Norm inequalities for the conjugate operator in two-weighted Lebesgue spaces

Abstract

In this article, first, we prove that weighted-norm inequalities for the M-harmonic conjugate operator on the unit sphere whenever the pair (u, v) of weights satisfies the A _p -condition, and udσ, vdσ are doubling measures, where dσ is the rotation-invariant positive Borel measure on the unit sphere with total measure 1. Then, we drive cross-weighted norm inequalities between the Hardy-Littlewood maximal function and the sharp maximal function whenever (u, v) satisfies the A _p -condition, and vdσ does a certain regular condition.

2000 MSC: primary 32A70; secondary 47G10.

1 Introduction

Let B be the unit ball of ℂⁿwith norm |z| = 〈z, z〉^1/2 where 〈, 〉 is the Hermitian inner product, S be the unit sphere and σ be the rotation-invariant probability measure on S.

For z ∈ B, ξ ∈ S, we define the $\mathcal{M}$-harmonic conjugate kernel K(z, ξ) by

$$iK\left(z,\xi \right)=2C\left(z,\xi \right)-P\left(z,\xi \right)-1,$$

where C(z, ξ) = (1 - 〈z, ξ〉)^-nis the Cauchy kernel and P(z, ξ) = (1 - |z|²)ⁿ/|1 - 〈z, ξ〉 |²ⁿis the invariant Poisson kernel [1].

For the kernels, C and P, refer to [2]. And for all f - A(B), the ball algebra, such that f(0) is real, the reproducing property of 2C(z, ξ) - 1 [2, Theorem 3.2.5] gives

$${\int }_{S}K\left(z,\xi \right)\mathsf{\text{Re}}f\left(\xi \right)d\sigma \left(\xi \right)=-i\left(f\left(z\right)-\mathsf{\text{Re}}f\left(z\right)\right)=\mathsf{\text{Im}}f\left(z\right).$$

For n = 1, the definition of K f is the same as the classical harmonic conjugate function and so we can regard K f as the Hilbert transform on the unit circle. The L^pboundedness property of harmonic conjugate functions on the unit circle for 1 < p < ∞ was introduced by Riesz in 1924 [3, Theorem 2.3 of Chapter 3]. Later, in 1973, Hunt et al. [4] proved that, for 1 < p < ∞, conjugate functions are bounded on weighted measured Lebesgue space if and only if the weight satisfies A _p -condition. Most recently, Lee and Rim [5] provided an analogue of that of [4] by proving that, for 1 < p < ∞, M-harmonic conjugate operator K is bounded on L^p(ω) if and only if the nonnegative weight ω satisfies the A _p (S)-condition on S; i.e., the nonnegative weight ω satisfies

$$\underset{Q}{sup}\frac{1}{\sigma \left(Q\right)}{\int }_Q\omega d\sigma {\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\omega }^{-1∕\left(p-1\right)}d\sigma \right)}^{p-1}:=A_p^{\omega }<\infty ,$$

where Q = Q(ξ, δ) = {η ∈ S : d(ξ, η) = |1 - 〈ξ, η〉|^1/2 < δ} is a non-isotropic ball of S.

To define the A _p (S)-condition for two weights, we let (u, v) be a pair of two non-negative integrable functions on S. For p > 1, we say that (u, v) satisfies two-weighted A _p (S)-condition if

$$\underset{Q}{sup}\frac{1}{\sigma \left(Q\right)}{\int }_{Q}ud\sigma {\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{v}^{-1∕\left(p-1\right)}d\sigma \right)}^{p-1}:=A_p<\infty ,$$

(1.1)

where Q is a non-isotropic ball of S. For p = 1, the A₁(S)-condition can be viewed as a limit case of the A _p (S)-condition as p ↘ 1, which means that (u, v) satisfies the A₁(S)-condition if

$$\underset{Q}{sup}\frac{1}{\sigma \left(Q\right)}{\int }_{Q}ud\sigma \left(\mathsf{\text{ess}}{sup}_Q{v}^{-1}\right):=A_1<\infty ,$$

where Q is a non-isotropic ball of S.

In succession of classical weighted-norm inequalities, starting from Muckenhoupt's result in 1975 [6], there have been extensive studies on two-weighted norm inequalities (for textbooks [7–10] and for related topics [11–17]). In [6], Muckenhoupt derives a necessary and sufficient condition on two-weighted norm inequalities for the Poisson integral operator. And then, Sawyer [18, 19] obtained characterizations of two-weighted norm inequalities for the Hardy-Littlewood maximal function and for the fractional and Poisson integral operators, respectively. As a result on two-weighted A _p (S)-condition itself, Neugebauer [20] proved the existence of an inserting pair of weights. Cruz-Uribe and Pérez [21] give a sufficient condition for Calderón-Zygmund operators to satisfy the weighted weak (p, p) inequality. More recently, Martell et al. [22] provide two-weighted norm inequalities for Calderón-Zygmund operators that are sharp for the Hilbert transform and for the Riesz transforms.

Ding and Lin [23] consider the fractional integral operator and the maximal operator that contain a function homogeneous of degree zero as a part of kernels and the authors prove weighted (L^p, L^q)-boundedness for those operators for two weights.

In [24], Muckenhoupt and Wheeden provided simple examples of a pair that satisfies two-weighted A _p (ℝ)-condition but not two-weighted norm inequalities for the Hardy-Littlewood maximal operator and the Hilbert transform. In this article, we prove the converse of the main theorem of [5] by adding a doubling condition for a weight function. And then by adding a suitable regularity condition on a weight function, we derive and prove a cross-weighted norm inequalities between the Hardy-Littlewood maximal function and the sharp maximal function.

