Convex set - in $\mathbb{R}^3$

Question

I need to prove $2$ sets, i hope you could help me: $x^2+y^2\leq z^2$ Is that a convex set? $S: 0\leq x_1\leq x_2\leq \cdots\leq x_n.$ I think about sum of convex sets - but not sure. So i woild say it is convex. Could anyone help?

Answer 1

The set $\{(x,y,z):x^2 + y^2 \leq z^2\}$ is not convex.

To see that the set is not convex, consider the points $(1,0,1)$ and $(1,0,-1)$, which are both in the set. However, the point $$ \frac 12 (1,0,1) + \frac 12 (1,0,-1) = (1,0,0) $$ is not in the set.

Your second set is indeed convex. We can show that this is the case as follows:

Suppose that $x = (x_1,\dots,x_n)$ and $y = (y_1,\dots,y_n)$ are elements of this set. Take any $s,t \geq 0$ with $s+t = 1$. For $i=1,\dots,n-1$, we have $$ s x_i + ty_i \leq sx_{i+1} + ty_{i} \leq s x_{i+1} + t y_{i+1} $$ and of course, $sx_1 + ty_1 \geq 0$. It follows that $sx + ty$ is an element of the set. So, the set is convex.