Uniform congergence Problem.

**younhock** · December 14th 2011, 01:39 AM

I have two question here:
1) Suppose $f_n = x^n$ , $\forall x\in [0,1] , n \in $\mathbb{N}$$ , how to find $A_\varepsilon \subseteq [0,1]$ such that $f_n$ is converge UNIFORMLY on $A_\varepsilon$ ?

2) Suppose $m(A) \nless \infty$ ,why there is no such $A_\varepsilon \subseteq A$ such that when $f_n = 1 _{[n,n+1]}$ , $f_n$ is converges uniformly on $A_\varepsilon$

**FernandoRevilla** · December 14th 2011, 01:49 AM

According to the Dini's Theorem, $f_n=x^n\to 0$ uniformly on any compact $K\subseteq [0,1)$ .

**younhock** · December 14th 2011, 04:36 AM

Originally Posted by FernandoRevilla

According to the Dini's Theorem, $f_n=x^n\to 0$ uniformly on any compact $K\subseteq [0,1)$ .

Is there any other approach ?

**FernandoRevilla** · December 14th 2011, 06:52 AM

Originally Posted by younhock

Is there any other approach ?

Yes, choose for example $K=[0,a)$ with $0<a<1$ , then for every $0<\epsilon<1$ and for $n$ positive integer such that $n>\log \epsilon/\log a$ we have $n>\log \epsilon/\log x$ for all $x\in K$ , or equivalently $|x^n|<\epsilon$ for all $x\in K$ .

**younhock** · December 14th 2011, 07:00 AM

Originally Posted by FernandoRevilla

Yes, choose for example $K=[0,a)$ with $0<a<1$ , then for every $0<\epsilon<1$ and for $n$ positive integer such that $n>\log \epsilon/\log a$ we have $n>\log \epsilon/\log x$ for all $x\in K$ , or equivalently $|x^n|<\epsilon$ for all $x\in K$ .

Oh, This is good. Thanks a lot.
How bout Question 2? I cant show that.

**vincisonfire** · December 16th 2011, 12:36 PM

Originally Posted by younhock

2) Suppose $m(A) \nless \infty$ ,why there is no such $A_\varepsilon \subseteq A$ such that when $f_n = 1 _{[n,n+1]}$ , $f_n$ is converges uniformly on $A_\varepsilon$

It seems it could be true or false depending on $A$ (if I understand the question properly).
It would converge uniformly to 0 on $[0,1]$ .
It would not converge uniformly (but pointwise) to 0 on $\bigcup\limits_{n=0}^\infty [n,n+2^{-n}]$ .

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