推 hau :謝謝！ 03/01 23:09
※ 編輯: ppia 來自: 125.231.219.112 (03/19 11:23)
※ 編輯: ppia 來自: 125.231.219.112 (03/19 11:25)
※ 引述《hau (小豪)》之銘言：
: x
: Let p > 1，f ≧ 0，f belongs to L^p((0,∞)) and F(x) = ∫ f(t) dt.
: 0
: Show that F(x) = o(x^(1/q)) as x→∞ where (1/p) + (1/q) = 1.
: 也就是要證明 lim F(x)/x^(1/q) = 0.
: x→∞
: 這題其實是兩小題，另一小題是證明 F(x) = o(x^(1/q)) as x→0.
: F(x) = o(x^(1/q)) as x→0 這情況簡單，
: 那逼近∞時，我試著加一個x的次方進去，再用Holder's inequality
: 但不行，也試過用Jeanson's inequality，不知是否是函數找的不好，也不行……
F(x)
Suppose lim sup ──── = c > 0. Put g(x)= f(x)/c.
x→∞ x^(1/q)
Then there exists an increasing sequence {x[n]} with x[n] → ∞
and
1/q x[n+1] 1/q
x[n] (1-ε[n]) < ∫ g(x) dx < x[n] (1+ε[n])
0
where ε[n] →0.
Moreover, since x[n] goes to infinity, we may assume x[n+1] > 2 x[n].
Put I[n] = [ x[n], x[n+1] ]. By the Holder inequality,
1/q 1/q 1/q 1/q
x[n+1] - x[n] - (ε[n+1] x[n+1] + ε[n] x[n] )
───(A)─── ───────(B)──────
p 1/p 1/q
< ∫g(x) dx < (∫g(x) dx) (x[n+1]-x[n]).
I[n] I[n]
Since x[n+1] > 2x[n], we have (x[n+1]-x[n])/x[n+1] > 1/2.
For (A), after appling the Mean Value Theorem we get:
(A) (x[n+1]-x[n])^(1/q) 1/q
──────── ≧ ────────── > (1/q)(1/2)
(x[n+1]-x[n])^(1/q) q x[n+1]^(1/q)
For (B), put δ[n] = max{ε[n],ε[n+1]}
(B) x[n+1] 1/q 1/q
──────── ≦ 2δ[n]( ──────) < 2δ[n] (1/2),
(x[n+1]-x[n])^(1/q) x[n+1]-x[n]
which goes to zero as n goes to infinity. Hence
p p
(1/q)/2 ≦ lim inf ∫ g(x) dx.
n →∞ I[n]
∞ p ∞ p
This implies ∫ g(x) dx = ∞ and thus ∫ f(x) dx = ∞.
0 0
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 125.231.231.25
※ 編輯: ppia 來自: 125.231.231.25 (02/20 22:03)