Re: [分析] bounded vs. convergent subsequence
假設 f_k monotone and uniformly bounded on [0,1]，則存在
逐點收斂之子序列. 大致步驟如下：
1 用對角線法，找出一個在有理數集上逐點收斂之子序列。仍將
此子序列記為 f_k。
2 不難看出，步驟一隻極限在有理數上遞增。將此極限延拓為一
[0,1] 上之遞增函數 f。延拓法可能不唯一，任取一個即可。
3 由遞增函數基本性質可知，f 在一可數集以外連續。
4 證明若 f 在 x 連續，則 f_k(x) 收斂至 f(x)。證明如下：
Given e, pick x_0 and x_1 such that x_0<x<x_1 and
f(x)-e<f(x_0)<f(x)<f(x_1)<f(x)+e
There is an N so that for all n>N, |f_n(x_0)- f(x_0)|<e
and |f_n(x_1)- f(x_1)|<e. In particulr, For all n>N,
f_n(x_0)> f(x)-2e and f_n(x_1)< f(x)+2e. This implies
|f_n(x)-f(x)|< 2e since f is monotone
5 用對角線法處理 f 不連續的點。
※ 引述《znmkhxrw (QQ)》之銘言：
: ※ 引述《GSXSP (Gloria)》之銘言：
: : 標題: [分析] bounded vs. convergent subsequence
: : 時間: Sat Jul 9 08:04:05 2016
: According to your repost, codomain of each f_k is R (set of real numbers).
: : Q1:
: : f_k (i) uniformly bounded for all i = 1, 2 ,...
: : Can we find a subsequence k_j such that
: : f_{k_j} (i) converge for all i=1,2,...
: The answer is positive. You can prove directly as your repost, or by general
: Arzela-Ascoli Theorem stated below.
: <Thm>
: Let X be a separable space and f_k(x) a sequence of functions from X to R.
: If f_k(x) is pointwisely bounded and locally equicontinuous,
: then there exists a uniformly convergent subsequence f_k_j(x)
: <Remark>
: 1.X is a separable space :=
: X contains a countable dense subset.
: 2.f_k(x) is pointwisely bounded :=
: for any a€X, there exists M>0 such that │f_k(a)│<= M for any k.
: 3.f_k(x) is locally equicontinuous :=
: for any a€X, there exists a neighborhood of a, say U_a, such that
: f_k(x) is equicontinuous on U_a.
: -------------------------------------------------------------------
: For your Q1：
: 1.X = set of natural numbers, of course a separable space
: 2.uniformly bounded implies pointwisely bounded
: 3.locally equicontinuous follows by X being a discrete metric space, since
: for each point, you may take delta=1 such that there's only itself inside.
: 4.from above, you obtain a uniformly convergent subsequence,
: of course convergent.
: : Q2:
: : f_k(x) uniformly bounded for all x \in [0,1]
: : Can we find a subsequence k_j such that
: : f_{k_j} (x) converge for all x \in [0,1]
: Negative, counterexample：f_k(x) = cos(nx), x€[-1,1]
: But there are two results.
: 1.f_k(x) doesn't have uniformly convergent subsequence.
: 2.f_k(x) doesn't have convergent subsequence.
: The former one is easier to be proved than the latter one, which relies on
: Real analysis. (I don't know if there's simpler proof or not)
: For some discussion, you may refer to #1FJlxeaW and #1GMPuLdT.
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