Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The title said it, I want to prove that $$ d(x+u, y + v) \le d(x, y) + d(u,v) $$ for every metric $d$. If the metric is induced by a norm, i.e. $d(x,y) := ||x-y||$, then this is easy. \begin{align*} d(x+u, y+v) & = ||x+u - (y+v)|| \\ & = ||x-y + u - v|| \\ & \le ||x-y||+||u-v|| \\ & = d(x,y) + d(u,v) \end{align*} But in the general case I have no idea how to get rid of the sums...

share|improve this question
    
What is the underlying space here? –  Sanchez Jan 22 '13 at 19:51
7  
$x+u$ isn’t meaningful in metric spaces in general; are you working specifically in $\Bbb R^n$? –  Brian M. Scott Jan 22 '13 at 19:51
add comment

2 Answers

up vote 4 down vote accepted

It isn’t even true in $\Bbb R$. The function

$$d:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto\left|\tan^{-1}x-\tan^{-1}y\right|$$

is a metric on $\Bbb R$. In this metric we have $d(-1,1)=\frac{\pi}2$, but $$\lim_{x\to\infty}d(x-1,x+1)=\lim_{x\to\infty}\left|\tan^{-1}(x-1)-\tan^{-1}(x+1)\right|=0\;,$$

even though $d(x,x)=0$ for all $x$.

More generally, suppose that it were true. Then you’d have $d(x+u,y+u)\le d(x,y)$ for all $x,y,u$, and hence $d(x,y)\le d\big((x+u)+(-u),(y+u)+(-u)\big)\le d(x+u,y+u)$ for all $x,y$, and $d$ would be translation-invariant. Clearly not all metrics on $\Bbb R^n$ are translation-invariant.

share|improve this answer
    
in what manner does your limit for $x\to \infty$ said that $d(x-1,x+1) \le d(x,x) + d(-1,1)$ could not hold? –  Stefan Jan 22 '13 at 20:07
    
@Stefan: It doesn’t: it shows that for sufficiently large $x$ the inequality $$d\big((x-1)+(-x),(x+1)+(-x)\big)\le d(x-1,x+1)+d(-x,-x)$$ cannot hold. –  Brian M. Scott Jan 22 '13 at 20:13
add comment

If I'm not mistaken, you can also take the pullback of the standard metric on $ℝ$ by $x ↦ x^3$, i.e. the metric $ℝ × ℝ → ℝ,\; (x,y) ↦ |x^3 - y^3|$, and choose $x=u=1$ and $y=v=-1$. Then the left hand side of the title inequality is $16$ and the right hand side is $2 + 2 = 4$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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