Let $f:[0,1] \rightarrow [0,1]$ be a twice continuously differentiable increasing function such that $f(0) = 0$ and $f(1) = 1$. There is a unique $q \in (0,1)$ such that $f''(x) > 0$ for all $x <q$ and $f''(x) < 0$ for all $x >q.$ It is also given that there is a unique unstable interior fixed point $\overline{p} \in (0,1)$ i.e., $f(\overline{p}) = \overline{p}$ and $f'(\overline{p}) > 1.$

One can show that the function $g(x)=1-f(x)$ has a unique interior fixed point $r \in (0,1)$ i.e., $g(r) = r.$ Can we conclude that $|g'(r)| > 1$?



You must log in to answer this question.

Browse other questions tagged .