# Harnack inequality for fractional Laplacian-type operators on hyperbolic
spaces

Jongmyeong Kim  Department of Mathematical Sciences, Seoul National
University, Seoul 08826, Republic of Korea<EMAIL_ADDRESS>,  Minhyun Kim
Fakultät für Mathematik, Universität Bielefeld, 33615 Bielefeld, Germany
<EMAIL_ADDRESS>and  Ki-Ahm Lee  Department of Mathematical
Sciences, Seoul National University, Seoul 08826, Republic of Korea & Korea
Institute for Advanced Study, Seoul 02455, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

The Krylov–Safonov theory for fully nonlinear nonlocal operators on hyperbolic
spaces of dimension three is established. Since the operators on hyperbolic
spaces exhibit qualitatively different behavior than those on manifolds with
nonnegative curvature, new scale functions are introduced which take the
effect of negative curvature into account.

The regularity theory in this work provides unified regularity results for
fractional-order and second-order operators in the sense that the regularity
estimates stay uniform as the fractional-order approaches 2. In the unified
regularity theory, the asymptotic behavior of the normalizing constant for the
fractional Laplacian plays a fundamental role. The dimension restriction has
been imposed to compute the explicit value of this constant by using the
Fourier analysis on hyperbolic spaces.

###### 2020 Mathematics Subject Classification:

35R01, 35B65, 35J60, 47G20.

##  1\. Introduction

In a celebrated series of papers [3, 4, 5] by Caffarelli and Silvestre, the
regularity theories such as Krylov–Safonov, Cordes–Nirenberg, and Evans–Krylov
theory are established for fractional-order operators on Euclidean spaces. The
most important feature is that the constants in regularity estimates do not
blow up and remain uniform as the order of operator converges to 2. It means
that the regularity theories for fractional-order and second-order operators
are unified. It has also been extended to the parabolic cases [24, 25, 26].

On the other hand, the regularity theory for the local operators on Riemannian
manifolds has been studied. In particular, the studies on the Harnack
inequalities, initiated by Yau [43] and Cheng–Yau [7], have been extended to
second-order divergence and non-divergence form operators. These are
extensions of the De Giorgi–Nash–Moser [31] and Krylov–Safonov Harnack
inequalities [2, 21, 42], respectively. See also [22, 23] for the parabolic
Harnack inequalities.

The natural extension is to obtain the regularity results for fractional-order
operators on Riemannian manifolds. Indeed, the Harnack inequality for nonlocal
operators on metric measure spaces with volume doubling property, which
includes Riemannian manifolds with nonnegative curvature, has been studied via
the Dirichlet form theory [6]. However, this approach does not provide the
unified regularity theory and is not appropriate for non-divergence form
operators. For nonlocal non-divergence form operators, the Krylov–Safonov
Harnack inequality on Riemannian manifolds with nonnegative curvature has
recently been established by the authors [19]. The result in [19] unifies the
Krylov–Safonov Harnack inequalities for local and nonlocal operators on
manifolds with nonnegative curvature as in the works of Caffarelli and
Silvestre.

In this paper, we continue to pursue the unified regularity theory for
fractional-order and standard local nonlinear operators on 3-dimensional
hyperbolic space $\mathbb{H}^{3}$ that has a nonnegative curvature. The study
on 3-dimensional hyperbolic spaces is geometrically meaningful especially in
general relativity. In this work, the dimension restriction has been imposed
to compute the explicit value of normalizing constant for the fractional
Laplacian, which is crucial for the unified regularity theory as in [3, 4, 5]
due to its asymptotic behavior as $\gamma$ approaches 1.

The fractional Laplacian $-(-\Delta_{\mathbb{H}^{n}})^{\gamma}$,
$\gamma\in(0,1)$, on hyperbolic spaces ${\mathbb{H}^{n}}$ is defined recently
in [1] by Banica, González, and Sáez to study extension problem, and the
explicit form of its kernel up to normalizing constant is provided. To the
best of authors’ knowledge, this normalizing constant is not known. It has
been observed [20] that finding the constant and studying its asymptotic
behavior are the most important steps when a scaling is not available. In this
regard, the first part of the paper is devoted to finding the explicit value
of the constant. Indeed, this can be done on hyperbolic spaces with curvature
$-\tau^{-2}$.

######  Theorem 1.1.

Let $\mathbb{H}^{3}_{\tau}$ be the 3-dimensional hyperbolic space with
curvature $-\tau^{-2}$. The fractional Laplacian
$-(-\Delta_{\mathbb{H}^{3}_{\tau}})^{\gamma}$ is given by

(1.1)
$-(-\Delta_{\mathbb{H}^{3}_{\tau}})^{\gamma}u(x)=\mathrm{p.v.}\int_{\mathbb{H}^{3}_{\tau}}(u(x^{\prime})-u(x))\mathcal{K}_{\gamma,\tau}(d_{\mathbb{H}^{3}_{\tau}}(x^{\prime},x))\,\mathrm{d}\mu_{\mathbb{H}^{3}_{\tau}}(x^{\prime}),\quad
x\in{\mathbb{H}^{n}}.$

Here, the kernel $\mathcal{K}_{\gamma,\tau}$ has an explicit form

(1.2)
$\mathcal{K}_{\gamma,\tau}(\rho)=C(3,\gamma)\frac{1/\tau}{\sinh\rho/\tau}\rho^{-1/2-\gamma}\frac{2K_{-3/2-\gamma}(\rho/\tau)}{\Gamma(3/2+\gamma)(2\tau)^{3/2+\gamma}},$

where $K_{\nu}$ denotes the modified Bessel function of the second kind and

$C(n,\gamma)=\frac{2^{2\gamma}\Gamma(\frac{n}{2}+\gamma)}{\pi^{n/2}|\Gamma(-\gamma)|}$

is the normalizing constant for the fractional Laplacian on Euclidean spaces
$\mathbb{R}^{n}$.

######  Remark 1.2.

  1. (i)

The novelty of Theorem 1.1 is that the normalizing constant has the same
asymptotic behavior with $1-\gamma$ as $\gamma\to 1$, which is crucial in the
upcoming regularity theory, as in [3, 4, 5].

  2. (ii)

It is natural to expect that $\mathcal{K}_{\gamma,\tau}$ converges to the
kernel of the fractional Laplacian on Euclidean space as curvature
$-\tau^{-2}$ approaches zero. Indeed, it follows from Lemma A.3 that

$\mathcal{K}_{\gamma,\tau}(\rho)\to C(3,\gamma)\rho^{-3-2\gamma},$

as $\tau\to\infty$.

  3. (iii)

The kernel $\mathcal{K}_{\gamma,\tau}$ decays exponentially as
$\rho\to\infty$, which is different from the behavior of the kernel on
manifolds with nonnegative curvature. The difference comes from the
exponential growth of the volume of balls in hyperbolic spaces, and this is
why the regularity theories on manifolds with negative and nonnegative
curvatures are dealt with separately.

  4. (iv)

In Theorem 1.1, the only 3-dimensional case is considered due to the
difficulty in some computation of integrals in, e.g., Section 4 and Section 5.
Such a restriction on dimensions is common on hyperbolic spaces. However, the
harmonic analysis in Section 3 is provided for general dimensions. Moreover,
most of the arguments in Section 6–Section 9 work in general dimensions with
minor modifications if Theorem 1.1 is true for general dimensions.

The proof of Theorem 1.1 is not as simple as in the case of Euclidean spaces.
In the Euclidean case, it follows from

(1.3)
$\displaystyle\begin{split}\widehat{(-\Delta)^{\gamma}u}(\xi)&=-\frac{1}{2}C(n,\gamma)\int_{\mathbb{R}^{n}}\frac{\widehat{u(\cdot+y)}(\xi)+\widehat{u(\cdot-y)}(\xi)-2\hat{u}(\xi)}{|y|^{n+2\gamma}}\,\mathrm{d}y\\\
&=-\frac{1}{2}C(n,\gamma)\left(\int_{\mathbb{R}^{n}}\frac{e^{i\xi\cdot
y}+e^{-i\xi\cdot y}-2}{|y|^{n+2\gamma}}\,\mathrm{d}y\right)\hat{u}(\xi)\\\
&=C(n,\gamma)\left(\int_{\mathbb{R}^{n}}\frac{1-\cos(\xi\cdot
y)}{|y|^{n+2\gamma}}\,\mathrm{d}y\right)\hat{u}(\xi)\\\
&=C(n,\gamma)\left(\int_{\mathbb{R}^{n}}\frac{1-\cos
y_{1}}{|y|^{n+2\gamma}}\,\mathrm{d}y\right)|\xi|^{2\gamma}\hat{u}(\xi)\end{split}$

and the definition
$\widehat{(-\Delta)^{\gamma}u}(\xi)=|\xi|^{2\gamma}\hat{u}(\xi)$ that

$C(n,\gamma)=\left(\int_{\mathbb{R}^{n}}\frac{1-\cos
y_{1}}{|y|^{n+2\gamma}}\,\mathrm{d}y\right)^{-1}=\frac{2^{2\gamma}\Gamma(\frac{n}{2}+\gamma)}{\pi^{n/2}|\Gamma(-\gamma)|}.$

The computation (1.3) relies heavily on the group structure of
$(\mathbb{R}^{n},+)$ that $\mathbb{H}^{n}_{a}$ does not have. Instead, the
gyrogroup structure of hyperbolic spaces, discovered by Ungar [35] and
developed as a tool for the formulation of special relativity as an
alternative to the use of Lorentz transformations to represent compositions of
velocities, can be used to perform a similar computation as in (1.3). See also
[39].

The aim of the next part of the paper is to establish the
Alexandrov–Bakelman–Pucci (or ABP for short) estimates, Krylov–Safonov Harnack
inequality, and Hölder estimates for solutions of nonlocal nonlinear elliptic
equations on 3-dimensional hyperbolic space, which are robust in the sense
that the estimates remain uniform as $\gamma$ approaches 1. Modeled on the
fractional Laplacian (1.1) with $\tau=1$, nonlinear operators of the
fractional Laplacian-type can be defined in the standard way. For a class
$\mathcal{L}_{0}$ of linear operators of the form

$Lu(x)=\mathrm{p.v.}\int_{{\mathbb{H}^{n}}}(u(x^{\prime})-u(x))\mathcal{K}(d_{{\mathbb{H}^{n}}}(x^{\prime},x))\,\mathrm{d}\mu_{\mathbb{H}^{n}}(x^{\prime}),\quad
x\in{\mathbb{H}^{n}},$

with kernels $\mathcal{K}$ satisfying

$\lambda\mathcal{K}_{\gamma}\leq\mathcal{K}\leq\Lambda\mathcal{K}_{\gamma},\quad
0<\lambda\leq\Lambda,$

the maximal and minimal operators are defined by

$\mathcal{M}^{+}u(x):=\mathcal{M}^{+}_{\mathcal{L}_{0}}u(x):=\sup_{L\in\mathcal{L}_{0}}Lu(x)\quad\text{and}\quad\mathcal{M}^{-}u(x):=\mathcal{M}^{-}_{\mathcal{L}_{0}}u(x):=\inf_{L\in\mathcal{L}_{0}}Lu(x),$

respectively. It is easy to see that these extremal operators are well-defined
at $x\in{\mathbb{H}^{n}}$ for any bounded function $u$ that is $C^{2}$ near
$x$ (see (5.2)). An operator $\mathcal{I}$ is said to be elliptic with respect
to $\mathcal{L}_{0}$ if

$\mathcal{M}^{-}_{\mathcal{L}_{0}}(u-v)(x)\leq\mathcal{I}(u,x)-\mathcal{I}(v,x)\leq\mathcal{M}^{+}_{\mathcal{L}_{0}}(u-v)(x)$

for every point $x\in{\mathbb{H}^{n}}$ and for all bounded functions $u$ and
$v$ which are $C^{2}$ near $x$.

The first step towards the Krylov–Safonov Harnack inequality and Hölder
estimate is showing the ABP-type estimate. The following theorem provides an
estimate on the distribution function of supersolutions to nonlocal nonlinear
operators. Throughout the paper, the functions $\frac{\sinh t}{t}$ and $t\coth
t$ are denoted by $\mathcal{S}(t)$ and $\mathcal{H}(t)$, respectively, as in
[42]. Moreover, we introduce a function
$\mathcal{T}(t)=t/\tanh^{-1}(\frac{1}{2}\tanh(t))$. To focus on the limiting
behavior as $\gamma\to 1$, we set $\tau=1$ from now on.

######  Theorem 1.3 (ABP-type estimate).

Let $n=3$, $\gamma_{0}\in(0,1)$, and assume $\gamma\in[\gamma_{0},1)$. Let
$u\in C^{2}(B_{5R})\cap L^{\infty}({\mathbb{H}^{n}})$ be a function on
${\mathbb{H}^{n}}$ satisfying $\mathcal{M}^{-}u\leq f$ in $B_{5R}$, $u\geq 0$
in ${\mathbb{H}^{n}}\setminus B_{5R}$, and $\inf_{B_{2R}}u\leq 1$. Let
$\mathscr{C}$ be a contact set defined by (6.1), then there is a finite
collection $\\{Q^{j}_{\alpha}\\}$ of dyadic cubes, with
$\mathrm{diam}(Q^{j}_{\alpha})\leq r_{0}$, such that
$Q^{j}_{\alpha}\cap\mathscr{C}\neq\emptyset$,
$\mathscr{C}\subset\cup_{j}\overline{Q}^{j}_{\alpha}$, and

(1.4) $|B_{R}|\leq\sum_{j}cF^{n}|Q^{j}_{\alpha}|,$

where $r_{0}$ is given by (5.7), and $c$ and $F$ are given by

$F=\mathcal{S}(7R)\bigg{(}\Lambda\mathcal{H}(7R)+\frac{R^{2}}{\mathcal{I}_{0}(R)}\max_{\overline{Q}^{j}_{\alpha}}f\bigg{)}_{+},$

and

(1.5)
$c=C\cosh^{n-1}(C\mathcal{T}(r_{0})^{2}r_{0}F)(C\mathcal{T}(r_{0})^{2}F)^{(n-1)\log\cosh(C\mathcal{T}(r_{0})^{2}r_{0}F)}\mathcal{T}(r_{0})^{2n}.$

The universal constant $C>0$ depends only on $n$, $\lambda$, $\Lambda$, and
$\gamma_{0}$.

The scale function $\mathcal{I}_{0}(R)$ in Theorem 1.3 will be introduced in
Section 5. The function $\frac{R^{2}}{\mathcal{I}_{0}(R)}$ corresponds to
$\frac{1}{2\gamma}R^{2\gamma}$ in the case of Euclidean spaces [3] or
manifolds with nonnegative curvature [19]. However, this function exhibits
qualitatively different behavior in the case of hyperbolic spaces, compared to
the simple scale function $\frac{1}{2\gamma}R^{2\gamma}$. This is because the
kernels $\mathcal{K}_{\gamma}$ decay exponentially as $\rho\to\infty$ while
those in the case of manifolds with nonnegative curvature decay polynomially.

It will be proved in Lemma 5.6 and Lemma 5.5 that $r_{0}\to 0$ and
$\mathcal{I}_{0}(R)\to 6$ as $\gamma\to 1$. Since $\lim_{t\to
0}\mathcal{T}(t)=2$, the dependence of $c$ in (1.5) on $R$ disappears in the
limit $\gamma\to 1$. As a consequence, the Riemann sum in (1.4) converges to
the integral

$C\int_{\mathscr{C}}\mathcal{S}^{n}(7R)\left(\Lambda\mathcal{H}(7R)+R^{2}f(x)\right)_{+}^{n}\,\mathrm{d}\mu_{\mathbb{H}^{n}}(x),$

with $C$ independent of $R$, which appears in [42, Theorem 1.2]. This implies
that Theorem 1.3 recovers the ABP estimate for second-order equations on
hyperbolic space as a limit. Moreover, Theorem 1.3 provides a new result even
for second-order operators because it covers fully nonlinear operators.
However, this paper provides the results for the case of dimension three only.

Finally, the Krylov–Safonov Harnack inequality and Hölder estimates for
solutions of nonlocal nonlinear equations on 3-dimensional hyperbolic space
are established. Similarly as in Theorem 1.3, we consider the case $\tau=1$.

######  Theorem 1.4 (Harnack inequality).

Let $n=3$, $\gamma_{0}\in(0,1)$ and assume $\gamma\in[\gamma_{0},1)$. If a
nonnegative function $u\in C^{2}(B_{7R})\cap L^{\infty}(\mathbb{H}^{3})$
satisfies

(1.6) $\mathcal{M}^{-}u\leq C_{0}\quad\text{and}\quad\mathcal{M}^{+}u\geq-
C_{0}\quad\text{in}~{}B_{7R},$

then

$\sup_{B_{\delta_{1}R}}u\leq
C\left(\inf_{B_{\delta_{1}R}}u+C_{0}\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\right)$

for some universal constants $\delta_{1}\in(0,1)$ and $C>0$ depending only on
$n$, $\lambda$, $\Lambda$, $R$, and $\gamma_{0}$.

Let us denote by $\|\cdot\|^{\prime}$ the non-dimensional norm in the
following theorem.

######  Theorem 1.5 (Hölder estimates).

Let $n=3$, $\gamma_{0}\in(0,1)$ and assume $\gamma\in[\gamma_{0},1)$. If $u\in
C^{2}(B_{7R})\cap L^{\infty}(\mathbb{H}^{3})$ satisfies (1.6), then

$\|u\|_{C^{\alpha}(\overline{B_{R}})}^{\prime}\leq
C\left(\|u\|_{L^{\infty}(\mathbb{H}^{3})}+C_{0}\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\right)$

for some universal constants $\alpha\in(0,1)$ and $C>0$ depending only on $n$,
$\lambda$, $\Lambda$, $R$, and $\gamma_{0}$.

Since $\mathcal{I}_{0}(R)\to 6$ as $\gamma\to 1$ and the universal constants
in Theorem 1.4 and Theorem 1.5 do not depend on $\gamma$, Theorem 1.4 and
Theorem 1.5 recover the classical results for second-order equations on
hyperbolic space as limits.

######  Corollary 1.6.

  1. (i)

If a nonnegative function $u\in C^{2}(B_{7R})\cap L^{\infty}(\mathbb{H}^{3})$
satisfies

(1.7) $\mathcal{M}^{-}(D^{2}u)\leq
C_{0}\quad\text{and}\quad\mathcal{M}^{+}(D^{2}u)\geq-
C_{0}\quad\text{in}~{}B_{7R},$

then

$\sup_{B_{\delta_{1}R}}u\leq
C\left(\inf_{B_{\delta_{1}R}}u+C_{0}R^{2}\right).$

  2. (ii)

If $u\in C^{2}(B_{7R})\cap L^{\infty}(\mathbb{H}^{3})$ satisfies (1.7), then

$\|u\|_{C^{\alpha}(\overline{B_{R}})}^{\prime}\leq
C\left(\|u\|_{L^{\infty}(\mathbb{H}^{3})}+C_{0}R^{2}\right).$

The universal constants $\delta,\alpha\in(0,1)$, and $C>0$ depend only on $n$,
$\lambda$, $\Lambda$, and $R$.

The main difficulties in establishing regularity results, Theorem 1.3, Theorem
1.4, and Theorem 1.5, arise from the effect of negative curvature. The volume
of ball in hyperbolic spaces behaves like that in Euclidean spaces when a
radius is small, while it grows exponentially as a radius gets bigger. Due to
this non-homogeneity of volume, the scaling property does not hold, making the
standard arguments for regularity results break. To overcome this difficulty,
we introduce new scale functions that take non-homogeneity into account and
provide some monotonicity properties (see Section 5).

Another difficulty arising from the non-homogeneity of volumes lies in the
dyadic ring argument in ABP estimates. In the ABP estimates, we find a dyadic
ring around a given contact point in which a supersolution is quadratically
close to a tangent paraboloid in a large portion of the ring. However, the
standard dyadic rings $B_{2^{-k}r}\setminus B_{2^{-(k+1)}r}$ no longer work in
the framework of hyperbolic spaces. It leads to introducing a hyperbolic
dyadic ring whose radii are determined by volumes of balls (see Section 6).
The hyperbolic dyadic ring turns out to be the natural “dyadic” ring in the
hyperbolic geometry.

For the construction of a barrier function, it is now standard to use the
distance function. However, computation is significantly different from the
standard one because of the hyperbolic structure. We observe in Section 7 how
the negative curvature of hyperbolic spaces affects the computations.

This paper is organized as follows. In Section 2, we recall several models for
hyperbolic spaces and Fourier analysis on these spaces. Using the Fourier
transform on hyperbolic spaces, the fractional Laplacian is defined as in [1].
Section 3 is devoted to the study of the gyrogroup structure of hyperbolic
spaces and some basic harmonic analysis. With these tools at hand, we prove
Theorem 1.1 in Section 4. In Section 5, new scale functions are introduced and
some monotonicity properties are studied. The regularity theory begins with
ABP-type estimates in Section 6. In this section Theorem 1.3 is proved. The
next step is the construction of a barrier function, and this is presented in
Section 7. This barrier function, together with the ABP estimates, is used to
obtain the so-called $L^{\varepsilon}$-estimates in Section 8. Theorem 1.4 and
Theorem 1.5 are proved in Section 9 and Section 10, respectively. In Appendix
A, some properties of special functions are collected.

##  2\. Preliminaries

In this section, we recall several models for the $n$-dimensional hyperbolic
spaces (see, e.g., [34, 30, 15]). Moreover, the fractional Laplacian is
defined as in [1] by using the Helgason Fourier transform on these spaces [14,
17, 33]. See also [15, 10]. Furthermore, some well-known properties on
hyperbolic spaces are collected.

Let us first recall the hyperboloid model

$\mathbb{H}^{n}_{\tau}=\\{(x_{0},\cdots,x_{n})\in\mathbb{R}^{n+1}:x_{0}^{2}-x_{1}^{2}-\cdots-
x_{n}^{2}=\tau^{2},x_{0}>0\\},$

with the metric induced by Lorentzian metric
$-\mathrm{d}x_{0}^{2}+\mathrm{d}x_{1}^{2}-\cdots\mathrm{d}x_{n}^{2}$ in
$\mathbb{R}^{n+1}$. The space $\mathbb{H}^{n}_{\tau}$ has a constant curvature
$-\tau^{-2}<0$. The internal product induced by the Lorentzian metric is
denoted by

$[x,x^{\prime}]=x_{0}x_{0}^{\prime}-x_{1}x_{1}^{\prime}-\cdots-
x_{n}x_{n}^{\prime},$

and the distance between two points $x$ and $x^{\prime}$ is given by

$d_{\mathbb{H}^{n}_{\tau}}(x,x^{\prime})=\tau\cosh^{-1}\left(\frac{[x,x^{\prime}]}{\tau^{2}}\right).$

Using the polar coordinates, $\mathbb{H}^{n}_{\tau}$ can also be realized as

$\mathbb{H}^{n}_{\tau}=\left\\{x=\tau(\cosh r,\sinh
r\omega)\in\mathbb{R}^{n+1}:r\geq 0,\omega\in\mathbb{S}^{n-1}\right\\}.$

Then, the metric and volume element are given by
$\mathrm{d}s^{2}=\tau^{2}(\mathrm{d}r^{2}+\sinh^{2}r\,\mathrm{d}\omega^{2})$
and
$\mathrm{d}\mu_{\mathbb{H}^{n}_{\tau}}=\tau^{n}\sinh^{n-1}r\,\mathrm{d}r\,\mathrm{d}\omega$,
respectively.

