---
abstract: |
    In this paper, we consider $\alpha$-harmonic functions in the half space $\mathbb{R}^n_+$: $$\left\{\begin{array}{ll}
    (-{\mbox{$\bigtriangleup$}})^{\alpha/2} u(x)=0,~u(x)>0, & \qquad x\in\mathbb{R}^n_+, \\
    u(x)\equiv0, & \qquad x\notin\mathbb{R}^{n}_{+}.
    \end{array}\right.
    \label{1}$$ We prove that all solutions of (\[1\]) have to assume the form $$u(x)=\left\{\begin{array}{ll}Cx_n^{\alpha/2}, & \qquad x\in\mathbb{R}^n_+, \\
    0, & \qquad x\notin\mathbb{R}^{n}_{+},
    \end{array}\right.
    \label{2}$$ for some positive constant $C$.
author:
- Wenxiong Chen Congming Li Lizhi Zhang Tingzhi Cheng
title: 'A Liouville theorem for $\alpha$-harmonic functions in $\mathbb{R}^n_+$'
---

[**Key words**]{} The fractional Laplacian, $\alpha$-harmonic functions, uniqueness of solutions, Liouville theorem, Poisson representation.\
\
\

Introduction
============

The fractional Laplacian in $R^n$ is a nonlocal pseudo-differential operator, assuming the form $$(-\Delta)^{\alpha/2} u(x) = C_{n,\alpha} \, \lim_{\epsilon {{\mbox{$\rightarrow$}}}0} \int_{\mathbb{R}^n\setminus B_{\epsilon}(x)} \frac{u(x)-u(z)}{|x-z|^{n+\alpha}} dz,
\label{Ad7}$$ where $\alpha$ is any real number between $0$ and $2$. This operator is well defined in $\cal{S}$, the Schwartz space of rapidly decreasing $C^{\infty}$ functions in $\mathbb{R}^n$. In this space, it can also be equivalently defined in terms of the Fourier transform $$\widehat{(-\Delta)^{\alpha/2} u} (\xi) = |\xi|^{\alpha} \hat{u} (\xi),$$ where $\hat{u}$ is the Fourier transform of $u$. One can extend this operator to a wider space of distributions.

Let $$L_{\alpha}=\{u: \mathbb{R}^n\rightarrow \mathbb{R} \mid \int_{\mathbb{R}^n}\frac{|u(x)|}{1+|x|^{n+\alpha}} \, d x <\infty\}.$$

Then in this space, we defined $(-\Delta)^{\alpha/2} u$ as a distribution by $$< (-\Delta)^{\alpha/2}u(x), \phi> \, = \, \int_{\mathbb{R}^n}  u(x) (-\Delta)^{\alpha/2} \phi (x) dx ,  \;\;\; \forall \, \phi \in C_0^{\infty}(\mathbb{R}^n) .$$

Let $$\mathbb{R}^n_+ = \{x=(x_1, \cdots, x_n\ \mid x_n > 0 \}$$ be the upper half space. We say that $u$ is $\alpha$-harmonic in the upper half space if $$\int_{\mathbb{R}^n}  u(x) (-\Delta)^{\alpha/2} \phi (x) dx = 0,  \;\;\; \forall \, \phi \in C_0^{\infty}(\mathbb{R}^n_+) .$$

In this paper, we consider the Dirichlet problem for $\alpha$-harmonic functions $$\left\{\begin{array}{ll}
(-{\mbox{$\bigtriangleup$}})^{\alpha/2} u(x)=0,~u(x)>0, & \qquad x\in\mathbb{R}^n_+, \\
u(x)\equiv0, & \qquad x\notin\mathbb{R}^{n}_{+},
\end{array}\right.
\label{1.1}$$ It is well-known that $$u(x)=\left\{\begin{array}{ll}Cx_n^{\alpha/2}, & \qquad x\in\mathbb{R}^n_+, \\
0, & \qquad x\notin\mathbb{R}^{n}_{+},
\end{array}\right.$$ is a family of solutions for problem (\[1.1\]) with any positive constant $C$.

A natural question is: [*Are there any other solutions?*]{}

Our main objective here is to answer this question and prove

Let $0<\alpha<2$, $u\in L_{\alpha}$. Assume $u$ is a solution of $$\left\{\begin{array}{ll}
(-{\mbox{$\bigtriangleup$}})^{\alpha/2} u(x)=0,~u(x)>0, & \qquad x\in\mathbb{R}^n_+, \\
u(x)\equiv0, & \qquad x\notin\mathbb{R}^{n}_{+}.
\end{array}\right.$$ then $$u(x)=\left\{\begin{array}{ll}Cx_n^{\alpha/2}, & \qquad x\in\mathbb{R}^n_+, \\
0, & \qquad x\notin\mathbb{R}^{n}_{+},
\end{array}\right.$$ for some positive constant $C$. \[mthm1\]

We will prove this theorem in the next section.

The Proof of the Liouville Theorem
==================================

In this section, we prove Theorem \[mthm1\]. The main ideas are the following.

We first obtain the Poisson representation of the solutions. We show that for $|x-x_r|<r$

$$u(x)=\int_{|y-x_r|>r}P_r(x-x_r,y-x_r)u(y)dy,
\label{P}$$

where  $x_r=(0, \cdots,0,r)$ , and $P_r(x-x_r,y-x_r)$ is the Poisson kernel for $|x-x_r|<r$ : $$\begin{aligned}
&&P_{r}(x-x_r,y-x_r)\nonumber\\&=&\left\{\begin{array}{ll}
\frac{\Gamma(n/2)}{\pi^{\frac{n}{2}+1}} \sin\frac{\pi\alpha}{2}
\left[\frac{r^{2}-|x-x_r|^{2}}{|y-x_r|^{2}-r^{2}}\right]^{\frac{\alpha}{2}}\frac{1}{|x-y|^{n}},\qquad& |y-x_r|>r,\\
0,& \text{elsewhere}.
\end{array}
\right.\end{aligned}$$

Then, for each fixed $x\in\mathbb{R}^n_+$, we evaluate first derivatives of $u$ by using (\[P\]). Letting $r \rightarrow \infty$, we derive $$\frac{\partial u}{\partial x_i}(x)=0,~~~~~i=1,2,\cdots,n-1.$$ and $$\frac{\partial u}{\partial x_n}(x)=\frac{\alpha}{2x_n}u(x).$$ These yield the desired results.

