Given a norm $\|\cdot\|$ on $\mathbb R^n$, the dual norm $\|\cdot\|_*$ is defined as $$\|z\|_* = \sup \{z^{\mathrm T} x : \|x\| \le 1\}\text.$$
My question is: what is the $\arg\max$ of this quantity?
This is important to me because I wanted to show that if $K$ is a norm cone then its dual is the norm cone of the dual norm. That is, if $K_{\|\cdot\|} = \{(x,r) : r \ge \|x\|\}$, then its dual cone is $K_* = K_{\|\cdot\|_*} = \{(y,t) : t \ge \|y\|_*\}$.
The set of maximizers in the definition of the dual norm equals the sub-differential of the dual norm at the given point.
Given a norm $\varphi : \mathbb{R}^{n} \to [0,\infty)$ with dual norm $\varphi^{*}$, recall that $\varphi^{*}$ is a convex function and, therefore, to each $z \in \mathbb{R}^{n}$, we can associate the sub-differential \begin{equation*} \partial \varphi^{*}(z) = \left\{ q \in \mathbb{R}^{n} \, \mid \, \forall z' \in \mathbb{R}^{n} \, \, \varphi^{*}(z') \geq \varphi^{*}(z) + q^{T} (z' - z) \right\}. \end{equation*} It is worth noting that $\partial \varphi^{*}(z)$ consists of vectors determining supporting hyperplanes at $z$ to the set $\{z' \in \mathbb{R}^{d} \, \mid \, \varphi^{*}(z') \leq \varphi^{*}(z)\}$ (when $z \neq 0$). Furthermore, since $\varphi^{*}$ is one-homogeneous, $\partial \varphi^{*}$ is zero-homogeneous (that is, $\partial \varphi^{*}(\alpha z) = \partial \varphi^{*}(z)$ for all $\alpha \in (0,\infty)$ and $z \in \mathbb{R}^{n}$). Therefore, at least geometrically (for $z \neq 0$), we may as well think of $\partial \varphi^{*}(z)$ as encoding the supporting hyperplanes at $z$ to the set $\{z' \in \mathbb{R}^{n} \, \mid \, \varphi^{*}(z') \leq 1\}$ by replacing $z$ by $\frac{z}{\varphi^{*}(z)}$.
Back to your question, it is a nice exercise in convex analysis to prove the following identity: \begin{equation*} \partial \varphi^{*}(z) = \{q \in \mathbb{R}^{n} \, \mid \, \varphi(q) = 1, \, \, q^{T} z = \varphi^{*}(z) \}. \end{equation*} In particular, if $q \in \partial \varphi^{*}(z)$, then \begin{equation*} q^{T} z = \varphi^{*}(z) = \max\left\{ p^{T} z \, \mid \, \varphi(p) \leq 1 \right\} \end{equation*} and these are all the maximizers.
Conversely, if $q \in \mathbb{R}^{n}$ satisfies $\varphi(q) = 1$, then there is a $z \in \mathbb{R}^{n}$ such that $\varphi^{*}(z) = q^{T} z$. This follows from the fact that $(\varphi^{*})^{*} = \varphi$; hence this holds if and only if $z \in \varphi^{*}(z) \partial \varphi(q)$.
To sum up, we see that the maximizers in the definition of the dual norm comprise the subdifferential of the dual norm at the corresponding point, which encodes the supporting hyperplanes to that point. This can be a useful way of thinking about the unit ball $\{p \in \mathbb{R}^{n} \, \mid \, \varphi(p) \leq 1\}$ and the dual unit ball $\{z \in \mathbb{R}^{n} \, \mid \, \varphi^{*}(z) \leq 1\}$, including questions like the strict convexity of the boundary (e.g. does the boundary contain a line segment?) as well as its differentiability (e.g. does the boundary have a corner a la the "taxi-cab norm"?).
Finally, I would be remiss not to mention that the previous considerations show that the following inequality holds: \begin{equation*} \forall q \in \mathbb{R}^{n}, \, \, \forall z \in \mathbb{R}^{n} \quad q^{T} z \leq \varphi(q) \varphi^{*}(z) \end{equation*} with equality if and only if $q \in \varphi(q) \partial \varphi^{*}(z)$ and $z \in \varphi^{*}(z) \partial \varphi(q)$. (Also, the discussion above applies to Finsler norms just as well as norms, modulo a few changes.)