Question on vector valued distributions: weak time derivative calculation Although I read all the similar posts here, I still can't find the answer in my question so I 'll try to pose it as clear as I can.
DEFINITION $1$: Let $f\in L^1_{loc}(I, X)$. Then $\;\;\langle T_f, \phi \rangle=\int_{I} f\phi$ for $\phi \in D(I)$, defines a  $X-$valued distribution on $I$, i.e $T_f \in D'(I,X)$
DEFINITION $2$: Let $T_f \in D'(I, X)$. We define the distributional derivative of $T_f$, $T_f' \in D'(I,X)$, by  $\langle T'_f,\phi \rangle=-\langle T_f,\phi' \rangle$, for $\phi \in D(I)$
DEFINITION $3$: Let $u \in L^1(0,T:;X)$. We say that $v \in L^1(0,T:;X)$ is the weak derivative of $u$ written $u'=v\;$ provided $\int_{0}^T \phi'(t)u(t)\; dt=-\int_{0}^T \phi(t)v(t) \;dt$ for all scalar test functions $\phi \in C^{\infty}_c(0,T)$
So, it is clear to me that the weak time derivative of $u$ is in fact the distributional derivative of $u$. I am a bit familiar to distribution theory but not in the $X-$valued one. However an analog should exist between these two and this is what I 'm trying to understand here.

Suppose now that $u \in L^2(0,T;H^1_0(U))$ with $u_t \in
 L^2(0,T;H^{-1}(U))$. If $\phi \in C^{\infty}_c(0,T;H^1_0)$ then what
  can we say about 
$(*)\int_{0}^T \int_U u_t(t,x) \phi (t,x) \;dxdt=$?

After some studying on this topic, I found that most calculations use the inner product as well as the dual pairing. More specifically, definition $3$ is equivalent to: 
$(**)\int_{0}^T (u(t),\psi'(t))_{L^2}=-\int_{0}^T \langle u_t(t),\psi(t) \rangle_{H^{-1},H^1}$ for $\psi \in C^{\infty}_c(0,T;H^1_0)$
If I understood right, $(*)$ using $(**)$ can be written as: 
$\int_{0}^T \int_U u_t(t,x) \phi (t,x) \;dxdt=-\int_{0}^T \int_U u(t,x) \phi_t (t,x) \;dxdt \Leftrightarrow \\ \int_{0}^T \langle u_t(t,\cdot),\phi(t,\cdot) \rangle_{H^{-1},H^1}=-\int_{0}^T (u(t,\cdot),\phi'(t,\cdot))_{L^2}$ 
Is the above correct?
I feel that what I 'm asking is quite elementary or silly but I've been stuck to definitions and expressions while all I want is to understand the explicit calculation of the distributional derivative of $u$.
Any help is much appreciated. Thanks in advance!
 A: Is the above correct?
I'd say yes, provided that $u\in L^2(0,T;L^2(U))$ instead of $L^2(0,T;H^{-1}(U))$. To see this, we could use the integration by parts formula:

Theorem (Dautray, p. 473 and p. 477). Let $V$ and $H$ be Hilbert spaces such that 
  
  
*
  
*$V$ is continuously and densely embedded into $H$
  
*$H$ is identified with $H'$
  
*$H'$ is  continuously and densely embedded into $V'$
(A) If $u\in L^2(a,b;V)$ and $u_t\in L^2(a,b;V')$, then $u\in C([a,b]; H)$.  
(B) If $u,v\in L^2(a,b;V)$ and $u_t,v_t\in L^2(a,b;V')$, then 
  $$\int_a^b\langle u_t(t),v(t)\rangle_{V'\times V}\;dt=(u(b),v(b))_H-(u(a),v(a))_H-\int_a^b\langle u(t),v_t(t)\rangle_{V\times V'}\;dt\tag{1}$$
  which has a sense by (A).

Therefore:


*

*If $u,v\in L^2(0,T;H_0^1(U))$ and $u_t,v_t\in L^2(0,T;H^{-1}(U))$, then 
\begin{align*}
\int_0^T\langle u_t(t),v(t)\rangle_{H^{-1}\times H_0^1}\;dt=(u(T),v(T))_{L^2}-(u(0),v(0))_{L^2}\\
-\int_0^T\langle u(t),v_t(t)\rangle_{H_0^1\times H^{-1}}\;dt\tag{2}
\end{align*}

*If we have the additional regularity $v\in C_c^\infty(0,T;H_0^1(U))\subset L^2(0,T;H_0^1(U))$, then 
$v_t\in C_c^\infty(0,T;H_0^1(U))\subset C_c^\infty(0,T;H^{-1}(U))\subset L^2(0,T;H^{-1}(U))$. In this case, the last equality becomes
$$\int_0^T\langle u_t(t),v(t)\rangle_{H^{-1}\times H_0^1}\;dt=-\int_0^T( u(t),v_t(t))_{L^2}\;dt.\tag{3}$$

*If we have the additional regularity $u_t\in L^2(0,T;L^2(U)) \subset L^2(0,T;H^{-1}(U))$, then the last equality becomes
$$\int_0^T(u_t(t),v(t))_{L^2}\;dt=-\int_0^T( u(t),v_t(t))_{L^2}\;dt,\tag{4}$$
which can be rewritten as
$$\int_0^T\int_U u_t(t,x)v(t,x)\;dx\;dt=-\int_0^T\int_U u(t,x)v_t(t,x)\;dx\;dt.\tag{5}$$
Thus $(5)$ is the same as $(4)$, which is a particular case of $(3)$, which is a particular case of $(2)$, which is a particular case of $(1)$.
