Dual norm and Legendre-Transformation of the norm

Question

In this post, the Legendre-Transformation of the norm on $\mathbb{R}^n$ was calculated using the dual norm.

I have two questions. First of all, I tried calculating the Legendre-Transformation myself: If $f(x)=|x|$, then $f^*(x)=\sup_{x\in\mathbb{R}^n} (x^Ty-|x|)$.

First $$ x^Ty-|x|\le |x|(|y|-1) $$ and thus $$ f^*(y)\le \begin{cases}+\infty \quad |y|>1\\ 0\quad\quad\, |y|\le 1 \end{cases} $$ and with the choices $x=0$ (if $|y|<1$) and $x=ne_i$ for $n\in\mathbb{N}$ and some $i\in\{1,...,n\}$ such that $y_i>1$ (in the case that $|y|>1$) respectively, the other inequality also follows.

So, I get the same thing, except with the norm and not the dual norm. Is there some mistake in my calculations or does the equivalence $$|y|\le 1 \Leftrightarrow |y|_*=\sup_{|z|\le 1}z^Ty\le 1 $$ actually hold?

I tried showing this in the following way:

If $|y|\le 1$, then $|y|_*\le \sup_{|z|\le 1}|y||z|\le 1$

and if $|y|_*\le 1$ then $|y|=y^T(\tfrac{y}{|y|})\le \sup_{|z|\le 1}z^Ty=|y|_*\le 1$.

This kind of surprised me and had me wondering if there is a mistake in my calculations. Could someone please help me out?

Answer 1

The confusion is that the norm $\|\cdot\|$ considered in the linked post any norm, and is not necessarily the euclidean norm (which we will write it $|\cdot|_2$). associated with the scalar product $(x,y) = x^Ty$. Therefore, the Cauchy-Schwarz inequality $x^Ty \leq \|x\|\|y\|$ does not hold.

Using the Cauchy-Schwartz inequality, you can indeed show that the dual norm of $|\cdot|_2$ is $|\cdot|_2$ itself, so your calculations are correct in the special case $\|\cdot\| = |\cdot|_2$.