Show that $(\mathcal{C}([0,1]),\|\cdot\|_{\infty})$ is isometric to $(\mathcal{C}([a,b]),\|\cdot\|_{\infty})$

Question

Here I think it is implict that we have to use the metric induced by the norm.
Also, an isometry here means a surjective function that preserves distances between points.

Well, I tried to find a isometric embedding that is also surjective. But I couldn't, and I think that the function that solves my problem isn't to intuitive (so I won't find it too easily) and so I came here asking for it… (hope this isn't duplicated).

My bounded creativity suggested me only $\varphi(f)(x)=f(x)b +(1-f(x))a$, but I think $\varphi$ is not surjective (and it doesn't even preserves distances hehe :/)

Help me, please!

Answer 1

You have the right idea, except you should apply it to the domain not the codomain of $f$, i.e., $$ \varphi(f)(x)=f(a+(b-a)x) $$ for each $f\in C([0,1])$.