Below is an exercise in Stein and Shakarchi's Functional Analysis.

A $d$-dimensional generalization of the identity for the Heaviside function is the identity $$\delta = \sum_{j=1}^d (\frac{\partial}{\partial x_j}) h_j,$$ with $h_j(x) = \frac{1}{A_d} \frac{x_j}{|x|^d}$, and $A_d = 2 \pi^{d/2}/\Gamma(d/2))$ denotes the area of the unit sphere in $\mathbb{R}^d$.

[Hint: When $d > 2$, write $\delta = \sum_{j=1}^d \frac{\partial}{\partial x_j}\left(\frac{\partial}{\partial x_j} C_d |x|^{−d+2}\right)$.]

Here both the Heaviside function and the Dirac delta should be treated as distributions. So the goal is to show that for any $\varphi \in \mathcal{S}$ we have $$-\frac{1}{A_d} \int_{\mathbb{R}^d} \frac{x_j}{|x|^d} \frac{\partial \varphi}{\partial x_j} dx = \varphi(0),$$ or when $d > 2$, $$C_d \int_{\mathbb{R}^d} \frac{1}{|x|^{d-2}} \Delta \varphi dx = \varphi(0),$$ where $\Delta$ is the Laplace operator.

I have trouble understanding how to deal with the singularity at the origin.

EDIT: For $d=2$, we can show $$\frac{1}{A_2} \int_{\mathbb{R}^2} \log|x| \Delta \varphi dx = \varphi(0)$$ by writing the Laplace operator in polar coordinates. In particular, we use can the fact $\int \frac{\partial^2 \varphi}{\partial \theta^2} d\theta = 0$.

So we should be good if we can show that $\int_{S^{d-1}} \Delta_{S^{d-1}}\varphi ds = 0$, where $\Delta_{S^{d-1}}\varphi$ is the last term in the Laplace operator in spherical coordinates in N dimensions. (It is easy to verify for $d=3$. But I don't know an easy way to show it is true for any $d$.)

  • $\begingroup$ What if you take the volume integral of the derivative with respect to the normal and turn it into a surface integral of the outward pointing normal of the function (generalized Greens/Stokes theorem)? You can then completely avoid the singularity at the origin. (Also hence the importance of the part about the area of the unit sphere.) $\endgroup$ Dec 23, 2022 at 23:06

1 Answer 1


We just need to integrate by parts twice and use the fact that $\nabla^2\left(\frac1{|x|^{d-2}}\right)=0$ for $\vec x\ne0$. Proceeding we have

$$ \begin{align} \int_{\mathbb{R}^d} \frac{1}{|\vec x|^{d-2}} \nabla^2\phi(\vec x)\,d^d x&=\underbrace{\int_{\mathbb{R}^d} \nabla\cdot \left(\frac{1}{|\vec x|^{d-2}} \nabla \phi(\vec x)\right)\,d^dx}_{=0\,\,\text{since}\,\,\phi \in C_C^\infty}-\int_{\mathbb{R}^d}\nabla \phi(\vec x)\cdot\nabla \left(\frac{1}{|\vec x|^{d-2}}\right)\,d^d x\\\\ &=-\lim_{\varepsilon\to 0}\int_{\mathbb{R}^d\setminus B(0,\varepsilon)}\nabla\cdot\left( \phi(\vec x)\nabla \left(\frac{1}{|\vec x|^{d-2}}\right)\right)\,d^d x\\\\ &=\lim_{\varepsilon\to 0}\int_{\partial B(0,\varepsilon)}\phi(\vec x)\frac{\partial}{\partial n}\left(\frac1{|\vec x|^{d-2}}\right)\,dS_d\\\\ &=K_d\phi(0) \end{align}$$

for some constant $K_d$. Can you finish now?

  • $\begingroup$ Super clear! Thank you! $\endgroup$ Dec 24, 2022 at 1:51
  • $\begingroup$ @PetraAxolotl Pleased to hear. And you're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Dec 24, 2022 at 2:38

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