Re: [分析] 複變一題

[keroro321 (日夕)]

※ 引述《young11539 (〝☆小小霈★”)》之銘言：
: If Ω is the punctured disk 0 < |z| < 1
: and if f is given by f(ζ) = 0 for |ζ| = 1
: f(0) = 1, show that all func. υ ≦ 0 in Ω
: where υ is subhamornic in Ω and
: limsup ν(z) ≦ f(ζ)
:  z→ζ
: where ζ are the boundary points

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證明主要是利用 Perron's method 得到的函數 u(x) , 證明

       " u(x) is bounded in Ω." (prove u=0 !)

然後再從 u(x) 的定義 , 得到本題的答案.
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需要用到書上 L. V. Ahlfors :"Complex Analysis," 3d ed. 此習題章節一些定理
                                               ￣￣￣
( 這章節介紹 Perron's method 來看 Dirichlet problem .)

Let B(f) be the class of functions υ which satisfy
   (i)  υ is subharmonic in Ω.
   (ii) limsup ν(z) ≦ f(ζ)  where ζ are the boundary(Γ) points of Ω .
        z→ζ

Let  u (z)= l.u.b. υ(z) for υ\in Ω .

以下提到用書上哪些定理 Chapter 6 > Section 4.2 >

Lemma1 → u is harmonic in Ω .

     如果 f 在一個 boundary point ζ 連續 :
Lemma2 → 在某些條件下  lim u(z) = f(ζ) .
                         z→ζ

     f :Γ→|R 連續函數
Theorem9 → 在什麼樣的 region 下 Dirichlet problem 可解 .
__


證明: lim u(z) = 0 for all |a|= 1 .
      z→a

      只要看 z=1 的情形的就好. (其他點相同做法)

      Let D = { |z| < 1 } , D(Γ) = { |z| = 1 }

      Let f:D(Γ)→|R  be a continuous function so that

                     f(1) = 0 , f(a)>0 for a≠1 .

      By Theorem9 , there exists a harmonic function F:D →|R such that

                     F|D(Γ) = f

      Because F is harmonic =>  F(0) > 0 !


      ------------------------------------------------
              F: harmonic in Ω                  D(Γ)∪{z=0}
                                                  ||
              F(1) = 0 , F(a) > 0  for all a \in  Γ － {z=1}

              So F is barrier at z=1 .
      ------------------------------------------------

      By Lemma2 ,  lim u(z) = f(1) = 0
                   z→1

─────────────────────
      Because (i)  ν(z) is subharmonic in Ω .
              (ii) limsup ν(z) ≦ f(ζ) ≦ 1 .  ( ζ \in Γ )
                    z→ζ

      Hence  ν(z) ≦ 1 for all z \in Ω .

      Note : g = 0 在 B(f) 裡面 ,

      so u(z) is "bounded" in Ω .

      既然已證明了 lim u(z) = 0  for all |a|=1
                   z→a

              而且 u(z) is bounded and harmonic in Ω={0 < |z| < 1}


              那麼 z=0 就是一個 "removable sigularity" .


              再由 Maximum Principle => u(z) = 0 in Ω .

              由 u(z) 定義 =>

                              ν(z)≦0 for all z in Ω .
                              ￣￣￣￣￣￣￣￣￣￣￣￣￣#


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