Existence of subsequences convergence with weak topology

Question

Let $\left\{ {{\varphi _n}} \right\}$ is the sequence bounded in ${L^\infty }\left( {0,\infty ;H_0^1\left( {0,1} \right)} \right)$. Is there exists $\varphi \in {L^\infty }\left( {0,\infty ;H_0^1\left( {0,1} \right)} \right)$ and subsequence $\left\{ {{\varphi _{{n_k}}}} \right\}$ such that ${\varphi _{{n_k}}} \to \varphi $ in weak star topology of ${L^\infty }\left( {0,\infty ;H_0^1\left( {0,1} \right)} \right)$ and for almost every $t \in \left( {0,\infty } \right)$ we have ${\varphi _{{n_k}}}\left( t \right) \to \varphi \left( t \right)$ weakly in ${H_0^1}\left(0,1 \right)$.

Answer 1

There will certainly be a weak-* convergent subsequence because $L^\infty(0, \infty; H)$ is the dual of the separable Banach space $L^1(0, \infty; H)$ and so its bounded sets are weak-* metrizable and relatively compact.

You can't expect an a.e. convergent subsequence though; this is not even true for $L^\infty(0, \infty; \mathbb{R})$ (consider $f_n(x) = e^{inx}$).