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S Apr 11, 2017 at 12:49 history bounty ended CommunityBot
S Apr 11, 2017 at 12:49 history notice removed CommunityBot
Apr 11, 2017 at 4:16 vote accept pxchg1200
Apr 10, 2017 at 1:49 comment added pxchg1200 Since $S_{0}^{1,2}(U)$ can be compact embedding in to $L^2(U)$, I think we can prove the following Poincare inequality for $p=2$, $$ \lambda_{1}\int_{U}|u(x)|^2dx\leq \int_{U}|D_{L}u(x)|^2dx $$ after variational calculus. the $\lambda_{1}$ means the first Dirichlet eigenvalues of $L$ which is strict positive.
Apr 8, 2017 at 13:52 answer added Henry.L timeline score: 2
Apr 6, 2017 at 3:05 comment added Deane Yang I have to concede that I appear to be wrong about this. I thought I knew how to do this, but I currently don't see how. If there is a global Poincaré inequality ($\|f\|_p \le C\|D_Lf\|_p$), then you would get the inequality.
Apr 3, 2017 at 15:09 comment added pxchg1200 It seems we can only obtain $$\|f\|_{L^{q}(U)}\leq C( \|D_{L}f\|_{L^{p}(U)}+\|f\|_{L^{p}(U)} )$$ will the sharp estimate
Apr 3, 2017 at 14:31 comment added pxchg1200 Do you mean use $\|D_{L}(\phi_{i}f) \|_{L^{p}(B_{i})}=\|(D_{L}\phi_{i})f+\phi_{i}D_{L}f\|_{L^{p}(B_{i}}$ ? then how can I combine it to $\|D_{L}f\|_{L^{p}(U)}$ ? @DeaneYang
Apr 3, 2017 at 14:25 comment added Deane Yang Use product rule for differentiation and Holder's inequality.
Apr 3, 2017 at 14:21 history edited pxchg1200 CC BY-SA 3.0
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Apr 3, 2017 at 14:01 history edited pxchg1200 CC BY-SA 3.0
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S Apr 3, 2017 at 10:50 history bounty started pxchg1200
S Apr 3, 2017 at 10:50 history notice added pxchg1200 Authoritative reference needed
S Oct 27, 2016 at 5:34 history bounty ended CommunityBot
S Oct 27, 2016 at 5:34 history notice removed CommunityBot
S Oct 19, 2016 at 4:16 history bounty started pxchg1200
S Oct 19, 2016 at 4:16 history notice added pxchg1200 Canonical answer required
Oct 18, 2016 at 3:15 comment added Deane Yang The set $B_R$ is still an open set, so if you cover $U$ with such "balls", you can still construct a partition of unity subordinate to that cover.
Oct 17, 2016 at 12:23 comment added Nate Eldredge Good point, I did not see the $c$ subscript.
Oct 17, 2016 at 5:25 comment added pxchg1200 @NateEldredge how does closed graph theorem works here? first you need $S_{c}^{1,p}(U)$ to be a Banach space in norm $\|\cdot\|=\left(\int_{U}|D_{L}u|^{p}dx\right)^{1/p}$
Oct 17, 2016 at 4:08 comment added pxchg1200 I wonder the subunit ball can cover $U$ or not,because the subunit ball is not like the open ball we used before. so I don't know how to use partition of unity, can you write the detail proof? @DeaneYang thank you very much!
Oct 17, 2016 at 3:45 comment added Deane Yang Nate, thanks. Didn't know I had such powers.
Oct 17, 2016 at 3:02 review Close votes
Oct 17, 2016 at 17:26
Oct 17, 2016 at 2:38 comment added Nate Eldredge @DeaneYang: You have enough reputation to vote to migrate...
S Oct 17, 2016 at 1:29 history suggested T. Amdeberhan CC BY-SA 3.0
Language errata corrected and clarifications added.
Oct 17, 2016 at 0:40 review Suggested edits
S Oct 17, 2016 at 1:29
Oct 17, 2016 at 0:32 comment added Deane Yang This should be migrated to math.stackexchange.com. But the idea is to use a partition of unity subordinate to a locally finite covering of $U$ by open balls to write $u$ as a sum of functions, each compactly supported on one of the balls, and apply Theorem 2.3 to each of these functions.
Oct 17, 2016 at 0:26 history asked pxchg1200 CC BY-SA 3.0