\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 322, pp. 1--4.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/322\hfil $L^p$-subharmonic functions in $\mathbb{R}^n$]
{$L^p$-subharmonic functions in $\mathbb{R}^n$}

\author[M. Damlakhi \hfil EJDE-2016/322\hfilneg]
{Moustafa Damlakhi}

\address{Moustafa Damlakhi \newline
Department of Mathematics,
College of Science, King Saud University,
P.O. Box 2455, Riyadh 11451, Saudi Arabia}
\email{damlakhi@ksu.edu.sa}


\thanks{Submitted August 8, 2016. Published December 20, 2016.}
\subjclass[2010]{31B05, 30D55}
\keywords{$L^p$-subharmonic functions; harmonic Hardy class}

\begin{abstract}
 We prove that if $u$ is an $L^p$-subharmonic function defined outside
 a compact set in $\mathbb{R}^n$, it is bounded above near infinity, in
 particular, if the subharmonic function $u$ is in $L^p(\mathbb{R}^n)$,
 $1\leq p<\infty $, then $u$  is non-positive. Some of the consequences of
 this property are obtained. We discuss the properties of subharmonic
 functions defined outside a compact set in $\mathbb{R}^n$ if they are also
 $L^p$ functions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks


\section{Introduction}

An upper semi-continuous function $u$,  taking the value infinity, and not
identically ($-\infty$) is called a subharmonic function in $\mathbb{R}^n$
if it has sub-mean value property. The properties of functions with
mean-value properties (called harmonic functions) are given in Axler \cite{a2}
analogous properties for subharmonic functions are also known. In this note,
we derive some properties of subharmonic functions on $\mathbb{R}^n$ when
they are also $L^p$ functions. For example, we show that a subharmonic 
$L^p$ function in $\mathbb{R}^n$ is non-positive.

From Anandam \cite{a1} it is easy to see that if $s(x)$ is a subharmonic
function defined outside a compact set in $\mathbb{R}^n$, then
$s(x)=v(x)+cu(x)+b(x)$ near infinity, where $v(x)$ is subharmonic on 
$\mathbb{R}^n$, $u(x)=\log | x| $ if $n=2$ and $u(x)=| x| ^{2-n}$ if $n>2$,
$c$ is constant and $b(x)$ is bounded harmonic function.
 We obtain some properties of $s(x)$ if it is in addition an $L^p$ function also.

In particular, we show that if the subharmonic function outside a compact
set is an $L^p$ function, then $s(x)$ tends to $0$ at infinity.

\section{Subharmonic functions in $L^p(\mathbb{R}^n)$}

In this note we consider $p<\infty $ and $n\geq 2$.

\begin{lemma} \label{lem2.1} 
Let $s\geq 0$ be a subharmonic function in $\mathbb{R}^n$. 
If $s\in L^p(\mathbb{R}^n),$ $p\geq 1$, then $s\equiv 0$.
\end{lemma}

\begin{proof} 
For $x_0\in\mathbb{R}^n$, let 
$B_{n}=\{ x:| x-x_0| =1\} $ and $\sigma _{n}$ be the
surface area of $B_{n}$. Since $\ s\geq 0$, $s^p$ is subharmonic and using
the polar coordinates for $x=(r,w)$ ,$| x-x_0| =r$.

By using the expression of the sub-mean-value property of $s^p$ we have
\[
s^p(x_0)\leq \frac{1}{\sigma _{n}}\int_{B_{n}}s^p(r,w)dw.
\]
From this inequality and using that
$s\in L^p(\mathbb{R}^n)$ we have
\[
\infty > \int_0^{\infty }\int_{B_{n}}s^p(r,w)r^{n-1}drdw
\geq  \int_0^{\infty }\sigma _{n}s^p(x_0)r^{n-1}dr.
\]

This is possible if and only if $s^p(x_0)=0$. Since $x_0$ is
arbitrary, $s^p\equiv 0$ in $\mathbb{R}^n$.
\end{proof}

\begin{theorem} \label{thm2.2} 
If $s$ is subharmonic function in $L^p(\mathbb{R}^n)$ with $p\geq1$, 
then $s\leq 0$.
\end{theorem}

\begin{proof} 
$s^{+}=\sup (s,0)$ is subharmonic and is in $L^p(\mathbb{R}^n)$. 
Hence by the Lemma \ref{lem2.1} $s^{+}\equiv 0$ and consequently $s\leq 0$ in
$\mathbb{R}^n$.
\end{proof}

\begin{corollary} \label{coro2.3}
Let $s$ be a subharmonic function in $L^p(\mathbb{R}^n)$,
 $1\leq p\leq \frac{n}{n-1}$. Then $s\equiv 0$.
\end{corollary}

\begin{proof} 
By the Theorem \ref{thm2.2}, $s\leq 0$ for all $p$, 1$\leq p<\infty $.

(1) Let $n=2$. Since $s$ is an upper bounded subharmonic function in
$\mathbb{R}^{2}$, it is a constant. If $s\in L^{\infty }(\mathbb{R}^{2})$, even
though Theorem \ref{thm2.2} does not hold, yet $s$ is also a constant in this case.

