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 $f(s)=cs-s^2/2$, $g(z)=-(z-c{)}^2/2$.

The lower conjugate uses an infimum, unlike the supremum defining a convex conjugate. For any $F$, $F_{\ast }(\tau )=-(-F{)}^{\ast }(-\tau )$ and $F(e)\le \tau :e-F_{\ast }(\tau )$. The inequality is informative only when the lower conjugate is finite. If the infimum is attained uniquely and the dual is differentiable, $D{F}_{\ast }(\tau )$ is its minimizing argument. This sign convention is useful for upper comparison-energy bounds.

The concave biconjugate is an upper envelope of $F$, and equals $F$ for a proper closed concave function under the usual duality hypotheses. It equals $-(-F{)}^{\ast \ast }$ in terms of the ordinary biconjugate. If $e=D{F}_{\ast }(\tau )$, differentiability and concavity of $F_{\ast }$ imply $F_{\ast }(\tau )+F_{\ast \ast }(e)=\tau :e$. For a finite differentiable dual the equality follows because $\tau$ minimizes $s:e-F_{\ast }(s)$.

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 $|h(z)|\le -2k_{-}|z-c|$.