PaataIvanisvili KentStateUniversity 2017 …pivanisv/bellman.pdfPaata Ivanisvili Kent State...

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Sharp estimates in harmonic analysis Paata Ivanisvili Kent State University 2017 University of California, Irvine Paata Ivanisvili Kent State University Sharp estimates

Transcript of PaataIvanisvili KentStateUniversity 2017 …pivanisv/bellman.pdfPaata Ivanisvili Kent State...

  • Sharp estimates in harmonic analysis

    Paata Ivanisvili

    Kent State University

    2017

    University of California, Irvine

    Paata Ivanisvili Kent State University Sharp estimates

  • Bellman function for extremal problems in BMO.(joint with N. Osipov, D. Stolyarov, P. Zatitskiy, V. Vasyunin)

    Paata Ivanisvili Kent State University Sharp estimates

  • Surface of minimal area k1 + k2 = 0.

    Paata Ivanisvili Kent State University Sharp estimates

  • Minimal concave function k1k2 = 0

    Paata Ivanisvili Kent State University Sharp estimates

  • Find the minimal concave function

    B(x , y) is minimal concave function on Ω with B|∂Ω = f .

    ProblemGiven a wire find B .

    Paata Ivanisvili Kent State University Sharp estimates

  • Projection

    Paata Ivanisvili Kent State University Sharp estimates

  • Foliation

    This we call foliation.

    Problem (equivalent)

    Given a wire find foliation.

    Paata Ivanisvili Kent State University Sharp estimates

  • Foliation

    This we call foliation.

    Problem (equivalent)

    Given a wire find foliation.

    Paata Ivanisvili Kent State University Sharp estimates

  • Concave envelope

    Paata Ivanisvili Kent State University Sharp estimates

  • Foliation

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle and Cup

    Demonstrate

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle and Cup

    DemonstratePaata Ivanisvili Kent State University Sharp estimates

  • Boundary value problem: homogeneous Monge–Ampère

    B(x , y) is minimal concave function on Ω with B|∂Ω = f .

    What if B(x , y) is minimal concave function on Ω \ D?

    Paata Ivanisvili Kent State University Sharp estimates

  • Boundary value problem: homogeneous Monge–Ampère

    B(x , y) is minimal concave function on Ω with B|∂Ω = f .What if B(x , y) is minimal concave function on Ω \ D?

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole

    Will not be minimal!

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole

    Will not be minimal!

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole: Ω \ D

    HessB ≤ 0 on Ω \ DB|∂Ω = f .B is minimal among all such solutions.det(Hess B) = 0 on Ω \ D;

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole: Ω \ D

    HessB ≤ 0 on Ω \ D

    B|∂Ω = f .B is minimal among all such solutions.det(Hess B) = 0 on Ω \ D;

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole: Ω \ D

    HessB ≤ 0 on Ω \ DB|∂Ω = f .

    B is minimal among all such solutions.det(Hess B) = 0 on Ω \ D;

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole: Ω \ D

    HessB ≤ 0 on Ω \ DB|∂Ω = f .B is minimal among all such solutions.

    det(Hess B) = 0 on Ω \ D;

    Paata Ivanisvili Kent State University Sharp estimates

  • Erasing a hole: Ω \ D

    HessB ≤ 0 on Ω \ DB|∂Ω = f .B is minimal among all such solutions.det(Hess B) = 0 on Ω \ D;

    Paata Ivanisvili Kent State University Sharp estimates

  • Evolution of the surface

    Paata Ivanisvili Kent State University Sharp estimates

  • Annulus and parabolic strips

    Ω \D

    Paata Ivanisvili Kent State University Sharp estimates

  • Left tangent lines 2D

    x1

    x2

    x2 = x21

    x2 = x21 + ε

    2

    Paata Ivanisvili Kent State University Sharp estimates

  • Left tangent lines 3D

    Paata Ivanisvili Kent State University Sharp estimates

  • Right tangent lines 2D

    x1

    x2

    x2 = x21

    x2 = x21 + ε

    2

    Paata Ivanisvili Kent State University Sharp estimates

  • Right tangent lines 3D

    Paata Ivanisvili Kent State University Sharp estimates

  • Cup

    x1

    x2

    x2 = x21

    x2 = x21 + ε

    2

    A0

    B0

    U1 U2

    Ωcup

    ΩRΩL

    Paata Ivanisvili Kent State University Sharp estimates

  • Cup

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle

    x1

    x2

    V

    x2 = x21

    x2 = x21 + ε

    2

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle and cup

    x1

    x2

    U−

    U+

    A

    B

    V

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle and cup

    Paata Ivanisvili Kent State University Sharp estimates

  • Angle and cup: evolution

    Paata Ivanisvili Kent State University Sharp estimates

  • Two cups

    Paata Ivanisvili Kent State University Sharp estimates

  • Two cups

    Paata Ivanisvili Kent State University Sharp estimates

  • Two cups: evolution

    Paata Ivanisvili Kent State University Sharp estimates

  • John–Nirenberg inequality and BMO space

    supI⊆[0,1]

