# Radial limits of solutions to elliptic partial differential equations

Paul M. Gauthier  Département de mathématiques et de statistique, Université
de Montréal, Montréal, Québec, Canada H3C3J7<EMAIL_ADDRESS>and  Mohammad Shirazi  Department of Mathematics and Statistics, McGill
University, Montréal, Québec, Canada, H3A 0B9<EMAIL_ADDRESS>To Professor Korevaar from the junior author, to Jaap from the senior author

###### Abstract.

For certain elliptic differential operators $L,$ we study the behaviour of
solutions to $Lu=0,$ as we tend to the boundary along radii in strictly
starlike domains in $\mathbb{R}^{n},n\geq 3.$ Analogous results are obtained
in other special domains. Our approach involves introducing harmonic line
bundles as instances of Brelot harmonic spaces and approximating continuous
functions by harmonic functions on appropriate subsets. These approximation
theorems on harmonic spaces yield interesting examples for approximation by
solutions of $Lu=0$ on some domains in $\mathbb{R}^{n}.$

First author was supported by NSERC (Canada) grant RGPIN-2016-04107.

##  1\. Introduction

On a domain $\Omega\subset\mathbb{R}^{n},\,n\geq 3,$ consider the following
partial differential operator

(1) $L=\sum_{i,j=1}^{n}a_{ij}\frac{\partial^{2}}{\partial x_{i}\partial
x_{j}}+\sum_{i=1}^{n}b_{i}\frac{\partial}{\partial x_{i}}+c.$

Following [14], we assume that $a_{ij}=a_{ji}\in C^{2,1},$ $b_{i}\in C^{1,1},$
$c\in C^{0,1}$ and $c\leq 0,$ where $C^{k,1}$ denotes the class of functions
on $\Omega$ which are $k$ times continuously differentiable, the
$k^{th}-$order partial derivatives being locally Lipschitz. We also assume
that the associated quadratic form $(a_{ij})$ is positive definite on
$\Omega,$ so $L$ is elliptic. A $C^{2}-$smooth solution of $Lu=0,$ on an open
subset $U\subset\Omega,$ will be called an $L-$harmonic function on $U$ and
solutions (in the sense of distributions) of $Lu\leq 0,$ which are lower
semicontinuous, will be called $L-$superharmonic functions on $U.$ We assume
that there is an $L-$superharmonic function $>0$ on $\Omega$ which is not
actually $L-$harmonic.

Let $\mathbb{B}^{n},$ and $S^{n-1}$ denote the unit ball, and the unit sphere
of $\mathbb{R}^{n},\,\geq 2,$ respectively. A point
$x\in\mathbb{R}^{n}\setminus\\{0\\}$ can be represented in polar coordinates
$x=(\theta,\rho),$ with $\theta\in S^{n-1}$ and $\rho\in(0,+\infty),$ via the
bijection $\mathbb{R}^{n}\setminus\\{0\\}\to
S^{m-1}\times(0,+\infty),\,x\mapsto(x/|x|,|x|).$

By a measurable set in a topological space, we mean, unless otherwise
specified, a Borel measurable set; by a measurable function between two
topological spaces, we mean a Borel measurable function and by a Borel measure
on a locally compact Hausdorff space, we mean a measure defined on the class
of Borel sets, such that the measure of compact sets is finite.

In the present paper, as a special case of our main result, Theorem 5.16, we
obtain the following result for $L-$harmonic functions.

######  Theorem 1.1.

Let $L$ be an operator of the form (1) on $\mathbb{B}^{n},\,n\geq 3,$
satisfying the above conditions. Let $F^{\prime}$ be a countable union of
closed (spherically) polar subsets of $S^{n-1}.$ Then, for every function
$\varphi$ continuous on $\mathbb{B}^{n},$ there is an $L-$harmonic function
$u$ on $\mathbb{B}^{n}$ such that

$\lim_{\rho\nearrow 1}\big{(}u-\varphi\big{)}((\theta,\rho))=0,\quad\textit{
for all }\theta\in F^{\prime}.$

Denote by $\mathbb{D}$ the unit disc in the complex plane $\mathbb{C}$ and by
$\mathbb{T}$ the unit circle $\\{z=e^{i\theta}:0\leq\theta<2\pi\\}.$ Theorem
1.1 can be considered as an analogue for solutions of elliptic partial
differential equations of the following approximation theorem of Bagemihl and
Seidel on the unit disc $\mathbb{D}$ (see formula (8) in [5]).

######  Theorem 1.2.

Let $E$ be an $F_{\sigma}$ set of first category on the unit circle
$\mathbb{T}.$ Then, for every function $\varphi$ continuous on the unit disc
$\mathbb{D},$ there is a harmonic function $u$ on $\mathbb{D}$ such that,

$\lim_{\rho\nearrow 1}\big{(}u-\varphi\big{)}\big{(}\rho
e^{i\theta}\big{)}=0,\quad\mbox{for all}\quad e^{i\theta}\in E.$

We remark that, since every set of first category is contained in an
$F_{\sigma}$ set of first category, we may drop the requirement that $F$ be an
$F_{\sigma}$ set.

Recently, we found an analogue of Theorem 1.2 for holomorphic functions of
several complex variables [18] (see also [24]). The proofs in this paper,
however, are independent of these results.

Some of the tools which we shall use were proved in the general context of
harmonic spaces in axiomatic potential theory. We bear in mind that, if $M$ is
a Riemannian manifold and $\mathcal{H}$ is the sheaf of harmonic functions on
$M,$ then $(M,\mathcal{H})$ is a harmonic space (see [6, p. xii]). These
spaces satisfy the axiom of polarity which states that a set $E\subset M$ is
polar provided that $E$ is thin at every point of $M.$

##  2\. Potential Theory

Brelot [9] developped an axiomatic potential theory on certain topological
spaces $\Omega.$ We shall restrict our attention to the case that $\Omega$ is
a (real, second countable) manifold. Suppose for a given manifold $\Omega,$ we
associate to every open subset $U\subset\Omega$ a vector space
$\mathcal{H}(U)$ of continuous functions on $U,$ which we call harmonic
functions, satisfying the following Axioms $1,2,3$ of Brelot.

Axiom 1 (sheaf axiom). If $U$ is an open subset of $\Omega$ and
$u:U\to\mathbb{R}$ is harmonic on $U,$ then the restriction of $u$ to every
open subset $V$ of $U$ is harmonic on $V$ and if, for every point $x\in U,$
the restriction of $u$ to some open neighbourhood of $x$ is harmonic, then $u$
is harrmonic on $U.$

Axiom 2. $\Omega$ has a basis of open sets which are regular for the Dirichlet
problem.

Axiom 3. If $u_{j}$ is an increasing sequence of harmonic functions on a
domain $U\subset\Omega$ and $u_{j}\to u,$ then either $u$ is harmonic on $U$
or $u\equiv+\infty.$

The pair $(\Omega,\mathcal{H})$, where $\mathcal{H}$ denotes the sheaf
$\mathcal{H}=\\{\mathcal{H}(U):U\subset\Omega\,\mbox{ is open}\\},$ is said to
be a Brelot (harmonic) space. On a harmonic space there is a natural notion of
superharmonic function and by a potential on a harmonic space, we mean a
superharmonic function $u\geq 0$ on $\Omega$ for which every harmonic minorant
$\geq 0$ vanishes identically [8]. A strong Brelot space is a Brelot space in
which there is a potential $>0.$

With the following additional domination axiom, it is possible to adapt the
greater part of classical potential theory to harmonic spaces. The original
statement of the domination axiom would require many additional definitions,
so we state an equivalent version for a strong Brelot space (see [11, p. 228].

Axiom D (domination axiom). For every locally bounded potential $p$ on
$\Omega$ and every relatively compact open set $U$ in $\Omega,$ the
(generalized) solution $H^{U}_{p}$ of the Dirichlet problem on $U$ with
boundary values $p$ is the greatest harmonic minorant of $p|_{U}.$

For the basic definitions and facts concerning potential theory, superharmonic
functions, potentials, thin sets and harmonic approximation on harmonic
spaces, see [9], [22] and [11] and for the particular case of potential theory
on Riemannian manifolds, we rely on [3] and the references therein. One of the
pleasant properties of superharmonic functions in harmonic spaces is the
following minimum principle (see [22, p. 428]).

######  Lemma 2.1.

A superharmonic function $\geq 0$ on a domain is either $>0$ or $\equiv 0.$

Notice that the constant function $0,$ is harmonic since $\mathcal{H}(\Omega)$
is a vector space. We can thus redefine a potential as a superharmonic
function $u\geq 0$ on $\Omega,$ whose greatest harmonic minorant is $0.$ Since
$0$ is harmonic, it is superharmonic and consequently $0$ is a potential,
since the greatest harmonic minorant of $0$ is itself. On Brelot spaces
$(\Omega,\mathcal{H})$ for which $0$ is not the only potential, the analogue
of classical potential theory can be developped much further. Suppose $u$ is a
potential other than $0$ on $\Omega$ and suppose $u(x)=0$ at some point
$x\in\Omega.$ Then, by Lemma 2.1, $u=0,$ which contradicts the choice of $u.$
Thus, if there is a potential $u$ which is not the constant $0,$ then $u>0.$
We have already named such a Brelot space a strong Brelot space. In other
words, a strong Brelot space is a Brelot space on which the constant $0$ is
not the only potential.

##  3\. Harmonic approximation on Riemannian manifolds

Let $M$ be a Riemannian manifold. A set $E\subset M$ is said to be thin at a
point $y$ if $y$ is not a limit point of $E$ or, when $y$ is a limit point of
$E$, if there is a superharmonic function $u$ on an open neighbourhood of $y$,
such that

$\liminf_{x\to y}u(x)>u(y),\quad x\in E.$

Let $E$ and $G$ be subsets of $M.$ We shall say that $E$ is nowhere thin in
$G$ if $E$ is thin at no point of $G.$ We recall the following simple lemma.

