Showing that a certain space is a Hilbert. I'm having trouble answering one question in my analysis class.
Let $\Omega \subset \mathbf{R}^{3}$, with $0\in\Omega$ and $\Omega$ bounded.
$$ W= \left\{\phi, \frac{\phi}{\lvert x\rvert} \in \mathbf{L}^2(\mathbf{R}^3-\overline\Omega),\phi \text{ is constant on }\partial\Omega, \nabla\phi   \in \mathbf{L}^2(\mathbf{R}^3-\overline\Omega)\right\}.$$
For $\phi$, $\psi$ in $W$, let
$$(\phi,\psi) =  \int_{\mathbf{R}^3-\Omega} \frac{\phi(x)\cdot\psi(x)}{\left\lvert x\right\rvert^2} \,dx + \int_{\mathbf{R}^3-\Omega} (\nabla\phi(x))\cdot(\nabla\psi(x)) \,dx.$$
Prove that $(W,(\cdot,\cdot))$ is a Hilbert space.
I can't figure out how to prove completeness. Any help would be appreciated.
 A: You have to assume something about $\Omega$ (e.g. for non-measurable $\Omega$, the assertion is probably wrong). Let us assume for simplicity that $\Omega$ is closed, since otherwise you will have difficulties to define $\nabla\varphi$.
Then a possible proof is as follows.
Let $\phi_n$ be a Cauchy sequence. Then $\nabla\phi_n$ is a Cauchy sequence in $X=L_2(\mathbb R^3-\Omega,\mathbb R^3)$, and $\phi_n$ is a Cauchy sequence in $Y=L_2(\mu)$ where $\mu$ is the weighted measure on $\mathbb R^3-\Omega$ with weight function $x\mapsto1/x$. Since $X$ and $Y$ are complete, there are $\psi\in X$, $\varphi\in Y$ with $\lVert\nabla\varphi_n-\psi\rVert_X\to0$, $\lVert\varphi_n-\varphi\rVert_Y\to0$.
Now on any bounded open subset $B\subseteq\mathbb R-\Omega$ the restriction $\phi_n|_B\in Z=W^{1,2}(B)$ is a Cauchy sequence, and since $W^{1,2}(B)$ is complete, there is $\Phi\in Z=W^{1,2}(B)$ with $\lVert\varphi_n|_B-\Phi\rVert_Z\to0$.
It follows that $\varphi|_B=\Phi$ and $\psi|_B=\nabla\Phi$, in particular $\psi|_B=\nabla\varphi|_B$. Since $B$ was arbitrary, it follows that $\psi=\nabla\varphi$.
Thus, $\varphi\in W$ and $\lVert\varphi_n-\varphi\rVert_W\to0$.