In the following, we use $C$ to denote various positive constants.

[*Step 1.*]{}

In this step, we obtain the Poisson representation (\[P\]) for the solutions of (\[1.1\]).

Let $$\hat{u}(x)=\left\{
\begin{array}{ll}
\int_{|y-x_r|>r}P_r(x-x_r,y-x_r)u(y)dy, & \qquad |x-x_r|<r, \\
u(x), & \qquad |x-x_r|\geq r.
\end{array}
\right.
\label{max}$$ We will prove that  $\hat{u}$ is  $\alpha$ -harmonic in  $B_r(x_r) $. The proof is similar to that in [@CL]. It is quite long and complex, hence for reader’s convenience, we will present it in the next section.

Let $w(x)=u-\hat{u}$ , then $$\left\{
\begin{array}{ll}
(-{\mbox{$\bigtriangleup$}})^{\alpha/2}w(x)=0, & \qquad |x-x_r|<r, \\
w(x)\equiv0, & \qquad |x-x_r|\geq r.
\end{array}
\right.$$

To show that  $w\equiv0$ ,we employ the following Maximum Principle.

(Silvestre, [@Si]) Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$, and assume that $v$ is a lower semi-continuous function on $\overline{\Omega}$ satisfying $$\left\{\begin{array}{ll}
(-{\mbox{$\bigtriangleup$}})^{\frac{\alpha}{2}}v\geq0,  &\mbox{in } \Omega,\\
v\geq0,&\mbox{on } \mathbb{R}^{n}\backslash\Omega.
\end{array}
\right.$$ then $v\geq0$ in $\Omega$.

Applying this lamma to both  $v=w$  and  $v=-w$ , we conclude that $$w(x)\equiv0.$$ Hence $$\hat{u}(x)\equiv u(x).$$ This verifies (\[P\]).

[*Step 2.*]{}

We will show that for each fixed  $x\in\mathbb{R}^n_+$ , $$\frac{\partial u}{\partial x_i}(x)=0,~~~~~i=1,2,\cdots,n-1.
\label{b1}$$ and $$\frac{\partial u}{\partial x_n}(x)=\frac{\alpha}{2x_n}u(x).
\label{b2}$$

From (\[b1\]), we conclude that  $u(x)=u(x_n)$, and this, together with (\[b2\]), immediately implies $$u(x)=Cx_n^{\alpha/2},$$ therefore $$u(x)=\left\{
\begin{array}{ll}
Cx_n^{\alpha/2}, & \qquad x\in\mathbb{R}^n_+, \\
0, & \qquad x\notin\mathbb{R}^n_+.
\end{array}
\right.$$ And this is what we want to derive.

Now, what left is to prove (\[b1\]) and (\[b2\]). Through an elementary calculation, one can derive that, for $i=1,2,\cdots,n-1,$ $$\begin{aligned}
\frac{\partial u}{\partial x_i}(x)&=&\int_{|y-x_r|>r}\left(\frac{-\alpha x_i}{r^2-|x-x_r|^2}+\frac{n(y_i-x_i)}{|y-x|^2}\right)P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\int_{|y-x_r|>r}\frac{-\alpha x_i}{r^2-|x-x_r|^2}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\&\quad&+\int_{|y-x_r|>r}\frac{n(y_i-x_i)}{|y-x|^2}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&:=&I_1+I_2.\label{b3}\end{aligned}$$

For each fixed  $x\in B_r(x_r)\subset\mathbb{R}^n_+$ and for any given  $\epsilon >0$ , we have $$\begin{aligned}
|I_1|&=&\left|\int_{|y-x_r|>r}\frac{-\alpha x_i}{r^2-|x-x_r|^2}P_r(x-x_r,y-x_r)u(y)dy\right|\nonumber\\
&\leq&\int_{|y-x_r|>r}\left|\frac{-\alpha x_i}{r^2-|x-x_r|^2}\right|P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\left|\frac{\alpha x_i}{r^2-|x-x_r|^2}\right|\int_{|y-x_r|>r}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\left|\frac{\alpha x_i}{r^2-|x-x_r|^2}\right|u(x)\nonumber\\
&\leq&\frac{C}{r}\nonumber\\
&<&\epsilon,~~~\text{for sufficiently large ~$r$~.}\label{b0}\end{aligned}$$

Here and in below, the letter  $C$  stands for various positive constants.

It is more delicate to estimate $I_2$. For each $R>0$, we divide the region $|y-x_r|>r$ into two parts: one inside the ball $|y|<R$ and one outside the ball.

$$\begin{aligned}
|I_2|&=&\left|\int_{|y-x_r|>r}\frac{n(y_i-x_i)}{|y-x|^2}P_r(x-x_r,y-x_r)u(y)dy\right|\nonumber\\
&\leq&\int_{|y-x_r|>r}\left|\frac{n(y_i-x_i)}{|y-x|^2}\right|P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\int_{\substack{|y-x_r|>r\\|y|>R}}\left|\frac{n(y_i-x_i)}{|y-x|^2}\right|P_r(x-x_r,y-x_r)u(y)dy\nonumber\\&\quad&+\int_{\substack{|y-x_r|>r\\|y|\leq R}}\left|\frac{n(y_i-x_i)}{|y-x|^2}\right|P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&:=&I_{21}+I_{22}.\label{b4}\end{aligned}$$

For the  $\epsilon>0$  given above, when  $|y|>R$ , we can easily derive $$\left|\frac{n(y_i-x_i)}{|y-x|^2}\right|\leq\frac{n}{|y-x|}\leq\frac{C}{R}<\epsilon,$$ for sufficiently large $R$. Fix this $R$, then $$\begin{aligned}
I_{21}&<&\int_{\substack{|y-x_r|>r\\|y|>R}}\epsilon P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&\leq&\epsilon\int_{|y-x_r|>r}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\epsilon u(x)\nonumber\\
&\leq&C\epsilon.\label{b5}\end{aligned}$$

