# Locally integrable functions versus distributions

We know that every locally integrable functions on $\mathbb R^n$ defines naturally a distribution. Moreover, every distribution is differentiable.

I think the following is right:

The fundamental solution to the Laplacian on $\mathbb R^n$ is locally integrable, hence defines a distribution. For example, in $n=3$, we have: $$\phi(x)=C_3|x|^{-1}$$ And $$\frac{\partial^2 \phi}{\partial x_j^2}=C_3'[-|x|^{-3}+3x_j^2|x|^{-5}],$$ which is still a distribution but fails to be locally integrable. Am I right?

• If the distributional derivative $\partial_{jj}\phi$ is represented by a measurable function, then so would be the same for the sum $\Delta \phi = \sum_{j=1}^{n} \partial_{jj} \phi$, which is not true since $\Delta \phi = \delta$. Commented Jul 16, 2017 at 8:55
• Thank you for your comment. So where was my calculation for the second derivative wrong? If I am correct, then $\Delta \phi(x)=0$ for $x\neq 0$. How can I recover the identity $\sum_j \partial_{jj}\phi=\delta$? Commented Jul 16, 2017 at 9:01

A bit of computation shows that, for any test function $\varphi \in C_c^{\infty}(\mathbb{R}^3)$ we have

$$\int_{\mathbb{R}^3} \frac{1}{|x|} \frac{\partial^2 \varphi}{\partial x_j^2}(x) \, dx = -\frac{4\pi}{3}\varphi(0) + \lim_{\epsilon \downarrow 0} \int_{\mathbb{R}^3\setminus B_{\epsilon}(0)} \varphi(x) \frac{\partial^2}{\partial x_j^2}\frac{1}{|x|} \, dx.$$

If you want to replicate this result, you can utilize the divergence theorem to compute both

$$\int_{\mathbb{R}^3\setminus B_{\epsilon}(0)} \operatorname{div}\left( \frac{1}{|x|} \, \mathrm{e}_j \frac{\partial \varphi}{\partial x_j} (x) \right) \, dx \qquad \text{and} \qquad \int_{\mathbb{R}^3\setminus B_{\epsilon}(0)} \operatorname{div}\left( \varphi (x) \, \mathrm{e}_j \frac{\partial}{\partial x_j} \frac{1}{|x|} \right) \, dx.$$

and then let $\epsilon \downarrow 0$.

Here, the failure of local integrability of $\partial_{jj} \frac{1}{|x|}$ near $x = 0$ is reflected in the emergence of the extra Dirac delta term $-\frac{4\pi}{3}\delta$. This is also where the distributional derivative departs from the ordinary differentiation.

Distributions are linear functionals on the space of test functions. When you write $\phi(x)=C|x|^{-1}$ is a distribution, what you mean is that it defines the distribution $$f\mapsto \int_{\Bbb R^3}\phi(x)f(x)\,dx.$$ (Here $f$ is smooth of compact support).

As a distribution, the second derivative of $\phi$ with respect to $x_i$ is the functional $$f\mapsto \int_{\Bbb R^3}\phi(x)\frac{\partial f}{\partial x_i^2}(x)\,dx.$$

If the support of $f$ does not contain the origin then integration by parts gives you $$\int_{\Bbb R^3}\phi(x)\frac{\partial f}{\partial x_i^2}(x)\,dx =\int_{\Bbb R^3}\psi(x)f(x)\,dx\tag{*}$$ with $$\psi(x)=C(-|x|^{-3}+3x_i^2|x|^{-5}).$$ What this means is that $\psi$ is the second derivative of $\phi$ (wrt $x_i$) on $\Bbb R^3-{(0,0,0)}$. The formula $(*)$ fails to define $\partial^2 f/\partial x_i^2$ for test functions $f$ whose support includes the origin, since $\psi$ is not $L^1$ in a neighbourhood of the origin.

To prove that $\phi$ is a fundamental solution of Laplace's equation what one must show is that $$\int_{\Bbb R^3}\phi(x)\left( \frac{\partial f}{\partial x_1^2}(x)+ \frac{\partial f}{\partial x_2^2}(x)+ \frac{\partial f}{\partial x_3^2}(x) \right)\,dx=f(0,0,0)$$ whenever $f$ is a test function. Try using spherical polars...