Operator norm composition operators

Question

Let $g\in C^1(\mathbb{R},[0,1])$ and consider the associated composition/Koopman operator on $L^2([0,1])$ defined for any $f\in L^2([0,1])$ by $$ C_g(f)(x) :=f\circ g(x). $$ Are there known bounds for the operator norm of $C_g$?

Answer 1

The operator $C_g$ may be unbounded. Let $g(x)=x^2.$ Then the function $f(x)=x^{-1/4}$ is belongs to $L^2(0,1),$ while $f\circ g(x)=x^{-1/2}$ doesn't.

As was observed by @daw the operator $C_g$ is well defined if the level sets $\{x\in [0,1]\,:\, g(x)=a\}$ have Lebesgue measure equal $0.$