Self-duality of a second-order cone
= Self-duality of a second-order cone
{title2=$\mathcal L^*=\mathcal L$}
For $(u,s),(v,t)$ in the <second-order cone>, the <Cauchy-Schwarz inequality> gives $u\cdot v+st\ge-\|u\|\|v\|+st\ge0$. Thus the cone is contained in its <dual cone>. Conversely, if $s<\|u\|$ and $u\ne0$, pairing $(u,s)$ with $(-u/\|u\|,1)$ is negative. If $u=0$ and $s<0$, use $(0,1)$. Every point outside the cone therefore lies outside its dual, proving that it is a <self-dual cone>.