# Can we prove the open mapping theorem using the maximum modulus principle?

Can we prove the open mapping theorem using Maximum Modulus Principle? I myself can prove the other way.

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Can you state the Maximum Modulus Principle for us, please? – Giuseppe Negro Mar 2 '13 at 18:46
Any non constant analytic function does not attains it maxima on interior point, if attain then it must be on the boundary. – Un Chien Andalou Mar 2 '13 at 20:45
– Cameron Buie Oct 24 '13 at 16:51
What do you mean by "prove ... using ..."? This can mean several different things. The answer below provides one interpretation. Another is to prove OMT using MMP and some other subset of knowledge that is generally considered more basic. So for instance, maybe the question means "show that OMT and MMP are logically equivalent with respect to ZFC". Or "show that OMT and MMP are logically equivalent with respect to ZFC + Cauchy's Theorem + Cauchy's Integral Theorem + elementary consequences of Cauchy's Theorem". The question needs clarification. – cantorhead May 13 at 13:09

If you plan to prove that $$\tag{1}(f\colon \Omega \to \mathbb{C}\ \text{satisfies the maximum modulus principle})\Rightarrow (f\ \text{is open}),$$ then you cannot because (1) is false.
Indeed there exist non-analytic functions which satisfy the maximum modulus principle, such as $$f(z)=\frac{1}{2\pi}\log \lvert z \rvert,\qquad z \in \Omega=\{\lvert z \rvert > 1\}.$$ This function $f$ coincides with its module and so it clearly cannot be open if we regard it as a function $\Omega\to \mathbb{C}$.
but we shall assume $f$ should be holomorphic, counter examples on non-analytic functions does not ensure – Un Chien Andalou Mar 3 '13 at 9:55