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vol. 55, no. 2 (2015), 119–126

Landau-type theorem for variable Lebesgue spaces

Lech Maligranda and Witold Wnuk

Summary. We describe, using elementary methods, the Köthe dual of variable Lebesgue spaces L

p(⋅)

, called also Nakano spaces, independenly for p(⋅) ∈ (1, ∞) and p(⋅) ∈ (0, 1). The case when p(⋅) ∈ [1, ∞] is also included.

Keywords Köthe dual spaces;

associate spaces;

Nakano spaces;

variable Lebesgue spaces

MSC 2010

46E30; 46B20; 46B42 Received: 2016-02-24, Accepted: 2016-03-11

Dedicated to Professor Henryk Hudzik on the occasion of his 70th birthday.

1. Introduction

In 1907, E. Landau [15] proved the reverse Hölder inequality: if (x n y n ) n =1 ∈ ℓ 1 for all sequences (x n ) ∈ ℓ p (1 < p < ∞), then ( y n ) ∈ ℓ q , where q is the conjugate exponent of p, that is 1/ p + 1/q = 1. Landau’s original proof is based on the classical Dini’s theorem that if a n ⩾ 0 and ∑ 1 a n = ∞, then ∑ 1 (a n /s n ) = ∞ and ∑ 1 (a n /s n 1+ε

) < ∞ for any ε > 0, where s n is the n-th partial sum s n = ∑

n

k =1 a k (details of his proof can be found, for example, in [3, I, pp. 529–531; 8, pp. 120–121; 12, pp. 1–2]). New proofs and generalizations were given in [5; 9; 10; 14; 16; 18; 19; 21; 23, pp. 237–238; 24–26]. Today we have more tools with which to check Landau’s theorem – the uniform boundedness principle (cf. [1, p. 179;

Lech Maligranda, Department of Engineering Sciences and Mathematics, Luleå University of Technology, SE-971 87 Luleå, Sweden (e-mail: lech.maligranda@ltu.se)

Witold Wnuk, Faculty of Mathematics and Computer Science, A. Mickiewicz University, Umultowska 87, 61-614 Poznań, Poland (e-mail: wnukwit@amu.edu.pl)

DOI 10.14708/cm.v55i2.1119 © 2015 Polish Mathematical Society

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4, p. 2524; 20, pp. 137–138]), or the general form of continuous linear functionals on ℓ p . In fact, Landau described the form of the Köthe dual space of ℓ p .

Let us recall that if E is an ideal in the space L 0 (µ) = L 0 (S , Σ, µ) of measurable func- tions with respect to the σ -finite measure space (S , Σ, µ) (i.e. E is a linear subspace, and

∣ f (s)∣ ⩽ ∣g(s)∣ µ-almost everywhere on S and f ∈ L 0 (µ), g ∈ E imply f ∈ E; additionally, E contains a function that is strictly positive µ-a.e.), then the Köthe dual (or associated space) E of E is defined as follows

E = {g ∈ L 0 (µ) ∶ f g ∈ L 1 (µ) for all f ∈ E}.

The space E coincides with the space of order continuous functionals F on E, i.e. f α ↓ 0 ⇒ F ( f α ) → 0 (see [28, Theorem 86.3]).

Our notation is standard. Let us only explain that χ A denotes the characteristic func- tion of the set A and f ∣ A denotes the restriction of the function f to A.

2. Results

We will devote attention to a special type of spaces E called variable (exponent) Lebesgue spaces, or Nakano spaces (see [6; 7, Chapter 3; 17, Chapter 3a; 22, pp. 281–284; 23, Chap- ter XI § 89]). For a measurable function p(⋅)∶ S → (0, ∞) we define the variable Lebesgue space as follows

L p (⋅) (µ) = { f ∈ L 0 (µ) ∶ ∃ λ >0 I p (⋅) (λ f ) = ∫

S ∣λ f (t)∣ p (t) d µ < ∞}.

The space L p (⋅) (µ) is an F -space (i.e. a complete metrizable topological vector space) with respect to an F -norm given by the formula

∥ f ∥ = inf {λ > 0 ∶ I p (⋅) ( f /λ) ⩽ λ}.

If p(t) ∈ [1, ∞) (µ-a.e.), then the topology generated by ∥ ⋅ ∥ is equivalent to the topology associated with the Luxemburg–Nakano norm

∥ f ∥ p (⋅) = inf {λ > 0 ∶ I p (⋅) ( f /λ) ⩽ 1}. (1) Applying Orlicz’s idea from [24] with slight modifications, we can easily obtain the following result (compare [6, Proposition 2.79] and [7, Theorem 3.2.13]).

