# Compactness in $C([0,1])$

I read in a paper (pp8) that the set $$A=\left\{f\in W^{1,1}(0,1):\sup |f|\leq C,\ \int_0^1|f'(t)|dt\leq M \right\}$$ is compact in $C([0,1])$, where $C$ and $M$ are fixed constants. I understand that $W^{1,1}(0,1)$ denotes in this situation the space of absolutely continuous functions. In order to apply the Arzela Ascoli Theorem, I would require uniform boundedness which is given, but to prove equicontinuity, a bound on $f'$ would have been nicer. I do not see how to use the second condition to prove equicontinuity. Could someone give a hint.

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Let $f_n$ such that $f_n(x)=0$ if $0\leq x\leq 1/2-1/n$ or $x\geq 1/2+1/n$, $f_n(1/2)=1$ and $f_n$ is piecewise linear. We can approximate these functions by smooth ones with compact support. The set $\{f_n,n\in\Bbb N\}$ is not compact in $C[0,1]$, unless I'm missing something. –  Davide Giraudo Nov 5 '12 at 8:54
@DavideGiraudo It seems to me that $\int_0^1 |f'_n(t)|\, dt \sim n$. –  Siminore Nov 5 '12 at 9:02
Sorry, I was computing the integral of $f_n$... –  Siminore Nov 5 '12 at 9:07

Let $$f_n(x):=\begin{cases} 0&\mbox{if }0\leq x\leq \frac 12-\frac 1n\mbox{ or }\frac 12+\frac 1n\leq x\leq 1;\\ 1&\mbox{if }x=\frac 12;\\ \mbox{piecewise linear}.& \end{cases}$$ We have $0\leq f_n(x)\leq 1$ for all $x$. This function is weakly differentiable, of weak derivative $n\chi_{(1/2-1/n,1/2)}-n\chi_{(1/2,1/2+1/n)}$, so $\int_{(0,1)}|f'_n(x)|dx=2$.
But $\{f_n,n\in\Bbb N\}$ is not relatively compact for the uniform norm as each subsequence converge pointwise to $\chi_{\{1,2\}}$, a discontinuous function, hence cannot converge uniformly.
An other way to see that is to note that this set is not equi-continuous at $1/2$.