A sobolev function $u$ that $u\in L^1$ but $\nabla u\in L^2$. Will it be $H^1$? Take $\Omega\subset \mathbb R^N$, open bounded, smooth boundary. Take $u_n\subset L^1$ a sequence of functions so that $u_n\to u$ strongly in $L^1$ and
$$
\sup_n\int_\Omega |\nabla u_n|^2dx<\infty,\,\,\int_\Omega |\nabla u|^2dx<\infty
$$
My question: Can we show that 
$$
\liminf \int_\Omega |\nabla u_n|^2dx\geq \int_\Omega |\nabla u|^2dx
$$
I got stuck on how to show that $u_n\in H^1(\Omega)$. Because $u_n$ is not in $L^2$, I don't know whether the embedding can work or not.
 A: The $L^2$-norm is weakly sequentially lower semicontinuous, then you can prove the desired inequality with the liminf provided you show that $\nabla u_n \rightharpoonup \nabla u$ in $L^2$. The fastest way to achieve this is probably to show that $\{u_n\}$ is bounded in $H^1(\Omega)$ (and then use the sequential Banach-Alaoglu theorem together with uniqueness of the weak limit, which will imply that $\nabla u_n$ weakly converges to the right stuff, i.e. $\nabla u$).
Then let me answer the question in the title: take $v \in L^1(\Omega)$ such that $\nabla v \in L^2(\Omega)$ and let's prove that $v \in H^1(\Omega)$. I am going to apply Poincare's inequality, so we really need the smoothness assumption on the domain (Lipschitz is all we need actually).
Let $v_k$ be a truncation of $v$, defined as follows: 
$$ v_k(x) = 
\begin{cases}
-k &\ \text{if}\ v(x) < -k \\
k &\ \text{if}\ v(x) > k \\
v(x) &\ \text{otherwise}.
\end{cases}
$$
Then we have that $v_k \in L^2(\Omega)$ and $|\nabla v_k| \le |\nabla v|$, indeed 
$$ \nabla v_k(x) = 
\begin{cases}
0 &\ \text{if}\ v(x) < -k \\
0 &\ \text{if}\ v(x) > k \\
\nabla v(x) &\ \text{otherwise}.
\end{cases}
$$
Now let $E$ be your favorite set of positive measure contained in $\Omega$ and let $$w_E := \frac{1}{\mathcal{L}^N(E)}\int_Ew(x)\,dx.$$ Then, by Poincare's inequality we get $$\int_{\Omega}|v_k(x) - (v_k)_E|^2\,dx \le C(\Omega,E)\int_{\Omega}|\nabla v_k(x)|^2\, dx \le C(\Omega,E)\int_{\Omega}|\nabla v(x)|^2\, dx$$ By Fatou's lemma we then get $$\int_{\Omega}|v(x) - v_E|^2\,dx \le C(\Omega,E)\int_{\Omega}|\nabla v(x)|^2\, dx < \infty.$$
This shows that if $\Omega$ is regular enough $$\{u : u \in L^1(\Omega), \nabla u \in L^2(\Omega)\} = H^1(\Omega).$$
