{"@context":{"@language":"en","Affiliation":"http:\/\/vivoweb.org\/ontology\/core#departmentOrSchool","AggregatedSourceRepository":"http:\/\/www.europeana.eu\/schemas\/edm\/dataProvider","Campus":"https:\/\/open.library.ubc.ca\/terms#degreeCampus","Creator":"http:\/\/purl.org\/dc\/terms\/creator","DateAvailable":"http:\/\/purl.org\/dc\/terms\/issued","DateIssued":"http:\/\/purl.org\/dc\/terms\/issued","Degree":"http:\/\/vivoweb.org\/ontology\/core#relatedDegree","DegreeGrantor":"https:\/\/open.library.ubc.ca\/terms#degreeGrantor","Description":"http:\/\/purl.org\/dc\/terms\/description","DigitalResourceOriginalRecord":"http:\/\/www.europeana.eu\/schemas\/edm\/aggregatedCHO","FullText":"http:\/\/www.w3.org\/2009\/08\/skos-reference\/skos.html#note","Genre":"http:\/\/www.europeana.eu\/schemas\/edm\/hasType","GraduationDate":"http:\/\/vivoweb.org\/ontology\/core#dateIssued","IsShownAt":"http:\/\/www.europeana.eu\/schemas\/edm\/isShownAt","Language":"http:\/\/purl.org\/dc\/terms\/language","Program":"https:\/\/open.library.ubc.ca\/terms#degreeDiscipline","Provider":"http:\/\/www.europeana.eu\/schemas\/edm\/provider","Publisher":"http:\/\/purl.org\/dc\/terms\/publisher","Rights":"http:\/\/purl.org\/dc\/terms\/rights","RightsURI":"https:\/\/open.library.ubc.ca\/terms#rightsURI","ScholarlyLevel":"https:\/\/open.library.ubc.ca\/terms#scholarLevel","Title":"http:\/\/purl.org\/dc\/terms\/title","Type":"http:\/\/purl.org\/dc\/terms\/type","URI":"https:\/\/open.library.ubc.ca\/terms#identifierURI","SortDate":"http:\/\/purl.org\/dc\/terms\/date"},"Affiliation":[{"@value":"Science, Faculty of","@language":"en"},{"@value":"Mathematics, Department of","@language":"en"}],"AggregatedSourceRepository":[{"@value":"DSpace","@language":"en"}],"Campus":[{"@value":"UBCV","@language":"en"}],"Creator":[{"@value":"Karslidis, Dimitrios","@language":"en"}],"DateAvailable":[{"@value":"2016-04-19T02:02:07","@language":"en"}],"DateIssued":[{"@value":"2016","@language":"en"}],"Degree":[{"@value":"Doctor of Philosophy - PhD","@language":"en"}],"DegreeGrantor":[{"@value":"University of British Columbia","@language":"en"}],"Description":[{"@value":"The main focus of this document is the small ball inequality. The small ball inequality is a functional inequality concerning the lower bound of the supremum norm of a linear combination of Haar functions supported on dyadic rectangles of a fixed volume. The sharp lower bound in this inequality, as yet unproven, is of considerable interest due to the inequality's numerous applications. We prove the optimal lower bound in this inequality under mild assumptions on the coefficients of a linear combination of Haar functions, and further investigate the lower bounds under more general assumptions on the coefficients. We also obtain lower bounds of such linear combinations of Haar functions in alternative function spaces such as exponential Orlicz spaces.","@language":"en"}],"DigitalResourceOriginalRecord":[{"@value":"https:\/\/circle.library.ubc.ca\/rest\/handle\/2429\/57662?expand=metadata","@language":"en"}],"FullText":[{"@value":"The small ball inequality with restricted coefficientsbyDimitrios KarslidisA thesis submitted in partial fulfilment of the requirements forthe degree ofDoctor of PhilosophyinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)The University of British Columbia(Vancouver)April 2016c\u00a9 Dimitrios Karslidis, 2016AbstractThe main focus of this document is the small ball inequality. The small ball inequal-ity is a functional inequality concerning the lower bound of the supremum norm ofa linear combination of Haar functions supported on dyadic rectangles of a fixedvolume. The sharp lower bound in this inequality, as yet unproven, is of consider-able interest due to the inequality\u2019s numerous applications. We prove the optimallower bound in this inequality under mild assumptions on the coefficients of a linearcombination of Haar functions, and further investigate the lower bounds under moregeneral assumptions on the coefficients. We also obtain lower bounds of such linearcombinations of Haar functions in alternative function spaces such as exponentialOrlicz spaces.iiPrefaceMuch of the following document is adapted from two of the author\u2019s research papers:[29] and [30]. In particular, Proposition 5.6 and Section 6.1, Section 6.2 form themain content of [29], On the signed small ball inequality with restricted coefficients,while Section 6.3, Section 6.4 and all of Chapter 7 are adapted from [30], Orlicz spacesbounds for special classes of hyperbolic sums. The first of these two manuscripts hasbeen accepted for publication and will appear in Indiana University MathematicsJournal and the second paper has been submitted.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Preliminary facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Statement of the small ball inequality . . . . . . . . . . . . . . . . . . 71.3 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Recent history of the small ball inequality . . . . . . . . . . . . . . . 143 Connections and applications . . . . . . . . . . . . . . . . . . . . . . . 193.1 Connection to probability theory . . . . . . . . . . . . . . . . . . . . 193.2 Connection to approximation theory . . . . . . . . . . . . . . . . . . 263.3 Connection to discrepancy theory . . . . . . . . . . . . . . . . . . . . 383.3.1 Conjectures in discrepancy theory . . . . . . . . . . . . . . . . 383.3.2 Proof of Theorem 3.17 . . . . . . . . . . . . . . . . . . . . . . 423.3.3 Hala\u00b4sz\u2019s proof of Conjecture 3.18 in d = 2 . . . . . . . . . . . 45iv4 The two-dimensional small ball inequality . . . . . . . . . . . . . . . 514.1 Temlyakov\u2019s proof of the two-dimensional small ball inequality . . . . 514.2 The sharpness of the small ball inequality in all dimensions . . . . . . 565 Elements of dyadic harmonic analysis . . . . . . . . . . . . . . . . . . 615.1 Khintchine\u2019s inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 615.2 The Littlewood-Paley inequalities . . . . . . . . . . . . . . . . . . . . 635.3 Connection of martingale differences with Haar functions . . . . . . . 676 Two proofs of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . 746.1 Motivation for the splitting condition . . . . . . . . . . . . . . . . . . 746.2 The first proof of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . 776.3 The second proof of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . 806.4 The sharpness of the small ball inequality with restricted coefficients . 847 Orlicz space bounds for special classes of hyperbolic sums . . . . . 897.1 Orlicz spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 907.2 The proof of Theorem 1.6 . . . . . . . . . . . . . . . . . . . . . . . . 937.3 Study of signed hyperbolic sums . . . . . . . . . . . . . . . . . . . . . 958 Conclusions and future projects . . . . . . . . . . . . . . . . . . . . . 102Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106vList of Figures1.1 A Haar function, hR, supported on a rectangle R. . . . . . . . . . . . 21.2 Examples of Haar functions. . . . . . . . . . . . . . . . . . . . . . . . 63.1 The function w used in the proof of the two-dimensional small ballprobability conjecture by Talagrand. . . . . . . . . . . . . . . . . . . 23viAcknowledgementsPrimarily, I would like to thank my supervisor Malabika Pramanik. No part of thisthesis would have been possible without her unwavering commitment and constantencouragement. Malabika has generously assisted me in all my scientific activities.She is always willing to give me advice and help me to navigate my way through theacademic world.I want to thank the members of my supervisory committee, Philip Loewen,Richard Froese and Akos Magyar for their valuable advice, their support, and theirtime. I would especially like to express my deep gratitude to Philip Loewen forsupporting me in all my academic pursuits during my time at UBC. I would like tothank Fok-Shuen Leung for his constant support and his valuable advice. I wantto express my gratitude to Dmitriy Bilyk for sharing his rich knowledge with me.We also have had many discussions on the small ball inequality from which I havebenefited a lot.I would also like to acknowledge the abundant support I received from my girl-friend, Pam Sargent, during the process of typing the thesis. She has helped memore than anybody else in drafting this document and has never ceased to believein my abilities as a researcher. \u03a3\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1i\u03c3\u03c4\u03c9 pio\u03bb\u03c5!Finally, I wish to thank my family, all of my friends and Idryma Paideias kaiEuropaikoy Politismou for their ample support all these years.viiChapter 1IntroductionIn this thesis, we focus on the small ball inequality, an inequality that conjecturesa lower bound for the supremum norm of any linear combination of Haar functions.The precise definition of Haar functions will be given in Section 1.1, roughly speakingthese are functions that take the value 1 on half of its support and \u22121 on the otherhalf. This inequality, which remains a conjecture in dimensions d \u2265 3, has importantapplications in several fields of mathematics. Here, we show that the conjecturedinequality is valid under some mild additional assumptions on the linear combinationsof Haar functions.To get an idea of what this inequality claims, first consider dyadic intervals in[0, 1]. These are intervals whose endpoints are consecutive dyadic rationals, i.e.,intervals of the form[m2n, m+12n)where n \u2265 0 and 0 \u2264 m < 2n. For example, [0, 14)and[14, 12)are dyadic intervals of [0, 1] of length 1\/4. Now, for a dyadic interval I,consider the function hI supported on I and taking the value \u22121 on the left half ofI, 1 on the right half of I, and zero outside of the interval. Finally, form a linearcombination of such functions hI , where the dyadic interval I is of a fixed length, i.e.,\u2211|I|=2\u2212n \u03b1IhI . For reasons described later on, one is often interested in determininghow small the largest possible magnitude of such linear combinations can be. Inother words, one seeks lower bounds of the supremum norm of functions of the form\u2211|I|=2\u2212n \u03b1IhI . We will explain in the next paragraph why this is reasonably simplewhen\u2211|I|=2\u2212n \u03b1IhI is considered. The present work focuses on lower bounds of the1supremum norm of the higher dimensional analogues of such functions. The small ballinequality is an inequality which conjectures a sharp lower bound for such quantities,as\u2211|R|=2\u2212n \u03b1RhR, where hR is a higher dimensional analogue of a function hI andits precise definition is given in (1.3).The key step in finding the largest possible magnitude of the function\u2211|I|=2\u2212n \u03b1IhIis to notice that dyadic intervals of a fixed length are disjoint. Therefore, the largestmagnitude is the largest of numbers |\u03b1I |. However, this is not the case in higherdimensions, and this creates a barrier in solving the higher dimensional problem.To further clarify the problem, consider a rectangle R, say with area |R| = 2\u2212n,in the unit square. Divide this rectangle into four subrectangles, Ri, i = 1, . . . , 4,where R1 corresponds to the upper left subrectangle, R3 corresponds to the lowerright subrectangle, and R2 and R4 correspond to the upper right and lower leftsubrectangles respectively. The length of each side of these subrectangles will behalf of the length of the corresponding side of the original rectangle. Consider afunction which takes the value +1 on R1 and R3, \u22121 on R2 and R4, and zero outsideof the rectangle R. Denote such functions by hR. A linear combination of suchfunctions corresponding to planar rectangles of area 2\u2212n are functions of the formH =\u2211|R|=2\u2212n \u03b1RhR. Letting the coefficients \u03b1R vary yields different correspondingfunctions H. In this setting, the small ball inequality conjectures a lower bound forthe largest value of H over all possible choices of coefficients in terms of n, for largevalues of n.\u22121+1+1\u22121Figure 1.1: A Haar function, hR, supported on a rectangle R.2The small ball inequality would be almost obvious if we only considered linearcombinations of functions which correspond to non-overlapping rectangles. When therectangles do overlap, the analysis is not straightforward. If we replace the planarrectangles by parallelepipeds, for example, the situation becomes much more difficultand delicate. In this thesis, we focus on obtaining lower bounds on the largest valueof H in terms of n, in any dimension.The formulation of the small ball inequality is given in Chapter 1. This chapter isintroductory in nature and discusses various versions of the small ball inequality, andwe give all of the preliminary facts needed to state the small ball inequality precisely.We also include the main results in Chapter 1 to better illustrate the connection ofthese results to the small ball inequality.In Chapter 2, we outline the history of this problem and progress to date.The main purpose of Chapter 3 is to highlight the connections between the smallball inequality and other areas of mathematics. Surprisingly, the validity of thesmall ball inequality is also a key to solving open problems in fields such as proba-bility theory, approximation theory and the theory of irregularities of distributions.These connections have been dealt with precisely in Chapter 3, but we provide somemotivation here in non-technical terms.Let the unit square T := [0, 1]2 represent the city of Vancouver. To each point ~t =(t1, t2) of T , we associate the price, X~t, of the property at that point. Since propertyprices in Vancouver are often high, it is natural to be interested in determining thelikelihood that the property prices in the entire city will fall below a prescribedvalue, say \u03b5 > 0. In other words, we are concerned with estimating P(sup~t\u2208T X~t < \u03b5)for sufficiently small \u03b5 > 0, where P is the probability measure of the underlyingspace. Under certain normality on the probability distribution of {X~t}, the small ballinequality will determine this probability. We give all the details of this connectionin Section 3.1 of Chapter 3.Now, suppose we have a collection of messages represented as binary strings of0\u2019s and 1\u2019s. Due to physical constraints, we would like to determine a subcollectionof the messages with the property that, given any message in the collection, thereis a member of that subcollection which recovers the given message with prescribed3accuracy. We obviously want to find the smallest subcollection that will do the job.This is related to the notion of entropy, and the small ball inequality plays a crucialrole in solving such problems. This is further explained in Section 3.2 of Chapter 3.In practice, there are many functions whose integrals cannot be written in termsof elementary functions. However, we may still be interested in estimating the areaunder such curves. One can, of course, approximate it by a Riemann sum, and itis natural to want to obtain a bound on the error in the approximation. It turnsout that how well-distributed the tags of the partition of the Riemann sum areaffects the error bound. The discrepancy function defined in Section 3.3 of Chapter3 provides a way of determining how far the distribution of the tags are from aperfectly randomized scenario. In a quantifiable sense that has been made precisein Section 3.3, minimizing the error bound in the numerical integration problemcorresponds to minimizing the discrepancy function, and the small ball inequalityplays a vital role in bounding the discrepancy function. This connection is presentedin Section 3.3 of Chapter 3.In Chapter 4, we present a detailed proof of the two-dimensional small ball in-equality. We also highlight the similarities between the arguments of this proof andthe proof, given in Chapter 3, of the two-dimensional lower bound of the discrepancyfunction\u2019s L\u221e-norm.In Chapter 5, we provide the background needed to understand the main re-sults of this thesis, presented in Chapter 6 and Chapter 7. In particular, we recallKhintchine\u2019s inequalities, a special case of the Littlewood-Paley inequalities for mar-tingales, since Khintchine\u2019s inequalities are used in the proof of Theorem 1.4. Wealso introduce the Littlewood-Paley inequalities for martingales and connect them tothe Littlewood-Paley inequalities for Haar functions, since such inequalities for Haarfunctions are used in the proofs of Theorem 1.3 and Theorem 1.6.Chapters 6 and 7 are primarily devoted to proving the main results. In particular,Theorem 1.3 and Theorem 1.4 are proved in Chapter 6. In Chapter 7, we first givethe necessary background information on exponential Orlicz spaces before movingon to proving Theorem 1.6. The reader will find this brief introduction to Orliczspaces helpful in understanding the proof of Theorem 1.6. We conclude Chapter 74with further investigation of hyperbolic sums in exponential Orlicz spaces.Chapter 8, the final chapter of the thesis, provides a summary of the results ofthis thesis and outlines some related future projects.1.1 Preliminary factsLet 1I(x) be the characteristic function of the interval I, i.e.,1I(x) =\uf8f1\uf8f2\uf8f31, x \u2208 I0, otherwise .Consider the collection of the dyadic intervals of [0, 1]:D ={I =[m2n,m+ 12n): m,n \u2208 Z, n \u2265 0, 0 \u2264 m < 2n}withD\u2217 = D \u222a {[\u22121, 1]}.(1.1)For every interval I \u2208 D, its left and right dyadic halves are denoted by Il and Irrespectively. We define the L\u221e-normalized Haar function, hI , corresponding to aninterval I as:hI(x) = \u22121Il +1Ir . (1.2)Note that h[\u22121,1](x) = 1[0,1)(x) for all x in [0, 1].It is easy to extend Haar functions to higher dimensions. We consider the familyof dyadic rectangles in dimension d \u2265 2:Dd = {R = R1 \u00d7 \u00b7 \u00b7 \u00b7 \u00d7Rd : Rj \u2208 D},i.e., every R \u2208 Dd is a Cartesian product of dyadic intervals. The Haar functionssupported on R are defined as a coordinate-wise product of one-dimensional Haarfunctions:hR(x1, . . . , xd) = hR1(x1) \u00b7 \u00b7 \u00b7hRd(xd), where R = R1 \u00d7 \u00b7 \u00b7 \u00b7 \u00d7Rd, Rj \u2208 D. (1.3)5(a) A Haar function sup-ported on the interval[0, 1).