Function spaces with variable exponents

Henning Kempka

September 22nd 2014

September 22nd 2014 · Henning Kempka 1 / 50 http://www.tu-chemnitz.de/

Outline1. Introduction & Motivation

First motivationSecond motivation

2. Variable exponent Lebesgue spacesDefinition of Lp(·)(Ω)Holders inequality

3. The Hardy-Littlewood maximal operator on Lp(·)(Ω)Basics about the maximal operatorBoundedness on Lp(·)(Ω)Interpolation in variable exponent Lebesgue spaces

4. Rubio de Francia extrapolationExtrapolation in variable Lebesgue spacesWavelet characterization of Lp(·)(Ω)

References

W. Orlicz: Uber konjugierte Exponentenfolgen. Studia Math. 3, (1931),200–212.O. Kovacik, J. Rakosnık: On spaces Lp(x) andW 1,p(x). Czechoslovak Math.J. 41(116)(1991), 592–618.

L. Diening: Maximal function on generalized Lebesgue spaces Lp(·).Mathematical Inequalities and Applications 7 no.2, (2004), 245–253.

L. Diening, P. Harjulehto, P. Hasto, M. Ruzicka: Lebesgue and Sobolevspaces with variable exponents, Lecture Notes in Mathematics2017,Springer, Heidelberg (2011).

D. V. Cruz-Uribe, A. Fiorenza: Variable Lebesgue spaces, Applied andNumerical Harmonic Analysis, Birkhauser/Springer, Heidelberg (2013).

Introduction & Motivation

Table of Contents

1. Introduction & MotivationFirst motivationSecond motivation

2. Variable exponent Lebesgue spaces

3. The Hardy-Littlewood maximal operator on Lp(·)(Ω)

4. Rubio de Francia extrapolation

Introduction & MotivationFirst motivation

Consider on R the function f(x) = |x|−1/3

f is well behaving, but f /∈ Lp(R) for every 1 ≤ p ≤ ∞I it either grows to quickly at the originI or it decays to slow at infinity

But: f ∈ L2([−2, 2]) and f ∈ L4(R \ [−2, 2])

Consider now g(x) = |x|−1/3 + |x− 1|−1/4:

Then g ∈ Lp([−2, 2]) for p < 3 and g ∈ Lq(R \ [−2, 2]) for q > 4But: We lost informatin on the local behaviour at the singularity x = 1!First solution: We can subdivide R even more: g ∈ L2([−1, 1/2]), g ∈ L3([1/2, 2]) and g ∈ L 9

2(R \ [−1, 2])

Consider now g(x) = |x|−1/3 + |x− 1|−1/4:

Then g ∈ Lp([−2, 2]) for p < 3 and g ∈ Lq(R \ [−2, 2]) for q > 4But: We lost informatin on the local behaviour at the singularity x = 1!First solution: We can subdivide R even more: g ∈ L2([−1, 1/2]), g ∈ L3([1/2, 2]) and g ∈ L 9

2(R \ [−1, 2])

Variable solutionIntroduce the exponent: p(x) = 9

2 −5/2

2|x|+1 , then

and p(0) = 2, p(1) = 113 and p(x)→ 9

2 for |x| → ∞.We have ∫

R|f(x)|p(x)dx <∞ and

∫R|g(x)|p(x)dx <∞.

Variable solutionIntroduce the exponent: p(x) = 9

2 −5/2

2|x|+1 , then

and p(0) = 2, p(1) = 113 and p(x)→ 9

2 for |x| → ∞.We have ∫

R|f(x)|p(x)dx <∞ and

∫R|g(x)|p(x)dx <∞.

Variable solution IIFurthermore, for the exponent: q(x) = 4− 2

2|x|+1

we have q(0) = 2, p(1) = 103 and p(x)→ 4 for |x| → ∞.

Then we have∫R|f(x)|q(x)dx <∞ and

∫R|g(x)|q(x)dx =∞.

Variable solution IIFurthermore, for the exponent: q(x) = 4− 2

2|x|+1

we have q(0) = 2, p(1) = 103 and p(x)→ 4 for |x| → ∞.

Then we have∫R|f(x)|q(x)dx <∞ and

∫R|g(x)|q(x)dx =∞.

Introduction & MotivationSecond motivation

Electrorheological fluidsI non-newtonian fluidsI change their viscosity dramatically (factor 1000) in reaction to an

electrical fieldI First observed in 1949 by WinslowI many applications:

I fast acting hydraulic valves & clutchesI ER brakes and shock absorbers

Electrorheological fluid, with and without electrical fieldSeptember 22nd 2014 · Henning Kempka 9 / 50 http://www.tu-chemnitz.de/

Electrorheological fluids IIThe model of these electrorheological fluids is connected via some nasty PDEs(with non standard growth conditions) to the

Dirichlet energy integral:∫

Ω|Du(x)|p(x)dx

where:I Du. . . symmetricpart of the gradientI p(x) = p(E(y, t)). . . is a function of the electrical field

Questions:I What is

∫Ω |f(x)|p(x)dx <∞?

