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Discrepancy theory and harmonic analysis
Dmitriy BilykUniversity of Minnesota
“Uniform distribution and quasi-Monte Carlo methods”Johann Radon Institute for Computational and Applied
Mathematics (RICAM)Linz, Austria
October 14, 2013
Dmitriy Bilyk Discrepancy and analysis
Geometric discrepancy
No rotations: discrepancy ≈ (log N)α
(K. Roth, W. Schmidt, van der Corput)
All rotations: discrepancy ≈ Nβ
(J. Beck, H. Montgomerry)
Pictures taken from Matousek’s “Geometric Discrepancy: An Illustrated Guide”
Dmitriy Bilyk Discrepancy and analysis
Discrepancy bounds in dimension d = 2
LOWER BOUND UPPER BOUND
Axis-parallel rectangles
L∞ log N log N
L2 log12 N log
12 N
Rotated rectangles
N1/4N1/4√
log N
Circles
N1/4N1/4√
log N
Convex Sets
N1/3N1/3 log4 N
Dmitriy Bilyk Discrepancy and analysis
Higher dimensions: d ≥ 3
LOWER BOUND UPPER BOUND
Axis-parallel boxes
L∞ (log N)d−1
2+η
(log N)d−1
L2 (log N)d−1
2 (log N)d−1
2
Rotated boxes
N12− 1
2d N12− 1
2d√
log N
Balls
N12− 1
2d N12− 1
2d√
log N
Convex Sets
N1− 2d+1 N1− 2
d+1 logc N
Dmitriy Bilyk Discrepancy and analysis
Discrepancy bounds in dimension d = 2
LOWER BOUND UPPER BOUND
Axis-parallel rectangles
L∞ log N log N
L2 log12 N log
12 N
Rotated rectangles
N1/4N1/4√
log N
Circles
N1/4N1/4√
log N
Convex Sets
N1/3N1/3 log4 N
Dmitriy Bilyk Discrepancy and analysis
Discrepancy function Lacunary Fourier series
DN(x) = #PN ∩ [0, x) − Nx1x2 f (x) ∼∑∞
k=1 ck sin nkx ,nk+1
nk> λ > 1
‖DN‖2 &√
log N ‖f ‖2 ≡√∑
|ck |2(Roth, ’54)
‖DN‖∞ & log N ‖f ‖∞ &∑|ck |
(Schmidt, ’72; Halasz, ’81) (Sidon, ’27)
Riesz product:∏(
1 + cfk)
Riesz product:∏(1 + cos(nkx + φk)
)‖DN‖1 &
√log N ‖f ‖1 & ‖f ‖2
(Halasz, ’81) (Sidon, ’30)
Riesz product:∏(
1 + i · c√log N
fk)
Riesz product:∏(1 + i · |ck |‖f ‖2
cos(nkx + θk))
Table : Discrepancy function and lacunary Fourier series
Dmitriy Bilyk Discrepancy and analysis
Sidon’s theorem and non-dense orbits
Sidon’s theorem:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that∣∣∣∣∑j
cj sin(2πnjx)
∣∣∣∣ & ε| log ε|−1∑j
|cj |.
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
Dmitriy Bilyk Discrepancy and analysis
Sidon’s theorem and non-dense orbits
Sidon’s theorem:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that∣∣∣∣∑j
cj sin(2πnjx)
∣∣∣∣ & ε| log ε|−1∑j
|cj |.
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
Dmitriy Bilyk Discrepancy and analysis
Origins: Erdos’ question
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
History:
Erdos posed the question in 1976.Khintchine (1926): ε2| log ε|−2.Pollington (1979), de Mathan (1980): ε4| log ε|−1.Katznelson (2001): ε2| log ε|−1.Akhunzhanov, Moschevitin (2004): ε2.
Dmitriy Bilyk Discrepancy and analysis
Origins: Erdos’ question
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
History:
Erdos posed the question in 1976.Khintchine (1926): ε2| log ε|−2.Pollington (1979), de Mathan (1980): ε4| log ε|−1.Katznelson (2001): ε2| log ε|−1.Akhunzhanov, Moschevitin (2004): ε2.
Dmitriy Bilyk Discrepancy and analysis
Connections
Another question of Erdos (1987):Let S = nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Define the graph G on Z: (n,m) is an edge iff |n −m| ∈ S.– What is the chromatic number of G ?
