# Infinite Blaschke product converges normally outside the accumulation points of $a_k$

Let $\Bbb D \subset \Bbb C$ be the unit circle. Let $a_k \in \Bbb D$ and $|a_k| \to 1$. Let $E \subset \partial \Bbb D$ be the set of accumulation points of $a_k$.

Define $B_k$ as follows: $B_k = \frac {\bar a_k}{|a_k|}\frac{a_k-z}{1-\bar a_k z}$ if $a_k \neq 0$, and $B_k=z$ if $a_k = 0$.

I want to show that if $\sum \limits_{k=1}^\infty (1-|a_k|)$ converges then $\prod \limits_{k=0}^\infty B_k$ converges unfiromly on every compact subset of $\Bbb C_\infty \setminus E$ to a meromorphic function.

Convergence to a meromorphic function is meaningful as convergence in the spherical metric.

What I know

Letting $B_k = 1+A_k$, we calculate that $A_k = (|a_k|-1)\frac{\bar a_k z+|a_k|}{|a_k|(1-\bar a_k z)}$ and $|A_k|\leq (1-|a_k|)\frac {1+|z|}{1-|z|}$, so on every compact subset of $\Bbb D$ we have by the M-test that $B(z)$ converges to an analytic function $B(z)$.

For compact subsets $K$ of $\Bbb C_\infty \setminus E$ that are not contained in $\Bbb D$, we can no longer use the M-test to deduce that $B(z)$ converges. How do we show that there is a uniform convergence on $K$ to a meromorphic function, then? If $K$ is disjoint from $\partial \Bbb D$ then by compactness it is at a positive distance from $\partial \Bbb D$ so we can find $r>1, R$ with $\frac r {|a_k|}<|z|<R$ on $K$ and for large enough $k$ values we know that on $K$ we have $|z|>\frac 1{|\bar a_k|}$ for all $k$, since $|a_k| \to 1$. We then note $\frac {1+|z|}{|1-|\bar a_kz||} =\frac {1+|z|}{|\bar a_k z|-1} \leq \frac{1+R}{r-1}$. So we can again use the M-test to deduce that the product converges normally to an analytic function when we omit some rational terms, so overall the product converges to a meromorphic $B(z)$. Hence the product is analytic on $\Bbb C_\infty \setminus \partial \Bbb D$.

On $\partial \Bbb D \setminus E$, how do I know that it is meromorphic?

On every such compact set $K$, by disjointness from $E$, there are only finitely many $a_k$ values in $K$.

To prove that the product is meromorphic, one approach is to find a finite number of terms in the product that except them the product is analytic, since then we multiply by those terms that are evidently meromorphic to obtain a meromorphic function.

## 1 Answer

I finally solved it: the set of accumulation points of $a_k$ is the same as the set of accumulation points of $\frac {1}{\bar a_k}$, so every compact set $K \subset \Bbb C \setminus E$ contains only finitely many points of the form $\frac {1}{\bar a_k}$, so when looking at the product after sufficiently many terms it does not contain any of them, so $|1-\bar a_k z|$ attains a positive minimum on $K$ and we can proceed as in the post with the M-test.