Timeline for Finding an optimal $p$ such that $u \in L^p$
Current License: CC BY-SA 3.0
5 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 21, 2013 at 7:29 | comment | added | Willie Wong | @math: translate along the $x$ axis. the $L^2$ of $y\partial_x \phi^\lambda_{\alpha\beta}$ remains invariant if you translate along the $x$ axis. And by choosing sufficiently fast decreasing supports, you can fit all of them inside $|x| < R$ for some finite $R$, making $x\partial_y \phi^\lambda_{\alpha\beta}$ bounded. | |
| Mar 21, 2013 at 0:00 | comment | added | mathenthusiast | on the $L^2$-norm of $y\partial_{x}\phi^{\lambda}_{\alpha\beta}$? What I am missing here? | |
| Mar 20, 2013 at 23:56 | comment | added | mathenthusiast | @Willie Wong Hi Willie, if I understand this correctly, in the last step you mean that we will get a function $u$ with countably many components of its support, with $u = \frac{1}{2^n}\phi^{\lambda}_{\alpha\beta}$ on each support, with $\lambda$ increasing and $n$ chosen suitably for each $\lambda$. This implies $u$ remains in $H^{2/3}$ and $\partial_{y}u$ in $L^2$. However, the way I am thinking, to get these disjoint supports, you have to push (translate) the original $\phi^{\lambda}_{\alpha\beta}$ with their supports in the plane. But does it readily follow that we still have a control | |
| Mar 19, 2013 at 21:20 | history | answered | Willie Wong | CC BY-SA 3.0 |