######  Lemma 3.1.

If $u$ is superharmonic in an open set $U\subset M$, then for each $y$ in $U$,

$\liminf_{x\to y}u(x)=u(y),\quad x\in U.$

The above lemma follows from the semi-continuity and mean-value properties for
superharmonic functions ([3, p. 69]) and implies that every open set in a
Riemannian manifold of dimension $n>1$ is nowhere thin in itself.

We shall say that a set $E\subset M$ is a polar set if there is a
superharmonic function $u$ on some open set $U$ such that $E\subset\\{x\in
U:u(x)=+\infty\\}$.

An easy argument shows that if $E$ is polar and closed in $M,$ then the open
set $M\setminus E$ is nowhere thin in $M.$ A polar set in $\mathbb{R}^{n}$ is
thin everywhere ([1, Theorem 7.2.2.]).

We recall the following interesting fact.

######  Lemma 3.2.

[6, Lemma A.1.6] If $M$ is a Riemannian manifold of dimension $m$ and $N$ is a
submanifold of dimension $n$, then $N$ is polar in $M$ if and only $n\leq
m-2.$

In particular, a point in a Riemannian manifold of dimension $n$ is polar if
and only if $n\geq 2$ and, since every subset of a polar set is polar, it
follows that there are no non-empty polar sets in $\mathbb{R}^{1}$ nor in
$S^{1}.$

Eucldian space $\mathbb{R}^{n}$ is a Riemannian manifold with the usual
distance $d_{n}$ as Riemannian metric. We consider the sphere $S^{n-1}$ as a
Riemannian manifold with the restriction of $d_{n}$ as Riemannian metric. This
induces a harmonic structure on $S^{n-1}$ and we shall sometimes add the
prefix “spherically” to the corresponding potential theoretic notions on
$S^{n-1},$ for example spherically thin or spherically polar sets.

If $\Omega$ is a locally compact Hausdorff space, we denote by $*_{\Omega},$
or simply $*$ if the context is clear, the ideal point of the one-point
compactification $\Omega^{*}=\Omega\cup\\{*\\}$ of $\Omega.$ We say that a
subset $X\subset\Omega^{*}$ is bounded if the ideal point $*_{\Omega}$ is not
in the closure of $X.$

For $F$ a closed subset of a Riemannian manifold $M.$ We define a hole of $F$
to be any bounded component of $M\setminus F.$ We denote by $\widehat{F}$ the
union of $F$ and all of its holes. Thus, $\widehat{F}$ is a closed set having
no holes.

We denote by $A(F)$ the family of continuous functions on $F$ which are
harmonic on the interiour $F^{\circ}.$ A closed set $F\subset M$ is said to be
a Mergelyan (respectively Carleman) set in $M$ if for every $u\in A(F)$, and
every $\epsilon,$ which is a positive constant (respectively, positive
continuous function on $F$) there exists a function $h$, harmonic on $M$, such
that $|h-u|<\epsilon$ on $F.$ The terminology is in honour of Mergelyan, who
showed (see [13]) that, if $K$ is a compact subset of $\mathbb{C}$ with
connected complement (i.e. $\widehat{K}=K$), then, for each function $f$
continuous on $K$ and holomorphic on $K^{\circ}$ and each positive constant
$\epsilon,$ there is a polynomial $p$ such that

$|f(z)-p(z)|<\epsilon,\quad\textit{for all}\quad z\in K$

and Carleman, who showed that, for each continuous function
$f(x),\,x\in\mathbb{R}$ and for each positive continuous function
$\epsilon(x),\,x\in\mathbb{R},$ there is an entire function
$g(z),z\in\mathbb{C},$ such that

$|f(x)-g(x)|<\epsilon(x),\,\quad\textit{ for all}\quad x\in\mathbb{R}.$

The following theorem from [4] gives a characterization of Mergelyan sets and
also of Carleman sets having no interior.

######  Theorem 3.3.

Let $F$ be a closed subset of a non-compact Riemannian manifold $M$. Then the
following are equivalent.

  * (a)

$F$ is a Mergelyan set.

  * (b)

All three of the following conditions hold:

(i) $M\setminus\widehat{F}$ and $M\setminus F^{\circ}$ are thin at the same
points of $F$.

(ii) $M^{\star}\setminus\widehat{F}$ is locally connected.

(iii) The holes of $F$ satisfy the long islands condition.

Moreover, if $F$ is a Mergelyan set and $F^{\circ}=\emptyset,$ then $F$ is a
Carleman set.

##  4\. Approximation on harmonic bundles

A continuous map $\phi:M\to N$ between two Riemannian manifolds $M$ and $N$ is
called a harmonic morphism (see [6, Def. 4.1.1]) if, for every open subset $V$
of $N$ and every harmonic function $u$ on $V,$ the composition $u\circ\varphi$
is harmonic on $\varphi^{-1}(V)$ (when non-empty). A simple example of a
harmonic morphism (see [6, Ch. 4]) is the natural projection $M\times N\to N$
from a Riemannian product.

The following quote is from [6, Ch. 12]. “The general problem of classifying
harmonic morphisms between manifolds of arbitrary dimensions remains far from
our reach at the present time; however, the problem becomes tractable when
they map from an $(n+1)$-dimensional to an $n$-dimensional manifold, so that
the regular fibres are one-dimensional.” This motivates us to consider a
harmonic line bundle, which we define to be a real line bundle
$(M,\,N,\,\pi,\,\mathbb{R}).$ Thus, for every point $x\in N,$ there is an open
neighbourhood $V_{x}\subset N$ and a homeomorphism
$\psi_{x}:\pi^{-1}(V_{x})\to V_{x}\times\mathbb{R}$ called a local
trivialization, such that $\pi_{x}\circ\psi_{x}=\pi,$ where $\pi_{x}$ is the
usual projection from $V_{x}\times\mathbb{R}$ onto $V_{x}.$ We assume that the
total space $M$ and the base space $N$ are Riemannian manifolds and that both
the projection $\pi$ and the local trivializations $\psi_{x}$ are harmonic
morphisms. We restrict our considerations to the cases where the base space
$N$ is either Euclidean space $\mathbb{R}^{n-1}$ or the unit sphere $S^{n-1}$
(in $\mathbb{R}^{n}$). Thus, locally, $M$ is homeomorphic to $\mathbb{R}^{n}.$
We warn the reader that the expression “harmonic line bundle” may have
different meanings in the literature.

The following result of Fuglede, furnishes an important property of harmonic
morphisms for two important projections.

######  Theorem 4.1.

[12, Theorems 2 and 3] For $n>1,$ let $\varphi:M\to N$ denote either the
“vertical projection”
$\mathbb{R}^{n}=\mathbb{R}^{n-1}\times\mathbb{R}\to\mathbb{R}^{n-1}$ or the
“radial projection” $\mathbb{R}^{n}\setminus\\{0\\}\to S^{n-1},$ given by
$x\to x/|x|.$ Then, $\varphi$ is a harmonic morphism and a set $E\subset N$ is
thin at a point $y\in N$ if and only if $\varphi^{-1}(E)$ is thin at some (and
hence every) point of the fiber $\varphi^{-1}(y)\subset M.$ Moreover, a set
$E\subset N$ is polar iff $\varphi^{-1}(E)$ is polar.

This is also in the book by Baird and Wood [6, Proposition. 4.3.7]. We remark
here that the case $\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ was first proved in [1,
Theorem 7.8.6].

######  Theorem 4.2.

Let $(M,\,N,\,\pi,\,\mathbb{R})$ be a harmonic line bundle with $M$ non-
compact. Let $F^{\prime}\subset N$ an $F_{\sigma}$ set, which we suppose to be
of first category if $\dim M=2$ and polar if $\dim M>2.$ Then, for every
continuous real-valued function $\varphi$ on $M,$ there is a harmonic function
$h$ on $M,$ such that, for all $x\in F^{\prime},$

$\big{(}h-\varphi\big{)}(m)\to 0,\quad\mbox{if}\quad
m\to*_{M},\quad\mbox{with}\quad m\in\pi^{-1}(x).$

###### Proof.

We first write $F^{\prime}$ as a union of closed polar sets if $n>2$, that is,
$\cup_{k}F^{\prime}_{k},$ where each $F_{k}^{\prime}$ is a closed polar set in
$N.$ For $n=2,$ we write $F^{\prime}$ as a union of nowhere dense closed sets,
that is, $\cup_{k}F^{\prime}_{k},$ where each $F_{k}^{\prime}$ is a nowhere
dense closed set in $N.$ We may assume that the $F_{j}^{\prime}$ are
increasing, $F_{j}^{\prime}\subset F_{j+1}^{\prime},\,j=1,2,\ldots.$

Let $(Q_{j})_{j=1}^{\infty}$ be a locally finite family of compact subsets of
$N,$ such that $(Q^{\circ}_{j})_{j=1}^{\infty}$ is an open cover of $N$ and
each $Q_{j}$ is contained in some $V_{j}=V_{x_{j}},$ for some $x_{j}\in N.$
For simplicity of notation, we denote $\psi_{x_{j}}$ by $\psi_{j}.$

Now, for each $k=1,2,\ldots,$ set

$K_{k}=\bigcup_{j=1}^{k}\psi_{j}^{-1}\big{\\{}Q_{j}\times[-m_{k},m_{k}]\big{\\}},$

where $m_{k},\,k=1,2,\ldots,$ is a strictly increasing sequence of natural
numbers, inductively chosen to increase so rapidly, that for each $k,$ and
each $j<k,$

(2)
$\psi_{j}^{-1}(Q_{j}\times\\{0\\})\cap\psi_{k}^{-1}(Q_{k}\times\mathbb{R})\subset\psi_{k}^{-1}(Q_{k}\times[-m_{k},m_{k}]).$

We claim that $(K_{k}^{\circ})_{k=1}^{\infty}$ is a cover of $M.$ To see this,
let $x\in M.$ Then, $\pi(x)$ is in some $Q_{j}^{\circ}\subset V_{j},$ so
$\psi_{j}(x)\in Q_{j}^{\circ}\times\mathbb{R}.$ Choose $k$ such that
$\psi_{j}(x)\in Q_{j}^{\circ}\times(-m_{k},m_{k}).$ Then
$x\in\psi^{-1}\\{Q_{j}^{\circ}\times(-m_{k},m_{k})\\}\subset K^{\circ}_{k},$
so $(K_{k}^{\circ})_{k=1}^{\infty}$ is indeed a cover of $M.$

Now, set

$F_{k}=\bigcup_{j=1}^{k}F_{kj},\quad\mbox{where}\quad
F_{kj}=\psi_{j}^{-1}\big{\\{}\big{(}F_{k}^{\prime}\cap
Q_{j}\big{)}\times\big{[}\mathbb{R}\setminus(-m_{k},m_{k})\big{]}\big{\\}}.$

We claim that each set $F_{kj}$ is closed in $M.$ To see this, we note that
$F_{kj}$ is a closed subset of $\pi^{-1}(V_{j}),$ which is contained in
$\pi^{-1}\big{(}F_{k}^{\prime}\cap Q_{j}\big{)}$ and the latter is closed in
$M.$ Thus, $F_{kj}$ is indeed closed in $M.$ Each $F_{kj}$ is also nowhere
dense, since it is a subset of the nowhere dense set
$\pi^{-1}(F_{k}^{\prime}\cap Q_{j}).$ For $n>2,$ it follows from Theorem 4.1
that the sets $F_{kj}$ are also polar in $M,$ since they are subsets of the
polar sets $\pi^{-1}(F_{k}^{\prime}\cap Q_{j}).$ It follows that each $F_{k}$
is closed, nowhere dense and if $n>2$ polar, being a finite union of such
sets.