Katznelson (2001):If ∃θ with inf ‖njθ‖ > δ, then chr(G ) < 1/δ.There exists a set A ⊂ N with upper densityd∗(A) & ε| log ε|−1 such that
(A− A) ∩ S = ∅.
(S = nj is not an intersective set).For H ⊂ N, define
δH = supd∗(A) : (A− A) ∩ H = ∅.
γH = infa0 : T (x) = a0 +∑n∈H
ah cos(2πnx) ≥ 0, T (0) = 1
Rusza (1984): γH ≥ δH .
Dmitriy Bilyk Discrepancy and analysis
Connections
Another question of Erdos (1987):Let S = nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Define the graph G on Z: (n,m) is an edge iff |n −m| ∈ S.– What is the chromatic number of G ?Katznelson (2001):If ∃θ with inf ‖njθ‖ > δ, then chr(G ) < 1/δ.
There exists a set A ⊂ N with upper densityd∗(A) & ε| log ε|−1 such that
(A− A) ∩ S = ∅.
(S = nj is not an intersective set).For H ⊂ N, define
δH = supd∗(A) : (A− A) ∩ H = ∅.
γH = infa0 : T (x) = a0 +∑n∈H
ah cos(2πnx) ≥ 0, T (0) = 1
Rusza (1984): γH ≥ δH .
Dmitriy Bilyk Discrepancy and analysis
Connections
Another question of Erdos (1987):Let S = nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Define the graph G on Z: (n,m) is an edge iff |n −m| ∈ S.– What is the chromatic number of G ?Katznelson (2001):If ∃θ with inf ‖njθ‖ > δ, then chr(G ) < 1/δ.There exists a set A ⊂ N with upper densityd∗(A) & ε| log ε|−1 such that
(A− A) ∩ S = ∅.
(S = nj is not an intersective set).
For H ⊂ N, define
δH = supd∗(A) : (A− A) ∩ H = ∅.
γH = infa0 : T (x) = a0 +∑n∈H
ah cos(2πnx) ≥ 0, T (0) = 1
Rusza (1984): γH ≥ δH .
Dmitriy Bilyk Discrepancy and analysis
Connections
Another question of Erdos (1987):Let S = nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Define the graph G on Z: (n,m) is an edge iff |n −m| ∈ S.– What is the chromatic number of G ?Katznelson (2001):If ∃θ with inf ‖njθ‖ > δ, then chr(G ) < 1/δ.There exists a set A ⊂ N with upper densityd∗(A) & ε| log ε|−1 such that
(A− A) ∩ S = ∅.
(S = nj is not an intersective set).For H ⊂ N, define
δH = supd∗(A) : (A− A) ∩ H = ∅.
γH = infa0 : T (x) = a0 +∑n∈H
ah cos(2πnx) ≥ 0, T (0) = 1
Rusza (1984): γH ≥ δH .
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Marshall Hall (Annals of Math., 48, 1947):any real number is representable as a sum of two continuedfractions with partial quotients at most 4.
Given any θ1, θ2, there exists α such that∣∣∣∣(α− θ1)− p
q
∣∣∣∣ > c
q2and
∣∣∣∣(α− θ2)− p
q
∣∣∣∣ > c
q2.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Marshall Hall (Annals of Math., 48, 1947):any real number is representable as a sum of two continuedfractions with partial quotients at most 4.
Given any θ1, θ2, there exists α such that∣∣∣∣(α− θ1)− p
q
∣∣∣∣ > c
q2and
∣∣∣∣(α− θ2)− p
q
∣∣∣∣ > c
q2.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Marshall Hall (Annals of Math., 48, 1947):any real number is representable as a sum of two continuedfractions with partial quotients at most 4.
Given any θ1, θ2, there exists α such that∣∣∣∣(α− θ1)− p
q
∣∣∣∣ > c
q2and
∣∣∣∣(α− θ2)− p
q
∣∣∣∣ > c
q2.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Lemma (Cassels (1956); Davenport (1964))
For all n > 0 there exists c = c(n) > 0 so that for any θ1, θ2, ..., θnthere exists α such that for all j = 1, ..., n∣∣∣∣
tan
(α− θj)−p
q
∣∣∣∣ > c
q2
for all p ∈ Z, q ∈ N.
c(n) ≈ 1/n2
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Lemma (Cassels (1956); Davenport (1964))
For all n > 0 there exists c = c(n) > 0 so that for any θ1, θ2, ..., θnthere exists α such that for all j = 1, ..., n∣∣∣∣
tan
(α− θj)−p
q
∣∣∣∣ > c
q2
for all p ∈ Z, q ∈ N.
c(n) ≈ 1/n2
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Lemma (Cassels (1956); Davenport (1964))
For all n > 0 there exists c = c(n) > 0 so that for any θ1, θ2, ..., θnthere exists α such that for all j = 1, ..., n∣∣∣∣ tan (α− θj)−
p
q
∣∣∣∣ > c
q2
for all p ∈ Z, q ∈ N.
c(n) ≈ 1/n2
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2for all θ ∈ Ω ???
