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The dimension-free estimate for the truncated maximal operator

  • Xudong Nie and Panwang Wang EMAIL logo
Published/Copyright: November 25, 2022
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Open Mathematics
From the journal Open Mathematics Volume 20 Issue 1

Abstract

We mainly study the dimension-free L p -inequality of the truncated maximal operator

M n a f ( x ) = sup t > 0 1 ∣ B a 1 ∣ ∫ B a 1 f ( x − t y ) d y ,

where B a 1 = { x : a ≤ ∣ x ∣ ≤ 1 } . When 0 ≤ a < 1 , we prove that ‖ M n a ‖ L p ( R n ) ≤ C ( p ) ‖ f ‖ L p ( R n ) for p > n / ( n − 1 ) . When a = 1 , we prove that ‖ M n 1 ‖ L p ( R n ) ≤ C ( p ) ‖ f ‖ L p ( R n ) for p ≥ 2 .

Keywords: maximal function; dimension-free estimate; Fourier multiplier
MSC 2010: 42B20; 42B25

1 Introduction

Let f ∈ S ( R n ) be a rapidly decreasing function defined on R n . For 0 ≤ a < 1 , define the truncated maximal operator

M n a f ( x ) = sup t > 0 1 ∣ B a 1 ∣ ∫ B a 1 f ( x − t y ) d y ,

where B a 1 ≔ { x : a ≤ ∣ x ∣ ≤ 1 } and ∣ B a 1 ∣ denote the volume of B a 1 . Conveniently, we denote the unite sphere S n − 1 by B 1 1 . Define

M n 1 f ( x ) = sup t > 0 1 ∣ B 1 1 ∣ ∫ B 1 1 f ( x − t y ) d σ ( y ) ,

where d σ ( y ) denotes the usual measure of the unite sphere and ∣ B 1 1 ∣ denotes its total mass. Obviously, M n 0 is the Hardy-Littlewood maximal operator and M n 1 is the sphere maximal operator. We mainly study the dimension-free L p -boundedness of the truncated maximal operators M n a for 0 ≤ a ≤ 1 . Stein first studied the sphere maximal operator M n 1 . In [1], he pointed out that if p > n / ( n − 1 ) , then

∥ M n 1 f ∥ L p ( R n ) ≤ C ( p , n ) ∥ f ∥ L p ( R n ) ,

where C ( p , n ) denotes a constant depending on p and n . The case for n = 2 was proved by Bourgain [2]. Later many authors tried to generalize the spherical maximal function to other hyper-surface. In [3], Sogge and Stein pointed out that if the Gaussian curvature of S does not vanish of infinite order at each point of S , then there exists a 2 < P 0 ( S ) < ∞ such that the corresponding surface maximal operator is strongly L p -bounded for p > p 0 ( S ) . Moreover, when the surface has at least one non-vanishing principal curvature everywhere in R n , then the maximal operator maps L p to L p for p > 2 [4]. More related works can be seen in [5,6, 7,8,9, 10,11,12, 13,14] and the references therein.

When a = 0 , M n a is the classical Hardy-Littlewood maximal operator. According to the result of [1], Stein [15] obtained that there exists a constant C ( p ) independent of n such that

(1) ∥ M n 0 f ∥ L p ( R n ) ≤ C ( p ) ∥ f ∥ L p ( R n )

for p > 1 . Analogously, many authors turned to focus on generalizing the dimension-free estimate for the maximal function to the general case. Bourgain [16] studied the maximal operator over central symmetric convex sets and obtained a similar result as (1) for p = 2 . Later, Bourgain [17] and Carbery [18] generalized this result for all p > 3 / 2 in different ways independently. For some special convex set such as q -balls, 1 ≤ q < ∞ , Müller [19] extended the L p bound to every p > 1 . Recently, Bourgain [20] showed that the strong type constant can be bounded by a constant, which is independent of the dimension for the Hardy-Littlewood maximal operator over cubes for all p > 1 . In this article, we will study the dimension-free estimate of the truncated maximal operator M n a , generalizing the results of Hardy-Littlewood maximal operators and sphere maximal operators proved in [1].

Theorem 1.1

Let f ∈ S ( R n ) . If 0 ≤ a < 1 , then there exists a constant C ( p ) independent of n and a such that

∥ M n a f ∥ L p ( R n ) ≤ ( 1 − a ) − 1 C ( p ) ∥ f ∥ L p ( R n ) ,

for p > 1 . When p > n / ( n − 1 ) ,

∥ M n a f ∥ L p ( R n ) ≤ C ( p ) ∥ f ∥ L p ( R n ) .

If a = 1 , then

(2) ∥ M n 1 f ∥ L p ( R n ) ≤ C ( p ) ∥ f ∥ L p ( R n ) ,

for p ≥ 2 .

2 The proof of main theorem for the case a = 1

By interpolation theorem, we only need to prove the case p = 2 . This result follows from the general L 2 -estimate on convolution maximal operators obtained by Bourgain [16].

