# Question on convex optimization and dual norms [closed]

I have the following questions about dual norms :

How do you prove that the dual of the dual norm is in fact the original norm? This is what I have so far:

If I have $||y||_*$ as the norm dual of $|| y ||$ then I know that $\\$

$||y||_*$ = $max_x \ x^Ty$ subject to $||x|| \leq 1$

In order to take the dual of this I first write the Lagrangian as follows:

$L(x,u) = - x^Ty + u*(||x|| -1)$

I rewrote this as:

$L(x,u) = - x^Ty + u*\sqrt{(\sum x_i^2)} \ - u$

Now, taking the dual of this by minimizing the Lagrangian we get the following :

$||y||_{**} = min_x L(x,u)$

I am not sure how to do this minimization. I would also like to confirm that all the former steps are correct. I understand that this is probably fairly simple - but I'm fairly new to this and any help would be very appreciated.

## closed as off-topic by Will Jagy, Pietro Majer, Carlo Beenakker, Andrey Rekalo, Ricardo AndradeOct 28 '13 at 11:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Questions about homework are generally off-topic. MathOverflow is for mathematicians to ask each other questions about their research." – Will Jagy, Pietro Majer, Carlo Beenakker, Andrey Rekalo
If this question can be reworded to fit the rules in the help center, please edit the question.

Thm.15.4 (Roc70). Let $f$ be a nonnegative convex function which vanishes at the origin. The polar $f^\circ$ of $f$ is then a nonnegative closed convex function which vanishes at the origin, and $f^{\circ\circ}=\text{cl}\:f$.
This theorem helps because if $f$ is a norm, then its polar $f^\circ$ is just the dual norm.