Let $V$ be a separable Hilbert space and define $X=L^2(0,T;V)$. Then $u\_m\to u$ weakly in $X$ means



> 
> for every $v\in X'=L^2(0,T;V')$
>  $$
> \int\_0^T\langle v(t),u\_m(t)\rangle\ dt\to\int\_0^T\langle v(t),u(t)\rangle\ dt\tag{1}
> $$
>  where $\langle\rangle$ is the dual pair $(V',V)$.
> 
> 
> 


If one assumes that



> 
> for every $v\in V'$
>  $$
> \int\_0^T\langle v,u\_m(t)\rangle\ dt\to\int\_0^T\langle v,u(t)\rangle\ dt\tag{2}
> $$
> 
> 
> 


then it is not generally true that $u\_m\to u$ weakly. The case $V=\mathbb{R}$ gives a counterexample. 


Would (2) together with $\{u\_m\}$ being bounded in $X$ be enough to imply (1)? (I don't have a way to test it.)


[Added: Are there any "handy" extra assumptions that would allow (2) imply (1)?]



Weak convergence in vector-valued Hilbert space
A subset $A\subset X'$ is called *total* if it is not contained in any proper closed subspace.
Note that $A$ is total in $X'$ iff its anihilator in $X$ is trivial.


Your space $X$ is a separable Hilbert space, thus closed bounded sets are compact and metrizable for the weak topology.
In particular, every bounded sequence has a weakly converging subsequence.
The following could be therefore proven easily by a sub-sub-sequence argument:


Assume $\{u\_m\}$ is bounded in $X$ and $A$ total in $X'$. Then $u\_m\to u$ weakly is equivalent to


$(\*)$ $\langle u\_m,v\rangle \to \langle u,v\rangle$ for every $v\in A$.


Note that if $A$ is not total in $X'$ then the condition $(\*)$ does not guarantee weak convergence (for example, take $u\_m=0$ for all $m$ and $u$ a nontrivial vector in the annihilator of $A$).


Now, back to our specific question. The set of constant functions is a proper closed subspace of $X'$, hence not total. Therefore the answer is "no".
What you are really seem to be asking is: what are good examples of total subsets of $X'$ that I can use? 
Given a total subset $B\subset V$, each of the sets
$$ \{ \chi\_{[a,b]}\cdot v\mid 0<a<b<1,~ v\in B\} $$
and 
$$ \{ f\cdot v\mid f\in C([0,1)),~v\in B\} $$ is total in $X'$ and might be suitable for your purposes. The proof is an exercise.