Throughout this article, Q denotes a non-isotropic ball of S induced by the non-isotropic metric d on S which is defined by d(ξ, η) = |1 - 〈ξ, η〉|^1/2. For notational simplicity, we denote ʃ _Q f dσ := f(Q) the integral of f over Q, and $\frac{1}{\sigma \left(Q\right)}{\int }_{Q}fd\sigma :=f_Q$ the average of f over Q. Also, for a nonnegative integrable function u and a measurable subset E of S, we write u(E) for the integral of u over E. We write Q(δ) in place of Q(ξ, δ) whenever the center ξ has no meaning in a context. For a positive constant s, sQ(δ) means Q(sδ). We say that a weight v satisfies a doubling condition if there is a constant C independent of Q such that v(2Q) ≤ Cv(Q) for all Q.

Theorem 1.1. Let 1 < p < ∞. If (u, v) satisfies two-weighted A _p' (S)-condition for some p' < p and udσ, vdσ are doubling measures, then there exists a constant C which depends on u, v and p, such that for all function f,

$${\int }_S{\left|Kf\right|}^{p}ud\sigma \le C{\int }_S{\left|f\right|}^{p}vd\sigma forallf\in L^p\left(v\right).$$

(1.2)

To prove the next theorem, we need a regularity condition for v such that for 1 ≤ p < ∞, we assume that for a measurable set E ⊂ Q and for σ(E) ≤ θσ(Q) with 0 ≤ θ ≤ 1, we get

$$v\left(E\right)\le \left(1-{\left(1-\theta \right)}^p\right)v\left(Q\right).$$

(1.3)

Let f ∈ L¹(S) and let 1 < p < ∞. The (Hardy-Littlewood) maximal and the sharp maximal functions M f, f^#p, resp. on S are defined by

$$\begin{array}{ll}\hfill Mf\left(\xi \right)& =\underset{Q}{sup}\frac{1}{\sigma \left(Q\right)}{\int }_Q\left|f\right|d\sigma ,\\ \hfill f^{\#p}\left(\xi \right)& =\underset{Q}{sup}{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|f-f_Q\right|}^{p}d\sigma \right)}^{1∕p},\end{array}$$

where each supremum is taken over all balls Q containing ξ. From the definition, the sharp maximal function f ↦ f^#pis an analogue of the maximal function M f, which satisfies f^#1(ξ) ≤ 2M f(ξ).

Theorem 1.2. Let 1 < p < ∞. If (u, v) satisfies two-weighted A _p (S)-condition and vdσ does (1.3), then there exists a constant C which depends on u, v and p, such that for all function f,

$${\int }_S{\left(Mf\right)}^{p}ud\sigma \le C\left({{\int }_S\left(f^{\#1}\right)}^{p}vd\sigma +{\int }_S{\left|f\right|}^{p}vd\sigma \right).$$

Remark. On the unit circle, we derive a sufficient condition for weighted-norm inequalities for the Hilbert transform for two weights.

The proofs of Theorem 1.1 will be given in Section 3. We start Section 2 by deriving some preliminary properties of (u, v) which satisfies the A _p (S)-condition. In Section 4, we prove Theorem 1.2.

2 Two-weight on the unit sphere

Lemma 2.1. If (u, v) satisfies two-weighted A _p (S)-condition, then for every function f ≥ 0 and for every ball Q,

$${\left(f_Q\right)}^{p}u\left(Q\right)\le A_p{\int }_Q{f}^{p}vd\sigma .$$

Proof. If p = 1 and (u, v) satisfies two-weighted A₁(S)-condition, we get, for every ball Q and every f ≥ 0,

$$\begin{array}{c}{f}_{Q}u(Q)=f(Q)u_Q\\ \le A_{1}f(Q)\frac{1}{\underset{Q}{\text{ess}\sup }{v}^{-1}}\\ \le A_1{\displaystyle {\int }_{Q}fv}d\sigma ,\end{array}$$

since $1/\underset{Q}{\text{ess}\sup }{v}^{-1}=\underset{Q}{\text{ess}\inf }v\le (\xi )$ for all ξ ∈ Q.

If 1 < p < ∞ and (u, v) satisfies two-weighted A _p (S)-condition, we have, for every ball Q and every f ≥ 0, using Holder's inequality with p and its conjugate exponent p/(p - 1),

$$\begin{array}{ll}\hfill f_Q& =\frac{1}{\sigma \left(Q\right)}{\int }_{Q}f{v}^{1∕p}{v}^{-1∕p}d\sigma \\ \le {\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{f}^{p}vd\sigma \right)}^{1∕p}{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{v}^{-1∕\left(p-1\right)}d\sigma \right)}^{\left(p-1\right)∕p}\end{array}$$

Hence,

$$\begin{array}{ll}\hfill {\left(f_Q\right)}^{p}u\left(Q\right)& =\frac{u\left(Q\right)}{\sigma \left(Q\right)}{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{v}^{-1∕\left(p-1\right)}d\sigma \right)}^{p-1}{\int }_Q{f}^{p}vd\sigma \\ \le A_p{\int }_Q{f}^{p}vd\sigma .\end{array}$$

Therefore, the proof is complete.

Corollary 2.2. If (u, v) satisfies two-weighted A _p (S)-condition, then

$${\left(\frac{\sigma \left(E\right)}{\sigma \left(Q\right)}\right)}^{p}u\left(Q\right)\le A_{p}v\left(E\right),$$

where E is a measurable subset of Q.

Proof. Applying Lemma 2.1 with f replaced by χ _E proves the conclusion.

