We have $f_k \rightharpoonup f$ in $L^2(0,T;W^{1,2}(\Omega))$, where the $f_k$ are said to satisfy the boundary condition $\nabla f_k \cdot\text{n}|_{\partial \Omega} = 0$ in the sense of traces. Further we have $\Delta f_k$ bounded in $L^p((0,T)\times \Omega)$ for some $p \geq 1$. Then it is claimed that $\nabla f \cdot \text{n}|_{\partial \Omega}=0$ in the sense of traces but I'm not sure if I understand this correctly. My idea is:
By the bound for $\Delta f_k$, it follows $\nabla f _k \rightharpoonup \nabla f$ in $L^2(0,T;W^{1,p}(\Omega))$. From the trace theorem we have $W^{1,p}(\Omega)$ embedded in $L^q(\partial \Omega)$ and may thus conclude
\begin{equation} \nabla f_k \rightharpoonup \nabla f \ \ \ \text{in } L^2(0,T;L^q(\partial \Omega)) \nonumber \end{equation}
and thus
\begin{equation} \nabla f_k \cdot \text{n} \rightharpoonup \nabla f \cdot \text{n} \ \ \ \text{in } L^2(0,T;L^q(\partial \Omega)). \nonumber \end{equation}
Hence,
\begin{equation} \|\nabla f \cdot \text{n}\|_{L^2(0,T;L^q(\partial \Omega))} \leq \liminf_{k \rightarrow \infty}\|\nabla f_k \cdot \text{n}\|_{L^2(0,T;L^q(\partial \Omega))} = 0, \end{equation}
which gives us the claimed statement.
My question is now, if I may argue like this? In particular, I'm not sure if this is what sense of traces means in this case, because we only consider the boundary of $\Omega$ and have in addition the time interval $(0,T)$. In the way stated above, don't we need $\nabla f(t) \cdot \text{n}|_{\partial \Omega}=0$ for all $t \in (0,T)$?
Thank you in advance for any help.