# Polyharmonic potential theory on the Poincaré disk

Massimo A. Picardello, Maura Salvatori, Wolfgang Woess Dipartimento di
Matematica Università di Roma “Tor Vergata” I-00133 Rome, Italy
<EMAIL_ADDRESS>Dipartimento di Matematica Università degli Studi
di Milano Via Saldini, 50, I-20133 Milano, Italy<EMAIL_ADDRESS>Institut für Diskrete Mathematik, Technische Universität Graz, Steyrergasse
30, A-8010 Graz, Austria<EMAIL_ADDRESS>
###### Abstract.

We consider the open unit disk $\mathbb{D}$ equipped with the hyperbolic
metric and the associated hyperbolic Laplacian $\mathfrak{L}\,$. For
$\lambda\in\mathbb{C}$ and $n\in\mathbb{N}$, a $\lambda$-polyharmonic function
of order $n$ is a function $f:\mathbb{D}\to\mathbb{C}$ such that
$(\mathfrak{L}-\lambda\,I)^{n}f=0$. If $n=1$, one gets $\lambda$-harmonic
functions. Based on a Theorem of Helgason on the latter functions, we prove a
boundary integral representation theorem for $\lambda$-polyharmonic functions.
For this purpose, we first determine $n^{\text{th}}$-order $\lambda$-Poisson
kernels. Subsequently, we introduce the $\lambda$-polyspherical functions and
determine their asymptotics at the boundary $\partial\mathbb{D}$, i.e., the
unit circle. In particular, this proves that, for eigenvalues not in the
interior of the $L^{2}$-spectrum, the zeroes of these functions do not
accumulate at the boundary circle. Hence the polyspherical functions can be
used to normalise the $n^{\text{th}}$-order Poisson kernels. By this tool, we
extend to this setting several classical results of potential theory: namely,
we study the boundary behaviour of $\lambda$-polyharmonic functions, starting
with Dirichlet and Riquier type problems and then proceeding to Fatou type
admissible boundary limits.

###### Key words and phrases:

Hyperbolic disk, hyperbolic Laplacian, polyharmonic functions, analytic
functionals, Riquier problem, Fatou type theorem

###### 2020 Mathematics Subject Classification:

31A30; 43A85, 58J32

The first author acknowledges support by MIUR Excellence Departments Project
awarded to the Department of Mathematics, University of Rome Tor Vergata,
MatMod@TOV. The second author acknowledges support as a visiting scientist at
TU Graz. The third author was supported by Austrian Science Fund project FWF
P31889-N35.

Dedicated to the memory of Alessandro Figà-Talamanca (1938-2023)

##  1\. Introduction

The aim of this article is to initiate a detailed study of the potential
theory associated with the polyharmonic, and more generally,
$\lambda$-polyharmonic functions for the hyperbolic Laplacian $\mathfrak{L}$,
that is, the solutions $f$ of $(\mathfrak{L}-\lambda\,I)^{n}f=0$, for
$n\in\mathbb{N}$ and $\lambda\in\mathbb{C}$.

A _polyharmonic function of order $n$_ on a Euclidean domain $D$ is a complex-
valued function $f$ ond $D$ that belongs to the kernel of the $n^{\text{th}}$
iterate of classical Euclidean Laplacian: $\Delta^{n}f\equiv 0$. The study of
polyharmonic functions goes back to work in the $19^{\text{th}}$ century, and
continues to be a very active topic. See e.g. the books by Aronszajn, Creese
and Lipkin [2] or by Gazzola, Grunau and Sweers [10]. A classical theorem of
Almansi [1] says that if the domain $D$ is star-like with respect to the
origin, then every polyharmonic function of order $n$ has a unique
decomposition

$f(z)=\sum_{k=0}^{n-1}|z|^{2k}\,h_{k}(z)\,,$

where each $h_{k}$ is harmonic on $D$, and $|z|$ is the Euclidean length of
$z\in D$. In particular, let the domain be the unit disk

$\mathbb{D}=\\{z=x+\mathfrak{i}\,y\in\mathbb{C}:|z|=\sqrt{x^{2}+y^{2}}<1\\}.$

Assume for the moment that in Almansi’s decomposition, each $h_{k}$ is non-
negative. Then it has an integral representation over the boundary
$\partial\mathbb{D}$ of the disk, that is, the unit circle, with respect to
the _Poisson kernel_

(1)
$P(z,\xi)=\frac{1-|z|^{2}}{|\xi-z|^{2}}\qquad(z\in\mathbb{D}\,,\;\xi=e^{\mathfrak{i}\,\phi}\in\partial\mathbb{D})$

Thus, Almansi’s decomposition on the disk reads as

(2)
$f(z)=\sum_{k=0}^{n-1}\int_{\partial\mathbb{D}}|z|^{2k}\,P(z,\xi)\,d\nu_{k}(\xi)\,,$

where $\nu_{0},\dots,\nu_{n-1}$ are non-negative Borel measures on the unit
circle. Without requiring non-negativity of the $h_{k}\,$, the result still
remains true, taking _analytic functionals_ (i.e., certain distributions)
$\nu_{k}$ instead of Borel measures: this follows from results by Helgason
[12], [13] which will be of crucial importance further below.

We now change our viewpoint and view $\mathbb{D}$ as the _hyperbolic_ or
_Poincaré disk_ with the hyperbolic length element and resulting metric

(3)
$ds=\frac{2\sqrt{dx^{2}+dy^{2}}}{1-|z|^{2}}\quad\text{and}\quad\rho(z,w)=\log\frac{|1-z\bar{w}|+|z-w|}{|1-z\bar{w}|-|z-w|}.$

We note that the Poisson kernel can be written as

(4)
$P(z,\xi)=e^{-\mathfrak{h}(z,\xi)}\quad\text{with}\quad\mathfrak{h}(z,\xi)=\lim_{w\to\xi}\Bigl{(}\rho(w,z)-\rho(w,0)\Bigr{)}\,,$

the _Busemann function_. The hyperbolic Laplace (or Laplace-Beltrami) operator
in the variable $z=x+\mathfrak{i}\,y$ is

(5)
$\mathfrak{L}=\frac{(1-|z|^{2})^{2}}{4}\,\Bigl{(}\partial_{x}^{2}+\partial_{y}^{2}\Bigr{)}\,.$

The harmonic functions for the two Laplacians on the disk clearly coincide,
but this is no more true for polyharmonic functions of higher order. While
there is an abundant ongoing literature on polyharmonic functions in the
Euclidean setting, we are not aware of an ample body of work for the
hyperbolic Laplacian, or more generally, for Laplace-Beltrami operators on
manifolds. A few references are, for example, Chung, Sario and Wang [6], Chung
[5] as well as Schimming and Belger [31] plus some of the citations in the
latter paper, and also Jaming [18].

The first main aim of this note is to provide an integral representation in
the spirit of (2) for _hyperbolically_ polyharmonic functions of order $n$.
More generally, we consider _$\lambda$ -polyharmonic_ functions of order $n$,
that is, solutions $f:\mathbb{D}\to\mathbb{C}$ of

$(\mathfrak{L}-\lambda\,I)^{n}f=0\,.$

Here, $I$ is the identity operator, and we are taking the $n^{\textrm{th}}$
iterate of $\,\mathfrak{L}-\lambda\,I\,$, where $\lambda\in\mathbb{C}$. If
$n=1$, we speak of a _$\lambda$ -harmonic function._ Considering $\lambda$ as
an “eigenvalue”, one should be careful with respect to the space on which the
operator acts. Indeed, we are _not_ referring to the action of $\mathfrak{L}$
as a self-adjoint operator on $L^{2}(\mathbb{D},\mathsf{area}_{h})$, where
$\mathsf{area}_{h}$ is the hyperbolic area measure of $\mathbb{D}$ and the
corresponding spectrum $(-\infty\,,-\frac{1}{4}\,]$ is continuous. The mapping

$\lambda(s)=s^{2}-\frac{1}{4}\,,\quad s\in\mathbb{C}\,,\;\Re(s)\geqslant 0$

maps the half-open half plane $\\{s\in\mathbb{C}:\Re(s)\geqslant
0\\}\setminus\\{\mathfrak{i}\,t:t<0\\}$ bijectively onto $\mathbb{C}$. We
write $s(\lambda)=\sqrt{\lambda+\frac{1}{4}}$ for the inverse mapping, where
the square root of $re^{\mathfrak{i}\,\phi}$ is
$\sqrt{r}\,e^{\mathfrak{i}\,\phi/2}$ for $r\geqslant 0$ and
$\phi\in(-\pi\,,\,\pi]$.

Here is our first main result.

######  Theorem 1.1.

Every $\lambda$-polyharmonic function $f:\mathbb{D}\to\mathbb{C}$ of order $n$
has a unique representation of the form

$f(z)=\begin{cases}\displaystyle\sum_{k=0}^{n-1}\int_{\partial\mathbb{D}}\mathfrak{h}(z,\xi)^{k}\,P(z,\xi)^{s(\lambda)+1/2}\,d\nu_{k}(\xi)\,,\quad&\text{if
}\;\lambda\neq-\frac{1}{4}\,,\\\\[14.0pt]
\displaystyle\sum_{k=0}^{n-1}\int_{\partial\mathbb{D}}\mathfrak{h}(z,\xi)^{2k}\,P(z,\xi)^{1/2}\,d\nu_{k}(\xi)\,,\quad&\text{if
}\;\lambda=-\frac{1}{4}\,,\end{cases}$

where $\nu_{0},\dots,\nu_{n-1}$ are analytic functionals on
$\partial\mathbb{D}$.

We postpone the precise definition of these functionals to §2, and we shall
also rescale the kernels in the integrals in a suitable manner to get the
(order $n+1$) $\lambda$-polyharmonic Poisson kernels
$P_{n}(z,\xi\,|\,\lambda)$, $n\geqslant 0$; see Proposition 2.6.

Our proof of Theorem 1.1 in §2 is inspired by related results obtained in a
discrete setting, mostly in recent work of Picardello and Woess [28] on
polyharmonic functions on general trees, that was preceded by a long paper by
Cohen et al. [7] who had used rather involved methods to obtain an integral
representation for polyharmonic functions on a regular tree with respect to
the standard graph Laplacian. (In [28], this is generalised and simplified.)
Further motivation for the present work came from Sava-Huss and Woess [30],
who studied the boundary behaviour of polyharmonic functions on regular trees.
There are many profound analogies between the hyperbolic disk and regular
trees. In the potential theoretic setting considered here, see the first part
of the note by Boiko and Woess [3] for an exposition of those analogies.111In
the formula for the hyperbolic Laplacian – which is (5) here – one of the two
squares is missing in [3].

The natural next goal is to study the asymptoptic behaviour of
$\lambda$-polyharmonic functions. For this purpose, but also by inherent
interest and for further possible applications, in §3, we introduce the family
of polyspherical functions $\Phi_{n}(z\,|\,\lambda)$, i.e., suitably
normalised $\lambda$-polyharmonic functions of order $n+1$ which only depend
on $r=|z|$. Here, the functions $\Phi_{0}(z\,|\,\lambda)$ for
$\lambda\in\mathbb{C}$ are the classical spherical functions of the Poincaré
disk. A major step, in itself of interest, is to determine the asymptotic
behaviour of $\Phi_{n}(z\,|\,\lambda)$ near $\partial\mathbb{D}$, that is, as
$r\to 1$ or equivalently, $R=\rho(z,0)\to\infty\,$; see Theorem 3.4.

This is important because, for the study of the boundary behaviour of
$\lambda$-polyharmonic functions one needs a suitable normalisation of the
polyharmonic kernels $P_{n}(z,\xi\,|\,\lambda)$, in order to compensate for
their growth or decay; this normalisation is then accomplished in §4; it
extends the classical case $n=0$ via the laborious computations of §3. Indeed,
when $n=0$, it is well-known that it is appropriate to normalise
$\lambda$-harmonic functions by the $\lambda$-spherical function, see e.g.
Michelson [27] and Sjögren [33], and for regular trees Korányi and Picardello
[21]. This cannot be done for $\lambda\in(-\infty\,,\,-\frac{1}{4})$, because
for these values of $\lambda$ the zeroes of the $\lambda$-spherical functions
accumulate at the boundary circle, while for all other values of $\lambda$,
there are no zeroes at all; see Remark 3.3.

So for arbitrary $n$, our normalisation consists in dividing by
$\Phi_{n}(z\,|\,\lambda)$, which is feasible since it follows from Theorem 3.4
that this function has no zeroes close to the boundary circle. In §4, we show
that the resulting normalised kernels are good approximate identities at the
boundary points, so that the classical convergence results hold for transforms
of functions and measures on $\partial\mathbb{D}$; see Proposition 4.6.

§5 is dedicated to another important issue: continuous extensions from
boundary data. We first limit attention to $n=0$ and show that for the
normalized kernel $P(z,\xi\,|\,\lambda)/\Phi_{0}(z\,|\,\lambda)$, the solution
of the Dirichlet problem with continuous boundary data is unique for any (even
complex) $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$ (Theorem
5.1). Then we extend the result to $n>0$ by formulating a suitable version of
the Riquier problem, adapted to the fact that the quotient of lower and higher
order polyspherical functions tends to zero at the boundary, and provide such
a solution (Corollary 5.4), which is inherently non-unique.

§6 answers another fundamental question on the asymptotic behaviour of
$\lambda$-polyharmonic functions, the Fatou theorem. Theorem 6.5 yields
admissible non-tangential convergence of the normalised transforms of measures
on $\partial\mathbb{D}$ for
$\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4}\,]$. For the critical
value $\lambda=-\frac{1}{4}$, we even have a wider approach region. Along the
classical guidelines, the proofs are based on maximal inequalities.

The last §7 is devoted to related examples (in the standard case $\lambda=0$),
discussions and open questions. In particular, we provide all details of an
example outlined to us by A. Borichev: a harmonic (indeed, analytic) function
$h(z)$ such that $h(z)/R$ is bounded but has no radial limits at the boundary,
as $R\to\infty$ where $R=\rho(z,0)\sim\Phi_{1}(z|0)$, the biharmonic spherical
function. In all the paper, for the reader’s benefit, we give most of the
details of the (sometimes lengthy) computations.

Acknowledgements. We acknowlege enlightening email conversations with
Alexander Borichev (Marseille), Fausto Di Biase (Pescara), Jean-Pierre Otal
(Toulouse) and Peter Sjögren (Göteborg).

##  2\. Integral representation

For each $\delta>0$, consider the space $\mathcal{H}(\mathbb{A}_{\delta})$ of
all holomorphic functions on the open annulus

$\mathbb{A}_{\delta}=\\{z\in\mathbb{C}:1-\delta<|z|<1+\delta\\}\,.$

The space is equipped with the topology of uniform convergence on compact
sets. The space $\mathcal{H}(\partial\mathbb{D})$ of analytic functions on the
unit circle consists of all functions $g:\partial\mathbb{D}\to\mathbb{C}$
which possess an extension in $\mathcal{H}(\mathbb{A}_{\delta})$ for some
$\delta=\delta(g)>0$. The topology on $\mathcal{H}(\partial\mathbb{D})$ is the
inductive limit of the topologies of $\mathcal{H}(\mathbb{A}_{\delta})$ as
$\delta\to 0$.

