question on fourier transform. I ask myself what
$$
   {\mathscr F}^{-1}( e^{it\xi} ({\mathscr F} \phi)'(\xi) )(s)
$$
is.  If it was just about 
$$
   {\mathscr F}^{-1}( e^{it\xi} ({\mathscr F} \phi)(\xi) )(s)
$$
it would be clear (a shift by $t$), the same is with
$$
   {\mathscr F}^{-1}(  ({\mathscr F} \phi)'(\xi) )(s),
$$
which gives a multiplication by $-is$. 
But what is about the combination of exponential and derivative? Any hints? Thanks, Eric

Why don't you just compute it?
$$
   {\mathscr F}^{-1}( e^{it\xi} ({\mathscr F} \phi)'(\xi) )(s)={\mathscr F}^{-1}( ({\mathscr F} \phi)'(\xi) )(s+t)=-i(s+t)\phi(s+t)
$$