(b) A Haar function sup-ported on the interval[0, 0.5).Figure 1.2: Examples of Haar functions.If we consider two distinct dyadic intervals, then either one will be strictly con-tained in the other, or they will be disjoint. Haar functions enjoy the followingproperties:h2R(x) = 1R(x), (1.4)\u222b[0,1]dhR(x)hR\u2032(x)dx = 0, whenever R 6= R\u2032, (1.5)\u222b[0,1]dhR(x)dx = 0, \u2200R \u2208 Dd. (1.6)LetHdn = {~r \u2208 Zd+ : |~r| := r1 + r2 + \u00b7 \u00b7 \u00b7+rd = n}withZd+ = {~r = (r1, . . . , rd) : rj \u2265 0 and rj \u2208 Z, j = 1, . . . , d},6andR~r = {R \u2208 Dd : Rj \u2208 D and |Rj| = 2\u2212rj , j = 1, . . . , d}.In other words, the set R~r is a collection of axis-parallel dyadic rectangles whose sidelength along the jth axis is 2\u2212rj . The rectangles in R~r are disjoint and partition thed-dimensional unit cube, [0, 1)d. Also, it will be useful to observe that the cardinalityof the set Hdn denoted by #Hdn is of order nd\u22121. This can be stated as #Hdn ' nd\u22121,which means that there exist positive constants C1(d) and C2(d) depending on thedimension d such that C1(d)nd\u22121 \u2264 #Hdn \u2264 C2(d)nd\u22121. This estimate agrees withour intuition, since each of the first d \u2212 1 coordinates of the vector ~r \u2208 Hdn canbe chosen in n different ways. The last coordinate is determined immediately oncethe first d \u2212 1 coordinates have been selected, since such ~r has prescribed l1-norm,|~r| = n.A function defined on [0, 1]d of the formf~r =\u2211R\u2208R~r\u03b1RhR with \u03b1R = \u00b11is called an ~r-function with parameter ~r \u2208 Hdn. These \u00b11-valued functions are alsoknown in the literature as generalized Rademacher functions. It can easily be verifiedthat ~r-functions are orthonormal with respect to the L2-norm.In all dimensions d, | \u00b7 | stands for the d-dimensional Lebesgue measure. Thesignum function of a real number x is defined bysgn(x) =\uf8f1\uf8f2\uf8f31, if x \u2265 0\u22121, if x < 0 . (1.7)1.2 Statement of the small ball inequalityAs mentioned earlier, the small ball inequality is a functional inequality concerningthe lower bound of the supremum norm of a sum of Haar functions supported ondyadic rectangles of a fixed volume. Such sums of Haar functions are also referred7to as hyperbolic sums.For brevity, let Adn = {R \u2208 Dd : |R| = 2\u2212n}, i.e., the set of all dyadic rectangleswhose d-dimensional volume is equal to 2\u2212n.Conjecture 1.1. (The small ball conjecture):In all dimensions d \u2265 2, for any choice of the coefficients \u03b1R, and all integersn \u2265 1, one has the following inequalitynd\u221222\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e\u2265 Cd2\u2212n\u2211R\u2208Adn|\u03b1R|, (1.8)where Cd is a constant depending only on the dimension d, not on n or the choice of{\u03b1R}.The critical feature of inequality (1.8) is the precise exponent of n occurring onthe left side. If we replace nd\u221222 with nd\u221212 , then we get the so-called \u201ctrivial bound\u201d:nd\u221212\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225L2\u2265 Cd2\u2212n\u2211R\u2208Adn|\u03b1R|. (1.9)If we compare the conjectured small ball inequality with inequality (1.9), noting thatthe L\u221e([0, 1]d)-norm is at least as large as the L2([0, 1]d)-norm, then we can see thatthe conjectured small ball inequality is better than (1.9) by a factor of\u221an.Proof of (1.9). We calculate the L2-norm of Hn =\u2211|R|=2\u2212n \u03b1RhR. Using the orthog-onality of Haar functions and the fact that \u2016hR\u201622 = 2\u2212n, Parseval\u2019s identity givesus\u2016Hn\u201622 =\u2211|R|=2\u2212n|\u03b1R|2\u2016hR\u201622 = 2\u2212n\u2211|R|=2\u2212n|\u03b1R|2\u2265 C2d2\u2212n(\u2211|R|=2\u2212n |\u03b1R|)22nnd\u22121,where in the last line, we used the Cauchy-Schwarz inequality and the fact that#{R \u2208 Dd : |R| = 2\u2212n} = 2n#Hdn ' 2nnd\u22121. Now, taking the square root of both8sides, we get\u2016Hn\u20162 \u2265 Cd( \u2211|R|=2\u2212n|\u03b1R|)2\u2212nn\u2212d\u221212 ,which is exactly (1.9).There is a simplified form of the small ball inequality which is of interest as well.If we consider coefficients {\u03b1R} from the set {\u22121, 1} in the small ball conjecture,then we obtain the signed small ball conjecture, i.e.,Conjecture 1.2. (The signed small ball conjecture) If \u03b1R = \u00b11 for every R \u2208 Adn,then we have \u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e\u2265 Cdn d2 , (1.10)where Cd is a constant depending only on the dimension d, not on n or the choice of{\u03b1R}.Note that when {\u03b1R} \u2282 {\u22121, 1}, then the trivial bound in (1.9) is given by\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225L2\u2265 Cdn d\u221212 . (1.11)In Section 4.2 we will prove that the conjectured signed small ball inequality, in-equality (1.10), is sharp. In other words, we will prove that there exists a positiveconstant Cd > 0 and a choice of the coefficients {\u03b1R} \u2282 {\u22121, 1} such that\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e\u2264 Cdn d2 .In particular, this fact will imply that if inequality (1.8) holds, then it is sharp aswell.91.3 Main resultsAccording to inequality (1.10), if one wants to attempt to solve the signed small ballconjecture, then one has to obtain a gain over the trivial bound given in (1.11), nd\u221212 ,by a factor of\u221an so that\u221an\u00b7n d\u221212 is the optimal lower bound in the signed small ballinequality. The best known gain over the trivial bound for general coefficients {\u03b1R} \u2282{\u22121, 1} for all d \u2265 3 was obtained by D. Bilyk, M. Lacey, and A. Vagharshakyanin [5]. We will state their result precisely in Chapter 2. In this work, we follow adifferent approach to that taken in [5]. Here, we impose a restriction on the collectionof coefficients and aim to get the optimal bound appearing in the signed small ballconjecture for this restricted collection of coefficients.To better understand the restriction we impose on the coefficients, let us observethat, for given n \u2265 1 and d \u2265 3, any choice of coefficients {\u03b1R}R\u2208Adn \u2282 {\u22121, 1} canbe generated by a map\u03b1\u2217 : Adn \u2192 {\u22121, 1}. (1.12)such that \u03b1\u2217(R) = \u03b1R for all R \u2208 Adn. SinceAdn \u2282n\u22c3r1=0A1r1 \u00d7n\u22c3r1=0Ad\u22121n\u2212r1 ,the map \u03b1\u2217 in (1.12) can be viewed as a restriction of a, not necessarily unique, map\u03b1 :n\u22c3r1=0A1r1 \u00d7n\u22c3r1=0Ad\u22121n\u2212r1 \u2192 {\u22121, 1}.The signed small ball conjecture states that inequality (1.10) must hold for thecollection of coefficients {\u03b1R := \u03b1(R) = \u03b1\u2217(R), R \u2208 Adn}. We will be interested incollections of coefficients generated by maps of the form:\u03b1 :n\u22c3r1=0A1r1 \u00d7n\u22c3r1=0Ad\u22121n\u2212r1 \u2192 {\u22121, 1}10with\u03b1(R1 \u00d7R2 \u00d7 \u00b7 \u00b7 \u00b7 \u00d7Rd) = \u03b11(R1) \u00b7 \u03b12(R2 \u00d7 \u00b7 \u00b7 \u00b7 \u00d7Rd), (1.13)where\u03b11 :n\u22c3r1=0A1r1 \u2192 {\u22121, 1} and \u03b12 :n\u22c3r1=0Ad\u22121n\u2212r1 \u2192 {\u22121, 1}.With a slight abuse of notation, we denote the values of a map in (1.13) by \u03b1R =\u03b1R1\u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd , and we letAsplit ={\u03b1R : R \u2208 Adn, \u03b1R = \u03b1R1 \u00b7 \u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd with \u03b1R1 , \u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd \u2208 {\u00b11}}.We will show that for any integer n \u2265 1, d \u2265 3 and any choice of coefficients in Asplit,Conjecture 1.2 holds, i.e.,Theorem 1.3. For all integers n \u2265 1, d \u2265 3 and all choices of coefficients {\u03b1R}R\u2208Adn \u2282Asplit we have that \u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e\u2265 Cdn d2 , (1.14)where Cd is a constant depending only on the dimension d, not on n or the choice of{\u03b1R}.Moreover, the above inequality is sharp.Theorem 1.4. For any integer n \u2265 1 and d \u2265 3, there exists a collection of coeffi-cients {\u03b1R}R\u2208Adn \u2282 Asplit such that\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e\u2264 Cdn d2 ,where Cd is a constant depending only on the dimension d.We have already seen that the L2-norm of the signed hyperbolic sum, Hn =\u2211|R|=2\u2212n \u03b1RhR with {\u03b1R} \u2282 {\u22121, 1}, can be calculated explicitly up to a multiplica-tive constant by using the orthogonality of Haar functions, i.e., \u2016Hn\u20162 ' n d\u221212 . It11turns out that, for any p < \u221e, the Lp-norm of the signed hyperbolic sum has thesame behaviour as its L2-norm, i.e., \u2016Hn\u2016p ' n d\u221212 , where the implicit constantsdepend on p and on the dimension d. We will see that this fact is a consequence ofthe Littlewood-Paley inequalities. In order to gain insight and to better understandthe signed small ball conjecture, we will also look for bounds of signed hyperbolicsums in function spaces which lie between Lp and L\u221e. Exponential Orlicz spaces,denoted by exp(La), are one type of such function spaces. These will be defined andfurther discussed in Chapter 7. Exponential Orlicz spaces have already appearedin the study of problems related to the small ball inequality. Based on the work ofBilyk et al. in [6], in which they prove an analogous result in dimension 2 for thediscrepancy function instead of hyperbolic sums, we conjecture the following:Conjecture 1.5. If \u03b1R = \u00b11 for every R \u2208 Adn, then for d \u2265 2 and n > 1 we have\u2225\u2225\u2225\u2225\u2225\u2225\u2211|R|=2\u2212n\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225exp(La)\u2265 C(d, a)n d\u221212 + 12\u2212 1a , 2 \u2264 a <\u221e, (1.15)where C(d, a) is a constant depending only on the dimension d and the scale ofintegrability a, not n or the choice of {\u03b1R}.The validity of this conjecture remains unknown in d \u2265 3. However, adjustingthe arguments given by V. Temlyakov in [49] shows that the conjecture is true indimension d = 2. We will also verify a weaker version of Conjecture 1.5 in alldimensions d \u2265 3 in Chapter 7. In particular, we will prove the following theorem:Theorem 1.6. For any integer n > 1 and any choice of coefficients \u03b1R \u2208 Asplit, wehave \u2225\u2225\u2225\u2225\u2225\u2225\u2211|R|=2\u2212n\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225exp(La)\u2265 C(d, a) n d\u221212 + 12\u2212 1a , 2 \u2264 a <\u221e (1.16)for d \u2265 3.Theorem 1.3 and Theorem 1.4 showcase the usage of the collection Asplit in prov-ing optimal bounds for the conjectured signed small ball inequality. Therefore, it12is natural to ask whether the conjectured inequalities (1.10) and (1.15) hold understructural assumptions on the coefficients. Such questions have been discussed inChapter 8.13Chapter 2Recent history of the small ballinequalityIn this chapter, we focus on outlining known results pertaining to the resolution ofthe small ball conjecture.In dimension d = 2, the small ball conjecture was first proved by M. Talagrand[47] in 1994, and Temlyakov [49] gave a second proof in 1995. However, in moregeneral dimensions d \u2265 3, the small ball conjecture remains unsolved.Remark 2.1. In what follows, the expression A & B means that there is a constantK such that A \u2265 KB. In our setting, K does not depend on n or {\u03b1R}.The first improvement over the trivial bound in higher dimensions was given indimension d = 3 by J. Beck [2]. In particular, the lower bound he obtained wasbetter than the trivial bound by a factor proportional to (log n)\u22121. In 2008, a bodyof work authored by Bilyk, Lacey, and Vagharshakyan [3, 4] made significant progresstoward understanding the small ball inequality. They proved the following theorem:Theorem 2.2. For d \u2265 3, there exists 0 < \u03b7(d) < 12such that, for all choices ofcoefficients \u03b1R and all non-negative integers n \u2265 1,nd\u221212\u2212\u03b7(d)\u2225\u2225\u2225\u2225\u2225 \u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e& 2\u2212n\u2211R\u2208Adn|\u03b1R|. (2.1)14In the proof of Theorem 2.2, no specific values of \u03b7(d) are given, though it is clearfrom the method of proof that \u03b7(d) must be small. The small ball conjecture saysthat (2.1) should hold with \u03b7 = 12.The proofs of the two-dimensional small ball inequality given by Talagrand andTemlyakov were fundamentally different. Temlyakov\u2019s proof used a duality argumentwhereas Talagrand\u2019s used a combinatorial argument. Here we give a rough outline ofTemlyakov\u2019s duality argument as similar techniques have been employed in provingTheorem 1.3 and Theorem 1.6.To begin the argument, Temlyakov constructed a function, called a test function,\u03a8 \u2208 L1([0, 1]2) such that\u2016\u03a8\u20161 \u2264 C, (2.2)where C is an absolute constant which does not depend on the choice of the coeffi-cients {\u03b1R} or the integer n \u2265 1. He further required that\u2329\u03a8,\u2211|R|=2\u2212n\u03b1RhR\u232a:=\u222b[0,1]d\u03a8 \u00b7\u2211|R|=2\u2212n\u03b1RhRdx \u2265 C2\u2212n\u2211|R|=2\u2212n|\u03b1R| (2.3)for all coefficients {\u03b1R} and every n \u2265 1. He then deduced the two-dimensionalsmall ball inequality from a simple application of Ho\u00a8lder\u2019s inequality to (2.3) whichexploited (2.2).The structure of the test function that Temlyakov constructed restricted theproof\u2019s validity to the two-dimensional small ball inequality. In particular, the testfunction that Temlyakov constructed was in the form of a Riesz product, i.e.,\u03a8 =n\u220fk=0(1 + fk), (2.4)where fk is the two-dimensional ~r-function defined byfk := f(k,n\u2212k) =\u2211R\u2208R~rsgn(\u03b1R)hR, ~r = (k, n\u2212 k).15In dimension d = 2, the product of Haar functions supported on dyadic rectangles ofa fixed volume is again a Haar function. This property, known as the Product Rule,allows one to write \u03a8\u22121 as a sum of Haar functions, and verify that \u03a8 satisfies (2.2)and (2.3). The precise formulation of the Product Rule is given in Lemma 3.23, statedin Section 3.3. Unfortunately, this property does not hold in dimensions d \u2265 3, andso Temlyakov\u2019s proof only applies in the two-dimensional case. Temlyakov\u2019s proofwill be presented in more detail in Chapter 4.The absence of the Product Rule in higher dimensions makes the general smallball conjecture a very challenging problem. However, Beck [2] and Bilyk, Lacey andVagharshakyan [3, 4] successfully modified Temlyakov\u2019s approach to yield partialresults in higher dimensions. To adapt Temlyakov\u2019s duality approach, they insteadconsidered the function\u03a8 =q\u220ft=1(1 + \u03c1\u02dcFt), (2.5)where q > 0 is a constant depending on n, \u03c1\u02dc > 0 is a constant depending on q, andFt =\u2211~r\u2208Atf~r. (2.6)Here, At is the collection of vectors whose first-coordinate depends on t in such away that {At}qt=1 partitions Hdn, i.e.,q\u2294t=1At = Hdn.Expanding the product in (2.5) allowed them to write\u03a8 = 1 + \u03a8sd + \u03a8q, (2.7)where \u03a8sd is a sum of a product of ~r-functions for which the Product Rule can beapplied, and \u03a8q is a sum of a product of ~r-functions for which the Product Rule fails.They then employed the Littlewood-Paley inequalities to obtain estimates on the Lp-norm of a product of ~r-functions for which the Product Rule fails. Combining these16estimates with combinatorial arguments, Bilyk, Lacey and Vagharshakyan provedthat\u2016\u03a8sd\u20161 \u2264 Cfor q = n\u03b7. As a result, they obtained\u3008Hn,\u03a8sd\u3009 =\u222b[0,1]dHn(x)\u03a8sd(x)dx&qn\u2212 d\u221212 \u00b7 2\u2212n\u2211R\u2208Adn|\u03b1R|.(2.8)Finally, they used Ho\u00a8lder\u2019s inequality to finish the proof of Theorem 2.2. Thistheorem gives an improvement over the trivial bound by a factor of q. Before theirwork, Beck had already used probabilistic and combinatorial methods to prove theL1-boundedness of \u03a8sd in dimension d = 3 for a value of q which was logarithmic inn and smaller than n\u03b7. Beck\u2019s work heavily influenced [3, 4].As mentioned in Chapter 1, the complete resolution of the small ball conjec-ture seems to be a difficult problem, but the signed small ball conjecture, stated asConjecture 1.2 on page 9, is thought to be more manageable, due to the additionalsimplifying assumptions on the coefficients {\u03b1R}. In the case of the signed small ballinequality, the trivial bound reduces to\u2225\u2225\u2225\u2225\u2225\u2225\u2211~r\u2208Hdnf~r\u2225\u2225\u2225\u2225\u2225\u2225L2& n d\u221212 , (2.9)whereHn =\u2211R\u2208Adn\u03b1RhR =\u2211~r\u2208Hdnf~r.The trivial bound can be obtained by using the orthogonality of generalizedRademacher functions and the fact that #Hdn, the cardinality of the set Hdn, is oforder nd\u22121. The exponent of the integer n appears naturally in the trivial bound(2.9). This comes from the volume constraint on dyadic rectangles, |R| = 2\u2212n, orequivalently |~r| = n, which reduces the number of \u201cfree\u201d parameters in the vector17~r \u2208 Hdn to d \u2212 1. Therefore, the total number of terms in the sum is of order nd\u22121.In [5], Bilyk, Lacey, and Vagharshakyan explicitly quantified the improvement overthe trivial bound, i.e., they proved the signed version of the Theorem 2.2, explicitlyproviding the range of values for \u03b7. In particular, they showed that, for every \u000f > 0,\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e\u2265 Cd \u00b7 n d\u221212 + 18d\u2212\u000f, (2.10)where \u03b1R \u2208 {\u00b11}. Under an additional assumption on the length of the first sideof the dyadic rectangle R, Bilyk, Lacey, I. Parissis and Vagharshakyan [7] made animprovement over (2.10) when d = 3. Specifically, they showed that\u2225\u2225\u2225\u2225\u2225 \u2211|R|=2\u2212n|R1|\u22652\u2212n\/2\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u221e& n 98 , (2.11)where \u03b1R \u2208 {\u00b11}, and the sum is taken over all dyadic rectangles R = R1\u00d7\u00b7 \u00b7 \u00b7\u00d7Rd \u2208Adn with the length of the first side R1 of R to be bounded below by 2\u2212n\/2.18Chapter 3Connections and applicationsAs indicated in the introduction, the small ball inequality lies at the interface ofmany areas of mathematics. This chapter, and the next, is devoted to a detailedanalysis of these connections. The material in these chapters has been adapted fromthe excellent survey written by Bilyk [10].3.1 Connection to probability theoryFor convenience, we start by recalling some definitions from probability theory.Definition 3.1. (a) If T is an arbitrary index set, a centred Gaussian process,g = {gt ; t \u2208 T}, is a stochastic process with the property that, for any positiveinteger m and for every choice of t1, . . . , tm \u2208 T , (gt1 , . . . , gtm) has a multivariatenormal distribution. That is, the joint density function of (gt1 , . . . , gtm) is givenbyf(~x) = {(2pi)m det(V)}\u22121\/2 exp{\u2212 12~xV\u22121~x\u2032}, ~x = (x1, . . . , xm) \u2208 Rm,where V = (vij), vij = E(gtigtj), is the m\u00d7m covariance matrix depending ont1, . . . , tm, and ~x\u2032 is the transpose of the vector ~x.