The Schwartz space, $S(\mathbb R): = \{f\in C^{\infty}(\mathbb R): \sup_{x\in \mathbb R} |(1+|x|)^{\alpha} D^{\beta}f(x)|< \infty , \forall \alpha, \beta \in \mathbb N \cup \{0\} \}$
Let $\phi \in \mathcal{S}$. Can we expect to find $a>0$ such that
$$| x^{r} \phi (x)| \leq |\phi (a)| \ (|x|>a, 1<r<2). $$
Edit: We take $1<r<2.$
No. Let $\phi\in\mathcal S$ be such that $\phi(x)=e^{-x}$ for $x>1$. Then, for any $r>1$, $x^r\,\phi(x)$ attains its maximum at $x=r$, and $\lim_{r\to\infty}x^r\,\phi(r)=\infty$. Given any $a>0$, you can allways find $r>a$ such that $x^r\phi(r)>\phi(a)$.
Suppose now that we limit the value of $r$ to lie in an interval $(1,R)$. Still it is not possible. Letting $x\to a$, we obtain the inequality $$ a^r\,|\phi(r)|\le|\phi(a)|,\quad 1<r<R. $$ Since we must have $|\phi(a)|\ne0$, it follows that $a^r\le1$ and $0<a\le1$. Then the inequality does not hold for any $\phi$ vanishing on $[-1,1]$.