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

Question

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 $\frac{d}{2}<Re(a)<d,$ and in the sence of distributional Fourier transforms if $0<Re(a)<d$.

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\ $I=\{z: 0<Re(z)<d-1\}$ 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. $\text{ }\blacksquare.$

Answer 1

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 $C_{d-a,d}C_{a,d} = (2\pi)^d.$

Answer 2

This is too complicated. There is a much simpler proof of the distributional Fourier transform with $0<\Re(a)<d$ 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.