Timeline for Second differential of total variation
Current License: CC BY-SA 4.0
11 events
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| Jun 10, 2022 at 7:50 | comment | added | Marko Rajkovic | @Dirk I have just checked and reached the same conclusion. I will check your book. I am aware of the "A pointwise characterization of the subdifferential of the total variation functional" by Bredies and Holler, which is a detailed extension of the above argument that locally $\partial TV(u)=-\text{div}(\frac{Du}{|Du|})$. Our first goal would be to, by observing smooth case, get a motivation for such local representation of the second differential. | |
| Jun 10, 2022 at 7:12 | comment | added | Dirk | I was wrong - Evans, Gariepy do not treat the differential of the TV seminorm (but kind of second derivatives of BV functions). Also the subdifferential is a bit more complicated - you can find a thorough discussion in the book "Mathematical Image Processing" (Secion 6.3.3) by Kristian Bredies and myself. We do not treat a "second differential" though. | |
| Jun 9, 2022 at 21:16 | comment | added | fedja | @MarkoRajkovic You understand correctly, but note that I changed the functional to something non-linear to get an interesting 1D example (the absolute value of the gradient in 2D and up is also non-linear and that's where all the trouble comes from). The point was that you will have no more luck getting rid of the derivatives in your formula than in the much simpler $\int u'v'$ 1D formula and there is no way to do it with the latter. | |
| Jun 9, 2022 at 20:40 | comment | added | Marko Rajkovic | @fedja If I see correctly, in 1D $TV(u)=\int u'$ so that $\partial TV(u)(v)=\int v'$ and $\partial^2 TV(u)(v,w)=0$. | |
| Jun 9, 2022 at 19:40 | comment | added | fedja | That's hopeless. Even if you are in 1D and your functional is $\int (f')^2$, then the first variation is $2\int f'u'=-2\int f''u$ but the second one is $\int u'v'$ and you cannot get rid of derivatives on both $u$ and $v$ no matter how hard you try: you can only kill one at the expense of increasing the order of the other one. | |
| Jun 9, 2022 at 19:05 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting
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| Jun 9, 2022 at 18:03 | comment | added | Marko Rajkovic | Thank you for the reference. I will give it a check. The expression is correct, but I want to be able to have it depending on v and w solely, and not their derivatives. | |
| Jun 9, 2022 at 18:00 | comment | added | Dirk | And what is wrong with you expression? It should be bilinear in (v,w) and it is. | |
| Jun 9, 2022 at 17:59 | comment | added | Dirk | There is something in "Measure theory and fine properties of functions" by Evans and Gariepy. if I remember correctly. | |
| Jun 9, 2022 at 17:41 | history | edited | Marko Rajkovic | CC BY-SA 4.0 |
specific case of interest
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| Jun 9, 2022 at 16:28 | history | asked | Marko Rajkovic | CC BY-SA 4.0 |