Dual norm equivalence?

Question

$\|\|$ is a norm in $R^n$, its dual norm is defined as

$\|s\|^*=max_{\|x\|=1}s^Tx$.

We denote $s^\#$ as any vector in the following set:

[Arg $max_x: \ \ s^Tx-\frac{1}{2}\|x\|^2$]

How to verify the claim $\|s^\#\|=\|s\|^*$?

Answer 1

Since $\|s\|^* = \displaystyle\max_{\|x\| = 1} s^Tx$, we know that $\|x\| = 1 \implies s^Tx \le \|s\|^*$.

Then, for any $x$, we have $\|\tfrac{1}{\|x\|}x\| = 1$, and so, $s^T(\tfrac{1}{\|x\|}x) \le \|s\|^*$, i.e. $s^Tx \le \|s\|^*\|x\|$.

Then, $s^Tx - \tfrac{1}{2}\|x\|^2 \le \|s\|^*\|x\| - \tfrac{1}{2}\|x\|^2 \le \tfrac{1}{2}(\|s\|^*)^2$. Equality only holds if $\|x\| = \|s\|^*$.

Therefore, if $s^\#$ maximizes $s^Tx - \tfrac{1}{2}\|x\|^2$, then $\|s^\#\| = \|s\|^*$.