Here's my question:
Consider the space $(\mathbb{R}^n, \| \cdot \|_\infty).$ For which vectors $v \in \mathbb{R}^n$ can we say that any linear functional $f: \langle v \rangle \to \mathbb{R}$ has a unique norm-preserving extension to the whole space?
My thought is that this question can be reduced to asking for which vectors we can find a sub-linear function $p$ such that $f$ is dominated by $p$ on $v$'s span. Hahn-Banach then guarantees that $f$ has an extension to the whole space, but that extension doesn't necessarily have to be unique, which I think is where this question gets tricky. At the moment, I'm trying to think about what further conditions we'd need to be guaranteed a unique extension. I'd really appreciate any help.
The problem is evidently invariant under scalar multiplication so we may suppose that both $\|v\|=1$ and $\|f\|=1$, in which case $f$ is necessarily given by $f(\lambda v)=\pm\lambda $. We may further replace $v$ by $-v$, if necessary, so as to be able to assume that $f(\lambda v)=\lambda $.
Next notice that a linear functional $g$, defined on the whole ${\bf R}^n$, has norm less than or equal to one, if and only if the affine hyperplane $$ \{x\in {\bf R}^n:g(x)=1\} $$ has empty intersection with the open unit ball $B(0, 1)$.
In order to produce a norm-preserving extension of our given functional $f$, we must therefore construct an affine hyperplane passing thru $v$, and leaving the whole of $B(0, 1)$ to the same side.
Noticing that $B(0, 1)$ is actually the cube $(-1,1)^n$, observe that if $v$ is one of the vertices of that cube, say $v = (\pm 1,\pm 1,\ldots , \pm1)$, then there are many ways to build a hyperplane thru $v$ not intercepting $B(0,1)$.