Let $\mathcal{C}$ be the class of all even Schwartz functions $f:\mathbb{R}\to\mathbb{R}$ satisfying the following conditions:
- The Fourier transform $\hat{f}$ is compactly supported;
- $f$ is non-increasing on $[0,\infty)$;
- $\|f\|_\infty\leq 1$.
Question: Does there exist a continuous function $F:[0,\infty)\to[0,\infty)$ such that:
- $F$ is non-increasing with $\displaystyle\lim_{x\to\infty}F(x)=0$;
- $\left\|f|_{\mathbb{R}\backslash[-r,r]}\right\|_\infty\leq F(r)$ for all $r\in[0,\infty)$ and every $f$ in $\mathcal{C}$?
Motivation: The question is motivated by the following observation. For a single fixed $f$ in $\mathcal{C}$, the family of rescaled functions $\{f(\frac{1}{t}\cdot)\}_{t>0}$ lies in $\mathcal{C}$. An element $f(\frac{1}{t}\cdot)$ in this family has Fourier transform $t\hat{f}(t\cdot)$, so that as $\text{supp}\hat{f}$ gets larger, the subset of $\mathbb{R}$ on which $f(\frac{1}{t}\cdot)$ is "small" also gets larger. The question above is an attempt at asking whether there is a uniform estimate for all functions in $\mathcal{C}$, not just those obtained by scaling.
Let $f$ a non-null function satisfying the hypothesis and define $f_a(x)=f(ax)$. For any $a>0$, $f_a$ also lies in $\mathcal{C}$.
We know that $f(1) > 0$ as $f$ cannot be compactly supported as well unless it is zero everywhere and $f$ is non-increasing on $\mathbb{R}^+$;
We have $f_a(\frac 1 a)=f(1)$. By letting $a \to 0$, $f_a$ will reach $f(1)>0$ at arbitrary large values. Hence we cannot find a continuous function decreasing to $0$ at infinity such that $f_a(r) \le F(r)$ for all $a>0,r \in \mathbb{R}$.
(As noted in the comments, $\left\|f|_{\mathbb{R}\backslash[-r,r]}\right\|_\infty=f(r)$)