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Endpoint estimates for spherical maximal operators acting on radial functions in hyperbolic spaces

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Abstract

In this article we investigate spherical maximal operators mα\mathfrak m^α of real order αα acting on radial functions in the hyperbolic space Hn\mathbb H^n. We show that for every n≥2n\geq 2, 1<p≤∞1<p\leq\infty and α>(1−n)/2α>(1-n)/2, mα\mathfrak m^α is bounded on Lradp(Hn)L^p_{\mathrm{rad}}(\mathbb H^n) if and only if \begin{align*} \begin{cases} α>1-n+\frac np, & 1<p\leq 2, α>\frac1p-\frac{n-1}2, & 2<p\leq\frac{2n}{n-1}, α\geq\frac{1-n}p, & p>\frac{2n}{n-1}. \end{cases} \end{align*} Moreover, we show that for 1<p≤2n/(n−1)1<p\leq 2n/(n-1) and \begin{align*} α= \begin{cases} 1-n+\frac np, & 1<p\leq 2, \frac1p-\frac{n-1}2, & 2<p\leq\frac{2n}{n-1}, \end{cases} \end{align*} mα\mathfrak m^α is bounded from Lradp,1(Hn)L^{p,1}_{\mathrm{rad}}(\mathbb H^n) to Lradp,∞(Hn)L^{p,\infty}_{\mathrm{rad}}(\mathbb H^n).

Introduction

In the 1970s, Stein [18] introduced the spherical maximal operators

Mα(f)(x):=sup⁡t>0∣1Γ(α)∫{y∈Rn:∣y∣<1}(1−∣y∣2)α−1f(x−ty) dy∣.\mathfrak{M}^{\alpha}(f)(x) := \sup_{t>0} \left| \frac{1}{\Gamma(\alpha)} \int_{\{y\in\mathbb{R}^{n}:|y|<1\}} (1-|y|^{2})^{\alpha-1}f(x-ty)\,dy \right|.

of complex order α\alpha, with a priori for Re⁡α>0\operatorname{Re}\alpha>0, on Rn\mathbb{R}^{n}. This definition can be analytically continued to Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2. In particular, when α=0\alpha=0, Mα\mathfrak{M}^{\alpha} recovers the standard spherical maximal operator M\mathfrak{M} defined by

M(f)(x):=sup⁡t>0∣∫{ω∈Rn:∣ω∣=1}f(x−tω) dω∣.\mathfrak{M}(f)(x) := \sup_{t>0} \left| \int_{\{\omega\in\mathbb{R}^{n}:|\omega|=1\}} f(x-t\omega)\,d\omega\right|.

where dωd\omega is the canonical surface measure of the unit sphere in Rn\mathbb{R}^{n}. Stein ([17 pp. 518–519] & [18 Theorem 2]) showed that for n≥2n\ge2, 1<p<∞1<p<\infty and α∈R\alpha\in\mathbb{R} with α>(1−n)/2\alpha>(1-n)/2, if

α>{1−n+np,1<p≤2,2−np,p>2,\alpha> \begin{cases} 1-n+\dfrac{n}{p}, & 1<p\le2,\\ \dfrac{2-n}{p}, & p>2, \end{cases}

then Mα\mathfrak{M}^{\alpha} is bounded on Lp(Rn)L^{p}(\mathbb{R}^{n}), and that if Mα\mathfrak{M}^{\alpha} is bounded on Lp(Rn)L^{p}(\mathbb{R}^{n}) for 1<p≤21<p\le2, then we must have

α>1−n+np.\alpha>1-n+\dfrac{n}{p}.

Stein’s result with α=0\alpha=0 indicates that for n≥3n\geq3 and p>n/(n−1)p>n/(n-1), M\mathcal{M} is bounded on Lp(Rn)L^p(\mathbb{R}^n), and that for n≥2n\geq2, M\mathcal{M} is bounded on Lp(Rn)L^p(\mathbb{R}^n) only if p>n/(n−1)p>n/(n-1). When n=2n=2, the same sufficient condition p>2=2/(2−1)p>2=2/(2-1) for the LpL^p-boundedness of M\mathcal{M} was obtained by Bourgain [2]. At the endpoint p=n/(n−1)p=n/(n-1), Bourgain [1] showed that Mα\mathcal{M}^{\alpha} is bounded from Lp,1(Rn)L^{p,1}(\mathbb{R}^n) to Lp,∞(Rn)L^{p,\infty}(\mathbb{R}^n) when n≥3n\geq3, and that the same restricted weak-type (p,p)(p,p) estimate holds for radial functions when n=2n=2. However, this 2-dimensional restricted weak-type estimate fails for general functions, as shown by Seeger, Tao and Wright [15]. For further results concerning LpL^p-boundedness of the operators Mα\mathcal{M}^{\alpha}, one can refer to [11, 12, 13, 20] and the references therein.

In the last decade, Ciaurri et al. [5, 6, 14] investigated Mα\mathcal{M}^{\alpha} acting on radial functions. They established the sharp range of α\alpha for Mα\mathcal{M}^{\alpha} to be bounded on Lradp(Rn)L^p_{\mathrm{rad}}(\mathbb{R}^n), the radial subspace of Lp(Rn)L^p(\mathbb{R}^n), as well as the corresponding endpoint result in terms of restricted weak-type estimates. More precisely, they showed that for n≥2n\geq2, 1<p≤∞1<p\leq\infty and α∈R\alpha\in\mathbb{R} with α>(1−n)/2\alpha>(1-n)/2, Mα\mathcal{M}^{\alpha} is bounded on Lradp(Rn)L^p_{\mathrm{rad}}(\mathbb{R}^n) if and only if

{α>1−n+np,1<p≤2,α>1p−n−12,2<p≤2nn−1,α≥1−np,p>2nn−1.\begin{cases} \alpha>1-n+\dfrac{n}{p}, & 1<p\leq2,\\ \alpha>\dfrac{1}{p}-\dfrac{n-1}{2}, & 2<p\leq\dfrac{2n}{n-1},\\ \alpha\geq\dfrac{1-n}{p}, & p>\dfrac{2n}{n-1}. \end{cases}

and that for 1<p≤2n/(n−1)1<p\leq2n/(n-1) and

α={1−n+np,1<p≤2,1p−n−12,2<p≤2nn−1,\alpha= \begin{cases} 1-n+\dfrac{n}{p}, & 1<p\leq2,\\ \dfrac{1}{p}-\dfrac{n-1}{2}, & 2<p\leq\dfrac{2n}{n-1}, \end{cases}

Mα\mathcal{M}^{\alpha} is bounded from Lradp,1(Rn)L^{p,1}_{\mathrm{rad}}(\mathbb{R}^n) to Lradp,∞(Rn)L^{p,\infty}_{\mathrm{rad}}(\mathbb{R}^n). See [14 Corollaries 1.5 and 1.6(b)].

In the 1980s, El Kohen [7] introduced on the hyperbolic space Hn\mathbb{H}^n the spherical maximal operators

mα(f)(x):=sup⁡t>0∣Mtα(f)(x)∣.\mathfrak{m}^{\alpha}(f)(x):=\sup_{t>0}\left|M_t^{\alpha}(f)(x)\right|.

of complex order α\alpha with Re⁡α>0\operatorname{Re}\alpha>0, where

Mtα(f)(x):=(sinh⁡t)2−nΓ(α)(cosh⁡t−1)α∫B(x,t)(cosh⁡t−cosh⁡d(x,y))α−1f(y) dy(1)M_t^{\alpha}(f)(x):=\frac{(\sinh t)^{2-n}}{\Gamma(\alpha)(\cosh t-1)^{\alpha}}\int_{B(x,t)}\left(\cosh t-\cosh d(x,y)\right)^{\alpha-1}f(y)\,dy \tag*{(1)}

is the spherical averaging operators of order α\alpha. Here d(⋅,⋅)d(\cdot,\cdot) is the geodesic distance, dxdx is the Riemannian measure and B(x,t)B(x,t) is the geodesic ball centered at xx with radius tt in Hn\mathbb{H}^n. The definition can be analytically continued to Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2. In particular, at order zero,

Mt0(f)(x)=∣Sn−1∣Avg⁡S(x,t)(f),(2)M_t^0(f)(x)=|\mathbb{S}^{n-1}|\operatorname{Avg}_{S(x,t)}(f), \tag*{(2)}

where S(x,t)S(x,t) is the geodesic sphere in Hn\mathbb{H}^n and Avg⁡S(x,t)\operatorname{Avg}_{S(x,t)} denotes its normalized surface average. El Kohen [7 Theorem 3] showed that for n≥2n\geq2, 1<p<∞1<p<\infty and α∈R\alpha\in\mathbb{R} with α>(1−n)/2\alpha>(1-n)/2, if

α>{1−n+np,1<p≤2,2−np,p>2,\alpha> \begin{cases} 1-n+\dfrac{n}{p}, & 1<p\leq2,\\ \dfrac{2-n}{p}, & p>2, \end{cases}

then mα\mathfrak{m}^{\alpha} is bounded on Lp(Hn)L^p(\mathbb{H}^n).

This result with α=0\alpha=0 yields that for n≥3n\geq3 and p>n/(n−1)p>n/(n-1), m0\mathrm{m}^{0} is bounded on Lp(Hn)L^{p}(\mathbb{H}^{n}). The case of n=2n=2 was later resolved by Ionescu [10]. More recently, Chen, Shen, Yan and the named author [4] gave the necessary condition for the LpL^{p}-boundedness of mα\mathrm{m}^{\alpha}, and improved the range of α\alpha in [7 Theorem 3] for p>2p>2. They showed that for n≥2n\geq2, 1<p<∞1<p<\infty and α∈R\alpha\in\mathbb{R} with α>(1−n)/2\alpha>(1-n)/2, if mα\mathrm{m}^{\alpha} is bounded on Lp(Hn)L^{p}(\mathbb{H}^{n}), then we must have

{α>1−n+np,1<p≤2,α≥1p−n−12,2<p≤2nn−1,α≥1−np,p>2nn−1.\begin{cases} \alpha>1-n+\dfrac{n}{p}, & 1<p\leq2,\\ \alpha\geq\dfrac{1}{p}-\dfrac{n-1}{2}, & 2<p\leq\dfrac{2n}{n-1},\\ \alpha\geq\dfrac{1-n}{p}, & p>\dfrac{2n}{n-1}. \end{cases}

and that if p>2p>2 and

α>{2−np−p−2ppn(pn−2),2<p≤pn,2−np−1ppn,p>pn\alpha> \begin{cases} \dfrac{2-n}{p}-\dfrac{p-2}{pp_{n}(p_{n}-2)}, & 2<p\leq p_{n},\\ \dfrac{2-n}{p}-\dfrac{1}{pp_{n}}, & p>p_{n} \end{cases}

with

pn={4,n=2,2(n+1)n−1,n≥3,p_{n}= \begin{cases} 4, & n=2,\\ \dfrac{2(n+1)}{n-1}, & n\geq3, \end{cases}

then mα\mathrm{m}^{\alpha} is bounded on Lp(Hn)L^{p}(\mathbb{H}^{n}). See [4 Theorems 1.1 and 1.2]. In addition, Zhang and the named author [19] investigated the related problems of maximal operators with the supremum restricted to lacunary radii.

