Factorization of the identity R. Lechner Joint work with N. J. - - PowerPoint PPT Presentation

Factorization of the identity

• R. Lechner

Joint work with N. J. Laustsen and P. F. X. Müller

• J. Kepler University, Linz

Bedlewo, July 19, 2016

Overview

1 Operators with large diagonal 2 Mixed-norm Hardy spaces Hp(Hq)

Overview

1 Operators with large diagonal 2 Mixed-norm Hardy spaces Hp(Hq)

A description of the problem class

• Let X be a Banach space and T : X → X a linear operator.

Find conditions on X and T such that the identity Id on X factors through T i.e. X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• The problem has finite dimensional (quantitative) and infinite

dimensional (qualitative) aspects.

• Classical examples for X include: ℓp (Pełczyński), ℓ∞ (Lindenstrauss),

L1 (Enflo-Starbird), Lp (Johnson-Maurey-Schechtman-Tzafriri), ℓp

n

(Bourgain-Tzafriri).

• Two parameters: Lp(Lq), 1 < p, q < ∞ (Capon).

A description of the problem class

• Let X be a Banach space and T : X → X a linear operator.

Find conditions on X and T such that the identity Id on X factors through T i.e. X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• The problem has finite dimensional (quantitative) and infinite

dimensional (qualitative) aspects.

• Classical examples for X include: ℓp (Pełczyński), ℓ∞ (Lindenstrauss),

L1 (Enflo-Starbird), Lp (Johnson-Maurey-Schechtman-Tzafriri), ℓp

n

(Bourgain-Tzafriri).

• Two parameters: Lp(Lq), 1 < p, q < ∞ (Capon).

A description of the problem class

• Let X be a Banach space and T : X → X a linear operator.

Find conditions on X and T such that the identity Id on X factors through T i.e. X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• The problem has finite dimensional (quantitative) and infinite

dimensional (qualitative) aspects.

• Classical examples for X include: ℓp (Pełczyński), ℓ∞ (Lindenstrauss),

L1 (Enflo-Starbird), Lp (Johnson-Maurey-Schechtman-Tzafriri), ℓp

n

(Bourgain-Tzafriri).

• Two parameters: Lp(Lq), 1 < p, q < ∞ (Capon).

A description of the problem class

• Let X be a Banach space and T : X → X a linear operator.

Find conditions on X and T such that the identity Id on X factors through T i.e. X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• The problem has finite dimensional (quantitative) and infinite

dimensional (qualitative) aspects.

• Classical examples for X include: ℓp (Pełczyński), ℓ∞ (Lindenstrauss),

L1 (Enflo-Starbird), Lp (Johnson-Maurey-Schechtman-Tzafriri), ℓp

n

(Bourgain-Tzafriri).

• Two parameters: Lp(Lq), 1 < p, q < ∞ (Capon).

A description of the problem class

• Let X be a Banach space and T : X → X a linear operator.

Find conditions on X and T such that the identity Id on X factors through T i.e. X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• The problem has finite dimensional (quantitative) and infinite

dimensional (qualitative) aspects.

• Classical examples for X include: ℓp (Pełczyński), ℓ∞ (Lindenstrauss),

L1 (Enflo-Starbird), Lp (Johnson-Maurey-Schechtman-Tzafriri), ℓp

n

(Bourgain-Tzafriri).

• Two parameters: Lp(Lq), 1 < p, q < ∞ (Capon).

A description of the problem class

• Let X be a Banach space and T : X → X a linear operator.

Find conditions on X and T such that the identity Id on X factors through T i.e. X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• The problem has finite dimensional (quantitative) and infinite

dimensional (qualitative) aspects.

• Classical examples for X include: ℓp (Pełczyński), ℓ∞ (Lindenstrauss),

L1 (Enflo-Starbird), Lp (Johnson-Maurey-Schechtman-Tzafriri), ℓp

n

(Bourgain-Tzafriri).