Lemma (Cassels (1956); Davenport (1964))
For all n > 0 there exists c = c(n) > 0 so that for any θ1, θ2, ..., θnthere exists α such that for all j = 1, ..., n∣∣∣∣ tan (α− θj)−
p
q
∣∣∣∣ > c
q2
for all p ∈ Z, q ∈ N. c(n) ≈ 1/n2
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2 · ψ(q)for all θ ∈ Ω ???
Examples:
Lacunary sequences, e.g. 2−jLacunary sets of finite order:
Lacunary of order M + 1 = union of a set S lacunary of orderM and lacunary sequences converging to to each point of SDirectional maximal function MΩ is bounded in Lp(R2) iff Ω isa finite union of lacunary sets of finite order (Bateman, 2007)E.g., Ω = 2−n1 + 2−n2 + ... + 2−nMnk∈N
Sets of Minkowski dimension 0 ≤ d ≤ 1
“Superlacunary” sets, e.g. 2−2n.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2 · ψ(q)for all θ ∈ Ω ???
Examples:
Lacunary sequences, e.g. 2−jLacunary sets of finite order:
Lacunary of order M + 1 = union of a set S lacunary of orderM and lacunary sequences converging to to each point of SDirectional maximal function MΩ is bounded in Lp(R2) iff Ω isa finite union of lacunary sets of finite order (Bateman, 2007)E.g., Ω = 2−n1 + 2−n2 + ... + 2−nMnk∈N
Sets of Minkowski dimension 0 ≤ d ≤ 1
“Superlacunary” sets, e.g. 2−2n.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2 · ψ(q)for all θ ∈ Ω ???
Examples:
Lacunary sequences, e.g. 2−j
Lacunary sets of finite order:
Lacunary of order M + 1 = union of a set S lacunary of orderM and lacunary sequences converging to to each point of SDirectional maximal function MΩ is bounded in Lp(R2) iff Ω isa finite union of lacunary sets of finite order (Bateman, 2007)E.g., Ω = 2−n1 + 2−n2 + ... + 2−nMnk∈N
Sets of Minkowski dimension 0 ≤ d ≤ 1
“Superlacunary” sets, e.g. 2−2n.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2 · ψ(q)for all θ ∈ Ω ???
Examples:
Lacunary sequences, e.g. 2−jLacunary sets of finite order:
Lacunary of order M + 1 = union of a set S lacunary of orderM and lacunary sequences converging to to each point of SDirectional maximal function MΩ is bounded in Lp(R2) iff Ω isa finite union of lacunary sets of finite order (Bateman, 2007)E.g., Ω = 2−n1 + 2−n2 + ... + 2−nMnk∈N
Sets of Minkowski dimension 0 ≤ d ≤ 1
“Superlacunary” sets, e.g. 2−2n.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2 · ψ(q)for all θ ∈ Ω ???
Examples:
Lacunary sequences, e.g. 2−jLacunary sets of finite order:
Lacunary of order M + 1 = union of a set S lacunary of orderM and lacunary sequences converging to to each point of SDirectional maximal function MΩ is bounded in Lp(R2) iff Ω isa finite union of lacunary sets of finite order (Bateman, 2007)E.g., Ω = 2−n1 + 2−n2 + ... + 2−nMnk∈N
Sets of Minkowski dimension 0 ≤ d ≤ 1
“Superlacunary” sets, e.g. 2−2n.
Dmitriy Bilyk Discrepancy and analysis
Question
Given Ω ⊂ [0, 1], does there exists α such that∣∣∣∣(α− θ)− p
q
∣∣∣∣ > c
q2 · ψ(q)for all θ ∈ Ω ???
