# Distributional Fourier Transform of $\frac{1}{|x|^a}$

Let $$h_a=\frac{\Gamma(\frac{a}{2})}{\pi^{\frac{a}{2}}}|x|^{-a}, x \in R^d$$ Then $$\hat{h_a}=h_{d-a}$$ in the sence of $$L^1+L^2-$$Fourier transforms if $$\frac{d}{2} and in the sence of distributional Fourier transforms if $$0.

This is Lemma $$4.1$$ in these lecture notes: http://www.math.ubc.ca/~ilaba/wolff/notes_march2002.pdf

I only managed to complete the details of the proof for $$d>2$$

How can i prove this for general $$d>0$$ with the same arguments i used for $$d>2$$?

Here is my complete proof for $$d>2$$

$$\underline{\text{Proof}}$$ Let $$h(x)=|x|^{-a}$$ where $$a\in (\frac{d}{2},d),$$ then $$h$$ is radial and\ $$h \in L^1(R^d)+L^2(R^d)$$ because $$h(x)=h_1+h_2$$ where $$h_1(x)=h(x)1_{\{|x|<1\}}(x) \in L^1(R^d)$$ $$h_2(x)=h(x)1_{\{|x| \geq 1\}}(x) \in L^2(R^d)$$\

Now recall that if $$f \in L^2(R^d)$$ and $$\phi_n \in \mathrm{S}$$ such that $$\phi_n \to^{L^2}f$$, then $$\hat{\phi_n} \to ^{L^2} \mathcal{F}(f)$$

Using this fact and simple changes of variables we can easily deduce that the $$L^2-$$Fourier transform $$\mathcal{F}$$ of a radial function $$f \in L^2(R^d)$$ is radial. Thus the $$L^1+L^2-$$Fourier transform of $$h$$ is radial.

Also using the previous fact, again we have $$\hat{h}(M\xi)=M^{-(d-a)}\hat{h}(\xi).$$ If $$\xi \in R^d$$ then $$\hat{h}(\xi)=\hat{h}(|\xi| \frac{\xi}{|\xi|})=|\xi|^{-(n-a)}\hat{h}( \frac{\xi}{|\xi|})$$ So $$\hat{h}(\xi)=c|\xi|^{-(d-a)}$$ where $$c=\hat{h}(x), \forall x \in \mathbb{S}^{d-1}$$

Using the Duality relation(which is true for $$L^2$$ functions by approximation)we have $$\int_{R^d}|x|^{-a}e^{-\pi |x|^2}dx=c\int_{R^d}|x|^{-(d-a)}e^{-\pi|x|^2}dx \text{ }(1)$$

Here we used the identity: $$\widehat{e^{-\pi|x|^2}}(\xi)=e^{-\pi|\xi|^2}$$ where $$e^{-\pi|x|^2} \in \mathrm{S}$$

Polar Coordinates formula and appropriate changes of variables to on both sides of the equation $$(1)$$ gives us: $$c=\frac{\Gamma(\frac{d-a}{2})\pi^{\frac{a}{2}}}{\Gamma(\frac{a}{2})\pi^{\frac{d-a}{2}}}$$

Hence $$\hat{h_a}=h_{d-a}.$$

Now for the general case, let $$\phi \in \mathrm{S},d>2$$ and $$A(z)=\int_{R^d}h_z\hat{\phi}$$ $$B(z)=\int_{R^d}h_{d-z}\phi$$\We will show that $$A(z),B(z)$$ are holomorphic in the strip\ $$I=\{z: 0 and agree everywhere on $$I.$$ Note that $$\frac{d}{2} < d-1$$

Recall that $$\Gamma(z)$$ is holomorphic define on the region $$\Omega=\{z:Re(z)>0\}$$ and has no zeroes, so the reciprocal gamma function $$\frac{1}{\Gamma}(z)$$ \ is holomorphic in $$\Omega.$$ So it suffices to show the holomorphy of the function $$G:I \to \Bbb{C}$$ where $$G(z)=\int_{R^d}|x|^{-z}\phi(x)dx$$

