# Is the integral $\frac{|u(x)|^2}{|x|^2}$ convergent?

Suppose $n \ge 3$ and $u \in C^\infty_c(\mathbb{R}^n)$. I know the Hardy's inequality $$\int_{\mathbb{R}^n} \frac{|u(x)|^2}{|x|^2} dx \le C_n \int_{\mathbb{R}^n} |\nabla u|^2 dx.$$ I would like to show the integral on the left hand side is convergent. Does the Hardy's inequality implicitly say the integral on the left hand side is convergent, or is the inequality well-defined because the integral on the left hand side is finite first?

Thank you.

Since $u$ is bounded in a neighbourhood of $0$, the finiteness of the left side is easy to show. It reduces to showing that $$\int_0^R r^{n-3}\; dr < \infty$$
For $n\ge 3$ you have that $\int_B\frac 1 {|x|^2}dx < \infty$, where $B$ is any compact set in $\mathbb R^n$.