Why is the weak limit of the derivatives the derivative of the weak limit here? In [1, chapter 8.2.1.b, p.466] the author uses the following argument:
Let $U \subset \mathbb{R}^N$ be an open, bounded domain with smooth boundary. Given a bounded sequence $(u_k)_{k \in \mathbb{N}}$ in $W^{1,q}(U)$ there exists a weakly convergent subsequence $(u_{k_j})_{j \in \mathbb{N}}$ and a function $u \in W^{1,q}(U)$ such that \begin{equation} \begin{cases} u_{k_j} \rightharpoonup u & \text{weakly in } L^q (U) \\ Du_{k_j} \rightharpoonup Du & \text{weakly in } L^q (U,\mathbb{R}^N) \end{cases} \end{equation}

My question is: how can he infer this rather than
\begin{equation} \begin{cases} u_{k_j} \rightharpoonup u & \text{weakly in } L^q (U) \\ Du_{k_j} \rightharpoonup Df & \text{weakly in } L^q (U,\mathbb{R}^N) ,\end{cases} \end{equation} where not necessarily $f=u$? 

I.e. how can he infer that the weak limit of the derivatives is the derivative of the weak limit?


[1] Evans, L. C.: Partial differential equations, Nr. 19 in Graduate
Studies in Mathematics. American Mathematical Society, Providence, RI,
2nd edition, 2010.

Because taking the weak derivative is a bounded linear operation, hence weakly continuous.
You can also prove it directly:
if $u_k$ converges weakly towards $u$ and $Du_k$ converges weakly towards $Df$, then
$$
\int Df \, \phi
\leftarrow
\int Du_k \, \phi
=
-\int u_k \, D\phi
\to
-\int u \, D \phi
=
\int Du \, \phi
$$
for all $\phi \in C_0^\infty(U)$.
