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

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.

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 $$H^s$$.

• Thank you very much for you answer. I have one small question, in the case $\theta<s$ does these hypothesis also implies that $u\in C([0,T]; H^{s'})$ for any $\theta\leq s'<s$? In other words, if we allow a little loss in space-regularity, is it posible to improve the the weak continuity in $H^s$ to strong continuity in $H^{s'}$ for $s'<s$?
– W2S
Sep 1, 2020 at 17:27
• Maybe if I add, for example, an extra hypothesis like $\partial_tu\in L^\infty(0,T;H^\alpha)$ with $\alpha\leq \theta$? Might that work?
– W2S
Sep 1, 2020 at 21:47