# Harmonic Bloch Space on the Real Hyperbolic Ball

A. Ers$\dot{\hbox{\i}}$n Üreyen  Department of Mathematics, Faculty of
Science, Eskişehir Technical University, 26470, Eskişehir, Turkey
<EMAIL_ADDRESS>
###### Abstract.

We study the Bloch and the little Bloch spaces of harmonic functions on the
real hyperbolic ball. We show that the Bergman projections from
$L^{\infty}(\mathbb{B})$ to $\mathcal{B}$, and from $C_{0}(\mathbb{B})$ to
$\mathcal{B}_{0}$ are onto. We verify that the dual space of the hyperbolic
harmonic Bergman space $\mathcal{B}^{1}_{\alpha}$ is $\mathcal{B}$ and its
predual is $\mathcal{B}_{0}$. Finally, we obtain an atomic decomposition of
Bloch functions as a series of Bergman reproducing kernels.

###### Key words and phrases:

real hyperbolic ball, hyperbolic harmonic function, Bloch space, Bergman
projection, atomic decomposition

###### 2010 Mathematics Subject Classification:

Primary 31C05; Secondary 46E22

This research is supported by Eskişehir Technical University Research Fund
under grant 23ADP054.

##  1\. Introduction

For $n\geq 2$ and $x,y\in\mathbb{R}^{n}$, let $\langle
x,y\rangle=x_{1}y_{1}+\dots+x_{n}y_{n}$ be the Euclidean inner product and
$\lvert x\rvert=\sqrt{\langle x,x\rangle}$ the corresponding norm. Let
$\mathbb{B}=\mathbb{B}_{n}=\\{x\in\mathbb{R}^{n}:\lvert x\rvert<1\\}$ be the
unit ball and $\mathbb{S}=\partial\mathbb{B}$ the unit sphere.

The hyperbolic ball is $\mathbb{B}$ equipped with the hyperbolic metric

$ds^{2}=\frac{4}{(1-\lvert x\rvert^{2})^{2}}\sum_{i=1}^{n}dx_{i}^{2}.$

For a $C^{2}$ function $f$, the hyperbolic (invariant) Laplacian $\Delta_{h}$
is defined by

$\Delta_{h}f(a)=\Delta(f\circ\varphi_{a})(0)\qquad(a\in\mathbb{B}),$

where $\Delta=\partial^{2}/\partial x_{1}^{2}+\dots+\partial^{2}/\partial
x_{n}^{2}$ is the Euclidean Laplacian and $\varphi_{a}$ is the involutory
Möbius transformation given in (7) that exchanges $a$ and $0$. Up to a factor
$1/4$, $\Delta_{h}$ is the Laplace-Beltrami operator associated with the
hyperbolic metric. A straightforward calculation shows

$\Delta_{h}f(a)=(1-\lvert a\rvert^{2})^{2}\Delta f(a)+2(n-2)(1-\lvert
a\rvert^{2})\langle a,\nabla f(a)\rangle,$

where $\nabla=\big{(}\partial/\partial x_{1},\dots,\partial/\partial
x_{n}\big{)}$ is the Euclidean gradient. We refer the reader to [16, Chapter
3] for details.

A $C^{2}$ function $f\colon\mathbb{B}\to\mathbb{C}$ is called hyperbolic
harmonic or $\mathcal{H}$-harmonic on $\mathbb{B}$ if $\Delta_{h}f(x)=0$ for
all $x\in\mathbb{B}$. We denote by $\mathcal{H}(\mathbb{B})$ the space of all
$\mathcal{H}$-harmonic functions equipped with the topology of uniform
convergence on compact subsets.

Let $\nu$ be the Lebesgue measure on $\mathbb{B}$ normalized so that
$\nu(\mathbb{B})=1$ and for $\alpha>-1$, let $d\nu_{\alpha}(x)=(1-\lvert
x\rvert^{2})^{\alpha}d\nu(x)$. For $0<p<\infty$, denote the Lebesgue classes
with respect to $d\nu_{\alpha}$ by $L^{p}_{\alpha}(\mathbb{B})$. The
$\mathcal{H}$-harmonic weighted Bergman space $\mathcal{B}^{p}_{\alpha}$ is
the subspace $L^{p}_{\alpha}(\mathbb{B})\cap\mathcal{H}(\mathbb{B})$. When
$p=2$, $\mathcal{B}^{2}_{\alpha}$ is a reproducing kernel Hilbert space and
for each $x\in\mathbb{B}$, there exists
$\mathcal{R}_{\alpha}(x,\cdot)\in\mathcal{B}^{2}_{\alpha}$ such that

(1)
$f(x)=\int_{\mathbb{B}}f(y)\overline{\mathcal{R}_{\alpha}(x,y)}\,d\nu_{\alpha}(y)\qquad(f\in\mathcal{B}^{2}_{\alpha}).$

The reproducing kernel $\mathcal{R}_{\alpha}$ is real-valued and the
conjugation above can be deleted.
$\mathcal{R}_{\alpha}(x,y)=\mathcal{R}_{\alpha}(y,x)$ and so
$\mathcal{R}_{\alpha}$ is $\mathcal{H}$-harmonic as a function of each
variable. We refer the reader to [15] and [16, Chapter 10] for details.

For $\alpha>-1$ and $\phi\in L^{1}_{\alpha}$, the Bergman projection operator
$P_{\alpha}$ is defined by

$P_{\alpha}\phi(x)=\int_{\mathbb{B}}\mathcal{R}_{\alpha}(x,y)\phi(y)\,d\nu_{\alpha}(y).$

In [19], estimates for the reproducing kernels have been obtained, and it is
shown that $P_{\gamma}:L^{p}_{\alpha}\to\mathcal{B}^{p}_{\alpha}$ is bounded
if and only $\alpha+1<p(\gamma+1)$. Further properties of the spaces
$\mathcal{B}^{p}_{\alpha}$ including their atomic decomposition have been
obtained in [20].

The purpose of this paper is to consider the $p=\infty$, i.e., the Bloch
space, case. For a $C^{1}$ function $f$, the hyperbolic gradient $\nabla^{h}$
is defined by

$\nabla^{h}f(a)=-\nabla(f\circ\varphi_{a})(0)=(1-\lvert a\rvert^{2})\nabla
f(a).$

The $\mathcal{H}$-harmonic Bloch space $\mathcal{B}$ consists of all
$f\in\mathcal{H}(\mathbb{B})$ such that

(2)
$p_{\mathcal{B}}(f)=\sup_{x\in\mathbb{B}}\lvert\nabla^{h}f(x)\rvert=\sup_{x\in\mathbb{B}}(1-\lvert
x\rvert^{2})\lvert\nabla f(x)\rvert<\infty.$

$p_{\mathcal{B}}$ is a seminorm and $\|f\|_{\mathcal{B}}=\lvert
f(0)\rvert+p_{\mathcal{B}}(f)$ is a norm on $\mathcal{B}$. The little Bloch
space $\mathcal{B}_{0}$ is the subspace consisting of functions $f$ with
$\lim_{\lvert x\rvert\to 1^{-}}\lvert(1-\lvert x\rvert^{2})\lvert\nabla
f(x)\rvert=0$.

The properties we state below for the $\mathcal{H}$-harmonic Bloch space
$\mathcal{B}$ are similar to the holomorphic or the harmonic case. However, we
would like to point out that there are differences between these and the
$\mathcal{H}$-harmonic case. For example, it is well known that polynomials
are dense in the holomorphic little Bloch space and similarly harmonic
polynomials are dense in the harmonic little Bloch space. However, this is not
true in the $\mathcal{H}$-harmonic case. In fact, when the dimension $n$ is
odd, there are not any non-constant $\mathcal{H}$-harmonic polynomials (see
Lemma 4.1). Besides, some elementary properties of harmonic (or holomorphic)
functions do not hold for $\mathcal{H}$-harmonic functions. For example, if
$f$ is harmonic, then the partial derivative $\partial f/\partial x_{i}$ and
the dilation $f_{r}(x)=f(rx)$ are also harmonic. However, neither of these are
true for $\mathcal{H}$-harmonic functions. Therefore, even if the final
results are similar, some of the proofs in the harmonic (or holomorphic) case
do not directly carry over to the $\mathcal{H}$-harmonic case.

Our first result is about projections onto $\mathcal{B}$ and
$\mathcal{B}_{0}$. Let $L^{\infty}(\mathbb{B})$ be the Lebesgue space of
essentially bounded functions, $C(\overline{\mathbb{B}})$ be the space of
functions continuous on $\overline{\mathbb{B}}$, and $C_{0}(\mathbb{B})$ be
its subspace consisting of functions vanishing on $\partial\mathbb{B}$.

######  Theorem 1.1.

For every $\alpha>-1$, $P_{\alpha}$ maps $L^{\infty}(\mathbb{B})$ boundedly
onto $\mathcal{B}$. It also maps $C(\overline{\mathbb{B}})$ and
$C_{0}(\mathbb{B})$ boundedly onto $\mathcal{B}_{0}$.

It has already been verified in [19, Theorem 1.5] that $P_{\alpha}\colon
L^{\infty}(\mathbb{B})\to\mathcal{B}$ is bounded and the main aspect of the
above theorem is the surjectivity. To achieve this we first characterize
$\mathcal{B}$ and $\mathcal{B}_{0}$ in terms of certain fractional
differential operators that are defined in Section 3. These operators are
compatible with $\mathcal{H}$-harmonic functions and the reproducing kernels,
and to understand the properties of $\mathcal{B}$, they are more adequate than
$\nabla^{h}$ or $\nabla$ used in (2).

We next consider the duality problem. For $1<p<\infty$, the dual of the
hyperbolic Bergman space $\mathcal{B}^{p}_{\alpha}$ can be identified with
$\mathcal{B}^{p^{\prime}}_{\alpha}$, where $p^{\prime}=p/(p-1)$ is the
conjugate exponent of $p$ (see [19, Corollary 1.4]). We complete the missing
$p=1$ case.

######  Theorem 1.2.

For $\alpha>-1$, the dual of $\mathcal{B}^{1}_{\alpha}$ can be identified with
$\mathcal{B}$ under the pairing

(3) $\langle f,g\rangle_{\alpha}=\lim_{r\to
1^{-}}\int_{r\mathbb{B}}f(x)g(x)\,d\nu_{\alpha}(x).$

More precisely, to each $\Lambda\in(\mathcal{B}^{1}_{\alpha})^{*}$, there
corresponds a unique $g\in\mathcal{B}$ with $\|g\|_{\mathcal{B}}$ equivalent
to $\|\Lambda\|$ such that $\Lambda(f)=\langle f,g\rangle_{\alpha}$.
Similarly, for every $\alpha>-1$ the dual of $\mathcal{B}_{0}$ can be
identified with $\mathcal{B}^{1}_{\alpha}$ under the pairing (3).

For an unbounded $g\in\mathcal{B}$ and $f\in\mathcal{B}^{1}_{\alpha}$, the
integral $\int_{\mathbb{B}}f(x)g(x)d\nu_{\alpha}(x)$ may not absolutely
converge; however the limit in (3) always exists. In the case of the
holomorphic Bloch space on the unit ball of $\mathbb{C}^{n}$, $g(z)=\log
1/(1-z_{1})$ is an unbounded Bloch function. On the other hand, it is not
obvious whether there exists an unbounded $\mathcal{H}$-harmonic Bloch
function. We give an example in Lemma 6.2.

