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Let $\{f_n\}$ be a sequence of monotonically increasing functions on $\mathbb{R}$.

Let $\{f_n\}$ be uniformly bounded on $\mathbb{R}$.

Then, there exists a subsequence $\{f_{n_k}\}$ pointwise convergent to some $f$.

Now, assume $f$ is continuous on $\mathbb{R}$.

Here, I want to prove that $f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$.

How do i prove this ?

I have proven that " $\forall \epsilon>0,\exists K\in\mathbb{N}$ such that $k≧K \Rightarrow \forall x\in\mathbb{R}, |f(x)-f_{n_k}(x)||<\epsilon \bigvee f_{n_k}(x) < \inf f + \epsilon \bigvee \sup f - \epsilon < f_{n_k}(x)$ ".

The argument is in the link below.

I don't understand why above statement implies "$f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$". Please explain me how..

Reference ; http://www.math.umn.edu/~jodeit/course/SP6S06.pdf

Thank you in advance!

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@Learer That's not the original question. It says "Prove $f_{n_k}\rightarrow f$ uniformly on $\mathbb{R}$" on p.167 Rudin PMA –  Katlus Dec 26 '12 at 5:55
    
@Learner Yes, so i'm asking that how does "$f_{n_k}$ converges uniformly over compacts on $\mathbb{R}$l" imply that " $f_{n_k}$ converges uniformly on $\mathbb{R}$". –  Katlus Dec 26 '12 at 6:07
    
It's not true. For example, consider $f_n(x)=I(x-n)$ where $I(\le0)=0$ and $I(>0)=1$. Reference –  Frank Science Feb 16 '13 at 10:53

1 Answer 1

It's absolutely my fault that i didn't even read (c) in the link. I extend the theorem in the link and my argument below is going to prove;

"If $K$ is a compact subset of $\mathbb{R}$ and $\{f_n\}$ is a sequence of monotonic functions on $K$ such that $f_n\rightarrow f$ pointwise on $K$, then $f_n\rightarrow f$ uniformly on $K$." (There may exist both $n,m$ such that $f_n$ is monotonically increasing while $f_m$ is monotonically decreasing)

Pf> Since $K$ is closed in $\mathbb{R}$, complement of $K$ is a disjoint union of at most countable open segments.

Let $K^C=\bigsqcup (a_i,b_i)$.

Define;

$$ g_n(x) = \begin{cases} f_n(x) &\text{if }x\in K \\ \frac{x-a_i}{b_i-a_i}f_n(b_i)+\frac{b_i-x}{b_i-a_i}f_n(a_i) & \text{if }x\in(a_i,b_i)\bigwedge a_i,b_i\in\mathbb{R} \\ f_n(b_i) &\text{if }x\in(a_i,b_i)\bigwedge a_i=-\infty \\ f_n(a_i) &\text{if }x\in(a_i,b_i)\bigwedge b_i=\infty \end{cases} $$

Then, $g_n$ is monotonic on $\mathbb{R}$ and $g_n\rightarrow g$ pointwise on $\mathbb{R}$ ans $g$ is a continuous extension of $f$.

Let $\alpha=\inf K$ and $\beta=\sup K$. Then, by the argument in the link, $g_n\rightarrow g$ uniformly on $[\alpha,\beta]$. Hence, $f_n\rightarrow f$ uniformly on $K$.

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