In this article we focus on mα\mathrm{m}^{\alpha} acting on radial functions, which depend only on d(x,0)d(x,0) for 00 being the origin in Hn\mathbb{H}^{n}. Our main result is the following.

Theorem 1.1. Suppose n≥2n\geq2, 1<p≤∞1<p\leq\infty and α∈R\alpha\in\mathbb{R} with α>(1−n)/2\alpha>(1-n)/2.

(i) mα\mathrm{m}^{\alpha} is bounded on Lradp(Hn)L_{\mathrm{rad}}^{p}(\mathbb{H}^{n}) if and only if

{α>1−n+np,1<p≤2,α>1p−n−12,2<p≤2nn−1,α≥1−np,p>2nn−1.\begin{cases} \alpha>1-n+\dfrac{n}{p}, & 1<p\leq2,\\ \alpha>\dfrac{1}{p}-\dfrac{n-1}{2}, & 2<p\leq\dfrac{2n}{n-1},\\ \alpha\geq\dfrac{1-n}{p}, & p>\dfrac{2n}{n-1}. \end{cases}

(ii) If

α={1−n+np,1<p≤2,1p−n−12,2<p≤2nn−1,\alpha= \begin{cases} 1-n+\dfrac{n}{p}, & 1<p\leq2,\\ \dfrac{1}{p}-\dfrac{n-1}{2}, & 2<p\leq\dfrac{2n}{n-1}, \end{cases}

then mα\mathrm{m}^{\alpha} is bounded from Lradp,1(Hn)L_{\mathrm{rad}}^{p,1}(\mathbb{H}^{n}) to Lradp,∞(Hn)L_{\mathrm{rad}}^{p,\infty}(\mathbb{H}^{n}).

The relations between α\alpha and pp in Theorem 1.1 are indicated in Figure 1 below.

The $(1/p,\alpha)$-plot for $n=2$

Figure 1. The (1/p,α)(1/p,\alpha)-plot for n=2n=2. mα\mathbf{m}^{\alpha} is bounded on Lradp(Hn)L_{\mathrm{rad}}^{p}(\mathbb{H}^{n}) if and only if (1/p,α)(1/p,\alpha) is on or above the solid segment OAOA (not containing AA), or strictly above the dashed folded segments ABCABC. In addition, if (1/p,α)(1/p,\alpha) is on the dashed folded segments ABCABC (containing AA), mα\mathbf{m}^{\alpha} is bounded from Lradp,1(Hn)L_{\mathrm{rad}}^{p,1}(\mathbb{H}^{n}) to Lradp,∞(Hn)L_{\mathrm{rad}}^{p,\infty}(\mathbb{H}^{n}).

We would like to mention that according to the arguments in [4], the threshold of the range of α\alpha in Theorem 1.1(i) corresponds to both the threshold of the Lebesgue index and the growth in time in the local smoothing estimates for wave equation with radial initial datum, obtained by Chen, Duong, Yan and the named author [3]. In addition, both the strong-type and the restricted weak-type boundedness coincide with its Euclidean counterpart [14 Corollaries 1.5 and 1.6(b)], indicating that the geometry at infinity induced by the negative curvature of Hn\mathbb{H}^{n} imposes no additional restriction on the LradpL_{\mathrm{rad}}^{p}-boundedness. Based on this observation, we conjecture that the sufficient condition for the general LpL^{p}-boundedness of mα\mathbf{m}^{\alpha} in [4 Theorems 1.1 and 1.2] can be sharpened so that there is no gap between them.

To show Theorem 1.1, we first derive the explicit radial kernel expressions of the operator Mtα\mathcal{M}_{t}^{\alpha} in terms of the Gauss hypergeometric function. We then decompose the operator mα\mathbf{m}^{\alpha} into local and global parts, with supremums taken over all small and large radii, respectively. The local part can be essentially dominated by the maximal operators with positive Euclidean radial kernels studied in [5, 6, 14]. For the global part, we continue splitting the supremum into unit intervals and use the exponential decay of the kernel to obtain summable estimates.

The rest of this article is organized as follows. Section 2 recalls basic facts about hyperbolic spaces and establishes the explicit radial kernel expressions and estimates. With these preparations, in Sections 3 we prove Theorem 1.1.

Preliminaries

In this section, we recall basic facts about the hyperbolic space Hn\mathbb{H}^{n} and obtain the explicit radial kernel expressions of the spherical maximal operators.

Hn as the hyperboloid

On R1+n\mathbb{R}^{1+n} we define the Minkowski metric

[x,y]=x0y0−x′⋅y′,(x0,x′), (y0,y′)∈R1+n.[x,y] = x_{0}y_{0} - x^{\prime} \cdot y^{\prime}, \qquad(x_{0},x^{\prime}),\ (y_{0},y^{\prime}) \in\mathbb{R}^{1+n}.

The hyperboloid model is

Hn:={x=(x0,x′)∈R1+n:[x,x]=1, x0>0}.\mathbb{H}^{n} := \{x = (x_{0},x^{\prime}) \in\mathbb{R}^{1+n} : [x,x] = 1,\ x_{0} > 0\}.

For x∈Hnx \in\mathbb{H}^{n}, its tangent space is

TxHn:={v∈R1+n:[v,x]=0}.T_{x}\mathbb{H}^{n} := \{v \in\mathbb{R}^{1+n} : [v,x] = 0\}.

The Riemannian metric g\mathbf{g} is given by the bilinear form

⟨v,w⟩x:=−[v,w],v,w∈TxHn.\langle v,w\rangle_{x} := -[v,w], \qquad v,w \in T_{x}\mathbb{H}^{n}.

This metric gives the complete simply connected space of constant sectional curvature −1-1. Its geodesic distance is given by

d(x,y):=arch⁡([x,y]).d(x,y) := \operatorname{arch}([x,y]).

Take 0:=(1,0,…,0)∈Hn\mathbf{0} := (1,0,\ldots,0) \in\mathbb{H}^{n}. The geodesic polar coordinates at 0\mathbf{0} are

x=(ch⁡R,sh⁡Rθ),R>0, θ∈Sn−1.x = (\operatorname{ch} R,\operatorname{sh} R\theta), \qquad R > 0,\ \theta\in S^{n-1}.

In these coordinates the Riemannian metric reads

g=dR2+(sh⁡R)2gSn−1.\mathbf{g} = dR^{2} + (\operatorname{sh} R)^{2}\mathbf{g}_{S^{n-1}}.

The induced Riemannian measure is therefore given by

dx=(sh⁡R)n−1dR dθ.(3)dx = (\operatorname{sh} R)^{n-1}dR\,d\theta. \tag*{(3)}

If x=(ch⁡R,sh⁡Rθ)x = (\operatorname{ch} R,\operatorname{sh} R\theta) and y=(ch⁡r,sh⁡rω)y = (\operatorname{ch} r,\operatorname{sh} r\omega), then we have the hyperbolic cosine law:

ch⁡d(x,y)=ch⁡Rch⁡r−sh⁡Rsh⁡r θ⋅ω.(4)\operatorname{ch} d(x,y) = \operatorname{ch} R\operatorname{ch} r - \operatorname{sh} R\operatorname{sh} r\,\theta\cdot\omega. \tag*{(4)}

Radial functions on Hn\mathbb{H}^{n} depend on R(x):=d(x,0)R(x) := d(x,\mathbf{0}) only. In other words, for such a function ff, there exists another function hh, namely the radial profile of ff, defined on (0,∞)(0,\infty), such that

f(x)=h(R(x))for x≠0.(5)f(x) = h(R(x)) \qquad\text{for } x \ne\mathbf{0}. \tag*{(5)}

In the sequel, we always identify ff with its radial profile. Note that in view of (2.1),

∫Hnf(x) dx=νn−1∫0∞f(r)(sh⁡r)n−1 dr.(6)\int_{\mathbb{H}^{n}} f(x)\,dx = \nu_{n-1}\int_{0}^{\infty} f(r)(\operatorname{sh} r)^{n-1}\,dr. \tag*{(6)}

where νn−1=∣Sn−1∣\nu_{n-1} = |S^{n-1}|.

Explicit kernel expressions for MtαM_t^\alpha

For Re⁡c>Re⁡b>0\operatorname{Re} c > \operatorname{Re} b > 0 and ∣z∣<1|z| < 1, let

F(a,b;c;z):=Γ(c)Γ(b)Γ(c−b)∫01(1−τ)c−b−1τb−1(1−zτ)−a dτ(7)F(a,b;c;z) := \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1 (1-\tau)^{c-b-1}\tau^{b-1}(1-z\tau)^{-a}\,\mathrm{d}\tau \tag*{(7)}

be the Gauss hypergeometric function. We further define

F(a,b;c;z):=F(a,b;c;z)Γ(c).(8)\mathbf{F}(a,b;c;z) := \frac{F(a,b;c;z)}{\Gamma(c)}. \tag*{(8)}

We need the following analytic facts about the functions FF and F\mathbf{F}.

Lemma 2.1. Suppose z∈Cz \in\mathbb{C} with ∣z∣<1|z| < 1.

  1. For Re⁡c>Re⁡b>0\operatorname{Re} c > \operatorname{Re} b > 0 the function F(a,b;c;z)F(a,b;c;z) is an entire function in aa.

  2. The function F(a,b;c;z)\mathbf{F}(a,b;c;z) is an entire function in cc.

  3. The functions FF and F\mathbf{F} satisfy

∣F(a,b;c;z)∣+∣F(a,b;c;z)∣≲{(1−∣z∣)Re⁡(c−a−b),a+b>c,1+log⁡11−∣z∣,a+b=c,1,a+b<c.|F(a,b;c;z)| + |\mathbf{F}(a,b;c;z)| \lesssim \begin{cases} (1-|z|)^{\operatorname{Re}(c-a-b)}, & a+b>c,\\ 1+\log\dfrac{1}{1-|z|}, & a+b=c,\\ 1, & a+b<c. \end{cases}

In particular, for Re⁡(c−a−b)>−1\operatorname{Re}(c-a-b)>-1, FF and F\mathbf{F} are both locally integrable in zz.

Proof. (1) follows readily from (7) and Morera’s Theorem.

To show (2), it follows from [8 (2) and (5) on pp. 56–57] that

F(a,b;c;z)=∑j=0∞Γ(a+j)Γ(a)Γ(b+j)Γ(b)zjj!Γ(c+j).\mathbf{F}(a,b;c;z)=\sum_{j=0}^{\infty}\frac{\Gamma(a+j)}{\Gamma(a)}\frac{\Gamma(b+j)}{\Gamma(b)}\frac{z^j}{j!\Gamma(c+j)}.

