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The unboundedness of the maximal directional Hilbert transform Marinelli, Alessandro 2018

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The Unboundedness of the MaximalDirectional Hilbert TransformbyAlessandro MarinelliM.Sc., Universita´ degli Studi di Padova, 2014A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)April 2018c© Alessandro Marinelli 2018AbstractIn this dissertation we study the maximal directional Hilbert transform op-erator associated with a set U of directions in the n-dimensional Euclideanspace. This operator shall be denoted by HU . We discuss in detail theproof of the (p, p)-weak unboundedness of HU in all dimensions n ≥ 2 andall Lebesgue exponents 1 < p < +∞ if U contains infinitely many directionsin Rn. This unboundedness result for HU is an immediate consequence ofa lower estimate for ‖HU‖Lp(Rn)→Lp,∞(Rn) that we prove if the cardinality ofU is finite. In this case, we show that ‖HU‖Lp(Rn)→Lp,∞(Rn) is bounded frombelow by√log (#U) up to a positive constant depending only on p and n,for any exponent p in the range 1 < p < +∞ and any n ≥ 2. These resultswere first proved by G. A. Karagulyan ([17]) in the case n = p = 2.iiLay SummaryThis dissertation is concerned with a mathematical object called the max-imal directional Hilbert transform. Associated with any collection U ofdirections in the plane or in the three-dimensional space, or also in the n-dimensional space with n > 3, there exists a maximal directional Hilberttransform for U . The results discussed prove an interesting, quantitativefact about this tranform if U contains only finitely many directions, butalso have a remarkable implication about its nature if U contains an infinitenumber of different directions. The proofs of these results combine classicalideas in harmonic analysis with tools and methods from Euclidean geometryand combinatorics.iiiPrefaceThe purpose of this dissertation is to present the proof of the (p, p)-weakunboundedness of the maximal directional Hilbert transform operator (de-noted by HU ) over an infinite set of directions U in Rn in all dimensionsn ≥ 2 and for all exponents 1 < p < +∞. The operator HU is defined asfollows:HUf(x) := supv∈U∣∣∣∣ lim→0 1pi∫{|t|>}f(x− νt)dtt∣∣∣∣.The aforementioned unboundedness is an immediate consequence of the es-timate‖HU‖Lp(Rn)→Lp,∞(Rn) ≥ Cp,n√log (#U), (0.0.1)that we will prove under the hypothesis that the cardinality of the set U isfinite.All of the work presented henceforth was conducted under the supervi-sion of Professor I.Laba and Professor M.Pramanik of the Department ofMathematics, University of British Columbia, between February 2016 andDecember 2017.The author studied the proof of the unboundedness for the same oper-ator, denoted HU , in the case p = n = 2 ([17]) and proved Lemma 2.1.2,that provides the fundamental step for the generalization of the result in[17] to all dimensions. Through this lemma, the author was able to provean unboundedness result for the aforementioned operator in the case n ≥ 2and 1 < p ≤ 2.Subsequently, in collaboration with I.Laba and M.Pramanik, we man-aged to find a stronger proof of unboundedness for the operator HU , thatboth strengthens the one previously obtained by the author and extends theunboundedness of HU to the dual range of exponents 2 ≤ p < +∞. This isan immediate consequence of a crucial upper estimate for the norm of HUthat is mainly due to Professor Laba and Professor Pramanik (see chapter4).Part of this work has been presented in the following publication: I. Laba,A. Marinelli, M. Pramanik. On the Maximal Directional Hilbert Transform.ivPreface2017. Preprint available at https://arxiv.org/abs/1707.01061.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . viList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . x1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The maximal directional Hilbert transform . . . . . . . . . . 11.1.1 A brief survey of the theory of the Hilbert transform 11.1.2 Main results . . . . . . . . . . . . . . . . . . . . . . . 31.2 Maximal directional operators in the literature: an overview 41.2.1 Convolution operators . . . . . . . . . . . . . . . . . . 41.2.2 The maximal directional function . . . . . . . . . . . 91.2.3 General maximal directional operators . . . . . . . . 211.2.4 HU in the literature . . . . . . . . . . . . . . . . . . . 222 The proof of Theorem 1.1.4 . . . . . . . . . . . . . . . . . . . 282.1 Some preliminary facts . . . . . . . . . . . . . . . . . . . . . 282.2 The proof of the theorem . . . . . . . . . . . . . . . . . . . . 322.3 The difference between the cases n = 2 and n > 2 . . . . . . 343 Tree systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 The proof of Proposition 2.1.1 . . . . . . . . . . . . . . . . . . 444.1 An important tree system . . . . . . . . . . . . . . . . . . . . 444.2 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . 494.3 The proof of the proposition . . . . . . . . . . . . . . . . . . 59viTable of Contents5 The geometric lemma . . . . . . . . . . . . . . . . . . . . . . . 636 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 676.1 Ongoing and future directions of research . . . . . . . . . . . 68Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70viiList of Figures1.1 The left half of the set E′ ∩ I is a lacunary sequence denotedby {yk}k∈N and converging to a. . . . . . . . . . . . . . . . . 111.2 The binary tree associated with the set Ω ={1, 12}. . . . . . 161.3 The binary tree associated with the set Ω ={2−k}k≥0. . . . . 171.4 The binary tree associated with the set Ω ={0, 14 ,12 ,34}. . . . 181.5 The ternary tree associated with the standard Cantor set viaBateman’s bijection procedure can be identified with the com-plete binary tree. . . . . . . . . . . . . . . . . . . . . . . . . . 191.6 If U is the set in Example (1.2.21), U ′ in the longest lacunarysequence with constant 1/2 and it has N + 1 elements. . . . . 252.1 The red sectors are the ones that fall within the half-planeΓuj : S1, . . . Sj . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2 Only 7 planes passing through the origin in R3 generate ahigh number of sectors. . . . . . . . . . . . . . . . . . . . . . 363.1 The nested supports of tree system functions. . . . . . . . . . 383.2 The double-indexing notation for the numbers 1, . . . 2m − 1allows us to represent them as vertices of a complete binarytree of height m. . . . . . . . . . . . . . . . . . . . . . . . . . 393.3 For a signed tree system, the signs of f(k)j are encoded in thebinary tree. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.4 The full binary tree with m = 3 and with the numbers tN . . 413.5 The permutation σ and the choice of lx. . . . . . . . . . . . . 424.1 A sequence of 4 indices represented as vertices of B4 . . . . . 514.2 Two sequences of three indices that belong to the same rayof B5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52viiiList of Figures4.3 The fixed vertices m1, . . .mk−1, n1, . . . nl−1 lie on the ray R′,connecting the root of the tree and the vertex max {mk−1, nl−1}.The innermost summation is over vertices mk = nl of heighth in the sub-tree with root max {mk−1, nl−1}). . . . . . . . . 585.1 The vertical line rx′ intersects all the planes at distinct points(in this case we have n = 3 and M = 3). . . . . . . . . . . . . 645.2 Again in the case n = 3, M = 3, we can pick distinct pointsQj on the vertical segment between two consecutive pointsPj . These points identify nonempty, disjoint sectors of R3. . . 65ixAcknowledgementsI would like to thank my supervisors, Professor Izabella Laba and Profes-sor Malabika Pramanik for their excellent and patient guidance during mystudies and the preparation of this dissertation.It has been a privilege for me to study at the University of BritishColumbia. I am grateful to this institution for its generous financial support.xChapter 1Introduction1.1 The maximal directional Hilbert transform1.1.1 A brief survey of the theory of the Hilbert transformLet us start with the following notation.Notation 1.1.1. We denote by S(Rn) the space of all C∞(Rn) functionswhose decay at infinity and the decay of all their derivatives at infinity isfaster than any polynomial decay (see [12, Definition 2.2.1]). In other words,f ∈ S(Rn) if ∀ α ∈ Nn (included the case α = (0, . . . 0)) and ∀ N ∈ N wehavesupRn{∣∣∣∂(α)f(x)∣∣∣ (1 + |x|)N} < +∞,or, equivalentlylim|x|→+∞∂(α)f(x)(1 + |x|)N = 0.It can be shown that S(Rn), referred to as the Schwartz space, is dense inLp(Rn) for any 1 ≤ p < +∞.Now we follow [12] and introduce the Hilbert transform on R as anoperator acting on functions in S(R) (see [12, Definition 5.1.1])):Definition 1.1.1. Let f ∈ S (R). We setHf(x) :=1pilim→0+∫{|t|>}f(x− t)dtt.The function Hf is called the Hilbert transform of f .It is easy to see that the previous definition is well-posed (it suffices tosplit the integral into two integrals over {|t| ≥ 1} and {1 ≥ |t| > }, respec-tively, and exploit the fact that the function 1/t is odd over the secondinterval).Before outlining the main properties of the operator H, we recall thedefinition of weak-Lp norm (we assume the reader to be familiar with the11.1. The maximal directional Hilbert transformnotion of strong-Lp norm). Henceforth, the notation |A| shall denote theLebesgue measure of a measurable subset A of Rn.Definition 1.1.2. Let 1 ≤ p < +∞. The space Lp,∞(Rn) is defined as theset of all measurable functions f on Rn such that‖f‖Lp,∞(Rn) := inf{C > 0 : |{x ∈ Rn : |f(x)| > α}| ≤(Cα)p∀α > 0}is finite. The term ‖f‖Lp,∞(Rn) is called the weak Lp-norm of f .We recall that an operator T defined on a dense subspace D of Lp(Rn)or Lp,∞(Rn), respectively, is called of (p, q)-strong or of (p, q)-weak type if,for any f ∈ D, one has ‖Tf‖Lq(Rn) ≤ Cp,q,n‖f‖Lp(Rn) or ‖Tf‖Lq,∞(Rn) ≤Cp,q,n‖f‖Lp(Rn), respectively.Theorem 1.1.1. Let f ∈ S (R) and 1 < p < +∞. Then:1. ‖Hf‖L1,∞(R) ≤ C‖f‖L1(R), where C is a constant independent of f .Therefore, by density H extends to a (1, 1)-weakly bounded linear op-erator on L1(R).2. ‖Hf‖Lp(R) ≤ Cp‖f‖Lp(R), where Cp is a constant depending on p butindependent of f . Therefore, by density H extends to a (p, p)-stronglybounded linear operator on Lp(R).The proofs of the previous results can be found in section 5.1.3 of [12].Let us now introduce the following multidimensional version of the operatorH.Definition 1.1.3. Let v be a unit vector in Rn, n ≥ 2. For f ∈ S (Rn), wesetHvf(x) :=1pilim→0+∫{|t|>}f(x− tv)dtt.The function Hvf is called the directional Hilbert transform of f along thedirection v.Remark 1.1.1. Some of the properties of the operator Hv can be easilyderived from the ones of H. Indeed, if e1 = (1, 0, . . . 0) is the first elementof the canonical basis of Rn, then there exists an orthogonal matrix O ∈On(Rn) such that Oe1 = v. It is easy to prove the identityHvf(x) = He1(f ◦O)(O−1x)21.1. The maximal directional Hilbert transformfor all x ∈ Rn (see [12]). Therefore, up to rotations in Rn, the study ofHvf can be reduced to that of He1f , which is obtained by applying theHilbert transform in the first variable followed by the identity operator inthe remaining variables. An easy consequence of this is the following result.Theorem 1.1.2. Let v be a unit vector of Rn and 1 < p < +∞. Then thereexists a constant Cp > 0 such that we have‖Hvf‖Lp(Rn) ≤ Cp‖f‖Lp(Rn) ,for all f ∈ S (Rn), where we can take Cp = ‖H‖Lp(R)→Lp(R). Finally, bydensity Hv extends to a (p, p)−strongly bounded linear operator on Lp(Rn).Now let us introduce the operator that we study in this dissertation.Definition 1.1.4. Let U be a collection of unit vectors in Rn, n ≥ 2. Forf ∈ S (Rn), we setHUf(x) := maxv∈U|Hvf | .The operator HU is called the maximal directional Hilbert transform of fover U .Through the rotation argument we used in Remark 1.1.1, it is straight-forward to see that, at least in the case of U with finite cardinality, HU is(p, p)−strongly bounded for all 1 < p < +∞. Indeed, if we denote by #Uthe cardinality of U , we have for f ∈ Lp(Rn):∥∥HUf∥∥p ≤ ∑v∈U‖Hvf‖p = #U‖He1f‖p .p,n #U‖f‖p , (1.1.1)where the notation .p,n and &p,n denote ≤ and ≥, respectively, up to amultiplicative constant depending on p and n. Nevertheless, there mightexist refinements of the the previous trivial bound that provide more infor-mation about the dependence of HU on #U . Moreover, the rough estimate(1.1.1) leaves open the possibility that there exist exponents p such thatHUis (p, p)-strongly (or weakly) bounded if #U = +∞. A partial answer tothese questions is given in the next section.1.1.2 Main resultsThe main result of this dissertation is the following theorem.Theorem 1.1.3. For any infinite set of unit vectors U in Rn with n ≥ 2,the operator HU cannot be extended to a bounded operator from Lp(Rn) toLp,∞(Rn) for any 1 < p < +∞.31.2. Maximal directional operators in the literature: an overviewIn other words, an inequality of the type‖HUf‖Lp,∞(Rn) ≤ Cp,n‖f‖Lp(Rn)for all f ∈ Lp (Rn) cannot be true for any constant depending only on 1 <p < +∞ and n if the set of unit vectors U in Rn is infinite. This is animmediate consequence of the quantitative estimate below:Theorem 1.1.4. For any 1 < p < +∞ and any n ≥ 2 there exists aconstant Cp,n such that, if U is a finite set of unit vectors in Rn, we have‖HU‖Lp(Rn)→Lp,∞(Rn) ≥ Cp,n√log (#U). (1.1.2)Indeed, the term√log (#U) clearly goes to infinity as #U → +∞.For n = 2 and p = 2, Theorems 1.1.3 and 1.1.4 were proved by G. A.Karagulyan in [17]. The proofs of these results follow the general outline ofKaragulyan’s argument, but extend it to all exponents 1 < p < +∞ and toall dimensions n ≥ 2.In the next chapter we widen our perspective on the operator HU andprovide more details about its properties and connections with