0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

\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*}

$\endgroup$
  • $\begingroup$ Ah! Collect the pairs of coordinates and substitute, thanks for solution! $\endgroup$ – Anton Fahlgren Dec 15 '15 at 17:59
  • $\begingroup$ @AntonFahlgren You're welcome ;-) $\endgroup$ – YYF Dec 15 '15 at 18:01
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.