[分析] 複變一題證明

看板Math作者 (聖潔之水)時間13年前 (2012/05/20 20:03), 編輯推噓0(001)
留言1則, 1人參與, 最新討論串1/1
Suppose that f is entire and that for each z, either |f(z)|≦1 or |f'(z)|≦1. Prove that f is a linear polynomial. Hint:Use a line integral to show |f(z)| ≦ A + |z| where A = max{1,|f(0)|} 後面的簡答是寫: z Let f(z) = f(z_0) + ∫ f'(w) dw, where the path of integration is along the z_0 ray from 0 to z, beginning at z_0, with z_0 = t_0 z, where t_0 = sup{t_1: |f(tz)|≦1, 0≦t≦t_1}. Clearly, then |f(z_0)|≦max{1,|f(0)|} and the integral is bounded by |z|. ========= 想問的是|f(z_0)|≦max{1,|f(0)|} 如果照t_0的假設 |f(0)|就會≦1, 是不是t_0的部分有錯 如果沒錯的話 可以分兩種情況1.|f(z_0)|≦1 2.|f'(z_0)|≦1 第二種情況要怎麼解決 z_0-δ z_0 我的想法是 |f(z_0)| = |f(0) + ∫ f'(w) dw + ∫ f'(w) dw| 0 z_0-δ ≦ |f(0)| + |B| + |C| where δ→0 |B| = |f(z_0-δ) - f(0)| = 0 不知道|C|要怎麼解決 z 然後 |∫ f'(w) dw| ≦|z| 這部分 是不是也要分兩種情況? z_0 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 42.79.57.74

05/21 09:55, , 1F
z0= t0*z ,t0=sup{t : |f(tz)|≦1 & 0≦t≦1 }
05/21 09:55, 1F
文章代碼(AID): #1FkDslhz (Math)