Re: [分析] 存在poly.ｐn->|x| pointwise

看板Math作者 (莉森)時間13年前 (2012/12/17 16:20), 編輯推噓1(1推 0噓 1→)
留言2則, 2人參與, 最新討論串3/3 (看更多)
※ 引述《lizzon (莉森)》之銘言： : Let h(x)=|x|. Show that : there is a sequence of polynomials {Ｐn} s.t. : Ｐn→h pointwise 小妹我參考樓上多位大大的方式，寫出以下推論，請各位高手幫小妹看有沒有錯 Let f is a continuous function on R By W.A.T, consider [-1,1], there is a sequence of poly. P s.t. for each ε>0 1,m |P (x)-f(x)|<ε for x￡[-1,1] whenever m≧N for some N 1,m 1 1 ˙ ˙ ˙ (*) [-n,n], there is a sequence of poly. P s.t. for each ε>0 n,m |P (x)-f(x)|<ε for x￡[-n,n] whenever m≧N for some N n,m n n ˙ ˙ ˙ Take P = P n n,Nn Claim: Pn→f pointwise on R ￣￣￣￣￣￣￣￣￣￣￣￣￣￣ Given x￡R and any ε>0 there is an integer n s.t. x￡[-n,n] then we have |Pn(x)-f(x)| = |P (x)-f(x)|<ε (By (*)) n,Nn Hence Ｐn→f pointwise ＃ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.25.147 ※ 編輯: lizzon 來自: 220.132.36.153 (12/17 18:08)
推
12/17 19:07, , 1F
你如何conclude的?
12/17 19:07, 1F
※ 編輯: lizzon 來自: 140.113.25.147 (12/17 22:08)
→
12/18 18:21, , 2F
從consider~claim之間的ε取成1/n
12/18 18:21, 2F
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文章代碼(AID): #1GpjN7qu (Math)
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文章代碼(AID): #1GpjN7qu (Math)