Suppose $\Omega = \mathbb{T}$ is the $1$D torus. Consider a sequence of functions $\{ u_{n} \}_{n \ge 1}$ such that this sequence is
- Cauchy in $L^{2}(\Omega)$,
- bounded in $H^{1}(\Omega)$.
Question 1: Is it true that there exist a subsequence $u_{n_{j}}$ converging to some $u$ strongly in $H^{1}(\Omega)$?
I think the answer is yes, but I would be very grateful if someone could verify.
My attempt:
The first fact implies that there exists a subsequence $u_{n_{j}}$ and some $u \in L^{2}(\Omega)$ such that $u_{n_{j}} \to u$ strongly in $L^{2}(\Omega)$. The second fact implies that there exists a subsequence denoted again by $u_{n_{j}}$ for convenience and some $v \in H^{1}(\Omega)$ such that $u_{n_{j}} \rightharpoonup ^{*} v$ (weak*) in $H^{1}(\Omega)$. But both convergences imply convergence in the sense of distributions. Since limits in $\mathcal{D}'$ are unique then we have $u = v$ in the distributional sense. However, since we know that $v \in H^{1}(\Omega)$, we must also have that $u \in H^{1}(\Omega)$. To see this last fact more concretely, $u=v$ implies that the distributional derivatives also coincide, i.e. $u' = v'$. But here $v' \in L^{2}(\Omega)$ and so $u' \in L^{2}(\Omega)$ too.
Question 2:
If this argument is correct then surely we could say that in general if $\{u_{n}\}$ is Cauchy in $L^{2}$, bounded in $H^{k}$ then there is a subsequence $u_{n_{j}}$ converging to some $u$ in $H^{k}$? This kind of seems too good to be true, but I don't have a great intuition for weak* convergence so my scepticism may be unfounded.