For a function $f$, recall the Fourier trnasformation $\widehat{f}(u)=\int_R e^{iux}f(x)dx$ (Maybe someone call it Fourier inversion, but it doesn't matter). Now let $T$ be a bounded self-adjoint operator from $L^2(R, dx)$ into itself. How to understand the notation $\widehat{\psi}(T)$, where $\psi$ is in the Schwart space?
Thanks!
If $F$ is a Borel function on $\mathbb{R}$, then you can define $F(T)$ using the spectral theorem. If $T$ has the spectral representation $$ Tf = \int \lambda dE(\lambda)f, $$ then $$ F(T)f = \int F(\lambda)dE(\lambda)f. $$ The domain of $F(T)$ consists of all $f$ for which $$ \int |F(\lambda)|^{2}d\|E(\lambda)f\|^{2} < \infty. $$ For your case, you want $\hat{\psi}(T)$. The function $\hat{\psi}$ is particularly well-behaved because $\psi$ is well-behaved. That should lead to a bounded operator.