= Inverse-flux quadratic bounds
{title2=$k_-(z-c)^2\le g(z)\le k_+(z-c)^2$}
Suppose $f(0)=0$, $f'(0)=c$, $h=(f')^{-1}$ is differentiable, and $k_-<h'/2<k_+<0$. Then $h(c)=g(c)=0$ for the <concave Legendre dual> $g$, and
$$
k_-\le\frac{h(z)}{2(z-c)}\le k_+\quad(z\ne c),\qquad k_-(z-c)^2\le g(z)\le k_+(z-c)^2.
$$
Integrate the derivative bounds from $c$ to $z$, then integrate $g'=h$. Multiplying the first inequality by a negative $z-c$ reverses both comparisons. A bound valid without cases is $|h(z)|\le-2k_-|z-c|$.
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