Maximum modulus principle: Difference between revisions
(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 is a domain (open connected subset). Let be a holomorphic function. The maximum principle states that if there exists a , such that for all , we have:
Then, is a constant function.
Facts used
Proof
Suppose is a nonconstant holomorphic function on a nonempty domain . We'll show that cannot have a maximum.
First, by the open mapping theorem, is an open map.
Also, observe that the 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 .