Suppose $f$ is holomorphic and $|f(\sin z)|<\infty.$ Then can I say $f(\sin z)=\sum_n^\infty a_n(\sin z)^n$ for $a_n\in\mathbb{C}$? Where does $|f(\sin z)|<\infty$ assumption play role here?
Edit: Domain of $f$ is not entire $\mathbb{C}$ plane.
Asked
Active
Viewed 31 times
0
-
By your data, if you meant $;f(\sin z);$ is bounded for all $;z\in\Bbb C;$ , then it is holomorphic and bounded and thus by Liouville's Theorem it is a constant map... – DonAntonio Nov 10 '20 at 21:09
-
@DonAntonio I am considering case when the domain is not the entire $\mathbb{C}$ plane. Thanks for pointing that out. – Nov 10 '20 at 21:11
-
Fine, then $;f(\sin z);$ is analytic as $;\sin z;$ is analytic at every single complex point, and thus a Taylor series exists for that function. – DonAntonio Nov 10 '20 at 21:13
-
@DonAntonio So we don't need to assume the boundedness? – Nov 10 '20 at 21:14
-
If $f$ is continuous, it is bounded on any compact subset of its domain. – Robert Israel Nov 10 '20 at 21:19
-
@ToInfinity Not at all. – DonAntonio Nov 10 '20 at 21:26
1 Answers
1
If $f$ is analytic in a neighbourhood of $0$, it has a Maclaurin series: there is $r > 0$ such that $$ f(z) = \sum_{n=0}^\infty a_n z^n \ \text{for}\ |z| < r $$
That implies $$ f(\sin(x)) = \sum_{n=0}^\infty a_n \sin(x)^n \ \text{whenever}\ |\sin(x)| < r $$
Since $\sin$ is continuous, there is some $\epsilon$ such that this is true when $|x| < \epsilon$.
The condition $|f(\sin(x))| < \infty$ plays no role. Indeed, without specifying for what $x$ this is true, it is essentially meaningless.
Robert Israel
- 448,999