######  Definition 2.1.

An _analytic functional_ $\nu$ on $\partial\mathbb{D}$ is an element of the
dual space of $\mathcal{H}(\partial\mathbb{D})$. We write

$\int_{\partial\mathbb{D}}g\,d\nu:=\nu(g)\,,\quad
g\in\mathcal{H}(\partial\mathbb{D})\,.$

A good way to understand the action of $\nu$ on
$\mathcal{H}(\partial\mathbb{D})$ is described in [8, p. 114], see Köthe [23]:
let $\nu_{n}=\nu(e^{-\mathfrak{i}\,n\phi})$, $n\in\mathbb{Z}$ be the Fourier
coefficients of $\nu$. Then

$\limsup_{|n|\to\infty}|\nu_{n}|^{1/|n|}\geqslant 1$

(and this characterises the analytic functionals). If
$g\in\mathcal{H}(\partial\mathbb{D})$ then the Fourier expansion
$g(e^{\mathfrak{i}\,\phi})=\sum_{n\in\mathbb{Z}}g_{n}\,e^{\mathfrak{i}\,n\phi}$
is such that

$\limsup_{|n|\to\infty}|g_{n}|^{1/|n|}<1.$

Then

(6)
$\int_{\partial\mathbb{D}}g\,d\nu=\sum_{n\in\mathbb{Z}}g_{n}\,\overline{\nu\,}_{\\!\\!n}\,.$

For more on analytic functionals, resp. hyperfunctions, see e.g. Hörmander
[16, Chapter IX] or Schlichtkrull [32].

It will be useful to write

$P(z,\xi\,|\,\lambda)=P(z,\xi)^{s(\lambda)+1/2}.$

The following results from an elementary and well-known computation.

######  Lemma 2.2.

For $\lambda\in\mathbb{C}$ and $\xi\in\partial\mathbb{D}$, the function
$z\mapsto P(z,\xi\,|\,\lambda)$ satisfies

$\mathfrak{L}P(\,\cdot\,,\xi\,|\,\lambda)=\lambda\,P(\,\cdot\,,\xi\,|\,\lambda)\,.$

We note here that we can write

(7) $P(z,\xi)=\frac{1-|z|^{2}}{(\xi-z)(1/\xi-\bar{z})}\,.$

In this form, for fixed $z\in\mathbb{D}$ and any $\lambda\in\mathbb{C}$, the
function $\xi\mapsto P(z,\xi\,|\,\lambda)$ is in
$\mathcal{H}(\mathbb{A}_{\delta})$ for $\delta=1-|z|$. Thus, as a function of
$\xi$ in the unit circle, it is in $\mathcal{H}(\partial\mathbb{D})$. We now
recall an important result.

######  Theorem 2.3 (Helgason [12], [13, Section V.6]).

For any $\lambda\in\mathbb{C}$, every $\lambda$-harmonic function $h$ for the
hyperbolic Laplacian on the Poincaré disk has a unique representation

$h(z)=\int_{\partial\mathbb{D}}P(z,\xi\,|\,\lambda)\,d\nu(\xi)\,,$

where $\nu=\nu^{h}$ is an analytic functional on $\partial\mathbb{D}$.

A very readable proof of this and several related results are contained in the
beautiful expository paper by Eymard [8].

######  Remark 2.4.

If the $\lambda$-harmonic function $h$ is positive real, then
$\lambda\geqslant-\frac{1}{4}$. Indeed, it is well known that positive
$\lambda$-harmonic functions exist precisely when
$\lambda\geqslant-\frac{1}{4}$; see e.g. Sullivan [35, Thm. 2.1].

For non-negative $h$, the functional $\nu^{h}$ is a non-negative Borel
measure. This follows from general Martin boundary theory, see e.g. Karpelevič
[19] or Taylor [36].

Before proving Theorem 1.1, we need to find suitable polyharmonic versions of
the Poisson kernel. That is, for each $n\in\mathbb{N}_{0}\,$, we want to have
a kernel of the form

(8)
$P_{n}(z,\xi\,|\,\lambda)=g_{n,\lambda}\bigl{(}-\mathfrak{h}(z,\xi)\bigr{)}\,P(z,\xi\,|\,\lambda)$

which satisfies

(9)
$(\mathfrak{L}-\lambda\,I)^{n}P_{n}(z,\xi\,|\,\lambda)=P(z,\xi\,|\,\lambda)\,.$

For this purpose, we shall use the following.

######  Lemma 2.5.

Let $f\in C^{2}(\mathbb{R})$ and set

$Q_{f}(z,\xi\,|\,\lambda)=f\bigl{(}-\mathfrak{h}(z,\xi)\bigr{)}\,P(z,\xi\,|\,\lambda)\,.$

Then

$(\mathfrak{L}-\lambda\,I)Q_{f}(z,\xi\,|\,\lambda)=Q_{g}(z,\xi\,|\,\lambda)\,,$

where $g=f^{\prime\prime}+2s\,f^{\prime}$ and $s=s(\lambda)\,$.

###### Proof.

Here (and frequently also later) we shall use the fact that the Busemann
function, hence also the Poisson kernel (as well as the Laplacian), are
rotation invariant:

(10)
$\mathfrak{h}(e^{\mathfrak{i}\,\alpha}z,e^{\mathfrak{i}\,\alpha}\xi)=\mathfrak{h}(z,\xi)\quad\text{for
all }\;z\in\mathbb{D}\,,\;\xi\in\partial\mathbb{D}.$

Thus, it is sufficient to consider $\xi=1$. Furthermore, from the Poincaré
disk model of the hyperbolic plane we can first pass to the upper half plane
model via the inverse Cayley transform, where in the new coordinates $(u,v)$
with $u\in\mathbb{R}$ and $v>0$, the hyperbolic Laplacian transforms into
$v^{2}(\partial^{2}u+\partial^{2}v)$ and the boundary point
$1\in\partial\mathbb{D}$ becomes $\mathfrak{i}\,\infty$. Then we make one more
change of variables, setting $w=\log v$ to obtain the _logarithmic model,_
where now $(u,w)\in\mathbb{R}^{2}$ and the hyperbolic Laplacian becomes

(11) $\mathfrak{L}=e^{2w}\partial^{2}_{u}+\partial^{2}_{w}-\partial_{w}\,.$

In these coordinates, the Busemann function and Poisson kernel at
$\mathfrak{i}\,\infty$ are

$\mathfrak{h}\bigl{(}(u,w),\mathfrak{i}\,\infty\bigr{)}=-w\quad\text{and}\quad
P\bigl{(}(u,w),\mathfrak{i}\,\infty\bigr{)}=e^{w}\,,$

and

$Q_{f}((u,w),\mathfrak{i}\,\infty|\lambda\bigr{)}=f(w)\,e^{(s+1/2)w}\,.$

The statement now follows by applying the Laplacian in the form of (11). ∎

######  Proposition 2.6.

For $\lambda\neq-\frac{1}{4}$ and $s=s(\lambda)$, the kernel

$P_{n}(z,\xi\,|\,\lambda)=\frac{1}{n!(2s)^{n}}\,\bigl{(}-\mathfrak{h}(z,\xi)\bigr{)}^{n}\,P(z,\xi)^{s+1/2}$

satisfies (9).

For $\lambda=-\frac{1}{4}$, where $s(\lambda)=0$, identity (9) holds for

$P_{n}(z,\xi|-\frac{1}{4})=\frac{1}{(2n)!}\,\mathfrak{h}(z,\xi)^{2n}\,P(z,\xi)^{1/2}.$

###### Proof.

In order to find a function $g_{n,\lambda}$ as in (8), we start of course with
$g_{0,\lambda}\equiv 1$. We proceed recursively, looking at each step for a
function $f_{n}=f_{n,\lambda}$ such that

(12)
$(\mathfrak{L}-\lambda\,I)\Bigl{[}f_{n,\lambda}\bigl{(}-\mathfrak{h}(z,\xi)\bigr{)}\,P(z,\xi\,|\,\lambda)\Bigr{]}=g_{n-1,\lambda}\bigl{(}-\mathfrak{h}(z,\xi\,|\,\lambda)\bigr{)}\,P(z,\xi\,|\,\lambda)\,.$

The function $f_{n}$ will then be replaced by the simpler
$g_{n}=g_{n,\lambda}$ which satisfies (8) before proceeding to $n+1$. By Lemma
2.5, $f_{n}$ must solve the differential equation

(13) $f_{n}^{\prime\prime}+2s\,f_{n}^{\prime}=g_{n-1}\,,\quad s=s(\lambda).$

The characteristic polynomial of (13) has roots $0$ and $-2s$, when
$\lambda\neq-\frac{1}{4}$. In the latter case, $0$ is a double root.

We start with $\lambda\neq-\frac{1}{4}$ and $n=1$. Since $g_{0}=1$, we are
looking for a special solution of (13) of the form $f_{1}(w)=A_{1,1}w$, whence
$A_{1,1}=1/(2s)$. We get $g_{1}(w)=f_{1}(w)=w/(2s)$, and going back to the
disc model, we obtain $P_{1}$ via (8), as proposed. We now prove by induction
on $n$ that by setting $g_{n}(w)=\dfrac{w^{n}}{n!(2s)^{n}}$ we obtain a
solution for $P_{n}\,$.

Suppose this is true for all orders up to $n-1$. The right hand side of (13)
is a polynomial of order $n-1$ in $w$. Hence there is a special solution of
the form $f_{n}(w)=\sum_{k=1}^{n}A_{n,k}w^{k}\,$. The coefficients $A_{n,k}$
are obtained as solutions of a system of linear equations, yielding a solution
of (12). However, by the induction hypothesis, the terms of order $k<n$ are
annihilated when applying $(\mathfrak{L}-\lambda\,I)^{n}$ to
$f_{n,\lambda}\bigl{(}-\mathfrak{h}(z,\xi)\bigr{)}\,P(z,\xi\,|\,\lambda)$, so
that for (9) we only need $g_{n}(w)=A_{n,n}w^{n}$. Inserting $f_{n}$ into
(13), comparison of the highest order coefficients yields
$\;2sn=\dfrac{w^{n-1}}{(n-1)!(2s)^{n-1}}$, which completes the induction step.

When $\lambda=-\frac{1}{4}$, the differential equation (13) simplifies to
$f_{n}^{\prime\prime}=g_{n-1}\,$. Here, we set $g_{n}=f_{n}\,$. Starting with
$g_{0}\equiv 1$, we integrate twice at each step and take just the highest
appearing power: $g_{1}(w)=w^{2}/2\,$, $g_{2}(w)=w^{4}/4!$, and so on, so that
$g_{n}(w)=w^{2n}/(2n)!$, as proposed. ∎

Note that when $\lambda=-\frac{1}{4}$, we even have
$(\mathfrak{L}-\lambda\,I)P_{n}(z,\xi|-\frac{1}{4})=P_{n-1}(z,\xi|-\frac{1}{4})$.

In the standard case $\lambda=0$, we just write $P_{n}(z,\xi)$ for
$P_{n}(z,\xi|0)$.

###### Proof of Theorem 1.1.

Along with the Poisson kernel, also the function $\xi\mapsto
P_{n}(z,\xi\,|\,\lambda)$ is in $\mathcal{H}(\mathbb{A}_{\delta})$ for
$\delta=1-|z|$, for every $n\in\mathbb{N}_{0}$, $\lambda\in\mathbb{C}$ and
$z\in\mathbb{D}$.

We claim that every $\lambda$-polyharmonic function $f$ of order $n$ has a
unique representation of the form

(14) $f(z)=\sum_{k=0}^{n-1}f_{k}(z)\quad\text{with}\quad
f_{k}(z)=\int_{\partial\mathbb{D}}P_{k}(z,\xi\,|\,\lambda)\,d\nu_{k}(\xi)\,,$

where $\nu_{0},\dots,\nu_{n-1}$ are analytic functionals on
$\partial\mathbb{D}$. Furthermore, when $\lambda\geqslant-\frac{1}{4}$ is
real, $\nu_{k}$ is a non-negative Borel measure if and only if

$(\mathfrak{L}-\lambda\,I)^{k}g_{k}\geqslant 0\,,\quad\text{where}\quad
g_{k}=f-(f_{n-1}+\dots+f_{k+1})\,.$

To prove this, we proceed by induction on $n$. For $n=1$, this is Theorem 2.3.
Suppose the statement is true for $n-1$. Let $f$ be $\lambda$-polyharmonic of
order $n$. Then $h=(\mathfrak{L}-\lambda\,I)^{n-1}f$ is $\lambda$-harmonic. By
Theorem 2.3, there is a unique analytic functional $\nu_{n-1}$ on
$\partial\mathbb{D}$ such that

$h(z)=\int_{\partial\mathbb{D}}P(z,\xi\,|\,\lambda)\,d\nu_{n-1}(\xi)\,,$

We set

$f_{n-1}(z)=\int_{\partial\mathbb{D}}P_{n-1}(z,\xi\,|\,\lambda)\,d\nu_{n-1}(\xi)\,.$

By Proposition 2.6,

$(\mathfrak{L}-\lambda\,I)^{n-1}f_{n-1}=h=(\mathfrak{L}-\lambda\,I)^{n-1}f\,.$

Thus, $f-f_{n-1}$ is $\lambda$-polyharmonic of order $n-1$, and we can apply
the induction hypothesis to that function in order to get the representation
of $f$. Uniqueness follows from Helgason’s Theorem 2.3. The statement on non-
negativity for real $\lambda\geqslant-\frac{1}{4}$ is a consequence of the
well known Poisson-Martin representation theorem for positive
$\lambda$-harmonic functions. Since (14) differs from the proposed result only
by the normalisation of the kernels $P_{n}\,$, the result is proved. ∎

The previous paper [28] on trees adopts a different method of proof, that
could also be used here: it consists in differentiating $P(z,\xi\,|\,\lambda)$
with respect to $\lambda$ instead of integrating a differential equation with
respect to $z$, which is inherent in the proof of Proposition 2.6. The
drawback is that this does not work at the critical value
$\lambda=-\frac{1}{4}$, while on the other hand, it can be applied to more
general domains as long as $\lambda$ belongs to the $L^{2}$-resolvent set of
the underlying Laplacian.

_From now on, we use the representation formula ( 14) for
$\lambda$-polyharmonic functions, instead of the one of the statement of
Theorem 1.1._

##  3\. Polyspherical functions

######  Definition 3.1.

For $\lambda\in\mathbb{C}$ and $n\in\mathbb{N}_{0}\,$, the $n^{\textrm{th}}$
_$\lambda$ -polyspherical funtion_ is

$\Phi_{n}(z\,|\,\lambda)=\int_{\partial\mathbb{D}}P_{n}(z,\xi\,|\,\lambda)\,d\xi\,,\quad
z\in\mathbb{D}\,,$

where $P_{n}$ is given by Proposition 2.6 and $d\xi=d\mathsf{m}(\xi)$ for the
normalized Lebesgue measure $\mathsf{m}$ on the unit circle.

The function $\Phi_{n}(z\,|\,\lambda)$ is $\lambda$-polyharmonic of order
$n+1$ and rotation invariant, i.e., it depends only on $|z|$. For $n=0$, we
recover the classical spherical functions $\Phi(z\,|\,\lambda)$, where we omit
the index $0\,$: for $z=r\,e^{i\phi}\in\mathbb{D}\,$,

(15) $\displaystyle\Phi(z\,|\,\lambda)$
$\displaystyle=\int_{\partial\mathbb{D}}P(z,\xi\,|\,\lambda)\,d\xi=\int_{\partial\mathbb{D}}P(z,\xi)^{s(\lambda)+1/2}\,d\xi$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\Bigl{(}\frac{1-r^{2}}{1+r^{2}-2r\cos\phi}\Bigr{)}^{s(\lambda)+1/2}d\phi\,.$

In particular, $\Phi(\,\cdot\,|\,0)\equiv 1$. The following is immediate from
Proposition 2.6.

