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I need help with the following problem, I have no idea how to proceed:

Let $u \colon \Omega \subseteq \mathbb{R}^N \to \mathbb{R}$ a continuous function, where $\Omega$ is open, connected and bounded. We define the set

$$\Gamma_u^+ = \{ y \in \Omega \colon (\exists p \in \mathbb{R}^N) \ (\forall x \in \Omega) \ u(x) \leq u(y) + \langle p,x-y \rangle\}$$ Prove that $f$ is concave iff $\Gamma_u^+ = \Omega$.

Any help will be appreciated.

yes
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    $p$ is an element of $\mathbb{R}^n$, right? Since you're taking $\langle p, x-y\rangle$. And when you said $f$, you mean $u$? Or am I confusing myself? – Vinícius Novelli Oct 14 '14 at 22:49
  • yes, thanks for the correction. – yes Oct 15 '14 at 00:01
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    Looks like you do not assume that $\Omega$ is a convex domain. How do you define a concave function, then? [What to do if $a,b\in \Omega$ but the line segment between them goes outside of $\Omega$?] –  Oct 15 '14 at 01:22
  • You can assume that $\Omega$ is convex but is easy to define the concept of convex/concave functions on non convex domains. We said that $f \colon \Omega \to \mathbb{R}$ is convex if for all $x,y \in \Omega$ such that $[x,y] \subseteq \Omega$ and all $t \in (0,1)$ we have that $f(tx + (1-t)y) \leq tf(x)+ (1-t)f(y)$. Thanks for your comment. – yes Oct 15 '14 at 16:07
  • Your set $\Gamma_u^+$ is the domain of the subdifferential of $-u$. Directly, if $u$ is concave then $-u$ is subdifferentiable on $\Omega$ since $\Omega$ is open (you can find this result in Rockafellar "Convex Analysis"). Conversely, I forsee a slight problem with your definition of concave function, but if $\Omega$ is convex you will find the result in the before mentioned book. – Neutral Element Nov 02 '14 at 08:23

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