Let $\Omega \subset \mathbb{R}^n$ a open and bounded set and $T>0$ fixed. Consider two functions $$u,v \in L^{\infty}(0,T; H_0^1(\Omega)).$$ Thus, in particular, for each $t \in [0,T]$, we have $u(t),v(t) \in H_0^1(\Omega)$ and we denote $u'=u_t$ and $v'=v_t$. Consider also $H_0^1(\Omega)$ with the inner product $(f,g)_{H_0^1}=(\nabla f, \nabla g)_{L^2}$, where $(\cdot, \cdot)_{L^2}$ denote the usual inner product of the $L^2(\Omega)$.
Question. Is true that $$\big(\nabla u(t), \nabla v'(t)\big)_{L^2}=\big(u(t), v(t)\big)_{H_0^1}?$$
I believe it is not true because it does not have control over $ v '(t)$.
No, for example $u(t,x)=v(t,x)=t\varphi(x)$ with $0 \neq \varphi \in C_c^\infty(\Omega)$ where $\Omega$ is a bounded domain in $\mathbb{R}^{n}$. Then, $v'(t,x)=\varphi(x)$ and $$(\nabla u(t),\nabla v'(t))_{L^2(\Omega)} = t\int_{\Omega}|\nabla \varphi(x)|^2 \, dx \neq t^2\int_\Omega | \nabla \varphi(x)|^2 \, dx=(u(t),v(t))_{H_0^1(\Omega)}.$$