How do you prove that the dual of the dual norm is the original norm?
This is what I have so far: If I have $\|y\|_* $ as the dual norm of $\|y\|$, then I know that $$ \|y\|_* = \qquad \max\limits_{x} \ x^Ty \quad \text{subject to} \quad \|x\| \le 1 $$ In order to take the dual of this I write the Lagrangian as follows: $$ L(x,u) = - x^Ty + u\cdot(\|x\| -1) $$ I rewrote this as $$ L(x,u) = - x^Ty - u + u \cdot \sqrt{\sum x_i^2} $$ Now, taking the dual of this by minimizing the Lagrangian we get $$ \|y\|_{**} = \min_{x} L(x,u). $$ I am not sure how to do this minimization.
By Hahn-Banach theorem in the answer above, $$\Vert y \Vert = \max_{x\not=0}\frac{x^T y}{\Vert x \Vert_*} = \Vert y \Vert_{**}$$ Done. I think it's enough if the Hahn-Banach theorem in the answer above is used in a correct way.