Cayley transform for unbounded operator M(f)(t)=t(f)(t)

Question

Let $M : C_c(\mathbb{R})\to L^2(\mathbb{R})$ be the unbounded, densely defined and symmetric operator defined by $M(f)(t)=t\cdot f(t)$. I want to determine the Cayley transform $$U=(M-iI)(M+iI)^{-1}$$ of $M$. I started with looking at $$U(M+iI)f(t)=(M-iI)f(t),$$ thus $$U(t\cdot f(t)+if(t))=t\cdot f(t)-if(t),$$ and from this it seems that $U$ is just the operation of complex conjugation (since $t$ and $f(t)$ must be real), but is there something I am missing?

All the best.

Answer 1

$M$ is multiplication by $t$, and $U$ is multiplication by $(t-i)/(t+i)$, which is a unitary multiplication operator on $L^2$.