• Two parameters: Lp(Lq), 1 < p, q < ∞ (Capon).

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Operators with large diagonal

• Let X be a Banach space and T : X → X a linear operator.
• Suppose that X has an unconditional basis (bn), and let b∗

n ∈ X∗ be so

that bn, b∗

m = 0, m = n, and bn, b∗ n = 1.

• We say that T has large diagonal (relative to (bn)) if

infn∈N |Tbn, b∗

n| > 0.

• For many Banach spaces X we know that the identity factors through
• perators T with large diagonal, i.e.

X

Id

• E
• X

X

T

X

P

• EP ≤ C.
• X = ℓp with the unit vector basis (Pełczyński)
• X = Lp (Andrew), X = Lp(Lq) (Capon) with the Haar basis

Can the identity operator on X be factored through each

• perator on X with large diagonal for all Banach spaces X

with an unconditional basis?

Answer:

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

There is an operator T on a Banach space X with an unconditional basis such that T has large diagonal, but the identity operator on X does not factor through T. Main ingredients for the proof:

• X is Gowers’ space with an unconditional basis (Gowers–Maurey).
• Fredholm theory.

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

The identity on mixed-norm Hardy spaces Hp(Hq), 1 ≤ p, q < ∞, factors through any operator T with large diagonal relative to the bi–parameter Haar basis. (1 < p, q < ∞ = Capon).

Can the identity operator on X be factored through each

• perator on X with large diagonal for all Banach spaces X

with an unconditional basis?

Answer:

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

There is an operator T on a Banach space X with an unconditional basis such that T has large diagonal, but the identity operator on X does not factor through T. Main ingredients for the proof:

• X is Gowers’ space with an unconditional basis (Gowers–Maurey).
• Fredholm theory.

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

The identity on mixed-norm Hardy spaces Hp(Hq), 1 ≤ p, q < ∞, factors through any operator T with large diagonal relative to the bi–parameter Haar basis. (1 < p, q < ∞ = Capon).

Can the identity operator on X be factored through each

• perator on X with large diagonal for all Banach spaces X

with an unconditional basis?

Answer:

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

There is an operator T on a Banach space X with an unconditional basis such that T has large diagonal, but the identity operator on X does not factor through T. Main ingredients for the proof:

• X is Gowers’ space with an unconditional basis (Gowers–Maurey).
• Fredholm theory.

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

The identity on mixed-norm Hardy spaces Hp(Hq), 1 ≤ p, q < ∞, factors through any operator T with large diagonal relative to the bi–parameter Haar basis. (1 < p, q < ∞ = Capon).

Can the identity operator on X be factored through each

• perator on X with large diagonal for all Banach spaces X

with an unconditional basis?

Answer:

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

There is an operator T on a Banach space X with an unconditional basis such that T has large diagonal, but the identity operator on X does not factor through T. Main ingredients for the proof:

• X is Gowers’ space with an unconditional basis (Gowers–Maurey).
• Fredholm theory.

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

The identity on mixed-norm Hardy spaces Hp(Hq), 1 ≤ p, q < ∞, factors through any operator T with large diagonal relative to the bi–parameter Haar basis. (1 < p, q < ∞ = Capon).

Can the identity operator on X be factored through each

• perator on X with large diagonal for all Banach spaces X

with an unconditional basis?

Answer:

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

There is an operator T on a Banach space X with an unconditional basis such that T has large diagonal, but the identity operator on X does not factor through T. Main ingredients for the proof:

• X is Gowers’ space with an unconditional basis (Gowers–Maurey).
• Fredholm theory.

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

The identity on mixed-norm Hardy spaces Hp(Hq), 1 ≤ p, q < ∞, factors through any operator T with large diagonal relative to the bi–parameter Haar basis. (1 < p, q < ∞ = Capon).

Can the identity operator on X be factored through each

• perator on X with large diagonal for all Banach spaces X

with an unconditional basis?

