# Uniform estimate on Schwartz functions with compactly supported Fourier transform

Let $$\mathcal{C}$$ be the class of all even Schwartz functions $$f:\mathbb{R}\to\mathbb{R}$$ satisfying the following conditions:

1. The Fourier transform $$\hat{f}$$ is compactly supported;
2. $$f$$ is non-increasing on $$[0,\infty)$$;
3. $$\|f\|_\infty\leq 1$$.

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.

• since $f$ is nonincreasing and even, with decay at infinity, isn't $f\ge 0$ and therefore $\left\|f|_{\mathbb{R}\backslash[-r,r]}\right\|_\infty = f(r)$? Commented Sep 14, 2020 at 8:39
• But can we choose $F$ to be independent of $f$? Commented Sep 14, 2020 at 8:41
• I'm aware of the meaning of your question (and don't know the answer yet), just thought that it was a lot of symbols to write something simple. In particular it was very clear that you were looking for a bound independent of $f$ Commented Sep 14, 2020 at 8:42
• I agree :) I'm trying to simplify it. Commented Sep 14, 2020 at 8:43

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, $$\left\|f|_{\mathbb{R}\backslash[-r,r]}\right\|_\infty=f(r)$$)

• Thanks for the answer. I realized that the question is not really the one I wanted to ask, as $f_a$ should not be a counter-example in the correct formulation. I'll ask again in a separate question soon. Commented Sep 14, 2020 at 12:53
• Sorry to hear that. I am curious to see the "real" one Commented Sep 14, 2020 at 15:16
• Here is the follow-up question: math.stackexchange.com/questions/3826750/…. Commented Sep 15, 2020 at 8:50