(b) A Brownian sheet, B, is a centred multi-parameter Gaussian process which19satisfies the covariance relationE(B(~s )B(~t ))=d\u220fj=1min(sj, tj),for ~s,~t \u2208 T := [0, 1]d, with ~s = (s1, . . . , sd) and ~t = (t1, . . . , td).The name \u201csmall ball inequality\u201d comes from probability theory, where one isinterested in determining the exact asymptotic behaviour of the small deviationprobability \u03c6(\u03b5) := \u2212 logP(\u2016B\u2016L\u221e([0,1]d) < \u03b5) as \u03b5\u2192 0+. In general, research in smalldeviation problems aims to uncover the asymptotic behaviour of the probability thata random process is accumulated in a small ball around the mean of the process insome norm.It is a known fact that in dimension d = 1 (when B is Brownian motion),lim\u03b5\u21920+\u03c6(\u03b5)\u03b5\u22122 =pi28, see [19]. In higher dimensions, the situation is more difficultand subtle, though upper bounds of the asymptotic behaviour of \u03c6(\u03b5) as \u03b5\u2192 0+ arecompletely understood. In particular, for all d \u2265 2 and \u03b5 > 0 small,\u03c6(\u03b5) \u2264 C\u03b52(log1\u03b5)2d\u22121. (3.1)This inequality was proved by T. Dunker, T. Ku\u00a8hn, M. Lifshits and W. Linde [18].Lower bounds of \u03c6(\u03b5) for small \u03b5 > 0 are not as well-understood as the upperbounds, though it is believed that they should have the same form.Conjecture 3.2. If d \u2265 2 and \u03c6(\u03b5) is the small deviation probability associated tothe Brownian sheet B, then for small \u03b5 > 0,C \u2032\u03b52(log1\u03b5)2d\u22121\u2264 \u03c6(\u03b5) \u2264 C\u03b52(log1\u03b5)2d\u22121, (3.2)where C and C \u2032 are absolute constants.Note that the right side of inequality (3.2) is equivalent to (3.1). Conjecture 3.2was shown to be true by Talagrand [47] in the two-dimensional case but remains20open in higher dimensions d \u2265 3.Remark 3.3. The two-sided inequality in Conjecture 3.2 is usually expressed morecompactly by writing \u03c6(\u03b5) ' 1\u03b52(log 1\u03b5)2d\u22121as \u03b5\u2192 0+.Though the precise lower bound appearing in Conjecture 3.2 has not been proven,there is a well-known lower bound on the L2-norm in dimensions d \u2265 2. Morespecifically, E. Csa\u00b4ki [17] has shown thatC\u03b52(log1\u03b5)2d\u22122\u2264 \u2212 logP(\u2016B\u2016L2([0,1]d) < \u03b5). (3.3)In fact, since \u2016B\u2016L2 \u2264 \u2016B\u2016L\u221e , we have\u03c6(\u03b5) \u2265 C\u03b52(log1\u03b5)2d\u22122, (3.4)where C is an absolute constant. Note that the lower bound of \u03c6(\u03b5) in (3.4) is smallerthan the lower bound stated in Conjecture 3.2 by a factor of log 1\u03b5for sufficiently small\u03b5 > 0 .There is a continuous analogue of the conjectured small ball inequality, which westate below, and we will see that the validity of this continuous analogue implies thetruth of Conjecture 3.2. Recall that D\u2217 = D \u222a {[\u22121, 1]}.Conjecture 3.4. In dimensions d \u2265 2, there exists an orthonormal basis of L2([0, 1]d),{uR}, enumerated using the index set Dd\u2217, such that, for any choice of coefficients\u03b1R,nd\u221222\u2225\u2225\u2225\u2225\u2225\u2225\u2211|R|=2\u2212n\u03b1R\u03b7R\u2225\u2225\u2225\u2225\u2225\u2225\u221e& 2\u2212 3n2\u2211|R|=2\u2212n|\u03b1R|, (3.5)where\u03b7R(x1, \u00b7 \u00b7 \u00b7 , xd) :=\u222b x10\u00b7 \u00b7 \u00b7\u222b xd0uR(y1, . . . , yd)dy1 \u00b7 \u00b7 \u00b7 dyd, (x1, \u00b7 \u00b7 \u00b7 , xd) \u2208 [0, 1]d.Conjecture 3.4 remains open in dimensions d \u2265 3, but Talagrand\u2019s [47] proof ofConjecture 3.2 in dimension d = 2 came from proving Conjecture 3.4 in d = 2. In21his work, he constructed an orthonormal basis {uR} of L2([0, 1]2), and correspondingfunctions \u03b7R, from a specific orthonormal basis {uI} of L2([0, 1]), and the corre-sponding functions \u03b7I . More precisely, in dimension 1, he divided a dyadic intervalI = [m \u00b7 2\u2212k, (m + 1) \u00b7 2\u2212k) into 4 successive dyadic subintervals of I of length 14|I|,i.e.,I =4\u22c3j=1Ij, Ij = [(4m+ j \u2212 1) \u00b7 2\u2212k\u22122, (4m+ j) \u00b7 2\u2212k\u22122), j = 1, . . . , 4. (3.6)Note that the centre of the dyadic interval I, denoted by c(I), is equal to(4m+ 2) \u00b7 2\u2212k\u22122. He then considered the functionuI(x) =1|I| 12 (\u22121I1(x) + 1I2(x) + 1I3(x)\u2212 1I4(x)), (3.7)and after a simple integration over a dyadic interval I, he showed that\u03b7I(x) =\uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3\u2212(x\u2212 (4m+ 2)2\u2212k\u22122 + 2\u2212k\u22121)|I|\u2212 12 , x \u2208 I1(x\u2212 (4m+ 2)2\u2212k\u22122)|I|\u2212 12 , x \u2208 I2 \u222a I3(2\u2212k\u22121 \u2212 (x\u2212 (4m+ 2)2\u2212k\u22122))|I|\u2212 12 , x \u2208 I40, otherwise.(3.8)The function \u03b7 can be rewritten in the equivalent form \u03b7I(x) = |I| 12w(x\u2212c(I)|I|), wherec(I) is the centre of the dyadic interval I. Here, w is a continuous non-constantfunction supported on the interval[\u221212, 12]with mean zero. The explicit formula ofw is given byw(x) =\uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3\u2212(x+ 12), \u221212\u2264 x \u2264 \u221214x, \u221214\u2264 x \u2264 14\u2212(x\u2212 12), 14\u2264 x \u2264 120, otherwise.22Figure 3.1: The function w used in the proof of the two-dimensional small ballprobability conjecture by Talagrand.In dimensions d \u2265 2, to a dyadic rectangle R = R1\u00d7\u00b7 \u00b7 \u00b7\u00d7Rd \u2208 Dd\u2217, he associatedthe functionuR(y1, . . . , yd) = uR1(y1) \u00b7 \u00b7 \u00b7uRd(yd) (3.9)so that {uR} formed an orthonormal basis for L2([0, 1]d). For this special choice of{uR}, \u03b7R =\u220fdj=1 \u03b7Rj(xj) where \u03b7Rj(xj) = |Rj|12w(xj\u2212c(Rj)|Rj |)and c(Rj) is the centreof the dyadic interval Rj, j = 1, . . . , d. Note that \u03b7R = 2\u2212n\/2\u220fdj=1 w(xj\u2212c(Rj)|Rj |)whenR \u2208 Adn.Proposition 3.5. Conjecture 3.4 implies Conjecture 3.2.The proof of this proposition has been adapted from [10].Proof of Proposition 3.5. The proof relies on Levy\u2019s [35] representation of a Brown-23ian sheet, B, in terms of a linear combination of independent N (0, 1) random vari-ables, i.e.,B(~s ) =\u2211k\u2208N\u03b3k\u03b7k(~s ), ~s \u2208 [0, 1]d, in distribution, (3.10)where {\u03b3k} are the independent N (0, 1) random variables,\u03b7k(~s ) =\u222b s10\u00b7 \u00b7 \u00b7\u222b sd0uk(y1, . . . , yd)dy1 \u00b7 \u00b7 \u00b7 dyd, ~s = (s1, \u00b7 \u00b7 \u00b7 , sd) \u2208 [0, 1]d,and {uk} is any orthonormal basis of L2([0, 1]d). Since there are countably manydyadic rectangles in [0, 1]d, there is a bijection between N and Dd\u2217. By a slight abuseof notation, we will suppress the bijection and now write \u03b3R and \u03b7R instead of \u03b3kand \u03b7k, respectively. We decompose the Brownian sheet B into two pieces:B = X~t + Y~t,whereX~t =\u2211|R|=2\u2212n\u03b3R(\u03c9)\u03b7R(~t ), Y~t =\u2211|R|6=2\u2212n\u03b3R(\u03c9)\u03b7R(~t ),with ~t \u2208 [0, 1]d. Note that X~t and Y~t are independent for each ~t \u2208 [0, 1]d. We willchoose the precise value of the integer n later. By Anderson\u2019s lemma, stated inLemma 3.6 below after the proof of this proposition, we get thatP(\u2016B\u2016L\u221e([0,1]d) < \u03b5) \u2264P\uf8eb\uf8ed\u2225\u2225\u2225\u2225 \u2211|R|=2\u2212n\u03b3R\u03b7R\u2225\u2225\u2225\u2225\u221e< \u03b5\uf8f6\uf8f8\u2264P\uf8eb\uf8edCn\u2212 d\u221222 2\u2212 3n2 \u2211|R|=2\u2212n|\u03b3R| < \u03b5\uf8f6\uf8f8\u2264P( \u2211|R|=2\u2212n|\u03b3R| < A),where A = \u03b5Cnd\u221222 23n2 and C is the implicit constant appearing in (3.5) which dependson the dimension d. Here, the conjectured small ball inequality (3.5) was used to24get the second inequality. Now, by applying Chebyshev\u2019s inequality, we can estimatethe distribution function of the sum of the absolute values of iid Gaussians {\u03b3R} asP( \u2211|R|=2\u2212n|\u03b3R| < A)\u2264 eAEe\u2212\u2211|R|=2\u2212n |\u03b3R|. (3.11)To calculate the quantity on the right side of (3.11), we first recall that the formulafor the moment generating function of the random variable |N (0, 1)| is given byM(t) = Ee|N (0,1)|t. A straightforward but tedious calculation then shows that M(t) =2et22\u222b t\u2212\u221ee(\u22121\/2)x2\u221a2pidx. Also, note that M(\u22121) \u2264 2e 12 \u00b7 0.16 \u2264 e 12 e\u2212C0 , where C0 is anabsolute constant larger than 1\/2. Note that such a constant can be found, since0.32 < e\u22121\/2 and e\u2212x is continuous at x = 1\/2. This, together with the fact that {\u03b3R}is a collection of K iid random variables, where K := #{R \u2208 Dd : |R| = 2\u2212n} =2n \u00b7#Hdn \u2265 Cd2nnd\u22121, gives us thateAEe\u2212\u2211|R|=2\u2212n |\u03b3R| = eA(M(\u22121))K \u2264 exp(1C\u03b5nd\u221222 23n2 \u2212 C12nnd\u22121), (3.12)where C1 =(C0 \u2212 12)Cd is a positive constant which depends only on the dimensiond. Now, for small \u03b5 > 0, if we let n be a positive integer satisfying12CC12\u2212n+12 (n+ 1)d2 \u2264 \u03b5 \u2264 12CC12\u2212n2 nd2 , (3.13)then the right side of inequality (3.13) implies that1C\u03b5nd\u221222 23n2 \u2264 12C12nnd\u22121. (3.14)This means thatexp(1C\u03b5nd\u221222 23n2 \u2212 C12nnd\u22121)\u2264 exp(\u2212 12C12nnd\u22121). (3.15)25Note that the left side of (3.13) implies that log 1\u03b5. n. Hence,P( \u2211|R|=2\u2212n|\u03b3R| < A)\u2264 exp(\u2212 12C12nnd\u22121)\u2264 exp(\u2212 12C1(2n2 )2ndn2d\u22121)\u2264 exp(\u2212 C\u2032\u03b52(log1\u03b5)2d\u22121),where C \u2032 is a positive constant. Therefore,P(\u2016B\u2016L\u221e([0,1]d) < \u03b5) \u2264 exp(\u2212 C\u2032\u03b52(log1\u03b5)2d\u22121).That is,\u03c6(\u03b5) & 1\u03b52(log1\u03b5)2d\u22121.The following lemma, which was used in the proof of Proposition 3.5, is a standardfact in probability theory. We state it here without proof.Lemma 3.6. (Anderson\u2019s Lemma [1]). Let Xt, Yt, t \u2208 T be independent centeredGaussian random processes. ThenP(supt\u2208T|Xt + Yt| < c)\u2264 P(supt\u2208T|Xt| < c).3.2 Connection to approximation theoryWe begin this section by recalling some notions from approximation theory whichwill be used throughout.Definition 3.7. (a) Let K\u02c6 be a subset of a metric space (K, \u03c1). The set K\u02c6 is calledan \u03b5-net for K, if for every x \u2208 K there exists y \u2208 K\u02c6 such that \u03c1(x, y) \u2264 \u03b5.(b) Points y1, . . . , ym \u2208 K are called \u03b5-separated if \u03c1(yi, yj) > \u03b5 for all i 6= j.26Note that if (K, \u03c1) is a compact metric space, then K contains a finite \u03b5-net forevery \u03b5 > 0.Definition 3.8. Let K\u02dc be a compact subset of a metric space (K, \u03c1). Then thecovering number of the set K\u02dc, N(\u03b5, K\u02dc), is the minimum of the cardinalities of \u03b5-netsfor K\u02dc, i.e.,N(\u03b5, K\u02dc) := min{N : \u2203{xk}Nk=1 \u2282 K\u02dc, K\u02dc \u2282N\u22c3k=1B(xk, \u03b5)},where B(xk, \u03b5) = {x \u2208 K\u02dc : \u03c1(x, xk) \u2264 \u03b5} is the ball centred at xk of radius \u03b5. The\u03b5-entropy of the set K\u02dc is then defined byH(\u03b5, K\u02dc) := log(N(\u03b5, K\u02dc)).Definition 3.9. Let (K, \u03c1) be a compact metric space. Given \u03b5 > 0, we define thepacking number, M(\u03b5,K), of the set K to beM(\u03b5,K) = max{m \u2208 N : \u2203 \u03b5-separated points y1, . . . , ym \u2208 K}.The study of covering numbers and entropies goes back A. Kolmogorov. Around1950, Kolmogorov developed a strong interest in problems in information theory. In-spired by Shannon\u2019s ideas, he introduced the notion of a covering number to measurethe size of a set, A, in a metric space. From the point of view of information theory,the entropy of A has the following usage. Consider the elements of A as \u201cmessages\u201dencoded in binary strings of 0\u2019s and 1\u2019s. Suppose that \u03b5 > 0 is a number smallenough so that any message x in A can be recovered with reasonable accuracy fromsome x\u2032 in A with \u03c1(x, x\u2032) \u2264 \u03b5. One should think of x as the original and x\u2032 as thetransmitted message. The quantity N(\u03b5, A) then captures the economic yet effectivecollection of transmittable messages.Kolmogorov used the concept of covering numbers to give a simpler proof ofVitushkin\u2019s theorem on superposition of functions. To do this, he found an upperbound for the \u03b5-entropy of W r\u221e(In), the class of functions of n variables defined on27the unit cube In in Rn and having uniformly bounded partial derivatives up to orderr. In particular, he showed thatlog(N(\u03b5,W r\u221e(In))) \u2264 C(1\u03b5)n\/r.This was the starting point of investigating the \u03b5-entropy of compact sets in functionspaces. We refer the interested reader to [32] for a thorough description of thedevelopment and history of covering numbers.The \u03b5-entropy of a set A also has applications in approximation theory, specif-ically, linear approximations. To present these concepts in an abstract setting, wemust first introduce the notion of the n-width of a set, A, in a normed space, anothertool invented by Kolmogorov to measure the size of A. Let X be a normed spaceequipped with a norm \u2016 \u00b7 \u2016, and let A be a subset of X. The n-width of the set A,dn(A), is defined to bedn(A) := infLnsupx\u2208Ainf\u03be\u2208Ln\u2016x\u2212 \u03be\u2016,where the outer infimum is taken over all n-dimensional subspaces Ln \u2282 X. We willrefer to the quantity inf\u03be\u2208Ln \u2016x\u2212\u03be\u2016 as the deviation of the approximation of x by Ln,supx\u2208A inf\u03be\u2208Ln \u2016x\u2212 \u03be\u2016 as the deviation of the approximation of the set A by Ln, anddn(A) as the optimal deviation of the approximation of A. For small \u03b5 > 0, in orderto ensure that dn(A) < \u03b5, n must be at least as large as some simple lower bounddepending on \u03b5, and is, roughly, H(\u03b5, A). More precisely, G. Lorentz [36] showedthatn \u2265 H(e2\u03b5, A)\u2212 C02 + log 1\u03b5+ log[H(e2\u03b5, A)\u2212 C0] ,where C0 is an absolute constant and A is a compact set in a separable Banach spaceX. From the above inequality, we see that, in order to see how n depends on \u03b5 forsmall \u03b5 > 0, one needs to understand how H depends on \u03b5.Here, we survey some results concerning the asymptotic behaviour of \u03c3p(\u03b5) :=H(\u03b5, B(MW p)), the \u03b5-entropy of a specific class of functions with bounded mixedderivatives. As we will see, this asymptotic behaviour can be determined explicitly if28one can prove the continuous version of the small ball conjecture. This topic gaineda lot of attention after the connection between the entropy of this class of functionswith bounded mixed derivatives and small deviation probabilities was discovered (seeTheorem 3.12).Next, we define a specific integration operator that will be used to give the precisedefinition of our space of functions with bounded mixed derivatives.Definition 3.10. The integration operator Td, acting on functions defined on theunit cube [0, 1]d, is defined by(Tdf)(x1, . . . , xd) :=\u222b x10\u00b7 \u00b7 \u00b7\u222b xd0f(y1, . . . , yd)dy1 \u00b7 \u00b7 \u00b7 dyd. (3.16)Let B\u221e and B(Lp) denote the unit ball in L\u221e([0, 1]d) and Lp([0, 1]d) respectively.Then, MW p([0, 1]d) := Td(Lp([0, 1]d)) is the space of functions on [0, 1]d with mixedderivatives \u2202df\u2202x1\u2202x2\u00b7\u00b7\u00b7\u2202xd in Lp, which, throughout, we will equip with the sup-norm,\u2016 \u00b7 \u2016\u221e. We then define the subset B(MW p) \u2282MW p by B(MW p) := Td(B(Lp)). Todefine the covering number of the set B(MW p), we must first show that B(MW p)is a precompact set with respect to the L\u221e-norm.Proposition 3.11. The closure of the set B(MW p), B(MW p)\u2016\u00b7\u2016\u221e, is a compact setwith respect to the L\u221e-metric for 1 < p <\u221e and for all d \u2265 1.Proof. To conclude that B(MW p)\u2016\u00b7\u2016\u221eis a compact set, we will use the Arzela`-Ascolitheorem. To do this, we need to show thatTd(B(Lp)) \u2282 B\u221e(0, c) (3.17)for some positive constant c, where B\u221e(0, c) is the ball centred at zero of radius cmeasured in L\u221e-norm, and that the family of functions {Td(f)}f\u2208B(Lp) is uniformlyequicontinuous, i.e.,\u2200\u03b5 > 0, \u2203\u03b4 > 0 such that |Td(f)(~s )\u2212 Td(f)(~t )| < \u03b5,\u2200~s,~t \u2208 [0, 1]d with (3.18)29\u2016~s\u2212 ~t \u20162 :=( d\u2211i=1|si \u2212 ti|2)1\/2< \u03b4 and \u2200f \u2208 B(Lp).First, for any f \u2208 B(Lp), we have that|Td(f)(~t )| \u2264\u222b t10\u00b7 \u00b7 \u00b7\u222b td0|f(y)|dy \u2264 \u2016f\u2016p \u2264 1,for every ~t \u2208 [0, 1]d. This shows that (3.17) holds.Now we will prove condition (3.18). To do this, we will show that|Td(f)(~s )\u2212 Td(f)(~t )| \u2264d\u2211i=1|ti \u2212 si|1\/q \u2200~t, ~s \u2208 [0, 1]d and \u2200f \u2208 B(Lp), (3.19)where q = pp\u22121 is the dual exponent of p, as (3.18) follows from (3.19) immediately.Suppose d \u2265 1 and ~t, ~s \u2208 [0, 1]d with ti 6= si for i = 1, . . . d. For any such ~s and ~t, wedefine the following sets:I1 := {i \u2208 {1, . . . , d} : si > ti},I2 := {i \u2208 {1, . . . , d} : si < ti}.Note that these sets depend on the choice of ~s,~t \u2208 [0, 1]d, and that I1\u222aI2 = {1, . . . , d}.Let [~0, ~s ] :=\u220fdi=1[0, si] and [~0,~t ] :=\u220fdi=1[0, ti]. Note that the corner that is furthestfrom the origin of each of these d-dimensional boxes is given by ~s and ~t respectively,and that the