Answer:

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

There is an operator T on a Banach space X with an unconditional basis such that T has large diagonal, but the identity operator on X does not factor through T. Main ingredients for the proof:

• X is Gowers’ space with an unconditional basis (Gowers–Maurey).
• Fredholm theory.

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

The identity on mixed-norm Hardy spaces Hp(Hq), 1 ≤ p, q < ∞, factors through any operator T with large diagonal relative to the bi–parameter Haar basis. (1 < p, q < ∞ = Capon).

Overview

1 Operators with large diagonal 2 Mixed-norm Hardy spaces Hp(Hq)

Mixed-norm Hardy spaces Hp(Hq)

• D = {[ k−1

2n , k 2n [ : n ≥ 0, 1 ≤ k ≤ 2n} denotes the dyadic intervals on

the unit interval,

• R = {I × J : I, J ∈ D} denotes the dyadic rectangles on the unit

square,

• (hI : I ∈ D) the L∞-normalized Haar system.
• hI×J(x, y) = hI(x)hJ(y) the L∞-normalized tensor product Haar

function, I × J ∈ R.

• Let f =

I×J∈R aI×JhI×J be a finite linear combination,

• Given 1 ≤ p, q < ∞, the space Hp(Hq) is the closure of

span{hR : R ∈ R} under the norm

• R∈R

aRhRHp(Hq) = 1 1

R∈R

a2

Rh2 R(x, y)

q/2 dy p/q dx 1/p .

Mixed-norm Hardy spaces Hp(Hq)

• D = {[ k−1

2n , k 2n [ : n ≥ 0, 1 ≤ k ≤ 2n} denotes the dyadic intervals on

the unit interval,

• R = {I × J : I, J ∈ D} denotes the dyadic rectangles on the unit

square,

• (hI : I ∈ D) the L∞-normalized Haar system.
• hI×J(x, y) = hI(x)hJ(y) the L∞-normalized tensor product Haar

function, I × J ∈ R.

• Let f =

I×J∈R aI×JhI×J be a finite linear combination,

• Given 1 ≤ p, q < ∞, the space Hp(Hq) is the closure of

span{hR : R ∈ R} under the norm

• R∈R

aRhRHp(Hq) = 1 1

R∈R

a2

Rh2 R(x, y)

q/2 dy p/q dx 1/p .

Mixed-norm Hardy spaces Hp(Hq)

• D = {[ k−1

2n , k 2n [ : n ≥ 0, 1 ≤ k ≤ 2n} denotes the dyadic intervals on

the unit interval,

• R = {I × J : I, J ∈ D} denotes the dyadic rectangles on the unit

square,

• (hI : I ∈ D) the L∞-normalized Haar system.
• hI×J(x, y) = hI(x)hJ(y) the L∞-normalized tensor product Haar

function, I × J ∈ R.

• Let f =

I×J∈R aI×JhI×J be a finite linear combination,

• Given 1 ≤ p, q < ∞, the space Hp(Hq) is the closure of

span{hR : R ∈ R} under the norm

• R∈R

aRhRHp(Hq) = 1 1

R∈R

a2

Rh2 R(x, y)

q/2 dy p/q dx 1/p .

Mixed-norm Hardy spaces Hp(Hq)

• D = {[ k−1

2n , k 2n [ : n ≥ 0, 1 ≤ k ≤ 2n} denotes the dyadic intervals on

the unit interval,

• R = {I × J : I, J ∈ D} denotes the dyadic rectangles on the unit

square,

• (hI : I ∈ D) the L∞-normalized Haar system.
• hI×J(x, y) = hI(x)hJ(y) the L∞-normalized tensor product Haar

function, I × J ∈ R.

• Let f =

I×J∈R aI×JhI×J be a finite linear combination,

• Given 1 ≤ p, q < ∞, the space Hp(Hq) is the closure of

span{hR : R ∈ R} under the norm

• R∈R

aRhRHp(Hq) = 1 1

R∈R

a2

Rh2 R(x, y)

q/2 dy p/q dx 1/p .

