Given Lorentz cone $L^{n+1}=\{(x,t)\in \mathbb R^{n+1}: ||x||_2 \leq t\}$
How can i proof the convexity and self duality? I tried to do it from definitions, but i couldn't solve it. Do i need more for the proof?
Given Lorentz cone $L^{n+1}=\{(x,t)\in \mathbb R^{n+1}: ||x||_2 \leq t\}$
How can i proof the convexity and self duality? I tried to do it from definitions, but i couldn't solve it. Do i need more for the proof?
Assume $\|y\|_2\le s$, $s\ge0$. Then $$ (x,y) + st \ge -\|x\|_2\|y\|_2 +st \ge0 \quad \forall \|x\|_2\le t, $$ hence $(y,s)$ in the dual cone of $L^{n+1}$. Conversely, let $(y,s)$ be in the dual cone. Then $$ (x,y)+st \ge 0\quad \forall \|x\|_2\le t. $$ Setting $x=-y$, $t=\|y\|$, shows $-\|y\|_2^2 + s\|y\|_2 \ge0$, hence $s\ge \|y\|_2$. This shows $(y,s)\in L^{n+1}$.