3 Proof of Theorem 1.1

In this section, we will prove the first main theorem. First, we derive the inequality between two sharp maximal functions of K f and f.

Lemma 3.1. Let f ∈ L¹(S). Then, for q > p > 1, there is a constant C_p,qsuch that (K f)^#p(ξ) ≤ C_p,qf^#q(ξ) for almost every ξ.

Proof. It suffices to show that for r ≥ 1 and q > 1, there is a constant C _rq such that (K f)^#r(ξ) ≤ C _rq f^#rq(ξ) for almost every ξ,

i.e., for Q = Q(ξ _Q , δ) a ball of S, we prove that there are constants λ = λ(Q, f) and C _rq such that

$${\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|Kf\left(\eta \right)-\lambda \right|}^{r}d\sigma \right)}^{1∕r}\le C_{r,q}{f}^{{\#}^q}\left({\xi }_Q\right).$$

(3.1)

Fix Q = Q(ξ _Q , δ) and write

$$\begin{array}{ll}\hfill f\left(\eta \right)& =\left(f\left(\eta \right)-f_Q\right){\chi }_{2Q}\left(\eta \right)+\left(f\left(\eta \right)-f_Q\right){\chi }_{S\backslash 2Q}\left(\eta \right)+f_Q\\ :=f_1\left(\eta \right)+f_2\left(\eta \right)+f_Q.\end{array}$$

Then, K f = K f₁ + K f₂, since K f _Q = 0.

For each z ∈ B, put

$$g\left(z\right)={\int }_S\left(2C\left(z,\xi \right)-1\right)f_2\left(\xi \right)d\sigma \left(\xi \right).$$

Then, g is continuous on B ∪ Q By setting λ = -ig(ξ _Q ) in (3.1), we shall drive the conclusion. By Minkowski's inequality, we split the integral in (3.1) into two parts,

$$\begin{array}{c}{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|Kf\left(\eta \right)+ig\left({\xi }_Q\right)\right|}^{r}d\sigma \left(\eta \right)\right)}^{1∕r}\\ \le {\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|K{f}_1\right|}^{r}d\sigma \right)}^{1∕r}+{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|K{f}_2+ig\left({\xi }_Q\right)\right|}^{r}d\sigma \right)}^{1∕r}:=I_1+I_2.\end{array}$$

(3.2)

We estimate I₁. By Holder's inequality, it is estimated as

$$\begin{array}{ll}\hfill I_1& \le {\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|K{f}_1\right|}^{rq}d\sigma \right)}^{1∕rq}\\ \le {\left(\frac{1}{\sigma \left(Q\right)}{\int }_S{\left|K{f}_1\right|}^{rq}d\sigma \right)}^{1∕rq}\le \frac{C_{rq}}{\sigma {\left(Q\right)}^{1∕rq}}{∥f_1∥}_{L^{rq}},\end{array}$$

since K is bounded on L^rq(S) (rq > 1). By replacing f₁ by f - f _Q , we get

$$\begin{array}{ll}\hfill {∥f_1∥}_{L^{rq}}& ={\left({\int }_{2Q}{\left|f-f_Q\right|}^{rq}d\sigma \right)}^{1∕rq}\\ \le {\left({\int }_{2Q}{\left|f-f_{2Q}\right|}^{rq}d\sigma \right)}^{1∕rq}+\sigma {\left(2Q\right)}^{1∕rq}\left|f_{2Q}-f_Q\right|.\end{array}$$

Thus, by applying Hölder's inequality in the last term of the above,

$$\begin{array}{ll}\hfill \sigma {\left(2Q\right)}^{1∕rq}\left|f_{2Q}-f_Q\right|& \le \frac{\sigma {\left(2Q\right)}^{1∕rq}}{\sigma \left(Q\right)}{\int }_Q\left|f-f_{2Q}\right|d\sigma \\ \le \frac{\sigma {\left(2Q\right)}^{1∕rq}\sigma {\left(Q\right)}^{1-1∕rq}}{\sigma \left(Q\right)}{\left({\int }_{2Q}{\left|f-f_{2Q}\right|}^{rq}d\sigma \right)}^{1∕rq}\\ =R_2^{1∕rq}{\left({\int }_{2Q}{\left|f-f_{2Q}\right|}^{rq}d\sigma \right)}^{1∕rq}\left(\mathsf{\text{by}}\left(4.2\right)\right).\end{array}$$

Hence,

$$I_1\le C_{rq}\left(1+R_2^{1∕rq}\right)f^{{\#}^{rq}}\left({\xi }_Q\right).$$

(3.3)

Now, we estimate I₂. Since f₂ ≡ 0 on 2Q, the invariant Poisson integral of f₂ vanishes on Q, i.e., $\underset{t_{\nearrow }1}{lim}{\int }_{S}P\left(t\eta ,\xi \right)f_2\left(\eta \right)d\sigma \left(\eta \right)=0$ whenever ξ ∈ Q. Thus, for almost all ξ ∈ Q,

$$iK{f}_2\left(\xi \right)={\int }_{S\backslash 2Q}\left(2C\left(\xi ,\eta \right)-1\right)f_2\left(\eta \right)d\sigma \left(\eta \right)=g\left(\xi \right)$$

and then, by Minkowski's inequality for integrals,

$$\begin{array}{ll}\hfill I_2& ={\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|iK{f}_2-g\left({\xi }_Q\right)\right|}^{r}d\sigma \right)}^{1∕r}\\ \le {\int }_{S\backslash 2Q}2\left|f_2\left(\eta \right)\right|{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{\left|C\left(\xi ,\eta \right)-C\left({\xi }_Q,\eta \right)\right|}^{r}d\sigma \left(\xi \right)\right)}^{1∕r}d\sigma \left(\eta \right).\end{array}$$

