This question is a follow-up to this post, but hopefully it's a better attempt at formulating the same idea. Roughly speaking, I would like to obtain a uniform estimate on the size of a certain class of Schwartz function "away" from the support of its Fourier transform.
More precisely, let $\mathcal{C}$ be the class of all even, non-negative Schwartz functions on $\mathbb{R}$ satisfying the following conditions:
- The Fourier transform $\hat{f}$ is compactly supported;
- $f$ is non-increasing on $[0,\infty)$;
- $\|f\|_\infty\leq 1$.
Question: For any $\epsilon>0$, does there exist a continuous function $F_\epsilon:[0,\infty)\to[0,\infty)$ such that:
- $F$ is non-decreasing with $\displaystyle\lim_{x\to\infty}F_\epsilon(x)=\infty$;
- $f(F_\epsilon(r_f))\leq\epsilon$ for every $f$ in $\mathcal{C}$,
where $r_f$ is any positive number such that $\text{supp}(\hat{f})\subseteq[-r_f,r_f]$?
Remark: Clearly when $\epsilon\geq 1$, there is no obstruction to the existence of $F_\epsilon$, so the question is really about $0<\epsilon\leq 1$.
The question is a generalization to the family $\mathcal{C}$ of the following example:
Example: Fix some $f$ in $\mathcal{C}$. Then the family of rescaled functions $\mathcal{C}^f:=\{f_t=f(t\cdot)\}_{t>0}$ is a subclass of $\mathcal{C}$. We know there exists some $C_f>0$ such that $f(C_f)\leq\epsilon$, hence $f_t(\frac{C_f}{t})\leq\epsilon$. Since $\hat{f}_t(\cdot) = \frac{1}{t}\hat{f}(\frac{1}{t}\cdot)$, the function $F_\epsilon(x)=\frac{C_f x}{r_f}$ works for the family $\mathcal{C}^f$.