regarding the concept of dual cone When studying the covex analysis, I am not clear about the concept of dual cone. In the following graph, $\mathcal{K}*$ was the dual cone. I marked two points, the read one is supposed to satisfy the inequality of $y^{T}x\leq 0$, where $x$ is represented as the green point. How can we guarantee that this inequatlity is satisfied, and where do thoes $90^{0}$ come from?

 A: For a fixed vector $y$, the set $\{x : y^Tx =  0\}$ is a line perpendicular to $x$, which I'll denote $y^\perp$ for brevity. (Dot product of two vectors is zero if and only if they are perpendicular.) This is the source of 90 degree angles you see on the diagram. The line $y^\perp$ divides the plane into two halfplanes. One of which, $\{x : y^Tx \ge  0\}$, is the one we are interested in. 
In order for $y$ to belong to the dual cone, we must have $y^Tx\ge 0$ for all $x\in K$. This means precisely that $K$ must be contained in the halfplane $\{x : y^Tx \ge  0\}$. (Consider this logic carefully). 
So, in order to test vector $y$ on being in $K^*$, we draw the line $y^\perp$   and check whether all of $K$ is on the same side of $y^\perp$  as $y$ itself. As usual, when studying a set, it helps to look at the boundary, i.e., the "edge cases". The vectors $y$  which  barely qualify for being in $K^*$ are those for which $y^\perp $ contains a side of cone $K$ (if you rotate such $y$ slightly in a wrong direction, the line $y^\perp$ will cross through $K$). This is why the boundary of $K^*$ is formed by half-lines perpendicular to the boundary of $K$.
