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:
$$\max\{|u|:||x||\leq 1\} \leq C(\max\{|g|:||x||=1\}+\max\{|f|:||x||<1\}).$$

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 $|x| \leq 1$. 
