17
$\begingroup$

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?

$\endgroup$
7
  • 2
    $\begingroup$ Can you prove this for two-dimensional normed spaces? $\endgroup$ Aug 19, 2018 at 0:36
  • $\begingroup$ @GeraldEdgar: no unfortunately ... $\endgroup$
    – user521337
    Aug 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 AA
    Aug 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 inner-product ... $\endgroup$
    – user521337
    Aug 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? $\endgroup$ Aug 29, 2018 at 14:14

2 Answers 2

3
$\begingroup$

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 $$ f(m+n) = -\sum_{k=1}^{-(m+n)} \mathbf{e}_k,\qquad \|f(m+n)\| = -m-n \\ f(m)+f(n) = \sum_{k=1}^{m} \mathbf{e}_k - \sum_{k=1}^{-n} \mathbf{e}_k =-\sum_{k=m+1}^{-n} \mathbf{e}_k\\ \|f(m)+f(n)\| = -n-m $$

added august 20
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 $\mathbb Z$.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Aug 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 $y$). $\endgroup$
    – YCor
    Aug 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. $\endgroup$
    – YCor
    Aug 29, 2018 at 14:08
3
$\begingroup$

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 $G$.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Uri Bader
    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. $\endgroup$
    – Uri Bader
    Aug 29, 2018 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.