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On some sets with minimal L 2 discrepancy Dmitriy Bilyk University of South Carolina, Columbia, SC MCQMC 2010 Warszawa, Polska August 20, 2010 Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis Discrepancy function Consider P N [0

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SLIDE 1

On some sets with minimal L2 discrepancy

Dmitriy Bilyk University of South Carolina, Columbia, SC MCQMC 2010 Warszawa, Polska August 20, 2010

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 2

Discrepancy function

Consider PN ⊂ [0, 1]d with #PN = N: DN(x) = ♯{PN ∩ [0, x)} − Nx1x2 . . . xd

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 3

L∞ estimates (star-discrepancy)

Theorem (Schmidt, 1972) For d = 2 we have DN∞ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 4

L∞ estimates (star-discrepancy)

Theorem (Schmidt, 1972) For d = 2 we have DN∞ log N d = 2, Lerch (1904); van der Corput: There exist PN ⊂ [0, 1]2 with DN∞ ≈ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 5

L∞ estimates (star-discrepancy)

Theorem (Schmidt, 1972) For d = 2 we have DN∞ log N d = 2, Lerch (1904); van der Corput: There exist PN ⊂ [0, 1]2 with DN∞ ≈ log N d ≥ 3, Halton (1960); Hammersely; Niederreiter; Faure There exist PN ⊂ [0, 1]d with DN∞ (log N)d−1

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 6

L∞ estimates (star-discrepancy)

Theorem (Schmidt, 1972) For d = 2 we have DN∞ log N d = 2, Lerch (1904); van der Corput: There exist PN ⊂ [0, 1]2 with DN∞ ≈ log N d ≥ 3, Halton (1960); Hammersely; Niederreiter; Faure There exist PN ⊂ [0, 1]d with DN∞ (log N)d−1 Theorem (DB, Lacey, Vagharshakyan, 2007) For d ≥ 3 there exists 0 < εd ≤ 1

2, such that

DN∞ (log N)

d−1 2 +εd Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 7

Lp estimates, 1 < p < ∞

Theorem DNp (log N)

d−1 2

Roth (p = 2) 1954, Schmidt (p = 2) 1977

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 8

Lp estimates, 1 < p < ∞

Theorem DNp (log N)

d−1 2

Roth (p = 2) 1954, Schmidt (p = 2) 1977 Theorem There exist PN ⊂ [0, 1]d with DNp ≈ (log N)

d−1 2

(Davenport; Roth; Halton, Zaremba; Chen, Skriganov)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 9

Van der Corput set

“Digit reversing” van der Corput set Denote the binary expansion of x ∈ [0, 1) by x =

  • i

xi · 2−i = 0.x1x2...xn... The van der Corput set Vn with 2n points is defined as: Vn = { ( 0.x1x2 . . . xn−1xn , 0.xnxn−1 . . . x2x1 ) : xi = 0, 1}

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 10

van der Corput set

van der Corput set with N = 23 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 11

van der Corput set

van der Corput set with N = 24 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 12

van der Corput set

van der Corput set with N = 25 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 13

van der Corput set

van der Corput set with N = 26 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 14

van der Corput set

van der Corput set with N = 27 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 15

van der Corput set

van der Corput set with N = 28 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 16

van der Corput set

van der Corput set with N = 29 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 17

van der Corput set

van der Corput set with N = 210 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 18

van der Corput set

van der Corput set with N = 211 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 19

van der Corput set

van der Corput set with N = 212 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 20

Van der Corput set

“Digit reversing” van der Corput set Denote the binary expansion of x ∈ [0, 1) by x =

  • i

xi · 2−i = 0.x1x2...xn... The van der Corput set Vn with 2n points is defined as: Vn = { ( 0.x1x2 . . . xn−1xn , 0.xnxn−1 . . . x2x1 ) : xi = 0, 1} Theorem (van der Corput) The set Vn satisfies with DVn∞ n ≈ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 21

Irrational lattice

Example Let α be an irrational number and let {x} denote the fractional part of x.