Examples:
Lacunary sequences, e.g. 2−jLacunary sets of finite order:
Lacunary of order M + 1 = union of a set S lacunary of orderM and lacunary sequences converging to to each point of SDirectional maximal function MΩ is bounded in Lp(R2) iff Ω isa finite union of lacunary sets of finite order (Bateman, 2007)E.g., Ω = 2−n1 + 2−n2 + ... + 2−nMnk∈N
Sets of Minkowski dimension 0 ≤ d ≤ 1
“Superlacunary” sets, e.g. 2−2n.
Dmitriy Bilyk Discrepancy and analysis
1. Cassels-Davenport iteration
Construct a sequence ...In−1 ⊃ In ⊃ ... with |In| → 0 so that∣∣∣∣(α− θ)− p
q
∣∣∣∣ ≥ c(n)
q2(∗)
for all α ∈ In, θ ∈ Ω, and R(n) ≤ q ≤ R(n + 1)—————————————————————————–
Let N(x) be the covering function of Ω (the minimal numberof intervals of length x needed to cover Ω).
Let F (x) ≈ x · N(x)
Theorem (DB, X. Ma, J. Pipher, C. Spencer)
There exists θ such that for all φ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(F−1
(1
q2
))for all p ∈ Z, q ∈ N.
Dmitriy Bilyk Discrepancy and analysis
1. Cassels-Davenport iteration
Construct a sequence ...In−1 ⊃ In ⊃ ... with |In| → 0 so that∣∣∣∣(α− θ)− p
q
∣∣∣∣ ≥ c(n)
q2(∗)
for all α ∈ In, θ ∈ Ω, and R(n) ≤ q ≤ R(n + 1)—————————————————————————–
Let N(x) be the covering function of Ω (the minimal numberof intervals of length x needed to cover Ω).
Let F (x) ≈ x · N(x)
Theorem (DB, X. Ma, J. Pipher, C. Spencer)
There exists θ such that for all φ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(F−1
(1
q2
))for all p ∈ Z, q ∈ N.
Dmitriy Bilyk Discrepancy and analysis
1. Cassels-Davenport iteration
Construct a sequence ...In−1 ⊃ In ⊃ ... with |In| → 0 so that∣∣∣∣(α− θ)− p
q
∣∣∣∣ ≥ c(n)
q2(∗)
for all α ∈ In, θ ∈ Ω, and R(n) ≤ q ≤ R(n + 1)—————————————————————————–
Let N(x) be the covering function of Ω (the minimal numberof intervals of length x needed to cover Ω).
Let F (x) ≈ x · N(x)
Theorem (DB, X. Ma, J. Pipher, C. Spencer)
There exists θ such that for all φ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(F−1
(1
q2
))for all p ∈ Z, q ∈ N.
Dmitriy Bilyk Discrepancy and analysis
1. Cassels-Davenport iteration
Construct a sequence ...In−1 ⊃ In ⊃ ... with |In| → 0 so that∣∣∣∣(α− θ)− p
q
∣∣∣∣ ≥ c(n)
q2(∗)
for all α ∈ In, θ ∈ Ω, and R(n) ≤ q ≤ R(n + 1)—————————————————————————–
Let N(x) be the covering function of Ω (the minimal numberof intervals of length x needed to cover Ω).
Let F (x) ≈ x · N(x)
Theorem (DB, X. Ma, J. Pipher, C. Spencer)
There exists θ such that for all φ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(F−1
(1
q2
))for all p ∈ Z, q ∈ N.
Dmitriy Bilyk Discrepancy and analysis
1. Cassels-Davenport iteration
Theorem (DB, X. Ma, J. Pipher, C. Spencer)
There exists α such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(F−1
(1
q2
))for all p ∈ Z, q ∈ N.
Ω finite: N = #Ω, F (x) ≈ Nx ,
F−1(F−1(1/q2)) ≈ 1
N2q2
Ω lacunary: N(x) ≈ log 1x , F (x) ≈ x · log 1
x , F−1(x) ≈ xlog(1/x) ,
F−1(F−1(1/q2)) ≈ 1
q2 log2 q
Ω “superlacunary”, F−1(F−1(1/q2)) ≈ 1q2(log log q)2
Dmitriy Bilyk Discrepancy and analysis
2. Trivial argument
Let h(q) be an increasing function of q such that∑ 1q·h(q) <∞. (h(q) = qε, h(q) = log1+ε q etc.)
For each q, cover Ω by N(δ(q)
)intervals Bk of length δ(q)
and remove the intervals Bk +(p/q − δ(q), p/q + δ(q)
).
