Re: [分析] 能否構造出滿足以下條件的函數序列
※ 引述《keroro321 (日夕)》之銘言：
: Is it possible to construct a sequence {f_i} of continuous functions
: such that ( f_i:|R—>|R )
: (i) f_i ≧ 0 for all i .
: (ii) { x ∣ lim inf f_n(x) = ∞ } = Q ( all rational numbers in |R )
: n->∞
: 先感謝各位板友的回應 !
Q_Q 構造看看:(概念寫一寫？)
首先 想像一個function
g(x) = { r(x) for some r > 0, if x in (n-0.1,n+0.1), n 是 0,1,2,3,4.....
{ 0, otherwise
把他想像成這樣好了：
/\ /\
/ \ / \
____/ \_______/ \______
0.9 1 1.1 1.9 2 2.1
然後我們現在想要用這種r,來當作以下的r唷！ 就是在中間peak是1,兩邊遞減到0.
構造開始！
因為有理數countable,不妨令count的順序q_1,q_2 .... 可以count完。
那想像 f_n(x) = { r(x), if x in (q-0.1^n,q+0.1^n) q in {q_1,q_2, ... q_n}
{ 0, otherwise
對有理數x,當n夠大 f_n(x) = 1,
對無理數y,如果存在有理數x使得, |y-x| < 0.1^n,
但是可以取到m>n,使|y-x| > 0.1^m, 讓f_m(y) = 0.
(這裡應該要對q_i的列舉順序動一下手腳)
=> 0 是 {f_m(y)}的一個limit point
(加上f_n>=0的條件)
=> liminf f_n(y) = 0
所以我們知道可以：
{ x ∣ lim inf f_n(x) = 1 } = Q ( all rational numbers in |R )
然後對同樣的函數乘以n，得到
{ x ∣ lim inf n*f_n(x) = ∞ } = Q ( all rational numbers in |R )
故得證.......
有錯請指正，謝謝
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◆ From: 140.112.90.67
※ 編輯: scan33scan33 來自: 140.112.90.67 (04/30 20:37)
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