I To which type of space does f belong?I Comparison to Lp(Ω)?I Further properties (inequalities, embeddings, . . . )

Electrorheological fluids IIThe model of these electrorheological fluids is connected via some nasty PDEs(with non standard growth conditions) to the

Dirichlet energy integral:∫

Ω|Du(x)|p(x)dx

where:I Du. . . symmetricpart of the gradientI p(x) = p(E(y, t)). . . is a function of the electrical field

Questions:I What is

∫Ω |f(x)|p(x)dx <∞?

I To which type of space does f belong?I Comparison to Lp(Ω)?I Further properties (inequalities, embeddings, . . . )

Variable exponent Lebesgue spaces

Table of Contents

1. Introduction & Motivation

2. Variable exponent Lebesgue spacesDefinition of Lp(·)(Ω)Holders inequality

Variable exponent Lebesgue spacesDefinition of Lp(·)(Ω)

For the rest Ω ⊂ Rn is an arbitrary but fixed open set.

Definition 1The class of variable exponents is

P(Ω) = p : Ω→ [1,∞] measurable .

For p ∈ P(Ω) and U ⊂ Ω we introduceI p−U = ess infx∈U p(x)

I p+U = ess supx∈U p(x)

I p− = p−ΩI p+ = p+

I Ω∞ = x ∈ Ω : p(x) =∞I Ω1 = x ∈ Ω : p(x) = 1I Ω∗ = x ∈ Ω : 1 < p(x) <∞

For the rest Ω ⊂ Rn is an arbitrary but fixed open set.

Definition 1The class of variable exponents is

P(Ω) = p : Ω→ [1,∞] measurable .

For p ∈ P(Ω) and U ⊂ Ω we introduceI p−U = ess infx∈U p(x)

I p+U = ess supx∈U p(x)

I p− = p−ΩI p+ = p+

I Ω∞ = x ∈ Ω : p(x) =∞I Ω1 = x ∈ Ω : p(x) = 1I Ω∗ = x ∈ Ω : 1 < p(x) <∞

Definition 2For p ∈ P(Ω) we define the convex modular

%p(·)(f) =

∫Ω\Ω∞

|f(x)|p(x)dx+ ess supx∈Ω∞

|f(x)|

and the variable exponent Lebesgue space Lp(·)(Ω) by

Lp(·)(Ω) = f : Ω→ C : %p(·)(f/λ) <∞ for some λ > 0.

Lemma 1 (Properties of the modular)

Let p ∈ P(Ω), then:(M1) For all f , we have %(f) ≥ 0 and %(|f |) = %(f);(M2) %(f) = 0⇐⇒ f(x) = 0 almost everywhere;(M3) If %(f) <∞, then |f(x)| <∞ for a.e. x ∈ Ω;(M4) % is convex, i.e.

%(αf + βg) ≤ α%(f) + β%(g) for all α, β ∈ [0, 1] with α+ β = 1;

(M5) If |f(x)| ≥ |g(x)| a.e, then %(f) ≥ %(g);(M6) Continuity property

If %(f/Λ) <∞ for some Λ > 0, then λ 7→ %(f/λ) is continuous anddecreasing on [Λ,∞)Further limλ→∞ %(f/λ) = 0.

Remark 1From the convexity (M4) of the modular it follows

%(αf) ≤ α%(f) for 0 < α < 1

%(αf) ≥ α%(f) for α > 1

Example 1

Let Ω = (1,∞), p(x) = x and f(x) = 1, then %(f) =∞ but for all λ > 1

%(f/λ) =

∫ ∞1

λ−xdx =1

λ log λ<∞.

Theorem 1For given p ∈ P(Ω), the space Lp(·)(Ω) is a normed vector space with theLuxemburg norm

‖f‖p(·) = infλ > 0 : %p(·)(f/λ) ≤ 1.

Proof.

I Lp(·)(Ω) is a vector spaceI (N1) ‖f‖p(·) ≥ 0 and ‖f‖p(·) = 0⇐⇒ f = 0

(N2) ‖αf‖p(·) = |α|‖f‖p for all α ∈ R(N3) ‖f + g‖p(·) ≤ ‖f‖p(·) + ‖g‖p(·)

Theorem 1For given p ∈ P(Ω), the space Lp(·)(Ω) is a normed vector space with theLuxemburg norm

‖f‖p(·) = infλ > 0 : %p(·)(f/λ) ≤ 1.

Proof.

I Lp(·)(Ω) is a vector spaceI (N1) ‖f‖p(·) ≥ 0 and ‖f‖p(·) = 0⇐⇒ f = 0

(N2) ‖αf‖p(·) = |α|‖f‖p for all α ∈ R(N3) ‖f + g‖p(·) ≤ ‖f‖p(·) + ‖g‖p(·)

Lemma 2Fix p ∈ P(Ω) then:

1. If ‖f‖p(·) ≤ 1, then %p(·)(f) ≤ ‖f‖p(·);2. If ‖f‖p(·) > 1, then %p(·)(f) ≥ ‖f‖p(·).

Proof.Remember:I %p(·)(f) =

∫Ω\Ω∞ |f(x)|p(x)dx+ ‖f‖L∞(Ω∞)

I ‖f‖p(·) = infλ > 0 : %p(·)(f/λ) ≤ 1I %p(·)(αf) ≤ α%p(·)(f) for 0 < α ≤ 1

Lemma 2Fix p ∈ P(Ω) then:

1. If ‖f‖p(·) ≤ 1, then %p(·)(f) ≤ ‖f‖p(·);2. If ‖f‖p(·) > 1, then %p(·)(f) ≥ ‖f‖p(·).

Proof.Remember:I %p(·)(f) =

∫Ω\Ω∞ |f(x)|p(x)dx+ ‖f‖L∞(Ω∞)

I ‖f‖p(·) = infλ > 0 : %p(·)(f/λ) ≤ 1I %p(·)(αf) ≤ α%p(·)(f) for 0 < α ≤ 1

Variable exponent Lebesgue spacesHolders inequality

Theorem 2 (Holders inequality)

Fix p ∈ P(Ω) and let p′ ∈ P(Ω) be defined by

p′(x)= 1 pointwise.