To estimate $I_{22}$, we employ the expression of the Poisson kernel. $$\begin{aligned}
I_{22}&=& C \int_{\substack{|y-x_r|>r\\|y|\leq R}}\left[\frac{r^2-|x-x_r|^2}{|y-x_r|^2-r^2}\right]^{\alpha/2}\frac{u(y)}{|x-y|^n}\left|\frac{n(y_i-x_i)}{|y-x|^2}\right|dy\nonumber\\
&\leq& C \int_{\substack{|y-x_r|>r\\|y|\leq R}}\left[\frac{2x_nr-|x|^2}{2y_nr-|y|^2}\right]^{\alpha/2}\frac{u(y)}{|x-y|^n}\frac{1}{|y-x|}dy\nonumber\\
&\leq&C\int_{\substack{|y-x_r|>r\\|y|\leq R}}\left[\frac{2x_nr-|x|^2}{2y_nr-|y|^2}\right]^{\alpha/2}\frac{u(y)}{|x-y|^{n+1}}dy\nonumber\\
&=&C\int_{\substack{|y-x_r|>r\\|y|\leq R, \, y_n>0}}\left[\frac{2x_nr-|x|^2}{2y_nr-|y|^2}\right]^{\alpha/2}\frac{u(y)}{|x-y|^{n+1}}dy\nonumber\\
&\leq&C_R \int_{\substack{|y-x_r|>r\\|y|\leq R, \, y_n>0}}\left[\frac{2x_nr-|x|^2}{2y_nr-|y|^2}\right]^{\alpha/2}\frac{1}{|x-y|^{n+1}}dy.\label{b6}\end{aligned}$$

Here we have used the fact that the $\alpha$-harmonic function $u$ is bounded in the region $$D_{R,r}=\left\{y=(y',y_n)\left|\right.|y-x_r|>r,~|y|<R,~y_n>0\right\}.$$ The bound depends on $R$, however is independent of $r$, since $D_{R, r_1} \subset D_{R, r_2}$ for $r_1>r_2$. For each such fixed open domain $D_{R,r}$, the bound of the $\alpha$-harmonic function $u$ can be derived from the interior smoothness ( see, for instance [@BKN] and [@FW]) and the estimate up to the boundary ( see [@RS]).

Set  $y=(y',y_n), \sigma=|y'|$, for fixed  $x$  and sufficiently large  $r$, we have

$$\begin{aligned}
\left[\frac{2x_nr-|x|^2}{2y_nr-|y|^2}\right]^{\alpha/2}&\leq&\frac{Cr^{\alpha/2}}{|2y_nr-|y|^2|^{\alpha/2}}\nonumber\\&=&\frac{Cr^{\alpha/2}}{|\sigma^2-2y_nr+y_n^2|^{\alpha/2}}\nonumber\\&=&\frac{Cr^{\alpha/2}}{|(y_n-r)^2+\sigma^2-r^2|^{\alpha/2}}.\label{b7}\end{aligned}$$

and $$\frac{1}{|x-y|^{n+1}}\leq\frac{C}{(1+|y|)^{n+1}}\leq\frac{C}{(1+|y'|)^{n+1}}=\frac{C}{(1+\sigma)^{n+1}}.\label{b8}$$

For convenience of estimate, we amplify the domain $D_{R,\,r}$ a little bit. Define $$\hat{D}_{R,\,r}=\left\{y=(y',y_n)\in\mathbb{R}^n_+\left|\right. |y-x_r|>r,~|y'|\leq R, 0<y_n<\bar{y}_n \right\}.$$ Here $\bar{y}_n$  satisfies $$(\bar{y}_n-r)^2+\sigma^2-r^2=0, \label{b9}$$ so that $\bar{y}=(y',\bar{y_n})\in\partial\hat{D}_{R,\,r}\cap\partial B_r(x_r)$. Then it is easy to see that $$D_{R,\,r}\setminus \hat{D}_{R,\,r}
\label{tian}$$

From (\[b9\]), for sufficiently large  $r$ (much larger than  $R$ ), we have $$\bar{y}_n= r - \sqrt{r^2-\sigma^2}.$$ Set $$y_n=r-s\sqrt{r^2-\sigma^2}.\label{b10}$$ Then for  $0<y_n<\bar{y}_n$, $$1<s<\frac{r}{\sqrt{r^2-\sigma^2}},\label{b11}$$ and $$dy_n=-\sqrt{r^2-\sigma^2}ds.\label{b12}$$

Continuing from the right side of (\[b6\]), we integrate in the direction of  $y_n$  first, and then integrate with respect to $y'$, setting $r$ sufficiently large (much larger than $R$), by(\[b6\]), (\[b7\]), (\[b8\]), (\[tian\]),( \[b10\]), (\[b11\]), and(\[b12\]), we derive $$\begin{aligned}
I_{22}&\leq&C\int_{\hat{D}_{R,\,r}}\left[\frac{2x_nr-|x|^2}{2y_nr-|y|^2}\right]^{\alpha/2}\frac{1}{|x-y|^{n+1}}dy\nonumber\\
&\leq&C\int_{|y'|<R}\int_0^{\bar{y}_n}\frac{r^{\alpha/2}}{|(y_n-r)^2+\sigma^2-r^2|^{\alpha/2}}dy_n\frac{1}{(1+\sigma)^{n+1}}dy'\nonumber\\
&\leq&C\int_0^R\int_0^{\bar{y}_n}\frac{r^{\alpha/2}}{|(y_n-r)^2+\sigma^2-r^2|^{\alpha/2}}dy_n\frac{1}{(1+\sigma)^{n+1}}\sigma^{n-2}d\sigma \qquad \label{b13}\\
&\leq&C\int_0^R\int_1^{\frac{r}{\sqrt{r^2-\sigma^2}}}\frac{r^{\alpha/2}}{[(s\sqrt{r^2-\sigma^2})^2-(r^2-\sigma^2)]^{\alpha/2}}\left|\sqrt{r^2-\sigma^2}\right|ds\frac{\sigma^{n-2}}{(1+\sigma)^{n+1}}d\sigma\nonumber\\
&=&C\int_0^Rr^{\alpha/2}(r^2-\sigma^2)^{\frac{1-\alpha}{2}}\int_1^{\frac{r}{\sqrt{r^2-\sigma^2}}}\frac{1}{(s^2-1)^{\alpha/2}}ds\frac{\sigma^{n-2}}{(1+\sigma)^{n+1}}d\sigma\nonumber\\
&\leq&C\int_0^Rr^{\alpha/2}(r^2-\sigma^2)^{\frac{1-\alpha}{2}}\int_1^{\frac{r}{\sqrt{r^2-\sigma^2}}}\frac{1}{(s-1)^{\alpha/2}}ds\frac{\sigma^{n-2}}{(1+\sigma)^{n+1}}d\sigma\label{b14}\\
&\leq&C\int_0^Rr^{\alpha/2}r^{1-\alpha}\int_1^{\frac{r}{\sqrt{r^2-\sigma^2}}}\frac{1}{(s-1)^{\alpha/2}}ds\frac{\sigma^{n-2}}{(1+\sigma)^{n+1}}d\sigma\label{b15}\\
&=&C\int_0^Rr^{1-\alpha/2}\left(\frac{r}{\sqrt{r^2-\sigma^2}}-1\right)^{1-\alpha/2}\frac{\sigma^{n-2}}{(1+\sigma)^{n+1}}d\sigma\nonumber\\
&=&C\int_0^Rr^{1-\alpha/2}\left(\frac{\sigma^2}{\left(r+\sqrt{r^2-\sigma^2}\right)\sqrt{r^2-\sigma^2}}\right)^{1-\alpha/2}\frac{\sigma^{n-2}}{(1+\sigma)^{n+1}}d\sigma\nonumber\\
&\leq&C\int_0^Rr^{1-\alpha/2}\left(\frac{1}{r^2}\right)^{1-\alpha/2}\frac{\sigma^{n-2+2-\alpha}}{(1+\sigma)^{n+1}}d\sigma\nonumber\\
&=&Cr^{\alpha/2-1}\int_0^R\frac{\sigma^{n-\alpha}}{(1+\sigma)^{n+1}}d\sigma\nonumber\\
&\leq&Cr^{\alpha/2-1}\nonumber\\
&=&\frac{C}{r^{1-\alpha/2}}.\label{b16}\end{aligned}$$

