## Completeness of H(\Omega)

$\displaystyle D(\Omega)=\{\overset{\rightharpoonup}{u}\in C^\infty_c(\Omega)| \nabla\cdot \overset{\rightharpoonup}{u}=0 \}$

$\displaystyle H(\Omega)$ is  the closure of $D(\Omega)$ under the norm of $H^1_0(\Omega)$

$\langle\displaystyle \overset{\rightharpoonup}{u}, \overset{\rightharpoonup}{v}\rangle_{H(\Omega)}=\int_{\Omega}\nabla\overset{\rightharpoonup}{u}\cdot \nabla\overset{\rightharpoonup}{v}dx$

$H(\Omega)$ is defined for arbitrary domain $\Omega\subset \mathbb{R}^3$ or $\mathbb{R}^2$, unless $\Omega=\mathbb{R}^2$

Let us prove $H(\Omega)$ is complete.

Simply take $u$ as $\overset{\rightharpoonup}{u}$, $u$ is understood as a vector function.

$\textbf{Proof:}$ $n=3$. Recall the Hardy inequality, for all $u\in C^\infty_c(\Omega)$

$\displaystyle \int_{\mathbb{R}^3}\frac{u^2(x)}{|x-y|^2}dx\leq 4\int_{\mathbb{R}^3}|\nabla u|^2dx$, $\forall y\in \mathbb{R}^3$, $\quad (1)$

Suppose $u_n$ is cauchy sequence in $H(\Omega)$, namely $\int_{\Omega}|\nabla u_n-\nabla u_m|^2dx\to 0$ as $n,m\to \infty$.

Since $D(\Omega)$ is dense in $H(\Omega)$, WLOG assume $u_n\in D(\Omega)$, $\forall n$.

Then apply $(1)$, we get,

$\displaystyle \int_{\Omega}\frac{(u_n(x)-u_m(x))^2}{|x-y|^2}dx\leq 4\int_{\Omega}|\nabla u_n-\nabla u_m|^2dx\to \infty$, $y$ is fixed.

This means $\displaystyle \frac{u_n(x)}{|x-y|}\to \frac{u(x)}{|x-y|}$ in $L^2(\Omega)$.

Apparently, $\nabla u_n$ should converge to some $v$ in $L^2(\Omega)$. Let us prove $\nabla u=v$.

For any $w\in D(\Omega)$

$\displaystyle \int_{\Omega}u_n\cdot\nabla wdx=-\int_{\Omega}\frac{u_n}{|x-y|}\cdot w\nabla|x-y| dx+\int_{\Omega}\frac{u_n}{|x-y|}\cdot\nabla(|x-y|w)$  $(2)$

Since $w\nabla|x-y|$ and $\nabla(|x-y|w)$ are both $C^\infty_c(\Omega)$ functions except a singular point $y$,

$\displaystyle \int_{\Omega}\frac{u_n}{|x-y|}\cdot w\nabla|x-y| dx\to \int_{\Omega}\frac{u}{|x-y|}\cdot w\nabla|x-y| dx$ as $n\to \infty$

$\displaystyle \int_{\Omega}\frac{u_n}{|x-y|}\cdot\nabla(|x-y|w)\to \int_{\Omega}\frac{u}{|x-y|}\cdot\nabla(|x-y|w)$ as $n\to \infty$

$(2)$ becomes

$\displaystyle \int_{\Omega}u_n\cdot\nabla wdx\to \int_{\Omega}u\cdot\nabla wdx$

However $\displaystyle \int_{\Omega}u_n\cdot\nabla wdx=-\int_{\Omega}\nabla u_n\cdot wdx \to -\int_{\Omega}v\cdot wdx$, which means

$\displaystyle \int_{\Omega}u\cdot\nabla wdx\to -\int_{\Omega}v\cdot wdx$

Then $\nabla u=v$ and it is easy to prove $v\in H(\Omega)$. And $u_n\to v$ in $H(\Omega)$

When $n=2$, use the same method by

$\displaystyle \int_{|x-y|>1}\frac{u^2(x)}{|x-y|^2\ln^2|x-y|}dx\leq 4\int_{|x-y|>1}|\nabla u|^2dx$, $\forall y\in \mathbb{R}^3$