If f ∈ Lp(·)(Ω) and g ∈ Lp′(·)(Ω), then f · g ∈ L1(Ω) and∫Ω|f(x)g(x)|dx ≤ Kp(·)‖f‖p(·)‖g‖p′(·) , where

Kp(·) =

p−− 1

)‖χΩ∗‖∞ + ‖χΩ∞‖∞ + ‖χΩ1‖∞.

Proof.

1. If ‖f‖p(·) = 0 or ‖g‖p′(·) = 0, then f · g = 0.Therefore: ‖f‖p(·) > 0 and ‖g‖p′(·) > 0.

2. Norm is homogenuous: We assume ‖f‖p(·) = 1 and ‖g‖p′(·) = 1.3. We consider the integral

∫|fg|dx on the sets Ω∞, Ω1 and Ω∗

I x ∈ Ω∞: p(x) =∞ and p′(x) = 1I x ∈ Ω1: p(x) = 1 and p′(x) =∞I x ∈ Ω∗: 1 < p(x) <∞

Youngs inequality: a · b ≤ ap(x)

p(x) + bp′(x)

p′(x)

Proof.

p(x) + bp′(x)

p′(x)

Proof.

p(x) + bp′(x)

p′(x)

Further properties of Lp(·)(Ω)

Banach spaces Lp(·)(Ω) is complete Banach spaces

Equivalent norm Introduce |||f |||p(·) = sup%p′(·)(g)≤1

∣∣∣∣∫Ωf(x)g(x)dx

∣∣∣∣, then

4‖f‖p(·) ≤ |||f |||p(·) ≤ 4‖f‖p(·)

Dual spaces(Lp(·)(Ω)

)′= Lp′(·) ⇐⇒ p ∈ L∞(Ω)

For 1 < p− ≤ p+ <∞ the spaces Lp(·)(Ω) are reflexiveSeperability Lp(·)(Ω) is seperable⇐⇒ p ∈ L∞(Ω)

Further properties of Lp(·)(Ω)

Banach spaces Lp(·)(Ω) is complete Banach spaces

Equivalent norm Introduce |||f |||p(·) = sup%p′(·)(g)≤1

∣∣∣∣∫Ωf(x)g(x)dx

∣∣∣∣, then

4‖f‖p(·) ≤ |||f |||p(·) ≤ 4‖f‖p(·)

Dual spaces(Lp(·)(Ω)

)′= Lp′(·) ⇐⇒ p ∈ L∞(Ω)

For 1 < p− ≤ p+ <∞ the spaces Lp(·)(Ω) are reflexiveSeperability Lp(·)(Ω) is seperable⇐⇒ p ∈ L∞(Ω)

Differences to Lp(Ω)Translation invariance If p(·) is not constant, then there exists an f ∈ Lp(·)(Ω)and an h ∈ Rn such that f(·+ h) /∈ Lp(·)(Ω).

Example 2

Let Ω = (−1, 1), 1 ≤ r < s <∞ and define

f(x) =

x−1/s, for 0 < x < 1

0, for − 1 < x ≤ 0p(x) =

r, for 0 < x < 1

s, for − 1 < x ≤ 0.

Then f ∈ Lp(·)(−1, 1) but f(·+ h) /∈ Lp(·)(−1, 1) for every h ∈ (0, 1).

Youngs convolution inequality If there exists a constant c > 0 such that for allf ∈ Lp(·)(Ω) and all g ∈ L1(Ω)

‖f ∗ g‖p(·) ≤ c‖f‖p(·)‖g‖1, then p(·) = const almost everywhere.

Differences to Lp(Ω)Translation invariance If p(·) is not constant, then there exists an f ∈ Lp(·)(Ω)and an h ∈ Rn such that f(·+ h) /∈ Lp(·)(Ω).

Example 2

Let Ω = (−1, 1), 1 ≤ r < s <∞ and define

f(x) =

x−1/s, for 0 < x < 1

0, for − 1 < x ≤ 0p(x) =

r, for 0 < x < 1

s, for − 1 < x ≤ 0.

Then f ∈ Lp(·)(−1, 1) but f(·+ h) /∈ Lp(·)(−1, 1) for every h ∈ (0, 1).

Youngs convolution inequality If there exists a constant c > 0 such that for allf ∈ Lp(·)(Ω) and all g ∈ L1(Ω)

‖f ∗ g‖p(·) ≤ c‖f‖p(·)‖g‖1, then p(·) = const almost everywhere.