Since for each j=0,1,…j=0,1,\ldots, 1/Γ(⋅+j)1/\Gamma(\cdot+j) is entire, for ∣z∣<1|z|<1 the above function is entire in cc.

To show (3), we apply [8 (2) and (5) on pp. 56–57] to derive

∣F(a,b;c;z)∣+∣F(a,b;c;z)∣≲∑j=0∞(1+j)Re⁡(a+b−c)−1∣z∣j.|F(a,b;c;z)|+|\mathbf{F}(a,b;c;z)|\lesssim\sum_{j=0}^{\infty}(1+j)^{\operatorname{Re}(a+b-c)-1}|z|^j.

This series is dominated by the desired estimate, so are F(a,b;c;z)F(a,b;c;z) and F(a,b;c;z)\mathbf{F}(a,b;c;z). □

Recall that for Re⁡α>0\operatorname{Re}\alpha>0, the spherical means MtαM_t^\alpha on Hn\mathbb{H}^n are defined by (1). We have the following explicit expression.

Lemma 2.2. Suppose n≥2n\geq2 and α∈C\alpha\in\mathbb{C}. For a smooth compactly supported radial function ff, Mtα(f)M_t^\alpha(f) can be analytically continued to Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2 and is also radial. We can write

Mtα(f)(R)=∫0∞Ktα(R,r)f(r)(sinh⁡r)n−1 dr,M_t^\alpha(f)(R)=\int_0^\infty K_t^\alpha(R,r)f(r)(\sinh r)^{n-1}\,\mathrm{d}r,

where for t,R,r>0t,R,r>0 away from t=∣R−r∣t=|R-r| and t=R+rt=R+r, KtαK_t^\alpha is given as follows:

  1. If t<∣R−r∣t<|R-r|, then Ktα(R,r)=0K_t^\alpha(R,r)=0.

  2. If ∣R−r∣<t<R+r|R-r|<t<R+r, then

Ktα(R,r)=2n−32νn−2Γ(n−12)(sinh⁡t)2−n(cosh⁡t−1)α(sinh⁡Rsinh⁡r)1−n2(cosh⁡t−cosh⁡(R−r))α+n−32K_t^\alpha(R,r)=2^{\frac{n-3}{2}}\nu_{n-2}\Gamma\left(\frac{n-1}{2}\right)\frac{(\sinh t)^{2-n}}{(\cosh t-1)^\alpha}(\sinh R\sinh r)^{\frac{1-n}{2}}\left(\cosh t-\cosh(R-r)\right)^{\alpha+\frac{n-3}{2}}
×F(3−n2,n−12;α+n−12;ch⁡t−ch⁡(R−r)2sh⁡Rsh⁡r).\times F\left(\frac{3-n}{2},\frac{n-1}{2};\alpha+\frac{n-1}{2};\frac{\operatorname{ch}t-\operatorname{ch}(R-r)}{2\operatorname{sh}R\operatorname{sh}r}\right).

(iii) If t>R+rt>R+r, then

Ktα(R,r)=2n−2νn−2(n−12)2(sh⁡t)2−nΓ(α)Γ(n−1)(ch⁡t−1)α(ch⁡t−ch⁡(R−r))α−1×F(1−α,n−12;n−1;2sh⁡Rsh⁡rch⁡t−ch⁡(R−r)).\begin{aligned} K_t^\alpha(R,r)=\frac{2^{n-2}\nu_{n-2}\left(\frac{n-1}{2}\right)^2(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)\Gamma(n-1)(\operatorname{ch}t-1)^\alpha}\left(\operatorname{ch}t-\operatorname{ch}(R-r)\right)^{\alpha-1} \\ &\times F\left(1-\alpha,\frac{n-1}{2};n-1;\frac{2\operatorname{sh}R\operatorname{sh}r}{\operatorname{ch}t-\operatorname{ch}(R-r)}\right). \end{aligned}

Moreover, for Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2, the values on t=∣R−r∣t=|R-r| and t=R+rt=R+r may be assigned arbitrarily.

Proof. Assume first Re⁡α>0\operatorname{Re}\alpha>0. In view of (1), (6) and (4), for x=(ch⁡R,sh⁡Rθ)x=(\operatorname{ch}R,\operatorname{sh}R\theta),

Mtα(f)(x)=(sh⁡t)2−nΓ(α)(ch⁡t−1)α∫0∞f(r)(sh⁡r)n−1×(∫Sn−1(ch⁡t−ch⁡Rch⁡r+sh⁡Rsh⁡r θ⋅ω)+α−1 dω) dr=νn−2(sh⁡t)2−nΓ(α)(ch⁡t−1)α∫0∞f(r)(sh⁡r)n−1×(∫−11(ch⁡t−ch⁡Rch⁡r+sh⁡Rsh⁡r w)+α−1(1−w2)n−32 dw) dr=νn−2(sh⁡t)2−nΓ(α)(ch⁡t−1)α∫0∞f(r)(sh⁡r)n−1(sh⁡Rsh⁡r)2−n×(∫ch⁡(R−r)ch⁡(R+r)(ch⁡t−v)+α−1(v−ch⁡(R−r))n−32(ch⁡(R+r)−v)n−32 dv) dr,\begin{aligned} M_t^\alpha(f)(x) &=\frac{(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)(\operatorname{ch}t-1)^\alpha}\int_0^\infty f(r)(\operatorname{sh}r)^{n-1}\\ &\quad\times\left(\int_{S^{n-1}}(\operatorname{ch}t-\operatorname{ch}R\operatorname{ch}r+\operatorname{sh}R\operatorname{sh}r\,\theta\cdot\omega)_+^{\alpha-1}\,d\omega\right)\,dr\\ &=\frac{\nu_{n-2}(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)(\operatorname{ch}t-1)^\alpha}\int_0^\infty f(r)(\operatorname{sh}r)^{n-1}\\ &\quad\times\left(\int_{-1}^1(\operatorname{ch}t-\operatorname{ch}R\operatorname{ch}r+\operatorname{sh}R\operatorname{sh}r\,w)_+^{\alpha-1}(1-w^2)^{\frac{n-3}{2}}\,dw\right)\,dr\\ &=\frac{\nu_{n-2}(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)(\operatorname{ch}t-1)^\alpha}\int_0^\infty f(r)(\operatorname{sh}r)^{n-1}(\operatorname{sh}R\operatorname{sh}r)^{2-n}\\ &\quad\times\left(\int_{\operatorname{ch}(R-r)}^{\operatorname{ch}(R+r)}(\operatorname{ch}t-v)_+^{\alpha-1}(v-\operatorname{ch}(R-r))^{\frac{n-3}{2}}(\operatorname{ch}(R+r)-v)^{\frac{n-3}{2}}\,dv\right)\,dr, \end{aligned}

where (A)+α−1(A)_+^{\alpha-1} denotes Aα−11A>0A^{\alpha-1}\mathbf{1}_{A>0}. Hence

Ktα(R,r)=νn−2(sh⁡t)2−nΓ(α)(ch⁡t−1)α(sh⁡Rsh⁡r)2−n×∫ch⁡(R−r)ch⁡(R+r)(ch⁡t−v)+α−1(v−ch⁡(R−r))n−32(ch⁡(R+r)−v)n−32 dv.(9)\begin{aligned} K_t^\alpha(R,r) &=\frac{\nu_{n-2}(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)(\operatorname{ch}t-1)^\alpha}(\operatorname{sh}R\operatorname{sh}r)^{2-n}\\ &\quad\times\int_{\operatorname{ch}(R-r)}^{\operatorname{ch}(R+r)}(\operatorname{ch}t-v)_+^{\alpha-1}(v-\operatorname{ch}(R-r))^{\frac{n-3}{2}}(\operatorname{ch}(R+r)-v)^{\frac{n-3}{2}}\,dv. \tag*{(9)} \end{aligned}

Now suppose Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2. (i) follows from analytic continuation and the fact that if ch⁡t<ch⁡(R−r)\operatorname{ch}t<\operatorname{ch}(R-r) then the integrand in (9) is zero. To see (ii), note that if ch⁡(R−r)<ch⁡t<ch⁡(R+r)\operatorname{ch}(R-r)<\operatorname{ch}t<\operatorname{ch}(R+r), letting

s:=v−ch⁡(R−r)ch⁡t−ch⁡(R−r)s:=\frac{v-\operatorname{ch}(R-r)}{\operatorname{ch}t-\operatorname{ch}(R-r)}

in the integral in (9), we have

Ktα(R,r)=2n−32νn−2(sh⁡t)2−nΓ(α)(ch⁡t−1)α(sh⁡Rsh⁡r)1−n2(ch⁡t−ch⁡(R−r))α+n−32×∫01(1−s)α−1sn−32(1−ch⁡t−ch⁡(R−r)2sh⁡Rsh⁡rs)n−32 ds.\begin{aligned} K_t^\alpha(R,r)=\frac{2^{\frac{n-3}{2}}\nu_{n-2}(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)(\operatorname{ch}t-1)^\alpha}(\operatorname{sh}R\operatorname{sh}r)^{\frac{1-n}{2}}\left(\operatorname{ch}t-\operatorname{ch}(R-r)\right)^{\alpha+\frac{n-3}{2}} \\ &\times\int_0^1 (1-s)^{\alpha-1}s^{\frac{n-3}{2}}\left(1-\frac{\operatorname{ch}t-\operatorname{ch}(R-r)}{2\operatorname{sh}R\operatorname{sh}r}s\right)^{\frac{n-3}{2}}\,ds. \end{aligned}

In view of (7),

Ktα(R,r)=2n−32νn−2Γ(n−12)(sh⁡t)2−n(ch⁡t−1)α(sh⁡Rsh⁡r)1−n2(ch⁡t−ch⁡(R−r))α+n−32×1Γ(α+n−12)F(3−n2,n−12;α+n−12;ch⁡t−ch⁡(R−r)2sh⁡Rsh⁡r).\begin{aligned} K_t^\alpha(R,r)=2^{\frac{n-3}{2}}\nu_{n-2}\Gamma\left(\frac{n-1}{2}\right)\frac{(\operatorname{sh}t)^{2-n}}{(\operatorname{ch}t-1)^\alpha}(\operatorname{sh}R\operatorname{sh}r)^{\frac{1-n}{2}}(\operatorname{ch}t-\operatorname{ch}(R-r))^{\alpha+\frac{n-3}{2}} \\ &\times\frac{1}{\Gamma\left(\alpha+\frac{n-1}{2}\right)}F\left(\frac{3-n}{2},\frac{n-1}{2};\alpha+\frac{n-1}{2};\frac{\operatorname{ch}t-\operatorname{ch}(R-r)}{2\operatorname{sh}R\operatorname{sh}r}\right). \end{aligned}

It remains to invoke (8) to obtain the desired expression. Then one uses Lemma 2.1(2) to analytically continue KtαK_t^\alpha to Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2.

