# Prove 'dual to the dual norm is the original norm' without using Hahn-Banach theorem?

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?

• 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. Commented Aug 30, 2020 at 7:44
• @iarbel84 yes I think you are right but I want to prove a more general case, not only l-p norm or something else. Commented Sep 1, 2020 at 11:57
• 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? Commented Sep 1, 2020 at 14:17
• @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. Commented Sep 2, 2020 at 8:20