Timeline for Classical Derivative, Weak Derivative and Integration by Parts
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2012 at 12:18 | vote | accept | Tatin | ||
| Dec 18, 2012 at 10:00 | comment | added | Daniel Spector | From my question and your response, I understood that you meant a weak derivative which was actually a function (since if you are studying this for the first time, the subtleties you encounter are often overlooked until later when you realize it is a game and a dance of being careful and having the right idea). My above response is in this context, and is a proof that they must agree, up to a set of Lebesgue measure zero. Implicit in my proof is a result that Sobolev functions are approximately differentiable, which uses essentially the Lebesgue differentiation theorem. | |
| Dec 18, 2012 at 2:47 | comment | added | Tatin | Let me say it again: 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 21:18 | history | edited | Daniel Spector | CC BY-SA 3.0 |
Used more math notation; added 8 characters in body
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| Dec 17, 2012 at 20:58 | comment | added | Bazin | According to the distribution definition of the weak derivative of an $L^1_{loc}$ function, you have with $\phi\in C^\infty_c(U)$ $$ \langle \nabla_w f,\phi\rangle_{\mathscr D'(U),\mathscr D(U)}=-\int f \nabla \phi dx. $$ On the other hand you assume that the function $f$ is differentiable on $U$ with gradient $\nabla f$. Now the formulation of your question is not really meaningful: $\nabla_w f$ is defined weakly whereas $\nabla f$ has a point wise definition. So the meaning of equality or difference on a set of positive measure does not mean anything. | |
| Dec 17, 2012 at 20:17 | history | answered | Daniel Spector | CC BY-SA 3.0 |