Can a polynomial function limited by x axis values be convex?

Question

Consider the given set:

$$ S = x ∈ E^2: x_2 − x_1^2 = 0, −1 \le x1 \le 1 $$

When I draw the set I find that it is a polynomial where the $$x_1$$ axis is cut at -1 and 1. The max value for $$x_2$$ is 1. It seems like a convex set to me but the book says it is not. Why is it not a convex set?

score 1 · Accepted Answer · answered Dec 12 '20 at 13:28

1

So, $a:=(\frac{1}{4},\frac{1}{2}) \in S$ and $b:=(\frac{-1}{4},\frac{1}{2}) \in S$ but $\frac{1}{2}(a+b) = (0, \frac{1}{2})$ which is not fulfilling $x_2-x_1^2=0$, hence not in $S$. So $\frac{1}{2}(a+b)$ is not in $S$, which renders $S$ is not konvex.

answered Dec 12 '20 at 13:28

Maksim

1,706

Is it possible to show this without giving x_1 and x_2 values? Purely algebracally – moli Dec 12 '20 at 13:43
@moli: Not sure what do you mean by pure algebracally ? – Maksim Dec 12 '20 at 13:48
I mean if we did not use real numbers like (1/4,1/2) could we still show this to be true? – moli Dec 12 '20 at 14:03

Can a polynomial function limited by x axis values be convex?

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