A Szemeredi-type theorem for subsets of the unit cube

A Szemerédi-type theorem for subsets of the unit cube
Vjekoslav Kovač (University of Zagreb)
Joint work with Polona Durcik (Caltech)
Supported by HRZZ UIP-2017-05-4129 (MUNHANAP)
Mathematical Institute, University of Bonn, July 3, 2020
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Ramsey theory
Search for patterns in large arbitrary structures
discrete vs. Euclidean
Ramsey (1920s) Erdős et al. (1973)
coloring theorems vs. density theorems
e.g., Van der Waerden’s theorem (1927) e.g., Szemerédi’s theorem (1975)
non-arithmetic patterns vs. arithmetic patterns
e.g., spherical conﬁgurations e.g., arithmetic progressions
This combination of choices: Cook, Magyar, and Pramanik (2015)
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Study of “large” measurable sets
We study large measurable sets A:
• for A ⊆ Rd this means
δ(A) := lim sp
R→∞
sp
x∈Rd
|A ∩ (x + [0, R]d)|
Rd
> 0
(the upper Banach density)
• for A ⊆ [0, 1]d this means
|A| > 0
(the Lebesgue meausure)
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An exemplary Euclidean density theorem
Δ = vertices of a non-degenerate n-dimensional simplex
Theorem (Bourgain (1986))
• Take A ⊆ Rn+1 measurable, δ(A) > 0. There exists λ0 = λ0(Δ, A) ∈ (0, ∞) such
that for every λ ≥ λ0 the set A contains an isometric copy of λΔ.
• Take δ ∈ (0, 1/2], A ⊆ [0, 1]n+1 measurable, |A| ≥ δ. Then
{λ ∈ [0, ∞) : A contains an isometric copy of λΔ}
contains an interval of length at least
(︀
exp(δ−C(Δ))
)︀−1
.
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Szemerédi’s theorem on Z
A question posed by Erdős and Turán (1936)
For a positive integer n ≥ 3 and a number 0 < δ ≤ 1/2 is there a positive integer N such
that each set S ⊆ {0, 1, 2, . . . , N − 1} with at least δN elements must contain a nontrivial
arithmetic progression of length n?
• n = 3 Roth (1953)
• n = 4 Szemerédi (1969)
• n ≥ 5 Szemerédi (1975)
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Szemerédi’s theorem on Z
Let N(n, δ) be the smallest such N
What are the best known bounds for N(n, δ)?
• N(3, δ) ≤ exp(δ−C) Heath-Brown (1987)
• N(4, δ) ≤ exp(δ−C) Green and Tao (2017)
• N(n, δ) ≤ exp(exp(δ−C(n))) Gowers (2001)
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Szemerédi’s theorem in [0, 1]d
Reformulation of Szemerédi’s theorem with the above bounds
For n ≥ 3 and d ≥ 1 there exists a constant C(n, d) such that for 0 < δ ≤ 1/2 and a
measurable set A ⊆ [0, 1]d with |A| ≥ δ one has
∫︁
[0,1]d
∫︁
[0,1]d
n−1∏︁
i=0
1A(x + iy) dy dx
≥
⎧
⎨
⎩
(︀
exp(δ−C(n,d))
)︀−1
when 3 ≤ n ≤ 4,
(︀
exp(exp(δ−C(n,d)))
)︀−1
when n ≥ 5.
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Gaps of progressions in A ⊆ [0, 1]d
What can be said about the following set?
gapsn(A) :=
{︀
y ∈ [−1, 1]d
: (∃x ∈ [0, 1]d
)
(︀
x, x + y, . . . , x + (n − 1)y ∈ A
)︀}︀
It contains a ball B(0, ϵ) for some ϵ > 0
A proof by Stromberg (1972):
• Assume that A is compact
• Find an open set U such that U ⊇ A and |U| ≤ (1 + 1/2n)|A|
• Take ϵ := dist(A, Rd U)/n
• For any y ℓ2 < ϵ take x ∈ A ∩ (A − y) ∩ · · · ∩ (A − (n − 1)y) =
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Number ϵ has to depend on “geometry” of A and not just on |A|
There is an obstruction already when we ask for much less
ℓ2
-gapsn(A) :=
{︀
λ ∈ [0, ∞) : (∃x, y)
(︀
x, x + y, . . . , x + (n − 1)y ∈ A, y ℓ2 = λ
)︀}︀
Does ℓ2-gapsn(A) contain an interval of length depending only on n, d, and |A|?
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No! Bourgain (1986):
A :=
{︀
x ∈ [0, 1]d
: (∃m ∈ Z)
(︀
m − 1/10 < ϵ−1
x 2
ℓ2 < m + 1/10
)︀}︀
.
• The parallelogram law:
x 2
ℓ2 − 2 x + y 2
ℓ2 + x + 2y 2
ℓ2 = 2 y 2
ℓ2 ,
• x, x + y, x + 2y ∈ A =⇒ m − 2/5 < 2 ϵ−1y 2
ℓ2 < m + 2/5
• |A| 1 uniformly as ϵ → 0+
One can switch attention to other patterns (spherical conﬁgurations) or ...
10/28

