Re: [分析] bounded vs. convergent subsequence

[znmkhxrw (QQ)]

※ 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.

--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 61.231.70.246
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1468226247.A.013.html
推 Vulpix      : 真的是只要double A，萬事都OK～
推 GSXSP       : 我看了#1GMPuLdT 所以L2, pointwise, 都不行
→ GSXSP       : how about adding the condition of monotone?
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
→ GSXSP       : 另外我其實不需要太強的收斂
→ GSXSP       : 不會說中文會上ptt嗎XD
→ GSXSP       : 我沒有Dini thm的條件, 只有f_k(x) monotone in x
→ GSXSP       : f_k(x) cts and monotone in x for all k
→ GSXSP       : 我也有在Mathoverflow上問 http://goo.gl/aBTUKW
→ GSXSP       : 裡面有我要做的 但我不是很確定到底怎麼樣的收斂夠
→ GSXSP       : 有人回in the sense of distributions可以
→ GSXSP       : 但我對那個不熟, 不知道適不適用我要證的存在性
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
→ GSXSP       : 他其實沒說Lp可。你給的例子也是Lp不收斂。
→ GSXSP       : converge in measure 的話怪怪的, 不曉得是什麼meas
→ GSXSP       : （我本來以為他是在講theory of distributions ）
→ GSXSP       : In measure再找子列a.e.的話就跟你的反例矛盾了
→ GSXSP       : 還有可能是weakly converge, 可是那是measure的收
→ GSXSP       : 斂 我不知道能不能弄成函數的。 你覺得那個泛函方程
→ GSXSP       : 怎麼樣的收斂才能構造出存在一解呢
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
→ GSXSP       : 嗯嗯. 不過問題太復雜感覺大家比較會不想詳細看
→ GSXSP       : 我再想想怎麼 rephrase
推 GSXSP       : 有f_k(x) increasing in x 的話我是不是可以用
→ GSXSP       : 這個Helly's selection theorem
→ GSXSP       : https://goo.gl/w2YXpq
→ GSXSP       : Michael說的應該是這個吧 pointwise+L1
→ GSXSP       : 但是not uniformly conv.的遞增函數例子是什麼呢？
→ GSXSP       : 哦, I_{(1-1/n,1]} 好像就不是uniformly conv了