Fix a compact $X\subset M.$ Since $(K_{k}^{\circ})_{k=1}^{\infty}$ is an
increasing open cover, there is a $k=k_{L}$ such that $L\subset
K_{k}^{\circ}\subset K_{k}.$ For $\ell>k,$

$F_{\ell}\cap
K_{k}=\left(\bigcup_{j=1}^{\ell}\psi_{j}^{-1}\big{\\{}\big{(}F_{\ell}^{\prime}\cap
Q_{j}\big{)}\times\big{[}\mathbb{R}\setminus(-m_{\ell},m_{\ell})\big{]}\big{\\}}\right)\cap\left(\bigcup_{j=1}^{k}\psi_{j}^{-1}\big{\\{}Q_{j}\times[-m_{k},m_{k}]\big{\\}}\right)=\emptyset,$

since
$\big{[}\mathbb{R}\setminus(-m_{\ell},m_{\ell})\big{]}\cap[-m_{k},m_{k}]=\emptyset.$
We have shown that, for every compact $X\subset M,$ there is a $k$ such that
$F_{\ell}\cap X=\emptyset,$ for $\ell>k,$ so the family
$(F_{\ell})_{\ell=1}^{\infty}$ is locally finite.

Set $F=\cup_{\ell=1}^{\infty}F_{\ell}.$ We now verify that the conditions in
part (b) of Theorem 3.3 are satisfied, so $F$ is a Mergelyan set.

We have that $F$ is the union of the countable family
$(F_{\ell})_{\ell=1}^{\infty}$ of sets which are closed, nowhere dense and if
$n>2$ polar. Moreover, $F$ itself is closed and nowhere dense, since this
family is locally finite. If $n>2,$ $F$ is also polar being a countable union
of polar sets.

We also claim $\widehat{F}=F$ ($F$ has no holes). To see this, note first
that, from the local trivializations it follows that, for each $x^{\prime}\in
N,$ the fibre $\pi^{-1}(x^{\prime})$ over $x^{\prime}$ is unbounded, connected
and closed in $M.$ Its closure in $M^{*}$ is thus
$\\{*\\}\cup\pi^{-1}(x^{\prime}),$ which is connected, because the closure of
a connected set is connected.

$M^{*}\setminus\pi^{-1}(F^{\prime})=\\{*\\}\cup\bigcup_{x^{\prime}\not\in
F^{\prime}}\pi^{-1}(x^{\prime})=\bigcup_{x^{\prime}\not\in
F^{\prime}}\left(\\{*\\}\cup\pi^{-1}(x^{\prime})\right)$

is therefore connected, being the union of a family of connected sets, sharing
a common point $\\{*\\}.$

We claim that $M^{*}\setminus\pi^{-1}(F^{\prime})$ is dense. Indeed, fix $x\in
M$ and choose $j$ so that $x^{\prime}=\pi(x)\in V_{j}.$ Then
$\psi_{j}(x)=(x^{\prime},t),$ for some $t\in\mathbb{R}.$ Let
$x_{i}^{\prime},\,i=1,2.\ldots,$ be a sequence in $V_{j}\setminus F^{\prime}$
tending to $x^{\prime}.$ Then, $(x_{i}^{\prime},t)\to(x^{\prime},t)$ in
$\psi_{j}(\pi^{-1}(V_{j})).$ Therefore, $\psi_{j}^{-1}((x^{\prime}_{i},t))\to
x$ in $M\setminus\pi^{-1}(F^{\prime}).$ Thus
$M^{*}\setminus\pi^{-1}(F^{\prime})$ is indeed dense. Since every set between
a connected set and its closure is also connected, it follows that
$M^{*}\setminus F$ is connected. Thus, $\widehat{F}=F$ as claimed.

Part $(i).$ We must check that $M\setminus\widehat{F}$ and $M\setminus
F^{\circ}$ are thin at the same points of $F.$ Since $\widehat{F}=F,$ we have
that $M\setminus\widehat{F}=M\setminus F$ and since $F^{\circ}=\emptyset,$ we
have that $M\setminus F^{\circ}=M.$ Thus, we must show that $M\setminus F$ and
$M$ are thin at the same ponts of $F.$ First note that every point of $F$ is
an interior point of $M,$ so $M$ is thin at no point of $F,$ by the remark
following Lemma 3.1.

To see that $M\setminus F$ is thin at no point of $F,$ we consider two cases.

Case I, $n=2.$ If a set is thin at a point $p$ in $\mathbb{R}^{2},$ then its
complement contains arbitrarily small circles centered at $p.$ In the local
trivializations (which are planar domains), $F$ contains no circles
whatsoever. Hence, the complement of $F$ in each local trivialization is non-
thin at every point of $F$ in this trivialization. Since the harmonic
isomophisms $\psi_{x}$ preserve thinness, $M\setminus F$ is non-thin at every
point of $F.$

Case II, $n>2.$ Since $\pi^{-1}(N\setminus F^{\prime})$ is a subset of
$M\setminus F,$ it is sufficient to show that $\pi^{-1}(N\setminus
F^{\prime})$ is non-thin at every point of $F.$ Let $y\in M,$ be an arbitrary
point in $F.$ By our definition, $y\in\pi^{-1}(F^{\prime}).$ Thus
$y\in\pi^{-1}(x),$ for some $x\in F^{\prime}.$ Now by Theorem 4.1,
$\pi^{-1}(N\setminus F^{\prime})$ is non-thin at $y$ if and only if
$N\setminus F^{\prime}$ is non-thin at $x.$ The latter follows from our
assumption on $F^{\prime}.$

Thus, in either case, the condition $(i)$ is fulfilled.

Part $(ii).$ Since $M\setminus F$ is connected and open in $M^{*}\setminus F,$
it follows that $M^{*}\setminus F$ is locally conneced at each point of
$M\setminus F.$ There remains only to check that $M^{*}\setminus F$ is locally
connected at the ideal point $*.$ Since the sets $(M^{*}\setminus F)\setminus
K_{k},\,k=1,2,\ldots,$ form a neighbourhood basis of the point $*$ in
$M^{*}\setminus F,$ it is sufficient to show that each $(M^{*}\setminus
F)\setminus K_{k}$ is connected. Now,

(3) $\big{(}M^{*}\setminus\pi^{-1}(F^{\prime})\big{)}\setminus
K_{k}\subset(M^{*}\setminus F)\setminus K_{k}$

and the left side is dense in the right side, so it is sufficient to show that
the left side is connected, because every set between a connected set and its
closure is also connected.

Consider first, for each $k,$

(4) $\big{(}M^{*}\setminus\pi^{-1}(F^{\prime})\big{)}\setminus
K_{k}=\bigcup_{x^{\prime}\not\in
F^{\prime}}\big{[}\\{*\\}\cup\big{(}\pi^{-1}(x^{\prime})\setminus
K_{k}\big{)}\big{]}.$

We have

$\pi^{-1}(x^{\prime})\setminus
K_{k}=\bigcup_{j=1}^{k}\psi_{j}^{-1}\big{(}\\{x^{\prime}\\}\times[-m_{k},m_{k}]\big{)}=\bigcup_{j=1}^{k}\alpha_{j},$

where each non-empty $\alpha_{j}$ is a compact arc in $\pi^{-1}(x^{\prime})$
or a point in $\pi^{-1}(x^{\prime}).$ The condition (2) imposed on the $m_{k}$
assures us that, $\alpha_{k}$ is non-empty, then each non-empty $\alpha_{j}$
meets $\alpha_{k}.$ Hence, the union $\alpha$ of the
$\alpha_{j},\,j=1,2,\ldots,k$ is either empty or a compact arc in
$\pi^{-1}(x^{\prime}).$ So

$\pi^{-1}(x^{\prime})\setminus K_{k}=\pi^{-1}(x^{\prime})\setminus\alpha$

is either all of $\pi^{-1}(x^{\prime})$ or two unbounded arcs, depending on
whether $\alpha$ is empty or not. In either case
$\pi^{-1}(x^{\prime})\setminus K_{k}$ is connected and unbounded and thus each
term on the right side of (4) is connected. Thus the right member of (4) is
connected, since it is a union of connected sets with a common point $*.$ This
concludes the proof of Part $(ii).$

Part $(iii).$ Note that $F$ has no holes, so there is nothing to verify.

We have shown that $F$ is a Mergelyan set, and since $F^{\circ}=\emptyset,$ it
is in fact a Carleman set in $M$, so for $\varphi$ continuous on $M,$ the
existence of the harmonic function $h$ on $M,$ satisfying the approximation
property

$|h(x)-\varphi(x)|\rightarrow 0,\quad\textit{ as }x\to*_{M},\quad
x\in\pi^{-1}(x^{\prime}),$

for all $x^{\prime}\in F^{\prime},$ follows from the definition of a Carleman
set (in $M$).