In the above, we derived (\[b13\]) by letting  $|y'|=\sigma$ . (\[b14\]) is valid because $$\begin{aligned}
\frac{1}{(s^2-1)^{\alpha/2}}&=&\frac{1}{(s+1)^{\alpha/2}}\frac{1}{(s-1)^{\alpha/2}}\nonumber\\&\leq&\frac{1}{(1+1)^{\alpha/2}}\frac{1}{(s-1)^{\alpha/2}}\nonumber\\&=&\frac{1}{2^{\alpha/2}}\frac{1}{(s-1)^{\alpha/2}}\nonumber\\&\leq&\frac{1}{(s-1)^{\alpha/2}}.\nonumber\end{aligned}$$ Since  $R$  is fixed and  $\sigma^2\leq R^2$ , when $r$  is sufficiently large ( much larger than  $R$ ), we have $r^2-\sigma^2>0$, and the value of  $(r^2-\sigma^2)^{\frac{1-\alpha}{2}}$  can be dominated by  $(r^2)^{\frac{1-\alpha}{2}}~$(i.e.  $r^{1-\alpha}$), this verifies (\[b15\]).

For the  $\epsilon>0$  given above and the fixed  $R$ , since  $0<\alpha<2$ , then by (\[b16\]) we can easily get $$I_{22}\leq C\frac{1}{r^{1-\alpha/2}}<\epsilon,\label{b17}$$ for sufficiently large  $r$ .

From (\[b3\]), (\[b0\]), (\[b4\]), (\[b5\]), and(\[b17\]), we derive $$\left|\frac{\partial u}{\partial x_i}(x)\right|<C\epsilon,$$ for sufficiently large  $R$  and much larger $r$.

The fact that  $\epsilon$  is arbitrary implies $$\left|\frac{\partial u}{\partial x_i}(x)\right| = 0.$$ This proves (\[b1\]).

Now, let’s prove (\[b2\]). Similarly, for fixed  $x\in B_r(x_r)\subset\mathbb{R}^n_+$ , through an elementary calculation, one can derive that $$\begin{aligned}
\frac{\partial u}{\partial x_n}(x)&=&\int_{|y-x_r|>r}\left(\frac{\alpha (r- x_n)}{r^2-|x-x_r|^2}+\frac{n(y_n-x_n)}{|y-x|^2}\right)P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\int_{|y-x_r|>r}\frac{\alpha (r- x_n)}{r^2-|x-x_r|^2}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\&\quad&+\int_{|y-x_r|>r}\frac{n(y_n-x_n)}{|y-x|^2}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&:=&J_1+J_2.\label{cc1}\end{aligned}$$

Similarly to  $I_2$, for sufficiently large  $r$, we can also derive $$|J_2|\leq C\epsilon,$$ for any  $\epsilon>0$ . That is $$J_2\rightarrow0, \quad\text{as}~~r\rightarrow\infty.\label{cc2}$$

Now we estimate  $J_1$. $$\begin{aligned}
J_1&=&\frac{\alpha (r- x_n)}{r^2-|x-x_r|^2}\int_{|y-x_r|>r}P_r(x-x_r,y-x_r)u(y)dy\nonumber\\
&=&\frac{\alpha (r- x_n)}{2x_nr-|x|^2}u(x).\end{aligned}$$ It follows that $$J_1\rightarrow\frac{\alpha}{2x_n}u(x),\quad \text{as}~~r\rightarrow\infty.\label{cc4}$$

By (\[cc1\]), (\[cc2\]), and(\[cc4\]), for each fixed  $x\in B_r(x_r)\subset\mathbb{R}^n_+$, letting  $r\rightarrow\infty$ , we arrive at $$\frac{\partial u}{\partial x_n}(x)=\frac{\alpha}{2x_n}u(x).$$ This verifies (\[b2\]), and hence completes the proof of Theorem \[mthm1\].\

$\hat{u}(x)$ is  $\alpha$ -harmonic in  $B_r(x_r)$ 
===================================================

In this section, we prove

$\hat{u}(x)$ defined by (\[max\]) in the previous section is $\alpha$-harmonic in $B_r(x_{r})$. \[thm3.1\]

The proof consists of two parts. First we show that $\hat{u}$ is harmonic in the average sense (Lemma \[lem3.1\]), then we show that it is $\alpha$-harmonic (Lemma \[lem3.2\]).

