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If i have $|f(z)|\leq|f(z^n)|$ in the unit disk, and $f$ is holomorphic, does it follow that $|f(z)|\leq \lim_{n\rightarrow\infty}|f(z^n)|=|f(0)|$?

TETRACTYS
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  • What do you think? Does $z^n \to 0$ imply $f(z^n) \to f(0)$? – Martin R Mar 30 '20 at 20:20
  • Well yes, but do I know that the inequality holds at the limit $f(0)$ if it holds everywhere else? I suppose that follows from the function being holomorphic? @Martin R – TETRACTYS Mar 30 '20 at 20:32
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    This has nothing to do with holomorphy. $f(z^n) \to f(0)$ because $f$ is continuous. And $|f(z)|\leq|f(z^n)|$ for all $n$ implies $\lim |f(z)|\leq \lim |f(z^n)|$, see for example https://math.stackexchange.com/q/1589689/42969. – Martin R Mar 30 '20 at 20:35
  • @Martin R Thanks that makes sense! – TETRACTYS Mar 30 '20 at 20:47
  • If the function is supposed to be holomorphic, not only do we have this (as @MartinR noted that depends only on the continuity of the function): we also have that $f$ is constant, from the maximum modulus principle, as you can see here https://math.stackexchange.com/questions/3566798/if-f-is-an-entire-function-satisfying-fz-leq-fz2-for-all-z-in-bbb/3566811#3566811 –  Mar 30 '20 at 23:11

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