What is this singular distribution $T(\phi)(x) = \int_{\mathbb{R}^n} \frac{\phi(y)}{|x-y|^n} \mathrm{d}y$?

Question

I want to study the behavior of this singular distribution $T \in \mathcal{D}'(\mathbb{R}^n)$, defined by

\begin{equation} T(\phi)(x) = \int_{\mathbb{R}^n} \frac{\phi(y)}{|x-y|^n} \mathrm{d}y, \tag{1} \end{equation}

with $\phi(x) \geq 0$, $\phi \in L^1(\mathbb{R}^n)$. I want to know if $T$ has a name, classified under what type of distributions, and its properties.

There are similar distributions. Namely if the denominator of the kernel was $(x-y)^n$ instead of $|x-y|^n$, then $T$ was just Riesz transform (Hilbert transform if $n=1$), and the integral can have a finite principal value.

In more general setting, if the kernel of the distribution $T(\phi)(x) = \int K_{\Omega}(x,y) \phi(y) \mathrm{d}y$ is of the form

\begin{equation} K_{\Omega}(x) = \frac{\Omega(x/|x|)}{|x|^n} \end{equation}

with the condition that $\Omega$ is zero mean value on unit sphere $\mathbb{S}^{n-1}$, then this is Calderon-Zygmund kernel, and I can relate the Calderon-Zygmund decomposition (Grafakos, section 5.2.1).

But the distribution of (1) is non of the above forms. becasue, its kernel $K_{\Omega}(x) = 1/|x|^n$ and $\Omega(x/|x|) =1 $ is not zero mean. I am wondering what are the name of such distributions, can they be studied by Calderon-Zygmund, or so they have their own properties? For instance, unless $\phi(0) = 0$, $T$ is always unbounded.

I am also looking for some reference suggestions. Thank you!

Answer 1

It is not a distribution on $\Bbb{R}^n$ because $$ \lim_{r \to 0}\int_{\mathbb{R}^n} \phi(x)\frac{1_{|x|> r}}{|x|^n} d^n x $$ diverges whenever $\phi(0) \ne 0$. What is a (tempered) distribution is $$<T,\phi>= \int_{\mathbb{R}^n} (\phi(x)-\phi(0) 1_{|x|<1})\frac{1}{|x|^n} d^n x $$ And $$T \ast \phi(y)= \int_{\mathbb{R}^n} (\phi(y-x)-\phi(y) 1_{|x|<1})\frac{1}{|x|^n} d^n x $$ The choice of $1_{|x|<1}$ is non-canonical but you can call it $$T=fp(\frac1{|x|^n})$$ (finite part) everyone will understand that you meant $$<fp(\frac1{|x|^n}),\phi>= \int_{\mathbb{R}^n} (\phi(x)-\phi(0) \psi(x))\frac{1}{|x|^n} d^n x=<T+c\delta ,\phi>$$ for some $\psi(0) = 1$

Answer 2

As you noted, unless $\phi(x) = 0$ this distribution isn't really a distribution since distributions need to be continuous linear functionals. But we can modify it by taking the action $$T_x(\phi) = \text{p.v.}\int_{\mathbb{R}^n} \frac{\phi(y)-\phi(x)}{|x-y|^n}dy$$ which we can consider as a family of distributions that are shifted by x from $$T_0(\phi) = \text{p.v.}\int_{\mathbb{R}^n} \frac{\phi(y)-\phi(0)}{|y|^n}dy$$ $T_0$ has some special properties, namely it is a homogeneous radial distribution of degree $-n$. In $\mathbb{R}^n$ another radial distribution with that homogeneity is the Dirac delta $\delta_0$ in $n$ dimensions. In some sense, $T_0$ can tell you what is happening with the test function everywhere except $0$, but $\delta_0$ can only tell you what is happening at $0$. In fact, any radial distribution with homogeneity $-n$ has to be some linear combination of $T_0$ and $\delta_0$