On Convolution Squares of Singular Measures

Transcription

1 On Convolution Squares of Singular Measures by Vincent Chan A thesis presented to the University of Waterloo in fulfillment of the thesis requirement for the degree of Master of Mathematics in Pure Mathematics Waterloo, Ontario, Canada, 2010 c Vincent Chan 2010

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3 I hereby declare that I am the sole author of this thesis. This is a true copy of the thesis, including any required final revisions, as accepted by my examiners. I understand that my thesis may be made electronically available to the public. iii

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5 Abstract We prove that if 1 > α > 1/2, then there exists a probability measure μ such that the Hausdorff dimension of its support is α and μ μ is a Lipschitz function of class α 1/2. v

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7 Acknowledgements I would like to deeply thank my supervisor, Kathryn Hare, for her excellent supervision; her comments were useful and insightful, and led not only to finding and fixing mistakes, but also to a more elegant way of presenting various arguments. A careful comparison of earlier drafts of this document and its current version would reveal the subtle errors detected by her keen eye. This thesis would not exist at all if it were not for Professor Thomas Körner, whom I thank for his help in clarifying points in his paper, which is the central focus in the pages to come. I am indebted to my friends and colleagues for their support and advice, particularly that of Faisal Al-Faisal, who, being in a similar situation to myself (including the bizarre sleep schedule), has become comparable to a brother in arms. Last but certainly not least, I am grateful for the support of my fiancée Vicki Nguyen, who, despite having no formal education in pure mathematics, suffered through reading dozens of pages while proofreading. She has never failed to provide encouragement when it was required most, and surely has kept me sane throughout the process. vii

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9 Contents 1 Preliminaries Introduction A tightness condition A Probabilistic Result 11 3 A Key Lemma 21 4 The Main Theorem ψ-lipschitz functions Density results APPENDICES 57 A Measure Theory 59 B Convergence of Fourier Series 63 C Probability Theory 65 D The Hausdorff Metric 67 Bibliography 74 ix

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11 Chapter 1 Preliminaries Unless otherwise stated, we will be working on T = R/Z = [ 1/2, 1/2], and only with positive, Borel, regular measures. We will use λ to denote the Lebesgue measure on T (and dx, dy, dt, and so forth to denote integration with respect to Lebesgue measure), and E = λ(e). Measures that are absolutely continuous or singular are taken with respect to λ in the case of T, and (left) Haar measure, should we require a more general setting. When speaking of measures living in an L p space or having other properties associated to functions, we mean the measure is absolutely continuous and we consider its Radon- Nikodym derivative to have this property. 1.1 Introduction The absolutely continuous measures form an ideal in the algebra of measures, so that for an absolutely continuous measure μ, μ μ is still absolutely continuous and so the Radon- Nikodym theorem gives rise to an associated function f L 1 such that μ μ = fλ. For any σ-finite measure μ, Lebesgue s Decomposition theorem decomposes μ as μ = μ ac + μ s, where μ ac is absolutely continuous and μ s is singular. It is thus natural to ask whether a similar situation holds with singular measures, that is, given a singular measure μ, does there exist a function f L 1 with μ μ = fλ? This is easily seen to be false, for if 1

12 μ = n a nδ xn is a discrete measure, then μ μ = n,m a na m δ xn+xm is also a discrete measure. To progress further, it may be helpful to appeal to an extension of Lebesgue s Decomposition theorem, which gives for any σ-finite measure μ three unique measures μ ac, μ cs and μ d which are absolutely continuous, continuous singular, and discrete, respectively, such that μ = μ ac + μ cs + μ d. Then we ask if we might recover the nice property of μ μ being absolutely continuous in the case of μ being a continuous singular measure. This is still not always the case, as there exist continuous singular measures on T where even their nth convolution power is singular for any n; for such an example we will need a preliminary definition. Definition 1.1. Let 1/2 a k 1/2 and define the trigonometric polynomials on T P N (t) = N (1 + 2a k cos 3 k t). k=1 It is a standard result that P N λ are probability measures which converge weak- in M(T) to a probability measure μ with 1 if n = 0, ˆμ(n) = M j=1 a k j if n = M j=1 ε j3 k j, ε j = ±1, 0 otherwise. This measure μ is called the Riesz product measure associated to {a k }. Lemma 1.2. If μ 1 and μ 2 are Riesz product measures, then so is μ 1 μ 2. Moreover, if μ i is associated to {a k,i }, then μ 1 μ 2 is associated to {a k,1 a k,2 }. Proof. We calculate the Fourier coefficients of the convolution: 1 if n = 0, ˆμ 1 μ 2 (n) = ˆμ 1 (n) ˆμ 2 (n) = M j=1 a k j,1a kj,2 if n = M j=1 ε j3 k j, ε j = ±1, 0 otherwise. (1.1) 2

13 Then we can define the trigonometric polynomials on T P N (t) = N (1 + 2a k,1 a k,2 cos 3 k t), k=1 noticing that 1/2 1/4 a k,1 a k,2 1/4 1/2. The weak- limit of P N λ produces a Riesz product measure whose coefficients satisfy (1.1), so we are finished by uniqueness. A standard result (see, for example, [6] Theorem 7.2.2) gives that a Riesz product measure μ is singular if and only if k=1 a2 k =. In particular, suppose μ is a Riesz product measure associated to the constant sequence {c} for some constant 0 < c 1/2. Certainly k=1 c2 =, so that μ is singular. Inductively applying Lemma 1.2, μ n = μ μ μ (the nth convolution power of μ) is a Riesz product measure associated to {c n }. Then k=1 c2n =, and so μ n remains singular. It is also not hard to show that Riesz product measures are continuous, in general. It should be remarked that the choice of frequencies 3 k is not essential, and Riesz product measures may be generalized further. In addition, the concept of Riesz product measures as a whole and the conclusion that there exist continuous singular measures whose convolution powers always remain singular can be extended to any infinite, compact, abelian group (see [6]). We now know it is impossible to hope the square convolution of a singular measure is absolutely continuous in general, even when restricting to the case of continuous singular measures. However, we may expect that are still some positive results; convolution behaves a smoothing operation, evidenced by the fact that both the continuous measures and the absolutely continuous measures form ideals in the space of measures. Indeed, a famous result due to Wiener and Wintner [16] produces a singular measure μ on T such that ˆμ(n) = o( n 1/2+ε ) as n for every ε > 0. By an application of the Hausdorff-Young inequality, such a measure can be shown to have the property that its square convolution μ μ is absolutely continuous, and in fact, μ μ L p (T) for every p 1. Other authors have generalized this result; Hewitt and Zuckerman [9] constructed a singular measure μ on any non-discrete, locally compact, abelian group G such that μ μ L p (G) for all p 1. It is worth noting that the condition of non-discrete cannot be 3

14 dropped, since otherwise all measures would be absolutely continuous. Using Rademacher- Riesz products, Karanikas and Koumandos [10] prove the existence of a singular measure with μ μ L 1 (G) for any non-discrete, locally compact group G. Shortly thereafter, Dooley and Gupta [3] produced a measure μ with μ μ L p (G) for every p 1 in the case of compact, connected groups and compact Lie groups using the theory of compact Lie groups. Saeki [15] took this concept further in a different direction by proving the existence of a singular measure μ on T with support having zero Lebesgue measure such that μ μ has a uniformly convergent Fourier series. This is an improvement on previous results; in this case the continuity of the partial Fourier sums for μ μ ensure that their limit, namely μ μ, is continuous. Then μ μ L (T) and subsequently μ μ L p (T) for p 1 since T is compact. Generalizing this, Gupta and Hare [7] show such a measure exists (now using Haar measure in place of Lebesgue measure) when replacing T with any compact, connected group. Körner [11] recently expanded on Saeki s work; to discuss how, we need the following definitions. Definition 1.3. For 1 β 0, we say a function f : T C is Lipschitz of class β 1, or simply β-lipschitz and write f Λ β if sup t T sup h β f(t + h) f(t) <. (1.2) h =0 Some references define a function to be Lipschitz of class β if there is a constant C with f(x) f(y) C x y β for every x, y T. These definitions are in fact equivalent, with a straightforward proof. Lemma 1.4. f Λ β if and only if f(x) f(y) C x y β for a constant C, for every x, y T. Proof. This is a simple matter of relabeling; use t = y and h = x y. 1 Functions with this property are also sometimes referred to as Hölder of class β. 4

