# Problem proving property quasi-convexity (quasi-concavity) & optima

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).

• Could you put down a definition of quasi-convexity you're using? Commented Oct 19, 2016 at 18:34
• @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. Commented Oct 19, 2016 at 18:37
• The property I want to prove provides a criterion to exclude quasi-convexity (this is a comment by the author of my textbook). Commented Oct 19, 2016 at 18:40

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...
• 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. Commented Oct 20, 2016 at 0:11
• I'll draw a picture, with the level set $L$ in pink and the sphere in blue. Commented Oct 20, 2016 at 0:25