- Let \(U \subset \mathbb { C }\) be an open set containing \(\overline { { \boldsymbol { D } } } ( \boldsymbol { P } , \boldsymbol { r } )\) . Prove that if \(f : U \to \mathbb { C }\) is a holomorphic function such that \(f\) is nowhere zero on \(\partial D ( P , r )\) and \(g : U \to \mathbb { C }\) is a holomorphic function sufficiently uniformly close to \(f\) on \(\partial D ( P , r )\) , then the number of zeros of \(f\) in \(D ( P , r )\) equals the number of zeros of g in \(D ( P , r )\) (counting multiplicity).
Solution
1. Boundary nonvanishing gives a positive approximation threshold.
Write \(D=D(P,r)\) and \(C=\partial D\). Continuity of \(f\) and compactness of \(C\) imply that \[ \delta=\min_{z\in C}|f(z)|>0. \] It suffices to require \(\sup_C|g-f|<\delta\). For such a \(g\), \[ |g(z)-f(z)|<\delta\leq|f(z)|\qquad(z\in C). \] In particular \(|g(z)|\geq|f(z)|-|g(z)-f(z)|>0\) on \(C\).
2. The two finite zero counts agree, with multiplicities.
Both \(f\) and \(g\) are holomorphic on the open neighborhood \(U\) of \(\overline D\). The strict boundary inequality in step 1 allows Rouché’s theorem to be applied to \(f\) and the perturbation \(g-f\). It gives the same number of zeros for \(f\) and \(f+(g-f)=g\) inside \(C\), counted with multiplicity [@SS03, Chapter 3, Theorem 4.3].
These counts are finite. Neither function is identically zero on the component containing the disk, since neither vanishes on its boundary. Infinitely many distinct zeros in \(\overline D\) would have an accumulation point in that compact subset of \(U\), contradicting the identity theorem. Every zero has finite multiplicity by its local Taylor expansion [@SS03]. Thus the equality is of precisely the finite multiplicity counts requested.