on sets of directions determined by subsets of ℝd

ON SETS OF DIRECTIONS DETERMINED BY SUBSETS OF Rd

By

ALEX IOSEVICH, MIHALIS MOURGOGLOU, AND STEVEN SENGER

Abstract. Given E ⊂ Rd , d ≥ 2, define

D(E) ≡{

x − y|x − y| : x, y ∈ E

}⊂ Sd−1

to be the set of directions determined by E . We prove that if the Hausdorff dimen-sion of E is greater than d − 1, then σ(D(E)) > 0, where σ denotes the surfacemeasure on Sd−1. In the process, we prove some tight upper and lower bounds forthe maximal function associated with the Radon-Nikodym derivative of the natu-ral measure on D. This result is sharp, since the conclusion fails to hold if E is a(d −1)-dimensional hyper-plane. This result can be viewed as a continuous analogof a recent result of Pach, Pinchasi, and Sharir ([22, 23]) on directions determinedby finite subsets of Rd . We also discuss the case when the Hausdorff dimension ofE is precisely d −1, where some interesting counter-examples have been obtainedby Simon and Solomyak ([25]) in the planar case. In response to the conjecturestated in this paper, T. Orponen and T. Sahlsten ([20]) have recently proved that ifthe Hausdorff dimension of E equals d −1 and E is rectifiable and is not containedin a hyper-pane, the Lebesgue measure of the set of directions is still positive.Finally, we show that our continuous results can be used to recover and, in somecases, improve the exponents for the corresponding results in the discrete settingfor large classes of finite point sets. In particular, we prove that a finite point setP ⊂ R

d , d ≥ 3, satisfying a certain discrete energy condition (Definition 3.1)determines � #P distinct directions.

1 Introduction

A large class of Erdos type problems in geometric combinatorics asks whether alarge set of points in euclidean space determines suitably large sets of geomet-ric relations or objects. For example, the classical Erdos distance problem askswhether N points in R

d , d ≥ 2, determine � N 2/d distinct distances, where hereand throughout, X � Y , with the controlling parameter N , means that for everyε > 0, there exists Cε > 0 such that X ≤ CεN εY . For a thorough description ofthese types of problems and recent results, see, for example, [1, 17, 21, 24, 26]and the references contained therein.

JOURNAL D’ANALYSE MATHEMATIQUE, Vol. 116 (2012)

DOI 10.1007/s11854-012-0010-x

355

356 ALEX IOSEVICH, MIHALIS MOURGOGLOU, AND STEVEN SENGER

Continuous variants of Erdos type geometric problems have also received muchattention in recent decades. Perhaps the best known of these is the Falconerdistance problem, which asks whether the Lebesgue measure of the distance set{|x − y| : x, y ∈ E} is positive, provided that the Hausdorff dimension of E ⊂ R

d ,d ≥ 2, is greater than d/2. For the best currently known results on this problem,see [5, 27, 3, 15, 18, 19]. Also see [4] for the closely related problem on finite pointconfigurations. For related problems under the assumption of positive Lebesguedensity, see, for example, [2, 8, 29].

In this paper, we study the sets of directions determined by subsets of euclideanspace. In the discrete setting, the problem of directions was studied in recent yearsby Pach, Pinchasi, and Sharir; see [22, 23]. In the latter paper, they proved thatif P is a set of n points in R

3, not all in a common line or plane, then the pairsof points of P determine at least 2n − 5 distinct directions if n is odd, and atleast 2n − 7 distinct directions if n is even. Our main result can be viewed as acontinuous variant of this result, where finite point sets are replaced with infinitesets of a given Hausdorff dimension. An explicit quantitative connection betweenour main result on directions (Theorem 1.2 below) and the work of Pach, Pinchasi,and Sharir is made in Section 3 below. We show that a finite set P satisfying the(d − 1 + ε)-adaptability assumption (see Definition 3.1 below) determines � #Pdistinct directions. In dimensions two and three, this result is weaker than then theresult of Pach, Pinchasi and Sharir described above. However, in dimensions fourand higher, our result gives, to the best of our knowledge, the only known bounds.

The problem of directions in the finite field setting was studied by the first listedauthor, Hannah Morgan, and Jonathan Pakianathan. See [13] and the referencescontained therein.

