Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 11 / 3

Use the curvature $-1$ hyperbolic metric $ds=2|dz|/(1-|z{|}^2)$. For the disk Möbius transformation ${\varphi }_w(z)=(z-w)/(1-\overline{w}z)$, set $r_w(z)=|{\varphi }_w(z)|$. The hyperbolic distance satisfiesThe triangle inequality givesThus convergence of the exponential sum is independent of the base point. Taking the base point zero, it is equivalent to the Blaschke condition ${\sum }_n(1-|z_n|)<\infty$. This condition implies that only finitely many terms lie in any compact subdisk, including finitely many repetitions of any point.

To construct the required holomorphic function, for $a\ne 0$ define the Blaschke factorFor $|z|\le R<1$, writing $a=|a|e^{i\alpha }$ givesConsequently, after the finitely many factors corresponding to small $|a|$ are removed, the sum of $|1-b_a|$ converges uniformly on each compact subdisk. Those factors are uniformly close to one, so their logarithms form an absolutely convergent series there. Their product tail is holomorphic and nonvanishing. Including the finitely many removed factors gives the Blaschke productwith exactly the requested zero multiplicities. Each finite product has modulus at most one in the disk, so the limit does too. Hence $f=B/2$ maps into the open disk and has exactly the prescribed zeros, proving the implication from the summability condition.

Let $f$ be the nonzero bounded holomorphic function in the zero-set condition. Choose $v$ with $f(v)\ne 0$ and compose with a disk Möbius transformation carrying zero to $v$. Call the resulting function $g$ and its zeros $a_n$, counted with multiplicity. Then $g(0)\ne 0$ and $|g|\le 1$.

For a radius $r$ with no zero on its circle, Jensen's formula givesTo see the formula directly, divide $g$ by its finitely many interior zero factors. The logarithm of the modulus of the resulting zero-free holomorphic function is harmonic near the closed disk and satisfies the mean value property. The circle average of $\log |r{e}^{i\theta }-a|$ for $|a|<r$ is $\log r$, obtained by expanding $\log (1-(a/r)e^{-i\theta })$. Subtracting its value at zero gives exactly the displayed zero sum.

Let $r\uparrow 1$ through zero-free circles. Increasing limits of the nonnegative summands giveSince $1-|a_n|\le -\log |a_n|$, the Blaschke condition follows for the transformed sequence. The hyperbolic metric identity and the base-point comparison proved above giveThis proves the reverse implication and completes the equivalence.

The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.