Define $M_{f} : L^2(X, \mu) \rightarrow L^2(X, \mu)$ to be $M_fh = fh$.
Assume the following
- $f, g: X \rightarrow [-1, 1]$ are measurable functions
- $U : L^2(X, \mu) \rightarrow L^2(X, \mu)$ is a unitary map such that $UM_fU^{-1} = M_g$
Prove that
- $UM_{\varphi \circ f}U^{-1} = M_{\varphi \circ g}$ for bounded measurable $\varphi : [-1, 1] \rightarrow \mathbb R$.
Here's what I have done so far. Consider the following proven facts:
Let $\varphi_n: [-1,1] \rightarrow \mathbb R$ be bounded measurable functions converging pointwise to $\varphi [-1, 1] \rightarrow \mathbb R$.
Suppose that $(\varphi_n)_{n \in \mathbb N}$ is uniformly bounded. Then for any $h \in L^2(X, \mu)$ $$\lim_{n \rightarrow \infty} M_{\varphi_n \circ f}h = M_{\varphi\circ f}h$$ where the convergence is in $L^2(X, \mu)$ norm.
Suppose $UM_{\varphi_1 \circ f}U^{-1} = M_{\varphi_1 \circ g}$ and $UM_{\varphi_2 \circ f}U^{-1} = M_{\varphi_2 \circ g}$ and let $c > 0$. Then $$UM_{(\varphi_1+\varphi_2) \circ f}U^{-1} = M_{(\varphi_1+\varphi_2) \circ g} \quad \text{and} \quad UM_{c\varphi_1 \circ f}U^{-1} = M_{c\varphi_1 \circ g}$$
Since $\varphi$ is bounded, there is a sequence of simple functions $(\varphi_n)_{n \in \mathbb N}$ such that $\varphi_n \nearrow \varphi$.
Hence, by the proven facts, it suffices to show that $UM_{\chi \circ f}U^{-1} = M_{\chi \circ g}$ for any characteristic function $\chi$.
This is where I am stuck.
I would appreciate any help or reference. If there's a different approach, that would be great, too.