intersection of these boxes is non-empty. In fact,[~0, ~s ]\u22c2[~0,~t ] =d\u220fi=1[0,min(ti, si)],wheremin(ti, si) = si, \u2200i \u2208 I2 and min(ti, si) = ti \u2200i \u2208 I1.30We have that[~0, ~s ] =([~0, ~s ]\u22c2[~0,~t ])\u22c3(\u22c3i\u2208I1Ai), (3.20)where\u22c3i\u2208I1 Ai is the complement of the set [~0, ~s ]\u22c2[~0,~t ] in [~0, ~s ], and each Ai dependson the choice of ~s and ~t. In fact, for i \u2208 I1, Ai = B1\u00d7\u00b7 \u00b7 \u00b7\u00d7Bd with Bj = [0, sj] \u2282 [0, 1],j 6= i, and Bi = (ti, si]. Moreover,[~0,~t] =([~0, ~s ]\u22c2[~0,~t ])\u22c3(\u22c3i\u2208I2A\u02dci), (3.21)where, for i \u2208 I2, A\u02dci = B\u02dc1\u00d7\u00b7 \u00b7 \u00b7\u00d7 B\u02dcd with B\u02dcj = [0, tj] \u2282 [0, 1], j 6= i, and B\u02dci = (si, ti].Note that the complement of the set [~0, ~s ]\u22c2[~0,~t ] in [~0,~t ] is\u22c3i\u2208I2 A\u02dci. Therefore,using relations (3.20), (3.21), and Ho\u00a8lder\u2019s inequality, we obtain|Td(f)(~s )\u2212 Td(f)(~t )| =\u2223\u2223\u2223\u2223 \u222b[~0,~s]f(y)dy \u2212\u222b[~0,~t]f(y)dy\u2223\u2223\u2223\u2223\u2264\u2211i\u2208I1\u222bAi|f(y)|dy +\u2211i\u2208I2\u222bA\u02dci|f(y)|dy\u2264\u2211i\u2208I1\u2016f\u2016p|ti \u2212 si|1\/q +\u2211i\u2208I2\u2016f\u2016p|ti \u2212 si|1\/q (\u2217)\u2264d\u2211i=1|ti \u2212 si|1\/q,(3.22)for any f \u2208 B(Lp), for all ~s,~t \u2208 [0, 1]d with si 6= ti, i = 1, . . . , d.From (3.22), we see that (3.19) holds for points ~s,~t \u2208 [0, 1]d with si 6= ti, i =1, . . . , d. Now, we will show that (3.19) holds for general points ~s,~t \u2208 [0, 1]d. Fix anytwo points ~s and ~t in [0, 1]d and consider the setI0 = {i \u2208 {1, . . . , d} : si = ti},i.e., I0 is the subset of indices in {1, . . . , d} for which the corresponding coordinates ofthe vectors ~s and ~t coincide. Let Ic0 denote the complement of the set I0 in {1, . . . , d}.31Now, letSI0 :=\u220fi\u2208I0[0, si], SIc0 :=\u220fi\u2208Ic0[0, si], and TIc0 =\u220fi\u2208Ic0[0, ti]be boxes corresponding to I0 and Ic0. We will let ySI0 , ySIc0, and yTIc0denote pointsbelonging to SI0 , SIc0 , and TIc0 respectively. Using Fubini\u2019s theorem, we get that|Td(f)(~s)\u2212 Td(f)(~t)| =\u2223\u2223\u2223\u2223 \u222bSI0[ \u222bSIc0f(y)dySIc0\u2212\u222bTIc0f(y)dyTIc0]dySI0\u2223\u2223\u2223\u2223\u2264\u222bSI0\u2223\u2223\u2223\u2223 \u222bSIc0f(y)dySIc0\u2212\u222bTIc0f(y)dyTIc0\u2223\u2223\u2223\u2223dySI0 .In order to estimate the absolute value of the difference of the two integrals in theabove inequality, we appeal to the case we considered previously. Now, for ySI0 \u2208 SI0 ,with the aid of (\u2217) in (3.22), we get that\u2223\u2223\u2223\u2223 \u222bSIc0f(y)dySIc0\u2212\u222bTIc0f(y)dyTIc0\u2223\u2223\u2223\u2223 \u2264\u2211i\u2208I1(\u222bSIc0|f(y)|pdySIc0)1\/p|ti \u2212 si|1\/q+\u2211i\u2208I2(\u222bTIc0|f(y)|pdyTIc0)1\/p|ti \u2212 si|1\/q.Here, I1 and I2 are subsets of Ic0 for which Ic0 = I1\u222aI2. Finally, integrating the aboveinequality with respect to ySI0 and applying Jensen\u2019s inequality gives|Td(f)(~s)\u2212 Td(f)(~t)| \u2264\u2211i\u2208I1\u2016f\u2016p|ti \u2212 si|1\/q +\u2211i\u2208I2\u2016f\u2016p|ti \u2212 si|1\/q\u2264 \u2016f\u2016p\u2211i\u2208Ic0|ti \u2212 si|1\/q \u2264d\u2211i=1|ti \u2212 si|1\/qfor any f \u2208 B(Lp) and any ~t, ~s \u2208 [0, 1]d. This completes the proof of the proposition.One branch of research in approximation theory is concerned with determiningthe asymptotic behaviour of \u03c3p(\u03b5) := log(N(\u03b5, p, d)) as \u03b5\u2192 0+, where32N(\u03b5, p, d) := min{N : \u2203{xk}Nk=1 \u2282 B(MW p), B(MW p) \u2282N\u22c3k=1(xk + \u03b5B\u221e)}is the covering number of the set B(MW p). Recall that this is the smallest num-ber of balls of radius \u03b5 needed to cover the set B(MW p), or, equivalently, the sizeof the smallest \u03b5-net of B(MW p). There is an interesting relation between thecovering number N(\u03b5, p, d) and the packing numberM(\u03b5, B(MW p)). Here, for con-venience, we will denote the packing number of B(MW p) by M(\u03b5, p, d) rather thanM(\u03b5, B(MW p)). In particular, it is easy to show thatM(2\u03b5, p, d) \u2264 N(\u03b5, p, d) \u2264M(\u03b5, p, d). (3.23)J. Kuelbs and V. Li [33] have established a deep connection between the asymp-totic behaviours of \u03c6(\u03b5) and \u03c3p(\u03b5). In particular, from their results, one can concludethe following:Theorem 3.12. Suppose d \u2265 2, b > 0, and \u03b5 > 0 is small. Then\u03c6(\u03b5) ' \u03b5\u22122(log1\u03b5)bif and only if \u03c32(\u03b5) ' \u03b5\u22121(log1\u03b5) b2.Here, the symbol \u201c ' \u201d is as in Remark 3.3. As mentioned in the previous section,the upper bound for \u03c6(\u03b5) has been proved for the special case when b = 2d\u2212 1. So,according to Theorem 3.12,\u03c32(\u03b5) \u2264 C\u03b5\u22121(log1\u03b5) 2d\u221212for small \u03b5 > 0, where C is an absolute constant. Thus, in the setting of approxima-tion theory, the small ball conjecture is equivalent to:33Conjecture 3.13. For all d \u2265 2 and \u03b5 > 0 small, we have that\u03c32(\u03b5) \u2265 C \u2032\u03b5\u22121(log1\u03b5) 2d\u221212,where C \u2032 is an absolute constant.Conjecture 3.13 was solved in dimension d = 2 by Kuelbs and Li [33]. They wereable to prove the conjecture in d = 2 by using Talagrand\u2019s result on the asymptoticbehaviour of \u03c6(\u03b5) in d = 2 and Theorem 3.12. Also, Temlyakov [50] gave a secondproof of Conjecture 3.13 in d = 2 by using the analogue of the small ball inequalityfor trigonometric polynomials. In particular, he showed that, in d = 2, for all1 < p <\u221e, and for \u03b5 > 0 small,\u03c3p(\u03b5) \u2265 Cp\u03b5\u22121(log1\u03b5) 32.Conjecture 3.13 remains open in d \u2265 3, though, using Theorem 3.12, it\u2019s not hardto see that Conjecture 3.13 would follow from a proof of Conjecture 3.4 in d \u2265 3.However, there is a way to deduce Conjecture 3.13 from Conjecture 3.4 withoutinvoking Theorem 3.12, which is what we do here.Proposition 3.14. If Conjecture 3.4 is true, then so is Conjecture 3.13.In order to prove Proposition 3.14, we will need a preliminary result. Considera bi-valued function \u03b2 : Adn \u2192 {\u22121, 1}, Adn = {R \u2208 Dd, |R| = 2\u2212n}. Let K = #Adn.Then, by defining \u03b2R := \u03b2(R) for R \u2208 Adn, we can associated \u03b2 with a vector~\u03b2 \u2208 {\u22121, 1}K whose \u201cRth\u201d coordinate is \u03b2R . It has already been noted that K =2n#Hdn ' 2nnd\u22121.Lemma 3.15. There is an integer n0 \u2265 1 such that, if n \u2265 n0 and K = 2n#Hdn,then there exists a set X \u2282 {\u22121, 1}K satisfying:1.\u2211|R|=2\u2212n |\u03b2\u2032R \u2212 \u03b2R| & K for any ~\u03b2 6= ~\u03b2\u2032 \u2208 X.2. log #X & K.34The proof of Lemma 3.15 will follow the proof of Proposition 3.14.Proof of Proposition 3.14. According to inequality (3.23), for a fixed small \u03b5 > 0,we need to produce a 2\u03b5-separated collection of functions F\u02dc = {F\u02dc~\u03b2} \u2282 B(MW 2) forwhich log #F\u02dc & 1\u03b5(log 1\u03b5) 2d\u221212.Now, to each such function \u03b2 or equivalently vector ~\u03b2 \u2208 {\u22121, 1}K , we can furtherassociate the function F\u02dc~\u03b2 given byF\u02dc~\u03b2 =c2n2 nd\u221212\u2211|R|=2\u2212n\u03b2R\u03b7R, (3.24)where c > 0 is a small constant and \u03b7R := TduR, as given in the statement ofConjecture 3.4. Moreover, F\u02dc~\u03b2 \u2208 B(MW 2) for every ~\u03b2 \u2208 {\u22121, 1}K . Indeed, since\u03b7R = TduR and {uR}R\u2208Adn is an orthonormal set of functions, we obtainF\u02dc~\u03b2 = Td(c2n2 nd\u221212\u2211|R|=2\u2212n\u03b2RuR),and \u2225\u2225\u2225\u2225 c2n2 nd\u221212\u2211|R|=2\u2212n\u03b2RuR\u2225\u2225\u2225\u222522=c22nnd\u221212n \u00b7#Hdn \u2264 1for c sufficiently small.Let X \u2282 {\u22121, 1}K be a set satisfying the conditions of Lemma 3.15. Then,F\u02dc = {F\u02dc~\u03b2}~\u03b2\u2208X is a 2\u03b5-separated collection of functions in B(MW 2) for whichlog #F\u02dc ' 1\u03b5(log1\u03b5) 2d\u221212as \u03b5\u2192 0+,for a suitable choice of n \u2265 n0. Indeed, let \u03b5 > 0 be any small number strictly lessthan 12n1\/20 2\u2212n0 , where n0 is the integer which appears in Lemma 3.15, and let n \u2265 n0be an integer for which n1\/22\u2212n \u2265 2\u03b5 \u2265 (n+ 1)1\/22\u2212(n+1). First, we show that F\u02dc is a2\u03b5-separated family with respect to the L\u221e-norm. From Conjecture 3.4, if ~\u03b2, ~\u03b2\u2032 \u2208 X35are such that ~\u03b2 6= ~\u03b2\u2032, then we get that\u2016F\u02dc~\u03b2 \u2212 F\u02dc~\u03b2\u2032\u2016\u221e =\u2225\u2225\u2225\u2225 c2n2 nd\u221212\u2211|R|=2\u2212n(\u03b2R \u2212 \u03b2\u2032R)\u03b7R\u2225\u2225\u2225\u2225\u221e&2\u2212n2 n\u2212 d\u221212 \u00b7 n\u2212 d\u221222 2\u2212 3n2\u2211|R|=2\u2212n|\u03b2R \u2212 \u03b2\u2032R|&n\u2212 2d\u221232 2\u22122n \u00b7 2n \u00b7 nd\u22121 = n1\/22\u2212n.(3.25)Hence, F\u02dc is a 2\u03b5-separated family with respect to the L\u221e-norm. Since2\u03b5 \u2265 (n+ 1)1\/22\u2212(n+1), we therefore have thatlog #F\u02dc = log #X & K & 2nnd\u22121& 2nn\u2212 12nd\u2212 12& 1\u03b5(log1\u03b5)d\u2212 12.Now, to prove Lemma 3.15, it will be useful to first introduce some notions fromcoding theory.In coding theory, a subset X \u2282 {\u22121, 1}K is called a binary code of length K, andwe can endow such a binary code with the Hamming distance. This is a distancefunction, dH , defined on X \u00d7X bydH(x, y) = #{j = 1, . . . , K : xj 6= yj},i.e., the Hamming distance between x and y is the number of components in which xand y do not coincide. It is easy to verify that (X, dH) is, in fact, a metric space. Theminimum Hamming distance of a binary code X is then defined to be the smallestHamming distance between its elements, minx,y\u2208X,x 6=ydH(x, y). So, in the languageof coding theory, to prove Lemma 3.15, we must find a binary code X of length36K = 2n#Hdn, such thatmin~\u03b2,~\u03b2\u2032\u2208X,~\u03b2 6=~\u03b2\u2032dH(~\u03b2, ~\u03b2\u2032) \u2265 K4, (3.26)andlog #X & K. (3.27)Since (3.26) tells us that\u2211|R|=2\u2212n |\u03b2\u2032R \u2212 \u03b2R| \u2265 2K4 & K for any ~\u03b2 6= ~\u03b2\u2032 \u2208 X, it\u2019sclear that (3.26) implies the first condition on X in Lemma 3.15.It is easy to show that there exists a binary code X satisfying (3.26). If thisbinary code X is equal to {\u22121, 1}K , then #X = 2K , and (3.27) follows immediately.In general, X will be a strict subset of {\u22121, 1}K , and so #X will be some fractionof 2K . Gilbert-Varshamov\u2019s theorem, stated below, provides us with a lower boundof this fraction, and this lower bound will be used in the proof of (3.27).Theorem 3.16. (Gilbert-Varshamov\u2019s theorem) Let A(m, k) denote the maximalsize of a binary code of length m and minimum Hamming distance at least k. ThenA(m, k) \u2265 2m\u2211k\u22121j=0(mj) .We will postpone the proof of Gilbert-Varshamov\u2019s theorem until we have com-pleted the proof of Lemma 3.15.Proof of Lemma 3.15. We will apply Gilbert-Varshamov\u2019s theorem with m = K andk = m4in order to show that log #X & K. First, note that there is a maximal codeX of length m and minimum Hamming distance m4, so Gilbert-Varshamov\u2019s theoremis applicable. Using Stirling\u2019s formula, m! ' 1\u221a2pim(me)m, we obtain(mm\/4)' 1\u221am(14)\u2212m4(34)\u22123m4.Applying Gilbert-Varshamov\u2019s theorem and recalling the fact that(mj)is an increas-37ing function of j when 0 \u2264 j < k, we get that#X \u2265 2m\u2211k\u22121j=0(mj) \u2265 2mk(mk) = 2mm\/4(mm\/4)' 1\u221am2m(14)m\/4(34)3m\/4& 1\u221amCm & bm\u221amCmbm& C\u02dcm,for a sufficiently large value of m, or, equivalently, for a big enough integer n. Here,C = 2(1\/4)1\/4(3\/4)3\/4 > 1, b is any number satisfying 1 < b < C, and C\u02dc = Cb. Hence,log #X & m.Finally, to conclude the section, we prove Gilbert-Varshamov\u2019s theorem.Proof of Gilbert-Varshamov\u2019s Theorem. First, note that, given any x \u2208 {\u22121, 1}m,#{y \u2208 X : dH(x, y) = j} =(mj)for j = 1, . . . ,m. If we let BH(x, k) denote theHamming ball centred at x of the radius k, i.e., BH(x, k) = {y | dH(x, y) < k},then #BH(x, k) =\u2211k\u22121j=0(mj). Let X be a maximal code of length m and minimumHamming distance at least k. Then\u22c3x\u2208X BH(x, k) = {\u22121, 1}m. As a result, weobtain2m = #\u22c3x\u2208XBH(x, k) \u2264\u2211x\u2208X#BH(x, k) = #Xk\u22121\u2211j=0(mj),which completes the proof.3.3 Connection to discrepancy theory3.3.1 Conjectures in discrepancy theoryDiscrepancy theory studies the deviation of the actual data from the ideal data. Oneof the most challenging problems in geometric discrepancy theory is, given a setPN = {p1, p2, . . . , pN} \u2282 [0, 1]d with cardinality #PN = N , to obtain a lower bound38on the discrepancy function,DN(x1, . . . , xd) = #{PN \u2229 [0, x1)\u00d7 \u00b7 \u00b7 \u00b7 \u00d7 [0, xd)} \u2212Nx1 \u00b7 \u00b7 \u00b7xd, (3.28)with respect to some norm. Relation (3.28) represents the difference between theactual and expected number of points in the box [0, x1) \u00d7 \u00b7 \u00b7 \u00b7 \u00d7 [0, xd). When thepoints p1, . . . , pN are uniformly distributed, these quantities are the same and soDN(x1, . . . , xd) = 0. Lower bounds of the discrepancy function with respect to theL\u221e-norm indicate the best case scenario for the distribution of points, i.e., howclose to uniformly distributed the distribution of the points can be. Motivation forstudying the behaviour of the discrepancy function comes from numerical integra-tion. In numerical integration, one attempts to approximate integrals of functionsf : [0, 1]d \u2192 R that cannot be computed analytically. Moreover, while doing this,one wants to minimize the error in their approximation. The Koksma-Hlawka in-equality connects the error, which appears in the process of the numerical integra-tion, with the discrepancy function. This inequality says that, for an N -point setP = {p1, . . . , pN} \u2282 [0, 1]d, and for all p \u2208 [1,\u221e],\u2223\u2223\u2223\u2223 \u222b[0,1]df(x)dx\u2212 1NN\u2211i=1f(pi)\u2223\u2223\u2223\u2223 \u2264 Vq(f)\u2016DN\u2016pN . (3.29)Here, q is the dual exponent of p, and Vq(f) is a quantity depending on f and p. Inthe one-dimensional case, d = 1, Vq(f) = \u2016f \u2032\u2016q when f is a smooth function. Theinterested reader can consult [31, 27] for the precise definition of Vq(f) in dimensionsd \u2265 2. From inequality (3.29), we see that an N -point set with smallest possibleLp-norm gives rise to the smallest error. Hence, we want to find an N -point setPN \u2282 [0, 1]d which minimizes the Lp-norm of DN . This motivates the study ofuniform lower bounds of the discrepancy function DN . The first quantitative resultin this direction, stated below, was obtained by K. F. Roth [39]. Note that we willprove Roth\u2019s theorem in the subsequent subsection.39Theorem 3.17. (Roth, 1954) In all dimensions d \u2265 2, for any N -point set PN \u2282[0, 1]d, one has\u2016DN\u20162 \u2265 Cd(logN) d\u221212 , (3.30)where Cd is an absolute constant that depends only on the dimension d.Furthermore, inequality (3.30) in Theorem 3.17 is sharp in the sense that thereexists an N -point set PN for which \u2016DN\u20162 \u2264 Cd(logN) d\u221212 . This fact was shownby Roth [40, 41], Chen [14] and Frolov [20]. The best known constant Cd in (3.30)was obtained by A. Hinrichs and L. Markhasin [26]. In general, the Lp-behaviourof the discrepancy function is well understood for 1 < p < \u221e. In fact, the Lp-behaviour of the discrepancy function is similar to the L2-behaviour, i.e., \u2016DN\u2016p \u2265C(d, p)(logN)d\u221212 , for 1 < p < \u221e. This was first shown by W. M. Schmidt [42,43]. He originally considered the values of p for which the dual exponent q is aneven integer using Roth\u2019s approach. One can also obtain such lower bounds on theLp-norm of DN for the full range of 1 < p < \u221e by using the Littlewood-Paleyinequalities. This will be explained in more detail after proving Roth\u2019s theoremand after introducing the Littlewood-Paley inequalities. In addition, examples of N -point sets with \u2016DN\u2016p \u2264 C(d, p)(logN) d\u221212 were constructed by Chen and Skriganov[15, 16] for p = 2 and by Skriganov [46] for p > 1. Determining the Lp-behaviour ofthe discrepancy functions for p = 1,\u221e is one of the most challenging open problemsin the field of geometric discrepancy.Conjecture 3.18. (L\u221e-Conjecture) For all d \u2265 2 and all PN \u2282 [0, 1]d with #PN =N ,\u2016DN\u2016\u221e \u2265 Cd(logN) d2 , (3.31)where Cd is an absolute constant depending only on the dimension d.Since L\u221e([0, 1]d) is continuously embedded in L2([0, 1]d), Roth\u2019s theorem imme-diately implies that\u2016DN\u2016\u221e \u2265 Cd(logN) d\u221212 . (3.32)Comparing (3.31) and (3.32), we see that there is a gap of order\u221alogN between theconjectured lower bound of the discrepancy function and the lower bound provided by40Roth\u2019s theorem. This gap was eliminated by Schmidt [42] in d = 2. More precisely,he showed that \u2016DN\u2016\u221e \u2265 C logN . One decade later, Hala\u00b4sz [25] gave another proofof the same result by combining the technique of using Riesz products with Roth\u2019sorthogonal function method. This is where the small ball inequality enters the areaof discrepancy theory. More precisely, Hala\u00b4sz\u2019s approach can be transferred to provethe small ball inequality in d = 2. After we prove Roth\u2019s theorem, we will analyzeHala\u00b4sz\u2019s proof and, in Chapter 4, show how his technique can be implemented toprove the small ball inequality in d = 2. The connection between problems indiscrepancy theory and the small ball inequality is based purely on the methods ofproof used. Currently, there is no indication that the results in discrepancy theorywill yield results concerning the small ball inequality, or vice versa.In 1989, by employing techniques using Riesz products, Beck [2] was able tomake some progress toward improving (3.32) in higher dimensions. Specifically, indimension d = 3, he showed that\u2016DN\u2016\u221e \u2265 C logN \u00b7 (log logN) 18\u2212\u03b5, for small \u03b5 > 0. (3.33)In other words, he improved the lower bound, logN , by a factor of order (log logN)18\u2212\u03b5,for some small \u03b5 > 0. In dimensions d \u2265 3, almost