# Weak convergence improved by Morrey embedding

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

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

$$$$\int _0^T (u_n(t, \eta(t,x)) - u(t, \eta(t,x)) ) \psi dt \rightarrow 0$$$$

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

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

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

• $W^{1,\infty}$ is the space of Lipschitz functions, so you don't need to apply Morrey embedding. – Michał Miśkiewicz Nov 16 '18 at 19:24
• 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 – jason paper Nov 17 '18 at 2:32