Let \(g\) be analytic for \(\abs{z} \le 1\) and \(\abs{g(z)} < 1\) for \(\abs{z} = 1\).
Show that \(g\) has a unique fixed point in \(\abs{z} < 1\).
What happens if we replace \(\abs{g(z)} < 1\) with \(\abs{g(z)} \le 1\) for \(\abs{z} = 1\)? Give an example if (a) is not true or give an proof if (a) is still true.
What happens if we simply assume that \(f\) is analytic for \(\abs{z} < 1\) and \(\abs{f(z)} < 1\) for \(\abs{z} < 1\)? Suppose that \(f(z) \not\equiv z\). Can \(f\) have more than one fixed point in \(\abs{z} < 1\)?
Hint: The map \(\psi_\alpha(z) = \frac{\alpha - z}{1 - \bar{\alpha} z}\) may be useful.