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I'm trying to solve the following problem I found in a convex analysis book but I don't know how to proceed:

Let $f: C \longrightarrow \mathbb{R}$ be a convex function over the convex set $C \subset \mathbb{R}^n$:

  1. If $f$ is bounded above on $C$ then $f$ is constant.

  2. If $(y;\alpha) \in O^+(\mathrm{epi}(f)) \cap \{-O^+(\mathrm{epi}(f))\}$ then $$f(x+\lambda y)=f(x)+\lambda \alpha, \forall x \in \mathbb C, \lambda \in \mathbb{R}$$ where $\mathrm{epi}(f)$ is the epigraph of $f$ and $O^+(\mathrm{epi}(f))$ is its recession cone.

I have no clue of how to solve point 1 (I've tried to prove it by contradiction but I don't advance), and for point 2, I know how to prove from the definitions of epigraph and recesion cone that $f(x+\lambda y)\leq f(x)+\lambda \alpha$ but I cannot prove the other inequality.

I'll appreciate any ideas or suggestions. Thank you.

EDITED: The definition of epigraph is $$\mathrm{epi}(f)=\{(x; \alpha) \in \mathbb{R}^{n+1}: f(x) \leq \alpha\}$$ And, for the recession cone, given a convex set $C \subset \mathbb{R}^n$, the recession cone of $C$ is the set $$O^+(C)=\{d \in \mathbb{R}^n: x+\lambda d \in C, \forall x \in C, \lambda \geq 0\}$$

Eparoh
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1 Answers1

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This is false. Consider $ x \mapsto |x|$ on $C:= [-1,1]$. Then this function is bounded, convex and non-constant.

J. De Ro
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  • Wow, it's true! So, the second part is also false? – Eparoh May 10 '20 at 15:15
  • I can't tell. I am not familiar with the notation in (2). – J. De Ro May 10 '20 at 15:19
  • It turns out the book was using some strange notation and the set $C$ is an affine variety and so that part is easy. For the other part, I've edited the question including the definitions. – Eparoh May 11 '20 at 07:38