I just noticed 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} => 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 $\mathbb{R}^2$.
Duality of cones is not the same as duality of normed vector spaces.
To get from the diamond to the square, observe that $$ B=\{y: \quad |x^T y|\leq 1\quad \forall |x|_{1}\leq 1\} $$ is a square, "created" from the diamond $D=\{x: |x|_{1}\leq 1\}$. The set $ D$ is the unit ball of $l_{1} $, whereas $ B $ is the unit ball of its dual space.