other maxi-mal directional operators. Moreover, we review the past and recent scientificliterature on these topics.The proof of Theorem 1.1.4 can be found in Chapter 3 and relies onseveral intermediate steps. The proofs of these results are treated in hesubsequent chapters.Remark 1.1.2. It is well known that the Lp-strong boundedness impliesthe Lp-weak boundedness. Consequently, Theorem 1.1.3 shows that HU isalso Lp-strongly unbounded.1.2 Maximal directional operators in theliterature: an overview1.2.1 Convolution operatorsIn modern harmonic analysis, the Hilbert transform H is an example of aconvolution operator with a principal value distribution (see, for instance[12]). The convolution of two functions, or of a tempered distribution anda function, is one of the most important operations in analysis and appears41.2. Maximal directional operators in the literature: an overviewnaturally in many contexts. If f and g are two L1(Rn) functions, we definethe convolution of f and g, denoted by f ∗ g, as the function:f ∗ g(x) :=∫Rnf(x− y)g(y)dt. (1.2.1)It is easy to check that the right-hand side of (1.2.1) is a well-defined functionon Rn up to a null set. The convolution of the functions f and g at the pointx provides the weighted average of f at x where the weighting is given byg. For example, if we denote by B1(O) the unit ball in Rn, by χA thecharacteristic function of a set A ⊂ Rn and, for r > 0, use the standardnotation gr := r−ng (·/r), we have:f∗ = (χB1(O))r(x) := r−n∫{|y|<r}f(x− y)dy = 1rn∫{|x−z|<r}f(z)dz,which is the usual integral average of the function f on the ball Br(x) up toa dimensional constant.Convolutions have been intensively studied in analysis. The action onfunctions of many important operators, for instance, is through a convolu-tion with a fixed kernel. Moreover, the convolution of a function with a fixeddensity is a smoothing operation that produces a certain approximation ofthe function (see for instance [12, Theorem 1.2.19, Theorem 2.3.20]).A fundamental tool in the study of convolution operators are the maxi-mal functions, the theory of which we are now briefly going to recall.Definition 1.2.1. Let f ∈ L1loc (Rn) and ωn be the notation for the Lebesguemeasure of the unit ball in Rn. We setMf(x) := supr>01ωnrn∫Br(x)|f(y)|dy,for all x ∈ Rn. The function Mf is called the Hardy-Littlewood maximalfunction of f .There are a few variants of the previous operator that it is useful todefine. For instance, for any x ∈ Rn let us consider the balls containingx instead of only the ones centered at x. For f ∈ L1loc(Rn), we define theuncentered Hardy-Littlewood maximal function:Mf(x) := supB ballB3x1|B|∫B|f(y)|dy.51.2. Maximal directional operators in the literature: an overviewAlternatively, we can use cubes instead of balls in both the centered anduncentered variant. It is easy to check that each one of these operators ispointwise bounded by a constant (depending only of n) times M.The following theorem summarizes the main properties of the maximalfunctions:Theorem 1.2.1. [12, Theorem 2.1.6] Let M be the operator defined inDefinition 1.2.1 and let 1 < p ≤ +∞. Then:1. M is of (1, 1)-weak type with norm 1 up to a dimensional constant,that is ‖Mf‖L1,∞(Rn) ≤ Cn‖f‖L1(Rn) for all f ∈ L1(Rn).2. M is of (p, p)-strong type with norm at most a constant depending onlyon p and n, that is ‖Mf‖Lp(Rn) ≤ Cp,n‖f‖Lp(Rn) for all f ∈ Lp(Rn).Therefore, for n = 1 and 1 < p < +∞, Theorem 1.1.1 and Theorem1.2.1 show that the operators M and H have the same properties in termsof (1, 1)-weak and (p, p)-strong boundedness.Maximal functions are crucial in the study of convolution operators be-cause they help in obtaining almost everywhere convergence for certain in-tegral averages of great importance in the subject of Fourier analysis. Tounderstand how, let us consider the following result:Theorem 1.2.2. [12, Theorem 2.1.14] Let K ∈ L1loc(Rn), 1 ≤ p ≤ +∞,1 ≤ q < +∞ and let us set K := −nK(·/) for  > 0. Let us supposethat the convolution K ∗ f is a well defined, measurable function for allf ∈ Lp(Rn) and let us set:T∗f := sup>0|K ∗ f | .Let D be a dense subspace of Lp(Rn) (for instance, D = C∞c (Rn), the space ofall smooth functions with compact support). Let us suppose that lim→0K∗fexists and is finite almost everywhere for all functions f ∈ D. Finally, letus suppose T∗ to be of (p, q)-weak type with norm Bp,q, that is‖T∗f‖Lq,∞(Rn) ≤ Bp,q,n‖f‖Lp(Rn) .Then the limit Tf := lim→0K ∗f exists almost everywhere and is finite forall functions f ∈ Lp(Rn) Moreover, we have‖Tf‖Lq,∞(Rn) ≤ Bp,q,n‖f‖Lp(Rn) .61.2. Maximal directional operators in the literature: an overviewThe point of the previous theorem is that the almost everywhere point-wise convergence of the family K ∗ f , with f in some dense subspace ofLp(Rn), and the (p, q)-weak boundedness of the maximal operator T∗ area sufficient condition for the existence of the almost everywhere pointwiselimit of Tf for all functions in Lp(Rn). This conclusion would be muchharder to prove without Theorem 1.2.2.An immediate application of the previous result is the famous differen-tiation theorem due to Lebesgue, which states that the average1/|Br|∫Br(x)f(y)dytends to f(x) as r → 0 for almost everywhere x ∈ Rn and for all f ∈ Lp(Rn).Indeed, if we set Trf := 1/|Br|∫Br(x)f(y)dy, we observe that Trf is well de-fined for all f ∈ L1(Rn), T∗ =M and Trf → f pointwise on Rn for, say, allf ∈ C∞c (Rn). But M is (p, p)-strongly bounded (Theorem 1.2.1), thereforethe result follows via Theorem 1.2.2.It can be shown that the Hardy-Littlewood maximal function of f con-trols the averages of f with respect to any radially decreasing L1(Rn) weight(see for instance [12, Theorem 2.1.10]). In other words, we can always con-trol from above the maximal operator sup>0|K ∗ f | withMf if the kernelK is summable and radially decreasing (or has a radially decreasing ma-jorant). Therefore, we get the almost everywhere pointwise convergence ofK ∗ f to f for any f in Lp(Rn) and 1 ≤ p < +∞ via Theorem 1.2.2.Nevertheless, the generality of Theorem 1.2.2 allow us to work withimportant convolution operators that do not fall into the previous category(for instance, operators whose kernels do not have a radially decreasingL1(Rn) majorant). The most relevant examples are the Fourier partial sumsof a square integrable function on the line or, equivalently, their integralversions:SNf(x) =∫ N−Nf̂(ξ)e2piiξxdξ, (1.2.2)for N ∈ N, where f̂ is the Fourier transform of f in L2(R). Indeed, it is easyto check that SNf is well defined for all f ∈ L2(R) and converges pointwiseto f for, say, all f in C∞c (R), which is a dense subspace of L2(R). Moreover,SNf(x) =sin (2piN(·))pi(·) ∗ f(x),71.2. Maximal directional operators in the literature: an overviewwith sin(2piN(·))pi(·) =(sin(2pi(·))pi(·))1/N, where we use the usual notaton gr :=r−ng (·/r) with r = 1/N . Nevertheless, it is well known that the func-tion sin(2pi(·))pi(·) is not summable on R. Carleson’s celebrated theorem (see [7],[13, Theorem 6.1.1] or [27, Theorem 7.1]), however, states that:∥∥∥ supN∈N|SNf |∥∥∥L2,∞(R)≤ C‖f‖L2(R) ,therefore we can still apply Theorem 1.2.2 and prove the almost everywhereconvergence of the Fourier series of any function in L2(R).Being the convolution operator with the simplest singular kernel, namely1/t, the Hilbert transform is the prototype of all singular integrals and itsstudy provided inspiration for many development of the subject (see [12],[33], [26]). Moreover, the operator H is deeply connected to the Fourierpartial sum operator (1.2.2). To see this, let us denote by g∨ the inverseFourier transform of the L2(Rn) function g. Operators of the type f →(mf̂)∨with m ∈ L∞(Rn) and f ∈ L2(Rn) are called Fourier multiplieroperators (see [12, Definition 2.5.11] or [33]). It is not hard to see that Hfcan be expressed as a Fourier multiplier operator with multiplier −isgn(·):Hf(x) =(−isgn(·)f̂)∨(x). (1.2.3)In particular, we have ‖Hf‖L2(R) = ‖f‖L2(R) via Parseval’s identity. There-fore, by density, H extends to an isometry of L2(R).On the other hand, it is easy to see that the study of SNf can be reducedto the one of the operatorTNf(x) =(χ(−∞,N)f̂)∨(x) =∫ N−∞f̂(ξ)e2piiξxdξ.But the multiplier of the operator TN is m(ξ) := 1/2− sgn(ξ −N)/2. Thismultiplier can be obtained from −isgn(·), the multiplier of the Hilbert trans-form, through a translation and a multiplicative constant. Consequently, bythe properties of the Fourier transform, TN behaves very similar to H. Inparticular, we can derive its (1, 1)-weak and (p, p)-strong boundedness on Rfor all N ∈ N and 1 < p < +∞ from the same properties of H (or the otherway around).81.2. Maximal directional operators in the literature: an overview1.2.2 The maximal directional functionFor f ∈ L1loc (Rn) we define the operatorMvf(x) := supr>012r∫ r−r|f(x− tv)|dt. (1.2.4)The function Mvf is called the maximal directional function of f alongthe direction v. Through the rotation argument used in Remark 1.1.1, wecan prove that Hv and Mv have the same (p, p)-strong boundedness for1 < p < +∞ on Rn. It might seem reasonable to think of the similarities(in terms of Lp-boundedness) between M and H, as well as between theirdirectional variants, as evidence of a deep likeness of the nature of the twooperators. Instead, as we shall see, a complete difference of behavior emergeswhen we study the analogue of HU for the maximal function, that is theoperator MUf(x) := maxν∈U |Mνf |. We call MU the maximal directionalfunction over the set U . Of course, if #U < +∞ it is trivial to show that,as we saw for HU , ∥∥MUf∥∥Lp(Rn) ≤ #U‖f‖Lp(Rn) . (1.2.5)But we will see that there are results of (p, p)-strong boundedness for theoperator MU even if #U = +∞. Theorem 1.1.3, instead, states that this isis not possible for HU .The operatorMU has been extensively studied over the past decades andthere are several papers that prove upper bounds for it. Before reviewingthe literature on this topic, let us give a few preliminary definitions:Definition 1.2.2. A sequence of real numbers {xk}k∈N converging to 0 issaid to be lacunary if there exists a number 0 < λ < 1 such that 0 < xk+1 ≤λxk for all k.Example of lacunary sequences are{2−k}k∈N ,{3−k}k∈N and, in general,{N−k}k∈N for a fixed N ∈ N.Some papers ([31] for instance) use the following, slightly more generaldefinition of lacunary sequence:Definition 1.2.3. A sequence {xk}k∈N of real numbers with xk k→ x (notnecessarily to 0) and xk 6= x for all k is called lacunary if there exists a0 < λ < 1 such that:|xk+1 − x||xk − x| < λ (1.2.6)for all k.91.2. Maximal directional operators in the literature: an overviewIs it clear that any sequence {xk}k∈N satisfying Definition 1.2.2 satisfiesDefinition 1.2.3 too (with x = 0), therefore we will use the latter throughoutthis dissertation.In what follows we use the notationdist(x,E) := infy∈E{|x− y|} ,where x ∈ R and E is a subset of R. We call dist(x,E) the distance betweenx and the set E.Definition 1.2.4. Let E and E′ be closed, zero measure sets in R. Thenwe say that E′ is a successor of E if there exists a constant C > 0 such that,for any x, y ∈ E′ with x 6= y, we have |y − x| ≥ Cdist(x,E).Definition 1.2.4 implies that we have|y − x| ≥ C max {dist(x,E) , dist(y,E)}for all x, y ∈ E′ with x 6= y. Roughly speaking, this means that the pointsof the set E′\E cannot lie too close to each other. For instance, in order tobetter understand the structure of a successor E′ of a set E, let us considerthe case E := {x0} (x0 ∈ R). If x 6= x0 and x ∈ E′, then we have dx :=dist(x,E) > 0. Now, Definition 1.2.4 implies that there are no points y ∈ E′such that |x− y| < Cdx. In other words, the neighborhood]x− Cdx , x+ Cdx[is contained in the complement of E′. The fact that dx → 0 as x → x0easily implies that a successor of {x0} must be contained in the union oftwo infinite monotonic sequences converging to x0 from above and frombelow, respectively. The discrete nature of E′ that emerges from the previousexample holds for more general sets E. Indeed, one of the consequencesof Definition 1.2.4 is that, if I is an open interval contained in Ec andhaving endpoints in E, then E′∩ I is contained in the union of two lacunarysequences converging to the endpoints of I. Indeed, let I =]a, b[ and supposethat there is a discrete infinite set of points of E′ in the left half of I. Thenwe can number them yk from the right to the left (see Figure 1.1 below) andwe have:|yk − a||yk+1 − a| =yk − ayk+1 − a = 1 +yk − yk+1yk+1 − a ≥≥ 1 + c |yk+1 − E|yk+1 − a = 1 + c > 1, (1.2.7)that is precisely the condition (1.2.6).101.2. Maximal directional operators in the literature: an overviewFigure 1.1: The left half of the set E′ ∩ I is a lacunary sequence denoted by{yk}k∈N and converging to a.Definition 1.2.5. Let a lacunary set of order 0 be a one-point set in R.Then, we define inductively a lacunary set of order k (k ≥ 1) to be a suc-cessor of a lacunary set of order k-1.Example 1.2.1. A lacunary set of order 1, therefore, is the successor of asingleton {x}, that is it is contained in the union of two lacunary sequencesconverging to x from the left and the right, respectively.Example 1.2.2. An example of lacunary set of order 2 isE′′ :={2−j + 2−i : i, j ∈ Z, i ≥ j} .Indeed, by setting E = {0} and E′ = {0, 2−j : j ∈ Z}, we have clearly thatthe lacunary sequence 2−j converges to 0 as j → +∞, therefore E′ is alacunary set of order 1. But, for i ≥ j, the numbers 2−j + 2−i lie between2−j and 2−j+1, and 2−j +2−i → 2−j as i→ +∞. This means that, if I is anopen interval that does not contain any element of E′ but with endpoints inE′ (say, I =]2−j , 21−j [), we have E′′ ∩ I = {2−j + 2−i : i ∈ Z, i ≥ j}, whichis clearly contained in a lacunary sequence.Example 1.2.3. By means of the same argument as that used in Example1.2.2, the setE′′′ :={2−k + 2−j + 2−i : i, j, k ∈ Z, i ≥ j ≥ k}.can easily be shown to be lacunary of order 3. And so on.Definition 1.2.6. We say that a set U ={uk = (u1k, u2k)}k∈N of vectors inR2 is L1-lacunary if there exists a number 0 < λ < 1 such that0 ≤ u2k+1u1k+1≤ λu2ku1k(1.2.8)for all k.111.2. Maximal directional operators in the literature: an overviewU is said to be L2-lacunary with node u∞ ∈ S1 if|uk+1 − u∞| ≤ 12|uk − u∞| (1.2.9)for all k.U is said to be L3-lacunary of order r if uk = e2piiθk for all k and thereexists a positive integer r such that {θk}k∈N is a lacunary set of order raccording to Definition 1.2.5.In short, a L1-lacunary set in the plane is a set of vectors whose slopesare a positive lacunary sequence converging to 0. It is not hard to see that aL2-lacunary set or a L3 lacunary set of order 1 can be covered by the unionof two L1-lacunary sets1.The case with U L1-lacunary set of directions in R2 was considered inseveral papers. An important result of (p, p)-strong boundedness for MUwas obtained by A. Nagel, E. M. Stein and S. Wainger in [28], whose maintheorem is actually a n-dimensional result (n ≥ 2) that we restate here:Theorem 1.2.3. [28, Theorem 1] Let γ(t) = (ta1 , . . . , tan) be a curve in Rnsuch that all the exponents aj are positive, and let {θk}k∈N be a lacunarysequence converging to 0. Let us consider the function:Ah,jf(x) :=1h∫ h0f(x− tγ(θj))dt.for f ∈ L1loc (Rn). Now, if we setM ∗f(x) := suph>0, j∈N|Ah,jf(x)| ,we have‖M ∗‖Lp(Rn)→Lp(Rn) .p 1. (1.2.10)for 1 < p < +∞.The (p, p)-boundedness of the operator M ∗ implies the one of the verysimilar operator MU , where U is the set of directions of the vectors γ(θj).1Indeed, we have from the condition (1.2.9) that uk → u∞ as k → +∞. Up to arotation of the set U , we can always assume that u∞ = (1, 0). Therefore, there areat most two sequences of unit vectors in U converging to (1, 0) from above and frombelow, respectively. It is not hard to prove that these two sequences satisfy a condition oflacunarity of type L1.121.2. Maximal directional operators in the literature: an overviewIndeed, by splitting the integral in (1.2.4) into two parts and using a changeof variables, it is easy to show thatMUf(x) ≤ 12(M ∗|f |(x) +M ∗∣∣f˜ ∣∣(−x)) ,where f˜(y) denotes the function f(−y).To see how Theorem 1.2.3 can be applied to L1-lacunary sets of direc-tions, let us consider a curve of the type (1, ta) and a lacunary sequence θkwith lacunarity constant λ. Then we have θak+1 ≤ λaθak for all k according toDefinition 1.2.2, which implies that the vectors (1, θak) form a L1-lacunaryset. Therefore, if we denote by U the set of directions of these vectors, wehave that MU is of (p, p)-strong type for 1 < p < +∞.A result of (p, p)-boundedness in dimension 2 for MU that extends theone that follows from [28] in the case of U L3-lacunary set is [31], due toSjo¨gren and Sjo¨lin. Roughly speaking, the idea is to iterate the argumentsof the proof in [28] in order to prove that‖MU‖Lp(R2)→Lp(R2) .p,r 1for 1 < p < +∞ and U L3-lacunary set of directions of order r (Corollary2.4, [31]).A more recent paper is [1], due to Alfonseca. In the context of L1-lacunary sets of directions, it generalizes the (p, p)-boundedness in [28] forthe directional maximal function MU (see [1, Theorem 5]) by showing thatit is possible to introduce new directions in the lacunary set U ⊂ S1 andstill get a bounded operator. More precisely:Theorem 1.2.4. [1, Theorem 5] Let {θk}k∈N be a positive lacunary se-quence converging to 0 and U0 be a L1-lacunary set of directions with slopes{θk}k∈N. Let Uk be any set of directions between two consecutive directionsof U0, for any k ∈ N. Finally, for all k ∈ N let us set U = ∪k≥0Uk,MUkf(x) := supθ∈Uk{Mθf(x)}and MU = supk≥0{MUk}. Then we have‖MU‖Lp(R2)→Lp(R2) .p supk≥0{‖MUk‖Lp(R2)→Lp(R2)},for any 1 < p < +∞.131.2. Maximal directional operators in the literature: an overviewTherefore, [1] generalizes the estimate for MU provided by (1.2.10) in[28]. Indeed, by setting Uk = ∅ for all k ≥ 1, U is trivially a L1-lacunary setof directions and‖MU‖Lp(R2)→Lp(R2) = ‖MU0‖Lp(R2)→Lp(R2) .p 1. (1.2.11)Nevertheless, [1] also generalizes Sjo¨gren and Sjo¨lin’s result [31]. To see this,let U be a L3-lacunary set of directions of order r ≥ 1 and let us proceed byinduction on r. The inequality (1.2.11) provides the verification for r = 1. Ifr > 1, Definition (1.2.6) implies that there exists a subset U ′ ⊂ U such thatthe elements of U between two neighboring directions of U ′ are contained intwo L1-lacunary sets of directions2. If we denote by {Uk}k≥1 the collectionsof elements of U between two neighboring directions of U ′, Theorem 1.2.4implies:‖MU‖Lp(R2)→Lp(R2) .p max{supk≥1{ ‖MUk‖p→p }, ‖MU ′‖p→p },which in turn implies ‖MU‖Lp(R2)→Lp(R2) .p,r 1 by induction.On the other hand, there are examples of sets of directions U in R2 forwhich there is no (p, p)-strong boundedness forMU . For example, if U = S1,the operator MU is not of (p, p)-strong type. This is a consequence of theexistence of the Besicovitch set3, that is a null set in the plane that containsa unit line segment in every direction (see [30]). Another important exampleof (p, p)-strong unboundedness for MU is when the slopes of the vectors ofU are a Cantor-type set: see [3] or [20] (see also the Example 1.2.8 below).Finally, the paper [2] by Bateman gives a necessary and sufficient condi-tion on the set U under which the inequality ‖MU‖p→p .p 1 holds, therebyproviding a result that completely characterizes the (p, p)-boundedness ofMU in the plane. The author uses the terminology of trees, that we verybriefly recall.A tree is an undirected two-dimensional graph with a root (or origin),which is the vertex from which the whole tree generates. This dissertationand the references we mention ([2], for instance) use binary trees, that istrees in which each vertex v has at most two children (called the 0th and1st children of v). In [2] and in what follows, the children of a vertex v aredenoted by c0(v) and c1(v). Accordingly, v will be called the parent of c0(v)and c1(v). Within the tree, we always move downward from a certain vertexto one of its children and along a certain path or ray, that is a sequence of2This is a consequence of the argument that has led to the inequality (1.2.7).3For the construction of the Besicovitch set, see for example [11] or [13].141.2. Maximal directional operators in the literature: an overviewparent - child couples of vertices where the child vertex of each couple agreeswith the parent vertex of the following one.If v is a vertex of a tree T , we will use the notation v ∈ T . The verticesalong the ray that connects the root of T with v are called the ancestorsof v. Their sequence in descending order is called the identifying sequenceof v. The height (or level) of v, denoted by h(v), is the number of parent -child couples of vertices in its identifying sequence4. If u is another vertexof the same tree, u is a descendant of v if they lie on the same ray andh(u) > h(v).The height of a tree T , denoted by h(T ), is defined as the supremumof the heights of the vertices of T :h(T ) := supv∈T{h(v)} .Finally, the sub-tree of T generated by the vertex v is the maximal sub-treeof T with v as its root.It turns out that there exists a way to encode certain subsets of theinterval [0, 1[ by binary trees via the following bijection:1. We identify the origin of our tree with the whole interval [0, 1[,2. Let I be a dyadic interval contained in [0, 1[ and let us identify it witha vertex v of a binary tree. Then we identify c0(v) with the left halfof I and c1(v) with the right half of I.In this way, if we denote by B the full, infinite binary tree, we have that anydyadic interval in [0, 1[ is associated with a specific vertex of B and thatB corresponds to the collection of all dyadic intervals of [0, 1[. The treeassociated with a generic Ω closed subset of [0, 1[, instead, will be obtainedthrough a certain trimming process applied to B. Let us recall, indeed,that [0, 1[ \Ω can be written as the union of countable many disjoint dyadicintervals, say {Ik}k∈N. Then we remove all sub-trees of B whose rootscorrespond to the intervals Ik. The residual tree obtained in this way isthe one we associate with the set Ω. This completes the definition of ourbijection between closed subsets of [0, 1[ and sub-trees of B, denote by Ω→TΩ.4The height of the root is always taken to be zero151.2. Maximal directional operators in the literature: an overviewExample 1.2.4. Let us set Ω :={1, 12}and let us construct its tree. Wehave[0, 1[ \Ω =[0,12[∪]12, 1[,The interval[0, 12[is dyadic, but]12 , 1[is not, therefore we have to write thelatter as a union of dyadic intervals. We have:]12, 1[=[34, 1[∪[58,34[∪ . . .In other words, we cover]12 , 1[starting from its right half: first the dyadicinterval[34 , 1[, then the dyadic interval half its size and next to it, that is[58 ,34[, then the dyadic interval half the size of the latter and next to it,and so on. Now let us construct TΩ: starting from the root, remove its leftchild (that is[0, 12[), the right child of its right child (that is[34 , 1[), theright child of its left child (that is[58 ,34[) and so on until we cover the wholeinterval]12 , 1[. The residual tree is shown in Figure 1.2 below.Figure 1.2: The binary tree associated with the set Ω ={1, 12}.Example 1.2.5. Now let us set Ω :={2−k}k≥0. We have:[0, 1[ \Ω =]12, 1[∪]14,12[∪]18,14[∪ . . . =+∞⋃k=1[ 12k,12k−1[.Unlike the previous example,[0, 12[is not contained in [0, 1[ \Ω, then theroot of TΩ must have two children. In order to find a cover of the interval[12 , 1[by disjoint dyadic intervals, we apply the same procedure used inExample 1.2.4, that is we split]12 , 1[into two parts of same length, keep[34 , 1[and repeat this argument for]12 ,34[. But we have to apply the same161.2. Maximal directional operators in the literature: an overviewprocedure to all other intervals]12k, 12k−1[: we remove from]12k, 12k−1[thedyadic interval Ik that covers its right half, than the dyadic interval half thesize of Ik and next to it and so on recursively. In this way, the structurethat emerges from the right children of the root will similarly develop fromthe right children of the left children of the root and so on, as we proceedleftward along the outer left edge of TΩ (see Figure 1.3 below).Figure 1.3: The binary tree associated with the set Ω ={2−k}k≥0.Crucial for the results in [2] is the concept of splitting number.Definition 1.2.7. Let T be a tree and v one of its vertices. We say that vis a splitting vertex if v has two children. We define the splitting numbersplit(R) of a path R along the tree T to be the number of splitting verticesalong R.In other words, split(R) is the number of forks along R.Definition 1.2.8. Let v be a vertex of a tree T , Tv be a sub-tree of Trooted at v (we write Tv ⊂ T ), and R be a path within Tv that begins atv (we denote this by R ∼ Tv). It is natural to define the splitting numbersplit(v) of the vertex v, denoted by split(v), assplit(v) := supTv⊂T{minR∼Tv{split(R)}}.Finally, we define the splitting number of the tree T , denoted by split(T ),assplit(T ) := supv∈T{split(v)} .171.2. Maximal directional operators in the literature: an overviewExample 1.2.6. For Ω as in Example 1.2.4, we have split(R) = 0 for allrays along TΩ, therefore split(v) = 0 for all v ∈ T and split(TΩ) = 0. If Ωis as in Example 1.2.5 and v lies on the outer left edge of TΩ, than we canfind rays R ∼ Tv with split(R) as large as we want, but split(v) is still =1. Therefore we have split(TΩ) = 1. Instead, let us set:Ω :={0,14,12,34}.In this case, it is not possible to remove the dyadic intervals of length 1/2or 1/4, therefore the root of TΩ has two children and four grandchildren.Through the leftward trimming argument used in Example 1.2.5, it is easyto see that four infinite leftward branches develop from these grandchildren(see Figure 1.4 below). This easily implies that split(root) = split(TΩ) = 2.Figure 1.4: The binary tree associated with the set Ω ={0, 14 ,12 ,34}.Subsequently the author introduces the following notion of lacunarity inthe framework of the theory of trees:Definition 1.2.9. Let T be a tree. Then, a sub-tree S ⊂ T is said to belacunary of order 0 if S consists of a single path along T . S is said to belacunary of order N if all splitting vertices of S lie along a lacunary tree oforder N − 1.Example 1.2.7. it can be shown that the set Ω :={2−j + 2−i : i ≥ j},proved to be lacunary of order 2 (see Example 1.2.2), is lacunary of order 2in Bateman’s sense, that is the tree TΩ is lacunary of order 2 according toDefinition 1.2.9. However, in this case TΩ is far more complex than the oneof the set{2−k}k≥0.181.2. Maximal directional operators in the literature: an overviewExample 1.2.8. It can be proved that C , the standard Cantor set, fallsunder the category of Theorem 1.2.6. Indeed, we can easily associate withC a tree according to Bateman’s bijection procedure, but a ternary oneinstead of binary. In other words, we split [0, 1[ into ternary intervals, whichcorrespond to the vertices of the full ternary tree, and associate any closedsubset Ω of [0, 1[ with the residual ternary tree that we get after removingthe vertices identified with the intervals of the ternary decomposition of[0, 1[ \Ω. In this way, it is easy to see that the tree associated with C canFigure 1.5: The ternary tree associated with the standard Cantor set viaBateman’s bijection procedure can be identified with the complete binarytree.be constructed from the root by adding only the 0-th and 2-th child toeach vertex and by proceeding recursively. But the corresponding tree canintuitively be identified with the full binary tree B (see Figure 1.5 above),which means that C is as lacunary of finite order as [0, 1[, that is it is notlacunary of finite order.Through Definition 1.2.9, Bateman proves the following result:Theorem 1.2.5. [2, Theorem 3 (A)] Let U be a set of direction in R2, letΩ ⊂ [0, 1[ be the set of their slopes and let us denote by TΩ the tree associatedwith Ω. If we have split(TΩ) = N , then the tree TΩ is lacunary of order Naccording to Definition 1.2.7 and we have‖MU‖Lp(R2)→Lp(R2) .p,N 1, (1.2.12)for any 1 < p < +∞. In particular, the previous bound does not depend onthe set U , but only on the order of its lacunarity.However, as we have seen, [31] proves the condition (1.2.12) under thehypothesis that the set U is L3-lacunary of order N . But the main result in[2] is the proof that the condition on U in Theorem 1.2.5 is also necessary.We need the following notion:191.2. Maximal directional operators in the literature: an overviewDefinition 1.2.10. We say that a closed subset Ω of [0, 1[ admits Kakeyasets if for each N ∈ N there is a collection RNΩ of rectangles of width 1 andeach pointed in a direction with slope in Ω such that:∣∣∣∣ ⋃R∈RNΩR∣∣∣∣ ≤ 1N∣∣∣∣ ⋃R∈RNΩ3R∣∣∣∣where 3R is the rectangle with the same width and center as R and threetimes the length.The previous condition turns out to be the one under which there is(p, p)-strong unboundedness for the maximal directional operator MU :Theorem 1.2.6. [2, Theorem 3 (B)] Let Ω ⊂ [0, 1[ be the set of slopes of acollection of unit vectors U in the plane. If split(TΩ) = +∞, then Ω admitsKakeya sets and MU is unbounded on Lp(R2)for any 1 < p < +∞.An example of result in n dimensions that generalizes the one proved in[28] is Carbery’s paper [6]. Let us consider the set of directions{(2k1u1, . . . 