Mixed-norm Hardy spaces Hp(Hq)

• D = {[ k−1

2n , k 2n [ : n ≥ 0, 1 ≤ k ≤ 2n} denotes the dyadic intervals on

the unit interval,

• R = {I × J : I, J ∈ D} denotes the dyadic rectangles on the unit

square,

• (hI : I ∈ D) the L∞-normalized Haar system.
• hI×J(x, y) = hI(x)hJ(y) the L∞-normalized tensor product Haar

function, I × J ∈ R.

• Let f =

I×J∈R aI×JhI×J be a finite linear combination,

• Given 1 ≤ p, q < ∞, the space Hp(Hq) is the closure of

span{hR : R ∈ R} under the norm

• R∈R

aRhRHp(Hq) = 1 1

R∈R

a2

Rh2 R(x, y)

q/2 dy p/q dx 1/p .

Mixed-norm Hardy spaces Hp(Hq)

• D = {[ k−1

2n , k 2n [ : n ≥ 0, 1 ≤ k ≤ 2n} denotes the dyadic intervals on

the unit interval,

• R = {I × J : I, J ∈ D} denotes the dyadic rectangles on the unit

square,

• (hI : I ∈ D) the L∞-normalized Haar system.
• hI×J(x, y) = hI(x)hJ(y) the L∞-normalized tensor product Haar

function, I × J ∈ R.

• Let f =

I×J∈R aI×JhI×J be a finite linear combination,

• Given 1 ≤ p, q < ∞, the space Hp(Hq) is the closure of

span{hR : R ∈ R} under the norm

• R∈R

aRhRHp(Hq) = 1 1

R∈R

a2

Rh2 R(x, y)

q/2 dy p/q dx 1/p .

Large diagonal relative to the bi-parameter Haar system

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

Let 1 ≤ p, q < ∞, δ > 0 and T : Hp(Hq) → Hp(Hq) be a linear operator with large diagonal, i.e. ThI×J, hI×J ≥ δ|I × J|. Then we have Hp(Hq)

Id

• E
• Hp(Hq)

Hp(Hq)

T

Hp(Hq)

P

• EP ≤ C/δ,

where C = 1 + 1/10000. By Capon’s bi–parameter construction (BI×J) and a random choice of signs εI×J, there exists a block basis b(ε)

I×J = K∈BI×J εK×LhK×L such that:

• (b(ε)

I×J) is equivalent to (hI×J),

• Tb(ε)

I×J = αI×Jb(ε) I×J + small error, with αI×J ≥ δ,

• The projection Qf =

I∈D f,b(ε)

I×J

b(ε)

I×J2 2

b(ε)

I×J is bounded on Hp(Hq).

Large diagonal relative to the bi-parameter Haar system

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

Let 1 ≤ p, q < ∞, δ > 0 and T : Hp(Hq) → Hp(Hq) be a linear operator with large diagonal, i.e. ThI×J, hI×J ≥ δ|I × J|. Then we have Hp(Hq)

Id

• E
• Hp(Hq)

Hp(Hq)

T

Hp(Hq)

P

• EP ≤ C/δ,

where C = 1 + 1/10000. By Capon’s bi–parameter construction (BI×J) and a random choice of signs εI×J, there exists a block basis b(ε)

I×J = K∈BI×J εK×LhK×L such that:

• (b(ε)

I×J) is equivalent to (hI×J),

• Tb(ε)

I×J = αI×Jb(ε) I×J + small error, with αI×J ≥ δ,

• The projection Qf =

I∈D f,b(ε)

I×J

b(ε)

I×J2 2

b(ε)

I×J is bounded on Hp(Hq).

