# Show that the following is a convex set

I've been banging my head against the wall trying to handle these proofs for two hours now, it seems very simple but I guess I need a hand starting out. I hope I at least know what to show:

Show that the set $S_1=\{(x_1,x_2), x_2 \geq x_1^2\}$ is convex. Intuitively it makes sense since this is the area above the parabola $x_2=x_1^2$. By the definition of convex sets I take points $(x_1,x_2)$ and $(x_3, x_4)$ such that $x_2 \geq x_1^2$ and $x_4 \geq x_3^2$, and I must show that $\lambda(x_1,x_2)+(1-\lambda)(x_3,x_4)$ is in the set $S_1$, i.e. $\lambda(x_2)+(1-\lambda)x_4 \geq (\lambda(x_1)+(1-\lambda)x_3)^2$, but I'm not getting there.

Thanks

\begin{align*} &\lambda x_2+(1-\lambda)x_4-[\lambda x_1+(1-\lambda)x_3]^2\\ &=\lambda x_2+(1-\lambda)x_4-(\lambda^2x_1^2+2\lambda(1-\lambda)x_1x_3+(1-\lambda)^2x_3^2)\\ &=\lambda[x_2-\lambda x_1^2]+(1-\lambda)[x_4-(1-\lambda)x_3^2]-2\lambda(1-\lambda)x_1x_3\\ &\geq\lambda(1-\lambda)x_1^2+(1-\lambda)\lambda x_3^2-2\lambda(1-\lambda)x_1x_3\\ &=\lambda(1-\lambda)(x_1-x_3)^2\\ &\geq0 \end{align*}

• Ah! Collect the pairs of coordinates and substitute, thanks for solution! – Anton Fahlgren Dec 15 '15 at 17:59
• @AntonFahlgren You're welcome ;-) – YYF Dec 15 '15 at 18:01

An idea (but perhaps not the simpler...): Your set $S$ is the set of points that are above all the tangents to the parabola. If you write the equation of such a tangent, at the point $(t,t^2)$, then it is $x_2=2tx_1-t^2$. So $(x_1,x_2)\in S$, iff you have $x_2\geq 2tx_1-t^2$ for all $t\in \mathbb{R}$. (It is easy to prove this, simply compute the discriminant of $t^2-2tx_1+x_2$, it has to be $\leq 0$). Now it is easy to finish.

• I'm gonna have a think about this one, thanks! – Anton Fahlgren Dec 15 '15 at 18:00