Gaps of 3-term progressions in other ℓp
norms
For p ∈ [1, ∞] deﬁne:
ℓp
-gapsn(A) :=
{︀
λ ∈ [0, ∞) : (∃x, y)
(︀
x, x + y, . . . , x + (n − 1)y ∈ A, y ℓp = λ
)︀}︀
Theorem (Cook, Magyar, and Pramanik (2015))
Take n = 3, p = 1, 2, ∞, d ≥ D(p), δ ∈ (0, 1/2], A ⊆ [0, 1]d measurable, |A| ≥ δ.
Then ℓp-gaps3(A) contains an interval of length depending only on p, d, and δ.
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Gaps of longer progressions in other ℓp
norms
Theorem (Durcik and K. (2020))
Take n ≥ 3, p = 1, 2, . . . , n − 1, ∞, d ≥ D(n, p), δ ∈ (0, 1/2], A ⊆ [0, 1]d measurable,
|A| ≥ δ. Then ℓp-gapsn(A) contains an interval of length at least
⎧
⎨
⎩
(︀
exp(exp(δ−C(n,p,d)))
)︀−1
when 3 ≤ n ≤ 4,
(︀
exp(exp(exp(δ−C(n,p,d))))
)︀−1
when n ≥ 5.
Modifying Bourgain’s example → sharp regarding the values of p
One can take D(n, p) = 2n+3(n + p) → certainly not sharp
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Quantities that detect progressions
σ(x) = δ( x p
ℓp − 1) = a measure supported on the unit sphere in the ℓp-norm
0
λ
(A) :=
∫︁
Rd
∫︁
Rd
n−1∏︁
i=0
1A(x + iy) dσλ(y) dx
0
λ
(A) > 0 =⇒ (∃x, y)
(︀
x, x + y, . . . , x + (n − 1)y ∈ A, y ℓp = λ
)︀
ϵ
λ
(A) :=
∫︁
Rd
∫︁
Rd
n−1∏︁
i=0
1A(x + iy)(σλ ∗ φϵλ)(y) dy dx
for a smooth φ ≥ 0 with
∫︀
Rd φ = 1
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Quantities that detect progressions
“The largeness/smoothness multiscale approach”
A variant of the approach introduced by Cook, Magyar, and Pramanik (2015)
ϵ
λ
(A) :=
∫︁
Rd
∫︁
Rd
n−1∏︁
i=0
1A(x + iy)(σλ ∗ φϵλ)(y) dy dx
λ ∈ (0, ∞), scale of largeness → detect APs with y ℓp = λ
ϵ ∈ [0, 1], scale of smoothness → the picture is blurred up to scale ϵ
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Proof strategy
0
λ
(A) = 1
λ
(A) +
(︀ ϵ
λ
(A) − 1
λ
(A)
)︀
+
(︀ 0
λ
(A) − ϵ
λ
(A)
)︀
• 1
λ
(A) = the structured part (“the main term”)
Controlled uniformly in λ using Szemerédi’s theorem (with the best known bounds)
as a black box
• ϵ
λ
(A) − 1
λ
(A) = the error part
Certain pigeonholing in λ is needed
Leads to some multilinear singular integrals
• 0
λ
(A) − ϵ
λ
(A) = the uniform part
Controlled uniformly in λ using Gowers uniformity norms
Leads to some oscillatory integrals
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Three propositions
Proposition handling the structured part
If λ ∈ (0, 1], δ ∈ (0, 1/2], A ⊆ [0, 1]d, |A| ≥ δ, then
1
λ
(A) ≥
⎧
⎨
⎩
(︀
exp(δ−E)
)︀−1
when 3 ≤ n ≤ 4,
(︀
exp(exp(δ−E))
)︀−1
when n ≥ 5.
This is essentially the analytical reformulation of Szemerédi’s theorem
16/28