As our final result we prove atomic decomposition of $\mathcal{H}$-harmonic
Bloch functions. Atomic decomposition of harmonic Bergman and Bloch functions
have been obtained in [4] (see also [3]). In the $\mathcal{H}$-harmonic case,
see [8] for atomic decomposition of Hardy spaces and [20] for Bergman spaces.

For $a,b\in\mathbb{B}$, the pseudo-hyperbolic metric is
$\rho(a,b)=\lvert\varphi_{a}(b)\rvert$. For $0<r<1$, let
$E_{r}(a)=\\{\,x\in\mathbb{B}:\rho(x,a)<r\,\\}$ be the pseudo-hyperbolic ball
with center $a$ and radius $r$. A sequence $\\{a_{m}\\}$ of points of
$\mathbb{B}$ is called $r$-separated if $\rho(a_{m},a_{k})\geq r$ for $m\neq
k$. An $r$-separated sequence is called an $r$-lattice if
$\bigcup_{m=1}^{\infty}E_{r}(a_{m})=\mathbb{B}$. Denote by $\ell^{\infty}$,
the space of bounded sequences with norm
$\|\\{\lambda_{m}\\}\|_{\ell^{\infty}}=\sup_{m\geq 1}\lvert\lambda_{m}\rvert$,
and by $c_{0}$ the subspace consisting of sequences that converge to $0$.

######  Theorem 1.3.

Let $\alpha>-1$. There is an $r_{0}<1/2$ depending only on $n$ and $\alpha$
such that if $\\{a_{m}\\}$ is an $r$-lattice with $r<r_{0}$, then for every
$f\in\mathcal{B}$ (resp. $\mathcal{B}_{0}$), there exits
$\\{\lambda_{m}\\}\in\ell^{\infty}$ (resp. $c_{0}$) such that

(4)
$f(x)=\sum_{m=1}^{\infty}\lambda_{m}\frac{\mathcal{R}_{\alpha}(x,a_{m})}{\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}}\qquad(x\in\mathbb{B}),$

where the series converges absolutely and uniformly on compact subsets of
$\mathbb{B}$ and the norm $\|\\{\lambda_{m}\\}\|_{\ell^{\infty}}$ is
equivalent to the norm $\|f\|_{\mathcal{B}}$.

By Lemma 7.1, the norm $\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}$
is equivalent to $(1-\lvert a_{m}\rvert^{2})^{-(\alpha+n)}$ and the theorem
remains true if one uses the representation

(5) $f(x)=\sum_{m=1}^{\infty}\lambda_{m}(1-\lvert
a_{m}\rvert^{2})^{\alpha+n}\,\mathcal{R}_{\alpha}(x,a_{m})\qquad(x\in\mathbb{B})$

instead of (4).

##  2\. Preliminaries

We denote positive constants whose exact values are inessential by the letter
$C$. The value of $C$ may be different in different occurrences. For two
positive expressions $X$ and $Y$, we write $X\lesssim Y$ to mean $X\leq CY$.
If both $X\leq CY$ and $Y\leq CX$, we write $X\sim Y$.

For $x,y\in\mathbb{B}$, we define

$[x,y]:=\sqrt{1-2\langle x,y\rangle+\lvert x\rvert^{2}\lvert y\rvert^{2}}.$

$[x,y]$ is symmetric, $[x,0]=1$, and if $y\neq 0$, then
$[x,y]=\bigl{\lvert}\lvert y\rvert x-y/\lvert y\rvert\bigr{\rvert}$. Therefore

(6) $[x,y]\geq 1-\lvert x\rvert\lvert y\rvert\qquad(x,y\in\mathbb{B}).$

Denote by $\mathcal{M}(\mathbb{B})$ the group of Möbius transformations that
preserve $\mathbb{B}$. For $a\in\mathbb{B}$, the canonical Möbius
transformation that exchanges $a$ and $0$ is given by

(7) $\varphi_{a}(x)=\frac{a\lvert x-a\rvert^{2}+(1-\lvert
a\rvert^{2})(a-x)}{[x,a]^{2}}\qquad(x\in\mathbb{B}).$

It is an involution, $\varphi_{a}^{-1}=\varphi_{a}$, and for all
$x\in\mathbb{B}$, the identity

(8) $1-\lvert\varphi_{a}(x)\rvert^{2}=\frac{(1-\lvert a\rvert^{2})(1-\lvert
x\rvert^{2})}{[x,a]^{2}}$

holds. The determinant of the Jacobian matrix of $\varphi_{a}$ satisfies ([16,
Theorem 3.3.1])

(9) $\lvert\det
J\varphi_{a}(x)\rvert=\frac{(1-\lvert\varphi_{a}(x)\rvert^{2})^{n}}{(1-\lvert
x\rvert^{2})^{n}}.$

The equality

(10) $[a,\varphi_{a}(x)]=\frac{1-\lvert a\rvert^{2}}{[x,a]}$

follows from (8) (see [13, Theorem 1.1] or [20, Lemma 2.1]).

For $a,b\in\mathbb{B}$, the pseudo-hyperbolic metric
$\rho(a,b)=\lvert\varphi_{a}(b)\rvert$ satisfies the equality

(11) $\rho(a,b)=\frac{\lvert a-b\rvert}{[a,b]}.$

The pseudo-hyperbolic ball $E_{r}(a)=\\{\,x\in\mathbb{B}:\rho(x,a)<r\,\\}$ is
also a Euclidean ball with (see [16, Theorem 2.2.2])

(12) $\text{center}=\frac{(1-r^{2})a}{1-\lvert
a\rvert^{2}r^{2}}\quad\text{and}\quad\text{radius}=\frac{(1-\lvert
a\rvert^{2})r}{1-\lvert a\rvert^{2}r^{2}}.$

For a proof of the following lemma, see [2, Lemma 2.1 and 2.2].

######  Lemma 2.1.

(i) For all $a,b\in\mathbb{B}$,

$\frac{1-\rho(a,b)}{1+\rho(a,b)}\leq\frac{1-\lvert a\rvert}{1-\lvert
b\rvert}\leq\frac{1+\rho(a,b)}{1-\rho(a,b)}.$

(ii) For all $a,b,x\in\mathbb{B}$,

$\frac{1-\rho(a,b)}{1+\rho(a,b)}\leq\frac{[x,a]}{[x,b]}\leq\frac{1+\rho(a,b)}{1-\rho(a,b)}.$

The hyperbolic metric on $\mathbb{B}$ is given by

$\beta(a,b)=\log\frac{1+\rho(a,b)}{1-\rho(a,b)}\qquad(a,b\in\mathbb{B}).$

Both metrics $\rho$ and $\beta$ are Möbius invariant.

Finally, we mention two integral estimates. Let $\sigma$ be the normalized
surface measure on $\mathbb{S}$. For a proof of the following lemma see [11,
Proposition 2.2].

######  Lemma 2.2.

Let $s>-1$ and $t\in\mathbb{R}$. For all $x\in\mathbb{B}$,

$\int_{\mathbb{S}}\frac{d\sigma(\zeta)}{\lvert
x-\zeta\rvert^{n-1+t}}\sim\int_{\mathbb{B}}\frac{(1-\lvert
y\rvert^{2})^{s}}{[x,y]^{n+s+t}}d\nu(y)\sim\begin{cases}\dfrac{1}{(1-\lvert
x\rvert^{2})^{t}},&\text{if $t>0$};\\\ 1+\log\dfrac{1}{1-\lvert
x\rvert^{2}},&\text{if $t=0$};\\\ 1,&\text{if $t<0$},\end{cases}$

where the implied constants depend only on $n,s,t$.

##  3\. Reproducing Kernels and Fractional Differential Operators

In this section we review the properties of the reproducing kernels and define
a family of differential operators $D^{t}_{s}$.

Denote by $H_{m}(\mathbb{R}^{n})$ the space of all homogeneous (Euclidean)
harmonic polynomials of degree $m$ on $\mathbb{R}^{n}$. It is finite
dimensional with $\text{dim}\,H_{m}\sim m^{n-2}$ $(m\geq 1)$. By homogeneity,
$q_{m}\in H_{m}(\mathbb{R}^{n})$ is determined by its restriction on
$\mathbb{S}$ which is called a spherical harmonic and the space of spherical
harmonics of degree $m$ is denoted by $H_{m}(\mathbb{S})$. Spherical harmonics
of different degrees are orthogonal on $L^{2}(\mathbb{S})$,

(13) $\int_{\mathbb{S}}q_{m}(\zeta)q_{k}(\zeta)\,d\sigma(\zeta)=0\qquad(m\neq
k,q_{m}\in H_{m}(\mathbb{S}),q_{k}\in H_{k}(\mathbb{S})).$

Point evaluation functionals are bounded on $H_{m}(\mathbb{S})$ and so for
every $\eta\in\mathbb{S}$, there exists $Z_{m}(\eta,\cdot)\in
H_{m}(\mathbb{S})$, called the zonal harmonic of degree $m$ with pole $\eta$,
such that for all $q_{m}\in H_{m}(\mathbb{S})$,

(14)
$q_{m}(\eta)=\int_{\mathbb{S}}q_{m}(\zeta)Z_{m}(\eta,\zeta)\,d\sigma(\zeta).$

$Z_{m}(\cdot,\cdot)$ is real valued, symmetric, and homogeneous of degree $m$
in each variable. On the diagonal, $Z_{m}(\zeta,\zeta)=\text{dim}\,H_{m}$, and
in general $\lvert Z_{m}(\eta,\zeta)\rvert\leq Z_{m}(\zeta,\zeta)$. Thus

(15) $\lvert Z_{m}(\eta,\zeta)\rvert\lesssim m^{n-2}\qquad(m\geq 1).$

For details we refer the reader to [1, Chapter 5].

For $q_{m}\in H_{m}(\mathbb{R}^{n})$, the solution of the
$\mathcal{H}$-harmonic Dirichlet problem on $\mathbb{B}$ with boundary data
$q_{m}|_{\mathbb{S}}$ is given by ([16, Theorem 6.1.1])

(16) $g(x)=S_{m}(\lvert x\rvert)q_{m}(x)\qquad(x\in\overline{\mathbb{B}}).$

That is, $g$ is $\mathcal{H}$-harmonic on $\mathbb{B}$, continuous on
$\overline{\mathbb{B}}$ and equals $q_{m}$ on $\mathbb{S}$. Here, the factor
$S_{m}(r)$ $(0\leq r\leq 1)$ is given by

(17)
$S_{m}(r)=\frac{F(m,1-\tfrac{1}{2}n;m+\tfrac{1}{2}n;r^{2})}{F(m,1-\tfrac{1}{2}n;m+\tfrac{1}{2}n;1)},$

where

(18) $F(a,b;c;z)=\sum_{k=0}^{\infty}\frac{(a)_{k}(b)_{k}}{(c)_{k}k!}\,z^{k}$

is the Gauss hypergeometric function. $S_{m}$ depends also on the dimension
$n$ but we do not write this for shortness. When the dimension $n$ is even the
hypergeometric series terminates and $S_{m}$ is a polynomial, but this is not
true in odd dimensions. $S_{m}(r)$ is a decreasing function of $r$, and is
normalized so that $S_{m}(1)=1$. $S_{0}\equiv 1$ and when $m\geq 1$, the
estimate ([15, Proposition I.6] or [18, Lemma 2.6])

(19) $1\leq S_{m}(r)\leq Cm^{n/2-1}\qquad(0\leq r\leq 1)$

holds, where $C=C(n)$ is a constant depending only on $n$.

