How to show $ \int_0^T\int_\Omega \frac{\partial }{\partial t}( \varphi^2 ) \,dx \,dt = \int_0^T \frac{d}{dt} (\int_\Omega \varphi^2 \, dx) \,dt $? $\Omega\subset \Bbb{R}^n$ is a bounded smooth domain.
$Q_T=\Omega\times [0,T]$.
$\mathring C^\infty(Q_T)=\{u\in C^\infty(Q_T) : u \text{ is zero near } \partial\Omega\times[0,T]\}$.
$\mathring W^{1,1}_2(Q_T)$ is closure of $\mathring C^\infty(Q_T)$ in $W^{1,1}_2(Q_T)$.  $W^{1,1}_2(Q_T)$ is Sobolov space, $u\in W^{1,1}_2(Q_T)$ means $u_t$ and $u_x$ belong to $L^2(Q_T)$.
For any $\varphi\in \mathring W^{1,1}_2(Q_T)$, how to show
$$
\int_0^T\int_\Omega \frac{\partial }{\partial t}( \varphi^2 ) \, dx \, dt = \int_0^T \frac{d}{dt} \left(\int_\Omega \varphi^2 \,dx\right)\, dt  
$$
 A: I will outline the proof but will let you to fill in the details for by the level of the proposition I presume you are able to.
Writing $\psi := \varphi^2$ to simplify notation, the first thing to note is that the outer integrals are equal
$$\int_0^T \left( \int_\Omega \frac{\partial \psi}{\partial t}\ dx \right) dt = \int_0^T \frac{d}{dt} \left( \int_\Omega \psi\ dx \right)\ dt$$
if and only if the inner integrals are equal almost everywhere with respect to Lebesgue measure
$$\int_\Omega \frac{\partial \psi}{\partial t}\ dx = \frac{d}{dt} \int_\Omega \psi\ dx\ \text{ a.e.}$$
That is, you need to show that the "integral with respect to $x$ and the derivative with respect to $t$ commutes". This is done through the Dominated Convergece Theorem on the usual Lebesgue measure space applied to the sequence of functions
$$f_n(x, t) := \frac{\psi\left(x, t + \frac{1}{n}\right) - \psi(x, t)}{\frac{1}{n}}$$
For this you should argue that this is indeed a sequence of Lebesgue measurable functions in $x$ and also
$$\lim_{n \to \infty} f_n = \frac{\partial \psi}{\partial t}$$
Next you argue that since $\psi$ vanishes when close enough to the boundary then also $\frac{\partial \psi}{\partial t}$ does, and since $\Omega$ is bounded and connected then $\left| \frac{\partial \psi}{\partial t} \right|$ attains its maximum.
Then by the Mean Value Theorem with respect to $t$ you should argue that fixed $t_0 \in [0, T]$, for all $n \in \mathbb{N}$ and $x \in \Omega$ it holds that
$$|f_n(x, t_0)| \leq \max_{\omega \in \Omega} \left| \frac{\partial \psi}{\partial t}(\omega, t_0) \right|$$
which is constant and therefore Lebesgue-integrable with respect to $x$ on a bounded set. Thus concluding the proof.
