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, $f_n(0)=0$;
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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 $n$.
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(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!
Best Answer
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.