On the togological dual of variable Lebesgue spaces

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On the togological dual of variable Lebesgue spaces

Jes´us Oc´ariz^1,2

Ongoing joint work with A. Amenta, J. Conde and D. Cruz-Uribe

1Universidad Aut´onoma de Madrid (UAM).

2Instituto de Ciencias Matem´aticas (ICMAT).

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Motivation of variable Lebesgue spaces

To make things simple, from now on we are going to work in an

unidimensional space. The concepts and results are easy generalized for higher dimensions.

Why do we need to introduce variable Lebesgue spaces?

Think about the function g(x) = |x|^−1/3.

It is extremely well-behaved, however g /∈ L^p(R) for every p.

Variable Lebesgue spaces will let us capture different kind of growths and will create a space where these functions could belong.

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Motivation of variable Lebesgue spaces

To make things simple, from now on we are going to work in an

unidimensional space. The concepts and results are easy generalized for higher dimensions.

Why do we need to introduce variable Lebesgue spaces?

Think about the function g(x) = |x|^−1/3.

It is extremely well-behaved, however g /∈ L^p(R) for every p.

Variable Lebesgue spaces will let us capture different kind of growths and will create a space where these functions could belong.

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Definition of variable Lebesgue spaces

Let I ⊂ Rⁿ interval and p : I → [1, +∞) be a measurable function.

How can we define the norm?

The na¨ıve candidate is

ρ(f ) :=

Z

I

|f (x)|^p(x)dx This is not a norm! It is a modular ⇒ Orlicz spaces!

Definition

The Luxembourg norm

kf k := inf {λ > 0 : ρ(f /λ) ≤ 1} .

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Definition of variable Lebesgue spaces

Let I ⊂ Rⁿ interval and p : I → [1, +∞) be a measurable function.

How can we define the norm?

The na¨ıve candidate is

ρ(f ) :=

Z

I

|f (x)|^p(x)dx This is not a norm! It is a modular ⇒ Orlicz spaces!

Definition

The Luxembourg norm

kf k := inf {λ > 0 : ρ(f /λ) ≤ 1} .

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Definition

The variable Lebesgue space L^p(·)(I) is the space of measurable functions such that kf k < ∞.

(L^p(·)(I), k · k) is a Banach space.

Could we have used another norm in the definition?

Yes. The Amemiya norm:

kf k_A:= inf

λ>0{λ + λρ(f /λ)} . Proposition

The Luxembourg and Amemiya norm are equivalents kf k ≤ kf k_A≤ 2kf k

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Definition

kf k_A:= inf

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Definition

kf k_A:= inf

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Definition

kf k_A:= inf

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Problem: The topological dual of this Banach space?

Given p(·), how much do we know about the properties of the Banach space L^p(·)(I)?

Separability?

Dense subsets?

Is reflexive?

What is its topological dual?

Two different cases and two different behaviors!

It depends on p(·), or more specifically on p₊= ess sup

x∈I

p(x)

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Problem: The topological dual of this Banach space?

Separability?

Dense subsets?

Is reflexive?

x∈I

p(x)

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Problem: The topological dual of this Banach space?

Separability?

Dense subsets?

Is reflexive?

x∈I

p(x)

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Problem: The topological dual of this Banach space?

Separability?

Dense subsets?

Is reflexive?

x∈I

p(x)

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Problem: The topological dual of this Banach space?

Separability?

Dense subsets?

Is reflexive?

x∈I

p(x)

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Problem: The topological dual of this Banach space?

Separability?

Dense subsets?

Is reflexive?

x∈I

p(x)

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Friendly case p

₊

< ∞

Definition

L^p(·)_b := {f ∈ L^p(·): f has compact support}

This is dense iff p+< ∞!

Proposition

L^p(·) is separable iff p+< ∞.

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Friendly case p

₊

< ∞

Definition

L^p(·)_b := {f ∈ L^p(·): f has compact support}

This is dense iff p+< ∞!

Proposition

L^p(·) is separable iff p+< ∞.

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Friendly case p

₊

< ∞

Proposition: H¨older inequality If p+< ∞,

Z

I

f (x)g(x)dx ≤ Ckf k_p(·)kgk_p0(·)

where p⁰(x) is the H¨older conjugate of p(x), that is, 1/p + 1/p⁰ = 1 L^p⁰^(·) is the dual

When p₊< ∞, we know the dual L^p⁰^(·)!

And, moreover, if p⁰₊< ∞, the space is reflexive.

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Friendly case p

₊

< ∞

Z

I

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Friendly case p

₊

< ∞

Z

I

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Bad examples of functions in L

^p(·)

Note the easy equivalence

f ∈ L^p(·)⇐⇒ ∃λ : ρ(f /λ) < ∞.

Example 1

Let I = [1, ∞) and p(x) = bxc, f (x) =

αj, x ∈ [j, j + α^−j_j ], 0, else.

On the one hand, f ∈ L^p(·). Indeed, if λ > 1, Z ∞

1

f (x) λ

p(x)

dx =

∞

X

j=1

α^j_j λ^jα^−j_j =

∞

X

j=1

1 λ^j < ∞.

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Bad examples of functions in L

^p(·)

Note the easy equivalence

f ∈ L^p(·)⇐⇒ ∃λ : ρ(f /λ) < ∞.

Example 2 Let

f₂(x) =

1, x ∈ A, 0, else.

The set A is defined so that the following holds: |A ∩ [j, j + 1]| = 1/2 for all j ∈ N. We choose that:

[0, 1/2] ⊂ A and [1/2, 1] ∩ A = ∅.

([1, 5/4] ∪ [3/2, 7/4]) ⊂ A and ([5/4, 3/2] ∪ [7/4, 2]) ∩ A = ∅.