two decades later, Bilyk, Lacey,Vagharshakyan [3, 4] improved the bound, (logN)d\u221212 , by a factor of order (logN)\u03b8by showing\u2016DN\u2016\u221e \u2265 Cd(logN) d\u221212 +\u03b8, (3.34)where \u03b8 is a small positive constant depending on the dimension d. Their workwas based on Beck\u2019s work and improved Beck\u2019s result. It is still unknown whetherConjecture 3.18 is sharp, though examples of sets have been constructed by J. H.Halton and J. M. Hammersley [23, 24] which obey the upper bound: \u2016DN\u2016\u221e \u2264Cd(logN)d\u22121. This leads to another L\u221e-conjecture:Conjecture 3.19. For all d \u2265 2 and all point sets PN in [0, 1]d with #PN = N wehave\u2016DN\u2016\u221e \u2265 Cd(logN)d\u22121, (3.35)41where Cd is an absolute constant.The L1-behaviour of the discrepancy function is another related open problem.Conjecture 3.20. (L1-Conjecture) For all d \u2265 2 and all N -point sets in [0, 1]d,\u2016DN\u20161 \u2265 Cd(logN) d\u221212 . (3.36)In dimension d = 2, Conjecture 3.20 was solved by Hala\u00b4sz [25]. Vagharshakyan[51] improved the constant Cd in Hala\u00b4sz\u2019s proof. It has also been conjectured that, for0 < p < 1, the Lp-norm of DN satisfies the lower bound analogous to that appearingin (3.36).3.3.2 Proof of Theorem 3.17Here, we prove that, for any N -point set in [0, 1]d and for all d \u2265 2, the L2-norm ofDN is of order (logN)d\u221212 . Having expanded the discrepancy function with respectto the orthogonal basis of Haar functions:DN =\u2211R\u2208Dd\u2217\u3008DN , hR\u3009\u2016hR\u201622hR,Roth\u2019s approach was to identify the terms of the expansion which do not contributesignificantly to \u2016DN\u20162. The following proposition quantifies what this means moreprecisely.Proposition 3.21. Let PN \u2282 [0, 1]d be a Npoint-set with #PN = N , and let n \u2208 Nbe such that 2n\u22122 \u2264 N < 2n\u22121. Then, for every ~r \u2208 Hdn, there exists an ~r-function,f~r, such that\u3008DN , f~r\u3009 :=\u222b[0,1]dDN \u00b7 f~rdx \u2265 cd > 0, (3.37)where the constant cd depends only on the dimension d.We will first assume that Proposition 3.21 holds to prove Theorem 3.17, and thenwe will prove Proposition 3.21 immediately afterward.42Proof of Theorem 3.17. Here, we use a duality argument and construct a test func-tion, F , such that \u2016F\u20162 ' log d\u221212 N and \u3008DN , F \u3009 & (logN)d\u22121. Using these twolower bounds, the Cauchy-Schwarz inequality will imply:\u2016DN\u20162 \u2265 \u3008DN , F \u3009\u2016F\u20162 & (logN)d\u221212 ,which is the estimate stated in Roth\u2019s theorem.Define F byF =\u2211~r\u2208Hdnf~r,where f~r is the ~r-function associated to ~r \u2208 Hdn which comes from Proposition 3.21.Using the orthogonality of ~r-functions and the fact that logN ' n, we get that\u2016F\u20162 =(\u2211~r\u2208Hdn\u2016f~r\u201622)1\/2= (#Hdn)1\/2 ' nd\u221212 ' (logN) d\u221212 .On the other hand, Proposition 3.21 ensures that\u3008DN , F \u3009 \u2265 cd#Hdn & nd\u22121 & (logN)d\u22121,which completes the proof.Remark 3.22. The same approach can, essentially, be used to prove the Lp lowerbounds of the discrepancy function. This will require us to estimate the Lp- norm ofthe function F , but, using the Littlewood-Paley inequalities, one can show that thishas the same behaviour as the L2-norm of F , i.e., \u2016F\u2016p ' (logN) d\u221212 . Now, togetherwith Ho\u00a8lder\u2019s inequality and Proposition 3.21, this will give us the estimate provedby Schmidt,\u2016DN\u2016p \u2265 \u3008DN , F \u3009\u2016F\u2016q & (logN)d\u221212 , (3.38)where q is the dual exponent of p \u2208 (1,\u221e). The Littlewood-Paley inequalities areneeded to replace the orthogonality argument used in calculating the L2-norm of F .43We refer the reader to Remark 5.8 for more details regarding the Lp-estimates of F .Proof of Proposition 3.21. First, we observe that the discrepancy function associatedto a set PN \u2282 [0, 1)d with #PN = N can be written in an equivalent form:DN(~x) =\u2211~p\u2208PN1[~p,~1)(~x)\u2212Nx1 \u00b7 \u00b7 \u00b7xd, (3.39)where ~p = (p1, . . . , pd) \u2208 PN \u2282 [0, 1)d and [~p,~1) = [p1, 1)\u00d7 \u00b7 \u00b7 \u00b7 \u00d7 [pd, 1). Fix a dyadicrectangle R = I1\u00d7\u00b7 \u00b7 \u00b7\u00d7 Id \u2208 Dd, and assume that the point ~p \u2208 PN is not containedin R. Therefore, there exists j \u2208 {1, . . . , d} such that pj is not contained in Ij. Asa result,\u222b[0,1]1[pj ,1)(xj)hIj(xj)dxj = 0, since either pj is to the right of Ij, in whichcase Ij \u2229 [pj, 1) = \u2205 and so\u222b[0,1]1[pj ,1)(xj)hIj(xj)dxj = 0, or pj is to the left of Ij and\u222b[0,1]1[pj ,1)(xj)hIj(xj)dxj =\u222b[0,1]hIj(xj)dxj = 0. Therefore,\u222b[0,1]d1[~p,~1) hR = 0, andwe get that \u2329 \u2211~p\u2208PN\u2229R=\u22051[~p,~1), hR\u232a= 0. (3.40)Also, since\u222bIjxjhIjdxj =|Ij |24, j = 1, . . . , d, we get\u3008x1 \u00b7 \u00b7 \u00b7xd, hR\u3009 = 4\u2212d|R|2. (3.41)Using (3.40) and (3.41), we get a crucial relation which will be essential for the restof the proof,\u3008DN , hR\u3009 = \u2212N4\u2212d|R|2, (3.42)for all dyadic rectangles R which do not contain points from PN .Recall that the setR~r = {R \u2208 Dd : Rj \u2208 D and |Rj| = 2\u2212rj , j = 1, . . . , d}is a collection of disjoint axis-parallel dyadic rectangles whose sides are of prescribedlengths. Now, for each ~r \u2208 Hdn, we are ready to define an ~r-function consisting oftwo parts, where the first part is a sum over the dyadic rectangles in R~r which do44not contain points from PN , and the second part is a sum over the dyadic rectanglesin R~r which contain points from PN . That is, we decompose f~r asf~r =\u2211R\u2208R~r:R\u2229PN=\u2205\u22121hR +\u2211R\u2208R~r:R\u2229PN 6=\u2205sgn\u3008DN , hR\u3009hR, (3.43)wheresgn(x) =\uf8f1\uf8f2\uf8f31, if x \u2265 0\u22121, otherwise .The inner product of the discrepancy function, DN , with the second term of thedecomposition in (3.43) is a positive number. Taking into account relation (3.42),this gives us\u3008DN , f~r\u3009 \u2265\u2211R\u2208R~r:R\u2229PN=\u2205N4\u2212d|R|2. (3.44)Since 2n\u22122 \u2264 N < 2n\u22121 and #R~r = 2n, the number of rectangles in R~r whichintersect the N -point set PN is at most 2n\u22121. Thus, we have#{R \u2208 R~r : R \u2229 PN = \u2205} \u2265 2n(1\u2212 12)\u2265 2n\u22121.Therefore, recalling that |R| = 2\u2212n, we get the desired estimate:\u3008DN , f~r\u3009 \u2265\u2211R\u2208R~r:R\u2229PN=\u2205N4\u2212d|R|2 \u2265 2n\u221212n\u22122 \u00b7 2\u22122n4d= cd.3.3.3 Hala\u00b4sz\u2019s proof of Conjecture 3.18 in d = 2Here, we give all the essential details of Hala\u00b4sz\u2019s proof of Schmidt\u2019s result for thediscrepancy function in d = 2,\u2016DN\u2016\u221e \u2265 C logN. (3.45)45Hala\u00b4sz\u2019s approach was to use a duality argument in combination with Riesz producttechniques and Roth\u2019s principle, Proposition 3.21. Using Riesz products turned outto be very fruitful in proving the two-dimensional small ball inequality, as we willsee in the next chapter.To present Hala\u00b4sz\u2019s proof, we need two auxiliary lemmas. The first lemma showshow Haar functions interact when multiplied.Lemma 3.23. (Product Rule) If R 6= R\u2032 \u2208 D2 are not disjoint dyadic rectanglesand |R| = |R\u2032|, thenhR \u00b7 hR\u2032 = \u00b11hR\u2229R\u2032 ,i.e., the product of two Haar functions is again a Haar function.We let the reader verify the result of this simple lemma. This lemma breaksdown in dimensions d \u2265 3, and the absence of the Product Rule in d \u2265 3 is a majorobstacle in solving the L\u221e-discrepancy conjecture.Lemma 3.24. Let f~r be any ~r-function with parameter s, i.e., |~r| := r1 + \u00b7 \u00b7 \u00b7+rd = s.Then, for some constant bd > 0,\u3008DN , f~r\u3009 \u2264 bdN2\u2212s.The interested reader can find the proof of the above lemma in [10].Hala\u00b4sz\u2019s proof. First, we construct a test function \u03a6 which satisfies two key estimates:\u2016\u03a6\u20161 . 1, (3.46)and|\u3008DN ,\u03a6\u3009| & logN. (3.47)Then, once we have shown that (3.46) and (3.47) hold, applying Ho\u00a8lder\u2019s inequalityin (3.47) will yield the conjectured estimate (3.45).The function \u03a6 will be a Riesz product, where the building blocks of the productwill be ~r-functions coming from Proposition 3.21. To be more precise, we fix n of46order logN , i.e., n ' logN , and we define the test function \u03a6 by\u03a6 :=n\u220fk=0(1 + \u03b3fk)\u2212 1, (3.48)where \u03b3 \u2208 (0, 1) is a small constant which will be determined later, andfk := f(k,n\u2212k) =\u2211R\u2208R~r\u03b1RhRis the two-dimensional ~r = (k, n\u2212 k)-function constructed in Proposition 3.21.We proceed to establish properties (3.46) and (3.47) for our test function \u03a6.Expanding the product in (3.48), we get\u03a6\u02dc :=n\u220fk=0(1 + \u03b3fk) = 1 + \u03a61 +n\u2211k=2\u03a6k, (3.49)where\u03a61 = \u03b3n\u2211k=0fk, (3.50)and\u03a6k = \u03b3k\u22110\u2264j1<\u00b7\u00b7\u00b7 0 for all k,and so \u03a6\u02dc is a positive function. In addition, with the aid of Lemma 3.23, we havethat\u222b[0,1]d\u03a6kdx = 0 for all k \u2265 1, since a finite product of ~r-functions is a linearcombination of Haar functions. Therefore,\u2016\u03a6\u02dc\u20161 =\u222b[0,1]d\u03a6\u02dcdx = 1. (3.52)Moreover, noticing that \u03a6 = \u03a6\u02dc\u2212 1, from the triangle inequality we obtain\u2016\u03a6\u20161 \u2264 \u2016\u03a6\u02dc\u20161 + 1 = 2, (3.53)47which is (3.46).Now, we concentrate on showing (3.47). Recall that, from Proposition 3.21,\u3008DN , fk\u3009 \u2265 cd for k = 0, . . . , n, for some constant cd < 1. From this it is easy to seethat\u3008DN ,\u03a61\u3009 \u2265 cd\u03b3 logN. (3.54)Hence, to complete the proof of (3.47), it suffices to show that, for some absoluteconstant c, we have\u3008DN ,n\u2211k=2\u03a6k\u3009 \u2264 2c\u03b32 logN, (3.55)for a small value of \u03b3 < 1 such that \u03b3 < min(12, cd2c). Indeed, since \u03a6 = \u03a61 +\u2211nk=2 \u03a6k,by applying the reverse triangle inequality, (3.54) and (3.55), we get|\u3008DN ,\u03a6\u3009| \u2265 \u3008DN ,\u03a61\u3009 \u2212 \u3008DN ,n\u2211k=2\u03a6k\u3009 \u2265 C(d, \u03b3) logN, (3.56)where C(d, \u03b3) = cd\u03b3\u2212 2c\u03b32. Note that C(d, \u03b3) > 0 since \u03b3 < cd2c . From the definitionof \u03a6k, k = 2, . . . , n, and Lemma 3.23, we can see that product of ~r-functions, fj1 \u00b7fj2 \u00b7 \u00b7 \u00b7 fjk , is again an ~r-function with ~r = (jk, n \u2212 j1). We rewrite the function \u03a6kwith respect to the parameter s as\u03a6k = \u03b3k2n\u2211s=n+1\u22110\u2264j1 n.It is easy to see that this rule does not hold in higher dimensions. For example,in d = 3, suppose R and R\u2032 are distinct non-disjoint rectangles of the same volumeand whose first sides coincide, i.e., R1 = R\u20321. Then the product of the Haar functionssupported on these dyadic rectangles is not a Haar function, since hR1 \u00b7 hR\u20321 = h2R1 =1R1 . The failure of this rule in higher dimensions makes the small ball inequality ahard problem in d \u2265 3.Proof of the small ball inequality for d = 2. Here, we solve the small ball conjecturefor d = 2 via a duality approach. To do this, we need to construct a test function,\u03a8, which satisfies the following two relations:\u2016\u03a8\u20161 \u2264 C, (4.2)where C is an absolute constant which does not depend on the choice of the coeffi-cients {\u03b1R} or the integer n \u2265 1, and\u2329\u03a8,\u2211|R|=2\u2212n\u03b1RhR\u232a:=\u222b[0,1]d\u03a8 \u00b7\u2211|R|=2\u2212n\u03b1RhRdx \u2265 C2\u2212n\u2211|R|=2\u2212n|\u03b1R|. (4.3)Once we construct such a test function, the small ball conjecture in d = 2 followsfrom a simple application of Ho\u00a8lder\u2019s inequality to (4.3) for p = 1 and q =\u221e.As in Hala\u00b4sz\u2019s proof, the test function \u03a8 is a Riesz product. Here, the building52blocks of \u03a8 are again ~r-functions, but this time they are defined byfk := f(k,n\u2212k) =\u2211|R|=2\u2212n:|R1|=2\u2212ksgn(\u03b1R)hR,and the test function \u03a8 is defined by\u03a8 =n\u220fk=0(1 + \u03b3fk). (4.4)Here, \u03b3 \u2208 (0, 1] is any constant. For simplicity, we choose the constant \u03b3 to be equalto 1, but we will see that the proof works for any positive constant \u03b3 \u2264 1.Now, we are ready to show that \u03a8 satisfies (4.2) and (4.3). First, since (1 +fk) isnonnegative for all k \u2208 {0, . . . , n}, we observe that the test function \u03a8 is nonnegative.Expanding the product in (4.4) gives\u03a8 = 1 + \u03a81 +n\u2211k=2\u03a8k, (4.5)where\u03a81 =n\u2211k=0fk, \u03a8k =\u22110\u2264j1<\u00b7\u00b7\u00b7 n, i.e., \u03a8k is a linear combinationof Haar functions supported on dyadic rectangles of volume strictly less than 2\u2212n.As a result, for every k \u2208 {2, . . . , n}, when we multiply \u03a8k with Hn we get a linearcombination of functions of the form hRhR\u2032 , where R = R1 \u00d7 R2, R\u2032 = R\u20321 \u00d7 R\u20322 aretwo-dimensional dyadic rectangles with |R| = 2\u2212n and |R\u2032| < 2\u2212n respectively. Since|R| 6= |R\u2032|, Ri 6= R\u2032i for either i = 1 or i = 2. Hence, hRihR\u2032i is a Haar function. Dueto the fact that the 1-dimensional integral of Haar functions is equal to zero, whenwe integrate the product \u03a8k \u00b7Hn over the unit square [0, 1]2, we therefore get (4.8).Now, we focus on the proof of (4.7). First, note that the function Hn can bewritten asHn =\u2211|R|=2\u2212n\u03b1RhR =\u2211~r\u2208H2n\u2211R\u2208R~r\u03b1RhR, (4.9)since \u222a~r\u2208H2nR~r = {R \u2208 D2 : |R| = 2\u2212n}. Using the above expression for Hn and theorthogonality of Haar functions, we have\u222b[0,1]2\u03a81Hndx =\u2211~r\u2208H2n\u2211R\u2208R~r\u222b[0,1]2|\u03b1R|h2Rdx = 2\u2212n\u2211|R|=2\u2212n|\u03b1R|, (4.10)which shows that (4.7) holds.From the proof of the two-dimensional small ball inequality we see that C2 = 1.If we had worked with the test function \u03a8 =\u220fnk=0(1 + \u03b3fk), \u03b3 < 1, then we wouldhave gotten that C2 = \u03b3 < 1. In Hala\u00b4sz\u2019s proof, the constant \u03b3 had a specific value.In particular, he used the test function \u03a6 =\u220fnk=0(1 + \u03b3fk) \u2212 1 = \u03a61 +\u2211nk=2 \u03a6k,where \u03a61 and \u03a6k, k \u2208 {2, . . . , n} were defined in (3.50) and (3.51) respectively. Thevalue of \u03b3 was chosen in such a way that the lower bound of \u3008DN ,\u2211nk=2 \u03a6k\u3009 wassmaller than the lower bound of \u3008DN ,\u03a61\u3009. Also, \u03a8 and \u03a6 differed in that Hala\u00b4sz\u2019s ~r-functions fk := f(k,n\u2212k) were chosen so that they had the property that \u3008DN , fk\u3009 \u2265 cdfor all k \u2208 {0, . . . , n}, whereas Temlyakov\u2019s didn\u2019t have any specific properties. The\u2018\u22121\u2019 term was included in the definition of \u03a6 for a technical reason. Specifically,including that term allowed Hala\u00b4sz to avoid calculating the integral\u222b[0,1]2DN(x)dx54when estimating the lower bound of \u3008DN ,\u03a6\u3009 :=\u222b[0,1]2DN(x)\u03a6(x)dx.In the proof of the two-dimensional small ball inequality we estimated the L\u221e-norm of Hn from below by estimating the integral\u3008\u03a8, Hn\u3009 :=\u222b[0,1]2\u03a8Hndx.The test function \u03a8 was defined as it was because it identifies small sets where Hn isvery large. As a result, the integral \u3008\u03a8, Hn\u3009 is a good estimate for the L\u221e-norm ofHn. Indeed, consider for simplicity that {\u03b1R} \u2282 {\u22121, 1}, i.e., sgn(\u03b1R) = \u03b1R. Then,in this case, Hn =\u2211~r\u2208H2n f~r =\u2211nk=0 fk and \u03a8 =\u220fnk=0(1 + fk)= 2n+1 1E, whereE = {~x \u2208 [0, 1]2 : fk(~x) = 1 for every k \u2208 {0, . . . , n}}.In other words, the function \u03a8 is supported on the set E where Hn is maximized. Asa consequence of this fact, we have that\u222b[0,1]2\u03a8Hndx =\u222bE\u03a8Hndx = \u2016Hn\u2016\u221e\u222bE\u03a8 =\u2016Hn\u2016\u221e, since \u2016\u03a8\u20161 =\u222bE\u03a8dx = 1.Remark 4.1. Temlyakov actually proved a stronger result. His proof carries throughif one replaces Hn with\u2211|R|\u22652\u2212n \u03b1RhR to give\u2225\u2225\u2225\u2225 \u2211|R|\u22652\u2212n\u03b1RhR\u2225\u2225\u2225\u2225\u221e& 2\u2212n\u2211|R|=2\u2212n|\u03b1R|.Riesz products had already been used by Sidon [44, 45] to study lacunary Fourierseries. Recall that an increasing sequence {\u03bbj}\u221ej=1 \u2282 N is called lacunary if thereexists q > 1 so that \u03bbj+1\/\u03bbj > q. Sidon used Riesz products to prove that if fhas a lacunary series, i.e., there exists a lacunary sequence \u039b such that the Fouriercoefficients of f ,f\u02c6(k) =\u222b[0,1]f(x)e\u22122piikxdx,55are supported on the sequence \u039b, then\u2016f\u2016\u221e &\u221e\u2211k=1|f\u02c6(k)|,and\u2016f\u20161 & \u2016f\u20162,where f is a bounded 1-periodic function. For the first inequality, he used the RieszproductPN(x) =N\u220fk=1(1 + cos(2pi\u03bbkx+ \u03b4k)),where \u03b4k was chosen so that ei\u03b4k = f\u02c6(k)\/|f\u02c6(k)|. He showed that \u2016PN\u20161 = 1 for everyN \u2208 N and \u2016f\u2016\u221e \u2265\u2223\u2223\u2223\u2223 \u222b[0,1] f(x)PN(x)\u2223\u2223\u2223\u2223 = 12 \u2211Nk=1 |f(\u03bbk)|. He completed the proofby letting N go to infinity. The same ideas were implemented for the proof of thesecond inequality.4.2 The sharpness of the small ball inequality inall dimensionsIn this section, we show that the small ball inequality is sharp, i.e., there exists aconstant Cd > 0 and a collection of {\u03b1R}, in fact {\u03b1R} \u2282 {\u22121, 1}, such that\u2225\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225\u221e\u2264 Cdn\u2212 d\u221222 \u00b7 2\u2212n\u2211R\u2208Adn|\u03b1R| = Cdn d2 .It suffices to show that the above inequality holds on average, i.e.,E\u2225\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225\u221e\u2264 Cdn d2 ,56where {\u03b1R(\u03c9)}R\u2208Adn are iid \u00b11-valued random variables on a probability space (\u2126,P)with P(\u03c9 \u2208 \u2126 : \u03b1R(\u03c9) = 1) = P(\u03c9 \u2208 \u2126 : \u03b1R(\u03c9) = \u22121) = 12 . Indeed, recall Markov\u2019sinequality: Let (X,\u03a3, \u00b5) be a measure space. If f \u2208 L1 we have that\u00b5({x : |f(x)| > a}) \u2264 \u2016f\u2016L1a,and for a = 2\u2016f\u2016L1 we have that\u00b5({x : |f(x)| > a}) \u2264 