Show that $f^{n}(0)=0$ for infinitely many $n\ge 0$.

Question

Let $f$ be an entire function. Suppose that for each $a\in \Bbb R$ there exists at least one coefficient $c_n$ in $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$ which is zero.

Then:

1. $f^{n}(0)=0$ for infinitely many $n\ge 0$.
2. $f^{2n}(0)=0$ for every $n\ge 0$.
3. $f^{2n+1}(0)=0$ for every $n\ge 0$.
4. $\exists k\ge 0$ such that $f^{n}(0)=0$ for all $n\ge k$.

My try:

Since $a\in \Bbb R$ is uncountable and $c_n$ is countable so there exists $b\in \Bbb R$ such that $c_n=0$ for infinitely many $n$ where $c_n=\dfrac{f^{n}(b)}{n!}\implies f^{n}(b)=0$.

So I feel that the correct options should be $1,4$.

But how can I show that $f^{n}(0)=0$ for infinitely many $n\ge 0$.

I only have $f^{n}(b)=0$ for infinitely many $n\ge 0$.But the question demands $f^{n}(0)=0$ for infinitely many $n\ge 0$. Please give some hints.

Answer 1

For each $n\geq 0$, let $E_n$ be the set of those $a\in\mathbb{C}$ for which $c_n=0$ in the power series expansion of $f$ about $a$. Then since $\mathbb{R}\subset \cup_{n=0}^{\infty}E_n$, at least one of the $E_n$ must be uncountable.

So choose some $n$ such that $E_n$ is uncountable, and write $$E_n=\bigcup_{j=1}^{\infty}E_n\cap \{|z|\leq j\}$$ Since $E_n$ is uncountable, at least one of the sets $E_n\cap\{|z|\leq j\}$ must be infinite, say $E_n\cap\{|z|\leq j_0\}$. But since $\{|z|\leq j_0\}$ is compact, it follows that $E_n\cap\{|z|\leq j_0\}$ has a limit point in $\{|z|\leq j_0\}$.

Therefore $f^{(n)}(z)=0$ on a set with a limit point, hence is identically zero. Hence $f^{(m)}=0$ for all $m\geq n$. This proves (4), and therefore proves (1) as well.

By looking at polynomials, you can show that (2) and (3) are not necessarily true.

Answer 2

For each $n$ define the open set $U_n=\{ a\in {\Bbb R} : f^{(n)}(a) \neq 0\}$. If $U_n$ is not dense then $f^{(n)}$ vanishes on some interval and we are in the polynomial case (case 4). So assume $U_n$ dense. Then by Baire the countable intersection $\bigcap_n U_n$ is dense in ${\Bbb R}$ which is not possible. So we are indeed in case 4.