Consider $V=H^1(\Omega)$ and a bounded sequence $f_n$ in $B = L^2((0,1),V)$, which is Hilbert with $(f,g)_B = \int_0^1(f,g)+(\nabla f, \nabla g)$.
Then, there exists $f \in B$ with, for all $g \in B$, the convergence $\int_0^1(f_n,g)+(\nabla f_n, \nabla g)\rightarrow \int_0^1(f,g)+(\nabla f, \nabla g)$.
Can we conclude that $\int_0^1(f_n,g)\rightarrow \int_0^1(f,g)$ and $\int_0^1 (\partial_i f_n, g)\rightarrow \int_0^1(\partial_i f, g)$, for all $g \in L^2(I,H)$?
This seems a very natural property for me to hold, but I couldn't get to prove it.
One could I guess try the weak convergence $(f_n,g)+(\nabla f_n, \nabla g)\rightarrow (f,g)+(\nabla f, \nabla g)$ for almost every time, but I think this is not attainable. On the other hand, a density argument where $p$ is substituted by functions with compact supports in space, doesn't seem to lead anywhere, as one cannot recover $B$ by such functions... Any idea here?
I guess that $H = L^2(\Omega)$ and $I = (0,1)$.
For every $g \in L^2(I;H)$, the mappings \begin{align*} V \in f &\mapsto \int_I (f,g)_H \, \mathrm{d}t,\\ V \in f &\mapsto \int_I (\partial f_i,g)_H \, \mathrm{d}t \end{align*} are linear and continuous from $V$ to $\mathbb R$. Since $f_n$ converges weakly in $V$ towards $f$, your desired convergences follow.