Complex Analysics/Maximum Principle

Introduction

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The maximum principle is a statement about holomorphic functions from the Complex Analysis. The magnitude   of a holomorphic function   cannot attain any strict local maxima within the domain of definition. Specifically, it asserts the following statement.

Statement

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Let   be a domain, and let   be holomorphic. If   has a local maximum in  , then   is constant. If   is bounded and   can be continuously extended to  , then   attains its maximum on  .

To prove this, we require a lemma that locally implies the conclusion.

Lemma

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Let   be open, and   be holomorphic. Let   be a local maximum point of  . Then   is constant in a neighborhood of  .

Proof of Lemma 1

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Let   be chosen such that   for all  . The Cauchy's integral formula'gives, for all  :

 

This allows us to establish the following estimation:

Proof of Lemma 2

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We derive the following estimation:

 

Proof of Lemma 3

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It follows that the inequality   must be an equality chain, implying

 .

Proof of Lemma 4

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Thus, we establish the constancy of   using the property:

  for all  ,

i.e.,   is constant on  .

Proof of Lemma 5

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If   is constant on  , then   must also be constant, where   is a constant.

Proof of Lemma 6

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Since   is holomorphic on  , the Cauchy-Riemann-Differential equation'apply:

 ,

and the following holds:

 .

Proof of Lemma 7

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Let   and  . Applying the chain rule to the partial derivatives, we obtain:

  and  .

Using the Cauchy-Riemann-Differential equation', replace the partial derivatives of   with those of  :

  and  , leading to:
  and  .

Proof of Lemma 8

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Squaring the above equations yields:

 ,
 .

Adding these equations gives:

 .

Proof of Lemma 9

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Factoring out   and  :

 .

Thus,

  or  .

Proof of Lemma 10

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With  , it follows that   since   and   are real-valued, implying  .

If  , then  , and  . By the Cauchy-Riemann-Differential equation,  . Thus,   is constant on  .

Proof

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Let   be a local maximum point of   in the domain  . Define   as the set of all   mapped to   (level set).

Proof 1: V is closed

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Since   is continuous, preimages of open sets are open, and preimages of closed sets are closed (in the relative topology of  ). Thus,   is closed in  .

Proof 2: V is open

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Using the lemma,   can also be represented as a union of open disks, and unions of open sets are open.

Proof 3: Connectivity

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Thus,   due to the connectivity of  , i.e.,   is constant.

Proof 4: G is bounded

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If   is bounded, then   is compact. Therefore, the continuous function   attains its maximum on  , say at  . If  , then   is constant on   (by the lemma) and hence on  , so   also attains its maximum on  . Otherwise,  , completing the proof.

See Also

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Page Information

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Translation and Version Control

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This page was translated based on the following source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Maximumprinzip

  • Date: 12/26/2024