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 UC is a domain (open connected subset). Let f:UC be a holomorphic function. The maximum modulus principle (sometimes called the maximum principle) states that if there exists a z0U, such that for all zU, we have:

|f(z)||f(z0)|

Then, f is a constant function.

Facts used

  1. Open mapping theorem

Proof

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

To prove: There does not exist any z0U with the property that |f(z)||f(z0)| for every zU>

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

Second, the modulus 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 aR or a=. Hence, there cannot be a maximum within U.