Can we have a map $f$, holomorphic on the open unit disk, such that $|f(z)|\rightarrow \infty$ as $|z|\rightarrow 1$? I think not, (at least I can't think of any such map), but I'd like to be able to prove this.


marked as duplicate by Daniel Fischer complex-analysis Nov 26 '16 at 13:12

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    $\begingroup$ Do you mean as $|z|\rightarrow 1$? 1/(z-1) blows up at one point, do you need it to blow up at all boundary points? $\endgroup$ – mathematician Nov 26 '16 at 8:20
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    $\begingroup$ To blow up at all boundary points consider the answer to this post $\endgroup$ – Mark Nov 26 '16 at 8:34

Such a function cannot exist. Suppose such a function exists. Let $z_1,\ldots,z_k$ be the zeros of $f$ in the unit disc with corresponding multiplicites $m_1,\ldots,m_k$. Then $$g(z) = \frac{(z - z_1)^{m_1}\cdots(z - z_n)^{m_k}}{f(z)}$$ is an analytic function in the unit disc such that $\lim_{\vert z \vert \to 1}g(z) = 0$. This contradicts the maximum modulus principle.

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    $\begingroup$ Note that we don't have $\lim\limits_{\lvert z\rvert \to 1} \lvert f(z)\rvert = +\infty$. That is not possible for holomorphic functions on the unit disk. Also, the denseness of dyadic rationals doesn't prove that the radial limit is $\infty$ at other points. $\endgroup$ – Daniel Fischer Nov 26 '16 at 11:26
  • $\begingroup$ @Daniel Fischer Why is it not possible to have holomorphic function that blows up at all boundary points? $\endgroup$ – Curious Dec 13 '16 at 1:48
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    $\begingroup$ @Curious See the edited answer. Such a function could have only finitely many zeros, and dividing those out would leave a zero-free holomorphic function that still blows up at all boundary points (since the domain is bounded). Then its reciprocal would violate the maximum modulus principle. $\endgroup$ – Daniel Fischer Dec 13 '16 at 9:10
  • $\begingroup$ @DanielFischer Excellent, thanks! $\endgroup$ – Curious Dec 13 '16 at 23:40

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