Generalizations of Lp(·)(Ω)Semimodular spaces On the vector space X there is defined a semimodular% : X → [0,∞] with some properties andX% = f ∈ X : %(f/λ) ≤ 1 for some λ > 0 is then a Semimodular space.Musielak-Orlicz spaces The space X(Ω) consists of all functions f such thatthere exists a λ > 0 with∫

(x,|f(x)|λ

)dx is finite.

The function Φ : Ω× [0,∞)→ [0,∞) is for almost every x ∈ Ω a Youngfunction Φ(x, ·).The function %Φ(f) =

∫Ω Φ(x, |f(x)|)dx defines a semimodular on the space

of measurable functions.If Φ(x, t) = χp(x)<∞(x)tp(x) + χp(x)=∞∞ · χ(1,∞)(t), thenX(Ω) = Lp(·)(Ω).

(x,|f(x)|λ

)dx is finite.

(x,|f(x)|λ

)dx is finite.

History of Lp(·)(Ω)

1. Early periodI 1931 Orlicz (he only wrote one paper about them and then studied Orlicz

spaces)I 1950 Nakano introduced modular spacesI 1961 TsenovI 1979 Sharapudinov, introduced the local log-Holder conditionI 1986 Zhikov applied Lp(·)(Ω) to problems in calculus of variations

2. Modern periodI 1991 Kovacik & Rakosnık, good overview of properties of Lp(·)(Ω)I ≥ 1990 A lot of papers on PDEs with non standard growth and the p(·)

Laplacian (Fan, Zhao, Harjulehto, Mingione,. . . )I ≥ 2000 Modelling of electrorheological fluids with Lp(·)(Ω) by RuzickaI 2004 Boundedness of the Hardy-Littlewood maximal operator (Diening)

History of Lp(·)(Ω)

1. Early periodI 1931 Orlicz (he only wrote one paper about them and then studied Orlicz

spaces)I 1950 Nakano introduced modular spacesI 1961 TsenovI 1979 Sharapudinov, introduced the local log-Holder conditionI 1986 Zhikov applied Lp(·)(Ω) to problems in calculus of variations

2. Modern periodI 1991 Kovacik & Rakosnık, good overview of properties of Lp(·)(Ω)I ≥ 1990 A lot of papers on PDEs with non standard growth and the p(·)

Laplacian (Fan, Zhao, Harjulehto, Mingione,. . . )I ≥ 2000 Modelling of electrorheological fluids with Lp(·)(Ω) by RuzickaI 2004 Boundedness of the Hardy-Littlewood maximal operator (Diening)

The Hardy-Littlewood maximal operator on Lp(·)(Ω)

Table of Contents

3. The Hardy-Littlewood maximal operator on Lp(·)(Ω)Basics about the maximal operatorBoundedness on Lp(·)(Ω)Interpolation in variable exponent Lebesgue spaces

The Hardy-Littlewood maximal operator on Lp(·)(Ω)Basics about the maximal operator

Definition 3Let f ∈ Lloc1 (Rn) then the Hardy-Littlewood maximal operatorMf is definedfor every x ∈ Rn by

Mf(x) = sup|Q|3x

∫Q|f(y)|dy,

where the supremum is taken over all cubes Q ∈ Rn containing x and sidesparallel to the axes.

Remark 2The cubes can be open or closed, they can be replaced by balls B or everythingcan be centered in x. All these definitions are up to constants the same.

Definition 3Let f ∈ Lloc1 (Rn) then the Hardy-Littlewood maximal operatorMf is definedfor every x ∈ Rn by

Mf(x) = sup|Q|3x

∫Q|f(y)|dy,

where the supremum is taken over all cubes Q ∈ Rn containing x and sidesparallel to the axes.

Remark 2The cubes can be open or closed, they can be replaced by balls B or everythingcan be centered in x. All these definitions are up to constants the same.

Theorem 3If 1 < p ≤ ∞ theM : Lp(Rn)→ Lp(Rn) is bounded, i.e.

‖Mf‖p ≤ c‖f‖p for all f ∈ Lp(Rn).

Proof.

easy M is strong (∞,∞), i.e.

‖Mf‖∞ ≤ ‖f‖∞.

hard M is weak (1, 1), i.e.

|x ∈ Rn :Mf(x) > t| ≤ c1‖f‖1t

ext. Marcinkiewicz interpolationSeptember 22nd 2014 · Henning Kempka 26 / 50 http://www.tu-chemnitz.de/

Proof.

‖Mf‖∞ ≤ ‖f‖∞.

|x ∈ Rn :Mf(x) > t| ≤ c1‖f‖1t

Proof.

‖Mf‖∞ ≤ ‖f‖∞.

|x ∈ Rn :Mf(x) > t| ≤ c1‖f‖1t

Proof.

‖Mf‖∞ ≤ ‖f‖∞.

|x ∈ Rn :Mf(x) > t| ≤ c1‖f‖1t

The Hardy-Littlewood maximal operator on Lp(·)(Ω)Boundedness on Lp(·)(Ω)

Conditions on p(·)Definition 4Let p ∈ P(Ω).

(i) p is said to be locally log Holder continuous, p ∈ C loclog(Ω), if there exists aconstant c0 such that

|p(x)− p(y)| ≤ c0

− log(|x− y|)for all x, y ∈ Ω with |x− y| < 1

(ii) p is log Holder continuous at infinity, p ∈ C∞log(Ω), if there existconstants c∞ and p∞ such that

|p(x)− p∞| ≤c∞

log(e+ |x|)for all x ∈ Ω.