    1|I |

    ∣∣∣∣{t ∈ I : ∣∣∣∣ϕ(t)− 1|I |∫Iϕ

    ∣∣∣∣ ≥ λ}∣∣∣∣ ≤ c1e−c2λ/‖ϕ‖BMO2([0,1])

    where

    ‖ϕ‖2BMO2([0,1])def= sup

    I⊆[0,1]

    1|I |

    ∫I

    ∣∣∣∣ϕ− 1|I |∫Iϕ

    ∣∣∣∣2 .QuestionFind sharp constants c1, c2 > 0.

    Paata Ivanisvili Kent State University Sharp estimates

  • John–Nirenberg inequality and BMO space

    supI⊆[0,1]

    1|I |

    ∣∣∣∣{t ∈ I : ∣∣∣∣ϕ(t)− 1|I |∫Iϕ

    ∣∣∣∣ ≥ λ}∣∣∣∣ ≤ c1e−c2λ/‖ϕ‖BMO2([0,1])where

    ‖ϕ‖2BMO2([0,1])def= sup

    I⊆[0,1]

    1|I |

    ∫I

    ∣∣∣∣ϕ− 1|I |∫Iϕ

    ∣∣∣∣2 .

    QuestionFind sharp constants c1, c2 > 0.

    Paata Ivanisvili Kent State University Sharp estimates

  • John–Nirenberg inequality and BMO space

    supI⊆[0,1]

    1|I |

    ∣∣∣∣{t ∈ I : ∣∣∣∣ϕ(t)− 1|I |∫Iϕ

    ∣∣∣∣ ≥ λ}∣∣∣∣ ≤ c1e−c2λ/‖ϕ‖BMO2([0,1])where

    ‖ϕ‖2BMO2([0,1])def= sup

    I⊆[0,1]

    1|I |

    ∫I

    ∣∣∣∣ϕ− 1|I |∫Iϕ

    ∣∣∣∣2 .QuestionFind sharp constants c1, c2 > 0.

    Paata Ivanisvili Kent State University Sharp estimates

  • Extremal problems on BMO: I & II

    Bε(x1, x2)def= sup

    ϕ : ‖ϕ‖BMO≤ε

    {∫ 10

    f (ϕ(t))dt :

    ∫ 10ϕ = x1,

    ∫ 10ϕ2 = x2,

    }

    Example: f (t) = 1(−∞,−λ)∪(λ,∞)(t) - John–Nirenberg inequality.

    Main results (IOSVZ):1. Bε is minimal concave function on Ω0 \ Ωε with B|∂Ω0 = f

    where Ωε = {(x1, x2) ⊂ R2 : x2 ≥ x21 + ε2}2. For each ε we find Bε (stationary algorithm).3. We described evolution of Bε, ε ≥ 0 (dynamical approach).

    The torsion of (t, t2, f (t)) changes sign finitely many times.

    Paata Ivanisvili Kent State University Sharp estimates

  • Extremal problems on BMO: I & II

    Bε(x1, x2)def= sup

    ϕ : ‖ϕ‖BMO≤ε

    {∫ 10

    f (ϕ(t))dt :

    ∫ 10ϕ = x1,

    ∫ 10ϕ2 = x2,

    }

    Example: f (t) = 1(−∞,−λ)∪(λ,∞)(t) - John–Nirenberg inequality.

    Main results (IOSVZ):1. Bε is minimal concave function on Ω0 \ Ωε with B|∂Ω0 = f

    where Ωε = {(x1, x2) ⊂ R2 : x2 ≥ x21 + ε2}2. For each ε we find Bε (stationary algorithm).3. We described evolution of Bε, ε ≥ 0 (dynamical approach).

    The torsion of (t, t2, f (t)) changes sign finitely many times.

    Paata Ivanisvili Kent State University Sharp estimates

  • Extremal problems on BMO: I & II

    Bε(x1, x2)def= sup

    ϕ : ‖ϕ‖BMO≤ε

    {∫ 10

    f (ϕ(t))dt :

    ∫ 10ϕ = x1,

    ∫ 10ϕ2 = x2,

    }

    Example: f (t) = 1(−∞,−λ)∪(λ,∞)(t) - John–Nirenberg inequality.

