Prove $\phi \in L_{loc}^1(\Re ^3)$ but $\phi \notin L^1(\Re ^3)$ for Poisson's Fundamental Solution in $\Re ^3$

Question

Let $\phi$ be the fundamental solution of Poisson’s equation in $\Re^3$:

$\phi(\vec x) = \frac{1}{4\pi}\frac{1}{|\vec x|}$ when $n \geq 3$

I am asked to prove that $\phi \in L_{loc}^1(\Re ^3)$ but $\phi \notin L^1(\Re ^3)$

Where $L_{loc}^1(\Re^3) = \{ \phi: \Re^3 \to \Re$ such that $\int_k |\phi(\vec x)d \vec x < \infty$ for any compact set $K \subset \Re^n\}$

So I need to show that $\int_k |\phi(\vec x)d \vec x $ is bounded on this compact set, but not on all of $\Re ^n$.

My thinking is that I want to show $\exists C>0$ s.t. $||\vec x|| \leq C \forall x \in k$

But I am not really sure how to do this

Answer 1

For a compact set $K$, find a $M>0$ so large that $K\subseteq\{|x|\leq M\}$, then $\displaystyle\int_{K}\dfrac{1}{|x|}dx\leq\int_{|x|\leq M}\dfrac{1}{|x|}dx=\omega_{2}\int_{0}^{M}\dfrac{1}{r}\cdot r^{2}dr<\infty$.

For general, note that $\displaystyle\int_{{\bf{R}}^{3}}\dfrac{1}{|x|}dx=\omega_{2}\int_{0}^{\infty}\dfrac{1}{r}\cdot r^{2}dr=\infty$, so $\dfrac{1}{|x|}\notin L^{1}({\bf{R}}^{3})$.