Definition 1.1. Given E ⊂ Rd , d ≥ 2, define

D(E) ={

x − y|x − y| : x, y ∈ E, and x �= y

}⊂ Sd−1

to be the set of directions determined by E .

Our main results are the following.

Theorem 1.2. Suppose E ⊂ Rd , d ≥ 2, has Hausdorff dimension greater

than d − 1. Denote by ν the probability measure on D(E) given by the relation∫h(ω)dν(ω) =

∫∫h(

x − y|x − y|

)dμ(x)dμ(y),

where μ is a Frostman measure on E, and let ν� denote the same measure corre-sponding to the set E� = {�x : x ∈ E}, where � ∈ O(d), the orthogonal group.

ON SETS OF DIRECTIONS DETERMINED BY SUBSETS OF Rd 357

Let ν�ε (ω) = ε−(d−1)ν�(B(ω, ε)), where B(ω, ε) is the ball of radius ε centered at

ω ∈ Sd−1. Then ν�ε (ω) = M�(ω)+R�ε (ω), where∫

sup0<ε<c |R�ε (ω)|dω � cs−(d−1),∫M�(ω)dω � 1, and there exists �0 ∈ O(d) such that

∫M�0 (ω)dω � 1.

In particular,

(1.1) σ(D(E)) > 0,

where σ denotes Lebesgue measure on S d−1.

Remark 1.3. Pertti Mattila recently pointed out to us that (1.1) follows from[19, Theorem 10.11]. Our method, which uses Fourier analysis, allows us to ob-tain more detailed information about the direction set measure and the associatedRadon-Nikodym derivative.

Remark 1.4. It is not difficult to check that if E is a (d − 1)-dimensionalLipschitz surface in R

d that is not contained in a (d − 1)-dimensional plane, thenσ(D(E)) > 0. It is reasonable to conjecture that the same conclusion holds if E

is merely a (d − 1)-dimensional rectifiable subset of Rd . We discuss the purelynon-rectifiable case in Subsection 1.1 below.

Remark 1.5. It is interesting to contrast Theorem 1.2 with the Besicovitch-Kakeya Conjecture (see e.g. [28] and the references contained therein), whichsays that any subset of R

d containing a unit line segment in every direction hasHausdorff dimension d . On the other hand, Theorem 1.2 says that having Haus-dorff dimension greater than d − 1 is sufficient for the set to contain endpointsof a segment of some length pointing in the direction of a positive proportion ofvectors in Sd−1.

1.1 Sharpness of the main results. In the following sense, Theorem 1.2cannot be improved. Suppose that E is contained in a (d − 1)-dimensional hyper-plane. Then σ(D(E)) = 0. It follows that the conclusion of Theorem 1.2 does nothold in general if the Hausdorff dimension of E is less than or equal to d − 1.

Another, very different, sharpness example comes from the theory of distancesets. Let Eq denote the q−d/s-neighborhood of q−1

(Z

d ∩[0, q]d), where Zd denotesthe standard integer lattice, and 0 < s < d . It is known that if qi is a sequence ofintegers given by q1 = 2, qi+1 > qi

i , then the Hausdorff dimension of E =⋂

i Eqi

is s; see, for example, [6, 7]. Observe that σ(D(Eq)) ≈ q−d(d−1)/s · qd , sincethe number of lattice points in [0, q]d , d ≥ 2, with relatively prime coordinatesis equal to qd (1 + o(1))/ζ (d), where ζ (t) is the Riemann zeta function. See, forexample, [16]. It follows that if s < d − 1, σ(D(Eq)) → 0 as q → ∞, and fromthis, it follows that σ(D(E)) = 0.


This example does not rule out the possiblility that s = d − 1, and one mightreasonably conjecture, consistent in spirit with the result due to Pach, Pinchasi,and Sharir stated above, that if the Hausdorff dimension of E is equal to d − 1,then (1.1) holds if and only E is not a subset of a single (d − 1)-dimensionalhyper-plane. However, this is not the case. A result due to Simon and Solomyak[25] shows that for every self-similar set of Hausdorff dimension one satisfying anadditional mild condition, the Lebesgue measure of D(E) is zero. In particular, ifE is the four-cornered Cantor set known as the Garnett set (see, e.g., [9]), then theHausdorff dimension of E is one and the Lebesgue measure of D(E) is zero. It isnot difficult to use Simon and Solomyak’s result to construct a set E of Hausdorffdimension d − 1 in R

d that is not contained in a hyperplane and for which theLebesgue measure of D(E) is zero.

