Why is $$r(\theta) = \frac{1}{\| \theta \|}$$ for $\theta \in S^{n-1}$, where $S^{n-1}$ is the $n-1$ dimensional unit spherical shell?
This is part of the following extract, which I'm a little lazy to type (so I'm putting a screenshot):
"since $r(\theta)$ is the largest number $r$ for which $r\theta \in K$" - why is this true, and how does it help?
By the way, we are working in $\mathbb{R}^n$, and using spherical polar coordinates.
Earlier in the notes, Prof. Ball states that the notation $|\cdot|$ is reserved for the Euclidean norm. So here he must be talking about something else. It should be the norm $\|\cdot\|_K$; in this specific instance he is talking about the norm given by $K=$ the just mentioned octahedron, i.e. the $\ell^1$ norm $\|x\|=\sum_1^n |x_i|$.
It seems that definitionally, $r(\theta)$ is the "radius of $K$ in direction $\theta$", namely the largest $r\ge0$ such that $r\theta \in K$. From $K = \{x\in \mathbb R^n : \|x\| \le 1\}$, it should be easy to see that all the points $r(\theta)\theta$ correspond to the boundary of the ball $$ \partial K = \partial \{x\in \mathbb R^n : \|x\| \le 1\} = \{x\in \mathbb R^n : \|x\|=1\}= \{ r(\theta)\theta : \theta\in S^{n-1}\} $$ Thus, letting $x=r(\theta)\theta$ be a general point of $\partial K$, on taking norms, $$ 1 = r(\theta)\|\theta\|$$ hence the result.
The context that would be missing from your question without the screenshot is that this is strongly dependent on the choice of $K$. if $\|x\|=|x|$ was indeed the euclidean norm, then the radius would be identically 1, which does indeed correspond to the unit ball in said norm.