$H^s$ boundedness and $L^2$ convergence imply limit is $H^s$ bounded

In a math text I'm reading, I've come across the following. We have a sequence $$\{u_n\} \subset H^s(\mathbb{R}), \qquad s \geq 4$$ and we know that $$\| u_n \|_{H^s} \leq C \qquad \text{uniformly}$$ $$u_n \to u \qquad \text{ in} \qquad L^2.$$ The authors then say that it follows from Fatou's lemma that the limit $u$ is in $H^s$. I'm having trouble seeing why.

I can see that, moving to the frequency side, Fatou's lemma says $$\int \langle \xi \rangle^{2s} \liminf \hat{u}_n^2 d \xi \leq \liminf \int \langle \xi \rangle^{2s} \hat{u}_n^2 d \xi \leq C,$$ but I don't see that $$\| u\|^2_{H^s} = \int \langle \xi \rangle^{2s} \hat{u}^2 d \xi \overset{??}{\lesssim} \int \langle \xi \rangle^{2s} \liminf \hat{u}_n^2 d \xi$$ which is what I would need to close the argument this way.

Is there another approach? Thanks!

Since $u \mapsto \hat u$ is an isometry, we have $\hat u_n \to \hat u$ in $L^2$. Taking a subsequence if necessary, we have $\hat u^2_n \to \hat u^2$ almost everywhere. Thus by Fatou's lemma, $$\int |\xi|^{2s} \hat u^2 d\xi = \int \liminf_{n\to \infty}\left( |\xi|^{2s} \hat{u_n}^2\right) d \xi \le \liminf_{n\to \infty} \int |\xi|^{2s} \hat{u}_n^2 d \xi<\infty$$