15 It is useful to note that the Λ β classes are nested downwards: Lemma 1.5. Suppose 1 β i 0 for i = 1, 2. If β 1 β 2, then Λ β1 Λ β2. Proof. Since 1 β i 0, for h T = [ 1/2, 1/2] we get that h β 1 h β 2. Thus if we certainly have implying Λ β1 Λ β2. sup t T sup t T sup h β 2 f(t + h) f(t) <, h =0 sup h β 1 f(t + h) f(t) <, h =0 We will also require the notion of Hausdorff dimension, which we recall here: Definition 1.6. Fix α 0 and let δ > 0. For a set E, define { } Hδ α (E) = inf E i α : E i E, E i δ, i=1 where E i can be taken to be intervals. Notice Hδ α (E) is monotone increasing as δ decreases, since fewer permissible collections of sets are taken in the infimum. Then we take i=1 H α (E) = lim Hα δ 0 + δ (E) = sup Hδ α (E), and we call H α (E) the α-hausdorff measure of E; it is standard that this is indeed a measure (see [4], for example). Of special interest is the case α = 1, in which case we recover the usual Lebesgue measure. It is easy to show that there is a critical value for α such that H α (E) = for α less than this critical value, and H α (E) = 0 for α greater than this critical value. Then we define the Hausdorff dimension of E by δ>0 dim H (E) = sup{α : H α (E) > 0} = sup{α : H α (E) = } = inf{α : H α (E) < } = inf{α : H α (E) = 0}. Heuristically, the Hausdorff dimension allows us to compare sets that are too sparse for the Lebesgue measure to be of use. 5

16 For our results, we will need to have a way of describing where a measure lives ; we make this formal by introducing the concept of the support of a measure. Definition 1.7. Let (X, T ) be a topological space, and μ be a Borel measure on (X, T ). Then the support of μ is defined to be the set of all points x in X for which every open neighborhood N x of x has positive measure: supp μ = {x X : μ(n x ) > 0, x N x T }. (1.3) Among other things, Körner demonstrated the existence of a probability measure μ whose support has a prescribed Hausdorff dimension (between 1 and 1/2) such that μ μ is a Lipschitz function. Our goal will be to prove one of his main results, which is the following: Theorem If 1 > α > 1/2, then there exists a probability measure μ such that dim H (supp μ) = α and μ μ = fλ where f Λ α 1/2. We will show that this theorem is indeed an extension of Saeki s work. Using Lemma 1.4, we can see that f Λ β yields a constant C such that f(x + h) f(x) C h β for any x and h, implying f(x + h) f(x) = o(((ln h 1 ) 1 ) (see Appendix B.1). By the Dini-Lipschitz test (see Appendix B.2), f has a uniformly convergent Fourier series. This says that the measure Körner produced has the same property as Saeki s, provided it were also singular. This fact will hold due to the Hausdorff dimension condition; indeed, the fact that dim H (supp μ) < 1 implies that supp μ = 0 by definition of the dimension and recalling H 1 is simply Lebesgue measure. Since we always have μ((supp μ) c ) = 0, this provides the necessary decomposition for μ to be singular. Consider now the condition 1 > α > 1/2. As we have seen, this ensures the constructed measure will be singular since sets of Hausdorff dimension less than 1 have zero Lebesgue measure; however the converse is not true. Indeed, it is well known that there exists a set of Hausdorff dimension 1 with zero Lebesgue measure. Then the case α = 1 could give still 6

17 give rise to singular measures. Since we are working on T which has Hausdorff dimension 1, monotonicity of Hausdorff dimension (easily seen from monotonicity of the Hausdorff measure) ensures we need not consider the case α > 1. We shall see soon that we also need α 1/2 for a positive result involving this Lipschitz condition, after proving a tightness condition. 1.2 A tightness condition In this section, we aim to show that the result in Theorem 4.19 is nearly the best we can hope for. In particular, we will show that if the support of a measure μ has Hausdorff measure dimension α and μ μ Λ β, then α 1/2 β. To begin, we mention a relationship between the Hausdorff dimension of the support of a measure and its Fourier coefficients. Definition 1.8. We define the s-energy of a finite, compactly supported measure μ on R n, denoted I s (μ), to be I s (μ) = dμ(x) dμ(y) x y s. Lemma 1.9. Suppose μ is a non-zero measure, 0 < η < 1, and Then dim H (supp μ) η. k =0 ˆμ 2 <. k 1 η Proof. By Theorem 2.2 in [8], with d = 1 and 0 < η < 1, we have for some constant b that ( I η (μ) b ˆμ(0) 2 + ) ˆμ(k) 2 <. k 1 η k =0 Then by Theorem 4.13 in [5], dim H (supp μ) η. Now, we develop a bound for portions of the l p norm of the Fourier coefficients of a β-lipschitz function. 7

18 Lemma If f Λ β and 0 < p 2, then ˆf(k) p C 1 n 1 p(β+ 1 2 ), for some constant C 1. n k 2n 1 Proof. Fix h [ 1/2, 1/2]. Let g(t) = f(t + h) f(t h). A simple calculation shows that the Fourier coefficients of g are given by ˆg(k) = 2i sin(2πkh) ˆf(k). By Parseval s identity, k ˆg(k) 2 = g 2, so that 4 ˆf(k) 2 sin 2 (2πkh) = f(t + h) f(t h) 2 dt. k Since f Λ β, there is a constant C such that f(t + h) f(t h) C (t + h) (t h) β = C 2h β, for all t T, so 4 k ˆf(k) 2 sin 2 (2πkh) (C 2h β ) 2 dt = 4 β C 2 h 2β 4C 2 h 2β. Thus for any n, n k 2n 1 ˆf(k) 2 sin 2 (2πkh) C 2 h 2β. As this holds for any h [ 1/2, 1/2], consider h = 1/(8n). Notice sin 2 ( kπ ) 1/2 for 4n n k 2n 1, so n k 2n 1 ˆf(k) 2 2C 2 h 2β = 2C 2 ( 1 8 )2β n 2β 2C 2 ( 1 8 )2 n 2β. (1.4) If p = 2, we are done; taking C 1 = 2C 2 ( 1 8 )2 = 1 32 C2, and (1.4) gives n k 2n 1 ˆf(k) 2 C 1 n 2β = C 1 n 1 2(β+ 1 2 ). 8

19 Otherwise, 0 < p < 2 and so 1 < 2/p, 1/(1 p/2) < ; notice they are conjugate indices. By Hölder s inequality with counting measure on { ˆf(k) p } n k 2n 1 and {1} n k 2n 1, we have ˆf(k) p p/2 ( ˆf(k) p ) 2/p 1 p/2 (1) 1/(1 p/2) n k 2n 1 = n k 2n 1 n k 2n 1 ˆf(k) 2 p/2 (2n) 1 p/2 n k 2n 1 (2C 2 ( 1 8 )2 n 2β ) p/2 (2n) 1 p 2 by (1.4) = C 1 n 1 p(β+1/2) as required, where C 1 = 2( C 8 )p is a constant. Lemma If μ is a measure with dim H (supp μ) = α and μ μ = fλ where f Λ β, then α 1/2 β. Proof. Since f Λ β, taking p = 1 in Lemma 1.10 yields ˆf(k) C 1 n 1 1(β+1/2) = C 1 n (1 2β)/2, n k 2n 1 for some constant C 1 depending on f. Since ˆf(k) = ˆμ(k) 2, we have ˆμ(k) 2 C 1 n 1 1(β+1/2) = C 1 n (1 2β)/2, and so if η > 0, we have n k 2n 1 n k 2n 1 ˆμ(k) 2 k 1 η n k 2n 1 ˆμ(k) 2 n 1 η C 1n (1 2β)/2 n 1 η = C 1 n (1+2β 2η)/2 for all n 1. In particular, for n = 2 j for each j 0, 2 j k 2 j+1 1 ˆμ(k) 2 k 1 η C 1(2 j ) (1+2β 2η)/2, 9

