Counterexamples of Arzèla Ascoli theorem for non-obeyed criteria

Solution 1:

1. 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.

2. 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 $n\ne m$.

3. Take $\Omega=[0,1]$, $f_n(x)=n$. Then the family is equicontinuous (any $\delta$ works for any $\varepsilon$!), but $\|f_n-f_m\|_\infty=|n-m|$.

Solution 2:

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 $\mathbb{R}$.