Three propositions
Proposition handling the error part
IF J ∈ N, λj ∈ (2−j, 2−j+1] for j = 1, 2, . . . , J, ϵ ∈ (0, 1/2], A ⊆ [0, 1]d measurable, then
J∑︁
j=1
⃒
⃒ ϵ
λj
(A) − 1
λj
(A)
⃒
⃒ ≤ ϵ−F
J1−2−n+2
.
Note the gain of J−2−n+2
over the trivial estimate
17/28

Three propositions
Proposition handling the uniform part
If d ≥ D(n, p), λ, ϵ ∈ (0, 1], A ⊆ [0, 1]d measurable, then
⃒
⃒ 0
λ
(A) − ϵ
λ
(A)
⃒
⃒ ≤ Gϵ1/3
.
Consequently, the uniform part can be made arbitrarily small by choosing a suﬃciently
small ϵ
Here is where the assumption p = 1, 2, . . . , n − 1, ∞ is needed
18/28

How to ﬁnish the proof?
Denote
ϑ :=
⎧
⎨
⎩
(︀
exp(δ−E)
)︀−1
when 3 ≤ n ≤ 4
(︀
exp(exp(δ−E))
)︀−1
when n ≥ 5
and choose
ϵ :=
(︁ ϑ
3G
)︁3
, J :=
⌊︀(︀
3ϑ−1
ϵ−F
)︀2n−2 ⌋︀
+ 1
Observe
2−J
≥
⎧
⎨
⎩
(︀
exp(exp(δ−C(n,p,d)))
)︀−1
when 3 ≤ n ≤ 4
(︀
exp(exp(exp(δ−C(n,p,d))))
)︀−1
when n ≥ 5
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How to ﬁnish the proof?
Take A ⊆ [0, 1]d such that |A| ≥ δ
By pigeonholing we ﬁnd j ∈ {1, 2, . . . , J} such that for every λ ∈ (2−j, 2−j+1] we have
⃒
⃒ ϵ
λ
(A) − 1
λ
(A)
⃒
⃒ ≤ ϵ−F
J−2−n+2
Now, I = (2−j, 2−j+1] is the desired interval!
Indeed, for λ ∈ I we can estimate
0
λ
(A) ≥ 1
λ
(A) −
⃒
⃒ ϵ
λ
(A) − 1
λ
(A)
⃒
⃒ −
⃒
⃒ 0
λ
(A) − ϵ
λ
(A)
⃒
⃒
≥ ϑ − ϵ−F
J−2−n+2
− Gϵ1/3
≥ ϑ − ϑ/3 − ϑ/3 > 0
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The error part
The most interesting part for us is the error part
We need to estimate
J∑︁
j=1
κj
(︀ ϵ
λj
(A) − 1
λj
(A)
)︀
for arbitrary scales λj ∈ (2−j, 2−j+1] and arbitrary complex signs κj, with a bound that is
sub-linear in J
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The error part
It can be expanded as
∫︁
Rd
∫︁
Rd
n−1∏︁
i=0
1A(x + iy)K(y) dy dx,
where
K(y) :=
J∑︁
j=1
κj
(︀
(σλj ∗ φϵλj )(y) − (σλj ∗ φλj )(y)
)︀
is a translation-invariant Calderón–Zygmund kernel
22/28

