# On a special type of normed linear spaces

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?

• Can you prove this for two-dimensional normed spaces? Aug 19, 2018 at 0:36
• @GeraldEdgar: no unfortunately ... Aug 19, 2018 at 0:38
• A naive comment: your condition on $V$ feels off the wall (but interesting!). Could you say more about it? Aug 19, 2018 at 1:42
• @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 inner-product ... Aug 19, 2018 at 3:38
• Hello @user521337 ... Can you provide the proof that in an inner product space, inequalities (Z) imply group homomorphism? Aug 29, 2018 at 14:14

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 $m>0, n \ge m$: $$f(m+n) = \sum_{k=1}^{m+n}\mathbf{e}_k,\quad \|f(n+m)\| = n+m, \\ f(n)+f(m) = \sum_{k=1}^{m}\mathbf{e}_k + \sum_{k=1}^{n}\mathbf{e}_k =\sum_{k=1}^m 2 \mathbf{e}_k+\sum_{k=m+1}^n \mathbf{e}_k, \\ \|f(n)+f(m)\| = 2m + n-m = n+m$$ 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 $$f(m+n) = -\sum_{k=1}^{-(m+n)} \mathbf{e}_k,\qquad \|f(m+n)\| = -m-n \\ f(m)+f(n) = \sum_{k=1}^{m} \mathbf{e}_k - \sum_{k=1}^{-n} \mathbf{e}_k =-\sum_{k=m+1}^{-n} \mathbf{e}_k\\ \|f(m)+f(n)\| = -n-m$$

A two-dimensional example.
$V = l^1_2$, two-dimensional $l^1$, 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 $$f(0) = (0,0),\\ f(k) = (1,k-1),\quad k\ge 1,\\ f(-k) = (-1,-k+1),\quad k\ge 1.$$ 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 non-homomorphism example in such a space? We would have to use a group other than $\mathbb Z$.

• 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 $0$. Aug 19, 2018 at 19:02
• 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.
– YCor
Aug 29, 2018 at 13:57
• 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 $y$).
– YCor
Aug 29, 2018 at 14:07
• @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. Aug 29, 2018 at 14:07
• 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.
– YCor
Aug 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 $$(*) \quad \forall~t\in [0,1],\quad \|tx+(1-t)y\|=1.$$ 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 $G$.

• It would be interesting to investigate $\mathbb Z^2$ or some nonabelian group. Aug 29, 2018 at 14:20
• @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. Aug 29, 2018 at 18:08
• 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. Aug 29, 2018 at 18:10