Why is the dual cone of $l^1$ is $l^\infty$?

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 $$\mathbb{R}^2$$.

• Your question is grossly unclear. Are you talking about dual norms or dual cones ? – dohmatob Jun 11 '16 at 19:47
• The question is edited. – Amin Jun 12 '16 at 3:51
• @Nurmister: The OP, Amin, is fairly active on the site, so I'd recommend proposing such corrections by Comments, which you have enough reputation to post. – hardmath Oct 3 '18 at 17:10
• Ah, I did not consider the OP's activity when making the edit -- yes, it would be better to leave tiny edits to an active OP. – Nurmister Oct 3 '18 at 17:17

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.