# The fourier transform of homogeneous distribution and related topics

When we have a distribuion $u\in \mathcal{D}'(R^n)$,and the restriction to $R^{n}\backslash{0}$ is homogeneous of degree a,we have $u \in \varphi'$ and $\widehat u$ is of degree(-n-a) in $R^{n}\backslash{0}$ .So it's easy to get when $-n< a <0$, $\widehat |x|^a=c_{a,n}|x|^{-n-a}$. my question is what's the result when $a>0$ ?Indeed, it's homogeneous of degree -n-a in$R^n$ and when a=n=1,we have $\widehat|x|=C p.v.|x|^{-2}$,where C is some constant. Futhermore,when we consider the wavefront set,it has the following:if u is a homogeneous distribuion in $R^{n}\backslash{0}$ ,then $(x,\xi) \in WF(u)\Leftrightarrow (\xi,-x) \in WF(\widehat u)$,where $x \neq 0$ , $\xi \neq 0$. Are there other interesting properties related to homogeneous distribuion ?

-

In the first place your presentation must be clarified. An homogeneous distribution $u$ of degree $a$ on $\mathbb R^n$ is characterized by $$\forall \lambda >0,\quad u(\lambda x)=\lambda^au(x),\qquad \text{i.e}\quad\langle u(x),\phi(x/\lambda\rangle\lambda^{-n}=\lambda^{a} \langle u(x),\phi(x\rangle,$$ which is also equivalent to the so-called Euler equation $x\cdot\partial_xu=au$. It is not difficult (but not trivial) to prove that an homogeneous distribution on $\mathbb R^n$ is a temperate distribution. Simple examples include in one dimension $pv(1/x)$ (degree $-1$), $H(x)$ (degree $0$), $(x+i0)^a$ (degree $a$), and in $n$ dimensions $\delta_0$ (degree $-n$), $\vert x\vert^a$ for $\Re a>-n$ (degree $a$),$\partial_x^\alpha \vert x\vert^a$ (degree $a-\vert \alpha\vert$). For $a\in \mathbb C$, $$\frac{x_+^a}{\Gamma(a+1)}\text{is an homogeneous distribution with degree a, entire with respect to a.}$$
On the other hand, you may consider homogeneous distributions on any cone of $\mathbb R^n$, e.g. on $\mathbb R^n\backslash 0$ and wonder if you can extend these distributions to homogeneous distributions on the whole $\mathbb R^n$. In particular the following result holds: if $u$ is an homogeneous distribution on $\mathbb R^n\backslash 0$ with degree $a$ which is not an integer $\le -n$, then there is a unique extension of $u$ to an homogeneous distribution on the whole $\mathbb R^n$.
Last but not least, if $u$ is an homogeneous distribution of degree $a$ on $\mathbb R^n$, it is temperate and its Fourier transform is indeed homogeneous on $\mathbb R^n$ with degree $-a-n$. In particular, you may consider $\vert x\vert ^a$ which is a radial distribution homogeneous of degree $a$ on $\mathbb R^n\backslash 0$; for $a$ avoiding integer values, it has a unique extension as described above as an homogeneous distribution with degree $a$ on $\mathbb R^n$ and its Fourier transform is also radial and homogeneous with degree $-a-n$, that is coincides with $c\vert \xi\vert^{-a-n}$ on $\mathbb R^n\backslash 0$.
@Bzain,thank you very much.your answer is really helpful.consider again $|x|^a$,where a is any positive integer,now we can't use the extension mathod to determine its Fourier transform.indeed, as you have mentioned above,it is coincides with $c|\xi|^{-a-n}$ on $R^n \backslash 0$,but can we write the explicit expression just as the case in one dimension. – user23078 Apr 26 '12 at 9:00
Well if a is an even positive integer, then you get an homogeneous polynomial in $x$ of degree $a$ and its Fourier transform is a linear combination of derivatives of order $a$ of he Dirac mass at 0. When $a$ is an odd positive integer, then the Fourier transform will be indeed radial and homogeneous with degree $-a-n$: you can see that as a sort of finite part with a formula such as $$\langle\hat u,\phi\rangle=c_{a,n}\int\vert \xi\vert^{-a-n} \bigl(\phi(\xi)-\sum_{0\le j\le N}\phi^{(j)}(0)\xi^j/j!\bigr) \chi(\xi)d\xi$$ – Bazin Apr 26 '12 at 19:30
yes,when a is an even positive integer,the case is easy and seems degenerate. i wanna understand in your last formula,is $\chi(\xi)$ standing for any cut-off funtion which vanish at 0 ? I'm really appreciated that if you can show me some references about this question.. – user23078 Apr 27 '12 at 2:55
To Huang Shanlin: the function $\chi$ is smooth with compact support and is 1 near 0, in such a way that you may add (I should have done this, but I wanted to insist on the way to get rid of the singularity) on the rhs of the above equality $$\int\vert \xi\vert^{-a-n}\phi(\xi)(1-\chi(\xi)) d\xi.$$ Best, Bazin. – Bazin Apr 30 '12 at 19:17
@Bazin,I found myself not completely understand it yet,I think your definition of $\hat u$ is not unique,since it depends on the choice of $\chi(\xi)$.so what's the problem behind this ? thanks in advance – user23078 Jun 20 '12 at 8:50