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I was wondering whether there is a way to show this identity

$$\pi \int_{\mathbb{R}^3} \frac{f(x)}{|x|} dx = \int_{\mathbb{R}^3} \frac{\widehat{f(x)}}{|x|^2} dx $$ without using distributions for $f \in \mathcal{S}(\mathbb{R}^n)$.

The reason I ask is the following: Obviously this identity makes perfect sense in the classical way, but the most obvious proof would use the fourier transform of $\frac{1}{|x|^2}$ which is clearly not defined in $L^1$ or $L^2$.

Thus, I would be interested, whether one can proof this identity with classical methods, i.e. no distribution theory, too?

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  • $\begingroup$ And why are you interested in that? $\endgroup$
    – Johannes Hahn
    Commented Nov 15, 2015 at 19:09
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    $\begingroup$ That does not answer my question. There are a lot of questions that can be stated without X, yet one/some/all/the best/the most famous/... proof(s) use X. Especially if X is unconnected to the original question, this is often considered a good thing because it reveals non-trivial connections between different parts of mathematics. Here on the other hand, you want to exclude a theory that is literally made for handling questions about fourier transforms. Why would you want to do that? Do you have a specific goal in mind? $\endgroup$
    – Johannes Hahn
    Commented Nov 15, 2015 at 19:18
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    $\begingroup$ @JohannesHahn it is not directly related to a research problem (as so many question on this page are), but it can be used to get the fundamental solution of Laplace's equation very straightforwardly by fourier transform methods without the use of distribution theory or tedious calculations and i considered this as intersting. $\endgroup$
    – user82546
    Commented Nov 15, 2015 at 19:32
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    $\begingroup$ @user82546 I suspect you can't get something for nothing. Either you use distribution theory (which is abstract but powerful) or you perform tedious calculations (which is elementary but ... tedious). An example of the latter is to mollify $f$ at 0, show the identity in this case, and then show that it converges in the right sense to the right limit. Distribution theory is created precisely to automate/abstract away this tedious process. $\endgroup$
    – Fan Zheng
    Commented Nov 15, 2015 at 21:33
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    $\begingroup$ A belated comment: I like this question, and the stated motivation. There are some calculations/questions in abstract harmonic analysis where it is -- for me at least -- more pleasing to use Plancherel and other $L^2$-techniques rather than $C^\infty$ or distributional techniques. Sometimes these suggest generalizations to locally compact groups that aren't Lie, and where Lie approximation seems a bit of a heavy machine $\endgroup$
    – Yemon Choi
    Commented Mar 12, 2017 at 18:36

2 Answers 2

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I think the constant in the equality you wrote down depends on how exactly you normalize the Fourier transform. Up to worrying about the constant, here's a proof that should work and may satisfy your criterion:

In dimension $d$, write $$ \int \frac{\widehat{f(\xi)}}{ |\xi|^{2}} d\xi = \int_0^\infty ds \int \hat{f}(\xi) e^{-s |\xi|^2} d\xi $$ Up to the normalization constant involving $(2 \pi)^d$ and justifying the convergence, the right hand side equals (by Plancharel) $$ \int_0^\infty ds \int f(x) e^{- \frac{|x|^2}{4 s} } (2s)^{-d/2} dx $$ Interchange the integrals and change the s variable such that $\tilde{s}^2 = (4s)^{-1}|x|^2$; the above equals $$ C \int \frac{f(x)}{|x|^{d - 2}} dx \int_0^\infty ~ e^{-\tilde{s}^2} \tilde{s}^{d - 3} d\tilde{s} $$ for an explicit constant $C$. Note that the power $d - 3 > -1$ since we are in dimension $d > 2$. The value of the $d\tilde{s}$ integral is then known and that will give your formula. (Equivalently, this answer gives a way to compute the Fourier transform of $|\xi|^{-2}$.)

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@Phil Isett's device also does apply to arbitrary $\int_{\mathbb R^n} f(x)/|x|^\alpha\;dx$ with $0< \Re(\alpha)<n$, as follows, disregarding constants and normalization of Fourier transforms: $$ \int_{\mathbb R^n} {f(x)\over |x|^\alpha}\;dx \;=\; {1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} f(x)\,e^{-t|x|^2}\,t^{\alpha/2}\;dx\;{dt\over t} $$ $$ = {1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-{1\over t}|x|^2}\,t^{{\alpha-n\over 2}}\;dx\;{dt\over t} $$ by Plancherel and by knowing how to take Fourier transform of Gaussians. Replace $t$ by $1/t$ to obtain (up to constants) $$ {1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-t|x|^2}\,t^{{n-\alpha\over 2}}\;dx\;{dt\over t} = {\Gamma({n-\alpha\over 2})\over \Gamma(\alpha/2)} \int_{\mathbb R^n} {\widehat{f}(x)\over |x|^{n-\alpha}}\;dx $$

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