Equivalent characterizations of the dual norm on finite dimensional vector spaces

Question

In their book on Convex Optimization, Boyd and Vandenberghe state that given a norm, $||\cdot||$, defined on $\mathbb{R}^n$, the dual norm is defined as $$||z||_*= \sup \{ z^Tx : ||x|| \leq 1 \}$$

In other places, I have encountered an equivalent characterization of the dual norm as $$||z||_*= \sup_{x \neq 0} \displaystyle\frac{z^Tx}{||x||}$$ I don't actually see how these two things are equivalent, even though this is said to be a simple one-liner.

In particular, what's confusing to me is that I would have argued the following, even though this seems to not be correct: By norm homogeneity, the set we're taking the supremum over is invariant to dilations. That is, for any $\alpha >0$, we have $$\displaystyle\frac{z^T(\alpha x)}{||\alpha x||} = \displaystyle\frac{\alpha (z^Tx)}{ |\alpha| ||x||} = \displaystyle\frac{z^Tx}{||x||}$$ and therefore we may find the supremum of the set by merely considering the $x$ values with some constant norm, for example, the unit norm: $$||z||_*= \sup_{x \neq 0} \displaystyle\frac{z^Tx}{||x||} = \sup_{x : ||x||=1} \displaystyle\frac{z^Tx}{||x||} = \sup \{ z^T x : ||x|| = 1\}$$.

Why is this logic incorrect?

Answer 1

You're correct. But note that in the BV definition, there is no benefit in taking $\|x\|<1$, so the definition may as well have stipulated that $\|x\|=1$.

Answer 2

Note that if $\|x\| < 1 $ and $z^T x \ge 0$, then $z^T {x \over \|x\|} \ge z^T x$. Hence $\sup \{z^T x : \|x \| \le 1 \} = \sup \{z^T x : \|x \|= 1 \}$.