Entire function with vanishing derivatives?

Solution 1:

Let $A_n$ be the set of points where $f^{(n)}(z)$ vanishes. Each $A_n$ is closed. By the Baire category theorem one of those sets must have a non empty interior which in turn implies that some derivative vanishes on an open ball. Therefore it is identically zero and $f$ is a polynomial.

Solution 2:

Let $n: [0,1] \to \{0,1,... \}$ be defined such that $n(x)$ is the smallest $k \ge 0$ such that $f^{(k)}(x) = 0$.

Since $[0,1]$ is uncountable, there must be some $k$ such that $Z=n^{-1} \{k\}$ has an infinite number of points. Since $[0,1]$ is compact, $Z$ has a limit point $p \in [0,1]$, and hence we have $f^{(k)}(z) = 0$ everywhere.

Elaboration: Let $p_n \in Z$ such that $p_n \to p$. Since $f^{(k)}(p_n) = 0$, the identify theorem (https://en.wikipedia.org/wiki/Identity_theorem) shows that $f^{(k)}(z) = 0$ everywhere from which it follows that $f$ is a polynomial (take the power series expansion around any point).

Solution 3:

we can also solve this without using baire category theorem..

we can use the fact that zeros of non zero analytic function is countable. Note that union of $A_{n}$ is $\mathbb{C}$ thus we must have some fix derivative identically 0. Now use power series and analytic continuation.