Dual norm of the dual norm is the primal norm

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.

The fact that dual to dual norm is equal to the original norm in case of finite-dimensional spaces is equivalent to the fact that the corresponding Banach space is reflexive. By James' theorem, a Banach space $B$ is reflexive if and only if every continuous linear functional on $B$ attains its maximum on the closed unit ball in $B$. That is surely true for finite-dimensional spaces with any norms, thus dual to dual norm must be equivalent to the original norm.

I've attempted to find a simple solution, take a look.

1. We are going to prove that $\|y\|_{**} \le \|y\|$. We have $$\|y\|_* = \max_{x\ne 0}\frac{x^Ty}{\|x\|}$$ and $$\|y\|_{**} = \max_{z\ne 0}\frac{z^Ty}{\|z\|_*}.$$ The latter equation can be rewritten as $$\|y\|_{**} = \max_{z\ne 0}\frac{z^Ty}{\max_{x\ne 0}\frac{x^Tz}{\|x\|}} = \max_{z\ne 0} \left(z^Ty\cdot \min_{x\ne 0}\frac{\|x\|}{x^Tz}\right) = \max_{z\ne 0} \min_{x\ne 0} \|x\|\frac{z^Ty}{z^Tx}.$$ By max-min inequality we obtain $$\|y\|_{**}\le \inf_{x\ne 0}\sup_{z\ne 0}\|x\|\frac{z^Ty}{z^Tx} = \inf_{x\ne 0}\|x\|\cdot\sup_{z\ne 0}\frac{z^Ty}{z^Tx} = \inf_{x\ne 0}\|x\|\cdot\left\{ \begin{array}{lc} \alpha, & y = \alpha x \\ +\infty, & y \ne \alpha x \end{array} \right. $$ The last conversion is easy to check: if $x$ and $y$ are linearly independent, we can find $z$ such that $(x,z)=0$ while $(y,z)\ne 0$. So on $$ \|y\|_{**} \le \inf_{x\ne 0} \alpha\cdot\|x\| = \inf_{\alpha\ne 0} \alpha\cdot\left\|\frac{y}{\alpha}\right\| = \|y\|. $$

2. By Hahn-Banach theorem, $$\|y\|=\max_{x\ne 0}\frac{x^Ty}{\|y\|_*},\ \ \ \|y\|_*=\max_{x\ne 0}\frac{x^Ty}{\|y\|_{**}} $$ Thus we can apply the same logic as in 1. to prove $\|y\|\le \|y\|_{**}$, finally obtaining $\|y\| = \|y\|_{**}$.

P. S. I have a strong feeling that there is something odd and unnecessary in this proof. But I can't see what. Can anyone help?

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.