Re: [分析] 絕對連續

看板Math作者 (紳士的野蠻要藏在袖子裡)時間12年前 (2013/09/11 04:26), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串2/2 (看更多)
造個反例: Let x_n = Σ 1/k^{3/2} k=1 to n. y_n = Σ 1/k^{3/2} + 1/2n^{3/2} k=1 to (n-1). where n\in N. Now define h(x) = + n^{-1/2} for any x \in [x_{n-1},y_n] - n^{-1/2} for any x \in [y_n,x_n] h is L^{\infty} => L^1. So x g(x):=∫ h(t) dt 0 is absolutely continuous, g(y_n)= n^{-1/2}1/2n^{3/2}=1/2n^2 and g(x_n)=0. This implies that \sqrt(g)(y_n)= 1/\sqrt(2)n and \sqrt(g)(x_n)=0. Therefore \sqrt(g) is not absolutely cont. Let f(x)=\sqrt(x) and g(x) defined as above. Here we have a counter example. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.247.0.115

09/12 14:32, , 1F
看到了! 謝謝
09/12 14:32, 1F
文章代碼(AID): #1IBu2DKK (Math)
文章代碼(AID): #1IBu2DKK (Math)