Maximum modulus principle and its corollary（Stein复分析）

Maximum modulus principle:

\quad I f f i s a n o n − c o n s t a n t a n d h o l o m o r p h i c f u n c t i o n i n a r e g i o n Ω , t h e n f c a n n o t a t t a i n a m a x i m u m i n Ω . If\quad f\quad is\quad a\quad non-constant\quad and\quad holomorphic\quad function\quad in\\ a\quad region\quad \Omega,\quad then\quad f\quad can\quad not\quad attain\quad a\quad maximum\quad in\quad \Omega. Iffisanon−constantandholomorphicfunctioninaregionΩ,thenfcannotattainamaximuminΩ.

Proof:

\quad S u p p o s e t h a t f a t t a i n s a m a x i m u m i n Ω , d e n o t e d b y ∣ f ( z 0 ) ∣ . S i n c e f i s h o l o m o r p h i c i n Ω , f i s a n o p e n m a p p i n g . T h e r e f o r e , i f D ⊂ Ω i s a s m a l l d i s c c e n t e r e d a t z 0 , i t s i m a g e f ( D ) i s o p e n a n d c o n t a i n s f ( z 0 ) . T h i s p r o v e s t h a t t h e r e a r e p o i n t s z ∈ D s u c h t h a t ∣ f ( z ) ∣ > ∣ f ( z 0 ) ∣ , a c o n t r a d i c t i o n . T h e p r o o f i s c o m p l e t e . Suppose\quad that\quad f\quad attains\quad a\quad maximum\quad in\quad\Omega,\quad denoted\quad by\\ |f(z_0)|.\quad Since\quad f\quad is\quad holomorphic\quad in\quad \Omega,\quad f\quad is\quad an\quad open\\ mapping.\quad Therefore,\quad if\quad D\sub \Omega\quad is\quad a\quad small\quad disc\quad centered\\ at\quad z_0,\quad its\quad image\quad f(D)\quad is\quad open\quad and\quad contains\quad f(z_0).\quad This\\ proves\quad that\quad there\quad are\quad points\quad z \in D\quad such\quad that\quad |f(z)|\gt|f(z_0)|\\,\quad a\quad contradiction.\\\quad The\quad proof\quad is\quad complete. SupposethatfattainsamaximuminΩ,denotedby∣f(z0)∣.SincefisholomorphicinΩ,fisanopenmapping.Therefore,ifD⊂Ωisasmalldisccenteredatz0,itsimagef(D)isopenandcontainsf(z0).Thisprovesthattherearepointsz∈Dsuchthat∣f(z)∣>∣f(z0)∣,acontradiction.Theproofiscomplete.

Corollary:

\quad S u p p o s e t h a t Ω i s a r e g i o n w i t h c o m p a c t c l o s u r e Ω ˉ . I f f i s h o l o m o r p h i c o n Ω a n d c o n t i n u o u s o n Ω ˉ , t h e n s u p z ∈ Ω ∣ f ( z ) ∣ ≤ s u p z ∈ Ω ˉ − Ω ∣ f ( z ) ∣ . Suppose\quad that\quad \Omega\quad is\quad a\quad region\quad with\quad compact\quad closure\quad \bar{\Omega}.\\If\quad f\quad is\quad holomorphic\quad on\quad \Omega \quad and\quad continuous\quad on\quad \bar{\Omega},\quad then\\sup_{z\in \Omega}|f(z)|\le sup_{z\in \bar{\Omega}-\Omega}|f(z)|. SupposethatΩisaregionwithcompactclosureΩˉ.IffisholomorphiconΩandcontinuousonΩˉ,thensupz∈Ω∣f(z)∣≤supz∈Ωˉ−Ω∣f(z)∣.

Proof:

\quad S i n c e f i s c o n t i n u o u s o n t h e c o m p a c t s e t Ω ˉ , t h e n ∣ f ( z ) ∣ a t t a i n s i t s m a x i m u m i n Ω ; b u t t h i s c a n n o t b e i n Ω i f f i s n o n − c o n s t a n t . I f f i s c o n s t a n t , t h e c o n c l u s i o n i s t r i v i a l . T h e p r o o f i s c o m p l e t e . Since\quad f\quad is\quad continuous\quad on\quad the\quad compact\quad set\quad \bar{\Omega},\quad then\\|f(z)|\quad attains\quad its\quad maximum\quad in\quad \Omega;\quad but\quad this\quad cannot\quad be\quad\\ in\quad \Omega\quad if\quad f\quad is\quad non-constant.\quad If\quad f\quad is\quad constant,\quad the\\ conclusion\quad is\quad trivial.\\\quad The\quad proof\quad is\quad complete. SincefiscontinuousonthecompactsetΩˉ,then∣f(z)∣attainsitsmaximuminΩ;butthiscannotbeinΩiffisnon−constant.Iffisconstant,theconclusionistrivial.Theproofiscomplete.