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.

  • $\begingroup$ since $f$ is nonincreasing and even, with decay at infinity, isn't $f\ge 0$ and therefore $\left\|f|_{\mathbb{R}\backslash[-r,r]}\right\|_\infty = f(r)$? $\endgroup$ Commented Sep 14, 2020 at 8:39
  • $\begingroup$ But can we choose $F$ to be independent of $f$? $\endgroup$
    – geometricK
    Commented Sep 14, 2020 at 8:41
  • $\begingroup$ 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 $f$ $\endgroup$ Commented Sep 14, 2020 at 8:42
  • $\begingroup$ I agree :) I'm trying to simplify it. $\endgroup$
    – geometricK
    Commented Sep 14, 2020 at 8:43

1 Answer 1


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

  • $\begingroup$ 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. $\endgroup$
    – geometricK
    Commented Sep 14, 2020 at 12:53
  • $\begingroup$ Sorry to hear that. I am curious to see the "real" one $\endgroup$
    – nicomezi
    Commented Sep 14, 2020 at 15:16
  • $\begingroup$ Here is the follow-up question: math.stackexchange.com/questions/3826750/…. $\endgroup$
    – geometricK
    Commented Sep 15, 2020 at 8:50

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