- Suppose f and g are holomorphic in a region containing the disc \(| z | \leq 1\) Suppose that f has a simple zero at \(z = 0\) and vanishes nowhere else in \(| z | \leq 1\) Let
\[ f _ {\epsilon} (z) = f (z) + \epsilon g (z). \]
Show that if ǫ is suficiently small, then
\(f _ { \epsilon } ( z )\) has a unique zero in \(| z | \leq 1\) , and
\(\mathrm { i f } \ z _ { \epsilon }\) is this zero, the mapping \(\epsilon \mapsto z _ { \epsilon }\) is continuous.
Solution
Because \(f\) has a simple zero at \(0\) and no other zero on \(|z|\le1\), choose \(0<r<1\) so small that \(0\) is the only zero of \(f\) in \(|z|\le r\). Since \(f\) is nonzero on \(|z|=r\), set \[ m=\min_{|z|=r}|f(z)|>0, \qquad M=\max_{|z|=r}|g(z)|. \] If \(|\epsilon|M<m\), then on \(|z|=r\), \[ |\epsilon g(z)|<|f(z)|. \] Rouché’s theorem implies that \(f_\epsilon=f+\epsilon g\) has exactly one zero in \(|z|<r\), counted with multiplicity. On the compact annulus \(r\le|z|\le1\), \(f\) has positive minimum \(m_1\), while \(g\) has finite maximum \(M_1\). If also \(|\epsilon|M_1<m_1\), then \(f_\epsilon\) has no zero there. Thus, for sufficiently small \(\epsilon\), \(f_\epsilon\) has a unique zero \(z_\epsilon\) in \(|z|\le1\).
To prove continuity, let \(\epsilon_n\to\epsilon\) within this small disc of parameters. The zeros \(z_{\epsilon_n}\) lie in the compact disc \(|z|\le r\). Any convergent subsequence has a limit \(z_*\). Since \[ f(z_{\epsilon_n})+\epsilon_n g(z_{\epsilon_n})=0, \] continuity gives \[ f(z_*)+\epsilon g(z_*)=0. \] By uniqueness of the zero for parameter \(\epsilon\), \(z_*=z_\epsilon\). Every convergent subsequence has the same limit, so the entire sequence converges to \(z_\epsilon\). Hence \(\epsilon\mapsto z_\epsilon\) is continuous.