The Discrepancy Function and the Small Ball Inequality in Higher Dimensions

Transcription

1 The Discrepancy Function and the Small Ball Inequality in Higher Dimensions Dmitriy Georgia Institute of Technology (joint work with M. Lacey and A. Vagharshakyan) 2007 Fall AMS Western Section Meeting University of New Mexico, Albuquerque, NM October 14, 2007

2 Geometric Discrepancy Let P N be an N point set in [0, 1] d of and let R [0, 1] d be a rectangle with sides parallel to the axis.

3 Geometric Discrepancy Let P N be an N point set in [0, 1] d of and let R [0, 1] d be a rectangle with sides parallel to the axis. D(P N, R) = {P N R} N vol (R)

4 Geometric Discrepancy Let P N be an N point set in [0, 1] d of and let R [0, 1] d be a rectangle with sides parallel to the axis. D(P N, R) = {P N R} N vol (R) Star-discrepancy: D(P N ) = sup D(P N, R) R

5 Geometric Discrepancy Let P N be an N point set in [0, 1] d of and let R [0, 1] d be a rectangle with sides parallel to the axis. D(P N, R) = {P N R} N vol (R) Star-discrepancy: D(P N ) = sup D(P N, R) R D(N) = inf P N D(P N )

6 Geometric Discrepancy Let P N be an N point set in [0, 1] d of and let R [0, 1] d be a rectangle with sides parallel to the axis. D(P N, R) = {P N R} N vol (R) Star-discrepancy: D(P N ) = sup D(P N, R) R D(N) = inf P N D(P N ) Can D(N) be bounded?

7 Geometric Discrepancy Let P N be an N point set in [0, 1] d of and let R [0, 1] d be a rectangle with sides parallel to the axis. D(P N, R) = {P N R} N vol (R) Star-discrepancy: D(P N ) = sup D(P N, R) R D(N) = inf P N D(P N ) Can D(N) be bounded? NO (van Aardenne-Ehrenfest; Roth)

8 Discrepancy function Enough to consider rectangles with a vertex at the origin

9 Discrepancy function Enough to consider rectangles with a vertex at the origin D N (x) = {P N [0, x)} Nx 1 x 2... x d

10 Discrepancy function Enough to consider rectangles with a vertex at the origin D N (x) = {P N [0, x)} Nx 1 x 2... x d Lower estimates for D N p???

11 Koksma-Hlawka Inequality f (x) dx 1 f (p) [0,1] d N p P N 1 N V (f ) D N

12 Koksma-Hlawka Inequality f (x) dx 1 f (p) [0,1] d N p P N 1 N V (f ) D N V (f ) is the Hardy-Krause variation of f

13 Koksma-Hlawka Inequality f (x) dx 1 f (p) [0,1] d N p P N 1 N V (f ) D N V (f ) is the Hardy-Krause variation of f V (f ) = d f [0,1] d x 1 x 2... x dx1 d... dx d

14 An example Consider a N N lattice

15 An example Consider a N N lattice

16 An example Consider a N N lattice D N (P) D N (Q) N 1 N = N

17 An example Consider a N N lattice D N (P) D N (Q) N 1 N = N Thus D N N 1 2

18 L p estimates, 1 < p < Theorem Roth (p=2), Schmidt D N p (log N) d 1 2

19 L p estimates, 1 < p < Theorem Roth (p=2), Schmidt Theorem There exist P N [0, 1] d with (Davenport, Roth, Frolov, Chen) D N p (log N) d 1 2 D N p (log N) d 1 2

20 L estimates Conjecture D N (log N) d 1 2

21 L estimates Conjecture D N (log N) d 1 2 Theorem (Schmidt) For d = 2 we have D N log N

22 L estimates Conjecture D N (log N) d 1 2 Theorem (Schmidt) For d = 2 we have D N log N d = 2 van der Corput: There exist P N [0, 1] 2 with D N log N

23 L estimates Conjecture D N (log N) d 1 2 Theorem (Schmidt) For d = 2 we have D N log N d = 2 van der Corput: There exist P N [0, 1] 2 with D N log N d 3 Halasz: There exist P N [0, 1] d with D N (log N) d 1