By Lemma 6.1.1 of [2], we get an upper bound such that

$$I_2\le C\delta {\int }_{S\backslash 2Q}\frac{\left|f_2\left(\eta \right)\right|}{{\left|1-⟨\eta ,{\xi }_Q⟩\right|}^{n+1∕2}}d\sigma \left(\eta \right),$$

(3.4)

where C is an absolute constant. Write $S\backslash 2Q={\bigcup }_{k=1}^{\infty }{2}^{k+1}Q\backslash 2^{k}Q$. Then, the integral of (3.4) is equal to

$$\begin{array}{l}\sum _{k=1}^{\infty }{\int }_{2^{k+1}Q\backslash 2^{k}Q}\frac{\left|f\left(\eta \right)-f_Q\right|}{{\left|1-⟨\eta ,{\xi }_Q⟩\right|}^{n+1∕2}}d\sigma \left(\eta \right)\\ \le \sum _{k=1}^{\infty }\frac{1}{2^{\left(2n+1\right)k{\delta }^{2n+1}}}{\int }_{2^{k+1}Q\backslash 2^{k}Q}\left|f-f_Q\right|d\sigma \\ \le \sum _{k=1}^{\infty }\frac{1}{2^{\left(2n+1\right)k{\delta }^{2n+1}}}\left({\int }_{2^{k+1}Q}\left|f-f_{2^{k+1}Q}\right|d\sigma +\sum _{j=0}^k{\int }_{2^{k+1}Q}\left|f_{2^{j+1}Q}-f_{2^{j}Q}\right|d\sigma \right).\end{array}$$

By Hölder's inequality, by (4.3),

$$\begin{array}{ll}\hfill {\int }_{2^{k+1}Q}\left|f-f_{2^{k+1}Q}\right|d\sigma & \le R_{2^{k+1}\delta }{\left(\frac{1}{\sigma \left(2^{k+1}Q\right)}{\int }_{2^{k+1}Q}{\left|f-f_{2^{k+1}Q}\right|}^{rq}d\sigma \right)}^{1∕rq}\\ \le R_{2^{k+1}\delta }{f}^{{\#}^{rq}}\left({\xi }_Q\right),\end{array}$$

(3.5)

Similarly, for each j,

$$\begin{array}{ll}\hfill {\int }_{2^{k+1}Q}\left|f_{2^{j+1}Q}-f_{2^{j}Q}\right|d\sigma & \le \frac{\sigma \left(2^{k+1}Q\right)}{\sigma \left(2^{j}Q\right)}{\int }_{2^{j}Q}\left|f-f_{2^{j+1}Q}\right|d\sigma \\ \le R_{2^{k-j+1}}{\int }_{2^{j+1}Q}\left|f-f_{2^{j+1}Q}\right|d\sigma \left(\mathsf{\text{by}}\left(4.2\right)\right)\\ \le R_{2^{k-j+1}}{R}_{2^{j+1}\delta }{f}^{{\#}^{rq}}\left({\xi }_Q\right)\left(\mathsf{\text{from}}\left(3.5\right)\mathsf{\text{with}}k=j\right)\\ =R_1{R}_{2^{k+2}\delta }{f}^{{\#}^{rq}}\left({\xi }_Q\right).\end{array}$$

Thus,

$$\sum _{j=0}^k{\int }_{2^{k+1}Q}\left|f_{2^{j+1}Q}-f_{2^{j}Q}\right|d\sigma \le \left(k+1\right)R_1{R}_{2^{k+2}\delta }{f}^{{\#}^{rq}}\left({\xi }_Q\right).$$

(3.6)

Since R _s increases as s ↗ ∞ and R₁ > 1, by adding (3.5) to (3.6), we have the upper bound as

$$\left(k+2\right)R_1{R}_{2^{k+2}\delta }{f}^{{\#}^{rq}}\left({\xi }_Q\right).$$

Eventually, the identity of $R_{2^{k+2}\delta }=R_1{2}^{2n\left(k+2\right)}{\delta }^{2n}$ yields that

$$I_2\le 2^{4n}C{R}_1^2\sum _{k=1}^{\infty }\frac{k+2}{2^k}{f}^{{\#}^{rq}}\left({\xi }_Q\right),$$

(3.7)

and therefore, combining (3.3) and (3.7), we complete the proof.

The main theorem depends on Marcinkiewicz interpolation theorem between two abstract Lebesgue spaces, which is as follows.

Proposition 3.2. Suppose (X, μ) and (Y, ν) are measure spaces; p₀, p₁, q₀, q₁are elements of [1, ∞] such that p₀ ≤ q₀, p₁ ≤ q₁and q₀ ≠ q₁and

$$\frac{1}{p}=\frac{1+t}{p_0}+\frac{t}{p_1},\frac{1}{q}=\frac{1-t}{q_0}+\frac{t}{q_1}\left(0<t<1\right).$$

If T is a sublinear map from$L^{p_0}\left(\mu \right)+L^{p_1}\left(\mu \right)$to the space of measurable functions on Y which is of weak-types (p₀,q₀) and (p₁,q₁), then T is of type (p, q).

Now, we prove the main theorem.

Proof of Theorem 1.1. Under the assumption of the main theorem, we will prove that (1.2) holds. We fix p > 1 and let f ∈ L^p(v).

