I wonder how we prove the statement in the title.
Actually, this is needed for solving the following problem.
Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be an entire function.
Suppose that for every $z_0 \in \mathbb{C}$ the power series expansion
$\sum_{k=0}^{\infty }a_k(z-z_0)^k$ of $f$ around $z_0$ satisfies that
there is $k \in \mathbb{N}$ s.t $a_k = 0$. Show that $f$ is a
polynomial.
I could show that $f$ can be written in a polynomial in each ball using the Identity Theorem, with the fact "an uncountable closed set contains an accumulation point."
However, I'm stuck there, and cannot proceed more from here to show that $f$ is indeed a polynomial on $\mathbb{C}$.
It seems like the Identity Theorem could be reused inductively here, but the problem occurs when setting a domain of two polynomials of two connected balls. (What I tried to show was any two connected balls eventually have the same polynomial expression of $f$ so that the statement could be automatically true.)
How should I show that the statement is true with the Identity Theorem, or is there an alternative way to do this?
Update with my work:
Let $P_0$ be the polynomial expression of $f$ on $B(z_0,r)$.
Assume that $P_0$ is entire (In other words, assume that it is defined on whole $\mathbb{C}$.)
Now, let $Z=\left \{ z\in \mathbb{C}:f(z)=P_0(z) \right \}$.
Observe that, $z_0 \in Z$, and in $Z \setminus \{z_0\}$, there is a sequence $\left \{ z_n \right \}_{n=2}^{\infty } $ s.t $z_n = z_0 + r/n$.
This shows that, $z_0$ is an accumulation point in $Z$, hence by the identity Thm, $f(z)=P_0(z)$ for any $z \in \mathbb{C}$.
Best Answer
If the polynomial has degree $n$ then $f^{(n+1)}$ is an entire function which vanishes on the given ball. By the identity theorem it vanishes everywhere. Hence, $f$ is a polynomial of degree at most $n$.