2
$\begingroup$

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 $[0, \infty)$.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.