Let $f$ be a function with compact support in $D\subset \mathbf{R}^2$. Is is known that $$-\Delta f=F^{-1}|x|^2 F f,$$ where $F$ is the Fourier transform. Which is the $n-$dimensional analogous formula?
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2this holds in any dimension, doesn't it? – Carlo Beenakker Feb 15 '14 at 20:45
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Essentially yes, but you have to worry about your class of functions and how Fourier transform is defined: multiplication by $|x|^2$ can bring you out of range of the Fourier transform. – Alexandre Eremenko Feb 16 '14 at 02:35
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When $f$ has compact support, its Fourier transform is in the Paley-Wiener space, and in particular is of rapid decay. Thus, multiplication by $|x|^2$ produces another function of rapid decay, certainly ok to apply Fourier transform... and the identity is correct, up to normalizing constant. – paul garrett Feb 16 '14 at 04:50
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@paulgarrett: if $f$ has compact support, its Fourier Transform is analytic, but it does not need to decay rapidly. If $f$ is smooth, its Fourier transform decays rapidly at infinity. – robjohn Feb 19 '14 at 12:10
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@robjohn, oop! Yes, indeed! :) – paul garrett Feb 19 '14 at 12:35
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As long as $f$ does not have any terrible singularities, it is a tempered distribution. Therefore we have $$ (f, -\Delta \varphi) = (f, F^{-1}|\xi|^2F\varphi) $$ when $\varphi$ is a Schwartz function (a function of rapid decay whose derivatives also decay rapidly). Thus $$ -\Delta f = F^{-1}|\xi|^2F f $$ where the operators $\Delta$, $F$, and $F^{-1}$ are interpreted in the sense of distributions. This is perhaps a vacuous statement, but the point is that, so long as $\Delta f$ is defined, you can write it in terms of the Fourier transform.
felipeh
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