Maximum modulus principle: Difference between revisions

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(New page: ==Statement== Suppose <math>U \subset \mathbb{C}</math> is a domain (open connected subset). Let <math>f:U \to \mathbb{C}</math> be a holomorphic function. The '''maximum principl...)
 
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* [[Open mapping theorem]]
* [[Open mapping theorem]]
==Proof==
Suppose <math>f</math> is a nonconstant holomorphic function on a nonempty domain <math>U</math>. We'll show that <math>f</math> cannot have a maximum.
First, by the [[open mapping theorem]], <math>f</math> is an open map.
Also, observe that the map <math>| \cdot |: \mathbb{C} \to [0,\infty)</math> is an open map. Thus, the composite map <math>|f|: U \to [0,\infty)</math> given by <math>z \mapsto |f(z)|</math> is also an open map. Thus, under this map, the image of <math>U</math> must be an open connected subset of <math>[0,\infty)</math> so it must be of the form <math>[0,a)</math> where <math>a \in \R</math> or <math>a = \infty</math>. Hence, there cannot be a maximum within <math>U</math>.

Revision as of 23:29, 18 April 2008

Statement

Suppose U⊂C is a domain (open connected subset). Let f:U→C be a holomorphic function. The maximum principle states that if there exists a z0∈U, such that for all z∈U, we have:

|f(z)|≤|f(z0)|

Then, f is a constant function.

Facts used

Proof

Suppose f is a nonconstant holomorphic function on a nonempty domain U. We'll show that f cannot have a maximum.

First, by the open mapping theorem, f is an open map.

Also, observe that the map |⋅|:C→[0,∞) is an open map. Thus, the composite map |f|:U→[0,∞) given by z↦|f(z)| is also an open map. Thus, under this map, the image of U must be an open connected subset of [0,∞) so it must be of the form [0,a) where a∈R or a=∞. Hence, there cannot be a maximum within U.