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Little brother: l

^p(·)

Let us think p : N → [1, +∞), and we can consider Definition

l^p(·) will denote the sequences of numbers such that

kf k := inf{λ > 0 :X

k

f (k) λ

p(k)

≤ 1} < ∞

Which one is ’easier’ to understand?

I p(k) = 1 + log(k).

I p(k) = k.

I p(k) = e^k.

The answer is the bigger the growth is, the ’easier’ !

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Little brother: l

^p(·)

l^p(·) will denote the sequences of numbers such that kf k := inf{λ > 0 :X

k

f (k) λ

p(k)

≤ 1} < ∞

I p(k) = 1 + log(k).

I p(k) = k.

I p(k) = e^k.

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Little brother: l

^p(·)

k

f (k) λ

p(k)

≤ 1} < ∞

I p(k) = 1 + log(k).

I p(k) = k.

I p(k) = e^k.

The answer is the bigger the growth is, the ’easier’ !

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Little brother: l

^p(·)

k

f (k) λ

p(k)

≤ 1} < ∞

I p(k) = 1 + log(k).

I p(k) = k.

I p(k) = e^k.

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Little brother: l

^p(·)

and l

^∞

Proposition: l^p(·) ⊆ l^∞

kf k∞≤ kf k_p(·)

Proposition: l^p(·) = l^∞

If p growths fast enough, that is, X

k

e^−p(k)< ∞

Then,

kf k_p(·)≤ Ckf k_∞

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Little brother: l

^p(·)

and l

^∞

k

e^−p(k)< ∞

Then,

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Little brother: l

^p(·)

and l

^∞

k

e^−p(k)< ∞

Then,

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Dual of Little brother: l

^p(·)

k

e^−p(k)< ∞

Then, l^p(·) is a renormalization of l^∞. And, consequently, its dual is equivalent to the dual of l^∞, that is, ba(N).

The p₊= ∞ appears!

This part is not separable, not reflexive...

We can find the little l^p(·) inside L^p(·). For example, considering the subspace of functions that are constant in the intervals (k, k + 1).

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Dual of Little brother: l

^p(·)

k

e^−p(k)< ∞

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Dual of Little brother: l

^p(·)

k

e^−p(k)< ∞

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Decomposition of the dual of the big brother: L

^p(·)

Proposition: the dual is bigger

(L^p(·))^∗ = L^p⁰^(·)⊕ Y where Y 6= 0.

The known part

L^p⁰^(·) comes from the subspace L^p_b⁰^(·).

This is what we can directly see, the function in compact intervals, where p₊ will be bounded.

Proposition: L^p_b⁰^(·)

f ∈ L^p_b⁰^(·) ⇐⇒ ρ(f /λ) < ∞ ∀λ > 0.

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Decomposition of the dual of the big brother: L

^p(·)

(L^p(·))^∗ = L^p⁰^(·)⊕ Y where Y 6= 0.

The known part

0

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Decomposition of the dual of the big brother: L

^p(·)

(L^p(·))^∗ = L^p⁰^(·)⊕ Y where Y 6= 0.

The known part

f ∈ L^p_b⁰^(·) ⇐⇒ ρ(f /λ) < ∞ ∀λ > 0.

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Decomposition of the dual of the big brother: L

^p(·)

(L^p(·))^∗ = L^p⁰^(·)⊕ Y where Y 6= 0.

The known part

0

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L

^∞

and L

^p⁰^(·)

Analogously as the discrete case, let us impose that Z

I

e^−p(x)dx < ∞

So, L^∞(I) ⊂ L^p⁰^(·). Note, that they are not the same (Example 1).

Proposition: L^∞

f ∈ L^∞⇐⇒ lim

k ρ((f − π_k(f ))/λ) < ∞ ∀λ > 0.

where,

πk(f )(x) :=







k k ≤ f (x) f (x) −k ≤ f (x) ≤ k

−k f (x) ≤ −k

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L

^∞

and L

^p⁰^(·)

I

e^−p(x)dx < ∞

Proposition: L^∞

f ∈ L^∞⇐⇒ lim

k ρ((f − π_k(f ))/λ) < ∞ ∀λ > 0.

where,

πk(f )(x) :=







k k ≤ f (x) f (x) −k ≤ f (x) ≤ k

−k f (x) ≤ −k

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L

^∞

and L

^p⁰^(·)

I

e^−p(x)dx < ∞

Proposition: L^∞

f ∈ L^∞⇐⇒ lim

k ρ((f − π_k(f ))/λ) < ∞ ∀λ > 0.

where,

πk(f )(x) :=







k k ≤ f (x) f (x) −k ≤ f (x) ≤ k

−k f (x) ≤ −k

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To see the L

^∞

at ’∞’

From the previous characterizations, we can deduce that L^p_b⁰^(·)⊂ L^∞. What happens in L^∞/L^p_b⁰^(·)?

Here, we have the pure finite additivity behaviour!

Hence,

(L^∞/L^p_b⁰^(·))^∗ = pba The dual of l^∞: change of language

The dual of l^∞ is ba(N), that can be decomposed as, ca + pba.

ca comes from the sequences from c₀₀ (L^p⁰^(·) is ca in the big case).

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To see the L

^∞

at ’∞’

Hence,

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To see the L

^∞

at ’∞’

Hence,

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Is that all?

Is L^∞ dense in L^p(·)?

If so, we would know the dual.

Proposition: not dense

L^∞ is not dense in L^p(·). Example 1 cannot be approximated by L^∞ functions.

Next step?

Maybe understand better l^p(·)(µ), to catch better what happens there at the ’triple’ infinity (at infinity, both the function and p are ’infinite’).

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Is that all?

Next step?

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Is that all?

Next step?

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Is that all?

Next step?

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On the togological dual of variable Lebesgue spaces