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I was trying to prove If f is holomorphic on the open unit disk and|f(z)|= 1 for|z|= 1, then f is a finite Blaschke product. In the proof, I saw Since|f(z)|→1 uniformly as|z|→1, there is an r <1 so that f is nonvanishing on the annulus r≤|z|<1. My question is why does |f(z)| tends to 1 uniformly? And why does f has a finite number of zeroes in the open unit disk?`

Krieg
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  • This question has been asked here before, I have answered one of its iterations recently. In my proof I do not use uniform continuity in any way. – SescoMath May 07 '20 at 05:15
  • It must have finitely many roots in the unit disk as otherwise it would be exactly $0$ by the identity theorem. – SescoMath May 07 '20 at 05:16
  • Thanks for clarifying the 2nd part. Can you tell me why the 1st part happens or provide a link of your answer? – Krieg May 07 '20 at 05:18
  • @Kreig See here: https://math.stackexchange.com/questions/3658683/problem-on-holomorphic-function-schwarzs-lemma/3658686#3658686 (upvote it if you like the answer lol). The gist of it is that since you know $f$ has finitely many roots in the disk, you can consider a finite Blashke product $g$ with the same roots and use the maximum modulus principle on $\vert f/g\vert$. This tells you that $f$ and $g$ are related by a phase, so $f$ is a Blashke product too. – SescoMath May 07 '20 at 05:20
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    The statement is not correct. You have to assume that $f$ is continuous on the closed disk. $f=0$ in the open disk and $f=1$ on the boundary gives a counter--example. – Kavi Rama Murthy May 07 '20 at 05:46

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