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Let D $\subset \mathbb{R^n}$ be an open convex domain and let f : D $\rightarrow \mathbb{R}$ be a map such that f has a locally strict maximum and a locally strict minimum.
Prove: The function f is neither quasi-convex nor quasi-concave.

I am trying to prove this property but my textbook gives very little information about quasi-convexity (quasi-concavity) at all to come up with an intelligent proof, so I have no idea where to start (or which properties to use).

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  • $\begingroup$ Could you put down a definition of quasi-convexity you're using? $\endgroup$
    – xyzzyz
    Commented Oct 19, 2016 at 18:34
  • $\begingroup$ @xyzzyz Of course. The map f is called quasi-convex if and only if the level set {x is an element of D : f(x) is less than or equal to alpha} is convex for every alpha in R. $\endgroup$
    – Anna D.
    Commented Oct 19, 2016 at 18:37
  • $\begingroup$ The property I want to prove provides a criterion to exclude quasi-convexity (this is a comment by the author of my textbook). $\endgroup$
    – Anna D.
    Commented Oct 19, 2016 at 18:40

1 Answer 1

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If $p$ is a locally strict maximum with $f(p) = r$, then for $\epsilon > 0$ sufficiently small $\{x: f(x) \le r-\epsilon\}$ contains a sphere around $p$ but not $p$ itself, so is not convex, and $f$ is not quasi-convex. Similarly...

EDIT:

enter image description here

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  • $\begingroup$ Thanks. Is the idea f is not convex (and not quasi-convex) since the open ball does not contain p (there is a 'hole' in the set)? How should I visualize this statement? $\endgroup$
    – Anna D.
    Commented Oct 19, 2016 at 19:21
  • $\begingroup$ I do not fully understand why this would be the case (open ball). $\endgroup$
    – Anna D.
    Commented Oct 19, 2016 at 20:12
  • $\begingroup$ If $\delta$ is small enough, the sphere $S_\delta = \{x: \|x - p\| = \delta\}$ is contained in $D$ and the maximum value of $f$ on $S_\delta$ is less than $r$. If $0 < \epsilon < r - \max_{S_\delta} f$, the level set $L = \{x: f(x) \le r - \epsilon\}$ contains $S_\delta$ but not $p$. If $x \in S_\delta$, so is $2 p - x$, and $(x + (2p-x))/2 = p \notin L$ while $p$ and $2p-x$ are in $L$, so $L$ is not convex. If $L$ is not convex, the function is not quasi-convex. $\endgroup$ Commented Oct 20, 2016 at 0:11
  • $\begingroup$ I'll draw a picture, with the level set $L$ in pink and the sphere in blue. $\endgroup$ Commented Oct 20, 2016 at 0:25
  • $\begingroup$ Thanks for the clarification and the visualization. It is a lot more clear now. $\endgroup$
    – Anna D.
    Commented Oct 22, 2016 at 20:23

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