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Any tips how to prove that the set

$$S = \{ x \in \mathbb{R^3} \mid x_1-x_2^2 \le x_3 \le x_1+x_2^2\}$$

is not convex? Any hint how to start would be very helpful.

Rodrigo de Azevedo
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lemonade
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  • Welcome to Maths SX! Perhaps using the equivalent condition $;|x_1-x_3|\le x_2^2;$ might simplify the problem? – Bernard Sep 15 '19 at 15:52
  • @Bernard uhh, not really. I just don't know what should I use here as a proof – lemonade Sep 15 '19 at 16:06
  • You might try to find two point in $S$ such that their middle point is not, to obtain a counterexample to convexity, for instance. – Bernard Sep 15 '19 at 16:09

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You can check that if $(x_1, x_2, x_3) \in S$, then $(x_1, -x_2, x_3) \in S$ using definition of $S$.

Thus $(x_1, 0, x_3) \in S$ if $S$ is convex and $(x_1, x_2, x_3) \in S$. This is because $(x_1, 0, x_3)$ lies on the line segment connecting $(x_1, x_2, x_3)$ and $(x_1, -x_2, x_3)$.

Note that $(1, 1, 0) \in S$ but $(1, 0, 0) \notin S$. Hence $S$ is not convex.

luxerhia
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