## Hardy’s inequality in higher dimension

Thm: Supppose ${u\in H^1(\mathbb{R}^n)}$ with ${n\geq 3}$ prove that

$\displaystyle \int_{\mathbb{R}^n}\frac{u^2}{|x|^2}dx\leq C(n)\int_{\mathbb{R}^n}|\nabla u|^2dx$

Proof: Suppose ${u\in C^\infty_c(\mathbb{R}^n)}$ firstly. We can find ball ${B_R(0)}$ large enough such that ${Supp. u\subset B_R}$, then we only need to prove the inequality within ${B_R}$.

Let

$\displaystyle I=\int_{B_R}\frac{u^2}{|x|^2}dx=\frac{-1}{2}\int_{B_R}x\cdot D\left(\frac{1}{|x|^2}\right)u^2dx$

Apply the Green identity on ${B_R\backslash B_\epsilon(0)}$,

$\displaystyle \int_{B_R\backslash B_\epsilon(0)}x\cdot D\left(\frac{1}{|x|^2}\right)u^2dx+\int_{B_R\backslash B_\epsilon(0)}\frac{1}{|x|^2}D(xu^2)dx=\int_{\partial B_\epsilon}\frac{1}{|x|^2}xu^2\cdot\nu dx$

where ${\nu}$ is the unit normal vector of ${\partial B_\epsilon}$

$\displaystyle \left|\int_{\partial B_\epsilon}\frac{1}{|x|^2}xu^2\nu dx\right|\leq \int_{B_\epsilon}\frac{u^2}{|x|^2}dx\leq \frac{1}{\epsilon}\leq \frac{1}{\epsilon}\sup u^2n w_n\epsilon^{n-1}$

Let ${\epsilon\rightarrow 0}$, we have

$\displaystyle -2I+\int_{B_R}\frac{1}{|x|^2}D(xu^2)dx=0$

$\displaystyle -2I+\int_{B_R}\frac{nu^2}{|x|^2}+\frac{1}{|x|^2}2uDu\cdot x=0$

$\displaystyle -2I+nI+2\int_{B_R}\frac{u}{|x|^2}Du\cdot xdx=0$

$\displaystyle I=\frac{2}{2-n}\int_{B_R}\frac{u}{|x|^2}Du\cdot xdx$

From Holder inequality, we have

$\displaystyle I\leq \left(\frac{2}{n-2}\right)^2\int_{B_R}|\nabla u|^2dx$

Next suppose ${u\in H^1(\mathbb{R}^n)}$. Since ${C^\infty_c(\mathbb{R}^n)}$ is dense in ${H^1(\mathbb{R}^n)}$, there exist ${u_k\subset C^\infty_c(\mathbb{R}^n)\rightarrow u}$ in ${H^1(\mathbb{R}^n)}$. From Fatou’s lemma

$\displaystyle \int_{\mathbb{R}^n}\frac{u^2}{|x|^2}dx\leq \lim\inf_{k\rightarrow \infty}\int_{\mathbb{R}^n}\frac{u_k^2}{|x|^2}dx\leq C(n)\lim\inf_{n\rightarrow \infty}\int_{B}|Du_k|^2=C(n)\int_{\mathbb{R}^n}|\nabla u|^2dx$

Remark: If ${n=2}$, we should prove that if ${u\in C^\infty_c(\Omega)}$, where ${\Omega=\{x|1\leq |x|<\infty\}}$

$\displaystyle \int_{\Omega}\frac{u^2}{|x|^2\ln^2|x|}dx\leq 4\int_{\Omega}|Du|^2dx$