Is $A$ convex if and only if $-\ln(i_{A})$ is convex?

Asked Dec 26 '15 at 15:59

Active Dec 27 '15 at 11:24

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Is it correct that we have : $A \subset \mathbb{R}^n$ is convex if and only if $-\ln(i_{A})$ is a convex function? where here $i_A$ takes the value 1 at $x \in A$ and $+\infty$ elsewhere.

I have found this statement a little bit weird since one knows that we usually compose with an nondecreasing convex function to construct convex functions where here -ln is nonincreasing convex.

edited Dec 27 '15 at 11:24

asked Dec 26 '15 at 15:59

user2015

Instead I propose A convex if and only if $ln(i_A)$ convex which can be easily proved using that $epi(ln(i_A))=A \times \mathbb{R}$ – user2015 Dec 26 '15 at 20:02
What's $A$? The question is rather unclear. – Benjamin Lindqvist Dec 27 '15 at 00:03
to Benjamin Lindqvist : Thank you for your response. $A$ is a subset of $\mathbb{R}^n$ – user2015 Dec 27 '15 at 11:25

Is $A$ convex if and only if $-\ln(i_{A})$ is convex?

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