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?

2$\begingroup$ Can you prove this for twodimensional normed spaces? $\endgroup$– Gerald EdgarAug 19, 2018 at 0:36

$\begingroup$ @GeraldEdgar: no unfortunately ... $\endgroup$– user521337Aug 19, 2018 at 0:38

$\begingroup$ A naive comment: your condition on $V$ feels off the wall (but interesting!). Could you say more about it? $\endgroup$– Wlod AAAug 19, 2018 at 1:42

1$\begingroup$ @WlodAA : well .... I can prove that if $(V,<,>)$ is an inner product space inducing the norm $.$ on $V$, then for every group $(G,*)$, every function $f:G \to V$ satisfying $f(x*y)\ge f(x)+f(y),\forall x,y\in G$, is a group homomorphism i.e. $f(x*y)=f(x)+f(y),\forall x,y\in G$ .... hence the question whether this property actually characterizes which norms come from an innerproduct ... $\endgroup$– user521337Aug 19, 2018 at 3:38

3$\begingroup$ Hello @user521337 ... Can you provide the proof that in an inner product space, inequalities (Z) imply group homomorphism? $\endgroup$– Gerald EdgarAug 29, 2018 at 14:14
2 Answers
Let's begin with an example, which supports the conjecture: not inner product, map from a group, satisfies the norm inequalities, but not a homomorphism.
$V = l^1$, the Banach space of sequences $\mathbf{x} = (x_1,x_2,x_3,\dots)$ with norm $\\mathbf x\ = \sum_{k=1}^\infty x_k$. This norm does not come from an inner product: parallelogram law fails. Write $\mathbf{e}_n = (0,\dots,0,1,0,\dots)$ with a $1$ in the $n$th coordinate, and all others $0$.
Group $G = (\mathbb Z, +)$.
Function $f : \mathbb Z \to l^1$ defined by: $$ f(0)=\mathbf{0}, \quad f(n) = \sum_{k=1}^n \mathbf{e}_k,\quad f(n) = \sum_{k=1}^n \mathbf{e}_k,\quad n>0 $$ Note that $f$ is not a homomorphism, since $f(2) = \mathbf{e}_1+\mathbf{e}_2 \ne 2 \mathbf{e}_1 = f(1)+f(1)$.
Now we check that $\f(m+n)\ = \f(m)+f(n)\$ for all $m,n \in \mathbb Z$. So the required inequality is actually equality.
Case $m=0$: we get $f(m+n) = f(0+n)= f(n)$ and $f(m)+f(n) = \mathbf{0}+f(n) = f(n)$, so $\f(m+n)\ = \f(m)+f(n)\$.
Case $n=0$: same.
Case $m>0, n \ge m$:
$$
f(m+n) = \sum_{k=1}^{m+n}\mathbf{e}_k,\quad \f(n+m)\ = n+m,
\\
f(n)+f(m) = \sum_{k=1}^{m}\mathbf{e}_k + \sum_{k=1}^{n}\mathbf{e}_k
=\sum_{k=1}^m 2 \mathbf{e}_k+\sum_{k=m+1}^n \mathbf{e}_k,
\\
\f(n)+f(m)\ = 2m + nm = n+m
$$
Case $m>0,0<n<m$: switch $n,m$ in the previous case.
Case $m<0,n<0$: switch signs in the previous two cases.
Case $m>0, 0>n \ge m$: then $0 \le m+n < m$ and
$$
f(m+n) = \sum_{k=1}^{m+n} \mathbf{e}_k,\qquad \f(m+n)\ = m+n
\\
f(m)+f(n) = \sum_{k=1}^{m} \mathbf{e}_k  \sum_{k=1}^{n} \mathbf{e}_k
=\sum_{k=n+1}^{m} \mathbf{e}_k,
\\
\f(m)+f(n)\ = m(n) = m+n
$$
Case $m>0, n<m$: then $n+m < 0$ and
$$
f(m+n) = \sum_{k=1}^{(m+n)} \mathbf{e}_k,\qquad \f(m+n)\ = mn
\\
f(m)+f(n) = \sum_{k=1}^{m} \mathbf{e}_k  \sum_{k=1}^{n} \mathbf{e}_k
=\sum_{k=m+1}^{n} \mathbf{e}_k\\
\f(m)+f(n)\ = nm
$$
added august 20
A twodimensional example.
$V = l^1_2$, twodimensional $l^1$, the space of ordered pairs $\mathbf{x}=(x_1,x_2)$ with norm $\\mathbf{x}\ = x_1+x_2$. This is sometimes known as the taxicab metric.
Then the map $f : \mathbb Z \to l^1_2$
defined by
$$
f(0) = (0,0),\\
f(k) = (1,k1),\quad k\ge 1,\\
f(k) = (1,k+1),\quad k\ge 1.
$$
satisfies $\f(n+m)\ = \f(n)+f(m)\$ for all $n,m \in \mathbb Z$.
Next we need to investigate normed spaces $V$ with the property
$\qquad\x+y\ = \x\+\y\ \Longrightarrow$ one of $x,y$ is a nonnegative multiple of the other.
This property of a norm is strictly weaker than "induced by an inner product". For example, $l^p$ with $1 < p < \infty$. Is there a nonhomomorphism example in such a space? We would have to use a group other than $\mathbb Z$.