To see (iii), note that if ch⁡t>ch⁡(R+r)\operatorname{ch}t>\operatorname{ch}(R+r), letting

s=v−ch⁡(R−r)2sh⁡Rsh⁡rs=\frac{v-\operatorname{ch}(R-r)}{2\operatorname{sh}R\operatorname{sh}r}

in the integral in (9), we have

Ktα(R,r)=2n−2νn−2(sh⁡t)2−nΓ(α)(ch⁡t−1)α(ch⁡t−ch⁡(R−r))α−1×∫01(1−s)n−32sn−32(1−2sh⁡Rsh⁡rch⁡t−ch⁡(R−r)s)α−1 ds.\begin{aligned} K_t^\alpha(R,r)=\frac{2^{n-2}\nu_{n-2}(\operatorname{sh}t)^{2-n}}{\Gamma(\alpha)(\operatorname{ch}t-1)^\alpha}(\operatorname{ch}t-\operatorname{ch}(R-r))^{\alpha-1} \\ &\times\int_0^1(1-s)^{\frac{n-3}{2}}s^{\frac{n-3}{2}}\left(1-\frac{2\operatorname{sh}R\operatorname{sh}r}{\operatorname{ch}t-\operatorname{ch}(R-r)}s\right)^{\alpha-1}\,ds. \end{aligned}

In view of (7), we obtain the desired expression. Then one uses Lemma 2.1(1) to analytically continue KtαK_t^\alpha to Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2.

Finally, for Re⁡α>(1−n)/2\operatorname{Re}\alpha>(1-n)/2, the expression in (ii) as well as Lemma 2.1(3) indicates that KtαK_t^\alpha is a locally integrable function on (0,∞)×(0,∞)(0,\infty)\times(0,\infty). Thus the values on these surfaces can be assigned arbitrarily. □

Lemmas 2.1(3) and 2.2(ii)(iii) immediately yield the following crucial estimates.

Lemma 2.3. Suppose α∈R\alpha\in\mathbb{R} with α>(1−n)/2\alpha>(1-n)/2. If ∣R−r∣<t<R+r|R-r|<t<R+r,

∣Ktα(R,r)∣≲(sh⁡t)2−n(ch⁡t−1)α(sh⁡Rsh⁡r)1−n2(ch⁡t−ch⁡(R−r))α+n−32(10)|K_t^\alpha(R,r)|\lesssim\frac{(\operatorname{sh}t)^{2-n}}{(\operatorname{ch}t-1)^\alpha}(\operatorname{sh}R\operatorname{sh}r)^{\frac{1-n}{2}}(\operatorname{ch}t-\operatorname{ch}(R-r))^{\alpha+\frac{n-3}{2}} \tag*{(10)}
×{(ch⁡(R+r)−ch⁡tsh⁡Rsh⁡r)α+n−32,1−n2<α<3−n2,1+log⁡2sh⁡Rsh⁡rch⁡(R+r)−ch⁡t,α=3−n2,1,α>3−n2.\times \begin{cases} \left(\dfrac{\operatorname{ch}(R+r)-\operatorname{ch}t}{\operatorname{sh}R\operatorname{sh}r}\right)^{\alpha+\frac{n-3}{2}}, & \dfrac{1-n}{2}<\alpha<\dfrac{3-n}{2},\\ 1+\log\dfrac{2\operatorname{sh}R\operatorname{sh}r}{\operatorname{ch}(R+r)-\operatorname{ch}t}, & \alpha=\dfrac{3-n}{2},\\ 1, & \alpha>\dfrac{3-n}{2}. \end{cases}

If t>R+rt>R+r,

∣Ktα(R,r)∣≲(sh⁡t)2−n(ch⁡t−1)α(ch⁡t−ch⁡(R−r))α−1(11)|K_t^\alpha(R,r)|\lesssim\frac{(\operatorname{sh}t)^{2-n}}{(\operatorname{ch}t-1)^\alpha}(\operatorname{ch}t-\operatorname{ch}(R-r))^{\alpha-1} \tag*{(11)}
×{(ch⁡t−ch⁡(R+r)ch⁡t−ch⁡(R−r))α+n−32,1−n2<α<3−n2,1+log⁡ch⁡t−ch⁡(R−r)ch⁡t−ch⁡(R+r),α=3−n2,1,α>3−n2.\times \begin{cases} \left(\dfrac{\operatorname{ch}t-\operatorname{ch}(R+r)}{\operatorname{ch}t-\operatorname{ch}(R-r)}\right)^{\alpha+\frac{n-3}{2}}, & \dfrac{1-n}{2}<\alpha<\dfrac{3-n}{2},\\ 1+\log\dfrac{\operatorname{ch}t-\operatorname{ch}(R-r)}{\operatorname{ch}t-\operatorname{ch}(R+r)}, & \alpha=\dfrac{3-n}{2},\\ 1, & \alpha>\dfrac{3-n}{2}. \end{cases}

We conclude this section by showing the following result, which holds for general Cc∞(Hn)C_c^\infty(\mathbb{H}^n) functions.

Lemma 2.4. For α>(1−n)/2\alpha> (1-n)/2, δ>0\delta> 0 and f∈Cc∞(Hn)f \in C^\infty_c(\mathbb{H}^n),

Mtα+δ(f)(x)≲Mtα(f)(x).\mathrm{M}_t^{\alpha+\delta}(f)(x) \lesssim\mathrm{M}_t^\alpha(f)(x).

Proof. In view of the proof of [7 Lemma 1],

Mtα+δ(f)(x)=1Γ(δ)∫01Mτ(t,u)α(f)(x)(sinh⁡τ(t,u)sinh⁡t)n−2uα(1−u)δ−1 du,\mathrm{M}_t^{\alpha+\delta}(f)(x)=\frac{1}{\Gamma(\delta)}\int_0^1\mathrm{M}_{\tau(t,u)}^\alpha(f)(x)\left(\frac{\sinh\tau(t,u)}{\sinh t}\right)^{n-2}u^\alpha(1-u)^{\delta-1}\,du,

where the function τ(t,u)≥0\tau(t,u)\geq0 is determined by

(sinh⁡τ(t,u)2)2=u(sinh⁡t2)2.\left(\sinh\frac{\tau(t,u)}{2}\right)^2=u\left(\sinh\frac{t}{2}\right)^2.

Note that sinh⁡τ(t,u)≲u1/2sinh⁡t\sinh\tau(t,u)\lesssim u^{1/2}\sinh t, whence

Mtα+δ(f)(x)≲Mtα(f)(x)∫01uα+n−22(1−u)δ−1 du≲Mtα(f)(x),\mathrm{M}_t^{\alpha+\delta}(f)(x)\lesssim\mathrm{M}_t^\alpha(f)(x)\int_0^1u^{\alpha+\frac{n-2}{2}}(1-u)^{\delta-1}\,du\lesssim\mathrm{M}_t^\alpha(f)(x),

as desired. □

Proof of Theorem 1.1

In this section, we decompose the maximal operator mα\mathrm{m}^\alpha into the local part

m0α(f)(R)=sup⁡0<t≤1∣Mtα(f)(R)∣\mathrm{m}_0^\alpha(f)(R)=\sup_{0<t\leq1}\left|\mathrm{M}_t^\alpha(f)(R)\right|

as well as the global parts

mTα(f)(R)=sup⁡T≤t≤T+1∣Mtα(f)(R)∣,T=1,2,….\mathrm{m}_T^\alpha(f)(R)=\sup_{T\leq t\leq T+1}\left|\mathrm{M}_t^\alpha(f)(R)\right|,\qquad T=1,2,\ldots.

We then separately consider their LradpL^p_{\mathrm{rad}}-boundedness.

The local part m0α\mathrm{m}_0^\alpha

The aim of this subsection is to show the following result.

Proposition 3.1. Suppose 1<p<∞1<p<\infty, α<1/p\alpha<1/p and

α≥{1−n+np,1<p≤2,1p−n−12,2<p≤2nn−1,1−np,p>2nn−1.(12)\alpha\geq \begin{cases} 1-n+\frac{n}{p}, & 1<p\leq2,\\ \frac{1}{p}-\frac{n-1}{2}, & 2<p\leq\frac{2n}{n-1},\\ \frac{1-n}{p}, & p>\frac{2n}{n-1}. \end{cases} \tag*{(12)}

Then m0α\mathrm{m}_0^\alpha is bounded from Lradp,1(Hn)L_{\mathrm{rad}}^{p,1}(\mathbb{H}^n) to Lradp,∞(Hn)L_{\mathrm{rad}}^{p,\infty}(\mathbb{H}^n). Furthermore, if p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, then m0α\mathrm{m}_0^\alpha is bounded on Lradp(Hn)L_{\mathrm{rad}}^p(\mathbb{H}^n).

To prove Proposition 3.1, we define the Euclidean radial kernel Etα(R,r)E_t^\alpha(R,r) as follows: (i) If t<∣R−r∣t<|R-r|, then Etα(R,r)=0E_t^\alpha(R,r)=0. (ii) If ∣R−r∣<t<R+r|R-r|<t<R+r, then

Etα(R,r)=(Rr)1−n2tn−2+2α(t2−(R−r)2)α+n−32{((R+r)2−t2)α+n−32Rr,1−n2<α<3−n2,1+log⁡(4Rr(R+r)2−t2),α=3−n2,1,α>3−n2.E_t^\alpha(R,r)=\frac{(Rr)^{\frac{1-n}{2}}}{t^{n-2+2\alpha}}\left(t^2-(R-r)^2\right)^{\alpha+\frac{n-3}{2}} \begin{cases} \frac{\left((R+r)^2-t^2\right)^{\alpha+\frac{n-3}{2}}}{Rr}, & \frac{1-n}{2}<\alpha<\frac{3-n}{2},\\[6pt] 1+\log\left(\frac{4Rr}{(R+r)^2-t^2}\right), & \alpha=\frac{3-n}{2},\\[6pt] 1, & \alpha>\frac{3-n}{2}. \end{cases}

(iii) If t>R+rt>R+r, then

Etα(R,r)=1tn−2+2α(t2−(R−r)2)α−1×{(t2−(R+r)2t2−(R−r)2)α+n−32,1−n2<α<3−n2,1+log⁡(t2−(R−r)2t2−(R+r)2),α=3−n2,1,α>3−n2.E_t^\alpha(R,r)=\frac{1}{t^{n-2+2\alpha}}\left(t^2-(R-r)^2\right)^{\alpha-1}\times \begin{cases} \left(\frac{t^2-(R+r)^2}{t^2-(R-r)^2}\right)^{\alpha+\frac{n-3}{2}}, & \frac{1-n}{2}<\alpha<\frac{3-n}{2},\\ 1+\log\left(\frac{t^2-(R-r)^2}{t^2-(R+r)^2}\right), & \alpha=\frac{3-n}{2},\\ 1, & \alpha>\frac{3-n}{2}. \end{cases}

We further define the associated maximal function

Eα(f)(R):=sup⁡t>0∫0∞Etα(R,r)∣f(r)∣rn−1 dr.\mathcal{E}^\alpha(f)(R):=\sup_{t>0}\int_0^\infty E_t^\alpha(R,r)|f(r)|r^{n-1}\,dr.

