Prove that $V$ is a closed subspace of $H^1$ Let $U \subset \mathbb{R}^n$ be connected. Consider the Sobolev space $H^1(U) = W^{1,2}(U) $ equipped with the usual inner product
$$ (u,v)_{H^1} = \int_U uv + DuDv \; dx  $$
Show that 
$$ V = \{ u \in H^1(U) : \int_U u \; dx = 0 \}. $$
is a closed subspace of $H^1(U)$, i.e $ \left( V , ( \cdot , \cdot)_{H^1} \right)$ is a Hilbert space.
Does anyone have any hints? I was thinking about defining a sequence $(u_n) \subset V$ such that $u_n \rightarrow u \in H^1$, and show that $u \in V$, but I got stuck. 
Thanks.

I have just an idea: I would consider $F \colon H^1 \to \mathbb{R}$, defined as $F(u) = \int_U u$. $F$ is a linear bounded (i.e. continuous) functional and $V = F^{-1}(0)$ is therefore closed being the inverse image of the closed set $\{0\}$.
EDIT I am sorry, I didn't notice $U$ is not bounded! This isn't working if $U$ is unbounded i guess...