2knun) : k ∈ Zn},with u fixed non-zero vector. The proof treats the case n = 3, but themethod is applicable in all other dimensions. The author takes a non-negative S(R) function ψ with compactly supported Fourier transform suchthat ψ(0) > 0 and, for f Schwartz function, considers the operatorTf := supk∈Znh>0∣∣∣(ψ̂(2kξ · hu)f̂(ξ))∨∣∣∣,where we use the notation 2kξ := (2k1ξ1, . . . 2knξn). Carbery proves in [6]that‖T‖Lp(Rn)→Lp(Rn) .p 1, (1.2.13)which, for n = 2, generalizes the result in [28]. Indeed, let us consider theset of vectors{(2k1u1, 2k2u2) : k ∈ Z2}or, equivalently, {(2ru1, u2) : r ∈ Z}.Up to a reflection of coordinates, we can suppose r ≤ 0 and obtain a set ofdirection that falls within the category [28] considers (it suffices to take thecurve γ(t) = (tu1, u2) with θk = 2k, k ≤ 0). Now, it is not hard to show thatT can be written as a maximal directional convolution operator. Indeed, if201.2. Maximal directional operators in the literature: an overviewwe set mk,h := h∣∣2ku∣∣ and denote by vk the direction of the vector 2ku, wehave(ψ̂(2kξ · hu)f̂(ξ))∨(x) = ∫R1mk,hψ( tmk,h)f(x− tνk)dt = ψmk,h ∗ f(x).Since ψ(0) > 0, there exists a constant C depending only on ψ such thatχ[0,1] ≤ Cψ on R. Therefore:M ∗f ≤ supk∈Znh>0{(χ[0,1])mk,h∗ |f |(x)}. supk∈Zn, h>0{ψmk,h∗ |f |(x)},for all f ∈ Lp (Rn), which implies (1.2.10).We now turn to the case of U generic finite set of directions. Upperbounds for the norm of MU depending on #U in R2 are due to Katz (see[18, Theorem 2] and [18, Proposition 2.1 (B)]). They are:‖MU‖L2(R2)→L2(R2) . log(#U) ,‖MU‖L2(R2)→L2,∞(R2) .√log(#U),‖MU‖Lp(R2)→Lp(R2) . log (#U)1/p, p > General maximal directional operatorsGeneral types of maximal directional operators have also been studied. Forexample, for n = 2 and U finite set of arbitrary directions, a few resultsstudy operators with general convolution kernels that include HU as a par-ticular case. Let K be a Calderon-Zygmund kernel with m = Kˇ Ho¨rmander-Mikhlin multiplier on R (that is m ∈ C∞(R\ {0}), ∣∣m(k)(ξ)∣∣ .k |ξ|−k; see[12], [26]), v be a unit vector and f be a Schwartz function. Let us setTvf(x) := lim→0+∫{|t|>}K(t)f(x− tv)dt, (1.2.14)TUf := maxv∈U|Tvf | . (1.2.15)It is clear that if we choose the Calderon-Zygmund kernel K(t) := 1t , we getTv =Hv and TU =HU .The paper [8] provides the estimate‖TU‖L2(R2)→L2(R2) . log(#U) , (1.2.16)and, in the case TU = HU and under a specific geometric condition on U ,proves its sharpness (see [8, Proposition 3.3] and estimate (1.2.22) below).211.2. Maximal directional operators in the literature: an overviewAnother example with U finite set of arbitrary directions in the plane is[9]. It provides the estimate (see [9, Theorem 1]):‖TU‖Lp(R2)→Lp(R2) .p log(#U) , (1.2.17)for 2 < p < +∞. Moreover, if U is a finite L2-lacunary set of directions,[9] provides an improvement of (1.2.17) in terms of a sharp upper estimate(see [9, Theorem 2]):‖TU‖Lp(R2)→Lp(R2) .p√log(#U) (1.2.18)for any 1 < p < +∞. Hence with respect [9, Theorem 1]), that is the esti-mate (1.2.17), we have both an improvement of the upper bound (√log(#U)instead of log(#U)) and an extension of the range of exponents p. Thesharpness of (1.2.18) follows from the condition (1.2.24) below.1.2.4 HU in the literatureWe now turn to the specific case of the maximal directional Hilbert transformHU . In the hypothesis #U = +∞ and n = 2, there is a fundamentalresult by Lacey and Li for U = S1 (see the paper [23]) with HU acting onfunctions smoothly restricted to a frequency annulus. In other words, theauthors consider a Schwartz function ζ with Fourier transform supported inthe annulus {1 ≤ |ξ| ≤ 2} and the operatorH ∗S1f := sup|v|=1|Hv(ζ ∗ f)| .The function ζ ∗ f is a smooth restriction of f to the frequency annulus{1 ≤ |ξ| ≤ 2} in the sense that, as it is well known, its Fourier transform isthe smoothly truncated localization ζ̂ f̂ of the function f̂ in the consideredannulus. The result proved in [23] is the following theorem.Theorem 1.2.7. [23, Theorem 1.1] For p > 2 the following estimates hold:∥∥H ∗S1∥∥L2(R2)→L2,∞(R2) . 1,∥∥H ∗S1∥∥Lp(R2)→Lp(R2) .p 1.There exists an analogue of the previous result for the maximal operatorM∗S1f(x) := sup|v|=1|Mν(ζ ∗ f)| ,221.2. Maximal directional operators in the literature: an overviewthat is∥∥M∗S1∥∥L2(R2)→L2,∞(R2) . 1. This is due to Bourgain (see [5] or [23,Proposition 3.1]). A sketch of the proof of the sharpness of the estimatesof weak square integrability for both H ∗S1 and M∗S1 , due to M. Christ, isincluded in [23].However, Theorem 1.2.7 is more difficult than Bourgain’s estimate forM∗ and it can be shown to imply Carleson’s theorem on the almost every-where convergence of Fourier series on the line ([7]).We can not omit to mention the paper [24], by the same authors of anddeeply related to [23]. It studies the case when the set of directions U istaken to be the range of a vector field for n = 2. In other words, we canconsider a vector field ~v : R2 → S1 and for, say, f ∈ S(R2), the operatorH~vf(x) :=1pilim→0+∫{|t|>}f(x− t~v(x))dtt.It arises naturally the problem of finding the conditions on the vectorfield ~v (for instance, a certain degree of regularity) under which there is(p, p)-strong (or weak) boundedness for the operator H~v.The paper [24], however, proves a result concerning a slightly differentversion of H~v, namely the truncated operatorH~v,f(x) :=1pilim′→0+∫{′<|t|<}f(x− t~v(x))dtt,where the parameter  is suitably determined by the smoothness propertiesof ~v. The motivation for the introduction of the operator H~v, lies in thefact that there are examples of smooth vector fields ~v for which H~v is notof (2, 2)-strong type (see for example [10], pag. 4).The results in [24], whose proofs largely use ideas and methods from theaforementioned paper [23] by the same authors, are difficult and deeply in-tertwined with profound questions about almost everywhere convergence ofFourier series. Indeed, the authors show that, assuming the vector field ~v tobe regular enough (say ‖~v‖C2(R2) ≤ 1), the bound∥∥H~v,~v∥∥L2(R2)→L2(R2) .1 implies the (2, 2)-strong boundedness of Carleson’s operator ([24, Propo-sition 2.4]).Moreover, [24] includes a survey of the known connections between thetheory of the Hilbert transforms along vector fields and other importantproblems and operators, such as the Kakeya maximal function. Referencesthat provide an historical perspective of the study of such problems, aswell as previous results with respect to the ones in [23] and [24], are alsomentioned (for example [32] and [25]).231.2. Maximal directional operators in the literature: an overviewFinally, there exist several generalizations and refinements of the tech-niques used in [23] and [24]. For instance, the paper [4] treats the case whenthe vector field ~v depends on just one coordinate. For other references andadditional results, the interested reader is referred to the papers [4] and therecent [14], [15] and [16].We now turn to the case of U finite. In this case, HU turns out toreveal a deep difference of behaviour with respect to MU in terms of (p, p)-boundedness.Let us start with upper estimates. In the case n = 2 and U finite set ofarbitrary directions, the inequality‖HU‖L2(R2)→L2(R2) . log(#U) , (1.2.19)is an immediate consequence of Menshov-Rademacher’s theorem (see [9])and relies on orthogonality. This theorem states that, if f1, . . . fN are or-thogonal functions in L2(R2), then the functionF := max1≤k≤N∣∣∣∣ ∑1≤j≤kfj∣∣∣∣satisfies the estimate‖F‖L2(R2) . log(N)∥∥∥∥ ∑1≤j≤Nfj∥∥∥∥L2(R2). (1.2.20)Now, it turns out thatHUf can be pointwise controlled by an operator of theform Ff = max1≤k<#U∣∣∣∑1≤j≤kfj∣∣∣, where the functions fk are of the type(χSk f̂)∨and the Sk are disjoint frequency sectors of the plane rooted at theorigin. The orthogonality of the fk (via Plancherel’s Theorem) provides theupper bound log(#U) for ‖HU‖L2(R2)→L2(R2) through the inequality (1.2.20).The more recent article [21] gives another proof of (1.2.19) ([21, Theorem1]) without using Rademacher-Menshov’s theorem.We have already mentioned the paper [9], that provides sharp upperestimates forHU depending on the geometric structure of U , namely (1.2.17)and (1.2.18). In terms of structural assumptions on U , the paper [9] studiesa third case. First we need the following definition.Definition 1.2.11. Let N be a positive integer. A finite set U ⊂ S1 with#U = N is a Vargas set with constant Q ifmax{#U ′ : U ′ is a L2-lacunary subset ofU}= Q logN.241.2. Maximal directional operators in the literature: an overviewThe Vargas sets with constants Q independent of N are simply referredto as Vargas sets.Example 1.2.9. Let us consider a uniformly distributed set of directions{e2piij2−N : j = 1, . . . 2N} or, equivalently, the set of equi-distributed pointsin the unit interval:U :={j2N: j = 1, . . . 2N}(1.2.21)It is not hard to see that U is a Vargas set with Q = 1. Indeed, if weconsider the subset U ′ consisting of the term 1 and the largest possiblelacunary sequence with constant 1/2, that is if we setU ′ :={vj :=2N−j2N: j = 0, 1, . . . N},we have #U ′ = N + 1 ≈ N , which is exactly the logarithm of #U (seeFigure 1.6).Figure 1.6: If U is the set in Example (1.2.21), U ′ in the longest lacunarysequence with constant 1/2 and it has N + 1 elements.Example 1.2.10. It can be proved that the N -truncations of Cantor setsare also Vargas sets, with Q = 4 (see [9], pag. 25).Vargas sets were first introduced in [19]. In [9], the authors explain theimportance of these sets in relation to, say, L1-lacunary sets of directions.Substantially, they are the two extreme cases in terms of maximum cardi-nality of their lacunary subsets (see the result proved in [9], subsection 7.1).The third result of [9] concerns the operator HU instead of the general oneTU . It states that, if U is a Vargas set with constant Q, then we have‖HU‖L2(R2)→L2,∞(R2) .Q√log(#U) log log(#U)6 ,251.2. Maximal directional operators in the literature: an overviewwhich is a slightly weaker bound with respect to estimate (1.2.18) becauseof the term log log(#U)6.Finally, we treat the case of upper estimates for n ≥ 2. The paper [21]provides an upper bound (see [21, Theorem 2]) for HU :‖HU‖L2(Rn)→L2(Rn) .n (#U)n−22n−2 .This estimate is sharp (see 1.2.23 below)Several lower bounds for the operator HU for specific types of sets U arealso known. In the case of U finite set of uniformly distributed directionsin the plane (U ={e2piij/#U : j = 1, . . .#U}), we have the aforementionedpaper [8], that gives us‖HUf‖L2(R2) & log(#U)‖f‖L2(R2) . (1.2.22)where f is the indicator function of the unit ball in R2. The estimate (1.2.22)follows from Proposition 3.3 of [8] and proves that (1.2.16) is sharp.The recent paper [10] proves that, in the case of U finite L3-lacunary5 setof directions of order D, for any 1 < p < +∞ there exist positive constantscp,D, Cp,D such thatcp,D√log(#U) ≤ ‖HU‖Lp(R2)→Lp(R2) ≤ Cp,D√log(#U).and always with U finite and under the condition that the directions ofU have equi-distributed slopes (that is slopes j/#U for j = 1, . . .