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Today during the qualifying exam I met this question:

Show that the maximum modulus principle implies the Liouville Theorem.

Well, this is my attempt:

It suffices to show that a bounded entire function can achieve its maximum modulus in complex plane. But I got messed up here. Can anyone give me some ideas?

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  • $\begingroup$ See also Second proof of proposition 6.3, pg. 58 of unc.edu/math/Faculty/met/complex.pdf $\endgroup$ – Jacob Bond Aug 12 '14 at 21:46
  • $\begingroup$ @Jacob Bond, Thank you very much! Can't believe I didn't think in that way... And, this proof also indicates that one cannot merely apply M.M.P to prove Liouville's Theorem. Riemann removable singularity theorem has to be used. $\endgroup$ – user167839 Aug 12 '14 at 21:51
  • $\begingroup$ @Jacob Bond, I just realized it's you.. I'm Qinfeng. How did you feel about the exam yesterday? $\endgroup$ – user167839 Aug 12 '14 at 22:02
  • $\begingroup$ @Jacob Bond, me too, I was hurt by the 1st problem, but felt good about the others. Well, good luck! $\endgroup$ – user167839 Aug 12 '14 at 22:06
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Show that the maximum modulus principle implies the Liouville Theorem.

The function

$$f(z) = \frac{z}{1+\lvert z\rvert}$$

which is a homeomorphism between $\mathbb{C}$ and the unit disk shows that the maximum modulus principle alone does not imply Liouville's theorem. We need some more properties of holomorphic functions.

It suffices to show that a bounded entire function can achieve its maximum modulus in complex plane. But I got messed up here. Can anyone give me some ideas?

Something closely related: Riemann's removable singularity theorem.

The Riemann sphere is compact, so every continuous function on the entire sphere ...

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  • $\begingroup$ Thanks Daniel. Actually I thought about the Riemann sphere during the exam, and I also noticed that $\infty$ is a removable singularity. But what could be implied here? I'm sorry I didn't get it. $\endgroup$ – user167839 Aug 11 '14 at 18:59
  • $\begingroup$ You get a holomorphic function defined on the entire sphere. In particular, that function is continuous. $\endgroup$ – Daniel Fischer Aug 11 '14 at 19:00
  • $\begingroup$ yeah, I noticed that. I mean, a continuous function $f$ on a compact manifold can achieve its maximum modulus on the boundary. Riemann sphere has no boundary, and maximum modulus can be obtained on $\infty$ if no extra result is considered. So how to derive a contradiction? $\endgroup$ – user167839 Aug 11 '14 at 19:03
  • $\begingroup$ @DanielFischer Thank you for pointing out my mistake! Apologies for misleading the OP. $\endgroup$ – angryavian Aug 11 '14 at 19:04
  • $\begingroup$ @user167839 $\lvert f\rvert$ must attain its maximum at some point on the sphere. If it attains the maximum at $\infty$ (and you don't know yet that it is constant), consider the function $g(z) = f(1/z)$. $\endgroup$ – Daniel Fischer Aug 11 '14 at 19:06

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