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Let $(M,g)$ be a complete Riemannian $m$-manifold, with bounded geometry and $m\geq2$. Suppose $Ric\geq(n-1)\kappa$. Let $B_p(r)$ be a geodesic open ball.

Q Can we find a constant $C=C(\kappa,r,m)$(independent on the point $p$), such that $$ \left(\frac{1}{Vol(B_p(r))}\int_{B_p(r)}\phi^{2^*}\right)^{\frac1{2^*}}\leq C \left(\frac{1}{Vol(B_p(r))}\int_{B_p(r)}|\nabla\phi|^2\right)^{1/2}, $$ for all compactly supported function $\phi$ on $B_p(R)$.

PS: I think it is proved in someone's paper or book, could anyone give me a reference?

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  • $\begingroup$ Federer-Fleming proved that the best local Sobolev constant is equal to the best constant for the local isoperimetric inequality. A local isoperimetric inequality is known to hold in a neighborhood of a point, where the constant depends on a lower bound of Ricci and a lower bound, as a function of $r < 0$, on the volume of a geodesic ball of radius $r$ centered at that point. Alas, I don't remember the references for this. $\endgroup$
    – Deane Yang
    Commented Jan 23, 2019 at 20:58
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    $\begingroup$ @DeaneYang Relative isoperimetric inequality for Ricci bounded below was proved by Buser. Buser's inequality is equivalent to a Poincare inequality. Ricci implies that the measure is doubling (Bishop–Gromov comparison theorem) and doubling plus Poincare implies the right Sobolev inequality. However, I am not sure if this gives the best constant as in the theorem stated in my answer. $\endgroup$
    – Piotr Hajlasz
    Commented Jan 23, 2019 at 23:44
  • $\begingroup$ @PiotrHajlasz, my guess is that the best constant corresponds to the constant curvature cases. The exponential factor seems to arise from the volume of a hyperbolic ball with radius $r$. But this is all speculation. $\endgroup$
    – Deane Yang
    Commented Jan 24, 2019 at 1:39

1 Answer 1

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The following result is Theorem 1.1 in:

P. Maheux, L. Saloff-Coste, Analyse sur les boules d'un opérateur sous-elliptique. Math. Ann. 303 (1995), 713–740.

Theorem. Let $(M,g)$ be a complete Riemannian $n$-manifold, such that $Ric\geq kg$ for some $k\in\mathbb{R}$. Let $1\leq p<n$ and $p^*=np/(n-p)$. Then there are constants $C=C(p)$ and $A=A(p)$ such that for every $r>0$ and $\phi\in C^\infty(B(x,r))$ we have $$ \left(\int_{B(x,r)}|\phi-\phi_{B(x,r)}|^{p^*}\right)^{1/p^*}\leq A e^{C(1+\sqrt{|k|}r)}r\, Vol_g(B(x,r))^{\frac{1}{p^*}-\frac{1}{p}} \left(\int_{B(x,r)}|\nabla\phi|^p\right)^{1/p}, $$ where $$ \phi_{B(x,r)}=\frac{1}{Vol_g(B(x,r))}\int_{B(x,r)}\phi. $$

It follows that under the above assumptions, if $\phi\in C_0^\infty(B(x,r))$, then $$ \left(\int_{B(x,r)}|\phi|^{p^*}\right)^{1/p^*}\leq A e^{C(1+\sqrt{|k|}r)}r\, Vol_g(B(x,r))^{\frac{1}{p^*}-\frac{1}{p}} \left(\int_{B(x,r)}|\nabla\phi|^p\right)^{1/p}. $$ (Perhaps with differnet costants $A$ and $C$.)

The key point is that the measure on manifolds with bounded Ricci curvature satisfies so called doubling condition and that such manifolds satisfy the Poincare inequality due to Buser. For more information about how the Poincaré inequality and doubling condition imply the Sobolev inequality in a very general setting see:

P. Hajłasz, P. Koskela, Sobolev met Poincaré. Mem. Amer. Math. Soc. 145 (2000), no. 688.

For a related question see; Sobolev and Poincare inequalities on compact Riemannian manifolds

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    $\begingroup$ Should Pekka be in the title, as opposed to being your co-author's name? $\endgroup$
    – mme
    Commented Jan 23, 2019 at 2:23
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    $\begingroup$ @MikeMiller Thank you for pointing this out. That is a hilarious typo. $\endgroup$
    – Piotr Hajlasz
    Commented Jan 23, 2019 at 2:25
  • $\begingroup$ @PiotrHajlasz, see my comment above. The theorem you quote appears to be a sharper version of the local Sobolev inequality known before. $\endgroup$
    – Deane Yang
    Commented Jan 23, 2019 at 21:02

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