- Let \(f _ { n } : D \to \mathbb { C } , n = 1 , 2 , 3 , . . . ,\) be a sequence of holomorphic functions on the unit disk D such that \(f _ { n } ^ { - 1 } ( 0 ) = \{ c _ { n } \}\) , where \(c _ { n } \in D\) . Suppose that \(f _ { n } \to f _ { 0 }\) uniformly, where \(f _ { 0 }\) is not constant.
Prove that \(f _ { 0 }\) has at most one zero in \(D\)
Can \(f _ { 0 }\) have no zeros? If so, give a necessary and sufficient condition on the \(c _ { n }\) for this to happen.
Solution
The uniform limit \(f_0\) of holomorphic functions is holomorphic on \(D\), and it is nonconstant, so its zeros are isolated.
1. (a) \(f_0\) has at most one zero in \(D\).
For a zero \(z_0\) of \(f_0\), take \(r>0\) with \(\overline{B(z_0,r)}\subset D\) containing no other zero of \(f_0\). By Hurwitz’s theorem, \(f_n\) has a zero in \(B(z_0,r)\) for all large \(n\). If \(f_0\) had two distinct zeros, disjoint such disks would give each large \(f_n\) two distinct zeros, while \(f_n^{-1}(0)=\{c_n\}\).
2. The zero in step 1 may be multiple: \(f_n(z)=\bigl(z-\frac1n\bigr)^2\) has \(f_n^{-1}(0)=\{\frac1n\}\) and converges uniformly on \(D\) to \(z^2\).
\(\abs{f_n(z)-z^2}=\abs{-\frac2nz+\frac1{n^2}}\le\frac3n\) on \(D\).
3. (b) Yes: \(f_n(z)=z-\bigl(1-\frac1n\bigr)\) has the single zero \(c_n=1-\frac1n\) and converges uniformly on \(D\) to \(z-1\), which has no zero in \(D\).
\(\abs{f_n(z)-(z-1)}=\frac1n\) for all \(z\).
4. (b) \(f_0\) has no zero in \(D\) if and only if \(\boxed{\abs{c_n}\to1}\).
If \(\abs{c_n}\not\to1\), a subsequence \(c_{n_k}\) converges to some \(c\in D\), and uniform convergence with continuity of \(f_0\) gives \(f_0(c)=\lim_kf_{n_k}(c_{n_k})=0\). Conversely, if \(f_0(z_0)=0\) for some \(z_0\in D\), Hurwitz’s theorem gives, for \(r\) as in step 1, a zero of \(f_n\) in \(B(z_0,r)\) for all large \(n\); that zero is \(c_n\), so \(\abs{c_n}\le\abs{z_0}+r<1\) for all large \(n\), and \(\abs{c_n}\not\to1\).
5. Q.E.D.
Steps 1 and 2 answer part (a), and steps 3 and 4 answer part (b).