A note on maximal operators related to Laplace-Bessel differential operators on variable exponent Lebesgue spaces

Definition 1

Given a function p ( ⋅ ) : R k , + n → [ 1 , ∞ ) , we say that p ( ⋅ ) is locally log-Hölder continuous, and denote this by p ( ⋅ ) ∈ P log ( R k , + n ) , if there exists a constant C 0 > 0 such that

We say that p ( ⋅ ) is log-Hölder continuous at infinity and denote this by p ( ⋅ ) ∈ P ∞ log ( R k , + n ) , if there exists a constant C ∞ such that

where p ∞ = lim x → ∞ p ( x ) > 1 .

Proposition 1

Let p ( ⋅ ) : R k , + n → [ 1 , ∞ ] be Lebesgue measurable and S denote the set of simple functions on R k , + n . If p + < ∞ , then S is a dense subset of L p ( ⋅ ) , γ ( R k , + n ) .

Theorem 1

[7] Let f be a function defined on R + n .

1. If f ∈ L 1 , γ ( R + n ) , then for every t > 0

   ∣ { x ∈ R + n : M γ f ( x ) > t } ∣ γ ≤ C t ∥ f ∥ L 1 , γ ( R + n ) ,

   where C is independent of x , r , t , and f .

2. If f ∈ L p , γ ( R + n ) , 1 < p ≤ ∞ , then M γ f ∈ L p , γ ( R + n ) and

   ∥ M γ f ∥ L p , γ ( R + n ) ≤ C ∥ f ∥ L p , γ ( R + n ) ,

   where C is independent of f .

Definition 2

[3] Let ℬ be a set of all Lebesgue measurable functions p ( ⋅ ) : R k , + n → ( 1 , ∞ ) . Then, the B -maximal operator M γ is bounded on L p ( ⋅ ) , γ ( R k , + n ) .

Theorem 2

[24] Let p ( ⋅ ) : R k , + n → ( 1 , ∞ ) be Lebesgue measurable with 1 < p − ≤ p + < ∞ . Then, the following conditions are equivalent:

1. p ( ⋅ ) ∈ ℬ ;

2. p ′ ( ⋅ ) ∈ ℬ ;

3. p ( ⋅ ) / q ∈ ℬ for some 1 < q < p − ;

4. ( p ( ⋅ ) / q ) ′ ∈ ℬ for some 1 < q < p − .

Proposition 2

[24] Let p ( ⋅ ) ∈ ℬ . We have a constant C > 0 so that for any B + ∈ B ,

Theorem 3

Let p ∈ P ∞ log ( R k , + n ) . Then,

where C is independent of f .

Proof

To prove the boundedness of the B -maximal operator on L p ( ⋅ ) , γ ( R k , + n ) , we will use the maximal function on a space of homogeneous type. Therefore, we will first introduce the fractional maximal function on a space of homogeneous type. Given a pseudo-metric space ( X , ρ ) and a positive measure μ . Then, we say that ( X , ρ , μ ) is a homogeneous-type space, if the measure μ satisfies the doubling condition

where C is independent of x and r > 0 , and B ( x , r ) = { y ∈ X : ρ ( x , y ) < r } .

Let ( X , ρ , μ ) be a homogeneous-type space and the maximal function defined as follows

It is well known that the maximal operator M μ is bounded on L p ( ⋅ ) ( X , μ ) [19, Corollary 1.6]. We shall use this result in the case X = R k , + n , ρ ( x , y ) = ∣ x − y ∣ , d μ ( x ) = ( x ′ ) γ d x . It is clear that this measure satisfies the doubling condition (1) (see [14]). Also in [3,7,24], it was proved that

From (5) and the boundedness of the maximal operator M μ on L p ( ⋅ ) ( X , μ ) , we obtain the boundedness of the B -maximal operator on L p ( ⋅ ) , γ ( R k , + n ) , i.e.,

Thus, the proof of theorem is completed.□

Theorem 4

Let p ( ⋅ ) ∈ P ( R k , + n ) . If p − = 1 , then B -maximal operator is not bounded on L p ( ⋅ ) , γ ( R k , + n ) variable exponent Lebesgue spaces.

