Timeline for Degree of a function in $H^{\frac{1}{2}}(\mathbb{S}^1,\mathbb{S}^1)$
Current License: CC BY-SA 3.0
6 events
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| Sep 4, 2013 at 15:01 | comment | added | Gatz' | Thank you for your help. But I still don't understand why a function $f \in H^{\frac{1}{2}}$ admits a weak derivative $\frac{\partial f}{\partial \theta}$ in $H^{-\frac{1}{2}}$. I believe it has something to do with the Fourier series characterisation of $H^s$ (and the duality between $H^{\frac{1}{2}}$ and $H^{-\frac{1}{2}}$), and I would be thankful for any good reference on the subject. | |
| Sep 3, 2013 at 15:52 | comment | added | Terry Tao | The functions $f$ and $\partial f/\partial\theta$ only need to lie in $H^{1/2}(S^1,C)$ and $H^{-1/2}(S^1,C)$ (rather than $H^{1/2}(S^1,S^1)$ and $H^{-1/2}(S^1,S^1)$) to define an inner product $\int_{S^1} \overline{f}\partial f/\partial \theta\ d\theta$, and this is immediate from the Fourier series characterisation of $H^s(S^1,C)$ (as given in the reference I gave above). | |
| Sep 3, 2013 at 5:24 | comment | added | Gatz' | Unfortunately, they do not explain why $f^{-1} \in H^{\frac{1}{2}}$ and $\frac{\partial f}{\partial \theta} \in H^{-\frac{1}{2}}$. I checked some of the references and they all say the same thing, arguing that the integral is well defined as a scalar product in the duality between $H^{\frac{1}{2}}$ and $H^{-\frac{1}{2}}$. I believe I missed something about that. | |
| Sep 3, 2013 at 4:58 | comment | added | Terry Tao | This survey of Brezis appears to comprehensively answer these sorts of questions: seminariostalca-santiago.bligoo.com.br/media/users/3/178983/… (see in particular Section 3). | |
| Sep 2, 2013 at 23:11 | review | First posts | |||
| Sep 2, 2013 at 23:56 | |||||
| Sep 2, 2013 at 22:52 | history | asked | Gatz' | CC BY-SA 3.0 |