I'm trying to understand the uniqueness result for renormalised solutions to the transport equation. A while back I tried to read the original source but could not get through it. Now I am trying to follow these notes.
I typed up a sketch of the bigger problem (First half of Prop 1.6) but I think its not so important for the point I'm stuck at. You can find it in an earlier version of this post.
The proof eventually leads to a Gronwall inequality, but only in the sense of distributions: $$ f' \le \|\nabla \cdot b\|_{L^\infty} f \quad \text{in } \mathcal D'$$ ($b$ is fixed with $b\in L^\infty, \nabla \cdot b\in L^\infty$.) By this I mean, for all test functions $\psi=\psi(t)\ge 0,$ $$ \int_0^\infty f\partial_t\psi + \|\nabla \cdot b\|_{L^\infty} f\psi \ d t \ge 0.$$
In addition, $f(0)=0$ is "obvious" formally in context. However, $f$ is in fact $$ f(t) =\int_{\mathbb R^n} w^2(t,x)\phi(t,x)\ dx $$ where $w\in L^\infty_{x,t}$ is a distributional solution to the transport equation $w_t + b\cdot\nabla w = 0$ with zero data, and $\phi\ge 0$ is a cutoff (satisfying certain bounds that allow us to reach the Grönwall.) I'm not sure how to rigourously get $f(0)=0$, or in what sense this should hold, because I don't know how $w$ attains the initial data. It is known that $w^2$ is also a distributional solution which might help (this is the renormalization property.)
Why is this enough to conclude that $f\equiv0$? Even the case $b=0$, i.e. if $f\ge 0$, $f'\le 0$ in $\mathcal D'$, and $f(0)=0$ in some appropriate sense. Can one conclude that $f = 0$? I do know how to solve e.g. $f'=0$ distributionally, and have checked the usual proofs of Grönwall, but was unable to conclude.
Thanks in advance for any pointers!