Prove that if \(f\) is holomorphic in the unit disc, bounded, and not identically zero, and \(z_1,z_2,\ldots\) are its zeros, with \(|z_k|<1\), then
\[ \sum_n(1-|z_n|)<\infty. \]
Hint. Use Jensen’s formula.
Solution
If \(0\) is a zero of \(f\), factor \(f(z)=z^m g(z)\) with \(g(0)\ne0\); this removes only finitely many zeros, so it suffices to treat the case \(f(0)\ne0\).
Let \(M=\sup_{\mathbb D}|f|\). For \(0<r<1\) such that no zero lies on \(|z|=r\), Jensen’s formula gives \[ \log |f(0)|+\sum_{|z_n|<r}\log\frac r{|z_n|} =\frac1{2\pi}\int_0^{2\pi}\log|f(re^{i\theta})|\,d\theta \le \log M. \] Hence \[ \sum_{|z_n|<r}\log\frac r{|z_n|}\le \log\frac M{|f(0)|}. \tag{1} \] Fix \(0<R<1\) and let \(r\uparrow1\) through radii larger than \(R\). From (1), \[ \sum_{|z_n|<R}\log\frac1{|z_n|}\le \log\frac M{|f(0)|}. \] Letting \(R\uparrow1\) and using monotone convergence for the nonnegative terms yields \[ \sum_n\log\frac1{|z_n|}<\infty. \] Finally, for \(0<t<1\), \(-\log t\ge1-t\). Therefore \[ \sum_n(1-|z_n|)\le \sum_n\log\frac1{|z_n|}<\infty. \]