Let $$z \in I$$ and define $$F(x,z)=|x|^{-z}\phi(x)$$ and let $$h_n \in \Bbb{C}$$ such that $$h_n \to 0$$ and $$|h_n| <\min\{1,\frac{d-Rez-1}{2}\}, \forall n \in \Bbb{N}$$. Then

$$\frac{F(x,z+h_n)-F(x,z)}{h_n} \to \frac{-\ln|x|}{x^z}\phi(x), \text{ }a.e$$ and also\ $$|\frac{F(x,z+h_n)-F(x,z)}{h_n}|= |\frac{1}{|h_n||x|^z}\bigg(\frac{1}{|x|^{h_n}}-1 \bigg)||\phi(x)|= \frac{|e^{-h_n \ln{|x|}}-1|}{|h_n||x|^{Re(z)}}|\phi(x)|$$ $$=\frac{|e^{-h_n \ln{|x|}}-1|}{|h_n||x|^{Re(z)}}|\phi(x)|1_{\{|x| \leq 1\}}+\frac{|e^{-h_n \ln{|x|}}-1|}{|h_n||x|^{Re(z)}}|\phi(x)|1_{\{|x|>1\}}$$

$$\textbf{(1)}$$ If $$|x| \leq 1$$ then,

$$\frac{|e^{-h_n \ln{|x|}}-1|}{|h_n||x|^{Re(z)}}1_{\{|x| \leq 1\}}|\phi(x)| \leq \frac{|\ln{|x|}||e^{|h_n ||\ln{|x|}|}}{|x|^{Re(z)}}1_{\{|x| \leq 1\}}|\phi(x)|=\frac{\ln{\frac{1}{|x|}}e^{|h_n| \ln{\frac{1}{|x|}}}}{|x|^{Re(z)}}1_{\{|x| \leq 1\}}|\phi(x)|$$ $$\leq \frac{1}{|x|^{Re(z)+1+|h_n|}}1_{\{|x| \leq 1\}}|\phi(x)|\leq \frac{1}{|x|^{Re(z)+1+\frac{d-1-Re(z)}{2}}}1_{\{|x| \leq 1\}}|\phi(x)| \in L^1(R^d)$$ since $$\phi$$ is bounded everywhere.

$$\textbf{(2)}$$ If $$|x|>1$$ then $$\frac{|e^{-h_n \ln{|x|}}-1|}{|h_n||x|^{Re(z)}}|\phi(x)|1_{\{|x|>1\}}\leq |x|^2|\phi(x)|1_{\{|x|>1\}} \in L^1(R^d)$$ since $$\phi \in \mathrm{S}.$$

Using $$\textbf{(1),(2)}$$ and the dominated convergence theorem we have that $$A(z),B(z)$$ are differentiable at $$z$$. So $$A,B$$ are differentiable at every $$z \in I$$ thus holomorphic in $$I$$ By $$\textbf{Proposition}$$, $$A(z)=B(z), \forall z \in (\frac{d}{2},d) \supset (\frac{d}{2},d-1)$$ Thus by identity theorem $$A(z)=B(z), \forall z \in I$$ If $$Re(z)>\frac{d}{2}$$ then $$h_a \in L^1(R^d)+L^2(R^d),$$ so its $$L^1+L^2$$ and distributional Fourier transforms coincide. $$\text{ }\blacksquare.$$

Here's a simple way to determine the form of $$\hat{h}_a$$:

The function $$h_a$$ satisfies $$x\cdot\nabla h_a = -a h_a$$. Taking the Fourier transform gives the equation $$i\nabla\cdot(ix\hat{h}_a) = -a \hat{h}_a.$$ The left hand side can be rewritten as $$-(\nabla\cdot x)\hat{h}_a - (x\cdot\nabla)\hat{h}_a.$$ Here, $$\nabla\cdot x = d$$ so we have a differential equation $$x\cdot\nabla\hat{h}_a = -(d-a)\hat{h}_a.$$ The solutions of this are $$\hat{h}_a = C_{a,d} |x|^{-(d-a)}.$$

But how to determine the constant $$C_{a,d}$$? Some thoughts:

Taking the Fourier transform again we get $$\hat{\hat{h}}_a = C_{d-a,a} C_{a,d} |x|^{-a}.$$ By the Fourier inversion theorem, $$\hat{\hat{h}}_a(x) = (2\pi)^d h(-x),$$ so $$C_{d-a,d}C_{a,d} = (2\pi)^d.$$

• +1...Thank you for your answer..your idea is slick..but can do you know how can i prove with the same arguments the theorem for d<2?..(I forgot to mention that i want a proof for $d<2$ with the same arguments i used for $d>2$.) – Marios Gretsas Aug 17 at 17:56

This is too complicated. There is a much simpler proof of the distributional Fourier transform with $$0<\Re(a) in this previous answer I gave: How to calculate $c_a$ where $\left(f\mapsto\int_{\mathbb{R}}\frac{f(t)-f(0)}{|t|^{a}}dt\right)=c_a\mathcal{F}_x(|x|^{-1+a})$

It is based on representing $$\frac{1}{|x|^a}$$ as a continuous superposition of Gaussians.

• +1 Thank you for your answer. – Marios Gretsas Aug 21 at 1:45