In the realm of rectifiable sets, we believe that Theorem 1.2 can be strengthenedas follows.

Conjecture 1.6. Let E ⊂ Rd , d ≥ 2, have Hausdorff dimension d − 1.

Suppose that E is rectifiable and is not contained in a hyper-plane. Then (1.1)holds.

Remark 1.7. After this paper was submitted, Conjecture 1.6 was resolved bya very nice argument due to Orponen and Sahlsten; see [20].

1.2 Structure of the paper. Theorem 1.2 is proved in Section 2. In Sec-tion 3, we describe an explicit connection between the main results of the paperand the discrete problems, such as those studied by Pach, Pinchasi, and Sharir.

2 Proof of Theorem 1.2

Let s be the Hausdorff dimension of E . Now, although the set D(E) is a subset ofthe (d −1)-dimensional sphere, in the arguments to follow it is convenient to workwith sets of slopes of line segments defined by pairs of points in the set E . For twopoints p and q with coordinates (p1, p2, . . . , pd ) and (q1, q2, . . . , qd ), respectively,we define the slope of the line segment between p and q as the (d − 1)-tuple{

p1 − q1

pd − qd,

p2 − q2

pd − qd, . . .

pd−1 − qd−1

pd − qd

}.

It is not difficult to see that if the (d −1)-dimensional Lebesgue measure of theset of slopes determined by E is positive, then (1.1) holds. With a slight abuse ofnotation, we refer to the set of slopes as D(E) as well. It is convenient to extract


two subsets from E separated from each other in at least one of the coordinates.To this end, we make the following construction.

Lemma 2.1. Suppose μ is a Frostman probability measure on E ⊂ Rd with

Hausdorff dimension greater than d − 1. Then there exist positive constants c1, c2

and subsets E1,E2 of E such that μ(E j ) ≥ c1 > 0, for j = 1, 2, and

max1≤k≤d

(inf{|xk − yk| : x ∈ E1, y ∈ E2}) ≥ c2 > 0.

Proof. We employ a stopping time argument. Define C0 to be the constantin the Frostman condition, μ(Br) ≤ C0rs. Let [0, 1]d be the unit cube in R

d , andsubdivide it into 4d smaller cubes, each of side length 1/4. Choose 2d collectionsof 2d sub-cubes, each such that no two cubes of the same collection touch eachother. Then by the pigeon-hole principle, at least one of them has measure greaterthan or equal to 1/2d . If there are two such cubes, Q1 and Q′

1, in the same collectionsuch that μ(Q1), μ(Q′

1) ≥ c/2d for some c > 0, then we are done. If not, thereexists a cube Q1 of side length 1/4 such that μ(Q1) ≥ 1/2d . We then repeat thesame procedure on the cube Q1. Now, either we have two cubes, Q2 and Q′

2, withμ(Q2) and μ(Q′

2) ≥ c/22d for some c > 0, which are in the same collection, orwe do not. If we do not, then again, there must be a cube Q2 with side length 1/42

such that μ(Q2) ≥ 1/22d . We can repeat this process, and at each stage check fortwo cubes from the same collection with the requisite measure. Let the integer n

depend on C0. If we fail to find two such cubes at the n-th iteration, we obtain acube Qn of side length 1/4n for which μ(Q) ≥ 1/2dn. By the Frostman measurecondition, there exists C0 > 1 such that 1/2dn ≤ μ(Q) ≤ C01/4sn, which is trueif n ≤ log2(C0)/(2s − d). So, picking n > log2(C0)/(2s − d) shows that it is onlytrue whenever s < d/2, and since s > d − 1, we have a contradiction. �

Let x and y be points in Rd with coordinates (x1, x2, . . . , xd ) and (y1, y2, . . . , yd ),

respectively. Apply Lemma 2.1 to E . Without loss of generality, let the sets E1 andE2 be separated in the d-th coordinate. Let μ1 and μ2 be restrictions of μ to thesets E1 and E2, respectively. Let t = (t1, t2, . . . , td−1). For slopes t ∈ [1/2, 1]d−1,define νε(t) to be the quantity

1εd−1 (μ× μ)

{(x, y) ∈ E1 × E2 : t1 − ε ≤ x1 − y1

xd − yd≤ t1 + ε, . . . ,

td−1 − ε ≤ xd−1 − yd−1

xd − yd≤ td−1 + ε

}.