Large diagonal relative to the bi-parameter Haar system

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

Let 1 ≤ p, q < ∞, δ > 0 and T : Hp(Hq) → Hp(Hq) be a linear operator with large diagonal, i.e. ThI×J, hI×J ≥ δ|I × J|. Then we have Hp(Hq)

Id

• E
• Hp(Hq)

Hp(Hq)

T

Hp(Hq)

P

• EP ≤ C/δ,

where C = 1 + 1/10000. By Capon’s bi–parameter construction (BI×J) and a random choice of signs εI×J, there exists a block basis b(ε)

I×J = K∈BI×J εK×LhK×L such that:

• (b(ε)

I×J) is equivalent to (hI×J),

• Tb(ε)

I×J = αI×Jb(ε) I×J + small error, with αI×J ≥ δ,

• The projection Qf =

I∈D f,b(ε)

I×J

b(ε)

I×J2 2

b(ε)

I×J is bounded on Hp(Hq).

Large diagonal relative to the bi-parameter Haar system

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

Let 1 ≤ p, q < ∞, δ > 0 and T : Hp(Hq) → Hp(Hq) be a linear operator with large diagonal, i.e. ThI×J, hI×J ≥ δ|I × J|. Then we have Hp(Hq)

Id

• E
• Hp(Hq)

Hp(Hq)

T

Hp(Hq)

P

• EP ≤ C/δ,

where C = 1 + 1/10000. By Capon’s bi–parameter construction (BI×J) and a random choice of signs εI×J, there exists a block basis b(ε)

I×J = K∈BI×J εK×LhK×L such that:

• (b(ε)

I×J) is equivalent to (hI×J),

• Tb(ε)

I×J = αI×Jb(ε) I×J + small error, with αI×J ≥ δ,

• The projection Qf =

I∈D f,b(ε)

I×J

b(ε)

I×J2 2

b(ε)

I×J is bounded on Hp(Hq).

Large diagonal relative to the bi-parameter Haar system

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

Let 1 ≤ p, q < ∞, δ > 0 and T : Hp(Hq) → Hp(Hq) be a linear operator with large diagonal, i.e. ThI×J, hI×J ≥ δ|I × J|. Then we have Hp(Hq)

Id

• E
• Hp(Hq)

Hp(Hq)

T

Hp(Hq)

P

• EP ≤ C/δ,

where C = 1 + 1/10000. By Capon’s bi–parameter construction (BI×J) and a random choice of signs εI×J, there exists a block basis b(ε)

I×J = K∈BI×J εK×LhK×L such that:

• (b(ε)

I×J) is equivalent to (hI×J),

• Tb(ε)

I×J = αI×Jb(ε) I×J + small error, with αI×J ≥ δ,

• The projection Qf =

I∈D f,b(ε)

I×J

b(ε)

I×J2 2

b(ε)

I×J is bounded on Hp(Hq).

Large diagonal relative to the bi-parameter Haar system

Theorem (N. J. Laustsen, R. L., P. F. X. Müller)

Let 1 ≤ p, q < ∞, δ > 0 and T : Hp(Hq) → Hp(Hq) be a linear operator with large diagonal, i.e. ThI×J, hI×J ≥ δ|I × J|. Then we have Hp(Hq)

Id

• E
• Hp(Hq)

Hp(Hq)

T

Hp(Hq)

P

• EP ≤ C/δ,

where C = 1 + 1/10000. By Capon’s bi–parameter construction (BI×J) and a random choice of signs εI×J, there exists a block basis b(ε)

I×J = K∈BI×J εK×LhK×L such that:

• (b(ε)

I×J) is equivalent to (hI×J),

• Tb(ε)

I×J = αI×Jb(ε) I×J + small error, with αI×J ≥ δ,

• The projection Qf =

I∈D f,b(ε)

I×J

b(ε)

I×J2 2

b(ε)

I×J is bounded on Hp(Hq).

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – ordering of R

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Figure: Order of the first 49 rectangles.

• If I ∈ D, then

I ∈ D is such that I ⊃ I and | I| = 2|I|.

• O⊳ (I × J) is the order

number of I × J.

