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Let $G:\mathbb R\to\mathbb R$ be a concave function, define $G_{\epsilon}: \mathbb R\to\mathbb R$ by

$$G_{\epsilon}(x)~~:=~~\max_{y\in [x-\epsilon, x+\epsilon]}G(y).$$

My question is the following: Is there some $h:\mathbb R\to \mathbb R$ s.t.

$$G_{\epsilon}(x)~+~h(x)(y-x)~-~\epsilon|h(x)|~~ \ge~~ G(y)\mbox{ for all } x, y \in\mathbb R?$$

PS: This question is related to the question of the following link

Generalization of concave envelope

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    $\begingroup$ Say, for $G(x)=-x^2$ such $h$ does not exist (for $x=y=0$ we get $h=0$, but it does not work either.) $\endgroup$
    – Fedor Petrov
    Commented Nov 4, 2016 at 14:48
  • $\begingroup$ @FedorPetrov Thanks a lot for the reply. Indeed, there is a mistake in my question: it should be $h(x)$ instead of $h$. I reformulated the question. $\endgroup$
    – CodeGolf
    Commented Nov 4, 2016 at 15:13
  • $\begingroup$ @FedorPetrov If $\epsilon=0$, it turns to prove that $G(x)+h(x)(y-x)\ge G(y)$ and taking $h$ to be the derivative of $G$ will work, i.e. $h(x)=G'(x)$. $\endgroup$
    – CodeGolf
    Commented Nov 4, 2016 at 15:13

1 Answer 1

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Yes, this is true. There are three cases.

1) $G_\varepsilon(x)=G(x+\theta)$ for some $\theta\in (-\varepsilon,\varepsilon)$. Then $G$ has a local, thus global, maximum at $x+\theta$, so we may take $h(x)=0$.

2) $G_\varepsilon(x)=G(x+\varepsilon)$. Then the left derivative of $G$ at a point $x+\varepsilon$ is non-negative. Denote it by $h$, the inequality follows from the fact that the graph of $G$ lies below the tangent line.

3) $G_\varepsilon(x)=G(x-\varepsilon)$, analogous to the case 2.

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