This arises in trying to understand the Uncertainty Principle. Suppose $\psi(x)$ is a Schwartz function on $\mathbb{R}^n$ such that $\|\psi\|_{2}=1$. Let $R_1$ denote the rectangle $[-1,1]^n$ in $\mathbb{R} ^n$. How can one show that $f(h),$ for $h$ a small parameter and $\epsilon > 0$ defined as
$$f(h):= h^{-n \epsilon}\int_{\mathbb{R}^n \setminus R_1} \left(\psi\left(\frac{x}{h^\epsilon}\right)\right)^2 dx$$
belongs to $O(h^\infty)$ as $h \to 0?$
I think this just follows from the definition of a Schwartz function. Fix $\epsilon > 0$ and let $I_h = [-\frac{1}{h^\epsilon},\frac{1}{h^\epsilon}]$. For $M \ge 2$, $|\psi(y)| \lesssim_M {|y|^{-M}}$ on the range $|y| \ge 1$ (say). Then, $\int_{\mathbb{R}^n\setminus I_h} \psi(y)^2 \lesssim_M \int_{\mathbb{R}^n\setminus I_h} \frac{1}{|y|^{2M}}dy \lesssim_M h^{\epsilon(2M-1)}$. As $M$ can be arbitrarily large, this shows $f(h) \in O(h^\infty)$.