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Let $\|\cdot\|$ be a norm on $\mathbb{R}^{n} .$ The associated dual norm, denoted $\|\cdot\|_{*},$ is defined as $$ \|z\|_{*}=\sup \left\{z^{\top} x \mid\|x\| \leq 1\right\} $$

I'm trying to prove $$\|x\|_{**} = \|x\|$$

here says we can prove it by Hahn-Banach theorem. i.e. $$\|y\|=\max _{x \neq 0} \frac{x^{T} y}{\|y\|_{*}}$$ But I think by definition, this is what we need to prove. So it seems says 'because this is correct, this is correct'. I think it proves nothing.

Are there any proof without using Hahn-Banach theorem?

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  • $\begingroup$ The original norm $p$ and dual norm $q$ must satisfy the relation $p+q=pq$. I think this decomposition must be unique for rational $p,q$ (not sure how to prove that), therefore the dual of the dual must be the original norm. And in the case the space is endowed with some $Q$ norm, $Q\succ 0$, then the dual norm will be $Q^{-1}$, therefore the bi-dual must be $Q$ again. $\endgroup$
    – iarbel84
    Aug 30, 2020 at 7:44
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    $\begingroup$ @iarbel84 yes I think you are right but I want to prove a more general case, not only l-p norm or something else. $\endgroup$
    – wz0919
    Sep 1, 2020 at 11:57
  • $\begingroup$ We covered $p$-norms and $Q$-norms. I agree that this is not general, but specific for $p,Q$, but what are the other norms you have in mind? $\endgroup$
    – iarbel84
    Sep 1, 2020 at 14:17
  • $\begingroup$ @iarbel84 Some matrix norm (spectral norm, trace norm,...) for example? For a specific norm maybe we can compute a concise expression of its dual norm, But for the general case the only expression is the definition perhaps. $\endgroup$
    – wz0919
    Sep 2, 2020 at 8:20

1 Answer 1

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