Suppose we have $u,v$ satisfying a ‘transport’ equation on $[0,1] \times [0,T]$; $$\partial_{t}u +v\partial_{x}u =0.$$ We assume further that $v \in L^{2}(0,T; L^{2})$, $\partial_{x}u \in L^{\infty}(0,T; L^{1})$ and also that $$ \int_{0}^{T}\int_{0}^{1} v \partial_{x}u \cdot f~dxdt \le C\label{A} \tag{A} $$ for any $f \in L^{1}(0,1)$.
Goal: show that $\|\partial_{t}u\|_{L^{1}_{t}W^{-1,2}_{x}} \le C$.
My attempt: Denote by $\langle \cdot, \cdot \rangle_{\ast}$ the action of a functional in $X^{\ast}=W^{-1,2}$ on an element of $X = W^{1, 2}_{0}$. We need to show that $$ \int_{0}^{T} \sup_{f : \|f\|_{X|=1}} |\langle \partial_{t}u, f \rangle_{\ast}| ~dt \le C. $$
Then taking $f \in X$ with $\|f\|_{X}=1$, $$ \begin{aligned} \langle \partial_{t}u, f \rangle_{\ast}(t) &= - (v \partial_{x}u,f)(t) \equiv - (v \partial_{x}u,f)_{L^{2}(0,1)}(t) \\ &= -\int_{0}^{1} v\partial_{x}u \cdot f~dx. \end{aligned} $$
Then it follows from (A) that $$ \int_{0}^{T} |\langle \partial_{t}u,f\rangle_{\ast}|~dt \le C. $$ Since $f$ was arbitrary we are done.
Is my attempt correct? I am just a bit unsure on whether I can treat the duality bracket as $L^{2}$ ( I think I can because of the isomorphism between $H^{1}_{0}$ and $H^{-1}$). And also just a bit curious if I am calculating this dual norm correctly..
I realise now that my attempt is not correct. Repeating the last computation:
$$\int_{0}^{T} |\langle \partial_{t}u, f \rangle_{\ast}|~dt = \int_{0}^{T} \left|\int_{0}^{1}v\partial_{x} u\cdot f~dx \right|~dt $$ which we can NOT bound because we only know that
$$ \left| \int_{0}^{T}\int_{0}^{1}v\partial_{x} u \cdot f~dx dt \right| \le C.$$ We need the modulus inside the time integral to actually conclude the argument.