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I would like to answer this question:

Let $\mathbb{E}:= \{z \in \mathbb{C}: |z| < 1\}$. Does there exist a holomorphic function $f: \mathbb{E} \to \mathbb{C}$ with $ f(0) = 2i$ and $\lim_{|z| \to 1}|f(z)| = 1$?

If we would replace $\lim_{|z| \to 1}|f(z)| = 1$ with the condition $|f(z)| = 1$ for all $z \in \mathbb{E}$, then I would know that the answer would be negative (identity theorem). Here I am unsure about whether or not the condition $\lim_{|z| \to 1}|f(z)| = 1$ changes something in regard to use the identity theorem.

I am thankful for some clarification.

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Since $\lim_{|z| \to 1}|f(z)| = 1$, there is an $r \in (0, 1)$ such that $|f(z)| \le 1.5$ for $r \le |z| < 1$.

The maximum modulus principle then implies that $|f(z)| \le 1.5$ for $|z| \le r$, so that $f(0) = 2i$ is not possible.

Remark: The same argument shows that if $f$ is holomorphic in the unit disk with $\lim_{|z| \to 1}|f(z)| = 1$ then $|f(z)| \le 1$ for all $z$.

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  • $\begingroup$ Just to make sure: You use the maximum modulus principle for bounded domains, correct? $\endgroup$
    – offline
    Jan 24 at 21:52
  • $\begingroup$ That is correct. $\endgroup$
    – Martin R
    Jan 24 at 21:58
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    $\begingroup$ You apply it to the closed disk with radius $r$. $\endgroup$
    – Martin R
    Jan 24 at 22:01
  • $\begingroup$ Now it's perfectly clear. Thank you! $\endgroup$
    – offline
    Jan 24 at 22:02

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