How to prove
$|-\Delta_{x} ((-x)^{\alpha} \phi(x))|\leq A_{j,\alpha}(1+|x|)^{-n-1}$
Hi all, i am reading Pseudo-differential Operators, singularities, applications. Y Egorov, on page 2. The inequality above is for $x\in\mathbb R^n$ and $\alpha$ multi index and $\phi$ in Schwartz spaces.
My attempt is for the simple case n=1 so $\alpha$ now is a natural number $m$, so $-\partial_{x}^{2}(-x^m\phi(x))$ then
$-\partial_{x}^{2}(-x^m\phi(x))=-\partial_{x}\partial_x (-x^m\phi(x))= -\partial_x (-mx^{m-1}\phi (x)-x^m\phi_x)= -\left(-m(m-1)x^{m-2}\phi(x)-mx^{m-1}\phi_x-mx^{m-1}\phi_x-x^m\phi_{xx} \right)= m(m-1)x^{m-2}\phi(x)-2mx^{m-1}\phi_x+x^{m}\phi_{xx}.$
Now, when i want to use that $\phi\in S$ or equivalently $\sup |x^{\alpha}D^{\beta}\phi(x)|<\infty$ the only thing that i have is the inequality above is finite, but the affirmation says in our case that $|-\partial_{x}^{2}(-x^m\phi(x))|< A_{1,m}(1+|x|)^{-2}$
How can to arrive in something like the main statement, please i will appreciate any hint, thank you
(Remark, of course i want to proof for the general case, but i think that i need to star with the simple case)
I think it boils down to checking the following two statements about Schwartz functions $\phi(x),\psi(x)$.
A slight generalization of 1. is that $p(x)\phi(x)$ is Schwartz whenever $p(x)$ is a polynomial. In any case, applying 1. and 2. sequentially to a given Schwartz function $\phi(x)$ (resp. $\psi(x):=(-x)^\alpha\phi(x)$) ensures that $f(x):= \Delta_x((-x)^\alpha \phi(x))$ is Schwartz. Then the estimate $|f(x)|\le C_{\alpha,n}(1+|x|)^{-(n+1)}$ follows from the definition of $f(x)$ being a Schwartz function.
I suspect the particular choice of exponent $(n+1)$ is chosen to justify the integration of $f(x)$ in a following step, and it may be unnecessarily distracting, since of course once you realize that $f(x)$ is Schwartz because of the properties 1. and 2., you know $|f(x)|\le C_{\alpha,N}(1+|x|)^{-N}$ for any $N$.