Strong and Weak Continuity in $L^p$ Spaces

Question

Set-Up

Let $\Omega \subseteq \mathbb{R}^n$ be a domain, $T>0$.

We find $u_m(x,t) : \Omega \times [0,T] \rightarrow \mathbb{R}^n$ which converge weakly to some $u \in L^{\infty}(0,T; L_{\sigma}^2)$.

Here, $L_{\sigma}^2 = \{ f \in L(\Omega)^n \ | \ \text{div}f = 0 \}$

We know that $||u_m||$ is uniform bounded, and $u_m(\cdot , t) := u(t)$ is continuous with respect to $t$ for all $m$. We also have that $u$ is weakly continuous in $L_{\sigma}^2$ with respect to $t$.

That is, $(u(t) , f) : \mathbb{R} \rightarrow \mathbb{R}$ is continuous, for all $f \in L^{2}_{\sigma}$, where $(\cdot , \cdot)$ is the inner-product on $L^2$. (See this previous question for a full explanation).

Claim

Our claim is that the inner-product $(u(t),u_m(t))$ is also continuous with respect to $t$, for all $m \in \mathbb{N}$.

That is, the integral $\int_{\Omega} u(t)u_m(t) \text{d}x$ is continuous with respect to $t$.

I believe that the above claim should hold, by the weak continuity of $u(t)$, and strong continuity of $u_m(t)$, but I cannot find an obvious way to show this... Does anyone know even how I might start? Thank you.

Answer 1

I assume (strong) continuity of $u_m$ means $u_m\in C((0,T);L^2_\sigma)$.

Continuity of $t\mapsto(u(t),u_m(t))$ in any $t_0\in(0,T)$ follows from the estimate

\begin{align*} &|(u(t_0),u_m(t_0))-(u(s),u_m(s))|\\ &\leq |(u(t_0),u_m(t_0))-(u(s),u_m(t_0))| + |(u(s),u_m(t_0))-(u(s),u_m(s))|\\ &\leq |(u(t_0),u_m(t_0))-(u(s),u_m(t_0))| + ||u(s)||_2\, ||u_m(t_0))-u_m(s)||_2\\ &\leq |(u(t_0),u_m(t_0))-(u(s),u_m(t_0))| + ||u||_{L^\infty(L^2)}\, ||u_m(t_0))-u_m(s)||_2. \end{align*}

Let $\epsilon>0$. The weak continuity of $u$ implies that the first term on the right-hand side above is less than $\epsilon$ for $|s-t_0|$ sufficiently small, and stong continuity of $u_m$ implies that the second term is also less than $\epsilon$ for $|s-t_0|$ sufficiently small.