Theorem
For any Ω ⊂ [0, 1] there exists α ∈ [0, 1) such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(1
q2 · h(q)
).
Ω has upper Minkowski dimension d : N(x) .(
1x
)d+ε,
F (x) . x1−d−ε, F−1(x) & x1
1−d−ε,
F−1(1/q2+ε) & q−2
1−d−ε
Dmitriy Bilyk Discrepancy and analysis
2. Trivial argument
Let h(q) be an increasing function of q such that∑ 1q·h(q) <∞. (h(q) = qε, h(q) = log1+ε q etc.)
For each q, cover Ω by N(δ(q)
)intervals Bk of length δ(q)
and remove the intervals Bk +(p/q − δ(q), p/q + δ(q)
).
Theorem
For any Ω ⊂ [0, 1] there exists α ∈ [0, 1) such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(1
q2 · h(q)
).
Ω has upper Minkowski dimension d : N(x) .(
1x
)d+ε,
F (x) . x1−d−ε, F−1(x) & x1
1−d−ε,
F−1(1/q2+ε) & q−2
1−d−ε
Dmitriy Bilyk Discrepancy and analysis
2. Trivial argument
Let h(q) be an increasing function of q such that∑ 1q·h(q) <∞. (h(q) = qε, h(q) = log1+ε q etc.)
For each q, cover Ω by N(δ(q)
)intervals Bk of length δ(q)
and remove the intervals Bk +(p/q − δ(q), p/q + δ(q)
).
Theorem
For any Ω ⊂ [0, 1] there exists α ∈ [0, 1) such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(1
q2 · h(q)
).
Ω has upper Minkowski dimension d : N(x) .(
1x
)d+ε,
F (x) . x1−d−ε, F−1(x) & x1
1−d−ε,
F−1(1/q2+ε) & q−2
1−d−ε
Dmitriy Bilyk Discrepancy and analysis
2. Trivial argument
Let h(q) be an increasing function of q such that∑ 1q·h(q) <∞. (h(q) = qε, h(q) = log1+ε q etc.)
For each q, cover Ω by N(δ(q)
)intervals Bk of length δ(q)
and remove the intervals Bk +(p/q − δ(q), p/q + δ(q)
).
Theorem
For any Ω ⊂ [0, 1] there exists α ∈ [0, 1) such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & F−1
(1
q2 · h(q)
).
Ω has upper Minkowski dimension d : N(x) .(
1x
)d+ε,
F (x) . x1−d−ε, F−1(x) & x1
1−d−ε,
F−1(1/q2+ε) & q−2
1−d−ε
Dmitriy Bilyk Discrepancy and analysis
3. Peres-Schlag method: more connections
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
Littlewood conjecture: for any (α, β) ∈ R2,
infq≥1
q · ‖qα‖ · ‖qβ‖ = 0.
Moschevitin,... 2010-2012:e.g. Bugeaud–Moschevitin 2011:the set of points (α, β) such that inf q · log2 q‖qα‖ · ‖qβ‖ > 0has Hausdorff dimension 2.
Einsiedler, Katok, Lindenstrauss (2006):the set of points (α, β) such that inf q · ‖qα‖ · ‖qβ‖ > 0 hasHausdorff dimension 0.
Dmitriy Bilyk Discrepancy and analysis
3. Peres-Schlag method: more connections
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
Littlewood conjecture: for any (α, β) ∈ R2,
infq≥1
q · ‖qα‖ · ‖qβ‖ = 0.
Moschevitin,... 2010-2012:e.g. Bugeaud–Moschevitin 2011:the set of points (α, β) such that inf q · log2 q‖qα‖ · ‖qβ‖ > 0has Hausdorff dimension 2.
Einsiedler, Katok, Lindenstrauss (2006):the set of points (α, β) such that inf q · ‖qα‖ · ‖qβ‖ > 0 hasHausdorff dimension 0.
Dmitriy Bilyk Discrepancy and analysis
3. Peres-Schlag method: more connections
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
Littlewood conjecture: for any (α, β) ∈ R2,
infq≥1
q · ‖qα‖ · ‖qβ‖ = 0.
Moschevitin,... 2010-2012:e.g. Bugeaud–Moschevitin 2011:the set of points (α, β) such that inf q · log2 q‖qα‖ · ‖qβ‖ > 0has Hausdorff dimension 2.
