In my work I stumbled upon a proposition (without proof, alas), which I can't really prove.
Suppose we have a family of functions $\left\{\phi_\epsilon (t,x,v)\right\}_{\epsilon\in(0,1]}$, and $M(v)$ - a good function (think $M(v) =e^{-|v|^2}$). $t>0$, $x,v\in \Bbb R^3$. Also, $\phi_\varepsilon\in C(\Bbb R_+; L^1_{loc}(\Bbb R^3\times \Bbb R^3))$.
We have a previous result: $\exists C>0$ such that $$\iint_{\Bbb R^3\times\Bbb R^3}(\phi_\epsilon(t,x,v))^2M(v)dv\,dx<C\quad\forall t>0,$$ i.e. this family is bounded in $L^\infty (\Bbb R_+;L^2(Mdv\,dx))$.
Basing on this, the authors of the article claim that another family of functions,$$\left\{2\phi_\epsilon(t,x,v)+\epsilon(\phi_\epsilon(t,x,v))^2\right\}_{\epsilon\in(0,1]}$$ is weakly relatively compact in $L^1_{loc}(dt\,dx; L^1(M dv))$.
Maybe the particular form of $\phi_\epsilon$ is useful here (the equation where it comes from), though I doubt it. I know that bounded sets in $L^p$ for $1<p<\infty$ are weakly precompact, but here I have $p=\infty$.
On a bright side, I have the continuous inclusion $L^2(Mdv)\subset L^1(Mdv)$, which allows to prove that the new family is included into $L^1_{loc}(dt\,dx; L^1(M dv))$. Banach–Alaoglu theorem and the fact that $(L^1)^\ast=L^\infty$ seems to prove this fact for weak-* topology. However, I don't see where the weak relative compactnees comes from.
Is there some comprehensible guide to dealing with such problems? Or maybe a theorem that I missed?
A simple sketch of the proof would be welcome, too.
edit
Upon further investigation I found a typo in the original formulation: apparently, they are talking about weak relative compactness (bolded it).
