# triangle inequality for a certain norm

Let $d$ be a metric on a (say real) vector space $E$, with the property $$d(x,x+cy)=|c|d(x,x+y)$$ for all $x,y\in E$ and scalars $c$. I am trying to prove that $x\mapsto d(x,0)$ defines a norm.

The triangle inequality gives me trouble (the others are easy): I need $d(x+y,0)\leq d(x,0)+d(y,0)$ for all $x,y\in E$.

Some thoughts: setting $y=\lambda x$ and $c=-1$ in the 'property' gives $d(x,(1-\lambda)x)=d(x,(1+\lambda)x)$, in particular $d(x,0)=d(x,2x)$.

The usual triangle inequality for $d$ gives $d(x+y,0)\leq d(x+y,x)+d(x,0)$, so also $2d(x+y,0)\leq d(x+y,x)+d(x+y,y)+d(x,0)+d(y,0)$.

I need something like translation invariance, or express $d(x+y,0)$ in terms of $d(y,0)$ ('get rid of the sum').

Some hints/suggestions? Thanks!

-
Are you sure your condition is sufficient? I get the idea ; you want the metric to be compatible with scalar multiplication, but maybe this is not enough, because none of your axioms allow compatibility with addition. – Patrick Da Silva Feb 9 '12 at 22:49
@PatrickDaSilva: I am sure insofar as this is an exercise in lecture notes (perhaps I should have added this in my question). I share your thoughts on addition compatibility...but perhaps there is some trick. – wildildildlife Feb 9 '12 at 23:07
Is this homework or just you trying to have fun? Perhaps you should add the homework tag if it is. – Patrick Da Silva Feb 10 '12 at 6:36
@PatrickDaSilva: it's not something I have to do or hand in, but I guess it can be regarded as homework. I added the tag. – wildildildlife Feb 10 '12 at 7:32
You say it's homework if you have to hand it in ; the point of this tag is that we want to know that we're helping you get a better grade. If it's your personal pleasure then the tag is not pertinent. – Patrick Da Silva Feb 10 '12 at 9:08

Be it homework or not – this is a very fine problem indeed!

It is enough to consider a plane $E$ with points $z=(x,y)$. Let $|z|:=\sqrt{x^2+y^2}$ denote the euclidean norm.

(i) Claim: For each line $g\subset E$ there is a constant $\lambda_g$ such that for all $z$, $w\in g$ one has $$d(z,w)=\lambda_g\ |z-w|\ .$$ Proof. Let $$g:\quad t\mapsto z(t)=a + t\ u\ ,\qquad |u|=1$$ be a parametrization of $g$ with euclidean arc length as parameter and put $$d^*(s,t):=d\bigl(z(s),z(t)\bigr)\ .$$ Then by the basic property of $d(\cdot,\cdot)$ we have for arbitrary $s$, $t\in{\mathbb R}$: $$d^*(s,t)=d^*\bigl(s,s+(t-s)\bigr)=|t-s|d^*(s,s+1)$$ and similarly $$d^*(s,t)=d^*(t,s)=d^*\bigl(t,t+(s-t)\bigr)=|t-s|d^*(t,t+1)\ .$$ For $t:=0$ the last two equations imply $|s|d^*(s,s+1)=d^*(s,0)=|s|d^*(0,1)$. It follows that $d^*(s,s+1)=d^*(0,1)=:\lambda_g$ for arbitrary $s\in{\mathbb R}$, whence $$d\bigl(z(s),z(t)\bigr)=d^*(s,t)=\lambda_g |t-s|=\lambda_g\bigl|z(t)-z(s)\bigr|\ ,$$ as claimed.

(ii) Claim: The scaling factor $\lambda_g$ depends only on the direction vector $u=(\cos\phi,\sin\phi)$ of $g$. In fact there is a continuous function $\lambda:\ S^1\to{\mathbb R}_{>0}$ such that $\lambda_g= \lambda(\phi)$.

Proof. It is enough to consider the following situation: Let $g_1$ be the line $y=1$ and $g_2$ be the line $y=-1$, and let $p_i:=\lambda_{g_i}$. We have to show that $p_1=p_2$. Consider for small $\phi >0$ the line $$g_\phi:\quad t \mapsto t \ (\cos\phi,\sin\phi)\ .$$ The line $g_\phi$ has a certain scaling factor $\lambda(\phi)$ and intersects $g_1$ at the point $b:=(\cot\phi, 1)$. Put $(0,0)=:o$, $(0,1)=:a$. Then we have by the triangle inequality for $d$: $$d(a,b)-d(o,a)\leq d(o,b)\leq d(o,a)+d(a,b)$$ or $$p_1\cot\phi - d(o,a)\leq \lambda(\phi)\ {1\over\sin\phi}\leq d(o,a)+p_1\cot\phi\ .$$ Multiplying by $\sin\phi>0$ we obtain $$p_1 \cos\phi -d(o,a)\sin\phi \leq\lambda(\phi)\leq p_1\cos\phi+d(o,a)\sin\phi\ ,$$ from which we conclude $\lim_{\phi\to 0+}\lambda(\phi)=p_1$. By symmetry (arguing on the left side) we also have $\lim_{\phi\to 0-}\lambda(\phi)=p_1$, so that in fact $\lim_{\phi\to0}\lambda(\phi)=p_1$. Using symmetry again it follows that $p_2=p_1$. Since the lines $g_1$, $g_2$ had nothing special about them it follows that any two parallel lines have the same scaling factor $\ \bigl(=:\lambda(\phi)\bigr)$; in particular the scaling factor of $g_0\colon\ y=0$ is also $\ =p_1$, which proves the continuity of the function $\lambda(\cdot)$. In particular, this function is bounded away from $0$ and $\infty$.

(iii) It follows from (i) and (ii) that the metric $d(\cdot,\cdot)$ is translation invariant. We now define the associated norm by $$\|z\|\ :=\ d(o,z)\ .$$ The identity $\|\alpha\ z\|=|\alpha|\ \|z\|$ is obvious, and $$\|z+w\|=d(0,z+w)\leq d(o,z)+d(z,z+w)=d(o,z)+d(o,w)=\|z\|+\|w\|$$ proves the triangle inequality for the norm.

-
Something's not clear in your proof of (i) : You show that $d^*(s,t)$ depends only on $d^*(s,s+1)=d^*(t,t+1)$ up to the scaling factor $|t-s|$, but nothing shows that $d^*(s,s+1) = d^*(0,1)$. It is not explicitly clear, and I must say I don't even see why it would hold. – Patrick Da Silva Feb 11 '12 at 14:22
@Patrick Da Silva: I have inserted an additional line. Hope it is clear now. – Christian Blatter Feb 11 '12 at 14:31
Thank you very much for your answer! Especially Claim 2 and its proof are quite clever. – wildildildlife Feb 12 '12 at 9:57