24 Conjectures Conjecture 1 D N (log N) d 1

25 Conjectures Conjecture 1 D N (log N) d 1 Conjecture 2 D N (log N) d 2

26 Roth s Orthogonal Function Method: Definitions Dyadic intervals are intervals of the form [k2 q, (k + 1)2 q ).

27 Roth s Orthogonal Function Method: Definitions Dyadic intervals are intervals of the form [k2 q, (k + 1)2 q ). For a dyadic interval I : h I = 1 Ileft + 1 Iright, Haar functions with L normalization.

28 Roth s Orthogonal Function Method: Definitions Dyadic intervals are intervals of the form [k2 q, (k + 1)2 q ). For a dyadic interval I : h I = 1 Ileft + 1 Iright, Haar functions with L normalization. In higher dimensions: for a rectangle R = I 1 I d h R (x 1,..., x d ) := h I1 (x 1 )... h Id (x d )

29 Roth s Orthogonal Function Method: Definitions All rectangles are in the unit cube in R d.

30 Roth s Orthogonal Function Method: Definitions All rectangles are in the unit cube in R d. Choose n appropriately so that n log 2 N and consider dyadic rectangles of volume 2 n.

31 Roth s Orthogonal Function Method: Definitions All rectangles are in the unit cube in R d. Choose n appropriately so that n log 2 N and consider dyadic rectangles of volume 2 n. H d n = {(r 1, r 2,..., r d ) N d : r r d = n}

32 Roth s Orthogonal Function Method: Definitions All rectangles are in the unit cube in R d. Choose n appropriately so that n log 2 N and consider dyadic rectangles of volume 2 n. H d n = {(r 1, r 2,..., r d ) N d : r r d = n} Definition For r H d n, call f an r function iff it is of the form f = ε R h R, ε R {±1}. R : R j =2 r j

33 Roth s Orthogonal Function Method: Proof For each r H d n, there exists f r such that D N, f r 1.

34 Roth s Orthogonal Function Method: Proof For each r H d n, there exists f r such that D N, f r 1. Construct the test function F = r H d n f r.

35 Roth s Orthogonal Function Method: Proof For each r H d n, there exists f r such that D N, f r 1. Construct the test function F = r H d n f r. { r H d n} n d 1 (log N) d 1.

36 Roth s Orthogonal Function Method: Proof For each r H d n, there exists f r such that D N, f r 1. Construct the test function F = r H d n f r. { r H d n} n d 1 (log N) d 1. F 2 (log N) d 1 2.

37 Roth s Orthogonal Function Method: Proof For each r H d n, there exists f r such that D N, f r 1. Construct the test function F = r H d n f r. { r H d n} n d 1 (log N) d 1. F 2 (log N) d 1 2. D N, F (log N) d 1.

38 Roth s Orthogonal Function Method: Proof For each r H d n, there exists f r such that D N, f r 1. Construct the test function F = r H d n f r. { r H d n} n d 1 (log N) d 1. F 2 (log N) d 1 2. D N, F (log N) d 1. Thus D N 2 D N, f r F 2 (log N) d 1 2

39 The Small Ball Inequality We are interested in non-trivial lower bounds for the hyperbolic sums of Haar functions: α R h R R =2 n

40 The Trivial Bound L 2 estimate n d 1 2 α R h R 2 n α R R =2 n R =2 n

41 The Trivial Bound L 2 estimate n d 1 2 α R h R 2 n α R R =2 n R =2 n holds with the L 2 norm on the left hand side

42 The Trivial Bound L 2 estimate n d 1 2 α R h R 2 n α R R =2 n R =2 n holds with the L 2 norm on the left hand side Cauchy-Schwarz inequality

43 The Trivial Bound L 2 estimate n d 1 2 α R h R 2 n α R R =2 n R =2 n holds with the L 2 norm on the left hand side Cauchy-Schwarz inequality every point is contained in n d 1 dyadic rectangles of area 2 n

