Let \(U\subset\mathbb C\) be open and contain the closure \(\overline D\) of a unit disk. Suppose holomorphic functions \(f_n:U\to\mathbb C\) converge uniformly on compact subsets of \(U\) to \(f\), and assume that \(f\) has no zero on \(\partial D\). Prove that for all sufficiently large \(n\), the functions \(f_n\) and \(f\) have the same number of zeros in \(D\), counted with multiplicity.
Boundary nonvanishing cannot be dropped. On the unit disk centered at zero, \(f_n(z)=z-(1-1/n)\) for \(n\geq2\) converges locally uniformly on the plane to \(f(z)=z-1\). Each \(f_n\) has one zero in the disk, while \(f\) has none there; its zero is on the boundary.
Solution
1. The limit is holomorphic and has a positive boundary modulus minimum.
The local uniform limit theorem makes \(f\) holomorphic on \(U\) [@SS03]. Since \(f\) is continuous and nonzero on the compact circle \(\partial D\), \[ \delta:=\min_{z\in\partial D}|f(z)|>0. \] In particular \(f\) is not identically zero on the component of \(U\) containing \(\overline D\). Its zeros in \(\overline D\) are finite in number: an infinite set of distinct zeros would have an accumulation point in this compact subset of \(U\), contradicting the identity theorem [@SS03]. Each zero has finite multiplicity by its Taylor expansion.
2. Uniform convergence on the boundary gives the same zero count.
1. This holds for every\(n\geq N\), as required.