We write p ∈ Plog(Ω) if 1/p ∈ C∞log(Ω) ∩ C loclog(Ω).September 22nd 2014 · Henning Kempka 27 / 50 http://www.tu-chemnitz.de/

Conditions on p(·)Definition 4Let p ∈ P(Ω).

(i) p is said to be locally log Holder continuous, p ∈ C loclog(Ω), if there exists aconstant c0 such that

|p(x)− p(y)| ≤ c0

− log(|x− y|)for all x, y ∈ Ω with |x− y| < 1

(ii) p is log Holder continuous at infinity, p ∈ C∞log(Ω), if there existconstants c∞ and p∞ such that

|p(x)− p∞| ≤c∞

log(e+ |x|)for all x ∈ Ω.

We write p ∈ Plog(Ω) if 1/p ∈ C∞log(Ω) ∩ C loclog(Ω).September 22nd 2014 · Henning Kempka 27 / 50 http://www.tu-chemnitz.de/

Lemma 3

(i) If p ∈ Plog(Ω), then p can be extented to p ∈ Plog(Rn) with the sameconstants c0, p∞, p

− and p+.(ii) If p ∈ P(Rn) with p+ <∞, then t.f.a.e.

(a) p ∈ Cloclog(Rn)

(b) There exists a constant c = c(n) such that for every cubeQ (or ballB) withx ∈ Q:

|Q|p(x)−p+Q ≤ c and |Q|p

−Q−p(x) ≤ c.

Lemma 3

(i) If p ∈ Plog(Ω), then p can be extented to p ∈ Plog(Rn) with the sameconstants c0, p∞, p

− and p+.(ii) If p ∈ P(Rn) with p+ <∞, then t.f.a.e.

(a) p ∈ Cloclog(Rn)

(b) There exists a constant c = c(n) such that for every cubeQ (or ballB) withx ∈ Q:

|Q|p(x)−p+Q ≤ c and |Q|p

−Q−p(x) ≤ c.

Theorem 4Let Ω ⊂ Rn and p ∈ Plog(Ω) with 1 ≤ p− ≤ p+ ≤ ∞, then

‖tχx∈Ω:Mf(x)>t‖p(·) ≤ c‖f‖p(·)

and if p− > 1, then

‖Mf‖p(·) ≤ c‖f‖p(·).

History of the theorem

2000 Michael Ruzicka conjectured in his book Electrorheological fluids:modeling and mathematical theory:

History of the theorem II2001 Lubos Pick and Michael Ruzicka provided a counterexample to the

boundedness, if p /∈ C loclog(Ω)

2002 His PhD student, Lars Diening, gave the first proof of the boundedness, forp+ <∞ and p is constant outside a large ball

≥2004 generalizations of the conditions used (p+ ≤ ∞ and C∞log(Ω) condition)2005 Andrei Lerner

p(x) = 2− a(1 + sin(log log(e+ |x|+ 1/|x|))) for small a > 0

Then p(·) is not continuous in x = 0 =⇒ p /∈ C loclog(Ω)

ButM : Lp(·)(Rn)→ Lp(·)(Rn) is bounded

History of the theorem II2001 Lubos Pick and Michael Ruzicka provided a counterexample to the

boundedness, if p /∈ C loclog(Ω)

2002 His PhD student, Lars Diening, gave the first proof of the boundedness, forp+ <∞ and p is constant outside a large ball

≥2004 generalizations of the conditions used (p+ ≤ ∞ and C∞log(Ω) condition)2005 Andrei Lerner

p(x) = 2− a(1 + sin(log log(e+ |x|+ 1/|x|))) for small a > 0

Then p(·) is not continuous in x = 0 =⇒ p /∈ C loclog(Ω)

ButM : Lp(·)(Rn)→ Lp(·)(Rn) is bounded

To prove Theorem 4

We prove Theorem 4 only in the special case 1 < p− ≤ p+ <∞, p ∈ C loclog(Ω)and |Ω| <∞.We need the following variable version of Jensens inequality:

Lemma 4Let p ∈ P(Ω) with p+ <∞ and p ∈ C loclog(Ω). Then there exists a constantc(p(·)) > 0 such that for all ‖f‖p ≤ 1

(Mf(x))p(x)

p− ≤ c(p(·))[(M|f(·)|

p(·)p−

)(x) + 1

]for all x ∈ Ω.

Proof.

1. Jensen inequality If q ∈ [1,∞] and f ∈ Lq(Ω), then for all Q ∈ Ω(1

∫Q|f(y)|dy

)q≤ 1

∫Q|f(y)|qdy since t 7→ tq is convex.

2. Set q(·) = p(·)p− , then q+ <∞ and q ∈ C loclog(Ω)

3. Let Qr = Q ⊂ Ω with x ∈ Q and l(Q) = r

r ≥ 1(

1|Q|∫Q|f(y)|dy

≤ c(p(·))

0 < r < 1(

1|Q|∫Q|f(y)|dy

≤ c(

1|Q|∫Q|f(y)|q(y)dy + 2

)4. Take supremum over all cubes Q with x ∈ Q.

Proof.

