Liouville's theorem for non constant functions

Question

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 $k$.

for $k=0$: There is such a function.
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 $f^k$? – José Carlos Santos Dec 10 '18 at 08:36 — José Carlos Santos, Dec 10 '18 at 08:36

Kavi Rama Murthy · Accepted Answer · 2018-12-10T10:00:02.797

1

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

edited Dec 10 '18 at 10:00

answered Dec 10 '18 at 08:39

Kavi Rama Murthy

311,013

Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that? – Sven Dec 10 '18 at 09:51
@SvenMath I have written the complete answer now. – Kavi Rama Murthy Dec 10 '18 at 10:00
Thank you:) Why is $|c_k| \leq \frac{1}{k!}$ – Sven Dec 10 '18 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 $1$. – Kavi Rama Murthy Dec 10 '18 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 Dec 10 '18 at 10:22
@SvenMath Picards theorem says if $f$ is a non-constant entire function then the range is either $\mathbb C$ or $\mathbb C \setminus {a}$ for some complex number $a$. – Kavi Rama Murthy Dec 10 '18 at 10:29
We didnt have that. I have to argue with Liovilles theorem? – Sven Dec 10 '18 at 10:56
@SvenMath $e^{if(z)}$ is a bounded entire function, hence constant. From this it follows that $f$ is also a constant. – Kavi Rama Murthy Dec 10 '18 at 11:54
Sorry for the late answer. How do get to $e^{if(z)}$ – Sven Dec 10 '18 at 14:15
Can you answer this, please? – Sven Dec 11 '18 at 15:20
@SvenMath $|e^{if(z)}|=e^{ Re, if(z)}=e^{- Im , f(z)}<1$. It is not a good idea to ask new questions in comments. Please post a new question. – Kavi Rama Murthy Dec 11 '18 at 23:14
Thank you for all your help:) I will do that:) – Sven Dec 12 '18 at 07:37

Liouville's theorem for non constant functions

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