$ d(u,v)=\int_0^1| u^\prime(x)-v^\prime(x) |^2 dx $ is a metric or not?

Question

The function defined by $ d(u,v)=\int_0^1| u^\prime(x)-v^\prime(x) |^2 dx $ defines a metric over the space $C([0,1]).$

I have proved the trivial thing i.e

$ d(u,v)\ge 0 $ and $ =0 \iff u=v $ and $d(u,v)=d(v,u)\ \forall\ u,v\ \in C([0,1]).$

But I am unable to prove the triangle inequality part. Please help me.

score 3 · Answer 1 · answered Sep 23 '15 at 01:32

3

Note that $d(1,0)=0$ but $1\neq 0$.

answered Sep 23 '15 at 01:32

Pedro

122,002

Okay, I got it. – user273279 Sep 23 '15 at 01:34
C[0,1] usually denotes the space of continuous real-valued functions with domain [0,1]. Some of these functions do not have a derivative at any point. – DanielWainfleet Sep 23 '15 at 02:57

score 0 · Answer 2 · answered Sep 23 '15 at 01:37

0

I think you have proved it wrong. Because if you will choose any constant function $u(x)=c_1$ and $ v(x)=c_2 $ s.t $c_1\neq c_2$ then $d(u,v)=0$ but $u\neq v.$ Hence, it is not a metric.

answered Sep 23 '15 at 01:37

Sachchidanand Prasad

2,770

score 0 · Answer 3 · answered Sep 23 '15 at 01:47

In addition to $d(1,0) = 0$, the triangle inequality is not satisfied.

$$ d(a,b) = ∫|a'-b'|^2 =∫|a'|^2 + ∫|b|^2 - 2∫a'b' = d(a,0) + d(b,0) - 2∫a'b'$$

So choosing $a=-b$ will result in

$$ d(a,b) = ∫|2b'|^2 =∫|a'|^2 + ∫|b'|^2 + 2∫|b'|^2 = d(a,0) + d(b,0) + 2d(b,0) \geq d(a,0) + d(b,0)$$

To check, try $a=x=-b$: then the statement reduces to $|1-(-1)|^2 = 4 \geq 2 = 1^2 + 1^2 $

$ d(u,v)=\int_0^1| u^\prime(x)-v^\prime(x) |^2 dx $ is a metric or not?

3 Answers3