12.So, when \u00b5 = P (P a probability measure) and f(\u03c9) =\u2225\u2225\u2225\u2211R\u2208Adn \u03b1R(\u03c9)hR\u2225\u2225\u2225\u221e we haveP({\u03c9 : |f(\u03c9)| > 2E(f)}) \u2264 12which implies thatP({\u03c9 : |f(\u03c9)| \u2264 2E(f)}) > 12.This implies that there exists \u03c9o \u2208 \u2126 such that|f(\u03c90)| =\u2225\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1R(\u03c90)hR\u2225\u2225\u2225\u2225\u2225\u2225\u221e\u2264 2E\uf8eb\uf8ed\u2225\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1R(\u03c9)hR\u2225\u2225\u2225\u2225\u2225\u2225\u221e\uf8f6\uf8f8. n d2 .To present the proof of the sharpness of the d-dimensional small ball inequality, weneed a preliminary lemma. This lemma is a standard fact in probability theory andwe state it here without a proof.Lemma 4.2. (Bernstein\u2019s inequality [28]) If Zi, i = 1, . . . , N , are independent andidentically distributed \u00b11-valued random variables with P(Zi = 1) = P(Zi = \u22121) =570.5 and (c1, . . . , cN) \u2208 RN , thenP(\u2223\u2223\u2223\u2223 N\u2211l=1ciZi\u2223\u2223\u2223\u2223 > \u03bb)\u2264 2 exp(\u2212 \u03bb24\u03c32),where \u03c32 =\u2211Ni=1 \u03c32i with \u03c32i = E((ciZi)2) and E(Zi) = 0, for i = 1, . . . , N .Theorem 4.3. The small ball inequality is sharp on average, i.e.,E\u2225\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225\u221e\u2264 Cdn d2 ,where {\u03b1R(\u03c9)}R\u2208Adn are iid \u00b11 valued random variables on a probability space (\u2126,P)with P(\u03b1R(\u03c9) = 1) = P(\u03b1R(\u03c9) = \u22121) = 12 .Proof. First, we observe that the function\u2211R\u2208Adn \u03b1R(\u03c9)hR(\u00b7) is constant on dyadiccubes of side length 2\u2212(n+1) for each \u03c9 \u2208 \u2126. This can be seen by noting that eachrectangle of volume 2\u2212n can be covered by the dyadic cubes of side length 2\u2212(n+1),and so each cube will lie in that part of the rectangle where the Haar function isconstant.As a consequence, the total number of such cubes is M = 2(n+1)d. If we denoteeach such cube by Qk, k = 1, . . .M , then we can create M random variablesXk(\u03c9) =\u2211R\u2208Adn\u03b1R(\u03c9)hR(xk),where xk \u2208 Qk for k = 1, . . . ,M . Applying Bernstein\u2019s inequality to Xk, for k =1, . . . ,M , we getP(|Xk| > t) \u2264 2 exp(\u2212 t24#Hdn). (4.11)Let \u03c8(t) = exp(t2) and define the following random variables:Yk(\u03c9) =Xk(\u03c9)Cnd\u221212,58where the constant C will be chosen later. Taking into account the fact that{\u03c9 : \u03c8(|Yk(\u03c9)|) > t} = {\u03c9 \u2208 \u2126 : |Yk(\u03c9)| > \u03c8\u22121(t)},we have thatE(\u03c8(|Yk|)) = \u222b \u221e0P(\u03c8(|Yk(\u03c9)|) > t)dt=\u222b \u221e0P(|Yk(\u03c9)| > \u03c8\u22121(t))dt=\u222b \u221e0P(|Xk(\u03c9)| > Cnd\u221212\u221alog t4#Hdn)=\u222b \u221e0min{1, 2 exp(\u2212C2nd\u22121 log t4#Hdn)}dt (by (4.11))\u2264\u222b \u221e0min{1, 2t\u2212K}dt,where #Hdn \u2264 Cdnd\u22121 and K = C24Cd. Now we choose K sufficiently large, or equiva-lently we choose C sufficiently large, so that\u222b \u221e0min{1, 2t\u2212K}dt \u2264 C \u2032,where C \u2032 is an absolute constant, so thatE(\u03c8(|Yk|)) \u2264 C \u2032. (4.12)Since \u03c8(t) = exp(t2) is a convex, increasing function on [0,\u221e) and\u03c8(supk=1,...,M{|Yk(\u03c9)|})= supk=1,...,M{\u03c8(|Yk(\u03c9)|)},59we get the following by applying Jensen\u2019s inequality:\u03c8(E(supk=1,...,M{|Yk(\u03c9)|})) \u2264 E(\u03c8( supk=1,...,M{|Yk(\u03c9)|}))\u2264 E(supk=1,...,M{\u03c8(|Yk(\u03c9)|)})\u2264 E(M\u2211k=1\u03c8(|Yk(\u03c9)|))\u2264 C \u2032M = C \u20322(n+1)d,where C \u2032 is an absolute constant, and the last line was obtained by using (4.12).Since \u03c8\u22121(t) =\u221alog t, we have thatE(supk=1,...,M{|Yk(\u03c9)|}) \u2264 (n+ 1) 12d 12\u221aC \u2032 log 2,which implies thatE(supk=1,...,M{|Xk(\u03c9)|}) \u2264 C \u2032\u2032n d\u221212 n 12d 12 = C \u2032\u2032n d2d 12 ,where C \u2032\u2032 is an absolute constant. Hence, we have shown thatE\u2225\u2225\u2225\u2225\u2225\u2225\u2211R\u2208Adn\u03b1R(\u03c9)hR\u2225\u2225\u2225\u2225\u2225\u2225\u221e. n d2 , (4.13)since supk=1,...,M{|Xk(\u03c9)|} = \u2225\u2225\u2225\u2211R\u2208Adn \u03b1R(\u03c9)hR\u2225\u2225\u2225\u221e , \u2200\u03c9 \u2208 \u2126.60Chapter 5Elements of dyadic harmonicanalysis5.1 Khintchine\u2019s inequalitiesKhintchine\u2019s inequalities allow us to, up to a constant, calculate the p-norms of alinear combination of iid random variables. We will use Khintchine\u2019s inequalities toprove Theorem 1.4.Let {Xi, 1 \u2264 i \u2264 N} be a collection of \u00b11-valued iid random variables on aprobability space (\u2126,P) such that P(Xi = 1) = P(Xi = \u22121) = 12 . Then, Khintchine\u2019sinequalities state that there exist constants Ap > 0 and Bp > 0 such thatAp[ N\u2211i=1|ai|2] 12\u2264\u2225\u2225\u2225\u2225 N\u2211i=1aiXi\u2225\u2225\u2225\u2225p\u2264 Bp[ N\u2211i=1|ai|2] 12, 1 < p <\u221e,for any finite sequence of real numbers (ai)Ni=1.The proofs of these inequalities rely on two key facts about the distributionfunction of the random variable SN =\u2211Ni=1 aiXi. In particular, the distribution61function of SN , dSN (\u03bb) = P({\u03c9 \u2208 \u2126 : |SN | > \u03bb}), satisfies the sub-Gaussian bounddSN([ N\u2211i=1a2i] 12\u03bb)\u2264 2e\u2212 12\u03bb2 , for all \u03bb > 0,and the Lp-norm of SN can be computed in terms of dSN :\u2016SN\u2016pp =\u222b \u221e0p\u03bbp\u22121dSN (\u03bb)d\u03bb, 0 < p <\u221e.Khintchine\u2019s inequalities can be extended to the case of complex-valued squaresummable sequences {aj}. When p = 2, the orthogonality of the random variablesXi, 1 \u2264 i \u2264 N , reduces Khintchine\u2019s inequalities to Parseval\u2019s identity:\u2016SN\u201622 =N\u2211i=1|ai|2.Moreover, Bp = 1 when 0 < p \u2264 2 and Ap = 1 for p \u2265 2, as the trivial estimates\u2016SN\u2016p \u2264 \u2016SN\u20162, 0 < p \u2264 2 and \u2016SN\u20162 \u2264 \u2016SN\u2016p, p \u2265 2 show respectively. For acomplete proof of these inequalities, we refer the interested reader to [21]. For theproof of the optimal values of Ap and Bp, we refer the reader to [22, 53].A direct application of Khintchine\u2019s inequalities shows that the p-norms of alinear combination of Rademacher functions are comparable, i.e., there exist positiveconstants Ci(p, q), i = 1, 2 such thatC1(p, q)\u2225\u2225\u2225\u2225 N\u2211k=0akrk\u2225\u2225\u2225\u2225p\u2264\u2225\u2225\u2225\u2225 N\u2211k=0akrk\u2225\u2225\u2225\u2225q\u2264 C2(p, q)\u2225\u2225\u2225\u2225 N\u2211k=0akrk\u2225\u2225\u2225\u2225p, q > p > 0. (5.1)Here, rk =\u2211I\u2208D:|I|=2\u2212k hI are Rademacher functions which are iid random variableson the probability space ([0, 1), | \u00b7 |).We would like to extend the above inequalities to more general functions. Morespecifically, we would like to extend them to\u2211Nk=0\u2211|I|=2\u2212k \u03b1IhI or, in the multivari-able case, to\u2211R\u2208Dd:|R|=2\u2212N \u03b1RhR, where the coefficients {\u03b1I} or {\u03b1R} depend not62only on the scale of the dyadic rectangles, but also on the rectangles themselves. Forthis reason, the Littlewood-Paley inequalities will be useful.5.2 The Littlewood-Paley inequalitiesIn this section, we will present the Littlewood-Paley inequalities for martingales.These inequalities can be considered as a generalization of Khintchine\u2019s inequalities.The special case of the Littlewood-Paley inequalities for Haar functions will be usedfor the proof of the higher dimensional analogue of inequalities (5.1). To introduce theLittlewood-Paley inequalities for martingales, we need some preliminary definitionsfrom probability theory.Suppose X \u2208 L1(\u2126,B,P) is an integrable real-valued random variable and letG \u2282 B a be sub-\u03c3-algebra. We define the conditional expectation of X given G to bea random variable, denoted by E(X|G), which satisfies(i) E(X|G) is G-measurable and integrable.(ii) For all G \u2208 G, we have \u222bGXdP =\u222bGE(X|G)dP.We mention only those properties conditional expectations possess that we will usethroughout this chapter.(1) Linearity: If X, Y \u2208 L1, and \u03b1, \u03b2 \u2208 R, we haveE(\u03b1X + \u03b2Y |G) = \u03b1E(X|G) + \u03b2E(Y |G).(2) E(E(X|G)) = E(X).(3) Product Rule: Let X, Y be random variables satisfying X, Y \u00b7X \u2208 L1. If Y \u2208 G,63i.e., Y is measurable with respect to G, thenE(XY |G) = Y E(X|G).(4) Smoothing: If G1 \u2282 G2 \u2282 B, thenE(E(X|G2)|G1) = E(X|G1).(5) If X \u2208 L1 is independent of the \u03c3-algebra G, thenE(X|G) = E(X).The proof of these properties can be found in [38].Definition 5.1. Suppose we are given integrable random variables {Xn}n\u22650 and \u03c3-algebras {Gn}n\u22650 which are sub-\u03c3-algebras of B. Then {(Xn,Gn)}n\u22650 is a martingaleif:(i) {Gn}n\u22650 is an increasing sequence of \u03c3-algebras, i.e.,G0 \u2282 G1 \u2282 G2 \u2282 \u00b7 \u00b7 \u00b7 \u2282 B.(ii) Xn is adapted, i.e., Xn \u2208 Gn, for each n \u2208 N0 = {0, 1, 2, . . .}.(iii) For 0 \u2264 m < n,E(Xn|Gm) = Xm.By the smoothing property, condition (iii) in the definition of a martingale isequivalent to E(Xn+1|Gn) = Xn. In many applications, Gn is the smallest \u03c3-algebragenerated by {Xi : 0 \u2264 i \u2264 n}, i.e. Gn = \u03c3(X0, . . . , Xn). Note that every martingale{(Xn,Gn)}n\u22650 can be written as Xn =\u2211nj=0 dj, where {(dj,Gj)}j\u22650 is a martingaledifference, i.e., dj \u2208 Gj with dj \u2208 L1 and E(dj+1|Gj) = 0 for j \u2265 0. Indeed, simplydefine d0 = X0 and dj = Xj\u2212Xj\u22121, j \u2265 1. We can see that a martingale is a natural64generalization of a sum of independent random variables. Indeed, suppose {Zn}n\u22650 isan independent sequence of integrable random variables such that E(Zn) = 0, n \u2265 0.Set Yn =\u2211ni=0 Zi and Gn = \u03c3(Z0, . . . , Zn). Then {(Yn,Gn)}n\u22650 is a martingale sinceE(Yn+1|Gn) = E(Yn|Gn) + E(Zn+1|Gn) = Yn + E(Zn+1) = Yn. Note that, in thiscase, {(Zi,Gi)}i\u22650 is a martingale difference. Moreover, we can create a martingalefrom a random variable X \u2208 L1 via the smoothing property. Specifically, if we takeXn = E(X|Gn), n \u2265 0, then {(Xn,Gn)}n\u22650 is a martingale.Now, if {(Xn,Gn)}n\u22650 is a martingale such that E(X2n) < \u221e, then {dj} is anorthogonal martingale difference, i.e.,E(didj) = 0, i 6= j.Indeed, using properties (2) and (3) of conditional expectation, if j > i, then we getE(didj) = E (E(didj|Gi))= E(diE(dj|Gi)) = 0.As an immediate consequence, we obtainE(X2n) = E( n\u2211j=0d2j), n \u2265 0,which is equivalent to\u2016Xn\u20162 =\u2225\u2225\u2225\u2225( n\u2211j=0d2j) 12\u2225\u2225\u2225\u22252. (5.2)The Littlewood-Paley inequalities are an Lp-version of (5.2), but they can also beviewed as an extension of Khintchine\u2019s inequalities to more general martingales.Theorem 5.2. (Littlewood-Paley inequalities) Let 1 < p < \u221e. There are positive65real numbers Mp and Np such that, if {(Xn,Gn)}n\u22650 is a martingale, thenMp\u2225\u2225\u2225\u2225( n\u2211j=0d2j) 12\u2225\u2225\u2225\u2225p\u2264 \u2016Xn\u2016p \u2264 Np\u2225\u2225\u2225\u2225( n\u2211j=0d2j) 12\u2225\u2225\u2225\u2225p, n \u2265 0. (5.3)If 1 < p <\u221e and supn\u22650 \u2016Xn\u2016p <\u221e then Xn converges to a random variable Xboth almost everywhere and with respect to the Lp-norm. Moreover, (5.3) becomesMp\u2225\u2225\u2225\u2225( \u221e\u2211j=0d2j) 12\u2225\u2225\u2225\u2225p\u2264 \u2016X\u2016p \u2264 Np\u2225\u2225\u2225\u2225( \u221e\u2211j=0d2j) 12\u2225\u2225\u2225\u2225p. (5.4)Here, the quantityS(X) =( \u221e\u2211j=0d2j) 12is called the square function of X.These inequalities were proven by D. Burkholder [12]. A major component ofBurkholder\u2019s proof of the Littlewood-Paley inequalities the following lemma.Lemma 5.3. If {dj} is a martingale difference and {\u03b5j} \u2282 {\u22121, 1}, then\u2225\u2225\u2225\u2225 n\u2211j=0\u03b5jdj\u2225\u2225\u2225\u2225p\u2264 cp\u2225\u2225\u2225\u2225 n\u2211j=0dj\u2225\u2225\u2225\u2225p, 1 < p <\u221e. (5.5)We will not prove inequality (5.5), but will show how to derive the Littlewood-Paley inequalities from Lemma 5.3.Proof of the Littlewood-Paley inequalities. Consider the Rademacher functionsrj =\u2211I\u2208D:|I|=2\u2212j hI , j = 0, . . . , n on [0, 1]. Using inequality (5.5), we immediatelyget thatc\u22121p \u2016Xn\u2016p \u2264\u2225\u2225\u2225\u2225 n\u2211j=0rj(t)dj\u2225\u2225\u2225\u2225p\u2264 cp\u2016Xn\u2016p, for every t \u2208 [0, 1]. (5.6)Raising both sides of the above inequality to the power p and integrating over the66interval [0, 1], we obtain(c\u22121p )p\u2016Xn\u2016pp \u2264\u222b[0,1]\u2225\u2225\u2225\u2225 n\u2211j=0rj(t)dj\u2225\u2225\u2225\u2225ppdt \u2264 (cp)p\u2016Xn\u2016pp. (5.7)Applying Fubini\u2019s Theorem and Khintchine\u2019s inequalities to the middle expressionof the above inequality, we have(c\u22121p )p\u2016Xn\u2016pp \u2264 Bpp\u2225\u2225\u2225\u2225( n\u2211j=0d2j) 12\u2225\u2225\u2225\u2225pp. (5.8)Thus,\u2016Xn\u2016p \u2264 Bpcp\u2225\u2225\u2225\u2225( n\u2211j=0d2j) 12\u2225\u2225\u2225\u2225p, (5.9)which is the right side of (5.3) with Np = Bpcp. The left side of (5.3) is provensimilarly.Also, Burkholder [13] found the values of the optimal constants Np and Mp oc-curring in the Littlewood-Paley inequalities:Np = max{p,pp\u2212 1}\u2212 1,Mp =(max{p,pp\u2212 1}\u2212 1)\u22121.5.3 Connection of martingale differences with HaarfunctionsRecall from (1.1) that D is the set of dyadic intervals of [0, 1], i.e.,D =\u221e\u22c3k=0A1k, (5.10)67where A1k is the collection of dyadic intervals of length 2\u2212k and D\u2217 = D\u222a [\u22121, 1]. Wewill be interested in the martingale:{(E(f |\u2206k),\u2206k)}k\u22650, (5.11)where \u2206k = \u03c3(A1k) and f \u2208 L1([0, 1]d). Note thatE(f |\u2206k) =\u2211I\u2208A1k\u3008f\u3009I 1I(x),where\u3008f\u3009I = 1|I|\u222bIf(x)dxis the average of f over I. In the next proposition, we relate the martingale differencedk = E(f |\u2206k)\u2212 E(f |\u2206k\u22121)to Haar functions.Proposition 5.4. For every f \u2208 L1([0, 1]) and for all k \u2265 1, we havedk(x) =\u2211I\u2208A1k\u22121\u3008f, hI\u3009|I| hI(x), x \u2208 [0, 1]. (5.12)Proof. First, we observe that, for every I \u2208 Ak\u22121, we have2\u3008f\u3009I = \u3008f\u3009Il + \u3008f\u3009Ir , (5.13)where Il and Ir are the left and right halves of the dyadic interval I. Also, we canexpress E(f |\u2206k) asE(f |\u2206k) =\u2211I\u2208A1k\u22121\u3008f\u3009Il 1Il +\u3008f\u3009Ir 1Ir (5.14)68and E(f |\u2206k\u22121), using (5.13), asE(f |\u2206k\u22121) =\u2211I\u2208A1k\u2212112(\u3008f\u3009Il + \u3008f\u3009Ir)(1Il +1Ir). (5.15)Subtracting (5.15) from (5.14), we getdk =\u2211I\u2208A1k\u2212112(\u2212\u3008f\u3009Il + \u3008f\u3009Ir)(\u22121Il +1Ir) =\u2211I\u2208A1k\u22121\u3008f, hI\u3009|I| hI , (5.16)which concludes the proof.It is well-known that the collection of Haar functions, {hI}I\u2208D\u2217 , is an uncondi-tional Schauder basis for Lp([0, 1]), 1 < p <\u221e. As a consequence, for f \u2208 Lp([0, 1])and 1 < p <\u221e, we get thatf =\u221e\u2211k=0dk (5.17)with respect to the Lp-norm. Furthermore, the square function of f is given byS(f) =(\u2223\u2223\u2223\u2223 \u222b 10f(x)dx\u2223\u2223\u2223\u22232 +\u2211I\u2208D|\u3008f, hI\u3009|2|I|2 .1I) 12. (5.18)If f has a Haar expansion,f =\u2211I\u2208D\u2217\u03b1IhI , (5.19)then its square function isS(f) =(\u2211I\u2208D\u2217|\u03b1I |2 1I) 12, (5.20)and (5.4) becomesMp\u2225\u2225\u2225\u2225( \u2211I\u2208D\u2217|\u03b1I |2 1I) 12\u2225\u2225\u2225\u2225p\u2264 \u2016f\u2016p \u2264 Np\u2225\u2225\u2225\u2225( \u2211I\u2208D\u2217|\u03b1I |2 1I) 12\u2225\u2225\u2225\u2225p(5.21)69for 1 < p <\u221e.For functions of the formf(~x) =\u2211R\u2208Dd\u2217\u03b1RhR(~x), ~x = (x1, . . . , xd) \u2208 [0, 1]d, (5.22)we define the product dyadic square functionSd(f) =( \u2211R\u2208Dd\u2217|\u03b1R|2 1R) 12. (5.23)Inequalities (5.21) hold even when the coefficients \u03b1I are elements of some Hilbertspace (see [52]). Using this fact, one can prove the d-dimensional version of inequal-ities (5.21):Theorem 5.5. (Product Littlewood-Paley Inequalities [8]) For functions of the formf =\u2211R\u2208Dd\u2217 \u03b1RhR, we have(Mp)d\u2016Sd(f)\u2016p \u2264 \u2016f\u2016p \u2264 (Np)d\u2016Sd(f)\u2016p, 1 < p <\u221e. (5.24)The next proposition compares different Lp-norms, p \u2208 (0,\u221e), of a function.In particular, when f is a special linear combination of Haar functions, the nextproposition shows that the norms \u2016f\u2016p are comparable for all p. The proof followsthe strategy outlined in Corollary 1.4 of [48, page 5-6], and we extend that corollaryto dimensions greater than one. Specifically, we establish the following:Proposition 5.6. Let f be a linear combination of Haar functions of the formf(~x) =\u2211R\u2208Adn\u03b1RhR(~x),such that the square function, Sd(f), is constant on [0, 1]d(i.e., Sd(f)(~x) = c(n, d)for every ~x \u2208 [0, 1]d, where the constant c(n, d) depends on the integer n \u2265 1 and onthe dimension d). Then there exist positive constants c1(p, q, d) and c2(p, q, d) such70thatc2\u2016f\u2016q \u2264 \u2016f\u2016p \u2264 c1\u2016f\u2016q for every 0 < p < q <\u221e. (5.25)Proof. First, we observe that the right side of inequality (5.25) is a simple conse-quence of Ho\u00a8lder\u2019s inequality (c1 = 1). Therefore, our goal is to prove thatc2\u2016f\u2016q \u2264 \u2016f\u2016p for every 0 < p < q <\u221e.We will show the above inequality by considering the following three cases:Case 1: 1 < p < q <\u221e.Case 2: 0 < p \u2264 1 < q <\u221e.Case 3: 0 < p < q \u2264 1.In Case 1, using the Littlewood-Paley inequalities and the fact that Sd(f) is inde-pendent of ~x, we get that\u2016f\u2016p \u2265 (Ap)d\u2016Sd(f)\u2016p = (Ap)d\u2016Sd(f)\u2016q \u2265(ApBq)d\u2016f\u2016q. (5.26)Thus, c2\u2016f\u2016q \u2264 \u2016f\u2016p with c2 =(ApBq)d.Now, we turn our attention to the proof of Case 2. First, choose \u03b1 \u2208 (0, 1) suchthat p > \u03b1q. Set r = p\u03b1qand s = rr\u22121 (i.e.,1r+ 1s= 1). Also, set b = 1 \u2212 \u03b1 so that\u03b1 + b = 1. Note that \u03b1r = pq< 1, and bs > 1 since (1\u2212 \u03b1) rr\u22121 > 1 if and only if1 > \u03b1r. Now, applying Ho\u00a8lder\u2019s inequality, we get that\u2016f\u2016qq =\u222b[0,1]d|f(x)|\u03b1q|f(x)|bqdx \u2264 \u2016f\u2016\u03b1q\u03b1rq\u2016f\u2016bqbsq. (5.27)But bs > 1 shows that bsq > q > 1, and Case 1 implies that there exists a positiveconstant c2 such that(c2)bq\u2016f\u2016bqbsq \u2264 \u2016f\u2016bqq . (5.28)Combining inequalities (5.27) and (5.28), we get that\u2016f\u2016qq \u2264 (c2)\u2212bq\u2016f\u2016\u03b1q\u03b1rq\u2016f\u2016bqq71which implies that\u2016f\u2016q(1\u2212b)q \u2264 (c2)\u2212bq\u2016f\u2016\u03b1q\u03b1rq.As a result, \u2016f\u2016q \u2264 (c2)\u2212 b\u03b1\u2016f\u2016\u03b1rq. Thus, Case 2 follows since \u03b1rq = p.Finally, we prove Case 3. Define the numbers \u03b1, b, r and s in the same way asin Case 2 and note that inequality (5.27) still holds. Since q \u2264 1 implies bsq \u2264 bs,using the right side of inequality (5.25), we get\u2016f\u2016bqbsq \u2264 \u2016f\u2016bqbs. (5.29)Now, putting together inequalities (5.27) and (5.29), we get that\u2016f\u2016qq \u2264 \u2016f\u2016\u03b1q\u03b1rq\u2016f\u2016bqbs. (5.30)Since bs > 1 \u2265 q, Case 2 gives us that \u2016f\u2016bqbs \u2264 (c2)\u2212bq\u2016f\u2016bqq . Combining thiswith inequality (5.30), we get that \u2016f\u2016qq \u2264 (c2)\u2212bq\u2016f\u2016\u03b1q\u03b1rq\u2016f\u2016bqq which is equivalent to\u2016f\u2016q(1\u2212b)q \u2264 (c2)\u2212bq\u2016f\u2016\u03b1qarq. As a consequence, \u2016f\u2016q \u2264 (c2)\u2212ba\u2016f\u2016arq. Since \u03b1rq = p,this completes the proof of Case 3.Remark 5.7. To summarize, the Littlewood-Paley inequalities are used in the proof of(5.25) for the case q > p > 1. In particular, the assumption that the square functionis constant on [0, 1]d permits the p-norms of the square function to be comparable.Combined with the Littlewood-Paley inequalities, this fact proves inequality (5.25)for q > p > 1. To prove (5.25) for the full range of the values of p and q, one needsto combine the proof of the special case q > p > 1 and a subtle treatment of Ho\u00a8lder\u2019sinequality. If we assume that the coefficients {\u03b1R} satisfy the inequalitiesc1 \u2264 |\u03b1R| \u2264 c2, for every R \u2208 Adn,where c1 and c2 are positive constants independent of R \u2208 Adn instead, then inequal-ities (5.25) still hold, as it is still true that the p-norms of the square function arecomparable.72Remark 5.8. It is clear that iff(~x) =\u2211R\u2208Adn\u03b1RhR(~x),with {\u03b1R}R\u2208Adn \u2282 {\u22121, 1}, then the square function, (Sdf), is independent of ~x \u2208[0, 1]d. More precisely,Sd(f)(~x) =[\u2211R\u2208Adn|\u03b1R|2 1R(~x)] 12=[#Hdn] 12 ,where #Hdn denotes the cardinality of the set Hdn. An immediate consequence of thisis that\u2016Fr1\u2016L1([0,1]d\u22121) & \u2016Fr1\u2016L2([0,1]d\u22121) & (n\u2212 r1)d\u221222 , r1 = 0, 1, . . . , n, (5.31)whereFr1(~x\u2032) =\u2211R\u2032\u2208Ad\u22121n\u2212r1\u03b1R\u2032hR\u2032(~x\u2032) (5.32)with ~x\u2032 = (x2, x3, . . . , xd) \u2208 [0, 1)d\u22121 and {\u03b1R\u2032} \u2282 {\u22121, 1}. Estimate (5.31) will becrucial in showing Lemma 6.4 and Theorem 1.3. Moreover, for r1 = 0, Fr1 is thefunction used in the proof of the lower bound of the Lp-norm of the discrepancyfunction when 1 < p <\u221e, and, if n ' logN , then \u2016Fr1\u2016p ' (logN)d\u221222 . In addition,defineAr1 =\u2211~r\u2032\u2208Hd\u22121n\u2212r1\u2211R\u2208R~r\u03b1RhR,with (r1, ~r\u2032) \u2208 Hdn, r1 = 0, 1, . . . , n, and {\u03b1R} \u2282 {\u22121.1}. Using the same reasoningas for Fr1 , we get\u2016Ar1\u20161 & \u2016Ar1\u20162 & (n\u2212 r1)d\u221222 , (5.33)r1 = 0, 1, . . . , n. These inequalities will be essential in proving Lemma 7.5.73Chapter 6Two proofs of Theorem 1.3The main objective of this chapter is to prove Theorem 1.3. In other words, weprove the signed small ball conjecture for all d \u2265 3 under an additional constrainton the coefficients {\u03b1R}R\u2208Adn . We will give two proofs of Theorem 1.3. The first onewill be based on a direct estimate of the L\u221e-norm of the hyperbolic sum Hn, andthe second one will be based on a duality argument which uses Riesz products. Thesecond proof is particularly important since the techniques used there can also beused to obtain lower bounds of hyperbolic sums in other function spaces, such asexponential Orlicz spaces. We will discuss this more in Chapter 7. However, beforeproving Theorem 1.3, we first motivate the additional constraint on the coefficientsthat appears in Theorem 1.3, the splitting condition.6.1 Motivation for the splitting conditionRecall that a coefficient \u03b1R \u2208 {\u22121, 1} satisfies the splitting condition if \u03b1R = \u03b1R1 \u00b7\u03b1R\u2032with \u03b1R1 , \u03b1R\u2032 \u2208 {\u22121, 1} and R = R1\u00d7R\u2032 = R1\u00d7R2\u00d7 \u00b7 \u00b7 \u00b7 \u00d7Rd. If one assumes thatthe coefficients of the signed hyperbolic sum satisfy the splitting condition, then thesigned hyperbolic sum Hn takes the formHn(~x) =n\u2211r1=0Br1(x1)Fr1(~x\u2032), ~x = (x1, ~x\u2032) \u2208 [0, 1]d, (6.1)74withBr1(x1) =\u2211|R1|=2\u2212r1\u03b1R1hR1(x1)andFr1(~x\u2032) =\u2211|R\u2032|=2\u2212(n\u2212r1)\u03b1R\u2032hR\u2032(~x\u2032), r1 = 0, 1, . . . , n.The functions {Br1}nr1=0 can be viewed as random variables defined on the probabilityspace (\u2126,B, P ) = ([0, 1),B([0, 1)), |\u00b7|), where |\u00b7| is the Lebesgue measure and B([0, 1))is the Borel \u03c3 -algebra on the interval [0, 1), as we will explain later. The power of thesplitting condition is that, when used to rewrite the hyperbolic sum as in (6.1), therandom variables {Br1}nr1=0 are independent, and the independence of these randomvariables will facilitate the computation of the supremum norm of Hn and lead tothe gain of n1\/2 over the trivial bound. We prove the independence of {Br1}nr1=0 afterthis preliminary discussion.To prove the independence of the generalized Rademacher functions {Br1}nr1=0,we will need to represent them in terms of iid binary random variables. To give allthe details of this representation, we need to recall some facts from probability theoryabout the dyadic sequence {di(x)}i\u22651 \u2282 {0, 1} associated to the dyadic expansion ofa point x \u2208 [0, 1),x =\u221e\u2211i=1di(x)2i. (6.2)By convention, we use the terminating dyadic expansion for the dyadic rationals{x = k2m: k,m \u2208 Z,m \u2265 0, 0 \u2264 k < 2m}.Fact 1: Each di is a random variable.Fact 2: The sequence {di, i \u2265 1} is iid. In other words, each di is identicallydistributed with P [di = 1] = P [di = 0] =12, and the di are independent.The proof of these facts can be found in [38, page 99-100].To each such dyadic sequence we associate a sequence of the iid\u00b11-valued randomvariables, {Xi}i\u22651, defined by75Xi =\uf8f1\uf8f2\uf8f3\u22121, if di = 01, if di = 1 ,where i \u2208 Z+ = {1, 2, . . .}. For every i \u2208 Z+, we let~Xi = (X1, . . . , Xi),~di = (d1, . . . , di),~bi = (b1, b2, . . . , bi) \u2208 {\u22121, 1}i,with the convention that {\u22121, 1}0 = {0}, ~d0 = 0, ~X0 = 0 and ~b0 = 0. Now, fixk \u2208 {0, 1, . . . , n} and a collection of numbers {\u03b1I}I\u2208D:|I|=2\u2212k \u2282 {\u22121, 1}, and definethe map:ak : {\u22121, 1}k \u2192 {\u03b1I}I\u2208D:|I|=2\u2212k , ak(~bk) = \u03b1I , (6.3)where I is the dyadic interval with |I| = 2\u2212k given byI =[k\u2211i=1bi + 12i+1,k\u2211i=1bi + 12i+1+ 2\u2212k).Observe that every x \u2208 I has the property that ~Xk(x) = ~bk. Note that a0(~b0) = \u03b1[0,1)and that this map is onto. Now, it is easy to verify thatBr1(x1) = ar1( ~Xr1(x1))Xr+1(x1), r1 = 0, 1, . . . , n. (6.4)Hence,Hn(~x) =n\u2211r1=0ar1( ~Xr1(x1))Xr1+1(x1)Fr1(~x\u2032), ~x = (x1, ~x\u2032) \u2208 [0, 1]d. (6.5)Proposition 6.1. The random variables {Bk = ak( ~Xk)Xk+1}nk=0 are iid binary ran-dom variables on the probability space ([0, 1),B([0, 1)), P = | \u00b7 |).76Proof. First, for fixed values of 0 \u2264 k \u2264 n and b0 \u2208 {\u22121, 1}, we show thatP (Xk+1ak( ~Xk) = b0) =12. We will then use this to show that the random vari-ables {ak( ~Xk)Xk+1}nk=0 are independent.Without loss of generality, we can assume that 0 < k \u2264 n. Fixing b0 \u2208 {\u22121, 1},we getP (Xk+1ak( ~Xk) = b0) =\u2211~bk\u2208{\u22121,1}kP(Xk+1ak( ~Xk) = b0, ~Xk = ~bk)=\u2211~bk\u2208{\u22121,1}kP(Xk+1 =b0ak(~bk), ~Xk = ~bk)= 2k12(12)k=12.Now, using the independence of the random variables {Xi}n+1i=1 , for any ~bn+1 \u2208{\u22121, 1}n+1, we get thatP (X1a0( ~X0) = b1, . . . , Xn+1an( ~Xn) = bn+1) = P (X1 = b\u02dc1, . . . , Xn+1 = b\u02dcn+1)=(12)n+1=n\u220fk=0P (Xk+1ak( ~Xk) = bk+1),where b\u02dc1, . . . , b\u02dcn+1 are \u00b11-valued deterministic constants depending on b1, . . . , bn+1.This last equality gives the independence of the random variables {ak( ~Xk)Xk+1}nk=0.6.2 The first proof of Theorem 1.3The next proposition gives us one of the main ingredients of the first proof of Theorem1.3: an explicit formula for the L\u221e-norm of a linear combination of iid \u00b11-valued77random variables.Proposition 6.2. Let (\u2126,F , P ) be a probability space and {Ui}ni=1 be a finite collec-tion of \u00b11-valued independent random variables for which P (Ui = \u00b11) > 0 for everyi = 1, . . . , n. If (ci)ni=1 is any finite deterministic sequence of real numbers, then\u2225\u2225\u2225\u2225\u2225n\u2211i=1ciUi\u2225\u2225\u2225\u2225\u2225\u221e=n\u2211i=1|ci|.Proof. It is clear that |\u2211ni=1 ciUi(\u03c9)| \u2264\u2211ni=1 |ci| for every \u03c9 \u2208 \u2126. So, to complete theproof, we need to find \u03c90 \u2208 \u2126 such that Ui(\u03c90) = sgn(ci), i = 1, . . . , n, where sgn isthe signum function defined in (1.7). Let A =\u22c2ni=1{\u03c9 \u2208 \u2126 : Ui(\u03c9) = sgn(ci)}. Usingthe independence of the random variables {Ui}ni=1, we get that P (A) =\u220fni=1 P (Ui =sgn(ci)) > 0, which shows that A is not a set of measure zero. Thus,\u2225\u2225\u2225\u2225\u2225n\u2211i=1ciUi\u2225\u2225\u2225\u2225\u2225\u221e=n\u2211i=1|ci|.The first proof of Theorem 1.3. Using the definition of the supremum norm and (6.5),we get\u2016Hn\u2016\u221e = sup~x\u2032\u2208[0,1]d\u22121supx1\u2208[0,1]\u2223\u2223\u2223\u2223\u2223n\u2211r1=0Fr1(~x\u2032)ar1(~Xr1(x1))Xr1+1(x1)\u2223\u2223\u2223\u2223\u2223 . (6.6)Now, if we can show that\u2016Hn\u2016\u221e \u2265n\u2211r1=0\u2016Fr1\u2016L1([0,1]d\u22121), (6.7)then we can use (5.31) to obtain the desired inequality. To obtain (6.7), we needto rewrite the inner supremum appearing on the right side in (6.6). Fix ~x\u2032 =(x2, x3, \u00b7 \u00b7 \u00b7 , xd) \u2208 [0, 1]d\u22121. Applying both Proposition 6.1 and Proposition 6.2 to78Ur1 = \u03b1r1( ~Xr1)Xr1+1 and cr1 = Fr1(x2, x3, \u00b7 \u00b7 \u00b7 , xd), r1 = 0, 1, . . . , n, we get thatsupx1\u2208[0,1]\u2223\u2223\u2223\u2223\u2223n\u2211r1=0Fr1(~x\u2032)ar1(~Xr1(x1))Xr1+1(x1)\u2223\u2223\u2223\u2223\u2223 =n\u2211r1=0|Fr1(~x\u2032)|. (6.8)Combining (6.6) and (6.8), we get\u2016Hn\u2016\u221e = sup~x\u2032\u2208[0,1]d\u22121n\u2211r1=0|Fr1(~x\u2032)|. (6.9)Inequality (6.7) easily follows since\u2016Hn\u2016\u221e \u2265\u2225\u2225\u2225\u2225\u2225n\u2211r1=0|Fr1|\u2225\u2225\u2225\u2225\u2225L1([0,1]d\u22121)=n\u2211r1=0\u2016Fr1\u2016L1([0,1]d\u22121).Now, using inequality (5.31), estimate (6.7) becomes\u2016Hn\u2016\u221e \u2265 cdn\u2211r1=0\u2016Fr1\u2016L2([0,1]d\u22121). (6.10)Using the orthogonality of Haar functions, we have that\u2016Fr1\u2016L2([0,1]d\u22121) = \u2016Sd\u22121(Fr1)\u2016L2([0,1]d\u22121) & (n\u2212 r1)d\u221222 . (6.11)Finally, putting inequalities (6.10) and (6.11) together, we obtain the desired result.Remark 6.3. The two basic components of the first proof are the following:\u2016Hn\u2016\u221e = sup~x\u2032\u2208[0,1]d\u22121n\u2211r1=0|Fr1(~x\u2032)|, (6.12)\u2016Fr1\u2016L1([0,1]d\u22121) & \u2016Fr1\u2016L2([0,1]d\u22121), (6.13)for r1 = 0, 1, . . . , n. According to Remark 5.7, if we consider coefficients {\u03b1R1\u03b1R\u2032}79in Hn for which \u03b1R1 \u2208 {\u22121, 1} and 0 < c1 \u2264 |\u03b1R\u2032| \u2264 c2 for all R1 \u00d7 R\u2032 \u2208 Adn,then inequality (6.13) still holds. Moreover, for the same collection of coefficients,the splitting condition ensures that equality (6.12) also remains true. Based onthese two facts, the small ball conjecture, Conjecture 1.1, can be proved for thisspecial collection of coefficients. Indeed, using (6.12), (6.13), and the Cauchy-Schwarzinequality, we obtain\u2016Hn\u2016\u221e &n\u2211r1=0\u2016Fr1\u2016L2([0,1]d\u22121)&n\u2211r1=0( \u2211~r\u2032\u2208Hd\u22121n\u2212r1\u2211R\u2032\u2208R~r\u2032|\u03b1R\u2032|22\u2212(n\u2212r1))1\/2& 2\u2212nn\u2212(d\u22122)\/2n\u2211r1=02r1\u2211~r\u2032\u2208Hd\u22121n\u2212r1\u2211R\u2032\u2208R~r\u2032|\u03b1R\u2032 |& 2\u2212nn\u2212(d\u22122)\/2n\u2211r1=0\u2211~r\u2032\u2208Hd\u22121n\u2212r1\u2211|R1|=2\u2212r1\u2211R\u2032\u2208R~r\u2032|\u03b1R1||\u03b1R\u2032|& 2\u2212nn\u2212(d\u22122)\/2\u2211|R|=2\u2212n|\u03b1R|,which shows the validity of Conjecture 1.16.3 The second proof of Theorem 1.3The second proof of Theorem 1.3 will use a duality argument. That is, for every col-lection of coefficients {\u03b1R} \u2282 Asplit, we will construct a test function \u03c8 \u2208 L1([0, 1]d)such that\u2016\u03c8\u2016L1 \u2264 C,80where C is a constant which does not depend on the choice of coefficients {\u03b1R} \u2282Asplit or the integer n \u2265 1. In addition, the test function \u03c8 will also satisfy\u3008\u03c8,Hn\u3009 :=\u222b[0,1]d\u03c8Hndx & nd2 ,where Hn =\u2211|R|=2\u2212n \u03b1RhR with {\u03b1R} \u2282 Asplit. Theorem 1.3 follows from the abovetwo inequalities by a simple application of Ho\u00a8lder\u2019s inequality to \u3008\u03c8,Hn\u3009. We willprove the two inequalities in Lemma 6.4. This lemma is also essential in provingTheorem 1.6. The construction of the test function \u03c8 will be similar to that inTemlyakov\u2019s proof of the two-dimensional small ball inequality, given in Section 4.1of Chapter 4.Lemma 6.4. For any n \u2265 1, d \u2265 3, and any choice of coefficients {\u03b1R}R\u2208Adn \u2282 Asplit,there exists a function \u03c8 \u2208 L1([0, 1]d) with \u2016\u03c8\u2016L1 = 1 such that\u3008\u03c8,Hn\u3009 =\u222b[0,1]d\u03c8Hndx & nd2 . (6.14)Here, the implicit constant depends only on the dimension d and not on the collectionof coefficients {\u03b1R}R\u2208Adn \u2282 Asplit or the integer n \u2265 1.Proof. Recall the expression of the functionHn(~x) =\u2211R\u2208Adn\u03b1RhR(~x) (6.15)given in (6.1) and define the test function \u03c8 as the Riesz product\u03c8(~x) =n\u220fr=0(1 +Br(x1)sgn(Fr(~x\u2032))). (6.16)Here, we have replaced r1 with r for our convenience. We claim that \u03c8 has theproperty that\u2016\u03c8\u2016L1([0,1]d) = 1 (6.17)81and satisfies (6.14). Indeed, we first observe that, for any ~x \u2208 [0, 1]d, \u03c8(~x) \u2265 0,since each factor in the product forming \u03c8 is nonnegative. Expanding the productin (6.16), we get\u03c8 = 1 + \u03c81 + \u03c82, (6.18)with\u03c81(~x) =n\u2211r=0Br(x1)sgn(Fr(~x\u2032)), (6.19)and\u03c82(~x) =n\u2211k=2\u22110\u2264s1<\u00b7\u00b7\u00b7 0, for every k \u2208 {1, . . . , 2(n+1)d}. To proceed, we will firstshow how (6.30) follows from (6.32), and then we will prove the claim.To see how (6.30) follows from the (6.32), let N = 2(n+1)d. Lemma 6.6 impliesthatE(sup1\u2264k\u2264N|Yk|).\u221an. (6.33)Combining (6.31) and (6.33) and using the fact that Zk is constant on the dyadiccube Qk for a fixed \u03c9 \u2208 \u2126, we get thatE\u2225\u2225\u2225 \u2211R\u2208Adn\u03b1RhR\u2225\u2225\u2225\u221e=E(sup1\u2264k\u2264N|Zk|)=E(sup1\u2264k\u2264NYkE|Zk|A). n d\u221212 E(sup1\u2264k\u2264NYk). n d\u221212\u221an.n d2 .Now we prove the claim. First, we estimate the distribution function of \u03c6(Yk) asfollows:P(\u03c6(Yk) > \u03bb) =P(Yk > \u03c6\u22121(\u03bb))=P(Yk >\u221a2 log \u03bb)=P(|Zk| > E|Xk|A\u221a2 log \u03bb)\u2264P(|Zk| > AC\u22121d nd\u221212\u221a2 log \u03bb).87Therefore, using Bernstein\u2019s inequality, stated in Lemma 4.2, we getP(\u03c6(Yk) > \u03bb) \u22642 exp(nd\u22121 log(\u03bb\u2212A2C\u22122d 2)4#Hdn)\u22642 exp(nd\u22121 log(\u03bb\u2212A2C\u22122d 2)4nd\u22121C \u2032d). 1\u03bbA2C\u22122d2C\u2032d. 1\u03bb2,for A chosen large enough. Now, applying the distribution identity,E(\u03c6(Yk)) =\u222b \u221e0P(\u03c6(Yk) > \u03bb)d\u03bb,we have thatE(\u03c6(Yk)) .