∎

##  5\. PDE’s

In this section $\Omega$ is a domain in $\mathbb{R}^{n},\,n>2$ and $L$ is an
elliptic partial differential operator of the form (1), satisfying the
conditions in the introduction on $\Omega.$ In the introduction, we defined
$L$-harmonic and $L$-superharmonic functions. If $L$ is defined on all of
$\mathbb{R}^{n}$ and $u$ is an $L$-harmonic function on $\mathbb{R}^{n},$ we
say that $u$ is an entire $L$-harmonic function. According to our earlier
definition, by an $L$-potential on a domain $\Omega\subset\mathbb{R}^{n},$ we
mean an $L$-superharmonic function $u\geq 0$ on $\Omega$ for which every
$L$-harmonic minorant $\geq 0$ vanishes identically [8]. Notice that the
constant function $0$ is an $L$-potential.

Denote by $\mathcal{H}_{L}(U)$ the real vector space of $L-$harmonic functions
on an open set $U\subset\Omega$ and by $\mathcal{H}_{L}$ the sheaf of
solutions $u$ of $Lu=0$ on open subsets of $\Omega,$ where we refrain from
including $\Omega$ in the notation for simplicity.

In Hervé [23] on page 245 (see also [9]), it is asserted that
$\mathcal{H}_{L}$ is a (Brelot) harmonic space. We shall denote the
corresponding potential theoretic notions by including the prefix $L.$ Thus,
we shall speak of $L-$harmonic functions, $L-$thinness, etc. When $L=\Delta,$
where $\Delta$ is the Laplace operator, we obtain classical potential theory.
Thus, $\Delta-$harmonic functions are classical harmonic functions,
$\Delta-$thinness is classical thinness, etc.

We wish to show the existence of an $L-$potential $>0.$ Our assumption that
$c\leq 0$ in equation (1) implies that the constant function $1$ is an
$L-$superharmonic function $>0,$ which is $L-$harmonic if and only if $c=0.$
In fact, our assumption that there is an $L-$superharmonic function $u\geq 0,$
which is not $L-$harmonic, implies the existence of an $L-$potential $>0.$
This follows immediately from the Riesz Representation Theorem [11, Theorem
2.2.2], which states that every $L-$superharmonic function $u\geq 0$ on a
domain $U$ has a unique representation as the sum $u=p+h$ of an $L$-potential
$p$ and an $L$-harmonic function $h.$ Since $u$ is not $L$-harmonic,
$p\not=0.$ Since $p$ is a potential, $p$ is an $L-$superharmonic function
$\geq 0.$ By Lemma 2.1, an $L-$superharmonic function $\geq 0$ on a domain
$\Omega$ is either $>0$ or $\equiv 0.$ Thus $p>0$ and we have confirmed the
existence of an $L-$potential $>0.$ Thus, under the assumptions in the
introduction, $\mathcal{H}_{L}$ is not only a Brelot space, but in fact a
strong Brelot space.

An additional important property of the Brelot space
$(\Omega,\mathcal{H}_{L})$ is that it satisfies Axiom D (see [22, p. 567]).

By an adjoint $L-$harmonic function $u$ on an open set $U,$ we mean a solution
$u$ of the adjoint equation $L^{*}u=0.$ The following axiom is important for
proving approximation results.

Axiom A* (quasi-analyticity). Every adjoint harmonic function $u$ on a
connected open set $U,$ which vanishes on a neighbourhood of some point of
$U,$ must vanish identically on $U.$

According to [2], the Brelot space $(\Omega,\mathcal{H}_{L})$ satisfies Axiom
A*.

The following convergence theorem can be found in [10, p. 17].

######  Theorem 5.1.

In a Brelot space, for a sequence $u_{n}$ of harmonic functions, locally and
uniformly bounded in an open set $U,$ pointwise convergence implies that the
limit is harmonic.

By a closed subset of $\Omega$ we mean a subset which is closed in the
topology of $\Omega.$ We say that a subset $X$ of $\Omega$ is $\Omega-$bounded
if it is relatively compact in $\Omega,$ that is, if its closure in the
topological space $\Omega$ is compact. If the context is clear, we shall omit
the prefex $\Omega.$ If $F$ is a closed subset of $\Omega,$ by a hole of $F,$
we mean a bounded component of $\Omega\setminus F.$ We denote by $\widehat{F}$
the union of $F$ with all of its holes. Clearly, $\widehat{F}$ has no holes
and if $K$ is a compact subset of $\Omega$ then $\widehat{K}$ is also compact.

The famous Mergelyan Theorem gives a complete characterization of those
compact subset $K$ of the complex plane $\mathbb{C},$ such that each function
continuous on $K$ and holomorphic on $K^{\circ}$ can be uniformly approximated
by polynomials. Since entire functions can be represented by power series,
approximation by polynomials on compact sets is equivalent to approximation by
entire functions. Thus, Mergelyan’s Theorem yieilds a characterization of
compact sets $K$ of the complex plane, for which every function $f\in C(K)$
satisfying the Cauchy-Riemann equation $\overline{\partial}f=0$ on
$K^{\circ},$ can be approximated by functions $g$ satisfying
$\overline{\partial}g$ on all on $\mathbb{C}.$ Thus, it is natural to call a
compact subset $K$ of $\Omega$ an $L$-Mergelyan set in $\Omega$ if for each
$u\in C(K)\cap\mathcal{H}_{L}(K^{\circ})$ and each $\epsilon>0,$ there is a
$v\in\mathcal{H}_{L}(\Omega)$ such that $|v-u|<\epsilon$ on $K.$ The following
theorem, which is a special case of a result of Gardiner, Goldstein and
GowriSankaran [14, Theorem. 4], characterizes $L-$Mergelyan sets of $\Omega$.

######  Theorem 5.2.

Let $L$ be as above and $K$ a compact subset of $\Omega.$ Then, the following
are equivalent.

(a) $K$ is an $L-$Mergelyan set in $\Omega.$

(b) $\Omega\setminus\widehat{K}$ and $\Omega\setminus K^{\circ}$ are $L-$thin
at the same points of $K.$

Results on approximation by global harmonic functions in the context of more
general harmonic spaces can be found in [7] and [21].

We need the following theorem of Hervé [22, Theorems 36.1 and 36.3].

######  Theorem 5.3.

A subset $E$ of $\Omega$ is $L-$thin at a point of $\Omega$ if and only if it
is $\Delta-$thin at that point and $E$ is $L$-polar if and only if it is
$\Delta-$polar.

This states that thinness for harmonic functions is the same as thinness for
$L-$harmonic functions. For this reason, for the remainder of this section, we
shall use the terms thin and polar, without the prefex $L$ or $\Delta,$
however, we should like to keep the prefex for $L-$Mergelyan sets.

######  Corollary 5.4.

For $L$ and $\Omega$ as above, if $K$ is an $L-$Mergelyan set in $\Omega$ and
$Q$ is a compact polar set in $\Omega,$ then $K\cup Q$ is an $L-$Mergelyan set
in $\Omega.$

###### Proof.

First of all, since $Q$ is polar,

(5) $\widehat{K\cup Q}=\widehat{K}\cup Q,\quad\mbox{and}\quad(K\cup
Q)^{\circ}=K^{\circ}.$

Suppose $\Omega\setminus(K\cup Q)^{\circ}$ is thin at a point $x\in K\cup Q.$
Since $(\Omega\cup Q)^{\circ}\subset\widehat{K\cup Q},$ it follows that
$\Omega\setminus\widehat{K\cup Q}$ is also thin at $x.$

Suppose conversely that $\Omega\setminus\widehat{K\cup Q}$ is thin at $x\in
K\cup Q.$ We wish to show that $\Omega\setminus(K\cup Q)^{\circ}$ is thin at
$x.$ If $x$ is in a hole of $K,$ then, indeed, $\Omega\setminus(K\cup
Q)^{\circ}=\Omega\setminus K^{\circ}$ is thin at $x.$ If $x$ is not in a hole
of $K,$ since $x\in K\cup Q,$ we have $x\in K\cup(Q\setminus\widehat{K}).$
Suppose $x\in Q\setminus\widehat{K}.$ Since adding a polar set does not cancel
thinness,

(6) $\big{[}\Omega\setminus\widehat{K\cup
Q}\big{]}\cup(Q\setminus\widehat{K})=\Omega\setminus\widehat{K}$

is thin at $x$ which is in $\Omega\setminus\widehat{K},$ which is absurd
because an open set cannot be thin at one of its points. Thus, $x\not\in
Q\setminus\widehat{K}$ and so $x\in K.$ By (6) we have
$\Omega\setminus\widehat{K}$ is thin at $x\in K.$ Since $K$ is a $L-$Mergelyan
set, this means that $\Omega\setminus K^{\circ},$ which by (5) is
$\Omega\setminus(K\cup Q)^{\circ},$ is thin at $x,$ which is what we wished to
prove.

∎

To replace the compact sets in Corollary 5.4 by closed sets, it will be
convenient to have the following form of the Tietze extension theorem.

######  Lemma 5.5.

If $X$ is a normal (topological) space, $0<c\leq+\infty$ and
$\phi:A\to(-c,+c)$ is a continuous function on a closed subset $A\subset X,$
then $\phi$ has a continuous extension $\Phi:X\to(-c,+c).$

###### Proof.

The Tietze theorem is usually stated for $c=+\infty,$ but this form is
equivalent, since $(-\infty,+\infty)$ and $(-c,+c)$ are homeomorphic. Now,
this is a simple consequence of [27, Theorem 8.2.11].