Every $\mathcal{H}$-harmonic function on $\mathbb{B}$ can be written as a
series of terms of the form (16). More precisely, for every
$f\in\mathcal{H}(\mathbb{B})$, there exists a unique sequence of polynomials
$q_{m}\in H_{m}(\mathbb{R}^{n})$ such that (see [7], [10], [12], [16, Theorem
6.3.1])

$f(x)=\sum_{m=0}^{\infty}S_{m}(\lvert x\rvert)q_{m}(x)\qquad(x\in\mathbb{B}),$

where the series converges absolutely and uniformly on compact subsets of
$\mathbb{B}$.

The hyperbolic Poisson kernel and its series expansion are given by [16,
Theorem 6.2.2]

(20) $P_{h}(x,\zeta)=\frac{(1-\lvert x\rvert^{2})^{n-1}}{\lvert
x-\zeta\rvert^{2(n-1)}}=\sum_{m=0}^{\infty}S_{m}(\lvert
x\rvert)Z_{m}(x,\zeta)\qquad(x\in\mathbb{B},\zeta\in\mathbb{S}).$

For the Bergman reproducing kernels $\mathcal{R}_{\alpha}(x,y)$ a closed
formula is not known, however, the following series expansion holds ([15,
Corollary III.5], [17, Theorem 5.3])

(21) $\mathcal{R}_{\alpha}(x,y)=\sum_{m=0}^{\infty}c_{m}(\alpha)S_{m}(\lvert
x\rvert)S_{m}(\lvert y\rvert)Z_{m}(x,y)\qquad(\alpha>-1,x,y\in\mathbb{B}),$

where the coefficients $c_{m}(\alpha)$ are determined by

(22)
$\frac{1}{c_{m}(\alpha)}=n\int_{0}^{1}r^{2m+n-1}S_{m}^{2}(r)(1-r^{2})^{\alpha}\,dr.$

A formula for the above integral is not known either. However, the estimate

(23) $c_{m}(\alpha)\sim m^{\alpha+1}\qquad(m\to\infty)$

holds (see ([15, Theorem III.6]) from which it follows that the series in (21)
converges absolutely and uniformly on $K\times\overline{\mathbb{B}}$ for every
compact $K\subset\mathbb{B}$.

It is clear from (22) that $c_{m}(\alpha)>0$. Using these coefficients we
define a family of fractional differential operators following [9] and [5].

######  Definition 3.1.

Let $s>-1$ and $s+t>-1$. If $f\in\mathcal{H}(\mathbb{B})$ has the series
expansion $f(x)=\sum_{m=0}^{\infty}S_{m}(\lvert x\rvert)q_{m}(x)$, then define

(24) $D^{t}_{s}f(x)=\sum_{m=0}^{\infty}\frac{c_{m}(s+t)}{c_{m}(s)}S_{m}(\lvert
x\rvert)q_{m}(x).$

The operator $D^{t}_{s}$ multiplies the $m^{\text{th}}$ term of the series
expansion of $f$ with the coefficient $c_{m}(s+t)/c_{m}(s)\sim m^{t}$ by (23).
Similar types of operators are frequently used in the theory of holomorphic
and harmonic Bergman spaces and act as differential operators of order $t$
(integral if $t<0$). For $\mathcal{H}$-harmonic functions slightly different
operators with multipliers $\Gamma(m+s)/\Gamma(m+\beta+s)$ are used in [14]
and Hardy-Littlewood inequalities are obtained.

######  Lemma 3.2.

For $f\in\mathcal{H}(\mathbb{B})$, the series in (24) absolutely and uniformly
converges on compact subsets of $\mathbb{B}$ and so
$D^{t}_{s}f\in\mathcal{H}(\mathbb{B})$. In addition,
$D^{t}_{s}:\mathcal{H}(\mathbb{B})\to\mathcal{H}(\mathbb{B})$ is continuous
when $\mathcal{H}(\mathbb{B})$ is equipped with the topology of uniform
convergence on compact subsets.

This lemma can be verified in the same way as [5, Theorems 3.1 and 3.2]. An
additional term $S_{m}(r)$ appears but it can easily be handled with the
estimate (19).

The operator $D^{t}_{s}$ is invertible with

(25) $D^{-t}_{s+t}D^{t}_{s}=D^{t}_{s}D^{-t}_{s+t}=\textrm{Id}.$

The role of $s$ is minor and one reason for its inclusion is to simplify the
action of $D^{t}_{s}$ on the reproducing kernel $\mathcal{R}_{s}$,

(26) $D^{t}_{s}\mathcal{R}_{s}(x,y)=\mathcal{R}_{s+t}(x,y).$

If $f\in\mathcal{H}(\mathbb{B})$ is also integrable, then $D^{t}_{s}$ can be
written as an integral operator.

######  Lemma 3.3.

Let $s>-1$, $s+t>-1$ and $f\in L^{1}_{s}(\mathbb{B})$.

  1. (i)

$D^{t}_{s}P_{s}f(x)=D^{t}_{s}\int_{\mathbb{B}}\mathcal{R}_{s}(x,y)f(y)d\nu_{s}(y)=\int_{\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)d\nu_{s}(y).$

  2. (ii)

If $f$ is also in $\mathcal{H}(\mathbb{B})$, then
$D^{t}_{s}f(x)=\int_{\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)d\nu_{s}(y).$

###### Proof.

For fixed $x\in\mathbb{B}$, the series in (21) uniformly converges for
$y\in\mathbb{B}$. Thus

$\displaystyle\int_{\mathbb{B}}\mathcal{R}_{s}(x,y)f(y)\,d\nu_{s}(y)$
$\displaystyle=\sum_{m=0}^{\infty}c_{m}(s)S_{m}(\lvert
x\rvert)\int_{\mathbb{B}}Z_{m}(x,y)S_{m}(\lvert y\rvert)f(y)\,d\nu_{s}(y)$
(27) $\displaystyle=:\sum_{m=0}^{\infty}c_{m}(s)S_{m}(\lvert
x\rvert)q_{m}(x).$

The function $q_{m}$ is in $H_{m}(\mathbb{R}^{n})$ and the series in (3)
absolutely and uniformly converges on compact subsets of $\mathbb{B}$. This
follows from (19) and (23), and the fact that $\lvert
Z_{m}(x,y)\rvert\lesssim\lvert x\rvert^{m}m^{n-2}$ by (15). Thus, the series
in (3) is the (unique) series expansion and by (24),
$D^{t}_{s}\int_{\mathbb{B}}\mathcal{R}_{s}(x,y)f(y)d\nu_{s}(y)=\sum_{m=0}^{\infty}c_{m}(s+t)S_{m}(\lvert
x\rvert)q_{m}(x)$. This series equals
$\int_{\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)d\nu_{s}(y)$ by the same
reasoning.

If $f\in L^{1}_{s}\cap\mathcal{H}(\mathbb{B})$, then $P_{s}f=f$ because the
reproducing property in (1) holds also for $f\in\mathcal{B}^{1}_{\alpha}$
([19, Lemma 7.1]). ∎

The following upper estimates of the reproducing kernels
$\mathcal{R}_{\alpha}$ have been obtained in [19, Theorem 1.2]. Here,
$\nabla_{x}$ means that the gradient is taken with respect to $x$.

######  Lemma 3.4.

Let $\alpha>-1$. There exists a constant $C=C(n,\alpha)>0$ such that for all
$x,y\in\mathbb{B}$,

  1. (a)

$\lvert\mathcal{R}_{\alpha}(x,y)\rvert\leq\dfrac{C}{[x,y]^{\alpha+n}}$,

  2. (b)

$\lvert\nabla_{x}\mathcal{R}_{\alpha}(x,y)\rvert\leq\dfrac{C}{[x,y]^{\alpha+n+1}}$.

These estimates lead to the following projection theorem (see [19, Theorem
1.1].

######  Lemma 3.5.

Let $1\leq p<\infty$ and $\alpha,\gamma>-1$. The operator $P_{\gamma}\colon
L^{p}_{\alpha}\to\mathcal{B}^{p}_{\alpha}$ is bounded if and only if
$\alpha+1<p(\gamma+1)$. In this case $P_{\gamma}f=f$ for $f\in
L^{p}_{\alpha}\cap\mathcal{H}(\mathbb{B})$.

##  4\. Elementary Properties of the Bloch Space

We first mention a few basic basic properties of $\mathcal{B}$ and
$\mathcal{B}_{0}$. The verifications are omitted as they are straightforward
and are similar to the holomorphic or the Euclidean harmonic case. The space
$\mathcal{B}$ is a Banach space with respect to the norm
$\|\cdot\|_{\mathcal{B}}$ and $\mathcal{B}_{0}$ is a closed subspace of
$\mathcal{B}$. The seminorm $p_{\mathcal{B}}$ is Möbius invariant, i.e.,
$p_{\mathcal{B}}(f\circ\psi)=p_{\mathcal{B}}(f)$ for all
$\psi\in\mathcal{M}(\mathbb{B})$. If $f\in\mathcal{B}$, then for every
$x,y\in\mathbb{B}$,

(28) $\lvert f(x)-f(y)\rvert\leq\frac{1}{2}\,p_{\mathcal{B}}(f)\beta(x,y).$

In particular, taking $y=0$ and using $\beta(x,0)\leq 1+\log 1/(1-\lvert
x\rvert)$ shows

(29) $\lvert f(x)\rvert\leq\|f\|_{\mathcal{B}}\Big{(}1+\log\frac{1}{1-\lvert
x\rvert}\Big{)}\qquad(f\in\mathcal{B},x\in\mathbb{B}).$

There are various results in [6] and [8] that show that $\mathcal{H}$-harmonic
functions can have different behaviours depending on whether the dimension $n$
is odd or even. We show here one more difference. Let $q_{m}\in
H_{m}(\mathbb{S})$. If the dimension $n$ is even, $S_{m}(\lvert x\rvert)$ is a
polynomial and the Poisson extension $P_{h}[q_{m}](x)=S_{m}(\lvert
x\rvert)q_{m}(x)$ is an $\mathcal{H}$-harmonic polynomial. This is not true
when the dimension $n$ is odd. In fact, in this case a non-constant polynomial
can not be $\mathcal{H}$-harmonic on $\mathbb{B}$.

######  Lemma 4.1.

In odd dimensions, there are no non-constant polynomials in
$\mathcal{H}(\mathbb{B})$.

