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Let $F:(a,b) \to [0,\infty)$ be a $C^1$, and let $\hat F$ be the (lower) convex envelope of $F$, i.e. $$ \hat F(x) = \sup \{ h(x) \mid \text{$h$ is convex on $[0, \infty)$}, h \le F \} \, . $$ $\hat F$ is a non-negative convex function.

Is $\hat F$ also $C^1$? Is it everywhere differentiable?

Does the answer change if $F$ is monotone?

Asaf Shachar
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1 Answers1

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If $F:(a,b) \to \Bbb R$ is bounded below (so that the convex envelope exists) and differentiable then its convex envelope $\hat F$ is continuously differentiable, i.e. a $C^1$ function. (The continuity of $F'$ is not needed for this conclusion.)

More generally one can show that if $F:(a,b) \to \Bbb R$ is bounded below, and differentiable at a point $c \in (a, b)$ then $\hat F$ is also differentiable at $c$.

This implies the desired conclusion: If $F$ is differentiable on $(a, b)$ then the same holds for $\hat F$. The derivative of a convex function is (weakly) increasing and cannot have jump discontinuities (because of Darboux's theorem). It follows that the derivative of $\hat F$ is continuous (see also Continuity of derivative of convex function).

In the following I'll write $G = \hat F$ for the convex envelope of $F$ (to save some keystrokes, and because $\hat{F}'$ renders ugly in MathJax).

Assume that$F$ is differentiable at $c$, but $G$ is not differentiable at $c$. Then $G_-'(c) < G_+'(c)$ where $G_-'$ and $G_+'$ denote the left and right derivative of $G$. Define $$ h(x) = \begin{cases} G(c) + G_-'(c)(x-c) & \text{ if } a < x \le c \, ,\\ G(c) + G_+'(c)(x-c) & \text{ if } c \le x < b \, . \end{cases} $$ Then $F(x) \ge G(x) \ge h(x)$ for all $x$.

Next show that $F(c) = G(c)$. Otherwise for sufficiently small $\epsilon$ the function $h_\epsilon$ obtained by replacing $h$ with a straight line segment on $[c-\epsilon, c+\epsilon]$ still satisfies $F(x) \ge h_\epsilon(x)$ for all $x$, but $h_\epsilon(c) > G(c)$. This contradicts the maximality of $G$.

It follows that $$ F(x) \ge \begin{cases} F(c) + G_-'(c)(x-c) & \text{ if } a < x \le c \\ F(c) + G_+'(c)(x-c) & \text{ if } c \le x < b \end{cases} $$ and that implies $$ F_-'(c) \le G_-'(c) < G_+'(c) \le F_+'(c) $$ in contradiction to the assumption that $F$ is differentiable at $c$.

Martin R
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  • Thanks. I really appreciate the fact that you give such good answers. This is not the first time you are helping me. Great answer. – Asaf Shachar Aug 29 '20 at 12:28
  • BTW, I think that your proof also implies the following: If $F(c)=G(c)$, and $F$ (and $G$) are differentiable at $c$, then $F'(c)=G'(c)$. Am I right? – Asaf Shachar Aug 29 '20 at 13:53
  • @AsafShachar: Generally, if $f$ and $g$ are differentiable functions with $f \ge g$ and $f(c) = g(c)$ then $f'(c) = g'(c)$. That is because the difference $h= f-g$ has a minimum at $x=c$. – Martin R Aug 29 '20 at 14:02
  • Yes, of course you are right! I actually thought on that reasoning myself a while ago, but then I didn't know that the convex envelope was differentiable, so things were more cumbersome with one-sided derivatives etc... – Asaf Shachar Aug 29 '20 at 14:31