The following result concerning the LradpL^p_{\mathrm{rad}}-boundedness of Eα\mathcal{E}^\alpha serves as a key ingredient in the proof of Proposition 3.1.

Lemma 3.2. Let 1<p<∞1<p<\infty, (1−n)/2<α<1/p(1-n)/2<\alpha<1/p. If (12) holds, then Eα\mathcal{E}^\alpha is bounded from Lp(R+,rn−1dr)L^p(\mathbb{R}_+,r^{n-1}dr) to Lp,∞(R+,rn−1dr)L^{p,\infty}(\mathbb{R}_+,r^{n-1}dr). Furthermore, if p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, then Eα\mathcal{E}^\alpha is bounded on Lp(R+,rn−1dr)L^p(\mathbb{R}_+,r^{n-1}dr).

Proof. This follows from the proofs of sufficiency part of [14 Theorems A and B(b)], see [14 Sections 3–4]. □

Now we present the

Proof of Proposition 3.1. Fix a radial function ff. We write

m0α(f)=1(0,2]m0α(f)+∑j=2∞1(j,j+1]m0α(f)=1(0,2]m0α(f1(0,3])⏟I1α(f)+∑j=2∞1(j,j+1]m0α(f1(j−1,j+2])⏟Ijα(f),(13)\begin{aligned} \mathfrak{m}_0^\alpha(f)=\mathbf{1}_{(0,2]}\mathfrak{m}_0^\alpha(f)+\sum_{j=2}^{\infty}\mathbf{1}_{(j,j+1]}\mathfrak{m}_0^\alpha(f) \\ &=\underbrace{\mathbf{1}_{(0,2]}\mathfrak{m}_0^\alpha(f\mathbf{1}_{(0,3]})}_{\mathfrak{I}_1^\alpha(f)} +\sum_{j=2}^{\infty}\underbrace{\mathbf{1}_{(j,j+1]}\mathfrak{m}_0^\alpha(f\mathbf{1}_{(j-1,j+2]})}_{\mathfrak{I}_j^\alpha(f)}, \tag*{(13)} \end{aligned}

where the second equality follows from Lemma 2.2(i).

We begin by estimating I1α(f)\mathfrak{I}_1^\alpha(f). Since

ch⁡a−ch⁡b=ea2(1−e−a+b)(1−e−a−b)\operatorname{ch}a-\operatorname{ch}b=\frac{e^a}{2}(1-e^{-a+b})(1-e^{-a-b})

Lemma 2.3 yields that for 0<t≤10<t\leq1, R≤2R\leq2 and r≤3r\leq3, we have

∣Ktα(R,r)∣≲Etα(R,r).|K_t^\alpha(R,r)|\lesssim E_t^\alpha(R,r).

Therefore, in view of Lemma 3.2, I1α\mathfrak{I}_1^\alpha satisfies the desired boundedness properties.

Fix j=2,3,…j=2,3,\ldots. It remains to estimate Ijα(f)\mathfrak{I}_j^\alpha(f). For 0<t≤10<t\leq1, R∈(j,j+1]R\in(j,j+1] and r∈(j−1,j+2]r\in(j-1,j+2], we have

ch⁡t−ch⁡(R−r)∼t2−(R−r)2,ch⁡(R+r)−ch⁡t∼sh⁡Rsh⁡r.\operatorname{ch}t-\operatorname{ch}(R-r)\sim t^2-(R-r)^2,\qquad\operatorname{ch}(R+r)-\operatorname{ch}t\sim\operatorname{sh}R\operatorname{sh}r.

Thus Lemma 2.3 yields

∣Ktα(R,r)∣(sh⁡r)n−1≲t2−n−2α(t2−(R−r)2)α+n−32,|K_t^\alpha(R,r)|(\operatorname{sh}r)^{n-1}\lesssim t^{2-n-2\alpha}\left(t^2-(R-r)^2\right)^{\alpha+\frac{n-3}{2}},

whence

m0α(f)(R)≲sup⁡0<t≤1t2−n−2α∫−tt(t2−r2)α+n−32∣f(R−r)∣ dr=sup⁡0<t≤1∫−11(1−r2)α+n−32∣f(R−tr)∣ dr.\begin{aligned} \mathfrak{m}_0^\alpha(f)(R)\lesssim\sup_{0<t\leq1}t^{2-n-2\alpha}\int_{-t}^{t}\left(t^2-r^2\right)^{\alpha+\frac{n-3}{2}}|f(R-r)|\,dr \\ &= \sup_{0<t\leq1}\int_{-1}^{1}(1-r^2)^{\alpha+\frac{n-3}{2}}|f(R-tr)|\,\mathrm{d}r. \end{aligned}

It follows from [14 Theorems A and B(b)] that IjαI_j^\alpha is bounded from Lp((j−1,j+2],dr)L^p((j-1,j+2],\mathrm{d}r) to Lp((j,j+1],dR)L^p((j,j+1],\mathrm{d}R) when α>1/p−(n−1)/2\alpha>1/p-(n-1)/2, and from Lp,1((j−1,j+2],dr)L^{p,1}((j-1,j+2],\mathrm{d}r) to Lp,∞((j,j+1],dR)L^{p,\infty}((j,j+1],\mathrm{d}R). These estimates in turn imply

∥Ijα(f)∥Lp((j,j+1],(shR)n−1dR)≲∥f∥Lp((j−1,j+2],(shr)n−1dr)for α>1p−n−12,\left\|I_j^\alpha(f)\right\|_{L^p((j,j+1],(\mathop{\rm sh}R)^{n-1}\mathrm{d}R)} \lesssim \|f\|_{L^p((j-1,j+2],(\mathop{\rm sh}r)^{n-1}\mathrm{d}r)} \qquad\text{for }\alpha>\frac{1}{p}-\frac{n-1}{2},

and

∥Ijα(f)∥Lp,∞((j,j+1],(shR)n−1dR)≲∥f∥Lp,1((j−1,j+2],(shr)n−1dr)for α≥1p−n−12.\left\|I_j^\alpha(f)\right\|_{L^{p,\infty}((j,j+1],(\mathop{\rm sh}R)^{n-1}\mathrm{d}R)} \lesssim \|f\|_{L^{p,1}((j-1,j+2],(\mathop{\rm sh}r)^{n-1}\mathrm{d}r)} \qquad\text{for }\alpha\geq\frac{1}{p}-\frac{n-1}{2}.

By the disjointness of (j,j+1](j,j+1] and finite overlapping of (j−1,j+2](j-1,j+2], we sum over j=2,3,…j=2,3,\ldots to obtain

∥∑j=2∞Ijα(f)∥Lradp(Hn)p=∑j=2∞∥Ijα(f)∥Lp((j,j+1],(shR)n−1dR)p≲∑j=2∞∥f∥Lp((j−1,j+2],(shr)n−1dr)p≲∥f∥Lradp(Hn)pfor α>1p−n−12.(14)\left\|\sum_{j=2}^{\infty}I_j^\alpha(f)\right\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}^p = \sum_{j=2}^{\infty}\left\|I_j^\alpha(f)\right\|_{L^p((j,j+1],(\mathop{\rm sh}R)^{n-1}\mathrm{d}R)}^p \lesssim \sum_{j=2}^{\infty}\|f\|_{L^p((j-1,j+2],(\mathop{\rm sh}r)^{n-1}\mathrm{d}r)}^p \lesssim \|f\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}^p \qquad\text{for }\alpha>\frac{1}{p}-\frac{n-1}{2}. \tag*{(14)}

while in view of [9 Proposition 1.4.9]

∥∑j=2∞Ijα(f)∥Lradp,∞(Hn)p≤∑j=2∞∥Ijα(f)∥Lp,∞((j−1,j],(shR)n−1dR)p≲∑j=2∞∥f∥Lp,1((j−1,j+2],(shr)n−1dr)p∼∑j=2∞(∫0∞(∫j−1j+21{∣f(r)∣>s}(shr)n−1 dr)1/pds)p≤(∫0∞(∑j=2∞∫j−1j+21{∣f(r)∣>s}(shr)n−1 dr)1/pds)p\begin{aligned} \left\|\sum_{j=2}^{\infty}I_j^\alpha(f)\right\|_{L^{p,\infty}_{\mathrm{rad}}(\mathbb{H}^n)}^p \leq \sum_{j=2}^{\infty}\left\|I_j^\alpha(f)\right\|_{L^{p,\infty}((j-1,j],(\mathop{\rm sh}R)^{n-1}\mathrm{d}R)}^p \lesssim \sum_{j=2}^{\infty}\|f\|_{L^{p,1}((j-1,j+2],(\mathop{\rm sh}r)^{n-1}\mathrm{d}r)}^p \\ &\sim \sum_{j=2}^{\infty}\left(\int_0^\infty\left(\int_{j-1}^{j+2}\mathbf{1}_{\{|f(r)|>s\}}(\mathop{\rm sh}r)^{n-1}\,\mathrm{d}r\right)^{1/p}\mathrm{d}s\right)^p \leq \left(\int_0^\infty\left(\sum_{j=2}^{\infty}\int_{j-1}^{j+2}\mathbf{1}_{\{|f(r)|>s\}}(\mathop{\rm sh}r)^{n-1}\,\mathrm{d}r\right)^{1/p}\mathrm{d}s\right)^p \end{aligned}
≲(∫0∞(∫0∞1{∣f(r)∣>s}(shr)n−1 dr)1/pds)p∼∥f∥Lradp,1(Hn)pfor α≥1p−n−12.(15)\lesssim \left(\int_0^\infty\left(\int_0^\infty\mathbf{1}_{\{|f(r)|>s\}}(\mathop{\rm sh}r)^{n-1}\,\mathrm{d}r\right)^{1/p}\mathrm{d}s\right)^p \sim \|f\|_{L^{p,1}_{\mathrm{rad}}(\mathbb{H}^n)}^p \qquad\text{for }\alpha\geq\frac{1}{p}-\frac{n-1}{2}. \tag*{(15)}

According to (13), these estimates yield the desired boundedness of m0αm_0^\alpha.

The global parts mTαm_{T}^{\alpha} (T=1,2,…T=1,2,\ldots)

The aim of this subsection is to show the following result.

Proposition 3.3. Suppose 1<p<∞1<p<\infty, T=1,2,…T=1,2,\ldots, α<1/p\alpha<1/p and (12) holds. Set

cp=(n−1)min⁡{1p,1p′}>0.c_p=(n-1)\min\left\{\frac{1}{p},\frac{1}{p'}\right\}>0.

Then

∥mTα(f)∥Lradp,∞(Hn)≲Te−cpT∥f∥Lradp,1(Hn).\left\|m_{T}^{\alpha}(f)\right\|_{L_{\mathrm{rad}}^{p,\infty}(\mathbb{H}^{n})}\lesssim T e^{-c_pT}\left\|f\right\|_{L_{\mathrm{rad}}^{p,1}(\mathbb{H}^{n})}.

