Show that there are no entire functions such that $\bigcup_{n = 0}^{\infty} \{ z \in \mathbb{C} : f^{(n)}(z) = 0 \} = \mathbb{R}$.
My attempt: So I tried this by contradiction. Suppose there is an entire function $g(z)$ that satisfied this claim. WLOG, suppose that $g(z)$ is non constant. Then, I picked $x \in \mathbb{R}$ and the least $j \in \mathbb{N}$ so that
$$ \big( \frac{\partial}{\partial z} \big)^{j} f(x) \neq 0. $$
Since $g$ is entire, I defined a taylor series expansion like this:
$$g(z) = \sum_{k = j}^{\infty} \frac{\big( \frac{\partial}{\partial z} \big)^{k} f(x)(z -x)^k}{k!}$$
I want to try to get that $g(z) = 0$ which would be my contradiction. I am not sure I am heading toward the right direction.
Can you please give some feedback if I did it right or not? Please give me hints only.
Thank you so much, guys!!
Best Answer
Contradiction: take $f(z)\equiv 0$ Then $f(z)$ is an entire function satisfying the given condition.
However, there would be no such non constant entire function. If there is, then by identity theorem, we get a contradiction.