Timeline for The characteristic (indicator) function of a set is not in the Sobolev space H¹
Current License: CC BY-SA 2.5
16 events
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| Oct 28, 2013 at 21:23 | comment | added | username | Maybe the last line of the argument could simply be: therefore $\chi_E(x)=\chi_E(x+h)$ a.e. for any $x$ and $h$, that is, $\chi\equiv0$ or $\chi\equiv1$ everywhere, both cases being forbidden | |
| Oct 28, 2013 at 20:43 | review | Suggested edits | |||
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| May 22, 2010 at 5:18 | comment | added | Pietro Majer | (erratum corrige: oh I wrote "Schwartz" instead of "Schwarz") | |
| May 19, 2010 at 6:19 | comment | added | Pietro Majer | Thank you, my pleasure. Your argument is more general in that also proves that chi_E is not W^1,1, nor W^1,1_loc; the argument I used at most covers W1,p for p>1. :) | |
| May 18, 2010 at 20:31 | comment | added | Tom Boardman | Aha! Found it in Dracorogna- never did follow calculus of variations as far as I should- nowhere to be found in my PDE books... You sir, have well and truly earned yourself a +1! Much more elegant than mine- which seems 'pretty sledgehammer to crack a nut' by comparison... | |
| May 18, 2010 at 20:02 | comment | added | Pietro Majer | Actually for L2 functions that inequality is a characterization of H1. This stuff is quite popular in calculus of variations and regularity. To prove the inequality, do it first for smooth functions with compact support (just write f(x+h)-f(x) as integral of the derivative in the direction h, then use Cauchy-Schwartz and Fubini. You will find C=||Df||_2^2). Then pass to the limit. An analog result holds with all p>1, to characterize W^{1,p}; with p=1 you get BV. Reference: I think Brezis (Fun.Anal.) is OK. | |
| May 18, 2010 at 19:34 | comment | added | Tom Boardman | My mistake! Chop it, bump it, mollify it and extend it by zero- I'm convinced you're still going to get a higher power of h than 2! Maybe I'm just being silly, but I swear I've never seen that inequality... Any chance you've got a reference for it? | |
| May 18, 2010 at 19:17 | comment | added | Pietro Majer | @Tom, x^2 is not in H1(R) :) | |
| May 18, 2010 at 19:14 | comment | added | Pietro Majer | (I tried to edit but had some problem with TeX). Anyway: just apply the first inequality to h/m where m is a positive integer, and make m steps to estimate the distance between E and E-h. Since m is arbitrary you conclude. Is it everything OK to you? This is quite an elementary proof. | |
| May 18, 2010 at 19:08 | comment | added | Tom Boardman | @ Pietro, I'm not sure about your first inequality- what if n=1 and f(x) is x^2 ? Trying h=1 for example seems to scupper things... | |
| May 18, 2010 at 18:39 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 18, 2010 at 18:16 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 18, 2010 at 18:09 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 18, 2010 at 17:58 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 18, 2010 at 17:41 | comment | added | Spencer | Any chance you could please just give extra clarification as to where the $m$'s come from? Many thanks. | |
| May 18, 2010 at 17:18 | history | answered | Pietro Majer | CC BY-SA 2.5 |