TITLE: On a special type of normed linear spaces
QUESTION [17 upvotes]: Let $(V,\|.\|)$ be a normed linear space such that for every group $(G,*)$, every function $f:G \to  V$ satisfying
$$
\|f(x*y)\|\ge \|f(x)+f(y)\|,\qquad\forall x,y\in G,\tag{Z}
$$ is a group homomorphism i.e. $f(x*y)=f(x)+f(y),\forall x,y\in G$. Then is it true that the norm on $V$ comes from an inner product?

REPLY [3 votes]: Not a full answer: the space should be strictly convex.
In fact, if you make your condition with the restricted demand that the group is $\mathbb{Z}$, then it is actually equivalent to the norm being strictly convex.
Recall that the norm is not strictly convex if there exist distinct $x,y\in V$ such that 
$$ (*) \quad \forall~t\in [0,1],\quad \|tx+(1-t)y\|=1. $$
Assuming this is the case we will be done by letting $f:\mathbb{Z}\to V$ be defined by
$$
f(n)=
\begin{cases}
nx & n\neq\pm 1 \\
y & n= \pm 1 
\end{cases}
$$
Assuming the norm is strictly convex,
using the observation that $f(-n)=-f(n)$ and scaling $x=f(1)$ to be of norm 1 it is enough to show by an induction on $n\in \mathbb{N}$ that for $y=f(n+1)-f(n)$ we have $(*)$.
This follows from the following two lines:
$$ \|f(n+1)\|\geq \|f(n)+f(1) \|=n+1 $$
$$ 1 =\|f(1)\| =\|f((n+1)+(-n))\| \geq \|f(n+1)-f(n)\| $$
from which you deduce first that $\|f(n+1)\|=n+1$ and then that 
both $x$ and $y$ are on the intersection of the unit sphere and the sphere of radius $n$ around $f(n+1)$.

Let us go back now to a general group $G$ and assume, in view of the above, that $V$ is strictly convex.
I claim that $g\mapsto |g|:=\|f(g)\|$ is a conjugation invariant seminorm on $G$ (recall that a seminorm on $G$ is a function $|\cdot|:G\to [0,\infty)$ satisfying for every $g,h\in G$, $|gh|\leq|g|+|h|$) which is homogeneous (that is it satisfies for every $n\in \mathbb{Z}$, $|g^n|=|n|\cdot |g|$).
The fact that $|\cdot |$ is a seminorm is easy:
$$ \|f(xy)\|\leq \|f(xy)-f(y)\|+\|f(y)\| \leq \|f(x)\|+\|f(y)\| $$
and the fact that it is homogeneous follows from the case $G=\mathbb{Z}$ discussed above.
The fact that $|\cdot |$ is conjugation invariant actually follows formally from the previous two facts: for $g,h\in G$, the inequality
$$ |ghg^{-1}|=|gh^ng^{-1}|/n \leq (|g|+|h^n|+|g^{-1}|)/n=|h|+2|g|/n$$
shows, by taking the limit on $n\to \infty$, that $|ghg^{-1}|\leq |h|$,
but substituting in this inequality $ghg^{-1}$ for $h$ and $g^{-1}$ for $g$ we get the reverse inequity.
Let my now make the remark that for a conjugation invariant seminorm $|\cdot |$ on a group $G$ it makes sense to define its kernel $K<G$ by
$$ K=\{g\in G\mid |g|=0 \} $$
and note that it is a normal subgroup and the seminorm descents to a well defined norm (that is, a seminorm with a trivial kernel) on the quotient group $G/K$. 
In our consideration we thus allowed to replace $G$ with $G/K$. Note that any group which admits an homogeneous norm is torsion free. In particular, we may assume this is the case for $G$.