######  Lemma 3.2.

For any $n\in\mathbb{N}\,$ and $\lambda\in\mathbb{C}$ and
$\xi\in\partial\mathbb{D}\,$,

$(\mathfrak{L}-\lambda\,I)^{n}\,\Phi_{n}(z\,|\,\lambda)=\Phi(z\,|\,\lambda)\,.$

In the specific case $\lambda=-\frac{1}{4}$ one even has

$(\mathfrak{L}-\lambda\,I)\,\Phi_{n}(z\,|\,-\\!\tfrac{1}{4})=\Phi_{n-1}(z\,|\,-\\!\tfrac{1}{4})\,.$

######  Remark 3.3.

For $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, the spherical
function $\Phi(r\,|\,\lambda)$ has no zeroes in $[0\,,\,1)$, while for real
$\lambda\in(-\infty\,,\,-\frac{1}{4})$, $\Phi(r\,|\,\lambda)$ has countably
many zeroes which accumulate at $1$.

For real $\lambda\in[-\frac{1}{4}\,,\,\infty)$, this is obvious because the
integrand in (15) is positive. For the other values of $\lambda$, to the best
of our knowledge, it seems that these facts are nowhere referred to in the
relevant literature on the hyperbolic Laplacian and its spherical functions.
Therefore, we add some explanations.

Expressing $\mathfrak{L}$ in polar coordinates leads to the differential
equation

$\frac{(1-r^{2})^{2}}{4}\,\Phi^{\prime\prime}(r\,|\,\lambda)+\frac{(1-r^{2})^{2}}{4r}\,\Phi^{\prime}(r\,|\,\lambda)-\lambda\,\Phi(r\,|\,\lambda)=0$

in the variable $r\in[0\,,1)$. We now substitute $z=\dfrac{1+r^{2}}{1-r^{2}}$
and write $\Psi(z\,|\,\lambda)=\Phi(r\,|\,\lambda)$. Then the above
differential equation transforms into

$(z^{2}-1)\,\Psi^{\prime\prime}(z\,|\,\lambda)+2z\,\Psi^{\prime}(z\,|\,\lambda)-\lambda\,\Psi(z\,|\,\lambda)=0\,.$

One sees that the latter is Legendre’s differential equation; see Hobson [17]
and, in particular, Hille [14], [15, equations (1) and (2)]. Recall that the
Legendre function $P_{a}$ with parameter $a\in\mathbb{C}$ solves
$(1-z^{2})P_{a}^{\prime\prime}(z)-2zP^{\prime}_{a}(z)+a(a+1)P_{a}(z)=0$, and

$P_{a}(z)=F\left(a+1,-a;1;\;\;\frac{1-z}{2}\right),$

where $F$ is Gauss’ hypergeometric function, and in our case,
$a(a+1)=\lambda$. We get

$\Phi(r\,|\,\lambda)=P_{a}(z)\quad\text{with}\quad
a=-\frac{1}{2}+s(\lambda)\quad\text{and}\quad
z=\frac{1+r^{2}}{1-r^{2}}\in[1\,,\infty).$

Compare with [8, identity (25)]. Note that the hyperbolic Laplacian in [8] is
4 times the one we are using here, and that $\Phi(z\,|\,\lambda)$ is the
function $\varphi_{0}(z,\mu)$ of [8], with $\mu=\tfrac{1}{2}+s(\lambda)$. Note
also that there are several known identities between hypergeometric functions
with different parameters. For example, Grellier and Otal [11] use a different
version, which coincides with the one given above, see e.g. Lebedev [24, p.
200].

It follows from old work of Mehler [26] and is explained in [14, pages 27–28,
identity (35)] that for $a=-\frac{1}{2}+b\,\mathfrak{i}\,$ with $b\neq 0$,
that is, for $\lambda\in(-\infty\,,\,-\frac{1}{4})$, the function $P_{a}(z)$
has countably many zeroes in $[1\,,\infty)$ which are such that the zeroes of
$\Phi(r\,|\,\lambda)$ accumulate at $1$. On the other hand, it is comprised in
[15, Theorem V and the page preceding it], that with $\Re(a)\neq-1/2$ the
function $P_{a}(z)$ has no zeroes in $[1\,,\infty)$.222We thank Jean-Pierre
Otal (Toulouse) for pointing us to the books by Lebedev and Hobson, which led
us to the PhD thesis of Hille [14] and its follow-up [15]. We also acknowledge
an exchange with Peter Sjögren (Göteborg) on the question how well this issue
of zeroes of the spherical functions is known in the community – apparently
not at all. ∎

For any fixed $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, we we
shall need the asymptotic behaviour of $\Phi_{n}(z\,|\,\lambda)$ when $|z|\to
1$, that is, when

$\rho(z,0)=\log\frac{1+|z|}{1-|z|}\to\infty\,.$

In view of Remark 3.3, we do not consider real
$\lambda\in(-\infty\,,\,-\frac{1}{4})$ because of the infinity of zeroes of
$\Phi(r\,|\,\lambda)$.

In the sequel, we shall adopt the following convention: in the formulas where
$0<r<1$ (resp. $z\in\mathbb{D}$ with $|z|=r$) appear, we always use capital
$R=\rho(r,0)=\rho(z,0)$, and we do the same in case of subscripts like
$r_{k}\leftrightarrow R_{k}\,$.

######  Theorem 3.4.

We have the following, as $R=\rho(z,0)\to\infty\,$.
(A)  If $\lambda\in\mathbb{C}\setminus(-\infty\,,-\frac{1}{4}\,]$ then

$\Phi_{n}(z\,|\,\lambda)\sim\frac{c(\lambda)}{n!\,\bigl{(}2s(\lambda)\bigr{)}^{n}}\,R^{n}\,\exp\Bigl{(}\bigl{(}s(\lambda)-1/2\bigr{)}R\Bigr{)}\,,$

where $c(\lambda)\neq 0$.
(B)  If $\lambda=-\frac{1}{4}$ then

$\Phi_{n}(z\,|\,\lambda)\sim\frac{2}{(2n+1)!\,\pi}\,R^{2n+1}\,\exp(-R/2)\,.$

In particular, for every $n\geqslant 1$ and
$\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$ there is
$0<r_{n,\lambda}<1$ such that

$\Phi_{n}(z\,|\,\lambda)\neq 0\quad\text{for all }\;z\in\mathbb{D}\;\text{
with }\;|z|\geqslant r_{n,\lambda}\,.$

For $n=0$, we set $r_{0,\lambda}=0$.

###### Proof.

For the parameter $0<r<1$, the even function

(16)
$P_{r}(\phi)=P(r,e^{\mathfrak{i}\,\phi})=\frac{1-r^{2}}{1+r^{2}-2r\cos\phi}\,,\quad\phi\in[-\pi\,,\pi]$

is strictly decreasing in $\phi\in[0\,,\pi]$. It attains its maximum in $0$
with value $P_{r}(0)=(1+r)/(1-r)=e^{R}$. We write

(17)
$\frac{P_{r}(\phi)}{P_{r}(0)}=\frac{1}{1+\tau^{2}\sin^{2}(\phi/2)}\quad\text{with}\quad\tau=\tau_{r}=\frac{2\sqrt{r}}{1-r}\sim
P_{r}(0)=e^{R}\;\text{ as }\;r\to 1\,.$

_Proof of_ (A). In this case, $\Re(s)>0$, where $s=s(\lambda)$. Then

$\displaystyle\frac{n!\,(2s)^{n}\,\Phi_{n}(r\,|\,\lambda)}{e^{(s+1/2)R}\,R^{n}}$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{P_{r}(\phi)^{s+1/2}\,\bigl{(}\log
P_{r}(\phi)\bigr{)}^{n}}{P_{r}(0)^{s+1/2}\,\bigl{(}\log
P_{r}(0)\bigr{)}^{n}}\,d\phi$
$\displaystyle=\frac{2}{\tau\pi}\int_{0}^{\pi/2}\frac{\tau}{(1+\tau^{2}\sin^{2}\phi)^{s+1/2}}\biggl{(}\frac{R-\log\bigl{(}1+\tau^{2}\sin^{2}\phi\bigr{)}}{R}\biggr{)}^{n}\,d\phi\,.$

We choose $0<a<\dfrac{2\Re(s)}{2\Re(s)+1}$ and decompose the last integral
into the two parts

$\displaystyle\int_{0}^{\tau^{-a}}$ plus
$\displaystyle\int_{\tau^{-a}}^{\pi/2}$.

As for the second integral,

(18)
$\biggl{|}\frac{R-\log\bigl{(}1+\tau^{2}\sin^{2}\phi\bigr{)}}{R}\biggr{|}\leqslant
1$

Furthermore,

(19) $\displaystyle(\tau^{2}\sin^{2}\phi)^{\Re(s)+1/2}$
$\displaystyle\geqslant(\tau^{2}\sin^{2}\tau)^{\Re(s)+1/2}$
$\displaystyle\sim\tau^{(1-a)(2\Re(s)+1)}\quad\text{for
}\;\phi\in[\tau^{-a}\,,\pi/2]\;\text{ as }\;\tau\to\infty\,,$

and we see that the second integral tends to $0$ by the choice of $a$.

As for the first integral in the decomposition, as $\tau$, hence $R$, tend to
infinity, we have

(20)
$(1+\tau^{2}\sin^{2}\phi)^{s+1/2}\sim(1+\tau^{2}\phi^{2})^{s+1/2}\quad\text{and}\quad\frac{R-\log\bigl{(}1+\tau^{2}\sin^{2}\phi\bigr{)}}{R}\sim
1-\frac{1+\tau^{2}\phi^{2}}{\log\tau}\,.$

Therefore, with the substitution $x=\tau\phi$, by dominated convergence as
$\tau\to\infty$, we obtain

$\displaystyle\frac{2}{\pi}\int_{0}^{\tau^{-a}}\frac{\tau}{(1+\tau^{2}\sin^{2}\phi)^{s+1/2}}$
$\displaystyle\biggl{(}\frac{R-\log\bigl{(}1+\tau^{2}\sin^{2}\phi\bigr{)}}{R}\biggr{)}^{n}$
$\displaystyle\sim\frac{2}{\pi}\int_{0}^{\tau^{-a}}\frac{\tau}{(1+\tau^{2}\phi^{2})^{s+1/2}}\biggl{(}1-\frac{\log\bigl{(}1+\tau^{2}\phi^{2}\bigr{)}}{\log\tau}\biggr{)}^{n}\,d\phi$
$\displaystyle=\frac{2}{\pi}\int_{0}^{\tau^{1-a}}\frac{1}{(1+x^{2})^{s+1/2}}\biggl{(}1-\frac{\log\bigl{(}1+x^{2}\bigr{)}}{\log\tau}\biggr{)}^{n}\,dx$
(21)
$\displaystyle\to\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{(1+x^{2})^{s+1/2}}\,dx=c(\lambda),$

recalling that $s=s(\lambda)$. This proves (A).
_Proof of_ (B). In this case, $s=0$, and

$\displaystyle\frac{(2n)!\,\Phi_{n}(r|-\tfrac{1}{4})}{e^{R/2}\,R^{2n}}$
$\displaystyle\sim\frac{2}{\pi}\,\int_{0}^{\pi/2}\frac{1}{(1+\tau^{2}\sin^{2}\phi)^{1/2}}\biggl{(}1-\frac{\log\bigl{(}1+\tau^{2}\sin^{2}\phi\bigr{)}}{\log\tau}\biggr{)}^{2n}\,d\phi\,.$

This time, we decompose the last integral into the two parts

$\displaystyle\int_{0}^{1/\log\tau}$ plus
$\displaystyle\int_{1/\log\tau}^{\pi/2}$.

By using the analogue of (20), the first integral is asymptotically equivalent
to

$\int_{0}^{1/\log\tau}\frac{1}{(1+\tau^{2}\,\phi^{2})^{1/2}}\biggl{(}1-\frac{\log\bigl{(}1+\tau^{2}\,\phi^{2}\bigr{)}}{\log\tau}\biggr{)}^{2n}\,d\phi\quad\text{as
}\;\tau\to\infty\,.$

We now substitute
$x=\dfrac{\log\bigl{(}1+\tau^{2}\,\phi^{2}\bigr{)}}{\log\tau}$, and observe
that the upper integration limit $1/\log\tau$ transforms into

$b_{\tau}=\frac{\log\bigl{(}1+(\tau/\log\tau)^{2}\bigr{)}}{\log\tau}\sim
2\,,\quad\text{as }\;\tau\to\infty\,.$

Thus, the latter integral becomes

$\frac{\log\tau}{2\tau}\int_{0}^{b_{\tau}}(1-x)^{2n}\frac{dx}{\bigl{(}1-\tau^{-x}\bigr{)}^{1/2}}\sim\frac{\log\tau}{\tau}\cdot\frac{1}{2n+1}\,.$

By (18), the second integral is bounded by

$\int_{1/\log\tau}^{\pi/2}\frac{d\phi}{\tau\sin\phi}=\frac{1}{\tau}\Bigl{(}-\log\,\tan\frac{1}{2\log\tau}\Bigr{)}\sim\frac{\log\tau}{\tau}\cdot\frac{\log(2\log\tau)}{\log\tau}\,,\quad\text{as
}\;\tau\to\infty\,.$

This leads to the asymptotic behaviour of (B). ∎

For $\lambda\in\mathbb{C}\setminus(-\infty\,,-\frac{1}{4}\,]$, we define the
_associated real eigenvalue_ $\lambda^{*}\in[-\frac{1}{4}\,,+\infty)$ by the
equation

(22) $s(\lambda^{*})=\Re\bigl{(}s(\lambda)\bigr{)}.$

We shall also need the function defined as

(23)
$|\Phi|_{n}(z\,|\,\lambda)=\int_{\partial\mathbb{D}}\bigl{|}P_{n}(z,\xi\,|\,\lambda)\bigr{|}\,d\xi\,,\quad
z\in\mathbb{D}\,.$

The same computations as in the proof of Theorem 3.4 with this modified
integrand lead to the following.

######  Proposition 3.5.

We have the following.
(A)  If $\lambda\in\mathbb{C}\setminus(-\infty\,,-\frac{1}{4}\,]$ and
$s=s(\lambda)$ then, as $R=\rho(z,0)\to\infty\,$

$|\Phi|_{n}(z\,|\,\lambda)\sim
C_{n}(\lambda)\,R^{n}\,\exp\Bigl{(}\bigl{(}\Re(s)-1/2\bigr{)}R\Bigr{)}\,,$

as $\;R=\rho(z,0)\to\infty\,$, where
$\;C_{n}(\lambda)=\dfrac{c(\lambda^{*})}{n!\,(2|s|)^{n}}\;$ with
$c(\lambda^{*})$ according to (3).
(B)  If $\lambda=-\frac{1}{4}$ then
$\;|\Phi|_{n}(z\,|\,\lambda)=\Phi_{n}(z\,|\,\lambda)\;$ with asymptotics given
by Theorem 3.4.B.

