If $f$ holomorphic such that $|f^{(n)}|\leq M$, then $f$ is a polynomial (verification)

Asked May 14 '21 at 22:32

Active May 14 '21 at 23:52

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I know a similar question was asked here, but I just need a clarification of the proof, I have no problem deleting my question after I get it.

If $f$ holomorphic over $\mathbb{C}$ such that $|f^{(n)}(z)|\leq M$, then $f$ is a polynomial $\leq n$ degree.

Letting $g(z)=f^{(n)}(z)$ from liouville theorem we get $f^{(n)}(z)=c$, $c \in \mathbb{C}$

My question: is $f$ a polynomial by definition since the $n$-th derivative is a constant ? and the proof is complete or do I have to prove something more ?

asked May 14 '21 at 22:32

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it's not by definition. You have to prove that constant $n^{th}$ derivative implies it is a polynomial (but the proof isn't too hard). You have to prove the existence of $a_0,\dots, a_n\in\Bbb{C}$ such that for all $z\in\Bbb{C}$, $f(z)=\sum_{k=0}^na_kz^k$. – peek-a-boo May 14 '21 at 22:39
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Use induction on $n$. – Kavi Rama Murthy May 14 '21 at 23:11
No, it's not by definition. See this answer. – vitamin d May 14 '21 at 23:14

If $f$ holomorphic such that $|f^{(n)}|\leq M$, then $f$ is a polynomial (verification)

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