If $u \in H^2_0(\Omega)$, is it true that $$u(x) \le C\mathrm{dist}(x,\partial \Omega)^2$$ as $x$ goes to the boundary?
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$\begingroup$ Any smoothness assumptions on the boundary? Also, in any case there is no hope for the lower bound (if $\asymp$ stands for the ratio being bounded between two positive constants). $\endgroup$– Mateusz KwaśnickiCommented Apr 2, 2019 at 21:35
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1$\begingroup$ As for the upper bound, $|u(x)| \le C \operatorname{dist}(x, \partial \Omega)^2$, this is not true even in dimension $1$: $(1-x^2)^{2-\delta}$ is in $H_0^2((-1, 1))$ if $\delta > 0$ is small enough. $\endgroup$– Mateusz KwaśnickiCommented Apr 2, 2019 at 21:38
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$\begingroup$ @MateuszKwaśnicki You're right: I've edited the question. We can assume whatever smoothness is necessary (although I hope none). $\endgroup$– user123672Commented Apr 2, 2019 at 21:38
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$\begingroup$ @MateuszKwaśnicki What do you mean by "if $\delta$ is small enough"? $\endgroup$– user123672Commented Apr 2, 2019 at 21:42
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2$\begingroup$ I bet the answer is "yes", although in an integral form rather than point-wise. I do not know the details, though. "Higher-order Hardy's inequality" (in Lipschitz domains) seems to be a good search phrase. $\endgroup$– Mateusz KwaśnickiCommented Apr 2, 2019 at 22:28
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