2.1. Theorem. Let p(t) ∈ (1, ∞) and let q(⋅)∶ S → (1, ∞) be such that p (t) 1 + q (t) 1 = 1 for all t ∈ S. Then the following statements are equivalent:

(i) f g ∈ L 1 (µ) for every f ∈ L p (⋅) (µ).

(ii) g ∈ L q (⋅) (µ).

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Proof. (i)⇒(ii). Since the measure space (S , Σ, µ) is σ -finite, it follows that there exists a sequence (S k ) of sets with finite positive measures and increasing to S. The function ∣g∣

determines a linear functional φ on L p (⋅) (µ) by φ( f ) = ∫ S f ∣g∣ d µ. Since φ is positive (i.e.

φ transforms nonnegative functions onto nonnegative numbers), then it is continuous (see [2, Theorem 5.19]). For fixed arbitrary η > 0, let

f (t) = (

∣g(t)∣

∥φ∥ p (⋅) + η )

q(t) p(t)

. Clearly,

q(t) p(t)

+ 1 = 1 p(t) − 1

+ 1 = p(t) p(t) − 1

= q(t), and so

1

∥φ∥ p (⋅) + η

f (t)∣g(t)∣ = (

∣g(t)∣

∥φ∥ p (⋅) + η )

q (t)

. We claim that

∫ S f (t) p (t) d µ ⩽ 1.

Indeed, otherwise we can find natural numbers N , k such that for A N ,k = {t∶ f (t) p (t) ⩽ N } ∩ S k

we obtain

1 < ∫

A

N , k

f (t) p (t) d µ = ρ < ∞.

Hence, by the convexity of (⋅) p (t) ,

A

N , k

( f (t)

ρ )

p (t)

d µ ⩽ 1 ρ ∫ A

N , k

f (t) p (t) d µ = 1, and so ∥ f χ A

N , k

∥ p (⋅) ⩽ ρ. The equality (2) implies

ρ = ∫ A

N , k

f (t) p (t) d µ = ∫ A

N , k

(

∣g(t)∣

∥φ∥ p (⋅) + η )

q (t)

d µ = ∫ A

N , k

1

∥φ∥ p (⋅) + η f ∣g∣ d µ

= 1

∥φ∥ p (⋅) + η φ( f χ A

N , k

) ⩽ 1

∥φ∥ p (⋅) + η

∥φ∥ p (⋅) ∥ f χ A

N , k

∥ p (⋅) ⩽

∥φ∥ p (⋅)

∥φ∥ p (⋅) + η ρ, a contradiction because ρ > 0. Since (2) holds, then f ∈ L p (⋅) (µ), which implies

∫ S (

∣g(t)∣

∥φ∥ p (⋅) + η )

q (t)

d µ = ∫

S

1

∥φ∥ p (⋅) + η

f ∣g∣ d µ = 1

∥φ∥ p (⋅) + η

φ( f ) < ∞,

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i.e. g ∈ L q (⋅) (µ). This means that (L p (⋅) ) ⊂ L q (⋅) . Moreover, ∥g∥ q (⋅) ⩽ ∥φ∥ p (⋅) + η, and so

∥g∥ q (⋅) ⩽ sup

∥ f ∥

p(⋅)

⩽1 ∫

S ∣ f (t)g(t)∣d µ = ∥g∥ p (⋅) .

(ii)⇒(i). Fix f ∈ L p (⋅) (µ) and let λ f , λ g > 0 be such that ∫ S (λ f ∣ f (t)∣) p (t) d µ < ∞ and

∫ S (λ g ∣g(t)∣) q (t) d µ < ∞. Remembering that p(t), q(t) > 1, by Young’s inequality for fixed t ∈ S, we obtain

uv ⩽ u p (t)

p(t) +

v q (t) q(t)

⩽ u p (t) + v q (t) for all u, v ⩾ 0.

Thus,

λ f λ g ∣ f (t)∣ ∣g(t)∣ ⩽ (λ f ∣ f (t)∣) p (t) + (λ g ∣g(t)∣) q (t)

for all t ∈ S, i.e. f g ∈ L 1 (µ). Moreover, ∥ f g∥ 1 ⩽ 2∥ f ∥ p (⋅) ∥g∥ q (⋅) and so L q (⋅) ⊂ (L p (⋅) ) with

∥g∥ p (⋅) ⩽ 2 ∥g∥ q (⋅) .

