推 wyob :感謝，還再努力研究第二題 02/08 10:58
※ 引述《wyob (Go Dolphins)》之銘言：
: 1.Let f be a continuous real-valued funtion defind on [a,b],and let
: b
: M= max│f(x)│.Show that lim (∫│f(x)│^n dx)^(1/n)=M
: x屬於[a.b] n→∞ a
: 2.Let f:(-1,2)→R be a real analytic function,
: If f(1/k)=0 for all k 屬於N,show that f is identically zero
: 第二題我自己解的怪怪的，因為好像沒有用到f定義在(-1,2)上
: 第一題該如何下手呢?
1.(1) If M = 0 => f(x) = 0 for all x
b
=> ∫ |f(x)|^n dx = 0 for all n
a
b
=> (∫ |f(x)|^n dx)^(1/n) = 0 for all n
a
b
=> lim (∫ |f(x)|^n dx)^(1/n) = 0 = M
n→∞ a
then we are done
(2) If M > 0
∵ f is continuous on [a,b]
∴ M exists and |f| is continuous on [a,b]
Let c belonging to [a,b] such that f(c) = M => |f(c)| = M
By definition of continuity of |f| at c ,
Given 0 < ε < M/2 , there exists δ > 0 such that
|x - c| < δ => ||f(x)| - |f(c)|| < ε
=> c - δ < x < c + δ => M - ε < |f(x)| < M + ε
c+δ c+δ b b
∴ ∫ (M-ε)^n dx ≦ ∫ |f(x)|^n dx ≦ ∫|f(x)|^n dx ≦ ∫ M^n dx
c-δ c-δ a a
n b n n
=> (M-ε) (2δ) ≦ ∫ |f(x)| dx ≦ M (b-a)
a
1/n b n 1/n 1/n
=> (M-ε)(2δ) ≦ (∫|f(x)| dx) ≦ M(b-a)
a
1/n b n 1/n 1/n
∵ limsup(M-ε)(2δ) ≦ limsup(∫|f(x)| dx) ≦ limsupM(b-a)
n→∞ n→∞ a n→∞
b n 1/n b n 1/n
=> M-ε ≦ limsup(∫|f(x)| dx) ≦ M => limsup(∫|f(x)| dx) = M
n→∞ a n→∞ a
1/n b n 1/n 1/n
and liminf(M-ε)(2δ) ≦ liminf(∫|f(x)| dx) ≦ liminfM(b-a)
n→∞ n→∞ a n→∞
b n 1/n b n 1/n
=> M-ε ≦ liminf(∫|f(x)| dx) ≦ M => liminf(∫ |f(x)| dx) = M
n→∞ a n→∞ a
b n 1/n
∴ lim (∫ |f(x)| dx) = M
n→∞ a
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