Lemma 2.1

Let K ∈ S ′ ( R n ) be a tempered distribution. Suppose K ^ ∈ L ∞ ( R n ) and is differentiable. Define the following quantities:

α j = sup 2 j ≤ ∣ ξ ∣ ≤ 2 j + 1 ∣ K ^ ( ξ ) ∣ and β j = sup 2 j ≤ ∣ ξ ∣ ≤ 2 j + 1 ∣ ⟨ ∇ K ^ ( ξ ) , ξ ⟩ ∣ .

for j ∈ Z . Then,

∥ sup t > 0 ∣ f ∗ K t ∣ ∥ L 2 ( R n ) ≤ C Γ ( K ) ∥ f ∥ L 2 ( R n ) ,

where

Γ ( K ) = ∑ j ∈ Z α j 1 / 2 ( α j + β j ) 1 / 2 .

Let M = 1 ∣ B n 1 ∣ d σ ( y ) and m ( ξ ) is the Fourier transform of M , where d σ ( y ) is the usual measure of unite sphere. Then,

m ( ξ ) = Γ n 2 π 1 2 Γ n − 1 2 ∫ − 1 1 e − 2 π i s ∣ ξ ∣ ( 1 − s 2 ) n − 3 2 d s ,

and

M n 1 f ( x ) = sup t > 0 ∫ R n f ∗ M t ( x ) .

Let P t ( t > 0 ) be the Poisson-semigroup on R n satisfying P t ^ ( ξ ) = e − t ∣ ξ ∣ . To derive (2), we take K ^ ( ξ ) = m ( ξ ) − e − n − 1 2 ∣ ξ ∣ . Stein [15] showed that there exists a constant C p independent of dimension n , such that

∥ sup t > 0 ∣ P t ∗ f ∣ ∥ L p ( R n ) ≤ C p ‖ f ‖ L p ( R n )

for all p > 1 . So it is sufficient to prove that

(3) ∥ sup t > 0 ∣ K t ∗ f ∣ ∥ L 2 ( R n ) ≤ C ‖ f ‖ L 2 ( R n ) .

To use Lemma 2.1, we will study the asymptotic properties of K ( ⋅ ) . Suppose n ≥ 3 , we give three estimates about m ( ξ ) .

Firstly, when ∣ ξ ∣ is large enough, we have

(4) ∣ m ( ξ ) ∣ = Γ n 2 π 1 2 Γ n − 1 2 ⋅ 1 2 π ∣ ξ ∣ ∫ − 1 1 e − 2 π i s ∣ ξ ∣ d d s ( 1 − s 2 ) n − 3 2 d s ≤ Γ n 2 π 1 2 Γ n − 1 2 ⋅ 1 2 π ∣ ξ ∣ ∫ − 1 1 d d s ( 1 − s 2 ) n − 3 2 d s ≤ C n 1 2 ∣ ξ ∣ − 1 .

When ∣ ξ ∣ is small enough, we obtain

(5) ∣ m ( ξ ) − 1 ∣ = Γ n 2 π 1 2 Γ n − 1 2 ∫ − 1 1 ( e − 2 π i s ∣ ξ ∣ − 1 ) ( 1 − s 2 ) n − 3 2 d s ≤ Γ n 2 π 1 2 Γ n − 1 2 ⋅ 2 π ∣ ξ ∣ ∫ − 1 1 ∣ s ∣ ( 1 − s 2 ) n − 3 2 d s ≤ C n − 1 2 ∣ ξ ∣ .

Finally, we obtain

(6) ∣ ⟨ ∇ m ( ξ ) , ξ ⟩ ∣ = Γ n 2 π 1 2 Γ n − 1 2 ∫ − 1 1 e − 2 π i s ∣ ξ ∣ ( − 2 π i s ∣ ξ ∣ ) ( 1 − s 2 ) n − 3 2 d s = Γ n 2 π 1 2 Γ n − 1 2 ∫ − 1 1 e − 2 π i s ∣ ξ ∣ d d s s ( 1 − s 2 ) n − 3 2 d s ≤ C n 1 2 ∫ 0 1 d d s s ( 1 − s 2 ) n − 3 2 d s ≤ C .

It is easy to know that

∣ e − n − 1 2 ∣ ξ ∣ − 1 ∣ ≤ n − 1 2 ∣ ξ ∣ ,

∣ e − n − 1 2 ∣ ξ ∣ ∣ ≤ n 1 2 ∣ ξ ∣ − 1 ,

and

∣ ⟨ ∇ e − n − 1 2 ∣ ξ ∣ , ξ ⟩ ∣ ≤ C .

Combining these two inequalities with (4)–(6), we have

∣ K ^ ( ξ ) ∣ ≤ C n 1 2 ∣ ξ ∣ − 1 ,

∣ K ^ ( ξ ) ∣ ≤ C n − 1 2 ∣ ξ ∣ ,

and

∣ < ∇ K ^ ( ξ ) , ξ > ∣ ≤ C .

