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Let $\varphi:\mathbb{R}^n \to [0,\infty)$ be compactly supported and $C^{\infty}$. Define $k_s(x) = |x|^{-\alpha}$ for $x \in \mathbb{R}^n$, where $0 < \alpha < n$. I know that, as tempered distributions,
$$ \mathcal{F}(k_\alpha) = k_{n-\alpha} $$ and $$ \mathcal{F}(\varphi \ast k_\alpha) = \mathcal{F}(\varphi)\mathcal{F}(k_\alpha). $$ Here $\mathcal{F}$ denotes the Fourier transform.

Question Is the following true? If yes, how can it be proved? \begin{align*} \int_{\mathbb{R}^n} | \mathcal{F}(\varphi \ast k_\alpha) |^2 dx = \int_{\mathbb{R}^n} | \mathcal{F}(\varphi)\mathcal{F}(k_\alpha) |^2 dx \end{align*}

In my application, we can assume $2(\alpha-n) > -n$ so that the integral on the right is is finite, in case that helps.

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  • $\begingroup$ The obstruction is that $|x|^{-s}$ is not in $L^2(\mathbb R^n)$, nor in $L^1(\mathbb R^n)$, for any $n$ or $s\in\mathbb R$... So, for example, $2(\alpha-n)<-n$ only assures good behavior at $\infty$, but not at $0$. $\endgroup$ – paul garrett Mar 6 '18 at 19:50
  • $\begingroup$ @paulgarrett Sorry, I had the inequality the wrong way. $\endgroup$ – LucasSilva Mar 6 '18 at 19:54
  • $\begingroup$ But I fear that there is no inequality that can guarantee local integrability at both $0$ and $\infty$... $\endgroup$ – paul garrett Mar 6 '18 at 19:56
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    $\begingroup$ @paulgarrett Perhaps you could correct me. For $|x| \leq 1$, we use $|\mathcal{F}(\varphi) \mathcal{F}(k_{\alpha}) |^2 \leq C |k_{n-\alpha}|^2 = C|x|^{2(\alpha - n)}$. For $|x| \geq 1$, we use $|\mathcal{F}(\varphi) \mathcal{F}(k_{\alpha}) |^2 \leq |\mathcal{F}(\varphi) k_{n-\alpha} |^2 \leq |\mathcal{F}(\varphi)|^2$. Doesn't that work? $\endgroup$ – LucasSilva Mar 6 '18 at 20:10
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    $\begingroup$ Oop: while shoveling snow to clear my head (!), I realized that you're right: all you need is local integrability at $0$, and the multiplication by the FT of a test function (or Schwartz) is sufficient at $\infty$. I guess it's also good to note that for distributions given by locally integrable functions "pointwise multiplication" makes sense (and agrees with the multiplication of tempered distributions by moderate-growth smooth functions, etc). $\endgroup$ – paul garrett Mar 6 '18 at 21:13
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The real question here is to show that $F(\varphi*u)=F(\varphi)\cdot F(u)$, as $L^2$ functions, for Schwartz functions $\varphi$ and suitable tempered distributions $u$, where $F(\varphi)\cdot F(u)$ is pointwise multiplication. For the latter to make sense, we need to have a prior condition on $u$ so that (for example) $F(u)$ is a locally integrable function, so has pointwise values a.e., and is completely described by those values (as opposed to Dirac $\delta$, for example). It suffices to have $u$ be in some Sobolev space, for example (all of which are inside the space of tempered distributions, and have locally $L^2$ Fourier transforms).

Then there is the question of a good definition/characterization of $\varphi*u$ for Schwartz functions and tempered distributions. One characterization is that it is a tempered distribution such that, for every Schwartz function $\psi$, $(\varphi*u)(\psi)=u(\varphi*\psi)$. For this to make sense, we need to know that $\varphi*\psi$ is Schwartz, which is indeed the case, by a variety of arguments.

Then, ignoring some signs (which disappear in the end), $$ F(\varphi*u)(\psi) \;=\; (\varphi*u)(F\psi) \;=\; u(\varphi*F\psi) \;=\; u(F(F\varphi\cdot \psi)) \;=\; (Fu)(F\varphi\cdot \psi) \;=\; (F\varphi\cdot Fu)(\psi) $$ by the usual characterization of multiplication of tempered distributions by Schwartz functions. This is the desired identity.

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  • $\begingroup$ I wonder if you can help me understand something a bit more. Your answer is really helpful. Now I definitely see that $F(\varphi \ast u) = F(\varphi) \cdot F(u)$ as tempered distributions, and that (with some condition on $u$), the RHS is an $L^2$ function. However, I don't see that this proves $F(\varphi \ast u)$ is an $L^2$ function. It shows that there exists an $L^2$ function $g$ such that $F(\varphi \ast u) = g$ as tempered distributions. With this in mind, I think the integral $\int |F(\varphi \ast u)|^2 dx$ I have written in the question does not really make sense. What do you think? $\endgroup$ – LucasSilva Mar 11 '18 at 16:00
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    $\begingroup$ Well, in one sense perhaps $F(\varphi * u)$ has not been proven to "be" an $L^2$ function, but in another sense we have proven that the distribution is given by integration-against-an-$L^2$-function. Since all the operations on distributions are (by design) compatible extensions of the corresponding operations on (integrate-against-) functions, operationally that distribution "is" an $L^2$ function. That is, it behaves in every respect like (integration-against-) an $L^2$ function. This is the criterion for equality of distributions... Is this addressing the issue? $\endgroup$ – paul garrett Mar 11 '18 at 16:08
  • $\begingroup$ Absolutely. What you wrote was exactly my understanding, I just wanted to confirm it. $\endgroup$ – LucasSilva Mar 11 '18 at 16:14
  • $\begingroup$ :) .............. $\endgroup$ – paul garrett Mar 11 '18 at 16:24

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