Let us also consider the Poincaré ball model
$\mathbb{B}^{n}_{t}=\\{y\in\mathbb{R}^{n}:|y|<t\\}$ with metric

(2.1) $\mathrm{d}s^{2}=\frac{4b^{2}}{(t^{2}-|y|^{2})^{2}}\,\mathrm{d}y^{2}.$

The volume measure is given by

(2.2)
$\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y)=\left(\frac{2b}{t^{2}-|y|^{2}}\right)^{n}\,\mathrm{d}y.$

If $b/t=\tau$, then the map defined by

(2.3) $\phi:(x_{0},x_{1},\cdots
x_{n})\in\mathbb{H}^{n}_{\tau}\mapsto\frac{t}{\tau+x_{0}}(x_{1},\cdots,x_{n})\in\mathbb{B}^{n}_{t},$

or equivalently, by

$\phi:\tau(\cosh r,\sinh r\omega)\in\mathbb{H}^{n}_{\tau}\mapsto
t\frac{\sinh\frac{r}{2}}{\cosh\frac{r}{2}}\omega\in\mathbb{B}^{n}_{t},$

is an isometry, and its inverse is given by

$\phi^{-1}:y\in\mathbb{B}^{n}_{t}\mapsto\left(\tau\frac{t^{2}+|y|^{2}}{t^{2}-|y|^{2}},\frac{2\tau
ty_{1}}{t^{2}-|y|^{2}},\cdots,\frac{2\tau
ty_{n}}{t^{2}-|y|^{2}}\right)\in\mathbb{H}^{n}_{\tau}.$

See [15, Chapter 8] for the proof. Therefore, there are several ways of
describing the distance function as follows:

$d_{\mathbb{B}^{n}_{t}}(0,y)=d_{\mathbb{H}^{n}_{\tau}}(0_{\tau},x)=\tau\cosh^{-1}\left(\frac{[0_{\tau},x]}{\tau^{2}}\right)=\tau\cosh^{-1}\left(\frac{t^{2}+|y|^{2}}{t^{2}-|y|^{2}}\right)=\tau\log\left(\frac{t+|y|}{t-|y|}\right),$

where $0_{\tau}=(\tau,0,\cdots,0)\in\mathbb{H}^{n}_{\tau}$. We shall write
$0_{\tau}=0$ if there is no confusion.

Note that the metric (2.1) tends to the Lebesgue measure $\mathrm{d}y$ as
$\tau\to\infty$, provided that $\tau=t/2$ and $b=t^{2}/2$. We will consider
this case in the sequel because we are interested in the convergence of the
normalizing constants for the fractional Laplacian in hyperbolic spaces to
those in $\mathbb{R}^{n}$.

Let us next recall the Helgason Fourier transform on ${\mathbb{B}^{n}_{t}}$.
The interested reader may consult [17, 14, 33, 15, 10]. The Helgason Fourier
transform is defined by

(2.4)
$\hat{u}(\lambda,\xi;t)=\int_{{\mathbb{B}^{n}_{t}}}u(y)e_{-\lambda,\xi;t}(y)\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y),\quad\lambda\in\mathbb{R},~{}\xi\in\mathbb{S}^{n-1},$

where

(2.5)
$e_{\lambda,\xi;t}(y)=\left(\frac{t^{2}-|y|^{2}}{|t\xi-y|^{2}}\right)^{\frac{n-1+i\lambda
t}{2}}$

is the eigenfunction with eigenvalue $-(\lambda^{2}+\frac{(n-1)^{2}}{t^{2}})$.
It is well known that the following inversion formula holds:

(2.6)
$u(x)=\int_{-\infty}^{\infty}\int_{\mathbb{S}^{n-1}}\hat{u}(\lambda,\xi;t)e_{\lambda,\xi;t}(x)\frac{\mathrm{d}\sigma(\xi)\,\mathrm{d}\lambda}{(t/2)^{n-1}|c(\lambda
t)|^{2}},$

where

$c(\lambda t)=\sqrt{2}(2\pi)^{n/2}\frac{\Gamma(\frac{i\lambda
t}{2})}{\Gamma(\frac{n-1+i\lambda t}{2})}$

is the Harish-Chandra coefficient. Moreover, the Plancherel formula holds:

(2.7)
$\int_{\mathbb{B}^{n}_{t}}|u(x)|^{2}\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(x)=\int_{-\infty}^{\infty}\int_{\mathbb{S}^{n-1}}|\hat{u}(\lambda,\xi;t)|^{2}\frac{\mathrm{d}\sigma(\xi)\,\mathrm{d}\lambda}{(t/2)^{n-1}|c(\lambda
t)|^{2}}.$

Note that all the formulas in this section can be written in terms of
$\mathbb{H}^{n}_{\tau}$ by using isometry (2.3). Furthermore, it is well known
[10] that (2.4), (2.6), and (2.7) converge to the standard Fourier transform,
inversion formula, and Plancherel formula on Euclidean spaces as $t\to\infty$.

We define the fractional Laplacian $(-\Delta_{{\mathbb{B}^{n}_{t}}})^{\gamma}$
by

(2.8)
$\widehat{(-\Delta_{\mathbb{B}^{n}_{t}})^{\gamma}u}=\left(\lambda^{2}+\frac{(n-1)^{2}}{t^{2}}\right)^{\gamma}\hat{u},$

which coincides with the definition in [1] when $a=t/2=1$. Similarly as in
[1], it has an integral representation

$\displaystyle-(-\Delta_{\mathbb{B}^{n}_{t}})^{\gamma}u(x)=\mathrm{p.v.}\int_{\mathbb{B}^{n}_{t}}(u(x^{\prime})-u(x))\mathcal{K}_{\gamma,t/2}(d_{\mathbb{B}^{n}_{t}}(x^{\prime},x))\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(x^{\prime}),$

with the singular kernel

(2.9)
$\mathcal{K}_{\gamma,t/2}(\rho)=\int_{-\infty}^{\infty}\left(\lambda^{2}+\frac{(n-1)^{2}}{t^{2}}\right)^{\gamma}k_{\lambda,t/2}(\rho)\,\mathrm{d}\lambda,$

where

$k_{\lambda,t/2}(x,x^{\prime})=k_{\lambda,t/2}(d_{{\mathbb{B}^{n}_{t}}}(x,x^{\prime}))=-\int_{\mathbb{S}^{n-1}}e_{\lambda,\xi;t}(x)e_{-\lambda,\xi;t}(x^{\prime})\,\frac{\mathrm{d}\sigma(\xi)}{(t/2)^{n-1}|c(\lambda
t)|^{2}}.$

The kernel (2.9) with the explicit value of constant will be investigated in
Section 4.

Let us now state some well-known results on hyperbolic spaces. The first one
is the volume doubling property that will be used frequently throughout the
paper.

######  Lemma 2.1.

For any $B_{r}\subset B_{R}\in\mathbb{H}^{n}$,

(2.10)
$\left(\frac{R}{r}\right)^{n}\leq\frac{|B_{R}|}{|B_{r}|}\leq\mathcal{D}\left(\frac{R}{r}\right)^{\log_{2}\mathcal{D}},$

where $\mathcal{D}=2^{n}\cosh^{n-1}(2R)$.

Lemma 2.1 is a direct consequence of the Bishop–Gromov inequality. See [40,
Theorem 18.8 and Corollary 18.11] for the first inequality and the second
inequality with $R=2r$ in (2.10), respectively. For the full inequality, we
find $k\in\mathbb{N}$ such that $R\in[2^{k-1}r,2^{k}r)$ and then iterate the
inequality.

The next result is the bound of Hessian of the squared distance. See [9, Lemma
3.12] for instance.

######  Lemma 2.2.

Fix $y\in{\mathbb{H}^{n}}$ and consider the distance function
$d_{\mathbb{H}^{n}}(\cdot,y)$. Then,

$D^{2}(d_{\mathbb{H}^{n}}^{2}(x,y)/2)(\xi,\xi)\leq\mathcal{H}(d_{\mathbb{H}^{n}}(x,y))|\xi|^{2}$

for all $\xi\in T_{x}{\mathbb{H}^{n}}$.

Let us close this section with the following generalization of Euclidean
dyadic cubes that will be used in the decomposition of the contact set and in
the Calderón–Zygmund technique.

######  Theorem 2.3 (Christ [8]).

There is a countable collection
$\\{Q_{\alpha}^{j}\subset{\mathbb{H}^{n}}:j\in\mathbb{Z},\alpha\in I_{j}\\}$
of open sets and constants $c_{1},c_{2}>0$ (with $2c_{1}\leq c_{2}$), and
$\delta_{0}\in(0,1)$, depending only on $n$, such that

  1. (i)

$|{\mathbb{H}^{n}}\setminus\cup_{\alpha}Q_{\alpha}^{j}|=0$ for each
$j\in\mathbb{Z}$,

  2. (ii)

if $i\geq j$, then either $Q_{\beta}^{i}\subset Q_{\alpha}^{j}$ or
$Q_{\beta}^{i}\cap Q_{\alpha}^{j}=\emptyset$,

  3. (iii)

for each $(j,\alpha)$ and each $i<j$, there is a unique $\beta$ such that
$Q_{\alpha}^{j}\subset Q_{\beta}^{i}$,

  4. (iv)

$\mathrm{diam}(Q_{\alpha}^{j})\leq c_{2}\delta_{0}^{j}$, and

  5. (v)

each $Q_{\alpha}^{j}$ contains some ball
$B(z_{\alpha}^{j},c_{1}\delta_{0}^{j})$.

The original statement of [8, Theorem 11] consists of six properties. As
mentioned in [8], the first five properties concern only the quasi-metric
space structure and the lsat property requires the space to be of homogeneous
type. Since the hyperbolic spaces are not of homogeneous type, the last
property—which is not needed in this work—cannot be included.

##  3\. The Möbius gyrogroup

In this section, the gyrogroup structure of the Poincaré ball model
${\mathbb{B}^{n}_{t}}$ is studied as a tool for analyzing kernels for the
fractional Laplacian. After first discovered in 1988 by Ungar [35], it has
been extensively studied in the context of abstract algebra, non-Euclidean
geometry, mathematical physics, and quantum information and computation [10,
11, 12, 36, 37, 38].

Let us first recall the Clifford algebra $\mathcal{C}\ell_{0,n}$ over
$\mathbb{R}^{n}$ before we define the gyrogroup addition. The Clifford algebra
$\mathcal{C}\ell_{0,n}$ is the associative real algebra generated by
$\mathbb{R}^{n}$ and $\mathbb{R}$ subject to the relation

(3.1) $x^{2}=-|x|^{2},\quad x\in\mathbb{R}^{n},$

where $|\cdot|$ denotes the standard Euclidean norm. For example,
$\mathcal{C}\ell_{0,1}$ and $\mathcal{C}\ell_{0,2}$ are isomorphic to the
field of complex numbers $\mathbb{C}$ and quaternions $\mathbb{H}$,
respectively. In the Clifford algebra, there is a unique anti-automorphism
satisfying $\bar{x}=-x$ and $\overline{xy}=\bar{y}\bar{x}$ for all
$x,y\in\mathbb{R}^{n}$. Clearly, it holds that $x\bar{x}=|x|^{2}=\bar{x}x$,
$|xy|^{2}=|x|^{2}|y|^{2}$, and $|x|=|\bar{x}|$. Moreover, it follows from the
relation (3.1) that

$x\bar{y}=\frac{1}{2}(x\bar{y}+y\bar{x})+\frac{1}{2}(x\bar{y}-y\bar{x})=\langle
x,y\rangle+\frac{1}{2}(x\bar{y}-y\bar{x}),$

where $\langle\cdot,\cdot\rangle$ is the inner product on $\mathbb{R}^{n}$.

The group of all conformal orientation preserving transformations of
${\mathbb{B}^{n}_{t}}$ consists of the mappings $K\varphi_{y}$, where $K\in
SO(n)$ and $\varphi_{x}$ is the Möbius transformation on
${\mathbb{B}^{n}_{t}}$. It is known [37, 38] that $\varphi_{x}$ is given by

$\varphi_{x}(y)=(x+y)\left(1+\frac{\bar{x}y}{t^{2}}\right)^{-1}=\frac{(1+\frac{2}{t^{2}}\langle
x,y\rangle+\frac{1}{t^{2}}|y|^{2})x+(1-\frac{1}{t^{2}}|x|^{2})y}{1+\frac{2}{t^{2}}\langle
x,y\rangle+\frac{1}{t^{4}}|x|^{2}|y|^{2}}.$

By defining the Möbius addition by

$x\oplus y=\varphi_{x}(y),$

the tuple $({\mathbb{B}^{n}_{t}},\oplus)$ forms a gyrogroup, that is, the
following properties hold:

  1. (i)

(Left identity) $0\oplus x=x$ for all $x\in{\mathbb{B}^{n}_{t}}$.

  2. (ii)

(Left inverse) For each $x\in{\mathbb{B}^{n}_{t}}$, there is $\ominus x=-x$
such that $(\ominus x)\oplus x=0$.

  3. (iii)

(Gyroassociative) $x\oplus(y\oplus z)=(x\oplus y)\oplus\mathrm{gyr}[x,y]z$ for
all $x,y,z\in{\mathbb{B}^{n}_{t}}$, where

$\mathrm{gyr}[x,y]z=\left(1+\frac{x\bar{y}}{t^{2}}\right)z\left(1+\frac{\bar{x}y}{t^{2}}\right)^{-1}$

is the gyration operator.

  4. (iv)

(Left loop property) $\mathrm{gyr}[x,y]=\mathrm{gyr}[x\oplus y,y]$ for all
$x,y\in{\mathbb{B}^{n}_{t}}$.

The gyrogroup $({\mathbb{B}^{n}_{t}},\oplus)$ is gyrocommutative, i.e.,
$x\oplus y=\mathrm{gyr}[x,y](y\oplus x)$ for all $x,y\in{\mathbb{B}^{n}_{t}}$.
Let us define the second operation $\boxplus$ so-called coaddition by

$x\boxplus y=x\oplus\mathrm{gyr}[x,\ominus y]y.$

It can be easily checked that the tuple $({\mathbb{B}^{n}_{t}},\boxplus)$
forms a gyrogroup with the left identity 0 and left inverse $\boxminus x=-x$.
We will use the abbreviation $x\ominus y=x\oplus(\ominus y)$ and $x\boxminus
y=x\boxplus(\boxminus y)$. Then we have

(3.2) $x\boxminus y=x\ominus\mathrm{gyr}[x,y]y.$

See [37, 38] for more properties of gyrogroups $({\mathbb{B}^{n}_{t}},\oplus)$
and $({\mathbb{B}^{n}_{t}},\boxplus)$. Among many properties, the following
cancellation laws will be used in Section 4.

######  Theorem 3.1.

[37, Theorem 2.15] Let $a,b\in{\mathbb{B}^{n}_{t}}$. Then $x=\ominus a\oplus
b$ is the unique solution of the equation $a\oplus x=b$, and $x=b\boxminus a$
is the unique solution of the equation $x\oplus a=b$.

In the rest of this section, we perform some computations on the
eigenfunctions $e_{\lambda,\xi;t}$ and the measures
$\mu_{{\mathbb{B}^{n}_{t}}}$ for future references.

######  Lemma 3.2.

For $\lambda\in\mathbb{R}$, $\xi\in\mathbb{S}^{n-1}$, and for
$y,z\in{\mathbb{B}^{n}_{t}}$,

$e_{-\lambda,\xi;t}(z\boxminus
y)=\left(\frac{\left|\xi-\frac{z}{t}\right|^{2}\left(1-\frac{|y|^{2}}{t^{2}}\right)\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}}{\left|\left(1-\frac{|z|^{2}|y|^{2}}{t^{4}}\right)\xi-\left(1-\frac{|y|^{2}}{t^{2}}\right)\frac{z}{t}+\left(1-\frac{|z|^{2}}{t^{2}}\right)\frac{y}{t}\right|^{2}}\right)^{\frac{n-1-i\lambda
t}{2}}e_{-\lambda,\xi;t}(z).$

###### Proof.

We use (3.2) and the formula

$1-\frac{|a\oplus
b|^{2}}{t^{2}}=\left(1-\frac{|a|^{2}}{t^{2}}\right)\left(1-\frac{|b|^{2}}{t^{2}}\right)\left|1+\frac{\bar{a}b}{t^{2}}\right|^{-2},$

which is given in [10, Equation (16)], to obtain

(3.3) $\displaystyle\begin{split}1-\frac{|z\boxminus
y|^{2}}{t^{2}}&=1-\frac{|z\oplus(-\mathrm{gyr}[z,y]y)|^{2}}{t^{2}}\\\
&=\left(1-\frac{|z|^{2}}{t^{2}}\right)\left(1-\frac{|y|^{2}}{t^{2}}\right)\left|1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right|^{-2}.\end{split}$

On the other hand, using (3.2) and the definition of gyration, we have

(3.4) $\displaystyle\begin{split}\left|\xi-\frac{z\boxminus
y}{t}\right|^{2}&=\left|\xi-\frac{z\ominus\mathrm{gyr}[z,y]y}{t}\right|^{2}\\\
&=\left|\xi-\frac{1}{t}(z-\mathrm{gyr}[z,y]y)\left(1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right)^{-1}\right|^{2}\\\
&=\left|\xi\left(1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right)-\frac{1}{t}(z-\mathrm{gyr}[z,y]y)\right|^{2}\left|1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right|^{-2}.\end{split}$

A further computation shows that

(3.5)
$\displaystyle\begin{split}&\left|\xi\left(1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right)-\frac{1}{t}(z-\mathrm{gyr}[z,y]y)\right|^{2}\\\
&=\left|\xi-\frac{z}{t}+\frac{1}{t}\left(1-\frac{\xi\bar{z}}{t}\right)\left(1+\frac{z\bar{y}}{t^{2}}\right)y\left(1+\frac{\bar{z}y}{t^{2}}\right)^{-1}\right|^{2}\\\
&=\left|\left(\xi-\frac{z}{t}\right)\left(1+\frac{\bar{z}y}{t^{2}}\right)+\left(1-\frac{\xi\bar{z}}{t}\right)\left(1+\frac{z\bar{y}}{t^{2}}\right)\frac{y}{t}\right|^{2}\left|1+\frac{\bar{z}y}{t^{2}}\right|^{-2}\\\
&=\left|\left(1-\frac{|z|^{2}|y|^{2}}{t^{4}}\right)\xi-\left(1-\frac{|y|^{2}}{t^{2}}\right)\frac{z}{t}+\left(1-\frac{|z|^{2}}{t^{2}}\right)\frac{y}{t}\right|^{2}\left|1+\frac{\bar{z}y}{t^{2}}\right|^{-2}.\end{split}$

Combining (3.3), (3.4), and (3.5), and recalling the definition of
eigenfunction (2.5), the lemma is proved. ∎

######  Lemma 3.3.

For $y,z\in{\mathbb{B}^{n}_{t}}$,

$\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(z\boxminus
y)=\left(\frac{1-\frac{|z|^{2}|y|^{2}}{t^{4}}}{\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}}\right)^{n-1}\,\mathrm{d}\mu_{\mathbb{B}^{n}_{t}}(z).$

Let us prove the following two lemmas in order to prove Lemma 3.3. The first
one is a computation of (3.2).

######  Lemma 3.4.

For $y,z\in{\mathbb{B}^{n}_{t}}$,

$z\boxminus
y=\bigg{(}1-\frac{|z|^{2}|y|^{2}}{t^{4}}\bigg{)}^{-1}\left(\left(1-\frac{|y|^{2}}{t^{2}}\right)z-\left(1-\frac{|z|^{2}}{t^{2}}\right)y\right).$

###### Proof.

A straightforward computation shows that

$\displaystyle z\boxminus y$
$\displaystyle=z\ominus\mathrm{gyr}[z,y]y=z\ominus\left(1+\frac{z\bar{y}}{t^{2}}\right)y\left(1+\frac{\bar{z}y}{t^{2}}\right)^{-1}$
$\displaystyle=\bigg{(}z-\left(1+\frac{z\bar{y}}{t^{2}}\right)y\left(1+\frac{\bar{z}y}{t^{2}}\right)^{-1}\bigg{)}\bigg{(}1-\frac{1}{t^{2}}\bar{z}\left(1+\frac{z\bar{y}}{t^{2}}\right)y\left(1+\frac{\bar{z}y}{t^{2}}\right)^{-1}\bigg{)}^{-1}$
$\displaystyle=\bigg{(}z+\frac{|z|^{2}}{t^{2}}y-y-\frac{|y|^{2}}{t^{2}}z\bigg{)}\left(1+\frac{\bar{z}y}{t^{2}}\right)^{-1}\bigg{(}1-\frac{|z|^{2}|y|^{2}}{t^{4}}\bigg{)}^{-1}\left(1+\frac{\bar{z}y}{t^{2}}\right)$
$\displaystyle=\left(\left(1-\frac{|y|^{2}}{t^{2}}\right)z-\left(1-\frac{|z|^{2}}{t^{2}}\right)y\right)\bigg{(}1-\frac{|z|^{2}|y|^{2}}{t^{4}}\bigg{)}^{-1}.$

∎

The next lemma provides the Jacobian determinant of $z\boxminus y$.

######  Lemma 3.5.

Let $J$ be the Jacobian matrix $\frac{\partial(z\boxminus y)}{\partial z}$.
Then,

$|J|=\left(1-\frac{|z|^{2}|y|^{2}}{t^{4}}\right)^{-(n+1)}\left(1-\frac{|y|^{2}}{t^{2}}\right)^{n}\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}.$

###### Proof.

Let $A=1-|z|^{2}|y|^{2}/t^{4}$. Using Lemma 3.4, the $(i,j)$-th entry, for
$i,j=1,\cdots,n$, is obtained as

$\frac{\partial(z\boxminus y)_{i}}{\partial
z_{j}}=\frac{1}{A^{2}}\left(1-\frac{|y|^{2}}{t^{2}}\right)\left(A\delta_{ij}+2\frac{z_{j}y_{i}}{t^{2}}+2\frac{|y|^{2}z_{i}z_{j}}{t^{4}}\right).$

We use the technique in [41, Lemma 5] to compute the Jacobian determinant
$|J|$. That is, we change $|J|$ into

$|J|=\frac{1}{A^{2n}}\left(1-\frac{|y|^{2}}{t^{2}}\right)^{n}\begin{vmatrix}~{}1&2z_{1}&\cdots&2z_{n}\\\
~{}0&A+2\frac{z_{1}y_{1}}{t^{2}}+2\frac{|y|^{2}z_{1}^{2}}{t^{4}}&\cdots&2\frac{z_{n}y_{1}}{t^{2}}+2\frac{|y|^{2}z_{1}z_{n}}{t^{4}}\\\
\vdots&\vdots&\ddots&\vdots\\\
~{}0&2\frac{z_{1}y_{n}}{t^{2}}+2\frac{|y|^{2}z_{n}z_{1}}{t^{4}}&\cdots&A+2\frac{z_{n}y_{n}}{t^{2}}+2\frac{|y|^{2}z_{n}^{2}}{t^{4}}\end{vmatrix}$

by adding zeroth row and zeroth column. For $i=1,\cdots,n$, we multiply the
first row of the above matrix by
$-\frac{y_{i}}{t^{2}}-\frac{|y|^{2}z_{i}}{t^{4}}$ and add it to the $i$-th
row. Then we obtain

$|J|=\frac{1}{A^{2n}}\left(1-\frac{|y|^{2}}{t^{2}}\right)^{n}\begin{vmatrix}1&2z_{1}&2z_{2}&\cdots&2z_{n}\\\
-\frac{y_{1}}{t^{2}}-\frac{|y|^{2}z_{1}}{t^{4}}&A&0&\cdots&0\\\
-\frac{y_{2}}{t^{2}}-\frac{|y|^{2}z_{2}}{t^{4}}&0&A&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
-\frac{y_{n}}{t^{2}}-\frac{|y|^{2}z_{n}}{t^{4}}&0&0&\cdots&A\end{vmatrix}.$

By multiplying the $j$-th column by
$\frac{1}{A}(\frac{y_{j}}{t^{2}}+\frac{|y|^{2}z_{j}}{t^{4}})$, $j=1,\cdots,n$,
and adding it to the first column, we arrive at

$|J|=\frac{1}{A^{2n}}\left(1-\frac{|y|^{2}}{t^{2}}\right)^{n}\begin{vmatrix}\frac{1}{A}\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}&2z_{1}&\cdots&2z_{n}\\\
0&A&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&A\end{vmatrix}=\frac{1}{A^{n+1}}\left(1-\frac{|y|^{2}}{t^{2}}\right)^{n}\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}.$

∎

We are now ready to prove Lemma 3.3.

