$L^\infty_\mathrm{loc}$ assumption in global existence for Boltzmann equation

Question

In short:

In P. Gérard's paper on the existence of global solutions to the Boltzmann equation from 1988 (or equivalently Cercignani's book), why are the stated assumptions (especially $A_n \in L^\infty_{loc}$) sufficient for proving velocity averaged compactness of the linearized collision kernels in Proposition 8 iii) ?

In particular, why is $\psi_n = \frac{A_n \ast f_n}{1 + \int f_n dv} \phi$ uniformly bounded in $L^\infty$ (here, $\phi \in L^\infty$ with compact support)?

Background information:

In addition to the famous paper by Lions and DiPerna "On the Cauchy Problem for Boltzmann Equations: Global Existence and Weak Stability" (Annals, 1989) there is the paper by P. Gérard "Solutions globales du problème de Cauchy pour l'équation de Boltzmann" (Séminaire Bourbaki, 1987-1988) which shows existence under the hypothesis that the kernel satisfies an $L^\infty_\mathrm{loc}$ assumption. This version of the proof of existence is also formulated in the book "The Mathematical Theory of Dilute Gases" (Cercignani, Illner, Pulvirenti, 1994).

The original DiPerna-Lions proof starts with the assumption that $A_n \in L^\infty$ then gets rid of this assumption completely in a second step. The Gérard approach tries to modify the first step in that it does only assume $A_n \in L^\infty_\mathrm{loc}$. This is already appropriate for most collision kernels, hence it's reasonable to stick to this assumption for the sake of a less technical proof (and skip the second part of DiPerna-Lions).

Incomprehensibly, it seems to me like the crucial step where DiPerna and Lions use $A_n \in L^\infty$ is not changed at all in the $L^\infty_\mathrm{loc}$ case (it's the step described above)! There is no comment from Cercignani or Gérard why the same argument should still work even without $L^\infty$.

Answer 1

In a way, this is a partial answer to this question. Indeed, I found a counterexample that shows that the above function $\psi_n$ is in general not bounded, if we choose $f_n$ and $A_n$ such that the required uniform bounds hold (ignoring the fact, that the $f_n$ are indeed solutions to truncated equations):

Let $A_n(x) = \sum_{k=0}^n 2^{3k} (\chi_k(x) + \chi_k(-x))$, where $\chi_k$ denotes the characteristic function of the interval $[2^k,2^k + 2^{-3k}]$, and suppose $f_n(x) = f(x) = \sum_k (\chi_k(x) + \chi_k(-x))$.

Then the $A_n$ are uniformly locally bounded and satisfy the uniform growth condition \begin{align} \sup_{n} \frac{1}{1+|\xi|^2} \int_{\xi_\ast \leq R} A_n(\xi - \xi_\ast) d\xi_\ast \to 0 \text{ as } \xi \to \infty. \end{align} Clearly, the second moments of $f$ are bounded. But $\int A(\xi) f(\xi) d\xi = (A \ast f)(0)$ is divergent.

The idea: On the one hand you know that the local $L^1$-norm of $A$ grows subquadratically at infinity. On the other hand the local $L^\infty$-norm of $A$ can still grow arbitrarily fast - you only need, that the sets, where $A$ is large, become small fast enough.

There doesn't seem to be any hope that the $\psi_n$ from my question are really bounded. Instead, I suspect, that this is rather some kind of gap in the proof and we need a completely different argument...

Answer 2

I thought if we replace $1+\int f^n \, d\xi$ by $1+A_n * f_n$, then the rest of the argument (passing to the limit etc.) will be just fine. If we go this way, we don't need to assume $A\in L^{\infty}(\mathbb{R}^d)$. It becomes also not necessary that $\varphi \in L^{\infty}((0,T)\times \mathbb{R}^d \times \mathbb{R}^d)$ is compactly supported.

cf. See (62) of the original paper by DiPerna and Lions