By Theorem 1.2, there is a constant C _p such that

$$\begin{array}{ll}\hfill {\int }_S{\left|Kf\right|}^{p}ud\sigma & \le {\int }_S{\left|M_u\left(Kf\right)\right|}^{p}ud\sigma \\ \le C_p{\int }_S{\left|{\left(Kf\right)}^{{\#}^1}\right|}^{p}ud\sigma \\ \le C_p{\int }_S{\left|f^{{\#}^q}\right|}^{p}ud\sigma \left(\mathsf{\text{by}}\mathsf{\text{Lemma}}3.1\mathsf{\text{with}}q>1,p∕q>1\right)\\ \le 2^p{C}_p{\int }_S{\left(M{\left|f\right|}^q\right)}^{p∕q}ud\sigma \left(\mathsf{\text{by}}\mathsf{\text{the}}\mathsf{\text{triangle}}\mathsf{\text{inequality}}\right).\end{array}$$

(3.8)

where M _u is the maximal operator with respect to udσ, the second inequality follows from the doubling condition of udσ.

Without loss of generality, we assume f ≥ 0. By Holder's inequality and by (1.1), we have

$$\begin{array}{ll}\hfill \frac{1}{\sigma \left(Q\right)}{\int }_{Q}d\sigma & \le {\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{f}^{p∕q}vd\sigma \right)}^{q∕p}{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{v}^{-1∕\left(p∕q-1\right)}d\sigma \right)}^{1-q∕p}\\ \le A_{p∕q}^{q∕p}{\left(\frac{1}{\sigma \left(Q\right)}{\int }_Q{f}^{p∕q}vd\sigma \right)}^{q∕p}{\left(\frac{\sigma \left(Q\right)}{v\left(Q\right)}\right)}^{q∕p}\mathsf{\text{for}}\mathsf{\text{all}}Q.\end{array}$$

Thus, if f _Q > λ, then

$$u\left(Q\right)\le A_{p∕q}{\lambda }^{p∕q}{\int }_Q{f}^{p∕q}vd\sigma \mathsf{\text{for}}\mathsf{\text{all}}Q.$$

(3.9)

Let E be an arbitrary compact subset of {ξ ∈ S: M f(ξ) > λ}. Since vdσ is a doubling measure, from (3.9), there exists a constant C _p,q such that

$$u\left(E\right)\le C_{p,q}{\lambda }^{p∕q}{\int }_S{f}^{p∕q}vd\sigma .$$

Thus, M f is of weak-type (L^p/q(vdσ),L^p/q(udσ)). Moreover,

$$\begin{array}{ll}\hfill {∥Mf∥}_{L^{\infty }\left(ud\sigma \right)}& \le {∥Mf∥}_{L^{\infty }}\left(\mathsf{\text{since}}ud\sigma \ll d\sigma \right)\\ \le {∥f∥}_{L^{\infty }}\\ ={∥f∥}_{L^{\infty }\left(ud\sigma \right)}\left(\mathsf{\text{since}}ud\sigma \ll d\sigma ,v>0\mathsf{\text{a}}\mathsf{\text{.e}}\mathsf{\text{.}}\mathsf{\text{by}}\left(1.1\right)\right).\end{array}$$

Now, by Proposition 3.2, M f is of type (L^r(vdσ), L^r(udσ)) for f > p/q.

Hence, the last integral of (3.8) is bounded by some constant times

$${\int }_S{\left|f\right|}^{qr}vd\sigma \left(\mathsf{\text{for}}\mathsf{\text{all}}r>p∕q\right).$$

Since q is arbitrary so that p/q > 0, we can replace qr by p with p > 1. Therefore, the proof is completed.

4 Proof of Theorem 1.2

Theorem 1.2 can be regarded as cross-weighted norm inequalities for the Hardy-Littlewood maximal function and the sharp maximal function on the unit sphere. For a single A _p -weight in ℝⁿ, refer to Theorem 2.20 of [8].

From Proposition 5.1.4 of [2], we conclude that when n > 1,

$$\frac{{\Gamma }^2\left(n∕2+1\right)}{2^{n-2}\Gamma \left(n+1\right)}{s}^{2n}\le \frac{\sigma \left(Q\left(s\delta \right)\right)}{\sigma \left(Q\left(\delta \right)\right)}\le \frac{2^{n-2}\Gamma \left(n+1\right)}{{\Gamma }^2\left(n∕2+1\right)}{s}^{2n},$$

and when n = 1,

$$\frac{2}{\pi }{s}^2\le \frac{\sigma \left(Q\left(s\delta \right)\right)}{\sigma \left(Q\left(\delta \right)\right)}\le \frac{\pi }{2}{s}^2$$

for any s > 0. Throughout the article, several kinds of constants will appear. To avoid confusion, we define the maximum ratio between sizes of two balls by

$$R_s:R_{s,n}=max\left(\frac{2^{n-2}\Gamma \left(n+1\right)}{{\Gamma }^2\left(n∕2+1\right)},\frac{\pi }{2}\right)s^{2n},$$

(4.1)

and thus, for every s > 0, for every δ > 0,

$$\sigma \left(sQ\left(\delta \right)\right)\le R_s\sigma \left(Q\left(\delta \right)\right).$$

(4.2)

Putting δ = 1 in (4.2), we get

$$\sigma \left(Q\left(s\right)\right)\le R_s.$$

(4.3)

To prove Theorem 1.2, we need some lemmas. The next result is a covering lemma on the unit sphere, related to the maximal function. Let f ∈ L¹(S) and let $t>{∥f∥}_{L^1\left(S\right)}$. We may assume ${∥f∥}_{L^1\left(S\right)}\ne 0$. Since {M f > t} is open, take a ball Q ⊂ {M f > t} with center at each point of {M f > t}. For such a ball Q,

$$\sigma \left(Q\right)\le \frac{1}{t}{\int }_Q\left|f\right|d\sigma .$$

(4.4)

Thus, to each ξ ∈ {M f > t} corresponds a largest radius δ such that the ball Q = Q(ξ, δ) ⊂ {M f > t} satisfies (4.4). Hence, we conclude the following simple covering lemma.

