The point is to apply integration by parts to show that $u$ is a weak solution to $\Delta u=0$ on $B_{R}(p)$. The computations are routine and standard in the literature, so we just skip it. ∎

The proof is rather standard and straightforward, which can be achieved by using separation of variables.

For the simplicity of notations, we denote

(6.43)

$$d\equiv m+n\geq 3.$$

Let $(r,\Theta)\in\mathbb{R}^{d}$ be the polar coordinate system in $\mathbb{R}^{d}$, so the Laplacian of $u$ can be written as

(6.44)

$$\Delta_{\mathbb{R}^{d}}u=\frac{\partial^{2}u}{\partial r^{2}}+\frac{d-1}{r}\cdot\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\cdot\Delta_{\mathbb{S}^{d-1}}u.$$

We make separation of variables on the punctured Euclidean space $\mathbb{R}^{d}\setminus\{0^{d}\}$. Let

(6.45)

$$\lambda_{j}\equiv j(j+d-2),\ j\in\mathbb{N},$$

be the spectrum of the unit round sphere $\mathbb{S}^{d-1}$. Correspondingly, let $\varphi_{j}\in C^{\infty}(\mathbb{S}^{d-1})$ satisfy

(6.46)

$$-\Delta_{\mathbb{S}^{d-1}}\varphi_{j}(\Theta)=\lambda_{j}\varphi_{j}(\Theta).$$

Then the function $u(r,\Theta)$ has the expansion along the fiber $\mathbb{S}^{d-1}$,

(6.47)

$$u(r,\Theta)=\sum\limits_{j=0}^{\infty}u_{j}(r)\cdot\varphi_{j}(\Theta).$$

Immediately, for each $j\in\mathbb{N}$, the coefficient function $u_{j}(r)$ solves the Euler-Cauchy equation,

(6.48)

$$u_{j}^{\prime\prime}(r)+\frac{d-1}{r}\cdot u_{j}^{\prime}(r)-\frac{1}{r^{2}}\cdot\lambda_{j}\cdot u_{j}(r)=0,$$

which has a general solution

(6.49)

$$u_{j}(r)=C_{j}\cdot r^{p_{j}}+C_{j}^{*}\cdot r^{q_{j}},$$

where $p_{j}=\frac{2-d+\sqrt{(d-2)^{2}+4\lambda_{j}}}{2}\geq 0$ and $q_{j}=\frac{2-d-\sqrt{(d-2)^{2}+4\lambda_{j}}}{2}<0$ solve the quadratic equation

(6.50)

$$w^{2}+(d-2)w-\lambda_{j}=0.$$

So it is obvious

	$\displaystyle p_{0}$	$\displaystyle=0,\quad q_{0}=2-d\leq-1,$
	$\displaystyle p_{j}$	$\displaystyle\geq p_{1}=1,$
(6.51)		$\displaystyle q_{j}$	$\displaystyle\leq q_{1}=1-d\leq-2,\quad j\in\mathbb{Z}_{+}.$

In the following, we will show that, given the growth condition (6.42) for $\mu_{p}\in(-1,1)\setminus\{0\}$, then for each $j\in\mathbb{N}$ and for each $r>0$, the coefficient $u_{j}(r)$ satisfies

(6.52)

$$|u_{j}(r)|\leq\frac{Q_{j}}{r^{\mu_{p}}},$$

where $Q_{j}\in\mathbb{R}$. In fact, so it follows from the expansion (6.47) that for each $j\in\mathbb{N}$,

(6.53)

$$u_{j}(r)=\int_{\mathbb{S}^{d-1}}u(r,\Theta)\cdot\varphi_{j}\dvol_{\mathbb{S}^{d-1}},$$

which implies

(6.54)

$$|u_{j}(r)|\leq|\varphi_{j}|_{L^{\infty}(\mathbb{S}^{d-1})}\cdot\int_{\mathbb{S}^{d-1}}\frac{1}{|x|^{\mu_{p}}}\dvol_{\mathbb{S}^{d-1}}.$$

Next, we will write the above integral in the polar coordinates $\Theta\equiv(\theta_{1},\ldots,\theta_{d-1})$ with $\theta_{1},\ldots,\theta_{d-2}\in[0,\pi]$ and $\theta_{d-1}\in[0,2\pi]$. Denote by $d\Theta\equiv d\theta_{1}\wedge d\theta_{2}\wedge\ldots\wedge d\theta_{d-1}$, then it is by elementary calculations that, $|x|=r^{m}\cdot\prod\limits_{k=1}^{d-m}|\sin\theta_{k}|$ and $\dvol_{\mathbb{S}^{d-1}}=\prod\limits_{k=1}^{d-2}(\sin^{d-k-1}\theta_{k})\cdot d\Theta$. Therefore,

(6.55)

$$\int_{\mathbb{S}^{d-1}}\frac{1}{|x|^{\mu_{p}}}\dvol_{\mathbb{S}^{d-1}}=\frac{1}{r^{\mu_{p}}}\int_{\mathcal{D}_{\Theta}}\frac{\prod\limits_{k=1}^{d-2}(\sin^{d-k-1}\theta_{k})}{\prod\limits_{k=1}^{d-m}|\sin\theta_{k}|^{\mu_{p}}}\cdot d\Theta,$$

where $\mathcal{D}_{\Theta}\equiv\{0\leq\theta_{1},\ldots,\theta_{d-2}\leq\pi,\ 0\leq\theta_{d-1}\leq 2\pi\}$. By assumption, $\mu_{p}\in(-1,1)\setminus\{0\}$, then $\frac{\prod\limits_{k=1}^{d-2}(\sin^{d-k-1}\theta_{k})}{\prod\limits_{k=1}^{d-m}|\sin\theta_{k}|^{\mu_{p}}}$ is integrable in $\mathcal{D}_{\Theta}$ and we denote

(6.56)

$$\mathcal{I}_{0}\equiv\int_{\mathcal{D}_{\Theta}}\frac{\prod\limits_{k=1}^{d-2}(\sin^{d-k-1}\theta_{k})}{\prod\limits_{k=1}^{d-m}|\sin\theta_{k}|^{\mu_{p}}}\cdot d\Theta.$$

Therefore, for each $j\in\mathbb{N}$, it holds that

(6.57)

$$|u_{j}(r)|\leq\frac{\mathcal{I}_{0}\cdot|\varphi_{j}|_{L^{\infty}(\mathbb{S}^{d-1})}}{r^{\mu_{p}}}\equiv\frac{Q_{j}}{r^{\mu_{p}}}$$

for all $r>0$.

Now we go back to the representation of $u_{j}(r)$ in (6.49) and we analyze the growth behavior of function as $r\ll 1$ and $r\gg 1$. Applying the assumption $\mu_{p}\in(-1,1)\setminus\{0\}$ and the gap obtained in (6.51), we have that, for each $j\in\mathbb{N}$, $C_{j}=C_{j}^{*}=0$. Therefore,

(6.58)

$$u\equiv 0\ \text{on}\ \mathbb{R}^{m+n}.$$

∎