$\begingroup$ Other observations. Suppose $V$ is a normed space and $f : G \to V$ satisfies the required norm inequality. Then $f(e) = 0$ where $e$ is the unit of the group; $f(x^{1}) = f(x)$; $\f(x^n)\ = n\f(x)\$ for $n \in \mathbb N$; an element of finite order maps to $0$. $\endgroup$ Aug 19, 2018 at 19:02

$\begingroup$ Gerald, I don't see why $\f(x^n)\=n\f(x)\$ (I don't see why the OP's hypothesis implies, for instance, $\f(xyz)\\ge \f(x)+f(y)+f(z)\$). However, it holds that $\f(x^2)\\ge 2\f(x)\$, and hence $\f(x^{2^n})\\ge 2^n\f(x)\$. In particular,it's unbounded when $f(x)\neq 0$, which forces $x$ to have infinite order. $\endgroup$– YCorAug 29, 2018 at 13:57

$\begingroup$ Consequently, $\f(xy)\\ge\f(x)+f(y)\=\f(y)\$ for all $x$ of finite order and all $y$, and in turn $\f(xy)\=\f(y)\$ for all $x$ of finite order and all $y$, and in particular, $f=0$ on the subgroup $T$ generated by torsion elements (still this is not enough to show that $f$ is $T$invariant, i.e. $f(xy)=f(y)$ for all $x$ of finite order and all $y$). $\endgroup$– YCorAug 29, 2018 at 14:07

$\begingroup$ @YCor ... I have not supplied the proof of the assertion $\f(x^n)\ = n\f(x)\$, but Uri has. And it is true that any element of finite order must map to $0$. I do not claim that $\f(xyz)\\ge \f(x)+f(y)+f(z)\$ except in the case $x,y,z$ are all powers of a single element. $\endgroup$ Aug 29, 2018 at 14:07

$\begingroup$ Yes (I also gave a proof that $f$ vanishes on elements of finite order in my first comment). I didn't read Uri's post carefully enough. $\endgroup$– YCorAug 29, 2018 at 14:08
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 $$ (*) \quad \forall~t\in [0,1],\quad \tx+(1t)y\=1. $$ 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 $g,h\in G$, $gh\leqg+h$) 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 $$ ghg^{1}=gh^ng^{1}/n \leq (g+h^n+g^{1})/n=h+2g/n$$ 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 $G$.

$\begingroup$ It would be interesting to investigate $\mathbb Z^2$ or some nonabelian group. $\endgroup$ Aug 29, 2018 at 14:20

$\begingroup$ @Gerald, I thought about $\mathbb{Z}^2$ a little bit with no success. However, I added to my answer a few remarks, in view of your discussion with Yves. $\endgroup$ Aug 29, 2018 at 18:08

$\begingroup$ Let me also add the speculation that (under mild assumptions on $V$) the image of any such function $f$ is contained in a one dimensional subspace, hence $f$ is indeed a homomorphism. $\endgroup$ Aug 29, 2018 at 18:10