###### Proof of Lemma 3.3.

By using (2.2), (3.3), and Lemma 3.5, we have

(3.6) $\displaystyle\begin{split}\mathrm{d}\mu_{\mathbb{B}^{n}_{t}}(z\boxminus
y)&=\left(1-\frac{|z\boxminus
y|^{2}}{t^{2}}\right)^{-n}\,\mathrm{d}(z\boxminus y)\\\
&=\left(1-\frac{|z|^{2}}{t^{2}}\right)^{-n}\left|1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right|^{2n}\left(1-\frac{|z|^{2}|y|^{2}}{t^{4}}\right)^{-(n+1)}\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}\,\mathrm{d}z.\end{split}$

Therefore, the desired result follows from

$\left|1-\frac{\bar{z}\mathrm{gyr}[z,y]y}{t^{2}}\right|=\left|1-\frac{|z|^{2}|y|^{2}}{t^{4}}\right|\left|1+\frac{\bar{z}y}{t^{2}}\right|^{-1}$

and (3.6). ∎

##  4\. Normalizing constant

In this section, we prove Theorem 1.1. Following the proof of [1, Lemma 2.1],
one can easily obtain that

(4.1)
$k_{\lambda,t/2}(\rho)=\frac{(-1)^{\frac{n+1}{2}}}{(2\pi)^{\frac{n+1}{2}}}\left(\frac{2}{t}\right)^{\frac{n-1}{2}}\left(\frac{\partial_{\rho}}{\sinh(2\rho/t)}\right)^{\frac{n-1}{2}}\cos(\lambda\rho)$

for $n=3$. Indeed, (4.1) holds for any $n\geq 3$ odd, and a similar formula
can be obtained for $n\geq 2$ even. However, let us focus on the case $n=3$ in
the sequel. As in [1, Theorem 2.4], we also have

(4.2)
$\mathcal{K}_{\gamma,t/2}(\rho)=\alpha_{\gamma}\left(\frac{\partial_{\rho}}{\sinh(2\rho/t)}\right)^{(n-1)/2}\left(\left(\frac{n-1}{t}\rho\right)^{-1/2-\gamma}K_{-1/2-\gamma}\left(\frac{n-1}{t}\rho\right)\right)$

for some constant $\alpha_{\gamma}$. The rest part of this section is devoted
to finding the constant $\alpha_{\gamma}$ in 3-dimensional case.

The idea is to follow the lines (1.3) as in the case of Euclidean spaces, in
light of the gyrogroup structure of ${\mathbb{B}^{n}_{t}}$. By utilizing
Theorem 3.1, Lemma 3.2, and Lemma 3.3, we have

(4.3) $\displaystyle\begin{split}\widehat{f(\cdot\oplus
y)}(\lambda,\xi)&=\int_{{\mathbb{B}^{n}_{t}}}f(x\oplus
y)e_{-\lambda,\xi;t}(x)\,\mathrm{d}\mu_{\mathbb{B}^{n}_{t}}(x)\\\
&=\int_{{\mathbb{B}^{n}_{t}}}f(z)e_{-\lambda,\xi;t}(z\boxminus
y)\,\mathrm{d}\mu_{\mathbb{B}^{n}_{t}}(z\boxminus y)\\\
&=\int_{{\mathbb{B}^{n}_{t}}}f(z)e_{-\lambda,\xi;t}(z)E(\lambda,\xi,y,z)\,\mathrm{d}\mu_{\mathbb{B}^{n}_{t}}(z)\end{split}$

and

(4.4) $\widehat{f(\cdot\ominus
y)}(\lambda,\xi)=\int_{{\mathbb{B}^{n}_{t}}}f(z)e_{-\lambda,\xi;t}(z)E(\lambda,\xi,-y,z)\,\mathrm{d}\mu_{\mathbb{B}^{n}_{t}}(z),$

where

$E(\lambda,\xi,y,z)=\left(\frac{\left|\xi-\frac{z}{t}\right|^{2}\left(1-\frac{|y|^{2}}{t^{2}}\right)\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}}{\left|\left(1-\frac{|z|^{2}|y|^{2}}{t^{4}}\right)\xi-\left(1-\frac{|y|^{2}}{t^{2}}\right)\frac{z}{t}+\left(1-\frac{|z|^{2}}{t^{2}}\right)\frac{y}{t}\right|^{2}}\right)^{\frac{n-1-i\lambda
t}{2}}\left(\frac{1-\frac{|z|^{2}|y|^{2}}{t^{4}}}{\left|1+\frac{\bar{z}y}{t^{2}}\right|^{2}}\right)^{n-1}.$

It is to be noted that $E$ in (4.3) and (4.4) are not independent of $z$,
which implies that the second equality in (1.3) already fails to hold in the
framework of hyperbolic spaces. We have obtained

(4.5)
$\displaystyle\begin{split}&\widehat{(-\Delta_{\mathbb{B}^{n}_{t}})^{\gamma}f}(\lambda,\xi)\\\
&=-\frac{1}{2}\int_{{\mathbb{B}^{n}_{t}}}\left(\widehat{f(\cdot\oplus
y)}(\lambda,\xi)+\widehat{f(\cdot\ominus
y)}(\lambda,\xi)-2\hat{f}(\lambda,\xi)\right)\mathcal{K}_{\gamma,t/2}(d)\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y)\\\
&=-\frac{1}{2}\int_{{\mathbb{B}^{n}_{t}}}f(z)e_{-\lambda,\xi;t}(z)\int_{{\mathbb{B}^{n}_{t}}}\left(E(\lambda,\xi,y,z)+E(\lambda,\xi,-y,z)-2\right)\mathcal{K}_{\gamma,t/2}(d)\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y)\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(z),\end{split}$

where $d$ denotes the distance $d_{{\mathbb{B}^{n}_{t}}}(0,y)$. However, even
though $E$ itself is not independent of $z$, it is plausible to guess that the
inner integral in the right-hand side of (4.5) is independent of $z$. Indeed,
we have the following result.

######  Lemma 4.1.

For any $\lambda\in\mathbb{R}$, $\xi\in\mathbb{S}^{2}$, and any
$z\in\mathbb{B}^{3}_{t}$,

(4.6)
$\displaystyle\begin{split}-\frac{1}{2}\int_{\mathbb{B}^{3}_{t}}\left(E(\xi,y,z)+E(\xi,-y,z)-2\right)\mathcal{K}_{\gamma,t/2}(d_{\mathbb{B}^{3}_{t}}(0,y))\,\mathrm{d}\mu_{\mathbb{B}^{3}_{t}}(y)\\\
=\alpha_{\gamma}2^{-3/2-3\gamma}\pi^{3/2}t^{2+2\gamma}|\Gamma(-\gamma)|\left(\lambda^{2}+\frac{4}{t^{2}}\right)^{\gamma}.\end{split}$

Lemma 4.1 actually finishes the proof of Theorem 1.1. Indeed, it follows from
(2.8), (4.5), and Lemma 4.1, that

(4.7)
$\alpha_{\gamma}=\frac{2^{3/2+3\gamma}}{\pi^{3/2}t^{2+2\gamma}|\Gamma(-\gamma)|}.$

Therefore, we obtain

$\mathcal{K}_{\gamma,t/2}(\rho)=C(3,\gamma)\frac{2/t}{\sinh(2\rho/t)}\rho^{-1/2-\gamma}\frac{2K_{-3/2-\gamma}(2\rho/t)}{\Gamma(3/2+\gamma)t^{3/2+\gamma}}$

from (4.2), (4.7), and (A.2). Recalling that $a=t/2$, we arrive at (1.2).

From now on, let us focus on proving Lemma 4.1. The proof of Lemma 4.1
consists of two parts: we will show that the imaginary part of the left-hand
side of (4.6) is zero in Lemma 4.2, and that the real part becomes the right-
hand side of (4.6), independently of $\xi$ and $z$.

######  Lemma 4.2.

Let $n=3$. For any $\lambda\in\mathbb{R}$, $\xi\in\mathbb{S}^{n-1}$, and any
$z\in{\mathbb{B}^{n}_{t}}$,

(4.8)
$\mathrm{Im}\left(\int_{{\mathbb{B}^{n}_{t}}}\left(E(\lambda,\xi,y,z)+E(\lambda,\xi,-y,z)-2\right)\mathcal{K}_{\gamma,t/2}(d_{{\mathbb{B}^{n}_{t}}}(0,y))\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y)\right)=0.$

###### Proof.

For any $T\in O(n)$, we have $E(\lambda,T\xi,y,Tz)=E(\lambda,\xi,T^{-1}y,z)$,
and hence

$\displaystyle\int_{{\mathbb{B}^{n}_{t}}}\left(E(\lambda,T\xi,y,Tz)+E(\lambda,T\xi,-y,Tz)-2\right)\mathcal{K}_{\gamma,t/2}(d_{{\mathbb{B}^{n}_{t}}}(0,y))\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y)$
$\displaystyle=\int_{{\mathbb{B}^{n}_{t}}}\left(E(\lambda,\xi,T^{-1}y,z)+E(\lambda,\xi,-T^{-1}y,z)-2\right)\mathcal{K}_{\gamma,t/2}(d_{{\mathbb{B}^{n}_{t}}}(0,y))\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y)$
$\displaystyle=\int_{{\mathbb{B}^{n}_{t}}}\left(E(\lambda,\xi,y,z)+E(\lambda,\xi,-y,z)-2\right)\mathcal{K}_{\gamma,t/2}(d_{{\mathbb{B}^{n}_{t}}}(0,y))\,\mathrm{d}\mu_{{\mathbb{B}^{n}_{t}}}(y).$

Thus, by taking $T\in O(n)$ that maps $\xi$ to $e_{1}$ and $z$ to a point in
$e_{1}e_{2}$-plane, we may assume that $\xi=e_{1}$ and
$z=(z_{1},z_{2},0,\cdots 0)$. In fact, we will prove

(4.9)
$\int_{\mathbb{S}^{n-1}}\mathrm{Im}\,E(\lambda,\xi,r\omega,z)\,\mathrm{d}\omega=0,$

from which (4.8) follows.

For notational convenience, let us write

$\mathrm{Im}\,E(\lambda,e_{1},y,z)=\left(\frac{A^{2}BD}{\left|Ae_{1}-B\frac{z}{t}+C\frac{y}{t}\right|^{2}\left|e_{1}+\frac{\bar{z}y}{t^{2}}\right|^{2}}\right)^{\frac{n-1}{2}}\sin\left(-\frac{\lambda
t}{2}\log\frac{BD\left|e_{1}+\frac{\bar{z}y}{t^{2}}\right|^{2}}{\left|Ae_{1}-B\frac{z}{t}+C\frac{y}{t}\right|^{2}}\right),$

where
$y=r(\omega_{1}\cos\theta,\omega_{1}\sin\theta,\omega^{\prime})\in\partial
B_{r}$ with $(\omega_{1},\omega^{\prime})\in\mathbb{S}^{n-2}$,
$\omega_{1}\in\mathbb{R}$, and $\omega^{\prime}\in\mathbb{R}^{n-2}$. Here, we
have set

$A=1-\frac{|z|^{2}r^{2}}{t^{4}},\quad B=1-\frac{r^{2}}{t^{2}},\quad
C=1-\frac{|z|^{2}}{t^{2}},\quad\text{and}\quad
D=\left|e_{1}-\frac{z}{t}\right|^{2}.$

To prove (4.9), let us first fix $(\omega_{1},\omega^{\prime})$ and set

$I:=\int_{\theta_{0}}^{2\pi+\theta_{0}}\mathrm{Im}\,E(\lambda,e_{1},r(\omega_{1}\cos\theta,\omega_{1}\sin\theta,\omega^{\prime}),z)\,\mathrm{d}\theta,$

where $\theta_{0}\geq 0$ is the smallest angle such that the function

$\Theta(\theta):=\frac{BD\left|e_{1}+\frac{r}{t^{2}}\bar{z}(\omega_{1}\cos\theta,\omega_{1}\sin\theta,\omega^{\prime})\right|^{2}}{\left|Ae_{1}-B\frac{z}{t}+C\frac{r}{t}(\omega_{1}\cos\theta,\omega_{1}\sin\theta,\omega^{\prime})\right|^{2}}$

attains its minimum. Let $\theta_{1}>\theta_{0}$ be the smallest angle such
that $\Theta$ attains its maximum, and $\theta_{2}>\theta_{1}$ the smallest
angle such that $\Theta$ attains its minimum. Then, we have
$\theta_{2}=\theta_{0}+2\pi$, and hence,

(4.10)
$I=\int_{\theta_{0}}^{\theta_{1}}\mathrm{Im}\,E(\lambda,e_{1},y,z)\,\mathrm{d}\theta+\int_{\theta_{1}}^{\theta_{2}}\mathrm{Im}\,E(\lambda,e_{1},y,z)\,\mathrm{d}\theta=:I_{0}+I_{1}.$

For $I_{i}$, $i=0,1$, let us consider the change of variables
$s=\Theta(\theta)$, $\theta_{i}\leq\theta\leq\theta_{i+1}$. We have

$s=\Theta(\theta)=\frac{X_{1}\omega_{1}\cos\theta+Y_{1}\omega_{1}\sin\theta+Z_{1}}{X_{2}\omega_{1}\cos\theta+Y_{2}\omega_{1}\sin\theta+Z_{2}},$

or equivalently,

$X\omega_{1}\cos\theta+Y\omega_{1}\sin\theta=Z,$

where

$\displaystyle X$
$\displaystyle:=X_{1}-X_{2}s:=2BD\frac{rz_{1}}{t^{2}}-\left(2AC\frac{r}{t}-2BC\frac{rz_{1}}{t^{2}}\right)s,$
$\displaystyle Y$
$\displaystyle:=Y_{1}-Y_{2}s:=2BD\frac{rz_{2}}{t^{2}}-\left(-2BC\frac{rz_{2}}{t^{2}}\right)s,\quad\text{and}$
$\displaystyle Z$
$\displaystyle:=Z_{2}s-Z_{1}:=\left(A^{2}-2AB\frac{z_{1}}{t}+B^{2}\frac{|z|^{2}}{t^{2}}+C^{2}\frac{r^{2}}{t^{2}}\right)s-\left(BD+\frac{r^{2}|z|^{2}}{t^{4}}BD\right).$

Let $W=\sqrt{X^{2}+Y^{2}}$, and let $\tilde{\theta}$ be the angle such that
$\cos\tilde{\theta}=\frac{X}{W}$ and $\sin\tilde{\theta}=\frac{Y}{W}$, then

(4.11)
$\theta=\tilde{\theta}+2k\pi\pm\arccos\frac{Z}{W\omega_{1}}\quad\text{for
some}~{}k\in\mathbb{Z},$

where the principal branch for $\arccos$ is taken to be $[0,\pi]$. Note that
$s=\Theta(\theta)$ increases on $[\theta_{0},\theta_{1}]$ and decreases on
$[\theta_{1},\theta_{2}]$. Since $Z_{2}>0$, $\arccos\frac{Z}{W\omega_{1}}$
decreases on $[\theta_{0},\theta_{1}]$ and increases on
$[\theta_{1},\theta_{2}]$. Thus, we have the minus sign in (4.11) on
$[\theta_{0},\theta_{1}]$ and the plus sign on $[\theta_{1},\theta_{2}]$. Let
us consider the case $\theta\in[\theta_{0},\theta_{1}]$, then

(4.12)
$\cos\theta=\frac{XZ+Y\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}\omega_{1}}\quad\text{and}\quad\sin\theta=\frac{YZ-X\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}\omega_{1}}.$

Using (4.12), we see that

(4.13)
$X_{1}\omega_{1}\cos\theta+Y_{1}\omega_{1}\sin\theta+Z_{1}=(X_{2}\omega_{1}\cos\theta+Y_{2}\omega_{1}\sin\theta+Z_{2})s=\frac{U-V\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}}s,$

where

$\displaystyle U$
$\displaystyle=-(X_{1}X_{2}+Y_{1}Y_{2})Z_{1}+(X_{1}^{2}+Y_{1}^{2})Z_{2}+\left((X_{2}^{2}+Y_{2}^{2})Z_{1}-(X_{1}X_{2}+Y_{1}Y_{2})Z_{2}\right)s,\quad\text{and}$
$\displaystyle V$ $\displaystyle=X_{1}Y_{2}-Y_{1}X_{2}.$

Thus, we have

(4.14)
$\displaystyle\begin{split}\left|e_{1}+\frac{r}{t^{2}}\bar{z}(\omega_{1}\cos\theta,\omega_{1}\sin\theta,\omega^{\prime})\right|^{2}&=\frac{(X_{1}\omega_{1}\cos\theta+Y_{1}\omega_{1}\sin\theta+Z_{1})}{BD}\\\
&=\frac{U-V\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{BDW^{2}}s.\end{split}$

Moreover, the equality

$\displaystyle(Y_{1}$ $\displaystyle
Z_{2}-Y_{2}Z_{1})\omega_{1}\cos\theta+(Z_{1}X_{2}-Z_{2}X_{1})\omega_{1}\sin\theta-(X_{1}Y_{2}-X_{2}Y_{1})\omega_{1}^{2}$
$\displaystyle=$
$\displaystyle~{}((Y_{1}Z_{2}-Y_{2}Z_{1})Y-(Z_{1}X_{2}-Z_{2}X_{1})X)\frac{\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}}$
$\displaystyle+\frac{(Y_{1}Z_{2}-Y_{2}Z_{1})XZ+(Z_{1}X_{2}-Z_{2}X_{1})YZ-(X_{1}Y_{2}-X_{2}Y_{1})W^{2}\omega_{1}^{2}}{W^{2}}$
$\displaystyle=$
$\displaystyle~{}((Y_{1}Z_{2}-Y_{2}Z_{1})Y-(Z_{1}X_{2}-Z_{2}X_{1})X)\frac{\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}}-\frac{(X_{1}Y_{2}-X_{2}Y_{1})(W^{2}\omega_{1}^{2}-Z^{2})}{W^{2}}$
$\displaystyle=$
$\displaystyle~{}\frac{U-V\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}}\sqrt{W^{2}\omega_{1}^{2}-Z^{2}},$

together with (4.13), yields that

(4.15)
$\displaystyle\begin{split}\mathrm{d}\theta&=\frac{(X_{2}\omega_{1}\cos\theta+Y_{2}\omega_{1}\sin\theta+Z_{2})^{2}}{(Y_{1}Z_{2}-Y_{2}Z_{1})\omega_{1}\cos\theta+(Z_{1}X_{2}-Z_{2}X_{1})\omega_{1}\sin\theta-(X_{1}Y_{2}-X_{2}Y_{1})\omega_{1}^{2}}\,\mathrm{d}s\\\
&=\frac{U-V\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}{W^{2}}\frac{1}{\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}\,\mathrm{d}s.\end{split}$

Therefore, combining (4.10), (4.14), and (4.15), and performing the same
computation on $[\theta_{1},\theta_{2}]$, we obtain

(4.16)
$\displaystyle\begin{split}I=&\int_{s_{0}}^{s_{1}}\frac{(ABD)^{n-1}s^{-\frac{n-1}{2}}}{\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}\left(\frac{W^{2}}{U-V\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}\right)^{n-2}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s\\\
&-\int_{s_{1}}^{s_{2}}\frac{(ABD)^{n-1}s^{-\frac{n-1}{2}}}{\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}\left(\frac{W^{2}}{U+V\sqrt{W^{2}\omega_{1}^{2}-Z^{2}}}\right)^{n-2}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s,\end{split}$

where $s_{i}=\Theta(\theta_{i})$, $i=0,1,2$. Since $n=3$, we have
$\omega_{1}=\sin\varphi$ and $\omega^{\prime}=\cos\varphi$, and hence

$\displaystyle\int_{\mathbb{S}^{2}}\mathrm{Im}\,E(e_{1},y,z)\,\mathrm{d}\omega$
$\displaystyle=\int_{0}^{\pi}\int_{s_{0}}^{s_{1}}\frac{(ABD)^{2}s^{-1}}{\sqrt{W^{2}\sin^{2}\varphi-Z^{2}}}\frac{2UW^{2}}{U^{2}-V^{2}(W^{2}\sin^{2}\varphi-Z^{2})}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s\sin\varphi\,\mathrm{d}\varphi$
$\displaystyle=4\int_{0}^{\pi/2}\int_{s_{0}}^{s_{1}}\frac{(ABD)^{2}s^{-1}}{\sqrt{W^{2}\sin^{2}\varphi-Z^{2}}}\frac{UW^{2}}{U^{2}-V^{2}(W^{2}\sin^{2}\varphi-Z^{2})}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s\sin\varphi\,\mathrm{d}\varphi.$

Note that $s_{0}$ and $s_{1}$ are solutions of $W^{2}\sin^{2}\varphi-Z^{2}=0$.
Since

(4.17)
$W^{2}-Z^{2}=-A^{2}B^{2}D^{2}\left(s-\frac{t-r}{t+r}\right)\left(s-\frac{t+r}{t-r}\right),$

Fubini’s theorem and change of variables $v=\sin\varphi$ lead us to

$\displaystyle\int_{\mathbb{S}^{2}}\mathrm{Im}\,E(e_{1},y,z)\,\mathrm{d}\omega$
$\displaystyle=\int_{\frac{t-r}{t+r}}^{\frac{t+r}{t-r}}\left(\int_{v_{0}}^{1}\frac{4(ABD)^{2}UW^{2}v}{(U^{2}-V^{2}(W^{2}v^{2}-Z^{2}))\sqrt{W^{2}v^{2}-Z^{2}}\sqrt{1-v^{2}}}\,\mathrm{d}v\right)\frac{1}{s}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s$
$\displaystyle=\int_{\frac{t-r}{t+r}}^{\frac{t+r}{t-r}}\left(\int_{v_{0}}^{1}\frac{4(ABD)^{2}Uv}{V^{2}W(\frac{U^{2}}{V^{2}W^{2}}+v_{0}^{2}-v^{2})\sqrt{v^{2}-v_{0}^{2}}\sqrt{1-v^{2}}}\,\mathrm{d}v\right)\frac{1}{s}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s,$

where $v_{0}=\frac{|Z|}{W}$. Note that
$R^{2}:=\frac{U^{2}}{V^{2}W^{2}}+v_{0}^{2}\geq 1$ since
$Z^{2}+\frac{U^{2}}{V^{2}}\geq Z^{2}\geq X^{2}+Y^{2}\geq W^{2}$. Moreover,
since

