sci.math.num-analysis #42803 (52 + 391 more) [1]--[1] From: Fragesteller [1] Newsgroups: sci.math.num-analysis,sci.math.research,sci.math.symbolic,sci.math [1] Friedrichs' Inequolity Question !!! Date: Fri Jun 12 15:05:40 EDT 1998 Hello ! It's well known, that for every function v(x) from the Sobolev-space H^1_0(\Omega) holds $$ \int_{\Omega} |v(x)|^2 dx \le C_f*\int_{\Omega} |\nabla v(x)|^2 dx $$ with a Friedrichs' constant C_f depending on the geometrical properties of region \Omega . It is rather obvious, the same inequolity stays hold, when the function v(x) is vanished only on a certain part of the boundary \partial\Omega . How can I find a proove of the last fact in the literature, probably in a case of convex set \Omega ? Thanks for eventual HELP ! S.R. End of article 42803 (of 42812) -- what next? [npq] sci.math.num-analysis #42804 (51 + 391 more) (1)--[1] From: rouben@math.umbc.edu (Rouben Rostamian) [1] Newsgroups: sci.math.num-analysis,sci.math.research,sci.math.symbolic,sci.math [1] Re: Friedrichs' Inequolity Question !!! Date: Fri Jun 12 17:52:28 EDT 1998 You don't need convexity. Connectedness and smooth boundary suffce. Here is a sketch for a proof. The books "Sobolev Spaces" by Adams "Variational inequalities" by Kinderlehrer and Stampacchia "Boundary value problems" by Lions and Magenes may have the details, but I am not sure. To prove your inequality by contradiction, suppose that there exists a sequence {v_n} of functions, which satisfy the boundary conditions and, as n -> infinity: \int v_n^2 = 1 (1) \int |\nabla v_n|^2 -> 0 (2) Observation 1: From (1) and (2) we see that {v_n} is bounded in H^1, therefore it has a weakly convergent subsequence, since H^1 is weakly compact. Without loss of generality, replace {v_n} by this convergent subsequence. Let v0 be the limit. The we have: v_n -> v0 weakly in H^1 (3) therefore by the compact imbedding of H^1 in L^2 (Rellich's Lemma) we have: v_n -> v0 strongly in L^2 This, together with (1) then implies that \int v0^2 = 1 (4) Observation 2: From (3) we see that v_n -> v0 in distributions. Therefore \nabla v_n -> \nabla v0 in distributions (5) But from (2) we see that \nabla v_n -> 0 in L^2 which also implies that \nabla v_n -> 0 in distributions. Then (5) implies that; \nabla v0 = 0 Therefore v0 is a constant. But v0 vanishes on part of the boundary, therefore v0 is identically zero. This contradicts (4). QED There are a lot of technical details hidden under the surface of the last two "Therefore"s. Refer to a good book for details. -- Rouben Rostamian End of article 42804 (of 42812) -- what next? [npq]