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I have a concave function $F:A \rightarrow \mathbb R$ and an interval $I=[\underline y,\bar y ]$ for which I consider the preimage

$$f^{-1}(I) := \{ x \in A \lvert f(x) \in I\} = \{ x \in A\lvert \ \underline y \leq f(x) \leq \bar y \}$$

my question is how to show that $f^{-1}(I)$ is convex if that is indeed the case?

Obviously for $x_1$ and $x_2$ in $f^{-1}(I)$ it must be the case that for $x_3 = \alpha x_1 + (1-\alpha)x_2$ that

$$f(x_3) = f(\alpha x_1 + (1-\alpha)x_2) \geq \alpha f(x_1) + (1-\alpha)f(x_2)\geq \underline y$$

but whether $f(x_3) \leq \bar y$ eludes me.

Jesper Hybel
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  • $f(x) = -x^{2}$ is the first thing that comes to mind for a concave function. Does this one check out? –  Dec 28 '20 at 18:34
  • Fair enough and thx. If you wanna make that into an answer I will accept accept your counterexample as an answer. – Jesper Hybel Dec 28 '20 at 18:36

1 Answers1

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The answer is no. Consider $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = -x^{2}$. We know that $f^{-1}([-4,-16]) = [-4,-2]\cup[2,4]$, which is disconnected and so not convex.

On the other hand, if $f$ is concave, then its super-level sets are convex. That is, $f^{-1}([c,\infty))$ is always convex, as you already noted.