Prove a cone $S=\{(x_1,x_2,x_3) \in \mathbb{R}^3 : x_1^2+x_2^2-x_3^2=0, x_3 \ge 0 \}$ is a convex set.

Question

I have to prove a cone $$S=\{(x_1,x_2,x_3) \in \mathbb{R}^3 : x_1^2+x_2^2-x_3^2=0, x_3 \ge 0 \}$$ is a convex set.

The definition of a convex set is known as

"A set $C$ is convex if for $x,y \in C$ and $\lambda \in [0,1]$ then $\lambda x + (1-\lambda)y \in C$".

I try to use this definition to prove S is a convex.

First, for $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3) \in C$ and $\lambda \in [0,1]$, let $z=\lambda x +(1-\lambda)y$. Then $$z_1^2+z_2^2-z_3^2 = \lambda^2 (x_1^2+x_2^2-x_3^2) +2\lambda(1-\lambda)(x_1y_1+x_2y_2-x_3y_3) +(\lambda(1-\lambda))^2(y_1^2+y_2^2-y_3^2). $$

Since $x,y \in S$ then $x_1^2+x_2^2-x_3^2=0, y_1^2+y_2^2-y_3^2=0$.

My problem is that I don't have any way to prove $x_1y_1+x_2y_2-x_3y_3=0$.

One way I have tried is that $$|x_1y_1+x_2y_2| \le \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}=x_3y_3. $$

Answer 1

I suggest the cone representation in spherical coordinates through $$ {x_1=R\sin\theta\cos\phi \\x_2=R\sin\theta\sin\phi \\x_3=R\cos\theta } $$ and prove the convexity using the new representation.