3.1 are explained in this paragraph. We suppress any pigeon hole factors. Suppose that (26) fails with ρ = 0 in the following slightly stronger sense: for half the circles C ∈C C δ t 1 1 C : µC >λ− δ− t >λδ (27) |{ δ }| for a ﬁxed choice of and t, and λ 100A . The factor t is motivated by (7). ≥ 0 t Following [6] we consider the set

Q = (C, C ,C ,C ):C satisﬁes (27),C,C ,C C , { 1 2 3 1 2 3 ∈Ct 19 λ dist(Cδ Cδ,Cδ Cδ) > for 1 i

C 3 Q min t . C | | |C| ∈C |C | Moreover, by the area estimate in Lemma 2.1,

2 1 1 C δ δ C δ Ct λδ λ− δ− t< µδ = C C t Cδ | ∩ | |C |√t C C ∈Ct for any C satisfying (27). We conclude that

3 3 3 1 2 1 1 λδ 2 t card(Q) λ− δ− t = . |C| δ2/√t |C| δ δ To ﬁnd an upper bound on card(Q) we assume that the majority of (C, C ,C ,C ) Q 1 2 3 ∈ satisfy C C1 and C C2 . Using Lemma 2.3 one can show that this is indeed the 2 ∈Ct 3 ∈Ct most signiﬁcant case, see the proof of Theorem 3.1 for details. Using (24) with η =0

and Lemma 3.1 we now obtain, choosing first C1, then C2,C3, and finally C (recall that λ t ) ≥ 3 2 3 2 2 3 t t card(Q) λ− , δ2 |C| δ δ δ2 C1 C1 C2 C2 C3 ∈C ∈Ct ∈Ct which agrees with our lower bound. To apply Lemma 3.1 note that Cδ Cδ is contained ∩ j δ xi x in an arc of length A0 t on C centered at x + r x −x sgn(r ri), see Lemma 2.1. Hence | i− | − definition (28) implies that ei(x, r) ej(x, r) >λ/20, as required. | − | We hope that the reader will not be sidetracked by the simple technicalities in the following proof. In order to make the heuristic argument above rigorous, we apply the pigeon hole principle several times. The various factors that arise by doing so as well as

η ρ the factors − in Proposition 2.1 can then be controlled by λ− .

20 1 Proof of Theorem 3.1: Let η = ρ/10 and choose , >c− as in Proposi- A⊂C |A| η |C| tion 2.1. Suppose (26) failed for half the circles in , i.e., for at least half the circles C A in A

δ 1 ρ 1 1 C : µA >cλ− − δ− λδ, (29) |{ δ ρ }| ≥ 2 with some dyadic λ depending on C. This will lead to a contradiction for suﬃciently large cρ. We will apply the pigeon hole principle to obtain a bound of this type that holds for a large number of circles and a ﬁxed choice of λ.Letν = ρ/100 and recall that kν ¯ aν = k∞=0 2− . We claim that there exists a dyadic λ so that

δ 1 ρ 1 1 C : µA >cλ¯− − δ− > λδ¯ (30) |{ δ ρ }| 2

1 1 ν for at least max( a− λ¯ , 1) many circles C . For if not, then the number of circles 2 ν |A| ∈A C for which (29) holds is 1 1 ν 1 < a− λ = , 2 ν |A| 2|A| λ=2 j 1 − ≤ contrary to our assumption. Since

C δ t µA C µA , δ ≤ δ /4 t 1 ≤ ≤ where the sum is taken over dyadic , t [δ, 1], and since ∈ ν ν = tν 4a2, t ≤ ν /4 t 1 /4 t 1 ≤ ≤ ≤ ≤ one easily sees that for each C satisfying (30) there are , t depending only on C so that

C δ t 2 1 ν 1 ρ 1 1 2 1 ν C : µA >c(4a )− λ¯− − δ− > (4a )− λδ¯ . (31) |{ δ ρ ν }| 2 ν

Indeed, given an x in the set appearing on the left–hand side of (30), we can ﬁnd , t depending on x so that

C 2 1 ν 1 ρ 1 At ¯− − µδ (x) >cρ(4aν)− λ δ− . (32)

21 For if this were not so, sum as above to get a contradiction. For the same reason, (30) implies that for some , t depending on C the measure of all x Cδ satisfying (32) has ∈ 1 2 1 ν ¯ to be > 2 (4aν)− λδ. Finally, there are ﬁxed values of an t so that (31) holds with 1 3 ν those , t for at least max( a− (λ¯ ) , 1) many circles. Let the set of those circles be 8 ν |A| 1 2 ν denoted by . Setting λ = a− λ¯, we conclude that there is a nonempty subset of B 8 ν B A satisfying

