Re: [分析] 存在poly.ｐn->|x| pointwise

[lizzon (莉森)]

※ 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
                          ＃


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◆ From: 140.113.25.147
※ 編輯: lizzon          來自: 220.132.36.153       (12/17 18:08)
推 znmkhxrw    :你如何conclude的?
※ 編輯: lizzon          來自: 140.113.25.147       (12/17 22:08)
→ lizzon      :從consider~claim之間的ε取成1/n