∫Q|f(y)|dy

)q≤ 1

r ≥ 1(

1|Q|∫Q|f(y)|dy

≤ c(p(·))

0 < r < 1(

1|Q|∫Q|f(y)|dy

≤ c(

1|Q|∫Q|f(y)|q(y)dy + 2

Proof.

∫Q|f(y)|dy

)q≤ 1

r ≥ 1(

1|Q|∫Q|f(y)|dy

≤ c(p(·))

0 < r < 1(

1|Q|∫Q|f(y)|dy

≤ c(

1|Q|∫Q|f(y)|q(y)dy + 2

Proof.

∫Q|f(y)|dy

)q≤ 1

r ≥ 1(

1|Q|∫Q|f(y)|dy

≤ c(p(·))

0 < r < 1(

1|Q|∫Q|f(y)|dy

≤ c(

1|Q|∫Q|f(y)|q(y)dy + 2

Proof of Theorem 4.

1. additional assumptions:I p ∈ Cloc

log(Ω)I 1 < p− ≤ p+<∞I |Ω| <∞

2. Since p+ <∞ andM is homogenuous it is enough to show:

%p(Mf) ≤ c(p(·)) for all f ∈ Lp(·)(Ω) with ‖f‖p(·) ≤ 1.

3. Set q(·) = p(·)p− and use

I (Mf)q(x)(x) ≤ c(p(·))((M|f(·)|q(·))(x) + 1)I M : Lp− → Lp− is bounded

Proof of Theorem 4.

log(Ω)I 1 < p− ≤ p+<∞I |Ω| <∞

Proof of Theorem 4.

log(Ω)I 1 < p− ≤ p+<∞I |Ω| <∞

The Hardy-Littlewood maximal operator on Lp(·)(Ω)Interpolation in variable exponent Lebesgue spaces

A positive interpolation result

Theorem 5 (Complex interpolation)

Let p0, p1 ∈ P(Ω), Θ ∈ (0, 1) and define

pΘ(x)=

1−Θ

p0(x)+

p1(x).

Then [Lp0(·)(Ω), Lp1(·)(Ω)

= LpΘ(·)(Ω).

Open question

This question for a Marcinkiewicz theorem in variable exponent Lebesguespaces was posed in 2004 by Diening, Hasto and Nekvinda.Question: Let T be a sublinear operator which is of weak type (π0(·), π0(·))and of weak type (π1(·), π1(·)). Is T then bounded from LπΘ(·) to LπΘ(·) with

πΘ(·)=

1−Θ

π0(·)+

π1(·)?

weak type (π(·), π(·)) means, there exist a constant such that∥∥ tχx∈Rn:|Tf(x)|>t∣∣Lπ(·)

∥∥ ≤ c∥∥f |Lπ(·)∥∥ ,

i.e. the operator T is bounded from Lπ(·) into Lπ(·),∞.

Open question

This question for a Marcinkiewicz theorem in variable exponent Lebesguespaces was posed in 2004 by Diening, Hasto and Nekvinda.Question: Let T be a sublinear operator which is of weak type (π0(·), π0(·))and of weak type (π1(·), π1(·)). Is T then bounded from LπΘ(·) to LπΘ(·) with

πΘ(·)=

1−Θ

π0(·)+

π1(·)?

weak type (π(·), π(·)) means, there exist a constant such that∥∥ tχx∈Rn:|Tf(x)|>t∣∣Lπ(·)

∥∥ ≤ c∥∥f |Lπ(·)∥∥ ,

i.e. the operator T is bounded from Lπ(·) into Lπ(·),∞.

Negative ResultIn general Marcinkiewicz Interpolation does not hold in the variable exponentsetting, ie.T . . . sublinear operatorT : Lπ0(·) → Lπ0(·),∞T : Lπ1(·) → Lπ1(·),∞Then in general it does not hold:T : Lπθ(·) → Lπθ(·) with 1/πθ(·) = (1− θ)/π0(·) + θ/π1(·)Idea for counterexample: Use usual Marcinkiewicz interpolation

H. Kempka, J. Vybıral: Lorentz spaces with variable exponents, Math. Nachr. 287no. 8-9 (2014), 938–954.

Negative ResultIn general Marcinkiewicz Interpolation does not hold in the variable exponentsetting, ie.T . . . sublinear operatorT : Lπ0(·) → Lπ0(·),∞T : Lπ1(·) → Lπ1(·),∞Then in general it does not hold:T : Lπθ(·) → Lπθ(·) with 1/πθ(·) = (1− θ)/π0(·) + θ/π1(·)Idea for counterexample: Use usual Marcinkiewicz interpolation

H. Kempka, J. Vybıral: Lorentz spaces with variable exponents, Math. Nachr. 287no. 8-9 (2014), 938–954.

Theorem 6 (Marcinkiewicz interpolation)

Let T be a sublinear operator which is bounded from Lp0 into Lq0,∞ and fromLp1 into Lq1,∞, where 0 < p0 6= p1 ≤ ∞ and 0 < q0 6= q1 ≤ ∞. Let0 < Θ < 1 and put

1−Θ

If p ≤ q,then T is also bounded from Lp into Lq .

