Prove that there are constants $0Let $f\in C^2_c(B(O,1))$ where $B(O,1)\subset \mathbb{R}^3$, $f\geq 0$ and $f=1$ on $B(O,1/2)$. Let $u$ be such that 
\begin{gather*}
-\Delta u = f(x) \text{ for } x\in\mathbb{R}^3 \\
u(x)\to 0 \text{ as } |x|\to\infty
\end{gather*}


*

*Represent $u$ using the fundamental solution to Laplace's equation.

*Prove that there are constants $0<A_1<A_2<\infty$ so that
$$ \dfrac{A_1}{|x|} \leq u(x) \leq \dfrac{A_2}{|x|} $$
for all $|x|\geq 2$.


I know that the fundamental solution when $n=3$  for the Poisson Equation (which is the inhomogenous form of Laplace Equation) is
$$\Phi(x)=\dfrac{1}{n(n-2)\alpha(n)}\cdot \dfrac{1}{|x|^{n-2}}
=\dfrac{1}{3\cdot 1\cdot \alpha(3)}\cdot \dfrac{1}{|x|} $$
where $\alpha(3)$ is the volume of a $3$ dimension ball.
How do I use this to represent $u$ as stated in (1)?
For (2), am I suppose to use Harnack's inequality with $r=A_1$ and $R=A_2$?
 A: The fundamental solution of Laplace's equation  in $\Bbb R^3$ is the (distributional) function $\Phi(x, x')$ satisfying
$-\nabla^2 \Phi(x, x') = \delta(x - x'); \tag 1$
it is given by
$\Phi(x, x') = \dfrac{1}{4\pi \vert x - x' \vert}. \tag 2$
Given the equation
$-\nabla^2 u(x) = f(x), \tag 3$
we have
$u(x) = \displaystyle \int_{\Bbb R^3} \Phi(x, x') f(x') \; dx' = \dfrac{1}{4\pi} \int_{\Bbb R^3} \dfrac{f(x')}{\vert x - x' \vert} \; dx'; \tag 4$
for
$f \in C_c^2(B(0, 1)), \tag 5$
the integrand of the rightmost integral vanishes if $x' \notin B(0, 1)$; therefore
$u(x) = \displaystyle \dfrac{1}{4\pi} \int_{B(0, 1)} \dfrac{f(x')}{\vert x - x' \vert} \; dx'; \tag 6$
now if
$\vert x \vert \ge 2, \; \vert x' \vert < 1, \tag 7$
$ \vert x - x' \vert \le \vert x \vert + \vert x' \vert < 2 \vert x \vert, \tag 8$
$\dfrac{1}{2 \vert x \vert} < \dfrac{1}{\vert x - x' \vert}; \tag 9$
then
$\dfrac{1}{4\pi}\dfrac{1}{2 \vert x \vert} \displaystyle \int_{B(0, 1)} f(x') \; dx' = \dfrac{1}{4\pi} \displaystyle \int_{B(0, 1)} \dfrac{f(x') }{2 \vert x \vert}\; dx' \le \dfrac{1}{4\pi} \displaystyle \int_{B(0, 1)} \dfrac{f(x') }{\vert x - x' \vert}\; dx' = u(x); \tag{10}$
if we set
$A_1 = \dfrac{1}{8\pi} \displaystyle \int_{B(0, 1)} f(x') \; dx', \tag{11}$ 
then
$\dfrac{A_1}{\vert x \vert} \le u(x); \tag{12}$
since $\vert x \vert \ge 2$, $1  / \vert x \vert \le 1/2$, so we also have
$\dfrac{1}{2} \vert x \vert \le \vert x \vert \vert 1 - \dfrac{1}{\vert x \vert} \vert = \vert \vert x \vert - 1 \vert < \vert \vert x \vert - \vert x' \vert \vert < \vert x - x' \vert \tag{13}$
or
$\dfrac{1}{\vert x - x' \vert} < \dfrac{2}{\vert x \vert}, \tag{14}$
whence
$u(x) = \dfrac{1}{4\pi} \displaystyle \int_{B(0, 1)} \dfrac{f(x') }{\vert x - x' \vert}\; dx' \le \dfrac{1}{4\pi} \displaystyle \int_{B(0, 1)} \dfrac{2f(x') }{\vert x \vert}\; dx' = \dfrac{1}{2\pi} \dfrac{1}{\vert x \vert} \displaystyle \int_{B(0, 1)} f(x') \; dx'; \tag{15}$
taking
$A_2 = \dfrac{1}{2\pi} \displaystyle \int_{B(0, 1)} f(x') \; dx',  \tag{16}$
we have
$u(x) \le \dfrac{A_2}{\vert x \vert}; \tag{17}$ 
combining (12) and (17),
$\dfrac{A_1}{\vert x \vert} \le u(x) \le \dfrac{A_2}{\vert x \vert}, \tag{18}$
the requisite estimate for $u(x)$, $\vert x \vert \ge 2$.