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 22

Irrational lattice

Example Let α be an irrational number and let {x} denote the fractional part of x. Define PN = i

N , {iα}

N−1

i=0

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 23

Irrational lattice

Example Let α be an irrational number and let {x} denote the fractional part of x. Define PN = i

N , {iα}

N−1

i=0

Theorem If the partial quotients of the continued fraction of α are bounded, then the discrepancy function of this set satisfies DN∞ ≈ log N.

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 24

Irrational lattice

Example Let α be an irrational number and let {x} denote the fractional part of x. Define PN = i

N , {iα}

N−1

i=0

Theorem If the partial quotients of the continued fraction of α are bounded, then the discrepancy function of this set satisfies DN∞ ≈ log N. In particular works for quadratic irrationalities α = u + √v.

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 25

Irrational lattice

Example Let α be an irrational number and let {x} denote the fractional part of x. Define PN = i

N , {iα}

N−1

i=0

Theorem If the partial quotients of the continued fraction of α are bounded, then the discrepancy function of this set satisfies DN∞ ≈ log N. In particular works for quadratic irrationalities α = u + √v. The idea goes as far back as 1904 (Lerch)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 26

Low discrepancy sets

The irrational (α = √ 2) lattice with N = 212 points Discrepancy ≈ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 27

Low discrepancy sets

The van der Corput set with N = 212 points Discrepancy ≈ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 28

Low discrepancy sets

Random set with N = 212 points Discrepancy ≈ √ N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 29

L2 bounds

Theorem (K. Roth) In dimension d = 2, for any N-point set PN ⊂ [0, 1]2,

  • DN
  • 2
  • log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 30

L2 bounds

Theorem (K. Roth) In dimension d = 2, for any N-point set PN ⊂ [0, 1]2,

  • DN
  • 2
  • log N

Standard sets fail to meet this bound For the van der Corput set and the irrational lattice, we have

  • DN
  • 2 ≈ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 31

Remedies

  • 1. Davenport’s reflection (symmetrization)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 32

Remedies

  • 1. Davenport’s reflection (symmetrization)
  • 2. Digital shifts (digit-scrambling)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 33

Remedies

  • 1. Davenport’s reflection (symmetrization)
  • 2. Digital shifts (digit-scrambling)
  • 3. Cyclic shifts (mod 1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 34

Remedies

  • 1. Davenport’s reflection (symmetrization)
  • 2. Digital shifts (digit-scrambling)
  • 3. Cyclic shifts (mod 1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 35

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • van der Corput set with N = 28 points

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 36

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • van der Corput set with N = 28 points

translated (mod 1) by 1/8

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 37

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • van der Corput set with N = 28 points

translated (mod 1) by 2/8

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 38

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • van der Corput set with N = 28 points

translated (mod 1) by 3/8

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 39

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • van der Corput set with N = 28 points

translated (mod 1) by 4/8

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 40

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • Theorem (K. Roth, 1979)

Eα

  • DVα

n

  • 2
  • log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 41

Remedy: Cyclic shifts

Define Vα

n =

  • (x + α) mod 1, y
  • : (x, y) ∈ Vn
  • Theorem (K. Roth, 1979)

Eα

  • DVα

n

  • 2
  • log N

Theorem (D.B., 2008) For α = 1 − k

2n , where

k =

  • 000111 ... 000111
  • 00001111 ... 00001111
  • 2

n1 n2 = 54 17

n1 digits n2 digits we have

  • DVα

n

  • 2
  • log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 42

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 43

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1)

  • [0,1]2 #{PN ∩ [0, x)}dx =
  • p∈Vn

(1 − p1)(1 − p2)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 44

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1)

  • [0,1]2 #{PN ∩ [0, x)}dx =
  • p∈Vn

(1 − p1)(1 − p2) = 2nE

  • 1 −

n

  • k=1

Xk · 2−k ·

  • 1 −

n

  • j=1

Xj · 2−n+j−1

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 45

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1)

  • [0,1]2 #{PN ∩ [0, x)}dx =
  • p∈Vn

(1 − p1)(1 − p2) = 2nE

  • 1 −

n

  • k=1

Xk · 2−k ·

  • 1 −

n

  • j=1

Xj · 2−n+j−1 where Xi are i.i.d with Pr(Xi = 0) = Pr(Xi) = 1

2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 46

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1)

