Let $C > 0$ be some constant and $L^2([-C,C])$ the square integrable functions on $[-C,C]$. Let $\delta > 0$ and let $M_{|\cdot |^\delta}$ denote the multiplication operator on $L^2(\mathbb R)$ defined by the given function, i.e. $M_{|\cdot |^\delta} \phi(x) = |x|^\delta \phi(x)$ with the usual domain $D(M_{|\cdot |^\delta}) = \{ \phi \in L^2(\mathbb R) \colon M_{|\cdot |^\delta} \phi \in L^2(\mathbb R) \}$. Let $\mathcal F$ be the Fourier transform.
Could it be true that $\mathcal F^{-1} M_{|\cdot |^\delta} \mathcal F$ leaves $L^2([-C,C])$ invariant, i.e. that $\mathcal F^{-1} M_{|\cdot |^\delta} \mathcal F \phi \in L^2([-C,C])$ for all $\phi \in L^2([-C,C])$?