We can extend the definition of the space L p (⋅) for functions p taking the values in [1, ∞] or in (0, ∞]. If we consider a measurable function p(⋅)∶ S → [1, ∞], then

I p (⋅) ( f ) = ∫

S ∖P

∣ f (t)∣ p (t) d µ + ess sup

t ∈P

∣ f (t)∣,

where P = {t ∈ S ∶ p(t) = ∞}, is a convex modular (cf. [6, p. 17] and [7, p. 74]; see also [16,22,23]), and with the norm (1) the space L p (⋅) is a Banach ideal space. The first ma- thematician who allowed this possibility was probably Sharapudinov [26] (Nakano in [23]

allowed it also but for a little different modular).

It is necessary to consider the sets P 1 = {t ∈ S ∶ p(t) = 1} and P 0 = S ∖ (P 1 ∪ P ).

If for t ∈ S we define the conjugate measurable function q(⋅)∶ S → [1, ∞] by

q(t) =

⎧ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎩

p (t)

p (t)−1 if t ∈ P 0 , 1 if t ∈ P ,

∞ if t ∈ P 1 ,

then Theorem 1 is still true. In fact, if L p (⋅) (µ) is generated by a function p taking values in [1, ∞], then

L p (⋅) (µ) = L p

0

(⋅) (µ 0 ) + L 1 (µ 1 ) + L ),

where p 0 means the restriction of p to P 0 and µ i , i ∈ {0, 1, ∞} are the restrictions of µ

to σ -algebras Σ ∩ P i = {A ∩ P i ∶ A ∈ Σ}. Hence, Theorem 2.1 and the classical result

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about the form of the dual space L 1 (µ) , as well as the representation theorem for order continuous functionals over L ), imply that f g ∈ L 1 (µ) for every f ∈ L p (⋅) (µ) if and only if g ∈ L q

0

(⋅) (µ 0 ) + L (µ 1 ) + L 1 ) = L q (⋅) (µ) (clearly q 0 denotes the restriction of q to P 0 ).

Now we investigate the spaces L p (⋅) (µ) generated by functions p(t) ∈ (0, 1). We have to consider two cases: the atomless and the purely atomic measures µ.

2.2. Theorem. Let p(t) ∈ (0, 1) and let µ be atomless. Then the following statements are

equivalent:

(i) f g ∈ L 1 (µ) for every f ∈ L p (⋅) (µ).

(ii) g = 0 µ-almost everywhere.

Proof. (i)⇒(ii). Let the sequence (S k ) have the same meaning as in the proof of The- orem 2.1. Put I n ,k = {t ∈ S k ∶ p(t) < p n }, where p n = 1 − 1

2 n . For every k there exists n k such that µ(I n ,k ) > 0 for n ⩾ n k . If n ⩾ n k and 0 ⩽ f ∈ L p

n

(µ∣ Σ∩I

n , k

), then

∫ I

n , k

f (t) p (t) d µ = ∫

I

n , k

∩{t∶ f (t)⩽1} f (t) p (t) d µ + ∫

I

n , k

∩{t∶ f (t)>1} f (t) p (t) d µ

⩽ µ(S k ) + ∫

I

n , k

f (t) p

n

d µ < ∞.

We have just shown that L p

n

(µ∣ Σ∩I

n , k

) ⊂ L p (⋅) (µ). Since µ∣ Σ∩I

n , k

is atomless and p n ∈ (0, 1) then, by the classical Day’s result, g∣ I

n , k

= 0 µ-a.e. for n ⩾ n k (see [2, Theorem 5.24]

or [16, p. 41]). Finally, g = 0 µ-a.e. because S = ⋃ n ,k=1 I n ,k .

The next result concerns variable Lebesgue spaces over purely atomic measures, and our proof of this result is inspired by [27, Lemma 4 and Theorem 7]. Let us recall that a measurable function is (almost everywhere) constant on an atom.

2.3. Theorem. Let p(t) ∈ (0, 1) and let µ be purely atomic. Assume that (S k ) k =1 is a sequence of atoms. If µ(S k ) = u k and p(S k ) = p k (k = 1, 2, . . .), then the following statements are equivalent:

(i) f g ∈ L 1 (µ) for every f ∈ L p (⋅) (µ).