\u222b 10d\u03bb+\u222b \u221e1\u03bb\u22122d\u03bb \u2264 C \u2032\u2032d .88Chapter 7Orlicz space bounds for specialclasses of hyperbolic sumsAs we have already mentioned, the Lp-bounds of the signed hyperbolic sums, Hn =\u2211|R|=2\u2212n \u03b1RhR with {\u03b1R} \u2282 {\u22121, 1}, are equivalent to the L2-bounds. Indeed,applying Proposition 5.6, there exist positive constants Ci(2, p, d), i = 1, 2, such thatC1(2, p, d)\u2016Hn\u20162 \u2264 \u2016Hn\u2016p \u2264 C1(2, p, d)\u2016Hn\u20162, 0 < p <\u221e.Using the orthogonality of Haar functions, we obtain \u2016Hn\u20162 = \u2016Sd(Hn)\u20162 ' n d\u221212 .Therefore, \u2016Hn\u2016p ' n d\u221212 for all p in (0,\u221e), where the implicit constants depend onthe value of p and the dimension d.In this chapter, we explore lower bounds of signed hyperbolic sums in exponentialOrlicz spaces. We have chosen to work with these spaces because, for 1 \u2264 p < \u221e,exponential Orlicz spaces can be considered as intermediate function spaces betweenLp and L\u221e, and the L\u221e-lower bounds of signed hyperbolic sums remain conjecturedfor all d \u2265 3 whereas the Lp-lower bounds are well-understood. Also, lower boundsof the discrepancy function have already been studied in exponential Orlicz spacesin dimension d = 2 [6]. We are interested in obtaining analogous lower bounds inthe case of signed hyperbolic sums. In dimension d = 2, we will adjust Temlyakov\u2019sproof of the two-dimensional small ball inequality for signed hyperbolic sums to the89framework of exponential Orlicz spaces, and show that the lower bounds obtained forthe discrepancy function can also be realized for signed hyperbolic sums. In higherdimensions, we will obtain lower bounds of signed hyperbolic sums in exponentialOrlicz spaces under the additional assumption that the coefficients \u03b1R are from theset Asplit.Before proving our main results, we first present the background material onOrlicz spaces needed to follow the proofs of the main theorems of this chapter,Theorem 1.6 and Theorem 7.4.7.1 Orlicz spacesFor p \u2265 1, Orlicz spaces are natural generalizations of Lp-spaces. To see this, notethat |x|p is a convex function for p \u2265 1, and a function f is in Lp(X,\u00b5) if\u222bX|f |p d\u00b5 <\u221e,whereas, roughly speaking, f is in the Orlicz space associated to a convex function\u03c6 : R\u2192 [0,\u221e) if \u222bX\u03c6(f) d\u00b5 <\u221e.More precisely, let \u03c6 : R \u2192 [0,\u221e] be a convex, even function with \u03c6(0) = 0and \u03c6 6\u2261 0. Given a finite measure space (X,M, \u00b5), one defines the Orlicz spaceassociated to the function \u03c6 asL\u03c6(X,\u00b5) ={f : X \u2192 R measurable : \u2203a > 0,\u222bX\u03c6(af)d\u00b5 <\u221e}.We endow Orlicz spaces with the so-called Luxembourg norm: for any measurablefunction f \u2208 L\u03c6(X,\u00b5), the norm of f is defined by\u2016f\u2016\u03c6 := inf{t > 0 :\u222bX\u03c6(f\/t)d\u00b5 \u2264 1}.90This is precisely the Minkowski functional associated to the setV ={f measurable :\u222bX\u03c6(f)d\u00b5 \u2264 1}.Note thatL\u03c6(X,\u00b5) ={f measurable : \u2016f\u2016\u03c6 <\u221e}.Such a function \u03c6 is increasing on R+ := [0,\u221e). Indeed, let 0 \u2264 s1 < s2, then theconvexity of \u03c6 implies that\u03c6(s1) \u2264 s2 \u2212 s1s2\u03c6(0) +s1s2\u03c6(s2) \u2264 \u03c6(s2).From the definition we can see that, to understand the composition of L\u03c6(X,\u00b5),we only need to know how \u03c6(x) grows as |x| \u2192 \u221e. More specifically, if \u03c6 \u223c h as|x| \u2192 \u221e, i.e.,C1h \u2264 \u03c6 \u2264 C2h, |x| \u2265M,for some positive constants C1, C2,M , then the two quantities \u2016f\u2016\u03c6 and \u2016f\u2016h areequivalent and we see that it suffices to know the asymptotic behaviour of \u03c6.Orlicz spaces coincide with Lp spaces for certain choices of \u03c6. If \u03c6(x) = |x|p (1 \u2264p <\u221e), then L\u03c6(X,\u00b5) = Lp(X,\u00b5) and \u2016f\u2016\u03c6 = \u2016f\u2016p, and if\u03c6(x) =\uf8f1\uf8f2\uf8f30, if |x| \u2264 1\u221e, otherwise ,then L\u03c6(X,\u00b5) = L\u221e(X,\u00b5) and the \u2016f\u2016\u03c6 = \u2016f\u2016\u221e. We will be interested in the Orliczspaces which are associated with functions of the form\u03c6a(x) =\uf8f1\uf8f2\uf8f3e|x|a \u2212 1, |x| \u2265 1(e\u2212 1)|x|, otherwise,where a > 0. These spaces are known as exponential Orlicz spaces and are denotedby exp(La).91To prove our main results, we will need an Orlicz space version of Ho\u00a8lder\u2019s in-equality, i.e., given \u03c6 as in the definition of L\u03c6, we want to find C > 0 and a function\u03c6\u2217 such that \u222b|fg|d\u00b5 \u2264 C\u2016f\u2016\u03c6\u2016g\u2016\u03c6\u2217 , (7.1)for all f \u2208 L\u03c6 and g \u2208 L\u03c6\u2217. Note that this requires that \u03c6\u2217 is an even convex functionwith \u03c6\u2217(0) = 0. To prove this generalization of Ho\u00a8lder\u2019s inequality, \u03c6\u2217 is taken to bethe function conjugate to \u03c6, i.e., \u03c6\u2217 : R\u2192 [0,\u221e] is the function defined by\u03c6\u2217(y) = sup{xy \u2212 \u03c6(x) : x \u2208 R}. (7.2)It follows from the definition that \u03c6\u2217 is convex, even and satisfies \u03c6\u2217(0) = 0. Anotherimmediate consequence of this definition is Young\u2019s inequality:xy \u2264 \u03c6(x) + \u03c6\u2217(y)for all x and y. Young\u2019s inequality implies that|f(x)g(x)| \u2264 \u03c6(f(x)) + \u03c6\u2217(g(x)), for all x \u2208 X. (7.3)We can now see that (7.1) holds for C = 2. Indeed, it suffices to show (7.1) for\u2016f\u2016\u03c6 = \u2016g\u2016\u03c6\u2217 = 1 and, in such a case, we obtain (7.1) simply by integrating bothsides of inequality (7.3).This generalization of Ho\u00a8lder\u2019s inequality will be applied to \u03c6a and \u03c6\u2217a to prove themain results of the chapter. Note that, by convention, the Orlicz spaces associatedto \u03c6\u2217a are denoted by L(logL)1a , and it can be verified that \u03c6\u2217a(y) \u223c |y|(log(1 + |y|))1aas |y| \u2192 \u221e.The next proposition deals with estimating Orlicz space norms of an indicatorfunction. A detailed proof can be found in [37].Proposition 7.1. Let E \u2208M. Then12\u00b5(E)(\u03c6a)\u22121(1\u00b5(E))\u2264 \u20161E \u2016\u03c6\u2217a \u2264 \u00b5(E)(\u03c6a)\u22121(1\u00b5(E)),92where (\u03c6a)\u22121 is the inverse of \u03c6a(x), x \u2265 0.Applying the above proposition to a probability measure gives us that\u20161E \u2016L(logL) 1a \u2264 P(E) \u00b7 (1\u2212 log(P(E))1a . (7.4)7.2 The proof of Theorem 1.6Before we proceed to the proof of Theorem 1.6, note that it follows from the definitionof exponential Orlicz spaces that, for each 1 < p <\u221e and a > 0, we have continuousembeddings L\u221e \u2282 exp(La) \u2282 Lp. Therefore,nd\u221212 . \u2016Hn\u20162 . \u2016Hn\u2016exp(La) (7.5)holds for any choice of coefficients {\u03b1R} \u2282 Asplit. In fact, (7.5) holds for any collection{\u03b1R} \u2282 {\u22121, 1}. We can see that the lower bound in (7.5) is better than the onegiven in Theorem 1.6 when a < 2, but when 2 \u2264 a < \u221e, inequality (1.16) from thestatement of Theorem 1.6 gives us an improvement over the lower bound in (7.5) bya factor of n12\u2212 1a for any collection of coefficients {\u03b1R} \u2282 Asplit. It is for this reasonthat Theorem 1.6 is stated only for a \u2265 2. As we will soon see, the proof of Theorem1.6 works for all a > 0.Proof of Theorem 1.6. From Lemma 6.4 there is a positive function \u03c8 such that\u2016\u03c8\u2016L1([0,1]d) = 1 and \u222b[0,1]d\u03c8Hndx & nd2 . (7.6)Also note that this test function can be written as\u03c8 = 2n+1 1E,whereE ={~x = (x1, ~x\u2032) \u2208 [0, 1]d : Br(x1) \u00b7 sgn(Fr(~x\u2032)) = 1 for all r = 0, . . . , n}.93Furthermore, |E| = 2\u2212(n+1) since \u222b[0,1]d\u03c8dx = 1. Applying Ho\u00a8lder\u2019s inequality forOrlicz spaces to (7.6), we get\u20162n+1 1E \u2016L(logL) 1a \u2016Hn\u2016exp(La) & nd2 . (7.7)Combining (7.4) with the fact that |E| = 12n+1, we obtain\u20162n+1 1E \u2016L(logL) 1a \u2264 Cd21an1a , n > 1. (7.8)Now, using (7.8), (7.7) becomes\u2016Hn\u2016exp(La) \u2265 (Cd)\u221212\u2212 1an d2\u2212 1a ,which completes the proof.From the proof of Theorem 1.6 we can see that the constant C(d, a) appearingin (1.16) must be equal to (Cd)\u221212\u22121a . Taking the limit of both sides of inequality(1.16) as the scale of integrability a approaches infinity, we get the signed small ballinequality when {\u03b1R} \u2282 Asplit.It is worth mentioning that bounds of this type have already appeared in thefield of irregularities of distribution of points (see [6]). In particular, Bilyk et al. [6]showed that, in d = 2, the discrepancy function,DN(x1, x2) := #(PN \u2229 [0, x1)\u00d7 [0, x2))\u2212Nx1x2,associated to the N -point set PN \u2282 [0, 1]2 satisfies the following lower bound\u2016DN\u2016exp(La) & (logN)1\u2212 1a , 2 \u2264 a <\u221e.They used a duality argument in the setting of exponential Orlicz spaces in theirproof which, in fact, provided the motivation for the proof of Theorem 1.6.One can also use a duality argument in framework of exponential Orlicz spacesto prove Conjecture 1.5 in d = 2. Indeed, recall Temlyakov\u2019s test function used in94the proof of the two-dimensional small ball inequality,\u03a8 =n\u220fk=0(1 + f(k,n\u2212k)) = 2n+1 1E,whereE = {~x \u2208 [0, 1]2 : f(k,n\u2212k) = 1, for all k \u2208 {0, 1, . . . , n}}.Adapting the proof of the two-dimensional small ball inequality to the signed hyper-bolic sum Hn =\u2211nk=0 f(k,n\u2212k), we have\u3008\u03a8, Hn\u3009 & n.Applying Ho\u00a8lder\u2019s inequality for Orlicz spaces as in the proof of Theorem 1.6, we get\u2016Hn\u2016exp(La) & n1\u2212 1a , 2 \u2264 a <\u221e.7.3 Study of signed hyperbolic sumsIn this section, we are concerned with exploring L\u221e and exp(La)-lower bounds of alinear combination of Haar functions with signed coefficients \u03b1R \u2208 {\u22121, 1}. To moti-vate our discussion, we first restate the signed small ball conjecture and Conjecture1.5 in an equivalent form.Conjecture 7.2. (Restatement of the signed small ball conjecture) There existsa vector ~\u00b5 = (\u00b50, . . . , \u00b5n) \u2208 {\u22121, 1}n+1 such that, for any choice of coefficients{\u03b2R} \u2282 {\u22121, 1} and any integer n \u2265 1, we have the inequality\u2225\u2225\u2225 \u2211~r\u2208Hdn\u00b5r1\u2211R\u2208R~r\u03b2RhR\u2225\u2225\u2225\u221e& n d2 , d \u2265 3. (7.9)By choosing \u00b5r1 = 1, r1 = 0, 1, . . . , n and \u03b2R = \u03b1R, R \u2208 Adn, it is immediatethat the small ball conjecture implies (7.9). To get the reverse implication, (7.9)implies the small ball conjecture, simply set \u03b2R = \u00b5r1\u03b1R for R = R1 \u00d7R\u2032 \u2208 Adn with|R1| = 2\u2212r1 , r1 = 0, . . . , n.95Conjecture 7.3. (Restatement of Conjecture 1.5) There exists a vector~\u00b5 = (\u00b50, . . . , \u00b5n) \u2208 {\u22121, 1}n+1 such that, for any choice of coefficients {\u03b2R} \u2282 {\u22121, 1}and any integer n > 1, we have the inequality\u2225\u2225\u2225 \u2211~r\u2208Hdn\u00b5r1\u2211R\u2208R~r\u03b2RhR\u2225\u2225\u2225exp(La)& n d\u221212 + 12\u2212 1a , 2 \u2264 a <\u221e, (7.10)for all d \u2265 3.The equivalence can be obtained by methods similar to those presented before.Using the techniques developed in the previous section, we will show the existenceof (n+ 1)-dimensional vectors ~\u00b5 and ~\u03bd \u2208 {\u22121, 1}n+1 which satisfy inequalities (7.9)and (7.10) respectively, though these vectors may depend on the choice of coefficients{\u03b1R} \u2282 {\u22121, 1}. Specifically, we show:Theorem 7.4. For any integer d \u2265 3 and any choice of coefficients {\u03b1R} \u2282 {\u22121, 1},there exist (n + 1)-dimensional vectors ~\u00b5 = (\u00b50, . . . , \u00b5n) \u2208 {\u22121, 1}n+1, n \u2265 1, and~\u03bd = (\u03bd0, . . . , \u03bdn) \u2208 {\u22121, 1}n+1, n > 1, such that\u2225\u2225\u2225 \u2211~r\u2208Hdn\u00b5r1\u2211R\u2208R~r\u03b1RhR\u2225\u2225\u2225\u221e& n d2 , (7.11)and \u2225\u2225\u2225 \u2211~r\u2208Hdn\u03bdr1\u2211R\u2208R~r\u03b1RhR\u2225\u2225\u2225exp(La)& n d\u221212 + 12\u2212 1a , (7.12)for 2 \u2264 a <\u221e.This theorem will be proved by employing a duality argument which uses a testfunction in the form of a Riesz product. Instead of working with the function\u2211~r\u2208Hdn\u00b5r1\u2211R\u2208R~r\u03b1RhR,we let \u00b5r1 : \u2126 \u2192 {\u22121, 1} be iid random variables with P(\u00b5r1 = \u00b11) = 12 , r1 =960, 1, . . . , n and consider the functionH~\u00b5(\u03c9, ~x) =\u2211~r\u2208Hdn\u00b5r1(\u03c9)\u2211R\u2208R~r\u03b1RhR(~x)defined on the probability space(\u2126\u00d7 [0, 1)d,P\u00d7 | \u00b7 |). Then, for a particular choiceof \u03c9 \u2208 \u2126, H~\u00b5(\u03c9, \u00b7) takes the form of the function inside the L\u221e and exp(La)-normin (7.9) and (7.10) respectively. The usage of these random variables combined with(5.33) will be crucial in getting the lower bounds in (7.11) and (7.12).The idea of introducing random coefficients for proving existence results is notnew; as we saw in Chapter 4, it was already used to prove the sharpness of theconjectured lower bound in the small ball inequality for all d \u2265 3. However, therandomization used in Chapter 4 required each coefficient in the collection {\u03b1R} tobe an iid \u00b11-valued random variable, and there were about nd\u221212n iid random coef-ficients needed for that proof, whereas the randomization used here is considerablymilder, employing just n+ 1 iid random variables.Just as we proved the lower bound appearing in Theorem 1.3 separately in Lemma6.4, we will prove the lower bound appearing in Theorem 7.4 in the following lemma.Lemma 7.5. Let d \u2265 3 and n \u2265 1. For each collection of coefficients \u03b1R \u2208 {\u22121, 1},there exists a positive test function \u03c8~\u00b5 \u2208 L1(\u2126\u00d7 [0, 1]d) such that\u2016\u03c8~\u00b5\u2016L1([0,1]d\u00d7\u2126) = 1 (7.13)andE~\u00b5E(\u03c8~\u00b5H~\u00b5) & nd2 , (7.14)where E~\u00b5 is the expected value with respect to the probability measure P and E(\u03c8~\u00b5H~\u00b5) =\u222b[0,1]d(\u03c8~\u00b5H~\u00b5)dx is the expected value with respect to the Lebesgue measure.Proof. We want to find a test function \u03c8~\u00b5 \u2208 L1(\u2126\u00d7 [0, 1]d) such that\u2016\u03c8~\u00b5\u2016L1(\u2126\u00d7[0,1]d) = 1 (7.15)97andE~\u00b5E(\u03c8~\u00b5H~\u00b5) & nd2 . (7.16)Note first thatH~\u00b5(\u03c9, ~x) =n\u2211r1=0\u00b5r1(\u03c9)Ar1(~x),whereAr1 =\u2211~r \u2032\u2208Hd\u22121n\u2212r1\u2211R\u2208R~r\u03b1RhR,with (r1, ~r\u2032) \u2208 Hdn. Consider the function \u03c8~\u00b5 defined by\u03c8~\u00b5(\u03c9, ~x) =n\u220fr=0(1 + \u00b5r(\u03c9)sgn(Ar(~x))) , (7.17)where we have replaced r1 with r for our convenience. Observe that, for all (\u03c9, ~x) \u2208\u2126\u00d7 [0, 1]d, \u03c8~\u00b5(\u03c9, ~x) \u2265 0, since each factor in the product is nonnegative. Expandingthe product in (7.17), we get\u03c8~\u00b5 = 1 + \u03c81 + \u03c82, (7.18)with\u03c81(\u03c9, ~x) =n\u2211r=0\u00b5r(\u03c9)sgn(Ar(~x)), (7.19)and\u03c82(\u03c9, ~x) =n\u2211k=2\u22110\u2264s1<\u00b7\u00b7\u00b7 1.This shows that there exists \u03c90 \u2208 \u2126 such that\u222b[0,1]d\u03c6a(H\u00b5(\u03c90, \u00b7)(2)\u22121C(d, a)nd2\u2212 1a)dx > 1,so \u2016H~\u00b5(\u03c90, \u00b7)\u2016exp(La) \u2265 12C(d, a)nd2\u2212 1a which completes the proof of Theorem 7.4.101Chapter 8Conclusions and future projectsIn Chapter 6 we gave two proofs of the signed small ball inequality under the addi-tional assumption that the coefficients are in Asplit. Imposing the splitting propertyallowed us to exploit the independence of the random variables {\u2211|R1|=2\u2212r1 \u03b1R1hR1 ,r1 = 0, . . . , n} defined on the probability space ([0, 1), | \u00b7 |). This, together with theLittlewood-Paley inequalities for Haar functions, enabled us to show that the signedsmall ball inequality holds in all dimensions d \u2265 3. In particular, it was shown that\u2016Hn\u2016\u221e &n\u2211r1=0\u2016Fr1\u20161 &n\u2211r1=0\u2016Fr1\u20162 &n\u2211r1=0(n\u2212 r1) d\u221222 & n d2 .It is natural to wonder if the signed small ball inequality might also hold forcoefficients which satisfy a different splitting property. That is, consider the collectionof coefficientsA\u2032split = {\u03b1R : R \u2208 Adn, \u03b1R = \u03b1R1\u00d7R2 \u00b7 \u03b1R3\u00d7\u00b7\u00b7\u00b7\u00d7Rd with \u03b1R1\u00d7R2 , \u03b1R3\u00d7\u00b7\u00b7\u00b7\u00d7Rd \u2208 {\u22121, 1}}.Problem 1: Given any n \u2265 1 and d \u2265 3, does the signed small ball inequality holdfor every collection of coefficients {\u03b1R} \u2282 A\u2032split?The new splitting property elevates the complexity of the problem. The pathswhich were taken in the proofs of Theorem 1.3 can not be followed here since, in this102case, the corresponding random variables generated by the new splitting property,\u2211|R\u2032|=2\u2212r1\u2212r2\u03b1R\u2032hR\u2032 , r1 \u2208 {0, . . . , n} and r2 \u2208 {0, . . . , n\u2212 r1}with R\u2032 = R1 \u00d7 R2, are not independent. This can be seen in the following way.First, observe that the signed small ball inequality is sharp when the coefficientsare restricted to the set A\u2032split. In other words, there exists a constant Cd > 0 andcoefficients {\u03b1R} \u2282 A\u2032split such that\u2225\u2225\u2225\u2225 \u2211|R|=2\u2212n\u03b1R1\u00d7R2\u03b1R3\u00d7\u00b7\u00b7\u00b7\u00d7RdhR\u2225\u2225\u2225\u2225\u221e\u2264 Cdnd\/2. (8.1)This can be shown using the same argument as that given in Theorem 1.4. If theaforementioned random variables were independent, then, as in the proof of Theorem1.3, we would get that\u2016Hn\u2016\u221e \u2265n\u2211r1=0n\u2212r1\u2211r2=0\u2016Fr1,r2\u2016L1([0,1]d\u22122), (8.2)where Fr1,r2 =\u2211|R\u2032\u2032|=2\u2212(n\u2212r1\u2212r2) \u03b1R\u2032\u2032hR\u2032\u2032 with R\u2032\u2032 = R3 \u00d7 \u00b7 \u00b7 \u00b7 \u00d7Rd. Since Proposition5.6 is also applicable to Fr1,r2 , \u2016Fr1,r2\u20161 & \u2016Fr1,r2\u20162, and this together with (8.2)shows that, for any collection of coefficients {\u03b1R} \u2282 A\u2032split,\u2016Hn\u2016\u221e \u2265n\u2211r1=0n\u2212r1\u2211r2=0\u2016Fr1,r2\u2016L2([0,1]d\u22122) &n\u2211r1=0n\u2212r1\u2211r2=0(n\u2212 r1 \u2212 r2)(d\u22123)\/2 & n(d+1)\/2.However, the above inequality contradicts the validity of (8.1).As shown in Remark 6.3, the techniques employed in the proofs of Theorem 1.3can be used to show that the d-dimensional small ball inequality holds for a specialfamily of coefficients {\u03b1R1 \u00b7 \u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd} with \u03b1R1 \u2208 {\u00b11} and c1 \u2264 |\u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd | \u2264 c2,for given positive constants c1 and c2. If we weaken the above condition so that wehave only an upper bound or only a lower bound, one might wonder if (1.8) still103holds. That is, letBsplit = {\u03b1R : R \u2208 Adn, \u03b1R = \u03b1R1 \u00b7 \u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd with \u03b1R1 \u2208 {\u00b11}}.Problem 2: Does the small ball inequality hold for all \u03b1R \u2208 Bsplit which satisfy0 < c \u2264 |\u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd|, or for all \u03b1R \u2208 Bsplit which satisfy |\u03b1R2\u00d7\u00b7\u00b7\u00b7\u00d7Rd | \u2264 c ?Under this weaker restriction on the coefficients, it is not known whether theLp-norms of such hyperbolic sums are comparable, as this cannot be deduced as adirect consequence of Remark 5.7.It would also be very interesting to know whether the conjectured lower bound ofsigned hyperbolic sums in exponential Orlicz spaces described in (1.15) is optimal.Problem 3: If d \u2265 2 and n > 1, is there a positive constant C(d, a) andcoefficients {\u03b1R} \u2282 {\u22121, 1} for which\u2225\u2225\u2225\u2225\u2225\u2225\u2211|R|=2\u2212n\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225exp(La)\u2264 C(d, a) n d\u221212 + 12\u2212 1a , 2 \u2264 a <\u221e ? (8.3)We would be able to prove the above inequality if we knew that, for any collectionof coefficients {\u03b1R} \u2282 {\u22121, 1} and d \u2265 2,\u2225\u2225\u2225\u2225\u2225\u2225\u2211|R|=2\u2212n\u03b1RhR\u2225\u2225\u2225\u2225\u2225\u2225exp(L2)\u2264 Cd n d\u221212 . (8.4)Indeed, first recall the useful interpolation inequality which says that, for any 0