Solution
= Solution
On $(0,1)$ the map is positive. Its maximum occurs at $x=1/\sqrt3$ and equals $2\lambda/(3\sqrt3)$. Thus if $\lambda<3\sqrt3/2$, then $f((0,1))\subset(0,1)$. Since $f$ is odd, also $f((-1,0))\subset(-1,0)$. Induction gives
$$
\boxed{\operatorname{sgn}x_n=\operatorname{sgn}x_0\quad\text{for }|x_0|<1.}
$$
The case $x_0=0$ is constant. These intervals are <positively invariant sets> for the discrete <dynamical system>.