Uniform bound in $L^\infty(0,T;X)$ implies pointwise a.e. convergence in $Y$ if $X \subset Y$ is compact?

Question

Suppose that $u_n \in L^\infty(0,T;X)$ is a uniformly bounded sequence, so $u_n \to u$ weakly-star to some $u$.

and that $X \subset Y$ is a compact embedding of Hilbert spaces. Does it follow that for a subsequence $$u_n(t) \to u(t)$$ strongly in $Y$?

The argument is, since $u_n$ is bounded, we have for a.e. $t$ that $$\lVert u_n(t) \rVert_X \leq C$$ bounded uniformly. So there is a subsequence $u_{n_j}(t) \to w(t)$ in $Y$ by the compact embedding for a.e. $t$. And then we can identify $u(t) = w(t)$, is that right?

Answer 1

This is not true even for $X=Y=\mathbb R$. Define $$ u_n(t) = sign(\sin(n t)) $$ on $(0,T):=(0,2\pi)$. This sequence converges weakly to $0$ in all $L^p$, $1\le p<\infty$, and weak-star to $0$ in $L^\infty$. Since $|u_n(t)|=1$ for almost all $t$ and all $n$, there is no subsequence that converges pointwise a.e. to zero.

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There are several problems in your argument. First, the map $t\mapsto u(t)$ is not continuous in $L^p$-spaces. Second, ignoring this, then you cannot go from

For all $t$ there is a subsequence $(u_{n_k})$ such that $u_{n_k}(t)$ converges.

to

There is a subsequence $(u_{n_k})$ such that $u_{n_k}(t)$ converges for all $t$.