Counterexamples of Arzèla Ascoli theorem for non-obeyed criteria I had an exam on functional analysis some time ago, and one of the questions I couldn't make any sense out was the following:
Let $\Omega\subset \mathbb{R}$ and $\{f_n\}$ a sequence of continuous functions from $\Omega$ to $\mathbb{R}$. If the following criteria are obeyed:


*

*$\exists M>0$ such that $||f_n||_{\infty}< M$ $\forall n\in\mathbb{N}$,

*$\Omega$ is compact,

*The sequence $\{f_n\}$ is uniform equicontinuous.


then the theorem of Arzèla Ascoli states that the sequence $\{f_n\}$ has a subsequence which converges in the $||.||_{\infty}$norm to a continuous function.
Show that the theorem is not true by stating counter examples in the cases:


*

*(1) and (2) are obeyed, but not (3),

*(1) and (3) are obeyed, but not (2),

*(2) and (3) are obeyed, but not (1).


I spend a lot of time thinking about this, but I couldn't think of any counter examples. For a non-compact subset of $\mathbb{R}$ I tried $(0,1)$ (as it is not closed) and for a bounded sequence I was thinking of $f_n = x^n$ but these didn't work. Can anyone help me with some counter examples and maybe a good way of thinking of them?
 A: *

*Let $\Omega=[0,1]$, $f_n(x)=x^n$. Then $\|f_n\|=1$ for all $n$. But the functions are not equicontinuous. If they were, there would exist $\delta>0$ such that $\delta<1$, for all $x,y\in(1-\delta,1]$, and for all $n$, 
$$
|y^n-x^n|<1/2.
$$
But if we take $n$ such that $(1-\delta/2)^n<1/2$, $y=1$ and $x=1-\delta/2$, then 
$$
|y^n-x^n|=1-(1-\delta/2)^n>1-1/2>1/2,
$$
a contradiction. So the family is not equicontinuous. 

*Let $\Omega=\mathbb R$, and define
$$
f_n(x)=\begin{cases}0,&\mbox{ if } x\not\in[n,n+1] \\ 2(x-n),&\mbox{ if } x\in[n,n+1/2]\\
2-2(x-n),&\mbox{ if } x\in[n+1/2,n+1] \end{cases}.
$$
Then $\|f_n\|\leq1$ for all $n$. Given $\varepsilon>0$, take $\delta=\varepsilon/2$. Then
$$
|f_n(x)-f_n(y)|<\varepsilon
$$
whenever $|x-y|<\delta$. So the family is equicontinuous. But $\|f_n-f_m\|_\infty=1$ if $n\ne m$. 

*Take $\Omega=[0,1]$, $f_n(x)=n$. Then the family is equicontinuous (any $\delta$ works for any $\varepsilon$!), but $\|f_n-f_m\|_\infty=|n-m|$.
A: For your second point, $(0,1)$ will work, but you'll see that any sequence of functions has a subsequence converging uniformly on any compact $K \subset (0,1)$; therefore what you'll be looking for is some nonuniformity at the endpoints $\{0,1\}$. 
An easier counterexample is to set $\Omega = \mathbb{R}$ and then use the sequence
$$
f_n(x) = \begin{cases}0 & x \leq n \\
x - n & n < x \leq n+1 \\
1 & x > n+1 \end{cases}
$$
The sequence converges pointwise to the zero function in the limit as $n \rightarrow \infty$, converges uniformly on compacts $K \subset \mathbb{R}$, but converges strictly nonuniformly on all of $\mathbb{R}$.
