Liouville's theorem for non constant functions












1












$begingroup$


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 $kgeq1$: There shouldn't be such a function, because $f(z)$ is not necessarily bounded, isn't it?






entire-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $f^{(k)}$ instead of $f^k$?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 8:36










  • $begingroup$
    Yes, $f^{(k)}$.Sorry for that
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:52


















1












$begingroup$


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 $kgeq1$: There shouldn't be such a function, because $f(z)$ is not necessarily bounded, isn't it?






entire-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $f^{(k)}$ instead of $f^k$?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 8:36










  • $begingroup$
    Yes, $f^{(k)}$.Sorry for that
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:52
















1












1








1





$begingroup$


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 $kgeq1$: There shouldn't be such a function, because $f(z)$ is not necessarily bounded, isn't it?






entire-functions







share|cite|improve this question











$endgroup$




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 $kgeq1$: There shouldn't be such a function, because $f(z)$ is not necessarily bounded, isn't it?






entire-functions




entire-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 9:16









mathreadler

14.9k72262




14.9k72262










asked Dec 10 '18 at 8:33









SvenMathSvenMath

357




357








  • 1




    $begingroup$
    Did you mean $f^{(k)}$ instead of $f^k$?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 8:36










  • $begingroup$
    Yes, $f^{(k)}$.Sorry for that
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:52
















  • 1




    $begingroup$
    Did you mean $f^{(k)}$ instead of $f^k$?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 8:36










  • $begingroup$
    Yes, $f^{(k)}$.Sorry for that
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:52










1




1




$begingroup$
Did you mean $f^{(k)}$ instead of $f^k$?
$endgroup$
– José Carlos Santos
Dec 10 '18 at 8:36




$begingroup$
Did you mean $f^{(k)}$ instead of $f^k$?
$endgroup$
– José Carlos Santos
Dec 10 '18 at 8:36












$begingroup$
Yes, $f^{(k)}$.Sorry for that
$endgroup$
– SvenMath
Dec 10 '18 at 9:52






$begingroup$
Yes, $f^{(k)}$.Sorry for that
$endgroup$
– SvenMath
Dec 10 '18 at 9:52












1 Answer
1






active

oldest

votes


















1












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:51










  • $begingroup$
    @SvenMath I have written the complete answer now.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:00










  • $begingroup$
    Thank you:) Why is $|c_k| leq frac{1}{k!}$
    $endgroup$
    – SvenMath
    Dec 10 '18 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 $1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 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?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 10:22













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:51










  • $begingroup$
    @SvenMath I have written the complete answer now.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:00










  • $begingroup$
    Thank you:) Why is $|c_k| leq frac{1}{k!}$
    $endgroup$
    – SvenMath
    Dec 10 '18 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 $1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 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?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 10:22


















1












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:51










  • $begingroup$
    @SvenMath I have written the complete answer now.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:00










  • $begingroup$
    Thank you:) Why is $|c_k| leq frac{1}{k!}$
    $endgroup$
    – SvenMath
    Dec 10 '18 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 $1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 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?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 10:22
















1












1








1





$begingroup$

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






share|cite|improve this answer











$endgroup$



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 10:00

























answered Dec 10 '18 at 8:39









Kavi Rama MurthyKavi Rama Murthy

60.5k42161




60.5k42161












  • $begingroup$
    Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:51










  • $begingroup$
    @SvenMath I have written the complete answer now.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:00










  • $begingroup$
    Thank you:) Why is $|c_k| leq frac{1}{k!}$
    $endgroup$
    – SvenMath
    Dec 10 '18 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 $1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 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?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 10:22




















  • $begingroup$
    Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 9:51










  • $begingroup$
    @SvenMath I have written the complete answer now.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:00










  • $begingroup$
    Thank you:) Why is $|c_k| leq frac{1}{k!}$
    $endgroup$
    – SvenMath
    Dec 10 '18 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 $1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 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?
    $endgroup$
    – SvenMath
    Dec 10 '18 at 10:22


















$begingroup$
Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
$endgroup$
– SvenMath
Dec 10 '18 at 9:51




$begingroup$
Sorry. It stands for the k-th derivative. I have to specify all Funktions. How can I do that?
$endgroup$
– SvenMath
Dec 10 '18 at 9:51












$begingroup$
@SvenMath I have written the complete answer now.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 10:00




$begingroup$
@SvenMath I have written the complete answer now.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 10:00












$begingroup$
Thank you:) Why is $|c_k| leq frac{1}{k!}$
$endgroup$
– SvenMath
Dec 10 '18 at 10:13




$begingroup$
Thank you:) Why is $|c_k| leq frac{1}{k!}$
$endgroup$
– SvenMath
Dec 10 '18 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 $1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 10:16




$begingroup$
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 $1$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 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?
$endgroup$
– SvenMath
Dec 10 '18 at 10:22






$begingroup$
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?
$endgroup$
– SvenMath
Dec 10 '18 at 10:22




















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