We have $f:[0,1]\to R$ continuous with $f(0)=0$ and $f(x)\ge \frac{x}{2}, \forall x \in[0,1]$. We have to show that $\forall a \in[0,1), \exists n$ natural number such that $[0,a] \subseteq \underbrace{(f\circ...\circ f)}_{n\text{ times}}([0,a])$
It's easy to prove by induction that $\underbrace{(f\circ...\circ f)}_{n\text{ times}}(x) \ge \frac{x}{2^n}$. Since the function is continous and $f(0)=0$, if we can find a number $x_0$ in $[0,a]$ for which exists a natural number $n_0$ such that $\underbrace{(f\circ...\circ f)}_{n_0\text{ times}}(x_0) \ge \frac{x_0}{2^{n_0}} \ge a$ we are done. Is this the right approach? What can I do to solve it?