= Solution
\b[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 $3/4\le1/2$.
The correct <inverse-flux quadratic bounds> are
$$
\boxed{\begin{aligned}
k_-(z-c)&\le h(z)/2\le k_+(z-c)&& (z\ge c),\\
k_+(z-c)&\le h(z)/2\le k_-(z-c)&& (z\le c),\\
k_-(z-c)^2&\le g(z)\le k_+(z-c)^2&& (z\in\mathbb R).
\end{aligned}}
$$
Indeed $h(c)=0$ and $h(z)/2=\int_c^z h'(r)/2\,dr$; 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 $g'=h$, integration once more gives the quadratic inequalities on both sides of $c$. A useful sign-independent consequence is $|h(z)|\le-2k_-|z-c|$.
Back to article page