Proof

For m ∈ N , take s m such that

For all m , since p − = 1 , the set

has positive measure. From Lebesgue differentiation theorem, for each χ E m ,

Particularly, if 0 < r ≤ R m , then there exists 0 < R m < 1 such that

Let B m = B + ( 0 , R m ) and define

Now, we will prove f m ∈ L p ( ⋅ ) , γ ( R k , + n ) and

First, note that since R m < 1 and − n − k − ∣ γ ∣ + 1 m + 1 < 0 ,

Then, we will use the equivalent definition of the B -maximal operator and averages over balls. Let x ∈ B m ∩ E m and take r = ∣ x ∣ ≤ R m . Then, we get

Let δ m = 2 − ( m + 1 ) . Then, we have

Therefore, since ∣ x ∣ − n − k − ∣ γ ∣ + 1 m + 1 is radially decreasing and from (3), since ∣ B + ( 0 , r ) ∩ E m ∣ γ ≥ ( 1 − 2 − ( n + k + ∣ γ ∣ ) ( m + 1 ) ) ∣ B + ( 0 , r ) ∣ γ , we get

If x ∉ B m ∩ E m , then this inequality is trivial. Therefore, ∥ M γ f m ∥ p ( ⋅ ) , γ ≥ C ( m + 1 ) ∥ f m ∥ p ( ⋅ ) , γ . Thus, when p − = 1 , we obtain that B -maximal operator is not bounded on L p ( ⋅ ) , γ ( R k , + n ) variable exponent Lebesgue spaces.□

Example 1

Let

and f ( x ) = ∣ x ∣ − 2 / 5 χ ( 0 , 1 ) ( x ) . Then, since ∣ x ∣ − 4 / 5 χ ( 0 , 1 ) ( x ) ∈ L 1 , γ ( R + ) , f ∈ L p ( ⋅ ) , γ ( R + ) . Moreover, M γ f ∉ L p ( ⋅ ) , γ ( R + ) . If x > 1 , then we have

Therefore, ϱ γ ( M γ f ) = ∞ , and thus, M γ f ∉ L p ( ⋅ ) , γ ( R + ) .

Theorem 5

Let 0 ≤ α < n + ∣ γ ∣ , f ∈ L p ( ⋅ ) , γ ( R k , + n ) , and p ( ⋅ ) : R k , + n → [ 1 , ∞ ) be such that 1 < p − ≤ p + < n + ∣ γ ∣ α and p ( ⋅ ) ∈ L H ( R k , + n ) . Define q ( ⋅ ) : R k , + n → [ 1 , ∞ ) by 1 p ( x ) − 1 q ( x ) = α n + ∣ γ ∣ . Then,

where C is a constant independent of f .

Proof

We use the fractional maximal function on a space of homogeneous type. Therefore, we will firstly introduce the fractional maximal function on a space of homogeneous type. Let ( X , ρ ) be a pseudo-metric space, ( X , ρ , μ ) be a homogeneous type space, and B ( x , r ) be as above. Then, the fractional maximal function is defined as:

The fractional maximal operator M μ β f satisfies the strong type inequality for 1 < p − < 1 β , 1 p ( x ) − 1 q ( x ) = β (see [25,26]), i.e.,

To complete the proof of Theorem 5, we will use the following statements. In the case X = R k , + n , ρ ( x , y ) = ∣ x − y ∣ , β = α n + ∣ γ ∣ , 0 ≤ α < n + ∣ γ ∣ , and d μ ( x ) = ( x ′ ) γ d x . It is clear that this measure satisfies the doubling condition (1).

It is well known that

and

for 1 ≤ k ≤ n and 1 ≤ i ≤ k (see [7]).

Now we will show that

Consider the definition of the B -fractional maximal function

Then, we can write

where

and χ B + ( 0 , r ) is the characteristic function of χ B + ( 0 , r ) ⊂ R k , + n . Note that T y χ B + ( 0 , r ) ( x ) = 0 for every y ∈ R k , + n ⧹ B + ( x , r ) , i.e., T y χ B + ( 0 , r ) ( x ) is supported in the ball B + ( x , r ) . Then, we have