# What is $\Delta\frac{1}{|\mathbf{x}|^2}$, as a distribution?

$\newcommand{\x}{\mathbf{x}}$Let $\x$ denote a vector in $\mathbb{R}^3$, $|\x|$ its magintude and $\Delta=\frac{\partial^2}{\partial x 2}+\frac{\partial^2}{\partial y 2}+\frac{\partial^2}{\partial z 2}$ the usual Laplacian. What is $$\Delta\frac{1}{|\x|^2}$$ in the sense of distribution? That is, what is $$\iiint_{\mathbb{R}^3}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x$$ for any test function $\varphi\in C^{\infty}_c(\mathbb{R}^3)$ (the subscript $c$ means compact support).

Attempt: I know how to do it for $\Delta\frac{1}{|\x|}$ (we get $\Delta\frac{1}{|\x|}=-4\pi\delta_0$, where $\delta_0$ is the delta distribution at $0$) and some others, and I am trying to reproduce the steps for this case but it does not seem to work. Let $\newcommand{\e}{\varepsilon}\e>0$ and consider the ball $B(0,\e)$ of radius $\e$ centered at $0$. Then, $$\iiint_{\mathbb{R}^3}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x= \iiint_{\mathbb{R}^3\setminus B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x+\iiint_{B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x$$ Then, we can see that $$\left|\iiint_{B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x\right|\leq \iiint_{B(0,\e)}\frac{1}{|\x|^2}|\Delta\varphi(\x)|d\x \leq C\int_0^\e\frac{1}{\rho^2}\rho^2 d\rho\to0$$ as $\e\to 0$. Now, since $\varphi$ has compact support, there is a compact subset $\Omega\subseteq \mathbb{R}^3$ that includes $B(0,\e)$ such that $\varphi$ and all its derivatives vanish on $\Omega^c$. Thus, $$\iiint_{\mathbb{R}^3\setminus B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x = \iiint_{\Omega\setminus B(0,\e)}\frac{1}{|\x|^2}\Delta\varphi(\x)d\x$$ Now by Green's Second Identity, $$\iiint_{\mathbb{R}^3\setminus B(0,\e)}\left(\frac{1}{|\x|^2}\Delta\varphi(\x)-\varphi(\x)\Delta\frac{1}{|\x|^2}\right)d\x = \iint_{\partial\Omega}\left(\frac{1}{|\x|^2}\frac{\partial\varphi(\x)}{\partial\mathbf{n}}-\varphi(\x)\frac{\partial}{\partial\mathbf{n}}\frac{1}{|\x|^2}\right)dS+ \iint_{\partial B(0,\e)}\left(\frac{1}{|\x|^2}\frac{\partial\varphi(\x)}{\partial\mathbf{n}}-\varphi(\x)\frac{\partial}{\partial\mathbf{n}}\frac{1}{|\x|^2}\right)dS$$ Since $\varphi$ and all its derivatives vanish on $\partial \Omega$, we are left with $$\iiint_{\mathbb{R}^3\setminus B(0,\e)}\left(\frac{1}{|\x|^2}\Delta\varphi(\x)-\varphi(\x)\Delta\frac{1}{|\x|^2}\right)d\x= \iint_{\partial B(0,\e)}\left(\frac{1}{|\x|^2}\frac{\partial\varphi(\x)}{\partial\mathbf{n}}-\varphi(\x)\frac{\partial}{\partial\mathbf{n}}\frac{1}{|\x|^2}\right)dS$$ Now usually we can work with the right-hand side to find what it does as $\e\to 0$. But here I don't see how...

• Maybe try using a chain rule: $\Delta(|x|^{-2}) = \Delta(|x|^{-1})^2$? – Neal Nov 14 '13 at 2:45

Taking Fourier transform, we have

$$\left( \Delta \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) = -|\xi|^{2} \left( \frac{1}{|x|^{2}} \right)^{\wedge}(\xi).$$