There exist a sublinear operator T and 0 < p0 6= p1 ≤ ∞, 0 < q0 6= q1 ≤ ∞and 0 < θ < 1 such thatI T : Lp0([0, 1])→ Lq0,∞([0, 1])

I T : Lp1([0, 1])→ Lq1,∞([0, 1])

I and p > q and T is not bounded from Lp([0, 1]) to Lq([0, 1]).September 22nd 2014 · Henning Kempka 38 / 50 http://www.tu-chemnitz.de/

1−Θ

I T : Lp1([0, 1])→ Lq1,∞([0, 1])

1−Θ

I T : Lp1([0, 1])→ Lq1,∞([0, 1])

CounterexampleUse the counterexample T to usual Marcinkiewicz from above and define T by

T f(x) :=

T (χ[0,1]f)(x− 1), if x ∈ [1, 2]

0 if x ∈ [0, 1).

π0(x) :=

p0, x ∈ [0, 1)

q0, x ∈ [1, 2]and π1(x) :=

p1, x ∈ [0, 1)

q1, x ∈ [1, 2]

thenI T is weak type (π0(·), π0(·))I T is weak type (π1(·), π1(·))I but not strong type (πθ(·), πθ(·))

CounterexampleUse the counterexample T to usual Marcinkiewicz from above and define T by

T f(x) :=

T (χ[0,1]f)(x− 1), if x ∈ [1, 2]

0 if x ∈ [0, 1).

π0(x) :=

p0, x ∈ [0, 1)

q0, x ∈ [1, 2]and π1(x) :=

p1, x ∈ [0, 1)

q1, x ∈ [1, 2]

thenI T is weak type (π0(·), π0(·))I T is weak type (π1(·), π1(·))I but not strong type (πθ(·), πθ(·))

Rubio de Francia extrapolation

Table of Contents

4. Rubio de Francia extrapolationExtrapolation in variable Lebesgue spacesWavelet characterization of Lp(·)(Ω)

Rubio de Francia extrapolationExtrapolation in variable Lebesgue spaces

Definition 5A weight w : Rn → (0,∞) belongs to the Muckenhoupt class Ap(1 < p <∞), if

∫Qw(x)dx

∫Qw(x)1−p′dx

)p−1

≤ C (1)

for all cubes Q ⊂ Rn.w is an A1 weight ifMw(x) ≤ Cw(x) for a.e. x ∈ Rn andA∞ =

⋃1≤p<∞Ap.

The smallest constant in (1) is denoted by Ap(w).

Remark 3

I Ap ⊂ Aq for p < q

I M : Lp(w)→ Lp(w) if, and only if, w ∈ Ap.September 22nd 2014 · Henning Kempka 41 / 50 http://www.tu-chemnitz.de/

∫Qw(x)dx

∫Qw(x)1−p′dx

)p−1

≤ C (1)

⋃1≤p<∞Ap.

Remark 3

∫Qw(x)dx

∫Qw(x)1−p′dx

)p−1

≤ C (1)

⋃1≤p<∞Ap.

Remark 3

Theorem 7Given Ω ⊂ Rn and a family of measureable functionsF = f, g such that for ap0 ≥ 1 and allw ∈ A1 and all (f, g) ∈ F∫

Ω|f(x)|p0w(x)dx ≤ c1

∫Ω|g(x)|p0w(x)dx,

where c1 only depends onA1(w).Let p ∈ Plog(Ω) with p0 ≤ p− ≤ p+ <∞, then

‖f‖p(·) ≤ c(p(·))‖g‖p(·) for all (f, g) ∈ F .

Theorem 7Given Ω ⊂ Rn and a family of measureable functionsF = f, g such that for ap0 ≥ 1 and allw ∈ A1 and all (f, g) ∈ F∫

where c1 only depends onA1(w).Let p ∈ Plog(Ω) with p0 ≤ p− ≤ p+ <∞, then

Proof.

I Set s(·) = p(·)p0

, then 1 ≤ s− ≤ s+ <∞ and s ∈ Plog(Ω).I We use

1. ‖f‖p0

p(·) = ‖|f |p0‖s(·)

2. ‖f‖s(·) ∼ sup‖h‖s′(·)≤1

∣∣∫Ωf(x)h(x)dx

∣∣3. ‖Mh‖s′(·) ≤ A‖h‖s′(·)

I Define the Rubio de Francia operator

Rh(x) =

∞∑k=0

Mkh(x)

(2A)k, whereMk =M · · ·M andM0 = id.

Proof.

I Set s(·) = p(·)p0

1. ‖f‖p0

p(·) = ‖|f |p0‖s(·)

2. ‖f‖s(·) ∼ sup‖h‖s′(·)≤1

∣∣3. ‖Mh‖s′(·) ≤ A‖h‖s′(·)

Rh(x) =

∞∑k=0

Mkh(x)

Proof.

I Set s(·) = p(·)p0

1. ‖f‖p0

p(·) = ‖|f |p0‖s(·)

2. ‖f‖s(·) ∼ sup‖h‖s′(·)≤1

∣∣3. ‖Mh‖s′(·) ≤ A‖h‖s′(·)

Rh(x) =

∞∑k=0

Mkh(x)

Proof.

I The Rubio de Francia operator

Rh(x) =

∞∑k=0

Mkh(x)

(2A)khas the properties:

(a) |h| ≤ Rh(b) ‖Rh‖s′(·) ≤ 2‖h‖s′(·)(c) M(Rh) ≤ 2ARh and therforeRh is an A1 weight.

