Prove that the normed spaces $(C[0,1], \| \cdot\|_2)$ and $(C[a,b], \|\cdot \|_2)$ where $\| \cdot\|_2$ is the Euclidean norm are isometric. Essentially I'm looking for a bijection $f: C[0,1]\to C[a,b]$ such that$$\|f(x) \|_2=\| x\|_2$$ I don't know how to go about finding this function, but I do know that it is possible.
$$\| x \|_2 = \sqrt{\int_a^b x(t)^2\,dt} 
$$
What if the norms in question were:
 $$\| x\|_{\infty}= \sup_{t \in [a,b]}|x(t)|$$
 $$\| x \|_1 = \int_a^b |x(t)| \, dt$$
How would these bijections look if asked to prove the "isometricity" of $$(C[0,1], \| \cdot\|_i) \text{     and    } (C[a,b], \|\cdot \|_j) i\neq j\ \ ; i,j\in{1,2,\infty}$$
 A: Consider a function $g\in C[0,1]$. Let $f[g]\in C[a,b]$ be defined by $f[g](x)=\frac{1}{\sqrt{b-a}}g(\frac{x-a}{b-a})$. You should be able to prove that's an isometry just by u-substitution!
The basic idea is, you translate/stretch the function so that it covers the new interval, then renormalize it to give it the same norm as before. You can apply the same idea to the other norms. Stretching it is the same, the only thing that differs is the renormalization factor. For example: the $\sup$ norm doesn't need renormalization.
A: Let $f(x)=y$ where $y(t) = x\left( \dfrac{t-a}{b-a} \right)\cdot\dfrac 1 {\sqrt{b-a}}$ for $a\le t\le b$.  Then
$$
\|y\|_2^2=\int_a^b (y(t))^2\,dt=\int_a^b \left(x\left( \frac{t-a}{b-a} \right)\right)^2 \frac{dt}{b-a} = \int_0^1 (x(u))^2\,du = \| x \|_2^2
$$
Let $g(x)=z$ where $z(t)= x\left( \dfrac{t-a}{b-a} \right)$  Then
$$
\|z\|_\infty = \sup_{a\le t\le b} |z(t)| = \sup_{0\le u\le 1} |x(u)| = \|z\|_\infty.
$$
In both of these, the substitution $u = \dfrac{t-a}{b-a}$ is used.
