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Let $F:[0,\infty) \to [0,\infty)$ be a $C^1$ function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $[0,1]$. Suppose also that $F|_{(1-\epsilon,1+\epsilon)}$ is strictly convex for some $\epsilon>0$.

Let $QF$ be the convex envelope of $F$. Does $QF$ have the same monotonicity properties as $F$?

i.e. is $QF$ strictly increasing on $[1,\infty)$, and strictly decreasing on $[0,1]$?

Asaf Shachar
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  • Do you mean the Moreau envelope (i.e. the result of infimal convolution with $|\cdot|^2 / 2$) or the bidual/biconjugate (the result of taking the Fenchel-Legendre conjugate twice)? – Zim Aug 28 '20 at 16:28
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    I meant for the largest convex function which does not exceed $F$. I think this is the second option you have mentioned. If you have any good reference on that topic, that would be great. – Asaf Shachar Aug 28 '20 at 16:52
  • It's not discussed much in Bauschke/Combettes. You might try Rockafellar/Wets, but I'm not really sure, sorry. – Zim Aug 29 '20 at 03:44
  • By the way, , I think these properties may imply that $F$ is quasiconvex (i.e. it has convex lower-level sets). – Zim Aug 29 '20 at 03:46
  • Since I wrote my answer it bothered me if the convexity near $x=1$ is actually needed. As it turns out, it is not needed, and even continuity is not needed. I have rewritten the answer accordingly. – Martin R Aug 30 '20 at 11:13

1 Answers1

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Your conjecture is almost correct. The convex envelope of $F$ is strictly decreasing on the (bounded) interval $[0, 1]$, and strictly increasing or identically zero on the (unbounded) interval $[1, \infty)$. $F$ need not even be continuous for this conclusion, and the strict convexity near $x=1$ is also not needed.

In the following, $\hat F$ denotes the (lower) convex envelope of $F$, that is $$ \hat F(x) = \sup \{ h(x) \mid \text{$h: \operatorname{dom}(F) \to \Bbb R$ is convex}, h \le F \} \, . $$

Then we have the following result for unbounded intervals:

Let $F: [a, \infty)\to \Bbb R$ be strictly increasing with $F(a) = 0$. Then $\hat F$ is identically zero or strictly increasing on $[a, \infty)$.

And for bounded intervals:

Let $I = [a, b]$ or $I = [a, b)$ and $F: I \to \Bbb R$ be strictly increasing with $F(a) = 0$. Then $\hat F$ is strictly increasing on $I$.

For a proof of the statement about unbounded intervals we distinguish two cases:

Case 1: $\liminf_{x \to \infty} F(x)/x = 0$. Then also $\liminf_{x \to \infty} G(x)/x = 0$, and that implies that $G$ is identically zero on $[a, \infty)$.

Case 2: $\liminf_{x \to \infty} F(x)/x > 0$. Then $F(x) > cx$ for some constant $c > 0$ and $x \ge x_1 > a$.

For $x_0 \in (a, x_1)$ set $m = \min(c, \frac{f(x_0)}{x_1 - x_0})$ and consider the function $h(x) = m(x-x_0)$. $h$ is convex with $h \le F$, so that $\hat F(x) \ge h(x) > 0$ on $(x_0, \infty)$.

Since $x_0$ can be arbitrarily close to $a$ it follows $\hat F(x) > 0$ on $(a, \infty)$. This implies that $\hat F$ is strictly increasing on $[a, \infty)$.

A similar reasoning as in the second case can be used to prove the statement about bounded intervals.

Martin R
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  • Thanks, this seems very nice. Just to be sure: what is your argument in case $1$ that $G=0$? Is it something in the following spirit: $G$ is convex --> it's above its tangents, and if there is a point $x>1$ where $G(x)>0$ then some tangent should have a strictly positive slope, thus $G$ should be super-linear at infinity, which is a contradiction. Was that what you had in mind, or did you have a different argument? I guess everything is easier if $G$ is $C^1$ (since then we don't need to be careful with distinguishing between one-sided derivatives...). – Asaf Shachar Aug 28 '20 at 18:12
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    @AsafShachar: That is it essentially, but you don't need the tangent or a derivative. If $G(y) > 0$ then $G$ lies above the line through $(1, 0)$ and $(y, G(y))$ for $x > y$ and that implies $\liminf_{x \to \infty} G(x)/x > 0$. – Alternatively, write down the convexity condition for $1 < y < x$ with fixed $y$ and see what happens for $x = x_n \to \infty$ with $\lim_{n \to \infty} G(x_n)/x_n = 0$. – Martin R Aug 28 '20 at 18:30
  • This is a really neat argument! I'm curious -- studying $\lim_{x\to+\infty} F(x)/x$ is closely related to the notion of supercoercivity. Did that play a role in how you came upon this argument? – Zim Aug 29 '20 at 03:53
  • @Zim: Unfortunately I am not familiar with the concept of supercoercivity, so it certainly did not play a role and I cannot say if it is somehow related. – What I did was to investigate under which conditions a non-negative convex function is identically zero, and that lead to considering that limes inferior. The difficult part was case 2, in other words to show that $G$ can not be zero only on some interval $[1, y]$. – Martin R Aug 29 '20 at 07:19
  • @MartinR Cool! Supercoercivity is usually a notion used to show existence of a minimizer, so I was surprised to see that limit appear in the argument. Anyways, thanks for explaining :) – Zim Aug 29 '20 at 15:07