#UA lower estimate for n ≥ 3, but only for p = 2, is provided in theaforementioned paper [21]. For a specific set U0 of directions in Rn, itobtains (Theorem 2) the sharp bound‖HU0‖L2(Rn)→L2(Rn) &n (#U0)n−22n−2 . (1.2.23)Finally, we turn to the case of lower estimates when U is a generic finiteset of directions in the plane. G.A.Karagulyan’s paper [17] shows that‖HU‖L2(R2)→L2,∞(R2) &√log(#U). (1.2.24)5The only difference between Sjo¨gren and Sjo¨lin’s introduction of the notion of lacu-narity of finite order and the approach used in [10] is that in [31] the authors give thedefinition of lacunary set of finite order in R (pag. 5 of [31])) and then extend it to S1through the identification of S1 with the interval [0, 1[ (see the definition of successor ofa set on pag. 15 of [31]). This is exactly the approach followed in this dissertation (Defi-nition 1.2.5 and the definition of L3-lacunarity, respectively). Instead, in [10] the authorsbypass the 1-dimensional case and work directly in S1 (pag. 2 and 3)261.2. Maximal directional operators in the literature: an overviewThis is a remarkable result that proves that HU is an unbounded operatoron L2(R2)as soon as U contains infinitely many directions. A very inter-esting aspect of the nature of HU emerges, then, in total contrast to MU ,as we saw (see [2, Theorem 3]).In this research landscape, the recent paper [22], whose results are ex-posed in this dissertation, provides the definitive proof that the unbounded-ness proved by Karagulyan regarding HU for any set of directions U in theplanar setting when #U = +∞ is actually inherent to the specific natureof the operator and endures even in the n-dimensional context. Indeed, amore general inequality than (1.2.24) holds, namely‖HU‖Lp(Rn)→Lp,∞(Rn) &p,n√log(#U), (1.2.25)for all n ≥ 2, all 1 < p < +∞ and all sets of directions U in Rn of finitecardinality (see [22, Theorem 1.1] or Theorem 1.1.4 below). The proof of the(p, p)-weak unboundedness of HU in the case #U = +∞ trivially followsfrom the divergence of the term√log(#U) as #U → +∞.27Chapter 2The proof of Theorem 1.1.4The proof of Theorem 1.1.4 relies on three main intermediate results thatwe will state here and prove in the following chapters.2.1 Some preliminary factsWe recall that if T : (X, ‖·‖X) → (Y, ‖·‖Y ) is a sublinear operator betweentwo normed or quasi-normed spaces6, the operator norm of T is defined as‖T‖X→Y := supf∈X\{0}‖Tf‖Y‖f‖X. (2.1.1)Therefore, as a direct consequence of the definition (2.1.1), to prove Theorem1.1.4 it suffices to find a function f ∈ Lp(Rn) such that‖HUf‖Lp,∞(Rn) ≥ Cp,n√log (#U) ‖f‖Lp(Rn) . (2.1.2)The proof of the inequality (2.1.2) has two components. First we constructa function f ∈ Lp(Rn) satisfying both a certain geometric condition in con-nection with the set U and the estimate‖f‖Lp(Rn) .p,n√log (#U). (2.1.3)Then we prove that this crucial geometric condition enables the function fto obey the lower estimate‖HUf‖Lp,∞(Rn) &p,n log (#U) , (2.1.4)which immediately implies Theorem 1.1.4 via the estimate (2.1.3).We can reduce the problem of proving the inequality (2.1.4) to a similarinequality involving an operator that turns out to be more convenient to6A quasi-norm ‖·‖ on a vector space X satisfies the norm axioms, except that thetriangle inequality is replaced by ‖x+ y‖ ≤ C(‖x‖+ ‖y‖) ∀ x, y ∈ X, for a certain con-stant C. It is easy to see that Lp,∞(Rn) is a quasi-normed space and that the constant Cdepends on p.282.1. Some preliminary factswork with rather than HU . To this end let v be a unit vector in Rn and letus setΓv := {x ∈ Rn : x · v > 0} . (2.1.5)Recalling that the operator H can be expressed as a multiplier operator withmultiplier −isgn(·) (see the equation (1.2.3)), for v ∈ U and f ∈ S(Rn) wehaveHvf = −2i(χΓv f̂)∨ − f. (2.1.6)To see this, we observe thatĤvf(ξ) = −isgn(ξ · v)f̂(ξ) = −i(χ{ξ· v > 0} − χ{ξ· v < 0})f̂(ξ)= −i(2χ{ξ· v > 0}f̂(ξ)− χ{ξ· v > 0}f̂(ξ)− χ{ξ· v < 0}f̂(ξ))= −i(2χΓujf̂(ξ)− f̂(ξ)).Therefore the equation (2.1.6) follows immediately. Consequently, if U isa set of directions in Rn and we use the notation Tvf :=(χΓv f̂)∨, it isimmediate that the boundedness of HU is equivalent to that of the operatorTUf := maxv∈U∣∣Tvf ∣∣.In particular, the inequality (2.1.2) will follow if we can prove the estimate(2.1.3) and that‖TUf‖Lp,∞(Rn) ≥ Cn,p log(#U) , (2.1.7)for any 1 < p <∞.Now let m be a large positive integer. We can assume without lossof generality that #U = 2m and study the norm ‖TU‖Lp(Rn)→Lp,∞(Rn) asfunction of m. Moreover, again without loss of generality, we can assume allvectors in U to belong to the first hyper-octant of Rn (that is, that all theircoordinates are strictly positive). Indeed, let us suppose that Theorem 1.1.4holds in this particular case and let U be a set of 2m generic unit vectors ofRn. It is easy to verify that‖TU‖Lp(Rn)→Lp,∞(Rn) =∥∥TR(U)∥∥Lp(Rn)→Lp,∞(Rn)for all orthogonal matrices R in On(R). Therefore, up to a rotation, it isalways possible to assume that each vector of U belongs to one of the 2n292.1. Some preliminary factshyper-octants of Rn (that is, that they do not belong to the union of then coordinate hyperplanes of Rn). Thus, we have that U is the union ofsets U ∩ H over all hyper-octants H of Rn. Consequently, there exists ahyper-octant H of Rn such that #(U ∩H) ≥ 2−n#U = 2m−n and, up to arotation, we can suppose H to be the first one. Finally, we have:‖TU‖p→p,∞ ≥ ‖TU∩H‖p→p,∞ &n,p√log (#(U ∩H))&√m− n &n√m.Last inequality holds because we can assume #U = m n. Therefore, oncethe inequality (2.1.7) holds in the particular case of U contained in the firsthyper-octant, it holds in the general one.Henceforth, we shall use the notation Q := [0, 1]n for the unit cube. Thefirst intermediate result consists of the introduction of the following notionand the subsequent lemma.Definition 2.1.1. Let m be a positive integer. We say that the sequence offunctions {fN}1≤N<2m defined on Rn is a signed tree system if the followingholds up to sets of measure 0:supp f(k+1)2j−1 ⊂{x ∈ Q : f (k)j (x) > 0}, (2.1.8)supp f(k+1)2j ⊂{x ∈ Q : f (k)j (x) < 0}. (2.1.9)where we identify f(k)j with fN if N = 2k + j − 1, for 1 ≤ j ≤ 2k andk = 0, 1, . . . ,m− 1 (see the double numbering notation (3.0.1) below).Lemma 2.1.1. (cf. [17]) There exists a permutation σ of the numbers{1, 2, . . . 2m − 1}, depending only on m, such that for every signed tree sys-tem f1, . . . , f2m−1 in Rn we havemax1≤l<2m∣∣∣∣∣l∑N=1fσ(N)(x)∣∣∣∣∣ ≥ 132m−1∑N=1|fN (x)| , (2.1.10)for all x ∈ Rn.We shall relate TU to the maximal operator on the left-hand side of(2.1.10) and use the previous inequality to prove estimate (2.1.7).The second intermediate result consists of the following simple general-ization of the geometric notion of sector in the plane and the subsequentproposition. We define a sector in Rn to be a portion of space delimited bya finite number of hyperplanes passing through the origin:302.1. Some preliminary factsDefinition 2.1.2. Let v1, v2, . . . vr unit vectors in Rn and r, s two positiveintegers with 1 ≤ s ≤ r. A sector in Rn is a nonempty set of the type{x ∈ Rn :{x · vj > 0 ∀ 1 ≤ j ≤ sx · vj < 0 ∀ s < j ≤ r}.Proposition 2.1.1. Let p0 be a fixed positive integer, let {XN}1≤N<2m bepairwise disjoint sectors in Rn, and let σ be the permutation defined throughthe condition (3.0.7) and used in Lemma 2.1.1. Then there exists a finitesequence {fN}1≤N<2m of smooth, L1(Rn) functions with compactly supportedFourier transforms satisfying the following properties:(a) supp f̂N ⊂ XN .(b)∥∥∥∑2m−1N=1 fN∥∥∥Lp(Rn)≤ C1√m, 1 ≤ p ≤ 2p0.(c)∣∣∣{x ∈ Q : max1≤l<2m∣∣∣∑lN=1 fσ(N)(x)∣∣∣ ≥ C2m}∣∣∣ > C3.with C1, C2, C3 > 0 depending only on p0 and n.We will set f :=∑2m−1N=1 fN , use (b) to prove the estimate (2.1.3) andboth (a) and (c) to prove estimate (2.1.7).The third and final intermediate result necessary to prove the estimate(2.1.7) consists of the following lemma, that uses the previously introducednotion of sector in Rn. We will use the notation (2.1.5) and, by a slightabuse of notation, also writeΓcv = {x ∈ Rn : x · v < 0} (2.1.11)Lemma 2.1.2. Let M ≥ 2 be an integer and U := {v1, v2, . . . vM} a collec-tion of distinct unit vectors in the first hyper-octant of Rn. Then there is anordering {u1, . . . uM} of U such that the following sectors are non-empty:S1 = Γu1 ∩ . . . ∩ ΓuM−1 ∩ ΓcuM ,S2 = Γu1 ∩ . . . ∩ ΓuM−2 ∩ ΓcuM−1 ∩ ΓcuM ,.........SM−2 = Γu1 ∩ Γu2 ∩ Γcu3 ∩ . . . ∩ ΓcuM ,SM−1 = Γu1 ∩ Γcu2 ∩ . . . ∩ ΓcuM .(2.1.12)312.2. The proof of the theoremMoreover, for all j = 1, 2, . . .M − 1 they satisfy the equationΓuj ∩M−1⋃k=1Sk =M−j⋃k=1Sk. (2.1.13)Lemma 2.1.2 will relate the set of directions U to the function f andis the foundamental ingredient for the passage to higher-dimensions of theresults in [17].2.2 The proof of the theoremSo let us suppose that U = {v1, . . . v2m} is a subset of unit vectors in thefirst hyper-octant of Rn and let us apply Lemma 4.3.1 with M = 2m. Thereexist nonempty, disjoint sectors S1, . . . S2m−1 of Rn satisfying:ΓuN ∩2m−1⋃r=1Sr =2m−N⋃r=1Sr ∀ N = 1, 2, . . . 2m − 1, (2.2.1)where u1, . . . u2m is a certain permutation of the vectors v1, . . . v2m . ClearlyTUf = maxuj∈U∣∣Tujf ∣∣, therefore we will work with the vectors {uj}1≤j<2m .Now let us fix 1 < p < +∞, let σ be the permutation defined through thecondition (3.0.7) and let us apply Proposition 2.1.1 to the sectors XN :=Sσ−1(N) for N = 1, . . . 2m − 1. There exists a collection {fN}1≤N<2m ofsmooth, L1(Rn) functions with compactly supported Fourier transforms sat-isfying (a) (that is supp f̂N ⊂ Sσ−1(N) for all N), (b) and (c). Then we setf :=2m−1∑N=1fN . (2.2.2)It is clear that (a) and (b) implysupp f̂ ⊂2m−1⋃N=1SN , ‖f‖Lp(Rn) .n,p√m,respectively. In particular, the second fact proves the estimate (2.1.3).Therefore, all we need to prove is the condition (2.1.4).We will estimate ‖TUf‖Lp(Rn) through the following crucial result.322.2. The proof of the theoremLemma 2.2.1. In the previous notation, we haveTUf = max1≤l<2m∣∣∣∣∣l∑N=1fσ(N)∣∣∣∣∣ , (2.2.3)where σ is, again, the permutation defined in (3.0.7).Proof. Since f̂ is supported in the union of the sectors {SN}1≤N<2m , by(2.2.1) we haveTurf =(χΓur2m−1∑N=1χSN f̂)∨=(2m−1∑N=1χΓur∩SN f̂)∨=(2m−r∑N=1χSN f̂)∨,for all r = 1, 2, . . . 2m−1. Therefore, if we denote by ϕr the function(χSr f̂)∨,the previous equalities imply thatTurf =2m−r∑N=1ϕN . (2.2.4)Consequently,TUf = max1≤r<2m∣∣∣∣∣2m−r∑N=1ϕN∣∣∣∣∣ . (2.2.5)Then the condition (2.2.3) follows if we prove that ϕN = fσ(N) for all N .But the sectors {SN}1≤N<2m are disjoint and, by construction, f̂σ(r) is theonly function whose support is contained in Sr. Therefore:ϕr =2m−1∑N=1(χSr f̂N)∨=(χSr f̂σ(r))∨=(f̂σ(r))∨= fσ(r). (2.2.6)The proof of the lemma is complete.At this point, in view of the equation (2.2.3), we apply Proposition2.1.1(c) and obtain∣∣∣∣∣{x ∈ Q : max1≤l≤2m−1∣∣∣∣∣l∑N=1fσ(N)(x)∣∣∣∣∣ ≥ Cm}∣∣∣∣∣= |{x ∈ Q : TUf(x) ≥ Cm}| > C ′,(2.2.7)332.3. The difference between the cases n = 2 and n > 2for suitable constants C,C ′ depending only on p and n. But an as easyconsequence of the estimate (2.2.7) is‖TUf‖Lp,∞(Q) &p,n m, (2.2.8)which clearly proves the condition (2.1.4). The proof of Theorem 1.1.4 isnow complete.2.3 The difference between the cases n = 2 andn > 2In [17], the specific characteristics of the geometry of the plane provide anatural and simple way to fulfill the condition (2.2.1). To see this, let ussuppose U to be a set of 2m unit vectors in the first quadrant of R2 and let{u1, . . . u2m} be its counterclockwise numbering. Let us consider the linesu⊥1 , . . . u⊥2m and the 2m−1 sectors that they determine in the 4-th quadrant.If we denote these sectors by S1, . . . S2m and we number them clockwise,it is easy to see that the two-dimensional version of the equation (2.2.1)immediately follows (see Figure 2.1).This fact is exactly the argument Karagulyan uses in [17]. However, thereis no natural analogue of this method in higher dimensions. For instance, the2m hyperplanes can determine a number of sectors much higher than 2m−1(see Figure 2.2 below) and the concepts of clockwise or counterclockwisenumbering make no sense. Therefore, there is no natural ordering for anycollection of 2m−1 sectors we can pick. In other words, it turns out that thecondition (2.2.1) is nontrivial in dimension n > 2. We still need 2m−1 sectorssatisfying the geometric condition (2.2.1) in order to relate the functions{fN}1≤N<2m to the set U , but in the n-dimensional case it is necessaryto prove their existence through a more delicate argument, namely Lemma2.1.2.342.3. The difference between the cases n = 2 and n > 2Figure 2.1: The red sectors are the ones that fall within the half-plane Γuj :S1, . . . Sj .352.3. The difference between the cases n = 2 and n > 2Figure 2.2: Only 7 planes passing through the origin in R3 generate a highnumber of sectors.36Chapter 3Tree systemsIn this chapter we prove Lemma 2.1.1, the first intermediate result for theproof Theorem 1.1.4. This fundamental lemma plays a crucial role in theproof of the estimate (2.1.4) and is based on the notion of Tree system. Treesystems, introduced by Karagulyan in [17], are finite sequences of functionswhose supports are nested in a way similar to Haar functions. Lemma 2.1.1was first proved by Nikishin and Ulyanov in [29] for finite sequences of Haarfunction, and by Karagulyan (see [17]) for general tree systems in R2. Below,we prove it in Rn for all n ≥ 2. The adjustments required to cover higherdimensions are minor, but since the proof in [17] contains some errors andnotational inconsistencies, we provide the complete argument.Let m be a large positive integer. We will consider finite sequences offunctions {fN}1≤N<2m where each fN is a complex-valued function sup-ported in the unit cube Q. We will use the following system of doublenumbering:f(k)j = fN , N = 2k + j − 1, 1 ≤ j ≤ 2k, k = 0, 1, . . .m− 1. (3.0.1)Thusf1 = f(0)1 ,f2 = f(1)1 , f3 = f(1)2 ,f4 = f(2)1 , f5 = f(2)2 , f6 = f(2)3 , f7 = f(2)4 ,etc.Definition 3.0.1. (a) We say that the sequence fN with N = 1, 2, . . . , 2m−1is a tree system if the following holds up to sets of measure 0:supp f(k+1)2j−1 ∩ supp f (k+1)2j = ∅,supp f(k+1)2j−1 ∪ supp f (k+1)2j ⊂ supp f (k)j .(3.0.2)37Chapter 3. Tree systems(b) We say that the sequence fN with N = 1, 2, . . . , 2m − 1 is a signedtree system if the following holds up to sets of measure 0:supp f(k+1)2j−1 ⊂{x ∈ Q : f (k)j (x) > 0}, (3.0.3)supp f(k+1)2j ⊂{x ∈ Q : f (k)j (x) < 0}. (3.0.4)for N = 2k + j − 1, 1 ≤ j ≤ 2k, k = 0, 1, . . . ,m− 1.Figure 3.1 shows the relations among the supports. Clearly, every signedtree system is a tree system.Figure 3.1: The nested supports of tree system functions.The nesting support properties of tree systems can be conveniently en-coded in a binary tree. Let us consider the full binary tree of height m andlet us denote it by Bm. Let us label each vertex as (k, j), where k is theheight of the vertex, so that k ranges from 0 to m − 1 and all vertices ofheight k are labelled lexicographically as (k, 1), (k, 2), . . . , (k, 2k). Given avertex (k, j), its parent can be identified as(k − 1, b j+12 c)if k ≥ 1, and itschildren can be identified as (k + 1, 2j − 1) and (k + 1, 2j) if k ≤ m − 2.In parallel to our double-numbering system N = 2k + j − 1, we will use asimilar double-labelling system for tree vertices, so that the vertex (k, j) willbe alternatively labelled by the number N(k, j) = 2k + j − 1 (see Figure 3.2below).38Chapter 3. Tree systemsFigure 3.2: The double-indexing notation for the numbers 1, . . . 