$\frac{\mathrm{d}}{\mathrm{d}v}\left(-\frac{VW}{U\sqrt{R^{2}-1}}\arctan\left(\frac{U}{VW\sqrt{R^{2}-1}}\sqrt{\frac{1-v^{2}}{v^{2}-v_{0}^{2}}}\right)\right)=\frac{v}{(R^{2}-v^{2})\sqrt{(v^{2}-v_{0}^{2})(1-v^{2})}},$

we obtain

$\int_{v_{0}}^{1}\frac{v}{(\frac{U^{2}}{V^{2}W^{2}}+v_{0}^{2}-v^{2})\sqrt{v^{2}-v_{0}^{2}}\sqrt{1-v^{2}}}\,\mathrm{d}v=\frac{\pi
V^{2}W^{2}}{2U\sqrt{U^{2}-V^{2}(W^{2}-Z^{2})}}.$

Therefore, we have

$\displaystyle\int_{\mathbb{S}^{2}}\mathrm{Im}\,E(e_{1},y,z)\,\mathrm{d}\omega$
$\displaystyle=\int_{\frac{t-r}{t+r}}^{\frac{t+r}{t-r}}\frac{2\pi(ABD)^{2}W}{\sqrt{U^{2}-V^{2}(W^{2}-Z^{2})}}s^{-1}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s.$

A simple computation shows that

$U^{2}-V^{2}(W^{2}-Z^{2})=\left((Z_{1}X_{2}-Z_{2}X_{1})^{2}+(Y_{1}Z_{2}-Y_{2}Z_{1})^{2}-(X_{1}Y_{2}-X_{2}Y_{1})^{2}\right)W^{2}.$

Furthermore, we have

$\displaystyle Z_{1}X_{2}-Z_{2}X_{1}$
$\displaystyle=-2ABD\left(C\frac{r}{t}\left(1+\frac{r^{2}\|z\|^{2}}{t^{4}}\right)-2\frac{rz_{1}}{t^{2}}\left(A-B\frac{z_{1}}{t}\right)\right),$
$\displaystyle Y_{1}Z_{2}-Y_{2}Z_{1}$
$\displaystyle=4ABD\frac{rz_{2}}{t^{2}}\left(A-B\frac{z_{1}}{t}\right),\quad\text{and}$
$\displaystyle X_{1}Y_{2}-X_{2}Y_{1}$
$\displaystyle=-4ABCD\frac{r^{2}z_{2}}{t^{3}},$

which yield that

$U^{2}-V^{2}(W^{2}-Z^{2})=4(ABD)^{4}W^{2}\left(\frac{rt}{t^{2}-r^{2}}\right)^{2}.$

Finally, the change of variables $u=\frac{t}{2}\log s$ yields that

$\displaystyle\int_{\mathbb{S}^{2}}\mathrm{Im}\,E(e_{1},y,z)\,\mathrm{d}\omega$
$\displaystyle=\pi\int_{\frac{t-r}{t+r}}^{\frac{t+r}{t-r}}\left(\frac{t}{r}-\frac{r}{t}\right)s^{-1}\sin\left(\frac{\lambda
t}{2}\log s\right)\,\mathrm{d}s$
$\displaystyle=\frac{2\pi}{t}\left(\frac{t}{r}-\frac{r}{t}\right)\int_{-d_{\mathbb{B}^{n}_{t}}(0,y)}^{d_{\mathbb{B}^{n}_{t}}(0,y)}\sin(\lambda
u)\,\mathrm{d}u=0,$

which proves (4.9). ∎

The proof of Lemma 4.2 for 2-dimensional case is simpler. Indeed, it follows
from (4.16) and (4.17) that

$I=\int_{\frac{t-r}{t+r}}^{\frac{t+r}{t-r}}\frac{ABDs^{-1/2}\sin\left(\frac{\lambda
t}{2}\log
s\right)}{\sqrt{-(s-\frac{t-r}{t+r})(s-\frac{t+r}{t-r})}}\,\mathrm{d}s=\int_{\log\frac{t-r}{t+r}}^{\log\frac{t+r}{t-r}}\frac{ABDe^{u/2}\sin\left(\frac{\lambda
t}{2}u\right)}{\sqrt{-(e^{u}-\frac{t-r}{t+r})(e^{u}-\frac{t+r}{t-r})}}\,\mathrm{d}u=0.$

The authors believe that Lemma 4.2 holds true for any dimension $n$, but could
not find how to compute the integral (4.16). We leave it for future
discussion.

Let us next prove Lemma 4.1.

###### Proof of Lemma 4.1.

The same computation as in Lemma 4.2 shows that

$\displaystyle\int_{\mathbb{S}^{2}}\mathrm{Re}\,E(\lambda,\xi,y,z)\,\mathrm{d}\omega$
$\displaystyle=\frac{2\pi}{t}\left(\frac{t}{r}-\frac{r}{t}\right)\int_{-d_{{\mathbb{B}^{n}_{t}}}(0,y)}^{d_{{\mathbb{B}^{n}_{t}}}(0,y)}\cos(\lambda
u)\,\mathrm{d}u$ $\displaystyle=\frac{4\pi}{\lambda
t}\left(\frac{t}{r}-\frac{r}{t}\right)\sin\left(\lambda
d_{\mathbb{B}^{n}_{t}}(0,y)\right),$

where $y=r\omega$. Thus, we have

$\displaystyle I:=$
$\displaystyle-\frac{1}{2}\int_{0}^{t}\int_{\mathbb{S}^{2}}\mathrm{Re}\left(E(\lambda,\xi,r\omega,z)+E(\lambda,\xi,-r\omega,z)-2\right)\mathcal{K}_{\gamma,t/2}\left(d_{\mathbb{B}^{n}_{t}}(0,y)\right)\frac{r^{2}}{(1-\frac{r^{2}}{t^{2}})^{3}}\,\mathrm{d}\omega\,\mathrm{d}r$
$\displaystyle=$ $\displaystyle~{}4\pi\int_{0}^{t}\left(1-\frac{1}{\lambda
t}\left(\frac{t}{r}-\frac{r}{t}\right)\sin\left(\lambda
d_{\mathbb{B}^{n}_{t}}(0,y)\right)\right)\mathcal{K}_{\gamma,t/2}\left(d_{\mathbb{B}^{n}_{t}}(0,y)\right)\frac{r^{2}}{(1-\frac{r^{2}}{t^{2}})^{3}}\,\mathrm{d}r.$

Using the change of variables

$\rho=\cosh^{-1}\left(\frac{t^{2}+r^{2}}{t^{2}-r^{2}}\right),\quad
r^{2}=t^{2}\frac{\cosh\rho-1}{\cosh\rho+1},\quad\mathrm{d}r=\frac{t}{\cosh\rho+1}\,\mathrm{d}\rho,$

we obtain

(4.18) $I=\frac{\pi t^{3}}{2}\int_{0}^{\infty}\left(1-\frac{2\sin(\lambda
t\rho/2)}{\lambda
t\sinh\rho}\right)\mathcal{K}_{\gamma,t/2}(t\rho/2)\sinh^{2}\rho\,\mathrm{d}\rho.$

Recall that the kernel $\mathcal{K}_{\gamma,t/2}$ is given by

(4.19)
$\displaystyle\begin{split}\mathcal{K}_{\gamma,t/2}(\rho)&=\alpha_{\gamma}\frac{1}{\sinh(2\rho/t)}\partial_{\rho}\left(\left(2\rho/t\right)^{-1/2-\gamma}K_{-1/2-\gamma}\left(2\rho/t\right)\right)\\\
&=(-\alpha_{\gamma})\frac{2/t}{\sinh(2\rho/t)}\left(2\rho/t\right)^{-1/2-\gamma}K_{-3/2-\gamma}\left(2\rho/t\right),\end{split}$

where we have used (A.2). By plugging (4.19) into (4.18), we have

$I=(-\alpha_{\gamma})\pi
t^{2}\int_{0}^{\infty}\left(\sinh\rho-\frac{2}{\lambda
t}\sin\left(\frac{\lambda
t}{2}\rho\right)\right)\rho^{-1/2-\gamma}K_{-3/2-\gamma}(\rho).$

Let $S^{k}_{\nu}$ and $L^{k}_{\nu}$ be indefinite integrals given in (A.3).
Then, for $\nu=3/2+\gamma$,

$I=(-\alpha_{\gamma})\pi
t^{2}\left[S^{1}_{\nu}(\rho)-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\frac{\lambda
t}{2}\right)^{2k}L^{2k+2}_{\nu}(\rho)\right]_{0}^{\infty}.$

Lemma A.1, Lemma A.2, and Lemma A.3 show that

$S^{1}_{\nu}-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\frac{\lambda
t}{2}\right)^{2k}L^{2k+2}_{\nu}\to 0$

as $\rho\to 0$, and

$S^{1}_{\nu}-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\frac{\lambda
t}{2}\right)^{2k}L^{2k+2}_{\nu}\to-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}\left(\frac{\lambda
t}{2}\right)^{2k}\frac{\sqrt{\pi}(2k+2)!}{2^{k+1}(k+1)!}2^{k-\nu}\Gamma(k-\gamma)$

as $\rho\to\infty$. Therefore, we arrive at

$\displaystyle I$
$\displaystyle=(-\alpha_{\gamma})2^{-\nu}\pi^{3/2}t^{2}\left[-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\left(\frac{\lambda
t}{2}\right)^{2k}(k-1-\gamma)(k-2-\gamma)\cdots(-\gamma)\Gamma(-\gamma)\right]$
$\displaystyle=(-\alpha_{\gamma})2^{-\nu}\pi^{3/2}t^{2}|\Gamma(-\gamma)|\sum_{k=0}^{\infty}\frac{\gamma(\gamma-1)\cdots(\gamma-k+1)}{k!}\left(\frac{\lambda
t}{2}\right)^{2k}$
$\displaystyle=(-\alpha_{\gamma})2^{-\nu}\pi^{3/2}t^{2}|\Gamma(-\gamma)|\left(1+\frac{\lambda^{2}t^{2}}{4}\right)^{\gamma}$
$\displaystyle=(-\alpha_{\gamma})2^{-3/2-3\gamma}\pi^{3/2}t^{2+2\gamma}|\Gamma(-\gamma)|\left(\lambda^{2}+\frac{4}{t^{2}}\right)^{\gamma},$

where we have used the binomial theorem for fractional exponent $\gamma$ in
the third equality. ∎

##  5\. Scale functions

Having the explicit form of kernels (1.2) at hand, we will proceed to the
regularity theory for fractional Laplacian-type operators on hyperbolic
spaces. For this purpose, some scale functions that are appropriate in the
framework of hyperbolic spaces are introduced in this section.

As mentioned in Section 1, it is of interest to study regularity estimates
whose constants do not blow up and stay uniform as the order of operator
approaches 2. To simplify computations and focus on the limit $\gamma\to 1$,
let us consider the the case $\tau=t/2=1$ and write
$\mathcal{K}_{\gamma,1}=\mathcal{K}_{\gamma}$. Let us define the scale
functions

(5.1)
$\displaystyle\begin{split}\mathcal{I}_{0}(R)&:=\mathcal{I}_{0,\gamma}(R):=\int_{B_{R}(x)}d_{{\mathbb{H}^{n}}}^{2}(z,x)\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z),\\\
\mathcal{I}_{\infty}(R)&:=\mathcal{I}_{\infty,\gamma}(R):=\int_{{\mathbb{H}^{n}}\setminus
B_{R}(x)}R^{2}\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z),\end{split}$

and
$\mathcal{I}(R):=\mathcal{I}_{\gamma}(R):=\mathcal{I}_{0,\gamma}(R)+\mathcal{I}_{\infty,\gamma}(R)$.
It follows from Theorem 1.1 and Lemma A.3 that
$\rho^{2}\mathcal{K}_{\gamma}(\rho)\sinh^{2}\rho\sim\rho^{1-2\gamma}$ as
$\rho\to 0$ and $\mathcal{K}_{\gamma}(\rho)\sinh^{2}\rho\sim\rho^{-1-\gamma}$
as $\rho\to\infty$, which guarantee the well-definedness of integrals in
(5.1). Moreover, we observe that

(5.2)
$|Lu(x)|\leq\Lambda\|u\|_{C^{2}(\overline{B_{R}})}\mathcal{I}_{0}(R)+2\Lambda\|u\|_{L^{\infty}({\mathbb{H}^{n}})}\mathcal{I}_{\infty}(R)<+\infty$

for any $L\in\mathcal{L}_{0}$ and any function $u\in
C^{2}(\overline{B_{R}(x)})\cap L^{\infty}({\mathbb{H}^{n}})$.

In 3-dimensional case, the scale functions (5.1) are represented explicitly by
using Bessel functions.

######  Lemma 5.1.

Let $n=3$. Then,

(5.3) $\displaystyle\mathcal{I}_{0}(R)=$
$\displaystyle~{}\frac{2^{\gamma}}{(1-\gamma)|\Gamma(-\gamma)|}R^{3-\gamma}\left(I_{1/2}K_{3/2+\gamma}+I_{3/2}K_{1/2+\gamma}\right)$
$\displaystyle~{}-\frac{2^{\gamma}}{(1-\gamma)(2-\gamma)|\Gamma(-\gamma)|}R^{3-\gamma}\left(I_{3/2}K_{1/2+\gamma}+I_{5/2}K_{-1/2+\gamma}\right)\quad\text{and}$
(5.4) $\displaystyle\mathcal{I}_{\infty}(R)=$
$\displaystyle~{}\frac{2^{\gamma}}{\gamma|\Gamma(-\gamma)|}R^{3-\gamma}\left(I_{1/2}K_{3/2+\gamma}+I_{3/2}K_{1/2+\gamma}\right),$

where $I_{\nu}$ and $K_{\nu}$ denote the modified Bessel function of the first
and second kind, respectively.

###### Proof.

The proof of (5.4) is very similar to that of (5.3), so let us prove (5.3)
only. Let $S^{k}_{\nu}$ and $C^{k}_{\nu}$ be functions defined by (A.3), then

$\displaystyle\mathcal{I}_{0}(R)$
$\displaystyle=|\mathbb{S}^{2}|\int_{0}^{R}\rho^{2}\mathcal{K}_{\gamma}(\rho)\sinh^{2}\rho\,\mathrm{d}\rho$
$\displaystyle=\frac{C(3,\gamma)|\mathbb{S}^{2}|}{\Gamma(3/2+\gamma)2^{1/2+\gamma}}\int_{0}^{R}\rho^{3/2-\gamma}K_{-3/2-\gamma}(\rho)\sinh\rho\,\mathrm{d}\rho$
$\displaystyle=\frac{2^{3/2+\gamma}}{\pi^{1/2}|\Gamma(-\gamma)|}\left(S^{3}_{3/2+\gamma}(R)-S^{3}_{3/2+\gamma}(0)\right),$

where $\nu=3/2+\gamma$. By using Lemma A.1 and the recurrence relations (A.1),
we obtain

$\displaystyle S^{3}_{3/2+\gamma}(R)=$
$\displaystyle~{}\frac{2^{-3/2}\pi^{1/2}}{(1-\gamma)}R^{3-\gamma}\left(I_{1/2}K_{3/2+\gamma}+I_{3/2}K_{1/2+\gamma}\right)$
$\displaystyle~{}-\frac{2^{-3/2}\pi^{1/2}}{(1-\gamma)(2-\gamma)}R^{3-\gamma}\left(I_{3/2}K_{1/2+\gamma}+I_{5/2}K_{-1/2+\gamma}\right),$

from which (5.3) follows. ∎

The scale functions $\mathcal{I}_{0}$ and $\mathcal{I}_{\infty}$ correspond to
$\frac{C(3,\gamma)|\mathbb{S}^{2}|}{2-2\gamma}R^{2-2\gamma}$ and
$\frac{C(3,\gamma)|\mathbb{S}^{2}|}{2\gamma}R^{2-2\gamma}$, respectively, in
the case of Euclidean spaces. While $R^{2-2\gamma}$ has a scaling property,
the scale functions $\mathcal{I}_{0}$ and $\mathcal{I}_{\infty}$ do not have
it. This is one of the main difficulties in the upcoming regularity theory.
However, we will observe that some monotonicity properties play a similar role
in the following sections. For example, $\mathcal{I}_{0}(R)$ is increasing and
$\mathcal{I}_{\infty}$ is decreasing by definition. Moreover, the following
lemma provides another monotonicity.

######  Lemma 5.2.

The function $\mathcal{I}_{0}(R)/R^{2-\gamma}$ is decreasing in $R$.

###### Proof.

Let us define the function

$\displaystyle F(R):=$
$\displaystyle\frac{(1-\gamma)|\Gamma(-\gamma)|}{2^{\gamma}}\frac{\mathcal{I}_{0}(R)}{R^{2-\gamma}}$
$\displaystyle=$ $\displaystyle
R\left(I_{1/2}K_{3/2+\gamma}+\frac{1-\gamma}{2-\gamma}I_{3/2}K_{1/2+\gamma}-\frac{1}{2-\gamma}I_{5/2}K_{-1/2+\gamma}\right).$

Using the relations (A.1) and (A.2), a straightforward calculation yields

$F^{\prime}(R)=-\gamma
I_{1/2}K_{3/2+\gamma}-(1-\gamma)I_{3/2}K_{1/2+\gamma}+I_{5/2}K_{-1/2+\gamma}.$

Thus, we conclude that

$F^{\prime}(R)\leq\left(-\gamma-(1-\gamma)+1\right)I_{5/2}K_{-1/2+\gamma}=0,$

by using the facts that $I_{\nu}$ is decreasing and $K_{\nu}$ is increasing
with respect to $\nu$ when $\nu>0$. ∎

An obvious consequence of Lemma 5.2 is the following.

######  Corollary 5.3.

The function $\mathcal{I}_{0}(R)/R^{2}$ is decreasing in $R$.

The following lemma shows a relation between two scale functions
$\mathcal{I}_{0}$ and $\mathcal{I}_{\infty}$. Let us mention that the function
$\mathcal{H}(t)=t\coth t$ naturally appears when the negative curvature is
involved as in Lemma 2.2.

######  Lemma 5.4.

For all $R>0$,

$\mathcal{I}_{\infty}(R)\leq\frac{1-\gamma}{\gamma}\mathcal{H}(R)\mathcal{I}_{0}(R).$

###### Proof.

Let us define the function

$F=\frac{\gamma|\Gamma(-\gamma)|}{2^{\gamma}}\left(\mathcal{I}_{\infty}-\frac{1-\gamma}{\gamma}\mathcal{H}\mathcal{I}_{0}\right)$

and show that $R^{-2+\gamma}F^{\prime}\leq 0$. Using Lemma 5.1 and the
relations (A.1) and (A.2), we have

$\displaystyle R^{-2+\gamma}F^{\prime}=$
$\displaystyle~{}2(1-\gamma)I_{1/2}K_{3/2+\gamma}+2I_{3/2}K_{1/2+\gamma}-2(1-\gamma)\mathcal{H}I_{1/2}K_{3/2+\gamma}$
$\displaystyle-\left(\mathcal{H}-\frac{R^{2}}{\sinh^{2}R}\right)\left(I_{1/2}K_{3/2+\gamma}+\frac{1-\gamma}{2-\gamma}I_{3/2}K_{1/2+\gamma}-\frac{1}{2-\gamma}I_{5/2}K_{-1/2+\gamma}\right).$

Using $\mathcal{H}I_{1/2}=RI_{-1/2}$ and (A.1) again, we obtain

$\mathcal{H}I_{1/2}K_{3/2+\gamma}=RI_{3/2}K_{-1/2+\gamma}+(1+2\gamma)I_{3/2}K_{1/2+\gamma}+I_{1/2}K_{3/2+\gamma}.$

Thus, in light of $\mathcal{H}-\frac{R^{2}}{\sinh^{2}R}\geq 0$ and
$\gamma\in(0,1)$, we have

$\displaystyle R^{-2+\gamma}F^{\prime}=$
$\displaystyle~{}(2-2(1-\gamma)(1+2\gamma))I_{3/2}K_{1/2+\gamma}-2(1-\gamma)RI_{3/2}K_{-1/2+\gamma}$
$\displaystyle-\left(\mathcal{H}-\frac{R^{2}}{\sinh^{2}R}\right)\left(I_{1/2}K_{3/2+\gamma}+\frac{1-\gamma}{2-\gamma}I_{3/2}K_{1/2+\gamma}-\frac{1}{2-\gamma}I_{5/2}K_{-1/2+\gamma}\right)$
$\displaystyle\leq$
$\displaystyle~{}-2\gamma(1-2\gamma)I_{3/2}K_{1/2+\gamma}-\left(\mathcal{H}-\frac{R^{2}}{\sinh^{2}R}\right)\left(I_{1/2}K_{3/2+\gamma}-I_{5/2}K_{-1/2+\gamma}\right).$

Moreover, since $I_{1/2}K_{3/2+\gamma}-I_{5/2}K_{-1/2+\gamma}\geq 0$, we use

$\mathcal{H}-\frac{R^{2}}{\sinh^{2}R}\geq\frac{R^{2}}{\sqrt{R^{2}+1}+1},$

to obtain

$\frac{R^{-2+\gamma}F^{\prime}}{I_{3/2}K_{1/2+\gamma}}\leq-2\gamma(1-2\gamma)-\frac{R^{2}}{\sqrt{R^{2}+1}+1}\left(\frac{I_{1/2}K_{3/2+\gamma}-I_{5/2}K_{-1/2+\gamma}}{I_{3/2}K_{1/2+\gamma}}\right).$

We now use bounds for ratios of modified Bessel functions

(5.5)
$\frac{I_{\nu+1/2}(R)}{I_{\nu-1/2}(R)}<\frac{R}{\sqrt{R^{2}+\nu^{2}}+\nu},\quad\nu\geq
0,$

and

(5.6)
$\frac{K_{\nu}(R)}{K_{\nu+1}(R)}\leq\frac{R}{\sqrt{R^{2}+(\nu-1/2)^{2}}+\nu+1/2},\quad\nu\geq
1/2.$

See [32, Theorem 3 and 6] for the proofs of these bounds. By using (5.5)–(5.6)
and monotonicity of $K_{\nu}$ with respect to $\nu$, we have

$\displaystyle\frac{I_{1/2}K_{3/2+\gamma}-I_{5/2}K_{-1/2+\gamma}}{I_{3/2}K_{1/2+\gamma}}$
$\displaystyle\geq\frac{(\sqrt{R^{2}+1}+1)(\sqrt{R^{2}+\gamma^{2}}+1+\gamma)}{R^{2}}-\frac{\sqrt{R^{2}+4}-2}{R}$
$\displaystyle\geq\frac{(\sqrt{R^{2}+1}+1)(\sqrt{R^{2}+\gamma^{2}}+1+\gamma)}{R^{2}}-\frac{\sqrt{R^{2}+1}+1}{R}$
$\displaystyle\geq(1+\gamma)\frac{\sqrt{R^{2}+1}+1}{R^{2}}.$

Therefore, we conclude that

$\frac{R^{-2+\gamma}F^{\prime}}{I_{3/2}K_{1/2+\gamma}}\leq-2\gamma(1-2\gamma)-(1+\gamma)=-(1+4\gamma)(1-\gamma)\leq
0.$

∎

Since the limit behavior of the scale functions as $\gamma\to 0$ are of
interest in the unified regularity theory, we also prepare the following
lemma.