Regarding the zeroes of the higher order polyspherical functions, we have
$P_{n}(0,\xi\,|\,\lambda)=\Phi_{n}(0\,|\,\lambda)=0$ for all
$\lambda\in\mathbb{C}$ and $n\geqslant 1$. For the following, recall that

$J_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(\cos\phi)^{n}\,d\phi=\begin{cases}\displaystyle\frac{1}{2^{n}}{n\choose
n/2}\,,&\text{if $n$ is even,}\\\\[11.38092pt] 0\,,&\text{if $n$ is
odd.}\end{cases}$

######  Lemma 3.6.

Let $n\geqslant 1$. Then we have the following as $r\to 0$.
(A)  If $\lambda\in\mathbb{C}\setminus\\{-\frac{1}{4}\\}$ and $s=s(\lambda)$,

$\Phi_{n}(r\,|\,\lambda)\sim\begin{cases}\dfrac{1}{\bigl{(}\frac{n}{2}!\bigr{)}^{2}\,(2s)^{n}}\,r^{n},&\text{if
$n$ is even,}\\\\[15.0pt]
\dfrac{1}{\frac{n+1}{2}!\,\frac{n-1}{2}!\,(2s)^{n-1}}\,r^{n+1},&\text{if $n$
is odd.}\end{cases}$

(B)  If $\lambda=-\frac{1}{4}$ then

$\Phi_{n}(r\,|\,\lambda)\sim\frac{1}{(n!)^{2}}\,r^{2n}.$

###### Proof.

In the sequel all $\mathfrak{o}(r^{k})$ are uniform in
$\phi\in(-\pi\,,\,\pi]$, as $r\to 0$. Short computations show that

$\displaystyle P(r,e^{\mathfrak{i}\,\phi})$
$\displaystyle=1+2r\,\cos\phi+\mathfrak{o}(r)\quad\text{and}\quad$
$\displaystyle\log P(r,e^{\mathfrak{i}\,\phi})$
$\displaystyle=2r\,\cos\phi+2r^{2}(\cos^{2}\phi-1)+\mathfrak{o}(r^{2})\,,$

so that

(24) $\displaystyle P(r,e^{\mathfrak{i}\,\phi})^{s+1/2}\,\bigl{(}\log
P(r,e^{\mathfrak{i}\,\phi})\bigr{)}^{n}$
$\displaystyle\qquad=\Bigl{(}1+(2s+1)\,r\,\cos\phi+\mathfrak{o}(r)\Bigr{)}$
$\displaystyle\qquad\quad\times\Bigl{(}2^{n}r^{n}\,\cos^{n}\phi+n2^{n+1}r^{n+1}(\cos^{n+1}\phi-\cos^{n-1}\phi)\Bigr{)}+\mathfrak{o}(r^{n+1})$
$\displaystyle\qquad=2^{n}r^{n}\,\cos^{n}\phi+2^{n}r^{n+1}\Bigl{(}(2s+1+n)\cos^{n+1}\phi-n\,\cos^{n-1}\phi\Bigr{)}+\mathfrak{o}(r^{n+1})$

Applying $\frac{1}{2\pi}\int_{-\pi}^{\pi}$ with respect to $\phi$, using the
above formula for $J_{n}\,$, and normalising as in Proposition 2.6, we obtain
the proposed asymptotic behaviour near $0$. ∎

While we do not know much about the zeroes of $\Phi_{n}(r\,|\,\lambda)$ in
general, we have the following for real eigenvalues.

######  Proposition 3.7.

For real $\lambda\geqslant-\frac{1}{4}$ and all $n\in\mathbb{N}$, we have
$\Phi_{n}(r\,|\,\lambda)>0$ for $0<r<1$.

###### Proof.

The statement is clear for even $n$, as well as for all $n$ when
$\lambda=-\frac{1}{4}$. Fix $r>0$ and consider

$F_{n}(s)=\int_{\partial\mathbb{D}}\bigl{(}\log
P(r,\xi)\bigr{)}^{n}\,P(r,\xi)^{s+1/2}\,d\xi\,,$

a function of $s\geqslant 0$. For $s=s(\lambda)$, one has
$\Phi_{n}(r\,|\,\lambda)=F_{n}(s)$ when $\lambda>-\frac{1}{4}$. Then for $s>0$

$\frac{d}{ds}F_{2n+1}(s)=F_{2n+2}(s)>0\,,$

so that $F_{2n+1}(s)$ is strictly increasing in $s\in[0\,,\,\infty)$. To
complete the proof it is enough to show that $F_{2n+1}(0)=0$, or equivalently,
that

$f_{2n+1}(z)=\int_{\partial\mathbb{D}}\bigl{(}\log
P(z,\xi)\bigr{)}^{2n+1}\,\sqrt{P(z,\xi)}\,d\xi=0\quad\text{for all
}\;z\in\mathbb{D}\,.$

We show this by induction on $n$. For $n=1$, Lemma 2.5 yields that
$(\mathfrak{L}+\frac{1}{4}I)f_{1}=0$ (recalling that $s=0$). Since $f_{1}$ is
radial, i.e., it depends only on $r=|z|$, it must be a constant multiple of
the spherical function $\Phi(z\,|\,-\frac{1}{4})$. Moreover, as $f_{1}(0)=0$,
that constant factor must be $0$, that is, $f_{1}\equiv 0$ on $\mathbb{D}$.
Now suppose that $n\geqslant 1$ and that we have already that $f_{2n-1}\equiv
0$ on $\mathbb{D}$. Once more, from Lemma 2.5, we get

$(\mathfrak{L}+\tfrac{1}{4}I)f_{2n+1}=2n(2n+1)\,f_{2n-1}\equiv 0\,$

so that also $f_{2n+1}$ must be a constant multiple of
$\Phi(z\,|\,-\frac{1}{4})$. As above, we get $f_{2n+1}\equiv 0$ on
$\mathbb{D}$. ∎

For $\lambda\in\mathbb{C}\setminus\mathbb{R}$, from Theorem 3.4 we know at
least that $\Phi_{n}(z\,|\,\lambda)\neq 0$ when $|z|$ is sufficiently close to
$1$.

##  4\. Polyharmonic kernels

######  Definition 4.1.

Let $n\in\mathbb{N}_{0}\,$ and $\lambda\in\mathbb{C}$. The _$n^{\text{th}}$
generalised Poisson transform_ of an analytic functional (or measure) $\nu$ on
$\partial\mathbb{D}$ is the function

$\operatorname{\boldsymbol{\Pi}}_{n,\lambda}\nu(z)=\int_{\partial\mathbb{D}}P_{n}(z,\xi\,|\,\lambda)\,d\nu(\xi)\,,\quad
z\in\mathbb{D}\,,$

where $P_{n}$ is given by Proposition 2.6. If $\nu$ is an absolutely
continuous measure with density function $g(\xi)$ with respect to the
normalised Lebesgue measure $\mathsf{m}$ on the circle, then we write
$\operatorname{\boldsymbol{\Pi}}_{n,\lambda}g(z)$ for the resulting transform.
When $n=0$, we just write

$\operatorname{\boldsymbol{\Pi}}_{\lambda}=\operatorname{\boldsymbol{\Pi}}_{0,\lambda}\quad\text{and}\quad\operatorname{\boldsymbol{\Pi}}=\operatorname{\boldsymbol{\Pi}}_{0,0}\,.$

Our aim is to consider boundary value problems for polyharmonic functions. The
most basic one is the Dirichlet problem: given a continuous function $g$ on
$\partial\mathbb{D}$, look for a harmonic function $h$ on $\mathbb{D}$ which
provides a continuous extension of $g$ to the closed disk
$\mathbb{D}\cup\partial\mathbb{D}$, the hyperbolic compactification. The
solution is unique and well known,
$h(z)=\operatorname{\boldsymbol{\Pi}}g(z)\,$, the classical Poisson transform.

If we consider $\lambda$-harmonic functions with
$0\neq\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, then we cannot
proceed in the same way, using $P(z,\xi)^{s(\lambda)+1/2}$. Indeed, if we take
$g\equiv 1$ on $\partial\mathbb{D}$, then its $\lambda$-Poisson transform is
$h(z)=\Phi(z\,|\,\lambda)$, which tends to $0$ as $|z|\to 1$ when $s=0$ or
$0<\Re(s)<1/2$, and to $\infty$ in absolute value when $\Re(s)>1/2$. It is
well-known that in this case one should normalise, and the natural candidate
is $\Phi(z\,|\,\lambda)$, see e.g. [27] and [21]. (Note that this
normalisation is not feasible when $\lambda\in(-\infty\,,\,-\frac{1}{4})$
because of the zeroes of the spherical function which accumulate at
$\partial\mathbb{D}$.)

######  Definition 4.2.

For $n\in\mathbb{N}_{0}\,$ and
$\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, the _normalised
polyharmonic kernel_ is

$\mathcal{K}_{n,\lambda}(z,\xi)=\frac{P_{n}(z,\xi\,|\,\lambda)}{\Phi_{n}(z\,|\,\lambda)}\,,\quad\xi\in\partial\mathbb{D}\,,\;z\in\mathbb{D}\,,\;|z|\geqslant
r_{n,\lambda}\,,$

where $r_{n,\lambda}$ is as in Theorem 3.4.

Thus,

$\int_{\partial\mathbb{D}}\mathcal{K}_{n,\lambda}(z,\xi)\,d\nu(\xi)=\frac{\operatorname{\boldsymbol{\Pi}}_{n,\lambda}\nu(z)}{\Phi_{n}(z\,|\,\lambda)}\,.$

######  Lemma 4.3.

There is a constant $\widetilde{C}(\lambda)$ for which the kernels of
Definition 4.2 satisfy

$\bigl{|}\mathcal{K}_{n,\lambda}(z,\xi)\bigr{|}\leqslant\widetilde{C}(\lambda)\,\mathcal{K}_{0,\lambda^{*}}(z,\xi)$

where $\lambda^{*}\geqslant-\frac{1}{4}$ is given by (22) and $|z|\geqslant
r_{n,\lambda}\,$.

###### Proof.

Note that
$|\mathfrak{h}(re^{\mathfrak{i}\,\alpha},\xi)|\leqslant\mathfrak{h}(r,1)=R$,
where (recall) $r=|z|$ and $R=\rho(z,0)=\rho(r,0)=\log\dfrac{1+r}{1-r}$. By
Theorem 3.4, as $r=|z|\to 1$,

$\bigl{|}\Phi(z\,|\,\lambda)\bigr{|}\sim\frac{|c(\lambda)|}{c(\lambda^{*})}\,\Phi(z\,|\,\lambda^{*}).$

Note also that by Theorem 3.4,

$\bigl{|}\Phi_{n}(z\,|\,\lambda)\bigr{|}\sim\begin{cases}\bigl{|}\Phi(z\,|\,\lambda)\bigr{|}\,R^{n}\,\dfrac{1}{n!\bigl{(}2|s(\lambda)|\bigr{)}^{n}}\,,&\text{if
}\;\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})\,,\\\\[12.0pt]
\bigl{|}\Phi(z\,|\,\lambda)\bigr{|}\,R^{2n}\,\dfrac{1}{(2n+1)!}\,,&\text{if
}\;\lambda=-\frac{1}{4}\;(=\lambda^{*})\,,\end{cases}$

as $r\to 1$. Therefore, there is $\widetilde{C}(\lambda)>0$ (depending on $n$)
such that

$\widetilde{C}(\lambda)\bigl{|}\Phi_{n}(z\,|\,\lambda)\bigr{|}\geqslant\begin{cases}\Phi(z\,|\,\lambda^{*})\,R^{n}\,,&\text{if
}\;\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})\,,\\\\[12.0pt]
\Phi(z\,|\,\lambda^{*})\,R^{2n}\,,&\text{if
}\;\lambda=-\frac{1}{4}\,,\end{cases}$

for all $r\geqslant r_{n,\lambda}\,$. It follows that
$\bigl{|}\mathcal{K}_{n,\lambda}(z,\xi)\bigr{|}$ is bounded above by

$\displaystyle\frac{P(z,\xi)^{\Re(s(\lambda))+1/2}R^{n}}{\bigl{|}\Phi_{n}(z\,|\,\lambda)\bigr{|}}$
$\displaystyle\leqslant\widetilde{C}(\lambda)\,\frac{P(z,\xi)^{s(\lambda^{*})+1/2}}{\Phi(z\,|\,\lambda^{*})},\quad\text{if
}\;\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4}\,]\,,$ and by
$\displaystyle\frac{P(z,\xi)^{1/2}R^{2n}}{\Phi_{n}(z|-\tfrac{1}{4})}$
$\displaystyle\leqslant\widetilde{C}(-\tfrac{1}{4})\,\frac{P(z,\xi)^{1/2}}{\Phi(z|-\tfrac{1}{4})},\quad\text{if
}\;\lambda=-\frac{1}{4}\,,$

which proves our claim. ∎

######  Remark 4.4.

(a) If $\lambda>-\frac{1}{4}$ then, by Lemma 3.6,
$\mathcal{K}_{n,\lambda}(z,\xi)$ is well-defined for all
$z\in\mathbb{D}\setminus\\{0\\}$. When $n$ is even then, in view of (24), it
extends continuously to $z=0$, and we can set $r_{n,\lambda}=0$ for even $n$.

However, when $n$ is odd, again in view of (24),
$\mathcal{K}_{n,\lambda}(z,\xi)$ has a pole at $z=0$ unless
$\xi=\pm\mathfrak{i}\,$. Thus, for Lemma 4.3 we need to work with
$r_{n,\lambda}=\varepsilon$ for odd $n$, where $\varepsilon>0$ is arbitrary
but fixed.
(b) If $\lambda=-\frac{1}{4}$ then $\mathcal{K}_{n,\lambda}(z,\xi)$ extends
continuosly to $z=0$ for all $n$, and we can always use
$r_{n,-\frac{1}{4}}=0$.

Note that $\mathcal{K}_{0,\lambda^{*}}(z,\cdot)\,d\mathsf{m}$ is a probability
measure on $\partial\mathbb{D}$ for each $z\in\mathbb{D}$.

Another look at the proof of Theorem 3.4 yields the following.

######  Lemma 4.5.

Let $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, $s=s(\lambda)$
and $n\in\mathbb{N}_{0}\,$. Then

$\displaystyle\lim_{r\to
1}\mathcal{K}_{n,\lambda}(r,e^{\mathfrak{i}\,\psi})=0\qquad\text{uniformly
for}$ $\displaystyle|\psi|\in$
$\displaystyle\begin{cases}[2\tau^{-a}\,,\,\pi]\,,&\text{if
}\;\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})\,,\quad\text{where
}\;0<a<\dfrac{2\Re(s)}{2\Re(s)+1}\,,\\\ [2(\log\tau)^{-a}\,,\,\pi]\,,&\text{if
}\;\lambda=-\frac{1}{4}\,,\quad\text{where }\;0<a<1\,.\end{cases}$

###### Proof.

In view of Lemma 4.3, we only need to prove this for $n=0$ and real
$\lambda\geqslant-\frac{1}{4}$. In this case, it is well known except maybe
for the fact that usually the lower bound for $|\psi|$ is required to be a
positive constant, while here it tends to $0$ as $r\to 1$.

First, look at case (A) of Theorem 3.4. With $\tau$ as in (17) and
$c(\lambda)$ as in (3), we have as $r\to 1$

$\frac{P(r,e^{\mathfrak{i}\,\psi}|\lambda)}{\Phi(r\,|\,\lambda)}\sim\frac{P_{r}(\psi)^{s+1/2}}{c(\lambda)\,P_{r}(0)^{s+1/2}\,e^{-R}}\sim\frac{1}{c(\lambda)}\,\frac{\tau}{\bigl{(}1+\tau^{2}\sin^{2}(\psi/2)\bigr{)}^{s+1/2}}.$

By (19) and (20), this tends to $0$ uniformly in the stated range.

In case (B) of Theorem 3.4, as $r\to 1$, that is, $\tau\to\infty$,

$\frac{P(r,e^{\mathfrak{i}\,\psi}|-\frac{1}{4})}{\Phi(r\,|\,-\frac{1}{4})}\sim\frac{\pi}{2}\,\frac{P_{r}(\psi)^{1/2}}{R\,e^{-R}\,P_{r}(0)^{1/2}}\sim\frac{\pi}{2}\,\frac{\tau/\log\tau}{\bigl{(}1+\tau^{2}\sin^{2}(\psi/2)\bigr{)}^{1/2}}.$

Again, this tends to $0$ uniformly in the stated range. ∎

The kernels are also rotation invariant:
$\mathcal{K}_{n,\lambda}(e^{\mathfrak{i}\,\alpha}z,e^{\mathfrak{i}\,\alpha}\xi)=\mathcal{K}_{n,\lambda}(z,\xi)$.
This fact and the last two lemmas yield the following by well-known methods.

