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Suppose $(X^m, g)$ is a closed Riemannian manifold of dimension $m$ with the following properties,

  1. There is a constant $\kappa$ such that $\kappa r^m \leq Vol(B(x, r)) \leq \kappa^{-1} r^m$ for every $r \in (0,1)$.

  2. For every $C^1$ function $f$, we have $(\int_X |f|^{\frac{2m}{m-2}})^{\frac{m-2}{m}} \leq C_S (\int_X |f|^2 + |\nabla f|^2)$.

  3. For every $C^1$ function $f$, we have $\int_X |f|^2 \leq C_P (\int_X |\nabla f|^2 + (\int_X f)^2)$.

Can we say that there is a positive constant $I$ depending only on $m, \kappa, C_S, C_P$ such that $\frac{Area(\partial \Omega)^m}{Vol(\Omega)^{m-1}} \geq I$ for every domain $\Omega$ satisfying $Vol(\Omega) \leq \frac{1}{2} Vol(X)$?

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mathoverflow.net/howtoask –  David Roberts Sep 19 '11 at 3:14
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I modified the problem. Now it should be clear. –  user17314 Sep 19 '11 at 16:31
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Thanks for rewriting it. I like the question. –  Deane Yang Sep 19 '11 at 17:36
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In 3) you forgot to add the assumption $\int_X f=0$. –  Rbega Sep 19 '11 at 17:46
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It is worth noting that this is essentially the same as asking whether the $L_2$ Sobolev constant determines a bound on the $L_1$ Sobolev constant. It is well known how to go in the opposite direction, but I don't believe I've ever seen any result in this direction. On the other hand, I don't believe I've seen any counterexample, either. –  Deane Yang Sep 19 '11 at 20:18
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