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 ${H\_0^1}\left(0,1 \right)$. 



Existence of subsequences convergence with weak topology
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 $f\_n(x) = e^{inx}$).