######  Lemma 5.5.

For any fixed $R>0$,

$\lim_{\gamma\to 1}\mathcal{I}_{0}(R)=6\quad\text{and}\quad\lim_{\gamma\to
1}\mathcal{I}_{\infty}(R)=0.$

###### Proof.

The modified Bessel functions $I_{\nu}$ and $K_{\nu}$ are elementary functions
when $\nu$ is a half-integer. For examples, we have

$\displaystyle I_{1/2}=\left(\frac{2}{\pi R}\right)^{\frac{1}{2}}\sinh R,$
$\displaystyle K_{1/2}=\left(\frac{\pi}{2R}\right)^{\frac{1}{2}}e^{-R},$
$\displaystyle I_{3/2}=\left(\frac{2}{\pi
R}\right)^{\frac{1}{2}}\left(-\frac{\sinh R}{R}+\cosh R\right),$
$\displaystyle
K_{3/2}=\left(\frac{\pi}{2R}\right)^{\frac{1}{2}}\left(1+\frac{1}{R}\right)e^{-R},$
$\displaystyle I_{5/2}=\left(\frac{2}{\pi
R}\right)^{\frac{1}{2}}\left(\left(1+\frac{3}{R^{2}}\right)\sinh
R-\frac{3}{R}\cosh R\right),$ $\displaystyle
K_{5/2}=\left(\frac{\pi}{2R}\right)^{\frac{1}{2}}\left(1+\frac{3}{R}+\frac{3}{R^{2}}\right)e^{-R}.$

Thus, by using

$\lim_{\gamma\to
1^{-}}(1-\gamma)|\Gamma(-\gamma)|=1\quad\text{and}\quad\lim_{\gamma\to
0}\gamma|\Gamma(-\gamma)|=1,$

a straightforward computation proves the lemma. ∎

In the work of Caffarelli and Silvestre [3], the quantity
$r_{0}=\rho_{0}2^{-1/(2-2\gamma)}R$, which is characterized by the relation
$(r_{0}/\rho_{0})^{2-2\gamma}=R^{2-2\gamma}/2$, plays a fundamental role. The
most important feature of this quantity is that it converges to 0 as
$\gamma\to 0$.

We define such quantity in a similar way in our framework. Since the scale
function $\mathcal{I}_{0}$ is strictly increasing, its inverse exists. Thus,
for given $R>0$, we define $r_{0}\in(0,R)$ by

(5.7) $r_{0}=\rho_{0}\mathcal{I}_{0}^{-1}(\mathcal{I}_{0}(R)/2)$

for some universal constant $\rho_{0}\in(0,1)$ that will be determined later.
Let us close the section with the following lemma.

######  Lemma 5.6.

For fixed $R>0$, $\lim_{\gamma\to 1}r_{0}=0$.

###### Proof.

Let us fix $R>0$ and suppose that $\lim_{\gamma\to 1}r_{0}\neq 0$. Then, since
$\tilde{r}:=\limsup_{\gamma\to 1}r_{0}\neq 0$, there is a sequence
$\gamma_{k}\to 1$ such that
$r_{0,k}:=\rho_{0}\mathcal{I}_{0,\gamma_{k}}^{-1}(\mathcal{I}_{0,\gamma_{k}}(R)/2)\in(0,R)$
converges to $\tilde{r}$. We have

(5.8)
$\mathcal{I}_{0,\gamma_{k}}(r_{0,k}/\rho_{0})=\frac{1}{2}\mathcal{I}_{0,\gamma_{k}}(R).$

By Lemma 5.5 and continuity of $\mathcal{I}_{0}$, the left-hand side of (5.8)
converges to 6, whereas the right-hand side of (5.8) converges to 3, which is
a contradiction. ∎

##  6\. Discrete ABP-type estimates

In this section, we provide the proof of Theorem 1.3. Throughout the section,
$u$ is assumed to be a supersolution given in Theorem 1.3. On Riemannian
manifolds, the distance squared function in construction of envelope was
suggested by Cabré in [2], and has been used by many in [21, 42, 22, 23]. More
precisely, for each $y\in B_{R}$, there is a unique paraboloid

$P_{y}(z)=c_{y}-\frac{1}{2R^{2}}d_{\mathbb{H}^{n}}^{2}(z,y)$

that touches $u$ from below, with a contact point $x\in B_{5R}$. The envelope
$\Gamma$ of $u$ is defined by

$\Gamma(z)=\sup_{y\in B_{R}}P_{y}(z),$

and the contact set is given by

(6.1) $\mathscr{C}=\\{x\in B_{5R}:u(x)=\Gamma(x)\\}.$

The first step toward to ABP-type estimates for nonlocal operators is to find
a ring around a given contact point in which supersolution $u$ is
quadratically close to the paraboloid in a large portion of the ring. In
Euclidean spaces [3], or more generally in Riemannian manifolds with
nonnegative curvature [19], the standard dyadic rings
$B_{2^{-k}r_{0}}\setminus B_{2^{-(k+1)}r_{0}}$ are used. However, these are
not appropriate within the framework of hyperbolic spaces due to the lack of
homogeneity of the volume of balls. We thus define $r_{k}$ recursively by

(6.2) $\frac{|B_{r_{k}}|}{|B_{r_{k-1}}|}=2^{-n},\quad k=1,2,\cdots,$

and a hyperbolic dyadic ring by $R_{k}=R_{k}(x)=B_{r_{k}}(x)\setminus
B_{r_{k+1}}(x)$. Note that we have from Lemma 2.1 that
$|B_{r_{k}}|/|B_{r_{k-1}}|\leq(r_{k}/r_{k-1})^{n}$, and hence $r_{k+1}\geq
r_{k}/2$. Using the hyperbolic dyadic rings, we will prove a series of lemmas
to deduce Theorem 1.3. For notational convenience, we shall write

$F(x):=\Lambda\mathcal{H}(7R)+\frac{R^{2}}{\mathcal{I}_{0}(R)}f(x)$

throughout the section.

######  Lemma 6.1.

Let $u$ be a supersolution given in Theorem 1.3. Then, there exists a
universal constant $C_{0}>0$, independent of $\gamma$ and $R$, such that for
each $x\in\mathscr{C}$ and $M_{0}>0$, there is an integer $k\geq 0$ satisfying

(6.3) $|G_{k}|\leq\frac{C_{0}}{M_{0}}F(x)|R_{k}|,$

where $G_{k}=R_{k}\cap\\{u>P_{y}+M_{0}(r_{k}/R)^{2}\\}$.

###### Proof.

Let $x\in\mathscr{C}$ and $M_{0}>0$ be given. Since $x$ minimizes the function
$u+\frac{1}{2R^{2}}d_{{\mathbb{H}^{n}}}^{2}(\cdot,y)$, we have
$\mathcal{M}^{-}u(x)\geq I_{1}+I_{2}+I_{3}$, where

$\displaystyle I_{1}$ $\displaystyle=\lambda\int_{B_{R}(x)\cup
B_{5R}}\delta\left(u+\frac{1}{2R^{2}}d_{{\mathbb{H}^{n}}}^{2}(\cdot,y),x,z\right)\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z),$
$\displaystyle I_{2}$ $\displaystyle=-\Lambda\int_{B_{R}(x)\cup
B_{5R}}\delta^{+}\left(\frac{1}{2R^{2}}d_{{\mathbb{H}^{n}}}^{2}(\cdot,y),x,z\right)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x))\,\mathrm{d}\mu(z),$
$\displaystyle I_{3}$
$\displaystyle=-\Lambda\int_{{\mathbb{H}^{n}}\setminus(B_{R}(x)\cup
B_{5R})}\delta^{-}(u,x,z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x))\,\mathrm{d}\mu(z),$

and $\delta(v,x,z)=(v(z)+v(\exp_{x}(-\exp_{x}^{-1}z))-2v(x))/2$ is the second
order incremental quotients. By the mean value theorem for integrals and Lemma
2.2, we obtain

(6.4) $I_{2}\geq-C\Lambda\mathcal{H}(7R)\frac{\mathcal{I}_{0}(R)}{R^{2}}.$

Since $u(x)\leq
u(x)+\frac{1}{2R^{2}}d_{{\mathbb{H}^{n}}}^{2}(x,y)\leq\inf_{B_{2R}}(u+\frac{1}{2R^{2}}d_{{\mathbb{H}^{n}}}^{2}(x,y)\leq
11/2<6$ and $u\geq 0$ in ${\mathbb{H}^{n}}\setminus B_{5R}$, we also have

(6.5) $I_{3}\geq-C\Lambda\frac{\mathcal{I}_{\infty}(R)}{R^{2}}.$

Let us now focus on $I_{1}$. Assume that (6.3) does not hold for all $k$.
Then, since

$\delta\left(u+\frac{1}{2R^{2}}d_{{\mathbb{H}^{n}}}^{2}(\cdot,y),x,z\right)\geq
M_{0}\left(\frac{r_{k}}{R}\right)^{2}\quad\text{on}~{}G_{k}$

and $\mathcal{K}$ is decreasing (which can be easily checked by calculating
$\mathcal{K}^{\prime}$ and using (A.2)), we have

$I_{1}\geq\lambda
M_{0}\sum_{k=1}^{\infty}\int_{G_{k}}\left(\frac{r_{k}}{R}\right)^{2}\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z)\geq\lambda
C_{0}\frac{F(x)}{R^{2}}\sum_{k=1}^{\infty}r_{k}^{2}\mathcal{K}_{\gamma}(r_{k})|R_{k}|.$

Since $r_{k+1}\geq r_{k}/2$ and $|R_{k}|=2^{n}|R_{k+1}|$, we obtain

$\displaystyle I_{1}$ $\displaystyle\geq\lambda
C_{0}\frac{F(x)}{R^{2}}\sum_{k=0}^{\infty}r_{k+1}^{2}\mathcal{K}_{\gamma}(r_{k+1})|R_{k+1}|$
$\displaystyle\geq\lambda
2^{-2-n}C_{0}\frac{F(x)}{R^{2}}\sum_{k=0}^{\infty}r_{k}^{2}\mathcal{K}_{\gamma}(r_{k+1})|R_{k}|$
$\displaystyle\geq\lambda
2^{-2-n}C_{0}\frac{F(x)}{R^{2}}\sum_{k=0}^{\infty}\int_{R_{k}}d_{{\mathbb{H}^{n}}}^{2}(z,x)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x))\,\mathrm{d}\mu(z)$
$\displaystyle=\lambda 2^{-2-n}C_{0}\frac{F(x)}{R^{2}}\mathcal{I}_{0}(r_{0}).$

Furthermore, by using Corollary 5.3 we have

(6.6) $I_{1}\geq\lambda
2^{-2-n}C_{0}\frac{F(x)}{R^{2}}\rho_{0}^{2}\mathcal{I}_{0}(r_{0}/\rho_{0})=\lambda
2^{-3-n}C_{0}\frac{F(x)}{R^{2}}\rho_{0}^{2}\mathcal{I}_{0}(R).$

Combining (6.4), (6.5), and (6.6), and then using (5.7) and Lemma 5.4,

$f(x)\geq\lambda
2^{-3-n}C_{0}\frac{F(x)}{R^{2}}\rho_{0}^{2}\mathcal{I}_{0}(R)-C\left(1+\gamma_{0}^{-1}\right)\Lambda\mathcal{H}(7R)\frac{\mathcal{I}_{0}(R)}{R^{2}}.$

By taking $C_{0}$ sufficiently large, we arrive at a contradiction. ∎

The next lemma shows that the function $\Gamma-P_{y}$ is $c$-convex with an
appropriate function $c$. See [13, 28] for the definition of $c$-convex
function. The proof is exactly the same with that of [19, Lemma 3.4] except
for the Hessian bound of distance squared function. That is, we use Lemma 2.2
instead of [19, Lemma 2.1].

######  Lemma 6.2.

Let $x\in\mathscr{C}$, $z\in{\mathbb{H}^{n}}$, and $y\in B_{R}$ be a vertex
point of a paraboloid $P_{y}$. Then,

$(\Gamma-P_{y})(z)\leq(1-t)(\Gamma-P_{y})(z_{1})+t(\Gamma-
P_{y})(z_{2})+\frac{1}{2R^{2}}t(1-t)\mathcal{H}(d_{{\mathbb{H}^{n}}}(y,z)+|\xi|)|\xi|^{2}$

for all $t\in(0,1)$, where $z_{1}=\exp_{z}(t\xi)$ and
$z_{2}=\exp_{z}((1-t)(-\xi))$.

Using Lemma 6.2, we show that the envelope is captured in a small ball near a
contact point by two paraboloids that are quadratically close to each other.

######  Lemma 6.3.

Under the setting of Lemma 6.1, there is a universal constant
$\varepsilon_{0}\in(0,1)$ such that if

$|\\{z\in R_{k}:\Gamma(z)>P_{y}(z)+h\\}|\leq\varepsilon_{0}|R_{k}|,$

then

$\Gamma\leq P_{y}+h+C\mathcal{H}(7R)\left(\frac{r_{k}}{R}\right)^{2}$

in $B_{\tilde{r}_{k+1}}(x)$, where
$\tilde{r}_{k+1}=\tanh^{-1}(\frac{1}{2}\tanh r_{k+1})$.

###### Proof.

Let us fix $z\in B_{\tilde{r}_{k+1}}(x)$ and set $D=\\{z\in
R_{k}:\Gamma(z)\leq P_{y}(z)+h\\}$. For $w\in R_{k}$, let us consider a
geodesic $c:\mathbb{R}\to{\mathbb{H}^{n}}$ passing through $w$ and $z$. Then,
$c(\mathbb{R})\cap R_{k}$ consists of two connected components
$c(t_{1},t_{2})$ and $c(t_{3},t_{4})$, where $t_{1}<t_{2}<t_{3}<t_{4}$ satisfy
$t_{4}-t_{3}=t_{2}-t_{1}$. We may assume that $w=c(t)\in c(t_{1},t_{2})$. We
define a map $\varphi_{z}:R_{k}\to R_{k}$ by
$\varphi_{z}(w)=c\left(-t+t_{1}+t_{4}\right)$, which is clearly one-to-one and
onto.

Among all the geodesics passing through the point $z$, let us consider
geodesics $c_{\perp}$ that are perpendicular to the geodesic joining $x$ and
$z$. Then $\cup\,c_{\perp}$ divides $R_{k}$ into two regions: let $A_{1}$ be
the smaller one and $A_{2}$ the bigger one. We have

(6.7) $|E|\leq|\varphi_{z}(E)|\quad\text{for any Borel set}~{}E\subset A_{1}.$

Indeed, we may assume that $z=0=(1,0)\in{\mathbb{H}^{n}}$ by using a global
isometry. Then the map $\varphi:=\varphi_{z}$ can be represented by

$\varphi(w)=(\cosh(r+C_{\theta}),\sinh(r+C_{\theta})(-\theta)),\quad w=(\cosh
r,\sinh r\theta),$

where
$C_{\theta}=d_{\mathbb{H}^{n}}(\varphi(w^{\ast}),0)-d_{\mathbb{H}^{n}}(w^{\ast},0)$,
with $w^{\ast}$, the intersection point of $\partial B_{r_{k+1}}(x)$ and the
geodesic segment joining 0 and $w$. Note that $\varphi$ is a smooth map
because it is a composition of smooth maps. Clearly, $C_{\theta}\geq 0$ if and
only if $w\in A_{1}$. Thus, we obtain

$\displaystyle|E|$ $\displaystyle=\iint{\bf
1}_{E}(w)\sinh^{n-1}r\,\mathrm{d}r\,\mathrm{d}\theta$
$\displaystyle\leq\iint{\bf
1}_{\varphi(E)}(\varphi(w))\sinh^{n-1}(r+C_{\theta})\,\mathrm{d}r\,\mathrm{d}\theta$
$\displaystyle=\iint{\bf
1}_{\varphi(E)}(\cosh\tilde{r},\sinh\tilde{r}(-\theta))\sinh^{n-1}\tilde{r}\,\mathrm{d}\tilde{r}\,\mathrm{d}\theta$
$\displaystyle=\iint{\bf
1}_{\varphi(E)}(\cosh\tilde{r},\sinh\tilde{r}\tilde{\theta})\sinh^{n-1}\tilde{r}\,\mathrm{d}\tilde{r}\,\mathrm{d}\tilde{\theta}=|\varphi(E)|,$

where we have used change of variables $\tilde{r}=r+C_{\theta}$ and
$\tilde{\theta}=-\theta$. This proves (6.7).

We next claim that

(6.8) $|R_{k}|\leq C|A_{1}|~{}\text{with}~{}C>0~{}\text{a universal
constant.}$

Let us first deduce the lemma assuming that (6.8) is true. If we show that
$\varphi_{z}(A_{1}\cap D)\cap D\neq\emptyset$, then there are points $w_{i}\in
A_{i}\cap D$, $i=1,2$, such that $\varphi_{z}(w_{1})=w_{2}$. Since
$\Gamma(w_{i})\leq P_{y}(w_{i})+h$, for $i=1,2$, the desired result follows
from Lemma 6.2. Assume to the contrary that $\varphi(A_{1}\cap D)\cap D$ is
empty. By (6.8), we have

$|A_{1}\cap D^{c}|\leq|R_{k}\cap D^{c}|\leq\varepsilon_{0}|R_{k}|\leq
C\varepsilon_{0}|A_{1}|.$

By taking $\varepsilon_{0}=(2C)^{-1}$, we obtain $|A_{1}\cap D|>|A_{1}|/2$.
Since $\varphi_{z}(A_{1}\cap D)\subset A_{2}\cap D^{c}$, it follows that

$\frac{1}{2}|A_{1}|<|A_{1}\cap D|\leq|\varphi_{z}(A_{1}\cap D)|\leq|A_{2}\cap
D^{c}|\leq|R_{k}\cap D^{c}|\leq\frac{1}{2}|A_{1}|,$

which is a contradiction.

From now on, we focus on the proof of (6.8). To this end, it is convenient to
use the Poincaré ball model $\mathbb{B}^{3}_{1}$ with $\tau=b=t=1$. Let
$\tilde{A}_{1}=\phi(A_{1})$ and $\tilde{R}_{k}=\phi(R_{k})$, where $\phi$ is
the isometry given by (2.3). Since we are concerned with volumes, we may
assume $\phi(z)=|\phi(z)|e_{1}$ so that $\tilde{A}_{1}$ is rotationally
symmetric with respect to $x_{1}$-axis. Let $\rho_{k}$ be such that
$r_{k}=d_{\mathbb{H}^{n}}(0,\rho_{k}e_{1})$. We observe that

(6.9) $\\{y\in\mathbb{B}^{3}_{1}:\rho_{k+1}<|y|<\rho_{k},~{}e_{1}\cdot
y/|y|>1/2\\}\subset\tilde{A}_{1}.$

Indeed, if we define $\tilde{A}_{1}^{\prime}$ in the same way as
$\tilde{A}_{1}$ with $z^{\prime}\in\partial B_{\tilde{r}_{k+1}}(x)$ instead of
$z\in B_{\tilde{r}_{k+1}}(x)$, then
$\tilde{A}_{1}\supset\tilde{A}_{1}^{\prime}$. Moreover, any geodesic that is
perpendicular to $x_{1}$-axis and passes through $\tilde{\rho}_{k+1}e_{1}$ is
contained in the sphere

(6.10)
$\bigg{(}x_{1}-\frac{1+\tilde{\rho}_{k+1}^{2}}{2\tilde{\rho}_{k+1}}\bigg{)}^{2}+x_{2}^{2}+x_{3}^{2}=\bigg{(}\frac{1-\tilde{\rho}_{k+1}^{2}}{2\tilde{\rho}_{k+1}}\bigg{)}^{2}.$

The $x_{1}$-coordinate of the intersection of the spheres (6.10) and
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=\rho_{k+1}^{2}$ is given by

$\frac{\tilde{\rho}_{k+1}}{1+\tilde{\rho}_{k+1}^{2}}(1+\rho_{k+1}^{2})=\frac{\tanh\tilde{r}_{k+1}}{\tanh
r_{k+1}}\rho_{k+1}=\frac{1}{2}\rho_{k+1},$

where we used

$d_{{\mathbb{H}^{n}}}(0,\rho
e_{1})=\cosh^{-1}\frac{1+\rho^{2}}{1-\rho^{2}}=\tanh^{-1}\frac{2\rho}{1+\rho^{2}}$

in the first equality. Note that the radius
$\tilde{r}_{k+1}=\tanh^{-1}(\frac{1}{2}\tanh r_{k+1})$ is chosen so that the
last equality holds. Therefore, (6.9) holds.

We now compute

$|A_{1}|=|\tilde{A}_{1}|\geq\int_{0}^{2\pi}\int_{\rho_{k+1}}^{\rho_{k}}\int_{0}^{\pi/3}\left(\frac{2}{1-\rho^{2}}\right)^{3}\rho^{2}\sin\varphi\,\mathrm{d}\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta=\pi\int_{\rho_{k+1}}^{\rho_{k}}\left(\frac{2}{1-\rho^{2}}\right)^{3}\rho^{2}\,\mathrm{d}\rho.$

Since

$|R_{k}|=|\tilde{R}_{k}|=4\pi\int_{\rho_{k+1}}^{\rho_{k}}\left(\frac{2}{1-\rho^{2}}\right)^{3}\rho^{2}\,\mathrm{d}\rho,$

(6.8) is proved with $C=4$. ∎

We define $\phi:{\mathbb{H}^{n}}\to B_{R}$ by a map assigning each point
$x\in{\mathbb{H}^{n}}$ a vertex point $y$ of the paraboloid $P_{y}$, where
$P_{y}$ is a paraboloid such that $\Gamma(x)=P_{y}(x)$, which is not
necessarily unique. Then, the flatness of $\Gamma$ obtained in Lemma 6.3
allows us to control the image of $\phi$, which can be understood as the image
of gradient of $\Gamma$.

######  Lemma 6.4.

Under the setting of Lemma 6.1, let $x\in\mathscr{C}$ and let $k$ be such that
(6.3) holds, and let $\varepsilon_{0}$ be the constant in Lemma 6.3. Then,

(6.11) $\left|\left\\{z\in
R_{k}:u(z)>P_{y}(z)+CF(x)(r_{k}/R)^{2}\right\\}\right|\leq\varepsilon_{0}|R_{k}|$

and

(6.12) $\phi\left(\overline{B(x,\tilde{r}_{k+1}/2)}\right)\subset
B(y,C\mathcal{S}(7R)\mathcal{T}(r_{k+1})F(x)r_{k}),$

where $\mathcal{T}(r)=r/\tanh^{-1}(\frac{1}{2}\tanh r)$ and $C>0$ is a
universal constant depending only on $n$, $\lambda$, $\Lambda$, and
$\gamma_{0}$.

###### Proof.

By taking $M_{0}=C_{0}F(x)/\varepsilon_{0}$ in Lemma 6.1, we obtain (6.11).
Moreover, by Lemma 6.3 we have

(6.13) $P_{y}\leq\Gamma\leq P_{y}+CF(x)\left(\frac{r_{k}}{R}\right)^{2}$

in $B_{\tilde{r}_{k+1}}(x)$, with a universal constant $C>0$.

To prove (6.12), let $z\in\overline{B(x,\tilde{r}_{k+1}/2)}$ and
$y_{\ast}\in\phi(z)$. We need to find a upper bound of
$d_{{\mathbb{H}^{n}}}(y_{\ast},y)$. Let $\xi_{1}=\exp_{z}^{-1}y_{\ast}$ and
$\xi_{2}=\exp_{z}^{-1}y$. Let us consider a family of geodesics

$c(s,t)=\exp_{z}(t(\xi_{1}+s(\xi_{2}-\xi_{1}))),$

and the Jacobi field $J$ along $c$. Then, by [16, Equation (1.8b)] (or see,
e.g. [18]), we have

$|J(1)|_{g(y_{\ast})}\leq\mathcal{S}(|\xi_{1}|)|J^{\prime}(0)|_{g(z)}\leq\mathcal{S}(7R)|\xi_{2}-\xi_{1}|_{g(z)}.$

Considering the curve $s\mapsto c(s,1)$, we obtain

$d_{{\mathbb{H}^{n}}}(y_{\ast},y)\leq\int_{0}^{1}|\partial_{s}c(s,1)|_{g(y_{\ast})}\,\mathrm{d}s\leq\mathcal{S}(7R)|\exp_{z}^{-1}y_{\ast}-\exp_{z}^{-1}y|_{g(z)}.$

By the Gauss lemma, we know that
$|\exp_{z}^{-1}y_{\ast}-\exp_{z}^{-1}y|_{g(z)}=R^{2}|\nabla
P_{y_{\ast}}(z)-\nabla P_{y}(z)|_{g(z)}$. Thus, it only remains to show that

(6.14) $R^{2}|\nabla P_{y_{\ast}}(z)-\nabla P_{y}(z)|_{g(z)}\leq
C\mathcal{T}(r_{k+1})F(x)r_{k}$

for some universal constant $C>0$.

