My problem is to prove the following identity: $$C_{\alpha}\int_{\mathbb R^n} \frac{1}{|x|^\alpha} \phi(x) dx = C_{n-\alpha}\int_{\mathbb R^n} \frac{1}{|x|^{n-\alpha}} \widehat{\phi}(x) dx$$ where $\phi:\mathbb R^n \to \mathbb C$ is on the Schwartz space , $0<\alpha <n$, $C_{\beta} = \Gamma(\beta/2)/\pi^{\beta/2}$ and $$\widehat{\phi}(x) = \int_{\mathbb R^n} \phi(y)e^{-2\pi i x\cdot y}dy.$$

In other words, how to calculate fourier transform of $1/|x|^\alpha$?


Denoting by $\widehat{\cdot}$ the Fourier transform in $\mathbb R^d$ and letting $$ f_\lambda(x)\equiv f(\lambda x), $$ then, by definition of the Fourier transform, \begin{align} \widehat {f_\lambda}(k)&=\int_{\mathbb R^n}e^{-ik\cdot x}f(\lambda x)dx\\ &=\int_{\mathbb R^n}e^{-ik\cdot x/\lambda}f(x)\lambda^{-d}dx\\ &=\lambda^{-d}\widehat f(k/\lambda). \end{align} Let $f(x)=F(|x|)$ be a radial function; if $M\in SO(d)$, then $f_M(x)\equiv f(Mx)=f(x)$ and hence, since $\widehat{ f_M}(k)=\widehat f(Mk)$, we have $\widehat {f_M}= \widehat f$, which in turn implies that also $\widehat f$ is radial. Now, suppose $f(x)=|x|^{-\alpha}$, for $0<\alpha<d$, then $$ f_\lambda(x) = \lambda^{-\alpha}f(x) $$ and, by the above remark, $$ \widehat f (k/\lambda)= \lambda^{d}\widehat {f_\lambda}(k) = \lambda^{d-\alpha}\hat f(k). $$ The only radial $\widehat f$ homogeneous of degree $-d+\alpha$ is $$ \widehat f (k) = |k|^{\alpha-d}, $$ up to constants.

Note that $f$ can be split as the sum of two pieces: let $B_a$ be the ball of radius $a$ centred at the origin, then \begin{align} f&\equiv u+v\\ &= \chi_{B_a}(x)\frac{1}{|x|^{\alpha}}+\left(1-\chi_{B_a}(x)\right)\frac{1}{|x|^{\alpha}}; \end{align} $u$ lies in $L^1(\mathbb R^d)$, for $0<\alpha<d$, and $v$ lies in $L^2(\mathbb R^d)$, for $\alpha>d/2$. This ensures $\widehat u \in L^\infty(\mathbb R^d)$ and $\widehat v\in L^2(\mathbb R^d)$ and that $\widehat f$ needs indeed to be a function in $L^1_{\text{loc}}(\mathbb R^d)$ from an abstract point of view. So $f,\widehat f\in L^1_{\text{loc}}(\mathbb R^d)$.

More generally, $f$ lies in $\mathscr S'(\mathbb R^d)$, which is stable under Fourier transform.

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    $\begingroup$ Nice trick! But there is something bothering me. Taking $f(x) = 1/|x|^\alpha$ we have that $\widehat{f}$ is a distribution on $S'$. I totally agree that it's homogeneous of degree $-d+ \alpha$ and invariant by action of $SO(n)$. But why does this implies that $\widehat{f}$ is of the form $c/|x|^{d-\alpha}$? I agree that if $\widehat{f}$ is a function then $\widehat{f}(x) = \widehat f(|x|e) =|x|^{-d+\alpha} \widehat f(e)$ where $e$ is a unitary vector, but i can't see why $\widehat f$ is a function. $\endgroup$ – Hugocito Jan 16 '16 at 23:34
  • $\begingroup$ @HugoC.Botós Good point, overlooked that. I have a strong belief that this statement also holds for tempered distribution, though. $\endgroup$ – Brightsun Jan 16 '16 at 23:55
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    $\begingroup$ @HugoC.Botós I added a remark concerning your observation. Let me know! $\endgroup$ – Brightsun Jan 17 '16 at 9:47
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    $\begingroup$ I just asked a question math.stackexchange.com/questions/1619691/… I belive there is a way of prove that without the restriction $2\alpha>d$. Thanks for making that observation. $\endgroup$ – Hugocito Jan 20 '16 at 15:34
  • $\begingroup$ Why f is a tempered distribution? $\endgroup$ – Senna Feb 22 at 20:01