20 so that k =0 ˆμ(k) 2 k 1 η = j=0 2 j k 2 j+1 1 ˆμ(k) 2 k 1 η C 1 (2 j ) (1+2β 2η)/2. Notice this converges whenever (1 + 2β 2η)/2 < 0, that is, whenever (1 + 2β)/2 > η. Then by Lemma 1.9, dim H (supp μ) η, for every η that satisfies (1 + 2β)/2 > η. Thus dim H (supp μ) (1 + 2β)/2, that is, α 1/2 β as required. j=0 Lemma 1.11 now gives us a motivation for the lower bound on α; taking β = 0 we see that for μ μ Λ 0, we must have dim H (supp μ) 1/2. Of course, this still leaves open the question of what happens in the case of α = 1/2; the boundary cases α = 1 and α = 1/2 are beyond the scope of this paper, as the proof relies on the strict inequalities. In Chapter 2, we prove a key lemma that provides us with a discrete measure satisfying useful boundedness properties which we help in our construction. The key to this lemma will be using a probabilistic argument, which will span most of the chapter. In Chapter 3, we utilize the key lemma from the previous chapter to construct an infinitely differentiable periodic function satisfying several key properties to be used in the main theorem. In Chapter 4, the main theorem will be proved, first by developing some complete metric spaces and applying the key lemma from the previous chapter in a density argument. Employing the Baire Category theorem will bridge the final gap towards the proof of the main theorem. 10

21 Chapter 2 A Probabilistic Result The key result that we will need is not easy to prove directly, so we will prove a probabilistic variant of it instead, which will imply the result we need. We proceed by proving the following result: Lemma 2.1. Suppose that 0 < Np 1 and m 2. Then if Y 1, Y 2,..., Y N are independent random variables with P(Y j = 1) = p, P(Y j = 0) = 1 p, it follows that ( N ) P Y j m j=1 2(Np)m. m! Proof. First note that Y j {0, 1} almost surely, so that their sum cannot exceed N with positive probability. That is, we may assume without loss of generality that m N. For 1 k N, define u k = ( N k ) p k = N! k!(n k)! pk. Since 0 < Np 1 and k, p > 0, we have (N k)p = Np kp 1. Then as k 1, u k+1 u k = N! (k+1)!(n k 1)! pk+1 N! k!(n k)! pk = 11 (N k)p k k

22 Since Y j takes on only 0 or 1 almost surely, ( N ) ( N N ) P Y j m = P Y j = k j=1 = k=m k=m j=1 ( ) N N p k (1 p) N k since Y j are independent k ( ) N N N p k = u k k k=m N u m 2 2u k m m since u k+1 1 u k 2 k=m k=m N! = 2 m!(n m)! pm 2(Np)m. m! Definition 2.2. For a set A R and a random variable X, we define the random variable δ X (A) by 1 if X(x) A, δ X (A)(x) = 0 otherwise. Lemma 2.3. If 1 > γ > 0 and ε > 0, there exists an M(γ, ε) 1 with the following property. Suppose that n 2, n γ N for sufficiently large n, and X 1, X 2,..., X N are independent random variables each uniformly distributed on Γ n = {r/n : 1 r n}. Then with probability at least 1 ε/n, N δ Xj ({r/n}) M(γ, ε) j=1 for all 1 r n. Proof. Since 1 γ > 0, there exists an integer M(γ, ε) 2 such that n 2 M(γ,ε)(1 γ) < ε 2 (2.1) 12

23 for all n 2. Fix r and set Y j = δ Xj ({r/n}). Note that Y j are independent random variables, as X j are. Since X j are uniformly distributed on Γ n, we get P(Y j = 1) = 1/n, P(Y j = 0) = 1 1/n. Since 1 > γ > 0 and n γ N, we have n n γ N, yielding Nn 1 1. Clearly 0 Nn 1. Then with p = n 1 and m = M(γ, ε) 2, Lemma 2.1 tells us that ( N ) ( N ) P δ Xj ({r/n}) M(γ, ε) = P Y j M(γ, ε) j=1 j=1 2(Nn 1 ) M(γ,ε) M(γ, ε)! 2nγM(γ,ε) n M(γ,ε) M(γ, ε)! 2n M(γ,ε)(1 γ) < ε n 2, where we used n γ N in the second inequality, and the final inequality follows from (2.1). Allowing r to vary, we get immediately ( N ) P δ Xj ({r/n}) M(γ, ε) for some 1 r n = j=1 ( N N ) P δ Xj ({r/n}) M(γ, ε) r=1 j=1 < N ε n 2 ε n, where we have used the fact that N n. Definition 2.4. Suppose P is a probability measure, X is a random variable with respect to the σ-algebra F, and G F is a sub-σ-algebra. If Y is a random variable with respect to the σ-algebra G such that for all A G, X dp = Y dp, A A 13

24 then we call Y the conditional expectation of X with respect to G, and denote it by E(X G). It is a consequence of the Radon-Nikodym Theorem that such a random variable always exists, and is unique almost surely. If X 1 is some random variable (with respect to the Borel σ-algebra B), we can define E(X X 1 ) := E(X σ(x 1 )). Recall σ(x 1 ) = {X 1 1 (B) : B B} is the smallest σ-algebra such that X 1 is a random variable. More generally, if X 1,..., X n are random variables, we define E(X X 1,..., X n ) := E(X σ(x 1,..., X n )). We will require two simple facts about conditional expectation. Proposition 2.5. (i) If X is G-measurable E(X G) = X. (ii) For any X, E(E(X G)) = E(X). Proof. Both parts are immediate from the definition. Lemma 2.6. Let δ > 0 and for j = 0, 1,..., n, let W j be a random variable that is measurable with respect to σ(x 1,..., X j ). Define Y j = W j W j 1 and suppose that E(e sy j X 0, X 1,..., X j 1 ) e a js 2 /2 for all s < δ and some a j 0. Suppose further that A N j=1 a j. Then provided that 0 x < Aδ, we have ( ) x 2 P( W N W 0 x) 2exp. 2A Proof. Suppose δ < s < δ. By definition of W j, we have e s(w N W 0 ) = e s(w N W N 1 ) e s(w N 1 W 0 ). Since W N 1 and W 0 are σ(x 1,..., X N 1 )-measurable, E(e s(w N W 0 ) X 0, X 1,..., X N 1 ) = e s(w N 1 W 0 ) E(e s(w N W N 1 ) X 0, X 1,..., X N 1 ) = e s(w N 1 W 0 ) E(e sy N X 0, X 1,..., X N 1 ) 14

25 e s(w N 1 W 0 ) e a N s 2 /2. By property of conditional expectation, taking expectation of both sides gives By induction, we get E(e s(w N W 0 ) ) E(e s(w N 1 W 0 ) )e a N s 2 /2. E(e s(w N W 0 ) ) N e a js 2 /2 e As2 /2, since A N j=1 a j. Then by Markov s inequality (see Appendix C.1), j=1 P(W N W 0 x) = P(e s(w N W 0 ) e sx ) E(e s(w N W 0 ) )e sx e As2 /2 e sx. Setting s = xa 1, 0 x < Aδ ensures that 0 xa 1 < δ, so that s < δ. Then by the above, ) P(W N W 0 x) e As2 /2 e sx = e Ax2 A 2 /2 e xa 1x = exp ( x2. 2A The same argument applies to the sequence W j, so that we also have ) P(W 0 W N x) exp ( x2. 2A The result follows. Lemma 2.7. Suppose φ : N R is a sequence with φ(n)(log n) 1/2 as n, and for any δ > 0, we have φ(n)n δ 0 as n. Suppose 1 > γ > 1/2, and the positive integer N = N(n) satisfies n γ N n 1/2+η for some η > 0, for all sufficiently large n. If ε > 0, there exists an M(γ) and an n 0 (φ, γ, ε) 1 with the following property. Suppose that n n 0 (φ, γ, ε), and n is odd. Suppose further that X 1, X 2,..., X N random variables each uniformly distributed on Γ n = {r/n T : 1 r n}. Then, if we write σ = N 1 N j=1 δ X j, we have σ σ({k/n}) n 1 φ(n)(log n)1/2 ε, Nn 1/2 15 are independent

