TITLE: Operator norm composition operators
QUESTION [0 upvotes]: 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 $C_g$?

REPLY [1 votes]: 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 $0.$
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