How to pass from $L^2(0,T;V')$ to $\mathcal{D}'\big(\Omega\times (0,T)\big)$?

Let $\mathcal{V}=\{v\in\mathcal{D}(\Omega)\times\mathcal{D}(\Omega)\mid\operatorname{div} v=0\}$, where $\Omega$ is a bounded open subset of $\mathbb{R}^2$ with smooth boundary. Set $V={\overline{\mathcal{V}}}^{H^1(\Omega)\times H^1(\Omega)}$. It is possible to show that
$$V=\{v\in H_0^1(\Omega)\times H_0^1(\Omega)\mid\operatorname{div} v=0\}.$$
Assume that $f\in L^2(0,T;V’)$ and define
$$F(t)=\int_0^tf(s)\;ds.\tag{1}$$

Question: Why $F$ belongs to $\mathcal{D}'(\Omega\times (0,T))$? And why the partial distributional derivative $\frac{\partial F}{\partial t}$ equals $f$ in $\mathcal{D}'(\Omega\times (0,T))$?

Motivation: Something like this is used in Temam’s book (p. 267) because he differentiate $F$ “in the distribution sense in $\Omega\times (0,\ell)$”, but I didn’t get it.

Solutions Collecting From Web of "How to pass from $L^2(0,T;V')$ to $\mathcal{D}'\big(\Omega\times (0,T)\big)$?"

Let us write $Q=\Omega\times(0,T)$.

Consider a function $v\in C(0,T;H^{-1}(\Omega))$ whose distributional derivative $v’$ in the sense of $\mathcal{D}'(0,T; H^{-1}(\Omega)$ belongs to $L^2(0,T; H^{-1}(\Omega))$.

The function $v$ can be seen as an element of $\mathcal{D}'(Q)$, defined by

\langle v,\varphi\rangle_{\mathcal{D}'(Q)}=\int_0^T\langle v(t),\varphi(\cdot,t)\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt,\quad\forall\ \varphi\in\mathcal{D}(Q).

Let $\partial_t v$ be the distributional derivative of $v$ with respect to $t$ in the sense of $\mathcal{D}'(Q)$. For any $\phi\in \mathcal{D}(\Omega)$ and $\theta\in \mathcal{D}(0,T)$,
\begin{align*}
\langle \partial_tv,\phi\theta\rangle_{\mathcal{D}'(Q)}
&=-\langle v,\phi\theta’\rangle_{\mathcal{D}'(Q)}\\
&=-\int_0^T\langle v(t), \phi\theta'(t)\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt=-\int_0^T\langle v(t)\theta'(t), \phi\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt\\
&=-\left\langle\int_0^T v(t)\theta'(t)\;dt, \phi\right\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}
=-\left\langle\langle v,\theta’\rangle_{\mathcal{D}'(0,T;H^{-1}(\Omega))} , \phi\right\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\\
&=\left\langle\langle v’,\theta\rangle_{\mathcal{D}'(0,T;H^{-1}(\Omega))} , \phi\right\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}
=\left\langle\int_0^T v'(t)\theta(t)\;dt, \phi\right\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\\
&=\int_0^T \langle v'(t)\theta(t), \phi\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt
=\int_0^T \langle v'(t), \phi\theta(t)\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt
\end{align*}
and thus
$$\langle \partial_tv,\varphi\rangle_{\mathcal{D}'(Q)}=\int_0^T \langle v'(t), \varphi(\cdot,t)\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt,\quad\forall\varphi\in\mathcal{D}(Q).$$
So, $v’$ can also be seen as an element of $\mathcal{D}'(Q)$, defined by
$$\langle v’,\varphi\rangle_{\mathcal{D}'(Q)}=\int_0^T\langle v'(t),\varphi(\cdot,t)\rangle_{H^{-1}(\Omega)\times H_0^1(\Omega)}\;dt,\quad\forall\ \varphi\in\mathcal{D}(Q)$$
and satisfying the equality $v’=\partial_t v$ em $\mathcal{D}'(Q)$.

Since (by extension) each element of $V’$ can be seen as an element of $[H^{-1}(\Omega)]^2$ (with the same norm), we conclude that $F\in C^([0,T];H^{-1}(\Omega))$ and thus $F,f\in [\mathcal{D}(Q)]^2$ according to the above identifcation. Moreover, if we write $f=(f_1,f_2)$, $F=(F_1,F_2)$ and $\partial_tF=(\partial_tF_1,\partial_tF_2)$ then
$$\partial_t F=(\partial_t F_1,\partial_t F_2)=(F_1′,F’_2)=(f_1,f_2)=f\quad\text{in}\quad[\mathcal{D}'(Q)]^2.$$

Probably, this is the meaning of “$\partial_tF=f$ in the distribution sense in $Q$” in the said book, and we are passing from $L^2(0,T;V’)$ to $[\mathcal{D}'(\Omega)]^2$.