Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 5 / 1 / 3 / f / Solution

The printed linear ordering is false for $z<c$. For example, $f(s)=-s^2/2$, $c=0$, $h(z)=-z$, $k_{-}=-3/4$, $k_{+}=-1/4$ satisfy the derivative hypothesis, but at $z=-1$ the printed lower bound would require $3/4\le 1/2$.

The correct inverse-flux quadratic bounds areIndeed $h(c)=0$ and $h(z)/2={\int }_c^z{h}^{\prime }(r)/2dr$; reversing the integration limits reverses the linear inequalities. Equivalently, for $z\ne c$, $k_{-}\le h(z)/(2(z-c))\le k_{+}$. Since $g(c)=0$ and $g^{\prime }=h$, integration once more gives the quadratic inequalities on both sides of $c$. A useful sign-independent consequence is $|h(z)|\le -2k_{-}|z-c|$.