Show that there exists $N \in \mathbb{N}$ such that for every $n \in \mathbb{N}$, one has $a_n = 0$, in other words show $f$ is a polynomial function.

Question

Let $\sum_{n=0}^{\infty}a_nz^n \in \mathbb{C}[[T]]$ be a power series over $\mathbb{C}$ with radius of convergence $R = \infty$, let $f: \mathbb{C} \to \mathbb{C}$ be the entire function given by $f(z) = \sum_{n=0}^{\infty}a_nz^n$. For every $z_0 \in \mathbb{C}$, let $$f(z) = \sum_{k=0}^\infty b_{k,z_0}(z-z_0)^k~~~\text{for } z \in \mathbb{C}$$ be the power series expansion of $f$ about $z_0$. Suppose for every $z_0 \in \mathbb{C}$, there exists $k \in \mathbb{N}$ (depending on $z_0$) such that $b_{k,z_0}=0$. Show that there exists $N \in \mathbb{N}$ such that for every $n \in \mathbb{N}$, one has $a_n = 0$, in other words show $f$ is a polynomial function.

I was trying to prove by contradiction that $f$ is not polynomial then $f(1/z)$ has essential singularity at $0$. But really have no idea how to continue... Any one has any idea.

score 2 · Accepted Answer · answered Aug 31 '18 at 10:37

2

For some $k$ $\{z:f^{(k)}(z)=0$ is uncountable (because $\mathbb C$ is uncountable ). This implies that this set has limit point, so $f^{(k)}\equiv 0$ and $f$ is a polynomial.

answered Aug 31 '18 at 10:37

Kavi Rama Murthy

311,013

Will be happy to provide more details if requested. – Kavi Rama Murthy Aug 31 '18 at 10:39
It is enough to consider all $z$ in the closed unit circle to get a finite limit point. – amsmath Aug 31 '18 at 10:43
... or a closed interval, as in https://math.stackexchange.com/a/995129/42969. – Martin R Aug 31 '18 at 10:44
I mean why for some $k$ the aforementioned set is uncountable. Why can we always find such a $k$. Thanks – Shweta Aggrawal Jun 11 '19 at 12:10
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Because you cannot write $\mathbb C$ as a countable union of countable sets. – Kavi Rama Murthy Jun 11 '19 at 12:14

Show that there exists $N \in \mathbb{N}$ such that for every $n \in \mathbb{N}$, one has $a_n = 0$, in other words show $f$ is a polynomial function.

1 Answers1