Let \(U\subset\mathbb C\) be connected and open, and let \(f_n:U\to\mathbb C\) be holomorphic with \(f_n^{-1}(0)=\{c_n\}\) for every \(n\geq1\). Suppose \(f_n\to f_0\) uniformly on \(U\), where \(f_0\) is nonconstant.
Prove that \(f_0\) has at most one zero in \(U\).
Can \(f_0\) have no zeros? Give a necessary and sufficient condition on \((c_n)\) for this to happen.
Solution
The zero-free case does occur. Its exact condition is that \((c_n)\) eventually leaves every compact subset of \(U\), equivalently that it has no subsequence converging to a point of \(U\).
1. The limit has at most one zero.
Uniform convergence implies compact-uniform convergence, so \(f_0\) is holomorphic [@SS03]. Since it is nonconstant on connected \(U\), its zeros are isolated. If it had distinct zeros \(a,b\), choose disjoint closed disks around them contained in \(U\), with no zero of \(f_0\) on either boundary. The minimum of \(|f_0|\) on the two boundaries is positive. For sufficiently large \(n\), uniform convergence gives \(|f_n-f_0|<|f_0|\) on both circles.
Rouché’s theorem then gives at least one zero of \(f_n\) in each disk, since \(f_0\) has one in each [@SS03]. The disjoint disks make these distinct, contradicting the single-point fiber \(f_n^{-1}(0)=\{c_n\}\). This proves (a).
2. A zero of the limit forces \(c_n\) to converge to it.
Suppose \(f_0(a)=0\). For every sufficiently small \(r>0\), the closed disk \(\overline{D(a,r)}\) lies in \(U\) and \(f_0\) has no zero on its boundary. As in step 1, Rouché’s theorem shows that \(f_n\) has a zero inside this disk for all sufficiently large \(n\). Its only zero is \(c_n\), so \(|c_n-a|<r\) eventually. Such radii can be chosen arbitrarily small; hence the entire sequence \(c_n\) converges to \(a\).
3. An interior subsequential limit of the zeros is a zero of \(f_0\).
Suppose \(c_{n_k}\to a\in U\). Choose a fixed compact disk \(K\subset U\) about \(a\); it contains \(c_{n_k}\) for all sufficiently large \(k\). Since \(f_{n_k}(c_{n_k})=0\), \[ |f_0(a)|\leq |f_0(a)-f_0(c_{n_k})| +\sup_{z\in K}|f_0(z)-f_{n_k}(z)|\longrightarrow0. \] The first term tends to zero by continuity of the fixed function \(f_0\), and the second by compact-uniform convergence. Thus \(f_0(a)=0\). Together with step 2, this proves that \(f_0\) is zero-free exactly when \((c_n)\) has no subsequence converging inside \(U\).
If the sequence does not eventually leave some compact set \(K\subset U\), infinitely many terms lie in \(K\); compactness yields an interior convergent subsequence. Conversely, an interior convergent subsequence eventually lies in a compact disk contained in \(U\). This proves the equivalent compact-escape formulation.
4. A nonconstant zero-free limit is possible under the stated uniform convergence.
Take \(U=D=\{|z|<1\}\) and \[ c_n=1-\frac1{n+1},\qquad f_n(z)=z-c_n, \qquad f_0(z)=z-1. \] Each \(f_n\) has exactly the one zero \(c_n\in D\), and \(\sup_{z\in D}|f_n(z)-f_0(z)|=1/(n+1)\to0\). The limit \(f_0\) is nonconstant and has no zero in \(D\); its only zero is the boundary point one. This supplies the example required in (b) and verifies the criterion.