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Improving estimates for discrete polynomial averaging operators
Vjekoslav Kovač (U. of Zagreb)
Joint work with
R. Han, M. T. Lacey, F. Yang (Georgia Tech), J. Madrid (UCLA)
Supported by HRZZ UIP-2017-05-4129 (MUNHANAP)
Zagreb Workshop on Operator Theory, Zagreb, June 29–30, 2020
1/12
What are averaging operators?
Finite averages
ANf := 1
N
∑︀N
k=1
Tkf; N ∈ N
Assumptions: X normed space, Tk : X → X isometries
Obvious property: AN X→X ≤ 1
More general assumptions: X, Y normed spaces, Tk : X → Y, Tk X→Y = 1
Again: AN X→Y ≤ 1
Can the constant 1 be improved? Sometimes
2/12
Norm-improving property
Finite averages
ANf := 1
N
∑︀N
k=1
Tkf; N ∈ N
Occasional additional property
AN X→Y ≤ c(N) < 1, c(N) → 0 as N → ∞
Appreciated in harmonic analysis, ergodic theory, etc.
We might want to find the optimal asymptotics of c(N)
All depends on the choice of Tk
3/12
Example: Translations on Z
1 ≤ p ≤ q ≤ ∞, ak ∈ Z
Tk : ℓp(Z) → ℓq(Z), (Tkf)(m) := f(m + ak)
Finite averages on Z
(ANf)(m) = 1
N
∑︀N
k=1
f(m + ak); N ∈ N, m ∈ Z
Norm-improving property now depends on the choice of numbers ak
Several features comes into play, coming from the Fourier analysis, number theory, etc.
4/12
Discrete polynomial averages
P: Z → Z polynomial with integer coefficients of degree d ≥ 2
Polynomial averages on Z
(ANf)(m) = 1
N
∑︀N
k=1
f(m + P(k)); N ∈ N, m ∈ Z
Theorem (R. Han, V. K., M. T. Lacey, J. Madrid, F. Yang (2019))
ANf ℓq(Z) ≤ C(P, p, q) N
−d( 1
p
− 1
q
)
f ℓp(Z)
in a certain range of exponents 1 ≤ p ≤ q ≤ ∞
Previously only the case P(x) = x2 was known — R. Han, M. T. Lacey, F. Yang (2019)
5/12
Trying out easy examples
Theorem
(ANf)(m) = 1
N
∑︀N
k=1
f(m + P(k))
ANf ℓq(Z) ≤ C(P, p, q) N
−d( 1
p
− 1
q
)
f ℓp(Z)
Take f = 1{1,2,3,...,2P(N)} =⇒ optimality of the decay N
−d( 1
p
− 1
q
)
Take f = 1{P(1),P(2),...,P(N)} =⇒ necessary condition d
q
≥ d−1
p
Take f = 1{0} =⇒ necessary condition d−1
q
≥ d
p
− 1
6/12
Range of exponents
Case d = 2
The theorem holds for
{︀
(p, q) : 1
q
≤ 1
p
,
2
q
>
1
p
,
1
q
>
2
p
− 1
}︀
When q = p the range specializes to 3
2
< p ≤ 2
The range in optimal modulo its boundary
Case d ≥ 3
The theorem holds for
{︀
(p, q) : 1
q
≤ 1
p
,
d2+d+1
q
>
d2+d−1
p
,
d2+d−1
q
>
d2+d+1
p
− 2
}︀
When q = p the range specializes to 2 − 2
d2+d+1
< p ≤ 2
No estimates outside the range
{︀
(p, q) : 1
q
≤ 1
p
,
d
q
≥ d−1
p
,
d−1
q
≥ d
p
− 1
}︀
When q = p this specializes to 2 − 1
d
≤ p ≤ 2
7/12
Range of exponents
(ANf)(m) = 1
N
∑︀N
k=1
f(m + P(k))
ANf ℓq(Z) ≤ C(P, p, q) N
−d( 1
