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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?

Thanks!

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  • $\begingroup$ 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. $\endgroup$ – SSepehr 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
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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.

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

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