Let $f:[0,1]\rightarrow \mathbb {R} $ a continuous function s.t. $f (0)=0$ and $\frac {x}{2}\leq f (x) $.
Show that for every $\alpha <1$ there exists $n\geq 1$ a positive integer s.t. $$[0,\alpha]\subseteq (f\circ f\circ...\circ f)([0,\alpha]) $$ where the function $f $ is composite for $n $ times.
I don't understand the requirement. If $[0,\alpha]\not\subset f ([0,\alpha]) $ then $f ([0,\alpha])\subseteq [0,1] $ because I have to compute $f\circ f $.
My idea to start the proof: Because $f $ is continuous on $[0,1] $ then $f$ is bounded and it can reach its bounds.
Take $f(x)=x/2$ and $\alpha=3/4$, the statement is not true. Can you double check that this is indeed the question asked?