  • [0,1]2 #{PN ∩ [0, x)}dx =
  • p∈Vn

(1 − p1)(1 − p2) = 2nE

  • 1 −

n

  • k=1

Xk · 2−k ·

  • 1 −

n

  • j=1

Xj · 2−n+j−1 where Xi are i.i.d with Pr(Xi = 0) = Pr(Xi) = 1

2

EXi · Xj = 1

4, but EX 2 j = 1 2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 47

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 48

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1) Theorem (Halton and Zaremba, 1968)

  • DVn
  • 2

2 = n2

82 + O(n)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 49

The integral

The integral of discrepancy is big 1 1 DVn(x)dx = n 8 + O(1) Theorem (Halton and Zaremba, 1968)

  • DVn
  • 2

2 = n2

82 + O(n)

  • DVα

n ≈

  • DVn − k

2 +

  • p∈Vn: p1≤k/2n

p2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

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SLIDE 50

Proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k),

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-51
SLIDE 51

Proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k), where fl(k) = #{0 ≤ j ≤ k : j = (jnjn−1 . . . j2j1)2 and jl = 1}

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-52
SLIDE 52

Proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k), where fl(k) = #{0 ≤ j ≤ k : j = (jnjn−1 . . . j2j1)2 and jl = 1} fl(k) =

  • 2l−1m

if k = 2lm + r, 0 ≤ r < 2l−1 2l−1m + 1 + r if k = 2lm + 2l−1 + r, 0 ≤ r < 2l−1

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-53
SLIDE 53

proof

fl(k) =

  • 2l−1m

if k = 2lm + r, 0 ≤ r < 2l−1 2l−1m + r + 1 if k = 2lm + 2l−1 + r, 0 ≤ r < 2l−1

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-54
SLIDE 54

proof

fl(k) =

  • 2l−1m

if k = 2lm + r, 0 ≤ r < 2l−1 2l−1m + r + 1 if k = 2lm + 2l−1 + r, 0 ≤ r < 2l−1 k =

n

  • i=1

ki · 2i−1 =

  • knkn−1 . . . kl+1 kl kl−1 . . . k2k1
  • 2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-55
SLIDE 55

proof

fl(k) =

  • 2l−1m

if k = 2lm + r, 0 ≤ r < 2l−1 2l−1m + r + 1 if k = 2lm + 2l−1 + r, 0 ≤ r < 2l−1 k =

n

  • i=1

ki · 2i−1 =

  • knkn−1 . . . kl+1
  • kl kl−1 . . . k2k1
  • 2

m

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-56
SLIDE 56

proof

fl(k) =

  • 2l−1m

if k = 2lm + r, 0 ≤ r < 2l−1 2l−1m + r + 1 if k = 2lm + 2l−1 + r, 0 ≤ r < 2l−1 k =

n

  • i=1

ki · 2i−1 =

  • knkn−1 . . . kl+1
  • kl kl−1 . . . k2k1
  • 2

m r

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-57
SLIDE 57

proof

fl(k) =

  • 2l−1m

if k = 2lm + r, 0 ≤ r < 2l−1 2l−1m + r + 1 if k = 2lm + 2l−1 + r, 0 ≤ r < 2l−1 k =

n

  • i=1

ki · 2i−1 =

  • knkn−1 . . . kl+1
  • kl kl−1 . . . k2k1
  • 2

m r fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-58
SLIDE 58

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-59
SLIDE 59

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-60
SLIDE 60

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k −
  • ki
  • Dmitriy Bilyk

Geometric Discrepancy and Harmonic Analysis

slide-61
SLIDE 61

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k − n

2

  • Dmitriy Bilyk

Geometric Discrepancy and Harmonic Analysis

slide-62
SLIDE 62

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k − n

2

  • +

O(1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-63
SLIDE 63

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k − n

2

  • +

O(1) + ???

