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In his book linear Analysis , Bollobás says

Given a metric $d$ on a vector space $X$, setting $||x||=d(x,0)$ defines a norm on $X$ iff $d(x,y)=d(x+z,y+z)$ and $d(\lambda x,\lambda y)=|\lambda|d(x,y)$ for all $x,y,z \in X$ and scalar $\lambda$.

The part where the distance is homogeneous and translation invariant implies that $||x||=d(x,0)$ is a norm is quite easy and I'm ready with that.

I'm stuck in the other side. In particular, I have no idea how to prove that the distance must be translation invariant (homogeneity is simple).

Any advice would be a great help.

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    I guess I would have set $|x-y| = d(x,y)$? – copper.hat Aug 19 '18 at 22:27
  • @copper.hat But this definition assumes that $d$ is translation-invariant, otherwise that norm is not well defined – Lorenzo Q Aug 19 '18 at 22:36
  • It can be a norm iff it is translation invariant, so I'm not sure what you are saying? – copper.hat Aug 19 '18 at 22:38
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    What I mean is if you start from an arbitrary metric $d$ on $X$, with $d(x+z,y+z)\neq d(x,y)$, then your definition yields two conflicting values for $|x-y|$: one from $d(x+z,y+z)$ and one from $d(x,y)$. – Lorenzo Q Aug 19 '18 at 22:42

1 Answers1

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The statement is not true.

Lets consider $\mathbb{R}^2$ as a metric space using the so-called "French railway distance" $\delta$. The distance $\delta(A,B)$ from a point $A$ to $B$ is the usual (euclidean) distance $\overline{AB}$ if they lie on the same ray from the origin $O$. Otherwise, $\delta(A,B)=\overline{AO}+\overline{OB}$.

We can write $\delta(A,B)=\begin{cases} |A|+|B|, & \text{if $A\neq \lambda B$} \\ |A-B|, & \text{if $A= \lambda B$ for some $\lambda$} \end{cases}$

I will show that $\delta$ is a distance function. In fact:

  1. $\delta(A,B) \geq 0\ \forall\ A,B\in \mathbb{R}^2$ and $\delta(C,C)=0\ \forall C \in \mathbb{R}^2$
  2. $\delta(A,B)=\delta(B,A)$. This is clear for any case.

  3. $\delta(A,B)\leq \delta(A,C)+\delta(C,B)$. Here we have a lot of cases:

    • $\fbox{$A\neq \lambda B$. $C\neq \lambda A$ and $C\neq \lambda B.$}$ Then $\delta(A,B)=|A|+|B|\leq |A|+|C|+|C|+|B|=\delta(A,C)+\delta(C,B)$.

    • $\fbox{$A\neq \lambda B$, $C\neq \lambda A$ but $C= \lambda B$ for some $\lambda$.}$ Then $\delta(A,B)=|A|+|B|=|A|+|B-C+C|\leq |A|+|C|+|B-C|=\delta(A,C)+\delta(C,B)$.

    • $\fbox{$A\neq \lambda B$, $C\neq \lambda B$ but $C= \lambda A$ for some $\lambda$.}$ As in the previous case, here inequality is also true by symmetry.
    • $\fbox{$A\neq \lambda B$, $C= \lambda_1 A$ for some $\lambda_1$ and $C= \lambda_2 B$ for some $\lambda_2$.}$ Cannot be possible.
    • $\fbox{$A= \lambda B$ for some $\lambda$, $C\neq \lambda_1 A$ (and then $C\neq \lambda_2 B$).}$ Then $\delta(A,B)=|A-B|\leq |A|+|B|\leq |A|+|C|+|B|+|C|=\delta(A,C)+\delta(C,B)$.
    • $\fbox{$A= \lambda B$ for some $\lambda$, $C= \lambda_1 A$ for some $\lambda_1$ (and then $C= \lambda_2 B$ for some $\lambda_2$).}$ Then $\delta(A,B)=|A-B|=|A-C+C-B|\leq |A-C|+|C-B|=\delta(A,C)+\delta(C,B)$.

And it is homogeneous, because clearly $\delta(\rho A,\rho B)=|\rho|\delta(A,B)$ for any case.

However, it is not translation invariant. If we take A=(3,0), B=(0,3), then $\delta(A,B)=6$ and translating by $C=(4,0)$ we have $A+C=(7,0)$ and $B+C=(4,3)$ so $\delta(A+C,B+C)=7+5=12.$

Now, we see that setting $||A||=d(A,0)$ defines a norm on $\mathbb{R}^2$. In fact:

i)$||A||=0$ iff $\delta(A,0)=0$ iff $A=0$.
ii)$||\lambda A||=\delta(\lambda A,0)=|\lambda A| = |\lambda||A| = |\lambda|\delta(A,0) =|\lambda|||A||$.
iii)$||A+B||=\delta(A+B,0)=|A+B|\leq |A|+|B|= \delta(A,0)+\delta(B,0)=||A||+||B||$.

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