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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.

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

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A somewhat more general theorem is available (though not entirely suitable for consumption by a beginner, but maybe it's ok). See Theorem 15.4 in Convex Analysis, by R. T. Rockafellar.

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.

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  • $\begingroup$ Is there another method to prove this? $\endgroup$ – Alice Oct 27 '13 at 22:17
  • $\begingroup$ To speak of polar functions, don't you need that the functions are positively homogeneous? $\endgroup$ – Dirk Oct 28 '13 at 7:27
  • $\begingroup$ Actually Dirk, on pg. 136, Rockafellar describes that how the notion of polars can be extended to the larger class of functions cited in the theorem above. That said, for the usual notion with positive homogeneity, Thm. 15.1 suffices. (also @Alice: the proof of Thm. 15.1 is quite simple, if you have access to the book, please check it out) $\endgroup$ – Suvrit Oct 28 '13 at 14:33

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