5
$\begingroup$

Let $N$ be a smooth manifold without boundary of dimension $n$. $M$ is a manifold with Lipschitz boundary if $M \subset N$, $M$ and $N$ are of the same dimension, and in the charts of $N$, the boundary of $M$ is Lipschitz.

The Nash inequality (sometimes also called $L^2$-type Nash inequality) reads $$ ||f||_2^{2+\frac{4}{n}} \le (C_1||f||_2^2+C_2||\nabla f ||_2^2)||f||_1^{\frac{4}{n}}$$ It is closely related to the Gagliardo-Nirenberg-Sobolev inequality $$||f||_p^2 \le C_1||f||_2^2+C_2||\nabla f||_2^2$$ where $\frac{1}{p}+\frac{1}{n}=\frac{1}{2}$.

The Sobolev inequality only makes sense if the dimension is at least $3$ and it holds for manifolds with Lipschitz boundary, see for example Aubin's book 'Nonlinear analysis on manifolds'.

If the dimension is at least $3$, the Nash and Sobolev inequality are equivalent to each other, see for example Carlsen, Kusuoka, Stroock 'Upper bounds for symmetric Markov transition functions'.

This means the only remaining case is dimension $2$. The Nash inequality holds for smooth manifolds without boundary, see for example Hebey 'Sobolev spaces on Riemmanian manifolds'.

Some papers like Humbert 'Best constants on the $L^2$-type Nash inequality' say that the Nash inequality also holds for manifolds with boundary, but I assume this means smooth boundary and the proof just says something like 'analogous to the Sobolev inequality'. His theorem also requires an injectivity radius bounded from below, which using the classic definition, does not make sense on manifolds with boundary (I had a question on injectivity radius on manifolds with boundary here recently, the upshot was that this is a nontrivial issue with multiple distinct reasonable definitions).

Edit: I forgot to mention that the manifolds are compact. Most of the theorems quoted also hold for noncompact manifolds with some suitable global curvature bounds but the compact case is all I'm interested in.

$\endgroup$
2
$\begingroup$

I think the answer to my question is yes, proved using a version of Stein's extension theorem. Theorem: Let $M$ be a manifold with Lipschitz boundary sitting inside $N$. Then there exists an extension operator E from functions on $M$ to functions on $N$ such that

a) $Ef|_{M}=f$, that is $E$ is an extension

b) $E: W^p_k(M) \rightarrow W^p_k(N)$ continuously for all $1 \le p \le \infty$ and all non-negative integral $k$ where $W^p_k$ is the Sobolev space of weak derivatives up to order $k$ lying in $L^p$

This is proved in Stein's book 'singular integrals and differentiability' for domains sitting in $\mathbb{R}^n$, the proof already uses charts, carrying it over to manifolds is fairly straightforward.

Using this theorem and the fact that the Nash inequality holds on the $N$, one can prove the Nash inequality on $M$ with a simple chain of inequalities. \begin{align*} &||f||_2^{2+\frac{4}{n}} \\ \le &||Ef||_2^{2+\frac{4}{n}} \\ \le & (C_1||Ef||_2^2+C_2||\nabla Ef ||_2^2)||Ef||_1^{\frac{4}{n}} \\ \le & \max(C_1,C_2) ||Ef||_{W_1^2}^2 ||Ef||_1^{\frac{4}{n}} \\ \le & \max(C_1,C_2)C^2 ||f||_{W_1^2}^2 ||f||_1^{\frac{4}{n}} \\ \le &(\max(C_1,C_2)C^2 ||f||_2^2+\max(C_1,C_2)C^2 ||\nabla f ||_2^2)||f||_1^{\frac{4}{n}} \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.