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Let $f: \mathbb R^n \rightarrow \mathbb R \cup \{-\infty,+\infty \}$ be a function. Let $epi f$ be the epigraph of $f$:

$$ epi(f)=\{(x,r)\in \mathbb R^n\times \mathbb R:f(x)\leq r \}. $$ Is it true that if $(x,r)\in conv (epi(f))$ and $s>r$, then $(x,s)\in conv(epi(f))$?

Thanks

Richard
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  • The answer to the question in the title is no: ${\rm conv(epi}(x\mapsto-x^2))$ is $\mathbb R^2$ and therefore not the epigraph of any function. –  Feb 16 '18 at 14:49
  • If it was be assumed that $f$ can be infinite, then it would be an epigraph of $f=-\infty$. – Richard Feb 16 '18 at 15:07
  • But you said $f:\mathbb R^n\to\mathbb R$, and $-\infty\not\in\mathbb R$. At least change it to be $f:\mathbb R^n\to\mathbb R\cup{-\infty,\infty}$. –  Feb 16 '18 at 15:51

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Second attempt to your question in the post:

Let be $(x,r)\in conv(epi(f))$ and $s>r$. If $s\geq f(x)$ or $r\geq f(x)$ its obvious, so let be $r<s<f(x)$. Then $conv(epi(f))$ is convex by definition of the convex hull and so for all $\lambda \in (0,1)$ there is $\lambda (x,r)+(1-\lambda)(x,f(x)) \in conv(epi(f))$. For $\lambda = \frac{s-f(x)}{r-f(x)}$ one get $\lambda (x,r)+(1-\lambda)(x,f(x)) = (x,s)$.

Duncan Idaho
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