$u_m\rightharpoonup u$ in $L^2(0,T;H)$ and $u'_m\rightharpoonup v$ in $L^2(0,T,H^*) \longrightarrow u'=v$ Asuume $H$ a Hilbert space, $u_m\rightharpoonup u$ in $L^2(0,T;H)$, and $u'_m\rightharpoonup v$ in $L^2(0,T, H^*)$. Does this imply $v=u'$? This is something I've been wrestling with and some online searching kind of gives a hand waving argument that it's true for $H_0^1(U)$. So maybe looking that this space would be a starting point? I'm just looking for a little bit of guidance on how to show something like this. 
 A: You must prove that
$$\int_0^T u(t)\varphi'(t)dt=-\int_0^Tv(t)\varphi(t) dt,\ \forall \varphi \in \mathcal{D}(0,T).$$
Note that these integrals are elements of $H$, since $u(t)\in H$ and $\varphi(t)\in \mathbb{R}$.
We know that $u_m\rightharpoonup u$ in $L^2(0,T;H)$ means 
$$\int_0^T (u_m(t),w(t))dt\to \int_0^T (u(t),w(t))dt\ \forall w\in L^2(0,T;H),$$
$(\cdot,\cdot)$ is the inner product of $H$. In this case, if $h\in H$ and $\varphi \in \mathcal{D}(0,T)$ we have $w=h\varphi'\in L^2(0,T;H)$, thus
$$
 \int_0^T (u_m(t)\varphi'(t),h )dt=\int_0^T (u_m(t),h\varphi'(t))dt\to \int_0^T (u(t),h\varphi'(t))dt=\int_0^T (u(t)\varphi'(t),h)dt\ \forall h\in H. \tag{1}\label{1}
$$
Since $H$ is a Hilbert space, from Riez representation theorem, we can identify $H\simeq H^\ast$ and so we have that $L^2(0,T;H^\ast)\simeq L^2(0,T;H)$. In this case, we have $u_m'\rightharpoonup v$  in $L^2(0,T;H)$( I'm not sure about this step).
As before, we can see that
$$ \int_0^T (u_m'(t)\varphi(t),h )dt=\int_0^T (u_m'(t),h\varphi(t))dt\to \int_0^T (v(t),h\varphi(t))dt=\int_0^T (v(t)\varphi(t),h)dt\ \forall h\in H.\tag{2}\label{2}
$$
On the other hand, we know that
$$\int_0^T u_m'(t)\varphi(t)dt=-\int_0^T u_m(t)\varphi'(t)dt,\ \forall \varphi\in \mathcal{D}(0,T).$$
Which implies that
$$ \int_0^T (u_m'(t)\varphi(t),h )dt=\left(\int_0^T u_m'(t)\varphi(t)dt,h\right)= \left(-\int_0^T u_m(t)\varphi'(t)dt,h\right)=-\int_0^T (u_m(t)\varphi'(t),h )dt,$$
$\forall h\in H.$
Using (\ref{1}) and \eqref{2} to passing to the limit in the last identity, we obtain 
$$\int_0^T (v(t)\varphi(t),h )dt=-\int_0^T (u(t)\varphi'(t),h )dt,\ \forall h\in H.$$
Since $h\in H$ is arbitrary, we can conclude that
$$\int_0^T v(t)\varphi(t)dt=-\int_0^T u(t)\varphi'(t)dt,\ \forall \varphi\in \mathcal{D}(0,T),$$
which concludes the proof.
A: I am not well familiar with the space $L^2(0,T;H)$ but for $H^1(\Omega)$ if $u_m\rightharpoonup u$ in $L^2(\Omega)$ and  $u_m'\rightharpoonup v$ in $L^2(\Omega)$ then you have: For every $f\in H^1(\Omega)$ $(f,u_m)_1=(f,u_m)+(f',u_m')\to (f,u)+(f',v)$. This means that for each functional in $f^*\in H^1(\Omega)$ we have that $\langle f^*,u_m\rangle $ converges ( we used Riesz representation theorem ). But then there is some element $\bar u\in H^1(\Omega)$ such that $u_m\rightharpoonup \bar u$ in $H^1(\Omega)$, which means that $u_m\rightharpoonup \overline u$ in $L^2(\Omega)$ and $u_m'\rightharpoonup \bar u'$ in $L^2(\Omega)$. As the weak limits are unique, it follows that $\bar u\equiv u$ and $\bar u'\equiv v$, i.e $u'=v$.
