Let \(f: \DD\to \DD\) with \(\ts{a_k}_{k\leq n}\) the zeros of \(f\) in \(\DD\). Show that \[ \abs{f(z)}\leq \prod_{k\leq n} \abs{ \psi_{a_k}(z) } .\]
Solution
Here \(\psi_a(z)\da{a-z\over1-\bar az}\), and the zeros \(a_k\) are listed with multiplicity. Define \(\Psi(z) \da \prod_{k\leq n} \psi_{a_k}(z)\) and \(g(z) \da f(z)/\Psi(z)\). Each zero of \(\Psi\) in \(\DD\) is a zero of \(f\) of at least the same order, so \(g\) has removable singularities and is holomorphic on \(\DD\). The claim is that \(\abs{g(z)} \leq 1\), which implies the result directly. Since \(\Psi\) is continuous on \(\overline\DD\) with \(\abs{\Psi} = 1\) on \(\abs{z} = 1\), \(m(r)\da\min_{\abs z=r}\abs{\Psi(z)}\to1\) as \(r\to1^-\). Fix \(z\in\DD\). For \(\abs z<r<1\) close enough to \(1\) that \(m(r)>0\), the maximum modulus principle on \(\abs\zeta\le r\) gives \[ \abs{g(z)}\le\max_{\abs\zeta=r}\abs{g(\zeta)} = \max_{\abs\zeta=r}{ \abs{f(\zeta)} \over \abs{\Psi(\zeta)} } \leq {1\over m(r)} \convergesto{r\to 1^-} 1 .\]