When does a functional from a vector's span to $\mathbb{R}$ have a unique norm-preserving extension to $\mathbb{R}^n$? 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.
 A: 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 $B(0,1)$.

The same goes for the case in which $v$ is in any edge of the cube, but of course not if $v$ is an interior point of one
of its faces.  Precisely, this means that $v$ must have one, and no more that one, coordinate with absolute value 1, so
voilà the
condition for the Hahn-Banach extension to be unique!