44 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) R =2 n α R h R

45 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) This conjecture is related to: R =2 n α R h R

46 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) This conjecture is related to: Irregularities of Distribution R =2 n α R h R

47 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) This conjecture is related to: Irregularities of Distribution Approximation Theory R =2 n α R h R

48 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) This conjecture is related to: Irregularities of Distribution Approximation Theory Probability Theory R =2 n α R h R

49 The Small Ball Conjecture and Discrepancy Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) R =2 n α R h R

50 The Small Ball Conjecture and Discrepancy Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) Conjecture 2 D N (log N) d 2 R =2 n α R h R

51 The Small Ball Conjecture and Discrepancy Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) Conjecture 2 D N (log N) d 2 R =2 n α R h R Notice that, in both conjectures, one gains a square root over the L 2 estimate.

52 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) R =2 n α R h R

53 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) R =2 n α R h R Random choice of sign shows this is sharp.

54 The Conjecture Small Ball Conjecture For dimensions d 2, we have 2 n R =2 n α R n 1 2 (d 2) R =2 n α R h R Random choice of sign shows this is sharp. Talagrand has proved this in d = 2. Another proof by Temlyakov.

55 Main result Theorem ( DB, Michael Lacey & Armen Vagharshakyan) In dimensions d 3 there is a η(d) > 0, such that for all choices of coefficients α R we have n d 1 2 η(d) α R h R 2 n α R (1) R =2 n R =2 n Previously known: d = 3 József Beck (1989): n η (log n) 1/8.

56 A New Result for the Discrepancy Function in higher dimensions Theorem (DB, M.Lacey, A.Vagharshakyan) For d 3 there is a choice of 0 < η(d) < 1 2 for which the following estimate holds for all collections P N [0, 1] d : D N (log N) d 1 2 +η(d).

57 A New Result for the Discrepancy Function in higher dimensions Theorem (DB, M.Lacey, A.Vagharshakyan) For d 3 there is a choice of 0 < η(d) < 1 2 for which the following estimate holds for all collections P N [0, 1] d : D N (log N) d 1 2 +η(d). η(d) 1 d 2

58 A New Result for the Discrepancy Function in higher dimensions Theorem (DB, M.Lacey, A.Vagharshakyan) For d 3 there is a choice of 0 < η(d) < 1 2 for which the following estimate holds for all collections P N [0, 1] d : D N (log N) d 1 2 +η(d). η(d) 1 d 2 Previously known: d = 3 József Beck (1989): (log N) η (log log N) 1/8.

59 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz Theorem In dimension 2, we have 2 n R =2 n α R R =2 n α R h R

60 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz Theorem In dimension 2, we have 2 n R =2 n α R The method of proof is by duality. R =2 n α R h R

61 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz Theorem In dimension 2, we have 2 n R =2 n α R The method of proof is by duality. Use Riesz products. R =2 n α R h R

62 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz Theorem In dimension 2, we have 2 n R =2 n α R The method of proof is by duality. Use Riesz products. R =2 n α R h R In Dimension 2, there is a Product Rule : If R = R, then, h R h R is either 0 1 R ±h R R.

63 Product Rule: Illustration

64 Product Rule Fails in higher dimensions!!!

65 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz H def = R =2 n α R h R

66 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz H def = f (k,n k) def = R =2 n α R h R R 1 =2 k, R 2 =2 n+k sgn(α R ) h R, 0 k n,

67 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz H def = f (k,n k) def = F def = R =2 n α R h R sgn(α R ) h R, 0 k n, R 1 =2 k, R 2 =2 n+k n ( ) 1 + f(k,n k) k=0

68 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz H def = f (k,n k) def = F def = F 0 R =2 n α R h R sgn(α R ) h R, 0 k n, R 1 =2 k, R 2 =2 n+k n ( ) 1 + f(k,n k) k=0 & F = 1 F 1 = 1