Furthermore, if p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, then the above estimate also holds with LradpL_{\mathrm{rad}}^{p} in place of Lradp,∞L_{\mathrm{rad}}^{p,\infty} and Lradp,1L_{\mathrm{rad}}^{p,1}.

Proof. We first write

mTα(f)≤1(1/10,∞)mTα(f1(1/10,∞))⏟J1α(f)+1(1/10,∞)mTα(f1(0,1/10])⏟J2α(f)+1(0,1/10]mTα(f)⏟J3α(f).(16)m_{T}^{\alpha}(f)\leq\underbrace{\mathbf{1}_{(1/10,\infty)}m_{T}^{\alpha}\bigl(f\mathbf{1}_{(1/10,\infty)}\bigr)}_{J_{1}^{\alpha}(f)}+\underbrace{\mathbf{1}_{(1/10,\infty)}m_{T}^{\alpha}\bigl(f\mathbf{1}_{(0,1/10]}\bigr)}_{J_{2}^{\alpha}(f)}+\underbrace{\mathbf{1}_{(0,1/10]}m_{T}^{\alpha}(f)}_{J_{3}^{\alpha}(f)}. \tag*{(16)}
  • Estimates for J1α(f)J_{1}^{\alpha}(f). We shall show that for α≥1/p−(n−1)/2\alpha\geq1/p-(n-1)/2

∥J1α(f)∥Lradp,∞(Hn)≲Te−cpT∥f∥Lradp,1(Hn),(17)\left\|J_{1}^{\alpha}(f)\right\|_{L_{\mathrm{rad}}^{p,\infty}(\mathbb{H}^{n})}\lesssim T e^{-c_pT}\left\|f\right\|_{L_{\mathrm{rad}}^{p,1}(\mathbb{H}^{n})}, \tag*{(17)}

and that for α>1/p−(n−1)/2\alpha>1/p-(n-1)/2 the above estimate holds with LradpL_{\mathrm{rad}}^{p} in place of Lradp,1L_{\mathrm{rad}}^{p,1}.

We first claim that for R≥1/10R\geq1/10 and α≥1/p−(n−1)/2\alpha\geq1/p-(n-1)/2,

sup⁡T≤t≤T+1∥Ktα(R,⋅)∥Lp′,∞([1/10,∞),(sinh⁡r)n−1dr)≲T1/p′e−cpT(sinh⁡R)−(n−1)/p(18)\sup_{T\leq t\leq T+1}\left\|K_{t}^{\alpha}(R,\cdot)\right\|_{L^{p',\infty}([1/10,\infty),(\sinh r)^{n-1}dr)} \lesssim T^{1/p'}e^{-c_pT}(\sinh R)^{-(n-1)/p} \tag*{(18)}

and that the same estimates holds with LpL^{p} in place of Lp′,∞L^{p',\infty} for α>1/p−(n−1)/2\alpha>1/p-(n-1)/2.

Once these estimates are prove, it follows from Lemma 2.2(i) and the Lorentz Hölder inequality (cf. [9], the proof of (v) on p. 58]) that for α≥1/p−(n−1)/2\alpha\geq1/p-(n-1)/2

J1α(f)(R)≤sup⁡T≤t≤T+1∥Ktα(R,⋅)∥Lp′,∞([1/10,∞),(sinh⁡r)n−1dr)×∥f∥Lp,1([max⁡{R−T−1,1/10},R+T+1],(sinh⁡r)n−1dr)≲T1/p′e−cpT(sinh⁡R)−(n−1)/p∥f∥Lp,1([max⁡{R−T−1,1/10},R+T+1],(sinh⁡r)n−1dr).\begin{aligned} J_{1}^{\alpha}(f)(R) &\leq\sup_{T\leq t\leq T+1}\left\|K_{t}^{\alpha}(R,\cdot)\right\|_{L^{p',\infty}([1/10,\infty),(\sinh r)^{n-1}dr)}\\ &\qquad{}\times\left\|f\right\|_{L^{p,1}([\max\{R-T-1,1/10\},R+T+1],(\sinh r)^{n-1}dr)}\\ &\lesssim T^{1/p'}e^{-c_pT}(\sinh R)^{-(n-1)/p} \left\|f\right\|_{L^{p,1}([\max\{R-T-1,1/10\},R+T+1],(\sinh r)^{n-1}dr)}. \end{aligned}

and for α>1/p−(n−1)/2\alpha>1/p-(n-1)/2

J1α(f)(R)≲T1/p′e−cpT(sinh⁡R)−(n−1)/p∥f∥Lp([max⁡{R−T−1,1/10},R+T+1],(sinh⁡r)n−1dr).J_{1}^{\alpha}(f)(R)\lesssim T^{1/p'}e^{-c_pT}(\sinh R)^{-(n-1)/p} \left\|f\right\|_{L^{p}([\max\{R-T-1,1/10\},R+T+1],(\sinh r)^{n-1}dr)}.

Note that these estimates imply, for j=0,1,…j=0,1,\ldots,

∥J1α(f)∥Lp((jT,(j+1)T],(sinh⁡R)n−1dR)≲Te−cpT×{∥f∥Lp,1([max⁡{(j−1)T−1,1/10},(j+2)T+1],(sinh⁡r)n−1dr),α≥1/p−(n−1)/2,∥f∥Lp([max⁡{(j−1)T−1,1/10},(j+2)T+1],(sinh⁡r)n−1dr),α>1/p−(n−1)/2.\begin{aligned} \left\|J_{1}^{\alpha}(f)\right\|_{L^{p}((jT,(j+1)T],(\sinh R)^{n-1}dR)} \lesssim T e^{-c_pT}\times \begin{cases} \left\|f\right\|_{L^{p,1}([\max\{(j-1)T-1,1/10\},(j+2)T+1],(\sinh r)^{n-1}dr)},& \alpha\geq1/p-(n-1)/2,\\ \left\|f\right\|_{L^{p}([\max\{(j-1)T-1,1/10\},(j+2)T+1],(\sinh r)^{n-1}dr)},& \alpha>1/p-(n-1)/2. \end{cases} \end{aligned}

Arguing analogously to (14) and (15) to sum over jj, we obtain the desired estimate (17).

To verify the claimed estimates (18) and its strong version, we apply Lemma 2.3 and the identity

ch⁡a−ch⁡b=ea2(1−e−a+b)(1−e−a−b)\operatorname{ch} a-\operatorname{ch} b=\frac{e^{a}}{2}\left(1-e^{-a+b}\right)\left(1-e^{-a-b}\right)

to obtain for R,r≥1/10R,r\ge1/10 and T≤t≤T+1T\le t\le T+1

∣Ktα(R,r)∣≲1{∣R−r∣<t}e−n−12t(sh⁡Rsh⁡r)1−n21{∣t−∣R−r∣∣,∣t−R−r∣≥14}⏟Kt,1α(R,r)+1{∣R−r∣<t}e−n−12t(sh⁡Rsh⁡r)1−n2ϕα(∣t−∣R−r∣∣)1{∣t−∣R−r∣∣≤14}⏟Kt,2α(R,r)+1{∣R−r∣<t}e−n−12t(sh⁡Rsh⁡r)1−n2ψα(∣t−R−r∣)1{∣t−R−r∣≤14}⏟Kt,3α(R,r),\begin{aligned} \left|K_{t}^{\alpha}(R,r)\right| \lesssim{}& \underbrace{\mathbf{1}_{\{|R-r|<t\}}e^{-\frac{n-1}{2}t}(\operatorname{sh}R\operatorname{sh}r)^{\frac{1-n}{2}}\mathbf{1}_{\{|t-|R-r||,|t-R-r|\ge\frac14\}}}_{K_{t,1}^{\alpha}(R,r)}\\ &+\underbrace{\mathbf{1}_{\{|R-r|<t\}}e^{-\frac{n-1}{2}t}(\operatorname{sh}R\operatorname{sh}r)^{\frac{1-n}{2}}\phi^{\alpha}(|t-|R-r||)\mathbf{1}_{\{|t-|R-r||\le\frac14\}}}_{K_{t,2}^{\alpha}(R,r)}\\ &+\underbrace{\mathbf{1}_{\{|R-r|<t\}}e^{-\frac{n-1}{2}t}(\operatorname{sh}R\operatorname{sh}r)^{\frac{1-n}{2}}\psi^{\alpha}(|t-R-r|)\mathbf{1}_{\{|t-R-r|\le\frac14\}}}_{K_{t,3}^{\alpha}(R,r)}, \end{aligned}

where

ϕα(v)={vα+n−32,1−n2<α<3−n2,1,α≥3−n2,ψα(v)={vα+n−32,1−n2<α<3−n2,log⁡1v,α=3−n2,1,α>3−n2.\phi^{\alpha}(v)= \begin{cases} v^{\alpha+\frac{n-3}{2}},&\frac{1-n}{2}<\alpha<\frac{3-n}{2},\\ 1,&\alpha\ge\frac{3-n}{2}, \end{cases} \qquad \psi^{\alpha}(v)= \begin{cases} v^{\alpha+\frac{n-3}{2}},&\frac{1-n}{2}<\alpha<\frac{3-n}{2},\\ \log\frac{1}{v},&\alpha=\frac{3-n}{2},\\ 1,&\alpha>\frac{3-n}{2}. \end{cases}

For α>(1−n)/2\alpha>(1-n)/2,

sup⁡T≤t≤T+1∥Kt,1α(R,⋅)∥Lp′([1/10,∞),(sh⁡r)n−1dr)≲T1p′e−cpT(sh⁡R)−n−1p.\sup_{T\le t\le T+1}\left\|K_{t,1}^{\alpha}(R,\cdot)\right\|_{L^{p'}([1/10,\infty),(\operatorname{sh}r)^{n-1}dr)} \lesssim T^{\frac{1}{p'}}e^{-c_pT}(\operatorname{sh}R)^{-\frac{n-1}{p}}.

In addition, for α≥1/p−(n−1)/2\alpha\ge1/p-(n-1)/2,

sup⁡T≤t≤T+1∥Kt,2α(R,⋅)∥Lp′,∞([1/10,∞),(sh⁡r)n−1dr)+sup⁡T≤t≤T+1∥Kt,3α(R,⋅)∥Lp′,∞([1/10,∞),(sh⁡r)n−1dr)≲e−cpT(sh⁡R)−n−1p,\begin{aligned} &\sup_{T\le t\le T+1}\left\|K_{t,2}^{\alpha}(R,\cdot)\right\|_{L^{p',\infty}([1/10,\infty),(\operatorname{sh}r)^{n-1}dr)}\\ &\quad+\sup_{T\le t\le T+1}\left\|K_{t,3}^{\alpha}(R,\cdot)\right\|_{L^{p',\infty}([1/10,\infty),(\operatorname{sh}r)^{n-1}dr)} \lesssim e^{-c_pT}(\operatorname{sh}R)^{-\frac{n-1}{p}}, \end{aligned}

and the above estimate holds with Lp′L^{p'} in place of Lp′,∞L^{p',\infty} for α>1/p−(n−1)/2\alpha>1/p-(n-1)/2. Therefore, the desired estimates (17) and its strong version are proved.

