Chun-Ju Lai

Modes of Convergence 整理

In Analysis on 2011/09/26 at 3:17 PM

之前王振男有教過, 最近需要用到, 所以我再整理後 把他打出來:
我們著眼於以下四種收斂方式:

AU:    f_n \to f almost uniformly
AE:    f_n \to f almost everywhere
M:    f_n \to f in measure
L^1:    f_n \to f in mean


一、Arbitrary measure spaces

\begin{array}{ccc}\text{AU}&\rightarrow&\text{AE}\\ \downarrow \\ \text{M}&\leftarrow&L^1\end{array}

1. 證成關係都只是 checking definitions
2. 反例
AU \not\to L^1 : 1/n \chi_{(0,n)}
AE \not\to M : \chi_{(n,n+1)}
\not\to AE : Typewritter seq
\not\to L^1 : n\chi_{[1/n,2/n]}

二、Finite measure spaces (X,M,\mu)

If \mu(X)<\infty, then
\begin{array}{ccc}\text{AU}&\leftrightarrow&\text{AE}\\ \downarrow \\ \text{M}&\leftarrow&L^1\end{array}
3. AE \to AU 為 Egoroff 定理

三、Dominated Cases

If |f_n|\leq g for some g in L^1, then

\begin{array}{ccc}\text{AU}&\leftrightarrow&\text{AE}\\ \downarrow \\ \text{M}&\leftrightarrow&L^1\end{array}

4. AE \to AU 為 modified Egoroff Theorem
5. M \to L^1 為 LDCT wrt conv. in measure


四、Existence of convergent subsequence

以 M \overset{s}{\to}AU 表
“若 f_n \to f in measure, 則 f_n 有 subseq g_n \to f almost uniformly “, 以此類推

證明先證 f_n 有 subseq g_n \to g almost uniformly
因此 g_n \to g in measure 且 g_n \to f in measure
便可說 g = h a.e. 得到 g_n \to f almost uniformly

由一、的結果可推得以下結果:

1. M \overset{s}{\to} AU
2. M \overset{s}{\to} AE
3. L^1 \overset{s}{\to} AU
4. L^1 \overset{s}{\to} AE

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