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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 $\mathbb{R}^2$.

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  • $\begingroup$ Your question is grossly unclear. Are you talking about dual norms or dual cones ? $\endgroup$ – dohmatob Jun 11 '16 at 19:47
  • $\begingroup$ The question is edited. $\endgroup$ – Amin Jun 12 '16 at 3:51
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    $\begingroup$ @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. $\endgroup$ – hardmath Oct 3 '18 at 17:10
  • $\begingroup$ 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. $\endgroup$ – Nurmister Oct 3 '18 at 17:17
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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.

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  • $\begingroup$ Here, I mean dual of cone. In fact, I would like to have a geometric interpretation of this point. $\endgroup$ – Amin Jun 11 '16 at 19:11
  • $\begingroup$ Then ask another question. The duality of vector spaces and cones is a different topic $\endgroup$ – Bananach Jun 12 '16 at 3:47
  • $\begingroup$ The question is edited. $\endgroup$ – Amin Jun 12 '16 at 3:51

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