Showing that the quotient of a function belonging to Schwartz Space by a strictly positive polynomial remains in the Schwartz Space

Question

Let $S\mathcal(R)$ be the Schwartz space. I want to show that if $f \in S\mathcal(R)$ then $f/P \in S\mathcal(R)$ where P is any strictly positive polynomial. I succeeded to show it for the case of $g(x)=e^{-x^{4}}/(1+x^{2})$ but i did not in the general case. Can someone help? Thank you.

Answer 1

Since $P$ is strictly positive, all you need to show is that $x^m D^k (f/P)$ is bounded on all of $\mathbb{R}$ for all $m$ and $k$; in particular, by continuity of everything in sight, it suffices to show that $x^m D^k (f/P)$ is bounded outside of some neighbourhood $[-\epsilon,\epsilon]$ of $0$.

Suppose that $P$ is of degree $n$, and write $P = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0$. Observe that $$ P(x) = a_n x^n \left(1+ \frac{a_{n-1}}{a_n} x^{-1} + \cdots + \frac{a_0}{a_n} x^{-n} \right) = a_n x^n (1 + g(x)), $$ where $g$ is continuous, indeed smooth, for $x \neq 0$, and satisfies $g(x) \to 0$ as $x \to \infty$. Then, for any $m \in \mathbb{N}$, when $x \neq 0$, $$ x^m \frac{f(x)}{P(x)} = x^m \frac{f(x)}{a_n x^n (1+g(x))} = \frac{x^{m-n}f(x)}{a_n(1+g(x))}, $$ which vanishes as $x \to \infty$ if $m < n$ and is bounded outside of $[-\epsilon,\epsilon]$ if $m \geq n$, precisely because $f \in S(\mathbb{R})$ and $\lim_{x \to \infty} g(x) = 0$.

Now, in general, along exactly the same lines, one can check that $$ P^{(k)}(x) = a_{n,k} x^{n-k}(1+g_k(x)), \quad a_{n,k} := n(n-1) \cdots (n-k+1)a_n, $$ where, once more, $g_k$ is smooth for $x \neq 0$ and satisfies $g_k(x) \to 0$ as $|x| \to +\infty$. Do you see how to show that all the derivatives of $f/P$ vanish quickly enough, using the quotient rule?