If i have a bounded sequence $\{u_n\}_n \in H^1(\Omega) \cap L^\infty(\Omega)$, then by the weak and weak-* compactness of the spaces, there exists subsequences and functions $u_1 \in H^1(\Omega)$, $u_2 \in L^\infty(\Omega)$ such that $u_n \rightharpoonup u_1$ in $H^1$, and $u_n \rightharpoonup u_2$ in $L^\infty$ weak-*. My question is, under what conditions will I get a limit with the same regularity as the sequence? In other words, when is $u_1 = u_2$?


It is easy to check that $u_n \rightharpoonup u_1$ in $H^1$ implies $u_n \rightharpoonup u_1$ in $L^2$. Similarly, $u_n \stackrel*\rightharpoonup u_2$ in $L^\infty$ implies $u_n \rightharpoonup u_2$ in $L^2$. By uniqueness of the weak limit, you have $u_1 = u_2$.

  • $\begingroup$ Thank you for the answer. However, it was my understanding that weak-* convergence in $L^\infty$ implies weak convergence in $L^1$. How come also in $L^2$? $\endgroup$ – Mattkb Mar 21 '18 at 12:17
  • $\begingroup$ You have $\int_\Omega u_n \, f \, \mathrm dx \to \int_\Omega u_2 \, f\, \mathrm{d}x$ for all $f \in L^1(\Omega)$. But $L^2(\Omega) \subset L^1(\Omega)$. Thus, this holds for all $f \in L^2(\Omega)$ and this is weak convergence in $L^2$. $\endgroup$ – gerw Mar 21 '18 at 12:20
  • $\begingroup$ Yes, i see. Thank you! $\endgroup$ – Mattkb Mar 22 '18 at 8:02
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    $\begingroup$ If you are pleased with the answer, you should accept it by clicking the checkmark. $\endgroup$ – gerw Mar 22 '18 at 11:33

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