3 ν a− (λ) (33) |B| ν |A| and ﬁxed numbers λ,, t [δ, 1] so that ∈ C δ t 6 3ν 1 ρ 1 C : µA c a− λ− − δ− >λδ (34) |{ δ ρ ν }| for all C . ∈B Case 1: λ 100A ≤ 0 t Here A0 is the constant from Lemmas 2.1 and 2.3. Then (34) implies that there is a circle C so that ∈A ρ C 2 δ t 6 3ν 1 1 t C : µA c a− λ− δ− >λδ |{ δ ρ ν }| and thus (recall η = ρ/10, ν = ρ/100)

C ρ δ t 6 1 1 η C : µA c a− λ− δ− t 2 − >λδ |{ δ ρ ν }| which contradicts our choice of , see (25), provided c is large. A ρ Case 2: λ>100A0 t Fix any C and let ∈B λ Q(C) = (C ,C ,C ):C ,C ,C C , dist(Cδ Cδ,Cδ Cδ) > { 1 2 3 1 2 3 ∈At i ∩ j ∩ 20 for 1 i

1 3 2 (C) 6 3ν ρ 1 t card(Q ) c a− λ− δ− . (36) ρ ν δ2 22 δ 2 By Lemma 2.1, all C C satisfy C Cδ δ and hence (34) implies ∈At | ∩ |∼ √t 2 6 3ν ρ C C δ At cρ aν− λ− < µδ t , i.e., (37) Cδ |A |√t 1 2 C 6 ρ 3ν 1 t c a− λ− δ− . |At| ρ ν δ2 Now ﬁx any C ,C C .LetR ,R be arcs of Cδ of thickness δ and length λ/5centered 1 2 ∈At 1 2 x x − j at ej = x r sgn(r rj) x x with j =1, 2, respectively. Lemma 2.1 implies that any − − | − j | C C with the properties ∈At

δ δ C Cδ = ,R C = for j =1, 2 ∩ ∅ j ∩ ∅

satisﬁes δ λ dist(Cδ C ,Cδ Cδ) > for j =1, 2. ∩ ∩ j 20 Since (34) implies that

C δ t 6 3ν 1 ρ 1 C (R R ):µA c a− λ− − δ− >λδ/2 |{ \ 1 ∪ 2 δ ρ ν }|

claim (36) follows from calculation (37) by choosing ﬁrst C1, then C2, and ﬁnally C3. Next we assert that the set

Q = (C, C ,C ,C ):C ,C,C ,C C , sgn(r r) = sgn(r r) 0 { 1 2 3 ∈B 1 2 3 ∈At 1 − 2 − = sgn(r r), (x x, x x) >λ/20 for 1 i

satisﬁes

1 3 2 6 3ν ρ 1 t card(Q ) c a− λ− δ− . (39) 0 |B| ρ ν δ2 First note that at least half the circles C C satisfy either sgn(¯r r) > 0 or sgn(¯r r) < ∈At − − 0. Thus one can add the condition

sgn(r r) = sgn(r r)=sgn(r r) 1 − 2 − 3 − 23 to the deﬁnition of Q(C) without violating (36). Furthermore, according to Lemma 2.1,

δ δ x xi (C) the set C Ci is centered at ei = x r sgn(r ri) x−x . Given any (C1,C2,C3) Q ∩ − − | − i| ∈ we therefore conclude from the conditions

dist(Cδ Cδ,Cδ Cδ) >λ/20, sgn(r r ) = sgn(r r ) i ∩ j ∩ − i − j

for 1 iλ/20, i − j −

and (39) follows from (36). We want to bound card(Q0) from above using Proposition 2.1

and Lemma 3.1. In order to do so, we need to specify the relative positions of (C1,C2,C3) for the majority of (C, C ,C ,C ) Q . This will be accomplished by applying the 1 2 3 ∈ 0 pigeon hole principle to the variables (x x, x x), (x x, x x), (x x, x x) 1 − 2 − 2 − 3 − 1 − 3 − 1 ν 1 ν 1 ν and ∆(C ,C ), ∆(C ,C ), x x , x x with weights a− λ ,a− λ ,a− λ and 1 2 2 3 | 1 − 2| | 2 − 3| ν 12 ν 13 ν 23 1 ν 1 ν 1 ν 1 ν jν aν− β1 ,aν− β2 ,aν− τ1 ,aν− τ2 , respectively (recall that aν = j 0 2− ). Indeed, the set ≥ Q = (C, C ,C ,C ) Q : λ (x x, x x) 2λ for 1 i

satisﬁes

7 ν card(Q) a− (λ λ λ β τ β τ ) card(Q ) ≥ ν 12 13 23 1 1 2 2 0 for a suitable choice of dyadic λ ,λ ,λ [λ/20, 1] and dyadic β ,τ ,β ,τ [δ, 1]. This 12 13 23 ∈ 1 1 2 2 ∈ is because any element of Q0 has to satisfy the conditions in (40) for some choice of those parameters. Hence, in view of (39) and (33),