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-64
SLIDE 64

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k − n

2

  • +

O(1) + ???

  • DVα

n ≈

  • DVn − k

2 +

  • p∈Vn: p1≤k/2n

p2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-65
SLIDE 65

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k − n

2

  • +

O(1) + ???

  • DVα

n ≈

n 8 − k 2 +

  • p∈Vn: p1≤k/2n

p2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-66
SLIDE 66

proof

  • p∈Vn: p1≤k/2n

p2 =

n

  • l=1

2−lfl(k) fl(k) =

n

  • i=l+1

ki · 2i−2 + kl ·

l−1

  • i=1

ki · 2i−1 + kl

  • p1≤k/2n

p2 =

n−1

  • l=1

n

  • i=l+1

ki · 2i−l−2 +

n

  • l=1

kl · 2−l +

n

  • l=2

l−1

  • i=1

ki · kl · 2i−l−1 ≈ 1 2

  • k − n

2

  • +

O(1) + n 8 ???

  • DVα

n ≈

n 8 − k 2 +

  • p∈Vn: p1≤k/2n

p2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-67
SLIDE 67

proof

We want:

n

  • i=1

ki = n 2+O(1) and S =

n

  • l=2

l−1

  • i=1

ki·kl·2i−l−1 = n 8+O(1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-68
SLIDE 68

proof

We want:

n

  • i=1

ki = n 2+O(1) and S =

n

  • l=2

l−1

  • i=1

ki·kl·2i−l−1 = n 8+O(1) k = (000111 ... 000111)2: S ≈ 13

108n

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-69
SLIDE 69

proof

We want:

n

  • i=1

ki = n 2+O(1) and S =

n

  • l=2

l−1

  • i=1

ki·kl·2i−l−1 = n 8+O(1) k = (000111 ... 000111)2: S ≈ 13

108n 13 108 ≈ 0.1203...

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-70
SLIDE 70

proof

We want:

n

  • i=1

ki = n 2+O(1) and S =

n

  • l=2

l−1

  • i=1

ki·kl·2i−l−1 = n 8+O(1) k = (000111 ... 000111)2: S ≈ 13

108n

k = (00001111 ... 00001111)2: S ≈ 19

136n

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-71
SLIDE 71

proof

We want:

n

  • i=1

ki = n 2+O(1) and S =

n

  • l=2

l−1

  • i=1

ki·kl·2i−l−1 = n 8+O(1) k = (000111 ... 000111)2: S ≈ 13

108n

k = (00001111 ... 00001111)2: S ≈ 19

136n

k =

  • 000111 ... 000111
  • 00001111 ... 00001111
  • 2

n1 n2 = 54 17

n1 digits n2 digits S = n 8 + O(1)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-72
SLIDE 72

Fourier coefficients DVα

n (n1, n2): (n1, n2) = 0

n1 = 0 OR n1 = 0, n2 ≡ 0 mod 2n

  • DVα

n (0, n2)

  • =
  • DVn(0, n2)
  • Dmitriy Bilyk

Geometric Discrepancy and Harmonic Analysis

slide-73
SLIDE 73

Fourier coefficients DVα

n (n1, n2): (n1, n2) = 0

n1 = 0 OR n1 = 0, n2 ≡ 0 mod 2n

  • DVα

n (0, n2)

  • =
  • DVn(0, n2)
  • n1 = 0, n2 = 2sm, s < n, m - odd
  • DVα

n (0, n2) −

DVn(0, n2)

  • ≤ 2s+1

2πn2 = 1 πm

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-74
SLIDE 74

Fourier coefficients DVα

n (n1, n2): (n1, n2) = 0

n1 = 0 OR n1 = 0, n2 ≡ 0 mod 2n

  • DVα

n (0, n2)

  • =
  • DVn(0, n2)
  • n1 = 0, n2 = 2sm, s < n, m - odd
  • DVα

n (0, n2) −

DVn(0, n2)