1 2 1 2

3 4 3 4

1 2 2

3 4 3 4

Figure 1. The first decomposition into four collections of four cubes each is shownwith a 1 in the cubes of the first collection, a 2 in the cubes of the second collection,etc. . .. In this case, the first decomposition is not enough, and a positive proportionof the mass is in the lower-right cube of the first collection. After the seconditeration, there are two shaded boxes, representing E1 and E2.

Since xd − yd is guaranteed to be more than c2 by Lemma 2.1, we can multiplyeach inequality through by the denominator to get

νε(t) ≈ 1εd−1

(μ1 ×μ2){(x, y) ∈ E1 ×E2 : (xd −yd )t1 −ε ≤ x1 −y1 ≤ (xd −yd )t1

+ ε, . . . , (xd − yd )td−1 − ε ≤ xd−1 − yd−1 ≤ (xd − yd )td−1 + ε}.Our plan is to show that

∫νε(t) is bounded above and below by a constant plus

an error term. For this, we write νε(t) as the sum of two terms. We prove that theintegral of first term in t1, . . . , td−1 is bounded above and below by two positiveconstants. Then we show that the integral of the modulus of the second secondterm is bounded above by a sufficiently small positive constant plus a constantmultiple of εs−(d−1). We then conclude that 1 − εs−(d−1) � νε(t) � 1 + εs−(d−1) for


a subset of [1/2, 1]d−1 of positive (d − 1)-dimensional Lebesgue measure. As wenoted above, this implies that σ(D(E)) > 0.

Let ψ : R → R be a smooth, even bump function whose support is containedin [−2,−1/2] ∪ [1/2, 2] and such that ψ(0) = 1. We have

νε(t) ≈ 1εd−1

∫∫ψ

((x1 − y1) − t1(xd − yd )

ε

)ψ

((x2 − y2) − t2(xd − yd )

ε

). . .

. . . ψ

((xd−1 − yd−1) − td−1(xd − yd )

ε

)dμ1(x)dμ2(y).

By Fourier inversion, this quantity equals

1εd−1

∫· · ·

∫ψ(λ1)ψ(λ2) · · · ψ(λd−1)e

2πiε λ1((x1−y1)−t1(xd −yd ))e

2πiε λ2((x2−y2)−t2(xd −yd )) · · ·

· · · e2πiε λd−1((xd−1−yd−1)−td−1(xd −yd ))dλ1dλ2 · · · dλd−1dμ1(x)dμ2(y)

=1εd−1

∫· · ·

∫ψ(λ1)ψ(λ2) · · · ψ(λd−1)

× μ1

(−λ1

ε,−λ2

ε, . . . ,

(t1λ1 + t2λ2 + · · · td−1λd−1

ε

))

× μ2

(−λ1

ε,−λ2

ε, ...,

(t1λ1 + t2λ2 + · · · td−1λd−1

ε

))dλ1dλ2 · · ·λd−1,

and by the change of variables λ ′j = λ j/ε, the latter is equal to

∫· · ·

∫ψ(ελ′

1)ψ(ελ′2) · · · ψ(ελ′

d−1)

× μ1(−λ′

1,−λ′2, . . . ,

(t1λ

′1 + t2λ

′2 + · · · td−1λ

′d−1

))× μ2

(−λ′1,−λ′

2, . . . ,(t1λ

′1 + t2λ

′2 + · · · td−1λ

′d−1

))dλ′

1dλ′2· · ·dλ′

d−1.

In what follows, we refer to λ′j as λ j , to simplify the exposition. We also define

λ = (λ1, λ2, . . . , λd−1) ∈ Rd−1.