3 2 3 25 ν 9ν 3ν 3ρ 3 t card(Q) c a− (β β τ τ ) λ − δ− |B| ρ ν 1 2 1 2 δ2 3 2 3 28 ν 10ν 4ν 3ρ 3 t c a− (β β τ τ ) λ − δ− . (41) |A| ρ ν 1 2 1 2 δ2 24 To bound card(Q) from above, ﬁrst observe that λ ,λ λ/20 >A implies by 12 23 ≥ 0 t C1 C2 Lemma 2.6 that C2 and C3 for any (C, C1,C2,C3) Q. Furthermore, in ∈Aβ1τ1 ∈Aβ2τ2 ∈ view of Lemma 2.3, the angles β1,β2 satisfy

τ β λ2 t2,τβ λ2 t2. (42) 1 1 ∼ 12 2 2 ∼ 23

Hence it follows from (24) of Proposition 2.1 and Lemma 3.1 that

card(Q) C : C Cj for j =1, 2, 3 and ≤ |{ ∈A ∈At C1 C1 C2 C2 C3 ∈A ∈Aβ1τ1 ∈Aβ2τ2 sgn(ri r) = sgn(rj r), (xi x, xj x) λij for 1 i

It is easy to see that (41) and (43) are incompatible for large cρ because ν and η are small compared with ρ and since τ ,τ 4t (indeed, if C ,C C then τ x x 1 2 ≤ 1 2 ∈Ct 1 ≤| 1 − 2|≤ x x + x x 4t), and β ,β (see (42) or Lemma 2.3). | 1 − | | − 2|≤ 1 2 ≥

4 Lp Lq estimates → It was shown in [9] and [11] that : Lp(R2) Lq(R2) provided ( 1 , 1 ) lies in the open M → p q 1 1 2 1 1 1 triangle with vertices (0, 0), ( 2 , 2 ) and ( 5 , 5 ) or on the half open segment (0, 0), ( 2 , 2 ) (which corresponds to Bourgain’s theorem). According to the examples in the introduction to [9] the triangle cannot be replaced by any strictly larger open set. However, the optimal inequalities on the shorter sides are unknown. The argument in [9] used both the combinatorial method from [6] and Fourier transform techniques, whereas [11] is based entirely on Fourier integral operator estimates. In this section we show how the purely

25 geometric/combinatorial argument above yields the full range of p q estimates just → stated. Since the technical details are very similar to those in the previous sections, we shall be very brief. By interpolation with (3) it suﬃces to show

f q, f 5 ,1 M ∞ 2 for every q<5. Using the argument from the ﬁrst part of the proof of Proposition 1.1 it is easy to see that this follows from the following theorem (with q(1 + σ)=5).

1 Theorem 4.1 Given σ>0 small, there exists with >c− for some constant A⊂C |A| σ |C| cσ depending only on σ and so that

δ 3 1 2 1 (1 σ) C : µA >c λ− 2 δ− (δ ) 2 − <λδ (44) |{ δ σ |C| }| for all C , 0 <λ 1. ∈A ≤

Before turning to the proof we give a heuristic discussion that parallels the one pre- ceding Theorem 3.1. Set σ = 0 and assume that (44) fails in the following sense. For half the circles C ∈C

C 3 1 δ t C : µC λ− 2 2 >λδ (45) |{ δ ≥ |C| }| with a ﬁxed choice of , t [δ, 1] and some ﬁxed λ A with A large. As before we ∈ ≥ t consider the set

Q = (C, C ,C ,C ):C satisﬁes (45),C,C ,C C , { 1 2 3 1 2 3 ∈Ct λ dist(Cδ Cδ,Cδ Cδ) > for 1 i

