The statement I am trying to prove:

Let $\{ f_n \}$ be a sequence of equicontinuous, real valued, uniformly bounded continuous functions on $\mathbb{R}$. Show that $\{ f_n \}$ has a convergent subsequence which converges uniformly on any bounded subset of $\mathbb{R}$ and pointwise on all of $\mathbb{R}$ to a continuous function. (Royden pg 210, problem 9).

I believe the first portion follows directly by Arzela Ascoli, once it is noted that if we are given some bounded subset $E$, we can look at $\overline{E}$ a compact set on which the family is equicontinuous and uniformly bounded.

For pointwise convergence on all of $\mathbb{R}$, the question seems to require some sort of diagonalization argument: Take some dense sequence of $\{x_i\}$ (the reals are separable) and write columns of $f_n$ applied to each individual point. There will be a convergent subsequence by Bolzano-Weirstrass and since the outputs are real and bounded by the uniform boundedness of the family. I am a bit fuzzy on this part: Repeating this literally pointwise and moving far enough diagonally, we can correct for different convergence "speeds" pluck out a single convergent subsequence for any given epsilon and have $\sup|f_n(x)-f(x)|<\epsilon$ for any given epsilon. Finally, continuity should follow from an $\epsilon/3$ argument and equicontinuity of the family.

However, in formalizing the above, I am getting caught up in the notation of that sort of argument and am hoping for a more elegant solution. Is there a better way? Am I just generally confused and that's why I'm having trouble writing things down formally?


  • $\begingroup$ If I'm not misunderstanding, in the first part of the question you're introducing different subsequences for different bounded subsets of $\mathbb{R}$. If you could manage to show that there exists one subsequence that converges uniformly on every bounded subset of $\mathbb{R}$, the pointwise convergence would follow immediately. $\endgroup$ – 1Emax Jan 29 '17 at 2:43
  • $\begingroup$ @SSepehr that confuses me a bit. I think it's not too hard to show that there is a subsequence converging on each bounded subset by just taking the closure, but I'm not sure how to formally extend each of these to a subsequence on $\mathbb{R}$ $\endgroup$ – qbert Jan 29 '17 at 2:56

By Arzela-ascoli, there's a subsequence $f_n^{(1)}$ that converges uniformly on $[-1, 1]$. Then there's a subsequence $f_n^{(2)}$ of $f_n^{(1)}$ which converges uniformly on $[-2, 2]$. Similarly we define $f_n^{(k)}$. Now the sequence $$f_1^{(1)}, f_2^{(2)}, f_3^{(3)}, \dots $$ clearly converges uniformly on every interval $[-m, m]$, so the first condition is satisfied.

For the second part, for every $x \in \mathbb{R}$, $x \in [-m, m]$ for some $m$. Thus the sequence converges at $x$ to some limit $L_x$. We'll define $f(x) = L_x$. It should be easy to see that $f$ is continuous everywhere.

  • $\begingroup$ Thank you. I overcomplicated this $\endgroup$ – qbert Jan 29 '17 at 23:45

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.