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
edited Nov 19 at 15:25
Empty
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asked Nov 18 at 18:55
J.Rodriguez
15310
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show 2 more comments
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
edited Nov 19 at 15:25
Empty
8,05742358
asked Nov 18 at 18:55
J.Rodriguez
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
Nov 18 at 19:10
@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
Nov 18 at 19:18
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
Nov 18 at 19:21
Yes.. $D$ can be any domain.
– J.Rodriguez
Nov 18 at 19:23
A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
Nov 18 at 19:25
|
show 2 more comments
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
edited Nov 19 at 15:25
Empty
8,05742358
asked Nov 18 at 18:55
J.Rodriguez
15310
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
edited Nov 19 at 15:25
Empty
8,05742358
asked Nov 18 at 18:55
J.Rodriguez
15310
edited Nov 19 at 15:25
Empty
8,05742358
asked Nov 18 at 18:55
J.Rodriguez
15310
edited Nov 19 at 15:25
Empty
8,05742358
edited Nov 19 at 15:25
Empty
8,05742358
edited Nov 19 at 15:25
Empty
8,05742358
8,05742358
asked Nov 18 at 18:55
J.Rodriguez
15310
asked Nov 18 at 18:55
J.Rodriguez
15310
asked Nov 18 at 18:55
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
Nov 18 at 19:10
@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
Nov 18 at 19:18
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
Nov 18 at 19:21
Yes.. $D$ can be any domain.
– J.Rodriguez
Nov 18 at 19:23
A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
Nov 18 at 19:25
|
show 2 more comments
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
Nov 18 at 19:10
@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
Nov 18 at 19:18
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
Nov 18 at 19:21
Yes.. $D$ can be any domain.
– J.Rodriguez
Nov 18 at 19:23
A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
Nov 18 at 19:25
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
Nov 18 at 19:10
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
Nov 18 at 19:10
@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
Nov 18 at 19:18
@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
Nov 18 at 19:18
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
Nov 18 at 19:21
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
Nov 18 at 19:21
Yes.. $D$ can be any domain.
– J.Rodriguez
Nov 18 at 19:23
Yes.. $D$ can be any domain.
– J.Rodriguez
Nov 18 at 19:23
A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
Nov 18 at 19:25
A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
Nov 18 at 19:25
|
show 2 more comments
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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
Nov 18 at 19:10
@MaikPickl but Liouville dont need that f is analytic in the entire complex plane?
– J.Rodriguez
Nov 18 at 19:18
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
Nov 18 at 19:21
Yes.. $D$ can be any domain.
– J.Rodriguez
Nov 18 at 19:23
A counterexample is usefull to but I dont have any intuition on this
– J.Rodriguez
Nov 18 at 19:25