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Extension versus representationThree examples

Measure TheoryMarczewski Centennial Conference

Bedlevo, September 2007

Guillermo P. Curbera Optimal domains and vector measures

Edward Marczewski

Optimal domains for classical operatorsand vector measures:

a new look at old problems

Guillermo P. Curbera

Universidad de SevillaSpain

Bedlevo, September 2007

Outline

1 Extension versus representationRepresentation theoremsExtension theorems

2 Three examplesThe Volterra operatorSobolev’s inequalityThe Hausdorff-Young inequality

Outline

Representation theoremsExtension theorems

Outline

The Riesz representation theorem

Frigyes Riesz (1909)

Let Λ: C([0, 1]) → C be a positive linear operator. Then,there exists a finite Borel measure

ν : B0([0, 1]) → R

such that:

∫f dν, f ∈ C([0, 1]).

Viewpoint: a representation theorem.

ν : B0([0, 1]) → R

such that:

∫f dν, f ∈ C([0, 1]).

ν : B0([0, 1]) → R

such that:

∫f dν, f ∈ C([0, 1]).

Viewpoint:

a representation theorem.

ν : B0([0, 1]) → R

such that:

∫f dν, f ∈ C([0, 1]).

The vector case

Bartle, Dunford, Schwartz (1955)

Let K be a compact Hausdorff space, X be a Banachspace, and T : C(K ) → X a weakly compact operator.Then, there exists a Borel (vector) measure

ν : B0(K ) → X

such that:

∫f dν, f ∈ C(K ).

The vector case

ν : B0(K ) → X

such that:

∫f dν, f ∈ C(K ).

The vector case

ν : B0(K ) → X

such that:

∫f dν, f ∈ C(K ).

Viewpoint:

a representation theorem.

The vector case

ν : B0(K ) → X

such that:

∫f dν, f ∈ C(K ).

Representation of operators I: Bochner

Let (Ω,Σ, µ) be a finite measure space and X be a Banachspace.

(Dunford, Pettis, Phillips (1940)Let T : L1(µ) → X be a weakly compact linear operator.Then, there exists g ∈ L∞(µ) such that

∫f ·g dµ, f ∈ L1(µ).

Then A ∈ Σ 7→ ν(A) := T (χA) ∈ X , is a vector measurewith a Bochner integrable density with respect to µ.

∫f ·g dµ, f ∈ L1(µ).

Representation of operators II: BDS

Let (Ω,Σ) be a measure space, X be a Banach space, andν : Σ → X a vector measure.

The BDS–integral:Let f : Ω → R be a measurable function. It is integrable withrespect to ν if... (Lebesgue type integration).

L1(ν) suitably normed is a Banach space (of classes) ofintegrable functions.

Integration operator: f ∈ L1(ν) 7→ Iν(f ) =∫

f dν ∈ X .

BDS–representation: Example

The Riemann–Liouville fractional integral of order α ∈ (0, 1):

Tα(f ) =1

Γ(α)

f (t)|x − t |α

dt , x ∈ [0, 1].

Tα : L∞([0, 1]) → Lp([0, 1]) continuous for all p ∈ [1,∞].

Tα is Bochner representable iff p < 1/α.

However, if νp(A) := Tα(χA) ∈ Lp([0, 1]), for A ∈ B0([0, 1]),then:

Tα(f ) = (BDS)−∫

f dνp, f ∈ L∞([0, 1]),

for all p ∈ [1,∞].

Tα(f ) =1

Γ(α)

f (t)|x − t |α

dt , x ∈ [0, 1].

Tα(f ) = (BDS)−∫

f dνp, f ∈ L∞([0, 1]),

for all p ∈ [1,∞].

Tα(f ) =1

Γ(α)

f (t)|x − t |α

dt , x ∈ [0, 1].

Tα(f ) = (BDS)−∫

f dνp, f ∈ L∞([0, 1]),

for all p ∈ [1,∞].

Tα(f ) =1

Γ(α)

f (t)|x − t |α

dt , x ∈ [0, 1].

Tα(f ) = (BDS)−∫

f dνp, f ∈ L∞([0, 1]),

for all p ∈ [1,∞].

Tα(f ) =1

Γ(α)

f (t)|x − t |α

dt , x ∈ [0, 1].

Tα(f ) = (BDS)−∫

f dνp, f ∈ L∞([0, 1]),

for all p ∈ [1,∞].

Outline

BDS–extension

The previous example has interesting features:

L∞([0, 1])Tα - Lp([0, 1])

L1(νp)Iνp

Moreover, the space L1(νp) is the largest Banach functionspace with order continuous norm for which such a factorizationexists (p 6= ∞).(Order continuous norm: fα ↓ 0 then ‖fα‖ ↓ 0.)