For the sake of simplicity we define the function

� (λ) = � (λ1, λ2, . . . , λd−1) := ψ(λ1)ψ(λ2) · · ·ψ(λd−1).

An easy calculation shows that �(λ) = ψ(λ1)ψ(λ2) · · · ψ(λd−1), which impliesthat �(0) = 1, since ψ(0) = 1. Also, note that � is continuous. By the preceding


discussion, one sees that

νε(t) =∫

· · ·∫

(�(ελ) − �(0))μ1 (−λ1, λ2, . . . , (t1λ1 + t2λ2 + ...td−1λd−1))

× μ2 (λ1, λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1)) dλ

+∫

· · ·∫μ1 (−λ1, λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1))

× μ2 (λ1, λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1)) dλ

=: Pε(t) + M (t).

We prove that if Pε(t) = Qε(t) + Rε(t), where Qε(t) and Rε(t) are error terms,then for sufficiently small η > 0,

∫M (t)dt ≈ 1,

∫|Qε(t)|dt � ηIs(μ), and

∫|Pε(t)|dt � εs−(d−1)Is(μ).

Let δ be a small positive real number, to be chosen later. We handle the error termPε by splitting it into two integrals, where the domains of integration are |λ| ≤ δ/ε

and |λ| > δ/ε respectively:

Pε =∫

· · ·∫

|λ|≤ δε

dλ1dλ2 · · · dλd−1 +∫. . .

∫|λ|>δ

ε

dλ1dλ2 · · · dλd−1 ≡ Qε(t) + Rε(t).

First we bound the L1 norm of the quantity Rε(t). Let ψ0 ∈ C∞0 (R) be such that

supp(ψ0) ⊂ [1/4, 2] and ψ0 = 1 in [1/2, 1].

∫[1/2,1]d−1

ψ0(t1) · · ·ψ0(td−1)|Rε(t)|dt1 · · · dtd−1

≤∫

[1/2,1]d−1

∫|λ|>δ/ε

|�(ελ) − �(0)|

× ∣∣μ1 (−λ1,−λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1))∣∣

×∣∣∣μ2 (−λ1,−λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1))

∣∣∣dλ1dλ2...dλd−1dt1 · · · dtd−1.

Applying the Cauchy-Schwarz inequlaity shows that the square of the expression


above

�∫

[1/2,1]d−1

∫|λ|>δ/ε

|�(ελ) − �(0)|ψ0(t1)ψ0(t2) · · ·ψ0(td−1)

∣∣μ1 (−λ1,−λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1))∣∣2 dλ1 · · · dλd−1dt1 · · · dtd−1∫

[1/2,1]d−1

∫|λ|>δ/ε

|�(ελ) − �(0)|ψ0(t1)ψ0(t2) · · ·ψ0(td−1)

∣∣∣μ2 (−λ1,−λ2, . . . , (t1λ1 + t2λ2 + · · · td−1λd−1))∣∣∣2 dλ1 . . . dλd−1dt1 · · · dtd−1

= A · B,

where A is the first integral, and B is the second. We break each of these integralsup into A j and B j , which are integrals over subsets which make up the wholeregion of integration. Define � j := {λ ∈ R

d−1 : |λ| > δ/ε, |λ j | ≈ |λ|}, andobserve that A ≤ ∑d−1

j =1

∫[1/2,1]d−1

∫� j

=∑d−1

j =1 A j .We now estimate the A j , while the estimates of the corresponding B j are iden-

tical. Without loss of generality, we may assume that j = 1 and λ1 > 0. Thesame proof works in the case λ1 < 0 with minor modifications. We introduce thechange of variables τi = ti , � j = −λ j , �d =

∑d−1k =1 tkλk, where 2 ≤ i ≤ d − 1 and

1 ≤ j ≤ d − 1. Note that the Jacobian can be considered to have positive sign;otherwise, we make a slightly different change of variables. Additionally, for therest of the proof, we continue writing our new variables τ i, � j as ti , λ j respectively,to stay consistent with our earlier notation. We now estimate the integral A1; theother pieces can be estimated in a similar fashion.