• O⊳ (

I × J) < O⊳ (I × J)

• O⊳ (I ×

J) < O⊳ (I × J)

• E.g., let I = J = [0, 1

2], then

• O⊳ (I × J) = 5,
• O⊳ (

I × J) = 3,

• and O⊳ (I ×

J) = 1.

Proof – case J = [0, 1]

Figure: Darkgray=past, lightgray=present, white=future.

• K0 × [0, 1] ∈ B

I×[0,1]

• K × [0, 1] ∈ BI×[0,1]

Proof – case J = [0, 1]

Figure: Darkgray=past, lightgray=present, white=future.

• K0 × [0, 1] ∈ B

I×[0,1]

• K × [0, 1] ∈ BI×[0,1]

[0, 1] K0 K K′ K K′′ K K′′′ K

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

The Gamlen-Gaudet construction (1973)

Figure: On the left side: construction of bI. On the right side: the corresponding index intervals I.

Proof – case J = [0, 1], I = [0, 1]

Figure: Darkgray=past, lightgray=present, white=future.

• [0, 1] × L0 ∈ B[0,1]×

J

• [0, 1] × L ∈ B[0,1]×J

Proof – case J = [0, 1], I = [0, 1]

Figure: Darkgray=past, lightgray=present, white=future.

• [0, 1] × L0 ∈ B[0,1]×

J

• [0, 1] × L ∈ B[0,1]×J

L0 L [0, 1] L′ L

Proof – case J = [0, 1], I = [0, 1]

Figure: Darkgray=past, lightgray=present.

• K0 × [0, 1] ∈ BI×[0,1]
• K1 × L1 ∈ B

I×J (⇒ K1 × [0, 1] ∈ B I×[0,1])

• K0 × L ∈ BI×J

Proof – case J = [0, 1], I = [0, 1]

Figure: Darkgray=past, lightgray=present.

• K0 × [0, 1] ∈ BI×[0,1]
• K1 × L1 ∈ B

I×J (⇒ K1 × [0, 1] ∈ B I×[0,1])

• K0 × L ∈ BI×J

L1 K0 L K1 L′

1

K0 L K′

1

L K′′

1

• The lightgray area is given by

BI×

J

• ∩

B

I×J

Proof – case J = [0, 1], I = [0, 1]

Figure: Darkgray=past, lightgray=present.

• K0 × [0, 1] ∈ BI×[0,1]
• K1 × L1 ∈ B

I×J (⇒ K1 × [0, 1] ∈ B I×[0,1])

• K0 × L ∈ BI×J

L1 K0 L K1 L′

1

K0 L K′

1

L K′′

1

• The lightgray area is given by

BI×

J

• ∩

B

I×J

Proof – case J = [0, 1], I = [0, 1]

Figure: Darkgray=past, lightgray=present.

• K0 × [0, 1] ∈ BI×[0,1]
• K1 × L1 ∈ B

I×J (⇒ K1 × [0, 1] ∈ B I×[0,1])

• K0 × L ∈ BI×J

L1 K0 L K1 L′

1

K0 L K′

1

L K′′

1

• The lightgray area is given by

BI×

J

• ∩

B

I×J

What does BI×J look like?

• BI×J =

K∈XI

• L∈YJ(K) K × L
• XI = {K0, K1, K2}

What does BI×J look like?

• BI×J =

K∈XI

• L∈YJ(K) K × L
• XI = {K0, K1, K2}

Yes, but how does BI×J REALLY look like?

• XI = {K0, K1, K2},
• BI×J =

K∈XI

• L∈YJ(K) K × L,
• J, J0, J1 ∈ D are such that J0 ∪ J1 = J and J0 ∩ J1 = ∅,
• BI×J in the top layer,
• BI×J0 and BI×J1 in the bottom layer.

Yes, but how does BI×J REALLY look like?

• XI = {K0, K1, K2},
• BI×J =

K∈XI

• L∈YJ(K) K × L,
• J, J0, J1 ∈ D are such that J0 ∪ J1 = J and J0 ∩ J1 = ∅,
• BI×J in the top layer,
• BI×J0 and BI×J1 in the bottom layer.