∎

For a closed subset $E$ of an open subset $U$ of $\Omega,$ $A(E)$ denotes the
set of all continuous functions on $E$ that are $L-$harmonic on the
$E^{\circ}.$ Moreover, $E$ is called an $L-$Carleman set in $U,$ if for each
$u\in A(E)$ and each continuous function $\epsilon:E\to(0,1],$ there exists
$v\in\mathcal{H}_{L}(U)$ such that $|v-u|<\epsilon$ on $E.$

By a chaplet $C$ in an open set $\Omega\subset\mathbb{R}^{n},$ we mean the
union of a family, locally finite in $\Omega,$ of pairwise disjoint,
connected, compact $L-$Mergelyan subsets of $\mathbb{R}^{n},$ contained in
$\Omega,$ whose complements in $\mathbb{R}^{n}$ are connected.

Note that $C$ is closed in $\Omega,$ since it is the union of a locally finite
family of closed subsets of $\Omega.$ The following theorem furnishes a method
for constructing interesting Carleman sets.

######  Theorem 5.6.

Let $\Omega$ be a domain in $\mathbb{R}^{n},\,n\geq 3,$ and $L$ an operator as
above in $\Omega.$ Let $F$ be a closed polar subset of $\Omega$ and $C$ a
chaplet in $\Omega.$ Then $E=F\cup C$ is an $L-$Carleman set in $\Omega.$

One way to prove the theorem would be to show that $L$ is the Laplace-Beltrami
operator for a Riemannian metric on $\Omega.$ The theorem would then simply be
special cases of Theorem 3.3. We, however, prefer to give a direct proof in
the next subsection.

###  5.1. Proof of Theorem 5.6

We need some construction and some lemmas before proving the theorem.

######  Lemma 5.7.

Given a chaplet $C$ in a domain $\Omega\subset\mathbb{R}^{n},$ there exists a
regular exhaustion

$\mathcal{K}_{1}\subset\mathcal{K}_{2}^{\circ}\subset\mathcal{K}_{2}\subset\mathcal{K}_{3}^{\circ}\subset\cdots,$

of $\Omega$ by compact $L-$Mergelyan subsets of $\Omega,$ such that
$C\cap\partial\mathcal{K}_{j}=\emptyset,\,j=1,2,\ldots.$

###### Proof.

Let us call the compact sets in the definition of the chaplet $C$ the parts of
$C.$ We may construct a regular exhaustion $(K_{j})_{j=1}^{\infty}$ of
$\Omega,$ whose members are smoothly bounded, such that, denoting by $C_{j}$
the union of those parts of $C$ which meet $\partial K_{j},$ we have
$C_{j}\subset K_{j+1}^{\circ},$ and $C_{j+1}\cap K_{j}=\emptyset,$ for
$j=1,2,\ldots.$ Now, for each $j=1,2,\ldots,$ let $V_{j}$ be a neighbourhood
of $C_{j},$ consisisting of the union of finitely many open balls of radius
$r_{j},$ centered at points of $C_{j},$ where $r_{j}$ are chosen decreasing to
zero so rapidly that $\overline{V}_{j}\cap(C\setminus C_{j})=\emptyset$ and
$\overline{V}_{j+1}\cap\partial K_{j}=\emptyset.$ Since each component of
$V_{j}$ meets $\partial K_{j},$ the sets $\mathcal{K}_{j}=K_{j}\setminus
V_{j}$ have the required properties.

∎

######  Lemma 5.8.

Let $K$ be a compact $L-$Mergelyan subset of a domain
$\Omega\subset\mathbb{R}^{n},$ with $\widehat{K}=K$ relative to $\Omega$ and
$Q\subset\Omega$ a compact $L-$Mergelyan subset of $\mathbb{R}^{n},$ with
$\widehat{Q}=Q$ relative to $\mathbb{R}^{n}$ and $K\cap Q=\emptyset.$ Then,
$K\cup Q$ is an $L-$Mergelyan subset of $\Omega.$

###### Proof.

Note that both $K$ and $Q$ are $L-$Mergelyan subsets of $\Omega,$ since the
property of being a $L-$Mergelyan subset of a domain is hereditary in the
sense that, if $\Omega_{1}\subset\Omega_{2}$ are domains in $\mathbb{R}^{n}$
and $Q\subset\Omega_{1}$ is a compact $L-$Mergelyan subset of $\Omega_{2},$
then it is also a $L-$Mergelyan subset of $\Omega_{1}.$

To conclude the proof, it is sufficient, as in the proof of Lemma 5.9, to show
that $(K\cup Q)^{\wedge}=K\cup Q$ in $\Omega.$ Thus, it is sufficient to show
that each component of $\Omega\setminus(K\cup Q)$ is unbounded in $\Omega.$

Let $U$ be a component of $\Omega\setminus K$ and set $Q_{U}=U\cap Q.$ We
claim the open subset that $U\setminus Q=U\setminus Q_{U}$ of
$\Omega\setminus(K\cup Q)$ is connected. To see this, fix $x\in U\setminus
Q_{U}.$ Since $U$ is (pathwise) connected, for an arbitrary $y\in U,$ we may
choose a path $\sigma_{y}$ in $U$ from $x$ to $y.$ If $\sigma_{y}\cap
Q_{U}=\emptyset,$ set $\Sigma_{y}=\sigma_{y}.$ Otherwise, since
$\widehat{Q}=Q$ in $\mathbb{R}^{n}$, noting that $Q_{U}$ is a compact subset
of $U,$ there exists $W,$ a neighbourhood of $Q_{U}$ in $U$ which is smoothly
bounded in $\mathbb{R}^{n},$ with $\overline{W}\subset
U\setminus(\\{x\\}\cup\\{y\\}).$ The set $W$ has finitely many components and
their closures are the components of $\overline{W}.$ Let
$\overline{W}_{1},\ldots,\overline{W}_{m}$ be those components which intersect
$\sigma_{y}.$ For each $j,$ let $X_{j}$ be the outer boundary of
$\overline{W}_{j},$ that is the boundary of the $\mathbb{R}^{n}-$unbounded
component of $\mathbb{R}^{n}\setminus W_{j}.$ For each $j,$ $\sigma_{x}$ has a
first point $p_{j}$ and a last point $q_{j}$ in $X_{j}.$ Let $\Sigma_{y}$ be
the connected subset of $U\setminus Q_{U}$ obtained from $\sigma_{y}$ by
replacing, for each $j,$ the portion $\sigma_{j}$ of $\sigma_{y}$ from $p_{j}$
to $q_{j}$ by the connected set $X_{j}.$ The set $\Sigma_{y}$ is a connected
subset in $U\setminus Q_{U}.$ Since

$U\setminus Q_{U}=U\setminus K=\bigcup_{y\in U\setminus K}\Sigma_{y}$

is the union of connected sets having the point $x$ in common, the open subset
$U\setminus Q_{U}$ of $\Omega\setminus(K\cup Q)$ is indeed connected as
claimed. Since each component $U$ of $\Omega\setminus K$ is unbounded in
$\Omega$ and $Q_{U}$ is bounded in $\Omega,$ each $U\setminus U_{Q}$ is
unbounded in $\Omega.$ Now each component of $\Omega\setminus(K\cup Q)$
intersects (and thus contains) such an $U\setminus U_{Q}.$ Thus each component
of $\Omega\setminus(K\cup Q)$ is unbounded in $\Omega.$

∎

We cannot strengthen the previous lemma by merely assuming that $K$ and $Q$
are disjoint Mergelyan sets in $\Omega.$ For example, if
$\Omega=\mathbb{R}^{n}\setminus\\{0\\},$
$S_{j}=\\{x\in\mathbb{R}^{n}:|x|=j\\},$ $K=S_{1}\cup S_{2}$ and $Q=S_{3},$
then $K$ and $Q$ are disjoint compact $L-$Mergelyan subsets of $\Omega.$
However, $K\cup Q$ is not an $L-$Mergelyan subset of $\Omega.$

The following lemma is a particular case of the previous one, but we shall
give a more direct proof.

######  Lemma 5.9.

The union of two disjoint compact $L-$Mergelyan sets in $\mathbb{R}^{n},$
having connected complements, is again an $L-$Mergelyan set in
$\mathbb{R}^{n}$.

###### Proof.

Let $K_{j},\,j=1,2$ be disjoint $L-$Mergelyan sets in $\mathbb{R}^{n},$ with
$\widehat{K}_{j}=K_{j},\,j=1,2.$ Since thinness is a local condition, it is
sufficient to show that $(K_{1}\cup K_{2})^{\wedge}=K_{1}\cup K_{2}.$ Thus, it
is sufficient to show that $X=\mathbb{R}^{n}\setminus(K_{1}\cup K_{2})$ is
connected. One easily shows this by verifying that, for a fixed $x\in X$, the
set of $y\in X,$ which can be connected to $x$ by a path in $X,$ is both open
and closed in $X.$

∎

######  Lemma 5.10.

Let $\Omega$ be a domain in $\mathbb{R}^{n},n\geq 3,$ $F$ a polar set in
$\Omega$ and $L,C,\mathcal{K}_{j},j=1,2,\ldots$ as above. For $j=1,2,\ldots,$
the sets $(\mathcal{K}_{j}\cup F\cup C)\cap\mathcal{K}_{j+1}$ are
$L-$Mergelyan sets in $\Omega.$

###### Proof.

Since the family of compacta whose union is $C$ is locally finite in $\Omega,$
by Lemma 5.9, $C\cap(\mathcal{K}_{j+1}\setminus\mathcal{K}_{j})$ is an
$L-$Mergelyan set in $\mathbb{R}^{n}.$ Then, by Lemma 5.8
$(\mathcal{K}_{j}\cup C)\cap\mathcal{K}_{j+1}$ is an $L-$Mergelyan set in
$\Omega.$ Finally, by Corollary 5.4 $(\mathcal{K}_{j}\cup F\cup
C)\cap\mathcal{K}_{j+1}$ is an $L-$Mergelyan set in $\Omega.$

∎

###### Proof.

(proof of Theorem 5.6) Note that $E$ is closed in $\Omega$ as it is the union
of two closed subsets of $\Omega.$ We may rename the compact sets
$(\mathcal{K}_{k})_{k=1}^{\infty}$ in the exhaustion obtained by Lemma 5.7
(for the chaplet $C$) such that $\mathcal{K}_{1}=\emptyset.$

