Schwartz space is closed under differentiation and multiplication by polynomials. In addition, if $f$ is a smooth function will all derivative bounded and $\psi$ is a Schwartz function, then $f\psi$ is also a Schwartz function. Why these statements are true?
[Math] Schwartz Space is closed under differentiation and multiplication by polynomials.
analysisfourier analysisfunctional-analysisschwartz-space
Best Answer
The first assertion is true because the finiteness of the suprema $\sup_t|t|^k|f^{(k)}(t)|$has to be check only for $k\geqslant 1$.
For multiplication by polynomials, we can notice that Schwartz space is a vector space, hence it suffices to do the proof for polynomials of the form $t^k$. One can argue by induction on $k$, or proceed directly from Leibnitz's formula.
Use Leibnitz' formula to compute the $n$-th derivative of the product. Then in order to control $\sup_t |t|^k|f^{(k)}(t)\psi^{(n-k)}(t)|$, use the fact that the function $t\mapsto \psi^{(n-k)}(t)$ is bounded .