# Generalization of Dirac Deltas as Derivative of Heaviside Functions in $\mathbb{R}^d$

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 $$d$$.)

• 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.) Dec 23, 2022 at 23:06

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?