  • Estimates for J2α(f)J_{2}^{\alpha}(f). We shall show that for α<1/p\alpha<1/p such that (12) holds,

∥J2α(f)∥Lradp′,∞(Hn)≲e−n−1p′T∥f∥Lradp,1(Hn),(19)\left\|J_{2}^{\alpha}(f)\right\|_{L_{\mathrm{rad}}^{p',\infty}(\mathbb{H}^{n})}\lesssim e^{-\frac{n-1}{p'}T}\left\|f\right\|_{L_{\mathrm{rad}}^{p,1}(\mathbb{H}^{n})}, \tag*{(19)}

and that for p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, the above estimate holds with Lradp′L_{\mathrm{rad}}^{p'} in place of Lradp′,∞L_{\mathrm{rad}}^{p',\infty} and Lradp,1L_{\mathrm{rad}}^{p,1}.

In view of Lemma 2.2(i), we further decompose

J2α(f)≤sup⁡T≤t≤T+11(t−12,t+12)∣Mtα(f1(0,110])∣⏟J2,1α(f)+sup⁡T≤t≤T+11(110,t−12]∣Mtα(f1(0,110])∣⏟J2,2α(f).\begin{aligned} J_{2}^{\alpha}(f)\le{}&\underbrace{\sup_{T\le t\le T+1}\mathbf{1}_{(t-\frac12,t+\frac12)}\left|M_{t}^{\alpha}\left(f\mathbf{1}_{(0,\frac{1}{10}]}\right)\right|}_{J_{2,1}^{\alpha}(f)}\\ &+\underbrace{\sup_{T\le t\le T+1}\mathbf{1}_{(\frac{1}{10},t-\frac12]}\left|M_{t}^{\alpha}\left(f\mathbf{1}_{(0,\frac{1}{10}]}\right)\right|}_{J_{2,2}^{\alpha}(f)}. \end{aligned}

For ∣t−R∣<1/2|t-R|<1/2 and r≤1/10r\le1/10, we have

ch⁡t−ch⁡(R−r)∼(t−R+r)eT,\operatorname{ch}t-\operatorname{ch}(R-r)\sim(t-R+r)e^{T},
∣ch⁡(R+r)−ch⁡t∣∼∣R+r−t∣eT,\left|\operatorname{ch}(R+r)-\operatorname{ch}t\right|\sim|R+r-t|e^{T},

sh⁡Rsh⁡r∼reT\operatorname{sh} R\operatorname{sh} r\sim r e^{T},

which implies

∣Kα(R,r)∣≲e−(n−1)TEtαt−T+2(R−T+2,r)\lvert K^\alpha(R,r)\rvert\lesssim e^{-(n-1)T}E_t^\alpha{}_{t-T+2}(R-T+2,r)

according to Lemma 2.3 and the definition of EtαE_t^\alpha. Therefore, in view of Lemma 3.2, for α<1/p\alpha<1/p satisfying (12)

∥J2,1α(f)∥Lp,∞((T−12,T+32),(sh⁡R)n−1dR)≲e−n−1p′T∥Eα(f1(0,110])∥Lp,∞((32,72),Rn−1dR)\left\|J_{2,1}^\alpha(f)\right\|_{L^{p,\infty}((T-\frac{1}{2},T+\frac{3}{2}),(\operatorname{sh}R)^{n-1}dR)} \lesssim e^{-\frac{n-1}{p'}T}\left\|E^\alpha\left(f\mathbf{1}_{(0,\frac{1}{10}]}\right)\right\|_{L^{p,\infty}((\frac{3}{2},\frac{7}{2}),R^{n-1}dR)}
≲e−n−1p′T∥f∥Lp,1((0,110],(sh⁡r)n−1dr),\lesssim e^{-\frac{n-1}{p'}T}\left\|f\right\|_{L^{p,1}((0,\frac{1}{10}],(\operatorname{sh}r)^{n-1}dr)},

and that for p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, the above estimate holds with LradpL_{\mathrm{rad}}^p in place of Lradp,∞L_{\mathrm{rad}}^{p,\infty} and Lradp,1L_{\mathrm{rad}}^{p,1}.

In addition, for t−R≥1/2t-R\geq1/2 and r≤1/10r\leq1/10, it follows from Lemma 2.3 that

∣Ktα(R,r)∣≲e−(n−1)T,\lvert K_t^\alpha(R,r)\rvert\lesssim e^{-(n-1)T},

which implies for all 1≤p≤∞1\leq p\leq\infty and α>(1−n)/2\alpha>(1-n)/2

∥J2,2α(f)∥Lp((110,T+12),(sh⁡R)n−1dR)≲e−n−1p′T∥f∥Lp((0,110],(sh⁡r)n−1dr).\left\|J_{2,2}^\alpha(f)\right\|_{L^p((\frac{1}{10},T+\frac{1}{2}),(\operatorname{sh}R)^{n-1}dR)} \lesssim e^{-\frac{n-1}{p'}T}\left\|f\right\|_{L^p((0,\frac{1}{10}],(\operatorname{sh}r)^{n-1}dr)}.

Therefore, the desired estimates (19) and its strong version are proved.

  • Estimates for J3α(f)J_3^\alpha(f). We shall show that for α<1/p\alpha<1/p such that (12) holds,

∥J3α(f)∥Lradp,∞(Hn)≲e−n−1pT∥f∥Lradp,1(Hn),(20)\left\|J_3^\alpha(f)\right\|_{L_{\mathrm{rad}}^{p,\infty}(\mathbb{H}^n)} \lesssim e^{-\frac{n-1}{p}T}\left\|f\right\|_{L_{\mathrm{rad}}^{p,1}(\mathbb{H}^n)}, \tag*{(20)}

and that for p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, the above estimate holds with LradpL_{\mathrm{rad}}^p in place of Lradp,∞L_{\mathrm{rad}}^{p,\infty} and Lradp,1L_{\mathrm{rad}}^{p,1}.

In view of Lemma 2.2(i), we further decompose

J3α(f)≤sup⁡T≤t≤T+11(0,110](R)∣Mtα(f1(t−12,t+12))∣+sup⁡T≤t≤T+11(0,110](R)∣Mtα(f1(0,t−12])∣.J_3^\alpha(f)\leq\sup_{T\leq t\leq T+1}\mathbf{1}_{(0,\frac{1}{10}]}(R)\left|M_t^\alpha\left(f\mathbf{1}_{(t-\frac{1}{2},t+\frac{1}{2})}\right)\right|+\sup_{T\leq t\leq T+1}\mathbf{1}_{(0,\frac{1}{10}]}(R)\left|M_t^\alpha\left(f\mathbf{1}_{(0,t-\frac{1}{2}]}\right)\right|.

For R≤1/10R\leq1/10 and ∣t−r∣<1/2\lvert t-r\rvert<1/2, we argue as in the estimate for J2,1α(f)J_{2,1}^\alpha(f) above to obtain

∣Ktα(R,r)∣(sh⁡r)n−1≲Etαt−T+2(R,r−T+2).\lvert K_t^\alpha(R,r)\rvert(\operatorname{sh}r)^{n-1}\lesssim E_t^\alpha{}_{t-T+2}(R,r-T+2).

Therefore, in view of Lemma 3.2, for α<1/p\alpha<1/p satisfying (12)

∥J3,1α(f)∥Lp,∞((0,110],(sh⁡R)n−1dR)≲∥Eα(f(⋅+T−2)1(32,72))∥Lp,∞((0,110),Rn−1dR)\left\|J_{3,1}^\alpha(f)\right\|_{L^{p,\infty}((0,\frac{1}{10}],(\operatorname{sh}R)^{n-1}dR)} \lesssim\left\|E^\alpha\left(f(\cdot+T-2)\mathbf{1}_{(\frac{3}{2},\frac{7}{2})}\right)\right\|_{L^{p,\infty}((0,\frac{1}{10}),R^{n-1}dR)}
≲∥f(⋅+T−2)∥Lp,1((32,72),rn−1dr)\lesssim\left\|f(\cdot+T-2)\right\|_{L^{p,1}((\frac{3}{2},\frac{7}{2}),r^{n-1}dr)}
≲e−n−1pT∥f∥Lradp,1(Hn),\lesssim e^{-\frac{n-1}{p}T}\left\|f\right\|_{L_{\mathrm{rad}}^{p,1}(\mathbb{H}^n)},

and that for p>2n/(n−1)p>2n/(n-1) and α=(1−n)/p\alpha=(1-n)/p, the above estimate holds with LradpL_{\mathrm{rad}}^p in place of Lradp,∞L_{\mathrm{rad}}^{p,\infty} and Lradp,1L_{\mathrm{rad}}^{p,1}.

In addition, for R≤1/10R\leq1/10 and t−r>1/2t-r>1/2, it follows from Lemma 2.3 that

∣Ktα(R,r)∣≲e−(n−1)T,\lvert K_t^\alpha(R,r)\rvert\lesssim e^{-(n-1)T},

which implies for all 1≤p≤∞1\leq p\leq\infty and α>(1−n)/2\alpha>(1-n)/2

∥J3,2α(f)∥Lp((0,110),(sh⁡R)n−1dR)≲e−n−1pT∥f∥Lradp(Hn).\left\|J_{3,2}^\alpha(f)\right\|_{L^p((0,\frac{1}{10}),(\operatorname{sh}R)^{n-1}dR)} \lesssim e^{-\frac{n-1}{p}T}\left\|f\right\|_{L_{\mathrm{rad}}^p(\mathbb{H}^n)}.

Therefore, the desired estimates (20) and its strong version are proved. Finally, we combine (16) with estimates (17)–(20) and there strong versions to conclude the proof of Proposition 3.3. □

Conclusion of the proof of Theorem 1.1

Proof of Theorem 1.1 (sufficiency part). Note that

mα(f)≤m0α(f)+∑T=1∞mTα(f).\mathbf{m}^{\alpha}(f) \le\mathbf{m}_{0}^{\alpha}(f) + \sum_{T=1}^{\infty} \mathbf{m}_{T}^{\alpha}(f).

Since for 1<p<∞1 < p < \infty and sufficiently small ε>0\varepsilon> 0, ∑T=1∞(T+1)e(ε−cp)T<∞\sum_{T=1}^{\infty}(T+1)e^{(\varepsilon-cp)T}<\infty, Propositions 3.1 and 3.3 imply the Lradp,1→Lradp,∞L_{\mathrm{rad}}^{p,1} \to L_{\mathrm{rad}}^{p,\infty}-boundedness of mα\mathbf{m}^{\alpha} whenever α<1/p\alpha< 1/p and (12) holds, as well as the LradpL_{\mathrm{rad}}^{p}-boundedness of mα\mathbf{m}^{\alpha} when p>2n/(n−1)p > 2n/(n-1) and α=(1−n)/p\alpha= (1-n)/p. Theorem 1.1(ii) is therefore proved.

