Timeline for Sobolev type inequalities involving affine metric
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2011 at 16:01 | comment | added | Deane Yang | You still need to figure out what you really mean by "makes the above integral affine invariant". As far as I can tell, it already is for an arbitrary function $f$. | |
| Aug 17, 2011 at 13:58 | comment | added | ShAs | I am intrested in the general case I was thinking taking special functions like affine invariant ones may still be interesting and make the computation easier and it makes the above integral affine invariant. | |
| Aug 17, 2011 at 3:35 | comment | added | Deane Yang | I'm still confused by whether you want to get an inequality for an arbitrary function $f$ on the hypersurface or for a specific function $f$ that you have in mind. If it is the latter, then you can take advantage of the specific properties of $f$. For example, if $f$ is defined in terms of the support function and the affine curvature, then the integrand can be written purely in terms of the support function and its derivatives up to order 3. You might want take advantage of this information. | |
| Aug 17, 2011 at 1:06 | comment | added | ShAs | I mean functions like affine support function. I wanted to have that integral as an affine invariant quantity. | |
| Aug 17, 2011 at 0:34 | comment | added | Deane Yang | But what does that mean? In general, if $x$ lies on the hypersurface, $Ax$ does not. And if you allow $f$ to be defined on all of $R^n$, then $f(x) = f(Ax)$ implies that $f$ is constant. | |
| Aug 16, 2011 at 22:43 | comment | added | ShAs | @ Deane Yang I mean $f(x)=f(Ax)$ for an affine transformation. | |
| Aug 16, 2011 at 22:11 | comment | added | ShAs | I assume the hypersurface to be convex. | |
| Aug 16, 2011 at 21:31 | comment | added | Deane Yang | What does "$f$ be affine invariant" mean? | |
| Aug 16, 2011 at 21:28 | comment | added | Deane Yang | If you don't assume the hypersurface to be convex, then the metric is not necessarily positive definite, In that case, I don't know much. If the hypersurface is assumed to be convex, then you are just trying to prove a Sobolev inequality for a Riemannian metric on the sphere, where metric happens to come from the affine structure. In that case, the usual way to prove a Sobolev inequality is to prove an isoperimetric inequality. So you want to figure out if the fact that the metric is the affine hypersurface metric implies an isoperimetric inequality. | |
| Aug 16, 2011 at 21:24 | history | edited | ShAs | CC BY-SA 3.0 |
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| Aug 16, 2011 at 20:19 | history | edited | ShAs | CC BY-SA 3.0 |
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| Aug 16, 2011 at 20:13 | answer | added | Igor Rivin | timeline score: 1 | |
| Aug 16, 2011 at 19:23 | comment | added | Shaoming Guo | maybe Poincare's inequality can give you some information | |
| Aug 16, 2011 at 19:03 | history | asked | ShAs | CC BY-SA 3.0 |