Showing a subset of $C([0,1])$ is compact. 
Let $${\cal F}=\left\{ f:\left[0,1\right]\to\mathbb{R} : \left|f\left(x\right)-f\left(y\right)\right|\le\left|x-y\right|\mbox{ and }{\displaystyle \int_{0}^{1}f\left(x\right)dx=1}\right\}.$$  Show that ${\cal F}$  is a compact subset of $C\left(\left[0,1\right]\right)$.

When I am trying to show a set is compact, I usually resort to the every open cover has a finite subcover definition.  But if this case, we are dealing with functions.  So I am having a difficulty "visualizing" what's going on. Any help or solutions would be appreciated.
Edit: I should mention that we are working with respect to the sup norm.
 A: According to Arzelà–Ascoli theorem you only have to show that $\mathcal F$ is


*

*equicontinuous, i.e. $(\forall x\in[0,1])(\forall \varepsilon>0)(\exists \delta>0)(\forall f\in\mathcal F) (\forall y) |y-x|<\delta \Rightarrow |f(y)-f(x)|<\varepsilon$;

*pointwise bounded;

*closed in $C[0,1]$.


Both equicontinuity and pointwise boundedness follow from the Lipschitz condition $|f(x)-f(y)|\le |x-y|$.
To show pointwise boundedness you can notice that $|f(x)-f(0)|\le |x|=x$, which means
$$f(0)-x \le f(x) \le f(0)+x.$$
If you apply integral $\int_0^1$ to the left inequality, you get $f(0)\le\int_0^1 (x+f(x))\,\mathrm{d}x=\frac32$. Now the right inequality implies
$f(x)\le \frac52$ for each $x$. 
(Thanks to Nate Eldredge, who pointed in his comment, that this was missing in my original answer.) 
To show that it is closed in sup-norm, you only have to show that if $f_n$ converges to $f$ uniformly and $f_n\in\mathcal F$, then the limit is in $\mathcal F$. 
We know that integral behaves well w.r.t. uniform convergence, see this questions. Proof of the fact that the condition $(\forall x,y\in [0,1])|f(x)-f(y)|\le |x-y|$ is preserved by uniform convergence is more-or-less standard. (In fact, in this part we only use pointwise convergence.)
