Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 9 / 2 / b

Let $m$ be the multiplicity at zero, and write $f(z)=z^{m}h(z)$ with $h(0)\ne 0$. The proof in solution gives $|f|\le 1$. For radii without boundary zeros, Jensen's formula saysThe absolute logarithmic hypothesis makes the left side tend to zero. The nonnegative zero sum increases to ${\sum }_n\log (1/|a_n|)$ by the monotone convergence theorem, so this series is finite. Hence ${\sum }_n(1-|a_n|)<\infty$, and the Blaschke condition supplies a Blaschke product $B$ with exactly these zeros and their multiplicities.

The quotient $g=f/B$ extends holomorphically across every zero because the multiplicities agree, and the extensions have nonzero values. Thus $f=Bg$ with $g$ zero-free and holomorphic. We will also need the stronger Blaschke factorization of a bounded holomorphic function, namely $|g|\le 1$. Let $B_N$ contain the factor $z^m$ and the first $N$ other zero factors. The quotient $f/B_N$ is holomorphic. On $|z|=R$ enclosing those finitely many zeros, its modulus is at most $1/{\min }_{|z|=R}|B_N(z)|$. The maximum modulus principle transfers this bound to the interior. Since the finite Blaschke product has boundary modulus one continuously, letting $R\uparrow 1$ gives $|f/B_N|\le 1$. Now let $N\to \infty$ using locally uniform convergence of the Blaschke product, first off the zeros and then by continuity at them. This proves $|g|\le 1$.