Maximum modulus principle: Difference between revisions

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==Facts used==
==Facts used==


* [[Open mapping theorem]]
# [[Open mapping theorem]]


==Proof==
==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.
'''Given''': <math>f</math> is a nonconstant holomorphic function on a nonempty domain <math>U</math>.  


First, by the [[open mapping theorem]], <math>f</math> is an open map.
'''To prove''': There does not exist any <math>z_0 \in U</math> with the property that <math>|f(z)| \le |f(z_0)|</math> for every <math>z \in U</math>>


Also, observe that the [[modulus 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>.
'''Proof''': First, by the [[open mapping theorem]] (fact (1)), <math>f</math> is an open map.
 
Second, the [[modulus 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>.

Latest revision as of 19:19, 12 September 2008

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis

This fact is an application of the following pivotal fact/result/idea: open mapping theorem
View other applications of open mapping theorem OR Read a survey article on applying open mapping theorem

Statement

Suppose is a domain (open connected subset). Let be a holomorphic function. The maximum modulus principle (sometimes called the maximum principle) states that if there exists a , such that for all , we have:

Then, is a constant function.

Facts used

  1. Open mapping theorem

Proof

Given: is a nonconstant holomorphic function on a nonempty domain .

To prove: There does not exist any with the property that for every >

Proof: First, by the open mapping theorem (fact (1)), is an open map.

Second, the modulus map is an open map. Thus, the composite map given by is also an open map. Thus, under this map, the image of must be an open connected subset of , so it must be of the form where or . Hence, there cannot be a maximum within .