Evelyn Freeman

2022-10-20

How to show that ${f}^{-1}\left(I\right)$ is convex if that is indeed the case?

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veirenca77

Expert

The answer is no. Consider$f:\mathbb{R}\to \mathbb{R}$ given by $f\left(x\right)=-{x}^{2}$. We know that ${f}^{-1}\left(\left[-4,-16\right]\right)=\left[-4,-2\right]\cup \left[2,4\right]$, which is disconnected and so not convex.
On the other hand, if $f$ is concave, then its super-level sets are convex. That is, ${f}^{-1}\left(\left[c,\mathrm{\infty }\right)\right)$ is always convex, as you already noted.

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