Nash inequality on a compact domain? - MathOverflow

Nash inequality on a compact domain?

2012-10-03T15:28:11Z

I have come across a few papers that make use of the Nash inequality for functions on a compact domain. Unfortunately, nobody cites a reference for the proof of this result. Is going from the classical Nash inequality on $\mathbb{R}^n$ to that on compact domains so trivial?

I'd really appreciate any references you know of.

EDIT: This is the statement I am looking for

Let $\mathcal{D} = \mathbb{T}^n$ be the unit square in $\mathbb{R}^n$ with periodic boundary conditions. There exist constants $C_1$ and $C_2$ such that such that for $f \in H^1(\mathcal{D})$ then $$||f||_{2}^{1 + \frac{n}{2}} \leq ||f||_1 \left(C_1||f||_{2}^2 + C_2||\nabla f||_2^2\right)^{\frac{n}{4}}$$

Answer by Luis Silvestre for Nash inequality on a compact domain?

2012-10-03T17:40:54Z

The original Nash inequality in $\mathbb R^d$ is $$\|\nabla f\|_2 \|f\|_1^{2/d} \geq c \|f\|_2^{1+2/d}$$ It is proved in this article: Nash, J. Continuity of solutions of parabolic and elliptic equations. Amer. J. Math. 80 1958 931–954.

The inequality is proved by the beginning of the paper. You can see it by the top of page 936. It is a very simple argument using the Fourier transform, so you can try to work it out in your periodic setting using Fourier series.

Note that the function $f \equiv 1$ fails the original Nash inequality in a compact domain. The extra term you have in yours is a correction for the compact case that is not necessary in the full space $\mathbb R^d$