Almost everywhere convergent subsequence in a Sobolev space Let $1 < p < n$ and let $p^\ast$ denote the Sobolev conjugate of $p$, i.e. let
$$
p^\ast := \frac{np}{n-p}.
$$
Denote by $\mathcal{D}^{1,p}(\mathbb{R}^n)$ the space
$$
\mathcal{D}^{1,p}\left(\mathbb{R}^n\right) := \left\{ u \in L^{p^\ast}(\mathbb{R}^n) : \nabla u \in L^p(\mathbb{R}^n) \right\}
$$
equipped with the norm
$$
\left\Vert u \right\Vert := \left\Vert \nabla u \right\Vert_{L^p(\mathbb{R}^n)}.
$$
Assume now that $v_m \rightharpoonup v$ weakly in $\mathcal{D}^{1,p}(\mathbb{R}^n)$ and that $(v_m)$ is uniformly bounded in $L^{p^\ast}(\mathbb{R}^n)$. After passing to a subsequence, can we assume that $v_m \to v$ almost everywhere point wise? 
Naturally my first idea was to try and use results about compact embeddings, but I am unaware of any such result involving this space $\mathcal{D}^{1,p}(\mathbb{R}^n)$.
Edit: Note that in this case we are working in the space $\mathcal{D}^{1,p}(\mathbb{R}^n)$ which is not the same as $W^{1,p}(\mathbb{R}^n)$. In particular, the embedding theorems for $W^{1,p}(\mathbb{R}^n)$ do not seem to apply here.
 A: First of all we need to understand what weak convergence in this space means. Observe that we have an isometric embedding $\mathcal{D}^{1,p}(\mathbb R^n) \hookrightarrow L^p(\mathbb R^n,\mathbb R^n)$ given by $u \mapsto \nabla u.$ By standard results about dual spaces of subspaces, we get that
$$ \mathcal{D}^{1,p}(\mathbb R^n)' \cong L^{p'}(\mathbb R^n,\mathbb R^n) / Y, $$
where
$$ Y = \left\{g \in L^{p'}(\mathbb R^n,\mathbb R^n) \middle| \int_{\mathbb R^n} g.\nabla u = 0 ,\ \forall u \in D^{1,p}(\mathbb R^n)  \right\}.$$
(Note I do not claim that $\mathcal{D}^{1,p}(\mathbb R^n)$ is complete in this argument $(\dagger)$.) From this it is easy to see that $v_m \rightharpoonup v$ weakly in $D^{1,p}(\mathbb R^n)$ if and only if for all $g \in L^{p'}(\mathbb R^n, \mathbb R^n),$
$$ \int_{\mathbb R^n} g.\nabla v_m \rightarrow \int_{\mathbb R^n} g.\nabla v. $$

Now the idea is that a.e. convergence is a local property, so it suffices to restrict to a bounded domain where we get compact embeddings into $L^p.$ For this fix $\Omega \subset \mathbb R^n$ a bounded domain, then extending by zero we get for any $g \in L^{p'}(\Omega,\mathbb R^n),$
$$ \int_{\Omega} g.\nabla v_m \rightarrow \int_{\Omega} g.\nabla v. $$
Now the restriction $(v_n|_{\Omega})$ defines a bounded sequence in $W^{1,p}(\Omega),$ where we have used Hölder and the uniform bound of $(v_n)$ in $L^{p^*}(\mathbb R^n).$ We know by the Rellich-Kondrachov compactness theorem that there exists $u \in W^{1,p}(\Omega)$ and a subsequence $(v_{n_k}|_{\Omega})$ such that $v_{n_k} \rightharpoonup u$ weakly in $W^{1,p}(\Omega)$ and $v_{n_k} \rightarrow u$ a.e. also. Then for any $g \in L^{p'}(\Omega,\mathbb R^n)$ we have,
$$ 0 = \lim_{m\rightarrow 0} \int_{\mathbb R^n} g.\nabla v_{n_k} - g.\nabla v = \int_{\mathbb R^n} g.\nabla(u-v). $$
Hence $\nabla(u-v) = 0$ a.e. in $\Omega,$ and so there is $\lambda_{\Omega} \in \mathbb R$ such that $v_{n_k} \rightarrow (v + \lambda_{\Omega})$ a.e. in $\Omega.$
Now take a compact exhaustion $\mathbb R^n = \bigcup_j \Omega_j$ and iteratively choose subsequences $v_{k,j}$ such that $v_{k,j} \rightarrow v + \lambda_i$ a.e. as $k \rightarrow \infty$ in $\Omega_i$ for all $i \leq j.$ Then observe by Fatou's lemma that,
$$ k_j^{p^*} |\Omega_j| \leq \liminf_{k \rightarrow \infty} \int_{\Omega_j} |v_{k,j} - v|^{p^*} \leq \sup_{m} \int_{\mathbb R^n} 2^{p^*}(|v_m|^{p^*} + |v|^{p^*}) < \infty, $$
so $k_j \rightarrow 0$ as $j \rightarrow \infty.$ Hence the diagonal sequence $v_{n_k} = v_{k,k}$ satisfies $v_{n_k} \rightarrow v$ a.e. in $\mathbb R^n.$
Now as the limit is unique, we see that every subsequence of $(v_m)$ has a further subsequence which converges a.e. to $v.$ Hence the entire sequence converges a.e. to $v.$

Added later $(\dagger)$: Writing $X = \mathcal{D}^{1,p}(\mathbb R^n)$ and $\lVert \cdot \rVert = \lVert \nabla \cdot \rVert_{L^p(\mathbb R^n,\mathbb R^n)},$ it is not clear whether $(X,\lVert\cdot\rVert)$ is complete (which is equivalent to its image being closed in $L^p(\mathbb R^n,\mathbb R^n)$). However I claim this is not necessary for the argument to work.
Let $\overline X$ be the completion of $X$ with respect to this norm, which can be identified with the closure of $X$ in $L^p(\mathbb R^n,\mathbb R^n)$ identified via this isometric embedding. Then the restriction map $\overline X' \ni f \mapsto f|_X$ defines an isometric isomorphism $X' \cong \overline X'.$ Also we have,
\begin{align*}
  Y &= \left\{ g \in L^{p'}(\mathbb R^n,\mathbb R^n) \middle| \int_{\mathbb R^n} g.\nabla u = 0, \ \forall u \in X \right\} \\
    &= \left\{ g \in L^{p'}(\mathbb R^n,\mathbb R^n) \middle| \int_{\mathbb R^n} g.\nabla u = 0, \ \forall u \in \overline X \right\}
\end{align*}
Identifying $L^{p'}(\mathbb R^n,\mathbb R^n) \cong L^p(\mathbb R^n,\mathbb R^n)',$ we have $Y = X^{\perp} = \overline X^{\perp}.$ Now the result about duality of subspaces is generally is stated for closed subspaces (see e.g. proposition 3.67 in section 3.16 of Introduction to Banach Spaces and Algebras by Allan & Dales), but in light of above we have,
$$ X' \cong \overline X' \cong L^p(\mathbb R^n,\mathbb R^n)' / \overline{X}^{\perp} \cong L^{p'}(\mathbb R^n,\mathbb R^n) / Y, $$
so it also holds even if $X$ is not complete.