69 Proof of Talagrand s Theorem, following Temlyakov, Schmidt and Halasz H def = f (k,n k) def = F def = F 0 R =2 n α R h R sgn(α R ) h R, 0 k n, R 1 =2 k, R 2 =2 n+k n ( ) 1 + f(k,n k) k=0 & H H, F = F = 1 F 1 = 1 n H, f (k,n k) = 2 n 1 k=0 R =2 n α R

70 Halasz s Proof of Schmidt s Theorem Theorem (Schmidt) For d = 2 we have D N log N Take n log 2 N and consider r = (k, n k)

71 Halasz s Proof of Schmidt s Theorem Theorem (Schmidt) For d = 2 we have D N log N Take n log 2 N and consider r = (k, n k) Use r functions such that D N, f r 1

72 Halasz s Proof of Schmidt s Theorem Theorem (Schmidt) For d = 2 we have D N log N Take n log 2 N and consider r = (k, n k) Use r functions such that D N, f r 1 F def = n k=0( 1 + cf(k,n k) ) 1

73 Halasz s Proof of Schmidt s Theorem Theorem (Schmidt) For d = 2 we have D N log N Take n log 2 N and consider r = (k, n k) Use r functions such that D N, f r 1 F def = n k=0( 1 + cf(k,n k) ) 1 F 1 2

74 Halasz s Proof of Schmidt s Theorem Theorem (Schmidt) For d = 2 we have D N log N Take n log 2 N and consider r = (k, n k) Use r functions such that D N, f r 1 F def = n k=0( 1 + cf(k,n k) ) 1 F 1 2 D N D N, F n + higher order terms

75 The Short Riesz Product Set q = n ɛ, this will be the length of our product.

76 The Short Riesz Product Set q = n ɛ, this will be the length of our product. Divide the integers {1, 2,..., n} into q disjoint intervals def I 1,..., I q, and let I t = { r N n : r 1 I t }.

77 The Short Riesz Product Set q = n ɛ, this will be the length of our product. Divide the integers {1, 2,..., n} into q disjoint intervals def I 1,..., I q, and let I t = { r N n : r 1 I t }. F t = r I t f r. ρ = q1/2 n (d 1)/2 ρ = true L 2 normalization. aqb n (d 1)/2 false normalization (b = 1/4, a << 1).

78 The Short Riesz Product Set q = n ɛ, this will be the length of our product. Divide the integers {1, 2,..., n} into q disjoint intervals def I 1,..., I q, and let I t = { r N n : r 1 I t }. F t = r I t f r. ρ = q1/2 n (d 1)/2 ρ = true L 2 normalization. aqb n (d 1)/2 false normalization (b = 1/4, a << 1). Ψ def = q (1 + ρ F t ). t=1

79 Definition Vectors r j N n, say that r 1,..., r J are strongly distinct iff there are no two vectors which are equal in any coordinate.

80 Definition Vectors r j N n, say that r 1,..., r J are strongly distinct iff there are no two vectors which are equal in any coordinate. Lemma The product of strongly distinct r functions is again an r function.

81 Definition Vectors r j N n, say that r 1,..., r J are strongly distinct iff there are no two vectors which are equal in any coordinate. Lemma The product of strongly distinct r functions is again an r function. Write Ψ = 1 + Ψ sd + Ψ sd Ψ sd = Sum of the strongly distinct products in the expansion of Ψ.

82 Definition Vectors r j N n, say that r 1,..., r J are strongly distinct iff there are no two vectors which are equal in any coordinate. Lemma The product of strongly distinct r functions is again an r function. Write Ψ = 1 + Ψ sd + Ψ sd Ψ sd = Sum of the strongly distinct products in the expansion of Ψ. We only need to show that Ψ sd 1 1.