    Main results (IOSVZ):1. Bε is minimal concave function on Ω0 \ Ωε with B|∂Ω0 = f

    where Ωε = {(x1, x2) ⊂ R2 : x2 ≥ x21 + ε2}2. For each ε we find Bε (stationary algorithm).3. We described evolution of Bε, ε ≥ 0 (dynamical approach).

    The torsion of (t, t2, f (t)) changes sign finitely many times.

    Paata Ivanisvili Kent State University Sharp estimates

  • Extremal problems on BMO: I & II

    Bε(x1, x2)def= sup

    ϕ : ‖ϕ‖BMO≤ε

    {∫ 10

    f (ϕ(t))dt :

    ∫ 10ϕ = x1,

    ∫ 10ϕ2 = x2,

    }

    Example: f (t) = 1(−∞,−λ)∪(λ,∞)(t) - John–Nirenberg inequality.

    Main results (IOSVZ):1. Bε is minimal concave function on Ω0 \ Ωε with B|∂Ω0 = f

    where Ωε = {(x1, x2) ⊂ R2 : x2 ≥ x21 + ε2}2. For each ε we find Bε (stationary algorithm).3. We described evolution of Bε, ε ≥ 0 (dynamical approach).

    The torsion of (t, t2, f (t)) changes sign finitely many times.

    Paata Ivanisvili Kent State University Sharp estimates

  • Other applications

    Theorem (Ivanisvili–Jaye–Nazarov)

    For any p > 1 and n ≥ 1 there exists A(p, n) > 1 such that∫Rn

    (supB3x

    1|B|

    ∫B|f (y)|dy

    )pdx ≥ A(p, n)

    ∫Rn|f (x)|pdx , ∀f ∈ Lp

    NO for centered maximal function p > nn−2 . f (x) = min{|x |n−1, 1}.

    Theorem (P.I.)

    supεI∈{−1,1}

    ∫ 10

    (∑I∈D

    εI 〈f , hI 〉hI)2

    + τ2f 2

    p/2 dx ≤ C ∫ 10|f |pdx

    Settles an open problem of Boros–Janakiraman–Volberg

    Paata Ivanisvili Kent State University Sharp estimates

  • Other applications

    Theorem (Ivanisvili–Jaye–Nazarov)

    For any p > 1 and n ≥ 1 there exists A(p, n) > 1 such that∫Rn

    (supB3x

    1|B|

    ∫B|f (y)|dy

    )pdx ≥ A(p, n)

    ∫Rn|f (x)|pdx , ∀f ∈ Lp

    NO for centered maximal function p > nn−2 .

    f (x) = min{|x |n−1, 1}.

    Theorem (P.I.)

    supεI∈{−1,1}

    ∫ 10

    (∑I∈D

    εI 〈f , hI 〉hI)2

    + τ2f 2

    p/2 dx ≤ C ∫ 10|f |pdx

    Settles an open problem of Boros–Janakiraman–Volberg

    Paata Ivanisvili Kent State University Sharp estimates

  • Other applications

    Theorem (Ivanisvili–Jaye–Nazarov)

    For any p > 1 and n ≥ 1 there exists A(p, n) > 1 such that∫Rn

    (supB3x

    1|B|

    ∫B|f (y)|dy

    )pdx ≥ A(p, n)

    ∫Rn|f (x)|pdx , ∀f ∈ Lp

    NO for centered maximal function p > nn−2 . f (x) = min{|x |n−1, 1}.

    Theorem (P.I.)

    supεI∈{−1,1}

    ∫ 10

    (∑I∈D

    εI 〈f , hI 〉hI)2

    + τ2f 2

    p/2 dx ≤ C ∫ 10|f |pdx

    Settles an open problem of Boros–Janakiraman–Volberg

    Paata Ivanisvili Kent State University Sharp estimates

  • Other applications

    Theorem (Ivanisvili–Jaye–Nazarov)

    For any p > 1 and n ≥ 1 there exists A(p, n) > 1 such that∫Rn

    (supB3x

    1|B|

    ∫B|f (y)|dy

    )pdx ≥ A(p, n)

    ∫Rn|f (x)|pdx , ∀f ∈ Lp

    NO for centered maximal function p > nn−2 . f (x) = min{|x |n−1, 1}.

    Theorem (P.I.)

    supεI∈{−1,1}

    ∫ 10

    (∑I∈D

    εI 〈f , hI 〉hI)2

    + τ2f 2

    p/2 dx ≤ C ∫ 10|f |pdx

    Settles an open problem of Boros–Janakiraman–Volberg

    Paata Ivanisvili Kent State University Sharp estimates

  • The end

    Thank you

    Paata Ivanisvili Kent State University Sharp estimates

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