[分析] 實變

[smartlwj (最後5天衝刺)]

1. Show that there is no function f such that both
         2                2
   f in L (R) and 1/f in L (R).

                   1
2. Suppose f,g in L [0 1] and |f(x)g(x)| ≧ 1 for any
                            1           1
   x in [0 1]. Prove that (∫|f(x)|dx)(∫|g(x)|dx) ≧ 1.
                            0           0

3. Let λ be the Lebesgue measure and let E be a Lebesgue

   measurable subset of [0 1] such that λ(E∩[a b]) ≧ (1/2)(b-a)

   for any 0 ≦ a < b ≦ 1. Show that λ(E) = 1.


請問這三題該怎麼做??

第一題我是想假設存在然後去証矛盾...可是我不會做 =.=
第二題沒有想法@@
第三題也沒有什麼想法去做...阿~~請教教我  謝謝


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◆ From: 140.114.34.117
推 goodGG      :1. Holder inequality => 1 in L^1 -><-
推 math1209    :2. Holder inequality. 3. Lebesgue 微分定理.
→ smartlwj    :第二題要怎麼用Holder?
→ smartlwj    :第一題ok了 感謝好gg大大
推 goodGG      :2. Consider √|f| and √|g|.
→ smartlwj    :懂了 謝謝好GG大和math大
推 math1209    :將 λ(E∩[a b]) 記為 ∫_S χ_E(t) dt, S = [a,x]
→ math1209    :So, ∫_S χ_E(t) dt ≧ 1/2 λ|S|.
→ math1209    :It implies (∫_S χ_E(t) dt)/|S|≧ 1/2.
→ math1209    :Let x tend to a, we finally ... you see.
→ keroro321   :第3題可以看看用以下方法 @@
→ keroro321   :suppose λ(E) = a <1 ,
→ keroro321   :V is any open set in [0,1],
→ keroro321   :λ(E∩[a,b]) ≧ (1/2)(b-a) implies
→ keroro321   :λ(E∩V) ≧ (1/2)λ(V)
→ keroro321   :Choose a closed set F contained in E such that
→ keroro321   :        λ(E-F) < (1-a)/4
→ keroro321   :let V=[0,1]-F , λ(V)≧1-λ(E)
→ keroro321   :λ(E)=λ(E∩F)+λ(E∩V)
→ keroro321   :     > a-(1-a)/4+(1-a)/2 = a+(1-a)/4
→ keroro321   :與假設矛盾 所以λ(E)只能是 1
推 math1209    :method of keroro321 is good.