How to prove the following set is convex?
$$\{(x,y,z)|x^2+y^2-z^2 \leqslant 0,z \geqslant 0\}$$
I try to make it by prove that $$[\theta x_1 + (1-\theta)x_2]^2 + [\theta y_1 + (1-\theta)y_2]^2 – [\theta z_1 + (1-\theta)z_2]^2 \leqslant 0$$,
but then stop at $$\theta^2(x_1^2+y_1^2-z_1^2)+(1-\theta)^2(x_2^2+y_2^2-z_2^2)-2\theta(1-\theta)(x_1x_2+y_1y_2-z_1z_2) \leqslant 0$$.
How can I continue or is there some other way to do this?
Best Answer
This set is the epigraph of the $2$-norm, which is a convex function. Hence, this set is convex.
By the way, this set is called the "second-order cone" or the "ice-cream cone". (More precisely, this is the ice-cream cone in $\mathbb R^3$.)