RBecause Eucildean space $\mathbb{R}^n$ has many good properties such as complete and reflexive or etc. And since it's finite dimensional, so all kinds of norms in $\mathbb{R}^n$ are equivalent, and in fact we have known that the dual space of $\mathbb{R}^n$ is isomorphic with $\mathbb{R}^n$ and the second dual is itself. So I have some questions which I think is correct based on those good properties of Euclidean space $\mathbb{R}^n$:
(1). Let $a\geq1$ and $||\cdot||_a$ be a norm in $\mathbb{R}^n$, so if the dual norm $||z||_{a^*}:=\sup_{||u||_a\leq1}\frac{|z^Tu|}{||u||_a}=||z||_b$ where $\frac{1}{a}+\frac{1}{b}=1$ (i.e. $a$ and $b$ are Hilbert conjugate) ?
(2). For the unit ball $B_a(0,1)$ with respect to norm $||\cdot||_a$, (we could consider it as a closed ball), we could know that the polar of it is also a unit ball in dual space, i.e. $B_a(0,1)^{\circ}=B_{a^*}(0,1)$, so if the equality $B_{a^*}(0,1)=B_b(0,1)$ holds or what should be the relation between these 2 unit balls?
(3). Based on the facts above, could we conclude the following equality:$$||u||_a=\sup_{||z||_{a^*}\leq1}\frac{|z^Tu|}{||z||_{a^*}}=\sup_{||z||_{b}\leq1}\frac{|z^Tu|}{||z||_{b}}$$which means the dual to the dual norm is the primal norm? It's easier to prove it in general case that we could use Hahn-Banach Theorem and some corollaries of it, but in Euclidean space, how could we prove the problem The dual norm of dual norm is the primal norm by using Separating Hyperplane Theorem? Thank you very much for everyone who contribute to the proof.