# Continuity of a function on a closed interval in $\mathbb{R}^3$

Is the following function continuous on $\Omega$ ? $$f(x)=\int_{\Omega}\frac{1}{|x-y|}dy,\quad x\in\Omega$$where $\Omega=[-a,a]^3$.

## 1 Answer

Hint:

First show the integral is convergent where

$$\lim_{\delta \to 0} \int_{B_\delta(x)} \frac{dy}{|x- y|} = 0$$

For $x$ and $x_0$ in $\Omega$, consider

$$|f(x) - f(x_0)| \leqslant \left|\int_{\Omega - (B_\delta(x_0)\cup B_\delta(x))}\left(\frac{1}{r} - \frac{1}{r_0} \right)\, dy \right|+ \left|\int_{B_\delta(x_0)}\left(\frac{1}{r} - \frac{1}{r_0} \right)\, dy \right| + \left|\int_{B_\delta(x)}\left(\frac{1}{r} - \frac{1}{r_0} \right)\, dy \right|$$

where $r = |x - y|$ and $r_0 = |x_0-y|.$

• I see. Thanks a lot – whereamI Sep 2 '17 at 3:07
• @whereamI: You're welcome and of course that first limit is easily shown by switching to a spherical coordinate system with origin at $x$ where $dy = r^2 \sin \theta \, dr \, d\theta \, d\phi$. – RRL Sep 2 '17 at 3:33
• By the way, can we also claim that $f(x)\in C^1(\Omega)$, i.e., continuous differentiable on $\Omega$ or $f(x)\in W^{1,\infty}(\Omega)$? – whereamI Sep 27 '17 at 0:46
• This integral is the potential at the point $x$ of a uniform distribution (charge, mass, etc.) throughout the region $\Omega$. We addressed the continuity, but it should also be the case that $f \in C^1$. Of course if $x$ is outside $\Omega$ then all of this is easy (take the derivative under the integral). If $x \in \Omega$, then ... – RRL Sep 27 '17 at 3:31
• With $x = (x_1,x_2,x_3)$ we can find $\frac{\partial f}{\partial x_1}$ as the limit as $h \to 0$ of $( f(x + (h,0,0)) - f(x) ) / h$ where we split $f$ into two integrals, one over the ball $B_\rho(x)$ and the other over $\Omega - B_\rho(x)$. If $\rho$ is sufficiently small then the first integral is arbitrarily small and the result ultimately is $\frac{\partial f}{\partial x_1} = \int_{\Omega} \frac{y_1 - x_1}{ |x - y|^3} \, dy$. Similar to the original question, you can show this is continuous. – RRL Sep 27 '17 at 3:39