Let $u_n: [0,T]\times \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a sequence with \begin{equation} u_n \rightharpoonup u \ \ \ \text{weakly star in } L^2(0,T;W^{1,\infty}(\mathbb{R}^3)) \end{equation}

and $\eta : [0,T]\times \mathbb{R}^3 \rightarrow \mathbb{R}^3$ continuous in time. Assume further $x \in \mathbb{R}^3$, $\psi \in L^2(0,T)$. Now it says

\begin{equation} \int _0^T (u_n(t, \eta(t,x)) - u(t, \eta(t,x)) ) \psi dt \rightarrow 0 \end{equation}

and I'm not sure if I understand why this is true. I think this follows from the Morrey embedding, which states that $W^{1,\infty }(\mathbb{R}^3) \subset C^{0,\alpha}(\mathbb{R}^3)$ is a compact embedding for $\alpha < 1$. What confuses me is that $\eta$ depends also on $t$ but I think this does not matter as the convergence in $C^0$ is uniform, we should have something like

\begin{equation} \int _0^T (u_n(t, \eta(t,x)) - u(t, \eta(t,x)) ) \psi dt \leq \int _0^T \sup_{y}|u_n(t, y) - u(t, y) | \psi dt \rightarrow 0 \end{equation}

although I'm not sure if this can be written like that. Is this argument correct?

  • $\begingroup$ $W^{1,\infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding. $\endgroup$ – Michał Miśkiewicz Nov 16 '18 at 19:24
  • $\begingroup$ But I think we need the compactness of the embedding to obtain the convergence uniformly in the second argument of $u_n$, right? Otherwise we only have the weak-star convergence and I don't think that this is enough $\endgroup$ – jason paper Nov 17 '18 at 2:32

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