How can function be assumed smooth in proving Hardy's inequality in Evans book?

Question

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.

Answer 1

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.