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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
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Any non constant analytic function does not attains it maxima on interior point, if attain then it must be on the boundary. –  El Angel Exterminador Mar 2 '13 at 20:45
    
Related. –  Cameron Buie Oct 24 '13 at 16:51

1 Answer 1

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 $\Omega\to \mathbb{C}$.

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but we shall assume $f$ should be holomorphic, counter examples on non-analytic functions does not ensure –  El Angel Exterminador Mar 3 '13 at 9:55
    
What I wanted to say is not that the open mapping theorem is false. Indeed it is true. I wanted to say that the property of maximum modulus alone does not imply the property of being an open mapping. –  Giuseppe Negro Mar 3 '13 at 11:32
    
Note that the converse implication is true: an open mapping (analytical or not) automatically satisfies the maximum modulus principle (see here). –  Giuseppe Negro Mar 3 '13 at 11:32
    
The conclusion is that, in principle, you can prove the open mapping theorem from the maximum modulus principle, of course. But this is not a natural thing to do. The contrapositive, that is, proving the maximum modulus principle from the open mapping theorem, is instead perfectly natural. –  Giuseppe Negro Mar 3 '13 at 11:36

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