Let $X=C[0,1]$ be the set of all continuous functions on the interval
$[0,1]$. Define:$$d_1(f,g)= \sup {\{ \left \lvert {f(t)-g(t)} \right \rvert : \ t \in [0,1]} \}$$
I want to expand this supremum metric of continuous functions defined on $[0,1]$ to the supremum metric of continuous functions on any interval $[a,b]$, that is:
$$d_2(f,g)= \sup {\{ \left \lvert {f(t)-g(t)} \right \rvert : \ t \in [a,b]} \}$$
The following hint was proposed to me:
There exists a bijection from $C[a,b]$ to $C[0,1]$. For every function $f(x)$ on $C[a,b]$ we can define a function $g(x)$ on $C[0,1]$ with:
$$g(x)=f \left(\frac{x-a}{b-a} \right)$$
I don't quite understand how this proves that if $d_1(f,g)$ is metric on $C[0,1]$, then $d_2$ is also a metric on $C[a,b]$. I imagine there is a certain theorem that connects these two.
Best Answer
You could show the following: If $(M,d_M)$ is a metric space and X some set and you have a bijection $h: X \to M$ then the function $d_X(x_1,x_2) := d_M(h(x_1),h(x_2))$ , $x_1,x_2 \in X$, is a metric on X.