When can I find a subsequence that converge weakly in $H(0,T, V)$ and pointwise a.e in $V$ for $t=0$ $?$ The following problem is based on finding solution of PDE's using Galerkin approximation method.
Let $\{u_{m}\}$ be a bounded sequence of functions in $H^{1}(0,T;L^{2})$ i.e.,
$$||u_{m}||_{H^{1}(0,T;L^{2})}\le M $$
that satisfies:
$$u_{m}(0,\cdot)=\sum^{m}_{k=1}(g,w_{k})w_{k}$$
$$u_{m}'(0,\cdot)=\sum^{m}_{k=1}(h,w_{k})w_{k}$$
where $g\in H_{o}^{1},h\in L^{2}$ and $\{w_{k}\}^{\infty}_{k=1}$ is an orthonormal basis of $L^{2}$.
Is there a subsequence $\{u_{m_{a}}\}^{\infty}_{a=1}$ such that $u_{m_{a}}$
converge weakly to an element $u\in H^{1}(0,T;L^{2})$ and $u_{m_{a}}(0,\cdot)$ converge a.e. pointwise to $g$ and $u_{m_{a}}'(0,\cdot)$ converge a.e. pointwise to $h$ ?
The idea I was trying is to choose a subsequence such that $u_{m_{j}}$
converge weakly to an element $u\in H^{1}(0,T;L^{2})$. This is possible because of Banach-Alaoglu Theorem.
Now we know that the sequence converge strongly to $g$ and $h$ in $L^{2}$ that is 
$$u_{m_{j}}(0,\cdot)\rightarrow g $$ in $L^{2}$.
$$u_{m_{j}}'(0,\cdot)\rightarrow h $$ in $L^{2}$.  
Does this fact follows from Parseval's identity?
Now since $$u_{m_{j}}(0,\cdot)\rightarrow g $$ we can choose a subsequence $\{u_{m_{j_{l}}}(0.\cdot)\}$ that converges a.e. to $g$.
Then choosing a subsequence of this last subsequence we can choose $\{u_{m_{j_{l_{r}}}}\}$ such that $\{u'_{m_{j_{l_{r}}}}(0.\cdot)\}$ that converges a.e. to $h$.
Does choosing this last sequence give the desire properties? 
 A: Is there a subsequence $\{u_{m_{a}}\}^{\infty}_{a=1}$ such that $u_{m_{a}}$
converge weakly to an element $u\in H^{1}(0,T;L^{2})$ and $u_{m_{a}}(0,\cdot)$ converge a.e. pointwise to $g$ and $u_{m_{a}}'(0,\cdot)$ converge a.e. pointwise to $h$?
Yes, and your approach looks good for me.
Does this fact follows from Parseval's identity?
If you define basis as a maximal set of linearly independent vectors, then you can say that this fact follows from the fact that $\{w_{k}\}^{\infty}_{k=1}$ is a basis of $L^{2}$. The proof (and the precise relation between this fact and the Persavel identity) can be found in the reference below.

Corollary 12.8 (A. W. Knapp): If $S$ is an orthonormal set in the Hilbert space $H$, then the following are equivalent:
(a) $S$ is maximal among orthonormal subsets of $H$.
(b) $\displaystyle u = \sum_{v_\alpha\in S}(u,v_\alpha)v_\alpha$ for all $u$ in $H$.
(c) $\displaystyle \|u\|^2 = \sum_{v_\alpha\in S}|(u,v_\alpha)|^2$ for all $u$ in $H$.
(d) $\displaystyle (u,v) = \sum_{v_\alpha\in S}(u,v_\alpha)\overline{(v,v_\alpha)}$ for all $u$ and $v$ in $H$.

Does choosing this last sequence give the desire properties?
Yes because you can define $\{u_{m_a}\}$ as follows:
$$u_{m_1}=u_{m_{j_{l_1}}},\quad u_{m_2}=u_{m_{j_{l_2}}},\quad u_{m_3}=u_{m_{j_{l_3}}} \cdots\quad  u_{m_a}=u_{m_{j_{l_a}}}\quad\cdots$$
