Power series of $f: \mathbb{C} \to \mathbb{C}$ with $f(\mathbb{R}) \subset \mathbb{R}$

Question

Let f be an entire function that satisfies the property $f(\mathbb{R}) \subset \mathbb{R}$.

Consider the power series of f: $$f(z) = \sum_{k=0}^\infty a_k z^k \ .$$ The question I'm asking myself is: Do all $a_k$ have to be real?

It seems like this should be true, because if at least one coefficient was not real, then one ought to be able to find some $x$ for which $f(x)$ is not real. However, I'm having difficulties proving this formally. Here is my attempt:

Let $x \in \mathbb{R}$. Then $$f(x) = \sum_{k=0}^\infty a_k x^k \ .$$ Therefore $$\Im(f(x)) = \Im(\sum_{k=0}^\infty a_k x^k) = \sum_{k=0}^\infty \Im(a_k) x^k = 0 \ .$$ Let $$g(z) = \sum_{k=0}^\infty \Im(a_k) z^k$$ which has a positive radius of convergence because f is analytic. Then $$g(\mathbb{R}) = 0$$ and since $\mathbb{R}$ has a limit point in $\mathbb{C}$, it follows by the identity theorem that $$g(z) = 0 \ \ \forall z \in \mathbb{C}$$ and therefore $$\Im(a_k) = 0 \ \ \forall k \ .$$

The points where I'm not quite sure is moving the $\Im( \dots )$ into the sum and applying the identity theorem by constructing $g$.

I appreciate any feedback on whether the proof holds (or perhaps whether or not the statement is true at all).

Answer 1

The statement is correct and your proof is also correct.

The function $v: \mathbb C \to \mathbb C$, defined by $v(z)=\Im (z)$, is continuous, hence, if $z_n \to z_0$, then $v(z_n) \to v(z_0)$. As a consequence we have: if $ \sum_{n=0}^{\infty}c_n$ is a convergent series, then

$$\Im(\sum_{n=0}^{\infty}c_n)=v(\sum_{n=0}^{\infty}c_n)=\sum_{n=0}^{\infty}v(c_n)=\sum_{n=0}^{\infty}\Im(c_n).$$

Answer 2

if an entire function is real on the real line, then all its derivatives at $z=0$ must be real (since they may be computed in $\mathbb{R}$). But:

$$ a_n = \frac1{n!}f^{(n)}(0) $$