sci.math #163118 (29 + 792 more) (1)--[1] From: Ray DeCampo Newsgroups: sci.math [1] Who put the W in Sobolev spaces? Date: Fri, 27 Sep 1996 17:44:34 -0400 Organization: Syracuse University Lines: 9 Message-ID: <324C4AC2.6A27@mailbox.syr.edu> NNTP-Posting-Host: sudial-74.syr.edu X-Mailer: Mozilla 2.02Gold (Win95; I) Thanks everybody for your responses to the question about the L in L^p. How about the W sometimes used for denoting Sobolev spaces? (I appreciate this a great deal. I am a grad student who hopes to be a prof someday and I'd like to be able to answer such questions myself!) Ray DeCampo End of article 163118 (of 163164) -- what next? [npq] sci.math #163128 (28 + 792 more) (1)--[1] From: lange@gpu5.srv.ualberta.ca (U Lange) [1] Re: Who put the W in Sobolev spaces? Ray DeCampo (rkdecamp@mailbox.syr.edu) wrote: : Thanks everybody for your responses to the question about : the L in L^p. How about the W sometimes used for denoting : Sobolev spaces? Actually there are the Symbols H and W for Sobolev spaces (and a famous proof that "H=W"). I am not sure, but IMHO it would make sense if "H" stands for "Hilbert" and "W" for "Weak derivative" (The "H"-definition of Sobolev spaces is based on the completion of the C^\infty-functions with respect to the Sobolev norms, while the "W"-definition is based on the integrability of the weak derivatives). -- Ulrich Lange Dept. of Chemical Engineering University of Alberta lange@gpu.srv.ualberta.ca Edmonton, Alberta, T6G 2G6, Canada End of article 163128 (of 163164) -- what next? [npq] sci.math #162672 (20 + 792 more) (1)--[1] From: ajordan1@aol.com (AJordan1) Newsgroups: sci.math [1] Help with the trace operator on sobolev spaces Date: 24 Sep 1996 15:05:31 -0400 Organization: America Online, Inc. (1-800-827-6364) Lines: 8 Sender: root@newsbf02.news.aol.com Message-ID: <529bdr$gnf@newsbf02.news.aol.com> Reply-To: ajordan1@aol.com (AJordan1) NNTP-Posting-Host: newsbf02.mail.aol.com I was on a seminar of distributions and Sobolev Spaces this summer. It ended with the trace theorem. I have no experience with PDE's but I would like to know some examples how to use the Trace Operator. It would be nice, if somebody can tell me an example or could send me the name of a book in witch I can find some. Thank's a lot! Andy End of article 162672 (of 163164) -- what next? [npq] sci.math #162717 (19 + 792 more) (1)--[1] From: lange@gpu3.srv.ualberta.ca (U Lange) [1] Re: Help with the trace operator on sobolev spaces AJordan1 (ajordan1@aol.com) wrote: : I was on a seminar of distributions and Sobolev Spaces this summer. It : ended with the trace theorem. I have no experience with PDE's but I would : like to know some examples how to use the Trace Operator. It would be : nice, if somebody can tell me an example or could send me the name of a : book in witch I can find some. Thank's a lot! What do you mean by "_use_ the trace operator"? I would think - apart from the proof of the trace theorem - the trace operator is never _explicitly_ "used". Normally you only use the fact that the trace operator is a _bounded_ operator. But this has very important consequences: The most relevant consequence of the trace theorem for PDEs is that it clarifies under which conditions it makes sense to impose boundary conditions to weak (distributional) solutions of a PDE. Then, it also tells you when (and in which sense!) the weak solution depends "continously" on the boundary conditions (Thus, the trace theorem is necessary to extend Hadamard's definition of a "well-posed problem" to the weak formulation of the PDE). A more practical application of the trace theorem is - for example - the numerical solution of a PDE with so-called "non-conforming Finite Elements": Using a standard triangulation for a circular domain, you replace the original circle by a polygon for the numerical solution. The error introduced by this replacement can be estimated by the trace theorem. -- Ulrich Lange Dept. of Chemical Engineering University of Alberta lange@gpu.srv.ualberta.ca Edmonton, Alberta, T6G 2G6, Canada End of article 162717 (of 163164) -- what next? [npq] sci.math.num-analysis #42975 (6 + 318 more) [1]--[1] From: "Teck Cheong Lim" Newsgroups: sci.math,sci.math.research,sci.math.num-analysis [1] H^s functions Date: Sun Jun 21 16:33:58 EDT 1998 Sobolev imbedding theorem says that functions in H^s(R^2) are continuous if s>1. Does anyone have an example of a function in H^1(R^2) that is not continuous? [ H^s(R^2) stands for L^2 functions whose weak derivatives up to order s are also in L^2] End of article 42975 (of 43063) -- what next? [npq] sci.math.num-analysis #42994 (5 + 318 more) (1)--[1] From: ullrich@hardy.math.okstate.edu Newsgroups: sci.math,sci.math.research,sci.math.num-analysis [1] Re: H^s functions Date: Tue Jun 23 11:33:41 EDT 1998 Yes. Say f is a function which is nice except near the origin, has compact support, and such that f(x) = log(log(1/|x|)) near the origin. It's certainly not continuous, and a little calculus shows it's in H^1(R^2). (The square of the gradient is 1/(x^2 * log(1/|x|)^2); integrate in polar coordinates.) If you're curious how I came up with that, it's clear that the question "must" be easy to answer if you look at the Fourier transform: Say p is the FT of f. Then f is in H^1 if and only if (*) Integral((1+|x|^2) * |p(x}|^2) < infinity. It "cannot" be difficult to determine whether this condition implies (**) Integral(|p|) < infinity. If (*) did imply (**) then this would give a theorem instead of a counterexample. Otoh given an example where (*) holds but (**) fails you can also get an example where p >= 0, and in that case the fact that (**) fails shows f must be unbounded near the origin. Or so you'd think. I came up with an obvious example where (*) holds but (**) fails: p(x) = 0 for |x| < 1, p(x) = 2^(-2n)/n for 2^n < |x| < 2^(n+1) . Then I realized it wasn't _quite_ as obvious as I'd thought at first that (**) false implies f unbounded, ie not _essentially_ bounded, near the origin (although it seems it "must" be so and it's probably easy) so I looked at how those things added up on the other side, smoothed things a little and the log(log(1/|x|)) popped out. David C. Ullrich -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading End of article 42994 (of 43063) -- what next? [npq] sci.math.research (moderated) #10331 (1 + 134 more) [1]--[1] From: chernoff@talinn.berkeley.edu (Paul R. Chernoff) [1] Multipliers from H^1 to L^2 Date: Mon Apr 27 12:47:51 EDT 1998 We work in 1 dimension. L^2 = L^2(R) and H^1 = the Sobolev space H^1(R). A complex valued function V is called a *multiplier* from H^1 to L^2 provided that, for every f in H^1, the product Vf is an L^2 function. By the closed graph theorem, it is easy to see that V is a multiplier iff there is a constant C such that, for every f in H^1, ||Vf||_2 \leq C( ||f||_2 + ||f'||_2 ) where f' is the derivative of f and the subscript denotes the L^2 norm. A trivial observation is that if V is the sum of an L^2 function and an L^{\infty} function, then V is such a multiplier. QUESTION: is the CONVERSE true? Or, if not, is there some simple character- ization of multipliers from H^1 to L^2? Note: It is easy to see that, if V is a multiplier, then V is *locally* square integrable. Moreover we have the integral bound \int_{-T}^{T} |V(x)|^2 dx = O(T) Both of these are consistent with V being in L^2 + L^{\infty}, but, of course, they do not prove this conclusion. --------------------------------------------------------------------- -- # Paul R. Chernoff chernoff@math.berkeley.edu # # Department of Mathematics # # University of California "Against stupidity, the gods themselves # # Berkeley, CA 94720 struggle in vain." -- Schiller # End of article 10331 (of 10845) -- what next? [npq] sci.math.research (moderated) #10333 (0 + 134 more) (1)--[1] From: "Gunter Bengel" [1] Re: Multipliers from H^1 to L^2 Date: Tue Apr 28 07:04:44 EDT 1998 Hardy's inequality states that \int_a^b |u(x)/x|^2 <= 4\int_a^b |u'(x)|^2 for u C-infinity with compact support in (a,b). So for a=-1, b=1 take a C-infinity-function g with compact support in (-1,1) g(x)/|x| is an H^1 - L^2 multiplier, but not locally L^2. Gunter End of article 10333 (of 10845) -- what next? [npq]