To this end, we prove that

(6.15)
$\left|\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\left(P_{y_{\ast}}-P_{y}\right)(c(t))\right|\leq
C\mathcal{T}(r_{k+1})F(x)\frac{r_{k}}{R^{2}}$

for all geodesics $c$, with unit speed, starting from $c(0)=z$. Suppose that
(6.15) does not hold for some $c$. We may assume that

$C\mathcal{T}(r_{k+1})F(x)\frac{r_{k}}{R^{2}}\leq\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\left(P_{y_{\ast}}-P_{y}\right)(c(t)),$

by considering $\tilde{c}(t)=c(-t)$ instead of $c(t)$ if necessary. Let
$\varepsilon>0$, then there is a $\delta>0$ such that if $|t|<\delta$, we have

(6.16)
$C\mathcal{T}(r_{k+1})F(x)\frac{r_{k}}{R^{2}}-\varepsilon\leq\frac{(P_{y_{\ast}}-P_{y})(c(t))-(P_{y_{\ast}}-P_{y})(c(0))}{t}\leq\frac{h(t)-h(0)}{t},$

where $h(t)=(\Gamma-P_{y})(c(t))$. Let $T>0$ be the first time such that
$c(T)\in\partial B_{3\tilde{r}_{k+1}/4}(x)$. Let $N$ be the least integer not
smaller than $T/\delta$, and let $0=t_{0}<t_{1}<\cdots<t_{N}=T$ be equally
distributed times. Then, $t_{i+1}-t_{i}=T/N\leq\delta$. By Lemma 6.2, we have

(6.17)
$\frac{h(t_{i})-h(t_{i-1})}{t_{i}-t_{i-1}}\leq\frac{h(t_{i+1})-h(t_{i})}{t_{i+1}-t_{i}}+\frac{\mathcal{H}(7R)}{2R^{2}}(t_{i+1}-t_{i-1}),\quad
i=1,2,\cdots,N-1.$

Thus, it follows from (6.16) and (6.17) that

(6.18)
$C\mathcal{T}(r_{k+1})F(x)\frac{r_{k}}{R^{2}}-\varepsilon\leq\frac{h(t_{i+1})-h(t_{i})}{T/N}+\frac{\mathcal{H}(7R)}{2R^{2}}\frac{2Ti}{N},\quad
i=1,2,\cdots,N-1.$

Summing up (6.18) for $i=1,2,\cdots,N-1$, we obtain

$N\left(C\mathcal{T}(r_{k+1})F(x)\frac{r_{k}}{R^{2}}-\varepsilon\right)\leq\frac{h(t_{N})-h(t_{0})}{T/N}+\frac{\mathcal{H}(7R)}{2R^{2}}\frac{2T}{N}\frac{N(N-1)}{2}.$

Since $c$ has a unit speed, we have $\tilde{r}_{k+1}/4<T<\tilde{r}_{k+1}$, and
hence

$\displaystyle C\mathcal{T}(r_{k+1})F(x)\frac{r_{k}}{R^{2}}-\varepsilon$
$\displaystyle\leq\frac{(\Gamma-P_{y})(c(T))-(\Gamma-
P_{y})(z)}{\tilde{r}_{k+1}/4}+\frac{\mathcal{H}(7R)}{2R^{2}}\tilde{r}_{k+1}$
$\displaystyle\leq\frac{(\Gamma-
P_{y})(c(T))}{\tilde{r}_{k+1}/4}+\frac{\mathcal{H}(7R)}{2R^{2}}\tilde{r}_{k+1}.$

Recalling that $\mathcal{T}(r_{k+1})=r_{k+1}/\tilde{r}_{k+1}$ and $r_{k+1}\geq
r_{k}/2$, and that $\varepsilon$ was arbitrary, we have

(6.19) $CF(x)\frac{r_{k}^{2}}{8R^{2}}\leq(\Gamma-
P_{y})(c(T))+\mathcal{H}(7R)\frac{r_{k}^{2}}{8R^{2}}.$

Since $c(T)\in\partial B_{3\tilde{r}_{k+1}/4}(x)\subset
B_{\tilde{r}_{k+1}}(x)$, the inequality (6.19) with sufficiently large
constant $C_{1}>0$ contradicts to (6.13). Therefore, we have proved (6.14),
which finishes the proof. ∎

We are now ready to prove a discrete ABP-type estimate, from which Theorem 1.3
follows.

######  Lemma 6.5.

Assume the same assumptions as in Theorem 1.3. There is a finite collection
$\mathcal{D}$ of dyadic cubes $\\{Q^{j}_{\alpha}\\}$, with diameters
$d_{j}\leq r_{0}$, such that the following hold:

  1. (i)

Any two different dyadic cubes in $\mathcal{D}$ do not intersect.

  2. (ii)

$\mathscr{C}\subset\bigcup\overline{Q}^{j}_{\alpha}$.

  3. (iii)

$|\phi(\overline{Q}^{j}_{\alpha})|\leq
C\cosh^{n-1}(C\mathcal{T}(r_{0})^{2}r_{0}F)(C\mathcal{T}(r_{0})^{2}F)^{(n-1)\log\cosh(C\mathcal{T}(r_{0})^{2}r_{0}F)}(\mathcal{T}(r_{0})^{2}F)^{n}|Q_{\alpha}^{j}|$.

  4. (iv)

$|B(z_{\alpha}^{j},2r_{0})\cap\\{u\leq\Gamma+C(\sup_{\overline{Q}_{\alpha}^{j}}F(x))(r_{0}/R)^{2}\\}|\geq\mu|Q^{j}_{\alpha}|$.

The constants $C>0$ and $\mu>0$ depend only on $n$, $\lambda$, $\Lambda$, and
$\gamma_{0}$.

###### Proof.

Let $c_{1}$, $c_{2}$, and $\delta_{0}$ be the constants given in Theorem 2.3,
which depend only on $n$. Let us fix the smallest integer $N$ such that
$c_{2}\delta_{0}^{N}\leq r_{0}$, then there are finitely many dyadic cubes
$Q_{\alpha}^{N}$ of generation $N$ such that
$\overline{Q}_{\alpha}^{N}\cap\mathscr{C}\neq\emptyset$ and
$\mathscr{C}\subset\cup_{\alpha}\overline{Q}_{\alpha}^{N}$. Whenever a dyadic
cube $Q_{\alpha}^{j}$ $(j\geq N)$ does not satisfy (iii) and (iv), we consider
all of its successors $Q_{\beta}^{j+1}\subset Q_{\alpha}^{j}$ instead of
$Q_{\alpha}^{j}$. Among these successors of $j+1$ generation, we only keep
those whose closures intersect $A$ and discard the rest. We prove that this
process must finish in a finite number of steps.

Assume to the contrary that the process produces an infinite sequence of
nested dyadic cubes $\\{Q^{j}_{\alpha}\\}_{j=N}^{\infty}$. Then, the
intersection of their closures is some contact point $x\in\mathscr{C}$. By
Lemma 6.4, there is a $k\geq 0$ such that (6.11) and (6.12) hold. Let $j\geq
N$ be such that $\delta_{0}\tilde{r}_{k+1}/2\leq
c_{2}\delta_{0}^{j}<\tilde{r}_{k+1}/2\leq r_{0}$, then it follows from Theorem
2.3 that

(6.20) $B(z_{\alpha}^{j},c_{1}\delta_{0}^{j})\subset
Q_{\alpha}^{j}\subset\overline{Q}_{\alpha}^{j}\subset B(x,\tilde{r}_{k+1}/2).$

Thus, it follows from (6.12) and (6.20) that

$|\phi(\overline{Q}_{\alpha}^{j})|\leq|\phi(\overline{B(x,\tilde{r}_{k+1}/2)})|\leq|B(y,C\mathcal{S}(7R)\mathcal{T}(r_{k+1})F(x)r_{k})|.$

Since $\mathcal{S}(7R)F(x)\leq F$ and $r_{k}\leq
2r_{k+1}=2\mathcal{T}(r_{k+1})\tilde{r}_{k+1}\leq
4\mathcal{T}(r_{0})c_{2}\delta_{0}^{j-1}$, we have

$|\phi(\overline{Q}_{\alpha}^{j})|\leq|B(z_{\alpha}^{j},C\mathcal{T}(r_{0})^{2}Fc_{1}\delta_{0}^{j})|.$

Therefore, by Lemma 2.1 we obtain

$|\phi(\overline{Q}_{\alpha}^{j})|\leq\mathcal{D}\left(C\mathcal{T}(r_{0})^{2}F\right)^{\log_{2}\mathcal{D}}|Q_{\alpha}^{j}|$

where
$\mathcal{D}=2^{n}\cosh^{n-1}(C\mathcal{T}(r_{0})^{2}c_{1}\delta_{0}^{j}F)$,
which shows that $Q_{\alpha}^{j}$ satisfies (iii).

If $z\in B(x,r_{k})$, then $d(z,z_{\alpha}^{j})\leq
d(z,x)+d(x,z_{\alpha}^{j})<r_{k}+c_{2}\delta_{0}^{j}\leq 2r_{0}$, which shows
that $B(x,r_{k})\subset B(z_{\alpha}^{j},2r_{0})$. Thus, by using (6.11), we
have

$\displaystyle|B(z_{\alpha}^{j},2r_{0})\cap\\{u\leq\Gamma+C(\sup_{\overline{Q}_{\alpha}^{j}}F(x))(r_{0}/R)^{2}\\}|$
$\displaystyle\geq|R_{k}\cap\\{u\leq P_{y}+CF(x)(r_{k}/R)^{2}\\}|$
$\displaystyle\geq(1-\varepsilon_{0})|R_{k}|$
$\displaystyle=(1-\varepsilon_{0})(2^{n}-1)|B_{r_{k+1}}|$
$\displaystyle\geq\mu|Q_{\alpha}^{j}|$

for some universal constant $\mu>0$. This proves that $Q_{\alpha}^{j}$ also
satisfies (iv), which yields a contradiction. Therefore, the process must stop
in a finite number of steps. ∎

##  7\. A barrier function

This section is devoted to a construction of a special barrier function, which
is a key ingredient together with the ABP-type estimates for the
Krylov–Safonov Harnack inequality. It is standard to use distance function to
construct a barrier function, but computations are significantly different
from the standard argument. We will observe how the negative curvature of
hyperbolic spaces comes into play. Let us begin with some inequalities.

######  Lemma 7.1.

Let $\alpha>0$ and $R_{0}>0$. Then,

(7.1) $(\cosh^{-1}(t\cosh R_{0}))^{-2\alpha}-R_{0}^{-2\alpha}\geq-2\alpha
R_{0}^{-2\alpha-2}\mathcal{H}(R_{0})\left(t-1\right)$

for all $t>1/\cosh R_{0}$. Moreover, for $\mathcal{H}(t)=t\coth t$,

(7.2) $\displaystyle\begin{split}&\frac{(\cosh^{-1}(t\cosh
R_{0}))^{-2\alpha-2}}{A^{2}\cosh^{2}R_{0}-1}-\frac{R_{0}^{-2\alpha-2}}{\sinh^{2}R_{0}}\\\
&\geq-\frac{1}{R_{0}^{2\alpha}}\left((2\alpha+2)+2\mathcal{H}(R_{0})\right)\mathcal{H}(R_{0})\frac{t-1}{R_{0}^{2}}\frac{1}{R_{0}^{2}\sinh^{2}R_{0}},\end{split}$

and

(7.3) $\displaystyle\begin{split}&(\cosh^{-1}(t\cosh
R_{0}))^{-2\alpha-1}\frac{t\cosh
R_{0}}{(t^{2}\cosh^{2}R_{0}-1)^{3/2}}-R_{0}^{-2\alpha-1}\frac{\cosh
R_{0}}{\sinh^{3}R_{0}}\\\ &\geq-
R_{0}^{-2\alpha}\left((2\alpha+1)\mathcal{H}(R_{0})-R_{0}^{2}+3\mathcal{H}^{2}(R_{0})\right)\mathcal{H}(R_{0})\frac{t-1}{R_{0}^{2}}\frac{1}{R_{0}^{2}\sinh^{2}R_{0}},\end{split}$

for all $t>1/\cosh R_{0}$.

###### Proof.

Since the function

$f(t):=(\cosh^{-1}(t\cosh(R_{0}))^{-2\alpha},\quad t>\frac{1}{\cosh R_{0}},$

is convex, (7.1) follows from the inequality $f(t)\geq
f(1)+f^{\prime}(1)(t-1)$. The inequalities (7.2) and (7.3) can be obtained
similarly by considering

$\displaystyle g(t)$ $\displaystyle:=\frac{(\cosh^{-1}(t\cosh
R_{0}))^{-2\alpha-2}}{t^{2}\cosh^{2}R_{0}-1}\quad\text{and}$ $\displaystyle
h(t)$ $\displaystyle:=(\cosh^{-1}(t\cosh R_{0}))^{-2\alpha-1}\frac{t\cosh
R_{0}}{(t^{2}\cosh^{2}R_{0}-1)^{3/2}},$

which are also convex functions. ∎

Using Lemma 7.1, we first construct a barrier function when $\gamma$ is
sufficiently close to 1.

######  Lemma 7.2.

Let $\delta\in(0,1)$. There are constants $\alpha>0$ and $\gamma_{0}\in(0,1)$,
depending only on $n$, $\lambda$, $\Lambda$, $\delta$, and $R$, such that the
function

$v(x)=\max\bigg{\\{}-\left(\frac{\delta}{20}\right)^{-2\alpha},-\left(\frac{d_{{\mathbb{H}^{n}}}(x,0)}{5R}\right)^{-2\alpha}\bigg{\\}}$

is a supersolution to

(7.4)
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v(x)+\Lambda\mathcal{H}(7R)\leq
0,$

for every $\gamma_{0}<\gamma<1$ and $x\in B_{5R}\setminus\overline{B}_{\delta
R/4}$.

###### Proof.

Fix $x$ and let $R_{0}:=d_{{\mathbb{H}^{n}}}(x,0)\in(\delta R/4,5R)$. We are
going to consider the coordinates centered at $x$. There is an isometry
$\varphi\in SO(1,n)$ such that $x=\varphi(0)$ and $0=\varphi(\cosh R_{0},\sinh
R_{0}e_{1})$ with $e_{1}\in\mathbb{S}^{n-1}$. Notice that 0 denotes
$(1,0,\cdots,0)\in{\mathbb{H}^{n}}$.

Let $z\in B_{R_{0}/2}(x)$, then $z=\varphi(\cosh r,\sinh r\omega)$ for some
$r\in[0,R_{0}/2)$ and $\omega\in\mathbb{S}^{n-1}$. By the hyperbolic law of
cosines, we have

$\displaystyle d_{\mathbb{H}^{n}}(z,0)$
$\displaystyle=d_{\mathbb{H}^{n}}(\varphi(\cosh r,\sinh r\omega),\varphi(\cosh
R_{0},\sinh R_{0}e_{1}))$ $\displaystyle=d_{\mathbb{H}^{n}}((\cosh r,\sinh
r\omega),(\cosh R_{0},\sinh R_{0}e_{1}))$ $\displaystyle=\cosh^{-1}(A-B)$

where $A=\cosh r\cosh R_{0}$ and $B=\sinh r\sinh R_{0}\omega_{1}$. Similarly,
we have

$d_{\mathbb{H}^{n}}(\exp_{x}(-\exp_{x}^{-1}z),0)=\cosh^{-1}(A+B).$

Thus, we obtain

$\delta(v,x,z)=-(5R)^{2\alpha}\frac{(\cosh^{-1}(A-B))^{-2\alpha}+(\cosh^{-1}(A+B))^{-2\alpha}-2R_{0}^{-2\alpha}}{2}.$

Since $(\cosh^{-1}(\cdot))^{-2\alpha}$ is convex at $A$, we obtain

$\displaystyle\delta\leq$
$\displaystyle-(5R)^{2\alpha}\left(\alpha(2\alpha+1)\frac{(\cosh^{-1}A)^{-2\alpha-2}}{(A^{2}-1)^{1/2}}+\alpha\frac{A(\cosh^{-1}A)^{-2\alpha-1}}{(A^{2}-1)^{3/2}}\right)B^{2}$
$\displaystyle-(5R)^{2\alpha}\left((\cosh^{-1}A)^{-2\alpha}-R_{0}^{-2\alpha}\right).$

Moreover, by applying Lemma 7.1 with $t=\cosh r$, we have

(7.5)
$\begin{split}\delta\leq&~{}\alpha(2\alpha+1)c_{\delta}\left((2\alpha+2+2\mathcal{H}(R_{0}))\mathcal{H}(R_{0})\frac{\cosh
r-1}{R_{0}^{2}}-1\right)\frac{\sinh^{2}r}{R_{0}^{2}}\omega_{1}^{2}\\\ &+\alpha
c_{\delta}\left(\left((2\alpha+1)\mathcal{H}(R_{0})-R_{0}^{2}+3\mathcal{H}^{2}(R_{0})\right)\frac{\cosh
r-1}{R_{0}^{2}}-1\right)\mathcal{H}(R_{0})\frac{\sinh^{2}r}{R_{0}^{2}}\omega_{1}^{2}\\\
&+2\alpha c_{\delta}\mathcal{H}(R_{0})\frac{\cosh r-1}{R_{0}^{2}},\end{split}$

where $c_{\delta}=(20/\delta)^{2\alpha}$.

Let us now compute

(7.6)
$\displaystyle\begin{split}\mathcal{M}^{+}v(x)\leq&~{}\int_{B(x,\frac{R_{0}}{2})}\left(\Lambda\delta^{+}(v,x,z)-\lambda\delta^{-}(v,x,z)\right)\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z)\\\
&~{}+\int_{{\mathbb{H}^{n}}\setminus
B(x,\frac{R_{0}}{2})}\left(\Lambda\delta^{+}(v,x,z)-\lambda\delta^{-}(v,x,z)\right)\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z)=:I_{1}+I_{2}.\end{split}$

We take $\alpha=\alpha(n,\lambda,\Lambda,R)>0$ sufficiently large so that

(7.7)
$\lambda(2\alpha+1)\fint_{\mathbb{S}^{n-1}}\omega_{1}^{2}\,\mathrm{d}\sigma-\Lambda\mathcal{H}(7R)>C_{1}\Lambda\mathcal{H}(7R),$

for some universal constant $C_{1}>0$ to be determined later. Then, we have

(7.8) $\begin{split}I_{1}\leq&~{}\Lambda\alpha
c_{\delta}\left((2\alpha+1)\mathcal{H}(R_{0})-R_{0}^{2}+3\mathcal{H}^{2}(R_{0})\right)\mathcal{H}(R_{0})\int_{B_{\frac{R_{0}}{2}}}\frac{\cosh
r-1}{R_{0}^{2}}\frac{\sinh^{2}r}{R_{0}^{2}}\omega_{1}^{2}\mathcal{K}_{\gamma}\,\mathrm{d}\mu(z)\\\
&~{}+\Lambda\alpha(2\alpha+1)c_{\delta}\left(2\alpha+2+2\mathcal{H}(R_{0})\right)\mathcal{H}(R_{0})\int_{B_{R_{0}/2}}\frac{\cosh
r-1}{R_{0}^{2}}\frac{\sinh^{2}r}{R_{0}^{2}}\omega_{1}^{2}\mathcal{K}_{\gamma}\,\mathrm{d}\mu(z)\\\
&~{}+\alpha
c_{\delta}\int_{B_{R_{0}/2}}\left(2\Lambda\mathcal{H}(R_{0})\frac{\cosh
r-1}{R_{0}^{2}}-\lambda(2\alpha+1+\mathcal{H}(R_{0}))\frac{\sinh^{2}r}{R_{0}^{2}}\omega_{1}^{2}\right)\mathcal{K}_{\gamma}\,\mathrm{d}\mu(z)\\\
=&~{}\alpha c_{\delta}\left(I_{1,1}+I_{1,2}+I_{1,3}\right).\end{split}$

We use (7.7) to estimate $I_{1,3}$ as follows:

(7.9)
$\displaystyle\begin{split}I_{1,3}&=\int_{B_{R_{0}/2}}\left(4\Lambda\mathcal{H}(R_{0})-4\lambda(2\alpha+1+\mathcal{H}(R_{0}))\cosh^{2}(r/2)\omega_{1}^{2}\right)\frac{\sinh^{2}\frac{r}{2}}{R_{0}^{2}}\mathcal{K}_{\gamma}\,\mathrm{d}\mu(z)\\\
&\leq\int_{0}^{R_{0}/2}\left(4\Lambda|\mathbb{S}^{n-1}|\mathcal{H}(7R)-4\lambda(2\alpha+1)\int_{\mathbb{S}^{n-1}}\omega_{1}^{2}\,\mathrm{d}\sigma\right)\frac{\sinh^{2}\frac{r}{2}}{R_{0}^{2}}\mathcal{K}_{\gamma}(r)\sinh^{n-1}r\,\mathrm{d}r\\\
&\leq-4C_{1}\Lambda\frac{\mathcal{H}(7R)}{R_{0}^{2}}\int_{0}^{R_{0}/2}|\mathbb{S}^{n-1}|\left(\frac{r}{2}\right)^{2}\mathcal{K}_{\gamma}(r)\sinh^{n-1}r\,\mathrm{d}r\\\
&=-C_{1}\Lambda\mathcal{H}(7R)\frac{\mathcal{I}_{0}(R_{0}/2)}{R_{0}^{2}}.\end{split}$

For $I_{1,1}$ and $I_{1,2}$, we observe that $\cosh r-1\leq Cr^{2}$ and
$\sinh^{2}r\leq Cr^{2}$ for $r\in[0,R_{0}/2]$, where $C$ is some constant
depending on $R$. Thus, by using Theorem 1.1 we obtain

(7.10) $\displaystyle\begin{split}I_{1,1}+I_{1,2}&\leq
C\Lambda\int_{0}^{R_{0}/2}\int_{\mathbb{S}^{n-1}}\frac{\cosh
r-1}{R_{0}^{2}}\frac{\sinh^{2}r}{R_{0}^{2}}\omega_{1}^{2}\mathcal{K}_{\gamma}(r)\sinh^{n-1}r\,\mathrm{d}\sigma\,\mathrm{d}r\\\
&\leq
C\Lambda\frac{C(3,\gamma)}{\Gamma(3/2+\gamma)2^{1/2+\gamma}}\frac{1}{R_{0}^{4}}\int_{0}^{R_{0}/2}r^{4-\nu}K_{-\nu-1}(r)\sinh
r\,\mathrm{d}r\\\ &\leq C\Lambda
C(3,\gamma)\frac{1}{R_{0}^{4}}S^{5}_{\nu}(R_{0}/2),\end{split}$

where $\nu=1/2+\gamma$ and $S^{5}_{\nu}$ is the function defined in (A.3).

