Timeline for The characteristic (indicator) function of a set is not in the Sobolev space H¹
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| Aug 30, 2010 at 0:20 | comment | added | Dorian | Tom what you have said here is false. See Evans chapter 5. If a function is in $W^{1,p}$ where $p > n$ in $\mathbb{R}^n$ then indeed what you said is true. So for one dimension, we do indeed have that $H^1 \subset C^0$. However this is not true in general. We only know there exists some weak derivative. We do not a priori know anything about almost everywhere differentiability of our function when $p < n$. | |
| May 22, 2010 at 5:28 | comment | added | Pietro Majer | of course it is not... W^{1,1}_loc(R) functions are continuous (or if you like, they have a continuous representant). W^1,1[a,b]=AC[a,b]. | |
| May 21, 2010 at 18:48 | comment | added | Tom Boardman | But isn't, say x^(-2) in W^{1,1}_loc(R)? | |
| May 21, 2010 at 15:57 | comment | added | Pietro Majer | I'm making a bit stronger statement: if f is in W^{1,1}_loc(R^n) and v is in R^n, then, for a.e. x in R^n, the function of 1 real variable u(t):=f(x+tv) is (in W^{1,1}_loc(R) hence) continuous : not just a.e. continuous (in particular this implies that f has a.e. partial derivatives, which is the theorem you referred to, I think). This way we can easily conclude. Actually this argument should be just a variation of yours. | |
| May 20, 2010 at 16:09 | comment | added | Tom Boardman | Is your alternative arument saying: "If it were in W^{1,1}_loc(R^n) it would be a.e cts wrt sections, but indicator functions are not QED" because if so, I'm not sure that indicator functions are in general not a.e. cts. | |
| May 20, 2010 at 13:31 | comment | added | Pietro Majer | Excuse me Tom, what do you mean ? | |
| May 19, 2010 at 16:18 | comment | added | Tom Boardman | But say \chi_([0,1]) in L^2(R) would be a.e. cts wouldn't it? Or am I dropping another x^2 in H^1 clanger? | |
| May 19, 2010 at 6:37 | comment | added | Pietro Majer | Another argument would be using the fact that any function in W^{1,1}_loc(R^n) has a.e. continuous 1 dimensional sections (in fact, for all v in R^n and a.e. x in R^n, the 1 variable function u(t):=f(x+ty) is W^{1,1}_loc(R).) de hoc satis. | |
| May 19, 2010 at 5:40 | comment | added | Pietro Majer | @Tom: By singularity I really mean blow up of the function. What theorem do you mean? Certainly not about differentiability a.e.; Rademacher's thm only works in W^1,infty. Oh, maybe by "differentiable a.e." you mean just having all partial derivatives a.e.; I see. | |
| May 18, 2010 at 18:41 | comment | added | Tom Boardman | @Spencer- It does. You just need to use the fact that there is an open ball which intersects with the set you are indicating in a set of measure zero (this is where your function 'goes up') and then let it 'roll down' over a subset of positive measure of the set you are indicating. | |
| May 18, 2010 at 17:53 | comment | added | Tom Boardman | @Pietro- if it's in H1 it will be a.e. differentiable- even the stuff with a dense set of singularities has that going for it. Don't worry- it's definitely a theorem ;) | |
| May 18, 2010 at 17:48 | comment | added | Pietro Majer | Just an objection: if f is differentiable in a point, it should continuous, hence locally bounded in that point, which is not quite true in H<sup>1</sup> if n>1 (such an f may have a dense set of singularities). | |
| May 18, 2010 at 17:39 | comment | added | Spencer | Does this arguement avoid assuming that the boundary of the set is of measure zero? I might be overcomplicating things but it's something that worried me last time I thought about the problem. | |
| May 18, 2010 at 17:10 | history | edited | Tom Boardman | CC BY-SA 2.5 |
added 1 characters in body
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| May 18, 2010 at 17:04 | history | answered | Tom Boardman | CC BY-SA 2.5 |