Application of mean value theorem on integral

Question

Let $f:[-1,1]\to \mathbb{R}$ be diferentiable function such that $F(1)=F(-1)=1$, where $F(x)=\int_0^x f(t) dt$. How to prove that $f'(c)>1$ for some $c\in (-1,1)$? By mean value theorem, we have that $F'(c)=f(c)=0$ for some $c\in (-1,1)$. But how to prove that $f'(c)>1$? Any hint is welcome. Thanks in advance.

Answer 1

Suppose $f'(x) \leq 1$ for all $x$. Let $G(x)=\frac {x^{2}} 2 -F(x)$. Then $G''(x)=1-f'(x) \geq 0$ so $G$ is a convex function . This gives $G(0)\leq \frac {G(1)+G(-1)} 2$. However, $G(0)=0$ , $G(1)=\frac 1 2 -1=-\frac1 2$ and $G(-1)=-\frac1 2$. So we end up with $0 \leq -\frac 1 2$ which is a contradiction.