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

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 $$g = T_\phi$$.

$$\newcommand{\R}{\mathbb{R}} \newcommand{\tto}{\underset{\text{onto}}\to}$$ 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 $$-\infty\le c (excluding the trivial case when $$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 $$g'_-(a) for some $$a$$ and $$b$$ such that $$c, 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 $$c). So, here we have $$g'_-(a)=g'_+(b)$$ for all $$a$$ and $$b$$ such that $$c. 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 $$c, the equation $$\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*}

• Thanks. Your answer only deals with convex functions with full domans (in the convex analytical sense), i.e for which $g(x) < \infty$ for all $x$. Would you mind extending your answer to extended convex functions as in the question ? Also, what do you mean by $-\infty\pm$ ? – dohmatob Sep 22 '19 at 12:42
• @dohmatob : I have now provided the requested modification, allowing $g$ to be defined on any convex subset of $\mathbb R$. As for $g(c+)$, it means $\lim_{x\downarrow c} g(x)$, as usual. Similarly defined is the left limit $g(d-)$. – Iosif Pinelis Sep 22 '19 at 15:27
• Thanks very much for your very detailed and instructive answer! – dohmatob Sep 22 '19 at 19:26
• As a supplement to this answer, let me add that if we define $U_C:=\{c-z \mid z \le c \in C\}$, then a $\phi$ which does the job is $\phi(x) = \psi^{-1}(x)= \begin{cases}z-x,&\mbox{ if }-x \in U_C,\\g^{-1}(z+x),&\mbox{ if }x \in U_C,\\+\infty,&\mbox{ else.}\end{cases}$ – dohmatob Sep 22 '19 at 22:34
• Oops! Apparently I forgot to save the added explicit expression for that $\phi$. Have done this now. – Iosif Pinelis Sep 23 '19 at 13:40