Timeline for Classical Derivative, Weak Derivative and Integration by Parts
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2012 at 12:18 | vote | accept | Tatin | ||
| Dec 18, 2012 at 2:46 | comment | added | Tatin | I'm sorry for the confusion. I have had the impression that the term "weak derivative" is used when the distributional derivative is a $L^1_{loc}$ function. I thought that this is a standard convention but, as it seems, I'm wrong. | |
| Dec 17, 2012 at 23:54 | comment | added | George Lowther | The definition of the weak derivative used here presupposes that it is locally integrable (by the looks of it), rather than just being a distribution. So $f$ must be of bounded variation on line segments and Lebesgue's differentiation theorem applies as mentioned by Terry Tao. | |
| Dec 17, 2012 at 20:47 | comment | added | Terry Tao | ... if one assumes Df is locally integrable (so that it becomes a distribution), then the answer is "no" in one dimension at least; see Proposition 25 of my notes terrytao.wordpress.com/2010/10/16/… . If instead one assumes that $D_w f$ is locally integrable, then the answer is again "no", from the Lebesgue differentiation theorem (Theorem 6, ibid). I haven't thought the higher dimensional case through, but perhaps one can leverage the 1D case using Fubini's theorem. | |
| Dec 17, 2012 at 20:41 | comment | added | Terry Tao | ... but $f$ is not assumed to lie in a Sobolev space with one degree of differentiability, but is instead merely assumed to lie in $L^1_{loc}$. In such a case, the weak derivative is a priori only defined as a distribution, which among other things means that hypothesis (3) is not well-formed without an additional hypothesis on either the weak derivative (to interpret it as a function or measure) or the strong derivative (to interpret it as a measure or distribution). ... | |
| Dec 17, 2012 at 20:17 | answer | added | Daniel Spector | timeline score: 4 | |
| Dec 17, 2012 at 18:42 | comment | added | Tatin | The way it is defined in the context of Sobolev spaces | |
| Dec 17, 2012 at 15:22 | comment | added | Daniel Spector | In what sense do we assume weak differentiability of $f$? As a distribution, measure, in a Sobolev space, etc.? | |
| Dec 17, 2012 at 12:10 | history | edited | Tatin | CC BY-SA 3.0 |
added 14 characters in body
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| Dec 17, 2012 at 9:59 | history | asked | Tatin | CC BY-SA 3.0 |