Complex Analysics/Maximum Principle
Introduction
editThe 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
editLet 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
editLet be open, and be holomorphic. Let be a local maximum point of . Then is constant in a neighborhood of .
Proof of Lemma 1
editLet 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
editWe derive the following estimation:
Proof of Lemma 3
editIt follows that the inequality must be an equality chain, implying
- .
Proof of Lemma 4
editThus, we establish the constancy of using the property:
- for all ,
i.e., is constant on .
Proof of Lemma 5
editIf is constant on , then must also be constant, where is a constant.
Proof of Lemma 6
editSince is holomorphic on , the Cauchy-Riemann-Differential equation'apply:
- ,
and the following holds:
- .
Proof of Lemma 7
editLet 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
editSquaring the above equations yields:
- ,
- .
Adding these equations gives:
- .
Proof of Lemma 9
editFactoring out and :
- .
Thus,
- or .
Proof of Lemma 10
editWith , it follows that since and are real-valued, implying .
If , then , and . By the Cauchy-Riemann-Differential equation, . Thus, is constant on .
Proof
editLet be a local maximum point of in the domain . Define as the set of all mapped to (level set).
Proof 1: V is closed
editSince 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
editUsing the lemma, can also be represented as a union of open disks, and unions of open sets are open.
Proof 3: Connectivity
editThus, due to the connectivity of , i.e., is constant.
Proof 4: G is bounded
editIf 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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- Date: 12/26/2024