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Brézis-Wainger Inequality on Riemannian Manifolds
Journal of Inequalities and Applications volume 2008, Article number: 715961 (2008)
The Brézis-Wainger inequality on a compact Riemannian manifold without boundary is shown. For this purpose, the Moser-Trudinger inequality and the Sobolev embedding theorem are applied.
There is no doubt that the Brézis-Wainger inequality (see ) is a very useful tool in the examination of partial differential equations. Namely, a lot of estimates to a solution of PDE are obtained with the help of the Brézis-Wainger inequality. Especially, the inequality is often applied in the theory of wave maps.
In this paper, we extend the Brézis-Wainger result onto a compact Riemannian manifold. We show the following theorem.
Let be an -dimensional compact Riemannian manifold, and , , for , where is a positive integer and is a real number. Then, and
where is a positive constant.
The proof relies on the application of a Moser-Trudinger inequality (see Theorem 2.2) and the Sobolev embedding theorem (see Theorem 2.1). Moreover, we will use the integral representation of a smooth function via the Green function (see ).
In order to make this paper more readable, we recall some definitions and facts from the theory of Sobolev spaces on Riemannian manifolds. In particular, we present useful inequalities and embeddings.
Let be a smooth, compact Riemannian manifold without boundary. We will denote by a space of smooth real functions. For and integer , we denote by the m th covariant derivative of . Next, for and for a fixed integer and a real , we set
where by we have denoted the Riemannian measure on the manifold .
We define the Sobolev space as a completion of with respect to .
We close this section stating the following results, which will be used in the proof of the main result.
Let be a smooth, compact Riemannian -manifold. Then, for any real numbers and any integers , if , then . Moreover, there exists a constant such that for all , the following inequality holds:
Theorem 2.2 (Moser-Trudinger Inequality ).
Let be a smooth, compact Riemannian -manifold and a positive integer, strictly smaller than . There exist a constant and such that for all with and , the following inequality holds:
3. Proof of the Main Result
In this section, we will prove the main result, that is, Theorem 1.1.
Let us notice that by the assumptions we have an embedding . First of all, we show the following lemma.
If , , and , then . Moreover, the following estimate holds:
Let us put
in Young's inequality
where . Then, we obtain an estimate
where is a constant from the Moser-Trudinger inequality (see Theorem 2.2).
Next, we can estimate
Subsequently, we can apply the Moser-Trudinger inequality to the right-hand side of the above inequality:
From this, the proof of Lemma 3.1 follows.
Now, we can go back to the proof of Theorem 1.1. First, we prove the theorem for such that and . Replacing by if necessary, we may suppose that
Let us recall (see ) that for , a compact Riemannian -manifold, there exists a Green function such that
(1)for any and any ,
where is a Riemannian volume of the manifold , and is the Laplace-Beltrami operator on a manifold;
(2)there exists a constant such that for any ,
where is a diagonal:
and is a Riemannian distance from to .
Let us define the map . Next, we apply the first property of the Green function to the map and to the point . Namely,
Subsequently, by the second property of the Green function we can estimate as follows:
Now, we will try to estimate the last term in the inequality (3.12). Let us notice that if , then . Next, by the Sobolev theorem (see Theorem 2.1),
and we have that .
Using elementary calculations, one can easily show the lemma.
There exist and a finite such that the following equality holds:
where is the exponent from the Sobolev theorem.
By Hölder's inequality with exponents , we can estimate the inequality (3.12) as follows:
where and is the exponent from Lemma 3.2. Finally, by Lemma 3.1 and the Sobolev theorem, we obtain
Let us stress that since , the constant does not depend on . We can rewrite the inequality (3.16) as follows:
Taking into account , we obtain
This finishes the proof of the inequality in the case such that and . Subsequently, one can easily obtain the inequality for such that .
Now, we prove the theorem for an arbitrary such that . We apply the density argument. Namely, for any , there exists such that
Next, we define
Such has zero-mean value. Moreover, the following inequality holds:
Finally, we take . This completes the proof.
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The author wishes to thank Professor Yuxiang Li for pointing out the paper . Moreover, the author thanks the referees for comments and invaluable suggestions.
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Cite this article
Górka, P. Brézis-Wainger Inequality on Riemannian Manifolds. J Inequal Appl 2008, 715961 (2008). https://doi.org/10.1155/2008/715961
- Partial Differential Equation
- Riemannian Manifold
- Sobolev Space
- Green Function
- Integral Representation