Yes, but how does BI×J REALLY look like?

• XI = {K0, K1, K2},
• BI×J =

K∈XI

• L∈YJ(K) K × L,
• J, J0, J1 ∈ D are such that J0 ∪ J1 = J and J0 ∩ J1 = ∅,
• BI×J in the top layer,
• BI×J0 and BI×J1 in the bottom layer.

And one more

• XI = {K0, K1, K2}
• I, J0, J1, J2, J3 ∈ D
• J0 ⊂ J1 ⊂ J2 ⊂ J3
• |J3| = 2|J2| = 4|J1| =

8|J0|

• BI×Jj in layer j
• The shaded vertical

plane connects the lines ℓ = {(x, y0) : x ∈ [0, 1)}

And one more

• XI = {K0, K1, K2}
• I, J0, J1, J2, J3 ∈ D
• J0 ⊂ J1 ⊂ J2 ⊂ J3
• |J3| = 2|J2| = 4|J1| =

8|J0|

• BI×Jj in layer j
• The shaded vertical

plane connects the lines ℓ = {(x, y0) : x ∈ [0, 1)}

Summary

The identity factors through operators with large diagonal (with respect to the bi–parameter Haar system) in the following Hardy spaces:

• Capon: Hp(Hq) for 1 < p, q < ∞,
• P. F. X. Müller: H1(H1),
• R. L., P. F. X. Müller: H1

n(H1 n),

• N. J. Laustsen, R. L., P. F. X. Müller: Hp(Hq) for 1 ≤ p, q < ∞.

Coming soon...

• Hp

n(Hq n).

Summary

The identity factors through operators with large diagonal (with respect to the bi–parameter Haar system) in the following Hardy spaces:

• Capon: Hp(Hq) for 1 < p, q < ∞,
• P. F. X. Müller: H1(H1),
• R. L., P. F. X. Müller: H1

n(H1 n),

• N. J. Laustsen, R. L., P. F. X. Müller: Hp(Hq) for 1 ≤ p, q < ∞.

Coming soon...

• Hp

n(Hq n).

Summary

The identity factors through operators with large diagonal (with respect to the bi–parameter Haar system) in the following Hardy spaces:

• Capon: Hp(Hq) for 1 < p, q < ∞,
• P. F. X. Müller: H1(H1),
• R. L., P. F. X. Müller: H1

n(H1 n),

• N. J. Laustsen, R. L., P. F. X. Müller: Hp(Hq) for 1 ≤ p, q < ∞.

Coming soon...

• Hp

n(Hq n).

Summary

The identity factors through operators with large diagonal (with respect to the bi–parameter Haar system) in the following Hardy spaces:

• Capon: Hp(Hq) for 1 < p, q < ∞,
• P. F. X. Müller: H1(H1),
• R. L., P. F. X. Müller: H1

n(H1 n),

• N. J. Laustsen, R. L., P. F. X. Müller: Hp(Hq) for 1 ≤ p, q < ∞.

Coming soon...

• Hp

n(Hq n).

Summary

The identity factors through operators with large diagonal (with respect to the bi–parameter Haar system) in the following Hardy spaces:

• Capon: Hp(Hq) for 1 < p, q < ∞,
• P. F. X. Müller: H1(H1),
• R. L., P. F. X. Müller: H1

n(H1 n),

• N. J. Laustsen, R. L., P. F. X. Müller: Hp(Hq) for 1 ≤ p, q < ∞.

Coming soon...

• Hp

n(Hq n).

• N. J. Laustsen, R. Lechner, and P. F. X. Müller.

Factorization of the identity through operators with large diagonal. ArXiv e-prints, September 2015. Richard Lechner and Paul F. X. Müller. Localization and projections on bi-parameter BMO.

• Q. J. Math., 66(4):1069–1101, 2015.