  • ≤ 2s+1

2πn2 = 1 πm Thus,

  • DVα0

n

− DVn

  • {n1=0,n2=0}
  • 2

2 n−1

  • s=0
  • m odd

1 m2 n = log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-75
SLIDE 75

Remedy: Davenport’s reflection

Davenport’s reflection principle

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-76
SLIDE 76

Remedy: Davenport’s reflection

Davenport’s reflection principle PN − → P ∪

  • 1 − x, y
  • : (x, y) ∈ PN
  • Dmitriy Bilyk

Geometric Discrepancy and Harmonic Analysis

slide-77
SLIDE 77

Remedy: Davenport’s reflection

Davenport’s reflection principle PN − → P ∪

  • 1 − x, y
  • : (x, y) ∈ PN
  • Irrational lattice: Davenport (1956)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-78
SLIDE 78

Remedy: Davenport’s reflection

Davenport’s reflection principle PN − → P ∪

  • 1 − x, y
  • : (x, y) ∈ PN
  • Irrational lattice: Davenport (1956)

van der Corput set: Chen, Skriganov (2003)

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-79
SLIDE 79

Fibonacci set

Example Let {bn}∞

n=1 be the Fibonacci numbers.

Define Fn =

  • k

bn ,

  • k bn−1

bn

bn−1

k=0

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-80
SLIDE 80

Fibonacci set

Example Let {bn}∞

n=1 be the Fibonacci numbers.

Define Fn =

  • k

bn ,

  • k bn−1

bn

bn−1

k=0

DFn∞ ≈ n ≈ log N

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-81
SLIDE 81

Fibonacci set

Example Let {bn}∞

n=1 be the Fibonacci numbers.

Define Fn =

  • k

bn ,

  • k bn−1

bn

bn−1

k=0

DFn∞ ≈ n ≈ log N Define the symmetrization of Fn: F′

n = {(p1, p2) ∪ (p1, 1 − p2) : (p1, p2) ∈ Fn}

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-82
SLIDE 82

Fibonacci set

Example Let {bn}∞

n=1 be the Fibonacci numbers.

Define Fn =

  • k

bn ,

  • k bn−1

bn

bn−1

k=0

DFn∞ ≈ n ≈ log N Define the symmetrization of Fn: F′

n = {(p1, p2) ∪ (p1, 1 − p2) : (p1, p2) ∈ Fn}

Theorem (DB, V.Temlyakov, R.Yu) For a symmetrized set F′

n

DF′

n2 ≈ (log N)1/2. Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-83
SLIDE 83

Fibonacci set

Theorem (DB, Temlyakov, Yu) DF′

n2

2 =

1 8b2

n bn−1

  • r=1

1 sin2

πbn−1r bn

  • · sin2

πr bn

+ 17 36 − 1 36b2

n

when bn is odd, DF′

n2

2 =

1 8b2

n bn−1

  • r=1

1 sin2

πbn−1r bn

  • · sin2

πr bn

+ 17 36 + 7 72b2

n

when bn is even.

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-84
SLIDE 84

Fibonacci set

Theorem (DB, Temlyakov, Yu) DF′

n2

2 =

1 8b2

n bn−1

  • r=1

1 sin2

πbn−1r bn

  • · sin2

πr bn

+ 17 36 − 1 36b2

n

when bn is odd, DF′

n2

2 =

1 8b2

n bn−1

  • r=1

1 sin2

πbn−1r bn

  • · sin2

πr bn

+ 17 36 + 7 72b2

n

when bn is even. 1 b2

n bn−1

  • r=1

1 sin2

πbn−1r bn

  • · sin2

πr bn

≈ 0.1193n.

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-85
SLIDE 85

Fibonacci set

1 b2

n bn−1

  • r=1

1 sin2

πbn−1r bn

  • · sin2

πr bn

≈ 0.1193n.

