I am studying first order systems of the form \begin{equation} L=\partial_t+K(t,x,D_x)\text{ where }D_x=-i\partial_x \end{equation} There is a change of variable and operator of our concern becomes \begin{equation} L_1=\big<D_x\big>^{-t}L\big<D_x\big>^{t} \end{equation} where $\big<\xi\big>=(1+|\xi|^2)^{1/2}$.
$L_1$ takes the form \begin{equation} L_1=\partial_t+K(t,x,D_x)+\log{\big<D_x\big>}+A \end{equation} where operator $A$ corresponds to lower order terms.
My question is about the form of $L_1$. How is it that $\big<D_x\big>^{-t}\partial_t\big<D_x\big>^{t}$ still gives a term $ \partial_t$? My calculations show only the term $\log{\big<D_x\big>}$.
Thank you in advance.
We should look at operators as if they are acting on functions. $\big<D_x\big>^{-t}\partial_t\big(\big<D_x\big>^{t}u(t,x)\big)=\big<D_x\big>^{-t}\big((\partial_t\big<D_x\big>^{t})u(t,x)+\big<D_x\big>^{t}\partial_tu(t,x)\big)\big)=\big<D_x\big>^{-t}\big(\big<D_x\big>^{t}\log\big<D_x\big>+\big<D_x\big>^{t}\partial_t\big)u(t,x)$