# Nash inequality on a compact domain?

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 $$||f||_{2}^{1 + \frac{n}{2}} \leq ||f||_1 \left(C_1||f||_{2}^2 + C_2||\nabla f||_2^2\right)^{\frac{n}{4}}$$

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Could you provide a precise statement of the inequality you want a reference for? – Deane Yang Oct 3 '12 at 16:50
In particular, if the functions on the compact domain are assumed to vanish on the boundary, then the inequality on $\mathbb{R}^n$ applies directly. – Deane Yang Oct 3 '12 at 16:58
@Deane: You're right I should have specified that the functions should be periodic. – RadonNikodym Oct 3 '12 at 17:19
You're effectively asking for the Nash inequality on a closed manifold. Cover the manifold by co-ordinate charts, fix a partition of unity subordinate to these charts. The inequality on the manifold then follows by applying the inequality on $\mathbb{R}^n$ on each co-ordinate chart to each compactly supported piece of the function and adding everything up. – Deane Yang Oct 3 '12 at 17:33
The Nash inequality is a particular case of the Gagliardo-Nirenberg inequalities, so you can look up the proofs for that. – Deane Yang Oct 3 '12 at 17:57

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 $\mathbb R^d$

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@Luis, thanks for your information. Yes, the original Fourier transform doesn't seem to translate very well to fourier series. There is another proof of the Nash inequality which I found in a paper of Stroock proved using estimates on the solution of the Heat equation. I'm attempting to adapt this one. – RadonNikodym Oct 3 '12 at 20:39

Sometimes one has to roll up his sleeves and get his hands dirty in analysis.

So here is the estimate you need:

$$\|f\|_2^2=(\sum_{|\xi|\le R}+\sum_{|\xi|>R}) |\hat f(\xi)|^2\ll_n R^n\sup_{\xi} |\hat f(\xi)|^2 + R^{-2}\sum_{|\xi|>R} |\xi|^2|\hat f(\xi)|^2$$

which in turn is bounded by

$$\|f\|_2^2\ll_n R^n \|f\|_1^2+R^{-2}\|\nabla f\|_2^2.$$

Now taking $R$ to be optimal a la Nash, you get

$$\|f\|_2\ll_n \|f\|_1^{\frac{2}{n+2}}\|\nabla f\|_2^{\frac{n}{n+2}}$$

which is even stronger than your inequality (note we can dispense with the term $\|f\|_2^2$ on the RHS.

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