Evans proved the Hardy's inequality, Hardy's inequality. In the first step, it is said that the function may be assumed smooth. I wonder why we can assume this function is smooth. I guess it may be due to density of smooth function. But if there is a sequence of smooth functions $u_m$ approaching to $u$ in $H^1(B(0,r))$, how can we assert by density: $\int_{B(0,r)}u^2/|x|^2=\lim_{m\rightarrow\infty}\int_{B(0,r)}u_m^2/|x|^2$ to conclude the final result.
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Right, the singular factor $1/|x|^2$ complicates passing to the limit. Let's relax it by introducing $\epsilon>0$ in the denominator: then there is no issue with $$\int_{B(0,r)}\frac{u^2}{|x|^2+\epsilon} = \lim_{m\rightarrow\infty}\int_{B(0,r)}\frac{u_m^2}{|x|^2+\epsilon}$$ Since $$ \int_{B(0,r)}\frac{u_m^2}{|x|^2+\epsilon} \le \int_{B(0,r)}\frac{u_m^2}{|x|^2} \le C \int_{B(0, r)} (|Du_m|^2 + u_m^2/r^2) $$ passing to the limit yields $$ \int_{B(0,r)}\frac{u^2}{|x|^2+\epsilon} \le C \int_{B(0, r)} (|Du|^2 + u^2/r^2) $$ This is true for every $\epsilon>0$, and $C$ does not depend on $\epsilon$, so we can let $\epsilon\to 0$ and conclude by the monotone convergence theorem.