Why is the weak limit of the derivatives the derivative of the weak limit here?

Question

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}

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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?

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

Answer 1

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 $\phi \in C_0^\infty(U)$.