In this problem, we discuss Blaschke products, which are bounded analogues in the disc of the Weierstrass products for entire functions.
Show that for \(0<|\alpha|<1\) and \(|z|\leq r<1\), the inequality
\[ \left|\frac{\alpha+|\alpha|z}{(1-\overline{\alpha}z)\alpha}\right| \leq \frac{1+r}{1-r} \]
holds.
Let \(\{\alpha_n\}\) be a sequence in the unit disc such that \(\alpha_n\neq 0\) for all \(n\) and
\[ \sum_{n=1}^{\infty}(1-|\alpha_n|)<\infty. \]
This condition holds when \(\{\alpha_n\}\) are the zeros of a bounded holomorphic function on the unit disc, by Problem 1. Show that the product
\[ f(z)=\prod_{n=1}^{\infty} \frac{\alpha_n-z}{1-\overline{\alpha_n}z} \frac{|\alpha_n|}{\alpha_n} \]
converges uniformly for \(|z|\leq r<1\) and defines a holomorphic function on the unit disc with precisely the zeros \(\alpha_n\). Show also that \(|f(z)|\leq 1\).
Solution
For part 1, write \(\rho=|\alpha|\). Then \[ \left|\frac{\alpha+\rho z}{(1-\overline\alpha z)\alpha}\right| \le \frac{|\alpha|+\rho|z|}{|\alpha|(1-|\alpha||z|)} =\frac{1+|z|}{1-\rho|z|} \le \frac{1+r}{1-r} \] for \(|z|\le r<1\).
Now put \[ b_\alpha(z)=\frac{\alpha-z}{1-\overline\alpha z}\frac{|\alpha|}{\alpha}. \] A direct calculation gives \[ 1-b_\alpha(z) =(1-|\alpha|)\frac{\alpha+|\alpha|z}{\alpha(1-\overline\alpha z)}. \] Thus, for \(|z|\le r<1\), \[ |1-b_{\alpha_n}(z)| \le \frac{1+r}{1-r}(1-|\alpha_n|). \tag{1} \] The assumed Blaschke condition makes the right-hand side summable. Hence \(\sum_n(1-b_{\alpha_n})\) converges uniformly on every closed disc \(|z|\le r<1\). The standard infinite-product criterion therefore implies that \[ \prod_{n=1}^\infty b_{\alpha_n}(z) \] converges uniformly on compact subsets to a holomorphic function \(f\), and the limit is nonzero at every point at which none of the factors vanishes. Consequently the zeros of \(f\) are precisely the \(\alpha_n\), with multiplicity.
For \(|z|<1\), \[ \left|\frac{\alpha-z}{1-\overline\alpha z}\right|<1, \] because \[ |1-\overline\alpha z|^2-|\alpha-z|^2 =(1-|\alpha|^2)(1-|z|^2)>0. \] Therefore every finite partial product has modulus at most \(1\), and passing to the locally uniform limit gives \(|f(z)|\le1\) on \(\mathbb D\).