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Let $s,s'\in\mathbb{R}$ satisfying $s'<s$. Consider $u=u(t,x)\in C(\mathbb{R},H^s(\mathbb{R}))$ and suppose that, as $t$ goes to infinity, we have the following weak convergence: $$ u(t,\cdot)\rightharpoonup u^*(x) \quad \hbox{in} \quad H^{s'}(\mathbb{R}), $$ where $s'<s$ and "$\rightharpoonup$" denotes the weak convergence in $H^{s'}$. If we additionally assume that $u(t,\cdot)$ is uniformly bounded in $H^s$, say $$ u\in L^{\infty}(\mathbb{R},H^s(\mathbb{R})), $$ does this imply that $u(t,\cdot)$ strongly converge to $u^*$ in $H^{s'}(K)$ for any $K\subset\mathbb{R}$ compact? Moreover, if we additionally assume that $u^*$ belongs to $H^s$ (recall that $s'<s$), does the previous hypotheses (weak convergence in $H^{s'}(\mathbb{R})$ and uniform boundedness in $H^s(\mathbb{R})$) imply strong convergence in $H^{r}(K)$ for all $s'\leq r<s$ and all $K\subset\mathbb{R}$ compact? In other words, if we have a function is converging in a very weak topology, but this function is also uniformly bounded in a stronger topology, does that implies that the function is locally-strongly converging in the topologies "in between" them (whenever "in between" has sense, like in this case)?

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