# Does weak convergence of $u_m(t)$ in $L^2(0,T;X)$ imply weak convergence of a subsequence of $u_m(t_0)$ in $X$ for a.e. $t_0$ in $[0,T]$?

In a book I'm reading (Navier Stokes Equations, by Constantin and Foias), the authors construct a sequence $u_m$ of functions in $L^2(0,T;V)$ which converge weakly to $u$ in $L^2(0,T;V)$. They then claim that $u_m(t_0)$ converges weakly to $u(t_0)$ for a.e. $t_0\in [0,T]$. Why is this true?

Here $V$ is the subset of $H^1_0(\Omega)$ ($\Omega\subset\mathbb{R}^n$ bounded, with nice boundary) whose elements are divergence-free; however, it seems unlikely to me that the definition of $V$ will matter for the question I am asking.

• As we can see in the @gerw 's answer, the result is false in general. But, as we can see in my answer, the result can be true in some cases. So, are you sure that the constructed sequence has no extra regularity? – Pedro Jan 10 '16 at 3:09

## 2 Answers

This is not true, even in the simplest case $V = \mathbb{R}$. To see this, fix $v \in V \setminus \{0\}$ and define $u_m(t) = v \, \sin(m \, t)$. Then, $u_m \rightharpoonup 0$ in $L^2(0,T;V)$, but it does not converge pointwise a.e.

One thing that ensures weak pointwise a.e. convergence, is Helly's selection theorem, see here and here.

I'd say that the definition of $V$ is important because we can prove a version of the desired result using the following theorem.

Theorem 1. Let $V$ be a Banach space such that $V'$ has the Radon-Nikodym property (for example, $V$ is Hilbert or it's separable reflexive) and $1\leq p<\infty$. If $$u_n\rightharpoonup u\quad\text{em}\quad L^p(0,T;V),$$ then $$\int_0^T\varphi u_n\;dt\rightharpoonup\int_0^T\varphi u\;dt\quad\text{em}\quad X,\qquad\forall\ \varphi\in C[0,T].$$

Theorem above implies the desired result provided that the sequence $\{u'_n\}$ of weak derivatives has sufficient regularity. More precisely:

Let $V$ be as in the hypothesis of Theorem 1. Assume that $$v_n\rightharpoonup v\quad\text{in}\quad L^2(0,T;V)$$ and that $\{v_n'\}$ is bounded in $L^b(0,T; V)$ for some $1<b<\infty$. Then $$v_n(s)\rightharpoonup v(s)\quad\text{in}\quad V,\qquad\forall\ s\in(0,T).$$

Proof:

Let $s\in(0,T)$ and $f\in V'$. We want to prove that

$$f(v_n(s))\to f(v(s))\quad\text{in}\quad\mathbb{R}.$$

Define $u_n=v_n-v$. Then $$u_n\rightharpoonup 0\quad\text{in}\quad L^2(0,T;V)\tag{1}$$ and we need to prove that $$f(u_n(s))\to 0\quad\text{in}\quad\mathbb{R}.$$

Set $\phi\in C^1([0,T];\mathbb{R})$ such that $\phi(0)=-1$ and $\phi(T)=0$.

Take $\lambda\in(0,1-s/T)$ (to be fixed later) and define $w_n(t)=u_n(s+\lambda t)$. Then $$u_n(s)=w_n(0)=\phi(t)w_n(t)\Big|_0^T=\int_0^T(\phi w_n)'\;dt=\underset{\alpha_n}{\underbrace{\int_0^T\phi' w_n\;dt}}+\underset{\beta_n}{\underbrace{\int_0^T\phi w_n'\;dt}}$$

Notice that $$\beta_n=\int_0^T\lambda\phi(t) u_n'(s+\lambda t)\;dt=\int_s^{s+\lambda T}\phi(\lambda^{-1}(\tau-s)) u_n'(\tau)\;d\tau.$$ Thus, by Hölder's inequality and by the boundedness of $\{v'_n\}$, \begin{aligned} \|\beta_n\|_{V}&\leq C\int_s^{s+\lambda T}\|u_n'(\tau)\|_V\;d\tau\\ &\leq C(\lambda T)^{1-\frac{1}{b}}\left(\int_s^{s+\lambda T}\|u_n'(\tau)\|_V^b\;d\tau\right)^{\frac{1}{b}}\\ &\leq C(\lambda T)^{1-\frac{1}{b}}\left(\int_0^T\|u_n'(\tau)\|_V^b\;d\tau\right)^{\frac{1}{b}}\\ %&= C(\lambda T)^{1-\frac{1}{b}}\|u_n'\|_{L^b(0,T;V)}\\ &\leq CT^{1-\frac{1}{b}}M\lambda^{1-\frac{1}{b}} \end{aligned} for some constant $M$. Therefore, given $\varepsilon>0$, we can choose $\lambda$ small enough such that $$\|\beta_n\|_{V}\leq\frac{\varepsilon}{2\|f\|},\quad \forall \ n\in\mathbb{N}.\tag{2}$$

Now, from $(1)$ we get $$u_n|_{(s,s+\lambda T)}\rightharpoonup 0\quad\text{em}\quad L^2(s,s+\lambda T;V)$$ and thus, by Theorem 1, $$\alpha_n=\int_0^T\phi' w_n\;dt=\lambda^{-1}\int_s^{s+\lambda T}\phi'(\lambda^{-1}(\tau-s)) u_n(\tau)\;d\tau\rightharpoonup 0\quad\text{in}\quad V$$ which implies $$f(\alpha_n)\to 0\quad\text{in}\quad\mathbb{R}.\tag{3}$$ From $(2)$ and $(3)$ we conclude that $f(u_n(s))=f(\alpha_n+\beta_n)\to 0$ as needed. $\blacksquare$