Timeline for Exercise 8.13 - Brezis
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 4, 2022 at 0:13 | comment | added | Yuval Peres | @IosifPinelis Thanks for explaining, indeed, we started from different definitions. | |
| May 3, 2022 at 3:21 | comment | added | Iosif Pinelis | @YuvalPeres : BTW, the definition of $W^{1,p}$ that you used in your answer differs from what I learned about 50 years ago as a student, which is the same as the one given in Wikipedia, which I think is the common one. I think the denseness of $C^1_c$ in $W^{1,p}$ is then a theorem, which of course shows that your definition is equivalent to what I think is the common one. | |
| May 3, 2022 at 3:11 | comment | added | Iosif Pinelis | @YuvalPeres : I think this is a well-known fact. E.g., Wikipedia has this: "In the one-dimensional problem it is enough to assume that the ${\displaystyle (k{-}1)}$-th derivative ${\displaystyle f^{(k-1)}}$ is differentiable almost everywhere and is equal almost everywhere to the Lebesgue integral of its derivative" (en.wikipedia.org/wiki/Sobolev_space#One-dimensional_case) | |
| May 3, 2022 at 3:06 | comment | added | Yuval Peres | @Iosif Pinelis That is correct, but this identity does not follow from the definition of the weak derivative in Sobolev space- it needs an additional argument or reference. Even after adding that, your proof is still likely to be shorter than mine. One place where this is discussed is the book by Evans and Gariepy, Theorem 4.9.1 | |
| May 3, 2022 at 2:41 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 3, 2022 at 0:40 | comment | added | Iosif Pinelis | @YuvalPeres : We can just re-define the function $u$ appropriately on a set of measure $0$ so as to make $u$ absolutely continuous -- and this is what I meant. I have now added the corresponding detail. I think this only simplifies the proof a bit. | |
| May 3, 2022 at 0:37 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 2, 2022 at 22:59 | comment | added | user253963 | I want to thank everyone | |
| May 2, 2022 at 22:14 | comment | added | Yuval Peres | The first identity in your answer, $$(D_hu)(x) =\frac{u(x+h)-u(x)}h =\int_0^1 ds\,u'(x+sh)$$ needs some justification, as $u$ is just a function in Sobolev space, and is formally just defined a.e. In particular, $u'$ is not a classical derivative. | |
| May 2, 2022 at 21:17 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |