Suppose $f$ is a smooth function compactly supported in some ball of radius $R$. Is $\|\widehat{f}\|_{L^{p}(\mathbb{R}^d)}$ comparable to $\|\widehat{f}\|_{L^{p}(B_{1/R})}$ where $B_{1/R}$ is any ball of radius $1/R$? Here by comparable I mean, does there exists an absolute constant $C$ such that $\|\widehat{f}\|_{L^{p}(\mathbb{R}^d)} \leq C\|\widehat{f}\|_{L^{p}(B_{1/R})}$?

I have heard that heuristically this is true by the uncertainty principle (that is, the Fourier transform of such a function is morally supported on the dual ball), but I do not know how to make this rigorous.

  • $\begingroup$ For $p=2$, this follows from the Amrein-Berthier theorem (google it, or google "uncertainty principle, harmonic analysis"). $\endgroup$ – user138530 Dec 18 '15 at 22:32

I have to confess I don't know the Amrein-Berthier theorem. But the answer to the question is no; whatever that theorem says about this must be somewhat different.

In one dimension, for simplicity, let $$f_A(t)=e^{iAt}f(t).$$Then $f_A$ is supported in $B$ as well. And $$\hat f_A(\xi)=\hat f(\xi-A),$$so $$||\hat f_A||_p=||\hat f||_p,$$while (assuming that $\hat f\in L^p$ to begin with)$$\lim_{A\to\infty}\int_K|\hat f_A|^p=0$$for any compact $K$.

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  • $\begingroup$ Excellent counter example. $\endgroup$ – Stella Biderman Dec 18 '15 at 22:50

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