2m−1 allowsus to represent them as vertices of a complete binary tree of height m.We will continue to use the notations and terminology of trees that wehave introduced in section 1.2.2. In particular, h(N) = k is to denote theheight of the vertex labelled with N and a ray of length l + 1 rooted at(k, j0) is a sequence of vertices {(k, j0), (k + 1, j1), . . . (k + l, jl)} where, foreach i = 1, . . . l, the vertex (k + i, ji) is a child of (k + i− 1, ji−1).Lemma 3.0.1. Let {fN}2m−1N=1 be a tree system. Then for (k, j) 6= (k′, j′),the supports of f(k)j and f(k′)j′ are either disjoint or nested. Moreover:(a) If (k, j) and (k′, j′) do not lie on the same tree ray (i.e. neithervertex is a descendant of the other), then the supports are disjoint.(b) If (k′, j′) is a descendant of (k, j), then supp f (k′)j′ ⊂ supp f (k)j .(c) For each x ∈ Q such that f (0)1 (x) 6= 0, there is a unique maximal rayR(x) = {f (0)1 , f (1)j1 , f(2)j2, . . . , f(k)jk} such that f (i)ji (x) 6= 0 for i = 1, . . . , k. Theray terminates at f(k)jkwhen both children of f(k)jktake value 0 at x.(d) If {fN}2m−1N=1 is a signed tree system, then we have the additionalproperty that R(x) encodes the sign of f (0)1 , f (1)j1 , . . . , f(k−1)jk−1 at x: it turnsleft at f(i)ji(i.e. goes to the left child f(i+1)2ji−1) if f(i)ji(x) > 0 and turns right(i.e. goes to the right child) if f(i)ji(x) < 0.Proof. Via notation (3.0.1), we will adapt the terminology of trees to thefunctions f(k)j , saying that f(k+1)2j−1 and f(k+1)2j are the (left and right, respec-tively) children of f(k)j , and that a function f(k′)j′ is a descendant of f(k)j ifk′ > k and the vertex (k, j) lies on the ray connecting (k′, j′) to the root (0, 1)of the tree. Other terminology related to binary trees can be adapted simi-larly. Therefore, each function f(k)j has two children (f(k+1)2j−1 and f(k+1)2j ) withmutually disjoint supports (up to sets of measure 0) contained in supp f(k)j .39Chapter 3. Tree systemsFigure 3.3: For a signed tree system, the signs of f(k)j are encoded in thebinary tree.Then (a) and (b) easily follow. The definition of ray of a binary tree implies(c) and (d) follows from condition (3.0.3) and (3.0.4), having identified theleft and right children of a vertex (k, j) as (k + 1, 2j − 1) and (k + 1, 2j)respectively.Remark 3.0.1. Note that if N = 2k+j−1, then 2k−1 +b j+12 c−1 = bN2 c =:N¯ . In particular, (3.0.3) and (3.0.4) are equivalent tosupp fN ⊂{x ∈ Q : (−1)j−1fN¯ (x) > 0}. (3.0.5)Now we are ready to prove the tree systems lemma, that we restate here:Lemma 3.0.2. (cf. [17]) There exists a permutation σ of the numbers{1, 2, . . . 2m − 1}, depending only on m, such that for every signed tree sys-tem f1, . . . , f2m−1 in Rn we havemax1≤l<2m∣∣∣∣∣l∑N=1fσ(N)(x)∣∣∣∣∣ ≥ 132m−1∑N=1|fN (x)| . (3.0.6)for all x ∈ Rn.Proof. We begin by defining the permutation σ of {1, . . . , 2m−1} as follows.For N = 2k + j − 1 as in (3.0.1), lettN = t(k)j :=2j − 12k+1∈ [0, 1] .Thus t1 =12 , t2 =14 , t3 =34 , t4 =18 , t5 =38 , t6 =58 , etc. Observe thata complete binary tree with m levels can be represented as a planar graphso that the vertex (k, j) has the x-coordinate t(k)j . Figure 3.4 illustrates thisfor m = 3.40Chapter 3. Tree systemsFigure 3.4: The full binary tree with m = 3 and with the numbers tNWe now rearrange the sequence {tN} ⊂ [0, 1] in increasing order. Specifi-cally, we define σ as the unique permutation of the numbers {1, 2, . . . 2m − 1}such thattσ(1) < tσ(2) < . . . < tσ(2m−1). (3.0.7)(In the example in Figure 7, we have σ(1) = 4, σ(2) = 2, σ(3) = 5, etc.).Clearly, σ depends only on m,Now it is easy to verify that, if f(k+1)2j−1 and f(k+1)2j are children of f(k)j ,thent(k+1)2j−1 = t(k)j −12k+2, t(k+1)2j = t(k)j +12k+2. (3.0.8)Iterating over tree levels from k + 1 to m, and using that∑k′i=k+2 2−i <∑∞i=k+2 2−i = 2−k−1 for any finite k′ ≥ k + 2, we see that whenever f (k′)j′ isa descendant of f(k)j , the corresponding numbers t(k′)j′ obeyt(k′)j′ ∈(t(k)j − 2−k−1, t(k)j + 2−k−1). (3.0.9)Now let x ∈ Q, and assume that f1(x) 6= 0 since otherwise there isnothing to prove because we would have fN (x) = 0 ∀ N . Definelx := max{h : 1 ≤ h < 2m, fσ(h)(x) < 0}. (3.0.10)It follow immediately from (3.0.10) that fσ(h)(x) ≥ 0 for all h > lx. Weclaim that, furthermore,fσ(h)(x) ≤ 0 ∀ h ≤ lx. (3.0.11)41Chapter 3. Tree systemsTo prove this, suppose for contradiction that h ≤ lx and fσ(h)(x) > 0. By(3.0.10), we cannot have h = lx. Letfσ(lx) = f(k)j , fσ(h) = f(s)i .Consider the ray R(x) defined in Lemma 3.0.1. If fσ(h) /∈ R(x), thenfσ(h)(x) = 0 and the claim is true. Conversely, let fσ(h) be a descendant offσ(lx). Since fσ(lx)(x) < 0, the ray R(x) turns right at fσ(lx), so that fσ(h)must be either f(k+1)2j or one of its descendants. Consequently, by (3.0.9)and then (3.0.8) we havetσ(h) > t(k+1)2j −12k+2= t(k)j = tσ(lx),which contradicts the assumption that h < lx and therefore tσ(h) < tσ(lx).Finally, consider the case when fσ(lx) is a descendant of fσ(h). If we havefσ(h)(x) > 0, then R(x) turns left at fσ(h), so that fσ(lx) must be eitherf(s+1)2i−1 or one of its descendants. Then, again by (3.0.9) and then (3.0.8), wehavetσ(lx) < t(s+1)2i−1 +12s+2= t(s)i = tσ(h),again contradicting our assumptions.Figure 3.5: The permutation σ and the choice of lx.42Chapter 3. Tree systemsTo recap, we have established the existence of a number lx such that{fσ(h)(x) ≤ 0 ∀ h ≤ lxfσ(h)(x) ≥ 0 ∀ h > lxLetS1 =lx∑N=1−fσ(N)(x), S2 =2m−1∑N=lx+1fσ(N)(x).Then S1, S2 ≥ 0,∑2m−1N=1 |fN (x)| = S1 + S2, andmax1≤l<2m∣∣∣∣∣l∑N=1fσ(N)(x)∣∣∣∣∣ = max(S1, S2 − S1).If S1 ≥ 13(S1 + S2), then (2.1.10) follows immediately. Suppose now thatS1 <13(S1 +S2). Then S2 >23(S1 +S2), and furthermore, 3S1 < S1 +S2 sothat 2S1 < S2. HenceS2 − S1 > S2 − S22=S22>13(S1 + S2),and (2.1.10) again follows.43Chapter 4The proof of Proposition2.1.1In this chapter we prove Proposition 2.1.1, that we restate below for thesake of completeness. We use the Definition 2.1.2 of sector in Rn.Proposition 4.0.1. Let p0 be a fixed positive integer, let {XN}1≤N<2m bepairwise disjoint sectors in Rn, and let σ be the permutation defined throughthe condition (3.0.7) and used in Lemma 2.1.1. Then there exists a finitesequence {fN}1≤N<2m of smooth, L1(Rn) functions with compactly supportedFourier transforms satisfying the following properties:(a) supp f̂N ⊂ XN .(b)∥∥∥∑2m−1N=1 fN∥∥∥Lp(Rn)≤ C1√m, 1 < p ≤ 2p0.(c)∣∣∣{x ∈ Q : max1≤l<2m∣∣∣∑lN=1 fσ(N)(x)∣∣∣ ≥ C2m}∣∣∣ > C3.with C1, C2, C3 > 0 depending only on p0 and n.Remark 4.0.1. in this chapter we shall make use of the notion of treesystem, introduced in Chapter 3 to prove the existence of the sequence{fN}1≤N<2m . However, we remark that this sequence is not a tree system.4.1 An important tree systemThe proof of Proposition 2.1.1 requires some preliminaries and a series oflemmas. We fix a Schwartz function φ on Rn such thatφ(x) > 0 ∀ x ∈ R,∫Rnφ(x)dx = 1,444.1. An important tree systemand supp φ̂ ⊂ Q. We set φl = lnφ(l·) for l ∈ N. We will continue to use thedouble-indexing notationN = 2k + j − 1, 1 ≤ j ≤ 2k, k = 0, 1, . . .m− 1,and the tree notation from Section 3. We will also use the notation Q (l, x)for the cube with sides parallel to the coordinate axes, side length equal tol and center x ∈ Rn.We define inductively a finite sequence of measurable subsets of Q whoseproperties are crucial for the construction of the functions of Proposition2.1.1. We set E1 := Q and, for N ≥ 2, N = 2k + j − 1 and N¯ := bN2 c,EN = E(k)j :={x ∈ EN¯ : (−1)j−1cos (x · p¯N¯ ) > 0}, (4.1.1)for a suitable choice of vectors p¯N ∈ Rn, N = 1, . . . 2m−1. The propertiesof the sets {EN}1≤N<2m are expressed in the following lemma.Lemma 4.1.1. Given p0 and {XN}1≤N<2m as in the Proposition 2.1.1, fora sufficiently large constant C  p0 there exists a choice of vectors p¯N ∈ Znand large constants lN such that:1. QN := Q(lN , p¯N ) ⊂ XN ,2. For all 1 ≤ h ≤ p0, and for all 2h-tuples 1 ≤ j1 ≤ . . . ≤ jh ≤ N, 1 ≤j′1 ≤ . . . ≤ j′h ≤ N such that #{r : jr = jmax} 6= #{r : j′r = jmax} withjmax = max {jh, j′h}, we have(Qj1 + . . .+Qjh) ∩ (Qj′1 + . . .+Qj′h) = ∅, (4.1.2)3. |EN | > 0 and∫EN|cos (2pix · p¯N )|dx > |EN |3 ,4. The functions gN := φlN ∗χEN obey 0 ≤ gN ≤ 1, supp ĝN ⊂ Q(lN , O),and ∥∥gN − χEN∥∥L1(Rn) + ∥∥gN − χEN∥∥L2p0(Rn) ≤ 2−Cm, (4.1.3)for all N = 1, . . . , 2m − 1.Proof. We first rewrite the condition 2 of the lemma in a form that will bemore convenient to work with. We haveQj1 + . . .+Qjh = Q(lj1 + . . .+ ljh , p¯j1 + . . .+ p¯jh) ,Qj′1 + . . .+Qj′h = Q(lj′1 + . . .+ lj′h , p¯j′1+ . . .+ p¯j′h).454.1. An important tree systemIf we set µ := #{r : jr = jmax} and ν := #{r : j′r = jmax}, and use thenotationlj :=∑jr<jmaxljr , lj′ :=∑j′r<jmaxlj′r ,then the condition 2 is equivalent to saying that(µ− ν)p¯jmax +∑jr<jmaxp¯jr −∑j′r<jmaxp¯j′r /∈ Q(lj + lj′ , O), (4.1.4)for any choice of indices {jr, j′r} such that µ 6= ν.We prove the lemma by induction on N . We set E1 := E(0)1 = Q andchoose l1 > 0 large enough so that (4.1.3) holds with N = 1. For anyp¯1 ∈ Zn we have∫Q|cos (x · p¯1)|dx ≥∫Qcos2 (x · p¯1)dx=∫Q1 + cos (2x · p¯1)2dx =|Q|2=|E1|2>|E1|3.This shows that condition 3 holds for N = 1. The reader will easily be ableto check that 0 ≤ gn ≤ 1 and that supp ĝ1 ⊂ supp φ̂l1 ⊂ [−l1, l1]n (sinceĝ1 = φ̂l1χ̂E1). Thus condition 4 holds for N = 1. Having chosen l1, wecan clearly find a translation vector p¯1 such that condition 1 holds. Finally,condition (4.1.4) is vacuous in this case (as there is only one cube availableso far) and hence µ = ν for any choice of the indices {jr, j′r}. This completesthe verification of the lemma for N = 1.Now, assume that N > 1 and that we have constructed p¯j , lj , Ej , gjobeying all conclusions of the lemma for j = 1, 2, . . . , N − 1. Define EN via(4.1.1) (note that this is possible since we have already constructed EN¯ andp¯N¯ ).Clearly EN is measurable, hence we can choose lN > 0 large enough sothat (4.1.3) holds. The properties 0 ≤ gN ≤ 1 and supp ĝN ⊂ [−lN , lN ]nfollow as in the case N = 1, establishing part 4.Next, for any p¯N ∈ Zn we observe that the periodic function x →cos (2pix · p¯N ) assumes positive (respectively negative) values on parallelstrips of thickness approximately 1/|p¯N |, namely1|p¯N |(k2− 14)< x · pˆN < 1|p¯N |(k2+14), (4.1.5)for k ∈ Z (pˆN being the direction of p¯N ). Since EN¯ is open relatively to Q,it is easy to see that the function cos (2pix · p¯N ) changes sign on EN¯ as long464.1. An important tree systemas |p¯N | is chosen large enough. Consequently, EN 6= ∅ and 0 < |EN | < |EN¯ |.Moreover: ∫EN|cos (x · p¯N )|dx ≥∫EN1 + cos (2x · p¯N )2dx=|EN |2+12∫ENcos (2x · p¯N ) dx.By the Riemann-Lebesgue lemma, χ̂EN (ξ)→ 0 as |ξ| → ∞. In particular,∫ENcos (2x · p¯N ) dx = Re χ̂EN (p¯N )→ 0 as |p¯N | → ∞,therefore the condition 3 is satisfied as long as |p¯N | is large enough.Now, since we have already picked the vectors p¯1, . . . p¯N−1 and the num-bers l1, . . . lN−1 in the previous steps of the construction, to prove part 2we must choose p¯N so that the condition (4.1.2) holds for all choices of theindices j1, . . . jh, j′1 . . . j′h. Equivalently, through condition (4.1.4), we musthavesp¯N /∈ −∑jr<Np¯jr +∑j′r<Np¯j′r + Q( ∑jr<Nljr +∑j′r<Nlj′r , O), (4.1.6)for s = ±1, . . . ,±p0. Substantially, we want the vectors sp¯N not to belongto finitely many disjoint, large cubes in Rn centered at certain points of Rn.This condition can easily be satisfied by picking |p¯N | large enough.A final choice of p¯N such that QN ⊂ XN completes the inductive stepand the proof of the lemma.The next lemma illustrates further, crucial properties of the sets EN .Lemma 4.1.2. Let EN be the sets defined in Lemma 4.1.1. Then the func-tions χN := χ(k)j := χEN form a tree system (see Definition 3.0.1(a)), and,in particular, obey the conclusions of Lemma 3.0.1(a)-(c). Moreover wehaveχ(k+1)2j−1 + χ(k+1)2j = χ(k)j . (4.1.7)Now, recalling the notation Bm for the complete binary tree of height m, letus denote by Bm,j0 the sub-tree of Bm rooted at the vertex j0 ∈ Bm. Then,if the height h(j0) of this vertex is equal to r ≤ m, we have∑j ∈Bm,j0h(j) = rχj = χj0 . (4.1.8)474.1. An important tree systemFinally we have2m−1∑N=1χN = m. (4.1.9)The equations (4.1.7), (4.1.8) and (4.1.9) hold up to sets of measure 0 onQ.Proof. Rewriting Lemma 4.1.1(3) in terms of j, k, we have E1 = Q andE(k+1)2j−1 :={x ∈ E(k)j : cos(x · p¯(k)j ) > 0},E(k+1)2j :={x ∈ E(k)j : cos(x · p¯(k)j ) < 0}.Therefore the sets E(k+1)2j−1 and E(k+1)2j are disjoint and contained in E(k)j .Consequently, the functions χ(k)j form a tree system. Since the set{x ∈ Q : cos(x · p¯(k)j ) = 0}has measure 0, we also have (4.1.7).Regarding (4.1.8), if h(j0) = m we have Bm,j0 = {j0} and there isnothing to prove. If h(j0) < r, then the parents of the indices j in Bm,j0with height r are all and only the ones of height r − 1. Then, according tothe condition (4.1.7), we have up to a null set:∑j∈Bm,j0h(j) = rχj =∑j′∈Bm,j0h(j′) = r−1χj′ . (4.1.10)Iterating the previous argument, (4.1.10) reduces to the characteristic func-tion of the root of Bm,j0 , that is χj0 , and (4.1.8) follows.Finally, iterating (4.1.7), we haveχQ = χ(0)1 = χ(1)1 + χ(1)2= (χ(2)1 + χ(2)2 ) + (χ(2)3 + χ(2)4 )= · · · =∑1≤j≤2kχ(k)j ,almost everywhere for every k = 1, 2, . . . ,m−1. Summing over k, we obtain(4.1.9).484.2. Preliminary results4.2 Preliminary resultsIn order to prove Proposition 2.1.1 and, in particular, part (b) for p = 2p0,we need to study the L2p0(Rn)-norm of the sum∑1≤N<2mfN . Since 2p0 isan even integer, we can expand the 2p0-power of the norm into a sum ofproducts of two terms, the first one of the type fj1 · · · fjp0 and the second oneof the typefj′1 · · ·fj′p0 for all indices j1, . . . jp0 and j′1 . . . j′p0 from 1 to 2m− 1.Namely: ∥∥∥∥ 2m−1∑N=1fN∥∥∥∥2p0L2p0(Rn)=∫Rn(2m−1∑N=1fN)p0( 2m−1∑N ′=1fN ′)p0dx=∑1≤j1...jp0≤2m−11≤j′1...j′p0≤2m−1∫Rnfj1 · · · fjp0fj′1 · · · fj′p0dx.