Let $(\epsilon_{k})_{k=0}^{\infty}$ be a sequence of positive numbers, such
that

$\epsilon_{k+1}+\epsilon_{k+2}+\cdots<\epsilon_{k},\quad\mbox{for}\quad
k=0,1,2,\ldots,$

and

$2\epsilon_{k-1}\leq\inf\\{\epsilon(y)\,:\,y\in(F\cup
C)\cap(\mathcal{K}_{k+1}\setminus\mathcal{K}_{k}^{\circ})\\},\quad\mbox{for}\quad
k=1,2,\ldots.$

Also, suppose $u\in A(E).$

Set $w_{0}=u$ on $(F\cup C)\cap\mathcal{K}_{2}.$ By Corollary 5.4, there is a
$v_{1}\in\mathcal{H}_{L}(\Omega),$ such that

$|v_{1}-w_{0}|<\epsilon_{2},\quad\textit{ on }\quad(C\cup
F)\cap\mathcal{K}_{2}.$

Note that on $(F\cup C)\cap\mathcal{K}_{2},$ one has

$|v_{1}-u|=|v_{1}-w_{0}|<\epsilon_{2}<\epsilon_{0}.$

We define a function $w_{1}$ in
$A\big{(}\mathcal{K}_{2}\cup(C\cap(\mathcal{K}_{3}\setminus\mathcal{K}_{2}))\cup(F\cap\partial\mathcal{K}_{3})\big{)},$
by setting $w_{1}=v_{1}$ on $\mathcal{K}_{2},$ and $w_{1}=u$ on
$(C\cap(\mathcal{K}_{3}\setminus\mathcal{K}_{2}))\cup(F\cap\partial\mathcal{K}_{3}).$
By the Tietze Lemma 5.5, extend $w_{1}$ to $\widetilde{w}_{1},$ from
$\mathcal{K}_{2}\cup(C\cap(\mathcal{K}_{3}\setminus\mathcal{K}_{2}))\cup(F\cap\partial\mathcal{K}_{3})$
continuously to $(\mathcal{K}_{2}\cup F\cup C)\cap\mathcal{K}_{3}$ such that

$|\widetilde{w}_{1}-u|<\epsilon_{2}\textit{ on }(F\cup
C)\cap(\mathcal{K}_{3}\setminus\mathcal{K}_{2}^{\circ}).$

We, however, keep denoting the extension $\widetilde{w}_{1}$ by $w_{1}$ to
simplify the notation. Noting that $w_{1}\in
A\big{(}\mathcal{K}_{2}\cup(C\cap(\mathcal{K}_{3}\setminus\mathcal{K}_{2}))\cup(F\cap\partial\mathcal{K}_{3})\big{)},$
by Lemma 5.10, there is a function $v_{2}\in\mathcal{H}_{L}(\Omega),$ such
that

$|v_{2}-w_{1}|<\epsilon_{3},\quad\textit{ on }\quad(\mathcal{K}_{2}\cup F\cup
C)\cap\mathcal{K}_{3}.$

Thus, $|v_{2}-v_{1}|<\epsilon_{3}$ on $\mathcal{K}_{2}$ and
$|v_{2}-u|<\epsilon_{3}+\epsilon_{2}<\epsilon_{1}$ on $(F\cup
C)\cap(\mathcal{K}_{3}\setminus\mathcal{K}_{2}^{\circ}).$

We proceed thus, by induction, to construct a sequence
$v_{k}\in\mathcal{H}_{L}(\Omega),$ such that

$|v_{k}-v_{k-1}|<\epsilon_{k+1}\textit{ on }\mathcal{K}_{k},$

and

$|v_{k}-u|<\epsilon_{k-1}\textit{ on }(F\cup
C)\cap(\mathcal{K}_{k+1}\setminus\mathcal{K}_{k}^{\circ}),$

for $k=1,2,\ldots.$

The sequence $(v_{k})_{k=1}^{\infty}$ is uniformly Cauchy on compact subsets
of $\Omega,$ and hence converges uniformly on compact subsets of $\Omega$ to a
function $v=\lim v_{k}$ in $\mathcal{H}_{L}(\Omega).$ Theorem 5.1, yields
$v\in\mathcal{H}_{L}(\Omega).$

Now suppose $y\in F\cup C.$ By exhaustion, there exists a fixed $k=1,2,\ldots$
such that $y\in(F\cup
C)\cap(\mathcal{K}_{k+1}\setminus\mathcal{K}_{k}^{\circ}).$ Therefore,

$|v(y)-u(y)|=|v_{k}(y)-u(y)+\sum_{j=k+1}^{\infty}(v_{j}-v_{j-1})(y)|\leq|v_{k}(y)-u(y)|+\sum_{j=k+1}^{\infty}|v_{j}(y)-v_{j-1}(y)|.$

Note that since $y\in\mathcal{K}_{k+1},$ it is in $\mathcal{K}_{j}$ for every
$j\geq k+1,$ hence $|v_{j}(y)-v_{j-1}(y)|<\epsilon_{j+1}.$ Also,

$|v_{k}(y)-u(y)|<\epsilon_{k-1}.$

Thus, the above yields

$|v(y)-u(y)|\leq\epsilon_{k-1}+\sum_{j=k+1}^{\infty}\epsilon_{j+1}\leq\epsilon_{k-1}+\epsilon_{k+1}\leq
2\epsilon_{k-1}\leq\epsilon(y).$

∎

###  5.2. Strictly Starlike Domains

We shall say that a domain $D\subset\mathbb{R}^{n}$ is strictly starlike with
respect to the origin $\bf 0$ if for every $p$ in $\overline{D}$, and every
$0\leq\lambda<1$, $\lambda p$ is in the interiour of $D$. We only consider
starlike domains with respect to the origin so we need not repeat “with
respect to the origin” again.

We first show that a domain $U$ is strictly starlike if and only if there is a
unique continuous function $R:S^{n-1}\to[0,+\infty],$ such that, in polar
coordinates,

$U=\\{x=(\theta,\rho):0\leq\rho<R(\theta),\,\theta\in S^{n-1}\\}.$

The “if part” is clear as the above $U$ is a strictly starlike domain. For the
the “only if part” we let $R(\theta)=sup\\{\rho:(\theta,\rho)\in U\\},$ for
$\theta\in S^{n-1}.$ The sup could well be $+\infty,$ for some values of
$\theta,$ as the domain is not necessarily bounded.

######  Lemma 5.11.

$R:S^{n-1}\to[0,+\infty]$ is continuous, where $[0,+\infty]$ has the usual
order topology.

###### Proof.

We first show that $R$ is lower semicontinuous in $S^{n-1}$. Suppose, to
obtain a contradiction, that there is a point $\theta\in S^{n-1},$ a finite
value $s<R(\theta)$ and a sequence $\theta_{j}\to\theta$ (in $S^{n-1}$) such
that $R(\theta_{j})<s.$ Since $U$ is open, it follows that no point in the
segment $\\{(\theta,r):s<r<R(\theta)\\}$ lies in $U,$ which contradicts the
definition of $R(\theta).$ Note that the argument is valid even if
$R(\theta)=+\infty.$

We now need to show the upper semicontinuity of $R$ in $S^{n-1}$. Fix
$\theta\in S^{n-1}.$ If $R(\theta)=+\infty,$ upper semicontinuity at $\theta$
is automatic. Suppose $R(\theta)<+\infty$ and suppose, to obtain a
contradiction, that there is a sequence $\theta_{j}\to\theta$ and
$\epsilon>0,$ such that $R(\theta_{j})>cR(\theta)+\epsilon.$ Then, for
$s=R(\theta)+\epsilon,$ $(\theta,s_{j})\in U.$ However, $(\theta,s)\not\in U.$
Therefore, $(\theta,s)\in\partial U.$ This contradicts the definition of
$R(\theta),$ as $U$ is a strictly starlike domain.

∎

Taking into account such a parametrization, we shall denote a strictly
starlike domain $U$ as $U=(S^{n-1},R).$ We wish to study the radial behaviour
of harmonic functions on strictly starlike domains and we begin with a theorem
on the radial limits of continuous functions.

######  Theorem 5.12.

Let $U=(S^{n-1},R)$ be a strictly starlike domain in $\mathbb{R}^{n},n\geq 2,$
and $\Psi:U\to(-\infty,+\infty)$ a continuous function. For $\theta\in
S^{n-1}$ we denote $\Psi^{*}(\theta)\in[-\infty,+\infty]$ the radial limit of
$h$ on the ray $\\{(\theta,r):0\leq r<R(\theta)\\}.$ Let $A\subset S^{n-1}$ be
the set of all $\theta\in S^{n-1}$ for which $\Psi^{*}(\theta)$ exists. Then
$A$ is a measurable set and the function $\Psi^{*}:A\to[-\infty,+\infty]$ is
of Baire class $0$ or $1.$

###### Proof.

If the radial limit along some ray is infinite then it is either $-\infty$ or
$+\infty.$ Indeed, suppose $\Psi(\theta,r)\to\infty$ as $r\to R(\theta),$ and
there are sequences $r^{+}_{j}\to R(\theta)$ and $r^{-}_{j}\to R(\theta),$
such that $\Psi(\theta,r^{+}_{j})\to+\infty$ and
$\Psi(\theta,r^{-}_{j})\to-\infty.$ By the intermediate value theorem, there
is a sequence $r^{\circ}_{j}\to R(\theta),$ such that
$\Psi(\theta,r^{\circ}_{j})=0.$ This contradicts the assumption that
$\Psi(\theta,r)\to\infty$ as $r\to R(\theta).$

Consider the homeomorphism

$h:U\longrightarrow\mathbb{B}^{n},\quad
x=(\theta,\rho)\longmapsto\big{(}\theta,\chi(0,\rho)\big{)},$

where $\chi$ denotes the chordal distance. Let $(r_{j})_{j=1}^{\infty}$ be a
sequence increasing to $1$ in $[0,1).$ We introduce the continuous functions

$R_{j}:S^{n-1}\to[0,+\infty),\quad\theta\mapsto|h^{-1}(\theta,r_{j})|,\quad
j=1,2,\ldots.$

Note that $0<R_{j}\leq R_{j+1},$ and $R_{j}\to R,$ as $j\to\infty.$ Then,
$\theta\in A$ if and only if

$\sup\\{\chi\big{(}\Psi(\theta,R_{j}(\theta)),\Psi(\theta,\rho)\big{)}:R_{j}(\theta)\leq\rho<R(\theta)\\}\to
0,\quad\mbox{as}\quad j\to\infty.$

Putting $\mathbb{Q}_{j}=\mathbb{Q}\cap[R_{j}(\theta),R(\theta)),$ this is
equivalent to

$\sup\\{\chi\big{(}\Psi(\theta,R_{j}(\theta)),\Psi(\theta,\rho)\big{)}:\rho\in\mathbb{Q}_{j}\\}\to
0,\quad\mbox{as}\quad j\to\infty.$

Thus,

$A=\bigcap_{i=1}^{\infty}\bigcup_{j=1}^{\infty}\bigcap_{\rho\in\mathbb{Q}_{j}}\big{\\{}\theta\in
S^{n-1}:\chi\big{(}\,\Psi(\theta,R_{j}(\theta)),\Psi(\theta,\rho)\,\big{)}\leq\frac{1}{i}\big{\\}},$

and since each function
$\chi\big{(}\,\Psi(\theta,R_{j}(\theta)),\Psi(\theta,\rho)\big{)}$ is
continuous in $\theta,$ it follows that $A$ is measurable and in fact an
$F_{\delta\sigma\delta}-$set.

