Timeline for Distributional derivative of non continuously differentiable functions
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2013 at 3:39 | answer | added | Phil Zaretzki | timeline score: 1 | |
| Jun 3, 2011 at 12:38 | answer | added | paul garrett | timeline score: 6 | |
| Jul 5, 2010 at 13:40 | answer | added | Andrey Rekalo | timeline score: 3 | |
| Jul 5, 2010 at 12:48 | comment | added | Harald Hanche-Olsen | @Willie: No offence. We're in total agreement, using different words. | |
| Jul 5, 2010 at 9:38 | comment | added | Willie Wong | @Harald: I hope I didn't cause any offence! I was just trying to observe that, granting all our comments are essentially contained in Rudin, unless the OP comes back and clarifies his thoughts, the question itself is not terribly meaningful. (In the sense that if our comments were on the mark, it is a case of "go back and re-read the book", and if our comments were off the mark, then god knows what the actual question is.) | |
| Jul 4, 2010 at 23:57 | comment | added | Harald Hanche-Olsen | @Willie: I never claimed my observations to be other than pedestrian. (And I am separated from all my books besides – it is no surprise that Rudin says the same things, so long as they are right.) Rather, I see my comments as probes to sharpen up the question a bit. | |
| Jul 4, 2010 at 23:19 | comment | added | Willie Wong | @Harald: and also, it seems our rather pedestrian observations are all contained in Rudin already. Example 6.14 states that "if $f$ is left-continuous of bounded variation in $\Omega\subset\mathbb{R}$, then $f'$ exists a.e. and is in $L^1$. In this case the distribution corresponding to $f'$ is equal to the distributional derivative of $f$ iff $f$ is absolutely continuous." This of course brings to mind the immediate example of the Cantor function, whose "classical derivative" can be identified with 0, and whose dist. der. is singularly continuous w.r.t. Lebesgue measure. | |
| Jul 4, 2010 at 22:48 | comment | added | Willie Wong | @Harald: Ah, I see. Thanks for the clarification. I am still a bit befuddled as to what the actual question is (which I tried to prod out of the OP from my previous two comments), though. (Side remark: if the function is differentiable everywhere [not just almost], couldn't we just use the fundamental theorem of calculus and integrations by parts to say that the distributional and classical derivatives "agree"? Or am I missing something obvious?) (Also, is $x^2\sin(x^{-2})$ differentiable at the origin?) | |
| Jul 4, 2010 at 20:11 | comment | added | Harald Hanche-Olsen | @Willie: Yes. But the distributional derivative is by definition just a distribution, so it is already meaningless to talk about its pointwise values. Anyway, after a bit more thought I suppose that the question must be about functions that are differentiable everywhere, yet not absolutely continuous. Perhaps the simplest example of which would look like $x^2\sin(x^{-2})$? | |
| Jul 4, 2010 at 19:51 | comment | added | Willie Wong | ... my point being that even in the original question, the classical derivative of a $C^1$ function only coincides with the distributional derivative in the sense that the classical derivative is a (perhaps preferred) representative in a equivalence class; or in other words one can choose another function as the distributional derivative such that it differs from the "classical derivative" on a set of measure zero. | |
| Jul 4, 2010 at 19:44 | comment | added | Willie Wong | If $f$ is merely absolutely continuous, doesn't its classical derivative only exist almost everywhere? | |
| Jul 4, 2010 at 18:09 | comment | added | Harald Hanche-Olsen | If $f$ is merely absolutely continuous, its classical and distributional derivatives still coincide, I believe? I don't have Rudin at hand, so I don't know what his example entails. | |
| Jul 4, 2010 at 17:03 | history | asked | shuhalo | CC BY-SA 2.5 |