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Given a function $f$ (in some suitable (EDIT: nice) function space) supported on a ball $B\subset\mathbb{R}^n$ of radius $R>0$, I have commonly heard people, who understand these things, say stuff like

"the Fourier transform $\hat{f}$ is morally supported on the dual ball"

where they mean by the dual ball, a ball $\tilde{B}$ of radius $R^{-1}$ whose center "does not matter".

  • What is precisely meant by this?

I would imagine it to mean something like: "there exists a uniform estimate of $|\hat{f}|$ outside the moral support giving some sort of nice decay". But in any such interpretation the center of that dual ball definitely does matter. It can't be that this function somehow has it's moral support on all possible translates of dual balls at the same time.

I understand there are several rigorous ways of expressing the uncertainty principle for Fourier transforms, one of which is the following:

Under the assumptions above,

$$\mathrm{osc}_{\tilde{B}} \hat{f} := \sup_{x,y\in \tilde{B}} |\hat{f}(x)-\hat{f}(y)|\le C\|\hat{f}\|_\infty$$

for some uniform constant $C>0$ and $\tilde{B}$ a ball of radius $R^{-1}$. This is a consequence of Bernstein's inequality and it means that $\hat{f}$ is roughly constant on balls of radius $R^{-1}$.

  • How is that related to $\hat{f}$ being morally supported on a ball of such radius?

Clearly this somehow doesn't fit to my above interpretation since here the center of the dual ball really does not matter since we're only talking about the function's oscillation, rather than it's size.

I would appreciate if somebody could order my thoughts and precisely tell me how these things are connected and what is meant by moral support.

Anonymous999
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  • The suitable function space is important here. If $f$ is badly discontinuous inside of $B$, its Fourier transform will have a heavy tail. The statement is true for very nice functions like Gaussians. The center does not matter because they talk about how spread-out $\hat f$ is. Also, it may help to observe that the transform of $f(\lambda x)$ is a multiple of $\hat f(\lambda^{-1}\xi)$. – 40 votes Jul 16 '13 at 19:42
  • Ok, assume $f\in C_0^\infty$. For example a real-valued bump function with support in $[-1,1]$. Then $\hat{f}(\xi)$ should be a function that's "big" in $[-1,1]$ with Schwartz decay as $|\xi|\rightarrow\infty$ right? What is confusing me is just that its big exactly in $[-1,1]$ and not in say $[-1+c,1+c]$ for any $c\in\mathbb{R}$. So the center does matter? The moral support should be $[-1,1]$ and not $[-1+c,1+c]$ for arbitrary $c$? What am I missing here? – Anonymous999 Jul 16 '13 at 20:11
  • Also I realize that if I now take that $f$ and multiply it by a modulating factor like $e^{i\lambda x}$ for some $\lambda\in\mathbb{R}$, the $\hat{f}$ will look the same just translated according to $\lambda$. Could that be part of what they mean by "the center doesn't matter"? – Anonymous999 Jul 16 '13 at 20:14
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    Of course, $R^{-1}$ is the same as $100R^{-1}$ in moral statements. Concerning smoothness - it is not enough to assume infinite differentiability, you need a quantitative smoothness assumption. Modulation will worsen the quantitative smoothness. Main thing is, people who know what they are talking about also know the context of their statements, sometimes implied but never stated. Without knowing this context you are wasting your time trying to make a precise statement out of nothing. – 40 votes Jul 16 '13 at 20:51
  • I guess you're right that I didn't provide enough context. I'll try and ask that question in a more precise context later. Thanks for your help. – Anonymous999 Jul 16 '13 at 20:59
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    One more remark: think of "A is morally B" as "Morally, A is B". They did not mean that there is a notion of "moral support". The meaning is that "the following statement (A is supported on B) is false when taken literally but true if suitably interpreted". – 40 votes Jul 16 '13 at 21:21

1 Answers1

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In this answer, I derive that $$ \|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2 $$ This relates the "mean-square" size of the support of $f$ and $\hat{f}$ and shows that they are inversely proportional or greater. That is, $$ \frac{\|xf\|_2}{\|f\|_2}\frac{\|\xi\hat{f}\|_2}{\|\hat{f}\|_2}\ge\frac{n}{4\pi} $$

This seems to fit with the ball of radius $R^{-1}$.

If we define the operator $S_rf(x)=r^{n/2}f(rx)$, then the mean-square size of the support of $S_rf$ is the mean square size of the support of $f$ divided by $r$. Furthermore, $$ \begin{align} \widehat{S_rf}(\xi) &=r^{n/2}\int f(rx)\,e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\\ &=r^{-n/2}\int f(x)\,e^{-2\pi ix\cdot\xi/r}\,\mathrm{d}x\\ &=r^{-n/2}\hat{f}(\xi/r)\\ &=S_{1/r}\hat{f}(\xi) \end{align} $$ That is, $S_r$ operating on $f$ divides its mean-square radius by $r$, but multiplies the mean-square radius of its Fourier Transform by $r$.

Moral Support

I think "morally supported" is partly a play on words as well as a heuristic guideline. I don't think those who use this term intend any precise definition for the "moral support" of a function, but rather they are promoting the idea that when you scale down the support of a function, you scale up the support of its Fourier Transform. Thus, when a function has a small support, its Fourier Transform should morally (that is, if everything else is going nicely in the universe) have a large support, and vice versa.

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robjohn
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  • Thanks a lot for your answer. This is not really what I meant, I knew these things and I'm still confused about what "moral support" means. I admit that my question is not really clear. – Anonymous999 Jul 16 '13 at 21:01
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    @Anonymous999: I've added to my answer what I think is meant by "moral support" and "morally supported". – robjohn Jul 16 '13 at 21:33