Analytic functions which are real on an infinite bounded subset of $\mathbb{R}$

Question

Let $f : \Bbb{C}\longrightarrow \Bbb{C}$ be an analytic function which is too real-valued ! To be more exact, For an infinite subset $A$ of $[0,1]$ we have $f(A)\subseteq \Bbb{R}$. Does it mean that for any $x\in \Bbb{R}$, $f(x)$ is real ?

Answer 1

Yes. Since $A$ is infinite in the compact space $[0,1]$, there is an accumulation point $c$ of $A$. I claim that the Taylor series coefficients about $c$ are real. If true, then (since $f$ is entire, its Taylor series about any point has infinite radius of convergence), $f$ is certainly real-valued on all of $\mathbb{R}$.

$n = 0$. Since $\mathbb{R}$ is closed in $\mathbb{C}$, $f$ is continuous, $f(A) \subset \mathbb{R}$, and $c \in \overline{A}$, we have $f(c) \in \overline{\mathbb{R}} = \mathbb{R}$.

$n = 1$: Since one knows the derivative exists, one can evaluate $f'(c) = \lim_{z \rightarrow c} \frac{f(z)-f(c)}{z-c}$ using any sequence $\{z_n\}$ which converges to $c$. In particular one can take $z_n \in A$ for all $n$, and this shows that $f'(c) \in \mathbb{R}$.

$n =2$: Put $c_0 = f(c)$. We already know that the linear approximation $f_1(z) = c_0 + c_1 (z-c)$ has real coefficients. By looking at a power series expansion, one sees that

$f''(c) = 2! \lim_{z \rightarrow c} \frac{f(z) - f_1(z)}{(z-c)^2}$.

Since $f_1(z)$ is a polynomial with real coefficients, it is real on all of $\mathbb{R}$, so once again evaluating the limit via a sequence with values in $A$ shows that it is real.

$n \geq 3$: Similarly, by induction we know that the $n-1$st Taylor polynomial has real coefficients. Subtract it off and compute

$f^{(n)}(c) = n! \lim_{z \rightarrow c} \frac{f(z) - f^{(n-1)}(z)}{(z-c)^{n}}$

along a sequence in $A$.