# Liouville's theorem for non constant functions

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?
• Did you mean $f^{(k)}$ instead of $f^k$? Commented Dec 10, 2018 at 8:36
• Yes, $f^{(k)}$.Sorry for that
– Sven
Commented Dec 10, 2018 at 9:52

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 $$|c_k| \leq \frac 1 {k!}$$.

• Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
– Sven
Commented Dec 10, 2018 at 9:51
• @SvenMath I have written the complete answer now. Commented Dec 10, 2018 at 10:00
• Thank you:) Why is $|c_k| \leq \frac{1}{k!}$
– Sven
Commented Dec 10, 2018 at 10:13
• 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 $1$. Commented Dec 10, 2018 at 10:16
• 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?
– Sven
Commented Dec 10, 2018 at 10:22