Mobius transformation without fixed points

Question

Let $T$ be a Mobius transformation of the unit disk $D=\{ |z|<1 \}$ onto itself, with no fixed points. Show that there exists $s \in \partial D$ such that $T^n(z) \rightarrow s$ as $n\rightarrow \infty$, for all $z\in D$.

Answer 1

We take two cases according to whether $T$ has one fixed point or two fixed points. (Since any non-identity Mobius transformation has one or two fixed points, these cases are exhaustive.) $$ $$ Case 1:

Suppose that $T$ has one fixed point. Let $f(z) = z+1$. Since all Mobius transformations with the same number of fixed points are conjugate, there exists a Mobius transformation $g$ so that $$T = g \circ f \circ g^{-1}.$$ The iterates of $T$ are given by $T^n = g \circ f^n \circ g^{-1}$. Fix $z \in D$. We claim that $T^n(z) \to g(\infty)$. To prove so, let $w_0 = g^{-1}(z)$, and define a sequence $\{w_n\}$ recursively by $w_{n+1} = f(w_n)$. We observe that for any $n \in \mathbb{N}$, $T^n(z) = g(w_n)$. By definition of $f$, $f^n(z) = z + n$. Hence, $f^n(z) \to \infty$ as $n \to \infty$. By continuity of $g$, $g(w_n) \to g(\infty)$. That is, $T^n(z) \to g(\infty)$.
It is immediate that $g(\infty) \in \partial D$; if it were in $\mathbb{C^*}-\overline{D}$, this would contradict continuity of $T$, whereas if it were in $D$, then this would contradict the assumption that the restriction of $T$ to $D$ has no fixed points.

$$ $$ Case 2:

Define $f$ by $f(z) = az$, where $a \in \mathbb{C} - \{0\}$ and $|a| < 1$. Since $f$ has two fixed points (namely $0$ and $\infty$), there exists a Mobius transformation $g$ so that $$T = g \circ f \circ g^{-1}.$$ The iterates of $T$ are given by $T^n = g \circ f^n \circ g^{-1}$. Since $|a| < 1$, $f^n(z) \to 0$ for any $z \in \mathbb{C}$. By a similar argument to that above, for any $z \in D$, we have $T^n(z) \to g(0)$.