1

Prove that if $d$ is a metric and $(X,d)$ is a metric space then $D(x,y)=(d(x,y))^2$ is also a metric on $X$.

I have problem of showing that $D(x,y)$ is well-defined and proving the triangle inequality to prove that $D(x,y)$ is a metric on $X$.

Please help me, thank you very much!

Math1000
  • 36,983
MATH14
  • 347
  • 2
    Thinking out loud a bit: $D(x,z)=(d(x,z))^2 \leq (d(x,y)+d(y,z))^2 = d(x,y)^2 + 2d(x,y) d(y,z) + d(y,z)^2 = D(x,y) + 2 d(x,y) d(y,z) + D(y,z)^2$. Can the inequality fail when the middle term is dropped? – Ian Sep 19 '15 at 23:15
  • Hmm this is true it holds because the middle term is positive so this is cleared now i just want to be sure about the well defining this metric – MATH14 Sep 19 '15 at 23:18
  • The problem is the other way around: it can happen that the middle term "saves" the inequality, and it would fail without this additional term. The examples in the answers demonstrate this point. – Ian Sep 19 '15 at 23:40
  • Yes i saw now that theoretically it changes nothing but as we find at least one example where the triangle inequality doesnt hold that means there something wrong – MATH14 Sep 19 '15 at 23:45
  • Also, typo earlier: at the end you get $D(x,z) \leq D(x,y)+D(y,z) + 2d(x,y)d(y,z)$. – Ian Sep 19 '15 at 23:53

2 Answers2

5

The assertion is false. Take an isosceles triangle with one angle equal to $120$ degrees. Then the sides are $1$, $1$, and $\sqrt{3}$. But $1^2+1^2\lt 3$.

André Nicolas
  • 507,029
3

$d(x,y)=\vert x-y\vert$ then $D(1,4)\leq D(1,3)+D(3,4)$ wont be correct. since $\vert 1-4\vert ^2=9> \vert 1-3\vert ^2+\vert 3-4\vert ^2=4+1=5$

R.N
  • 4,318
  • triangle inequality is convese, isnot it? – R.N Sep 19 '15 at 23:37
  • infact you have to have $\vert 1-4\vert ^2=9\leq \vert 1-3\vert ^2+\vert 3-4\vert ^2=4+1=5$ which isnot correct – R.N Sep 19 '15 at 23:38
  • Hmm so we found an example for the metric d(x,y) for which the triangle inequality doesnt hold so does it imply that D(x,y) is not a metric since the third condition doesnt hold ? – MATH14 Sep 19 '15 at 23:41
  • you are quite right – R.N Sep 19 '15 at 23:44