###### Proof.

Suppose that $p$ is a polynomial of degree $M\geq 1$ and
$p\in\mathcal{H}(\mathbb{B})$. Then $p=\sum_{m=0}^{M}p^{(m)}$, where $p^{(m)}$
is a homogeneous polynomial of degree $m$ and by [1, Theorem 5.7], each
$p^{(m)}$ can be written in the form

$p^{(m)}(x)=\sum_{j=0}^{[m/2]}\lvert x\rvert^{2j}q_{m-2j}^{(m)}(x),$

with $q_{m-2j}^{(m)}\in H_{m-2j}(\mathbb{R}^{n})$. Thus

(30) $p(x)=\sum_{m=0}^{M}\sum_{j=0}^{[m/2]}\lvert
x\rvert^{2j}q_{m-2j}^{(m)}(x)=\sum_{k=0}^{M}\sum_{j=0}^{[(M-k)/2]}\lvert
x\rvert^{2j}q_{k}^{(k+2j)}(x),$

where we make the change of index $k=m-2j$. On the boundary $\mathbb{S}$,

$p(\zeta)=\sum_{k=0}^{M}\sum_{j=0}^{[(M-k)/2]}q_{k}^{(k+2j)}(\zeta)=:\sum_{k=0}^{M}Q_{k}(\zeta)\qquad(\zeta\in\mathbb{S}),$

with $Q_{k}\in H_{k}(\mathbb{S})$, and since $p$ is $\mathcal{H}$-harmonic,
$p=P_{h}[p|_{\mathbb{S}}]$ and so

(31) $p(x)=\sum_{k=0}^{M}S_{k}(\lvert x\rvert)Q_{k}(x)\qquad(x\in\mathbb{B}).$

Because $p$ is non-constant, there exist $1\leq i\leq M$ and
$\eta\in\mathbb{S}$ such that $Q_{i}(\eta)\neq 0$. For $0\leq r\leq 1$, we
compute the integral

$I=\int_{\mathbb{S}}p(r\zeta)Z_{i}(\eta,\zeta)\,d\sigma(\zeta)$

in two ways, using (30) and then (31). First, by (30),

$I=\sum_{k=0}^{M}\sum_{j=0}^{[(M-k)/2]}r^{k+2j}\int_{\mathbb{S}}q_{k}^{(k+2j)}(\zeta)Z_{i}(\eta,\zeta)\,d\sigma(\zeta)=r^{i}\sum_{j=0}^{[(M-i)/2]}r^{2j}q_{i}^{(i+2j)}(\eta),$

by (13) and (14). Next, by (31),

$I=\sum_{k=0}^{M}S_{k}(r)r^{k}\int_{\mathbb{S}}Q_{k}(\zeta)Z_{i}(\eta,\zeta)\,d\sigma(\zeta)=r^{i}S_{i}(r)Q_{i}(\eta).$

Combining these we see that
$S_{i}(r)Q_{i}(\eta)=\sum_{j=0}^{[(M-i)/2]}r^{2j}q_{i}^{(i+2j)}(\eta)$ for all
$0<r\leq 1$. Because $Q_{i}(\eta)\neq 0$, this shows that $S_{i}(r)$ is a
polynomial of $r$. This is a contradiction because when the dimension $n$ is
odd, $S_{i}$ is not a polynomial for $i\geq 1$ since its hypergeometric series
do not terminate. ∎

Let $q_{m}\in H_{m}(\mathbb{R}^{n})$. When the dimension $n$ is even,
$P_{h}[q_{m}|_{\mathbb{S}}](x)=S_{m}(\lvert x\rvert)q_{m}(x)$ is a polynomial
and so is in $\mathcal{B}_{0}$. It is also in $\mathcal{B}_{0}$ when the
dimension is odd. This follows from the following two elementary facts about
hypergeometric series. First,

(32) $\frac{d}{dz}F(a,b;c;z)=\frac{ab}{c}F(a+1,b+1;c+1;z)$

and second, if $\Re\\{c-a-b\\}>0$, then $F(a,b;c;z)$ uniformly converges and
so is bounded on the closed disk $\\{z:\lvert z\rvert\leq 1\\}$.

######  Lemma 4.2.

If $q_{m}\in H_{m}(\mathbb{R}^{n})$, then
$P_{h}[q_{m}|_{\mathbb{S}}](x)=S_{m}(\lvert x\rvert)q_{m}(x)$ is in
$\mathcal{B}_{0}$.

###### Proof.

The case $n=2$ is obvious and we assume $n\geq 3$. By (32),

$\displaystyle\frac{\partial}{\partial x_{i}}S_{m}(\lvert x\rvert)q_{m}(x)={}$
$\displaystyle
2x_{i}\frac{m(1-\tfrac{1}{2}n)}{m+\tfrac{1}{2}n}\,\frac{F(m+1,2-\tfrac{1}{2}n;m+\tfrac{1}{2}n+1;\lvert
x\rvert^{2})}{F(m,1-\tfrac{1}{2}n;m+\tfrac{1}{2}n;1)}\,q_{m}(x)$
$\displaystyle+S_{m}(\lvert x\rvert)\frac{\partial}{\partial x_{i}}q_{m}(x).$

The hypergeometric function in the first term is bounded since
$\Re\\{c-a-b\\}=n-2>0$. Since the second term is also bounded, the result
follows. ∎

We next show that Bergman projections of polynomials are in $\mathcal{B}_{0}$.

######  Lemma 4.3.

$P_{\alpha}p\in\mathcal{B}_{0}$ for every polynomial $p$ and $\alpha>-1$.

###### Proof.

We can assume $p$ is homogeneous. By [1, Theorem 5.7] again, $p$ can be
written in the form $p=q_{m}+\lvert x\rvert^{2}q_{m-2}+\cdots+\lvert
x\rvert^{2k}q_{m-2k}$, where $k=[m/2]$ and $q_{j}\in H_{j}(\mathbb{R}^{n})$.
Thus, it suffices to show that $P_{\alpha}(\lvert
x\rvert^{k}q_{j})\in\mathcal{B}_{0}$ for every $k\geq 0$ and $q_{j}\in
H_{j}(\mathbb{R}^{n})$. Now, using the uniform convergence of the series in
(21), integrating in polar coordinates and then using (13) and (14), we obtain

$\displaystyle P_{\alpha}$ $\displaystyle\bigl{(}\lvert
x\rvert^{k}q_{j}\bigr{)}(x)=\int_{\mathbb{B}}\mathcal{R}_{\alpha}(x,y)\lvert
y\rvert^{k}q_{j}(y)\,d\nu_{\alpha}(y)$
$\displaystyle=\sum_{m=0}^{\infty}c_{m}(\alpha)S_{m}(\lvert
x\rvert)\int_{0}^{1}nr^{n-1}S_{m}(r)r^{k+m+j}(1-r^{2})^{\alpha}\int_{\mathbb{S}}Z_{m}(x,\zeta)q_{j}(\zeta)\,d\sigma(\zeta)dr$
$\displaystyle=c_{j}(\alpha)S_{j}(\lvert
x\rvert)q_{j}(x)\int_{0}^{1}nr^{n-1}S_{j}(r)r^{k+2j}(1-r^{2})^{\alpha}\,dr$
$\displaystyle=CS_{j}(\lvert x\rvert)q_{j}(x),$

which belongs to $\mathcal{B}_{0}$ by Lemma 4.2. ∎

######  Remark 4.4.

The above proof shows also that for every polynomial $p$ and $\alpha>-1$,
$P_{\alpha}p\in\text{span}\bigl{\\{}S_{m}(\lvert x\rvert)q_{m}(x)\,|\,q_{m}\in
H_{m}(\mathbb{R}^{n}),m=0,1,2\dotsc\bigr{\\}}$.

##  5\. Projections onto the Bloch and the Little Bloch Space

The aim of this is section is to prove Theorem 1.1. The main issue is to show
that the projection $P_{\alpha}\colon L^{\infty}\to\mathcal{B}$ is onto. To
achieve this we first characterize $\mathcal{B}$ and $\mathcal{B}_{0}$ in
terms of the differential operators $D^{t}_{s}$.

######  Lemma 5.1.

For every $\alpha>-1$, $P_{\alpha}\colon L^{\infty}(\mathbb{B})\to\mathcal{B}$
is bounded. In addition, if $f\in C(\overline{\mathbb{B}})$, then
$P_{\alpha}f\in\mathcal{B}_{0}$.

###### Proof.

The estimate in Lemma 3.4(b) together with Lemma 2.2 immediately implies that
$P_{\alpha}$ maps $L^{\infty}(\mathbb{B})$ boundedly into $\mathcal{B}$ (see
[19, Theorem 1.5]). That $P_{\alpha}$ maps $C(\overline{\mathbb{B}})$ into
$\mathcal{B}_{0}$ follows from the Stone-Weierstrass theorem and Lemma 4.3. ∎

We next obtain two estimates. One is similar to Lemma 2.2, and the other to
[20, Lemma 4.3], but include an extra term $\beta(x,y)$, the hyperbolic
distance between $x$ and $y$.

######  Lemma 5.2.

Let $s>-1$ and $t>0$.

  1. (i)

There exists a constant $C=C(n,s,t)>0$ such that for all $x\in\mathbb{B}$,

$\int_{\mathbb{B}}\frac{\beta(x,y)\,(1-\lvert
y\rvert^{2})^{s}}{[x,y]^{n+s+t}}\,d\nu(y)\leq\frac{C}{(1-\lvert
x\rvert^{2})^{t}}.$

  2. (ii)

Given $\varepsilon>0$, there exists $0<r_{\varepsilon}<1$ such that for all
$r$ with $r_{\varepsilon}<r<1$ and all $x\in\mathbb{B}$,

$\int_{\mathbb{B}\backslash E_{r}(x)}\frac{\beta(x,y)\,(1-\lvert
y\rvert^{2})^{s}}{[x,y]^{n+s+t}}\,d\nu(y)<\frac{\varepsilon}{(1-\lvert
x\rvert^{2})^{t}}.$

###### Proof.

The proof is similar to the proof of [20, Lemma 4.3], requiring only a minor
modification. For $0\leq r<1$, let

$I_{r}(x)=(1-\lvert x\rvert^{2})^{t}\int_{\mathbb{B}\backslash
E_{r}(x)}\frac{\beta(x,y)\,(1-\lvert
y\rvert^{2})^{s}}{[x,y]^{n+s+t}}\,d\nu(y),$

where for $r=0$, $\mathbb{B}\backslash E_{0}(x)=\mathbb{B}$. In the integral
make the change of variable $y=\varphi_{x}(z)$. Note that
$\varphi_{x}^{-1}(\mathbb{B}\backslash
E_{r}(x))=\mathbb{B}\backslash\mathbb{B}_{r}$. Employing (8), (9) and (10),
and the fact that $\beta(x,\varphi_{x}(z))=\beta(0,z)$ by the Möbius-
invariance of $\beta$, we obtain that

$I_{r}(x)=\int_{\mathbb{B}\backslash\mathbb{B}_{r}}\frac{\beta(0,z)\,(1-\lvert
z\rvert^{2})^{s}}{[x,z]^{n+s-t}}\,d\nu(z).$

Next, using $\beta(0,z)\leq 1+\log 1/(1-\lvert z\rvert)$ and integrating in
polar coordinates shows

$I_{r}(x)\leq\int_{r}^{1}n\tau^{n-1}\Big{(}1+\log\frac{1}{1-\tau}\Big{)}\,(1-\tau^{2})^{s}\int_{\mathbb{S}}\frac{d\sigma(\zeta)}{\lvert\tau
x-\zeta\rvert^{n+s-t}}\,d\tau.$

Estimating the inner integral with Lemma 2.2 in three cases and using the
inequality $1-\tau^{2}\lvert x\rvert^{2}\geq 1-\tau^{2}$, we see that

$\int_{\mathbb{S}}\frac{d\sigma(\zeta)}{\lvert\tau x-\zeta\rvert^{n+s-t}}\leq
Cg(\tau):=\begin{cases}\dfrac{1}{(1-\tau^{2})^{1+s-t}},&\text{if
$1+s-t>0$};\\\ 1+\log\dfrac{1}{1-\tau^{2}},&\text{if $1+s-t=0$};\\\
1,&\text{if $1+s-t<0$},\end{cases}$

where $C$ depends only on $n,s,t$. Thus

$I_{r}(x)\leq
C\int_{r}^{1}n\tau^{n-1}\Big{(}1+\log\frac{1}{1-\tau}\Big{)}\,(1-\tau^{2})^{s}g(\tau)\,d\tau.$

Because $s>-1$ and $t>0$, in all the three cases, the above integral is finite
when $r=0$. This proves both parts of the lemma. ∎

We prove one more estimate. In the lemma below we consider the integral over
$r\mathbb{B}$ for $0<r\leq 1$, not just $\mathbb{B}$. This generality will be
needed later.

