Given convex l.s.c. function $f$, find decreasing convex function $\phi$ such that $f(x) \equiv \sup_y x\phi(y)-\phi(-y)$

Question

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

Question

• 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.

Examples

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

Observations

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 $g = T_\phi$.

Answer 1

$\newcommand{\R}{\mathbb{R}} \newcommand{\tto}{\underset{\text{onto}}\to}$ Let us answer the reformulated question: given a convex function $g\colon C\to\R$, when is it possible to find a decreasing convex function $\phi\colon\R\to\R$ such that \begin{equation} \phi\circ(-\phi^{-1})=g? \tag{1} \end{equation} Here $C$ is a convex subset of $\R$, that is, an interval with some endpoints $c$ and $d$ such that $-\infty\le c<d\le\infty$ (excluding the trivial case when $c=d$).

First here, note that any convex function from $\R$ to $\R$ is continuous. So, the generalized inverse $\psi:=\phi^{-1}$ of a decreasing convex function $\phi\colon\R\to\R$ is just a regular inverse, and the function $\psi=\phi^{-1}$ is also decreasing and convex.

Also, (1) implies that $\psi=\phi^{-1}$ is defined only on $C$, so that we may write $\phi\colon\R\tto C$ (rather than $\phi\colon\R\to\R$) and $\psi=\phi^{-1}\colon C\tto\R$, where $\tto$ means that the map is onto. So, again in view of (1), we may write
\begin{equation} g=\phi\circ(-\phi^{-1})\colon C\tto C. \tag{1.00} \end{equation} Moreover, because the function $\phi\colon\R\tto C$ is decreasing and convex, we see that necessarily \begin{equation*} C=(c,d)\text{ for }d=\infty\text{ and some}\ c\in[-\infty,\infty). \tag{1.0} \end{equation*}

Next, (1) can be rewritten as \begin{equation} \psi\circ g=-\psi, \tag{1a} \end{equation} which implies $\psi\circ g\circ g=-\psi\circ g=\psi$ and hence \begin{equation} g\circ g=\text{id}_C, \tag{2} \end{equation} where $\text{id}_C$ is the identity map of $C$; this simple observation is crucial.

In particular, $g$ is one-to-one. Since $g$ is convex and l.s.c., it follows that $g$ is continuous. Therefore and because $g$ is one-to-one, we see that either $g$ is increasing on $C$ or decreasing on $C$. The convexity of $g$ also implies the existence of the left and right derivatives $g'_-(x)$ and $g'_+(x)$ at any point $x\in(c,d)$.

Consider first the case when $g$ is increasing on $C$. Then the identity $g(g(x))=x$ for $x\in C$ implies $g'_-(g(x))g'_-(x)=1$ and $g'_+(g(x))g'_+(x)=1$ for all $x\in(c,d)$. So, if $g'_-(a)<g'_+(b)$ for some $a$ and $b$ such that $c<a\le b<d$, then $g'_-(g(a))>g'_+(g(b))$, which latter contradicts the convexity of $g$ (which implies that $g'_-(x)\le g'_+(y)$ for all $x$ and $y$ such that $c<x\le y<d$). So, here we have $g'_-(a)=g'_+(b)$ for all $a$ and $b$ such that $c<a\le b<d$. Thus, if the function $g$ is increasing on $C$, it must be affine. It is then easy to see that the condition $g\circ g=\text{id}_C$ implies $g=\text{id}_C$. So, in this case the only solution of equation (1a) is $\psi=0$, which is not a decreasing function.

It remains to consider the case when $g\colon C\tto C$ is decreasing and convex. Then $g(c+)=d$ and $g(d+)=c$, and $g$ is a bijection of $C$. Since $g$ is continuous and $c<d$, the equation \begin{equation} g(z)=z \tag{3} \end{equation} has a unique root $z\in(c,d)$.

Since $g$ is decreasing and convex, we also have $g'_+(g(x))g'_-(x)=1$ for all $x\in(c,d)$. Hence, $g'_+(z)g'_-(z)=1$. Also, $g'_-(z)\le g'_+(z)\le0$. Therefore, \begin{equation} g'_-(z)\le-1. \tag{4} \end{equation}

For $x\in C$, let now \begin{equation*} \psi(x):=\left\{ \begin{aligned} z-x&\text{ if }x\ge z,\\ g(x)-z&\text{ if }x\le z. \end{aligned} \right. \tag{5} \end{equation*} This definition is valid, because, in view of (3), $z-x=0=g(x)-z$ if $x=z$. Next, $\psi$ is decreasing, since $g$ is decreasing.

Further, $\psi$ is obviously convex on $[z,\infty)\cap C$, and $\psi$ is convex on $(-\infty,z]\cap C$ -- because $g$ is convex. Also, in view of (4), $\psi'_+(z)=-1\ge g'_-(z)=\psi'_-(z)$. So, $\psi$ is convex on $C$.

Moreover, (i) in view of (1.0), $\psi(d-)=\psi(\infty-)=-\infty$ and (ii) in view of (1.00), $\psi(c+)=g(c+)-z=d-z=\infty$, so that $\psi\colon C\tto\R$ and hence $\phi=\psi^{-1}\colon\R\tto C$.

Finally, recalling that $g$ is decreasing and (2) holds, it is straightforward to check that (1a) holds for the so-defined $\psi$.

Thus, for any l.s.c. convex function $g\colon C\to\R$, the following two conditions are equivalent to each other:

(I) there exists a decreasing convex function $\phi\colon\R\to\R$ such that (1) holds;

(II) $C$ is as in (1.0) and $g$ is a decreasing involution of $C$ -- that is, $g$ is a decreasing bijection of $C$ satisfying condition (2).

One may also note that, if condition (II) holds and if $\psi$ is given by (5), then for $\phi=\psi^{-1}$ and all real $y$ we have \begin{equation*} \phi(y)=\left\{ \begin{aligned} z-y&\text{ if }y\le0,\\ g^{-1}(z+y)&\text{ if }y\ge0. \end{aligned} \right. \end{equation*}