On the other hand, using the fact that $v$ is bounded and Lemma 5.4, we obtain

(7.11) $I_{2}\leq
C\Lambda\frac{\mathcal{I}_{\infty}(R_{0}/2)}{(R_{0}/2)^{2}}\leq
C\Lambda\frac{1-\gamma}{\gamma}\mathcal{H}(7R)\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}.$

Thus, (7.6), (7.8), (7.9), (7.10), (7.11), and Corollary 5.3 yield

(7.12) $\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v(x)\leq
C\alpha\Lambda\left(-C_{1}+C(3,\gamma)\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\frac{1}{R_{0}^{4}}S^{5}_{\nu}(R_{0}/2)+\frac{1-\gamma}{\gamma}\right)\mathcal{H}(7R).$

Recall from Lemma 5.5 that $\mathcal{I}_{0}(7R)\to 6$ as $\gamma\to 1$. Since
$C(3,\gamma)\to 0$ as $\gamma\to 1$, we can make the second and third terms in
(7.12) be as small as we want, by choosing $\gamma_{0}$ close to 1. Therefore,
the proof is finished by assuming that we have taken $\alpha$ sufficiently
large so that (7.4) holds. ∎

In the following lemma, we construct a barrier function for any
$\gamma\in(\gamma_{0},1)$ for given $\gamma_{0}\in(0,1)$.

######  Lemma 7.3.

Given $\gamma_{0}\in(0,1)$ and $\delta\in(0,1)$, there exist universal
constants $\alpha>0$ and $\kappa\in(0,1/4]$, depending only on $n$, $\lambda$,
$\Lambda$, $\delta$, and $\gamma_{0}$, such that the function

$v(x)=\max\bigg{\\{}-\left(\frac{\kappa\delta}{20}\right)^{-2\alpha},-\left(\frac{d_{\mathbb{H}^{n}}(x,0)}{5R}\right)^{-2\alpha}\bigg{\\}}$

is a supersolution to

(7.13)
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v(x)+\Lambda\mathcal{H}(7R)\leq
0,$

for every $\gamma_{0}<\gamma<1$ and $x\in B_{5R}\setminus\overline{B}_{\delta
R/4}$.

###### Proof.

Let $\gamma_{1}$ and $\alpha_{1}$ be the $\gamma_{0}$ and $\alpha$ in Lemma
7.2, respectively. When $\gamma\in[\gamma_{1},1)$, the desired result holds
with $\alpha_{1}$ and $\kappa=1/4$.

Let us now assume $\gamma\in(\gamma_{0},\gamma_{1})$. For $x$ with
$R_{0}:=d_{\mathbb{H}^{n}}(x,0)\in(\delta R/4,5R)$, we know that $v\in
C^{2}(B(x,R_{0}/2))$ and that $\delta^{+}(v,x,z)$ is bounded for
$z\in{\mathbb{H}^{n}}\setminus B(x,R_{0}/2)$. Thus, we have

(7.14) $\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v(x)\leq
C\mathcal{H}(7R)-\lambda\int_{{\mathbb{H}^{n}}}\delta^{-}(v,x,z)\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z).$

If we take $\alpha=\max\\{\alpha_{1},n/2\\}$, then the function
$\delta^{-}(-(d_{\mathbb{H}^{n}}(\cdot,0)/5R)^{-2\alpha},x,z)$ is not
integrable. Therefore, the last integral in (7.14) can be made arbitrarily
large, by taking $\kappa$ small. In particular, we choose $\kappa$ so that
(7.13) holds. ∎

######  Corollary 7.4.

Let $\delta\in(0,1)$ and assume $0<\gamma_{0}\leq\gamma<1$. Then, there is a
function $v_{\delta}$ such that

$\begin{cases}v_{\delta}\geq 0&\text{in}~{}M\setminus B_{5R},\\\
v_{\delta}\leq 0&\text{in}~{}B_{2R},\\\
\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v_{\delta}+\Lambda\mathcal{H}(7R)\leq
0&\text{in}~{}B_{5R}\setminus\overline{B}_{\delta R/4},\\\
\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v_{\delta}\leq
C\Lambda\mathcal{H}(7R)&\text{in}~{}B_{5R},\\\ v\geq-C&\text{in}~{}B_{5R},\\\
\end{cases}$

for some universal constant $C>0$, depending only on $n$, $\lambda$,
$\Lambda$, $\delta$, and $\gamma_{0}$.

###### Proof.

Let $\alpha$ and $\kappa$ be the constants given in Lemma 7.3, and define a
function $v_{\delta}(x)=\psi(d_{\mathbb{H}^{n}}^{2}(x,0)/R^{2})$, where $\psi$
is a smooth and increasing function on $[0,\infty)$ such that

$\psi(t)=\left(\frac{3^{2}}{5^{2}}\right)^{-\alpha}-\left(\frac{t}{5^{2}}\right)^{-\alpha}\quad\text{if}~{}t\geq(\kappa\delta)^{2}.$

We already know from Lemma 7.3 that
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v_{\delta}+\Lambda\mathcal{H}(7R)\leq
0$ in $B_{5R}\setminus\overline{B}_{\delta R/4}$. Finally, for
$x\in\overline{B}_{\delta R/4}$, we have $|\delta(v_{\delta},x,z)|\leq
C\mathcal{H}(7R)d_{\mathbb{H}^{n}}(x,z)^{2}/R^{2}$ for $z\in B_{R}(x)$ and
$|\delta(v_{\delta},x,z)|\leq C$ for $z\in{\mathbb{H}^{n}}\setminus B_{R}(x)$
with a uniform constant $C>0$. Therefore, we conclude
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v_{\delta}\leq
C\Lambda\mathcal{H}(7R)$ in $B_{5R}$, with the help of Lemma 5.4. ∎

##  8\. $L^{\varepsilon}$-estimate

In this section, we prove the so-called $L^{\varepsilon}$-estimate, which
forms a basis for the proofs of Harnack inequality and Hölder estimates. It
connects a pointwise estimate to an estimate in measure.

######  Lemma 8.1.

Assume $0<\gamma_{0}\leq\gamma<1$, and let $\delta\in(0,1)$. If $u\in
C^{2}(B_{7R})$ is a nonnegative function on ${\mathbb{H}^{n}}$ satisfying
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}u\leq\varepsilon_{\delta}$
in $B_{7R}$ and $\inf_{B_{2R}}u\leq 1$, then

$\frac{|\\{u\leq M_{\delta}\\}\cap B_{\delta R}|}{|B_{7R}|}\geq\mu_{\delta},$

where $\varepsilon_{\delta}>0$, $\mu_{\delta}\in(0,1)$, and $M_{\delta}>1$ are
universal constants depending only on $n$, $\lambda$, $\Lambda$, $\delta$,
$R$, and $\gamma_{0}$.

###### Proof.

Let $v_{\delta}$ be the barrier function constructed in Corollary 7.4 and
define $w=u+v_{\delta}$. Then $w$ satisfies $w\geq 0$ in $M\setminus B_{5R}$,
$\inf_{B_{2R}}w\leq 1$, and
$\mathcal{M}^{-}w\leq\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\varepsilon_{\delta}+\mathcal{M}^{+}v_{\delta}$
in $B_{5R}$. By applying Theorem 1.3 to $w$ with its envelope $\Gamma_{w}$, we
have

$|B_{R}|\leq\sum_{j}cF^{n}|Q^{j}_{\alpha}|,$

where

$F=\mathcal{S}(7R)\bigg{(}\Lambda\mathcal{H}(7R)+\frac{R^{2}}{\mathcal{I}_{0}(R)}\bigg{(}\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\varepsilon_{\delta}+\max_{\overline{Q}^{j}_{\alpha}}\mathcal{M}^{+}v_{\delta}\bigg{)}\bigg{)}_{+}$

and
$c=C\cosh^{n-1}(C\mathcal{T}(r_{0})^{2}r_{0}F)(C\mathcal{T}(r_{0})^{2}F)^{(n-1)\log\cosh(C\mathcal{T}(r_{0})^{2}r_{0}F)}\mathcal{T}(r_{0})^{2n}$.
By Corollary 5.3, we obtain

$F\leq\mathcal{S}(7R)\bigg{(}\varepsilon_{\delta}+\Lambda\mathcal{H}(7R)+\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\max_{\overline{Q}^{j}_{\alpha}}\mathcal{M}^{+}v_{\delta}\bigg{)}_{+}.$

Since
$\Lambda\mathcal{H}(7R)+\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v_{\delta}\leq
0$ in $B_{5R}\setminus\overline{B}_{\delta R/4}$ and
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}v_{\delta}\leq
C\Lambda\mathcal{H}(7R)$ in $B_{5R}$, we have

$|B_{R}|\leq
C\varepsilon_{\delta}^{n}\sum_{\overline{Q}_{\alpha}^{j}\cap\overline{B}_{\delta
R/4}=\emptyset}|Q_{\alpha}^{j}|+C\sum_{\overline{Q}_{\alpha}^{j}\cap\overline{B}_{\delta
R/4}\neq\emptyset}|Q_{\alpha}^{j}|$

for some universal constant $C>0$, depending on $R$. By taking
$\varepsilon_{\delta}>0$ sufficiently small, we have

$|B_{7R}|\leq C\sum_{\overline{Q}_{\alpha}^{j}\cap\overline{B}_{\delta
R/4}\neq\emptyset}|Q_{\alpha}^{j}|.$

By using Lemma 6.5 (iv), we obtain

$|B_{7R}|\leq C\sum_{\overline{Q}_{\alpha}^{j}\cap\overline{B}_{\delta
R/4}\neq\emptyset}|B(z_{\alpha}^{j},2r_{0})\cap\\{w\leq\Gamma_{w}+C\\}|.$

Whenever $\overline{Q}_{\alpha}^{j}\cap\overline{B}_{\delta
R/4}\neq\emptyset$, the ball $B(z_{\alpha}^{j},2r_{0})$ is contained in
$B_{\delta R}$ if we have taken $\rho_{0}=\delta/4$. Indeed, for $z\in
B(z_{\alpha}^{j},2r_{0})$

$d_{\mathbb{H}^{n}}(z,0)\leq
d_{\mathbb{H}^{n}}(z,z_{\alpha}^{j})+d_{\mathbb{H}^{n}}(z_{\alpha}^{j},z_{\ast})+d_{\mathbb{H}^{n}}(z_{\ast},0)\leq
2r_{0}+r_{0}+\delta R/4<\delta R,$

where $z_{\ast}$ is a point in
$\overline{Q}_{\alpha}^{j}\cap\overline{B}_{\delta R}$. By taking a subcover
of $\\{B(z_{\alpha}^{j},2r_{0})\\}$ with finite overlapping and using
$v_{\delta}\geq-C$ in $B_{5R}$, we arrive at

$|B_{7R}|\leq C|\\{u\leq M_{\delta}\\}\cap B_{\delta R}|$

for some $M_{\delta}>1$. Taking $\mu_{\delta}=1/C$ finishes the proof. ∎

Lemma 8.1, together with the Calderón–Zygmund technique developed in [2],
provides the following $L^{\varepsilon}$-estimate. As in [2], we fix
$\delta=\frac{2c_{1}}{c_{2}}\delta_{0}$ and
$\delta_{1}=\delta_{0}(1-\delta_{0})/2\in(0,1)$. Let $k_{R}$ be the integer
satisfying

$c_{2}\delta_{0}^{k_{R}-1}<R\leq c_{2}\delta_{0}^{k_{R}-2},$

which is the generation of a dyadic cube whose size is comparable to that of
some ball of radius $R$.

######  Lemma 8.2.

Assume $0<\gamma_{0}\leq\gamma<1$. Let $\varepsilon_{\delta}$, $\mu_{\delta}$,
and $M_{\delta}$ be the constants in Lemma 8.1. Let $u\in C^{2}(B_{7R})$ be a
nonnegative function on ${\mathbb{H}^{n}}$ satisfying
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}u\leq\varepsilon_{\delta}$
in $B_{7R}$ and $\inf_{B_{\delta_{1}R}}u\leq 1$. If $Q_{1}$ is a dyadic cube
of generation $k_{R}$ such that $\inf_{x\in
Q_{1}}d_{\mathbb{H}^{n}}(x,0)\leq\delta_{1}R$, then

$|\\{u>M_{\delta}^{i}\\}\cap Q_{1}|\leq(1-c_{\delta})^{i}|Q_{1}|.$

for all $i=1,2,\cdots$. As a consequence, we have

$|\\{u>t\\}\cap Q_{1}|\leq Ct^{-\varepsilon}|Q_{1}|,\quad t>0,$

for some universal constants $C>0$ and $\varepsilon>0$.

######  Corollary 8.3 (Weak Harnack inequality).

Assume $0<\gamma_{0}\leq\gamma<1$. If $u\in C^{2}(B_{2R})$ is a nonnegative
function satisfying $\mathcal{M}^{-}u\leq C_{0}$ in $B_{2R}$, then

$\left(\fint_{B_{R}}u^{p}\,\mathrm{d}\mu(z)\right)^{1/p}\leq
C\left(\inf_{B_{R}}u+C_{0}\frac{R^{2}}{\mathcal{I}_{0}(R)}\right),$

where $p>0$ and $C>0$ are universal constants depending only on $n$,
$\lambda$, $\Lambda$, $R$, and $\gamma_{0}$.

See, e.g., [2, Theorem 8.1] for the proof of Corollary 8.3.

##  9\. Harnack inequality

The purpose of this section is to prove the Krylov–Safonov Harnack inequality
by using Lemma 8.2. A simple scaling argument will provide Theorem 1.4.

######  Theorem 9.1.

Assume $0<\gamma_{0}\leq\gamma<1$. If a nonnegative function $u\in
C^{2}(B_{7R})$ satisfies

$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}u\leq\varepsilon_{0}\quad\text{and}\quad\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}u\geq-\varepsilon_{0}\quad\text{in}~{}B_{7R}$

and $\inf_{B_{\delta_{1}R}}u\leq 1$, then

$\sup_{B_{\delta_{1}R/4}}u\leq C,$

where $\varepsilon_{0}>0$ and $C>0$ are universal constants depending only on
$n$, $\lambda$, $\Lambda$, $R$, and $\gamma_{0}$.

###### Proof.

Let $\varepsilon$ and $\varepsilon_{\delta}$ be the constants given in Lemma
8.2, and let $t>0$ be the minimal value such that the following holds:

$u(x)\leq
h_{t}(x):=t\bigg{(}\frac{1}{\mathcal{D}}\bigg{(}1-\frac{d_{\mathbb{H}^{n}}(x,z_{0})}{\delta_{1}R}\bigg{)}^{\log_{2}\mathcal{D}}\bigg{)}^{-1/\varepsilon}\quad\text{for
all}~{}x\in B_{\delta_{1}R},$

where for $\mathcal{D}=2^{n}\cosh^{n-1}(2\delta_{0}R)$. Since
$\sup_{B_{\delta_{1}R/4}}u\leq
t\mathcal{D}^{-\frac{1}{\varepsilon}\log_{2}(3/8)}$, we can conclude the
theorem once we show that $t\leq C$ for some universal constant $C$.

Let $x_{0}\in B_{\delta_{1}R}$ be a point such that $u(x_{0})=h_{t}(x_{0})$.
Let $d=\delta_{1}R-d_{{\mathbb{H}^{n}}}(x_{0},0)$, $r=d/2$, and
$A=\\{u>u(x_{0})/2\\}$, then we have

$u(x_{0})=h_{t}(x_{0})=t\mathcal{D}^{1/\varepsilon}\left(\frac{2r}{\delta_{1}R}\right)^{-\frac{1}{\varepsilon}\log_{2}\mathcal{D}}.$

We apply Lemma 8.2 to $u$ to obtain

$|A\cap Q_{1}|\leq C\left(\frac{u(x_{0})}{2}\right)^{-\varepsilon}|Q_{1}|\leq
Ct^{-\varepsilon}\frac{1}{\mathcal{D}}\left(\frac{r}{R}\right)^{\log_{2}\mathcal{D}}|Q_{1}|,$

where $Q_{1}$ is the unique dyadic cube of generation $k_{R}$ that contains
the point $x_{0}$.

We will show that there is a small constant $\theta>0$ such that

(9.1) $|A^{c}\cap Q_{2}|\leq\frac{1}{2}|Q_{2}|,$

where $Q_{2}\subset Q_{1}$ is the dyadic cube of generation $k_{\theta r/14}$
containing the point $x_{0}$, provided that $t$ is large. However, when $t$ is
sufficiently large, we also have

$|A\cap Q_{2}|\leq|A\cap
Q_{1}|\leq\frac{C}{t^{\varepsilon}\mathcal{D}}\left(\frac{r}{R}\right)^{\log_{2}\mathcal{D}}|B(z,c_{2}\delta_{0}^{k_{R}})|\leq\frac{C}{t^{\varepsilon}}|B(z,c_{1}\delta_{0}^{k_{r\theta/14}})|\leq\frac{C}{t^{\varepsilon}}|Q_{2}|<\frac{1}{2}|Q_{2}|,$

where $B(z,c_{1}\delta_{0}^{k_{\theta r/14}})$ is a ball contained in $Q_{2}$.
This contradicts to (9.1) and will lead us to a conclusion that $t$ is
uniformly bounded.

Let us now focus on proving (9.1). For every $x\in B(x_{0},\theta r)$, we have

$u(x)\leq h_{t}(x)\leq t\left(\frac{1}{\mathcal{D}}\left(\frac{d-\theta
r}{\delta_{1}R}\right)^{\log_{2}\mathcal{D}}\right)^{-1/\varepsilon}=\left(1-\frac{\theta}{2}\right)^{-\frac{1}{\varepsilon}\log_{2}\mathcal{D}}u(x_{0}).$

We define a function

$v(x):=\left(1-\frac{\theta}{2}\right)^{-\frac{1}{\varepsilon}\log_{2}\mathcal{D}}u(x_{0})-u(x).$

Since we will apply Lemma 8.2, we need a function which is nonnegative on the
whole space. Thus, we apply Lemma 8.2 to $w:=v^{+}$ in $B(x_{0},7(\theta
r/14))$. For $x\in B(x_{0},7(\theta r/14))$, we have

(9.2)
$\displaystyle\begin{split}\mathcal{M}^{-}w(x)&\leq\mathcal{M}^{-}v(x)+\mathcal{M}^{+}v_{-}(x)\\\
&\leq-\mathcal{M}^{+}u(x)+\Lambda\int_{{\mathbb{H}^{n}}\setminus
B(x_{0},\theta
r)}v^{-}(z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x))\,\mathrm{d}\mu(z)\\\
&\leq\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\varepsilon_{0}+\Lambda\int_{{\mathbb{H}^{n}}\setminus
B(x_{0},\theta
r)}(u(z)-(1-\theta/2)^{-\frac{1}{\varepsilon}\log_{2}\mathcal{D}}u(x_{0}))^{+}\mathcal{K}_{\gamma}(d_{{\mathbb{H}^{n}}}(z,x))\,\mathrm{d}\mu(z).\end{split}$

To compute the last integral in (9.2), we introduce another auxiliary function

$g_{\beta}(x):=\beta\left(1-\frac{d_{{\mathbb{H}^{n}}}(x,0)^{2}}{R^{2}}\right)^{+},$

with the largest number $\beta>0$ satisfying $u\geq g_{\beta}$. From the
assumption $\inf_{B_{\delta_{1}R}}u\leq 1$, we have
$(1-\delta_{1}^{2})\beta\leq 1$. Let $x_{1}\in B_{R}$ be a point where
$u(x_{1})=g_{\beta}(x_{1})$. Since

$\displaystyle\int_{\mathbb{H}^{n}}\delta^{-}(u,x_{1},z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))\,\mathrm{d}\mu(z)$
$\displaystyle\leq\int_{\mathbb{H}^{n}}\delta^{-}(g_{\beta},x_{1},z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))\,\mathrm{d}\mu(z)$
$\displaystyle\leq C\mathcal{H}(7R)\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}},$

we obtain that

$\displaystyle\varepsilon_{0}$
$\displaystyle\geq\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}u(x_{1})$
$\displaystyle\geq\lambda\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\int_{\mathbb{H}^{n}}\delta^{+}(u,x_{1},z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))\,\mathrm{d}\mu(z)-C\Lambda\mathcal{H}(7R).$

Thus, we have

(9.3)
$\displaystyle\begin{split}&\int_{\mathbb{H}^{n}}(u(z)-c)^{+}\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))\,\mathrm{d}\mu(z)\\\
&\leq\int_{\mathbb{H}^{n}}\delta^{+}(u,x_{1},z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))\,\mathrm{d}\mu(z)\leq
C\mathcal{H}(7R)\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}},\end{split}$

where $c:=1/(1-\delta_{1}^{2})$.

If $u(x_{0})\leq c$, then we find an upper bound
$t=u(x_{0})(\delta_{1}R/d)^{-\frac{1}{\varepsilon}\log_{2}\mathcal{D}}\leq
c\delta_{1}^{-\frac{1}{\varepsilon}\log_{2}\mathcal{D}}$, which finishes the
proof. Otherwise, it follows from (9.2) and (9.3) that

$\displaystyle\mathcal{M}^{-}w(x)$
$\displaystyle\leq\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\varepsilon_{0}+\Lambda\int_{{\mathbb{H}^{n}}\setminus
B(x_{0},\theta
r)}(u(z)-c)_{+}\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x))\,\mathrm{d}\mu(z)$
$\displaystyle\leq\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\varepsilon_{0}+\Lambda
M\int_{{\mathbb{H}^{n}}\setminus B(x_{0},\theta
r)}(u(z)-c)_{+}\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))\,\mathrm{d}\mu(z)$
$\displaystyle\leq\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\varepsilon_{0}+CM\mathcal{H}(7R)\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}},$

where

$M=\sup\left\\{\frac{\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x))}{\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(z,x_{1}))}:x\in
B(x_{0},\theta r/2),x_{1}\in B_{R},z\in{\mathbb{H}^{n}}\setminus
B(x_{0},\theta r)\right\\}.$

Let $d=d_{\mathbb{H}^{n}}(z,x)$ and $d_{1}=d_{\mathbb{H}^{n}}(z,x_{1})$ for
the sake of brevity. If $d\geq d_{1}$, then by [27, Theorem 3.1] we have

$\frac{\mathcal{K}_{\gamma}(d)}{\mathcal{K}_{\gamma}(d_{1})}=\frac{\sinh
d_{1}}{\sinh
d}\left(\frac{d_{1}}{d}\right)^{1/2+\gamma}\frac{K_{3/2+\gamma}(d)}{K_{3/2+\gamma}(d_{1})}\leq\frac{e^{d_{1}}\sinh
d_{1}}{d_{1}}\frac{d}{e^{d}\sinh d}.$

Since the function $s\mapsto e^{s}\sinh s/s$ is increasing, we obtain
$\mathcal{K}_{\gamma}(d)/\mathcal{K}_{\gamma}(d_{1})\leq 1$. When $d<d_{1}$,
by [27, Theorem 3.1] again we have

$\frac{\mathcal{K}_{\gamma}(d)}{\mathcal{K}_{\gamma}(d_{1})}\leq\frac{e^{d_{1}}\sinh
d_{1}}{e^{d}\sinh
d}\left(\frac{d_{1}}{d}\right)^{2+2\gamma}\leq\frac{e^{d_{1}}\sinh
d_{1}}{e^{d}\sinh d}\left(\frac{d_{1}}{d}\right)^{4}.$

Since $d_{1}/d\to 1$ as $d\to\infty$, so is
$\mathcal{K}_{\gamma}(d)/\mathcal{K}_{\gamma}(d_{1})$. Thus, for any cases the
ratio $\mathcal{K}_{\gamma}(d)/\mathcal{K}_{\gamma}(d_{1})$ is bounded by a
universal constant which is independent of $\gamma$. By using Corollary 5.3,
we arrive at

$\frac{(\theta r/2)^{2}}{\mathcal{I}_{0}(\theta
r/2)}\mathcal{M}^{-}_{\mathcal{L}_{0}}w\leq
C\frac{\mathcal{I}_{0}(R)/R^{2}}{\mathcal{I}_{0}(\theta r/2)/(\theta
r/2)^{2}}\mathcal{H}(7R)\leq C$

in $B(x_{0},7(\theta r/14))$.

