I have the following problem (considering the vector space structure of $\mathbb{R}^n$):

Let $d: \mathbb{R}^n \times \mathbb{R}^n \to [0,\infty)$ be defined by: $$d(x,y)=\begin{cases}\| x-y \|, \mbox{if $x$ and $y$ are linearly dependent}\\ \|x\|+\|y\|, \mbox{otherwise} \end{cases}$$

Prove that $(\mathbb{R}^n,d)$ is a metric space.

I had no difficulties prooving that $d$ satisfies the propoerties of a metric function but the triangle inequality one.

For proving that $\forall x,y,z \in \mathbb{R}^n$ the triangle inequality holds, I divided by cases:

  1. $x,y$ are linearly independent but $x,z$ not;
  2. $x,y$ L.D. and $x,z$ L.D.;
  3. $x,y$ L.I but $x,z$ L.D.; and
  4. $x,y,z$ L.I.

I could prove all except 3. I have

$d(x,y)=\|x\|+\|y\|$, and $d(x,z)+d(z,y)=\|x-z\|+\|z\|+\|y\|$

I tried to use that $\vert \|a\|-\|b\| \vert \leq \|a+b\|$, but I couldn't...

Could you help me? Thanks in advance.

OBS.: I can't find a basic proof here to: prooving that the following are metrics (provided that $d$ is) $d_1(x,y)=\min\{1,d(x,y)\}; d_2(x,y)=\frac{d(x,y)}{1+d(x,y)}$; (I got stuck prooving triangle inequality too)

  • 1
    $\begingroup$ What does "L.D." mean in this context? $\endgroup$ – kimchi lover Aug 27 '18 at 21:32
  • $\begingroup$ @kimchilover I'd guess at linearly dependent $\endgroup$ – Tom Collinge Aug 27 '18 at 22:02
  • $\begingroup$ Linearly dependent! I'm sorry.. $\endgroup$ – Robson Aug 27 '18 at 22:46

For (3), if $x, z$ are linearly dependent then for some $\alpha$, $z = \alpha x$
And if $x, y$ are linearly independent then so are $ z, y$
So, $d(x, z) + d(z, y) = ||x - z|| + ||z|| + ||y||$
$= ||(1 - \alpha) x|| + ||z|| + ||y|| = |1-\alpha|.||x|| + ||\alpha x|| + ||y||$
$ = (|1 -\alpha| + |\alpha|) ||x|| + ||y||$
Which for any value of $ \alpha$ is $ \ge ||x|| + ||y|| = d(x, y)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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