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Let $f_k : \mathbb R \to \mathbb R$ be an equicontinuous function sequence.

If for every $k$, $f_k(0)=0$ then the sequence $\langle f_k \rangle$ has a convergent subsequence.

$\langle f_k \rangle$ may not necessarily have a uniformly convergent subsequence (consider the counterexample $f_k(x)= \frac x k$) so the question is whether even in this case there is a pointwise convergent subsequence.

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bellcircle
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2 Answers2

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This is a partial answer.

Lemma (Rudin PMA Them 7.23) If $\{f_n\}$ is a pointwise bounded sequence of real or complex valued functions on a countable set $E$ then it has a subsequence $\{f_{n_k}\}$ that converges pointwise on $E$.

For proof of this lemma, see the first part of this proof for Arzela Ascoli theorem, in which you may just replace the rational numbers with any countable set in any metric space.

So if you can show $\{f_n\}$ is pointwise bounded (which I believe is true but I don't know how to prove), then there exists a subsequence $\{f_{n_k}\}$ that converges pointwise on $\Bbb Q$. So it must converge on $\Bbb R$: for any $x\in\Bbb R$, given any $\epsilon>0$, find $\delta>0$ so that $|f(s)-f(t)|<\epsilon/3$ for all $f\in \{f_k\}$ and $|s-t|<\delta$. Now find $r\in\Bbb Q$ such that $|x-r|<\delta$, find $K$ such that for any $k,k'\ge K$, $|f_{n_{k'}}(r)-f_{n_k}(r)|<\epsilon/3$. Then for any $k,k'\ge K$ $$|f_{n_{k'}}(x)-f_{n_{k}}(x)|\le |f_{n_{k'}}(x)-f_{n_{k'}}(r)|+|f_{n_{k'}}(r)-f_{n_k}(r)|+|f_{n_{k}}(r)-f_{n_{k}}(x)|<\epsilon.$$ So $\{f_{n_k}(x)\}$ is a Cauchy sequence and thus convergent.

So the crucial part is to show $\{f_{n}\}$ is pointwise bounded, and of course the key condition here is $f_n(0)=0$, but for now I still have no clue.

Vim
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  • Pointwise boundedness can be proved from equicontinuity. Thanks! – bellcircle Jul 16 '17 at 08:26
  • @bellcircle indeed. I just realised we can prove, by equicontinuity and $f(0)=0$, that on any $[0,G]$, ${f}$ must be uniformly bounded because: given any $\epsilon>0$, once we divide $[0,G]$ uniformly into $N$ subintervals where $N$ is so large that each subinterval has length less than the corresponding $\delta$, then $$\sup_{f\in{f_n},x\in[0,G]}|f(x)|\le (N+1)\epsilon.$$ which follows from the triangle inequality. – Vim Jul 16 '17 at 08:33
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The Arzelà–Ascoli theorem may be applied here along with an additional diagonal argument to deal with the fact that $\mathbb{R}$ is $\sigma-$compact but not compact.

  • Let $g_k = f_k \upharpoonright [-1, 1]$ so it is an equicontinuous sequence of real-valued functions defined on a closed and bounded interval. For the hypotheses of Arzelà–Ascoli theorem to hold it remains to show that it is uniformly bounded. But this follows easily from equicontinuity and the fact that $f_k(0) = 0$ for each $k \in \mathbb{N}$.

    We can now apply the theorem to obtain a subsequence $g_{k_n^{(1)}}$ of $g_k$ which is uniformly convergent so $f_{k_n^{(1)}}$ converges uniformly on $[-1, 1]$.

  • Suppose for the sake of induction step that we have chosen a subsequence $f_{k_n^{(s-1)}}$ of $f_k$ which is uniformly convergent on $[-(s-1), s-1]$ where $s \in \mathbb{N}$. In the manner similar to the above we prove that the sequence $g_{k_n^{(s-1)}} := f_{k_n^{(s-1)}} \upharpoonright [-s, s]$ satisfies the hypotheses of the Arzelà–Ascoli theorem, so we pick a uniformly convergent subsequence $g_{k_n^{(s)}},$ hence $f_{k_n^{(s)}}$ converges uniformly on $[-s, s]$.

  • Finally, let $f_{k_n} = f_{k_n^{(n)}}$. It follows from the construction that $f_{k_n}$ is a subsequence of $f_k$ which converges uniformly on every interval $[-s, s]$, so it converges compactly on $\mathbb{R}$.

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Adayah
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