p
− 1
q
)
f ℓp(Z)
Best explained visually
8/12
Sketch of the proof in the case d ≥ 3
Regard AN as “projections” of
Higher-dimensional “universal” polynomial averages
(̃︀ANf)(m1, m2, . . . , md) := 1
N
∑︀N
k=1
f(m1 + k, m2 + k2, . . . , md + kd);
N ∈ N, (m1, m2, . . . , md) ∈ Zd
We turn to proving
Theorem
̃︀ANf ℓq(Zd) d,p,q N
− d(d+1)
2
( 1
p
− 1
q
)
f ℓp(Zd)
9/12
Sketch of the proof in the case d ≥ 3
Write (̃︀ANf)(m1, m2, . . . , md) as
1
N
N∑︁
k=1
∫︁
Td
̂︀f(t1, t2, . . . , td)e2πi((m1+k)t1+(m2+k2)t2+···+(md+kd)td)
dt1dt2 · · · dtd
=
∫︁
Td
̂︀f(t1, t2, . . . , td)SN(t1, t2, . . . , td)e2πi(m1t1+m2t2+···+mdtd)
dt1dt2 · · · dtd,
where
SN(t1, t2, . . . , td) := 1
N
∑︀N
k=1
e2πi(kt1+k2t2+···+kdtd); (t1, t2, . . . , td) ∈ Td = (R/Z)d are
the normalized exponential sums
10/12
Sketch of the proof in the case d ≥ 3
Take q = p
Apply the Hausdorff–Young inequality twice to ̃︀ANf = −1
(︀
( f)SN
)︀
:
̃︀ANf ℓp (Zd) ≤ ̂︀fSN Lp(Td) ≤ ̂︀f Lp (Td) SN Ls(Td) ≤ f ℓp(Zd) SN Ls(Td),
where 1
s
= 1
p
− 1
p
= 2
p
− 1
Theorem (Vinogradov’s mean value conj. – J. Bourgain, C. Demeter, L. Guth (2016))
SN Ls(Td) ≤ C(d, s)N− d(d+1)
2s for d ≥ 3 and s > d(d + 1)
11/12
Thank you for your attention!
12/12

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Improving estimates for discrete polynomial averaging operators

  • 1. Improving estimates for discrete polynomial averaging operators Vjekoslav Kovač (U. of Zagreb) Joint work with R. Han, M. T. Lacey, F. Yang (Georgia Tech), J. Madrid (UCLA) Supported by HRZZ UIP-2017-05-4129 (MUNHANAP) Zagreb Workshop on Operator Theory, Zagreb, June 29–30, 2020 1/12
  • 2. What are averaging operators? Finite averages ANf := 1 N ∑︀N k=1 Tkf; N ∈ N Assumptions: X normed space, Tk : X → X isometries Obvious property: AN X→X ≤ 1 More general assumptions: X, Y normed spaces, Tk : X → Y, Tk X→Y = 1 Again: AN X→Y ≤ 1 Can the constant 1 be improved? Sometimes 2/12
  • 3. Norm-improving property Finite averages ANf := 1 N ∑︀N k=1 Tkf; N ∈ N Occasional additional property AN X→Y ≤ c(N) < 1, c(N) → 0 as N → ∞ Appreciated in harmonic analysis, ergodic theory, etc. We might want to find the optimal asymptotics of c(N) All depends on the choice of Tk 3/12
  • 4. Example: Translations on Z 1 ≤ p ≤ q ≤ ∞, ak ∈ Z Tk : ℓp(Z) → ℓq(Z), (Tkf)(m) := f(m + ak) Finite averages on Z (ANf)(m) = 1 N ∑︀N k=1 f(m + ak); N ∈ N, m ∈ Z Norm-improving property now depends on the choice of numbers ak Several features comes into play, coming from the Fourier analysis, number theory, etc. 4/12