######  Lemma 5.3.

Let $s>-1$ and $t>0$. There exists a constant $C=C(n,s,t)>0$ such that for all
$0<r\leq 1$ and $f\in\mathcal{B}$,

$(1-\lvert
x\rvert^{2})^{t}\,\Bigl{\lvert}\int_{r\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)\,d\nu_{s}(y)\Bigr{\rvert}\leq
C\|f\|_{\mathcal{B}}.$

###### Proof.

We write

$\displaystyle\int_{r\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)d\nu_{s}(y)$
$\displaystyle=\int_{r\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(x)d\nu_{s}(y)$
$\displaystyle+\int_{r\mathbb{B}}\mathcal{R}_{s+t}(x,y)\big{(}f(y)-f(x)\big{)}d\nu_{s}(y)=:h_{1,r}(x)+h_{2,r}(x).$

Integrating in polar coordinates gives

$h_{1,r}(x)=f(x)\int_{0}^{r}n\tau^{n-1}(1-\tau^{2})^{s}\int_{\mathbb{S}}\mathcal{R}_{s+t}(x,\tau\zeta)\,d\sigma(\zeta)\,d\tau.$

Now, by the mean-value theorem for $\mathcal{H}$-harmonic functions, the inner
integral is $\mathcal{R}_{s+t}(x,0)$ and by (21),
$\mathcal{R}_{s+t}(x,0)=c_{0}(s+t)$ for all $x\in\mathbb{B}$ because
$Z_{m}(x,0)=0$ for $m\geq 1$, $Z_{0}\equiv 1$ and $S_{0}\equiv 1$. Therefore,
using also (29), we obtain

(33) $\displaystyle\begin{split}\bigl{\lvert}h_{1,r}(x)\bigr{\rvert}&\leq
c_{0}(s+t)\lvert
f(x)\rvert\int_{0}^{1}n\tau^{n-1}(1-\tau^{2})^{s}\,d\tau=C\lvert f(x)\rvert\\\
&\leq C\|f\|_{\mathcal{B}}\Bigl{(}1+\log\frac{1}{1-\lvert
x\rvert}\Bigr{)},\end{split}$

with $C$ depending only on $n,s,t$. This shows $(1-\lvert
x\rvert^{2})^{t}\lvert h_{1,r}(x)\rvert\lesssim\|f|_{\mathcal{B}}$.

Next, by (28) and Lemma 3.4(a),

$\lvert h_{2,r}(x)\rvert\lesssim
p_{\mathcal{B}}(f)\int_{r\mathbb{B}}\frac{\beta(x,y)\,d\nu_{s}(y)}{[x,y]^{n+s+t}}\lesssim\|f\|_{\mathcal{B}}\int_{\mathbb{B}}\frac{\beta(x,y)\,d\nu_{s}(y)}{[x,y]^{n+s+t}}\lesssim\frac{\|f\|_{\mathcal{B}}}{(1-\lvert
x\rvert^{2})^{t}},$

where in the last inequality we use Lemma 5.2(i). This proves the lemma. ∎

######  Proposition 5.4.

Let $s>-1$, $t>0$ and $f\in\mathcal{H}(\mathbb{B})$.

  1. (i)

$f\in\mathcal{B}$ if and only if $(1-\lvert x\rvert^{2})^{t}D^{t}_{s}f(x)\in
L^{\infty}(\mathbb{B})$. In this case,

$\|f\|_{\mathcal{B}}\sim\|(1-\lvert
x\rvert^{2})^{t}D^{t}_{s}f(x)\|_{L^{\infty}},$

where the implied constants depend only on $n,s,t$, and are independent of
$f$. In addition,

(34) $P_{s}\bigl{[}(1-\lvert x\rvert^{2})^{t}D^{t}_{s}f(x)\bigr{]}=f.$

  2. (ii)

$f\in\mathcal{B}_{0}$ if and only if $(1-\lvert
x\rvert^{2})^{t}D^{t}_{s}f(x)\in C_{0}(\mathbb{B})$.

This proposition verifies the onto part of Theorem 1.1 and completes its
proof. It also shows that for every $s>-1$ and $t>0$, $\|(1-\lvert
x\rvert^{2})^{t}D^{t}_{s}f(x)\|_{L^{\infty}}$ is a norm on $\mathcal{B}$
equivalent to $\|f\|_{\mathcal{B}}$. In the rest of the paper we mostly employ
these norms as they are easier to work with than $\|\cdot\|_{\mathcal{B}}$.

###### Proof.

Suppose $f\in\mathcal{B}$. Then $f\in L^{1}_{s}\cap\mathcal{H}(\mathbb{B})$ by
(29) and by Lemma 3.3(ii) we have
$D^{t}_{s}f(x)=\int_{\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)d\nu_{s}(y)$. That
$\|(1-\lvert
x\rvert^{2})^{t}D^{t}_{s}f(x)\|_{L^{\infty}}\lesssim\|f\|_{\mathcal{B}}$
follows now from Lemma 5.3.

Suppose now that $(1-\lvert x\rvert^{2})^{t}D^{t}_{s}f(x)\in
L^{\infty}(\mathbb{B})$. We first show (34) which holds because $D^{t}_{s}f\in
L^{1}_{s+t}(\mathbb{B})\cap\mathcal{H}(\mathbb{B})$ and by Lemma 3.3(ii) and
(25),

$P_{s}\bigl{[}(1-\lvert
x\rvert^{2})^{t}D^{t}_{s}f(x)\bigr{]}(x)=\int_{\mathbb{B}}\mathcal{R}_{s}(x,y)D^{t}_{s}f(y)\,d\nu_{s+t}(y)=D^{-t}_{s+t}(D^{t}_{s}f)(x)=f(x).$

It now follows from Lemma 5.1 that $f\in\mathcal{B}$ and
$\|f\|_{\mathcal{B}}\leq\|P_{s}\|\,\|(1-\lvert
x\rvert^{2})^{t}D^{t}_{s}f(x)\|_{L^{\infty}}$.

(ii) Suppose $f\in\mathcal{B}_{0}$. For $\varepsilon>0$, pick
$r>r_{\varepsilon}$ with $r_{\varepsilon}$ is as given in Lemma 5.2 (ii). As
in the proof of Lemma 5.3, we write
$D^{t}_{s}f(x)=\int_{\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(y)d\nu_{s}(y)$ in the
form

$\displaystyle D^{t}_{s}f(x)={}$
$\displaystyle\int_{\mathbb{B}}\mathcal{R}_{s+t}(x,y)f(x)\,d\nu_{s}(y)+\int_{\mathbb{B}\backslash
E_{r}(x)}\mathcal{R}_{s+t}(x,y)\big{(}f(y)-f(x)\big{)}\,d\nu_{s}(y)$
$\displaystyle+\int_{E_{r}(x)}\mathcal{R}_{s+t}(x,y)\big{(}f(y)-f(x)\big{)}\,d\nu_{s}(y)=:h_{1}(x)+h_{2}(x)+h_{3}(x).$

We have $(1-\lvert x\rvert^{2})^{t}h_{1}(x)\in C_{0}(\mathbb{B})$ by (33).
Next, applying (28), Lemma 3.4(a) and then Lemma 5.2(ii) shows that for some
constant $C=C(n,s,t)$,

$\lvert h_{2}(x)\rvert\leq Cp_{\mathcal{B}}(f)\int_{\mathbb{B}\backslash
E_{r}(x)}\frac{\beta(x,y)\,(1-\lvert
y\rvert^{2})^{s}}{[x,y]^{n+s+t}}\,d\nu(y)<Cp_{\mathcal{B}}(f)\frac{\varepsilon}{(1-\lvert
x\rvert^{2})^{t}}.$

Thus $(1-\lvert x\rvert^{2})^{t}\lvert h_{2}(x)\rvert\lesssim\varepsilon$.

To estimate $h_{3}$ note that for $y\in E_{r}(x)$, by the mean-value
inequality,

$\lvert f(y)-f(x)\rvert\leq\lvert y-x\rvert\sup_{z\in E_{r}(x)}\lvert\nabla
f(z)\rvert$

and by (11), $\lvert y-x\rvert=\rho(x,y)[x,y]<r[x,y]\lesssim r(1-\lvert
x\rvert^{2})$ since by part (ii) of Lemma 2.1, $[x,y]\sim[x,x]=1-\lvert
x\rvert^{2}$. Therefore

$\lvert f(y)-f(x)\rvert\lesssim(1-\lvert x\rvert^{2})\sup_{z\in
E_{r}(x)}\lvert\nabla f(z)\rvert\lesssim\sup_{z\in E_{r}(x)}(1-\lvert
z\rvert^{2})\lvert\nabla f(z)\rvert,$

where the last inequality follows from Lemma 2.1(i). Hence, by Lemma 2.2,

$\displaystyle(1-\lvert x\rvert^{2})^{t}\lvert h_{3}(x)\rvert$
$\displaystyle\lesssim\sup_{z\in E_{r}(x)}(1-\lvert z\rvert^{2})\lvert\nabla
f(z)\rvert\,(1-\lvert
x\rvert^{2})^{t}\int_{\mathbb{B}}\frac{d\nu_{s}(y)}{[x,y]^{n+s+t}}$
$\displaystyle\lesssim\sup_{z\in E_{r}(x)}(1-\lvert z\rvert^{2})\lvert\nabla
f(z)\rvert.$

Now, by (12), for $z\in E_{r}(x)$ we have $\lvert
z\rvert\geq\dfrac{\bigl{\lvert}\lvert x\rvert-r\bigr{\rvert}}{1-r\lvert
x\rvert}$ and the right-hand side tends to $1$ as $\lvert x\rvert\to 1^{-}$.
Since $f\in\mathcal{B}_{0}$, this shows that $\lim_{\lvert x\rvert\to
1^{-}}(1-\lvert x\rvert^{2})^{t}\lvert h_{3}(x)\rvert=0$; and we conclude that
$(1-\lvert x\rvert^{2})^{t}D^{t}_{s}f(x)\in C_{0}(\mathbb{B})$.

Conversely, if $(1-\lvert x\rvert^{2})^{t}D^{t}_{s}f(x)\in C_{0}(\mathbb{B})$,
then $f\in\mathcal{B}_{0}$ by Lemma 5.1 and (34). ∎

The next corollary is the $\mathcal{H}$-harmonic counterpart of the fact that
harmonic polynomials are dense in the harmonic little Bloch space.

######  Corollary 5.5.

$\text{span}\bigl{\\{}S_{m}(\lvert x\rvert)q_{m}(x)\,|\,q_{m}\in
H_{m}(\mathbb{R}^{n}),m=0,1,2\dotsc\bigr{\\}}$ is dense in $\mathcal{B}_{0}$
and therefore $\mathcal{B}_{0}$ is separable.

###### Proof.

Pick any $\alpha>-1$ and $t>0$. If $f\in\mathcal{B}_{0}$, then
$\phi(x)=(1-\lvert x\rvert^{2})^{t}D^{t}_{\alpha}f(x)$ is in
$C_{0}(\mathbb{B})$ and $P_{\alpha}\phi=f$. Since $\phi$ can be approximated
in $C_{0}(\mathbb{B})$ by polynomials, the result follows from Remark 4.4. ∎

##  6\. Duality

We begin with writing the pairing in (3) as an absolutely convergent integral.