Let $$\varepsilon^{(r)}_{\alpha}(x)=\left\{\begin{array}{ll}
0, &|x|<r.\\
\frac{\Gamma(n/2)}{\pi^{\frac{n}{2}+1}} \sin\frac{\pi\alpha}{2}\frac{r^\alpha}{(|x|^2-r^2)^{\frac{\alpha}{2}}|x|^n}, &|x|>r.
\end{array}
\right.$$

We say that $u$ is $\alpha$-harmonic in the average sense (see [@L]) if for small $r$, $$\varepsilon^{(r)}_{\alpha}\ast u(x)=u(x).$$

Let $$\begin{aligned}
&&P_{r}(x-x_r,y-x_r)\nonumber\\&=&\left\{\begin{array}{ll}
\frac{\Gamma(n/2)}{\pi^{\frac{n}{2}+1}} \sin\frac{\pi\alpha}{2}
\left[\frac{r^{2}-|x-x_r|^{2}}{|y-x_r|^{2}-r^{2}}\right]^{\frac{\alpha}{2}}\frac{1}{|x-y|^{n}},\qquad& |y-x_r|>r,|x-x_r|<r\\
0,& \text{elsewhere}.
\end{array}
\right.\end{aligned}$$

Let $u(x)$ be any measurable function outside $B_r(x_{r})$ for which $$\int_{\mathbb{R}^n}\frac{|u(z)|}{(1+|z-x_{r}|)^{n+\alpha}}dz<\infty.
 \label{d1}$$ Let $$\hat{u}(x)=\left\{\begin{array}{ll}
\int_{|y-x_{r}|>r}P_{r}(y-x_{r},x-x_{r})u(y)dy,& |x-x_{r}|<r,\\
u(x),& |x-x_{r}|\geq r.
\end{array}
\right.$$ Then $\hat{u}(x)$ is $\alpha$-harmonic in the average sense in $B_{r}(x_{r})$, i.e. for sufficiently small $\delta$, we have $$(\varepsilon_\alpha^{(\delta)}\ast\hat{u})(x)=\hat{u}(x),\quad   |x-x_{r}|<r,$$ where $\ast$ is the convolution. \[lem3.1\]

**Proof.**

The outline is as follows.

*i)* Approximate $u$ by a sequence of smooth, compactly supported functions $\{u_k\}$, such that $u_k(x) {{\mbox{$\rightarrow$}}}u(x)$ and $$\int_{|z-x_{r}|>r}\frac{|u_k(z)-u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^\frac{\alpha}{2}}dz {{\mbox{$\rightarrow$}}}0.
\label{d2}$$

This is possible under our assumption (\[d1\]).

*ii)* For each $u_k$, find a signed measure $\nu_k$ such that $supp\,\nu_k\subset B^c_r(x_{r})$ and $$u_k(x)=U_{\alpha}^{\nu_k}(x), \;\; |x-x_{r}|>r.$$ Then $$\hat{u}_k(x)=U_{\alpha}^{\nu_k}(x), \quad |x-x_{r}|<r.$$

*iii)* It is easy to see that $\hat{u}_k(x)$ is $\alpha$-harmonic in the average sense for $|x-x_{r}|<r$. That is, for each fixed small $\delta>0$, $$(\varepsilon_\alpha^{(\delta)}\ast\hat{u}_k)(x)=\hat{u}_k(x).$$ By showing that as $k {{\mbox{$\rightarrow$}}}\infty$ $$\varepsilon_\alpha^{(\delta)}\ast\hat{u}_k {{\mbox{$\rightarrow$}}}\varepsilon_\alpha^{(\delta)}\ast\hat{u},$$ and $$\hat{u}_k {{\mbox{$\rightarrow$}}}\hat{u},$$ we arrive at $$(\varepsilon_\alpha^{(\delta)}\ast\hat{u})(x)=\hat{u}(x),\quad  |x-x_{r}|<r.$$

Now we carry out the details.

*i)* There are several ways to construct such a sequence $\{u_k\}$. One is to use the mollifier. Let $$u|_{B_k(x_{r})}(x)=\left\{\begin{array}{ll}
u(x),&|x-x_{r}|<k,\\
0,&|x-x_{r}|\geq k,
\end{array}
\right.$$ and $$J_\epsilon(u|_{B_k(x_{r})})(x)=\int_{\mathbb{R}^n}j_\epsilon(x-y)u|_{B_k(x_{r})}(y)dy.$$

For any $\delta>0$, let $k$ be sufficiently large (larger than $r$) such that $$\int_{|z-x_{r}|\geq k}
\frac{|u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz<\frac{\delta}{2}.$$ For each such $k$, choose $\epsilon_k$ such that $$\int_{B_{k+1}\backslash B_r}
\frac{|u_k(z)-u|_{B_k(x_{r})}(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz
<\frac{\delta}{2},$$ where $u_k=J_{\epsilon_k}(u|_{B_k(x_{r})})$. It then follows that $$\begin{aligned}
&\quad&\int_{|z-x_{r}|>r}\frac{|u_k(z)-u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz\\
&\leq&\int_{B_{k+1}(x_{r})\backslash B_r(x_{r})}\frac{|u_k(z)-u|_{B_k(x_{r})}(z)|+
|u|_{B_k(x_{r})}(z)-u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz\\
&&+\int_{|z-x_{r}|>k+1}\frac{|u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz\\
&=&\int_{B_{k+1(x_{r})}\backslash B_r(x_{r})}\frac{|u_k(z)-u|_{B_k(x_{r})}(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz+
\int_{|z-x_{r}|\geq k}\frac{|u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}dz\\
&<&\frac{\delta}{2}+\frac{\delta}{2}=\delta.\end{aligned}$$

Therefore, as $k {{\mbox{$\rightarrow$}}}\infty$, $$\int_{|z-x_{r}|>r}\frac{|u_k(z)-u(z)|}{|z-x_{r}|^n(|z-x_{r}|^2-r^2)^\frac{\alpha}{2}}dz {{\mbox{$\rightarrow$}}}0.
\label{w1}$$