(4.2.1)Now we rewrite the block of functionsfj1 · · · fjp0fj′1 · · · fj′p0 , (4.2.2)in another form. Indeed, the same function may occur more than once inthe block if two or more of the indices j1, . . . jp0 and j′1, . . . j′p0 have the samevalue. Let us denote by m1 < . . . < mk and n1 < . . . < nl, respectively, thetwo increasing sequences of indices that occur in the block, for some integernumbers 1 ≤ k, l ≤ p0. Then it is easy to see that there exist integer numbersµ1, . . . µk and ν1, . . . νl such that any term fmj occurs exactly µj times andfnj exactly νj times in the block, for all j. Moreover, by construction wemust have µ1 + . . . + µk = ν1 + . . . + νk = p0. In the proof of Proposition2.1.1, we will work with the block (4.2.2) re-expressed in the formfµ1m1 · · · fµkmkfν1n1 · · · fνlnl , (4.2.3)and call the numbers µ1, . . . µk and ν1, . . . νl the multiplicities of m1, . . . ,mkand n1, . . . nl, respectively.Through the previous considerations and the compact notation µ¯ :=(µ1, . . . µk) and ν¯ := (ν1, . . . νk) for the multiplicities, we can rewrite thesum (4.2.1) in the form∑1≤ k,l,≤ p0∑µ¯,ν¯µ1+...µk=p0ν1+...νl=p0Cµ¯,ν¯∑1≤m1<···<mk<2m1≤n1<···<nl<2m∫Rnfµ1m1 · · · fµkmkfν1n1 · · · fνlnl dx,494.2. Preliminary resultswhere Cµ¯,ν¯ =(p0µ¯)(p0ν¯)in multinomial notation7.The first two sums above involve only p0 and not m, therefore we obtain(b) of the Proposition 2.1.1 as soon as we show that∣∣∣∣∣ ∑1≤m1<···<mk<2m1≤n1<···<nl<2m∫Rnfµ1m1 · · · fµkmkfν1n1 · · · fνlnl dx,∣∣∣∣∣ .p0 mp0 , (4.2.4)for all k, l, µ¯ and ν¯ such that µ1 + . . .+ µk = ν1 + . . .+ νk = p0. A delicateargument, that requires several lemmas and definitions, turns out to benecessary for the optimal book-keeping of the terms in the sum (4.2.4).Let p0 and m be as in Proposition 2.1.1, and let us consider three integernumbers h, k, l such that 1 ≤ k, l ≤ p0, 1 ≤ h ≤ m. We will use the compactnotation m¯ := (m1, . . .mk) for the sequence m1, . . .mn of indices / verticesof Bm as in sum (4.2.4). Moreover, if R is a ray of Bm and all the verticesm1 . . .mk lie on R, we will write m¯ ∈ R.Let p1 be an integer number with 1 ≤ p1 ≤ p0. We will associate asequence of positive integer numbers µ1, . . . µk such thatA(µ¯) := µ1 + . . .+ µk = p1 ≤ p0, (4.2.5)with a sequence of indices m¯. We will say that the index / vertex mj hasp1-multiplicity µj for all j = 1, . . . k (or that m¯ has p1-multiplicity µ¯) andwrite mp1(m¯) = µ¯. When p1 = p0, or when p1 is clear from the context, wewill simply use the name multiplicities, as in the argument that leads to thesum in the estimate (4.2.4).Example 4.2.1. Let us set, for instance, m = 4, p0 = p1 = 3, k = 2 and letus consider Figure 4.1 below. We want to assign a 3-multiplicity µ¯ = (µ1, µ2)to, say, the sequence (m2,m4) = (5, 15). We must have µ1 + µ2 = 3, thenwe can consider the numbers 2 and 1, respectively, and assign them to theindices m2 and m4. In this case, we say that 5 has multiplicity 2, 15 hasmultiplicity 1 and write m3(5, 15) = (2, 1).Finally, we set the notation r ∧ s := max {r, s} for r, s positive integers.Having set up these preliminary notations, we need the following definition.Definition 4.2.1. Let p0, p1,m, h, k, l be as above and let µ¯ = (µ1, . . . µk) amultiplicity vector such that A(µ¯) = p1. We setMh, µ¯(p1) :={m¯ ∈ Nk : 1 ≤ m1 < · · · < mk < 2h, mp1(m¯) = µ¯}.7We use the notation(p0µ¯)= p0!µ1!···µk! . For any fixed choice of µ¯ and m1, . . .mk, theblock fµ1m1 · · · fµkmk occurs exactly(p0µ¯)times in the expansion of the sum (4.2.1).504.2. Preliminary resultsFigure 4.1: A sequence of 4 indices represented as vertices of B4Let 1 ≤ p2 ≤ p0 and ν¯ = (ν1, . . . νl) be a p2-multiplicity vector. We setMh, µ¯, ν¯(p1, p2) := {(m¯, n¯) ∈Mh, µ¯ (p1)×Mh, ν¯ (p2) : h(mk ∧ nl) = h} .In other words, Mh,µ¯,ν¯(p1, p2) is the collection of all pairs of strictly in-creasing sequences of indices (m¯, n¯) in the range 1, 2, . . . 2h − 1, of lengthk and l and multiplicities µ¯ and ν¯, respectively, such that either h(mk) orh(nl) is h. For simplicity of notation, the parameters p1, p2 will be sup-pressed if they are clear from the context or not relevant in an argument ora calculationIn the process of summation of the various blocks of functions in theestimate (4.2.4), a special role is played by two subclasses of the family ofvectors Mh, µ¯, ν¯(p1, p2):Definition 4.2.2. Let t p0, p1, p2,m, h, k, l, µ¯ and ν¯ as in Definition 4.2.1.Let us recall the notation Bh for the complete binary tree of height h andthat we write R ∼ Bh if R is a ray of Bh. We set:M′h, µ¯, ν¯(p1, p2) := {α¯ = (m¯, n¯) ∈Mh,µ¯,ν¯(p1, p2) : m¯, n¯ ∈ R, R ∼ Bh}M′′h, µ¯, ν¯(p1, p2) := {α¯ = (m¯, n¯) ∈Mh,µ¯,ν¯(p1, p2) : mk = nl, µk = νl} .Example 4.2.2. Let us set, for instance, m = h = 5, p0 = p1 = p2 = 3 andk = l = 3, so that we have µ¯ = ν¯ = (1, 1, 1), and let 1¯ be the notation forthis vector. Let us consider the vertices in Figure 4.2 below. We have thatα¯ := (m1,m2,m3, n1, n2, n3) ∈ M′′5,1¯,1¯(3, 3) because the vertices 1, 2, 5, 10and 20 lie on the same ray, m3 = n3 = 20 and the multiplicities coincide. If514.2. Preliminary resultsone of these vertices was 4, instead, α¯ would not even belong toM′5,1¯,1¯(3, 3)because its vertices would not lie on the same ray.Now let m,h, p0, p1, p2 be as above, but let us set k = l = 2, µ¯ =(2, 1) and ν¯ = (1, 2). If we set α¯ := (2, 10, 1, 5), The vector α¯ belongs toM′r, µ¯, ν¯(3, 3) \M′′r, µ¯, ν¯(3, 3). Indeed, the vertices of α¯ lie on the same ray, butm3(m2) = m3(10) = 1 6= 2 = m3(5) = m3(n2).Figure 4.2: Two sequences of three indices that belong to the same ray ofB5Before embarking on the proof of the various steps that are necessary toprove Proposition 2.1.1, we need to set up a few other multi-index notationsNotation 4.2.1. Let h, p0, p1, p2, µ¯ and ν¯ as in Definition (4.2.1) and letα¯ ∈ Mh, µ¯, ν¯(p1, p2). Let {gN}1≤N<2m , {p¯N}1≤N<2m and {EN}1≤N<2m bethe system of functions, the vectors and the sets defined in Lemma Preliminary resultsWe set fN (x) := e2piix·p¯N gN (x) for all N = 1, 2, . . . 2m − 1 and:Fα¯(x) := fµ1m1 · · · fµkmkfν1n1 · · · fνlnl (x),vα¯ :=∑1≤j≤kµj p¯mj −∑1≤j≤lνj p¯mj ,Eα¯ := Em1 ∩ . . . ∩ Emk ∩ En1 ∩ . . . ∩ Enl ,Gα¯(x) := e2piivα¯·xχEα¯(x).Through the previous notation, we can identify the whole blockfµ1m1 · · · fµkmkfν1n1 · · · fνlnlwith the vector α¯, that is with two sequences of vertices of the tree Bh,counted with multiplicity. We will use the notations (4.2.1) throughout therest of this chapter.The last notational conventions we need are the usual big O notation,that is X = O(Y ) (X,Y complex numbers) to mean that |X| ≤ c |Y | forc suitable positive constant, and the following one for quantities X and Ydepending on m. We will write X ∼= Y if |X − Y | ≤ c12−c2m, where themultiplicative constant c1 and the exponent c2 will always be sufficientlylarge, may depend on p0 and n, and may change from line to line but re-main independent of m. In our applications, c2 will depend on the largeconstant C from Lemma 4.1.1(4). Assuming that C >> p0 was chosen largeenough, we will always be able to ensure that c2 >C10 .Now we are ready to state and prove the following lemmas.Lemma 4.2.1. Let p1, p2, h, k, l as in Definition (4.2.1) (in particular, letA(µ¯) and A(µ¯) be p1 and p2, respectively). Let C  p0 be the same constantof Lemma 4.1.1. Then we have:1.∫RnFα¯(x)dx ∼=∫RnGα¯(x)dx,2. Eα¯ 6= ∅ if and only if α¯ ∈M′h, µ¯, ν¯ . That is, if there exists a ray Rα¯ ofBh such that m¯, n¯ ∈ R.534.2. Preliminary resultsProof. We have∫RnFα¯(x)dx =∫Rne2piivα¯·x(gµ1m1 · · · gµkmkgν1n1 · · · gνlml)(x)dx=∫Em1e2piivα¯·x(gµ1−1m1 · · · gµkmkgν1n1 · · · gνlml)(x)dx+∫Rne2piip¯m1·x(gm1(x)− χEm1(x))Fα¯−e1(x)dx = A+B.(4.2.6)Let us start with the term B. We recall that the function gN is of thetype φlN ∗ χEN with φ Schwartz function and φl = lnφ(l·). But ‖gN‖∞ .‖χEN ‖∞ = 1, thus ‖Fα¯−e1‖∞ . 1. Then, from Lemma 4.1.1(4) we have|B| .∫Rn∣∣(gm1(x)− χEm1(x)∣∣ dx ≤ 12Cm .We now turn to the term A. We iterate the argument in (4.2.6) and peelaway a factor gN on every iteration. After A(µ¯) +A(ν¯) = p1 +p2 iterations,we have ∣∣∣∣∫RnFα¯(x)dx −∫RnGα¯(x)dx∣∣∣∣ . p1 + p22Cm .But p1 + p2 ≤ 2p0, therefore we have (p1 + p2)2−Cm = O( p02Cm)and (1)follows.We now turn to (2). The fact that the functions {χEN }1≤N<2m are atree systems (see Lemma 4.1.2) implies that, for any two indices N < N ′,the sets EN and EN ′ are either disjoint or nested. Their intersection isnonempty precisely when EN ⊇ EN ′ , that is if and only if the vertex N isan ancestor of the vertex N ′ on Bh. Therefore, Eα¯ is nonempty if and onlyif the indices in α¯ are all ancestors or descendants of one another. But thisoccurs precisely when they all lie on a ray of Bh, that is if α¯ ∈M′h,µ¯,ν¯ .Lemma 4.2.1 has the following important consequence: for any 1 ≤p1, p2 ≤ p0 and any multiplicity vectors µ¯, ν¯ such that A(µ¯) = p1 andA(ν¯) = p2, we have∑α¯∈Mh, µ¯, ν¯∫RnFα¯(x)dx ∼=∑α¯∈Mh, µ¯, ν¯∫RnGα¯(x)dx ∼=∑α¯∈M′h, µ¯, ν¯∫RnGα¯(x)dx.Indeed, by Lemma 4.2.1(1), the first sum is equal to the second one up toan additive term of size O(2−Cm)times the cardinality of Mm,µ¯,ν¯ , which is544.2. Preliminary resultsO(22p0m). Therefore, the size of the error is O(2−(C−2p0)m), as small as wewant as long as the constant C in Lemma 4.1.1 is taken  p0. Finally, theidentity between the second sum and the third one is a direct consequenceof Lemma 4.2.1(2).The following lemma shows the importance of the subfamily M′′m, µ¯, ν¯ .Lemma 4.2.2. Let p1, p2, h, k, l as in Definition (4.2.1) (in particular, letA(µ¯) and A(µ¯) be p1 and p2, respectively). Then we have∫RnGα¯(x)dx ∼= 0,unless mk = nl and µk = νl, that is unless α¯ ∈M′′h, µ¯, ν¯ .Proof. We repeat the argument used to prove Lemma 4.2.1(1), but from theterms Gα¯ back to the terms Fα¯:∫RnGα¯(x)dx =∫Rne2piivα¯·xχEα¯(x)dx=∫Rne2piivα¯·xχEm1(x)χEα¯−e¯1(x)dx=∫Rne2piivα¯·xgm1(x)χEα¯−e¯1(x)dx+∫Rne2piip¯m1·x(gm1(x)− χEm1(x))Gα¯−e1(x)dx = A′ +B′,(4.2.7)where we used the fact thatGα¯(x) = e2piip¯m1χEm1(x)Gα¯−e1(x).Using Lemma 4.1.1(4), we have B′ ∼= 0. On the other hand,A′ =∫Rne2piip¯m1 ·xgm1(x)e2piivα−e1·xχEα¯−e¯1(x)dx =∫Rnfm1(x)Gα¯−e1(x)dx.Iterating the argument in (4.2.7) A(µ¯) +A(ν¯) = p1 + p2 times, we have∫RnGα¯(x)dx ∼=∫RnFα¯(x)dx.Now we apply Plancherel’s theorem:∫RnFα¯(x)dx =∫Rnf̂ (µ1)m1 ∗ · · · ∗ f̂ (µk)mk f̂(ν1)n1 ∗ · · · ∗ f̂ (νl)nl dξ,554.2. Preliminary resultswhere we use the notationsf̂(µj)mj := f̂mj ∗ · · · ∗ f̂mj︸ ︷︷ ︸µjtimes, f̂(νj)nj := f̂nj ∗ · · · ∗ f̂nj︸ ︷︷ ︸νjtimes.But the term f̂(µ1)m1 ∗ · · · ∗ f̂ (µk)mk f̂ (ν1)n1 ∗ · · · ∗ f̂ (νl)nl is supported in the set(µ1Qm1 + . . .+ µkQmk) ∩ (ν1Qn1 + . . .+ νlQnl).Therefore, according to Lemma 4.1.1(2), the previous intersection is emptyunless mk = nl and µk = νl, that is unless α¯ ∈M′′h, µ¯, ν¯ .Now, for h, µ¯, ν¯, p1, p2 as in Lemma 4.2.2, we setS(h, µ¯, ν¯) :=∑α¯∈M′h, µ¯, ν¯∫RnGα¯(x)dx.Moreover, let us set µ¯j := (µ1, . . . µk−j) and ν¯j′:=(ν1, . . . νl−j′)for 1 ≤ j <k and 1 ≤ j′ < l.The next lemma establishes the crucial property of the sums S(h, µ¯, ν¯).Lemma 4.2.3. S(h, µ¯, ν¯) obeys the recursion formulaS(h, µ¯, ν¯) ∼=h−1∑h1=1S(h1, µ¯1, ν¯1) , (4.2.8)where, according to the notation (4.2.5), we have A(µ¯1) = p1 − µk andA(ν¯1) = p2 − νl. Moreover,S(h, µ¯, ν¯) ∼={0 if µ¯ 6= ν¯O(hk−1)if µ¯ = ν¯(4.2.9)Proof. Let us first prove the formula (4.2.8). By Lemma 4.2.2 we have:S(h, µ¯, ν¯) ∼=∑α¯∈M′′h, µ¯, ν¯∫RnGα¯(x)dx. (4.2.10)A vector α¯ ∈M′′h, µ¯, ν¯(p1, p2) has the formα¯ = (m1, . . .mk−1,mk, n1, . . . nl−1,mk) .564.2. Preliminary resultsLet us use the notation α¯′ for its first k − 1 and l − 1 entries of α¯, namelyα¯′ := (m1, . . .mk−1, n1, . . . nl−1) .The sum (4.2.10) can now be expressed as a sum over all vectors α¯′ insome family M′h1, µ¯1, µ¯1(p1 − µk, p2 − νl), with h1 < h, times a sum over allvertices mk of height h on the binary tree Bm:∑α¯∈M′′h, µ¯, ν¯∫RnGα¯(x)dx =h−1∑h1=1∑α¯′∈M′h1, µ¯1, ν¯1∑mk:h(mk)=h∫RnGα¯′(x)χEmk(x)dx=h−1∑h1=1∑α¯′∈M′h1, µ¯1, ν¯1∫RnGα¯′(x) ∑mk :h(mk)=hχEmk(x)dx.The expression in brackets equals the function χE′α by the condition (4.1.9)of Lemma 4.1.2 (see Figure 4.3 below for an illustration of this process ofsummation). Therefore we haveS(h, µ¯, ν¯) ∼=h−1∑h1=1∑α¯′∈M′h1, µ¯1, ν¯1∫RnGα¯′(x)dx =h−1∑h1=1S(h1, µ¯1, ν¯1) .Now let us turn to the condition (4.2.9). If µ¯ 6= ν¯ (including the casek 6= l, that is if µ¯ and ν¯ have a different number of entries), let 0 ≤ r ≤ k∧ lbe the smallest index such that µk−t 6= νl−t. If t > 0, by iterating theformula (4.2.8) we haveS(h, µ¯, ν¯) ∼=h−1∑h1=1S(h1, µ¯1, ν¯1) ∼= h−1∑h1=1h1−1∑h2=1S(h2, µ¯2, ν¯2)∼= . . . ∼=h−1∑h1=1h1−1∑h2=1· · ·ht−1−1∑ht=1S(ht, µ¯t, ν¯t) .We can then apply Lemma 4.2.2 with the parameters h, k, l, µ¯, ν¯ replaced byht, k − t, l − t, µ¯t, ν¯t, and with p1, p2 replaced by p1 − µ1 − . . . − µk−t andp2 − ν1 − . . .