# Equicontinuity, compactness and uniform convergence.

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 $$f$$.

#### Attempt

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.

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

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 $$f$$