  • 5. Discrete polynomial averages P: Z → Z polynomial with integer coefficients of degree d ≥ 2 Polynomial averages on Z (ANf)(m) = 1 N ∑︀N k=1 f(m + P(k)); N ∈ N, m ∈ Z Theorem (R. Han, V. K., M. T. Lacey, J. Madrid, F. Yang (2019)) ANf ℓq(Z) ≤ C(P, p, q) N −d( 1 p − 1 q ) f ℓp(Z) in a certain range of exponents 1 ≤ p ≤ q ≤ ∞ Previously only the case P(x) = x2 was known — R. Han, M. T. Lacey, F. Yang (2019) 5/12
  • 6. Trying out easy examples Theorem (ANf)(m) = 1 N ∑︀N k=1 f(m + P(k)) ANf ℓq(Z) ≤ C(P, p, q) N −d( 1 p − 1 q ) f ℓp(Z) Take f = 1{1,2,3,...,2P(N)} =⇒ optimality of the decay N −d( 1 p − 1 q ) Take f = 1{P(1),P(2),...,P(N)} =⇒ necessary condition d q ≥ d−1 p Take f = 1{0} =⇒ necessary condition d−1 q ≥ d p − 1 6/12
  • 7. Range of exponents Case d = 2 The theorem holds for {︀ (p, q) : 1 q ≤ 1 p , 2 q > 1 p , 1 q > 2 p − 1 }︀ When q = p the range specializes to 3 2 < p ≤ 2 The range in optimal modulo its boundary Case d ≥ 3 The theorem holds for {︀ (p, q) : 1 q ≤ 1 p , d2+d+1 q > d2+d−1 p , d2+d−1 q > d2+d+1 p − 2 }︀ When q = p the range specializes to 2 − 2 d2+d+1 < p ≤ 2 No estimates outside the range {︀ (p, q) : 1 q ≤ 1 p , d q ≥ d−1 p , d−1 q ≥ d p − 1 }︀ When q = p this specializes to 2 − 1 d ≤ p ≤ 2 7/12
  • 8. Range of exponents (ANf)(m) = 1 N ∑︀N k=1 f(m + P(k)) ANf ℓq(Z) ≤ C(P, p, q) N −d( 1 p − 1 q ) f ℓp(Z) Best explained visually 8/12
  • 9. Sketch of the proof in the case d ≥ 3 Regard AN as “projections” of Higher-dimensional “universal” polynomial averages (̃︀ANf)(m1, m2, . . . , md) := 1 N ∑︀N k=1 f(m1 + k, m2 + k2, . . . , md + kd); N ∈ N, (m1, m2, . . . , md) ∈ Zd We turn to proving Theorem ̃︀ANf ℓq(Zd) d,p,q N − d(d+1) 2 ( 1 p − 1 q ) f ℓp(Zd) 9/12
  • 10. Sketch of the proof in the case d ≥ 3 Write (̃︀ANf)(m1, m2, . . . , md) as 1 N N∑︁ k=1 ∫︁ Td ̂︀f(t1, t2, . . . , td)e2πi((m1+k)t1+(m2+k2)t2+···+(md+kd)td) dt1dt2 · · · dtd = ∫︁ Td ̂︀f(t1, t2, . . . , td)SN(t1, t2, . . . , td)e2πi(m1t1+m2t2+···+mdtd) dt1dt2 · · · dtd, where SN(t1, t2, . . . , td) := 1 N ∑︀N k=1 e2πi(kt1+k2t2+···+kdtd); (t1, t2, . . . , td) ∈ Td = (R/Z)d are the normalized exponential sums 10/12
  • 11. Sketch of the proof in the case d ≥ 3 Take q = p Apply the Hausdorff–Young inequality twice to ̃︀ANf = −1 (︀ ( f)SN )︀ : ̃︀ANf ℓp (Zd) ≤ ̂︀fSN Lp(Td) ≤ ̂︀f Lp (Td) SN Ls(Td) ≤ f ℓp(Zd) SN Ls(Td), where 1 s = 1 p − 1 p = 2 p − 1 Theorem (Vinogradov’s mean value conj. – J. Bourgain, C. Demeter, L. Guth (2016)) SN Ls(Td) ≤ C(d, s)N− d(d+1) 2s for d ≥ 3 and s > d(d + 1) 11/12
  • 12. Thank you for your attention! 12/12