26 and σ({k/n}) M(γ) N for all 1 k n, with probability at least 1/2. Proof. Set M(γ) = M(γ, 1/4) coming from Lemma 2.3. Without loss of generality, assume M(γ) 3. Fix r for now, and define Y 1, Y 2,..., Y N as follows. If j 1 v=1 δ X v ({u/n}) M(γ) for all u with 1 u n, set Y j = 2j 1 n j 1 + δ 2Xj ({r/n}) + 2 δ Xv+Xj ({r/n}). Otherwise, set Y j = 0. Take X 0 = W 0 = 0 and W j = jm=1 Y m; notice Y j = W j W j 1. Suppose first that j 1 v=1 δ X v ({u/n}) M(γ) for all u with 1 u n. Observe that if s is a fixed integer, then X j + s/n and 2X j are uniformly distributed over Γ n (using addition modulo 1), since n is odd and X j are independent. Then by uniform distribution, E(Y j ) = 2j 1 n = 2j 1 n v=1 j 1 + E(δ 2Xj ({r/n})) + 2 E(δ Xv+Xj ({r/n})) v=1 + 1 j 1 n n = 2j j 1 n n v=1 On the other hand, if it is not the case that j 1 v=1 δ X v ({u/n}) M(γ) for all u with 1 u n, then Y j = 0 by construction so that E(Y j ) = 0 automatically. = 0. We wish to make use of Lemma 2.6, and so we bound E(e sy j X 0, X 1,..., X j 1 ). Notice as n γ N and γ < 1, we have n > n γ N. Then (2N 1)/n 2N/n 2. Suppose first that j 1 v=1 δ X v ({u/n}) M(γ) for all u with 1 u n. Then Y j 2j 1 n 2N 1 n j 1 + δ 2Xj ({r/n}) + 2 δ Xv+Xj ({r/n}) v= M(γ) M(γ) 3M(γ), 16

27 where we have used the fact that M(γ) 3. Setting Z j = Y j + (2j 1)/n, we notice that Z j = 0 if any of δ Xv+X j ({r/n}) = 0 for 1 v j. As X j are uniformly distributed and by subadditivity, we thus get P(Z j = 0) j/n. Since E(Y j ) = 0, ( ) E(e sy j (sy j ) k s k k ) = E = E(Yj ) k! k! k=0 k=0 s k k s k = 1 + E(Yj ) 1 + k! k! E Y j k k=2 k=2 s k = 1 + P(Z j = 0) k! E( Y j k Z j = 0) + P(Z j = 0) 1 + k=2 k=2 s k k! E( Y j k Z j = 0) + j n k=2 k=2 s k k! E( Y j k Z j = 0). s k k! E( Y j k Z j = 0) where the third line follows from the partition formula, and the final inequality follows since P(Z j = 0) 1 and P(Z j = 0) j/n. Since Z j = Y j + (2j 1)/n, in the case Z j = 0 we have Y j = (2j 1)/n. Otherwise, our earlier approximation gives Y j 3M(γ). Thus, E(e sy j ) k=2 k=2 ( s 2j 1 n )k k! ( s 2N 1 n )k k! + j n + N n ( s 3M(γ)) k k=2 k! ( s 3M(γ)) k since 1 j N. Notice if we have a constant 0 a 1, then k=2 k=2 a k k! a 2 2 = k a2. k=1 For sufficiently large n, we have 0 s 2N 1 1. If s < 1, we also get 0 s 3M(γ) n 3M(γ) 1, so by our quick result above, E(e sy j ) 1 + ( s 2N 1 ) 2 + N n n ( s 3M(γ)) N 2 n 2 s2 + 9M(γ) 2 N n s2 1 + (4 + 9M(γ) 2 )Nn 1 s M(γ) 2 Nn 1 s 2 exp ( 20M(γ) 2 Nn 1 s 2 /2 ), k! 17

28 where we have used M(γ) 2 9 in the second to last inequality, and the fact that 1+x e x for every real x in the last inequality. Define a j = 20M(γ) 2 N n > 0. If it is not the case that j 1 v=1 δ X v ({u/n}) M(γ) for all u with 1 u n, then by definition of Y j, we get automatically Define E(e sy j ) = E(e 0 ) = E(1) = 1 exp ( 20M(γ) 2 Nn 1 s 2 /2 ). A = N a j = j=1 N 20M(γ) 2 Nn 1 = 20M(γ) 2 N 2 n 1. Set δ = 1 3M(γ) and x = εφ(n)(log n) 1/2 Nn 1/2. j=1 To apply Lemma 2.6, we must check that 0 x < Aδ for sufficiently large n. Indeed, φ(n)(log n) 1/2 as n, while ε > 0 and Nn 1/2 > 0 for every n, so that x 0 for sufficiently large n. In order to show x < Aδ, we must have εφ(n)(log n) 1/2 Nn 1/2 < 20M(γ) 2 N 2 n 1 1 3M(γ) 3 ε 20M(γ) φ(n)(log n)1/2 n 1/2 < N. By hypothesis, N is such that there exists η > 0 with N > n 1/2+η for all sufficiently large n. For this value of η > 0, we get φ(n)n η/2 < 20M(γ) 3ε of φ. Then we easily get that for sufficiently large n, for sufficiently large n by property 3 ε 20M(γ) φ(n)(log n)1/2 n 1/2 < n η/2 (log n) 1/2 n 1/2 < n 1/2+η < N, as required. Then by Lemma 2.6, ) ) P( W N W 0 x) 2 exp ( x2 = 2 exp ( ε2 φ(n) 2 (log n)n 2 n 1 2A 40M(γ) 2 N 2 n ) 1 = 2 exp ( ε2 40M(γ) 2 φ(n)2 log n. Notice ε2 40M(γ) 2 < 0 is a constant, and we are given φ(n) 2 log n as n. Then, we may choose n 0 (φ, γ, ε) 1 such that for all n n 0, P( W N W 0 εφ(n)(log n) 1/2 Nn 1/2 ) 1 4. (2.2) 18

29 For the rest of the proof, we assume n satisfies this condition. Recall M(γ) = M(γ, 1/4), as given by Lemma 2.3. By that lemma, we have that with probability at least 1 1/(4n), N δ Xv ({r/n}) M(γ), (2.3) v=1 for all 1 r n. Then for every 1 r n and for every 1 j N, j δ Xv ({r/n}) v=1 N δ Xv ({r/n}) M(γ), v=1 with probability at least 1 1/(4n). Then by construction, By definition of W j, Notice while W N W 0 = Y j = 2j 1 n j 1 + δ 2Xj ({r/n}) + 2 δ Xv+Xj ({r/n}). v=1 ( ) N 2j 1 j 1 + δ 2Xj ({r/n}) + 2 δ Xv+X n j ({r/n}). j=1 N j=1 ( 2j 1 ) = N 2 n n, ( ) j 1 N δ 2Xj ({r/n}) + 2 δ Xv+Xj ({r/n}) = j=1 v=1 v=1 N v=1 j=1 N δ Xv+Xj ({r/n}). (2.4) Combining (2.2) and (2.4), we get that with probability at least 1 1/(4n) 1/4 1/2, we have N W N W 0 = v=1 j=1 Writing σ = N 1 N j=1 δ X j, this becomes N δ Xv+Xj ({r/n}) N 2 n < εφ(n)(log n)1/2 Nn 1/2. σ σ({r/n}) n 1 φ(n)(log n)1/2 < ε, Nn 1/2 19

