Let $f: \mathbb R \rightarrow (-\infty,+\infty]$ be a lower-semicontinuous convex function.


  • Under what futher conditions does there exists a convex decreasing function $\phi: \mathbb R \rightarrow \mathbb R$ such that $f(x) = \sup_y x\phi(y)-\phi(-y)$ for all $x \in \mathbb R$ ?
  • Construct such a $\phi$ explicitly.


$f(x) = -2\sqrt{x}$ if $ x \ge 0$ and $f(x)=+\infty$ else, then one may take $\phi(x) \equiv e^{-x}$.


Given a decreasing function $\phi$, define its (pseudo-)inverse by $\phi^{-1}(x):=\inf\{t \in \mathbb R \mid \phi(t) \le x\}$, with the usual convention that $\inf\emptyset = +\infty$. Let $T_\phi(x) := \phi(-\phi^{-1}(x))$, for all $x \in \mathbb R$. Then, with the change of variable $z := \phi(y)$ one has

$$ \sup_y x\phi(y)-\phi(-y)=\sup_zxz-\phi(-\phi^{-1}(z)) := T_\phi^*(x), $$ where $T_\phi^*$ is the convex conjugate of $T_\phi$. This suggests that one should consider the following problem rather:

Reformulated problem. Given a convex l.s.c function $g$, find decreasing convex function $\phi$ such that $g = T_\phi$.


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