# The Arzelà–Ascoli theorem fails on a half-open interval

Can we find an example:

(1) $\lbrace f_n \rbrace_n$ is a family of real-valued functions defined on $[0,1)$ such that this family is uniformly bounded and equicontinuous, $f_n(0)=0$;

~~~ Uniformly bounded: $|f_n(x)|\leq M \ \ \ \forall n$ and for a fixed $M>0$, Equicontinuity for $\lbrace f_n \rbrace _n$: for any small $\varepsilon >0$, there is a $\delta > 0$ such that whenever $|x-y|< \delta$, then $|f_n(x)-f_n(y)|< \varepsilon$ for all $n$. ~~~

(2) this family does not have a uniformly convergent subsequence on $\mathcal{C}[0,1)$.

This is actually to find a counterexample for Arzela-Ascoli theorem on the space $\mathcal{C}[0,1)$ where $[0,1)$ is not compact, and all other conditions are satisfied.

What if an additional assumption is added: $f_n(1^-)=1$? Thanks a lot!

• Just to make sure we're on the same page, is equicontinuity a uniform property for you (for all $\varepsilon > 0$ there is a $\delta > 0$ such that for all $x$ ...) or a pointwise property (for all $x$, for all $\varepsilon > 0$, there is a $\delta > 0$ ...)? – Daniel Fischer Nov 14 '13 at 20:34
• equicontinuity for {f_n}_n: for any small ε>0, there is a δ>0 such that whenever |x-y|<δ, then |f_n(x)-f_n(y)|<ε for all n. – user108871 Nov 14 '13 at 20:37
• yes, uniform property, just as what I wrote. Thanks. – user108871 Nov 14 '13 at 20:46

Okay, so it's uniform equicontinuity. Then there cannot be such an example, since all $f_n$ are uniformly continuous, hence can be continuously extended to $[0,1]$, and the family of extended functions is still uniformly bounded and uniformly equicontinuous. Then by Ascoli's theorem, it contains a uniformly (on $[0,1]$) convergent subsequence.

If we worked with pointwise equicontinuity instead, then

$$f_n(x) = \begin{cases}0 &, x \leqslant 1 - \frac1n\\ n(x-1)+1 &, x \geqslant 1 - \frac1n \end{cases}$$

would give an example.

The family $\{ f_n : n\in\mathbb{Z}^+\}$ is uniformly bounded, and equicontinuous in all $x \in [0,1)$ (in a neighbourhood of each point $x \in [0,1)$, only finitely many $f_n$ are not constant). It converges pointwise to $0$ (and locally uniformly), but not uniformly.

• Yes, this works perfectly. Thank you very much! – user108871 Nov 14 '13 at 20:58
• perfectly clear in this case, but I've difficult to find out what doesn't work in your counterexample in the case that the family of functions are defined over $[0,1]$ (and so Ascoli-Arzelà has to work properly). There is a problem when $n \to \infty$? thanks in advance! – Riccardo Nov 14 '13 at 21:51
• @RicPed If the functions were defined on $[0,1]$, the family would not be equicontinuous in every point, namely not in $1$. – Daniel Fischer Nov 14 '13 at 21:54
• mmh, ok I think I'm starting following your reasoning. It is the fact that for a fixed $\delta$, for every $y$ such that $|y-1| \leq \delta$ there is a $\bar{n}$ s.t. $f_n(y)=0$ $\forall n > \bar{n}$? – Riccardo Nov 14 '13 at 22:21
• @RicPed Right. At $1$, there is no $\delta$ for any $\varepsilon < 1$. – Daniel Fischer Nov 14 '13 at 22:25