Let \(f\) be holomorphic on a neighborhood of the closed unit disk \[ \overline{B_1(0)}=\{z:|z|\le1\}. \] Suppose \[ \max_{|z|=1}|f(z)|\le1. \] Prove that there exists \(z\) with \(|z|\le1\) such that \[ f(z)=z. \]
Solution
For each integer \(m\ge1\), set \[ \alpha_m\coloneqq1+\frac1m \] and \[ g_m(z)\coloneqq f(z)-\alpha_m z. \]
1. For every \(m\ge1\), there exists \(z_m\) with\(\abs{z_m}<1\) such that\[ f(z_m)=\alpha_m z_m. \]
On the unit circle, \[ \abs{f(z)}\le1<\alpha_m=\abs{\alpha_m z}. \] Hence, by Rouché’s theorem, the functions \(g_m(z)=f(z)-\alpha_m z\) and \(-\alpha_m z\) have the same number of zeros in the open unit disk, counted with multiplicity. The latter has exactly one zero there. Therefore \(g_m\) has a zero \(z_m\) with \(\abs{z_m}<1\), and \(g_m(z_m)=0\) is exactly \(f(z_m)=\alpha_m z_m\).
2. There are a subsequence \((z_{m_k})\) and a point \(z\) with\(\abs{z}\le1\) such that\[ z_{m_k}\longrightarrow z. \]
Every \(z_m\) lies in the closed unit disk, which is compact. Thus the sequence \((z_m)\) has a convergent subsequence whose limit remains in the closed unit disk.
3. The limit point \(z\) from step 2 satisfies\[ f(z)=z. \]
Since \(f\) is holomorphic on a neighborhood of the closed unit disk, it is continuous there. Using step 1 along the subsequence from step 2, \[ \begin{aligned} f(z) &=\lim_{k\to\infty}f(z_{m_k})\\ &=\lim_{k\to\infty}\alpha_{m_k}z_{m_k}\\ &=1\cdot z=z, \end{aligned} \] because \(\alpha_m\to1\).
4. There exists \(z\) with \(\abs{z}\le1\) and\[ \boxed{f(z)=z}. \]
Step 2 gives \(\abs{z}\le1\), and step 3 gives the fixed-point identity.
5. Q.E.D.
Step 4 is the required conclusion.