Continuity in $H^\theta$ implies weak convergence in $H^s$

Question

Let us consider $\theta,s\in\mathbb{R}$, and the standard Sobolev spaces $H^s(\mathbb{R})$ and $H^\theta(\mathbb{R})$, with no apriori relation between $s$ and $\theta$. Consider a function $$ u\in C([0,T],H^\theta(\mathbb{R}))\cap L^\infty([0,T],H^s(\mathbb{R})). $$ Finally, consider a sequence of times $t_n\to0$. Does the continuity of $u$ with values in $H^\theta$ implies $$ u(t_n)\rightharpoonup u(0) \quad\hbox{in} \quad H^s(\mathbb{R})? $$ Of course, this is trivial if $\theta\geq s$ . I am wondering what about the case $\theta<s$.

Answer 1

Yes, it is true also in the case $\theta < s$.

Note that (see Lemma II.5.9 in "Mathematical Tools for the Navier-Stokes Equations and Related Models") $$u \in C([0,T];H^\theta) \cap L^\infty(0,T;H^s) = C([0,T];H^s_{weak}).$$

That means $t \mapsto \langle \psi, u(t) \rangle$ is continuous for all $\psi \in (H^s)'$. That means for $t_n \to 0$ we have

$$\langle \psi, u(t_n) \rangle \to \langle \psi, u(0) \rangle,$$

for all $\psi \in (H^s)'$. In other words,

$$u(t_n) \rightharpoonup u(0)$$

in $H^s$.