(2) Let $n\geq 3$. Since $-s$ a is positive supharmonic function, by Riesz
decomposition $-s=u+v$ where $u$ is a potential and $h\geq0$ is harmonic and
hence constant. Since $u\leq -s$, $u\in L^p(\mathbb{R}^n)$.

Now, if $B$ is the unit ball in $\mathbb{R}^n$, $n\geq 3$, we define the
function
\[
\vartheta (x)=\begin{cases}
1&\text{if }x\in B  \\
|x| ^{2-n}&\text{if }x\in\mathbb{R}^n- B
\end{cases}
\]
Then $\vartheta (x)$ is a potential and
$u(x)\geq (\underset{x\in\overline{B}}{\inf }u(x))\vartheta (x) \in \mathbb{R}^n$
Consequently, $\vartheta (x)\in L^p(\mathbb{R}^n)$, but this would imply
\[
\int_1^{\infty } (r^{2-n})^pr^{n-1}\, dr<\infty .
\]
This is not possible if $p(2-n)+n-1\geq-1$ which means that if
$p\leq \frac{n}{n-2},u\equiv 0$ and hence $s$ is constant.
Thus, for all $n\geq2$, $s\equiv A$, a constant. Since $s\in L^p(\mathbb{R}^n)$,
$A\equiv 0$ when $p<\infty $.
\end{proof}

\begin{corollary} \label{coro2.4}
If $s$ is a subharmonic function in $L^p(\mathbb{R}^n)$, $p\geq 1$, which 
is associated measure $\mu $ in a local Riesz representation, $\mu $ does not 
charge points; that is $\mu (\{x\} )=0$ for every $x \in \mathbb{R}^n$.
\end{corollary}

\begin{proof} 
In view of the above corollary, we assume $n\geq 3$. By
Theorem \ref{thm2.2}, it follows that $u=-s$ is a potential. Since $\mu $ is the
measure associated with the subharmonic function $s$, it is always non-positive.
If we Suppose that it is strictly negative at a point, it leads to
a contradiction. In fact, we assume that $\mu (\{ x_0\})=\alpha<0$ for some 
$x_0$ in $\mathbb{R}^n$.

Let $B=\{ x:| x-x_0| <1\} $. Since $u(x)\geq -\alpha | x-x_0| ^{2-n}$ in $B$ 
and since $u$  is in $L^p(B)$, we should have $p<n/(n-2)$.

In $\mathbb{R}^n-B$. Since 
$u(x)\geq (\min_{x\in \overline{B}}| x-x_0| ^{2-n})$  and since 
$u \in L^p(\mathbb{R}^n)$. We should have $p>n/(n-2)$ as in the proof 
of Corollary \ref{coro2.3}.

Thus, for any choice of $p\geq 1$, $u\notin L^p(\mathbb{R}^n)$. This
contradiction shows that $\mu (\{ x_0\} )=0$.
\end{proof}

Recall that a $C^{\infty }$ function $q(x)$ in an open set in 
$\mathbb{R}^n $ is called a quasiharmonic function if $\Delta q=-1$.

\begin{corollary}[{\cite[pp. 120-122]{s1}}]  \label{coro2.5}
 Let $s$ be subharmonic in $\mathbb{R}^n$ such that $\Delta s=A$, a constant 
(with $\Delta $ in the sense of distributions). 
Suppose  $s\in L^p( \mathbb{R}^n)$, $p\geq1$. Then $s\equiv 0$.
\end{corollary}

\begin{proof}  
Since $s$ is subharmonic, $A=\Delta s\geq 0$. 
Suppose $A>0$. Note by the Theorem \ref{thm2.2}. $s\leq 0$. Since
\[
\Delta s=A,\quad \Delta \Big( s(x)-\frac{A| x|^{2} }{2n}\Big) =0
\]
and hence there exists a harmonic function $h(x)\in\mathbb{R}^n$ such that
$s(x)=\frac{A}{2n}| x| ^{2}$ $+h(x)$ a.e.

If two subharmonic functions are equal a.e., they are equal every where;
hence $s(x)=\frac{A}{2n}| x| ^{2}$ $+h(x)$.
Since $h(x)\leq s(x)\leq 0$, $h$ is a constant which leads to a
contradiction since  $s\leq 0$ and $A| x| ^{2}$ tends to
$\infty $. Hence $\Delta s=0$.
Thus $s$ is a harmonic function in $L^p(\mathbb{R}^n)$. In this case,
the Theorem \ref{thm2.2}. implies that $s\equiv 0$.
\end{proof}

\section{$L^p$ Subharmonic function outside a compact set in $\mathbb{R}^n$}

Let $u$ be subharmonic function outside a compact set in $\mathbb{R}^n$.
We say that $u$  extends subharmonically in $\mathbb{R}^n$ if there
exists a subharmonic function $V$ in $\mathbb{R}^n$, such that $V$ is not
bounded from above by a harmonic function in $\mathbb{R}^n$ and $V=u$
outside a compact set.