Einsiedler, Katok, Lindenstrauss (2006):the set of points (α, β) such that inf q · ‖qα‖ · ‖qβ‖ > 0 hasHausdorff dimension 0.
Dmitriy Bilyk Discrepancy and analysis
3. Peres-Schlag method: more connections
Peres–Schlag 2010:Let nj ⊂ N be lacunary with
nj+1
nj> 1 + ε.
Then there exists θ ∈ [0, 1] such that for all j ∈ N
‖njθ‖ & ε| log ε|−1.
Littlewood conjecture: for any (α, β) ∈ R2,
infq≥1
q · ‖qα‖ · ‖qβ‖ = 0.
Moschevitin,... 2010-2012:e.g. Bugeaud–Moschevitin 2011:the set of points (α, β) such that inf q · log2 q‖qα‖ · ‖qβ‖ > 0has Hausdorff dimension 2.
Einsiedler, Katok, Lindenstrauss (2006):the set of points (α, β) such that inf q · ‖qα‖ · ‖qβ‖ > 0 hasHausdorff dimension 0.
Dmitriy Bilyk Discrepancy and analysis
3. Peres-Schlag method
Theorem (DB, Ma, Pipher, Spencer)
Assume that for every Q ∈ N we have
1/δ(Q)∑q=Q
q · F (δ(q)) . 1. Then
there exists α such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & δ(q)
for all p ∈ Z, q ∈ N.
Ω lacunary of order M: N(x) ≈ logM 1x , F (x) ≈ x · logM 1
x ,
δ(q) ≈ 1
q2 logM+1 q
Dmitriy Bilyk Discrepancy and analysis
3. Peres-Schlag method
Theorem (DB, Ma, Pipher, Spencer)
Assume that for every Q ∈ N we have
1/δ(Q)∑q=Q
q · F (δ(q)) . 1. Then
there exists α such that for all θ ∈ Ω∣∣∣∣(α− θ)− p
q
∣∣∣∣ & δ(q)
for all p ∈ Z, q ∈ N.
Ω lacunary of order M: N(x) ≈ logM 1x , F (x) ≈ x · logM 1
x ,
δ(q) ≈ 1
q2 logM+1 q
Dmitriy Bilyk Discrepancy and analysis
A. Erdos-Turan inequality
For a sequence ω = (ωn) ⊂ [0, 1]:
DN(ω) = supx⊂[0,1]
∣∣∣∣#n ≤ N : ωn ≤ x − Nx
∣∣∣∣
Theorem (Erdos-Turan)
For any sequence ω ⊂ [0, 1] we have
DN(ω) .N
m+
m∑h=1
1
h
∣∣∣∣∣N∑
n=1
e2πihωn
∣∣∣∣∣for all natural numbers m.
For the Kronecker sequence ω = nα, we have∣∣∣∣∣N∑
n=1
e2πihnα
∣∣∣∣∣ ≤ 1
| sinπhα|.
1
‖hα‖.
Dmitriy Bilyk Discrepancy and analysis
A. Erdos-Turan inequality
For a sequence ω = (ωn) ⊂ [0, 1]:
DN(ω) = supx⊂[0,1]
∣∣∣∣#n ≤ N : ωn ≤ x − Nx
∣∣∣∣Theorem (Erdos-Turan)
For any sequence ω ⊂ [0, 1] we have
DN(ω) .N
m+
m∑h=1
1
h
∣∣∣∣∣N∑
n=1
e2πihωn
∣∣∣∣∣for all natural numbers m.
For the Kronecker sequence ω = nα, we have∣∣∣∣∣N∑
n=1
e2πihnα
∣∣∣∣∣ ≤ 1
| sinπhα|.
1
‖hα‖.
Dmitriy Bilyk Discrepancy and analysis
A. Erdos-Turan inequality
For a sequence ω = (ωn) ⊂ [0, 1]:
DN(ω) = supx⊂[0,1]
∣∣∣∣#n ≤ N : ωn ≤ x − Nx
∣∣∣∣Theorem (Erdos-Turan)
For any sequence ω ⊂ [0, 1] we have
DN(ω) .N
m+
m∑h=1
1
h
∣∣∣∣∣N∑
n=1
e2πihωn
∣∣∣∣∣for all natural numbers m.
For the Kronecker sequence ω = nα, we have∣∣∣∣∣N∑
n=1
e2πihnα
∣∣∣∣∣ ≤ 1
| sinπhα|.
1
‖hα‖.