30 and (2.3) becomes σ({r/n}) M(γ) N. Allowing r to take values from 1 to n, the result follows. Lemma 2.8. Suppose φ, γ, and N are as in Lemma 2.7. If ε > 0, there exists an M(γ) and an n 0 (φ, γ, ε) 1 with the following property. Suppose that n n 0 (φ, γ, ε), and n is odd. Then we can find N points x j {r/n T : 1 r n} (not necessarily distinct) such that, writing μ = N 1 N j=1 δ xj, we have and for all 1 k n. μ μ({k/n}) n 1 φ(n)(log n)1/2 ε, Nn 1/2 μ({k/n}) M(γ) N Proof. This follows immediately from Lemma 2.7, since an event with positive probability must have at least one occurrence. 20

31 Chapter 3 A Key Lemma In this chapter, we will take steps to convert the discrete measure generated by Lemma 2.8 into an incredibly well-behaved function, being a periodic, positive, infinitely differentiable function satisfying several key properties; this procedure will be completed in four lemmas. We will write 1 A for the indicator function of the set A. Lemma 3.1. Suppose φ, γ, and N are as in Lemma 2.7. If ε > 0, there exists an M(γ) and an n 0 (φ, γ, ε) 1 with the following property. Suppose that n n 0 (φ, γ, ε), and n is odd. Then we can find N points x j {r/n T : 1 r n} (not necessarily distinct) such that, writing we have: g = n N N 1 [xj (2n) 1,x j +(2n) 1 ), j=1 (i) g g 1 ε φ(n)(log n)1/2 n 1/2 N. (ii) For each t T, if 0 < h < 1/n, then h 1 g g(t + h) g g(t) 4ε φ(n)(log n)1/2 n 3/2. N 21

32 (iii) g(t) nm(γ) N (iv) g 1 = 1. for all t T. Proof. By Lemma 2.8, we can find N points (not necessarily distinct) such that, writing we have and x j {r/n T : 1 r n} μ = N 1 N j=1 δ xj, μ μ({k/n}) n 1 φ(n)(log n)1/2 ε, Nn 1/2 μ({k/n}) M(γ) N for all 1 k n. Observe that as δ xj is supported on {x j }, for any a, b and any x T we have δ xj 1 [a,b) (x) = As convolution respects addition, 1 [a,b) (x y) dδ xj (y) = 1 [a,b) (x x j ) = 1 [xj +a,x j +b)(x). nμ 1 [ (2n) 1,(2n) 1 ) = n N N δ xj 1 [ (2n) 1,(2n) 1 ) = n N j=1 N 1 [xj (2n) 1,x j +(2n) 1 ) = g. j=1 Since 1 [ (2n) 1,(2n) 1 ) is symmetric almost everywhere, 1 [ (2n) 1,(2n) 1 ) 1 [ (2n) 1,(2n) 1 ) = 1 [ (2n) 1,(2n) 1 )(t x)1 [ (2n) 1,(2n) 1 )(t) dt = 1 [x (2n) 1,x+(2n) 1 )(t)1 [ (2n) 1,(2n) 1 )(t) dt 0 if x < 1/n, 1[ (2n) = 1,x+(2n) 1 )(t) dt if 1/n x 0, 1[x (2n) 1,(2n) 1 )(t) dt if 0 x 1/n, 0 if x > 1/n, 22

33 0 if x > 1/n, = 1/n x if x 1/n, = max {0, 1/n x }. Writing Δ n = max {0, 1 n x }, the previous calculations give g g = (nμ 1 [ (2n) 1,(2n) 1 )) (nμ 1 [ (2n) 1,(2n) 1 )) = μ μ nδ n. In particular, we have g g(r/n) = (μ μ) nδ n (r/n) = n max {0, 1 n (r/n) t } d(μ μ)(t). Since μ is supported on {k/n}, 1 k n, so is μ μ. Then the above integral is zero except possibly when t = r 0 /n for r 0 Z, i.e. nt Z. Now, Δ n ((r/n) t) is non-zero only if 1 n (r/n) t > 0, yielding 1 > r nt. But nt Z, implying nt = r. Therefore, g g(r/n) = n(1 n (r/n) (r/n) )μ μ({r/n}) = nμ μ({r/n}). (3.1) It is routine to check that for x j {r/n T : 1 r n}, we have 1 [xi (2n) 1,x i +(2n) 1 ) 1 [xj (2n) 1,x j +(2n) 1 )(x) 0 if x / [x i + x j 1/n, x i + x j + 1/n], = x (x i + x j ) + 1/n if x [x i + x j 1/n, x i + x j ], x + (x i + x j ) + 1/n if if x [x i + x j, x i + x j + 1/n]. Notice x i + x j is of the form r/n as well, so that this convolution is continuous and piecewise linear on each interval [(r 1)/n, r/n], for 1 r n. Then g g, being a linear combination of such convolutions, will be continuous 1 and piecewise linear on each 1 Continuity can also be obtained by noticing g L 2 (T), and L 2 (T) L 2 (T) C(T). 23

34 interval [(r 1)/n, r/n], for 1 r n. Thus, g g attains a local maximum at points in {r/n T : 1 r n}. By property of the chosen μ, we have By (3.1), this becomes μ μ({r/n}) n 1 φ(n)(log n)1/2 ε. Nn 1/2 g g(r/n) 1 ε φ(n)(log n)1/2 n 1/2. N Since g g has local maxima on {r/n T : 1 r n}, we get verifying part (i). g g 1 = max g g(r/n) 1 εφ(n)(log n)1/2 n 1/2, 1 r n N To verify part (ii), we need to check the claimed inequality for each t T. We have two cases. Case 1: t = r/n, for some 1 r n. Suppose 0 < h < 1/n. Since g g is linear on [r/n, (r + 1)/n], g g(r/n + h) can be expressed as the convex combination: Then g g ( r + h) = (1 nh)(g g) ( ) ( r n n + nh(g g) r+1 ) n. h 1 g g(t + h) g g(t) = h 1 g g( r n + h) g g( r n ) = h 1 (1 nh)(g g)( r r+1 ) + nh(g g)( ) g g( r ) n n n = h 1 n h g g( r+1 n ) g g( r n ) n( g g( r+1) 1 + g g( r ) 1 ) n n n (ε φ(n)(log n)1/2 n 1/2 + ε φ(n)(log ) n)1/2 n 1/2 N N = 2ε φ(n)(log n)1/2 n 3/2. N 24

35 The argument is similar if 1/n < h < 0. Case 2: t = r/n, for any 1 r n. Suppose 0 < h < 1/n. Then there exists r/n, for some 1 r n, within a distance of 1/n of both t and t + h. By Case 1, h 1 g g(t + h) g g(t) h 1 ( g g(t + h) g g( r ) + g g( r ) g g(t) ) n n Cases 1 and 2 together give part (ii). 4ε φ(n)(log n)1/2 n 3/2. N Fix t T. We have that [t (2n) 1, t + (2n) 1 ) intersects {r/n T : 1 r n} exactly once, say at k/n. Then as μ is supported on {r/n}, by property of μ, completing part (iii). By definition of g, it is easy to see that verifying (iv). g(t) = nμ 1 [ (2n) 1,(2n) 1 )(t) = n 1 [ (2n) 1,(2n) 1 )(t y) dμ(y) g 1 = = n N = n N = nμ({k/n}) nm(γ) N, n N N 1 [xj (2n) 1,x j +(2n) 1 )(t) dt j=1 N (x j + (2n) 1 ) (x j (2n) 1 ) j=1 N j=1 1 n = n N N n = 1, We next smooth the function obtained in the previous lemma: 25

