- Let Ω be a bounded open subset of \(\mathbb { C } ,\) and \(\varphi : \Omega \to \Omega\) a holomorphic function. Prove that if there exists a point \(z _ { 0 } \in \Omega\) such that
\[ \varphi (z _ {0}) = z _ {0} \quad \mathrm{and} \quad \varphi^ {\prime} (z _ {0}) = 1 \]
then \(\varphi\) is linear.
[Hint: Why can one assume that \(z _ { 0 } = 0 ?\) Write \(\varphi ( z ) = z + a _ { n } z ^ { n } + O ( z ^ { n + 1 } )\) near 0, and prove that if \(\varphi _ { k } = \varphi \circ \cdots \circ \varphi\) (where \(\varphi\) appears k times), then \(\varphi _ { k } ( z ) =\) \(z + k a _ { n } z ^ { n } + O ( z ^ { n + 1 } )\) . Apply the Cauchy inequalities and let \(k \to \infty\) to conclude the proof. Here we use the standard O notation, where \(f ( z ) = O ( g ( z ) )\) as \(z \to 0\) means that \(| f ( z ) | \leq C | g ( z ) |\) for some constant C as \(| z | \xrightarrow { } 0 . ]\)
Connectedness is needed. For \(\Omega=\mathbb D\cup D_1(3)\), the map equal to \(z\) on \(\mathbb D\) and to the constant \(3\) on \(D_1(3)\) is holomorphic from \(\Omega\) to \(\Omega\), fixes \(0\) with derivative \(1\), and is not linear.
Solution
Translating \(\Omega\) by \(-z_0\), assume \(z_0=0\), so \(\varphi(0) = 0\) and \(\varphi'(0) = 1\). Since \(\Omega\) is bounded, there is \(M\) with \(\abs z\le M\) for all \(z \in \Omega\), and there is \(r > 0\) with \(\overline{D}_r(0) \subset \Omega\). For \(k\ge1\) let \(\varphi_k\) be the \(k\)-fold iterate of \(\varphi\); since \(\varphi(\Omega)\subseteq\Omega\), \(\abs{\varphi_k}\le M\) on \(\Omega\).
1. If \(\varphi(z) = z + a_n z^n + O(z^{n+1})\) with \(n\ge2\), then \(\varphi_k(z) = z + k a_n z^n + O(z^{n+1})\) for every \(k\ge1\).
By induction on \(k\); the case \(k = 1\) is the hypothesis. If the claim holds for \(k\), then \[\varphi_{k+1}(z) = \varphi_k(z) + a_n \varphi_k(z)^n + O(\varphi_k(z)^{n+1}) = z + (k+1) a_n z^n + O(z^{n+1}),\] because \(\varphi_k(z)^n=z^n+O(z^{n+1})\).
2. Every Taylor coefficient of \(\varphi\) at \(0\) of order \(n\ge2\) vanishes.
Suppose not, and let \(n \ge 2\) be the least order with \(a_n\ne0\), so that \(\varphi(z) = z + a_n z^n + O(z^{n+1})\). By step 1, \(\varphi_k^{(n)}(0)/n! = k a_n\), and the Cauchy inequalities on \(\abs z=r\) give \(\abs{k a_n} \le M/r^n\) for every \(k \ge 1\). Letting \(k \to \infty\) gives \(a_n = 0\), a contradiction.
3. Q.E.D.
By step 2, \(\varphi(z)=z\) near \(0\), so \(\varphi(z)=z\) on the connected component of \(\Omega\) containing \(0\) by the identity theorem. Undoing the translation, \(\varphi(z)=z\) there; in particular \(\varphi\) is linear when \(\Omega\) is connected.