Differential operators in $\Bbb R^n$

Question

Put $P_j=\frac{\partial}{\partial \xi_j}$ et $Q_j=2 i \xi_j$ with$\xi=\left(\xi_1, \ldots, \xi_n\right)$ et $x=\left(x_1, \ldots, x_n\right)$. How to prove :

1. $\exp \left(\sum_{j=1}^n x_j P_j\right)(f)(\xi)=f(\xi+x)$.
2. $\exp \left(\sum_{j=1}^n x_j Q_j\right)(f)(\xi)=\exp (2 i(x, \xi)) f(\xi)$. where $(x, \xi)=\sum_{j=1}^n x_j \xi_j$ and $i^2=-1$. Thanks

Answer 1

I want first to change your notations, sticking to the usual variables $x,\xi$ in the phase space. As a general statement about pseudo-differential operator with a symbol $a(x,\xi)$, I wish to write $$ \bigl(\text{Op}(a) f\bigr)(x)=\int e^{2iπ x\cdot \xi} a(x,\xi) \hat f(\xi) \, d\xi. $$ Let $y\in \mathbb R^n$ be given. Then we have $$ f(x+y)=\int e^{2iπ (x+y)\cdot \xi}\hat f(\xi) \, d\xi = \int e^{2iπ x\cdot \xi}e^{2iπ y\cdot \xi}\hat f(\xi) \, d\xi= \bigl(\operatorname{Op}(a_y) f\bigr)(x), \quad a_y(x,\xi)=e^{2iπ y\cdot \xi}, $$ which is your first formula.

Let $\eta\in \mathbb R^n$ be given. Then we have $$ e^{2iπ x\cdot \eta}f(x)=\int e^{2iπ x\cdot \xi}e^{2iπ x\cdot \eta}\hat f(\xi) \, d\xi=\bigl(\text{Op}(b_\eta) f\bigr)(x),\quad b_\eta(x,\xi)=e^{2iπ x\cdot \eta}, $$ and this is your second formula.