######  Proposition 4.6.

Let $n\in\mathbb{N}_{0}\,$ and
$\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$. For a measurable
function $g:\partial\mathbb{D}\to\mathbb{C}$, resp. a complex Borel measure
$\nu$ on $\partial\mathbb{D}$, let

$f(z)=\operatorname{\boldsymbol{\Pi}}_{n,\lambda}g(z)\,,\quad\text{resp.}\quad
f(z)=\operatorname{\boldsymbol{\Pi}}_{n,\lambda}\nu(z)\,.$

Then the following properties hold.

  1. (i)

If $g\in C(\partial\mathbb{D})$ then
$\;\;\displaystyle\lim_{z\to\xi}\frac{f(z)}{\Phi_{n}(z\,|\,\lambda)}=g(\xi)\;\;$
for all $\xi\in\partial\mathbb{D}$ and uniformly as $|z|\to 1$.

  2. (ii)

If $g\in L^{p}(\partial\mathbb{D})$ ($1\leqslant p<\infty$) then
$\;\;\displaystyle\lim_{r\to
1}\frac{f(r\xi)}{\Phi_{n}(r\,|\,\lambda)}=g(\xi)\;\;$ for almost every
$\xi\in\partial\mathbb{D}$ and in $L^{p}(\partial\mathbb{D})$.

  3. (iii)

If $g\in L^{\infty}(\partial\mathbb{D})$ then $\;\;\displaystyle\lim_{r\to
1}\frac{f(r\xi)}{\Phi_{n}(r\,|\,\lambda)}=g(\xi)\;\;$ in the weak*-topology of
$L^{\infty}(\partial\mathbb{D})$.

  4. (iv)

If $\nu$ is a finite Borel measure on $\partial\mathbb{D}$ then the measures
$\;\;\dfrac{f(r\xi)}{\Phi_{n}(r\,|\,\lambda)}\,d\xi\;\;$ converge to $\nu$ in
the weak*-topology, as $r\to 1$.

The classical reference is Zygmund [37, chapter 17, Theorems 1.20 & 1.23]. See
also Stein and Weiss [34, §I.1, Theorems 1.18 & 1.25 and §II.2] and Garnett
[9, Theorem 3.1]. In these references, the results are presented for the upper
half space, resp. half plane and carry over to the disk model of the
hyperbolic plane.

##  5\. Dirichlet and Riquier problem at infinity

Reconsider statement (i) of Proposition 4.6. It says that for any
$g\in\mathcal{C}(\partial\mathbb{D})$, the function

(25)
$f(z)=\frac{\operatorname{\boldsymbol{\Pi}}_{n,\lambda}g(z)}{\Phi_{n}(z\,|\,\lambda)}$

is $\lambda$-polyharmonic of order $n+1$ which provides a continuous extension
of $g$ to $\\{z\in\overline{\mathbb{D}}:|z|\geqslant r_{n,\lambda}\\}\,$. When
$n\geqslant 1$ in (25) then we cannot expect that the given $f$ is the unique
function with this property. Indeed, for example also

$\lim_{z\to\xi}\frac{f(z)+\Phi_{k}(z\,|\,\lambda)}{\Phi_{n}(z\,|\,\lambda)}=g(\xi)\quad\text{when
}\;0\leqslant k<n\,.$

However, when $n=0$ and we are considering $\lambda$-harmonic functions, this
is the solution of the $\lambda$-Dirichlet problem, valid on all of
$\overline{\mathbb{D}}$ since $\Phi(\cdot\,|\,\lambda)$ has no zeroes in
$\mathbb{D}$. It is well-known to be unique when $\lambda=0$. The extension to
_real_ $\lambda\geqslant-\frac{1}{4}$ with normalisation by
$\Phi(z\,|\,\lambda)$ is well understood via the maximum principle applied to
the kernels $\mathcal{K}_{0,\lambda}(z,\xi)$ of Definition 4.2, which are
probability kernels with respect to $\mathsf{m}$ for real $\lambda$. When
$\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$ is _complex_ , we
can still prove uniqueness, via a different technique in place of the standard
one.

######  Theorem 5.1.

Let $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$ and
$g\in\mathcal{C}(\partial\mathbb{D})$. Then
$h(z)=\operatorname{\boldsymbol{\Pi}}_{\lambda}g(z)$ is the _unique_
$\lambda$-harmonic function for which

(26) $\lim_{\mathbb{D}\ni
z\to\xi}\frac{h(z)}{\Phi(z\,|\,\lambda)}=g(\xi)\quad\text{for every
}\;\xi\in\partial\mathbb{D}.$

###### Proof.

We know that the given function $h$ is a solution of the $\lambda$-Dirichlet
problem, i.e. it satisfies (26). In order to show uniqueness, it is enough to
prove that the constant function $0$ is the only solution, when $g\equiv 0$ on
$\partial\mathbb{D}$. In other words, we assume that $h$ is a
$\lambda$-harmonic function on $\mathbb{D}$ with

$\lim_{\mathbb{D}\ni z\to\xi}\frac{h(z)}{\Phi(z\,|\,\lambda)}=0\quad\text{for
every }\;\xi\in\partial\mathbb{D}\,,$

and we have to show that $h\equiv 0$.

Let
$\overset{\mathchoice{}{}{\mbox{\raisebox{-2.15277pt}[0.0pt][0.0pt]{$\scriptstyle\frown$}}}{\mbox{\raisebox{-1.50696pt}[0.0pt][0.0pt]{$\scriptscriptstyle\frown$}}}}{h}$
be the spherical average of $h$ around $0$,

$\overset{\mathchoice{}{}{\mbox{\raisebox{-2.15277pt}[0.0pt][0.0pt]{$\scriptstyle\frown$}}}{\mbox{\raisebox{-1.50696pt}[0.0pt][0.0pt]{$\scriptscriptstyle\frown$}}}}{h}(z)=\frac{1}{2\pi}\int_{-\pi}^{\pi}h(|z|e^{\mathfrak{i}\,\phi})\,d\phi\,.$

In particular,
$\overset{\mathchoice{}{}{\mbox{\raisebox{-2.15277pt}[0.0pt][0.0pt]{$\scriptstyle\frown$}}}{\mbox{\raisebox{-1.50696pt}[0.0pt][0.0pt]{$\scriptscriptstyle\frown$}}}}{h}(0)=h(0)$.
Then
$\overset{\mathchoice{}{}{\mbox{\raisebox{-2.15277pt}[0.0pt][0.0pt]{$\scriptstyle\frown$}}}{\mbox{\raisebox{-1.50696pt}[0.0pt][0.0pt]{$\scriptscriptstyle\frown$}}}}{h}$
is also $\lambda$-harmonic, and it is rotation-invariant. Now, up to
multiplication with constants, $\Phi(z\,|\,\lambda)$ is the unique
$\lambda$-harmonic function which is rotation-invariant. See e.g. [8] for this
well-known fact. Therefore

$\overset{\mathchoice{}{}{\mbox{\raisebox{-2.15277pt}[0.0pt][0.0pt]{$\scriptstyle\frown$}}}{\mbox{\raisebox{-1.50696pt}[0.0pt][0.0pt]{$\scriptscriptstyle\frown$}}}}{h}(z)=h(0)\,\Phi(z\,|\,\lambda)\,.$

On the other hand, the function on the closed disk which is
$h/\Phi(\cdot\,|\lambda)$ in the interior and $0$ on the boundary is
continuous, whence uniformly continuous, so that

$\lim_{|z|\to 1}\frac{h(z)}{\Phi(z\,|\,\lambda)}=0\quad\text{uniformly in
}\;z.$

Since $\Phi(z\,|\,\lambda)$ only depends on $|z|$, also

$\lim_{|z|\to
1}\frac{\overset{\mathchoice{}{}{\mbox{\raisebox{-2.15277pt}[0.0pt][0.0pt]{$\scriptstyle\frown$}}}{\mbox{\raisebox{-1.50696pt}[0.0pt][0.0pt]{$\scriptscriptstyle\frown$}}}}{h}(z)}{\Phi(z\,|\,\lambda)}=0.$

We conclude that $h(0)=0$.

Now let $z_{0}\in\mathbb{D}$ be arbitrary. Then there is an isometry $\gamma$
of the Poincaré disk (a Möbius transform) such that $\gamma\,0=z_{0}\,$. The
isometries commute with the hyperbolic Laplacian, whence also the function
$h_{\gamma}(z)=h(\gamma z)$ is $\lambda$-harmonic. If $|z|\to 1$ then also
$|\gamma z|\to 1$. Therefore also $h(\gamma z)/\Phi(\gamma z\,|\,\lambda)\to
0$ as $|z|\to 1$. Now Theorem 3.4 implies

$\frac{\Phi(\gamma
z\,|\,\lambda)}{\Phi(z\,|\,\lambda)}\sim\exp\Bigl{(}(s-1/2)\bigl{(}\rho(\gamma\,z,o)-\rho(z,o)\bigr{)}\Bigr{)}\quad\text{as
}\;|z|\to 1\,.$

This is bounded, since

$\bigl{|}\rho(\gamma\,z,o)-\rho(z,o)\bigr{|}=\bigl{|}\rho(z,\gamma^{-1}o)-\rho(z,o)\bigr{|}\leqslant\rho(\gamma^{-1}o,o).$

We infer that

$\frac{h_{\gamma}(z)}{\Phi(z\,|\,\lambda)}=\frac{h(\gamma z)}{\Phi(\gamma
z\,|\,\lambda)}\,\frac{\Phi(\gamma z\,|\,\lambda)}{\Phi(z\,|\,\lambda)}\to
0\quad\text{as }\;|z|\to 1\,.$

We can now apply the above argument to $h_{\gamma}$ and its spherical average,
and conclude that $h(z_{0})=0$. This is true for every $z_{0}\in\mathbb{D}$. ∎

The next Lemma is in preparation of the Riquier problem for
$\lambda$-polyharmonic functions.

######  Lemma 5.2.

For $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, let $f$ be
$\lambda$-polyharmonic of order $n+1$ on $\mathbb{D}$ and such that the
$\lambda$-harmonic function $h=(\mathfrak{L}-\lambda\,I)^{n}f$ satisfies

$\lim_{\mathbb{D}\ni
z\to\xi}\frac{h(z)}{\Phi(z\,|\,\lambda)}=g(\xi)\quad\text{for every
}\;\xi\in\partial\mathbb{D}\,$

where $g\in\mathcal{C}(\partial\mathbb{D})$. Then

$f(z)=\operatorname{\boldsymbol{\Pi}}_{n,\lambda}g(z)+f_{*}(z)\,,$

where $f_{*}$ is $\lambda$-polyharmonic of order $n$.

###### Proof.

This is very similar to the inductive argument in the proof of Theorem 1.1. It
follows from Theorem 5.1 that
$h=\operatorname{\boldsymbol{\Pi}}_{\lambda}g=:h_{g}\,$. Write
$f_{g}=\operatorname{\boldsymbol{\Pi}}_{n,\lambda}g\,$. By Lemma 5.2,

$(\mathfrak{L}-\lambda\,I)^{n}f_{g}=h_{g}=(\mathfrak{L}-\lambda\,I)^{n}f\,.$

Therefore $(\mathfrak{L}-\lambda\,I)^{n}f_{*}=0$. ∎

In the setting of the Euclidean Laplacian $\Delta$ on a bounded domain, the
Riquier problem asks for solutions of $\Delta^{n}f=0$ with prescribed boundary
data $g_{k}=\Delta^{k}f$ for $k=0,\dots,n-1$. This is not applicable to the
hyperbolic setting of $(\mathfrak{L}-\lambda\,I)^{n}$, even when $\lambda=0$,
because in any case, the quotient of lower and higher order polyspherical
functions tends to zero at the boundary. For this reason, we propose a
different formulation.

######  Definition 5.3.

Let $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$ and
$g_{0}\,,\dots\,,g_{n-1}\in\mathcal{C}(\partial\mathbb{D})$. Then a solution
of the associated _Riquier problem at infinity_ is a polyharmonic function

$f=f_{0}+\dots+f_{n-1}$

of order $n$, where each $f_{k}$ is $\lambda$-polyharmonic of order $k+1$ and

$\lim_{z\to\xi}\frac{(\mathfrak{L}-\lambda\,I)^{k}f_{k}(z)}{\Phi(z\,|\,\lambda)}=g_{k}(\xi)\quad\text{for
every }\;\xi\in\partial\mathbb{D}\,.$

Note that by Remark 3.3, the denominator in the last quotient is always non-
zero.

######  Corollary 5.4.

A solution of the Riquier problem as in Definition 5.3 is given by

$f_{k}(z)=\operatorname{\boldsymbol{\Pi}}_{k,\lambda}g_{k}(z)\,,\quad
k=0,\dots,n-1\,.$

One also has for every $\xi\in\partial\mathbb{D}$ and
$k\in\\{0\,,\dots,n-1\\}$

$\lim_{z\to\xi}\frac{f_{k}(z)}{\Phi_{k}(z\,|\,\lambda)}=g_{k}(\xi)\quad\text{and}\quad\lim_{z\to\xi}\frac{f_{j}(z)}{\Phi_{k}(z\,|\,\lambda)}=0\;\text{
for }\,j<k.$

##  6\. A Fatou theorem for polyharmonic functions

######  Definition 6.1.

(i) For $0<\delta\leqslant\pi$, consider the arc
$B_{\delta}=\\{e^{\mathfrak{i}\,\phi}\colon|\phi|<\delta\\}\subset\partial\mathbb{D}\,$,
and for $\zeta=e^{\mathfrak{i}\,\alpha}$ consider the rotated arc
$B_{\delta}(\zeta)=\zeta
B_{\delta}=\\{e^{\mathfrak{i}\,\phi}\colon|\phi-\alpha|<\delta\\}$ with
measure (normalised arc length)
$\mathsf{m}\bigl{(}B_{\delta}(\zeta)\bigr{)}=\delta/\pi$. The
_Hardy–Littlewood maximal operator_ is defined on functions $g\in
L^{1}(\partial\mathbb{D})$ as

$\mathcal{M}g(\zeta)=\sup\left\\{\frac{1}{\mathsf{m}\bigl{(}B_{\delta}(\zeta)\bigr{)}}\int_{B_{\delta}(\zeta)}|g(\xi)|\,d\xi\;\colon\;0<\delta\leqslant\pi\right\\}.$

(ii) Let $[0,\zeta]$ be the line segment with endpoints $0$ and
$\zeta=e^{\mathfrak{i}\,\alpha}$ in the unit disc $\mathbb{D}$, and for
$a\geqslant 0$ consider the _admissible region_ , or _tubular domain_

$\Gamma_{a}(\zeta)=\Bigl{\\{}z\in\mathbb{D}\colon\rho\bigl{(}z,[0,\zeta]\bigr{)}\leqslant
a\Bigr{\\}}.$

The _non-tangential (tubular) maximal operator of width $a\geqslant 0$_ is
defined on functions $g\in L^{1}(\partial\mathbb{D})$ as

$\mathfrak{M}_{a}^{(n,\lambda)}g(\zeta)=\sup\left\\{\,\biggl{|}\int_{\partial\mathbb{D}}\mathcal{K}_{n,\lambda}(z,\xi)\,g(\xi)\,d\xi\biggr{|}\;\colon\;z\in\Gamma_{a}(\zeta)\,,\;|z|\geqslant
r_{n,\lambda}\right\\},$

where $\mathcal{K}_{n,\lambda}$ is the kernel introduced in Definition 4.2 and
$r_{n,\lambda}$ is as in Theorem 3.4.
(iii) With the same ingredients as in (ii), the _enlarged-admissible region_
is

$\Gamma_{(a)}(\zeta)=\Bigl{\\{}z\in\mathbb{D}\colon\rho\bigl{(}z,[0,\zeta]\bigr{)}\leqslant
a+\log\rho(z,0)\Bigr{\\}}.$

The _extended maximal operator of width $a$_ is defined on functions $g\in
L^{1}(\partial\mathbb{D})$ as

$\mathfrak{M}_{(a)}^{(n,\lambda)}g(\zeta)=\sup\left\\{\,\biggl{|}\int_{\partial\mathbb{D}}\mathcal{K}_{n,\lambda}(z,\xi)\,g(\xi)\,d\xi\biggr{|}\colon
z\in\Gamma_{(a)}(\zeta)\,,\;|z|\geqslant r_{n,\lambda}\right\\}.$

Both $\Gamma_{a}(\zeta)$ and $\Gamma_{(a)}(\zeta)$ are (Euclidean) convex
subsets of $\mathbb{D}$ which touch the boundary $\partial\mathbb{D}$ only at
$\zeta$. In the Euclidean metric, the “tube” $\Gamma_{a}(\zeta)$ is conical at
$\zeta$ with Stolz angle $2\arctan(\sinh a)$. On the other hand, the boundary
curve of the larger domain $\Gamma_{(a)}(\zeta)$ is tangent to
$\partial\mathbb{D}$ at $\zeta$.