Lemma 4.1 (Covering lemma on S). Let f ∈ L¹(S) be non-trivial. Then, for$t>{∥f∥}_{L^1\left(S\right)}$, there is a collection of balls {Q _t,j } such that

(i) $\left\{\xi \in S:Mf\left(\xi \right)>t\right\}\subset {\bigcup }_j{Q}_{t,j,}$

(ii) $\sigma \left(Q_{t,j}\right)\le t^{-1}\left|f\right|\left(Q_{t,j}\right),$

where each Q _t,j has the maximal radius of all the balls that satisfy (ii) in the sense that if Q is a ball that contains some Q _t,j as its proper subset, then σ(Q) > t^-1 ʃ _Q |f| dσ holds.

Now, we are ready to prove Theorem 1.2.

Proof of Theorem 1.2. Fix 1 < p < ∞. We may assume f^#1∈ L^p(v) and f ∈ L^p(v), otherwise, Theorem 1.2 holds clearly. Since v satisfies the doubling condition, we have ||M f|| _{Lp (v)}≤ C||f^#1|| _{Lp (v)}. Combining this with f^#1∈ L^p(v), we have ||M f||_Lp(v)< ∞.

Suppose that f is non-trivial and we may assume that f ≥ 0. Let

$$t>max\left(2,2\underset{2}{\overset{2}{R}},R_3\right){∥f∥}_{L^1\left(S\right)}.$$

For ε > 0, E _t be a compact subset of {M f > t} such that u({M f > t}) < u(E _t ) + e^-tε. Indeed, since u is integrable, u dσ is a regular Borel measure absolutely continuous with respect to σ.

Suppose {Q_t,j} is a collection of balls having the properties (i) and (ii) of Lemma 4.1. Since {Q _t,j } is a cover of a compact set E _t , there is a finite subcollection of {Q _t,j }, which covers E _t . By Lemma 5.2.3 of [2], there are pairwise disjoint balls, $Q_{t,j_1},Q_{t,j_2},...,Q_{t,j_{\ell }}$ of the previous subcollection such that

$$\begin{array}{ll}\hfill E_t& \subset \bigcup _{k=1}^{\ell }3Q_{t,j_k},\\ \hfill \sigma \left(E_t\right)& \le R_3\sum _{k=1}^{\ell }\sigma \left(Q_{t,j_k}\right),\end{array}$$

where ℓ may depend on t. To avoid the abuse of subindices, we rewrite $Q_{t,j_k}$ as Q _t,j . Let us note that from the maximality of Q _t,j ,

$$t>\frac{1}{\sigma \left(2Q_{t.j}\right)}{\int }_{2Q_{t,j}}fd\sigma \ge \frac{\sigma \left(Q_{t.j}\right)}{\sigma \left(2Q_{t.j}\right)}t\ge R_2^{-1}t.$$

(4.5)

Fix $Q_0=2Q_{kt,j_0}$, where κ^-1 = 2R₂. (Here, κ < 1/2, since R₂ > 1.) Let λ > 0 that will be chosen later. From the definition of the sharp maximal function, there are two possibilities: either

$$Q_0\subset \left\{f^{{\#}^1}>\lambda t\right\}\mathsf{\text{or}}{Q}_0\not\subset \left\{f^{{\#}^1}>\lambda t\right\}.$$

(4.6)

In the first case, since Q_t,j's are pairwise disjoint,

$$\sum _{\left\{j:Q_{t,j}\subset Q_0\subset \left\{f^{{\#}^1}>\lambda t\right\}\right\}}v\left(Q_{t,j}\right)\le v\left(\left\{f^{{\#}^1}>\lambda t\right\}\right),$$

and also,

$${\displaystyle \sum _{\underset{Q_0\subset \{f^{{\#}^1}>\text{λ}t\}}{Q_0}}{\displaystyle \sum _{\{j:Q_{t,j}\subset Q_0\}}v(Q_{t,j})\le v(\{f^{{\#}^1}>\text{λ}t\})}}.$$

(4.7)

In the second case,

$$\frac{1}{\sigma \left(Q_0\right)}{\int }_{Q_0}|f-f_{Q_0}|d\sigma \le \lambda t.$$

(4.8)

Since $2^{-1}t>{∥f∥}_{L^1\left(S\right)}$, by (4.5), taking $f_{Q_0}\le R_{2}kt=2^{-1}t$ into account, we have

$$\begin{array}{ll}\hfill \sum _{\left\{j:Q_{t,j}\subset Q_0\not\subset \left\{f^{{\#}^1}>\lambda t\right\}\right\}}\left(t-t∕2\right)\sigma \left(Q_{t,j}\right)& \le \sum _{\left\{j:Q_{t,j}\subset Q_0\not\subset \left\{f^{{\#}^1}>\lambda t\right\}\right\}}{\int }_{Q_{t,j}}f-f_{Q_0}d\sigma \\ \le \sum _{\left\{j:Q_{t,j}\subset Q_0\not\subset \left\{f^{{\#}^1}>\lambda t\right\}\right\}}{\int }_{Q_{t,j}}|f-f_{Q_0}|d\sigma \\ \le {\int }_{Q_0}|f-f_{Q_0}|d\sigma \\ \le \lambda t\sigma \left(Q_0\right)\left(\mathsf{\text{by (4}}\mathsf{\text{.8)}}\right).\end{array}$$

