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I know that for a dependent-mesh norm Nitsche' method for boundary conditions is coercive. But how to prove this for the usual norms for $H^1$?

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    $\begingroup$ I don't think it is possible to prove anything useful in this way, since you cannot uniformly control the jump terms without mesh-dependency in your norm. Coercivity is after all only one of many components for the analysis. And don't forget that Nitsche's method is nonconforming, i.e., the solutions are not sought for in the usual H^1 plus boundary conditions. $\endgroup$ – Christian Waluga Dec 20 '15 at 9:06
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    $\begingroup$ To put this another way: If it were possible to show coercivity with respect to the standard $H^1$ norm, nobody would have bothered with mesh-dependent norms. $\endgroup$ – Christian Clason Dec 20 '15 at 15:24
  • $\begingroup$ To Christian Waluga. Can you give me more explanations, pléiade ? Thank you $\endgroup$ – Mohamed Cheddadi Jan 20 '16 at 0:39

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