I just noticed somewhere in Convex Optimization that the dual cone of $l^1$ is $l^\infty$! (A diamond in $\mathbb{R}^2$ for $l^1$ is a square in $\mathbb{R}^2$ for $l^\infty$.) In fact I cannot imagine that. Can you please explain it geometrically by the definition of the dual cone? [Ref. Convex Optimization book, Stephen Boyd]
$K = \{(x,t): \Vert x\Vert_1 \le t\} \Rightarrow K^* = \{(x,t): \Vert x\Vert_\infty \le t\}$
Definition:
$K$ is a cone, then the dual cone is : $K^* = \{y: x^T y \geq 0 \ \text{for all} \ x \in K\}$
I would be glad if you have any comment about that. For simplicity you can discuss about that in $\mathbb{R}^2$.
The key is the dual relationship $\|x\|_\infty = \max_{\|z\|_1 \le 1} z^T x$.
Note \begin{eqnarray} K^* &=& \{ (y,s) | x^T y + st \ge 0 \text{ for all } (x,t) \in K \} \\ &=& \{ (y,s) | -x^T y + s \ge 0 \text{ for all } (-x,1) \in K \}\\ &=& \{ (y,s) | x^T y \le s \text{ for all } \|x\|_1 \le 1 \}\\ &=& \{ (y,s) | \max_{\|x\|_1 \le 1 }x^T y \le s \}\\ &=& \{ (y,s) | \|y\|_\infty \le s \}\\ \end{eqnarray}