By the definition of λd , t1 =(λd + t2λ2 + · · · td−1

)λd−1/λ1, and therefore

A1 ≤∫

[1/2,1]d−2ψ0(t2) . . .ψ0(td−1)

∫�1

∫|λd |�|λ|

ψ0

(λd + t2λ2 + · · · + td−1λd−1

λ1

)

× |�(ελ) − �(0)| ∣∣μ1(λ1, . . . , λd )∣∣2 dλd

dλ1

λ1dλ2 . . . dλd−1dt2 . . . dtd−1.

Defining λ′ := (λ1, λ2, . . . , λd−1, λd ) = (λ, λd ) ∈ Rd , one can easily deduce

that �1 × {λd : λd � |λ|} ⊂ {λ : |λ′| > δ/ε}. Hence,

A1 �∫

· · ·∫

|λ|>δ/ε

∣∣μ1(λ1, . . . , λd )∣∣2 1

|λ′|dλ1 · · · dλd ≡ I1.

Now, we use the energy integral bounds in polar coordinates. Recall that since theHausdorff dimension of E is s (see [18]), the energy integral of μ, which is givenby

Is(μ) :=∫ ∞

0

∫Sn−1

∣∣μ(ξ )∣∣2

|ξ |d−sdξ =

∫ ∞

0

∫Sn−1

∣∣μ(rθ)∣∣2 rd−1

rd−sdrdθ,


is finite. Rewriting the integral I1 in polar coordinates, we have

I1 =∫ ∞

δ/ε

∫Sn−1

rd−s

∣∣μ1(rθ)∣∣2

rrd−1

rd−sdrdθ =

∫ ∞

δ/ε

∫Sn−1

∣∣μ1(rθ)∣∣2

rs−(d−1)

rd−1

rd−sdrdθ

≤( εδ

)s−(d−1)Is(μ),

which concludes the proof of

(2.1)∫

|Rε(t)| dt � εs−(d−1).

By the continuity of �, for a sufficiently small positive constant η, we can findδ > 0 such that

(2.2)∫

|Qε(t)|dt � ηIs(μ).

It remains only to prove that, for the main term,

(2.3)∫

M (t)dt ≈ 1.

The argument for the upper bound is similar to the one used in the proof of(2.2), but is simpler, since we do not use the continuity of �. It remains only toprove the lower bound. To this end, by the same change of variables made in theproof of (2.1), we can reduce case to proving a lower bound for the modulus of theintegral∫∫

�′1

ψ0

(λd + t2λ2 + · · · + td−1λd−1

λ1

)μ1(λ1, . . . , λd )μ1(λ1, . . . , λd )

dλ1 . . . dλd

|λ′| ,

where�′1 :=

{λ ∈ R

d−1 : |λ1| ≈ |λ|}. Moreover, since

λd + t2λ2 + · · · + td−1λd−1

λ1∈ supp(ψ0) ⊂

[14, 2

],

we can reduce the case even more, by observing that the domain of integrationcan be considered as an appropriate d-dimensional sector S. By readjusting theconstants, we may assume that the sector is of angle π/3. Then, denoting by �the rotation by π/3, we can write

5∑i =0

∫�iS

μ1(λ1, . . . , λd )μ2(λ1, . . . , λd )dλ1 · · · dλd

|λ′|=∫μ1(λ1, . . . , λd )μ2(λ1, . . . , λd )

dλ1 · · · dλd

|λ′| .


Since E1,E2 ⊂ [0, 1]d are separated by the stopping time argument employedpreviously, one can assume that∫

μ1(λ1, . . . , λd )μ2(λ1, . . . , λd )dλ1 . . . dλd

|λ′| ≈∫∫

dμ1(x)dμ2(y)|x − y| ≈ 1.

Thus, there exists i0 ∈ {0, 1, . . . , 5} such that∫�i0S

μ1(λ1, . . . , λd )μ2(λ1, . . . , λd )dλ1 · · · dλd

|λ′| � 1,

and writing �i0 := �0, we obtain, by a change of variables,

(2.4)∫S

μ1(�−10 λ′)μ2(�−1

0 λ′)dλ′

|λ′| � 1.

