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We have $f(z)=u(x,y)+iv(x,y)$ is analytic and $u(x,y)\leq u_0$ for all point $(x,y)$ in $xy$ plane. We have to show that $u(x,y)$ is constant throughout the plane.

My try: I think Liouville's theorem may be helpful but dont know how to proceed with it.Thank you.

MatheMagic
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  • $e^f$ is bounded entire, so you can apply Liouville – user160738 Mar 07 '17 at 13:19
  • sorry sir..will you please elaborate ? – MatheMagic Mar 07 '17 at 13:20
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    You know that $e^z$ is analytic, and so $e^{f(z)}$ is also analytic. But for any $z$, $|e^z|=e^{\Re z}$, and you are given that $\Re f$ is bounded by $u_0$. So $e^f$ is also bounded, by $e^{u_0}$ (it's of course bounded below by $0$ – user160738 Mar 07 '17 at 13:24
  • @user160738:That means by Liouville's theorem $e^{f}$ is constant but how can we conclude u is constant? – MatheMagic Mar 07 '17 at 13:27
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    Observe that the image is contained in a half-plane. Geometrically what you can do, shift and rotate it to upper half plane and send it to unit disk! hence you have a bounded map form $\mathbb{C}$ to the unit disk. Now apply Liouville's theorem. – user300 Mar 07 '17 at 13:27
  • Actually, I think this question is a duplicate: http://math.stackexchange.com/questions/229312/an-entire-function-whose-real-part-is-bounded-must-be-constant – user160738 Mar 07 '17 at 13:32

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Observe that your real part of $f$ is bounded that is $u(x,y) \leq u_{0}$.

Let us consider $g(x,y) := e^{f(x,y)}$ , observe that $g$ is analytic on whole complex plane as $f$ is analytic in the whole complex plane and also $g$ is bounded by $e^{u_{0}}$ , so applying Liouville's theorem to $g$ , $g$ must be constant thus implying $f$ to be constant.

BAYMAX
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