Re: [分析] 實變sequential convergene/compactness

看板Math作者 (紳士的野蠻要藏在袖子裡)時間12年前 (2013/05/31 10:12), 編輯推噓0(002)
留言2則, 1人參與, 最新討論串2/2 (看更多)
※ 引述《BA0954016 (YES)》之銘言: : 1. Let 1<p<∞ and f_0 belong to L^p(R). For each natural number n, define : f_n(x)=f_0(x-n) for all x. Define f=0 on R. Show that {f_n}-->f in L^p(R). Is : this true for p=1? pf. Let g\in L^q(R) where 1/p+1/q=1. Then for any \epsilon > 0, there is M > 0 s.t. ||g\chi_{|x|>M}||_q \leq \epsilon. Therefore |< f_n, g >|\leq \epsilon ||f_n||_p + ||g||_q ||f_n\chi_{|x|>M}||_p where ||f_n\chi_{|x|>M}||_p -> 0 as n -> \infty. Done. When p=1, this statement fails by taking f_0 as a positive function and g=1 as a constant function. : 2. Let [a,b] be a nondegenerate closed, bounded interval. In the Banach space : C[a,b], normed by the maximum norm, find a bounded sequence that fails to : have any strongly convergent subsequence. Counter example. \chi_{(2^{-n},2^{-n+1})} (say, a=0,b=1 in our case.) : 感謝各位實變高手能給我一些解答或方向! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 76.127.164.190

05/31 22:40, , 1F
2. 可以取 sin(nx) 或 cos(nx), 用反證法證沒有
05/31 22:40, 1F

05/31 22:41, , 2F
收斂子數列
05/31 22:41, 2F
文章代碼(AID): #1Hg0SCRo (Math)
文章代碼(AID): #1Hg0SCRo (Math)