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Let $(M,g)$ be a closed, Riemannian manifold of dimension $n>4$. Let $K$ be the best constant for the Sobolev inequality

$||u||^2_p \leq K \int_{{\Bbb{R}}^n} (\Delta u)^2 dx,$

where $p=\frac{2n}{n-4}$. Here we are assuming that $u \in H^2(\Bbb{R}^n)$, so constants are excluded. Then by work in Djadli et al. we have it that there exists a constant $B$ such that for any $\epsilon$ greater than zero, we have it that

$||u||_p^2 \leq (K+\epsilon) \int_M (\Delta u)^2 + B(|\nabla u|^2 + u^2) dv_g,$

for all $u \in H^2(M)$. My question is whether or not there is a $H^3(M)$ generalization of this result -- something like the following: for every $\epsilon > 0$ there exists a constant $B$ such that

$||u||_q^2 \leq (M + \epsilon) \int_M |\nabla \Delta u|^2 + B((\Delta u)^2 + |\nabla u|^2 + u^2) dv_g,$

for all $u \in H^3(M)$, where $q = \frac{2n}{n-6}$, $n>6$, and $M$ is the best constant for the Sobolev inequality on $\Bbb{R}^n$,

$||u||^2_q \leq M \int_{{\Bbb{R}}^n} |\nabla \Delta u|^2 dx.$

I tried to modify the proof given by Djadli et al., but it breaks down because $\nabla \Delta u$ will involve second order derivatives of the metric tensor.

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If $M$ is a closed manifold, then your first inequality is false. Just set $u$ equal to a nonzero constant function. –  Deane Yang Jun 27 '11 at 17:18
    
Deane: Right. There is another hypothesis needed. –  Viktor Bundle Jun 27 '11 at 20:28
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I have not checked the details but I believe that what you want is true and using the same proof. You just need to compute $[\nabla, \Delta]$ carefully to check that it is a zeroth order operator involving only the curvature tensor. –  Deane Yang Jun 27 '11 at 22:40
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Also useful is the observation that $|\Delta|\nabla f|| \le |\Delta\nabla f|$, so you can get the inequality you want by applying the second displayed equation with $u$ replaced by $|\nabla u|$. –  Deane Yang Jun 28 '11 at 9:02

1 Answer 1

The desired inequality is correct. It is easily generalized from Aubin's proof (which is given in Lee&Parker's "Yamabe Problem") of the classical Sobolev inequality for Riemannian manifolds. One simply use a partition of unity argument to transfer the $\Bbb{R}^n$ result to the compact manifold. This requires numerous applications of the Cauchy-Schwarz inequality and the Cauchy inequality after one expands the derivatives of the product of the function with the partition functions. The key idea is that you can estimate all of the second, first, and zero order terms in a crude fashion, because the third order term is the only one that needs to be estimated carefully. This process seems like it could be generalized easily to get similiar Sobolev inequalities for $H^k(M)$ where $k>3$.

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This should work but I find it much easier and cleaner to work without co-ordinates and use formulas for commuting covariant derivatives (which is curvature more or less by definition), as well as the fact that the metric is covariantly constant. –  Deane Yang Jun 28 '11 at 9:03

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