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Suppose we have a sequence of distinct complex numbers $\{a_n\}$ such that $a_n\rightarrow 1$ and the $a_n$ satisfy the Blaschke condition $\sum (1-|a_n|)<\infty$. Does there exist a Blaschke product $B$ such that $B(a_n)=0$ for each $n$ and $B(1)=1$?

I know that there is a Blaschke product $B$ for which $B(a_n)=0$ for all $n$, and my question is about what control we have over choosing certain boundary values of $B$. Can we ever guarantee that $B(1)$ exists, and if so, can we choose its value?

Thank you for your help.

user122916
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  • Does your series $\sum_n (1 - |a_n|)$ converge hust conditionally, or absolutely? Or do you think it matters? – user2566092 Aug 06 '15 at 00:09
  • Does your summation series converge or just have a bounded set of finite partial sums? – DanielWainfleet Aug 06 '15 at 06:07
  • What do you mean by $B(1)$? Since the zeros of $B$ accumulate at $z=1$, there is no chance that $B$ is (can be extended to) a holomorphic function in a neighbourhood of $1$. So you must specify in what way you want to consider boundary values of $B$. You may be interested in http://math.stackexchange.com/questions/5331 – mrf Aug 06 '15 at 11:36
  • @mrf A Blaschke product has the form $B(z)=\displaystyle\prod_{n=1}^\infty \frac{\overline{a_n}}{|a_n|}\frac{a_n-z}{1-\overline{a_n}z}$. If each $a_n$ is real, then since $\frac{\overline{a_n}}{|a_n|}\frac{a_n-z}{1-\overline{a_n}z}=-1$ for all $n$, $B(1)$ doesn't exist. But if instead $B(z)=\displaystyle\prod_{n=1}^\infty \frac{z-a_n}{1-\overline{a_n}z}$ is the interpolating Blaschke product, then $B(1)=1$. So an infinite Blaschke product in this scenario need not be holomorphic in a neighborhood of $1$. But can we choose a Blaschke product for which $B(1)$ exists? – user122916 Aug 07 '15 at 05:48

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