Indeed, for any $\varphi \in \mathcal{S}(\Bbb{R}^{3})$,

\begin{align*} \langle (\Delta |x|^{-2})^{\wedge}, \varphi \rangle &= \langle \Delta |\xi|^{-2}, \hat{\varphi}(\xi) \rangle = \langle |\xi|^{-2}, \Delta \hat{\varphi}(\xi) \rangle = \langle |\xi|^{-2}, - (|x|^{2}\varphi)^{\wedge}(\xi) \rangle \\ &= - \langle (|x|^{-2})^{\wedge}(\xi), |\xi|^{2}\varphi(\xi) \rangle = - \langle |\xi|^{2} (|x|^{-2})^{\wedge}(\xi), \varphi(\xi) \rangle. \end{align*}

But we have

\begin{align*} \left( \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) &= \int_{\Bbb{R}^{3}} \frac{e^{-i\xi\cdot x}}{|x|^{2}} \, dx = \int_{0}^{\infty} \int_{S^{2}} e^{-i\xi\cdot r \omega} d\sigma_{\omega} dr \\ &= \int_{0}^{\infty} \int_{0}^{2\pi} \int_{0}^{\pi} e^{-i|\xi| r \cos\phi} \sin\phi \, d\phi d\theta dr \\ &= 2\pi \int_{0}^{\infty} \int_{-1}^{t} e^{i|\xi| r t}\, dt dr \qquad (t = -\cos\phi) \\ &= \frac{4\pi}{|\xi|} \int_{0}^{\infty} \frac{\sin |\xi|r}{r} \, dt = \frac{2\pi^{2}}{|\xi|}. \end{align*}

So it follows that

$$\left( \Delta \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) = -|\xi|^{2} \left( \frac{1}{|x|^{2}} \right)^{\wedge}(\xi) = -2\pi^{2} |\xi|$$

and hence

$$\Delta \frac{1}{|x|^{2}} = -2\pi^{2} \sqrt{-\Delta} \delta_{0}.$$

• Thank you very much for this answer. But what does $\sqrt{-\Delta}$ mean? – Spenser Nov 14 '13 at 21:04
• @Spenser, It is a pseudo-differential operator defined by $$\sqrt{-\Delta} u = \{ |\xi|\hat{u}(\xi) \}^{\vee}.$$ Since differential operators in the space-side becomes multiplication operators by polynomials, it makes sense to call it something similar to differential operator, hence pseudo-differential operator. It can also be defined by utilizing functional calculus. – Sangchul Lee Nov 14 '13 at 21:26

The relevant distribution theory here was surely known long before L. Schwartz, but he certainly fully legitimized it, and a bit of further work by A. Grothendieck on holomorphic vector-valued functions (e.g., distribution-valued functions) makes the following viewpoint completely rigorous. Gelfand-Shilov's "Generalized Functions" volume I does such things, and many more complicated ones, at length.

For $\Re(s)>2$, on $\mathbb R^n$, $\Delta |x|^s = s(s+n-2)\cdot |x|^{s-2}$, noting that the right-hand side is locally integrable, hence, gives a distribution by integration-against-it.

This relation also gives the meromorphic continuation of the distribution $|x|^s$, by dividing through by $s(s+n-2)$ and replacing $s$ by $s+2$, and induction: $|x|^s=(\Delta |x|^{s+2})/(s+2)(s+n)$. Since $|x|^{s+2}$ is locally integrable for $\Re(s)>-n$, for $n>2$ there is in fact no pole at $s=-2$, and the first pole is at $s=-n$.

The first pole of that meromorphic family of distributions at $s=-n$ can be shown to be a constant multiple (depending on dimension $n$) of Dirac $\delta$ by applying $|x|^s$ to $e^{-\pi |x|^2}$ (for example) and looking at the numerical meromorphic continuation... and regularizing a general $f$ by $f-f(0)\cdot e^{-\pi |x|^2}$...

For $n=3$, the first pole is at $s=-3$, with residue a multiple of $\delta$. The relation above shows that there is no pole at $s=-4$ for $n=3$, so we'd imagine that the value is a constant multiple of $|x|^{-4}$. However this is not a locally integrable function! At this point, one might think of J. Hadamard's "finite part" trick, which is to integrate by parts and ignore the terms that blow up. In fact, as Riesz showed a few years later, this is the same outcome as by analytic continuation. So one has a choice about how to present that distribution. Yes, for $f$ vanishing sufficiently at $0$, it is just (a constant multiple of) integration against $|x|^{-4}$.