Maximum modulus principle - Revision history

Vipul at 19:19, 12 September 2008

2008-09-12T19:19:58Z

← Older revision		Revision as of 19:19, 12 September 2008
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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>.

Vipul: 7 revisions

2008-05-18T19:16:26Z

7 revisions

← Older revision	Revision as of 19:16, 18 May 2008
(No difference)

Vipul: /* Proof */

2008-04-26T19:24:24Z

Proof

← Older revision		Revision as of 19:24, 26 April 2008
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	First, by the [[open mapping theorem]], <math>f</math> is an open map.		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>.		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>.

Vipul at 19:20, 26 April 2008

2008-04-26T19:20:22Z

← Older revision		Revision as of 19:20, 26 April 2008
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	{{basic fact}}		{{basic fact}}
			{{application of\|open mapping theorem}}
	==Statement==		==Statement==

Vipul at 19:16, 26 April 2008

2008-04-26T19:16:10Z

← Older revision		Revision as of 19:16, 26 April 2008
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			{{basic fact}}
	==Statement==		==Statement==

Vipul at 13:53, 19 April 2008

2008-04-19T13:53:30Z

← Older revision		Revision as of 13:53, 19 April 2008
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	==Statement==		==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 principle''' states that if there exists a <math>z_0 \in U</math>, such that for all <math>z \in U</math>, we have:		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 modulus principle''' (sometimes called the '''maximum principle''') states that if there exists a <math>z_0 \in U</math>, such that for all <math>z \in U</math>, we have:

	<math>\|f(z)\| \le \|f(z_0)\|</math>		<math>\|f(z)\| \le \|f(z_0)\|</math>

Vipul: Maximum principle for holomorphic functions moved to Maximum modulus principle

2008-04-19T13:52:14Z

Maximum principle for holomorphic functions moved to Maximum modulus principle

← Older revision	Revision as of 13:52, 19 April 2008
(No difference)

Vipul at 23:29, 18 April 2008

2008-04-18T23:29:11Z

← Older revision		Revision as of 23:29, 18 April 2008
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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>.

Vipul: New page: ==Statement== Suppose $U \subset \mathbb{C}$ is a domain (open connected subset). Let $f:U \to \mathbb{C}$ be a holomorphic function. The '''maximum principl...

2008-04-18T23:05:38Z

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...

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 principle''' states that if there exists a <math>z_0 \in U</math>, such that for all <math>z \in U</math>, we have:

<math>|f(z)| \le |f(z_0)|</math>

Then, <math>f</math> is a constant function.

==Facts used==

* [[Open mapping theorem]]