83 Methods Conditional expectation arguments

84 Methods Conditional expectation arguments L p estimates

85 Methods Conditional expectation arguments L p estimates Littlewood-Paley inequalities with sharp constants in p (Burkholder; Wang; R. Fefferman, J. Pipher)

86 Methods Conditional expectation arguments L p estimates Littlewood-Paley inequalities with sharp constants in p (Burkholder; Wang; R. Fefferman, J. Pipher) Exponential moments ( hyperbolic Chang-Wilson-Wolf, subgaussian estimates)

87 Methods Conditional expectation arguments L p estimates Littlewood-Paley inequalities with sharp constants in p (Burkholder; Wang; R. Fefferman, J. Pipher) Exponential moments ( hyperbolic Chang-Wilson-Wolf, subgaussian estimates) Analysis of coincidences

88 The Crucial Lemma of Beck Lemma Beck Gain: We have the estimate f r f s p 3/2 n 3/2 p r s N n r 2 =s 2

89 Restricted inequality in all dimensions Theorem ( DB, Michael Lacey & Armen Vagharshakyan) In dimensions d 3 there is a η(d) > 0, such that for all choices of coefficients α R {±1} we have α R h R n d 1 R =2 n For d = 3, η = 1 10 ε. 2 +η(d) (2)

90 Restricted inequality in all dimensions Theorem ( DB, Michael Lacey & Armen Vagharshakyan) In dimensions d 3 there is a η(d) > 0, such that for all choices of coefficients α R {±1} we have α R h R n d 1 R =2 n For d = 3, η = 1 10 ε. In higher dimensions, η(d) 1 d. 2 +η(d) (2)

91 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator.

92 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator. Define Ball s def = {f : f s + MD f s 1}

93 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator. Define Ball s def = {f : f s + MD f s 1} Define N s (ɛ) := least number of L balls of radius ɛ needed to cover Ball s

94 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator. Define Ball s def = {f : f s + MD f s 1} Define N s (ɛ) := least number of L balls of radius ɛ needed to cover Ball s Conjecture For d 2, one has the estimate log N 2 (ɛ) 1 ɛ ( ) log 1 d 1/2 ɛ

95 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator. Define Ball s def = {f : f s + MD f s 1} Define N s (ɛ) := least number of L balls of radius ɛ needed to cover Ball s Conjecture For d 2, one has the estimate log N 2 (ɛ) 1 ɛ Upper bound is known ( ) log 1 d 1/2 ɛ

96 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator. Define Ball s def = {f : f s + MD f s 1} Define N s (ɛ) := least number of L balls of radius ɛ needed to cover Ball s Conjecture For d 2, one has the estimate log N 2 (ɛ) 1 ɛ Upper bound is known ( ) log 1 d 1/2 ɛ Small Ball inequality for smooth wavelets implies the lower bound (Talagrand)

97 Covering Numbers of Mixed Derivative Spaces Let MD = D x1 D xd be the mixed derivative operator. Define Ball s def = {f : f s + MD f s 1} Define N s (ɛ) := least number of L balls of radius ɛ needed to cover Ball s Conjecture For d 2, one has the estimate log N 2 (ɛ) 1 ɛ Upper bound is known ( ) log 1 d 1/2 ɛ Small Ball inequality for smooth wavelets implies the lower bound (Talagrand) Small Ball Inequality for Haars implies a bound for N 1 (ɛ)

98 Small Ball Problem for the Brownian Sheet Let B : [0, 1] d R be the Brownian Sheet: d EB(s)B(t) = min{s j, t j } j=1

99 Small Ball Problem for the Brownian Sheet Let B : [0, 1] d R be the Brownian Sheet: d EB(s)B(t) = min{s j, t j } j=1 Theorem (Kuelbs, Li) log P( B C[0,1] d < ɛ) ɛ 2 ( log 1 ɛ ) β iff log N 2 (ɛ) ɛ 1 ( log 1 ɛ ) β/2

100 Small Ball Problem for the Brownian Sheet Let B : [0, 1] d R be the Brownian Sheet: d EB(s)B(t) = min{s j, t j } j=1 Theorem (Kuelbs, Li) log P( B C[0,1] d < ɛ) ɛ 2 ( log 1 ɛ ) β iff log N 2 (ɛ) ɛ 1 ( log 1 ɛ ) β/2 Small Ball Problem For d 2, we have log P( B C[0,1] d < ɛ) ɛ 2 ( log 1 ɛ ) 2d 1