Dmitriy Bilyk Discrepancy and analysis
A. Erdos-Turan inequality
Lemma
For the sequence ω = nα we have
DN(ω) .N
m+
m∑h=1
1
h · ‖hα‖
for all natural numbers m.
Assume that α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q)
for each p ∈ Z, q ∈ N.
We have
m∑h=1
1
h‖hα‖. ψ(2m) log m +
m∑h=1
ψ(2h) log h
h
m∑h=1
1
h‖hα‖. log2 m + ψ(m) +
m∑h=1
ψ(h)
h
Dmitriy Bilyk Discrepancy and analysis
A. Erdos-Turan inequality
Lemma
For the sequence ω = nα we have
DN(ω) .N
m+
m∑h=1
1
h · ‖hα‖
for all natural numbers m.
Assume that α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q)
for each p ∈ Z, q ∈ N.
We have
m∑h=1
1
h‖hα‖. ψ(2m) log m +
m∑h=1
ψ(2h) log h
h
m∑h=1
1
h‖hα‖. log2 m + ψ(m) +
m∑h=1
ψ(h)
h
Dmitriy Bilyk Discrepancy and analysis
A. Erdos-Turan inequality
Lemma
For the sequence ω = nα we have
DN(ω) .N
m+
m∑h=1
1
h · ‖hα‖
for all natural numbers m.
Assume that α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q)
for each p ∈ Z, q ∈ N.
We have
m∑h=1
1
h‖hα‖. ψ(2m) log m +
m∑h=1
ψ(2h) log h
h
m∑h=1
1
h‖hα‖. log2 m + ψ(m) +
m∑h=1
ψ(h)
h
Dmitriy Bilyk Discrepancy and analysis
B. Partial quotients
For the Kronecker sequence ω = nα
DN(ω) .m+1∑j=1
aj ,
where α = a0 + 1a1+ 1
a2+ 1...
= [a0; a1, a2, a3, ...],
pnqn
= [a0; a1, a2, ..., an], and qm ≤ N < qm + 1.
If α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q) , then an+1 ≤ ψ(qn).
Ω “superlacunary”, e.g. Ω =
2−2m
:
F−1(F−1(1/q2)) ≈ 1q2(log log q)2 , i.e. ψ(q) = (log log q)2
One can deduce that
DN(ω) . log N · (log log N)2,
while Erdos-Turan would only yield log2 N.
Dmitriy Bilyk Discrepancy and analysis
B. Partial quotients
For the Kronecker sequence ω = nα
DN(ω) .m+1∑j=1
aj ,
where α = a0 + 1a1+ 1
a2+ 1...
= [a0; a1, a2, a3, ...],
pnqn
= [a0; a1, a2, ..., an], and qm ≤ N < qm + 1.
If α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q) , then an+1 ≤ ψ(qn).
Ω “superlacunary”, e.g. Ω =
2−2m
:
F−1(F−1(1/q2)) ≈ 1q2(log log q)2 , i.e. ψ(q) = (log log q)2
One can deduce that
DN(ω) . log N · (log log N)2,
while Erdos-Turan would only yield log2 N.
Dmitriy Bilyk Discrepancy and analysis
B. Partial quotients
For the Kronecker sequence ω = nα
DN(ω) .m+1∑j=1
aj ,
where α = a0 + 1a1+ 1
a2+ 1...
= [a0; a1, a2, a3, ...],
pnqn
= [a0; a1, a2, ..., an], and qm ≤ N < qm + 1.
If α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q) , then an+1 ≤ ψ(qn).
Ω “superlacunary”, e.g. Ω =
2−2m
:
F−1(F−1(1/q2)) ≈ 1q2(log log q)2 , i.e. ψ(q) = (log log q)2
One can deduce that
DN(ω) . log N · (log log N)2,
while Erdos-Turan would only yield log2 N.
Dmitriy Bilyk Discrepancy and analysis
B. Partial quotients
For the Kronecker sequence ω = nα
DN(ω) .m+1∑j=1
aj ,
where α = a0 + 1a1+ 1
a2+ 1...
= [a0; a1, a2, a3, ...],
pnqn
= [a0; a1, a2, ..., an], and qm ≤ N < qm + 1.
If α satisfies∣∣∣α− p
q
∣∣∣ > 1q2·ψ(q) , then an+1 ≤ ψ(qn).