######  Lemma 6.1.

Let $\alpha>-1$, $f\in\mathcal{B}^{1}_{\alpha}$ and $g\in\mathcal{B}$. For
every $t>0$,

$\langle f,g\rangle_{\alpha}=\lim_{r\to
1^{-}}\int_{r\mathbb{B}}f(x)g(x)\,d\nu_{\alpha}(x)=\int_{\mathbb{B}}f(x)(1-\lvert
x\rvert^{2})^{t}D^{t}_{\alpha}g(x)\,d\nu_{\alpha}(x).$

###### Proof.

Since the reproducing property in (1) holds also in $\mathcal{B}^{1}_{\alpha}$
and $f\in\mathcal{B}^{1}_{\alpha}\subset\mathcal{B}^{1}_{\alpha+t}$,

$\lim_{r\to 1^{-}}\int_{r\mathbb{B}}f(x)g(x)\,d\nu_{\alpha}(x)=\lim_{r\to
1^{-}}\int_{r\mathbb{B}}\int_{\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)f(y)\,d\nu_{\alpha+t}(y)g(x)\,d\nu_{\alpha}(x),$

which, after changing the order of the integrals (possible since for $\lvert
x\rvert\leq r$, the functions $\mathcal{R}_{\alpha+t}(x,y)$ and $g(x)$ are
bounded), equals

$\lim_{r\to 1^{-}}\int_{\mathbb{B}}f(y)(1-\lvert
y\rvert^{2})^{t}\int_{r\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)g(x)\,d\nu_{\alpha}(x)\,d\nu_{\alpha}(y).$

The term $(1-\lvert
y\rvert^{2})^{t}\bigl{\rvert}\int_{r\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)g(x)d\nu_{\alpha}(x)\bigr{\rvert}$
is bounded by a constant independent of $r$ and $y$ by Lemma 5.3 and the fact
that $\mathcal{R}_{\alpha+t}$ is symmetric. Thus, by the dominated convergence
theorem, we can push the limit into the integral and obtain

$\lim_{r\to
1^{-}}\int_{r\mathbb{B}}f(x)g(x)\,d\nu_{\alpha}(x)=\int_{\mathbb{B}}f(y)(1-\lvert
y\rvert^{2})^{t}\int_{\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)g(x)\,d\nu_{\alpha}(x)\,d\nu_{\alpha}(y).$

This gives the desired result since the inner integral is $D^{t}_{\alpha}g(y)$
by Lemma 3.3(ii). ∎

###### Proof of Theorem 1.2.

For $g\in\mathcal{B}$, define
$\Lambda_{g}\colon\mathcal{B}^{1}_{\alpha}\to\mathbb{C}$ by
$\Lambda_{g}(f)=\langle f,g\rangle_{\alpha}$. Pick some $t>0$. By Lemma 6.1
and Proposition 5.4,

(35) $\lvert\langle
f,g\rangle_{\alpha}\rvert\leq\|f\|_{\mathcal{B}^{1}_{\alpha}}\|(1-\lvert
x\rvert^{2})^{t}D^{t}_{\alpha}g\|_{L^{\infty}}\lesssim\|f\|_{\mathcal{B}^{1}_{\alpha}}\|g\|_{\mathcal{B}},$

and so $\Lambda_{g}\in(\mathcal{B}^{1}_{\alpha})^{*}$ and
$\|\Lambda_{g}\|\lesssim\|g\|_{\mathcal{B}}$.

Conversely, let $\Lambda\in(\mathcal{B}^{1}_{\alpha})^{*}$. Pick
$\gamma>\alpha$. Then, by Lemma 3.5, $\Lambda\circ
P_{\gamma}\in(L^{1}_{\alpha})^{*}$ and by the Riesz representation theorem
there exists $\psi\in L^{\infty}(\mathbb{B})$ with
$\|\psi\|_{L^{\infty}}=\|\Lambda\circ P_{\gamma}\|$ such that for all $\phi\in
L^{1}_{\alpha}(\mathbb{B})$,

$(\Lambda\circ
P_{\gamma})\phi=\int_{\mathbb{B}}\phi(y)\psi(y)\,d\nu_{\alpha}(y).$

When $\phi=f\in\mathcal{B}^{1}_{\alpha}$, we have $P_{\gamma}f=f$ and so

$\displaystyle\Lambda(f)$
$\displaystyle=\int_{\mathbb{B}}P_{\gamma}f(y)\psi(y)\,d\nu_{\alpha}(y)=\int_{\mathbb{B}}\int_{\mathbb{B}}\mathcal{R}_{\gamma}(y,x)f(x)\,d\nu_{\gamma}(x)\psi(y)\,d\nu_{\alpha}(y)$
(36)
$\displaystyle=\int_{\mathbb{B}}f(x)\int_{\mathbb{B}}\mathcal{R}_{\gamma}(y,x)\psi(y)\,d\nu_{\alpha}(y)\,d\nu_{\gamma}(x),$

where we can change the order of the integrals because $\psi\in L^{\infty}$,
and by Lemma 3.4(a) and Lemma 2.2,
$\int_{\mathbb{B}}\lvert\mathcal{R}_{\gamma}(y,x)\rvert\,d\nu_{\alpha}(y)\lesssim(1-\lvert
x\rvert^{2})^{-(\gamma-\alpha)}$. Let

$g(x):=P_{\alpha}\psi(x)=\int_{\mathbb{B}}\mathcal{R}_{\alpha}(x,y)\psi(y)\,d\nu_{\alpha}(y).$

By Theorem 1.1, $g$ is in $\mathcal{B}$ and
$\|g\|_{\mathcal{B}}\lesssim\|\psi\|_{L^{\infty}}=\|\Lambda\circ
P_{\gamma}\|\lesssim\|\Lambda\|$. Further, by Lemma 3.3(i),

$D^{\gamma-\alpha}_{\alpha}g(x)=D^{\gamma-\alpha}_{\alpha}\int_{\mathbb{B}}\mathcal{R}_{\alpha}(x,y)\psi(y)\,d\nu_{\alpha}(y)=\int_{\mathbb{B}}\mathcal{R}_{\gamma}(x,y)\psi(y)\,d\nu_{\alpha}(y).$

Hence by (36) and the symmetry of $\mathcal{R}_{\gamma}$,

$\Lambda(f)=\int_{\mathbb{B}}f(x)(1-\lvert
x\rvert^{2})^{\gamma-\alpha}D^{\gamma-\alpha}_{\alpha}g(x)\,d\nu_{\alpha}(x),$

which shows that $\Lambda=\Lambda_{g}$.

To see the uniqueness of $g$, note that for $x_{0}\in\mathbb{B}$,
$\mathcal{R}_{\alpha}(x_{0},\cdot)$ is bounded on $\mathbb{B}$ and so belongs
to $\mathcal{B}^{1}_{\alpha}$. In addition, if $g\in\mathcal{B}$, then
$\mathcal{R}_{\alpha}(x_{0},\cdot)g$ is in $L^{1}_{\alpha}(\mathbb{B})$ by
(29). Thus

(37)
$\displaystyle\begin{split}\Lambda_{g}(\mathcal{R}_{\alpha}(x_{0},\cdot))&=\langle\mathcal{R}_{\alpha}(x_{0},\cdot),g\rangle_{\alpha}=\lim_{r\to
1^{-}}\int_{r\mathbb{B}}\mathcal{R}_{\alpha}(x_{0},x)g(x)d\nu_{\alpha}(x)\\\
&=\int_{\mathbb{B}}\mathcal{R}_{\alpha}(x_{0},x)g(x)d\nu_{\alpha}(x)=g(x_{0}),\end{split}$

by the reproducing property. Hence, if $g_{1}\neq g_{2}$, then
$\Lambda_{g_{1}}\neq\Lambda_{g_{2}}$. We conclude that to each
$\Lambda\in(\mathcal{B}^{1}_{\alpha})^{*}$, there corresponds a unique
$g\in\mathcal{B}$ with $\|g\|_{\mathcal{B}}\sim\|\Lambda\|$ and
$\Lambda=\Lambda_{g}$.

We next show that $\mathcal{B}_{0}^{*}$ can be identified with
$\mathcal{B}^{1}_{\alpha}$ for any $\alpha>-1$. For
$f\in\mathcal{B}^{1}_{\alpha}$, define
$\Lambda_{f}\colon\mathcal{B}_{0}\to\mathbb{C}$ by $\Lambda_{f}(g)=\langle
f,g\rangle_{\alpha}$. By (35), $\Lambda_{f}\in\mathcal{B}_{0}^{*}$ and
$\|\Lambda_{f}\|\lesssim\|f\|_{\mathcal{B}^{1}_{\alpha}}$. Suppose now that
$\Lambda\in\mathcal{B}_{0}^{*}$. By Theorem 1.1, $\Lambda\circ P_{\alpha}\in
C_{0}(\mathbb{B})^{*}$ and by the Riesz representation theorem, there exists a
complex Borel measure $\mu$ on $\mathbb{B}$ with
$\lvert\mu\rvert(\mathbb{B})=\|\Lambda\circ P_{\alpha}\|$ such that for all
$\phi\in C_{0}(\mathbb{B})$,

$(\Lambda\circ P_{\alpha})\phi=\int_{\mathbb{B}}\phi(y)\,d\mu(y).$

Pick a $t>0$. For $g\in\mathcal{B}_{0}$, let $\phi(x)=(1-\lvert
x\rvert^{2})^{t}D^{t}_{\alpha}g(x)$. Then by Proposition 5.4, $\phi\in
C_{0}(\mathbb{B})$ with $\|\phi\|_{L^{\infty}}\sim\|g\|_{\mathcal{B}}$ and
$P_{\alpha}\phi=g$. Thus

$\Lambda(g)=(\Lambda\circ
P_{\alpha})\phi=\int_{\mathbb{B}}\phi(y)\,d\mu(y)=\int_{\mathbb{B}}(1-\lvert
y\rvert^{2})^{t}D^{t}_{\alpha}g(y)\,d\mu(y).$

Further, since $D^{t}_{\alpha}g\in L^{1}_{\alpha+t}$, by the reproducing
property, $D^{t}_{\alpha}g=P_{\alpha+t}(D^{t}_{\alpha}g)$. Inserting this into
the above equation and using the symmetry of $\mathcal{R}_{\alpha+t}$ we
obtain

(38) $\displaystyle\begin{split}\Lambda(g)&=\int_{\mathbb{B}}(1-\lvert
y\rvert^{2})^{t}\int_{\mathbb{B}}\mathcal{R}_{\alpha+t}(y,x)D^{t}_{\alpha}g(x)\,d\nu_{\alpha+t}(x)\,d\mu(y)\\\
&=\int_{\mathbb{B}}\int_{\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)(1-\lvert
y\rvert^{2})^{t}\,d\mu(y)(1-\lvert
x\rvert^{2})^{t}D^{t}_{\alpha}g(x)\,d\nu_{\alpha}(x),\end{split}$

where we can change the order of the integrals since $(1-\lvert
x\rvert^{2})^{t}D^{t}_{\alpha}g(x)$ is bounded, and by Lemma 3.4(a) and Lemma
2.2,
$\int_{\mathbb{B}}\lvert\mathcal{R}_{\alpha+t}(y,x)\rvert\,d\nu_{\alpha}(x)\lesssim(1-\lvert
y\rvert^{2})^{-t}$. Let

$f(x):=\int_{\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)(1-\lvert
y\rvert^{2})^{t}\,d\mu(y).$