According to three inequalities mentioned earlier and Lemma 2.1, we obtain

∥ sup t > 0 ∣ f ∗ K t ∣ ∥ L 2 ( R n ) ≤ C ‖ f ‖ L 2 ( R n ) .

3 The proof of main theorem for the case 0 < a < 1

For 0 ≤ a < 1 , we have

(7) M n a f ( x ) = sup t > 0 1 ∣ B a 1 ∣ ∫ B a 1 f ( x − t y ) d y ≤ ∣ B 0 1 ∣ ∣ B a 1 ∣ sup t > 0 1 ∣ B 0 1 ∣ ∫ B 0 1 ∣ f ( x − t y ) ∣ d y ≤ 1 1 − a M n 0 ( ∣ f ∣ ) ( x ) .

It follows from (1) and (7) that

‖ M a f ‖ L p ( R n ) ≤ 1 1 − a C ( p ) ‖ f ‖ L p ( R n ) .

The general case has been proved. It remains to consider the special case p > n / ( n − 1 ) . We introduce the following operator:

M n , m a f ( x ) = sup t > 0 ∫ a ≤ ∣ y ∣ ≤ 1 f ( x − t y ) ∣ y ∣ m d y ∫ a ≤ ∣ y ∣ ≤ 1 ∣ y ∣ m d y .

Noting that

(8) ∫ a ≤ ∣ y ∣ ≤ 1 f ( x − t y ) ∣ y ∣ m d y ∫ a ≤ ∣ y ∣ ≤ 1 ∣ y ∣ m d y = ∫ a 1 ∫ S n − 1 f ( x − t ρ θ ) ρ m + n − 1 d θ d ρ ∫ a 1 ∫ S n − 1 ρ n + m − 1 d θ d ρ .

It implies

M n , m a f ( x ) ≤ M n 1 ( ∣ f ∣ ) ( x ) .

From this, it follows that

(9) ∥ M n , m a f ∥ L p ( R n ) ≤ C ( p , n ) ‖ f ‖ L p ( R n ) .

For τ ∈ O ( n ) , define

M k , τ a f ( x ) = sup t > 0 ∫ a ≤ ∣ y 1 ∣ ≤ 1 f ( x − t τ ( y 1 , 0 ) ) ∣ y 1 ∣ n − k d y 1 ∫ a ≤ ∣ y 1 ∣ ≤ 1 ∣ y 1 ∣ n − k d y 1 ,

where ( y 1 , 0 ) ∈ R k × R n − k . Then

M k , τ a f ( x ) = sup t > 0 ∫ a ≤ ∣ y 1 ∣ ≤ 1 f τ ( τ − 1 x − t ( y 1 , 0 ) ) ∣ y 1 ∣ n − k d y 1 ∫ a ≤ ∣ y 1 ∣ ≤ 1 ∣ y 1 ∣ n − k d y 1 = M n , k a f τ ( τ − 1 x ) ,

where f τ ( ⋅ ) = f ( τ ⋅ ) . Hence,

(10) ∥ M k , τ a f ∥ L p ( R n ) ≤ A ( p , k ) ‖ f ‖ L p ( R n ) .

Since

1 ∣ B a 1 ∣ ∫ B a 1 f ( x − t y ) d y = ∫ a ≤ ∣ y ∣ ≤ 1 f ( x − t y ) d y ∫ a ≤ ∣ y ∣ ≤ 1 d y = ∫ O ( n ) ∫ a ≤ ∣ y 1 ∣ ≤ 1 f ( x − t τ ( y 1 , 0 ) ) ∣ y 1 ∣ n − k d y 1 d τ ∫ O ( n ) ∫ a ≤ ∣ y 1 ∣ ≤ 1 ∣ y 1 ∣ n − k d y 1 d τ ,

we obtain

(11) M n a f ( x ) ≤ ∫ O ( n ) M k , τ a ( ∣ f ∣ ) ( x ) d τ ,

where d τ is the normalized Harr measure on the orthogonal group O ( n ) .

It follows from (10) and (11) that

(12) ‖ M n a f ‖ L p ( R n ) ≤ A ( p , k ) ‖ f ‖ L p ( R n ) .

Since n > p / ( p − 1 ) , we take k to be the smallest integer greater than p / ( p − 1 ) . Our theorem follows from (12).

  1. Funding information: The research was supported by the Hebei Province introduced overseas student support projects (Grant No. C20190365).

  2. Author contributions: All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors state no conflict of interests.

  4. Data availability statement: Not applicable

  5. Code availability: Not applicable

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Received: 2022-01-15
Revised: 2022-10-09
Accepted: 2022-11-08
Published Online: 2022-11-25

© 2022 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  149. Efficient fixed-point iteration for generalized nonexpansive mappings and its stability in Banach spaces
https://doi.org/10.1515/math-2022-0529
Keywords for this article
maximal function; dimension-free estimate; Fourier multiplier
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