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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 $L^\infty(0,T;L^p(\mathbb R^n))$ $(1<p<\infty)$, 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.

Thanks for your help.

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    $\begingroup$ 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? $\endgroup$ – gerw Jun 19 at 6:51
  • $\begingroup$ 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. $\endgroup$ – Airlast Jun 19 at 16:42
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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 $L^\infty(0,T;L^p(\mathbb R^n))$ $(1<p<\infty)$ 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 $$ \sup_{\varepsilon>0} \|u^\varepsilon\|_{L^\infty(0,T;L^1(\mathbb R^n))} \leq C \text{ and } \sup_{\varepsilon>0} \|u^\varepsilon_n\|_{L^\infty(0,T;L^p(\mathbb R^n))} \leq C(n)$$

The authors claim this entails the desired weak compactness.

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.

Is this way ok, or is there another way to interpret what the authors claim.

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