The following fact is well-known (see, for instance, [22, Corollary of Lemma
1.1]).

######  Proposition 6.2.

The Hardy–Littlewood maximal operator is of weak type $(1,1)$ and strong type
$(p,p)$ for $1<p\leqslant\infty$.

Our aim is to prove the following.

######  Proposition 6.3.

For every $a\geqslant 0$ and $n\in\mathbb{N}_{0}$, there is a constant
$C(a,\lambda)>0$ such that for all $g\in L^{1}(\partial\mathbb{D})$

$\displaystyle\mathfrak{M}_{a}^{(n,\lambda)}\,g$ $\displaystyle\leqslant
C(a,\lambda)\,\mathcal{M}g\,,\quad\text{if
}\;\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})\,,$
$\displaystyle\mathfrak{M}_{(a)}^{(n,\lambda)}\,g$ $\displaystyle\leqslant
C(a,\lambda)\,\mathcal{M}g\,,\quad\text{if }\;\lambda=-\frac{1}{4}\,.$

Note that the inequality in the critical case is stronger, since the
underlying domain is enlarged. By Lemma 4.3, it is sufficient to prove
Proposition 6.3 for $n=0$ (i.e., for $\mathfrak{L}-\lambda\,I$) and real
$\lambda\geqslant-\frac{1}{4}$, showing that

(27) $\displaystyle\mathfrak{M}_{a}^{(0,\lambda)}\,g$ $\displaystyle\leqslant
C(a,\lambda)\,\mathcal{M}g\,\quad\text{if
}\;\lambda>-\frac{1}{4}\,,\quad\text{and}\quad$
$\displaystyle\mathfrak{M}_{(a)}^{(0,\lambda)}\,g$ $\displaystyle\leqslant
C(a,\lambda)\,\mathcal{M}g\,\quad\text{if }\;\lambda=-\frac{1}{4}\,.$

That is, we only need to work with the standard spherical functions
$\Phi(z\,|\,\lambda)$.

In this case, the proof is practically folklore when $\lambda>-\frac{1}{4}$,
see [37], [20] or [27], while the extended version for $\lambda=-\frac{1}{4}$
is contained in a note by Sjögren [33]. However, the extent to which the proof
is folklore is such that it is hard to find a simple version for the
hyperbolic Laplacian, and the note [33] is also not very easily accessible.
Therefore, for the reader’s convenience, we include a simple proof (more
direct via partial integration than typical versions which decompose the unit
circle in countably many arcs).

###### Proof of Proposition 6.3.

1) As mentioned, it is sufficient to work with $n=0$ and $\lambda$ real. Then
$\mathcal{K}_{n,\lambda}(z,\xi)\,d\xi$ is a probability measure, and we have
rotation invariance:
$\;\mathcal{K}_{n,\lambda}(e^{\mathfrak{i}\,\alpha}z,e^{\mathfrak{i}\,\alpha}\xi)=\mathcal{K}_{n,\lambda}(z,\xi)\,$.
Thus, it is enough to prove the inequality at $\zeta=1$.
2) For the rest of this proof, we set

(28) $t=s(\lambda)+1/2\quad\text{and}\quad R^{*}=\begin{cases}1\,,&\text{if
}\;\lambda>-\frac{1}{4}\,,\\\\[5.69046pt] R\,,&\text{if
}\;\lambda=-\frac{1}{4}\,.\end{cases}$

Thus $t>1/2$ when $\lambda>-\frac{1}{4}$, and $t=1/2$ when
$\lambda=-\frac{1}{4}$. We can subsume both admissible regions $\Gamma_{a}(1)$
and $\Gamma_{(a)}(1)$ under

$\bigl{\\{}z=r\,e^{\mathfrak{i}\,\alpha}\in\mathbb{D}:\rho\bigl{(}z,[0,1]\bigr{)}\leqslant
a+\log R^{*}\bigr{\\}}\,,$

where, as always, $R=\log\dfrac{1+r}{1-r}\,$. Elementary computations with
hyperbolic distance yield for $z=r\,e^{\mathfrak{i}\,\alpha}$ with $r<1$ and
$|\alpha|\leqslant\pi$ that for any $a\geqslant 0$,

(29) $\rho\bigl{(}z,[0,1]\bigr{)}\leqslant a+\log
R^{*}\iff\underbrace{\frac{1-r^{2}}{2r}\,\sinh(a+\log
R^{*})}_{\displaystyle=:K_{r}}\geqslant\begin{cases}|\sin\alpha|\,,&\text{if
}\;|\alpha|<\pi/2\\\ 1\,,&\text{if }\;|\alpha|\geqslant\pi/2\,.\end{cases}$

If $|\alpha|\geqslant\pi/2\,$, this is the same as $R\leqslant a+\log
R^{*}\,$. Thus, there is $r_{a}\in(0\,,\,1)$ such that when $r\geqslant r_{a}$
then the second of the above cases is excluded, and that $\sinh(a+R_{a})>1$,
where $R_{a}=\rho(r_{a},0)$.
3) In view of (17), we need to show that for $z=r\,e^{\mathfrak{i}\,\alpha}$
in the range of (29),

(30)
$\frac{e^{R}}{2\pi\,R^{*}}\int_{-\pi}^{\pi}|g(e^{\mathfrak{i}\,\phi})|\,\frac{1}{\bigl{(}1+\frac{4r}{(1-r)^{2}}\sin^{2}\frac{\phi-\alpha}{2}\bigr{)}^{t}}\,d\phi\leqslant
C(a,\lambda)\,\mathcal{M}g(1).$

If $r\in[0\,,\,r_{a})$ then the kernel of the above integral is bounded, so
that the estimate is immediate with a suitable value of $C(a,\lambda)$. So we
now consider the case $r\geqslant r_{a}\,$, where we know that
$|\alpha|<\pi/2$, and $\alpha\to 0$ as $r\to 1$. For $\phi\in(-\pi\,,\,\pi]$,
we have $|\phi-\alpha|/2\leqslant 3\pi/4$, whence there is
$\kappa\in(0\,,\,1)$ such that
$|\sin\frac{\phi-\alpha}{2}|\geqslant\kappa\,|\frac{\phi-\alpha}{2}|$, as well
as $|\sin\alpha|\geqslant\kappa\,|\alpha|$. We may assume that $r_{a}$ is such
that $2K_{r}\leqslant\pi$ for $r\geqslant r_{a}\,$. Now the left hand side of
(30) is bounded above by

$\frac{e^{R}}{2\pi\,R^{*}}\int_{-\pi}^{\pi}|g(e^{\mathfrak{i}\,\phi})|\,\frac{1}{\bigl{(}1+\frac{r}{\kappa^{2}(1-r)^{2}}(\phi-\alpha)^{2}\bigr{)}^{t}}\,d\phi\,.$

We decompose the last integral into
$\int_{|\phi|<2K_{r}/\kappa}+\int_{|\phi|\geqslant 2K_{r}/\kappa}\,$: for the
first part,

$\displaystyle\frac{e^{R}}{2\pi\,R^{*}}$
$\displaystyle\int_{-2K_{r}/\kappa}^{2K_{r}/\kappa}|g(e^{\mathfrak{i}\,\phi})|\,\frac{1}{\bigl{(}1+\frac{r}{\kappa^{2}(1-r)^{2}}(\phi-\alpha)^{2}\bigr{)}^{t}}\,d\phi$
$\displaystyle\leqslant\frac{e^{R}}{2\pi\,R^{*}}\int_{-2K_{r}/\kappa}^{2K_{r}/\kappa}|g(e^{\mathfrak{i}\,\phi})|\,d\phi\leqslant\frac{4K_{r}\,e^{R}}{2\pi\,\kappa\,R^{*}}\,\mathcal{M}g(1)\leqslant\frac{4e^{a}}{2\pi\,\kappa\,r_{a}}\,\mathcal{M}g(1)\,.$

For the second part, we set

$H(\phi)=\int_{0}^{\phi}\bigl{(}|g(e^{\mathfrak{i}\,\psi})|+|g(e^{-\mathfrak{i}\,\psi})|\bigr{)}\,d\psi\,,\quad\phi\in[-\pi\,,\,\pi].$

Then

$H(\phi)\leqslant 2\phi\,\mathcal{M}g(1)\,.$

Observe that, by (29) and the choices of $r$ and $\kappa$, we have
$|\alpha|\leqslant K_{r}/\kappa$, so that $|\phi|\geqslant 2K_{r}/\kappa$
implies $|\phi-\alpha|\geqslant|\phi|/2$. We get, with
$c_{\kappa}=(2\kappa)^{2t}/(r_{a}\,\pi)$, and using partial integration

$\displaystyle\frac{e^{R}}{2\pi\,R^{*}}$
$\displaystyle\int_{2K_{r}/\kappa\leqslant|\phi|\leqslant\pi}|g(e^{\mathfrak{i}\,\phi})|\,\frac{1}{\bigl{(}1+\frac{r}{\kappa^{2}(1-r)^{2}}(\phi-\alpha)^{2}\bigr{)}^{t}}\,\,d\phi$
$\displaystyle\leqslant
c_{\kappa}\,\frac{(1-r)^{2t-1}}{R^{*}}\,\int_{2K_{r}/\kappa}^{\pi}\bigl{(}|g(e^{\mathfrak{i}\,\phi})|+|g(e^{-\mathfrak{i}\,\phi})|\bigr{)}\frac{1}{\phi^{2t}}\,d\phi$
$\displaystyle=c_{\kappa}\,\frac{(1-r)^{2t-1}}{R^{*}}\left(\frac{H(\pi)}{\pi^{2t}}-\frac{H(2K_{r}/\kappa)}{(2K_{r}/\kappa)^{2t}}+2t\int_{2K_{r}/\kappa}^{\pi}\frac{H(\phi)}{\phi^{2t+1}}\,d\phi\right)$
$\displaystyle\leqslant
c_{\kappa}\,\frac{(1-r)^{2t-1}}{R^{*}}\left(\frac{2\pi}{\pi^{2t}}+4t\int_{2K_{r}/\kappa}^{\pi}\frac{1}{\phi^{2t}}\,d\phi\right)\mathcal{M}g(1)$
$\displaystyle\leqslant\begin{cases}c_{\kappa}\left(\dfrac{2}{\pi^{2t-1}}+\dfrac{4t}{2t-1}\Bigl{(}\dfrac{\kappa}{\sinh
a}\Bigr{)}^{2t-1}\right)\mathcal{M}g(1)\,,&\text{if }\;t>1/2,\\\\[14.22636pt]
c_{\kappa}\left(\dfrac{2+2\log\pi}{R_{a}}+2\right)\mathcal{M}g(1)\,,&\text{if
}\;t=1/2\,,\end{cases}$

because $-\log(2K_{r}/\kappa)\leqslant R$ by the choices of $r_{a}$ and
$\kappa$. This concludes the proof. ∎

As a consequence, we have the following convergence theorem (in the discrete
setting of trees, see [21, Theorem 1, Theorem 3] for $\lambda$-harmonic
functions, and [30, Theorem 4.6] for regular trees and $\lambda$-polyharmonic
functions).

######  Definition 6.4.

A function $f\colon\mathbb{D}\to\mathbb{C}$ _converges admissibly, ( $\equiv$
non-tangentially)_ resp. _enlarged-admissibly_ at a boundary point
$\xi\in\partial\mathbb{D}$ if, for every $a\geqslant 0$, the limit

$\lim_{\Gamma_{a}(\xi)\ni
z\to\xi}f(z)\,,\quad\text{resp.}\quad\lim_{\Gamma_{(a)}(\xi)\ni z\to\xi}f(z)$

exists and is finite.

######  Theorem 6.5 (Fatou theorem for $\lambda$-polyharmonic functions).

Let $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$ and $\nu$ a
Borel measure on $\partial\mathbb{D}$. Then the normalised
$\lambda$-polyharmonic function

$\frac{\operatorname{\boldsymbol{\Pi}}_{n,\lambda}\nu(z)}{\Phi_{n}(z\,|\,\lambda)}\qquad(\,|z|\geqslant
r_{n,\lambda}\,)$

converges admissibly at $\mathsf{m}$-almost every point in
$\partial\mathbb{D}$. The limit is the Radon–Nykodim derivative
$d\nu^{\text{ac}}/d\mathsf{m}$, where (recall) $\mathsf{m}$ is the normalised
arc length measure, and $\nu^{\text{ac}}$ is the absolutely continuous part of
$\nu$.

If $\lambda=-\frac{1}{4}$, convergence is even enlarged-admissible.

###### Proof.

It is well-known that we can assume $\nu$ to be absolutely continuous with
respect to $\mathsf{m}$. We prove this fact for the sake of completeness: if
$\nu$ is singular with respect to $\mathsf{m}$, then there is an
$\mathsf{m}$-null set $E\subset\partial\mathbb{D}$ such that
$\nu(\mathbb{D}\setminus E)=0$. For every $\varepsilon>0$, let
$\\{A_{j}\subset\partial\mathbb{D}\colon j=1,\dotsc,n\\}$ be a finite
collection of open arcs covering $E$ with
$\mathsf{m}\bigl{(}\bigcup_{j=1}^{n}A_{j}\bigr{)}<\varepsilon\,$. Let
$U=\bigcup_{j=1}^{n}A_{j}$. Then

$\left|\frac{f(z)}{\Phi_{n}(z\,|\,\lambda)}\right|\leqslant\int_{U}\frac{\bigl{|}P(z,\xi\,|\,\lambda)\,\mathfrak{h}(z,\xi\,|\,\lambda)^{n}\bigr{|}}{|\Phi_{n}(z\,|\,\lambda)|}\,d\nu(\xi)\,,$

which tends to $0$ when $z\to\zeta\in\partial\mathbb{D}\setminus U$ by Lemma
4.5.

