# On DiPerna-Lions compactness' arguments.

In DiPerna Lions, Ordinary differential equations, transport theory and Sobolev spaces (1989), the authors used topological arguments that remains obscure to me. Page 515 the authors used an approximation $$u^\varepsilon$$ of the transport equation, which for our purpose have no importance. I separate my doubt in $$3$$ points:

1. The authors claim (middle of p. 515) that for $$u^\varepsilon$$ uniformly bounded in $$L^\infty(0,T;L^p(\mathbb R^n))$$ $$(1, we can extract a subsequence which converges weakly in $$L^\infty(0,T;L^p(\mathbb R^n))$$. Up to this point I understand the argument if by weak convergence the authors mean weak-* convergence on the dual of $$L^1(0,T;L^q(\mathbb R^n))$$ with $$q$$ the conjugate of $$p$$, being exactly $$L^\infty(0,T;L^p(\mathbb R^n))$$ (Diestel and Uhl, Vector measures, IV.1 Theorem 1).

2. The most delicate point to me is now. They claim if $$u^\varepsilon$$ is uniformly bounded in $$L^\infty(0,T;L^\infty(\mathbb R^n))$$, we can extract a subsequence which converges weak-* to some $$u$$ that is according to the paper in $$L^\infty(0,T;L^\infty(\mathbb R^n))$$. In such a case we have that $$L^\infty(0,T;L^\infty(\mathbb R^n))$$ is isometric to a proper subspace of $$L^1(0,T;L^1(\mathbb R^n))'$$ thus I can presume that $$u^\varepsilon$$ converges (up to a subsequence) weak-* to a continuous linear form $$\psi$$ on $$L^1(0,T;L^1(\mathbb R^n))$$ but not an element on $$L^\infty(0,T;L^\infty(\mathbb R^n))$$ except if someone show me a bounded subset in $$L^\infty(0,T;L^\infty(\mathbb R^n))$$ is weakly* sequentially close. On the other side we also have that $$L^\infty(0,T;L^\infty(\mathbb R^n))$$ is a proper subspace of $$L^\infty((0,T)\times\mathbb R^n)$$ (matter of measurability) thus I can presume that $$u^\varepsilon$$ may converge (up to a sbsequence) weak-* to $$u$$ belonging to $$L^\infty((0,T)\times\mathbb R^n)$$. We can also deduce that $$\Psi = u$$ in distribution. But up to this moment I cannot conclude that I have a representent in $$L^\infty(0,T;L^\infty(\mathbb R^n))$$. Can somebody help me to explain this assertion?

3. I have similar question on the case $$p=1$$, but I leave this for later.

• Ad 1: I would also think that the author means weak-* convergence. Ad 2: Without knowing the book, I would guess that (a) the author is not aware of the difference between $L^\infty(0,T;L^\infty(\mathbb R^n))$ and $L^\infty((0,T)\times \mathbb R^n)$ or (b) he uses a different definition of $L^\infty(0,T;L^\infty(\mathbb R^n))$ (in which measurability is replaced by weak-* measurability) [in this case, the space coincides with $L^\infty((0,T)\times\mathbb R^n)$]. Is a weak-* convergent subsequence in $L^\infty(L^\infty)$ really needed? – gerw Jun 19 at 6:51
• Thanks again gerw! I was not aware on weak-* measurable version and with your indication I found in Leoni "A first course in Sobolev space" (2nd Edition) the space $L_w^{\infty}(X,L^\infty(\mathbb R^n))$ of weak*-measurable function (Def. 8.16) identify by the Riesz representation (Theorem 8.17) to the space $L^1(0,T;L^1(\mathbb R^n))$. I will pursue reading the paper but I thing it is enough. – Airlast Jun 19 at 16:42

I finished by point 3. I promized. The paper is disponible here for more information. And I recal that $$u^\varepsilon$$ are regular engouh (continuous) so that measurability in any subsequent space is valid.
The case $$p=1$$. The authors say the sequence $$u^\varepsilon$$ uniformely bounded in $$L^\infty(0,T;L^1(\mathbb R^n))$$ is weakly relatively compact in $$L^\infty(0,T;L^1_{loc}(\mathbb R^n))$$ and there is an accumulation point in $$L^\infty(0,T;L^1(\mathbb R^n))$$.
To prove this assertion they approach $$u^\varepsilon$$ by a sequence $$u^\varepsilon_n$$ uniformely bounded in $$\varepsilon$$ (not in $$n$$) in $$L^\infty(0,T;L^p(\mathbb R^n))$$ $$(1 and belonging to $$L^\infty(0,T;L^1(\mathbb R^n))$$ which satisfies $$\lim_{n\to \infty} \sup_{\varepsilon>0}\|u^\varepsilon - u^\varepsilon_n \|_{L^\infty(0,T;L^1(\mathbb R^n))} =0$$ so that we have in particular $$\sup_{\varepsilon>0} \|u^\varepsilon\|_{L^\infty(0,T;L^1(\mathbb R^n))} \leq C \text{ and } \sup_{\varepsilon>0} \|u^\varepsilon_n\|_{L^\infty(0,T;L^p(\mathbb R^n))} \leq C(n)$$
First I don't now what is the weak convergence in $$L^\infty(0,T;L^1_{loc}(\mathbb R^n))$$ but I can imagine they understand the convergence again $$L^1(0,T;C_c(\mathbb(\mathbb R^n))$$. What I can also presume is that $$u^\varepsilon$$ weakly* converges in $$L^\infty_w(0,T;\mathcal M(\mathbb R^n))$$ as the dual of $$L^1(0,T;C_0(\mathbb R^n))$$ where $$\mathcal M(\mathbb R^n)$$ is the space of finite Radon measures on $$\mathbb R^n$$. Note $$\mu$$ the limit. I would show that $$\mu$$ can be identify to an element of $$L^\infty(0,T;L^1(\mathbb R^n))$$. To that we may by the relation $$\langle u^\varepsilon,\varphi\rangle = \langle u^\varepsilon -u^\varepsilon_n,\varphi\rangle + \langle u^\varepsilon_n,\varphi \rangle$$ for all $$\varphi\in L^1(0,T;C_c(\mathbb R^n))$$ and assuming that $$u^\varepsilon_n$$ converges weakly* to $$u_n$$ in $$L^\infty(0,T;L^p(\mathbb R^n)$$. Thus we can choose $$\delta>0$$ and $$n$$ large enough such that $$|\langle \mu,\varphi\rangle| \leq \delta + |\langle u_n,\varphi \rangle|$$ for all $$\varphi$$ with norm less than $$1$$. We can separate the variable and we have $$|\langle \mu(t),\varphi\rangle| \leq \delta + |\langle u_n(t),\varphi \rangle| \qquad a.e. t\in(0,T)$$ for all $$\varphi\in C_c(\mathbb R^n)$$ no-negative and less than one and then I presume considering a negligeable set $$A$$ by measure theory (some regularity) we can prove that $$\mu(t)(A)$$ is as small as we want thus of measure null and conclude that a.e. $$t \in(0,T)$$, $$\mu(t)$$ as a density w.r.t. Lebesgue measure. Finally we construct a function $$u:(0,T) \to L^1(\mathbb R^n)$$ in $$L^\infty_w(0,T;\mathcal M(\mathbb R^n))$$ limit of the $$u^\varepsilon$$ in the sense above mentioned but again measurability in $$L^1$$ is not obvious.