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I have to show by using Liouville's theorem, whether there are non-constant entire functions such that:

$$ |f^k(z)| \leq 1 \forall z \in \mathbb{C}$$ and fixed $k$.

  1. for $k=0$: There is such a function.
  2. for $k\geq1$: There shouldn't be such a function, because $f(z)$ is not necessarily bounded, isn't it?
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    $\begingroup$ Did you mean $f^{(k)}$ instead of $f^k$? $\endgroup$ Commented Dec 10, 2018 at 8:36
  • $\begingroup$ Yes, $f^{(k)}$.Sorry for that $\endgroup$
    – Sven
    Commented Dec 10, 2018 at 9:52

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If $f^{k}$ stands for the $k-$ th power then $|f^{k}(z)| \leq 1$ is same as $|f(z)| \leq 1$ provided $k \neq 0$. Hence $f$ is a constant if $k \neq 0$. If $f^{k}$ stands for the $k-$ th derivative then any polynomial of degree at most $k$ with leading coefficient small enough ; the answer is $f(z)=\sum_{j=0}^{k} c_j z^{j}$ with $|c_k| \leq \frac 1 {k!}$.

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  • $\begingroup$ Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that? $\endgroup$
    – Sven
    Commented Dec 10, 2018 at 9:51
  • $\begingroup$ @SvenMath I have written the complete answer now. $\endgroup$ Commented Dec 10, 2018 at 10:00
  • $\begingroup$ Thank you:) Why is $|c_k| \leq \frac{1}{k!}$ $\endgroup$
    – Sven
    Commented Dec 10, 2018 at 10:13
  • $\begingroup$ When you calculate the k-th derivative you will get just one term, namely $k! c_k$; this must be bounded in absolute value by $1$. $\endgroup$ Commented Dec 10, 2018 at 10:16
  • $\begingroup$ I see:) Can i ask you another example. Is there a entire non-constant function, where $ f( \mathbb{C})$ is in the upper half plane? $\endgroup$
    – Sven
    Commented Dec 10, 2018 at 10:22

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