Let $X$ be a compact topological space and let $K\subset \mathbb{R}^n$ be compact. Let $F = \{f_n\}_{n \ge 1}$ be a sequence of continuous functions where $f_n : X \to K$ that converges pointwise to $f$.

Show that if $F$ is equicontinuous, then $f$ is continuous and $f_n$ converges uniformly to $f$.


a) It is enough to show that the convergence is uniform, since the Uniform Limit Theorem then implies that $f$ is continuous, correct?

b) Since the convergence is pointwise, $F$ must be pointwise bounded, correct? (If it's not, then I have a big misunderstanding...)

c) By Arzela's theorem, since $X$ is compact,$F$ pointwise bounded and equicontinuous, $F$ has a uniformly convergent subsequence, correct?

d) Since $F$ has a uniformly convergent subsequence, can I conclude that it converges uniformly to $f$? Why? Why not?

Any help is highly appreciated.

  • $\begingroup$ A standard sub-subsequence argument should do the trick $\endgroup$ Apr 25, 2022 at 13:16
  • 1
    $\begingroup$ @EvangelopoulosF. So all steps are correct, and I only need to justify step d? $\endgroup$
    – JustANoob
    Apr 25, 2022 at 14:07
  • $\begingroup$ Pointwise convergence always implies pointwise boundedness. But since all $f_n$ map into $K$ (which is bounded), $F$ is even uniformly bounded. $\endgroup$
    – Paul Frost
    Apr 25, 2022 at 23:04

1 Answer 1


Hi you are right with point b) and c).

let's show that f is continuous

if $F$ is equicontinuous let $\epsilon >0$ let $x>0$

there exist $\delta >0$ such that $\forall n \in N$ $|x-y|< \delta =>|f_n(x)-f_n(y)|< \frac{\epsilon}{3}$

So let $y$ such that $|x-y|< \delta$

by pointwise convergence let $n \in N $ such that $|f_n(y)-f(y)|<\frac{\epsilon}{3}$

and $|f_n(x)-f(x)|<\frac{\epsilon}{3}$

then $|f(x)-f(y)|< |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$

then f is continuous.

As mentionned in the comment to conclude you can use this sub-subsequence argument :

let's show that $f_n$ uniformly converge to $f$

We assume that $f_n$ don't uniformly converge to $f$ So there exist $\epsilon$ and there exist $\phi$ such that $\forall n \in N ~~ |f_{\phi(n)}-f|_{\infty}> \epsilon$ but you can apply arzela's theorem to the subsequence $(f_{\phi(n)})$ because the subsequence verify the theorem's conditions. So there exist $\psi$ such that $(f_{\phi(\psi(n))})$ uniformly converge to $f$

but as a subsequence of $(f_{\phi(n)})$

$\forall n \in N$ $|f_{\phi(\psi(n))}-f|_{\infty }> \epsilon $

This is a contradiction hence $f_n$ uniformly converge to $f$


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