Concave Legendre dual ($g(z)={\inf }_s(zs-f(s))$)

For a differentiable strictly concave function $f$ whose derivative is a bijection of $\mathbb{R}$, defineThe unique minimizing point is $h(z)$, and $g^{\prime }=h$ follows by comparing minimizers at adjacent arguments, even when $h$ is not differentiable. This dual is concave and equals $-(-f{)}^{\ast }(-z)$ in terms of the convex conjugate. If $f^{\prime \prime }<0$, then $g^{\prime \prime }=1/f^{\prime \prime }(h)$. For $f(s)=cs-s^2/2$, $g(z)=-(z-c{)}^2/2$.

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