Let \(g\) be analytic for \(|z|\leq 1\) and \(|g(z)| < 1\) for \(|z| = 1\).
Show that \(g\) has a unique fixed point in \(|z| < 1\).
What happens if we replace \(|g(z)| < 1\) with \(|g(z)|\leq 1\) for \(|z|=1\)? 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 \(|z| < 1\) and \(|f(z)| < 1\) for \(|z| < 1\)? Suppose that \(f(z) \not\equiv z\). Can f have more than one fixed point in \(|z| < 1\)?
Hint: The map \(\displaystyle{\psi_{\alpha}(z)=\frac{\alpha-z}{1-\bar{\alpha}z}}\) may be useful.
Solution
Goal: Let \(g\) be analytic on \(|z| \le 1\) with \(|g(z)| < 1\) for \(|z| = 1\). (a) Show \(g\) has a unique fixed point in \(|z| < 1\). (b) What changes if \(|g(z)| \le 1\) on \(|z| = 1\)? (c) If \(f\) is analytic on \(|z| < 1\) with \(|f(z)| < 1\) and \(f \not\equiv z\), can \(f\) have more than one fixed point in \(|z| < 1\)?
1. (a) Setup: apply Rouch'e to \(g(z) - z\) on \(|z| = 1\).
1.1. On \(|z| = 1\), \(|g(z)| < |z| = 1\).
given \(|g(z)| < 1\) on \(|z| = 1\).
1.2. \(g(z) - z\) and \(-z\) have the same number of zeros in \(|z| < 1\).
Rouch'e’s theorem with \(f(z) = -z\) and \(g\) as perturbation: \(|g(z)| < |{-z}|\) on \(|z| = 1\) by 1, and \((g - z) = (-z) + g\).
1.3. \(g\) has exactly one fixed point in \(|z| < 1\).
\(-z\) has exactly one zero (at \(z = 0\), simple) in \(|z| < 1\); by 2, \(g(z) - z\) has exactly one zero (counting multiplicity) there.
2. (b) With \(|g(z)| \le 1\) on \(|z| = 1\), the conclusion of (a) may fail.
example \(g(z) = \tfrac{z+1}{2}\): \(|g(z)| \le \tfrac{|z|+1}{2} = 1\) on \(|z| = 1\) (equality at \(z = 1\)), but the only fixed point solves \(z = \tfrac{z+1}{2}\), i.e. \(z = 1\), which lies on \(|z| = 1\), not in \(|z| < 1\). So \(g\) has no fixed point in the open disk.
3. (c) No: a non-identity analytic map \(f: D \to D\) has at most one fixed point in \(D\).
3.1. Recall the Schwarz–Pick inequality: for \(a, b \in D\), \(\rho(f(a), f(b)) \le \rho(a, b)\) for the hyperbolic distance \(\rho\), with strict inequality if \(f\) is not an automorphism.
standard form of Schwarz–Pick.
3.2. If \(f\) had two distinct fixed points \(a, b \in D\), then \(f\) must be an automorphism.
\(\rho(a, b) = \rho(f(a), f(b)) \le \rho(a, b)\); equality in 1 forces \(f \in \Aut(D)\).
3.3. A non-identity automorphism of \(D\) has exactly one fixed point in \(D\).
writing \(\psi(z) = e^{i\theta}\frac{z - \alpha}{1 - \bar\alpha z}\), the fixed-point equation is a quadratic with product of roots of modulus \(|\alpha| < 1\), so exactly one root lies in \(D\) (or, for \(\alpha = 0\), the only fixed point in \(D\) is \(0\) unless \(\psi\) is the identity).
3.4. Conclusion for (c).
if \(f \not\equiv z\) had two fixed points, 2 makes \(f\) a non-identity automorphism, contradicting 3; hence at most one fixed point.
4. Q.E.D.
1 proves (a); 2 answers (b) with an example; 3 answers (c).
Solution: Part 1
Use Rouché: if \(\abs{f(z)} < 1\) is strict when \(\abs{z} = 1\), then consider \(F(z) \da f(z) - z\). Write the big part as \(M(z) = z\) and the small as \(m(z) = f(z)\), then on \(\abs{z} = 1\) \[ \abs{m(z)} = \abs{f(z)} < 1 = \abs{z} = \abs{M(z)} ,\] so \(M(z)\) and \(m(z) + M(z) = f(z) - z\) have the same number of zeros in \(\DD\) – precisely one.
Solution: Part 2
There is still a unique fixed point. Use the Brouwer fixed point theorem: since \(g\) is holomorphic on \(\bar{\DD}\), it is in particular continuous. By the Brouwer fixed point theorem, every continuous map \(\bar{\DD} \to \bar{\DD}\) has a fixed point. If \(g\) is nonconstant, then the fixed point is unique by Schwarz: without loss of generality one can assume \(f(0) = 0\) by composing with a Blaschke factor. Apply Schwarz to \(f\), then if \(f(a) = a\) we have the equality clause and \(f(z) = \lambda z\). Since \(a = f(a) = \lambda a\), \(\lambda = 1\) and \(f\) is the identity. If \(g\) is constant, then \(\abs{g(z)} < 1\) on \(\abs{z} = 1\) forces \(g\equiv 0\).
Solution: Part 3
Note that there is a major difference between self maps to \(\DD\) versus \(\bar{\DD}\). By the argument in part 2, if \(f(z)\) is not the identity then \(f\) can have at most one fixed point. Moreover, not every map \(f:\DD\to\DD\) need have a fixed point: consider \[ g: \HH &\to \HH \\ z &\mapsto z+1 .\] Now conjugate with the Cayley map \(C:\HH\to \DD\) to define \(f\da Cg\inverseof{C}:\DD\to \DD\) which has no fixed points at all.