The error part
If d = 1 and K(y) is a truncation of 1/y, then this becomes the (dualized and truncated)
multilinear Hilbert transform,
∫︁
R
∫︁
[−R,−r]∪[r,R]
n−1∏︁
i=0
fi(x + iy)
dy
y
dx
• When n ≥ 4, no Lp-bounds uniform in r, R are known
• Tao (2016) showed a bound of the form o(J), where J = log(R/r)
• Durcik, K., and Thiele (2016) showed a bound of the form O(J1−ϵ
)
23/28

The error part — the actual proof (g = st. Gauss., h(l)
= ∂lg)
̃︀Λα,αk,...,αn−1
k,l,a,b
:=
∫︁ b
a
∫︁
(Rd)2(n−k−1)
⃒
⃒
⃒
∫︁
(Rd)3
k h(l)
tαk
(y − pk)h(l)
tα
(pk + · · · + pn−1) dpk dx dy
⃒
⃒
⃒
×
(︁ n−1∏︁
j=k+1
gtαj (pj)gtαj (uj − pj)
)︁
dpk+1 · · · dpn−1 duk+1 · · · dun−1
dt
t
Λα,αk,...,αn−1
k,l,a,b
:=
∫︁ b
a
∫︁
(Rd)2(n−k)
⃒
⃒
⃒
∫︁
Rd
k h(l)
tαk
(y − pk) dy
⃒
⃒
⃒gtα(pk + · · · + pn−1)
×
(︁ n−1∏︁
j=k+1
gtαj (pj)gtαj (uj − pj)
)︁
dpk · · · dpn−1 duk+1 · · · dun−1 dx
dt
t
k :=
k∏︁
i=0
∏︁
(rk+1,...,rn−1)∈{0,1}n−k−1
1A
(︁
x + iy +
n−1∑︁
s=k+1
rs(i + s − k)us
)︁
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The error part — the actual proof
By the mathematical induction on k = 1, 2, . . . , n − 1 we prove:
• ̃︀Λα,α1,...,αn−1
1,l,a,b
1 (k = 1)
• Λα,αk,...,αn−1
k,l,a,b
, ̃︀Λα,αk,...,αn−1
k,l,a,b
(ααk · · · αn−1)2
(︁
log b
a
)︁1−2−k+1
(k ≥ 2)
One can observe:
• Induction basis k = 1 is the entangled (twisted) paraproduct associated with a cube
(= entangled singular Brascamp–Lieb), telescoping + positivity needed only
• LHS of the desired estimate is bounded by a superposition of the forms Λ for
k = n − 1
• Induction step consists of telescoping + Cauchy–Schwarzing
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An open problem
Problem
Prove or disprove: if n ≥ 4, p = 1, 2, . . . , n − 1, d ≥ D (n, p), A ⊆ Rd measurable,
δ(A) > 0, then ∃λ0 = λ0(n, p, d, A) ∈ (0, ∞) such that for every λ ≥ λ0 one can ﬁnd
x, y ∈ Rd satisfying x, x + y, . . . , x + (n − 1)y ∈ A and y ℓp = λ.
Its diﬃculty ←→ lack of our understanding of boundedness
properties of multilinear Hilbert transforms
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Other patterns
Groups of allowed symmetries play a major role. Compare:
corner: (x, y), (x + s, y), (x, y + s)
Durcik, K., and Rimanić (2016)
isosceles right triangle: (x, y), (x + s, y), (x, y + t) with s ℓ2 = t ℓ2
essentially Bourgain (1986)
square (cube, box): (x, y), (x + s, y), (x, y + t), (x + s, y + t) with s ℓ2 = t ℓ2
Lyall and Magyar (2016, 2019), Durcik and K. (2018, 2020)
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Thank you for your attention!
28/28

A Szemeredi-type theorem for subsets of the unit cube

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