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

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