Thus,

$$\sum _{\left\{j:Q_{t,j}\subset Q_0\not\subset \left\{f^{{\#}^1}>\lambda t\right\}\right\}}\sigma \left(Q_{t,j}\right)\le 2\lambda \sigma \left(Q_0\right).$$

(4.9)

In (4.9), take a small λ > 0 such that

$$2\lambda <1.$$

(4.10)

(Note that the condition (4.10) enables us to use (1.3).) Thus, (4.9) can be written as

$$\sum _{\left\{j:Q_{t.j}\subset Q_0\not\subset \left\{f^{{\#}^1}>\lambda t\right\}\right\}}v\left(Q_{t,j}\right)\le \left(1-{\left(1-2\lambda \right)}^p\right)v\left(2Q_{\kappa t,j_0}\right).$$

Adding up all possible Q₀'s in the second case of (4.6), we get

$${\displaystyle \sum _{\underset{Q_0\not\subset \{f^{{\#}^1}>\lambda t\}}{Q_0}}{\displaystyle \sum _{\{j:Q_{t,j}\subset Q_0\}}v(Q_{t,j})\le \left(1-{(1-2\lambda )}^p\right){\displaystyle \sum _{k}v(2Q_{\kappa t,k})}}}.$$

(4.11)

Since $\left\{Mf>t\right\}\subset \left\{Mf>R_2^{-1}t\right\}$ and $\sigma \left(2Q_{t,j}\right)\le R_2{t}^{-1}{\int }_{2Q_{t,j}}fd\sigma$ holds (4.5), we can construct the collection of balls $\left\{Q_{R_2^{-1}t,j}\right\}$ which covers $\left\{Mf>R_2^{-1}t\right\}$ with maximal radius just the same way as Lemma 4.1, so that 2Q _t,j is contained in $\left\{Q_{R_2^{-1}t,i}\right\}$ for some i. Recall that $R_2^{-1}kt=2^{-1}{R}_2^{-2}t>{∥f∥}_{L^1\left(S\right)}$, hence, (4.11) turns into

$${\displaystyle \sum _{\underset{Q_0\not\subset \{f^{{\#}^1}>\lambda t\}}{Q_0}}{\displaystyle \sum _{\{j:Q_{t,j}\subset Q_0\}}v(Q_{t,j})\le \left(1-{(1-2\lambda )}^p\right)}{\displaystyle \sum _{k}v(Q_{R_2^{-1}\kappa t,k})}}.$$

(4.12)

Combining (4.7) and (4.11), we summarize

$$\sum _{j}v\left(Q_{t,j}\right)\le v\left(\left\{f^{{\#}^1}>\lambda t\right\}\right)+\sum _k\left(1-{\left(1-2\lambda \right)}^p\right)v\left(Q_{R_2^{-1}\kappa t,k}\right).$$

(4.13)

Now, put

$$\begin{array}{ll}\hfill {\alpha }_v\left(t\right)& =\sum _{j}v\left(Q_{t,j}\right),\\ \hfill {\beta }_u\left(t\right)& =u\left(E_t\right).\end{array}$$

(4.14)

Then,

$$\begin{array}{ll}\hfill {\beta }_u\left(t\right)& \le {\int }_{{\cup }_{j}3Q_{t,j}}ud\sigma \le \sum _j{\int }_{3Q_{t,j}}ud\sigma \\ \le A_p\sum _j{\int }_{3Q_{t,j}}vd\sigma \left(\mathsf{\text{by Corollary 2}}\mathsf{\text{.2 with }}E=Q=3Q_{t,j}\right)\\ \le A_p\sum _j{\int }_{Q_{R_3^{-1}t,j}}vd\sigma \\ =A_p{\alpha }_v\left(R_3^{-1}t\right),\end{array}$$

(4.15)

where the fourth inequality follows from the fact that $3Q_{t,j}\subset Q_{R_3^{-1}t,i}$ for some i. Indeed, we can construct $Q_{R_3^{-1}t,i}$ as before, since $R_3^{-1}t>{∥f∥}_{L^1\left(S\right)}$.

Eventually, putting the constant $e_p={\int }_0^{\infty }{t}^p{e}^{-t}dt$, and $N=max\left(2,2\underset{2}{\overset{2}{R}},R_3\right)$ (which depends only on n), we have

$$\begin{array}{ll}\hfill {\int }_S|Mf{|}^{p}ud\sigma & \le {\int }_0^{\infty }p{t}^{p-1}{\beta }_u\left(t\right)dt+e_p\epsilon \\ \le {\int }_0^{N\parallel f{\parallel }_{L^1\left(S\right)}}p{t}^{p-1}{\beta }_u\left(t\right)dt+A_p{\int }_{N\parallel f{\parallel }_{L^1\left(S\right)}}^{\infty }p{t}^{p-1}{\alpha }_v\left(R_3^{-1}t\right)dt+e_p\epsilon \left(\mathsf{\text{by (4}}\mathsf{\text{.15)}}\right)\\ :=I+II+e_p\epsilon .\end{array}$$

The first term I is dominated by

$$\begin{array}{ll}\hfill N^p\parallel u{\parallel }_{L^1\left(S\right)}\parallel f{\parallel }_{L^1\left(S\right)}^p& \le N^p\parallel f{\parallel }_{L^{p\left(v\right)}}^p\parallel u{\parallel }_{L^1\left(S\right)}{\left({\int }_S{v}^{-1∕\left(p-1\right)}d\sigma \right)}^{p-1}\\ \le N^p{A}_p\parallel f{\parallel }_{L^{p\left(v\right)}}^p\left(\mathsf{\text{since }}\left(u,v\right)\in A_p\left(S\right)\right),\end{array}$$

where the first inequality follows from Hölder's inequality for ${∥f∥}_{L^1\left(S\right)}^o$.