\begin{proposition} \label{prop3.1}
Let $u$ be an $L^p$ subharmonic function outside
a compact set. Then $u$ cannot be extended subharmonically in $\mathbb{R}^n$.
\end{proposition}

\begin{proof} 
Let $V$ be subharmonic function in $\mathbb{R}^n$ not
bounded from above by a harmonic function in $\mathbb{R}^n$ such that $V=u$
outside a compact set. Then for large $r$, the function $s$ defined as
\[
s=\begin{cases}
u & \text{if }|x| \geq r\\
D_{r}u &\text{if } | x| <r
\end{cases}
\]
Where $D_r u$ is the Dirichlet solution in $|x| <r$ with
boundary values $u$, is subharmonic in $\mathbb{R}^n$
and $s\geq V$.

If $u(x)\in L^p$ in $|x| \geq r$, $s(x)$ is in the
harmonic Hardy class $h^p$ in $|x| <r$ (see \cite[page 103]{a2}) 
and hence there exists a  harmonic function $H(x)$ in $| x| <r$ such that 
$| s| ^p<H$.
Then
\[
\int_{| x| <r }| s(x)| ^p\ dx\leq
c_{n}H(0),
\]
for a constant $c_{n}$. That is $s(x)\in L^p$ in 
 $| x| <r$. Which implies that $s\in L^p(\mathbb{R}^n)$ since 
$s(x)=u(x)$ in $| x| \geq r$. By Theorem \ref{thm2.2}, $s\leq 0$
and hence $V\leq 0$ in $\mathbb{R}^n$, a contradiction.
\end{proof}

\begin{corollary} \label{coro3.2} 
Let $u(x)$ be subharmonic in an open set $w$
containing $| x| \geq r$ in $\mathbb{R}^n$. Suppose 
$u\in L^p(w)$ for some $p\geq 1$. Then $u(x)$ is upper bounded in $| x| \geq r$.
\end{corollary}

\begin{proof} 
By hypothesis $u^{+}(x)$ is an $L^p$ subharmonic function
in an open set containing $| x| \geq r$.

(1) In $\mathbb{R}^{2}$, if $u^{+}$ is not upper bounded in 
$|x| \geq r$ , it can be extended subharmonically in $\mathbb{R}^{2}$
(see \cite[Corollary 1]{a1}). This is a contradiction with 
Proposition \ref{prop3.1},
since $u^{+}\in L^p$ in $| x| \geq r$.
This means that $u^{+}(x)$ and hence $u(x)$ is upper bounded in 
$|x| \geq r$.

(2) In $\mathbb{R}^n$, $n\geq 3$, there exists a subharmonic function $s(x)$
in $\mathbb{R}^n$ and some $\alpha \leq 0$ such that 
$u^{+}(x)=s(x)-\alpha| x| ^{2-n}$ in $| x| \geq r$  (see
\cite[Theorem 1]{a1}). Hence $s(x)\geq \alpha | x| ^{2-n}$ in $| x| \geq r$.

Let $M(R,s)$ denote the the mean-value of $s(x)$ in $| x| =R$.
If $\lim_{R\to \infty } M(R,s)=\infty $, then 
$\lim_{R\to \infty } M(R,u^{+})=\infty $. Hence $u^{+}$
can be extended subharmonically in $\mathbb{R}^n$ 
(see \cite[Theorem 2]{a1}), a contradiction; thus $\lim_{R\to \infty }M(R,s)=\infty $.

When $\lim_{R\to \infty } M(R,s)$ is finite, $s$ has a
harmonic majorant $h$ in $\mathbb{R}^n$. 
Since $h$ is lower bounded, it is a constant $c$ and $c\geq 0$. 
(We remark in passing that $c$ here can be
chosen as $0$ if $p>\frac{n}{n-2}$, see Corollary \ref{coro2.3}).
 Hence $u^{+}(x)$ is bounded in $| x| \geq r$, and consequently $u(x)$ is
upper bounded by $c-\alpha | x| ^{2-n}$ in $|x| \geq r$. Thus, in all 
cases $u(x)$ is upper bounded in  $| x| \geq r$.
\end{proof}

\begin{remark} \label{rmk} \rm
In particular, we deduce that if $h$ is an $L^p$
subharmonic function defined outside a compact set in $\mathbb{R}^n$, then
$h$  tends to $0$ at at infinity; if $h$ is a harmonic function
defined outside a compact set in $\mathbb{R}^n$, $n\geq 3$. 
Tending to $0$ at infinity, then $h$ is in $L^p$ in $| x| \geq r$ for
large $r$ if $p> \frac{n}{n-2}$.
\end{remark}

\begin{thebibliography}{0}

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\bibitem{a2} S. Axler, P. Bourdon, W. Ramey;
\emph{Harmonic function theory}, Springer Verlag, New York. 1992.

\bibitem{s1} L. Sario, M. Nakai, C. Wang, L. O. Chung;
\emph{Classification theory of Riemannian manifolds}, Springer Verlag,
 L. N. 605, (1977).

\end{thebibliography}

\end{document}