Let $Q_{2}\subset Q_{1}$ be the dyadic cube of generation $k_{\theta r/14}$
containing the point $x_{0}$. Then by Lemma 8.2, we have

$\displaystyle|\\{u<u(x_{0})/2\\}\cap Q_{2}|$
$\displaystyle=|\\{w>\left(\left(1-\theta/2\right)^{-\gamma}-1/2\right)u(x_{0})\\}\cap
Q_{2}|$
$\displaystyle\leq\frac{C|Q_{2}|}{\left(\left(1-\theta/2\right)^{-\gamma}-1/2\right)^{\varepsilon}u(x_{0})^{\varepsilon}}\left(\inf_{B(x_{0},\delta_{1}\theta
r/14)}w+C\right)^{\varepsilon}.$

We can make the quantity $(1-\theta/2)^{-\gamma}-1/2$ bounded away from 0 by
taking $\theta>0$ sufficiently small. Recalling that
$w(x_{0})=((1-\theta/2)^{-\gamma}-1)u(x_{0})$, we obtain

$|\\{u<u(x_{0})/2\\}\cap Q_{2}|\leq
C|Q_{2}|\left(((1-\theta/2)^{-\gamma}-1)^{\varepsilon}+\left(\frac{C}{u(x_{0})}\right)^{\varepsilon}\right).$

We choose a constant $\theta>0$ sufficiently small so that

$C\left((1-\theta/2)^{-\gamma}-1\right)^{\varepsilon}\leq\frac{1}{4}.$

If $t>0$ is sufficiently large so that $C(C/u(x_{0}))^{\varepsilon}<1/4$, then
we arrive at (9.1). Therefore, $t$ is uniformly bounded and the desired result
follows. ∎

##  10\. Hölder estimates

In this section, the following Hölder regularity result is proved. Theorem 1.5
follows from simple scaling and covering arguments.

######  Lemma 10.1.

Assume $0<\gamma_{0}\leq\gamma<1$. There is a universal constant
$\varepsilon_{0}$ such that if $u\in C^{2}(B_{7R})$ is a function such that
$|u|\leq\frac{1}{2}$ in $B_{7R}$ and

$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}u\geq-\varepsilon_{0}\quad\text{and}\quad\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}u\leq\varepsilon_{0}\quad\text{in}~{}B_{7R},$

then $u\in C^{\alpha}$ at $0\in{\mathbb{H}^{n}}$ with an estimate

$|u(x)-u(0)|\leq CR^{-\alpha}d_{\mathbb{H}^{n}}(x,0)^{\alpha},$

where $\alpha\in(0,1)$ and $C>0$ are universal constants depending only on
$n$, $\lambda$, $\Lambda$, $R$, and $\gamma_{0}$.

###### Proof.

Let $R_{k}:=7\cdot 4^{-k}R$ and $B_{k}:=B_{R_{k}}$. It suffices to construct
an increasing sequence $\\{m_{k}\\}_{k\geq 0}$ and a decreasing sequence
$\\{M_{k}\\}_{k\geq 0}$ such that $m_{k}\leq u\leq M_{k}$ in $B_{k}$ and
$M_{k}-m_{k}=4^{-\alpha k}$. We initially choose $m_{0}=-1/2$ and $M_{0}=1/2$
for the case $k=0$. Let us assume that we have sequences up to $m_{k}$ and
$M_{k}$ and find $m_{k+1}$ and $M_{k+1}$.

For $x\in B_{2R_{k+1}}$, let $Q_{1}$ be a dyadic cube of generation
$k_{R_{k+1}/7}$. In $Q_{1}$, either $u>(M_{k}+m_{k})/2$ or
$u\leq(M_{k}+m_{k})/2$ in at least half of the points in measure. We assume

(10.1) $|\\{u>(M_{k}+m_{k})/2\\}\cap Q_{1}|\geq\frac{1}{2}|Q_{1}|.$

A function defined by

$v(x):=\frac{u(x)-m_{k}}{(M_{k}-m_{k})/2}$

satisfies $v\geq 0$ in $B_{k}$ by the induction hypothesis. To apply Lemma
8.2, let us consider a function $w:=v^{+}$, which satisfies

(10.2) $|\\{w>1\\}\cap Q_{1}|\geq\frac{1}{2}|Q_{1}|$

by (10.1). Since $\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}v\leq
2\varepsilon_{0}/(M_{k}-m_{k})$ in $B_{7R}$, we have

$\displaystyle\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\mathcal{M}^{-}w$
$\displaystyle\leq\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}(\mathcal{M}^{-}v+\mathcal{M}^{+}v^{-})$
$\displaystyle\leq\frac{2\varepsilon_{0}}{M_{k}-m_{k}}\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}+\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\mathcal{M}^{+}v^{-}$

in $B_{3R_{k+1}}$. By Lemma 5.2, we have

$\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\frac{\mathcal{I}_{0}(7R)}{(7R)^{2}}\leq\left(\frac{R_{k+1}}{7R}\right)^{\gamma}=4^{-(k+1)\gamma}<4^{-k\gamma_{0}}.$

Thus, we obtain

$\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\mathcal{M}^{-}w\leq
2\varepsilon_{0}+\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\mathcal{M}^{+}v^{-},$

by assuming $\alpha<\gamma_{0}$.

For $\mathcal{M}^{+}v^{-}$, we use an inequality
$v(z)\geq-2((d_{\mathbb{H}^{n}}(z,0)/R_{k})^{\alpha}-1)$,
$z\in{\mathbb{H}^{n}}\setminus B_{k}$, which follows from the definition of
$v$ and the properties of sequences $M_{k}$ and $m_{k}$. Then, for any
$x_{0}\in B_{3R_{k+1}}$, we have

$\displaystyle\mathcal{M}^{+}v^{-}(x_{0})$
$\displaystyle\leq\Lambda\int_{{\mathbb{H}^{n}}\setminus
B_{k}}v^{-}(z)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(x_{0},z))\,\mathrm{d}\mu(z)$
$\displaystyle\leq 2\Lambda\int_{{\mathbb{H}^{n}}\setminus
B_{k}}\left(\left(\frac{d_{\mathbb{H}^{n}}(z,0)}{R_{k}}\right)^{\alpha}-1\right)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(x_{0},z))\,\mathrm{d}\mu(z).$

Since $d_{\mathbb{H}^{n}}(z,0)\leq 4d_{\mathbb{H}^{n}}(z,x_{0})$, we obtain

(10.3) $\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\mathcal{M}^{+}v^{-}\leq
2\Lambda\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\int_{{\mathbb{H}^{n}}\setminus
B(x_{0},R_{k+1})}\left(\left(\frac{d_{\mathbb{H}^{n}}(z,x_{0})}{R_{k+1}}\right)^{\alpha}-1\right)\mathcal{K}_{\gamma}(d_{\mathbb{H}^{n}}(x_{0},z))\,\mathrm{d}\mu(z).$

Let $I$ be the right-hand side of (10.3). By the dominated convergence
theorem, we know that $I$ converges to 0 as $\alpha\to 0$ for each $\gamma$.
Let $\alpha_{\gamma}>0$ be the constant such that $I\leq\varepsilon_{0}$
whenever $\alpha\leq\alpha_{\gamma}$. Since $I$ is continuous with respect to
$\alpha$ and $\gamma$, $\alpha_{\gamma}$ is chosen continuously. Thus, the
quantity $\min_{\gamma\in[\gamma_{0},1]}\alpha_{\gamma}$ is positive and
depends on $\gamma_{0}$ (not on $\gamma$). By choosing
$\alpha=\min_{\gamma\in[\gamma_{0},1]}\alpha_{\gamma}$, we obtain

$\frac{R_{k+1}^{2}}{\mathcal{I}_{0}(R_{k+1})}\mathcal{M}^{-}w\leq
3\varepsilon_{0}$

in $B(x,7(R_{k+1}/7))$ for $x\in B_{2R_{k+1}}$. Therefore, by Lemma 8.2 and
(10.2), we have

$\displaystyle\frac{1}{2}|Q_{1}|\leq|\\{w>1\\}\cap Q_{1}|\leq
C|Q_{1}|\left(w(x)+3\varepsilon_{0}\right)^{\varepsilon},$

or equivalently, $\theta\leq w(x)+3\varepsilon_{0}$ for some universal
constant $\theta>0$. By taking $\varepsilon_{0}<\theta/6$, we arrive at
$w\geq\theta/2$ in $B_{2R_{k+1}}$. Thus, if we set $M_{k+1}=M_{k}$ and
$m_{k+1}=M_{k}-4^{-\alpha(k+1)}$, then

$M_{k+1}\geq u\geq
m_{k}+\frac{M_{k}-m_{k}}{4}\theta=M_{k}-\left(1-\frac{\theta}{4}\right)4^{-\alpha
k}\geq m_{k+1}$

in $B_{k+1}$.

When (10.1) does not hold, a similar proof can be made by using
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{+}u\geq-\varepsilon_{0}$
instead of
$\frac{(7R)^{2}}{\mathcal{I}_{0}(7R)}\mathcal{M}^{-}u\leq\varepsilon_{0}$. ∎

##  Appendix A Special functions

The equation

$\rho^{2}\frac{\mathrm{d}^{2}y}{\mathrm{d}\rho^{2}}+\rho\frac{\mathrm{d}y}{\mathrm{d}\rho}-(\rho^{2}+\nu^{2})y=0$

is called the modified Bessel’s equation, and its solutions are given by

$I_{\nu}(\rho)=\sum_{j=0}^{\infty}\frac{1}{j!\Gamma(\nu+j+1)}\left(\frac{\rho}{2}\right)^{2j+\nu}\quad\text{and}\quad
K_{\nu}(\rho)=\frac{\pi}{2}\frac{I_{-\nu}(\rho)-I_{\nu}(\rho)}{\sin\nu\pi}.$

They are called modified Bessel functions of the first and second kind,
respectively. They satisfy the recurrence relations

(A.1) $K_{\nu+1}-K_{\nu-1}=\frac{2\nu}{R}K_{\nu},\quad
I_{\nu-1}-I_{\nu+1}=\frac{2\nu}{R}I_{\nu},$

and the following system of first-order differential equations:

(A.2) $\begin{cases}I_{\nu}^{\prime}&=I_{\nu-1}-\frac{\nu}{R}I_{\nu},\\\
I_{\nu}^{\prime}&=I_{\nu+1}+\frac{\nu}{R}I_{\nu},\end{cases}\quad\text{and}\quad\begin{cases}K_{\nu}^{\prime}&=-K_{\nu-1}-\frac{\nu}{R}K_{\nu},\\\
K_{\nu}^{\prime}&=-K_{\nu+1}+\frac{\nu}{R}K_{\nu}.\end{cases}$

Using (A.1) and (A.2), we compute the indefinite integrals

(A.3)
$\displaystyle\begin{split}S_{\nu}^{k}(\rho)&=\int\rho^{k-\nu}K_{-\nu}(\rho)\sinh\rho\,\mathrm{d}\rho,\\\
C_{\nu}^{k}(\rho)&=\int\rho^{k-\nu}K_{-\nu}(\rho)\cosh\rho\,\mathrm{d}\rho,\end{split}$

for $k=0,1,\cdots$, and $\nu\neq k+1/2$, and

(A.4) $L_{\nu}^{k}(\rho)=\int\rho^{k-\nu}K_{-\nu}(\rho)\,\mathrm{d}\rho,$

for $k=0,2,4,\cdots$, and $\nu\neq k+1/2$.

######  Lemma A.1.

For $k=0,1,\cdots$, and for $\nu\neq k+1/2$, the indefinite integrals
$S_{\nu}^{k}$ and $C_{\nu}^{k}$ are given by

(A.5)
$S_{\nu}^{k}=\sqrt{\frac{\pi}{2}}\sum_{j=0}^{k}\frac{(-1)^{j}k!/(k-j)!\rho^{k+3/2-\nu}}{(k+1-2\nu)\cdots(k+1+j-2\nu)}\left(K_{\nu-j}I_{(-1)^{j}/2}+K_{\nu-j-1}I_{(-1)^{j+1}/2}\right)$

and

(A.6)
$C_{\nu}^{k}=\sqrt{\frac{\pi}{2}}\sum_{j=0}^{k}\frac{(-1)^{j}k!/(k-j)!\rho^{k+3/2-\nu}}{(k+1-2\nu)\cdots(k+1+j-2\nu)}\left(K_{\nu-j}I_{(-1)^{j+1}/2}+K_{\nu-j-1}I_{(-1)^{j}/2}\right),$

respectively.

###### Proof.

We know from [29, 10.43.3 and 10.39.1] that (A.5) and (A.6) hold true for
$k=0$. For $k\geq 1$, we use $S_{\nu}^{0}$ to see that

$\displaystyle S_{\nu}^{k}$
$\displaystyle=\int\rho^{k}\partial_{\rho}S_{\nu}^{0}\,\mathrm{d}\rho=\rho^{k}S_{\nu}^{0}-k\int\rho^{k-1}\frac{\rho^{1-\nu}}{1-2\nu}(K_{\nu}\sinh\rho+K_{\nu-1}\cosh\rho)\,\mathrm{d}\rho$
$\displaystyle=\rho^{k}S^{0}_{\nu}-\frac{k}{1-2\nu}S_{\nu}^{k}-\frac{k}{1-2\nu}C_{\nu-1}^{k-1},$

which yields

(A.7)
$S_{\nu}^{k}=-\frac{k}{k+1-2\nu}C_{\nu-1}^{k-1}+\sqrt{\frac{\pi}{2}}\frac{\rho^{k+3/2-\nu}}{k+1-2\nu}\left(K_{\nu}I_{1/2}+K_{\nu-1}I_{-1/2}\right).$

Similarly, by using $C_{\nu}^{0}$ we obtain

(A.8)
$C_{\nu}^{k}=-\frac{k}{k+1-2\nu}S_{\nu-1}^{k-1}+\sqrt{\frac{\pi}{2}}\frac{\rho^{k+3/2-\nu}}{k+1-2\nu}\left(K_{\nu}I_{-1/2}+K_{\nu-1}I_{1/2}\right).$

The lemma follows by solving the recurrence relations (A.7) and (A.8). ∎

For $L_{\nu}^{k}$, we need the modified Struve functions. The functions

$\displaystyle\mathbf{H}_{\nu}(\rho)$
$\displaystyle=\sum_{j=0}^{\infty}\frac{(-1)^{j}}{\Gamma(j+3/2)\Gamma(\nu+j+3/2)}\left(\frac{\rho}{2}\right)^{2j+\nu+1}\quad\text{and}$
$\displaystyle\mathbf{L}_{\nu}(\rho)$
$\displaystyle=\sum_{j=0}^{\infty}\frac{1}{\Gamma(j+3/2)\Gamma(\nu+j+3/2)}\left(\frac{\rho}{2}\right)^{2j+\nu+1}$

are called the modified Struve functions and solve the modified Struve’s
equation

$\frac{\mathrm{d}^{2}y}{\mathrm{d}\rho^{2}}+\frac{1}{\rho}\frac{\mathrm{d}y}{\mathrm{d}\rho}-\left(1+\frac{\nu^{2}}{\rho^{2}}\right)y=\frac{(\rho/2)^{\nu-1}}{\sqrt{\pi}\Gamma(\nu+1/2)}.$

######  Lemma A.2.

For $k=0,1,\cdots$, and for $\nu\neq k+1/2$, the indefinite integral
$L_{\nu}^{2k}$ is given by

$\displaystyle L^{2k}_{\nu}=$
$\displaystyle-\sum_{j=0}^{k-1}\frac{(2k)!(k-j)!}{2^{j}k!(2k-2j)!}\rho^{2k-j-\nu}K_{-\nu+1+j}$
$\displaystyle+\frac{\pi^{\frac{1}{2}}\Gamma(1/2-\nu+k)}{2^{\nu+1-k}}\frac{(2k)!}{2^{k}k!}\rho\left(K_{-\nu+k}{\bf
L}_{-\nu+k-1}+K_{-\nu+k-1}{\bf L}_{-\nu+k}\right).$

###### Proof.

By using the integration by parts, we obtain the recurrence relation

(A.9)
$L_{\nu}^{2k}=-\int\rho^{2k-1}\partial_{\rho}(\rho^{1-\nu}K_{1-\nu})\,\mathrm{d}\rho=-\rho^{2k-\nu}K_{1-\nu}+(2k-1)L_{\nu-1}^{2k-2}.$

Therefore, the lemma follows from (A.9) and

(A.10)
$L^{0}_{\nu}=\int\rho^{-\nu}K_{-\nu}\,\mathrm{d}\rho=\pi^{\frac{1}{2}}2^{-\nu-1}\Gamma(1/2-\nu)\rho\left(K_{-\nu}{\bf
L}_{-\nu-1}+K_{-\nu-1}{\bf L}_{-\nu}\right).$

For (A.10), see, e.g., [29, 10.43.2]. ∎

Let us close the section by collecting some asymptotic behavior of the
modified Bessel functions and modified Struve functions.

######  Lemma A.3.

The asymptotic behavior of the modified Bessel functions are given by

$\displaystyle I_{\nu}(\rho)$
$\displaystyle\sim\frac{1}{\Gamma(\nu+1)}\left(\frac{\rho}{2}\right)^{\nu},\quad\nu\neq-1,-2,\cdots,$
$\displaystyle K_{\nu}(\rho)$
$\displaystyle\sim\frac{1}{2}\Gamma(\nu)\left(\frac{\rho}{2}\right)^{-\nu},\quad\mathrm{Re}\,\nu>0,$

as $\rho\to 0$, and

$\displaystyle I_{\nu}(\rho)$
$\displaystyle\sim\frac{e^{\rho}}{\sqrt{2\pi\rho}},$ $\displaystyle
K_{\nu}(\rho)$ $\displaystyle\sim\sqrt{\frac{\pi}{2\rho}}e^{-\rho},$

as $\rho\to\infty$. The asymptotic behavior of the modified Struve function
$\mathbf{L}_{\nu}$ is given by

$\displaystyle{\bf L}_{\nu}(\rho)$
$\displaystyle\sim\frac{(\rho/2)^{\nu+1}}{\Gamma(3/2)\Gamma(3/2+\nu)}\quad\text{as}~{}\rho\to
0,~{}\text{and}$ $\displaystyle{\bf L}_{\nu}(\rho)$
$\displaystyle\sim\frac{e^{s}}{\sqrt{2\pi s}}\quad\text{as}~{}\rho\to\infty.$

For further properties of special functions, the reader may consult the book
[29].

## Acknowledgement

The research of Jongmyeong Kim is supported by the National Research
Foundation of Korea(NRF) grant funded by the Korea
government(MSIT)(2016R1E1A1A01941893). The research of Minhyun Kim is funded
by the Deutsche Forschungsgemeinschaft (GRK 2235/2 2021 - 282638148). The
research of Ki-Ahm Lee is supported by the National Research Foundation of
Korea(NRF) grant funded by the Korea government(MSIP):
NRF-2020R1A2C1A01006256.

## References

  * [1] V. Banica, M. d. M. González, and M. Sáez.  Some constructions for the fractional Laplacian on noncompact manifolds.  Rev. Mat. Iberoam., 31(2):681–712, 2015.
  * [2] X. Cabré.  Nondivergent elliptic equations on manifolds with nonnegative curvature.  Comm. Pure Appl. Math., 50(7):623–665, 1997.
  * [3] L. Caffarelli and L. Silvestre.  Regularity theory for fully nonlinear integro-differential equations.  Comm. Pure Appl. Math., 62(5):597–638, 2009.
  * [4] L. Caffarelli and L. Silvestre.  The Evans-Krylov theorem for nonlocal fully nonlinear equations.  Ann. of Math. (2), 174(2):1163–1187, 2011.
  * [5] L. Caffarelli and L. Silvestre.  Regularity results for nonlocal equations by approximation.  Arch. Ration. Mech. Anal., 200(1):59–88, 2011.
  * [6] Z.-Q. Chen, T. Kumagai, and J. Wang.  Elliptic Harnack inequalities for symmetric non-local Dirichlet forms.  J. Math. Pures Appl. (9), 125:1–42, 2019.
  * [7] S. Y. Cheng and S. T. Yau.  Differential equations on Riemannian manifolds and their geometric applications.  Comm. Pure Appl. Math., 28(3):333–354, 1975.
  * [8] M. Christ.  A $T(b)$ theorem with remarks on analytic capacity and the Cauchy integral.  Colloq. Math., 60/61(2):601–628, 1990.
  * [9] D. Cordero-Erausquin, R. J. McCann, and M. Schmuckenschläger.  A Riemannian interpolation inequality à la Borell, Brascamp and Lieb.  Invent. Math., 146(2):219–257, 2001.
  * [10] M. Ferreira.  Harmonic analysis on the Möbius gyrogroup.  J. Fourier Anal. Appl., 21(2):281–317, 2015.
  * [11] M. Ferreira and G. Ren.  Möbius gyrogroups: a Clifford algebra approach.  J. Algebra, 328:230–253, 2011.
  * [12] T. Foguel and A. A. Ungar.  Gyrogroups and the decomposition of groups into twisted subgroups and subgroups.  Pacific J. Math., 197(1):1–11, 2001.
  * [13] W. Gangbo and R. J. McCann.  The geometry of optimal transportation.  Acta Math., 177(2):113–161, 1996.
  * [14] I. M. Gelfand, S. G. Gindikin, and M. I. Graev.  Selected topics in integral geometry, volume 220 of Translations of Mathematical Monographs.  American Mathematical Society, Providence, RI, 2003.  Translated from the 2000 Russian original by A. Shtern.
  * [15] V. Georgiev.  Semilinear hyperbolic equations, volume 7 of MSJ Memoirs.  Mathematical Society of Japan, Tokyo, second edition, 2005.  With a preface by Y. Shibata.
  * [16] N. Gigli.  Second order analysis on $(\mathscr{P}_{2}(M),W_{2})$.  Mem. Amer. Math. Soc., 216(1018):xii+154, 2012.
  * [17] S. Helgason.  Geometric analysis on symmetric spaces, volume 39 of Mathematical Surveys and Monographs.  American Mathematical Society, Providence, RI, second edition, 2008.
  * [18] J. Jost.  Riemannian geometry and geometric analysis.  Universitext. Springer, Cham, seventh edition, 2017.
  * [19] J. Kim, M. Kim, and K.-A. Lee.  Harnack inequality for nonlocal operators on manifolds with nonnegative curvature.  arXiv preprint arXiv:2101.xxxxx, 2021.
  * [20] M. Kim and K.-A. Lee.  Regularity for fully nonlinear integro-differential operators with kernels of variable orders.  Nonlinear Anal., 193:111312, 27, 2020.
  * [21] S. Kim.  Harnack inequality for nondivergent elliptic operators on Riemannian manifolds.  Pacific J. Math., 213(2):281–293, 2004.
  * [22] S. Kim, S. Kim, and K.-A. Lee.  Harnack inequality for nondivergent parabolic operators on Riemannian manifolds.  Calc. Var. Partial Differential Equations, 49(1-2):669–706, 2014\.