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
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).
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 , kf (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 ) < ∞,
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 ∞ (µ ∞ ),
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
about the form of the dual space L 1 (µ) ∗ , as well as the representation theorem for order continuous functionals over L ∞ (µ ∞ ), 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 (µ ∞ ) = 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 , kf (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 , kf (t) p
nd µ < ∞.
We have just shown that L p
n(µ∣ Σ∩I
n , k) ⊂ L p (⋅) (µ). Since µ∣ Σ∩I
n , kis 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
ku k p
k
−1
< ∞.
Proof. (i)⇒(ii). Put b k = ∣g(S k )∣ and suppose, to the contrary of (ii), that b
p
nn
kku n
kp
nk
−1
> k 2 for some subsequence (n k ).
Let
f (S n
k) = 1 b n
ku n
kand f (S j ) = 0 for j ∉ {n 1 , n 2 , . . . }.
Then
∫ S
f (t)
p (t)
d µ =
∞
∑
k =1
f (S n
k) p
nku n
k=
∞
∑
k =1
1 b n
kp
nk
u n
kp
nk
−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
ku 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
ku 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
ku k <
1 M
. Therefore,
( 1
j x k )
p
k∣g(S k )∣ p
ku k p
k= ( 1 j
x k )
p
ku k ∣g(S k )∣ p
ku 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
ku k p
k= ( 1
j x k )
p
ku k ∣g(S k )∣ p
ku k p
k−1
⩽ M(
1 j
x k )
p
ku 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
ku 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