Following Brightsun solution we have $\widehat f (\xi) = k/|x|^{n-\alpha}$. Let $g(x) = e^{-\pi x^2}$, we have $\widehat g = g$. Since $$ \int_{\mathbb R^n} \widehat{f}(x)g(x)dx = \int_{\mathbb R^n} f(x)\widehat{g}(x)dx,$$ we have $$ \int_{\mathbb R^n} \frac{k}{|x|^{n-\alpha}}e^{-\pi x^2}dx = \int_{\mathbb R^n} \frac{1}{|x|^\alpha}e^{-\pi x^2}dx.$$ In polar coordinates we have: $$\int_{\mathbb R^n} \frac{1}{|x|^{\alpha}}e^{-\pi x^2}dx = \sigma(\mathbb S^{n-1})\int_{0}^{\infty} \frac{e^{-\pi r^2}r^{n-1}}{r^{\alpha}} = \sigma(S^{n-1})\int_{0}^{\infty} e^{-\pi r^2}r^{n-1-\alpha} dr,$$ where $\sigma$ is the measure on the sphere. Taking $u = \pi r^2$ we obtain $$\sigma(S^{n-1})\int_{0}^{\infty} e^{-\pi r^2}r^{n-1-\alpha} dr = \frac{\sigma(S^{n-1})}{\pi^{\alpha/2}}\int_{0}^{\infty}u^{\alpha/2 -1} e^{-u}du = \frac{\sigma(S^{n-1})\Gamma(\alpha/2)}{\pi^{\alpha/2}}.$$ Analogously, we can deduce $$\int_{\mathbb R^n} \frac{k}{|x|^{n-\alpha}}e^{-\pi x^2}dx = \frac{k\sigma(S^{n-1})\Gamma((n-\alpha)/2)}{\pi^{(n-\alpha)/2}}$$ and we have $$k=\frac{\Gamma(\alpha/2)\pi^{(n-\alpha)/2}}{\Gamma((n-\alpha)/2)\pi^{\alpha/2}}.$$ Therefore, the identity $$C_{\alpha}\int_{\mathbb R^n} \frac{1}{|x|^\alpha} \phi(x) dx = C_{n-\alpha}\int_{\mathbb R^n} \frac{1}{|x|^{n-\alpha}} \widehat{\phi}(x) dx$$ holds.

  • $\begingroup$ Nice! Do you think this method also resolves the controversy in math.stackexchange.com/questions/1341370/… ? I didn't think of it to be honest... $\endgroup$ – Brightsun Jan 17 '16 at 1:07
  • $\begingroup$ I think it works fine with direct calculation in the case $d>3$. I belive that de critical case in the identity in my question works with use $log |x|$ in the place of $1/|x|^{\alpha} $ in the case that $d=\alpha $. That would prove the result to d=2. $\endgroup$ – Hugocito Jan 17 '16 at 1:50
  • $\begingroup$ Why are we able to use $$ \int_{\mathbb R^n} \widehat{f}(x)g(x)dx = \int_{\mathbb R^n} f(x)\widehat{g}(x)dx, $$ when $f(x)=|x|^{-\alpha}$ does not satisfy $$ \sup_{x \in R^d}|x^a\left(\frac{d}{dx}\right)^bf(x) | < \infty$$ specifically for $a>\alpha$ and $b=0$. In Stein's Fourier Analysis text, both $f$ and $g$ need to satisfy that sup condition in order to use the identity in question. $\endgroup$ – alpastor Oct 9 at 2:15
  • $\begingroup$ This identity works in several levels of generality. You can take $f$ and $g$ in $L^1$ and the identity is true by Lemma 8.25 of Folland's Real Analysis book (2nd edition). You also can use $f$ and $g$ in $L^2$ or $f\in S'$ and $g \in S$. $\endgroup$ – Hugocito Oct 9 at 8:21
  • $\begingroup$ In the last case you have to write the identity in the language of tempered distribution. $\endgroup$ – Hugocito Oct 9 at 8:22

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