How can I find the difference between weakly converge in Sobolev spaces and all orders weak derivative weakly converge in $L^p$

Question

$U$ is a open subset in $\mathbb{R}^n$, $u_k$ is a sequence in $H^1(U)$ (Sobolev space).

1. $u_k$ weakly converges to $u$ in $H^1(U)$.
2. $u_k$ and $Du_k$ weakly converge to $v$ and $Dv$ in $L^2(U)$. How to show that (1) implies (2), and (2) implies (1)?

I can't find the way to prove them. In other words, what is the difference between weakly converge in Sobolev spaces and the all order weak derivative weak converge in $L^p(U)$? Could someone give me some details, thank you!

Answer 1

If $u_k \rightharpoonup u$ in $H^1(U)$, then $u_k\rightharpoonup u$ and $Du_k\rightharpoonup Du$ in $L^2(U)$ as $id:H^1\to L^2$ and $D:H^1\to L^2$ are linear and continuous.

Now let $u_k\rightharpoonup u$ and $Du_k\rightharpoonup Du$ in $L^2(U)$. Let $f\in H^1(U)^*$. Since $H^1(U)$ is a Hilbert space there is $w\in H^1(U)$ such that $$\int_\Omega \nabla w\cdot \nabla v +wv\ dx = f(v)$$ for all $v\in H^1(U)$. Using this representation it is easy to prove $f(u_k)\to f(u)$.