Let us now define the measure μ�0j on the set E�0

j := {�−10 x : x ∈ E j } for

j = 1, 2 by∫

f (x)dμ�0j (x) =

∫f (�−1

0 x)dμ j (x). Taking f (x) = e−2πix·y shows that

μ�0j (y) = μ j (�−1

0 y); thus, (2.4) may be written as

(2.5)∫S

μ�01 (λ′)μ�0

2 (λ′)dλ′

|λ′| � 1.

The estimate (2.5) provides us with the lower bound we were seeking for themeasures μ�0

j instead of μ j . This allows us to prove that the set of directionsdetermined by the sets E�0

1 and E�02 has positive Lebesgue measure. To this end,

without loss of generality we define the measure ν�0 supported on closed cube[1/2, 1]d−1 by∫

g(t)dν�0(t) =∫

g(

x1 − y1

xd − yd,

x2 − y2

xd − yd, · · · , xd−1 − yd−1

xd − yd

)dμ�0

1 (x)dμ�02 (y),

and ν�0ε (t) either by

(2.6)1εd−1μ

�0 × μ�0

{(x, y) ∈ E�0

1 × E�02 : t1 − ε

≤ x1 − y1

xd − yd≤ t1 + ε, . . . , td−1 − ε ≤ xd−1 − yd−1

xd − yd≤ td−1 + ε

},

which, by a stopping time argument, is comparable to

1εd−1

μ�01 × μ�0

2

{(x, y) ∈ E�0

1 × E�02 : |x1 − y1 − (xd − yd )t1|

≤ ε, . . . , |xd−1 − yd−1 − (xd − yd )td−1| ≤ ε

}.


We follow mutatis mutandis the preceding argument for νε(t), and in view of(2.5) we prove that

∫M�0 (t)dt ≈ 1,

∫ |Q�0ε (t)|dt � η, and

∫ |R�0ε (t)|dt � εs−(d−1).

As a result, since η > 0 is sufficiently small, we have

(2.7) 1 − εs−(d−1) �∫ν�0ε (t)dt � 1 + εs−(d−1).

By Lebesgue’s decomposition, dν�0 = dν�0ac + dν�0

s , where ν�0ac � Ld−1 and

ν�0s ⊥ Ld−1. Moreover, letting f be the Radon-Nikodym derivative of ν�0

ac withrespect to Ld−1, we can write dν�0

ac = fdLd−1. Since ν�0 is a finite Borel measure,limε→0 ν

�0 (B(t, ε)/Ld−1(B(t, ε))) = f (t), for Ld−1-a.e. t. By (2.7) and Fatou’slemma,

(2.8)∫

lim infε→0

ν�0 (B(t, ε))Ld−1(B(t, ε))

dt ≤∫

lim infε→0

ν�0 (B(t, ε))Ld−1(B(t, ε))

dt =∫

f (t)dt � 1,

where B(t, ε) is the (d − 1)-dimensional ball with center at t and of radius ε. Notethat f < ∞ Ld−1-a.e.

Let us now define for an appropriate constant C0 > 0,

ν�0∗ (t) := supε<C0

ν�0 (B(t, ε))Ld−1(B(t, ε))

.

Observe that ν�0∗ (t) provides us with a dominating function for ν�0ε (t), and by

our previous arguments for the upper bounds,∫ν�0∗ (t)dt � 1. Fatou’s lemma

for ν�0∗ (t) − ν�0ε (t) and (2.7) show that

1 � lim supε→0

∫ν�0 (B(t, ε))Ld−1(B(t, ε))

dt ≤∫

lim supε→0

ν�0 (B(t, ε))Ld−1(B(t, ε))

dt =∫

f (t)dt,

which, in view of (2.8), gives

(2.9)∫

f (t)dt ≈ 1.

Let F be the support of ν�0ac and note that Ld−1(F) = Ld−1([1/2, 1]d−1),

ν�0�F = ν�0ac , and ν�0�F � Ld−1. Thus for every Borel set B in F, ν�0�F(B) =∫

B f (t)dt, from which setting B = F in conjunction with (2.9) implies that the setof directions determined by E�0

1 and E�02 contains a set of positive Lebesgue mea-

sure. Since, by definition, E�01 and E�0

2 determine the same number of directionswith E1 and E2, the proof of Theorem 1.2 is complete.


