# Dual to the dual norm is the original norm (?)

I have the following questions about dual norms :

How do you prove that the dual of the dual norm is in fact the original norm? This is what I have so far:

If I have $\|y\|_*$ as the norm dual of $\|y\|$ then I know that $\\$

$\|y\|_* = \max\limits_{\large{x}} \ x^Ty$ subject to $\|x\| \leq 1$

In order to take the dual of this I write the Lagrangian as follows:

$$L(x,u) = - x^Ty + u*(\|x\| -1)$$

I rewrote this as:

$$L(x,u) = - x^Ty + u*\sqrt{\Big(\sum x_i^2\Big)} \ - u$$

Now, taking the dual of this by minimizing the Lagrangian we get the following :

$$\|y\|_{**} = \min_{\large{x}} L(x,u)$$

I am not sure how to do this minimization. I understand that this is probably fairly simple -but as I'm fairly new to this and any help would be very appreciated.

• In your problem statement it appears you are assuming that $\|y\|$ is the Euclidean two-norm. Is that right? If so, notice that $\|y\|_* = \|y\|$ and your claim follows immediately. To see this, notice that you can replace your inequality constraint with equality, and that $x=y/\|y\|$ then clearly is the maximizer. – user7530 Oct 28 '13 at 4:55
• No. ||x|| is any general norm. If you look at the last bullet point at this link then they state the the same thing (dual of the dual norm is primal) but don't prove it. I understand how to prove it if it was the Euclidean norm. – Alice Oct 28 '13 at 5:37
• In that case, there is a mistake in the rewriting: $\|x\|$ is not generally $\sqrt{\sum x_i^2}$. – user7530 Oct 28 '13 at 13:14
• You have to use the Hahn-Banach theorem. – Stephen Montgomery-Smith Nov 4 '13 at 23:39

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.
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 $\|y\| = \|y\|_{**}$.
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.