Let $X: L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ be a multiplication operator, i.e. $f(x) \mapsto xf(x)$ for $f \in L^2$. Multiplication operators are known to be self-adjoint on some dense subset of $L^2$, and so by Stone's theorem on one-parameter unitary groups $X$ is the infinitesimal generator of some one-parameter group of operators $U(t)$ such that $U(t) = e^{itX}$.
I am interested in explicitly finding the action of $e^{itX}$ on an $L^2$ function $f$. In an analogy with the heat semigroup, I think this can be determined by thinking of $e^{itX}$ as the solution to the differential equation $$\partial_t f = ixf$$ but I don't think this would make sense since $f$ is only assumed to be a function of $x$.
How can one find and/or describe how $e^{itX}$ acts on $L^2$ functions? Can anything be said about $e^{tx}$?
Not sure what the differential equation angle gives you. Maybe someone else can answer that. But the action is very easy to describe.
If $X$ is the map $f(x) \mapsto xf(x)$, then $U(t)$ is the map $f(x) \mapsto e^{itx}f(x)$. It’s still an action by pointwise multiplication, but now it’s a one-parameter ($t$) family of unitary operators on $L^2$.