*ii)* For each $u_k$, there exists a signed measure $\psi_k$ such that $$u_k(x)=U_{\alpha}^{\psi_k}(x).$$ Indeed, let $\psi_k(x)= C (-{\mbox{$\bigtriangleup$}})^{\frac{\alpha}{2}}u_k(x)$, then $$\begin{aligned}
U_{\alpha}^{\psi_k}(x)&=&\int_{\mathbb{R}^n}\frac{C}{|x-y|^{n-
 \alpha}}(-{\mbox{$\bigtriangleup$}})^{\frac{\alpha}{2}}u_k(y)dy\\
 &=&\int_{\mathbb{R}^n}(-{\mbox{$\bigtriangleup$}})^{\frac{\alpha}{2}}\left[\frac{C}{|x-y|^{n-
 \alpha}}\right]u_k(y)dy\label{d3}\\
 &=&\int_{\mathbb{R}^n}\delta(x-y)u_k(y)dy=u_k(x).
 \end{aligned}$$ Here we have used the fact that $\frac{C}{|x-y|^{n-\alpha}}$ is the fundamental solution of $(-{\mbox{$\bigtriangleup$}})^{\alpha/2}$.

Let $\psi_k|_{B_r(x_{r})}$ be the restriction of $\psi_k$ on $B_r(x_{r})$ and $$\tilde{\psi}_k(y)=\int_{|x-x_{r}|<r}P_r(y-x_r, x-x_r)\psi_k|_{B_r(x_{r})}(x)dx,$$ we have $$U_{\alpha}^{\tilde{\psi}_k}(x)=U_{\alpha}^{\psi_k|_{B_r(x_{r})}}(x), \;\; |x-x_{r}|> r,$$ and $supp\,\tilde{\psi}_k\subset B^c_r(x_{r}).$ Here we use the fact (see (1.6.12$'$) [@L]) that $$\frac{1}{|z-x|^{n-\alpha}}=\int_{|y-x_{r}|>r}\frac{P_r(y-x_r,x-x_r)}{|z-y|^{n-\alpha}}dy,\qquad
 \:|x-x_{r}|<r,\: |z-x_{r}|>r.
 \label{Po}$$

Let $\nu_k=\psi_k-\psi_k|_{B_r(x_{r})}+\tilde{\psi}_k$, then $supp\,\nu_k\subset B^c_r(x_{r})$, and $$U_{\alpha}^{\nu_k}(x)=U_{\alpha}^{\psi_k}(x)+U_{\alpha}^{\tilde{\psi}_k}(x)
-U_{\alpha}^{\psi_k|_{B_r(x_{r})}}(x)=U_{\alpha}^{\psi_k}(x),\quad |x-x_{r}|>r.$$ That is $$u_k(x) = U_{\alpha}^{\nu_k}(x) ,\quad |x-x_{r}|>r.$$

Again by (\[Po\]), we deduce $$\hat{u}_k(x) = U_{\alpha}^{\nu_k}(x) , \quad |x-x_{r}|<r.$$ In this case $\hat{u}_k$ is $\alpha$-harmonic (in the sense of average) in the region $|x-x_{r}|<r$ (see [@L]).

*iii)*For each fixed $x$, we first have $$\hat{u}_k(x) {{\mbox{$\rightarrow$}}}\hat{u}(x).$$ In fact, by (\[w1\]), $$\begin{aligned}
\hat{u}_k(x)-\hat{u}(x)&=&\int_{|y-x_{r}|>r}P_r(y-x_{r},x-x_{r})[u_k(y)-u(y)]dy\\
&=&C\int_{|y-x_{r}|>r}\frac{(r^2-|x-x_{r}|^2)^{\frac{\alpha}{2}}[u_k(y)-u(y)]}{(
|y-x_{r}|^2-r^2)^{\frac{\alpha}{2}}|x-y|^n}dy\\&{{\mbox{$\rightarrow$}}}&0.\end{aligned}$$ Next, we show that, for each fixed $\delta>0$ and fixed $x$, $$(\varepsilon_\alpha^{(\delta)}\ast\hat{u}_k)(x) {{\mbox{$\rightarrow$}}}(\varepsilon_\alpha^{(\delta)}\ast\hat{u})(x).
 \label{d8}$$

Indeed, $$\begin{aligned}
 &&(\varepsilon_\alpha^{(\delta)}\ast\hat{u}_k)(x)- (\varepsilon_\alpha^{(\delta)}\ast\hat{u})(x)\\
&=&C\int_{|y-x|>\delta}\frac{\delta^\alpha[\hat{u}_k(y)-
\hat{u}(y)]}{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n}dy\\
&=&C\{\int_{\substack{|y-x|>\delta\\ |y-x_{r}|<r-\eta}}
\frac{\delta^\alpha[\hat{u}_k(y)-
\hat{u}(y)]}{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n}dy\\
&&+\int_{\substack{|y-x|>\delta \\r-\eta<|y-x_{r}|<r}}
\frac{\delta^\alpha[\hat{u}_k(y)-
\hat{u}(y)]}{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n}dy\\
&&+
\int_{\substack{|y-x|>\delta \\|y-x_{r}|>r}}
\frac{\delta^\alpha[\hat{u}_k(y)-
\hat{u}(y)]}{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n}dy\}\\
&=&C(I_1+I_2+I_3).\end{aligned}$$

For each fixed $x$ with $|x-x_{r}|<r$, choose $\delta$ and $\eta$ such that $$B_\delta(x)\cap B^c_{r-2\eta}(x_{r})=\emptyset.$$

It follows from (\[w1\]) that as $k {{\mbox{$\rightarrow$}}}\infty$, $$I_3=\int_{\substack{|y-x|>\delta \\
                      |y-x_{r}|>r}}\frac{\delta^\alpha[u_k(y)-
u(y)]}{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n}dy~ {{\mbox{$\rightarrow$}}}~ 0.$$

$$\begin{aligned}
I_2&=&\int_{\substack{ |y-x|>\delta \\
                      r-\eta<|y-x_{r}|<r}}
\frac{\delta^\alpha \int_{|z-x_{r}|>r}P_r(z-x_{r},y-x_{r})[u_k(z)-u(z)]dz}
{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n}dy\\
&=&C\delta^\alpha\int_{|z-x_{r}|>r}\frac{u_k(z)-u(z)}{(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}
\int_{\substack{ |y-x|>\delta \\
                      r-\eta<|y-x_{r}|<r}}
\frac{(r^2-|y-x_{r}|^2)^{\frac{\alpha}{2}}dy}
{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n|z-y|^n}dz\\
&=&C\delta^\alpha\int_{|z-x_{r}|>r}\frac{u_k(z)-u(z)}{(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}
\cdot I_{21}(x,z)dz.\end{aligned}$$

