# Geometric interpretation of the dual cone of $l^1$ is $l^\infty$? [duplicate]

This question already has an answer here:

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 $\mathbb{R}^2$.

## marked as duplicate by Community♦Oct 13 '16 at 23:23

The key is the dual relationship $\|x\|_\infty = \max_{\|z\|_1 \le 1} z^T x$.
• @AminJaili: If you mean $K \subset \mathbb{R}^2$ then since $\|x\|_1 = \|x\|_\infty = |x|$, we see that $K=K^*$. I think it is easier to first look for geometric inspiration in the context of the unit balls in $l_1, l_\infty$ rather than the cones. The geometric intuition is that a convex compact set can be described as a collection of points or the intersection of all half planes containing the set. This is the essence of duality. – copper.hat Jun 12 '16 at 15:06
• @AminJaili: No, you are confusing the cones with the unit balls. If $K \subset \mathbb{R}^2$, then it is defined by $\{(x,t)| |x| \le t \}$ which doesn't distinguish between $l_1, l_\infty$ norms (since they are the same on $\mathbb{R}$). Convince yourself of the duality of the unit balls first before looking at cone duality. In some sense they are the same thing, but you need to develop intuition first. – copper.hat Jun 12 '16 at 15:30