[機統] Converge in L^p norm
本來是這樣:
Xn → X in L^p
Yn → Y in L^q
then XnYn → XY in L^1 for 1/p + 1/q = 1
我想問: 不要 1/p + 1/q = 1 這個條件是否也可以
例如說 p=q=1.
他證明是有用到Holder inequality
不過這樣是否也可:
proof:
Case1 X = 0 and Y = 0
2
∥Xn Yn∥ ≦ ∥Xn∥ ∥Yn∥ → 0 as n → ∞
1 1 1
hence ∥Xn Yn∥ → 0
1
Case2 Xn = X for all n and Y = 0
X_M (w) = min { X(w) , M }
2
∥X_M Yn∥ ≦ ∥X_M∥ ∥Yn∥ → 0 as n → ∞ for all M > 0
1 1 1
=> Given ε> 0 exist N s.t.
2 2
∥X_M Yn∥ < ε for all n ≧ N
1
∵ | X_M Yn | → |X Yn| increasingly a.s.
=> E|X_M Yn | → E|X Yn| increasingly
∴ ∥X Yn∥ = lim ∥X_M Yn∥ ≦ lim ε = ε for all n ≧ N
1 M→∞ 1 M→∞
hence ∥X Yn∥ → 0
1
General case
XnYn - XY = (Xn-X)(Yn-Y) + X(Yn-Y) + Y(Xn-X) → 0
case1 case2 case2
Hence XnYn → XY in L^1
這樣是否正確?
而且這樣好像 p , q ,1 這三個數字可以換成很多別的數字?
(用math1209大貼過的 a＋b ＝ c 且 c/r ＝ a/p ＋ b/q,
c a b
∥fg∥ ≦ ∥f∥ ∥g∥ .
r p q
好像很多數字都可以，另外我想問一下這個不等式在哪裡找得到或是證明阿?
a,b,c,r,p,q都要是正整數嗎?)
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※ 編輯: GSXSP 來自: 218.168.28.250 (02/28 16:51)
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