About weak convergence in Sobolev space $W^{1,p}(U)$ The usual definition for a sequence $u_k $ converge weakly to $u$ in $W^{1,p} (U)$ is that $ u_k \rightharpoonup u$ in $L^p (U)$ and ${u_k}_ {x_i } \rightharpoonup u_{x_i }$ in $L^p (U)$ for all $i$.
But if I think $W^{1,p } (U)$ itself as a Banach space, weak convergence means for any bounded linear operator $f$ on it, $f(u_k ) \to f(u)$.
Are the two definitions equivalent? It's easy to get the first one from the second, but I have trouble seeing how the inverse hold.
Any help is appreciated. Many thanks in advance.
 A: Any linear functional $f$ on $W^{1,p}(U)$, $1 \le p < \infty$ can be identified with $(v_0,\ldots,v_n) \in L^{q}(U)^{n+1}$, $1/q + 1/p = 1$, via
$$f(u) = \int_U u \, v_0 + \sum_{i=1}^n u_{x_i} \, v_i \, \mathrm{d}x.$$
This shows the equivalency of your conditions.
And this identification can be shown by using
$T : W^{1,p}(U) \to L^p(U)^{n+1}$,
$$u \mapsto (u, u_{x_1}, \ldots, u_{x_n}).$$
This $T$ is isometric (if the norms are defined accordingly). Hence, a bounded, linear functional $f$ on $W^{1,p}(U)$ can be identified with bounded functional on a closed subspace of $L^p(U)^{n+1}$. Now, extend it by Hahn-Banach and identify with an element from $L^q(U)^{n+1}$.
A: A complete answer was already given. In this post I will consider the case $1<p<\infty$ (in which case we can exploit the reflexivity and get a stronger result).
Let $X$ be a normed space over $\mathbb K$, $X'$ the dual of $X$, $(x_k)$ a sequence in $X$ and $x\in X$.
Definition. We say that $(x_k)$ converges weakly to $x$ in $X$ if
$$f(x_k)\to f(x)\quad\text{in}\quad\mathbb K,\quad \quad\forall \ f\in X'.$$
Notation: If $(x_k)$ converges weakly to $x$ in $X$, we write
$$u_k\rightharpoonup u\quad\text{in}\quad X.$$

Theorem. Let $1<p<\infty$, $(u_k)\subset W^{1,p}(\Omega)$ and $u\in W^{1,p}(\Omega)$. The following assertions are equivalent.
  (a) $u_k\rightharpoonup u$ in $W^{1,p}(\Omega)$.
  (b) $u_k\rightharpoonup u$ in $L^p(\Omega)$ and ${u_k}_{x_i}\rightharpoonup u_{x_i}$ in $L^p(\Omega)$ for all $i=1,...,n$.
  (c) $u_k\rightharpoonup u$ in $L^p(\Omega)$ and $({u_k}_{x_i})$ is weakly convergent in $L^p(\Omega)$ for all $i=1,...,n$.
  (d) $u_k\rightharpoonup u$ in $L^p(\Omega)$ and $(\nabla {u_k})$ is bounded in $[L^p(\Omega)]^n$.

Proof.
$\mathbf{(a)\Rightarrow(b)}$. For $i=1,...,n$, define $f_i:W^{1,p}(\Omega)\to L^p(\Omega)$ by $f_i(v)=v_{x_i}$. Note that each $f_i$ is continuous because
$$\|f_i(v)\|_{L^p}^p=\|v_{x_i}\|_{L^p}^p\leq \|v\|_{L^p}^p+\sum_{i=1}^n\left\|v_{x_i}\right\|_{L^p}^p=\|v\|_{W^{1,p}}^p$$
As continuous linear operators preserve weak convergence,
$${u_k}_{x_i}=f_i(u_k)\ \rightharpoonup f_i(u)=u_{x_i}\quad\text{in}\quad L^p(\Omega),\qquad\forall \ i=1,...,n.$$
Analogously, $f_0:W^{1,p}(\Omega)\to L^p(\Omega)$ given by $f_0(v)=v$ is continuous and thus we also have
$$u_k=f_0(u_k)\ \rightharpoonup f_0(u)=u\quad\text{in}\quad L^p(\Omega).$$
$\mathbf{(b)\Rightarrow(c)}$. Trivial.
$\mathbf{(c)\Rightarrow(d)}$. Every weakly convergent sequence is norm bounded. Thus, $({u_k}_{x_1}), ..., ({u_k}_{x_n})$ are all bounded in $L^p(\Omega)$.
$\mathbf{(d)\Rightarrow(a)}$.
As $(u_k)$, $({u_k}_{x_1}), ..., ({u_k}_{x_n})$ are bounded in $L^p(\Omega)$, we conclude that $(u_k)$ is bounded in $W^{1,p}(\Omega)$. Let $(u_{k_m})$ be a subsequence of $(u_k)$. Since $W^{1,p}(\Omega)$ is reflexive, there exist a subsequence $(u_{k_{m'}})$ of $(u_{k_m})$ and a function $w\in W^{1,p}(\Omega)$ such that
$$u_{k_{m'}}\rightharpoonup w\quad\text{in}\quad W^{1,p}(\Omega).$$
It follows from the implication $\text{(a)}\Rightarrow\text{(b)}$ that
$$u_{k_{m'}}\rightharpoonup w\quad\text{in}\quad L^p(\Omega).$$
It follows from the uniqueness of the weak limit that $w=u$. In short: every subsequence of $(u_k)$ has a subsequence which converges to $u$. This implies that the entire sequence converges weakly to $u$. $\square$
Source: Le Dret (Proposition 1.8, p. 16).
