Three-dimensional Hardy bound

[The divergence of the radial weight] On ${\mathbb{R}}^3\setminus \{0\}$ one has $\mathrm{div}(x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2)={\sum }_{i=1}^3{\mathrm{\partial }}_i(x_i\mathrm{\mid }x{\mathrm{\mid }}^{-2})=3\mathrm{\mid }x{\mathrm{\mid }}^{-2}-2\mathrm{\mid }x{\mathrm{\mid }}^{-2}=\mathrm{\mid }x{\mathrm{\mid }}^{-2}$, because the derivative of $\mathrm{\mid }x{\mathrm{\mid }}^{-2}$ in the direction $x$ is $-2\mathrm{\mid }x{\mathrm{\mid }}^{-4}{\sum }_i{x}_i^2=-2\mathrm{\mid }x{\mathrm{\mid }}^{-2}$.

[Integration by parts on the punctured space] Let $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\};\mathbb{C})$ and put $W(x):=\mathrm{\mid }u(x){\mathrm{\mid }}^{2}x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2$. Then $W$ is $C^1$ on ${\mathbb{R}}^3\setminus \{0\}$ and compactly supported there, and step 1.1 gives $\mathrm{div}W=\mathrm{\nabla }(\mathrm{\mid }u{\mathrm{\mid }}^2)\cdot x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2+\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2$. Since ${\int }_{{\mathbb{R}}^3}\mathrm{div}W=0$ by [F1], and $\mathrm{\nabla }(\mathrm{\mid }u{\mathrm{\mid }}^2)=2\mathrm{Re}(\bar{u}\mathrm{\nabla }u)$, we obtain ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^{2}dx=-2\mathrm{Re}{\int }_{{\mathbb{R}}^3}\bar{u}(x)x\cdot \mathrm{\nabla }u(x)\mathrm{\mid }x{\mathrm{\mid }}^{-2}dx$.

[Hardy bound away from the origin] With $u$ as in step 2.1, the absolute value of the right-hand integral is at most $\int (\mathrm{\mid }u\mathrm{\mid }\mathrm{/}\mathrm{\mid }x\mathrm{\mid })\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$ by [F2], while the left-hand side is $\mathrm{\parallel }u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2^2$; both quantities are finite by [F6]. Hence $\mathrm{\parallel }u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2^2\le 2\mathrm{\parallel }u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$, and if the factor is zero the conclusion is immediate, and otherwise division by its positive value yields $\mathrm{\parallel }u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$, which is (a).

[Removing the puncture] Let $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{C})$ and use the radial $\chi =1-{\chi }_0\in C^{\mathrm{\infty }}({\mathbb{R}}^3;[0,1])$ of [F7], with $\chi =0$ for $\mathrm{\mid }x\mathrm{\mid }\le 1$, $\chi =1$ for $\mathrm{\mid }x\mathrm{\mid }\ge 2$ and $\mathrm{\mid }\mathrm{\nabla }\chi \mathrm{\mid }\le C$; put ${\chi }_{\epsilon }(x):=\chi (x\mathrm{/}\epsilon )$. Then $u_{\epsilon }:={\chi }_{\epsilon }u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\};\mathbb{C})$, because its support is contained in the closed set $\{\mathrm{\mid }x\mathrm{\mid }\ge \epsilon \}\cap \mathrm{supp}u$ and hence stays a positive distance from the origin, so step 3.1 applies: $\mathrm{\parallel }{u}_{\epsilon }\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }\mathrm{\nabla }{u}_{\epsilon }{\mathrm{\parallel }}_2$. As $\epsilon \downarrow 0$ the functions $u_{\epsilon }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ converge to $u\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ pointwise on $\{\mathrm{\mid }x\mathrm{\mid }>0\}$ and are dominated by $\mathrm{\mid }u\mathrm{\mid }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }\in L^2$ by [F6], so $\mathrm{\parallel }{u}_{\epsilon }\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2\to \mathrm{\parallel }u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2$ by [F5] applied to the squared moduli. Moreover $\mathrm{\nabla }{u}_{\epsilon }={\chi }_{\epsilon }\mathrm{\nabla }u+u\mathrm{\nabla }{\chi }_{\epsilon }$ with $\mathrm{\parallel }{\chi }_{\epsilon }\mathrm{\nabla }u-\mathrm{\nabla }u{\mathrm{\parallel }}_2^2\le {\int }_{\mathrm{\mid }x\mathrm{\mid }<2\epsilon }\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2\to 0$ and, with $M:=\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$, $\mathrm{\parallel }u\mathrm{\nabla }{\chi }_{\epsilon }{\mathrm{\parallel }}_2^2\le C^2{\epsilon }^{-2}{\int }_{\mathrm{\mid }x\mathrm{\mid }\le 2\epsilon }{M}^{2}dx\le C^2{M}^2\frac{4\pi (2\epsilon {)}^3}{3{\epsilon }^2}=O(\epsilon )\to 0$; hence $\mathrm{\parallel }\mathrm{\nabla }{u}_{\epsilon }{\mathrm{\parallel }}_2\to \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$, and passing to the limit in the display gives (b).

[Density extension to $W^{1,2}$] Let $u\in W^{1,2}({\mathbb{R}}^3;\mathbb{C})$ and choose $u_n\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{C})$ with $\mathrm{\parallel }{u}_n-u{\mathrm{\parallel }}_{W^{1,2}}\to 0$ by [F3]. Step 4.1 applied to $u_n-u_m$ gives $\mathrm{\parallel }{u}_n\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }-u_m\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }\mathrm{\nabla }(u_n-u_m){\mathrm{\parallel }}_2$, so $(u_n\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid })$ is Cauchy in $L^2$; let $w\in L^2$ be its limit, so $\mathrm{\parallel }w{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$ because $\mathrm{\parallel }\mathrm{\nabla }{u}_n{\mathrm{\parallel }}_2\to \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$. By [F4] some subsequence satisfies $u_{n_k}\to u$ and $u_{n_k}\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }\to w$ pointwise a.e.; on the conull set where $u_{n_k}\to u$ and $\mathrm{\mid }x\mathrm{\mid }>0$ we have $u_{n_k}(x)\mathrm{/}\mathrm{\mid }x\mathrm{\mid }\to u(x)\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, so $w=u\mathrm{/}\mathrm{\mid }\cdot \mathrm{\mid }$ a.e. and (c) follows.