So we can assume that $\nu$ is absolutely continuous with respect to the
normalised Lebesgue measure $\mathsf{m}$, with $g=d\nu/d\mathsf{m}\in
L^{1}(\partial\mathbb{D},\mathsf{m})$. The rest of the proof is the continuous
analogue of [21, Theorems 1 and 3] and [30, Theorem 4.6]. In brief, we can
find a sequence $(g_{k})$ in $\mathcal{C}(\partial\mathbb{D})$ such that
$\|g-g_{k}\|<1/2^{k}$, and by propositions 6.2 and 6.3,

$\sum_{k}\mathsf{m}\bigl{[}\mathfrak{M}_{a}^{n,\lambda}(g-g_{k})\geqslant\varepsilon\bigr{]}<\infty\,.$

By the Borel-Cantelli Lemma,

$\lim_{k\to\infty}\mathfrak{M}_{a}^{n,\lambda}(g-g_{k})(\xi)=0\quad\text{for
$\mathsf{m}$-almost every }\;\xi\in\partial\mathbb{D}.$

We can now apply Proposition 4.6(i) to each of the $g_{k}$ to get the proposed
convergence at all those points $\xi\in\partial\mathbb{D}\,$. ∎

Let us call a $\lambda$-polyharmonic function $f$ _regular_ if all the
analytic functionals $\nu_{k}$ in the boundary representation of Theorem 1.1,
or better (equivalently) formula (14), are complex Borel measures on
$\partial\mathbb{D}$. The fact that
$\Phi_{k}(r\,|\,\lambda)/\Phi_{n}(r\,|\,\lambda)\to 0$ for $k<n$, when $r\to
1$, yields the following.

######  Corollary 6.6.

For $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-\frac{1}{4})$, let $f$ be a
regular $\lambda$-polyharmonic function of order $n+1$ and $\nu_{n}$ the
highest-order representing measure of $f$ in (14). Then
$f(z)/\Phi_{n}(z\,|\,\lambda)$ converges admissibly (resp. enlarged-
admissibly, if $\lambda=-\frac{1}{4}$) at almost every
$\xi\in\partial\mathbb{D}$. The limit function is the Radon–Nykodim derivative
$d\nu_{n}^{\text{ac}}/d\mathsf{m}\,$.

Compare the following with [31].

######  Corollary 6.7.

For $\lambda\in\mathbb{C}\setminus(-\infty\,,\,-1/4)$, the only radial
$\lambda$-polyharmonic functions of order $n+1$ on $\mathbb{D}$ are the linear
combinations of the $\lambda$-polyspherical functions
$\Phi_{0},\dotsc,\Phi_{n}$.

##  7\. Examples, complements, open problems

In this section, we study examples of (hyperbolically) harmonic and bi-
harmonic functions which are not regular. We focus on the case $\lambda=0$,
where of course $s(0)+1/2=1$.

First, for $0\leqslant r<1$ and $\xi\in\partial\mathbb{D}$,

$P(r,\xi)=\frac{1-r^{2}}{1+r^{2}-r\xi-r/\xi}=\sum_{n\in\mathbb{Z}}r^{|n|}\,\xi^{n}.$

Thus, for any fixed $z=r\,e^{\mathfrak{i}\,\alpha}\in\mathbb{D}$, the Fourier
expansion of the Poisson kernel in the boundary variable
$\xi=e^{\mathfrak{i}\,\phi}$ is

$P(r\,e^{\mathfrak{i}\,\alpha},e^{\mathfrak{i}\,\phi})=\sum_{n\in\mathbb{Z}}r^{|n|}\,e^{-\mathfrak{i}\,n\alpha}\,e^{\mathfrak{i}\,n\phi}.$

Next, we want do determine the Fourier expansion of $-\mathfrak{h}(z,\xi)=\log
P(z,\xi)$. By (7),

$\log
P(r,\xi)=\log(1-r^{2})-\log(\xi-r)-\log\Bigl{(}\frac{1}{\xi}-r\Bigr{)}=\log(1-r^{2})+\sum_{0\neq
n\in\mathbb{Z}}\frac{r^{|n|}}{|n|}\,\xi^{n}\,,$

so that we have the Fourier expansion

$\log
P(r\,e^{\mathfrak{i}\,\alpha},e^{\mathfrak{i}\,\phi})=\log(1-r^{2})+\sum_{0\neq
n\in\mathbb{Z}}\frac{r^{|n|}}{|n|}\,e^{-\mathfrak{i}\,n\alpha}\,e^{\mathfrak{i}\,n\phi}.$

Now we can write

$P(r,\xi)\log P(r,\xi)=\sum_{n\in\mathbb{Z}}r^{|n|}\,d_{|n|}(r)\,\xi^{n}\,,$

since the coefficients of $\xi^{n}$ and $\xi^{-n}$ must coincide. For
$n\geqslant 0$,

$\displaystyle r^{n}\,d_{n}(r)$ $\displaystyle=r^{n}\log(1-r^{2})+\sum_{0\neq
k\in\mathbb{Z}}r^{|n-k|}\frac{r^{|k|}}{|k|}$
$\displaystyle=r^{n}\log(1-r^{2})+\underbrace{\sum_{k=1}^{\infty}\frac{r^{n+2k}}{k}}_{\displaystyle=-r^{n}\log(1-r^{2})}+\sum_{k=1}^{n}\frac{r^{n}}{k}+\sum_{k=n+1}^{\infty}\frac{r^{2k-n}}{k}\,,$

whence

(31)
$d_{n}(r)=\sum_{k=1}^{n}\frac{1}{k}+\sum_{k=1}^{\infty}\frac{r^{2k}}{k+n}\,.$

Note that $d_{n}(r)$ is in fact a function of $r^{2}$. We have

$d_{n}(r)-d_{n-1}(r)=\sum_{k=0}^{\infty}\frac{(1-r^{2})r^{2k}}{k+n}\geqslant
0\,,$

so that

$d_{n}(r)=d_{0}(r)+(1-r^{2})\sum_{k=0}^{\infty}\frac{r^{2k}}{k}\sum_{m=1}^{n}\frac{k}{k+n}\,,$

whence

(32) $\log\frac{1}{1-r^{2}}=d_{0}(r)\leqslant
d_{n}(r)\leqslant\displaystyle\bigl{(}1+n(1-r^{2})\bigr{)}\,\log\frac{1}{1-r^{2}}\,.$

We conclude that the Fourier expansion of the biharmonic Poisson kernel is

$P_{1}(r\,e^{\mathfrak{i}\,\alpha},e^{\mathfrak{i}\,\phi})=P(r\,e^{\mathfrak{i}\,\alpha},e^{\mathfrak{i}\,\phi})\,\log
P(r\,e^{\mathfrak{i}\,\alpha},e^{\mathfrak{i}\,\phi})=\sum_{n\in\mathbb{Z}}r^{|n|}\,d_{|n|}(r)\,e^{-\mathfrak{i}\,n\alpha}\,e^{\mathfrak{i}\,n\phi}.$

######  Definition 7.1.

Let $h(z)$ be a harmonic function on $\mathbb{D}$ and let $\nu^{h}$ be the
analytic functional on $\partial\mathbb{D}$ in its Poisson representation. The
_associated biharmonic function_ is

$f_{h}(z)=\int_{\partial\mathbb{D}}P_{1}(z,\xi)\,d\nu^{h}(\xi)\,.$

Now, let us start with an analytic, whence harmonic function

(33)
$h(z)=\sum_{n=0}^{\infty}h_{n}\,z^{n}\,,\quad\limsup_{n\to\infty}|h_{n}|^{1/n}\leqslant
1$

on $\mathbb{D}$. We compute the Fourier coefficients
$\nu_{n}^{h}=\nu^{h}(e^{-\mathfrak{i}\,n\phi})$ of the corresponding analytic
functional $\nu^{h}$, that is

$h(z)=\int_{\partial\mathbb{D}}P(z,\xi)\,d\nu^{h}(\xi)=\sum_{n\in\mathbb{Z}}r^{|n|}\,e^{-\mathfrak{i}\,n\alpha}\,\overline{\nu_{n}^{h}}\,,$

where $z=r\,e^{-\mathfrak{i}\,\alpha}$. Comparison with

$h(z)=\sum_{n=0}^{\infty}h_{n}\,r^{n}\,e^{\mathfrak{i}\,n\alpha}$

yields

(34) $\nu_{n}^{h}=\begin{cases}\overline{h\,}_{\\!-n}\,,&\text{if
}\;n\leqslant 0\,,\\\ 0\,,&\text{if }\;n>0\,.\end{cases}$

Then the associated biharmonic function is

$f_{h}(z)=\sum_{n=0}^{\infty}h_{n}\,d_{n}(|z|)\,z^{n}\,.$

Below we shall use the fact that (32) implies

(35)
$\left|f_{h}(z)-\log\frac{1}{1-r^{2}}h(z)\right|\leqslant(1-r^{2})\sum_{n=1}^{\infty}n\,|h_{n}|\,r^{n}$

with $\log\frac{1}{1-r^{2}}\sim R$ as $r=|z|\to 1$. Moreover, the real part

$\Re
h(z)=\sum_{n=0}^{\infty}h_{n}\,r^{n}\,\frac{h_{n}e^{\mathfrak{i}\,n\alpha}+\overline{h\,}_{\\!n}e^{-\mathfrak{i}\,n\alpha}}{2}$

is harmonic, and the Fourier coefficients of the corresponding analytic
functional $\nu^{\Re h}$ are

$\nu_{n}^{\Re h}=\begin{cases}\Re h_{0}\,,&\text{if }\;n=0,\\\\[3.0pt]
h_{n}/2\,,&\text{if }\;n>0,\\\\[3.0pt] \overline{h\,}_{\\!-n}/2\,,&\text{if
}\;n<0\,.\end{cases}$

(Indeed, every real harmonic function arises as the real part of an analytic
function.) It follows that

$f_{\Re h(z)}=-\int_{\partial\mathbb{D}}P_{1}(z,\xi)\,d\nu^{\Re h}(\xi)=\Re
f_{h}(z)\,.$

######  Discussion 7.2.

Euclidean and hyperbolic harmonic functions coincide. We know that if a
harmonic function $h$ is bounded then its representing analytic functional is
in fact a measure with bounded density with respect to the Lebesgue measure on
$\partial\mathbb{D}$. Euclidean biharmonic functions are of the form
$h_{1}(z)+r^{2}h_{2}(z)$ with $h_{1}\,,h_{2}$ harmonic. The Euclidean
biharmonic kernel is $r^{2}P(z,\xi)$, and no normalisation is in place. In
this context, Mazalov [25] provides an example of a _bounded_ biharmonic
function which has no radial limits at the boundary. The function is of the
form $(1-r^{2})\,\Re h(z)$ where $h(z)$ is as in (33) with a lacunary sequence
of coefficients $h_{n}\geqslant 0$. It is bounded, while $h(z)$ is not
bounded. What would be an analogue in the hyperbolic case ? Since the
biharmonic kernel needs to be normalised by $\Phi_{1}(z|0)\sim
R\sim|\log(1-r)|$ as $R\to\infty$ (recall that $R=\rho(z,0)$ and $r=|z|$), we
would look for a hyperbolically biharmonic function $f(z)$ such that
$f(z)/|\log(1-r)|$ is bounded but has no radial limits.

Now suppose that $h(z)$ is such that $h_{n}\geqslant 0$ and such that
$f(z)/|\log(1-r)|$ is bounded. Then we have by (32) for $|z|=r$

$|h(z)|\leqslant
h(r)\leqslant\sum_{n=0}^{\infty}h_{n}\frac{d_{n}(r)}{d_{0}(r)}r^{n}\leqslant\frac{f_{h}(r)}{|\log(1-r)|}\,.$

Hence also the harmonic function $h(z)$ is bounded, the (common) representing
analytic functional of $h(z)$ and $f_{h}(z)$ is a measure with bounded density
and $f_{h}(z)/|\log(1-r)|$ has admissible limits almost everywhere on
$\partial\mathbb{D}$ by Theorem 6.5. It is also worth mentioning that
$f_{h}(z)/|\log(1-r)|-h(z)$ has admissible limit $0$ almost everywhere.∎

######  Question 7.3.

Let $\nu$ be an analytic functional on $\partial\mathbb{D}$ and

$f(z)=\operatorname{\boldsymbol{\Pi}}_{1,0}\nu(z)=\int_{\mathbb{D}}P_{1}(z,\xi)\,d\nu(\xi)$

be such that $f(z)/R$ is bounded. Is it true that $\nu$ must be a measure with
bounded density ?

On an example of Borichev.333We acknowledge literature hints of Fausto Di
Biase (Pescara) which led us to the work of Borichev. We are particularly
grateful to Alexander Borichev (Marseille) who indicated this clever example
to us.
The following interesting example is closely related to the results and
methods in the paper by Borichev et al. [4].

######  Proposition 7.4.

There is a harmonic function $h(z)$ on $\mathbb{D}$ such that
$h(z)/|\log(1-r)|$ is bounded, but has no radial limits at any point of
$\partial\mathbb{D}$.

Of course, this function is also $\mathfrak{L}$-biharmonic, so it is an
example of a biharmonic function in the sense of Discussion 7.2.

We now provide the details of the proof of Proposition 7.4.

The set $K=\bigl{\\{}\frac{t+1}{6}\,e^{3\pi
t\mathfrak{i}\,}:t\in[0\,,\,1]\bigr{\\}}$ is a spiral winding one and a half
times around the origin with radius $r$ varying from $1/6$ to $1/3$. Using
Runge’s approximation theorem as for example stated by Rudin [29, Thm. 13.7],
one finds a homogeneous polynomial

(36)
$p(z)=\sum_{j=1}^{s}p_{j}\,z^{j}\quad\text{with}\quad|p(z)-5/3|<2/3\;\text{
for all }\;z\in K\,.$

It is of course harmonic, and on $\mathbb{D}$,

$|p(z)|\leqslant B\,|z|\,,\quad\text{where}\quad B=\sum_{j=1}^{s}|p_{j}|\,.$

We construct the function

(37) $h(z)=\sum_{k=1}^{\infty}k!\,p\bigl{(}z^{2^{k!}}\bigr{)}\,,\quad
z\in\mathbb{D}\,.$

Note that $p(0)=0$.

