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
Best Answer
\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*}