[分析] 一題Stein複變上的習題

[Bourbaki (大狐狸)]

p.281 prob 3

Suppose Ω is a simply connected domain that excludes the three roots of the
polynomials 4z^3 - (g_2)z - (g_3). For ρ belonging Ω and ρ fixed, define
        ω             dz
I(ω)=∫ -------------------------------      ω belonging Ω
       ρ [4z^3 - (g_2)z - (g_3)]^(1/2)

then I has an inverse given by p(z+α) for some α
that is I(p(z+α))=z for approprite α.

                d
他的hint說證明----I(p(z+α)) = +1 or -1 再利用p是偶函數
               dz
              ^^^^^^^^^^^^^^^^^^^^^^^^^這裡有做到

但是我不知道怎麼用偶函數這個條件來讓I(p(z+α))=z
請強者指教
謝謝^^

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◆ From: 140.112.25.15
※ 編輯: Bourbaki        來自: 140.112.25.15        (05/17 20:03)
→ keroro321   :也許這個建議會對你有幫助,先從ρ附近來看.
→ keroro321   :由Ω的定義,知道I(ω) is well-defined.!!
→ keroro321   :ρ fixed,方程式p(z)=ρ has two roots (z_0,z_1).
→ keroro321   :(I(p(z_i)))'=I'(p(z_i))*p'(z_i)  (i=0,1)
→ keroro321   :Becaus p is even (p' is odd), we can choose a
→ keroro321   :root z_2 (z_0 or -z_0=z_1 (mod M)) such that
→ keroro321   :(I(p(z_2)))'= 1, 從連續性知道
→ keroro321   :存在 a nbd U of z_2 使得 p(U)包含於Ω 並且
→ keroro321   :(I(p(z)))'= 1 , So I(p(z))= z+c(cons.)!! (z屬於U)
→ keroro321   :I(p(z_2))=I(ρ)= 0 =z_2+c , hence c=-z_2
→ keroro321   :I(p(z+z_2)) = z , (z+z_2 屬於U)
→ keroro321   :所以在 ρ 附近做到了,之後應該可以從類似證明
→ keroro321   :open connected in |R^n => path connected
→ keroro321   :證到整個 Ω.