######  Lemma 7.5.

The series defining $h(z)$ converges absolutely in $\mathbb{D}$, and

$\sup_{z\in\mathbb{D}}\frac{|h(z)|}{\bigl{|}\log(1-|z|)\bigr{|}}<\infty\,.$

###### Proof.

Consider

$f(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n}=-\log(1-z).$

For real $r\in[0\,,\,1)$, the sequence $(r^{n}/n)$ is decreasing, whence

$\displaystyle f(r)$
$\displaystyle=\sum_{l=0}^{\infty}\sum_{n=2^{l}}^{2^{l+1}-1}\frac{r^{n}}{n}\geqslant\sum_{l=0}^{\infty}2^{l}\frac{r^{2^{l+1}-1}}{2^{l+1}-1}\geqslant\frac{1}{2}\sum_{l=1}^{\infty}r^{2^{l}}=\frac{1}{2}\sum_{k=1}^{\infty}\sum_{l=k!}^{(k+1)!-1}r^{2^{l}}$
$\displaystyle\geqslant\frac{1}{2}\sum_{k=1}^{\infty}\bigl{(}(k+1)!-k!\bigr{)}r^{2^{(k+1)!-1}}\geqslant\frac{1}{2}\sum_{k=1}^{\infty}\bigl{(}1-\frac{1}{k+1}\bigr{)}(k+1)!\,r^{2^{(k+1)!}}\geqslant\frac{1}{4}\sum_{k=2}^{\infty}k!\,r^{2^{k!}}\,.$

Therefore

$\sum_{k=1}^{\infty}k!\,|z|^{2^{k!}}\leqslant|z|^{2}+4\bigl{|}\log(1-|z|)\bigr{|}\leqslant
c\,\bigl{|}\log(1-|z|)\bigr{|}\,.$

Now

$|h(z)|\leqslant\sum_{j=1}^{s}|p_{j}|\sum_{k=1}^{\infty}k!\,|z|^{j\,2^{k!}}\leqslant
c\sum_{j=1}^{s}|h_{j}|\bigl{|}\log(1-|z|^{j})\bigr{|}\leqslant
c\,B\,\bigl{|}\log(1-|z|)\bigr{|},$

as stated. ∎

######  Lemma 7.6.

For every $\alpha\in(-\pi\,,\pi]$,

$\limsup_{r\to
1}\frac{|h(r\,e^{\mathfrak{i}\,\alpha})|}{\bigl{|}\log(1-r)\bigr{|}}\geqslant
1\,.$

###### Proof.

We fix $\alpha$ and $N\in\mathbb{N}$ and can find $r_{N}=r_{N}(\alpha)$ such
that

$w_{N}=r_{N}\,e^{\mathfrak{i}\,\alpha\,2^{N!}}\in K\,.$

Thus,

$1/6\leq r_{N}\leqslant
1/3\,,\quad\bigl{|}p(w_{N})-5/3|<2/3\,,\quad\text{and}\quad\Re\bigl{(}p(w_{N})\bigr{)}>1.$

Now we choose

$z_{N}=r_{N}^{1/2^{N!}}\,e^{\mathfrak{i}\,\alpha}\,,$

so that $|z_{N}|\to 1$, and we consider

$h(z_{N})-N!\,p\bigl{(}z_{N}^{2^{N!}}\bigr{)}=\sum_{k=1}^{N-1}k!\,p\bigl{(}z_{N}^{2^{k!}}\bigr{)}+\sum_{k=N+1}^{\infty}k!\,p\bigl{(}z_{N}^{2^{k!}}\bigr{)}.$

We can estimate the first sum by

$\left|\sum_{k=1}^{N-1}k!\,p\bigl{(}z_{N}^{2^{k!}}\bigr{)}\right|\leqslant
B\sum_{k=1}^{N-1}k!\leqslant 2B\,(N-1)!=\frac{2B}{N}N!\,.$

and, by setting $x_{N}=(1/3)^{N!}$ and using that $2^{n}\geqslant n$, the
second sum is majorized as follows:

$\displaystyle\left|\sum_{k=N+1}^{\infty}k!\,p\bigl{(}z_{N}^{2^{k!}}\bigr{)}\right|$
$\displaystyle\leqslant
B\,N!\sum_{k=N+1}^{\infty}\frac{k!}{N!}\,x_{N}^{2^{(k!/N!)-1}}\leqslant
B\,N!\sum_{k=N+1}^{\infty}\frac{k!}{N!}\,x_{N}^{(k!/N!)-1}$
$\displaystyle\leqslant
B\,N!\sum_{m=N+1}^{\infty}m\,x_{N}^{m-1}=B\,N!\,\underbrace{\frac{(N+1)x_{N}^{N}-Nx_{N}^{N+1}}{(1-x_{N})^{2}}}_{\displaystyle\to
0\,,\;\text{ as }\;N\to\infty}\,.$

It follows that

$\Bigl{|}h(z_{N})-N!\,\underbrace{p\bigl{(}z_{N}^{2^{N!}}}_{\displaystyle
p(w_{n})}\bigr{)}\Bigr{|}\leqslant N!\,\epsilon_{N}\,,\quad\text{where
}\;\epsilon_{N}\to 0\;\text{ as }\;N\to\infty\,.$

Now note that $1<\log(1/r_{N})<2$. Using this and the fact that for $x>0$ one
has $x-x^{2}/2\leqslant 1-e^{-x}\leqslant x$, one obtains that

$1-|z_{N}|=1-e^{-2^{-N!}\log(1/r_{N})}\begin{cases}<2\cdot 2^{-N!}\\\
>\frac{1}{2}\cdot 2^{-N!}.\end{cases}$

We deduce that $\bigl{|}\log(1-|z_{N}|)\bigr{|}\sim N!\,$, whence

$\left|\frac{h(z_{N})}{\bigl{|}\log(1-|z_{N}|)\bigr{|}}-p(w_{N})\right|\to
0\quad\text{ as }\;N\to\infty\,.$

Since $|p(w_{N})|\geqslant\Re\bigl{(}p(w_{N}))>1$, the statement is proved. ∎

######  Lemma 7.7.

$\qquad\displaystyle\lim_{N\to\infty}\sup_{|z|=1-2^{-N!\sqrt{N}}}\frac{h(z)}{\bigl{|}\log(1-|z|)\bigr{|}}=0\,.$

###### Proof.

It is sufficient to show that

$\lim_{N\to\infty}\frac{h_{0}\bigl{(}1-2^{-N!\sqrt{N}}\bigr{)}}{N!\sqrt{N}}=0\,,\quad\text{where}\quad
h_{0}(z)=\sum_{k=1}^{\infty}k!\,z^{2^{k!}}.$

Similarly as above, using the fact that
$(1-2^{-N!\sqrt{N}})^{2^{N!\sqrt{N}}}\leqslant 1/e$ and setting
$y_{N}=e^{-N!}$, we find

$\displaystyle h_{0}\bigl{(}1-2^{-N!\sqrt{N}}\bigr{)}$
$\displaystyle\leqslant\sum_{k=1}^{N}k!\,+\sum_{k=N+1}^{\infty}k!\,\bigl{(}\frac{1}{e}\bigr{)}^{2^{k!-N!\sqrt{N}}}$
$\displaystyle\leqslant
2\,N!+N!\sum_{k=N+1}^{\infty}\frac{k!}{N!}\,(y_{N})^{(k!/N!)-1}y_{N}^{1-\sqrt{N}}$
$\displaystyle\leqslant
2\,N!+N!\sum_{m=N+1}^{\infty}m\,(y_{N})^{m-1}y_{N}^{1-\sqrt{N}}$
$\displaystyle=2\,N!+N!\underbrace{\frac{(N+1)y_{N}^{N+1-\sqrt{N}}-Ny_{N}^{N+2-\sqrt{N}}}{(1-y_{N})^{2}}}_{\displaystyle\to
0\,,\;\text{ as }\;N\to\infty}.$

Divided by $N!\sqrt{N}$, this tends to $0$. ∎

Lemmas 7.5, 7.6 and 7.7 prove Proposition 7.4. With very small adaptations,
one verifies that the same statement holds for the harmonic function $\Re
h(z)$. We conclude with the following observation, which should not look too
surprising. (Recall that for $\lambda=0$, $\Phi_{2}(z|0)\sim R^{2}/2$, as
$r=|z|\to 1$.)

######  Lemma 7.8.

The biharmonic function $f_{h}$ associated with the function $h$ of (37) is
such that $f_{h}(z)/R^{2}$ is bounded and has no radial limits at any point of
$\partial\mathbb{D}$.

###### Proof.

Similarly to Lemma 7.5, for $0\leqslant r<1$,

$\displaystyle\frac{1}{1-r}\log\frac{1}{1-r}$
$\displaystyle=\sum_{n=1}^{\infty}\left(\sum_{j=1}^{n}\frac{1}{j}\right)r^{n}\geqslant\sum_{k=1}^{\infty}\left(\sum_{n=2^{k!}}^{2^{(k+1)!}-1}\sum_{j=1}^{n}\frac{1}{j}\right)r^{2^{(k+1)!}}$
$\displaystyle\stackrel{{\scriptstyle(*)}}{{\geqslant}}\tilde{c}\sum_{k=1}^{\infty}(k+1)!\,2^{(k+1)!}\,r^{2^{(k+1)!}}.$

For $(*)$, see below. We get via (35)

$\left|f_{h}(z)-\log\frac{1}{1-r^{2}}\,h(z)\right|\leqslant(1-r^{2})\sum_{j=1}^{N}j\,|p_{j}|\sum_{k=1}^{\infty}k!\,2^{k!}\,r^{j2^{k!}}\leqslant\widetilde{C}\log\frac{1}{1-r}$

for a suitable constant $\widetilde{C}>0$. The statement follows by dividing
by $\bigl{(}\log\frac{1}{1-r}\bigr{)}^{2}$ and applying Proposition 7.4.

Let us prove $(*)$:

$\displaystyle Q$
$\displaystyle:=\frac{1}{(k+1)!\,2^{(k+1)!}}\sum_{n=2^{k!}}^{2^{(k+1)!}-1}\sum_{j=1}^{n}\frac{1}{j}$
$\displaystyle\phantom{:}=\frac{1}{(k+1)!\,2^{(k+1)!}}\,2^{k!}\sum_{j=1}^{2^{k!}}\frac{1}{j}\;+\;\frac{1}{(k+1)!\,2^{(k+1)!}}\sum_{j=2^{k!}+1}^{2^{(k+1)!}-1}\frac{2^{(k+1)!}-j}{j}.$

As $k\to\infty$, the first term behaves like

$\frac{2^{k!}\,k!\,\log 2}{(k+1)!\,2^{(k+1)!}}\to 0\,,$

while the second term is

$\frac{1}{(k+1)!}\sum_{j=2^{k!}+1}^{2^{(k+1)!}-1}\Bigl{(}\frac{1}{j}-\frac{1}{2^{(k+1)!}}\Bigr{)}\to\log
2\,.$

Thus, the quotient $Q$ is bounded below by some $\tilde{c}>0$ for all $k$. ∎

## References

  * [1] Almansi, E.: _Sull’integrazione dell’equazione differenziale $\Delta^{2n}=0$._ Annali di Matematica, Serie III 2 (1899) 1–59.
  * [2] Aronszajn, N., Creese, T. M., and Lipkin, L. J.: _Polyharmonic Functions._ Oxford Math. Monographs, Oxford University Press, New York, 1983.
  * [3] Boiko, T., and Woess, W.: _Moments of Riesz measures on Poincaré disk and homogeneous tree – a comparative study._ Expositiones Math. 33, no. 3 (2015) 353–374.
  * [4] Borichev, A., Lyubarskiĭ, Yu., Malinnikova, E., and Thomas, P.: _Radial growth of functions in the Korenblum space._ Algebra i Analiz 21, no.6 (2009) 47–65; reprinted in St. Petersburg Math. J. 21, no. 6. (2010) 877–891.
  * [5] Chung, L. O.: _Biharmonic and polyharmonic principal functions._ Pacific J. Math. 86, no. 2 (1980) 437–445.
  * [6] Chung, L. O., Sario, L., and Wang, C.: _Riemannian manifolds with bounded Dirichlet finite polyharmonic functions._ Ann. Scuola Norm. Sup. Pisa Cl. Sci. 27, no. 4 (1973) 809–814.
  * [7] Cohen, J. M., Colonna, F., Gowrisankaran, K., and Singman, D.: _Polyharmonic functions on trees._ Amer. J. Math. 124, no. 5 (2002) 999–1043.
  * [8] Eymard, P.: _Le noyau de Poisson et la théorie des groupes._ Symposia Mathematica 22 (1977) 107–132.
  * [9] Garnett, J. B.: _Bounded Analytic Functions._ Revised first edition. Graduate Texts in Mathematics 236. Springer, New York, 2007.
  * [10] Gazzola, F., Grunau, H-Ch., and Sweers, G. : _Polyharmonic Boundary Value Problems._ Lecture Notes in Mathematics 1991, Springer, Berlin, 2010.
  * [11] Grellier, S., and Otal, J-P.: _Bounded eigenfunctions in the real hyperbolic space._ Int. Math. Res. Not. 62 (2005) 3867–3897.
  * [12] Helgason, S.: _Eigenspaces of the Laplacian; integral representations and irreducibility._ J. Functional Analysis 17 (1974), 328–353.
  * [13] Helgason, S.: _Geometric Analysis on Symmetric Spaces._ Mathematical Surveys and Monographs 39. American Mathematical Society, Providence, RI, 1994.
  * [14] Hille, E.: _Some problems concerning spherical harmonics._ Ark. för Mat., Astron. och Fys. 13, No. 17, 76 p. (1919).
  * [15] Hille, E.: _On the zeros of Legendre functions._ Ark. för Mat., Astron. och Fys. 17, No. 22, 16 p. (1923).
  * [16] Hörmander, L.: _The Analysis of Linear Partial Differential Operators. I._ Second edition, Springer-Verlag, Berlin, 1990.
  * [17] Hobson, E. W.: _The Theory of Spherical and Ellipsoidal Harmonics._ Cambridge Univ. Press., 1931.
  * [18] Jaming, Ph.: _Harmonic functions on classical rank one balls._ Boll. Unione Mat. Ital. Sez. B 4, no. 3 (2001) 685–702.
  * [19] Karpelevič, F. I.: _The geometry of geodesics and the eigenfunctions of the Beltrami-Laplace operator on symmetric spaces._ Trudy Moskov. Mat. Obšč. 14 (1965) 48–185 (Russian); translated as Trans. Moscow Math. Soc. 1965 (1967) 51–199.
  * [20] Korányi, A.: _Boundary behavior of Poisson integrals on symmetric spaces._ Trans. Amer. Math. Soc. 140 (1969) 393–409.
  * [21] Korányi, A., and Picardello, M. A.: _Boundary behaviour of eigenfunctions of the Laplace operator on trees._ Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 13, no. 3 (1986) 389–399.
  * [22] Korányi, A., and Vagi, S.: _Singular integrals on homogeneous spaces and some problems of classical analysis._ Ann. Scuola Norm. Sup. Pisa Cl. Sci. 25 (1971) 575–618.
  * [23] Köthe, G.: _Die Randverteilungen analytischer Funtionen._ Math. Z. 57 (1952) 13–33.
  * [24] Lebedev, N. N.: _Special Functions and their Applications._ Revised English edition Translated and edited by Richard A. Silverman. Prentice-Hall, Inc., Englewood Cliffs, N.J., 1965.
  * [25] Mazalov, M. Ya.: _On the existence of angular boundary values of polyharmonic functions in the ball._ J. Math. Sci. (N.Y.) 234, no. 3, (2018) 362–368.
  * [26] Mehler, F. G.: _Ueber eine mit den Kugel- und Cylinderfunctionen verwandte Function und ihre Anwendung in der Theorie der Elektricitätsvertheilung._ Math. Annalen 18 (1881) 161–194.
  * [27] Michelson, H. L.: _Fatou theorem for eigenfunctions of the invariant differential operators on symmetric spaces._ Trans. Amer. Math. Soc. 177 (1973) 257–274.
  * [28] Picardello, M. A., and Woess, W.: _Boundary representations of $\lambda$-harmonic and polyharmonic functions on trees._ Potential Analysis 51, no. 4 (2019) 541–561.
  * [29] Rudin, W.: _Real and Complex Analysis._ Third edition. McGraw-Hill Book Co., New York, 1987.
  * [30] Sava-Huss, E., and Woess, W.: _Boundary behaviour of $\lambda$-polyharmonic functions on regular trees._ Ann. Mat. Pura Appl. (4) bf 200, no. 1 (2021) 35–50.
  * [31] Schimming, R., and Belger, M.: _Polyharmonic radial functions on a Riemannian manifold._ Math. Nachr. 153 (1991) 207–216.
  * [32] Schlichtkrull, H.: _Hyperfunctions and Harmonic Analysis on Symmetric Spaces._ Progress in Mathematics 49. Birkhäuser, MA, 1984.
  * [33] Sjögren, P.: _Une remarque sur la convergence des fonctions propres du laplacien à valeur propre critique._ In _Théorie du Potentiel (Orsay, 1983)_ , 544–548, Lecture Notes in Math. 1096, Springer, Berlin, 1984.
  * [34] Stein, E. M., and Weiss, G.: _Introduction to Fourier Analysis on Euclidean Spaces._ Princeton University Press, Princeton, N.J., 1971.
  * [35] Sullivan, D.: _Related aspects of positivity in Riemannian geometry._ J. Differential Geom. 25, no. 3 (1987) 327–351.
  * [36] Taylor, J. C.: _The Martin compactification associated with a second order strictly elliptic partial differential operator on a manifold M._ In: _Topics in Probability and Lie Groups: Boundary Theory,_ 153–202. CRM Proceedings Lecture Notes 28, American Mathematical Society, Providence, RI, 2001.
  * [37] Zygmund, A.: _Trigonometric Series_ , 2nd edition. Cambridge Univ. Press, New York, 1959.