Prove the following map is an isometry.

Question

I have a map $$T :C[0,1] \to C[a,b]$$ such that $$T(f)=f \circ s$$ where $s=a+t(b-a)$ is a homeomorphism from $[0,1] \to [a,b]$. ($C[0,1]$ and $C[a,b]$ are the spaces of continuous functions on $[0,1]$ and $[a,b]$ respectively). Prove $$T(f)=f \circ s$$ is an isometry on the metric $$d(f,g)=\sup_{t\in[a,b]}\bigl|f(t)-g(t)\bigr|.$$ So for $f,g \in C[0,1]$ I have $$d_{C[0,1]}(T(f),T(g))=d_{C[0,1]}(f \circ s,g \circ s)$$ Now I can't proceed from there, I was thinking that the fact that $s$ is a homeomorphism gives $f \circ s=f$ since the domain and range of $s$ are similar enough but I am not sure if this is true. Can you help me proceed?

Answer 1

If the distance in $C[a,b]$ is $d(f,g)=\sup_{t\in[a,b]}\bigl|f(t)-g(t)\bigr|$, then that's just because$$\sup_{t\in[0,1]}\bigl|f(s(t))-g(s(t))\bigr|=\sup_{t\in[a,b]}\bigl|f(t)-g(t)\bigr|.$$This follows from the fact that $s$ is a bijection.