Then $f\in\mathcal{H}(\mathbb{B})$ and by Fubini’s theorem and the estimate in
the previous line,

$\|f\|_{L^{1}_{\alpha}}\leq\int_{\mathbb{B}}(1-\lvert
y\rvert^{2})^{t}\int_{\mathbb{B}}\lvert\mathcal{R}_{\alpha+t}(x,y)\rvert\,d\nu_{\alpha}(x)\,d\lvert\mu\rvert(y)\lesssim\lvert\mu\rvert(\mathbb{B}).$

Thus $f\in\mathcal{B}^{1}_{\alpha}$ with
$\|f\|_{\mathcal{B}^{1}_{\alpha}}\lesssim\lvert\mu\rvert(\mathbb{B})=\|\Lambda\circ
P_{\alpha}\|\lesssim\|\Lambda\|$, and $\Lambda=\Lambda_{f}$ by (38).
Uniqueness of $f$ can be verified in the same way as the previous part. We
note that for $x_{0}\in\mathbb{B}$, $\mathcal{R}_{\alpha}(x_{0},\cdot)$ is in
$\mathcal{B}_{0}$ because
$\lvert\nabla\mathcal{R}_{\alpha}(x_{0},\cdot)\rvert$ is bounded by Lemma
3.4(b). Using also the fact that $\mathcal{R}_{\alpha}(x_{0},\cdot)$ is
bounded, we see that $f\mathcal{R}_{\alpha}(x_{0},\cdot)$ is in
$L^{1}_{\alpha}$ and obtain $\langle
f,\mathcal{R}_{\alpha}(x_{0},\cdot)\rangle_{\alpha}=f(x_{0})$ as in (37). ∎

We finish this section by verifying that there exists an unbounded
$\mathcal{H}$-harmonic Bloch function.

######  Lemma 6.2.

There exists an unbounded function in $\mathcal{B}$.

###### Proof.

Let $e_{1}=(1,0,\dots,0)\in\mathbb{S}$ and $\phi(x)=(1-\lvert
x\rvert^{2})^{n-1}P_{h}(x,e_{1})$, where $P_{h}$ is the hyperbolic Poisson
kernel in (20). It is obvious that $\phi\in L^{\infty}(\mathbb{B})$ and so the
Bergman projection

$f(x):=P_{0}\phi(x)=\int_{\mathbb{B}}\mathcal{R}_{0}(x,y)P_{h}(y,e_{1})(1-\lvert
y\rvert^{2})^{n-1}\,d\nu(y)$

is in $\mathcal{B}$ by Theorem 1.1. To see that $f$ is unbounded we find its
series expansion. Note that by the integral representation of $D^{t}_{s}$ in
Lemma 3.3(ii) (with $s=n-1$ and $t=-(n-1)$) we have
$f(x)=D^{-(n-1)}_{n-1}P_{h}(x,e_{1})$. Therefore, by the series expansion of
$P_{h}$ in (20),

$f(x)=D^{-(n-1)}_{n-1}P_{h}(x,e_{1})=\sum_{m=0}^{\infty}\frac{c_{m}(0)}{c_{m}(n-1)}S_{m}(\lvert
x\rvert)Z_{m}(x,e_{1}).$

Observe that when $x=re_{1}$, all the terms in the above series are positive.
We have $Z_{m}(re_{1},e_{1})=r^{m}Z_{m}(e_{1},e_{1})\sim r^{m}m^{n-2}$ $(m\geq
1)$, $c_{m}(0)/c_{m}(n-1)\sim m^{-(n-1)}$ by (23), and $S_{m}(r)\geq 1$ by
(19). Thus

$f(re_{1})\gtrsim 1+\sum_{m=1}^{\infty}\frac{r^{m}}{m}$

which tends to $\infty$ as $r\to 1^{-}$. ∎

##  7\. Atomic Decomposition

Throughout the section we employ Proposition 5.4 and use any one of the
equivalent norms $\|(1-\lvert
x\rvert^{2})^{t}D^{t}_{\alpha}f(x)\|_{L^{\infty}}$ $(\alpha>-1,t>0)$ for the
Bloch space $\mathcal{B}$.

######  Lemma 7.1.

For every $\alpha>-1$ and $a\in\mathbb{B}$, the kernel
$\mathcal{R}_{\alpha}(\cdot,a)$ is in $\mathcal{B}_{0}$. In addition, there
exists $C=C(n,\alpha)>0$ such that for all $a\in\mathbb{B}$,

(39) $\frac{1}{C\,(1-\lvert
a\rvert^{2})^{\alpha+n}}\leq\|\mathcal{R}_{\alpha}(\cdot,a)\|_{\mathcal{B}}\leq\frac{C}{(1-\lvert
a\rvert^{2})^{\alpha+n}}.$

###### Proof.

Pick some $t>0$. For fixed $a\in\mathbb{B}$, by (26),
$D^{t}_{\alpha}\mathcal{R}_{\alpha}(x,a)=\mathcal{R}_{\alpha+t}(x,a)$ which is
bounded by Lemma 3.4(a) and the inequality $[x,a]\geq 1-\lvert a\rvert$ by
(6). Thus $(1-\lvert x\rvert^{2})^{t}D^{t}_{\alpha}\mathcal{R}_{\alpha}(x,a)$
is in $C_{0}(\mathbb{B})$ and $\mathcal{R}_{\alpha}(x,a)\in\mathcal{B}_{0}$.
Further,

$(1-\lvert
x\rvert^{2})^{t}\bigl{\lvert}\mathcal{R}_{\alpha+t}(x,a)\bigr{\rvert}\lesssim\frac{(1-\lvert
x\rvert^{2})^{t}}{[x,a]^{\alpha+t+n}}\lesssim\frac{1}{(1-\lvert
a\rvert^{2})^{\alpha+n}},$

again by $[x,a]\geq 1-\lvert x\rvert$ and $[x,a]\geq 1-\lvert a\rvert$, which
gives the second inequality in (39). The first inequality follows from [19,
Lemma 6.1] which shows that when $x=a$, $\mathcal{R}_{\alpha+t}(a,a)\sim
1/(1-\lvert a\rvert^{2})^{\alpha+t+n}$, and the fact that
$\mathcal{R}_{\alpha+t}(\cdot,a)$ is continuous. ∎

######  Lemma 7.2.

Suppose $\alpha>-1$ and $\\{a_{m}\\}$ is $r$-separated for some $0<r<1$. Then
the operator $T=T_{\\{a_{m}\\},\alpha}\colon\ell^{\infty}\to\mathcal{B}$
mapping $\lambda=\\{\lambda_{m}\\}$ to

$T\lambda(x)=\sum_{m=1}^{\infty}\lambda_{m}\frac{\mathcal{R}_{\alpha}(x,a_{m})}{\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}}\qquad(x\in\mathbb{B}),$

is bounded. The above series converges absolutely and uniformly on compact
subsets of $\mathbb{B}$. In addition, if $\lambda\in c_{0}$, then
$T\lambda\in\mathcal{B}_{0}$.

###### Proof.

We first verify that $\sum_{m=1}^{\infty}(1-\lvert
a_{m}\rvert^{2})^{\alpha+n}<\infty$. To see this, note that the balls
$E_{r/2}(a_{m})$ are disjoint, and for fixed $r$,
$\nu(E_{r/2}(a_{m}))\sim(1-\lvert a_{m}\rvert^{2})^{n}$ by (12). Also, for
$y\in E_{r/2}(a_{m})$, we have $(1-\lvert y\rvert^{2})\sim(1-\lvert
a_{m}\rvert^{2})$ by Lemma 2.1(i). Thus

$\sum_{m=1}^{\infty}(1-\lvert
a_{m}\rvert^{2})^{\alpha+n}\sim\sum_{m=1}^{\infty}\int_{E_{r/2}(a_{m})}(1-\lvert
y\rvert^{2})^{\alpha}\,d\nu(y)\leq\int_{\mathbb{B}}(1-\lvert
y\rvert^{2})^{\alpha}\,d\nu(y)<\infty.$

To see that the series absolutely and uniformly converges on compact subsets
of $\mathbb{B}$, suppose $\lvert x\rvert\leq R<1$. Then
$\lvert\mathcal{R}_{\alpha}(x,a_{m})\rvert\leq C$ for all $m$ by Lemma 3.4 and
(6). Using also Lemma 7.1, we obtain

$\sum_{m=1}^{\infty}\lvert\lambda_{m}\rvert\frac{\lvert\mathcal{R}_{\alpha}(x,a_{m})\rvert}{\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}}\lesssim\|\lambda\|_{\ell^{\infty}}\sum_{m=1}^{\infty}(1-\lvert
a_{m}\rvert^{2})^{\alpha+n}<\infty.$

Next, we pick some $t>0$. By the continuity of $D^{t}_{\alpha}$ in Lemma 3.2,
we can push $D^{t}_{\alpha}$ into the series and using (26), Lemma 3.4(a),
Lemma 7.1, and the fact that $[x,y]\sim[x,a_{m}]$ for $y\in E_{r/2}(a_{m})$ by
Lemma 2.1, we obtain

$\displaystyle\big{\lvert}D^{t}_{\alpha}(T\lambda)(x)\big{\rvert}$
$\displaystyle\lesssim\|\lambda\|_{\ell^{\infty}}\sum_{m=1}^{\infty}\frac{(1-\lvert
a_{m}\rvert^{2})^{\alpha+n}}{[x,a_{m}]^{\alpha+t+n}}\sim\|\lambda\|_{\ell^{\infty}}\sum_{m=1}^{\infty}\int_{E_{r/2}(a_{m})}\frac{(1-\lvert
y\rvert^{2})^{\alpha}}{[x,y]^{\alpha+t+n}}d\nu(y)$
$\displaystyle\leq\|\lambda\|_{\ell^{\infty}}\int_{\mathbb{B}}\frac{(1-\lvert
y\rvert^{2})^{\alpha}}{[x,y]^{\alpha+t+n}}d\nu(y)\lesssim\frac{\|\lambda\|_{\ell^{\infty}}}{(1-\lvert
x\rvert^{2})^{t}},$

where in the last inequality we use Lemma 2.2. Hence $T\lambda\in\mathcal{B}$
and $\|T\lambda\|_{\mathcal{B}}\lesssim\|\lambda\|_{\ell^{\infty}}$.

Finally, suppose $\lambda\in c_{0}$. For $\varepsilon>0$, let $M$ be such that
$\sup_{m\geq M}\lvert\lambda_{m}\rvert<\varepsilon$. Then

$T\lambda(x)=\sum_{m=1}^{M-1}\lambda_{m}\frac{\mathcal{R}_{\alpha}(x,a_{m})}{\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}}+\sum_{m=M}^{\infty}\lambda_{m}\frac{\mathcal{R}_{\alpha}(x,a_{m})}{\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}}=:h_{1}(x)+h_{2}(x).$

By Lemma 7.1, $h_{1}$ is in $\mathcal{B}_{0}$, and by the previous paragraph
$\|h_{2}\|_{\mathcal{B}}\lesssim\sup_{m\geq M}\lvert\lambda_{m}\rvert$. Thus
$\limsup_{\lvert x\rvert\to 1^{-}}(1-\lvert x\rvert^{2})^{t}\lvert
D^{t}_{\alpha}(T\lambda)(x)\rvert\lesssim\varepsilon$ and $T\lambda$ is in
$\mathcal{B}_{0}$. ∎

Following [4], we associate with an $r$-lattice $\\{a_{m}\\}$ the following
disjoint partition $\\{E_{m}\\}$ of $\mathbb{B}$. Let
$E_{1}=E_{r}(a_{1})\backslash\bigcup_{m=2}^{\infty}E_{r/2}(a_{m})$ and for
$m=2,3,\dots$, inductively define

$E_{m}=E_{r}(a_{m})\backslash\biggl{(}\bigcup_{k=1}^{m-1}E_{k}\,\bigcup\bigcup_{k=m+1}^{\infty}E_{r/2}(a_{k})\biggr{)}.$

The following properties hold: (i) $E_{r/2}(a_{m})\subset E_{m}\subset
E_{r}(a_{m})$, (ii) The sets $E_{m}$ are disjoint, (iii)
$\bigcup_{m=1}^{\infty}E_{m}=\mathbb{B}$.

