A function is 0 if is almost bounded











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I am having problems with this excercise, if $f$ is analytic in $D subset mathbb{C}$ and:



$$|f(cos(z))| leq m|z|^n$$ for some $m,n in mathbb{N}$ and $z in D$. Then show that $f equiv 0$.



Please some hint could be very useful.






complex-analysis analytic-functions







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  • By Liouvilles theorem $f(cos(z))$ is a polynomial of at most degree $n$. Now use the periodicity of $cos$ to show that $fcirc cos$ has infinitely many roots.
    – Maik Pickl
    2 days ago










  • @MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
    – J.Rodriguez
    2 days ago












  • Hmm, I just saw you’re function is not defined on the whole of the complex plane. Are you sure about that? What is $D$ in your exercise?
    – Maik Pickl
    2 days ago










  • Yes.. $D$ can be any domain.
    – J.Rodriguez
    2 days ago










  • A counterexample is usefull to but I dont have any intuition on this
    – J.Rodriguez
    2 days ago















up vote
0
down vote

favorite












I am having problems with this excercise, if $f$ is analytic in $D subset mathbb{C}$ and:



$$|f(cos(z))| leq m|z|^n$$ for some $m,n in mathbb{N}$ and $z in D$. Then show that $f equiv 0$.



Please some hint could be very useful.






complex-analysis analytic-functions







share|cite|improve this question
























  • By Liouvilles theorem $f(cos(z))$ is a polynomial of at most degree $n$. Now use the periodicity of $cos$ to show that $fcirc cos$ has infinitely many roots.
    – Maik Pickl
    2 days ago










  • @MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
    – J.Rodriguez
    2 days ago












  • Hmm, I just saw you’re function is not defined on the whole of the complex plane. Are you sure about that? What is $D$ in your exercise?
    – Maik Pickl
    2 days ago










  • Yes.. $D$ can be any domain.
    – J.Rodriguez
    2 days ago










  • A counterexample is usefull to but I dont have any intuition on this
    – J.Rodriguez
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am having problems with this excercise, if $f$ is analytic in $D subset mathbb{C}$ and:



$$|f(cos(z))| leq m|z|^n$$ for some $m,n in mathbb{N}$ and $z in D$. Then show that $f equiv 0$.



Please some hint could be very useful.






complex-analysis analytic-functions







share|cite|improve this question















I am having problems with this excercise, if $f$ is analytic in $D subset mathbb{C}$ and:



$$|f(cos(z))| leq m|z|^n$$ for some $m,n in mathbb{N}$ and $z in D$. Then show that $f equiv 0$.



Please some hint could be very useful.






complex-analysis analytic-functions




complex-analysis analytic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









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8,03242358










asked 2 days ago









J.Rodriguez

15310




15310












  • By Liouvilles theorem $f(cos(z))$ is a polynomial of at most degree $n$. Now use the periodicity of $cos$ to show that $fcirc cos$ has infinitely many roots.
    – Maik Pickl
    2 days ago










  • @MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
    – J.Rodriguez
    2 days ago












  • Hmm, I just saw you’re function is not defined on the whole of the complex plane. Are you sure about that? What is $D$ in your exercise?
    – Maik Pickl
    2 days ago










  • Yes.. $D$ can be any domain.
    – J.Rodriguez
    2 days ago










  • A counterexample is usefull to but I dont have any intuition on this
    – J.Rodriguez
    2 days ago


















  • By Liouvilles theorem $f(cos(z))$ is a polynomial of at most degree $n$. Now use the periodicity of $cos$ to show that $fcirc cos$ has infinitely many roots.
    – Maik Pickl
    2 days ago










  • @MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
    – J.Rodriguez
    2 days ago












  • Hmm, I just saw you’re function is not defined on the whole of the complex plane. Are you sure about that? What is $D$ in your exercise?
    – Maik Pickl
    2 days ago










  • Yes.. $D$ can be any domain.
    – J.Rodriguez
    2 days ago










  • A counterexample is usefull to but I dont have any intuition on this
    – J.Rodriguez
    2 days ago
















By Liouvilles theorem $f(cos(z))$ is a polynomial of at most degree $n$. Now use the periodicity of $cos$ to show that $fcirc cos$ has infinitely many roots.
– Maik Pickl
2 days ago




By Liouvilles theorem $f(cos(z))$ is a polynomial of at most degree $n$. Now use the periodicity of $cos$ to show that $fcirc cos$ has infinitely many roots.
– Maik Pickl
2 days ago












@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
2 days ago






@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
2 days ago














Hmm, I just saw you’re function is not defined on the whole of the complex plane. Are you sure about that? What is $D$ in your exercise?
– Maik Pickl
2 days ago




Hmm, I just saw you’re function is not defined on the whole of the complex plane. Are you sure about that? What is $D$ in your exercise?
– Maik Pickl
2 days ago












Yes.. $D$ can be any domain.
– J.Rodriguez
2 days ago




Yes.. $D$ can be any domain.
– J.Rodriguez
2 days ago












A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
2 days ago




A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
2 days ago















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