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Let $\mathcal{M}$ be a compact smooth hypersurface in $\mathbb{R}^{n}$ with affine metric $h$ and affine volume measure $d\mu_{h}$. If $$\int_{\mathcal{M}}|\nabla f|^2_{h}d\mu_{h}$$ is small what can I say about $f~$? Sort of Sobolev type inequalities? Here $|.|_{h}^2$ denotes the norm with respect to $h$.

Edit: I can let $f$ be affine invariant also.

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maybe Poincare's inequality can give you some information –  Shaoming Guo Aug 16 '11 at 19:23
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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. –  Deane Yang Aug 16 '11 at 21:28
    
What does "$f$ be affine invariant" mean? –  Deane Yang Aug 16 '11 at 21:31
    
I assume the hypersurface to be convex. –  ShAs Aug 16 '11 at 22:11
    
@ Deane Yang I mean $f(x)=f(Ax)$ for an affine transformation. –  ShAs Aug 16 '11 at 22:43

1 Answer 1

Check out the extensive oevre of Yang, Zhang, Lutwak (Deane Yang is active here, and will shed more light, I hope).

Sharp affine Lp Sobolev inequalities Lutwak, D Yang… - Journal of Differential Geometry, 2002

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Alas, although we do prove affine Sobolev inequalities, I don't offhand have any immediate answer for the question above. I don't see any obvious connection between the integral above and our affine invariant Sobolev "norms". If I find the time, I'll try to think about it more. –  Deane Yang Aug 16 '11 at 21:13

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