Finally, the function $\Psi^{*}:A\to[-\infty,+\infty]$ is of Baire class $0$
or $1,$ since it is the limit of a sequence of continuous functions; that is,

$\lim_{j\to\infty}\Psi\big{(}\theta,R_{j}(\theta)\big{)}=\Psi^{*}(\theta),\quad\mbox{for
all}\quad\theta\in A.$

∎

Let $U=(S^{n-1},R)$ be a strictly starlike domain in $\mathbb{R}^{n},$ and let
$Q_{1},Q_{2},\ldots$ be a regular exhaustion of $U$ by compact sets of the
form

$Q_{k}=\\{(\theta,\rho):\rho\leq R_{k}(\theta)\\},$

where $(R_{k})_{k=1}^{\infty}$ is a strictly increasing sequence of smooth
functions $R_{k}:S^{n-1}\to(0,+\infty)$ converging to the function $R.$ Note
that the $Q_{k}^{\prime}$s are smoothly bounded domains in $U.$ We wish to
associate to the exhaustion $Q_{1},Q_{2},\ldots,$ a class of closed subsets of
$U.$

Let $F^{\prime}$ be a countable union of closed subsets $F^{\prime}_{k}\subset
S^{n-1},k=1,2,\ldots,$ which are spherically polar. We may assume that
$F^{\prime}_{1}\subset F_{2}^{\prime}\subset\cdots.$ Consider the closed sets

$F_{k}=\\{(\theta,\rho):\,\theta\in
F_{k}^{\prime},\,\,R_{k}(\theta)\leq\rho<R(\theta)\\},\quad\,k=1,2,\dots.$

and set $F=F_{1}\cup F_{2}\cup\ldots.$ We say that such a set $F$ is a
stalactite associated to the exhaustion $Q_{1},Q_{2},\ldots$ in the strictly
starlike domain $U.$

######  Lemma 5.13.

Stalactites in a strictly starlike domain $U$ are polar subsets of $U.$

###### Proof.

This follows from Theorem 4.1.

∎

######  Lemma 5.14.

Let $U$ be an open subset of $\mathbb{R}^{n},$ $X$ be a smooth closed
hypersurface or a smoothly bordered domain in $U$ and $x$ be a point of
$\partial X$. Then, $U\setminus X$ is not thin at $x.$

###### Proof.

Because of the smoothness assumption, $U\setminus X$ contains a cone, whose
vertex is $x.$ Thus, by the classical Poincaré-Zaremba cone condition (see
e.g. [19]), $U\setminus X$ is not thin at $X.$

∎

Theorem 5.6 gave us an explicit method for constructing Carleman sets. We
shall now invoke this method to obtain one of our main results. Let
$F^{\prime}$ be as above and let $E^{\prime}_{1}\subset
E^{\prime}_{2}\subset\cdots,$ be an increasing sequence of compact subsets of
$S^{n-1}\setminus F^{\prime},$ such that $m(S^{n-1}\setminus
E^{\prime}_{j})<1/j,$ where $m$ is Hausdorff $(n-1)-$measure on $S^{n-1}.$ We
define closed spherical caps $C_{j}$ as follows

$C_{j}=\left\\{\big{(}\theta,R_{j}(\theta)\big{)}:\theta\in
E^{\prime}_{j}\right\\},$

which are subsets of the closed hypersurfaces $\partial Q_{j}.$ Set
$C=\cup_{j=1}^{\infty}C_{j}.$ Note that by Lemma 5.14, $C$ is a chaplet in
$U.$

######  Theorem 5.15.

Let $U=(S^{n-1},R)$ be a strictly starlike domain in $\mathbb{R}^{n},n\geq 3,$
$L$ be an operator of the form (1) on $U,$ $F$ be a stalactite in $U$ and
$C=C_{1}\cup C_{2}\cup\cdots$ be as above. Then the set

$E=F\cup C,$

is an $L-$Carleman set in $U.$

###### Proof.

This is a consequence of Lemma 5.13, Lemma 5.14 and Theorem 5.6.

∎

For a strictly starlike domain $U=(S^{n-1},R)$ in $\mathbb{R}^{n},$ a function
$u:U\to[-\infty,+\infty]$ and a point $\theta\in S^{n-1},$ denote by
$C(u,\theta)$ the radial cluster set

$C(u,\theta)=\\{w\in[-\infty,+\infty]:\textit{ there exists a sequence
}r_{j}(\theta)\nearrow R(\theta),\,\textit{ such that }u(r_{j},\theta)\to
w\\}.$

######  Theorem 5.16.

Suppose $U=(S^{n-1},R)$ is a strictly starlike domain in $\mathbb{R}^{n},n\geq
3,$ and let $L$ be an operator of the form (1) on $U.$ Moreover, let
$F^{\prime}$ be a countable union of compact (spherically) polar subsets of
$S^{n-1}.$ Then, for every function $\varphi$ continuous on $U,$ there is an
$L-$harmonic function $h$ on $U$ such that,

$\lim_{\rho\nearrow
R(\theta)}\big{(}h-\varphi\big{)}((\theta,\rho))=0,\quad\mbox{for
all}\quad\theta\in F^{\prime}$

and

$C(h,\theta)=[-\infty,+\infty],\quad\mbox{for a.e.}\quad\theta\in
S^{n-1}\setminus F^{\prime}.$

###### Proof.

We may express $F^{\prime}$ in the form $F^{\prime}=\cup_{j}F^{\prime}_{j},$
where $F_{1}^{\prime}\subset F_{2}^{\prime}\subset\cdots$ are closed polar
sets. We may assume that the sequence $(F_{j}^{\prime})_{j=1}^{\infty}$ is
infinite, because if $F^{\prime}=F_{1}^{\prime}\cup\cdots\cup F_{k}^{\prime},$
we may set $F_{j}^{\prime}=F_{k}^{\prime},$ for all $j\geq k.$ To the compact
sets $F_{j}^{\prime}\subset S^{n-1}$ we associate the closed sets
$F_{j}\subset U$ as above, for $j=1,2,\ldots.$ The set

$E=F\cup C$

is an $L-$Carleman set in $U,$ by Theorem 5.15.

Let $(c_{j})_{j=1}^{\infty}$ be a dense sequence in $(-\infty,+\infty).$ We
define a continuous function $\psi$ on $E,$ by setting $\psi=\varphi$ on $F$
and $\psi=c_{j}$ on $C_{j}$ for $j=1,2,\ldots.$

On $U,$ consider the continuous function $\rho:U\to(0,+\infty),$ defined as
$\rho(x)=|x|$ and the defining function $R:U\to(0,+\infty],$ for $U,$ which by
Lemma 5.11 is also continuous and define the function
$\epsilon:U\to(0,+\infty),$ by setting $\epsilon(x)=\chi(\rho(x),R(x)),$ which
is continuous, since
$(0,+\infty)\times(0,+\infty)\subset\overline{\mathbb{C}}\times\overline{\mathbb{C}}$
and the spherical distance $\chi$ is continuous on the Riemann sphere
$\overline{\mathbb{C}}.$ Of course $\epsilon(x)=\epsilon(\theta,\rho)$ tends
to zero, as $\rho\to R(\theta),$ for all $\theta\in S^{n-1}.$

Finally, by the definition of a $L-$Carleman set, there is an $L-$harmonic
function $h$ on $U,$ such that $|\varphi(x)-h(x)|<\epsilon(x),$ for all $x\in
E.$

∎

##  6\. Some special domains for $\Delta$-harmonic functions

###  6.1. Radial domains

By a polar set in the sphere $S^{n},$ we mean a polar set in the harmonic
space $S^{n},$ as an $n$-dimensional Riemannian manifold. There is no polar
set in $S^{1},$ because of the remark following Lemma 3.2.

By Theorem 4.1, if we consider $\pi:\mathbb{R}^{n}\setminus\\{0\\}\to
S^{n-1},\,x\mapsto x/|x|$ (i.e. the radial projection), then $E^{\prime}$ in
$S^{n-1}$ is a spherically polar set if and only if $E=\pi^{-1}(E^{\prime})$
is polar in $\mathbb{R}^{n}\setminus\\{0\\}.$

Let $U^{\prime}$ be a domain in $S^{n-1}$ and
$r:U^{\prime}\to[0,+\infty),\,R:U^{\prime}\to(0,+\infty]$ be continuous
functions, with $r<R.$ We call the domain

$U=(U^{\prime},r,R)=\\{(\theta,\rho)\in S^{n-1}\times\mathbb{R}^{+}:\theta\in
U^{\prime},\,r(\theta)<\rho<R(\theta)\\}$

a radial line domain.

