Implications of weak convergence on the Lebesgue space to Sobolev space Suppose $\Omega$ is a bounded smooth domain in $\mathbb{R}^N$, $1<p<\infty$. We define the Sobolev space
$$
W^{1,p}(\Omega)=\{u:\Omega\to\mathbb{R} : ||u||_{W^{1,p}(\Omega)}=||u||_{L^p(\Omega)}+||\nabla u||_{L^p(\Omega)}<\infty\}.
$$
Let $u_n\to u$ in $L^p(\Omega)$ weakly and $\nabla u_n \to \nabla u$ weakly in $L^p(\Omega)$.
Then is it true that $u_n\to u$ weakly in $W^{1,p}(\Omega)$? 
I have a hint that the map
$$
T:W^{1,p}(\Omega)\to L^p(\Omega)\times L^p(\Omega)
$$ 
defined by $T(u)=(u,\nabla u)$ is an isometry. But unable to prove.
Please help with proper explanation!
Thanks in advance.
 A: The hint is not quite correct, because $\nabla u$ is an $\mathbb R^N$-valued function. If we define $L^p(\Omega,\mathbb R^N)$ to be the space of measurable functions $v : \Omega \rightarrow \mathbb R^N$ such that $|v| \in L^p(\Omega,\mathbb R^N)$ equipped with the norm $ \lVert v \rVert_{L^p(\Omega,\mathbb R^N)} := \lVert |v| \rVert_{L^p(\Omega)},$ this we can show the map,
$$ T : W^{1,p} \rightarrow L^p(\Omega) \times L^p(\Omega,\mathbb R^N)$$
sending $u \mapsto (u,\nabla u)$ is an isometric embedding. Here if we define the norm on the product space as $\lVert (u,v) \rVert = \lVert u \rVert_{L^p(\Omega)} + \lVert v \rVert_{L^p(\Omega,\mathbb R^N)},$ then it's somewhat tautological.

To prove the result, there are a few subtle technical points we need to address. Following the hint, what we want to say is that $T(u_n)$ converges weakly to $T(u)$ because of our hypotheses and hence $u_n$ converges in $W^{1,p}$ using the isometry. However, it's not totally obvious why this is okay.
First of all, if $X,Y$ are Banach spaces then we get a natural isomorphism $(X \times Y)^* \simeq X^* \times Y^*,$ by sending $f \mapsto (f|_{X \times \{0\}},f|_{\{0\} \times Y}).$ So if $\phi \in \left(L^p(\Omega) \times L^p(\Omega,\mathbb R^N)\right)^*,$ viewing $\phi = (\phi_1,\phi_2)$ we get,
$$ \phi(T(u_n)) = \phi_1(u_n) + \phi_2(\nabla u_n) \rightarrow \phi_1(u) + \phi_2(\nabla u) = \phi(T(u)),$$
as $n \rightarrow \infty$ by assumption that they converge weakly.
Now as $T$ is an isometric embedding, we can define an inverse map,
$$ S : \mathrm{Im}\, T \rightarrow W^{1,p}. $$
Since $T$ is an isometry, this is well-defined and bounded such that $ST = \mathrm{id}.$ So if $\psi \in (W^{1,p}(\Omega))^*,$ we get $\psi \circ S$ is an element of $(\mathrm{Im}\, T)^*.$ By Hahn-Banach, this extends to a linear function on $\phi$ on $L^p(\Omega) \times L^p(\Omega,\mathbb R^N).$ Then we get,
$$ \psi(u_n) = \psi(ST(u_n)) = \phi(T(u_n)) \rightarrow \phi(T(u)) = \psi(ST(u)) = \psi(u) $$
as $n \rightarrow \infty.$ Hence we get $u_n$ converges weakly to $u$ in $W^{1,p}(\Omega).$
