Equivalent of $\ln\ln(N(\epsilon))$ where $N(\varepsilon)$ is the minimum of balls for covering $A$. 
Let $E=\mathcal{C}^0([0,1],\mathbb{R})$ with the uniform convergence and $$A=\{f\in E\ |\ f(0) = 0\text{ and } \forall x,y\in[0,1]\ |f(x) - f(y)| \leqslant |x-y|\} $$
For $\varepsilon >0$ denote by $N(\varepsilon)$ the minimum number of balls of radius $\epsilon$ necessary for covering $A$.
Give an equivalent of $\ln\ln(N(\epsilon)$ when $\varepsilon \rightarrow 0$.

(ENS oral) 
My attempt :
Let $B$ the unit open ball,
If $B \subset \bigcup_{n=1}^{N}B(x_n,\varepsilon)$ we have 
$$
vol(B)<\sum_{n=1}^{N}vol(B(x_n,\varepsilon))<N\varepsilon^nvol(B)
$$
So $N\ge \frac{1}{\varepsilon^n}$.
Then I tried to build a covering of $N$ balls,
We choose $x_1 \in B$ , if $B\subset B(x_1,\varepsilon)$ we stop, if not we choose $x_2\in B$ such that $\vert\vert x_2-x_1\vert\vert > \varepsilon$
It seems we have an induction here, 
Assume we have construct $x_1,\cdots,x_{n-1} \in B$ with  $\vert\vert x_i-x_j\vert\vert > \varepsilon$ for $i\ne J$. If $B\subset B(x_j,\varepsilon)$ we stop. If not we choose $x_n\in B$ such that $\vert\vert x_n-x_i\vert\vert > \varepsilon$ for $i<n$.
Unfortunataly I don't see how can I continue, I wish I Had a frame.
 A: For the purposes of asymptotics, it suffices to consider the case $\epsilon = 1/n$. Consider the set $PL(\epsilon)$ piecewise linear functions such that $f(0)=0$ and on each interval $[(k-1)\epsilon, k\epsilon]$ the derivative $f'$ is either $\equiv 1$ or $\equiv -1$. There are $2^n$ such functions. I claim that for every $f\in A$ there is $g\in PL(\epsilon)$ such that $\|f-g\|\le  \epsilon$. 
The definition of $g$ is inductive. Let $g(0)=0$. For $k=1,\dots,n$ let $g(k\epsilon)$ be one of the numbers $g((k-1)\epsilon)\pm \epsilon$ that is closer to $f(k\epsilon)$. If they are equidistant, take either. 
Now let's check that $|g(k\epsilon)-f(k\epsilon)|\le \epsilon$ for all $k$, again by induction. The base case is clear. Assuming $|f-g|$ is at most $\epsilon$ at $(k-1)\epsilon$, we have $$|f(k\epsilon) - g((k-1)\epsilon)|\le 2\epsilon$$ The  choice of $g(k\epsilon)$, according to the definition, brings the difference $|f(k\epsilon) - g(k\epsilon)|$ down to at most $\epsilon$. 
I'll leave it for you to check that $|f(x)-g(x)|\le \epsilon $ holds for all $x$ (which is true); of course you may be satisfied with the straightforward bound $|f(x)-g(x)|\le 2\epsilon $. 
For the lower bound, it suffices to note that any two elements of $PL(\epsilon)$ are at distance $\ge \epsilon$ from each other. Hence, $N(\epsilon/2)\ge 2^n$.
In conclusion, $$\lim_{\epsilon\to 0}\frac{\ln \ln N(\epsilon)}{ \ln(1/\epsilon)}=1 $$