(ii) sup k ∈N ∣g(S k )∣ p

k

u k p

k

−1

< ∞.

Proof. (i)⇒(ii). Put b k = ∣g(S k )∣ and suppose, to the contrary of (ii), that b

p

n

n

kk

u n

k

p

n

k

−1

> k 2 for some subsequence (n k ).

Let

f (S n

k

) = 1 b n

k

u n

k

and f (S j ) = 0 for j ∉ {n 1 , n 2 , . . . }.

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Then

∫ S

f (t)

p (t)

d µ =

k =1

f (S n

k

) p

nk

u n

k

=

k =1

1 b n

k

p

n

k

u n

k

p

n

k

−1 <

k =1

1 k 2

< ∞,

i.e. f ∈ L p (⋅) (µ), but ∣ f (S n

k

)g(S n

k

)∣µ(S n

k

) = f (S n

k

)b n

k

u n

k

= 1 for all k, and hence f g ∉ L 1 (µ), a contradiction.

(ii)⇒(i). Let sup k ∈N ∣g(S k )∣ p

k

u k p

k

−1

< M < ∞ and fix f ∈ L p (⋅) (µ). Put x k = ∣ f (S k )∣. We can choose a natural number j such that

∫ S ( 1 j

f (t))

p (t)

d µ = ∑

k

( 1 j

x k )

p

k

u k <

1 M

. Therefore,

( 1

j x k )

p

k

∣g(S k )∣ p

k

u k p

k

= ( 1 j

x k )

p

k

u k ∣g(S k )∣ p

k

u k p

k

−1

< 1.

Hence, 1 j x k ∣g(S k )∣u k < 1, and we obtain 1

j

x k ∣g(S k )∣u k ⩽ ( 1

j x k )

p

k

∣g(S k )∣ p

k

u k p

k

= ( 1

j x k )

p

k

u k ∣g(S k )∣ p

k

u k p

k

−1

⩽ M(

1 j

x k )

p

k

u k . Thus

∫ S ∣ f (t)g(t)∣ d µ = j ∑

k

1 j

x k ∣g(S k )∣u k ⩽ jM ∑

k

( 1

j x k )

p

k

u k < j, i.e. f g ∈ L 1 (µ).

If (S , Σ, µ) is a σ -finite measure space, then there exists a sequence (A k ) of µ-atoms such that for S 0 = ⋃ k A k the measure µ∣ Σ ∩(S∖S

0

) is atomless, or else µ(Σ ∩ (S ∖ S 0 )) = {0}.

Considering a Σ-measurable function p(⋅)∶ S → (0, ∞), we can separate three sets:

P <1 = {t ∈ S ∶ p(t) < 1}, P 1 = {t ∈ S ∶ p(t) = 1}, P >1 = {t ∈ S ∶ p(t) > 1}.

Putting together the theorems presented above we obtain the following result.

2.4. Theorem. Let p(⋅)∶ S → (0, ∞) be a Σ-measurable function. For a σ -finite measure µ

the following statements are equivalent:

(i) f g ∈ L 1 (µ) for every f ∈ L p (⋅) (µ).

(ii) g χ P

>1

∈ L q (⋅) (µ∣ Σ∩P

>1

), where q(⋅)∶ P >1 → (1, ∞) satisfies p (t) 1 + q (t) 1 = 1 for t ∈ P >1 ; mo- reover, g χ P

1

∈ L (µ∣ Σ∩P

1

) and sup k ∣g(A k ∩ P <1 )∣ p (A

k

∩P

<1

) µ(A k ∩ P <1 )

p (A

k

∩P

<1

)−1

<

∞ and g∣ (P

<1

)∖S

0

= 0 µ-a.e.

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Finally, we should mention that quite recently in a Kamińska–Kubiak paper [13, Ap- pendix A] appeared the proof of the Köthe duality of the Musielak–Orlicz spaces L M (Mu- sielak–Orlicz spaces include Orlicz spaces and variable Lebesgue spaces) in the case when the function M∶ S × [0, ∞) → [0, ∞] is convex in the second variable, which means that the space L M is a Banach space. The proof in [13] uses a Hudzik–Maligranda [11] result on Orlicz functions characterizing equality in Young’s inequality.

Acknowledgements Witold Wnuk acknowledges the support of The National Center of

Science, Poland, Grant no. N N 201 605340.

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