To show the sufficiency part of Theorem 1.1(i), we fix 1<p<∞1 < p < \infty,

α>{1−n+np,1<p≤2,1p−n−12,2<p≤2nn−1,1−np,p>2nn−1,(21)\alpha> \begin{cases} 1-n+\dfrac{n}{p}, & 1<p\le2,\\ \dfrac{1}{p}-\dfrac{n-1}{2}, & 2<p\le\dfrac{2n}{n-1},\\ \dfrac{1-n}{p}, & p>\dfrac{2n}{n-1}, \end{cases} \tag*{(21)}

and choose 1<p0<p<p1<∞1 < p_{0} < p < p_{1} < \infty sufficiently close to pp such that α\alpha satisfies (21) as well as α<1/p\alpha< 1/p at p0p_{0} and p1p_{1}. Interpolation between the Lradq,1→Lradq,∞L_{\mathrm{rad}}^{q,1} \to L_{\mathrm{rad}}^{q,\infty}-boundedness of mα\mathbf{m}^{\alpha} at q=p0q=p_{0} and q=p1q=p_{1} (cf. [9 Theorem 1.4.19]) yields the LradpL_{\mathrm{rad}}^{p}-boundedness of mα\mathbf{m}^{\alpha}. Thus all the desired LradpL_{\mathrm{rad}}^{p}-boundedness of mα\mathbf{m}^{\alpha} for 1<p<∞1 < p < \infty under the additional restriction α<1/p\alpha< 1/p is proved. At p=∞p=\infty, it follows directly from (2) that m0\mathbf{m}^{0} is bounded on L∞(Hn)L^{\infty}(\mathbb{H}^{n}). Finally, we apply Lemma 2.4 to remove the additional restriction α<1/p\alpha< 1/p for 1<p<∞1 < p < \infty and α=0\alpha=0 at p=∞p=\infty. This proves the sufficiency part of Theorem 1.1(i), and thus the whole sufficiency part of Theorem 1.1 is concluded. □

Proof of Theorem 1.1 (necessity part). We shall construct three counterexamples to show that the necessity conditions for the LradpL_{\mathrm{rad}}^{p}-boundedness of mα\mathbf{m}^{\alpha} are α>1−n+n/p\alpha> 1-n+n/p, α>1/p−(n−1)/2\alpha> 1/p-(n-1)/2 and α≥(1−n)/p\alpha\ge(1-n)/p, respectively.

  • The condition α>1−n+n/p\alpha> 1-n+n/p. Let 0<δ<1/1000 < \delta< 1/100 be a fixed sufficiently small number to be determined later. For 0<ε<1/1000 < \varepsilon< 1/100 we set

fε:=1[ε(1+δ),ε(1+2δ)].f_{\varepsilon}:=\mathbf{1}_{[\varepsilon(1+\delta),\varepsilon(1+2\delta)]}.

For 5ε≤R≤1/105\varepsilon\le R \le1/10, we choose t=R−εt=R-\varepsilon. If r∈supp⁡fεr\in\operatorname{supp}f_{\varepsilon}, then ∣R−r∣<t<R+r|R-r|<t<R+r,

sinh⁡Rsinh⁡r∼Rε,cosh⁡t−cosh⁡(R−r)∼Rδε.\sinh R\sinh r\sim R\varepsilon,\qquad\cosh t-\cosh(R-r)\sim R\delta\varepsilon.

Since F(a,b;c;0)=1F(a,b;c;0)=1, by taking δ=δα,n\delta=\delta_{\alpha,n} sufficiently small such that the hypergeometric function term in Lemma 2.2(ii) is uniformly positive and in view of Lemma 2.2(ii), we have

Ktα(R,r)≳R1−n−αεα−1,K_{t}^{\alpha}(R,r)\gtrsim R^{1-n-\alpha}\varepsilon^{\alpha-1},

whence for RR as above

Mtα(fε)(R)≳R1−n−αεα−1∫ε(1+δ)ε(1+2δ)rn−1 dr∼(εR)α+n−1.M_{t}^{\alpha}(f_{\varepsilon})(R)\gtrsim R^{1-n-\alpha}\varepsilon^{\alpha-1}\int_{\varepsilon(1+\delta)}^{\varepsilon(1+2\delta)}r^{n-1}\,dr\sim\left(\frac{\varepsilon}{R}\right)^{\alpha+n-1}.

This, together with the fact that ∥fε∥Lradp(Hn)∼εn\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)} \sim\varepsilon^n, implies

∥mαfε∥Lradp(Hn)≳∥fε∥Lradp(Hn)εp(α+n−1)−n∫5ε110Rn−1−p(n−1+α) dR∼∥fε∥Lradp(Hn)×{1,α>1−n+np,log⁡1ε,α=1−n+np,εp(n−1−np+α),α<1−n+np.\begin{aligned} \|\mathbf{m}^{\alpha}f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)} \gtrsim\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}\varepsilon^{p(\alpha+n-1)-n}\int_{5\varepsilon}^{\frac{1}{10}}R^{n-1-p(n-1+\alpha)}\,\mathrm{d}R \\ &\sim\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}\times \begin{cases} 1, & \alpha>1-n+\frac{n}{p},\\ \log\frac{1}{\varepsilon}, & \alpha=1-n+\frac{n}{p},\\ \varepsilon^{p(n-1-\frac{n}{p}+\alpha)}, & \alpha<1-n+\frac{n}{p}. \end{cases} \end{aligned}

Thus α>1−n+n/p\alpha>1-n+n/p is necessary for the LradpL^p_{\mathrm{rad}}-boundedness of mα\mathbf{m}^{\alpha}.

  • The condition α>1/p−(n−1)/2\alpha>1/p-(n-1)/2. Let δ>0\delta>0 be a fixed sufficiently small number to be determined later. For 0<ε<δ/20<\varepsilon<\delta/2 we set

fε(r):=(r−1)−1/p(log⁡1r−1)−11(1+ε,1+δ)(r).f_\varepsilon(r):=(r-1)^{-1/p}\left(\log\frac{1}{r-1}\right)^{-1}\mathbf{1}_{(1+\varepsilon,1+\delta)}(r).

If r∈supp⁡fεr\in\operatorname{supp}f_\varepsilon, 5/4≤R≤3/25/4\leq R\leq3/2 and t=R−1t=R-1, then ∣R−r∣<t<R+r|R-r|<t<R+r,

sinh⁡Rsinh⁡r∼1,cosh⁡t−cosh⁡(R−r)∼r−1≤δ.\sinh R\sinh r\sim1,\qquad\cosh t-\cosh(R-r)\sim r-1\leq\delta.

Since F(a,b;c;0)=1F(a,b;c;0)=1, by taking δ=δα,n\delta=\delta_{\alpha,n} sufficiently small such that the hypergeometric function term in Lemma Lemma 2.2(ii) is uniformly positive and in view of Lemma Lemma 2.2(ii), we have

Ktα(R,r)≳(r−1)α+n−32,K_t^\alpha(R,r)\gtrsim(r-1)^{\alpha+\frac{n-3}{2}},

whence for RR as above and α≤1/p−(n−1)/2\alpha\leq1/p-(n-1)/2

∣Mtα(fε)(R)∣≳∫εδsα−1p+n−12−1log⁡1s ds→∞as ε→0+.|M_t^\alpha(f_\varepsilon)(R)|\gtrsim\int_\varepsilon^\delta\frac{s^{\alpha-\frac{1}{p}+\frac{n-1}{2}-1}}{\log\frac{1}{s}}\,\mathrm{d}s\to\infty\qquad\text{as }\varepsilon\to0^+.

On the other hand, ∥fε∥Lradp(Hn)≲1\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}\lesssim1 uniformly in ε\varepsilon. Thus α>1/p−(n−1)/2\alpha>1/p-(n-1)/2 is necessary for the LradpL^p_{\mathrm{rad}}-boundedness of mα\mathbf{m}^{\alpha}.

  • The condition α≥(1−n)/p\alpha\geq(1-n)/p. Let 0<δ<1/1000<\delta<1/100 be a fixed sufficiently small number to be determined later. For 0<ε<δ/20<\varepsilon<\delta/2 we set

fε:=1[1+δε2,1+δε].f_\varepsilon:=\mathbf{1}_{[1+\frac{\delta\varepsilon}{2},1+\delta\varepsilon]}.

For ε≤R≤2ε\varepsilon\leq R\leq2\varepsilon, we choose t=1−R+2δεt=1-R+2\delta\varepsilon. If r∈supp⁡fεr\in\operatorname{supp}f_\varepsilon, then ∣R−r∣<t<R+r|R-r|<t<R+r,

sinh⁡Rsinh⁡r∼ε,cosh⁡t−cosh⁡(R−r)∼δε.\sinh R\sinh r\sim\varepsilon,\qquad\cosh t-\cosh(R-r)\sim\delta\varepsilon.

Since F(a,b;c;0)=1F(a,b;c;0)=1, by taking δ=δα,n\delta=\delta_{\alpha,n} sufficiently small such that the hypergeometric function term in Lemma Lemma 2.2(ii) is uniformly positive and in view of Lemma Lemma 2.2(ii), we have

Ktα(R,r)≳εα−1,K_t^\alpha(R,r)\gtrsim\varepsilon^{\alpha-1},

whence for RR as above

Mtα(fε)(R)≳εα.M_t^\alpha(f_\varepsilon)(R)\gtrsim\varepsilon^\alpha.

This, together with the fact that ∥fε∥Lradp(Hn)∼ε1/p\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}\sim\varepsilon^{1/p}, implies

∥mαfε∥Lradp(Hn)≳∥fε∥Lradp(Hn)εα−1/p(∫ε2εRn−1 dR)1/p∼∥fε∥Lradp(Hn)εα+n−1p.\|\mathbf{m}^{\alpha}f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)} \gtrsim\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}\varepsilon^{\alpha-1/p} \left(\int_\varepsilon^{2\varepsilon}R^{n-1}\,\mathrm{d}R\right)^{1/p} \sim\|f_\varepsilon\|_{L^p_{\mathrm{rad}}(\mathbb{H}^n)}\varepsilon^{\alpha+\frac{n-1}{p}}.

When p=∞p = \infty, the above estimates are understood by the convention 1/∞=01/\infty= 0. Thus α≥(1−n)/p\alpha\ge(1-n)/p is necessary for the LradpL^p_{\mathrm{rad}}-boundedness of mα\mathbf{m}^{\alpha}.

Declaration of AI usage

During preparation of the manuscript, the author used GPT-6 Astra for assistance with concrete computations and language editing. The author takes responsibility for the mathematical arguments and the final content of the paper.

Acknowledgements

The author would like to thank Naijia Liu and Professor Lixin Yan for helpful discussions.

References

YUNXIANG WANG, DEPARTMENT OF MATHEMATICS, SUN YAT-SEN UNIVERSITY, GUANGZHOU, 510275, P.R. CHINA

Email address: [email protected]

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