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Suppose that $f$ is an entire function and that there is a bounded sequence of real numbers $a_1, a_2, ... $ such that $f(a_n)$ is real for all $n$. Show that $f(x)$ is real for all real $x$.

Thoughts so far: Since $a_n$ is a bounded sequence of real numbers, we now that the set ${a_n}$ has an accumulation point. Now, since we're given that the function is entire, my first thought was to apply Liouville's Theorem somehow, but I can't get it to work out. A hint would be much appreciated.

Context: I'm studying for a qual, so just a hint at this point would be most helpful.

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  • $\begingroup$ I think you can use the Cauchy Riemann relation to prove that the imaginary part of the function has to be $0$. (Cauchy Riemann equation is equivalent to Laplace's equation) $\endgroup$ – mastrok Aug 25 '15 at 3:51
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Hint. $g(z) = \overline{f(\bar{z})}$ is also an entire function such that $f(a_n) = g(a_n)$ for all $n$. What can you conclude from the identity theorem?

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  • $\begingroup$ Awesome, so we can conclude that $g = f$, and thus $f(x)$ is real for all real $x$, since this is clearly true for $g$. My questions is, is it obvious that $g$ is entire? Or is this a well-known fact? $\endgroup$ – user19817 Aug 25 '15 at 4:59
  • $\begingroup$ @m.deslauriers There are many ways to see that. You may look at the Cauchy-Riemann equation. Or, you can expand $f$ as power series and check that $g$ has the McLaurin series with coefficients conjugated. $\endgroup$ – Sangchul Lee Aug 25 '15 at 7:09
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Hint:Since a real and bounded sequence has a limit point.Let us call it 'a'.It is easy to conclude that f(a+∆)=0 for small ∆.By Identity theorem we have f(x)=0 identically.

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