C 3 which will again satisfy card(Q) minC t . Note that for any C satisfying (45) ≥|C| ∈C |C | 2 3 1 C δ δ C δ 2 2 Ct λδ λ− µδ = C C t |C| ≤ Cδ | ∩ | |C |√t C C ∈Ct 26 and thus 3 3 1 λδ card(Q) λ− 2 2 . |C| |C| δ2/√t On the other hand, assuming as above that the majority of (C, C ,C ,C ) Q satisfy 1 2 3 ∈ C C1 and C C2 (it will follow from Lemma 2.3 that this is the most signiﬁcant 2 ∈Ct 3 ∈Ct case), 2 3 card(Q) λ− . δ2 C1 C1 C2 C2 C3 ∈C ∈Ct ∈Ct In view of Proposition 2.1,

1 3 4 1 2 Cj 3 2 t 4 t |C | |C| δ δ for j =1, 2 and thus

1 3 2 1 2 2 5 2 t 3 2 − card(Q) 2 λ . |C| δ δ δ 3 3 2 4 Comparing the upper and lower bounds yields λ t , which contradicts our assumption on λ. Proof of Theorem 4.1: Let η = σ/20 , ν = σ/100, and choose as in Proposition 2.1. A Assuming that (44) fails for at least half the circles in we obtain as in the proof of A Theorem 3.1 that for ﬁxed λ,, t [δ, 1] ∈ C 3 1 δ t 6 3ν 1 2 (1 σ) C : µA c a− λ− 2 δ− (δ ) 2 − >λδ (46) |{ δ σ ν |C| }| 3 ν for all C , where a− (λ) . ∈B |B| ν |A| Case 1: λ 100A ≤ 0 t On the one hand, using Lemma 2.1 and (46), we obtain 2 6 3ν 1 2 1 (1 σ) C C δ 2 2 At cσaν− λ− (δ ) − µδ t . (47) |C| δ |A |√t C On the other hand, (24) implies that

1 3 2 (1+σ) 1 2 1 2 t C 2 (1 σ) η t − − . |A | |C| δ δ 27 As the reader will easily verify, this contradicts (47) for large cσ since λ t and since ν and η are small compared with σ.

Case 2: λ>100A0 t With Q deﬁned as in (40),

7 ν card(Q) aν− (λ12λ13λ23β1τ1β2τ2) card(Q0) 3 1 3 ν 7 3ν ν 6 3ν 1 √t 2 1 (1 σ) c− a− (λ) a− λ (β τ β τ ) c a− λ− 2 (δ ) 2 − |C| η ν ν 1 1 2 2 σ ν δ2 |C|

for a suitable choice of dyadic λ ,λ ,λ [λ/20, 1] and dyadic β ,τ ,β ,τ [δ, 1]. 12 13 23 ∈ 1 1 2 2 ∈ Indeed, this follows from the same reasoning as in the proof of Theorem 3.1 if one uses (46) instead of (34). On the other hand, applying Proposition 2.1 and Lemma 3.1 as in the upper bound (43) and using (42), i.e., τ β λ2 t2,τβ λ2 t2, we conclude that 1 1 ∼ 12 2 2 ∼ 23

1 3σ 1 3σ 1 4 + 4 1 4 + 4 2 3 2 3 β1 τ1 2 η 3 (1 σ) β2 τ2 2 η card(Q) c β− 4 − c β− |C| η δ δ 1 |C| η δ δ 2 2 3 (1 σ) 1 4 − (λ λ λ )− |C| δ2 12 13 23 3 1 3σ η η 1 3σ 3 3σ 1+ 2 (1 σ) 2 4 + 4 4 + 4 2 1 δ− − − (λ12λ23t ) (τ1τ2β1− β2− ) (λ12λ13λ23)− |C| 1 3σ 2 +2ν 4 6 2 3 (1 σ) 1 + 3σ η η 1 + 3σ 3 +4ν t − 2 δ− (δ ) 2 − t 2 2 (τ τ β− β− ) 4 4 λ− 2 , |C| |C| 1 2 1 2

where we have used λ ,λ ,λ λ/20 in the last step. Since β ,β and 12 13 23 ≥ t 1 2 ≥ τ1,τ2 4t, and since ν and η are small compared with σ, the lower and upper estimate ≤ for card(Q) will contradict each other for large cσ.

Acknowledgement : I thank Thomas Wolﬀfor his interest in this work, his encourage- ment, and his ﬁnancial support, as well as for helpful comments on a preliminary version of this paper.

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Author’s Address : 253–37 Caltech, Pasadena, CA 91125, U.S.A. [email protected] Current address : School of Mathematics, Institute for Advanced Study, Olden Lane, Princeton, NJ 08540, U.S.A.