36 Lemma 3.2. Suppose φ, γ, and N are as in Lemma 2.7. If ε > 0, there exists an M(γ) and an n 0 (φ, γ, ε) 1 with the following property. Suppose that n n 0 (φ, γ, ε), and n is odd. Then we can find a positive, infinitely differentiable function f such that: (i) f f 1 ε φ(n)(log n)1/2 n 1/2 N. (ii) (f f) 4ε φ(n)(log n)1/2 n 3/2 N. (iii) f nm(γ) N. (iv) f 8n2 M(γ) N. (v) f(t) dt = 1. T (vi) supp f can be covered by N intervals of length 2/n. Proof. Take M(γ), n 0 (φ, γ, ε), and g as in Lemma 3.1. Let K : R R be an infinitely differentiable positive function such that K(x) = 0 for x 1/2, K (x) 8 for all x, and K(x) dx = 1. (Such a function does exist: as an example, take R K(x) = R k(x) k(x) dx, where exp ( ) 1 if x < 1 1 4x k(x) =, otherwise. After some work, one can show that K(x) is infinitely differentiable and K (x) 7.2 everywhere. It is trivial that it is non-negative and the support lies in ( 1/2, 1/2).) Define K n : T R by K n (t) = nk(nt) for 1/2 t < 1/2. Then K n and K n K n are also positive and lie in L 1 (T), so for each n, we get that ( ) 2 1/2 K n K n 1 = K n 2 1 = nk(nt) dt 1/2 ( 2 ( n/2 2 1/2 = K(x) dx) = K(x) dx) = 1, n/2 1/2 26

37 since K(x) is supported on ( 1/2, 1/2) and n/2 1/2 for every n, where equality holds since we have positive functions. Set f = g K n. Then for s T, (f f 1)(s) = (g K n ) (g K n )(s) 1 = (g g)(s t)(k n K n )(t) dt K n K n (t) dt (g g)(s t) 1 (K n K n )(t) dt and so verifying part (i). f f 1 g g 1 K n K n 1 ε φ(n)(log n)1/2 n 1/2, N Fix t T. By Lemma 3.1, if 0 < h < 1/n, then Then h 1 g g(t + h) g g(t) 4ε φ(n)(log n)1/2 n 3/2. N h 1 f f(t + h) f f(t) = h 1 (g g) (K n K n )(t + h) (g g) (K n K n )(t) = h 1 (g g)(t + h y)(k n K n )(y) dy (g g)(t y)(k n K n )(y) dy h 1 g g(t + h y) g g(t y) (K n K n )(y) dy 4ε φ(n)(log n)1/2 n 3/2 (K n K n )(y) dy N = 4ε φ(n)(log n)1/2 n 3/2. N 27

38 Since g, K n L 1 and K n is infinitely differentiable, f = g K n is infinitely differentiable. Then f f is infinitely differentiable, and so the following limit exists: for each t T, and hence verifying part (ii). (f f) (t) = lim h 1 f f(t + h) f f(t) 4ε φ(n)(log n)1/2 n 3/2, h 0 N (f f) 4ε φ(n)(log n)1/2 n 3/2, N Recall g(t) nm(γ) for each t T, so that for s T, N nm(γ) f(s) = g K n (s) g(s t) K n (t) dt N K n(t) dt = nm(γ) N, and hence verifying part (iii). have f nm(γ) N, Since K n is infinitely differentiable, (g K n ) = g K n 2. By choice of K, for s T we f (s) = (g K n ) (s) = g K n(s) = verifying part (iv). 1/2 1/2 = n2 M(γ) N g(s t) n 2 K (nt) dt nm(γ) N 1/2 1/2 K (x) dx n2 M(γ) N g(s t) K n(t) dt 1/2 1/2 1/2 1/2 n 2 K (nt) dt 8 dx = 8n2 M(γ), N Since g and K n are positive, f(t) dt = T (g K n )(t) dt = g 1 K n 1 = 1, verifying part (v). 2 Indeed, we may consider the Fourier transform of each function: for each r Z we have ((g K n ) )ˆ(r) = irgˆ K n (r) = irˆg(r) ˆK n (r) = ˆg(r) ˆK n(r) = (g K n)ˆ(r). 28

39 By construction, the support of g can be covered by N intervals of length 1/n. K is supported on ( 1/2, 1/2), so that K n is supported on ( 1/(2n), 1/(2n)). Then the support of f = g K n is covered the sum of these supports, so is covered by N intervals of length 2/n, for part (vi). In our proof, we will not need such precise estimates on the bounds, so we modify the previous lemma and introduce a connection with the parameter α. Lemma 3.3. Suppose that 1/2 > α 1/2 > β > 0. If ε > 0, there exists an n 1 (α, β, ε) 1 with the following property. Suppose that n n 0 (φ, γ, ε), and n is odd. Then we can find a positive, infinitely differentiable function f such that: (i) f f 1 εn β /2. (ii) (f f) εn 1 β. (iii) f εn. (iv) f εn 2. (v) f(t) dt = 1. T (vi) supp f can be covered by less than εn α /2 intervals of length 2/n. (vii) ˆf(r) ε for all r = 0. Proof. Without loss of generality, assume ε < 1, for it only puts a more severe restriction on the bounds. Set κ = (α + β + 1/2)/2, φ(n) = log(n), and N = n κ. Notice φ : N R is a sequence with φ(n)(log n) 1/2 = (log n) 3/2 as n and for any η > 0, we have φ(n)n η = (log n)n η 0 as n. We have α > κ > β + 1/2, so since 1 > α and β > 0, we have 1 > κ > 1/2. N is a function of n taking values in the positive integers. Since κ = (α + β + 1/2)/2, we define η = (α (β + 1/2))/2 > 0 so that κ = β + 1/2 + η and κ = α η. For sufficiently large n we get N(n) = n κ n β+1/2 n (β/2)+(1/2). 29

40 For every n, n κ n κ = N. By Lemma 3.2, there exists an M(κ) and an n 0 (φ, κ, ε) 1 with the following property. If n n 0 and n is odd, we can find a positive, infinitely differentiable function f such that: (i) f f 1 ε φ(n)(log n)1/2 n 1/2 N. (ii) (f f) 4ε φ(n)(log n)1/2 n 3/2 N. (iii) (iv) f nm(κ) N. f 8n2 M(κ) N. (v) f(t) dt = 1. T (vi) supp f can be covered by N intervals of length 2/n. This f is the function we want in the lemma. Choose n 1 (α, β, ε) n 0 sufficiently large so that if n n 1 is odd, then 4(log n) 3/2 n η < ε, (3.2) and 8M(κ) n κ 1 < ε. (3.3) Using our particular values, 4(log n) 3/2 n η < ε 1 turns (i) into f f 1 ε (log n)3/2 n 1/2 n κ ε (log n)3/2 n 1/2 n β η β (log n)3/2 = εn εn β /2. n η We also get f f 1 < ε 2, since 4(log n) 3/2 n η < ε. This will be useful in proving part (vii) later. (ii) becomes, after another application of (3.2), (f f) 4ε (log n)3/2 n 3/2 n κ Now using (3.3), (iii) becomes 4ε (log n)3/2 n 3/2 n β η 1 β 4(log n)3/2 = εn εn 1 β. n η f nm(κ) n κ εn, 30

41 and (iv) becomes f 8n2 M(κ) n κ εn 2. (v) gives what we need immediately. (vi) tells us that supp f can be covered by n κ = n α η intervals of length 2/n. Our choice of n satisfying (3.2) will ensure that n η < ε/2 provided n > exp(1/4) 1.28, which we will assume. Thus, supp f can be covered by less than n α η < εn α /2 intervals of length 2/n. For r = 0, verifying part (vii). ˆf(r) = ˆf f(r) 1/2 = ˆf f(r) ˆ1(r) 1/2 f f 1 1/2 < ε, For the final step of this chapter, we will impose a periodicity condition on the smooth function we generate. Lemma 3.4. Suppose that 1/2 > α 1/2 > β > 0. There exists an integer k(α, β) such that given any ε > 0, there exists an m 1 (k, α, β, ε) 1 with the following property. Suppose that m > m 1 (k, α, β, ε), and m is odd. Then we can find a positive, infinitely differentiable, periodic function F with period 1/m with the following properties: (i) F F 1 ε. (ii) (F F ) εm k(1 β). (iii) F εm k. (iv) F εm 2k+1. (v) F (t) dt = 1. T 31