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-86
SLIDE 86

Remedy: Digital shift

Digital shift

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-87
SLIDE 87

Remedy: Digital shift

Digital shift Define Vn,σ =

  • 0.x1x2...xn−1xn , 0.(xn ⊕ σn)...(x2 ⊕ σ2)(x1 ⊕ σ1)
  • Dmitriy Bilyk

Geometric Discrepancy and Harmonic Analysis

slide-88
SLIDE 88

Remedy: Digital shift

Digital shift Define Vn,σ =

  • 0.x1x2...xn−1xn , 0.(xn ⊕ σn)...(x2 ⊕ σ2)(x1 ⊕ σ1)
  • σk = 0, 1,

k = 1, ..., n

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-89
SLIDE 89

Remedy: Digital shift

Digital shift Define Vn,σ =

  • 0.x1x2...xn−1xn , 0.(xn ⊕ σn)...(x2 ⊕ σ2)(x1 ⊕ σ1)
  • σk = 0, 1,

k = 1, ..., n ”⊕” is addition mod 2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-90
SLIDE 90

Remedy: Digital shift

Digital shift Define Vn,σ =

  • 0.x1x2...xn−1xn , 0.(xn ⊕ σn)...(x2 ⊕ σ2)(x1 ⊕ σ1)
  • σk = 0, 1,

k = 1, ..., n ”⊕” is addition mod 2 σk ≈ n/2

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-91
SLIDE 91

Remedy: Digital shift

Digital shift Define Vn,σ =

  • 0.x1x2...xn−1xn , 0.(xn ⊕ σn)...(x2 ⊕ σ2)(x1 ⊕ σ1)
  • σk = 0, 1,

k = 1, ..., n ”⊕” is addition mod 2 σk ≈ n/2 Chen & Skriganov; Niederreiter; Pillichshammer, Larcher, Faure, Kritzer, etc

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-92
SLIDE 92

exp(Lα) estimates: B., Lacey, Parissis, Vagharshakyan

Theorem (DB, Lacey, Parissis, Vagharshakyan 2008) For any N-point set PN ⊂ [0, 1]2 we have DNexp(Lα) (log N)1−1/α , 2 ≤ α < ∞ .

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-93
SLIDE 93

exp(Lα) estimates: B., Lacey, Parissis, Vagharshakyan

Theorem (DB, Lacey, Parissis, Vagharshakyan 2008) For any N-point set PN ⊂ [0, 1]2 we have DNexp(Lα) (log N)1−1/α , 2 ≤ α < ∞ . Theorem (Faure, Pillichshammer 2008; DB, Lacey, Parissis, Vagharshakyan 2008) For 1 < p < ∞, the digit-shifted van der Corput set satisfies DNp p1/2(log N)1/2,

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-94
SLIDE 94

exp(Lα) estimates: B., Lacey, Parissis, Vagharshakyan

Theorem (DB, Lacey, Parissis, Vagharshakyan 2008) For any N-point set PN ⊂ [0, 1]2 we have DNexp(Lα) (log N)1−1/α , 2 ≤ α < ∞ . Theorem (Faure, Pillichshammer 2008; DB, Lacey, Parissis, Vagharshakyan 2008) For 1 < p < ∞, the digit-shifted van der Corput set satisfies DNp p1/2(log N)1/2, Theorem The digit-scrambled van der Corput set satisfies DNexp(Lα) (log N)1−1/α , 2 ≤ α < ∞ .

Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-95
SLIDE 95

BMO estimates: B., Lacey, Parissis, Vagharshakyan

Dyadic Chang-Fefferman BMO1,2: functions f on [0, 1]2 for which f BMO := sup

U⊂[0,1]2

  • |U|−1

R∈D2 R⊂U

f , hR2 |R| 1/2 < ∞ . Theorem For any N-point set PN ⊂ [0, 1]2 we have DNBMO (log N)

1 2 Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis

slide-96
SLIDE 96

BMO estimates: B., Lacey, Parissis, Vagharshakyan

Dyadic Chang-Fefferman BMO1,2: functions f on [0, 1]2 for which f BMO := sup

U⊂[0,1]2

  • |U|−1

R∈D2 R⊂U

f , hR2 |R| 1/2 < ∞ . Theorem For any N-point set PN ⊂ [0, 1]2 we have DNBMO (log N)

1 2

Theorem The van der Corput set satisfies DNBMO (log N)

1 2 Dmitriy Bilyk Geometric Discrepancy and Harmonic Analysis