######  Lemma 7.3.

Suppose $\alpha>-1$, $t>0$, and $\\{a_{m}\\}$ is an $r$-lattice. If
$\\{E_{m}\\}$ is the associated sequence defined above, then the operator
$U=U_{\\{a_{m}\\},\alpha,t}\colon\mathcal{B}\to\ell^{\infty}$,

$Uf=\Big{\\{}D^{t}_{\alpha}f(a_{m})\,\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}\,\nu_{\alpha+t}(E_{m})\Big{\\}}_{m=1}^{\infty}$

is bounded. In addition, if $f\in\mathcal{B}_{0}$, then $Uf\in c_{0}$.

###### Proof.

Because $E_{r/2}(a_{m})\subset E_{m}\subset E_{r}(a_{m})$ and $r$ is fixed, by
Lemma 2.1 and (12), $\nu_{\alpha+t}(E_{m})\sim(1-\lvert
a_{m}\rvert^{2})^{\alpha+t+n}$. Combining this with Lemma 7.1 shows

$\lvert
D^{t}_{\alpha}f(a_{m})\rvert\,\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}\,\nu_{\alpha+t}(E_{m})\sim(1-\lvert
a_{m}\rvert^{2})^{t}\lvert D^{t}_{\alpha}f(a_{m})\rvert.$

Thus $\|Uf\|_{\ell^{\infty}}\lesssim\|f\|_{\mathcal{B}}$. If
$f\in\mathcal{B}_{0}$, then $(1-\lvert x\rvert^{2})^{t}D^{t}_{\alpha}f(x)\in
C_{0}(\mathbb{B})$ and so $Uf$ is in $c_{0}$, since $\lim_{m\to\infty}\lvert
a_{m}\rvert=1$. ∎

###### Proof of Theorem 1.3.

Pick some $t>0$ and define the operators $U\colon\mathcal{B}\to\ell^{\infty}$
and $T\colon\ell^{\infty}\to\mathcal{B}$ as above. We show that there exists a
constant $C=C(n,\alpha,t)$ such that $\|I-TU\|_{\mathcal{B}\to\mathcal{B}}\leq
Cr$, where $I$ is the identity operator. This implies that
$\|I-TU\|_{\mathcal{B}\to\mathcal{B}}<1$ when $r$ is sufficiently small, $TU$
is invertible, and hence $T$ is onto. In the little Bloch case replacing
$\mathcal{B}$ with $\mathcal{B}_{0}$ and $\ell^{\infty}$ with $c_{0}$, we
obtain $T\colon c_{0}\to\mathcal{B}_{0}$ is onto.

Let $f\in\mathcal{B}$. In the calculations below we suppress constants that
depend only on $n,\alpha,t$, and be careful that they do not depend on $r$ or
$f$. Note that

$TUf(x)=\sum_{m=1}^{\infty}D^{t}_{\alpha}f(a_{m})\mathcal{R}_{\alpha}(x,a_{m})\,\nu_{\alpha+t}(E_{m})$

and the series converges absolutely and uniformly on compact subsets of
$\mathbb{B}$. By continuity we can push $D^{t}_{\alpha}$ into the series, and
using (26) obtain

$\displaystyle D^{t}_{\alpha}(TUf)(x)$
$\displaystyle=\sum_{m=1}^{\infty}D^{t}_{\alpha}f(a_{m})\mathcal{R}_{\alpha+t}(x,a_{m})\,\nu_{\alpha+t}(E_{m})$
$\displaystyle=\sum_{m=1}^{\infty}\int_{E_{m}}D^{t}_{\alpha}f(a_{m})\mathcal{R}_{\alpha+t}(x,a_{m})\,\nu_{\alpha+t}(y).$

Further, since $D^{t}_{\alpha}f\in L^{1}_{\alpha+t}(\mathbb{B})$, by the
reproducing property,

$D^{t}_{\alpha}f(x)=\int_{\mathbb{B}}\mathcal{R}_{\alpha+t}(x,y)D^{t}_{\alpha}f(y)d\nu_{\alpha+t}(y)=\sum_{m=1}^{\infty}\int_{E_{m}}\mathcal{R}_{\alpha+t}(x,y)D^{t}_{\alpha}f(y)d\nu_{\alpha+t}(y).$

Thus

$\displaystyle D^{t}_{\alpha}(I-TU)f(x)$
$\displaystyle=\sum_{m=1}^{\infty}\int_{E_{m}}\bigl{(}\mathcal{R}_{\alpha+t}(x,y)-\mathcal{R}_{\alpha+t}(x,a_{m})\bigr{)}D^{t}_{\alpha}f(y)\,d\nu_{\alpha+t}(y)$
$\displaystyle+\sum_{m=1}^{\infty}\int_{E_{m}}\bigl{(}D^{t}_{\alpha}f(y)-D^{t}_{\alpha}f(a_{m})\bigr{)}\mathcal{R}_{\alpha+t}(x,a_{m})\,d\nu_{\alpha+t}(y)$
$\displaystyle=:h_{1}(x)+h_{2}(x).$

We first estimate $h_{1}$. Pick $y\in E_{m}$. Since $E_{m}\subset
E_{r}(a_{m})$, a convex set, by the mean value inequality

$\bigl{\lvert}\mathcal{R}_{\alpha+t}(x,y)-\mathcal{R}_{\alpha+t}(x,a_{m})\bigr{\rvert}\leq\lvert
y-a_{m}\rvert\sup_{z\in
E_{r}(a_{m})}\lvert\nabla_{z}\mathcal{R}_{\alpha+t}(x,z)\rvert.$

By (11), $\lvert y-a_{m}\rvert=\rho(y,a_{m})[y,a_{m}]$ and since
$\rho(y,a_{m})<r<1/2$, we have $[y,a_{m}]\sim[y,y]=1-\lvert y\rvert^{2}$ by
Lemma 2.1(ii), with the suppressed constants not depending on $r$. Thus
$\lvert y-a_{m}\rvert\lesssim r(1-\lvert y\rvert^{2})$. Similarly, for all
$x\in\mathbb{B}$ and $z\in E_{r}(a_{m})$, we have
$[x,z]\sim[x,a_{m}]\sim[x,y]$ by Lemma 2.1(ii). Hence, by Lemma 3.4(b),

$\bigl{\lvert}\nabla_{z}\mathcal{R}_{\alpha+t}(x,z)\bigr{\rvert}\lesssim\frac{1}{[x,z]^{\alpha+t+n+1}}\sim\frac{1}{[x,y]^{\alpha+t+n+1}}.$

We conclude that for all $x\in\mathbb{B}$ and $y\in E_{m}$,

(40)
$\bigl{\lvert}\mathcal{R}_{\alpha+t}(x,y)-\mathcal{R}_{\alpha+t}(x,a_{m})\bigr{\rvert}\lesssim
r\frac{1-\lvert
y\rvert^{2}}{[x,y]^{\alpha+t+n+1}}\lesssim\frac{r}{[x,y]^{\alpha+t+n}},$

where in the last inequality we use $[x,y]\geq 1-\lvert y\rvert$ by (6). Thus

$\displaystyle\lvert h_{1}(x)\rvert$ $\displaystyle\lesssim
r\sum_{m=1}^{\infty}\int_{E_{m}}\frac{(1-\lvert y\rvert^{2})^{\alpha+t}\lvert
D^{t}_{\alpha}f(y)\rvert}{[x,y]^{\alpha+t+n}}\,d\nu(y)\lesssim
r\|f\|_{\mathcal{B}}\int_{\mathbb{B}}\frac{(1-\lvert
y\rvert^{2})^{\alpha}}{[x,y]^{\alpha+t+n}}\,d\nu(y)$ $\displaystyle\lesssim
r\|f\|_{\mathcal{B}}\frac{1}{(1-\lvert x\rvert^{2})^{t}},$

where the last inequality follows from Lemma 2.2.

To estimate $h_{2}$, note first that by the reproducing property, we have

$D^{t}_{\alpha}f(y)-D^{t}_{\alpha}f(a_{m})=\int_{\mathbb{B}}\bigl{(}\mathcal{R}_{\alpha+t}(y,z)-\mathcal{R}_{\alpha+t}(a_{m},z)\bigr{)}D^{t}_{\alpha}f(z)\,d\nu_{\alpha+t}(z).$

Therefore, by (40) with the symmetry of $\mathcal{R}_{\alpha+t}$, and Lemma
2.2,

$\lvert D^{t}_{\alpha}f(y)-D^{t}_{\alpha}f(a_{m})\rvert\lesssim
r\int_{\mathbb{B}}\frac{(1-\lvert z\rvert^{2})^{\alpha+t}\lvert
D^{t}_{\alpha}f(z)\rvert}{[y,z]^{\alpha+t+n}}\,d\nu(z)\lesssim
r\|f\|_{\mathcal{B}}\frac{1}{(1-\lvert y\rvert^{2})^{t}}.$

Hence, using also Lemma 3.4, the fact that $[x,a_{m}]\sim[x,y]$ for all
$x\in\mathbb{B}$ and $y\in E_{m}$, and finally Lemma 2.2, we deduce

$\displaystyle\lvert h_{2}(x)\rvert$ $\displaystyle\lesssim
r\|f\|_{\mathcal{B}}\sum_{m=1}^{\infty}\int_{E_{m}}\frac{(1-\lvert
y\rvert^{2})^{\alpha}}{[x,a_{m}]^{\alpha+t+n}}\,d\nu(y)\sim
r\|f\|_{\mathcal{B}}\sum_{m=1}^{\infty}\int_{E_{m}}\frac{(1-\lvert
y\rvert^{2})^{\alpha}}{[x,y]^{\alpha+t+n}}\,d\nu(y)$
$\displaystyle=r\|f\|_{\mathcal{B}}\int_{\mathbb{B}}\frac{(1-\lvert
y\rvert^{2})^{\alpha}}{[x,y]^{\alpha+t+n}}\,d\nu(y)\lesssim
r\|f\|_{\mathcal{B}}\frac{1}{(1-\lvert x\rvert^{2})^{t}}.$

Thus, $\|(1-\lvert x\rvert^{2})^{t}D^{t}_{\alpha}(I-TU)f(x)\|_{L^{\infty}}\leq
Cr\|f\|_{\mathcal{B}}$ and the proof is completed. ∎

To see that the representation (5) can be used instead of (4), the only change
needed is to replace $\|\mathcal{R}_{\alpha}(\cdot,a_{m})\|_{\mathcal{B}}$
with $(1-\lvert a_{m}\rvert^{2})^{-(\alpha+n)}$ in the definitions of $T$ and
$U$. The proofs of the Lemmas 7.2 and 7.3 become simpler; and $TU$ and the
proof of Theorem 1.3 remain the same.

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