Re: [分析] bounded vs. convergent subsequence
※ 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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原來你會說中文XDD
你講的是這個吧 Dini定理
https://en.wikipedia.org/wiki/Dini%27s_theorem
不過要加上"連續"這個假設
※ 編輯: znmkhxrw (61.231.70.246), 07/12/2016 00:46:00
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看了你的連結與下面與那篇下面的回應 我有幾點想說
1.Michael的"in the sense of distributions"其實就是converge in measure
但是我覺得奇怪的是 他提到說1.converge in measure 可
2.converge uniformly 不可
3.converge in L^p 可
就我所知，實變中關於converge in measure且有關子列的，是conv. in measure
則有收斂子列(a.e.)，而converge in L^p implies conv. in measure
然後就是母列收斂的話，要滿足什麼條件才能L^p收斂
但是你問題是母列不收斂，要滿足什麼條件才能讓子列收斂，我是還蠻好奇M大回應
的詳細版本
2.另外一個回應是說你問的問題比你原問題更加限制(我是不知道這邊限制是指條件更多
還是更少 更少當然更難)
直接看書上是countable set,你要推廣到uncountable或是compact set
那個人好像回說這個問題也可以試試看用A-A定理去看
不過問題點就在於，A-A定理的兩個精神
(a) pointwisely bdd：用對角線法 造出收斂子列
但！只是適用於可數集
(b) locally equi.cont.：如果不是可數集，就是用等連續來連接附近的點
至於為什麼連的到，正是separable的性質 dense subset
使可數個點就可以大約代表整個集合
而這個誤差 就用等連續來彌補
因此 沒有等連續的話 我個人不知道有什麼其他A-A版本可以幫助你
※ 編輯: znmkhxrw (61.231.70.246), 07/12/2016 12:15:21
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我說的 可 不可 可 是從他的語意推敲的 基於什麼條件下是怎樣的可或不可
所以你提出的疑問我自然也有 才說期待他的詳細陳述到底是怎樣
你那個問題我目前也沒有答案耶 不如改成直接問這個問題吧?
※ 編輯: znmkhxrw (42.72.14.91), 07/12/2016 15:29:24
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