[分析] 沒有均勻收斂子列
今天看到課本後面的題目
請造一個連續函數列f_n(x)€C[a,b]
且f一致有界
但是f不存在均勻收斂子列
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這例子我知道怎麼造
只是我在想,一樣的條件下,是否造得出f_n(x)不存在"逐點"收斂子列
也就是說是否造的出一串f_n,使得:
f_n(x)€C[a,b],f一致有界,但所有f_n(x)的子列f_n(k)(x)都存在一點c€[a,b]使得
lim f_n(k)(c)不存在
k→inf
試了一段時間試不出來
所以想證一定可以找的到"逐點"收斂子列
目前只知道for all c€[a,b] , f_n(c)有界
by Bolzano-Weierstrass 定理 存在一個收斂子列 f_n(k)(c)
可是這個子列的選取與c有關係
如果c是有限個,只要把f排排站(R^n的B-W的證法)
然後取對角線就可以找到共同的子列了
可是現在c€[a,b]
所以這個方法到這邊就卡住了@@
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你是說 f_n(x) = sin(1/x) €C(0,1] , for all n (這邊先不care原題要是interval)
然後claim f_n(x)的任何子列都不會是逐點收斂列???
※ 編輯: znmkhxrw 來自: 1.171.13.124 (09/19 19:23)
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可是,Let f_n(k)(x) is a subsequence of f_n(x)
hence f_n(k)(x) = sin(1/x) (因為f_n(x)一開始就跟n無關了)
而這個函數處處逐點收斂阿= = 因為k跑到無限大還是一樣是sin(1/x) for all x
※ 編輯: znmkhxrw 來自: 1.171.13.124 (09/19 19:29)
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