BDS–extension

L∞([0, 1])Tα - Lp([0, 1])

L1(νp)Iνp

BDS–extension

L∞([0, 1])Tα - Lp([0, 1])

L1(νp)Iνp

BDS–extension

L∞([0, 1])Tα - Lp([0, 1])

L1(νp)Iνp

BDS–extension

More precisely, if there exists a Banach function space E withorder continuous norm, and an operator T : E → X whichextends Tα

L∞([0, 1])Tα - Lp([0, 1])

then E ⊂ L1(νp) and the integration operator Iνp extends T .

BDS–extension

More precisely, if there exists a Banach function space E withorder continuous norm, and an operator T : E → X whichextends Tα

L∞([0, 1])Tα - Lp([0, 1])

then E ⊂ L1(νp) and the integration operator Iνp extends T .

Extension versus representation

The Riesz theorem:

C([0, 1])Λ

L1(ν)Iν

The BDS theorem:

C(K )T

L1(ν)Iν

The Riesz theorem:

C([0, 1])Λ

L1(ν)Iν

The BDS theorem:

C(K )T

L1(ν)Iν

The Riesz theorem:

C([0, 1])Λ

L1(ν)Iν

The BDS theorem:

C(K )T

L1(ν)Iν

General situation

Theorem

T : E → X linearE Banach function spaceX Banach space

such thatTfn → Tf weakly in X

if fn ↑ f ∈ E

ν(A) := TχA

is σ–additiveE → L1(ν)Integration operator Iν

extends T

L1(ν) is the optimal domain for T (with order continuousnorm).

General situation

Theorem

if fn ↑ f ∈ E

ν(A) := TχA

extends T

General situation

Theorem

if fn ↑ f ∈ E

ν(A) := TχA

extends T

The Volterra operatorSobolev’s inequalityThe Hausdorff-Young inequality

Outline

The Volterra integral operator

Defined by

f 7−→ Vf (x) :=

0f (t) dt , x ∈ [0, 1].

[V , L∞] =

f : V |f | ∈ L∞([0, 1])

[V , L1] =

f : V |f | ∈ L1([0, 1])

[V , L∞] = L1([0, 1]).

[V , L1] = L1((1− t)dt).

Defined by

f 7−→ Vf (x) :=

0f (t) dt , x ∈ [0, 1].

[V , L∞] =

f : V |f | ∈ L∞([0, 1])

[V , L1] =

f : V |f | ∈ L1([0, 1])

[V , L∞] = L1([0, 1]).

[V , L1] = L1((1− t)dt).

Defined by

f 7−→ Vf (x) :=

0f (t) dt , x ∈ [0, 1].

[V , L∞] =

f : V |f | ∈ L∞([0, 1])

[V , L1] =

f : V |f | ∈ L1([0, 1])

[V , L∞] = L1([0, 1]).

[V , L1] = L1((1− t)dt).

Defined by

f 7−→ Vf (x) :=

0f (t) dt , x ∈ [0, 1].

[V , L∞] =

f : V |f | ∈ L∞([0, 1])

[V , L1] =

f : V |f | ∈ L1([0, 1])

[V , L∞] = L1([0, 1]).

[V , L1] = L1((1− t)dt).

Optimal Lp-domain

Let 1 < p < ∞.

QUESTION: What can we say about

[V , Lp] =

f : V |f | ∈ Lp([0, 1])?

ANSWER: Since Lp = (L1, L∞)1/p′,p, then

[V , Lp] =([V , L1], [V , L∞]

)1/p′,p

Optimal Lp-domain

Let 1 < p < ∞.

[V , Lp] =

f : V |f | ∈ Lp([0, 1])?

[V , Lp] =([V , L1], [V , L∞]

)1/p′,p

Optimal Lp-domain

Let 1 < p < ∞.

[V , Lp] =

f : V |f | ∈ Lp([0, 1])?

[V , Lp] =([V , L1], [V , L∞]

)1/p′,p

Optimal X–domain

The result is valid more generally:

For X rearrangement invariant (i.e. interpolation) space on[0, 1] we define the optimal domain

[V , X ] =

f : V |f | ∈ X.

Since X = (L1, L∞)ρ, then

[V , X ] =([V , L1], [V , L∞]

For Volterra convolution operators (under certainconditions on φ):

f 7−→ Vφf (x) :=

0f (t)φ(x − t) dt , x ∈ [0, 1].

Optimal X–domain

[V , X ] =

f : V |f | ∈ X.

[V , X ] =([V , L1], [V , L∞]

f 7−→ Vφf (x) :=

0f (t)φ(x − t) dt , x ∈ [0, 1].