On the other hand,

$$\begin{array}{ll}\hfill II& \le A_p{C}_p{\int }_0^{\infty }p{t}^{p-1}v\left(\left\{f^{{\#}^1}>R_3^{-1}t\right\}\right)dt\left(\mathsf{\text{by Lemma 4}}\mathsf{\text{.3}}\right)\\ =A_p{C}_p{R}_3^p{\int }_0^{\infty }p{t}^{p-1}v\left(\left\{f^{{\#}^1}>t\right\}\right)dt\left(\mathsf{\text{by the change of variable}}\right)\\ =A_p{C}_p{R}_3^p{\int }_S|f^{{\#}^1}{|}^{p}vd\sigma .\end{array}$$

Hence,

$${\int }_S|Mf{|}^{p}ud\sigma \le N^p{A}_p\parallel f{\parallel }_{L^p\left(v\right)}^p+A_p{C}_p{R}_3^p{\int }_S|f^{{\#}^1}{|}^{p}vd\sigma +e_p\epsilon .$$

The first and the last integrals are independent of ε. Letting ε ↘ 0, therefore, the proof is complete after accepting Lemma 4.3.

Lemma 4.2. Let α _v be defined in (4.14). Then, for every q ≥ p and every r > 0,

$${\int }_0^r{t}^{q-1}{\alpha }_v\left(t\right)dt<\infty .$$

Proof. For a positive real number r, we set

$$I_r={\int }_0^{r}q{t}^{q-1}{\alpha }_v\left(t\right)dt.$$

(4.16)

Since ${\sum }_{j}v\left(Q_{t,j}\right)\le {\int }_{\left\{Mf>t\right\}}vd\sigma$, we have

$$I_r\le {\int }_0^{r}q{t}^{q-1}{\int }_{\left\{Mf>t\right\}}vd\sigma dt.$$

We note that I _r is finite, since p ≥ p₀ and it is bounded by

$$\frac{q{r}^{q-p}}{p}{\int }_0^{r}p{t}^{p-1}{\int }_{\left\{Mf>t\right\}}vd\sigma dt\le \frac{q{r}^{q-p}}{p}\parallel Mf{\parallel }_{L^p\left(v\right)}^p<\infty ,$$

since M f ∈ L^p(v). Therefore, the proof is complete.

Now, filling up next lemma, we finish the proof of Theorem 1.2.

Lemma 4.3. Under the same assumption as Theorem 1.2, if α _v is defined in (4.14), then there is a constant C _p such that

$${\int }_0^{\infty }{t}^{p-1}{\alpha }_v\left(t\right)dt\le C_p{\int }_0^{\infty }{t}^{p-1}v\left(\left\{f^{{\#}^1}>t\right\}\right)dt.$$

Proof. Recall (4.13), i.e.,

$${\alpha }_v\left(t\right)\le v\left(\left\{f^{{\#}^1}>\lambda t\right\}\right)+\left(1-{\left(1-2\lambda \right)}^p\right){\alpha }_v\left(R_2^{-1}\kappa t\right).$$

By integration, it follows that

$$\begin{array}{l}{\int }_0^r{t}^{p-1}{\alpha }_v\left(t\right)dt\\ \le {\int }_0^r{t}^{p-1}v\left(\left\{f^{\#1}>\lambda t\right\}\right)dt\\ +\left(1-{\left(1-2\lambda \right)}^p\right){\int }_0^r{t}^{p-1}{\alpha }_v\left(R_2^{-1}kt\right)dt\\ ={\int }_0^r{t}^{p-1}v\left(\left\{f^{\#1}>\lambda t\right\}\right)dt\\ +\left(1-{\left(1-2\lambda \right)}^p\right)R_2^p{k}^{-p}{\int }_0^{R_2^{-1}kr}{t}^{p-1}{\alpha }_v\left(t\right)dt\\ \le {\int }_0^r{t}^{p-1}v\left(\left\{f^{\#1}>\lambda t\right\}\right)dt\\ +2^p{R}_2^{2p}\left(1-{\left(1-2\lambda \right)}^p\right){\int }_0^r{t}^{p-1}{\alpha }_v\left(t\right)dt\left(\mathsf{\text{since}}k=2^{-1}{R}_2^{-1},R_2^{-1}k<1\right),\end{array}$$

(4.17)

where the equality is due to the change of variable.

Take a small λ so that

$$\begin{array}{ll}\hfill 2^p{R}_2^{2p}\left(1-{\left(1-2\lambda \right)}^p\right)& <1∕2,\\ \hfill 2\lambda & <1,\end{array}$$

where the second inequality comes from (4.10). Then, by Lemma 4.2, (4.17) can be written as

$$\begin{array}{ll}\hfill \frac{1}{2}{\int }_0^r{t}^{p-1}{\alpha }_v\left(t\right)dt& \le {\int }_0^r{t}^{p-1}v\left(\left\{f^{{\#}^1}>\lambda t\right\}\right)dt\\ ={\lambda }^{-p}{\int }_0^{\lambda r}{t}^{p-1}v\left(\left\{f^{{\#}^1}>t\right\}\right)dt,\end{array}$$

where the equality is caused by the change of variable.

Finally, letting r ↗ ∞, we obtain

$${\int }_0^{\infty }{t}^{p-1}{\alpha }_v\left(t\right)dt\le 2{\lambda }^{-p}{\int }_0^{\infty }{t}^{p-1}v\left(\left\{f^{{\#}^1}>t\right\}\right)dt.$$

Therefore, the proof is complete.