Proof $\int\psi(x)P(x)dx=((Id-\mathcal{L})P)(0)$

Question

I have a question on the Friz's book "A couse on rough path", exercise 13.8, I can not understant the following formula for $\psi$, Proof for any $\psi$ integrating to 1 one can find a differential operator $\mathcal{L}$ of order $\alpha$ with constant cofficients and without constant term such that $\int\psi(x)P(x)dx=((Id-\mathcal{L})P)(0)$ for all polynomials $P$ of degree $\alpha$.