Noting that in the ring $r-\eta<|y-x_{r}|<r$, we have $$|x-y|>\eta+\delta.$$ It then follows that $$\begin{aligned}
&&I_{21}(x,z)\nonumber\\
&\leq&\frac{1}{(2\eta\delta+\eta^2)^{\frac{\alpha}{2}}(\eta+\delta)^n}
\int_{r-\eta<|y-x_{r}|<r}\frac{(r^2-|y-x_{r}|^2)^{\frac{\alpha}{2}}dy}{|z-y|^n}
\nonumber\\
&=&C\int_{r-\eta}^r (r^2-\tau^2)^{\frac{\alpha}{2}}
\left\{\int_{S_\tau} \frac{1}{|z-y|^n} d\sigma_y\right\}d\tau\nonumber\\
&=&C\int_{r-\eta}^r (r^2-\tau^2)^{\frac{\alpha}{2}}
\left\{\int_0^\pi \frac{\omega_{n-2}(\tau\sin\theta)^{n-2}\tau d\theta}
{(\tau^2+|z-x_{r}|^2-2\tau|z-x_{r}|\cos\theta)^{\frac{n}{2}}}\right\}d\tau\nonumber\\
&=&C\int_{r-\eta}^r (r^2-\tau^2)^{\frac{\alpha}{2}}\frac{1}{\tau^n}
\int_0^\pi \frac{\tau^{n-1}\sin^{n-2}\theta d\theta}{(
(\frac{|z-x_{r}|}{\tau})^2-2\frac{|z-x_{r}|}{\tau}\cos\theta+1)^{\frac{n}{2}}}d\tau
\label{d5}\\
&=&C\int_{r-\eta}^r \frac{(r^2-\tau^2)^{\frac{\alpha}{2}}}{\tau}
\frac{d\tau}{(\frac{|z-x_{r}|}{\tau})^{n-2}((\frac{|z-x_{r}|}{\tau})^2-1)}
\int_{0}^\pi\sin^{n-2}\beta d\beta \label{d6}\\
&<&\frac{Cr^{n-1}}{|z-x_{r}|^{n-2}}\int_{r-\eta}^r
\frac{(r^2-\tau^2)^{\frac{\alpha}{2}}}{|z-x_{r}|^2-\tau^2} d\tau\nonumber\\
&=&\frac{Cr^{n-1}}{|z-x_{r}|^{n-2}}\cdot J.\nonumber\end{aligned}$$ In the above, to derive (\[d6\]) from (\[d5\]), we have made the following substitution (See Appendix in [@L]): $$\frac{\sin\theta}{\sqrt{(\frac{|z-x_{r}|}{\tau})^2-2\frac{|z-x_{r}|}{\tau}\cos\theta+1}}
=\frac{\sin\beta}{\frac{|z-x_{r}|}{\tau}},$$

To estimate the last integral $J$, we consider

*(a)*For $r<|z-x_{r}|<r+1$, $$J\leq\int_{r-\eta}^r\frac{(r+\tau)^{\frac{\alpha}{2}-1}}{(r-
\tau)^{1-\frac{\alpha}{2}}}d\tau\leq C_{\alpha,r}.$$

*(b)*For $|z-x_{r}|\geq r+1$, obviously, $$J\sim\frac{1}{|z-x_{r}|^2}, \mbox{ for $|z-x_{r}|$ large}.\nonumber$$

In summary, $$I_{21}(x,z)\sim\left\{\begin{array}{ll}
1,  &\mbox{ for $|z-x_{r}|$ near r},\\
|z-x_{r}|^n,&\mbox{ for $|z-x_{r}|$ large}.
\end{array}
\right.
\nonumber$$

Therefore, by (\[w1\]), as $k {{\mbox{$\rightarrow$}}}\infty$, $$I_2=\delta^\alpha\int_{|z-x_{r}|>r}\frac{u_k(z)-u(z)}{(|z-x_{r}|^2 -r^2)^{\alpha/2}} I_{21}(x,z)dz ~{{\mbox{$\rightarrow$}}}~ 0.$$

Now what remains is to estimate $$I_1=\delta^\alpha\int_{|z-x_{r}|>r}\frac{u_k(z)-u(z)}{(|z-x_{r}|^2-r^2)^{\frac{\alpha}{2}}}
I_{11}(x,z)dz,$$ where $$I_{11}(x,z)=\int_{\substack{|y-x|>\delta \\
                      |y-x_{r}|< r-\eta}}
 \frac{(r^2-|y-x_{r}|^2)^{\frac{\alpha}{2}}dy}
 {(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|x-y|^n|z-y|^n}.$$

$$\begin{aligned}
 I_{11}(x,z)
&\leq&\frac{r^\alpha}{\delta^n}\int_{\substack{ |y-x|>\delta \\
                      |y-x_{r}|< r-\eta}}
 \frac{dy}{(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}|z-y|^n}\\
&\leq&\frac{r^\alpha}{\delta^n(|z-x_{r}|-r+\eta)^n}
\int_{\delta<|y-x|<2r}\frac{dy}
 {(|x-y|^2-\delta^2)^{\frac{\alpha}{2}}}\\
 &=&\frac{r^\alpha}{\delta^n(|z-x_{r}|-r+\eta)^n}
\int_\delta^{2r}
\frac{\omega_{n-1}\tau^{n-1}d\tau}{(\tau^2-\delta^2)^{\frac{\alpha}{2}}}\\
&\leq&\frac{C}{|z-x_{r}|^n}.\end{aligned}$$

By (\[w1\]), as $k {{\mbox{$\rightarrow$}}}\infty$, we have $I_1 {{\mbox{$\rightarrow$}}}0.$ This verifies (\[d8\]) and hence completes the proof.

