Isometry between $C[a,b]$ and $C[0,1]$

Question

I need to find isometry between two spaces of continuous functions $C[a,b]$ and $C[0,1]$. That means to find function $ \phi\colon C[a,b] \longrightarrow C[0,1] $ which is bijection and $d_{\infty}(f,g)=d_{\infty}(\phi(f),\phi(g))$. I know that there is bijection between $[a,b]$ and $[0,1]$ $y=(b-a)x +a$. I have idea to every function $g(y) $ from $C[a,b]$ join function $g((b-a)x +a)$ from $C[0,1]$. But I don't know how to prove that this is bijection. I would accept any other way of solving this.

Answer 1

Simply define $\phi(f)(x)=f\bigl((b-a)x+a\bigr)$. Its inverse is $\phi^{-1}(f)(x)=f\left(\frac{x-a}{b-a}\right)$. Besides\begin{align}d_\infty\bigl(\phi(f),\phi(g)\bigr)&=\sup_{x\in[0,1]}\bigl|f\bigl((b-a)x+1\bigr)-g\bigl((b-a)x+1\bigr)\bigr|\\&=\sup_{x\in[a,b]}\bigl|f(x)-g(x)\bigr|\\&=d_\infty(f,g).\end{align}

Answer 2

Define $\psi:[0,1]\to[a,b]$ by $\psi(x)=(b-a)x+a$, and define $\phi:C([a,b])\to C([0,1])$ by $\phi f=f\circ\psi$. That $\phi$ is a bijection follows from the fact that $\psi$ is a bijection: Define $\rho:C([0,1])\to C([a,b])$ by $\rho f=f\circ \psi^{-1}$. Then $\rho\phi f=f\circ(\psi\circ\psi^{-1})=f$ and $\phi\rho f=f\circ(\psi^{-1}\circ\psi)=f$.

That $\phi$ is an isometry can be shown as follows: since $\phi$ is linear, it suffices to show that $\|\phi f\|=\|f\|$ for all $f\in C([a,b])$. For such $f$, note that $$\{|f(x)|:x\in[a,b]\}=\{|f(\psi(x))|:x\in [0,1]\}=\{|\phi f(x)|:x\in[0,1]\},$$ so taking supremums yields $\|f\|=\|\phi f\|$.