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Let $f$ be an entire function, which takes real values on the real axis and has no zeros. Suppose $f$ is bounded for $|\operatorname{Im} z| > a > 0$ where $a>0$. Is $f$ a constant?


I would be able to conclude this with Liouville's theorem if I knew that $\operatorname{Im}z$ is bounded in the strip $|\operatorname{Im} z|\le a$, but I don't see how to prove this.

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  • $\begingroup$ Show that $f$ is bounded on the strip $|\text{Im} z| \le a$ too, so it's bounded everywhere, then apply Liouville's theorem. $\endgroup$ – Alex M. Dec 12 '15 at 20:47
  • $\begingroup$ Thanks Alex M. for your answer. The problem is precisely that we can not show that the function is bounded for |Im z|<a. $\endgroup$ – delrio Dec 14 '15 at 0:33
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    $\begingroup$ entire functions are constructed by : starting from polynomials, constructing new functions by addition and multiplication and also converging infinite sums and products (which allows $f \to e^f$) thus if you can prove the theorem for polynomials and by induction following each of these construction rules, you'll get your theorem. another way would be to find a trick (like applying the the maximum or minimum modulus to $\cos(f)$) directly showing that the function has to be constant, but it will be a less general approach $\endgroup$ – reuns Feb 10 '16 at 11:04
  • $\begingroup$ i am sure this is a duplicate q. $\endgroup$ – DanielWainfleet Feb 10 '16 at 13:23
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    $\begingroup$ Possible duplicate of Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 >1$? $\endgroup$ – user147263 Feb 13 '16 at 0:28
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I found here and in that book p.153-155 that this would be a counter-example :

$$f(z) = \int_0^\infty \frac{e^{z t}}{t^t}dt$$

which is an entire function bounded by $ \frac{1}{|c|}$ on any ray starting from the origin to infinity with an angle $\pm e^{i (\pi/2 \pm c)}$. Thus it is unbounded only on a small strip around the imaginary axis.

(I'm not 100% convinced of his proof yet)

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    $\begingroup$ The present question requires also that the function have no zeros. But $\exp f(z)$ seems to do the trick. $\endgroup$ – Noam D. Elkies Feb 11 '16 at 5:09

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