Let \(\Omega\) be a star-shaped domain of \(\mathbb{C}\). Let \(\{f_n\}_{n=1}^\infty\) be a sequence of injective analytic functions on \(\Omega\) that converges uniformly on every compact subset of \(\Omega\) to a function \(f\). Prove that:
\(f\) is analytic on \(\Omega\), and
\(f\) is either injective or a constant.
Solution
Since \(f_n\to f\) uniformly on compact subsets, the locally uniform limit theorem for holomorphic functions implies that \(f\) is holomorphic. For completeness, if \(T\Subset\Omega\) is any triangle, then \[ \int_{\partial T}f(z)\,dz =\lim_{n\to\infty}\int_{\partial T}f_n(z)\,dz=0, \] so Morera’s theorem gives holomorphicity.
Assume that \(f\) is not constant. Suppose, toward a contradiction, that \(f(a)=f(b)\) for distinct \(a,b\in\Omega\). Since \(\Omega\) is star-shaped it is connected, and since \(f\) is nonconstant, the zeros of \[ g(z)=f(z)-f(a) \] are isolated. Choose a small closed disk \(D\) around \(b\) whose closure lies in \(\Omega\), which does not contain \(a\), and such that \(b\) is the only zero of \(g\) in \(D\).
Set \[ g_n(z)=f_n(z)-f_n(a). \] Then \(g_n\to g\) uniformly on \(D\). By Hurwitz’s theorem, for all sufficiently large \(n\), \(g_n\) has a zero \(b_n\) in \(D\). But \[ g_n(b_n)=0 \quad\Longrightarrow\quad f_n(b_n)=f_n(a). \] Injectivity of \(f_n\) forces \(b_n=a\), impossible because \(a\notin D\). Hence \(f\) is injective. Therefore the limit is either injective or constant.