42 (vi) We can find a finite collection of intervals I such that I supp F and I α < ε. I I I I (vii) ˆF (r) ε for all r = 0. (viii) h β F F (t + h) F F (t) ε for all t, h T with h = 0. Proof. Since 1/2 > α 1/2 > β > 0, we can find α 1 and β 1 such that 1/2 > α 1/2 > α 1 1/2 > β 1 > β > 0. For an explicit choice, let α 1 = 3 4 α ( β + 1 2), β1 = 1 4 Since 1/2 > α 1/2 > β > 0, clearly β 1 > 0 and ( ) α β. 4 α = β (α 1 2 β) > β 1, while and Moreover, α 1 1 = (α 1) 1(α 1 β) < β 1 = 1(α 1 β) + β > β, 4 2 α 1 = α 1(α 1 β) < α. 4 2 Choose an integer k such that k(β 1 β) 1 (as β 1 > β) and k(α α 1 ) > 1 α (as α > α 1 ); then k is determined entirely by α and β alone. By Lemma 3.3, if ε > 0 there exists an n 1 (α 1, β 1, ε) 1 such that if for some odd m, n = m k > n 1 (note that n is odd since m is), then we can find a positive, infinitely differentiable function f such that: (i) f f 1 εm kβ 1 /2. (ii) (f f) εm k(1 β1). (iii) f εm k. 32

43 (iv) f εm 2k. (v) f(t) dt = 1. T (vi) supp f can be covered by less than εm kα 1 /2 intervals of length 2m k. (vii) ˆf(r) ε for all r = 0. Extend f by periodicity to [ m/2, m/2]. Setting F (t) = f(mt) for t [ 1/2, 1/2], we get that F is a positive, infinitely differentiable, periodic function with period 1/m, since f is a function on T = [ 1/2, 1/2] with endpoints identified. We have F F (s) = F (s t)f (t) dt = = m/2 1/2 1/2 m/2 1/2 f(m(s t))f(mt) dt f(ms u)f(u) 1 1/2 m du = f(ms u)f(u) du = f f(ms), by translation invariance of Lebesgue measure. Then F has the following properties: (i) F F 1 εm kβ 1 /2. (ii) (F F ) εm k(1 β 1)+1. (iii) F εm k. (iv) F εm 2k+1. (v) F (t) dt = 1. T (vi) supp F can be covered by less than εm kα 1+1 /2 intervals of length 2m k 1. (vii) ˆF (r) ε for all r = 0. As we chose k such that k(β 1 β) 1, we get 1 kβ 1 kβ, and so (i) gives F F 1 εm kβ 1 /2 εm 1 kβ 1 /2 εm kβ /2 ε, verifying (i). We have kept the inequality F F 1 εm kβ /2 as it is instrumental in proving (viii) later. 33

44 Similarly, we have by choice of k that k(1 β) k(1 β 1 ) + 1, and so (ii) gives (F F ) εm k(1 β1)+1 εm k(1 β), verifying (ii). (vi) tells us that there is a finite collection of intervals I (in fact, with at most εm kα1+1 /2 elements) such that I supp F, and I I I α ε mkα 1+1 (2m k 1 ) α = ε2 α 1 m 1 α+k(α1 α). 2 I I Since α 1 < α and k is fixed satisfying k(α α 1 ) > 1 α, there exists an m 1 (k, α, β, ε) 1 such that if m m 1, then 2 α 1 m 1 α+k(α1 α) < 1, so that I α < ε, I I verifying (vi). We have already proven part (ii), namely that (F F ) εm k(1 β). Then since F F is smooth, the Mean Value theorem yields h 1 F F (t + h) F F (t) (F F ) εm k(1 β). Thus if h m k, h β F F (t + h) F F (t) = h 1 β h 1 F F (t + h) F F (t) ε h 1 β m k(1 β) ε, since (1 β) > 0. If h m k, then the inequality we showed in the proof of (i) above gives h β F F (t + h) F F (t) h β 2 F F 1 ε h β m kβ ε, since β < 0. This verifies part (viii). The remaining parts (iii), (iv), (v), and (vii) follow directly from their counterparts (iii), (iv), (v), and (vii). 34

45 Chapter 4 The Main Theorem Our method of proof will, in fact, result in a slightly more general version than what we need for Theorem For the sake of generality, we will introduce the concept of ψ- Lipschitz functions in the following section, and develop a complete metric space to work in so that we may eventually apply a Baire Category argument. 4.1 ψ-lipschitz functions Lemma 4.1. (i) Consider the space F of non-empty closed subsets of T. If we set { } d F (E, F ) = max sup inf e f, sup inf e f, e E f F f F e E then (F, d F ) is a complete metric space. d F is known as the Hausdorff metric. (ii) Consider the space E consisting of ordered pairs (E, μ) where E F and μ is a probability measure with supp μ E and ˆμ(r) 0 as r. If we take d E ((E, μ), (F, σ)) = d F (E, F ) + sup ˆμ(r) ˆσ(r), r Z then (E, d E ) is a complete metric space. 35

46 (iii) Consider the space G consisting of those (E, μ) E such that μ μ = f μ λ with f μ continuous. If we take d G ((E, μ), (F, σ)) = d E ((E, μ), (F, σ)) + f μ f σ, then (G, d G ) is a complete metric space. Proof. (i) It is a standard result that the Hausdorff metric defined with usual distance is a complete metric (see Appendix D.3). (ii) If (E, μ), (F, σ) E with E = F and μ = σ, then it is clear that d E ((E, μ), (F, σ)) = 0. Conversely, if d E ((E, μ), (F, σ)) = d F (E, F ) + sup r Z ˆμ(r) ˆσ(r) = 0, then E = F by (i) and μ = σ by the uniqueness theorem. Since d F is a metric, d E ((E, μ), (F, σ)) = d E ((F, σ), (E, μ)) for any (E, μ), (F, σ) E. Triangle inequality follows directly from the usual triangle inequality on and part (i). Thus, (E, d E ) is indeed a metric space. To see completeness, let (E n, μ n ) be a Cauchy sequence in (E, d E ). Then {E n } is Cauchy in (F, d F ), so by completeness, we can find a set E F such that d F (E n, E) 0 as n. Since the space of probability measures is weak- compact, there exists a convergent subsequence, say μ nk μ w-, μ a probability measure. Since T is compact, this implies ˆμ nk (r) ˆμ(r) for every fixed r. By hypothesis, ˆμ nk (r) 0 as r, and by the Cauchy condition, as k, l, and thus ˆμ(r) 0 as r. sup ˆμ nk (r) ˆμ nl (r) 0 r Z Suppose x supp μ. By definition, for every neighborhood U x of x, we have μ(u x ) > 0. Since μ n μ w-, there exists an integer N x sufficiently large that n N x implies μ n (U) μ(u) for every open set U. In particular, for every neighborhood U x of x, this gives μ n (U x ) μ(u x ) > 0, so that x supp μ n if n N x. Let ε > 0. By property of Hausdorff metric, (see Appendix D.4), we have that there exists N ε such that n N ε implies E n E ε. Combined with the previous paragraph, this gives x supp μ n E n E ε for n max{n x, N ε }. Since this holds for arbitrary ε and E is closed, we have that x E, and so supp μ E. 36

