Proving that $(\operatorname{sech}(x))^a$ is a Schwartz function for every a>0

Question

I wanna show that, for every $a>0$ , the function $h_a$ defined as

\begin{equation} h_a(x) = \frac{1}{(\cosh(x))^a} \end{equation}

is a Schwartz function, where $x \in \mathbb{R}$.

Remembering that a function $f$ is a Schwartz function if satisfies, for every $m,q = 0,1,2,\dots $, the propertie that

\begin{equation} \lim_{\vert x \vert \rightarrow \infty} (1 + \vert x \vert)^m\vert f^{(q)}(x)\vert = 0 \end{equation}

where of course $f^{(q)} = \frac{d^q}{dx^q}f(x)$.

Answer 1

$h_a$ is clearly $C^\infty$. For the behavior at infinity, observe that $$ 0<(\operatorname{sech}x)^a<2^a\,e^{-a|x|}. $$ This gives you the desired estimates for $h_a$. You will have to do some work for the derivatives.