What is the right Lebesgue space $L^p_{loc}(\mathbb{R}^3)$ to which the function $\frac{1}{\vert x\vert}$ belongs?

Question

In dimension $3$, what is the right Lebesgue space $L^p_{loc}(\mathbb{R}^3)$ to which the function $\frac{1}{\vert x\vert}$ belongs?

There is a general rule that involves the dimension of $\mathbb{R}^n$, the function $\frac{1}{\vert x\vert^{\alpha}}$ and the right $L^p_{loc}(\mathbb{R}^n)$ to which $\frac{1}{\vert x\vert^{\alpha}}$ belongs?

Answer 1

Let $B(0,1)$ denote the unit ball in $R^n$ and $\omega_n = |B(0,1)|$. Then the integral

$$ \int_{B(0,1)} \frac{1}{|x|^\alpha}dx = \omega_n\int_0^1 \frac{1}{r^\alpha}r^{n-1}dr = \omega_n\int_0^1\frac{1}{r^{\alpha-n+1}}dr $$

converges if and only if $\alpha-n+1<1$, i.e. $\alpha < n$.

From this you can deduce what you are interested in, namely that $\frac{1}{|x|^\alpha} \in L_{\rm loc}^p(\mathbb{R}^n)$ if and only if $\alpha p < n$.

Answer 2

The only problem is the origin. It is continuous, and therefore bounded everywhere else on any compact set not containing the origin. By a change of variable, we are interested in the $p$ for which the integral below exists, $$\int_{0}^{\rho} \int_{-\pi}^{\pi} \int_{0}^{2 \pi} \left( \frac{1}{r} \right)^{p} r^{2} \sin(\phi) d \theta d \phi dr$$ We find that whenever $p < 3$, this is, $$4 \pi \frac{\rho^{3-p}}{3-p}$$ And moreover, the integral doesn't exist when $3 \geq p$. And so $\frac{1}{|x|} \in {L_{loc}^{p}}(\mathbb{R}^{3})$ whenever $p<3$.