Weak convergence in $L^{2}(0,T;H^{-1}(\Omega))$ What does weak convergence in $L^{2}(0,T;H^{-1}(\Omega))$ means?
$\Omega$ is open, bounded, has boundary smooth  and etc...
 A: Notation: For an arbitrary linear functional $g:V\to\mathbb{R}$, $\langle g, v\rangle$ is the value of $g$ at $v\in V$.

The meaning of
  $$u_n\rightharpoonup u\quad\text{in}\quad L^2(0,T;H^{-1}(\Omega))\tag{A}$$
  is
  $$\int_0^T\langle u_n(s),w(s)\rangle\ ds\to\int_0^T\langle u(s),w(s)\rangle\ ds\quad\text{in}\quad \mathbb{R}, \quad \forall\ w\in L^2(0,T;H_0^1(\Omega))\tag{B}$$
  that implies (by taking $w(s)=\phi(s)v$)
  $$\int_0^T\langle u_n(s),\phi(s)v\rangle\ ds\to\int_0^T\langle u(s),\phi(s)v\rangle\ ds\quad\text{in}\quad \mathbb{R}, \quad \forall\ (\phi,v)\in C_0^\infty(0,T)\times H_0^1(\Omega).$$
  The weak convegence $(A)$ also implies
  $$\int_0^Tu_n(s)v(s)\ ds\rightharpoonup\int_0^Tu(s)v(s)\ ds\quad\text{in}\quad H^{-1}(\Omega),\quad\forall\ v\in C_0^\infty(0,T)\tag{C}.$$

Let me (try to) explain it. Initially, consider the more general case
$$u_n\rightharpoonup u\quad\text{in}\quad L^p(0,T;X)\tag{1}.$$
According to the usual definition, $(1)$ means
$$\langle \Lambda,u_n\rangle\to \langle\Lambda,u\rangle\quad\text{in}\quad \mathbb{R},\quad \forall\ \Lambda\in \Big(L^p(0,T;X)\Big)'.\tag{2}$$
Theorems 23.28 and 23.29 in Kuttler's book shows that, for $1/p+1/q=1$, the mapping
\begin{align}
\theta:L^q(0,T;X')&\longrightarrow  \Big(L^p(0,T;X)\Big)'\\
g&\longmapsto\theta_g
\end{align}
is a bijective linear isometry, where $\theta_g$ is the mapping
\begin{align}
\theta_g:L^p(0,T;X)&\longrightarrow\mathbb{R}\\
f&\longmapsto\int_0^T\langle g(s),f(s)\rangle\ ds.
\end{align}
So, given $\Lambda\in \Big(L^p(0,T;X)\Big)'$, there exists an unique $h_\Lambda\in L^q(0,T,X')$ such that
$$\langle \Lambda,f\rangle=\int_0^T\langle h_\Lambda(s),f(s)\rangle\ ds,\quad\forall\ f\in L^p(0,T;X).\tag{3}$$
From $(2)$ and $(3)$, it follows that
$$\int_0^T\langle h_\Lambda(s),u_n(s)\rangle\ ds\to\int_0^T\langle h_\Lambda(s),u(s)\rangle\ ds\quad\text{in}\quad \mathbb{R}, \quad \forall\ \Lambda\in \Big(L^p(0,T;X)\Big)'$$
Notice that, if $\Lambda$ runs through all $\Big(L^p(0,T;X)\Big)'$ then $h_\Lambda$ runs through all $L^q(0,T;X')$. So, $(1)$ means

$$\int_0^T\langle h(s),u_n(s)\rangle\ ds\to\int_0^T\langle h(s),u(s)\rangle\ ds\quad\text{in}\quad \mathbb{R}, \quad \forall\ h\in L^q(0,T;X').$$

As a particular case, we conclude that the meaning of $(A)$ is
$$\int_0^T\langle h(s),u_n(s)\rangle\ ds\to\int_0^T\langle h(s),u(s)\rangle\ ds\quad\text{in}\quad \mathbb{R}, \quad \forall\ h\in L^2(0,T;(H^{-1}(\Omega))').\tag{4}$$
Since $H_0^1(\Omega)$ is reflexive (that is, the canonical mapping from $H_0^1(\Omega)$ to $(H^{-1}(\Omega))'$ is a bijective linear isometry), we conclude that:


*

*Given $h\in L^2(0,T;(H^{-1}(\Omega))')$, there exists $w\in L^2(0,T;H_0^{1}(\Omega))$ such that
$$\langle h(s),v\rangle=\langle v,w(s)\rangle,\quad \forall\ (s,v)\in (0,T)\times H^{-1}(\Omega);$$

*Given $w\in L^2(0,T;H_0^{1}(\Omega))$, there exists $h\in L^2(0,T;(H^{-1}(\Omega))')$ such that
$$\langle h(s),v\rangle=\langle v,w(s)\rangle,\quad \forall\ (s,v)\in (0,T)\times H^{-1}(\Omega).$$


So, $(B)$ is obtained from $(4)$.
Item $(C)$ is also obtained from $(4)$ as explained here.
A: We have that $u_n \rightharpoonup u$ in $L^2(0,T;H^{-1}(\Omega))$ if and only if
$$\int_0^T \langle u_n(t), w(t) \rangle_{H^{-1}(\Omega), H^1_0(\Omega)} \to \int_0^T \langle u(t), w(t) \rangle_{H^{-1}(\Omega), H^1_0(\Omega)}$$
for all $w \in L^2(0,T;H^1_0(\Omega))$. 
Notice that the dual space of $L^2(0,T;H^1_0(\Omega))$ is $L^2(0,T;H^{-1}(\Omega)).$
