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Feb 23, 2017 at 16:00 vote accept Chee Han
Sep 26, 2016 at 19:15 comment added Chee Han Now that I think of it, my subsequent question sounds really dumb and obvious.
Sep 26, 2016 at 15:50 comment added Chee Han I will check out the paper later today, thanks for the reference! Does a similar result holds for $\mathcal{F}$ instead of $\partial\mathcal{D}$ thou? Actually, now that I know the embedding is compact, any weakly convergent sequence in $H^1(\mathcal{D})$ maps to a strongly convergent sequence in $L^2(\partial\mathcal{D})$; does that implies that the sequence is also strongly convergent in $L^2(\mathcal{F})$, where $\mathcal{D}, \mathcal{F}$ are defined as above in my question?
Sep 26, 2016 at 10:17 history answered Hannes CC BY-SA 3.0