Extension of uniform limit in Arzela Ascoli Suppose, I have a sequence $f_n([0, \infty))$ of functions such that they are equicontinuous and uniformly bounded. So, I can get a uniformly convergent subsequence $f_{n_k}$ over $[0,T]$. I want the uniform limit to get extended over the whole $[0, \infty)$ such that it remains continuous. So, I extract a uniformly convergent sub-sequence of $f_{n_k}$ over $[0,2T]$ (to make the limit continuous). Keep reapeating this process. 
Is this a way to construct a uniform limit for the whole $[0, \infty)$.     
 A: That can not be done in general. Let $f: \mathbb R \to \mathbb R$ to be a nontrivial continuous function with compact support. Then the sequence of functions $f_n : [0,\infty) \to \mathbb R$ defined by 
$$f_n(x):= f(x-n)$$
are uniformly bounded and equicontinuous. But the convergence to the zero function cannot be made uniform. Indeed, there is $\epsilon>0$ such that 
$$\sup_{x\in \mathbb R} |f_n(x) - f_m(x)| > \epsilon$$
for all $n \in \mathbb N$. Thus this sequence cannot have uniform limit in $[0,\infty)$. 
There is a standard way to construct a limit in some sense. Let $N \in \mathbb N$. Then let $(f_{n_k})$ be a subsequence of $(f_n)$ so that $f_{n_k}$ converges uniformly to $f_1$ on $[0,1]$. Denote the subsequence by $f^1_n$. Then pick another subsequence of $(f^2_n)$ of $(f^1_n)$ so that it converges to $f_2$ in $[0, 2]$. As $f^2_n$ is a sebsequence of $f^1_n$, $f_1 = f_2$ on $[0,1]$. 
Then we can do that for every $[0,N]$ and we come up with a "seqeuence of subsequences" $(f^N_n)$ and a function $f : [0,\infty) \to \mathbb R$ so that $f^N_n$ converges as $n\to \infty$ to $f$ in $[0,N]$. 
Then the sequence $g_n := f^n_n$ has the property that $g_n \to f$ on each bounded intervals. This is called the diagonal process. 
