Let $\{u_n\}_{n\in\mathbb{N}}$ a bounded sequence in $H_0^1({\Omega})$ for some $\Omega\subset\mathbb{R}^n$ (which is "appropriate" for the following theorems). Since $H_0^1({\Omega})$ is compactly embedded in $L^2(\Omega)$ by the Rellich-Kondrashev theorem, there exists a subsequence that converges strongly in $L^2(\Omega)$. $H_0^1({\Omega})$ is also reflexive so by the Banach-Alaoglu theorem there also exists a subsequence that converges weakly in $H_0^1({\Omega})$.

Why should these two subsequences have the same limit? I would say that since $H_0^1({\Omega})\subset L^2(\Omega)$ we have $L^2(\Omega)^*\subset H_0^1({\Omega})^*$ so weak convergence in $H_0^1({\Omega})$ implies weak convergence in $L^2(\Omega)$, and the weak and strong limit must coincide. But what tells me that the two subsequences, to begin with, are the same? After all those two theorems prove that there exists a subsequence such that..., so they may be as well be two different subsequences. I think I have to go deeper and take a further subsequence of the subsequences, but I don't know how to formalize it without making a mess.

Note: I used $H_0^1({\Omega})$ and $L^2(\Omega)$ as an example but the same reasoning applies to any two Sobolev spaces connected by the Rellich-Kondrashev theorem.


You cannot infer that the subsequences have the same limit, e.g., choose any $u \in H_0^1(\Omega) \setminus \{0\}$ and consider $$u_n = (-1)^n \, u$$.

Then, $\{u_{2n}\}$ converges strongly in $L^2(\Omega)$ and $\{u_{2n+1}\}$ converges weakly in $H_0^1(\Omega)$ and the limits are different.

However, any subsequence which converges strongly in $L^2(\Omega)$ converges weakly in $H_0^1(\Omega)$ and vice versa. This can be seen by a subsequence-subsequence argument.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.