In the proof of the existence of weak solutions to the NSE (Navier-Stokes Equations by Constantin and Foias, Chapter 8), the following argument is made:

Let $u_m$ converges weakly to $u$ in $L^2(0,T;V)$ where $$ V=\overline{\{f\in (C_0^\infty(\Omega))^n\mid \nabla\cdot f=0\}}^{H^1(\Omega)} $$ By extracting a subsequence, we may assume that $u_m(t_0)$ converges to $u(t_0)$ weakly in $V$ for all $t_0\in[0,T]\setminus E$ for some $E$ of Lebesgue measure $0$.

I don't understand why this is true. The statement would not be true when $V=\mathbb{R}$ according to the comments to a previous question. Would anyone point me to the related theorems in known references (or sketch the proof)?

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    $\begingroup$ Weak convergence in $L^2$ alone is not enough for convergence a.e. of a subsequence (think of $\sin(mt)v$, for example). There must be something more to this particular $u_m$, what is it? $\endgroup$ – Jean Duchon Feb 4 '16 at 8:40

To provide some context, first let $\Omega$ be a bounded domain in $\mathbb{R}^3$ with smooth boundary.

The popular way of constructing weak solutions is by Galerkin approximation with the eigenfunctions of the Stokes operator (due to Hopf in 1951, or see Ladyzhenskaya's book Chapter 6, Section 6). A second way is to solve the mollified equations $$ u_t + \text{div} (u \otimes u_\rho) - \Delta u + \nabla p = f, $$ where $u_\rho$ is the mollification of $u$ with an appropriate test function. This is what Leray did for the unbounded case, but it is possible in a bounded domain as well.

In either case, you obtain a sequence of functions $u^{(N)}$ which is bounded in $L^\infty_t L^2_x(\Omega \times (0,T))$ and $L^2_t H^1_{x,0}(\Omega \times (0,T))$. By boundedness, there is a subsequence $u^{(N)}$ converging in an appropriate sense (weak-$\ast$ in the former case, weak in the latter).

Others have remarked that the above modes of convergence are not enough to obtain a subsequence converging pointwise a.e. (indeed, for the target space $\mathbb{R}$, weak convergence in $L^2(0,T)$ not does imply pointwise convergence a.e.).

To obtain pointwise a.e. convergence in time, we can prove strong convergence in a space like $L^2_t L^2_x(\Omega \times (0,T))$. To do this, consider the Aubin-Lions lemma with $p=2$ and the triple $$ H_0^1 \overset{\text{cpt}}\hookrightarrow L^2 \hookrightarrow H^{-3}.$$ The compactness of the first embedding follows from the boundedness of $\Omega$ and the Rellich-Kondrachov theorem. Aubin-Lions tells us that $$ W = \{ f \in L^2_tH_0^1(\Omega \times (0,T)) : f_t \in L^2_tH_x^{-3}(\Omega \times (0,T)) \} \overset{\text{cpt}}\hookrightarrow L^2_tL^2_x(\Omega \times (0,T)).$$ I choose to bound the derivative in $H_x^{-3}(\Omega)$ because I want the use of the Sobolev embedding theorem in $n=3$. The nice thing about Aubin-Lions is that you can take as negative a Sobolev space as you want.

To meet the requirements of Aubin-Lions, we remark that the sequence $u^{(N)}$ is already bounded in $L^2_tH^1_{x,0}$. It remains to bound the derivative in $L^2_tH_x^{-3}$. Something like the following should be attainable from the equation: $$ || \partial_t u^{(N)} ||_{L^2_t H_x^{-3}} \lesssim ||u^{(N)}||_{L^\infty_tL^2_x}^2 + ||u^{(N)}||_{L^2_t H^1_x} + ||f^{(N)}||_{L^2_t H^{-1}_x} \lesssim 1. $$ Aubin-Lions then tells us you can choose a subsequence such that \begin{align} u^{(N)} \to u &\text{ in } L^2_tL^2_x(\Omega \times (0,T)) \\ u^{(N)} \overset{\ast}\rightharpoonup u &\text{ in } L^\infty_tL^2_x(\Omega \times (0,T)) \\ u^{(N)} \rightharpoonup u &\text{ in } L^2_tH^1_x(\Omega \times (0,T)). \end{align} The strong convergence gives you pointwise a.e. convergence on the time interval, which is what you asked for.

Please note: I have essentially given you a poor rewording of the argument in Seregin's lecture notes (see Chapter 5), and I am mostly just rerecording it here. There is a preprint of his notes which comes up on Google, if you want.

For the case $\Omega = \mathbb{R}^3$, the Aubin-Lions lemma about compactness is not available (at least, not globally). In such case, you can look at what Leray did in his original 1934 paper, which is available here in English and French.

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  • $\begingroup$ You detailed answer helps a lot. Thank you! $\endgroup$ – Jack Mar 7 '16 at 15:39

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