For a non-singular transformation $T$ the operator $f\mapsto f\circ T$ is an isometry on $\mathcal L^\infty(X,\mu)$

Question

Let $(X,\mu)$ be a measure space and $T:X\to X$ be a non-singular transformation, that is, $\mu(T^{-1}(A))=0$ if and only if $\mu(A)=0.$ Now consider a map $\Phi:\mathcal L^\infty(X,\mu) \to \mathcal L^\infty(X,\mu)$ be defined by $\Phi(f)=f\circ T$ for all $f\in\mathcal L^\infty(X,\mu)$. Now I want to show that $\Phi$ is a linear isometry.
Now for linearity, let $\lambda\in \mathbb C$ and $f,g\in \mathcal L^\infty(X,\mu)$. Then we have, for any $x\in X$, \begin{align*}\Phi(f+\lambda g)(x) &= (f+\lambda g)(T(x))\\ &=f(T(x))+\lambda (g(T(x)))\\ &=\Phi(f)(x)+\lambda\Phi(g)(x).\end{align*} So I have proved the linearity. Also I proved that $\Phi$ is an isometry on $\mathcal L^2(X,\mu)$ and also got here when $T$ is measure preserving but I am unable to show that $\Phi$ is an isometry on $\mathcal L^\infty(X,\mu)$ when $T$ is non-singular, that is, I want to show $\|f\|_\infty=\|f\circ T\|_\infty.$ Please help me to solve this. Thanks in advance.

Answer 1

I think non-singular transformations are supposed to be bijective. Otherwise, the statement is false. Counter-example: Let $X=[0,1]$ with Lebesgue measure, $Tx=\frac x 2$, $f(x)=0$ for $x \le \frac1 2$ and $f(x)=2$ for $x>\frac 1 2$. Then, $\|\Phi(f)\|=0$ since $f(T(x))=0$ for all $x$. But $\|f\|=2$.

Suppose $|f(x)| \le M$ whenver $x \in E$ where $\mu(E^{c})=0$. Then, $|f(T(x))|\leq M$ whenever $Tx \in E$ i.e. whenever $x\in T^{-1}(E).$ It follows that $|f\circ T|\leq M$ a.e. It follows from the definition of $L^{\infty}$ norm that $\|\Phi(f)\|=\|f\circ T|\leq \|f\|$. The reverse inequality follows in a similar way.