Maximum Principle for Poisson Equation For a smooth $u(x)$, $x \in \mathbb{R}^n$, satisfying:
$\Delta u = -f$ for $||x||<1$ ,    $u=g$ on $||x||=1$
I want to show that there exists a constant $C$ such that:
$$\max\{|u|:||x||\leq 1\} \leq C(\max\{|g|:||x||=1\}+\max\{|f|:||x||<1\}).$$
 A: Well, implicitly, $A = \|f\|_\infty$ and $B = \|g\|_\infty$ must exist. 
Thus, let us consider
$$u_+(x) = B + A (1-|x|^2)$$
It is a simple matter to check that $\Delta(u_+ - u) \leq 0$ where $|x| < 1$ and that $u_+ - u \geq 0$ on $|x|=1$. The maximum principle tells you the rest. 
You can analogously construct
$$u_-(x) = -B - A (1-|x|^2)$$
Using these as super and subsolutions for your problem, you can see that 
$$|u(x)| \leq A + B$$
where $|x| \leq 1$. 
A: If $u$ is a smooth function, then also $f$ and $g$ are smooth. Defined the Newtonian Potential $v$ of $f$ by $$v(x)=\int_B \Gamma(x-y)f(y)dy$$
where $B$ is the unit open ball about the origin and $\Gamma$ is the fundamental sollution of the Laplace equation. Consider the solution of the auxiliary problem $$
 \left\{ \begin{array}{rl}
 -\Delta w=0 &\mbox{ in $B$} \\
  w=v-g &\mbox{ in $\partial B$}
       \end{array} \right.
$$
By unicity we conclude that $u=v-w$. Also, from the maximum principle, we have that $$\|w\|_{L^\infty(\overline{B})}\leq \|v\|_{L^\infty(B)}+\|g\|_{L^\infty(\partial B)}$$
From here, you have two things to do:
1 - Show that there exist a constan $C_1$, such that $\|v\|_{L^\infty(B)}\leq C_1\|f\|_{L^\infty(B)}$
2 - Use the representation of $u$ in terms of $v$ and $w$ to conclude.
