推 thisday :請問g是哪一個? 12/18 16:19
※ 編輯: bineapple 來自: 123.192.216.62 (12/18 16:31)
推 thisday :感謝^^ 12/18 16:34
※ 引述《thisday (小綱)》之銘言：
: If f(z) is analytic in |z|≦ 1 and satisfies |f| = 1 on |z| = 1
: show that f(z) is rational.
: 我有想用Blaschke product
Let a represent any root of f with order m in |z|<1.
Consider f(z)Π[(1-a*z)/(z-a)]^m := g(z)
a
where a* is the conjugate of a, then |g|=1 on |z|=1, and g has no root
in |z|<1, so by applying maximum modulus principle to g and 1/g we
get that g is a constant since |z|≦1 is compact, this shows that
f is rational.
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※ 編輯: bineapple 來自: 123.192.216.62 (12/17 05:01)