1
$\begingroup$

In the book Navier Stokes Equations by Constantin and Foias, the folloiwng 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 $0$.

For what conditions on a Banach space $X$ does one have the following general result?

If $(u_m)$ converges weakly in $L^2(0,T;X)$ for some Banach space $X$, then it has a subsquence $u_{m_k}$ such that $u_{m_k}(t)$ converges weakly in $X$ for almost every $t\in[0,T]$.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ As stated, this is not even true for X=R. $\endgroup$
    – Michael Renardy
    Commented Feb 2, 2016 at 2:21
  • $\begingroup$ According to the comment of Michael Renardy, you have to specify more explicitly what you mean by "weak convergence". Perhaps this should mean $L^2$-convergence of $\varphi \circ u_m$ for all $\varphi\in X'$? $\endgroup$
    – Jochen Wengenroth
    Commented Feb 2, 2016 at 9:07
  • $\begingroup$ @JochenWengenroth: For "$u_m$ converges weakly in $Z:=L^2(0,T;X)$" I mean $f(u_m)$ converges for every $f$ in the dual of $Z$. $\endgroup$
    – user14319
    Commented Feb 4, 2016 at 2:50
  • $\begingroup$ Hence, Michael Renardy's comment applies. $\endgroup$
    – Jochen Wengenroth
    Commented Feb 4, 2016 at 8:37

0

Reset to default

You must log in to answer this question.