Re: [分析] 實變一題請教

看板Math作者 (QQ)時間13年前 (2012/11/19 18:45), 編輯推噓2(202)
留言4則, 3人參與, 最新討論串2/2 (看更多)
※ 引述《kkk3388 (haha)》之銘言: : 我想三天了 : 拜託版上高手 解答一下 : (1){g_n} be integrable on E and positive , : which converges to an integrable function g : (2){f_n} be measurable functions on E such that |f_n|<g_n, : and fn converges to f. : If lim Sgn = Sg, then Sf=limSfn : n->infinity ~(->這個符號是積分) : <hint>use fatou`s lemma : 這是中山大學2011博班考題 : 萬分感激 謝謝 │f_n│< g_n => -g_n < f_n < g_n => { 0 < f_n + g_n < 2g_n ---(1) { 0 > f_n - g_n > -2g_n ---(2) by fatou's lemma for (1), ∫liminf (f_n + g_n) ≦ liminf∫(f_n + g_n) ---(1') ~~~~~~~~~~~~~~~~~~~~ ↓(equals) ∫(f + g) for (2), ∫limsup (f_n - g_n) ≧ limsup∫(f_n - g_n) ---(2') ~~~~~~~~~~~~~~~~~~~~ ↓(equals) ∫(f - g) since ∫f_n is bounded by <lemma>(see p.s.) (1') = liminf∫(f_n + g_n) = (liminf∫f_n) +∫g (2') = limsup∫(f_n - g_n) = (limsup∫f_n) -∫g Hence ∫(f + g) ≦ (liminf∫f_n) +∫g =>∫f ≦ liminf∫f_n and ∫(f - g) ≧ (limsup∫f_n) - ∫g =>∫f ≧ limsup∫f_n Hence lim∫f_n = ∫f # <lemma> if a_n is bounded, b_n is convergent sequence with finite value L then liminf(a_n + b_n) = (liminf a_n) + L and limsup(a_n + b_n) = (limsup a_n) + L ----------------------------------------------------- 有錯請指證~ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 1.171.8.55 ※ 編輯: znmkhxrw 來自: 1.171.8.55 (11/19 18:50)

11/19 20:51, , 1F
長的蠻像LDCT的XD
11/19 20:51, 1F

11/19 21:20, , 2F
了解了 謝謝 感恩
11/19 21:20, 2F

11/19 22:48, , 3F
對阿~一開始想去湊有沒有一個general的g去壓
11/19 22:48, 3F

11/19 22:48, , 4F
不過失敗QQ 後來拆拆看就有了XD
11/19 22:48, 4F
※ 編輯: znmkhxrw 來自: 1.171.7.198 (11/27 00:59)
文章代碼(AID): #1GgWt3Ka (Math)
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文章代碼(AID): #1GgWt3Ka (Math)