3 Some connections between continuous and discreteaspects of the problem at hand

In this section, we appeal to a conversion mechanism developed in [12, 11, 14]to deduce a Pach-Pinchasi-Sharir type result from Theorem 1.2. In the aforemen-tioned papers, the conversion mechanism was used in the context of distance sets.However, as we demonstrate, the idea is quite flexible and lends itself to a varietyof applications.

Definition 3.1. Let P be a set of n points contained in [0, 1]d , d ≥ 2. Definethe measure dμs

P(x) = n−1 · nd/s · ∑p∈P χBn−1/s (p)(x)dx, where χBn−1/s (p)(x) is thecharacteristic function of the ball of radius n−1/s centered at p. We say that P is s-adaptable if P is n−1/s-separated and Is(μP) =

∫∫ |x − y|−sdμsP(x)dμs

P(y) < ∞.This is equivalent to the statement n−2 ∑

p�=p′∈P |p − p′|−s � 1.

To put it simply, s-adaptability means that a discrete point set P can be thick-ened into a set which is uniformly s-dimensional in the sense that its energy inte-gral of order s is finite. Unfortunately, as is shown in [14], there exist finite pointsets which is not s-adaptable for certain ranges of the parameter s. The point is thatthe notion of Hausdorff dimension is much more subtle than the simple “size” es-timate. This turns out to be a serious obstruction to efforts to convert “continuous”results into “discrete analogs”.

Theorem 3.2. Suppose that for arbitrarily small ε > 0, P is a (d − 1 + ε)-adaptable set in [0, 1]d , d ≥ 2, consisting of n points. Then

(3.1) #D(P) � n.

Moreover, there exists D′(P) ⊂ D(P) such that #D′(P) ≥ #D(P)/2 and the ele-ments in D′(P) are n−(d−1)/(d−1+ε)-separated.

As noted in the Introduction, in dimensions 2 and 3 this result is much weakerthan what is already known. Another weakness of this result is that it holds onlyfor s-adaptable sets. However, in dimensions four and higher, Theorem 3.2 appearsto give a new result in the discrete setting.

Proof of Theorem 3.2. Thicken each point of P by n−1/s, where s >

d − 1. Let EP denote the resulting set. Then σ(D(EP)) � n−(d−1)/s · #D(P). Bythe adaptability assumption and the proof of Theorem 1.2, we see that #D(P) �n(d−1)/s. This proves (3.1), since we may take s arbitrarily close to d − 1. Notethat since Theorem 1.2 does not hold for s = d − 1, we cannot replace (3.1) with

(3.2) #D(P) ≥ C#P.


To see that there exists a subset of the direction set which is n−(d−1)/(d−1+ε)-separated, we recall that σ(D(EP)) > 0; so we can break D(EP) up into pieces,each with Lebesgue measure n−(d−1)/(d−1+ε) and each of which contains a repre-sentative from D(P). Then, by a simple pigeon-hole argument, we see that at least#D(P)/2d−1 of these must be separated. �

Remark 3.3. One should take note that the separation statement is relativelyunique to continuous techniques. Most standard discrete results say nothing aboutthe separation of distinct elements in a given set. For example, in [22], there isa sharp lower bound on the number of distinct directions determined by a set ofpoints in R

3, but there are no guarantees on the separation or distribution of thesedirections on S2.

Acknowledgments. The authors are deeply grateful to Pertti Mattila for sev-eral very helpful remarks which significantly improved this paper.

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Alex IosevichDEPARTMENT OF MATHEMATICS

UNIVERSITY OF ROCHESTER

ROCHESTER, NY 14627 USAemail: [email protected]

Mihalis MourgoglouDEPARTEMENT DE MATH EMATIQUES D’ORSAY

UNIVERSITE PARIS-SUD 11ORSAY, FRANCE

email: [email protected]

Steven SengerDEPARTMENT OF MATHEMATICS

UNIVERSITY OF MISSOURI

COLUMBIA, MO 65201 USA

CURRENT ADDRESS

DEPARTMENT OF MATHEMATICAL SCIENCES

UNIVERSITY OF DELAWARE

NEWARK, DE 19716 USAemail: [email protected]

(Received September 23, 2010 and in revised form March 2, 2011)

on sets of directions determined by subsets of ℝd

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