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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 $1/|x|^\alpha$?

Hugo
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2 Answers2

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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.

Brightsun
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    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. – Hugo Jan 16 '16 at 23:34
  • @HugoC.Botós Good point, overlooked that. I have a strong belief that this statement also holds for tempered distribution, though. – Brightsun Jan 16 '16 at 23:55
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    @HugoC.Botós I added a remark concerning your observation. Let me know! – Brightsun Jan 17 '16 at 09:47
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    I just asked a question http://math.stackexchange.com/questions/1619691/why-does-fourier-transform-of-1-x-alpha-is-a-function I belive there is a way of prove that without the restriction $2\alpha>d$. Thanks for making that observation. – Hugo Jan 20 '16 at 15:34
  • Why f is a tempered distribution? – Senna Feb 22 '19 at 20:01
  • @Brightsun What is your definition of the Fourier transform? $e^{-xy}$ or $e^{-2\pi I ixy}$ – Hermi Apr 07 '20 at 18:13
  • @BobOakley I guess I used $e^{-ikx}$ judging from the second equation in the post... – Brightsun Apr 09 '20 at 07:31
  • @Brightsun Thx. I have another question. How to get the Fourier transform of the $1/|x|^2$ for 2 dimension? Or can we get a similar formula that $\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$? – Hermi Apr 09 '20 at 07:37
  • @BobOakley Once you have the answer as a function of $\xi$ for generic $D$ you can obtain the answer for $D=2$ by analytic continuation. Letting $D=2+\epsilon$, expand for small $\epsilon$ and you should find a $\xi$-independent pole $1/\epsilon$, plus a finite term proportional to $\log|\xi|$. – Brightsun Apr 09 '20 at 12:29
  • @Brightsun Sorry, I am confused. What is your $D$? And how to find such $\xi$? – Hermi Apr 09 '20 at 16:06
  • @Hermi The Fourier transform of $1/|x|^2$ in dimension $2$ is computed here https://math.stackexchange.com/questions/3723136/the-fourier-transform-of-1-p3 – LL 3.14 Dec 14 '23 at 16:23
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Following Brightsun solution we have $\widehat f (\xi) = k/|\xi|^{n-\alpha}$. 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 $$ \sigma(S^{n-1})\int_{0}^{\infty} e^{-\pi r^2}r^{n-1-\alpha} dr = \frac{\sigma(S^{n-1})}{2\pi^{(n-\alpha)/2}} \int_{0}^{\infty}u^{\frac{n-\alpha}{2} -1} e^{-u}du = \frac{\sigma(S^{n-1})\Gamma(\frac{n-\alpha}{2})}{2\pi^{(n-\alpha)/2}}.$$ Analogously, we can deduce $$\int_{\mathbb R^n} \frac{k}{|x|^{n-\alpha}}e^{-\pi x^2}dx = \frac{k\sigma(S^{n-1})\Gamma(\alpha/2)}{2\pi^{\alpha/2}}$$ and we have $$ k = \frac{\Gamma((n-\alpha)/2)\pi^{\alpha/2}}{\Gamma(\alpha/2)\pi^{(n-\alpha)/2}}.$$ 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.

Pass Stoneke
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Hugo
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  • Nice! Do you think this method also resolves the controversy in http://math.stackexchange.com/questions/1341370/fundamental-solution-to-the-laplace-equation-by-fourier-transform ? I didn't think of it to be honest... – Brightsun Jan 17 '16 at 01:07
  • 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. – Hugo Jan 17 '16 at 01:50
  • 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. – alpastor Oct 09 '19 at 02:15
  • 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 $g \in S$. – Hugo Oct 09 '19 at 08:21
  • In the last case you have to write the identity in the language of tempered distribution. – Hugo Oct 09 '19 at 08:22
  • I think it is noteworthy that you are using a different definition of the Fourier transformation than Brightsun does in his comment, thus coming to a different conclusion what the exact constant should be. – Pass Stoneke Dec 18 '23 at 13:38