I am trying to prove that the second-order cone defined as $C = \{(x,t): || x ||_2 \leq t, t\geq 0\}$ is a cone and is convex. I want to use the definition of convexity. Here is what I have so far:
Defn. of convexity: For any $x_1, x_2 \in C, \theta x_1 + (1-\theta)x_2 \in C$.
Taking 2 points in the set of second-order cone:
$\\ (x_1, t_1); \|x_1\|_2 \leq t_1, t_1\geq 0.$
$\\ (x_2, t_2); \|x_2\|_2 \leq t_2, t_2\geq 0.$
Multiplying $(x_1, t_1), (x_2, t_2)$ by $\theta, (1-\theta)$ respectively, and adding, I get:
$\theta ||x_1||_2 + (1-\theta)||x_2||_2 \leq \theta t_1 + (1-\theta)t_2$
How do I proceed from here to prove that the second-order cone is convex? I saw the following post regarding Convexity of a second order cone but it looks like an affine transformation is applied to the second-order cone.
Best Answer
We want to show $\|\theta x_1 + (1-\theta)x_2\| \leq\theta t_1 + (1-\theta)t_2$. By the triangle inequality, $\|\theta x_1 + (1-\theta)x_2\| \leq \|\theta x_1\| + \|(1-\theta)x_2\| = \theta \|x_1\| + (1-\theta)\|x_2\|\leq \theta t_1 + (1-\theta) t_2$, where the last inequality follows from the assumption that the two points lie in the cone.