$$\lim_{r\rightarrow0}\frac{1}{r^\alpha}\left[u(x)-\varepsilon^{(r)}_{\alpha}\ast u(x)\right]
=c(-{\mbox{$\bigtriangleup$}})^\frac{\alpha}{2}u(x).
\label{c0}$$

where $c=\frac{\Gamma(n/2)}{\pi^{\frac{n}{2}+1}} \sin\frac{\pi\alpha}{2}$. \[lem3.2\]

**Proof.** $$\begin{aligned}
&&\frac{1}{r^\alpha}\left[u(x)-\varepsilon^{(r)}_{\alpha}\ast u(x)\right]\nonumber\\
&=&\frac{1}{r^\alpha}u(x)- c\int_{|y-x|>r}\frac{u(y)}{(|x -y|^2-r^2)^{\frac{\alpha}{2}}|x-y|^n}dy\nonumber\\
&=&c\int_{|y-x|>r}
\frac{u(x)-u(y)}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}|x-y|^n}dy.
\label{c1}\end{aligned}$$ Here we have used the property that $$\int_{|y-x|>r}\varepsilon^{(r)}_{\alpha}(x-y)=1.$$

Compare (\[c1\]) with $$(-{\mbox{$\bigtriangleup$}})^\frac{\alpha}{2}u(x)=\lim_{r\rightarrow0}\int_{|y-x|>r}
\frac{u(x)-u(y)}{|x-y|^{\alpha+n}}dy.$$ One may expect that $$\lim_{r\rightarrow0}\int_{|y-x|>r}
\frac{u(x)-u(y)}{|x-y|^{\alpha+n}}dy=\lim_{r\rightarrow0}\int_{|y-x|>r}
\frac{u(x)-u(y)}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}|x-y|^n}dy.$$

Indeed, consider $$\begin{aligned}
&&\int_{|y-x|>r}
\frac{u(x)-u(y)}{|x-y|^{n}}\left(\frac{1}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{|x-y|^{\alpha}}\right)dy\nonumber\\
&=&\int_{r<|y-x|<1}\frac{u(x)-
u(y)}{|x-y|^{n}}\left(\frac{1}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{|x-y|^{\alpha}}\right)dy\nonumber\\
&&+\int_{|y-x|\geq1}\frac{u(x)-
u(y)}{|x-y|^{n}}\left(\frac{1}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{|x-y|^{\alpha}}\right)dy\nonumber\\
&=&I_{1}+I_{2}.\label{c2}\end{aligned}$$

It is easy to see that as $r\rightarrow0$, $I_2$ tends to zero. Actually, same conclusion is true for $I_1$. $$\begin{aligned}
I_1&=&\int_{r<|y-x|<1}\frac{\nabla u(x)(y-x)+O(|y-x|^2)}{|x-
y|^{n}}\left(\frac{1}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{|x-y|^{\alpha}}\right)dy\nonumber\\
\label{c2.5}\\
&\leq&C\int_{r<|y-x|<1}\frac{|x-
y|^2}{|x-y|^n}\left(\frac{1}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{|x-y|^{\alpha}}\right)dy\label{c3}\\
&=&C\int_r^1\frac{\tau^2}{\tau^n}\left(\frac{1}{(\tau^
2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{\tau^{\alpha}}\right)\tau^{n-1}d\tau\label{c4}\\
&\leq&C\int_1^\infty\left(\frac{1}{r^\alpha(s^
2-1)^{\frac{\alpha}{2}}}
-\frac{1}{r^\alpha s^\alpha}\right)sr^2ds\label{c5}\\
&=&Cr^{2-\alpha}\int_1^\infty\left(\frac{s^\alpha-(s^
2-1)^{\frac{\alpha}{2}}}{(s^
2-1)^{\frac{\alpha}{2}}s^\alpha}\right)sds.\label{c6}\end{aligned}$$ Equation (\[c2.5\]) follows from the Taylor expansion. Due to symmetry, we have $$\int_{r<|y-x|<1}\frac{\nabla u(x)(y-x)}{|x-
y|^{n}}\left(\frac{1}{(|x-y|^2-r^2)^{\frac{\alpha}{2}}}
-\frac{1}{|x-y|^{\alpha}}\right)dy=0$$ and get (\[c3\]). By letting $|y-x|=\tau$ and $\tau=rs$ respectively, one obtains (\[c4\]) and (\[c5\]). It is easy to see that the integral in (\[c6\]) converges near 1. To see that it also converges near infinity, we estimate $$s^\alpha-(s^2-1)^{\frac{\alpha}{2}}.$$

Let$f(t)=t^{\alpha/2}$. By the *mean value theorem*,

$$\begin{aligned}
f(s^2)-f(s^2-1)
&=&f^\prime(\xi)(s^2-(s^2-1))\\
&=&\frac{\alpha}{2}\xi^{\frac{\alpha}{2}-1}\sim s^{\alpha-2}, \mbox{ for $s$ sufficiently large.}\end{aligned}$$

This implies that $$\frac{s^\alpha-(s^2-1)^{\frac{\alpha}{2}}}{(s^2-1)^{\frac{\alpha}{2}}s^\alpha}s
\sim\frac{s^{\alpha-2}s}{(s^2-1)^{\frac{\alpha}{2}}s^\alpha}
\sim\frac{1}{s^{1+\alpha}}.$$ Now it is obvious that (\[c6\]) converges near infinity. Thus we have $$\int_1^\infty\left(\frac{s^\alpha-(s^
2-1)^{\frac{\alpha}{2}}}{(s^
2-1)^{\frac{\alpha}{2}}s^\alpha}\right)sds<\infty.$$ Since $0<\alpha<2$, as $r\rightarrow0$, (\[c6\]) goes to zero, i.e. $I_1$ converges to zero. Together with (\[c1\]) and (\[c2\]), we get (\[c0\]). This proves the lemma.

[CL]{}

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[*Authors’ Addresses and E-mails:*]{}

Wenxiong Chen

Department of Mathematical Sciences

Yeshiva University

New York, NY, 10033 USA

wchen@yu.edu

Congming Li

Department of Applied Mathematics

University of Colorado,

Boulder CO USA

cli@clorado.edu

Lizhi Zhang

School of Mathematics and Information Science

Henan Normal University

azhanglz@163.com

Tingzhi Cheng

Department of Mathematics

Shanghai JiaoTong University

nowitzki1989@126.com.