Optimal X–domain

[V , X ] =

f : V |f | ∈ X.

[V , X ] =([V , L1], [V , L∞]

f 7−→ Vφf (x) :=

0f (t)φ(x − t) dt , x ∈ [0, 1].

Outline

Sobolev’s classical inequality

Theorem (1938)

Let Ω ⊂ Rn be a bounded domain and let 1 ≤ p < n. Thereexist a constant C > 0 such that

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω),

where q := npn−p .

Note: p < npn−p ⇒ ‖u‖p ≤ ‖u‖ np

Theorem (1938)

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω),

Theorem (1938)

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω),

Refining Sobolev’s inequality

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω)

For a fixed norm in the left hand side, finding a smaller norm inright hand side.

Optimal problem:

For a fixed norm in the left hand side, find the smallestnorm in right hand side.

‖u‖Lq(Ω) ≤ C

‖ |∇u| ‖Lp(Ω)

, u ∈ C10(Ω)

For a fixed norm in the left hand side,

finding a smaller norm inright hand side.

Optimal problem:

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω)

Optimal problem:

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω)

Optimal problem:

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω)

Optimal problem:

‖u‖Lq(Ω) ≤ C ‖ |∇u| ‖Lp(Ω), u ∈ C10(Ω)

Optimal problem:

An example

Fix Lp–norm on the LHS (n′ < p < ∞).

Optimality within Lebesgue norms: the Lnp

n+p –norm isoptimal on the RHS, that is,

‖u‖Lp(Ω) ≤ C ‖ |∇u| ‖L

npn+p (Ω)

More general (rearrangement invariant, r.i.) norms: the

n+p ,p–norm is optimal on the RHS, that is,

‖u‖p ≤ C ‖ |∇u| ‖ npn+p ,p,

sharper than the classical Sobolev’s inequality, since

‖ |∇u| ‖ npn+p ,p ≤ ‖ |∇u| ‖ np

An example

‖u‖Lp(Ω) ≤ C ‖ |∇u| ‖L

npn+p (Ω)

‖u‖p ≤ C ‖ |∇u| ‖ npn+p ,p,

‖ |∇u| ‖ npn+p ,p ≤ ‖ |∇u| ‖ np

An example

‖u‖Lp(Ω) ≤ C ‖ |∇u| ‖L

npn+p (Ω)

‖u‖p ≤ C ‖ |∇u| ‖ npn+p ,p,

‖ |∇u| ‖ npn+p ,p ≤ ‖ |∇u| ‖ np

An example

‖u‖Lp(Ω) ≤ C ‖ |∇u| ‖L

npn+p (Ω)

‖u‖p ≤ C ‖ |∇u| ‖ npn+p ,p,

‖ |∇u| ‖ npn+p ,p ≤ ‖ |∇u| ‖ np

An example

‖u‖Lp(Ω) ≤ C ‖ |∇u| ‖L

npn+p (Ω)

‖u‖p ≤ C ‖ |∇u| ‖ npn+p ,p,

‖ |∇u| ‖ npn+p ,p ≤ ‖ |∇u| ‖ np

Reduction to one variable

Theorem (Edmunds, Kerman, Pick, 2000)

Let X , Y be r.i. spaces on [0,1]. TFAE:

‖u‖X(Ω) ≤ C‖ |∇u| ‖Y (Ω), for all u ∈ C10(Ω)

‖Tf‖X ≤ K‖f‖Y , for all f ∈ X

where T is the kernel operator associated with Sobolev’sinequality, defined for f : [0, 1] → R by

t ∈ [0, 1] 7−→ Tf (t) :=

tf (s)s

1n−1 ds

t ∈ [0, 1] 7−→ Tf (t) :=

tf (s)s

1n−1 ds

t ∈ [0, 1] 7−→ Tf (t) :=

tf (s)s

1n−1 ds

Optimality & one-variable optimality

Given a r.i. space X on [0,1] (a r.i. space X (Ω) on Ω).

Which is the largest r.i. function space Y (Ω) such that

‖u‖X(Ω) ≤ C‖ |∇u| ‖Y (Ω)?

i.e. the weakest size condition on |∇u| so that u ∈ X (Ω)?

Which is the largest r.i. function space Y such that

T : Y → X continuously?

‖u‖X(Ω) ≤ C‖ |∇u| ‖Y (Ω)?

Optimal domains and vector measure

Given a r.i. function space X, which is the largest Banachfunction space Y such that

T : Y −→ X continuously?

The optimal domain [T , X ] =

f : T |f | ∈ X

Associated vector measure: νX (A) := T (χA).