Ω “superlacunary”, e.g. Ω =
2−2m
:
F−1(F−1(1/q2)) ≈ 1q2(log log q)2 , i.e. ψ(q) = (log log q)2
One can deduce that
DN(ω) . log N · (log log N)2,
while Erdos-Turan would only yield log2 N.
Dmitriy Bilyk Discrepancy and analysis
Directional Discrepancy: Setting
PN – a set of N points in [0, 1]d
R ⊂ [0, 1]d – a measurable set.
D(PN ,R) = ]PN ∩ R − N · vol (R)
Let Ω ⊂ [0, π/2)
AΩ = rectangles pointing in directions of Ω
DΩ(N) = infPN
supR∈AΩ
|D(PN ,R)|
Dmitriy Bilyk Discrepancy and analysis
All rotations
Ω = [0, π/2)
Theorem (J. Beck, 1987-88)
N1/4 . DΩ(N) . N1/4√
log N
Dmitriy Bilyk Discrepancy and analysis
No rotations: axis-parallel rectangles
Ω = 0
Theorem (Lerch, 1904; Schmidt, 1972)
DΩ(N) ≈ log N
Theorem (L2: Roth, 1954; Davenport, 1956)
D(2)Ω (N) ≈
√log N
Ω finite −→ same (Beck-Chen, Chen-Travaglini)
Dmitriy Bilyk Discrepancy and analysis
No rotations: axis-parallel rectangles
Ω = 0
Theorem (Lerch, 1904; Schmidt, 1972)
DΩ(N) ≈ log N
Theorem (L2: Roth, 1954; Davenport, 1956)
D(2)Ω (N) ≈
√log N
Ω finite −→ same (Beck-Chen, Chen-Travaglini)
Dmitriy Bilyk Discrepancy and analysis
Irrational Lattice
Example
Let α ∈ [0, π]. Denote by PαN the lattice 1√NZ2 rotated by α and
intersected with [0, 1]2.
Theorem
If tanα is badly approximable, i.e.∣∣∣∣tan(α)− p
q
∣∣∣∣ & 1
q2
for all p ∈ Z, q ∈ N, then the discrepancy of PαN with respect toaxis-parallel rectangles satisfies
D(PαN) ≈ log N.
Dmitriy Bilyk Discrepancy and analysis
Directional Discrepancy
DΩ(PαN ) . supθ∈Ω
Dc√N (n · tan(α− θ))
Theorem (DB, Ma, Pipher, Spencer)
Lacunary directions: DΩ(N) . log3 N
Lacunary of order M: DΩ(N) . logM+2 N
“Superlacunary” sequence
DΩ(N) . log N · (log log N)2
Ω has upper Minkowski dimension 0 ≤ d < 1
DΩ(N) . Nd
d+1 +ε.
Dmitriy Bilyk Discrepancy and analysis
Directional Discrepancy
DΩ(PαN ) . supθ∈Ω
Dc√N (n · tan(α− θ))
Theorem (DB, Ma, Pipher, Spencer)
Lacunary directions: DΩ(N) . log3 N
Lacunary of order M: DΩ(N) . logM+2 N
“Superlacunary” sequence
DΩ(N) . log N · (log log N)2
Ω has upper Minkowski dimension 0 ≤ d < 1
DΩ(N) . Nd
d+1 +ε.
Dmitriy Bilyk Discrepancy and analysis
Critical dimension
If Ω has upper Minkowski dimension 0 ≤ d ≤ 1:
DΩ(N) . Nd
d+1+ε. (∗)
but for Ω = [0, 2π), i.e. all rotations,
DΩ(N) . N1/4√
log N
So (∗) is interesting only for dd+1 <
14 , i.e.
d <1
3.
Dmitriy Bilyk Discrepancy and analysis
Critical dimension
If Ω has upper Minkowski dimension 0 ≤ d ≤ 1:
DΩ(N) . Nd
d+1+ε. (∗)
but for Ω = [0, 2π), i.e. all rotations,
DΩ(N) . N1/4√
log N
So (∗) is interesting only for dd+1 <
14 , i.e.
d <1
3.
Dmitriy Bilyk Discrepancy and analysis
Critical dimension
If Ω has upper Minkowski dimension 0 ≤ d ≤ 1:
DΩ(N) . Nd
d+1+ε. (∗)
but for Ω = [0, 2π), i.e. all rotations,
DΩ(N) . N1/4√
log N
So (∗) is interesting only for dd+1 <
14 , i.e.
d <1
3.
Dmitriy Bilyk Discrepancy and analysis