Remark 4If one uses Rubio de Francia interpolation in the constant exponent case andA1 ⊂ Ap we get the following version:If there exists a p0 ≥ 1 such that for all w ∈ Ap0∫

where c1 only depends on Ap0(w).Let p ∈ Plog(Ω) with 1 < p− ≤ p+ <∞, then

Remark 4If one uses Rubio de Francia interpolation in the constant exponent case andA1 ⊂ Ap we get the following version:If there exists a p0 ≥ 1 such that for all w ∈ Ap0∫

where c1 only depends on Ap0(w).Let p ∈ Plog(Ω) with 1 < p− ≤ p+ <∞, then

Rubio de Francia extrapolationWavelet characterization of Lp(·)(Ω)

Wavelet bases in Rn

Let D = Qjk = 2−j([0, 1)n + k) : j ∈ Z and k ∈ Zn be dyadic cubes.A finite set of functions Ψ = ψ1, . . . , ψL ⊂ L2(Rn) is called anorthonormal Waveletbasis, ifΨl

Q = ψlQjk(x) = 2jn2 ψl(2jx− k) : j ∈ Z, k ∈ Zn, 1 ≤ l ≤ L is an

ONB of L2(Rn).Define:

WΨf =

L∑l=1

∑Q∈D

∣∣∣⟨ f, ψlQ⟩∣∣∣2 |Q|−1χQ

∞∑j=−∞

∑k∈Zn

L∑l=1

∣∣∣⟨ f, ψlQjk⟩∣∣∣2 2jnχjk

Wavelet bases in Rn

Let D = Qjk = 2−j([0, 1)n + k) : j ∈ Z and k ∈ Zn be dyadic cubes.A finite set of functions Ψ = ψ1, . . . , ψL ⊂ L2(Rn) is called anorthonormal Waveletbasis, ifΨl

Q = ψlQjk(x) = 2jn2 ψl(2jx− k) : j ∈ Z, k ∈ Zn, 1 ≤ l ≤ L is an

ONB of L2(Rn).Define:

WΨf =

L∑l=1

∑Q∈D

∣∣∣⟨ f, ψlQ⟩∣∣∣2 |Q|−1χQ

∞∑j=−∞

∑k∈Zn

L∑l=1

∣∣∣⟨ f, ψlQjk⟩∣∣∣2 2jnχjk

A wavelet system is called admissible, if there exists an p0 ∈ (1,∞) such that

C−1w ‖f |Lp0(w)‖ ≤ ‖WΨf |Lp0(w)‖ ≤ Cw ‖f |Lp0(w)‖

for all w ∈ Ap0 .

Remark 5Admissible wavelet systems are:I Haar waveletsI Spline waveletsI Daubechies waveletsI Wavelets out of Multi resolution analysis

A wavelet system is called admissible, if there exists an p0 ∈ (1,∞) such that

C−1w ‖f |Lp0(w)‖ ≤ ‖WΨf |Lp0(w)‖ ≤ Cw ‖f |Lp0(w)‖

for all w ∈ Ap0 .

Remark 5Admissible wavelet systems are:I Haar waveletsI Spline waveletsI Daubechies waveletsI Wavelets out of Multi resolution analysis

Theorem 8Let p ∈ Plog(Rn) with 1 < p− ≤ p+ <∞ and Ψ be an admissible waveletsystem. Then for all f ∈ Lp(·)(Rn) we have

‖f‖p(·) ∼

∥∥∥∥∥∥∥ L∑l=1

∑Q∈D

∣∣∣⟨ f, ψlQ⟩∣∣∣2 |Q|−1χQ

1/2∥∥∥∥∥∥∥p(·)

Proof.Apply Theorem 7 and Remark 4 on

F1 = (|f |,WΨf) : f ∈ L∞ and suppf compact andF2 = (WΨf, |f |) : f ∈ L∞ and suppf compact.

Further use: L∞ ∩ suppf compact is dense in Lp(·)(Rn).September 22nd 2014 · Henning Kempka 48 / 50 http://www.tu-chemnitz.de/

Theorem 8Let p ∈ Plog(Rn) with 1 < p− ≤ p+ <∞ and Ψ be an admissible waveletsystem. Then for all f ∈ Lp(·)(Rn) we have

‖f‖p(·) ∼

∥∥∥∥∥∥∥ L∑l=1

∑Q∈D

∣∣∣⟨ f, ψlQ⟩∣∣∣2 |Q|−1χQ

1/2∥∥∥∥∥∥∥p(·)

Proof.Apply Theorem 7 and Remark 4 on

F1 = (|f |,WΨf) : f ∈ L∞ and suppf compact andF2 = (WΨf, |f |) : f ∈ L∞ and suppf compact.

Further use: L∞ ∩ suppf compact is dense in Lp(·)(Rn).September 22nd 2014 · Henning Kempka 48 / 50 http://www.tu-chemnitz.de/

Further applications of Rubio de Francia extrapolation

I Boundedness of the fractional maximal and the Riesz potential operatoron Lp(·)(Rn)

I Boundedness of Calderon-Zygmund integral operators on Lp(·)(Rn)

I Vector-valued maximal inequality, i.e. the maximal operator is bounded onLp(·)(`q).

Thank you for your attention!

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