Let \(f_n\to f\) uniformly on compact subsets of an open connected set \(\Omega\subset\mathbb C\), where every \(f_n\) is analytic and \(f\) is not identically zero. Show that if \(f(w)=0\), then there are points \(z_n\to w\) such that \(f_n(z_n)=0\) for all sufficiently large \(n\).
Solution
1. The limit function \(f\) is analytic on \(\Omega\), and its zero at \(w\)is isolated.
Uniform convergence on compact subsets of analytic functions implies that the limit is analytic. Since \(\Omega\) is connected and \(f\) is not identically zero, the identity theorem implies that the zeros of \(f\) are isolated.
2. Choose radii\[ r_1>r_2>\cdots>0, \qquad r_k\longrightarrow0, \]such that every closed disk\[ \overline{D(w,r_k)}\subset\Omega \]and \(w\) is the only zero of \(f\) in each disk.
By 1, there is \(r>0\) such that \[ \overline{D(w,r)}\subset\Omega \] and \(f\) has no zero in \(\overline{D(w,r)}\setminus\{w\}\). Any strictly decreasing sequence \(r_k\to0\) with \(r_k<r\) has the stated properties.
3. For each \(k\), there is an integer \(N_k\) such that every\(n\ge N_k\) has at least one zero in \(D(w,r_k)\).
On the circle \[ C_k=\{z:|z-w|=r_k\} \] the continuous function \(|f|\) has a positive minimum \[ m_k:=\min_{C_k}|f|>0. \] Compact-uniform convergence gives \(N_k\) such that, for \(n\ge N_k\), \[ |f_n(z)-f(z)|<m_k\le |f(z)| \qquad(z\in C_k). \] Rouché’s theorem therefore says that \(f_n\) and \(f\) have the same number of zeros in \(D(w,r_k)\), counted with multiplicity. Since \(f(w)=0\), that number is positive. Hence \(f_n\) has at least one zero in the disk.
4. The thresholds \(N_k\) may be chosen strictly increasing.
Replace the sequence from 3 recursively by \[ N_1'=N_1, \qquad N_k'=\max\{N_k,N_{k-1}'+1\} \quad(k\ge2). \] The property in 3 remains true with \(N_k'\) in place of \(N_k\), and the new sequence is strictly increasing. Rename it \(N_k\).
5. There are zeros \(z_n\) of \(f_n\), for all sufficiently large \(n\),such that \(z_n\to w\).
For \(n\ge N_1\), let \(k(n)\) be the largest integer \(k\) with \[ N_k\le n. \] By 3, choose a zero \(z_n\) of \(f_n\) satisfying \[ |z_n-w|<r_{k(n)}. \] Because the strictly increasing sequence \(N_k\) tends to infinity, \(k(n)\to\infty\) as \(n\to\infty\). Since \(r_k\to0\), \[ |z_n-w|<r_{k(n)}\longrightarrow0. \] Thus \(z_n\to w\), as required.