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edited Jun 27 2011 at 20:30

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.

changed reference

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edited Jun 27 2011 at 14:30

Viktor Bundle
999●1●8

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 $p=\frac{2n}{n-4}$. Then by work in Ledoux 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 Ledoux Djadli et al., but it breaks down because $\nabla \Delta u$ will involve second order derivatives of the metric tensor.

tex

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edited Jun 27 2011 at 1:13

Viktor Bundle
999●1●8

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 $p=\frac{2n}{n-4}$. Then by work in Ledoux 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 $q = \frac{2n}{n-6}$, $n>5$, 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 Ledoux et al., but it breaks down because $\nabla \Delta u$ will involve second order derivatives of the metric tensor.