######  Theorem 6.1.

Let $U\subset\mathbb{R}^{n},n\geq 3,$ be a radial line domain and
$F^{\prime}\subset U^{\prime}$ an $F_{\sigma}$ set, which we assume to be
spherically polar. Then, for every function $\varphi$ continuous on $U,$ there
is a harmonic function $h$ on $U$ such that,

$\big{(}h-\varphi\big{)}(\theta,\rho)\to 0,\quad\mbox{as}\quad\rho\downarrow
r(\theta),\quad\mbox{or}\quad\rho\uparrow R(\theta)\quad;\quad\textit{ for all
}\quad\theta\in F^{\prime}.$

###### Proof.

This is not an immediate consequence of Theorem 4.2, because $U$ is not a
trivial harmonic bundle, since the fibre over a point $\theta\in S^{n-1}$ is
not the whole line $(-\infty,+\infty),$ but rather an open interval
$\big{(}r(\theta),R(\theta)\big{)}$ in $[0,+\infty].$ However, the only
property of the projection $\pi$ which we needed in the proof of Theorem 4.2
was that the pull back of polar, which is the case here.

We write $F$ as a union $F^{\prime}=F_{1}^{\prime}\cup
F_{2}^{\prime}\cup\cdots,$ of closed polar sets as before. Let
$(r_{k})_{k=1}^{\infty}$ be a decreasing and $(R_{l})_{l=1}^{\infty}$ be an
increasing sequence of continuous functions on $U^{\prime}$ such that
$r_{k}\downarrow r$ and $R_{k}\uparrow R.$ We set

$F_{k}=\big{\\{}(\theta,\rho):\theta\in
F_{k}^{\prime},\,\,\rho\in\big{(}r(\theta),R(\theta)\big{)}\setminus\big{(}r_{k}(\theta),R_{k}(\theta)\big{)}\big{\\}}$

and $F=\cup_{k=1}^{\infty}F_{k}.$ Then, as in the proof of Theorem 4.2, the
set $F$ is a Carleman set for harmonic approximation in $U.$ The rest of the
proof is like that of Theorem 4.2 but much simpler.

∎

######  Theorem 6.2.

Suppose $U=(S^{n-1},R)$ is a strictly starlike domain in $\mathbb{R}^{n},n\geq
3,$ and let $F^{\prime}$ be an $F_{\sigma}$ subset of $S^{n-1},$ which is
spherically polar. Then, for every function $\varphi$ continuous on $U,$ there
is a function $h$ harmonic on $U$ such that,

$\lim_{\rho\nearrow
R(\theta)}\big{(}h-\varphi\big{)}((\theta,\rho))=0,\quad\mbox{for
all}\quad\theta\in F^{\prime}.$

###### Proof.

The proof is analogous to the proof of Theorem 6.1 since, again the pull back
of nowhere dense sets and polar sets are respectively nowhere dense and polar.
The theorem is also covered by Theorem 5.16.

∎

###  6.2. Vertical line domains

Given a domain $U^{\prime}\subset\mathbb{R}^{n-1},$ by a vertical line domain
$U=(U^{\prime},c,d),$ with base $U^{\prime}$, we mean a domain
$U\subset\mathbb{R}^{n}$ of the form

$U=\\{(p^{\prime},y)\in\mathbb{R}^{n-1}\times\mathbb{R}=\mathbb{R}^{n}:p^{\prime}\in
U^{\prime},\,c(p^{\prime})<y<d(p^{\prime})\\},$

where $c,d:U^{\prime}\to[-\infty,+\infty]$ are continuous functions, with
$c<d.$ Note that $U^{\prime}$ is the projection $\pi(U)$ of $U.$

######  Theorem 6.3.

Let $U=(U^{\prime},c,d)$ be a vertical line domain and $F^{\prime}\subset
U^{\prime}$ an $F_{\sigma}$ set, which we suppose to be polar. Then, for every
$\varphi$ continuous on $U,$ there is a harmonic function $h$ on $U,$ such
that, for all $x^{\prime}\in F^{\prime},$

$\big{(}h-\varphi\big{)}(x^{\prime},y)\to 0,\quad\mbox{if}\quad y\downarrow
c(x^{\prime})\quad\mbox{or}\quad y\uparrow d(x^{\prime}).$

###### Proof.

We first write $F^{\prime}$ as a union of closed polar sets, that is,
$\cup_{k}F^{\prime}_{k},$ where each $F_{k}^{\prime}$ is a closed polar set in
$U^{\prime}.$

Let $(c_{k})_{k=1}^{\infty}$ and $(d_{k})_{k=1}^{\infty}$ be sequences of
continuous functions on $U^{\prime},$ with the sequence
$(c_{k})_{k=1}^{\infty}$ strictly decreasing to $c,$ $(d_{k})_{k=1}^{\infty}$
strictly increasing to $d$ and $c_{1}<d_{1}.$ For $k=1,2,\ldots,$ set

$F_{k}=\\{(x^{\prime},y):x^{\prime}\in F_{k}^{\prime},\,c(x^{\prime})<y\leq
c_{k}(x^{\prime})\\}\cup\\{(x^{\prime},y):x^{\prime}\in
F_{k}^{\prime},\,d_{k}(x^{\prime})\leq y<d(x^{\prime})\\}$

and $F=\cup_{k=1}^{\infty}F_{k}.$

It is straightforward, using Theorem 3.3 to show that $F$ is a Mergelyan set
and the rest of the proof follows the pattern we have established.

∎

###  6.3. Upper domains.

Let us call a domain $U\subset\mathbb{R}^{n},\,n>1,$ an upper domain if, for
each $x=(x_{1},\ldots,x_{n-1},x_{n})\in U,$ every point
$(x_{1},\ldots,x_{n-1},y),\,y>x_{n},$ is also in $U.$ Writing
$\mathbb{R}^{n}=\mathbb{R}^{n-1}\times\mathbb{R},$ let
$\pi:\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ be the associated projection. For
$x\in\mathbb{R}^{n},$ we write $x=(p^{\prime},x_{n}),$ where
$\pi(x)=p^{\prime}=(p_{1},\dots,p_{n-1}).$ If $U\subset\mathbb{R}^{n}$ is an
upper domain, we denote $U^{\prime}=\pi(U),$ which is a domain in
$\mathbb{R}^{n-1}.$ For $p^{\prime}\in U^{\prime},$ set

(7)
$L(p^{\prime})=\inf\left\\{x_{n}:(p^{\prime},x_{n})=(p_{1},\ldots,p_{n-1},x_{n})\in
U\right\\}.$

The upper domain $U$ can be described as

$\\{x=(p^{\prime},y)\in U^{\prime}\times\mathbb{R}:L(p^{\prime})<y\\}.$

######  Lemma 6.4.

If $U$ is an upper domain, the function $p^{\prime}\mapsto L(p^{\prime})$
mapping $U^{\prime}$ into $[-\infty,+\infty)$ is upper semicontinuous.

###### Proof.

That $L(p)<+\infty$ follows from the connectivity of $U.$ Suppose, to obtain a
contradiction, that there is a point $p^{\prime}\in U^{\prime},$ a finite
value $y>L(p^{\prime})$ and a sequence $p^{\prime}_{j}\to p^{\prime}$ such
that $L(p^{\prime}_{j})>y.$ Since $U$ is open, it follows that no point in the
segment $\\{(p^{\prime},t):L(p^{\prime})<t<y\\}$ lies in $U,$ which
contradicts the definition of $L(p^{\prime}).$

∎

An upper domain $U$ will be called a strictly upper domain if, for every
$x=(x_{1},\ldots,x_{n-1},x_{n})\in\partial U,$ every point
$(x_{1},\ldots,x_{n-1},y),\,y>x_{n},$ is in $U.$ For example, in
$\mathbb{R}^{2},$ the open upper half-space with the vertical segment
$\\{(x,y):0<y\leq 1\\}$ deleted is an upper domain, but not a strictly upper
domain.

######  Lemma 6.5.

If $U$ is a strictly upper domain, then the function $p^{\prime}\mapsto
L(p^{\prime}):U^{\prime}\to[-\infty,+\infty)$ is continuous.

###### Proof.

In view of the previous lemma, we need only show lower semicontinuity. Fix
$q^{\prime}\in U^{\prime}.$ If $L(q^{\prime})=-\infty,$ lower semicontinuity
at $q^{\prime}$ is automatic. Suppose $L(q^{\prime})>-\infty$ and suppose, to
obtain a contradiction, that there is a sequence $p^{\prime}_{j}\to
q^{\prime}$ and $\epsilon>0,$ such that
$L(p^{\prime}_{j})<L(q^{\prime})-\epsilon.$ Then, there is a
$y_{n}<L(q^{\prime})$ and a sequence $p^{\prime}_{j}\to q^{\prime}$ with
$(p^{\prime}_{j},y_{n})\in U.$ Therefore, $(q^{\prime},y_{n})\in\partial U.$
By strictness, $\big{(}q^{\prime},L(q^{\prime})\big{)}\in U,$ which
contradicts the definition of $L(q^{\prime}).$

∎

We have the following particular case of Theorem 6.3.

######  Corollary 6.6.

Let $U\subset\mathbb{R}^{n},n\geq 2,$ be a strictly upper domain and
$F^{\prime}\subset U^{\prime}$ an $F_{\sigma}$ set, which we suppose to be
polar. Then, for every $\varphi$ continuous on $U,$ there is a harmonic
function $h$ on $U,$ such that, for all $x^{\prime}\in F^{\prime},$

$\big{(}h-\varphi\big{)}(x^{\prime},y)\to 0,\quad\mbox{if}\quad y\downarrow
L(x^{\prime})\quad\mbox{or}\quad y\uparrow+\infty.$

###### Proof.

We may let $U=(U^{\prime},c,d),$ where $c=L,$ defined by (7), and $d=+\infty,$
then apply Theorem 6.3.

∎

## Acknowledgment

The junior author would like to thank the senior author for introducing the
problem and for his generous support. Also he is grateful to Dr. Dmitry
Jakobson and Dr. Javad Mashreghi for the financial support, and hospitality,
during his postdoct studies at McGill University.

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