− νk−t, respectively. We conclude that S(ht, µ¯t, ν¯t) ∼= 0. Thisimplies S(h, µ¯, ν¯) ∼= 0, since there are at most ht = O(mp0) terms to sum.On the other hand, if µ¯ = ν¯ (which implies k = l) we can iterate theformula (4.2.8) k times and find thatS(h, µ¯, ν¯) ∼=h−1∑h1=1h1−1∑h2=1· · ·hk−1−1∑hk=1S(hk, µ1, µ1) .574.2. Preliminary resultsFigure 4.3: The fixed vertices m1, . . .mk−1, n1, . . . nl−1 lie on the ray R′,connecting the root of the tree and the vertex max {mk−1, nl−1}. The in-nermost summation is over vertices mk = nl of height h in the sub-tree withroot max {mk−1, nl−1}).584.3. The proof of the propositionBut for any h1 we haveS(h1, µ1, µ1) =∑m1 :h(m1)=h1∫RnχEm1(x)dx =∫RnχE1(x)dx = 1.Therefore we getS(h, µ¯, ν¯) ∼=h−1∑h1=1h1−1∑h2=1· · ·hk−2−1∑hk−1=11 = O(hk−1).4.3 The proof of the propositionNow we are ready to prove Proposition 2.1.1.Proof. The condition (a) follows immediately from Lemma 4.1.1(1):supp f̂N ⊆ QN ⊂ XN .Next, we turn to (b). We can prove it for p = 2p0 and p = 1, then useinterpolation between these two cases. As we observed, for p = 2p0 itsuffices to prove the estimate (4.2.4), that is∣∣∣∣∣ ∑1≤m1<···<mk<2m1≤n1<···<nl<2m∫Rnfµ1m1 · · · fµkmkfν1n1 · · · fνlnl dx,∣∣∣∣∣ .p0 mp0 ,for fixed k, l ≤ p0 and fixed multiplicities µ¯, ν¯ such that A(µ¯) = A(ν¯) = p0.According to the notation we set in Definition 4.2.1, we can rewrite theprevious sum in the formm∑h=1∑α¯∈Mh,µ¯,ν¯∫RnFα¯(x)dx.Lemma 4.2.3 yieldsm∑h=1∑α¯∈Mh,µ¯,ν¯∫RnFα¯(x)dx ∼=m∑h=1S(h, µ¯, ν¯) ∼={0 if µ¯ 6= ν¯O(mk)if µ¯ = ν¯In both cases, S(m, µ¯, ν¯) = O(mp0) and (b) follows for p = 2p0.594.3. The proof of the propositionWe now turn to the case p = 1. To simplify the notation, let us sete(p¯N · x) := e2piix·p¯mN . We have from the condition (4.1.3):∥∥∥∥ 2m−1∑N=1fN∥∥∥∥L1(Rn)=∥∥∥∥ 2m−1∑N=1e(p¯N ·)gN∥∥∥∥L1(Rn)≤∥∥∥∥ 2m−1∑N=1e(p¯N ·)χEN∥∥∥∥L1(Rn)+2m−1∑N=1‖gN − χEN ‖L1(Rn)∼=∥∥∥∥ 2m−1∑N=1e(p¯N ·)χEN∥∥∥∥L1(Rn)But the term∑2m−1N=1 e(p¯N ·)χEN is supported in Q. Therefore by Ho¨lder’sinequality, the condition (4.1.3) and Proposition 2.1.1(b) for p = 2p0, wehave:∥∥∥∥ 2m−1∑N=1e(p¯N ·)χEN∥∥∥∥L1(Rn)≤∥∥∥∥ 2m−1∑N=1e(p¯N ·)χEN∥∥∥∥L2p0(Rn)≤∥∥∥∥ 2m−1∑N=1fN∥∥∥∥L2p0(Rn)+∥∥∥∥ 2m−1∑N=1e(p¯N ·)(χEN − gN)∥∥∥∥2p0∼=∥∥∥∥ 2m−1∑N=1fN∥∥∥∥L2p0(Rn).p0√m.Then (b) follows for p = 1.Finally, we prove (c). We definef˜N (x) := cos (2pip¯N ·x)χEN (x),for N = 1, 2, . . . 2m − 1. The next lemma establishes the properties of thefunctions f˜N .Lemma 4.3.1. We have:1.∥∥f˜N − Re fN∥∥Lp(Rn) ∼= 0 ∀ 1 < p ≤ 2p0,2.{f˜N}N=1,2,...2m−1 is a signed tree system.Proof. We start with (1). Let 1 < p ≤ 2p0. We have∥∥f˜N − Re fN∥∥pLp(Rn) = ∫Rn∣∣cos (2pip¯N ·x)(χEN(x)− gn(x))∣∣pdx≤∫Rn|χEN(x)− gn(x)|pdx ∼= 0,(4.3.1)604.3. The proof of the propositionby interpolating between the cases p = 2p0 and p = 1, through the condition(4.1.3).Concerning (2), we observe that f˜N (x) and cos (2pip¯N · x) have the samesign in the set EN . Therefore, by (4.1.1) we obtainsupp f˜N = EN ={x ∈ EN¯ : (−1)j−1cos (2pix · p¯N¯ ) > 0}={x ∈ EN¯ : (−1)j−1f˜N¯ (x) > 0}={x ∈ Q : (−1)j−1f˜N¯ (x) > 0},which is exactly the condition (3.0.5).The fact that{f˜N}N=1,2,...2m−1 is a signed tree system allows us to useLemma 2.1.1. If σ is the permutation defined in (3.0.7), via the triangleinequality and the inequality (2.1.10) we havemax1≤ l<2m∣∣∣∣ l∑N=1fσ(N)(x)∣∣∣∣ ≥ max1≤ l<2m∣∣∣∣ l∑N=1Re fσ(N)(x)∣∣∣∣≥ max1≤ l<2m∣∣∣∣∣l∑N=1f˜σ(N)(x)∣∣∣∣∣− E(x) ≥ 132m−1∑N=1∣∣f˜N (x)∣∣ − E(x),(4.3.2)for all x ∈ Rn, whereE(x) :=2m−1∑j=1∣∣∣f˜j(x)− Re fj(x)∣∣∣ .Therefore, if we set f˜(x) :=∑2m−1N=1∣∣f˜N (x)∣∣, the estimate (4.3.2) says thatE + max1≤ l<2m∣∣∣∣ l∑N=1fσ(N)∣∣∣∣ ≥ f˜ ,on Rn, which implies∣∣∣{E ≥ cm2}∣∣∣ + ∣∣∣∣∣Q ∩{max1≤ l<2m∣∣∣∣ l∑N=1fσ(N)∣∣∣∣ ≥ cm2}∣∣∣∣∣ ≥ ∣∣∣Q ∩ {f˜ ≥ cm}∣∣∣ ,for any constant c. But the size of the set{E ≥ cm2 } is small because, byChebyshev’s inequality and estimate (4.3.1), we have∣∣∣∣{x ∈ Q : E(x) > 1}∣∣∣∣ ≤ ‖E‖pLp(Rn) ∼= 0.614.3. The proof of the propositionTherefore we can prove Proposition 2.1.1(c) working on f˜ and estimating∣∣∣Q ∩ {f˜ ≥ cm}∣∣∣ from below.First, by Lemma 4.1.1(3) we have∫Q∣∣f˜N (x)∣∣dx = ∫EN|cos (2pix · p¯N )| dx > |EN |3.Then, by using Lemma 4.1.2, integrating equation (4.1.9) on Q and summingover all N , we get∫Q2m−1∑N=1∣∣f˜N (x)∣∣dx ≥ 132m−1∑N=1|EN | = m3. (4.3.3)Therefore, the identity (4.3.3) and the condition (4.1.9) yield∫Qf˜(x)dx ≥ m3,f˜(x) ≤2m−1∑N=1χN (x) = m,(4.3.4)for a. e. x ∈ Q. The final claim is that the two conditions in (4.3.4) imply∣∣∣{x ∈ Q : f˜(x) ≥ m4}∣∣∣ > c0 (4.3.5)for a suitable constant c0 > 0. We havem∣∣∣{x ∈ Q : f˜(x) ≥ m4}∣∣∣+ m4≥∫Q∩{f˜≥m4 }f˜(x)dx +∫Q∩{f˜<m4 }f˜(x)dx=∫Qf˜(x)dx ≥ m3,which implies ∣∣∣{x ∈ Q : f˜(x) ≥ m4}∣∣∣ ≥ 13− 14> 0. (4.3.6)Thus the claim (4.3.5) is verified and the condition (4.3.6) proves (c) via theinequality (4.3.2). The proof is complete.62Chapter 5The geometric lemmaIn this chapter we provide the proof of Lemma 2.1.2. In what follows westill use the notations (2.1.5) and (2.1.11).Lemma 5.0.1. Let M ≥ 2 be an integer and U := {v1, v2, . . . vM} a collec-tion of distinct unit vectors in the first hyper-octant of Rn (that is all theircoordinates are strictly positive). Then there is an ordering {u1, . . . uM} ofU such that the following sectors are non-empty:S1 = Γu1 ∩ . . . ∩ ΓuM−1 ∩ ΓcuM ,S1 = Γu1 ∩ . . . ∩ ΓuM−2 ∩ ΓcuM−1 ∩ ΓcuM ,.........SM−2 = Γu1 ∩ Γu2 ∩ Γcu3 ∩ . . . ∩ ΓcuM ,SM−1 = Γu1 ∩ Γcu2 ∩ . . . ∩ ΓcuM .(5.0.1)Moreover, for all j = 1, 2, . . .M − 1 they satisfy the equationΓuj ∩M−1⋃k=1Sk =M−j⋃k=1Sk. (5.0.2)Proof. Let us denote by v⊥1 , . . . v⊥M the hyperplanes in Rn orthogonal to theunit vectors v1, . . . vM , respectively. For x′ ∈ Rn−1 let us denote by rx′ thevertical line passing through the point (x′, 0) ∈ Rn:rx′ :={(x′, t) ∈ Rn : t ∈ R} = {(x′, 0) + ten ∈ Rn : t ∈ R} .It is not hard to show that the line rx′ intersects v⊥1 , . . . v⊥M at distinct pointsfor almost every x′ ∈ Rn−1. Indeed, for any j ∈ {1, . . .M} the vectorvj is in the first hyper-octant of Rn, therefore its n-th coordinate v(n)j isstricty positive. Consequently, for any x′ ∈ Rn−1 we can find a t such thatrx′(t) · vj = x′ · v′j + tv(n)j = 0 and prove the existence of the intersectionpoint. However, for two or more of the vectors v1, . . . vM these points couldcoincide.63Chapter 5. The geometric lemmaTo show that for almost every x′ ∈ Rn−1 this is not the case, let usconsider all possible intersections v⊥j1 ∩v⊥j2 for 1 ≤ j1 < j2 ≤M . The vectorsvj are distinct in the first hyper-octant and, therefore, pairwise linearlyindependent. Consequently, the sets v⊥j1 ∩ v⊥j2 are linear subspaces of Rn ofdimension n − 2 and their n-dimensional Lebesgue measure is 0. Now, ifwe denote by pi : Rn → Rn−1 the projection on the first n − 1 coordinates(pi(x) = (x1, . . . xn−1)), the (n − 1)-dimensional Lebesgue measure of theprojections of these subspaces through pi is also 0. Therefore, for any x′ ∈Rn−1 in the complement of the union of these projections, the intersectionpoints between rx′ and the hyperplanes v⊥1 , . . . v⊥M must be all distinct.Now, for a fixed x′ ∈ Pc, let t1 < t2 < · · · < t2m be the real num-bers such that the intersection points with the hyper-planes are rx′(t1) =(x′, t1), . . . rx′(tM ) = (x′, tM ). Let us denote them by P1, . . . PM , respec-tively. It is clear that there exists a permutation8 σ˜ of the numbers 1, 2, . . .Msuch that Pj ∈ v⊥σ˜(j) for all j. In other words, if we imagine to move along theline rx′ in the positive t direction, there exists a permutation σ˜ of {1, 2, . . .M}such that v⊥σ˜(1) is the first hyperplane the line rx′ intersects (at, say, P1), v⊥σ˜(2)is the second hyperplane the line rx′ intersects (at, say, P2) and so on untilthe last hyperplane v⊥σ˜(M), being intersected at PM (see Figure 5.1 below).Let us set u⊥j := v⊥σ˜(j) for all j.Figure 5.1: The vertical line rx′ intersects all the planes at distinct points(in this case we have n = 3 and M = 3).8It is not hard to see that the permutation σ˜ depends only on the connected componentof the set Pc in which we take the point x′64Chapter 5. The geometric lemmaNow let us choose M − 1 real numbers τ1, τ2, . . . τM−1 such that tj <τj < tj+1 for all j = 1, . . .M − 1 and consider the pointsQ1 := rx′(τ1), . . . QM−1 := rx′(τM−1).Geometrically, the point Qj lies on the vertical segment PjPj+1 for j =1, . . .M − 1. Therefore, Q1 lies above u⊥1 but below u⊥2 , . . . u⊥M , Q2 liesabove u⊥1 and u⊥2 but below u⊥3 , . . . u⊥M and so on until QM−1, that liesabove u⊥1 , . . . u⊥M−1 but below u⊥M (see Figure 5.2 below). In other words,for all j = 1, . . .M − 1 we haveu1 · ~OQj > 0, . . . uj · ~OQj > 0,uj+1 · ~OQj < 0, . . . uM · ~OQj < 0.The existence of these points show that, if we define the sectors S1, . . . SM−1as in (5.0.1), by construction it follows thatQj ∈ SM−j ,for all j = 1, . . .M − 1. Therefore the sectors S1, . . . SM−1 are nonempty.Figure 5.2: Again in the case n = 3, M = 3, we can pick distinct points Qjon the vertical segment between two consecutive points Pj . These pointsidentify nonempty, disjoint sectors of R3.It remains to prove the condition (5.0.2). We haveΓuj ∩M−1⋃k=1Sk =M−1⋃k=1Γuj ∩ Γu1 ∩ Γu2 ∩ . . . ∩ ΓuM−k ∩ ΓcuM−k+1 ∩ . . . ∩ ΓcuM .65Chapter 5. The geometric lemmaIf j ≤ M − k, Γuj is among the half-spaces that contain Sk. Then we haveSk ⊂ Γuj and Sk ∩ Γuj = Sk. If, instead, j > M − k, we have Sk ⊂ Γcuj andSk ∩ Γuj = ∅. ThereforeM−1⋃k=1Γuj ∩ Sk =M−j⋃k=1Γuj ∩ Sk,which is the condition (5.0.2). The proof is complete.66Chapter 6ConclusionIn this dissertation we studied the norm ‖HU‖Lp(Rn)→Lp,∞(Rn) and its be-havior in connection with the structural properties of the set U and, inparticular, with #U if U is finite. The remarkable fact about the lowerestimate (1.2.25) is that it holds for all finite sets of directions U in all di-mensions. As we have pointed out at the end of section 1.2.4, this generalityenabled us to derive Theorem 1.1.3, that clearly sets HU apart from othermaximal directional operators for which there exist results of (p, p)-strongboundedness even if the set of directions has infinite cardinality (see section1.2).The results in this dissertations complete and generalize the ones inKaragulyan’s paper [17] in at least two ways. First, Karagulyan provesestimate (2.1.3) only for p = 2 through the orthogonality in L2(R2), that isin the simplest case. Indeed, if{XN}1≤N<2m are disjoint sectors in R2, thefunctions{fN}1≤N<2m that we obtain from Proposition 2.1.1 are orthogonalto each other and we have∥∥∥∥ 2m−1∑N=1fN∥∥∥∥2L2(R2)=2m−1∑N=1‖fN‖2L2(R2) .But the right-hand side of the previous identity can easily be proved to be. m, which proves estimate (2.1.3) for p = n = 2.Instead, as we have seen in Chapter 4, the proof of the inequality (2.1.3)requires much more work. In particular, it is necessary to strengthen thedisjointness of the supports of the Fourier transforms of the functions fN andto use a delicate argument involving repeated applications of Plancherel’stheorem.The second important contribution with respect to the paper [17] besidesthe trivial adaptation from the planar setting is, as we have seen in Chap-ter 2, the treatment of the condition (2.2.1) in higher dimensions. Unlikethe two-dimensional case, where the fulfillment of this condition is triviallyproved through the clockwise ordering of the sectors, a slightly more delicateproof is necessary in dimension n > 2. The argument is provided by Lemma2.1.2.676.1. Ongoing and future directions of research6.1 Ongoing and future directions of researchHere we briefly mention some possible further lines of research and openproblems in the theory of the maximal Hilbert transform HU . With theweak and strong Lp-unboundedness of HU when #U = +∞ having beenestablished, several results show that the growth rate of ‖HU‖Lp(Rn)→Lp(Rn)and ‖HU‖Lp(Rn)→Lp,∞(Rn), as functions of #U , depends on certain propertiesof the set U . The term√log(#U) provides a minimum growth rate for theLp,∞(Rn)-norms of the operator HU that is valid for all exponents 1 < p <+∞, for all sets of directions U and in all dimensions n ≥ 2, but the abovenorms can grow, as #U increases, faster than√log(#U). For instance, [8]shows that‖HU‖L2(R2)→L2(R2) & log(#U) (6.1.1)if U is a set uniformly distributed directions in the plane. By coupling thecondition (6.1.1) with the inequality (1.2.19), we conclude that, under thisgeometric condition on U , we have ‖HU‖L2(R2)→L2(R2) ≈ log(#U).Instead, let us consider U to be, say, L1 lacunary (see Definition 1.2.8).For instance, let us consider the finite setU :={vk|vk| : vk =(1, 2−k−100), k = 1, . . .#U}. (6.1.2)On one hand, the inequality (1.2.25) implies that‖HU‖Lp(R2)→Lp,+∞(R2) &p√log(#U).On the other hand, as the remarkable result in the paper [10] shows, wehave‖HU‖Lp(R2)→Lp(R2) .p√log(#U),which implies that‖HU‖Lp(R2)→Lp(R2) ≈ ‖HU‖Lp(R2)→Lp,∞(R2) ≈p√log(#U)for any 1 < p < +∞. More generally, if U is a L3 lacunary set of order r,the inequality (1.2.25) and [10] show that‖HU‖Lp(R2)→Lp(R2) ≈ ‖HU‖Lp(R2)→Lp,∞(R2) ≈p,r√log(#U).Therefore, in this case, the growth rate of the norm ofHU perfectly matchesour bound√log(#U).686.1. Ongoing and future directions of researchTo sum up, what clearly seems to emerge from these and other exam-ples is that the growth rate of ‖HU‖Lp(Rn)→Lp(Rn) and ‖HU‖Lp(Rn)→Lp,∞(Rn)depends on the geometric properties of the set of directions U . Therefore,in the planar setting as well as in a multidimensional space, the problem ofunderstanding exactly how arises naturally. At least for n = 2, the idealsituation would be a result in the spirit of Bateman’s (see [2]), establish-ing the necessary and sufficient conditions on U , in terms of its geometricproperties, for a certain growth rate of the (p, p)-norms of HU .69Bibliography[1] A. Alfonseca. Strong type inequalities and an almost-orthogonalityprinciple for families of maximal operators along directions in R2. J.London Math. Soc. (2) 67, no. 1: 208–218, 2003.[2] M. Bateman. Kakeya sets and directional maximal operators in theplane. Duke Math. J. 147, no. 1: 55–77, 2009.[3] M. Bateman and N.H. Katz. Kakeya sets in Cantor directions. 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