Free subgroups of piecewise linear homeomorphisms of the circle For convenience, I recall the definition of piecewise linear homeomorphism:

A homeomorphism $f$ of the real line $\mathbb{R}$ is called piecewise linear if there is an increasing sequence of real numbers $x_i$ parametrized by $i \in \mathbb{Z}$ such that $\lim\limits_{i \to \pm \infty} x_i =\pm \infty$ and such that the restriction of $f$ to each interval $[x_i,x_{i+1}]$ coincides with an affine map. If such a homeomorphism satisfies $f(x+1)=f(x)+1$ for all $x$, then it induces a homeomorphism of the circle $\mathbb{S}^1 \simeq \mathbb{R} / \mathbb{Z}$. Such a homeomorphism of $\mathbb{S}^1$ is called a piecewise linear homeomorphism of the circle.

The group of orientation-preserving piecewise linear homeomorphisms is denoted by $PL_+$.
In his document Groups acting on the circle, Etienne Ghys showed that $PL_+([0,1])$ does not contain any non-abelian free group (theorem 4.6). Then, he precises that the result does not hold for $PL_+(\mathbb{S}^1)$ and moreover that it is very easy to find two piecewise linear homeomorphisms of $PL_+(\mathbb{S}^1)$ generating a free subgroup of rank two using ping-pong lemma.
However, I tried to find such a pair of homeomorphisms and I did not find that so easy... Do you see a nice argument making the remark easy?
 A: Example. For convenience, I will use the group of even integers to define the quotient circle: $S^1= R/2Z$. This does not affect anything. Start with an odd PL map $f: [-1, 1]\to [-1, 1]$ which fixes $-1, 0, 1$, is linear on the four intervals $\pm [0, 10^{-2}], \pm [10^{-2}, 1]$, and satisfies
$$
f(\pm 10^{-2})= \pm (1- 10^{-2}). 
$$
The map $f$ clearly descends to a PL homeomorphism $h_1: S^1\to S^1$ (since we form $S^1$ by identifying the end-points of the interval $[-1,1]$. Now, every rotation of $S^1$ is the projection of the translation of $R$ and, hence, is "linear". Now, let $T$ denote the order 4 rotation of $S^1$ (it lifts to the translation by $0.5$ of the real line) and define
$$
h_2= T h_1 T^{-1}. 
$$ 
Then the group generated by $h_1, h_2$ is free of rank 2. This is a direct application of the ping-pong argument on the circle. 
A: I finally found a solution: 
It is known that $\left( \begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix} \right)$ and $\left( \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} \right)$ generate a free subgroup of rank two in $SL_2(\mathbb{Z})$; on the other hand, $SL_2(\mathbb{Z})$ acts naturally on $\mathbb{R}P^1 \simeq \mathbb{S}^1$. It leads to introduce
$$f(x)= \left\{ \begin{array}{cl} \frac{3}{2}x & \text{if} \ 0 \leq x \leq \frac{1}{3} \\ x+ \frac{1}{6} & \text{if} \ \frac{1}{3} \leq x \leq \frac{2}{3} \\ \frac{2}{3} x + \frac{1}{3} & \text{if} \ \frac{2}{3} \leq x \leq 1 \end{array} \right. \ \text{and} \ g(x)= \left\{ \begin{array}{cl} x+ \frac{2}{3} & \text{if} \ 0 \leq x \leq \frac{1}{3} \\ 3x-1 & \text{if} \ \frac{1}{3} \leq x \leq \frac{2}{3} \\ \frac{1}{3}x + \frac{1}{3}  & \text{if} \ \frac{2}{3} \leq x \leq 1 \end{array} \right..$$
Now it is sufficient to apply ping-pong lemma to $f^2$ and $g^2$ on $\left(0, \frac{1}{3} \right) \cup \left( \frac{2}{3} , 1 \right)$ and $\left( \frac{1}{3}, \frac{2}{3} \right)$.
In fact, it is readable directly on the graphs of $f$ and $g$:

