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 $[0, \infty)$.


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.

  • $\begingroup$ Confusion. I am not claiming that under equi-continuity and uniform boundedness pointwise convergence becomes uniform convergence. I want to know whether the statement of Arzela-Ascoli true for non-compact domains also. $\endgroup$ – Anonymous Nov 14 '14 at 11:41
  • $\begingroup$ @user148951 I have edited the answer. The theorem does not hold on noncompact domain. $\endgroup$ – user99914 Nov 14 '14 at 11:46
  • $\begingroup$ I am not saying that the theorem holds on non-compact domains, my point is I want to extend the continuous limit $X(.)$ over the whole interval by expressing it as increasing union of compact domains then using Arzela-Ascoli over nested subsequences. In summary, still I won't be having any subsequence whose limit is $X$. But, the definition of $X(.)$ over the whole interval is well-defined. Does this make sense. That's why I wrote extension of uniform limit, not arzela-ascoli over non-compact. $\endgroup$ – Anonymous Nov 14 '14 at 13:17
  • $\begingroup$ @user148951: I edited the answer again. Please have a look. $\endgroup$ – user99914 Nov 14 '14 at 21:07

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