Let $a$ be a point of $S^{n-1}$ fixed by a parabolic transformation $\phi$ of $B^{n}$ (conformal ball model). One has to show that if $x$ is in $\bar{B^{n}}$ , then
$$lim_{m \rightarrow \infty} \phi^{m}(x)=a$$
I have tried taking the convergent subsequence using compactness of $\bar{B^{n}}$ but could not prove the required. Any kind of help would be appreciated.
If there exists a point $x\in\overline{B^n}$ such that $\lim_{m\rightarrow\infty}\phi^m(x)\neq a$ then there is some positive real number $r$ so that $d_{\overline{B^n}}(\phi^m(x),a)>r$ for infinitely many $m>0$. Enumerate the corresponding exponents $m_i$. This gives us a sequence of points $(\phi^{m_i}(x))_{i>0}$ in the complement of the open $r$-ball about $a$. This complement is compact, so there is a convergent subsequence, with limit $z\neq a$. By continuity of $\phi$, the point $z$ must be fixed, contradicting the fact that $\phi$ is parabolic.