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Today I was told by my tutor that on closed manifolds like $\mathbb{S}^2$ (we can stick to $\mathbb{S}^2$ here) the following Sobolev inequality holds (if the right-hand side exists, the left-hand side does too.)

$$||f||_p^2 \le a || \nabla f||^2_2 + b ||f||^2_2$$

for any $p \in (2, \infty)$ and some $a,b \ge 0$ depending on $p$. Note, that the $p-norms$ are taken w.r.t. the surface measure on the sphere.

I know that the Sobolev inequality enables us to lift $f \in L^2$ to some $f \in L^p$ but here he claimed that it is possible to lift it to all $p \in (2, \infty)$ which is rather unbelievable. I did not manage to show this, but was wondering if anybody of you knows how to do this?

user167575
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That's preculiar to $\mathbb S^2$, or more precisely, two dimensional compact Riemannian manifolds $M$ only. Recall that for $p<2$, the conjugate $p^*$ is given by

$$p^* = \frac{2p}{2-p}.$$ The usual Sobolev inequality is

$$\begin{split} \| u\|_{L^{p^*}} &\le C(p)(\|u\|_p + \| \nabla u\|_p) \\ &\le \tilde C(p) (\|u\|_2 +\|\nabla u\|_2) \end{split}$$

by applying Holder inequality as $p<2$. Note that for all $q\in (2, \infty)$, there is $p<2$ so that $q = p^*$. Thus the fabulous one is really the usual one with Holder inequality.

One important remark is that the constant blows up to $\infty$ as $p\to \infty$: If not, all $W^{1, 2}$ functions will have uniformly $L_p$ bound for all $p$ and that would imply $W^{1, 2} (M)\subset L^\infty(M)$, which is not true.

  • just a question, you said that $| u|{L^{p^*}} \le C(p)(|u|_p + | \nabla u|_p)$ is the standard sobolev inequality, but isn't $| u|{L^{p^*}} \le C(p) | \nabla u|_p $ actually the standard one and you just added an additional term or do we really have to add there the term $||u||_p$? Cause I couldn't find your version in the wikipedia article https://en.wikipedia.org/wiki/Sobolev_inequality – user167575 Oct 15 '15 at 22:28
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    On a manifold, you have to use the Sobolev imbedding on manifold. You can find it in Aubin's Nonlinear analysis on manifold. One can see how that $|u|_p$ comes out at the right: One need to apply the usual inequality to each coordinate chart, so we use a partition of unity ${\phi_i}$ and apply the usual one to each $\phi_iu $ (which is now of compact support). So $|\phi_i u|_q\le C |\nabla(\phi_i u)|_p$. Using product rule on the right hand side, one sees that $|u|_p$ would come out. @user167575 –  Oct 16 '15 at 01:14
  • thank you that was very helpful. By the way you said that $W^{1,2}(M) $ cannot be a subset of the essentially bounded functions. Did you have a particular example in mind that shows that this cannot hold? Cause I thought about it and could not find one. – user167575 Oct 16 '15 at 12:26
  • @user167575 See here: http://math.stackexchange.com/questions/222418/example-of-a-discontinuous-and-bounded-function-for-the-limiting-case-w1-n –  Oct 16 '15 at 14:03