47 Weak convergence gives ˆμ nk (r) ˆμ(r) for each fixed r, thus d E ((E nk, μ nk ), (E, μ)) 0 as k. Since a Cauchy sequence with a convergent subsequence must converge to the same limit as the subsequence, d E ((E n, μ n ), (E, μ)) 0, completing the proof. (iii) That (G, d G ) is a metric follows a similar proof as in (ii). To see completeness, let (E n, μ n ) be a Cauchy sequence in (G, d G ). Then (E n, μ n ) is Cauchy in (E, d E ), so by completeness, we can find (E, μ) E such that d E ((E n, μ n ), (E, μ)) 0 as n. Since f μn is Cauchy in the uniform norm, f μn converges uniformly to some continuous function f. By properties of weak convergence, μ μ = fλ, so (E, μ) G and d G ((E n, μ n ), (E, μ)) 0. Definition 4.2. Let 1 > α > 1/2 and suppose that ψ : R + R + is a strictly increasing, continuous function with ψ(t) t α 1/2 for all t 0, ψ(0) = 0. We call ψ a generalized α-power function. For our uses, ψ need only be defined on [0, 1/2] since we are on the torus and will be using non-negative values. Definition 4.3. Suppose ψ is a generalized α-power function. We say that f : T C is ψ-lipschitz if sup ψ( h ) 1 f(t + h) f(t) <. t,h T,h =0 In this case, we denote this value by ω ψ (f); notice ω ψ (f) 0. In the particular case of ψ(t) = t β, this definition reduces to the usual definition of Lipschitz of class β, recall (1.2). We denote by Λ ψ the collection of all ψ-lipschitz functions. Analogously to Lipschitz functions (see Lemma 1.4), we have a result linking this definition to another form which may be more useable in certain circumstances, with the same proof. Lemma 4.4. f Λ ψ if and only if f(x) f(y) Cψ( x y ) for a constant C, for every x, y T. 37

48 We develop some basic properties of operations under ω ψ which will be used later. Lemma 4.5. Let ψ be a generalized α-power function. Suppose f, g Λ ψ, c is a constant, and F L 1 (T). Then the following hold: (i) Λ ψ -translation invariance: c + f Λ ψ and ω ψ (f + c) = ω ψ (f). (ii) Λ ψ -positive homogeneity: cf Λ ψ and ω ψ (cf) = c ω ψ (f). (iii) Λ ψ -addition: f + g Λ ψ and ω ψ (f + g) ω ψ (f) + ω ψ (g). (iv) Λ ψ -multiplication: fg Λ ψ and ω ψ (fg) ω ψ (f) g + ω ψ (g) f. (v) Λ ψ -convolution: f F Λ ψ and ω ψ (f F ) ω ψ (f) F 1. Proof. (i) and (ii) are immediate from the definition. (iii) This follows by triangle inequality. (iv) If f, g Λ ψ, we have sup ψ( h ) 1 (fg)(t + h) (fg)(t) t,h T,h =0 sup ψ( h ) 1 ( f(t + h) f(t) g(t + h) + f(t) g(t + h) g(t) ) t,h T,h =0 sup ψ( h ) 1 f(t + h) f(t) g(t + h) + sup ψ( h ) 1 f(t) g(t + h) g(t) t,h T,h =0 t,h T,h =0 sup ψ( h ) 1 f(t + h) f(t) g + sup ψ( h ) 1 f g(t + h) g(t) t,h T,h =0 t,h T,h =0 = ω ψ (f) g + ω ψ (g) f <, so that fg Λ ψ and ω ψ (fg) ω ψ (f) g + ω ψ (g) f. (v) If f Λ ψ and F L 1 (T), we have for fixed t, h T, h = 0: ψ( h ) 1 (f F )(t + h) (f F )(t) = ψ( h ) 1 f(t + h y)f (y) dy f(t y)f (y) dy 38

49 ψ( h ) 1 f(t + h y) f(t y) F (y) dy ω ψ (f) F (y) dy = ω ψ (f) F 1 <, so that taking the supremum over permissible t and h yields f F Λ ψ and ω ψ (f F ) ω ψ (f) F 1. There is also a nice relationship between the ω ψ value of a function and the magnitude of its derivative, provided f is continuously differentiable. Lemma 4.6. Let ψ be a generalized α-power function. If f : T C has a continuous derivative, then f Λ ψ and ω ψ (f) f. Proof. If f : T C has continuous derivative, then f has a maximum on T, so f <. By the Mean Value theorem, for any t, h T, h = 0, we have h 1 f(t + h) f(t) f, so then h ψ( h ) yields ψ( h ) 1 f(t + h) f(t) h 1 f(t + h) f(t) f. Taking the supremum over t, h T with h = 0, we get so that f Λ ψ and sup ψ( h ) 1 f(t + h) f(t) f < t,h T,h =0 ω ψ (f) f. Lemma 4.7. Let 1 > α > 1/2 and suppose that ψ is a generalized α-power function. Consider the space L ψ consisting of those (E, μ) G such that μ μ = f μ λ with f μ Λ ψ. If we take d ψ ((E, μ), (F, σ)) = d G ((E, μ), (F, σ)) + ω ψ (f μ f σ ), then (L ψ, d ψ ) is a complete metric space. 39

50 Proof. If (E, μ), (F, σ) L ψ with E = F and μ = σ, then it is clear that d ψ ((E, μ), (F, σ)) = 0. Conversely, if d ψ ((E, μ), (F, σ)) = d G ((E, μ), (F, σ)) + ω ψ (f μ f σ ) = 0, then E = F and μ = σ since d G is a metric. Since d G is a metric and by Λ ψ -positive homogeneity, d ψ ((E, μ), (F, σ)) = d ψ ((F, σ), (E, μ)) for any (E, μ), (F, σ) L ψ. Triangle inequality follows directly from Λ ψ -addition and the fact that d G is a metric. Thus, (L ψ, d ψ ) is indeed a metric space. If (E n, μ n ) is a Cauchy sequence in (L ψ, d ψ ), then it is Cauchy in (G, d G ). Thus there exists (E, μ) G such that d G ((E n, μ n ), (E, μ)) 0 as n. By definition of G, we may write μ n μ n = f n λ and μ μ = fλ with f n and f continuous. The condition with d G above gives f n f uniformly. If m n, we have for any t, h T, h = 0 that ψ( h ) 1 (f f n )(t + h) (f f n )(t) ψ( h ) 1 (f f m )(t + h) (f f m )(t) + ψ( h ) 1 (f m f n )(t + h) (f m f n )(t) ψ( h ) 1 (f f m )(t + h) (f f m )(t) + d ψ ((E n, μ n ), (E m, μ m )) ψ( h ) 1 (f f m )(t + h) (f f m )(t) + sup d ψ ((E p, μ p ), (E q, μ q )). p,q n Allowing m, the first term limits to 0 and thus ψ( h ) 1 (f f n )(t + h) (f f n )(t) sup d ψ ((E p, μ p ), (E q, μ q )) p,q n for all t, h T, h = 0. Thus f f n Λ ψ and so Λ ψ -addition gives f Λ ψ, implying (E, μ) L ψ. Moreover, the above calculation gives ω ψ (f f n ) sup d ψ ((E p, μ p ), (E q, μ q )) p,q n so that d ψ ((E n, μ n ), (E, μ)) 0 as n, and hence (L ψ, d ψ ) is a complete metric space. 40

51 Lemma 4.8. Let 1 > α > 1/2 and suppose that ψ is a generalized α-power function. Consider the set {(E, μ) L ψ : μ μ = f μ λ, f μ is infinitely differentiable}. Let M ψ denote the closure of this set with respect to the d ψ metric. Then (M ψ, d ψ ) is a complete metric space. Proof. By definition, (M ψ, d ψ ) is a closed subspace of the complete metric space (L ψ, d ψ ) (by Lemma 4.7), and hence is complete. 4.2 Density results Let 1 > α > 1/2 and suppose that ψ is a generalized α-power function. Let H n be the subset of (M ψ, d ψ ) consisting of those (E, μ) M ψ such that we can find a finite collection of closed intervals I with I E and I α+1/n < 1 n. I I I I Notice that H n H n+1. Our goal will be to show that H n are dense in (M ψ, d ψ ) which will be Lemma Lemma 4.9. Let 1 > α > 1/2 and suppose that ψ is a generalized α-power function. Suppose further that n 1, g : T R is a positive, infinitely differentiable function with g(t) dt = 1 T and H is a closed set with H supp g. Then, given ε > 0, we can find a positive, infinitely differentiable function f : T R with T f(t) dt = 1 and a closed set E supp f such that (E, fλ) H n and d ψ ((E, fλ), (H, gλ)) < ε. 41