Prove $H^1(\Omega)$ is complete given that $L_2(\Omega)$ is complete Given that $L_2(\Omega)$ is complete, prove that $H^1(\Omega)$ is complete. Hint: Assume that $\|v_j-v_i\|\to 0$ as $i,j\to\infty$. Show that there are $v,w_k$ such that $\|v_j-v\|\to 0,\|\partial v_j/\partial x_k -w_k\|\to 0$ and that $w_k=\partial v/\partial x_k$ in the sense of weak derivative.
So far I did,
Since $L_2(\Omega)$ is complete we get $$\|v_j-v\|_{L_2}\to 0$$ where $v\in L_2(\Omega).$
Let $\{v_i\}$ be a Cauchy sequence in $H^1(\Omega)$, then 
\begin{align*}
&\|v_j-v_i\|_{H^1}\to 0\\
\implies\,&\|v_j-v_i\|_{L_2}+\|\nabla v_j-\nabla v_i\|_{L_2}\to 0.
\end{align*}
which gives us $$\|v_j-v_i\|_{L_2}\to 0$$
and $$\|\nabla v_j-\nabla v_i\|_{L_2}\to 0.$$
Now I am stuck, any help would be greatly appreciated. Thanks in advance.
 A: If $v_j$ is a Cauchy sequence in $H^1$, then $v_j$ and $\nabla v_j$ are Cauchy sequences in $L^2(\Omega)$. By completeness of $L^2$, we can assume that 
$$ v_j \to v, \quad \nabla v_j \to V \quad \text{in } L^2. $$
It's left to show that $v$ is weakly differentiable and $\nabla v = V$. 
Fix $\varphi \in C_c^\infty(\Omega, \mathbb{R}^n)$. By the definition of the weak derivative, 
$$ \int_\Omega \nabla v_j \cdot \varphi = - \int_\Omega v_j \operatorname{div} \varphi $$
for each $j=1,2,3,\ldots$. Taking the limit (both $\varphi$ and $\operatorname{div} \varphi$ are fixed functions in $L^2$)
$$ \int_\Omega V \cdot \varphi = - \int_\Omega v \operatorname{div} \varphi. $$
This holds for arbitrary $\varphi \in C_c^\infty(\Omega, \mathbb{R}^n)$, thus $V$ is the weak gradient of $v$. 
This proves that $v$ is an element of $H^1(\Omega)$ and $v_j \to v$ in $H^1(\Omega)$. 

This proof can be also phrased in an alternative way. If we choose to view the elements of $H^1(\Omega)$ as pairs of functions $(v, \nabla v)$, then the space $H^1(\Omega)$ is the following subspace of the Banach space $X = L^2(\Omega) \times L^2(\Omega,\mathbb{R}^n)$: 
$$ H^1(\Omega) = \left\{ (v,V) \in L^2(\Omega) \times L^2(\Omega,\mathbb{R}^n) : \forall_{\varphi \in C_c^\infty(\Omega, \mathbb{R}^n)} \int_\Omega V \cdot \varphi = - \int_\Omega v \operatorname{div} \varphi \right\}. $$
For each $\varphi \in C_c^\infty(\Omega, \mathbb{R}^n)$ denote 
$$ l_\varphi \colon X \to \mathbb{R}, \quad l_\varphi(v) = \int_\Omega V \cdot \varphi + \int_\Omega v \operatorname{div} \varphi. $$
Observe that this is a bounded functional on $X$, so its kernel $\ker l_\varphi$ is a closed subspace of $X$. Thus the intersection 
$$ H^1(\Omega) = \bigcap_{\varphi \in C_c^\infty(\Omega, \mathbb{R}^n)} \ker l_\varphi $$
is also a closed subspace of $X$. Since $X$ is complete, so is $H^1(\Omega)$. 
Both reasonings apply to arbitrary Sobolev spaces $W^{1,p}(\Omega)$, but the latter shows something more in the case $p>1$. Since $L^p(\Omega) \times L^p(\Omega,\mathbb{R}^n) \cong L^p(\Omega, \mathbb{R}^{n+1})$ is a reflexive Banach space and $W^{1,p}(\Omega)$ is its closed subspace, $W^{1,p}(\Omega)$ is also reflexive (this is a consequence of the Hahn-Banach theorem). 
