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On a compact (without boundary) Riemannian manifold (eg. some surface in $\mathbb{R}^n$), I'm looking for a result like $$\lVert \nabla u\rVert_{L^2}^2 \leq \epsilon\lVert u\rVert_{H^1}^2+\epsilon^{-1}\lVert u\rVert_{L^2}^2$$ for $u \in H^2,$ where $\epsilon$ is of my choosing.

Does this or something similar hold (eg. instead, the inequality can have the LHS in $H^1$ norm and the $H^1$ norm on the RHS can be the $H^2$ norm.)

I saw something like this in Adams but this requires the domain to satisfy some unpleasant conditions. Thanks.

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If I understood correctly, what you are looking for is interpolation inequalities. The question is more suitable for MSE. – timur Jan 7 '13 at 19:20
If $\|u\|_{H^1} = \|\nabla u\|_{L^2} + \|u\|_{L^2}$, I don't see how this could be true. – Deane Yang Jan 7 '13 at 20:00
If you put the $H^2$ norm on the right hand side instead of the $H^1$ norm, this is Ehrling's lemma, which is well known. – Michael Renardy Jan 7 '13 at 20:03
But if you mean to use the $H^2$ norm on the RHS, then this is indeed an interpolation inequality. It indeed is probably in Adams for function on a domain in $R^n$, but this can be transferred to a manifold using co-ordinate charts and a partition of unity. – Deane Yang Jan 7 '13 at 20:05
Chris, any domain with smooth boundary satisfies the cone condition. You would cover the manifold by open sets that are diffeomorphic to a ball and apply the result in Adams to the function multiplied by the appropriate cutoff function restricted to one of the open sets and pulled back to the ball. – Deane Yang Jan 7 '13 at 20:24

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