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Symmetrization of Functions and the Best Constant of 1DIM Sobolev Inequality
Journal of Inequalities and Applications volume 2009, Article number: 874631 (2009)
Abstract
The best constants of Sobolev embedding of into for and are obtained. A lemma concerning the symmetrization of functions plays an important role in the proof.
1. Introduction
Let be a Sobolev space which consists of the functions whose derivatives up to vanish at , that is,
where denotes th derivative of in a distributional sense. The purpose of this paper is to investigate the best constant of Sobolev inequality
where and . The result is as follows.
Theorem 1.1.
The best constant of inequality (1.2) is
where satisfies and in (1.5) is the unique solution of the equation
satisfying .
For special values of , the solution of (1.6) can be written explicitly, and is expressed as follows.
Corollary 1.2.
One has
The best constants and were recently obtained by Oshime [1]. This paper gives an alternative proof which simplifies the derivation process of and and computes further constant . To compute these constants, the following lemma with respect to the symmetrization of functions plays an important role.
Lemma 1.3.
Let be an integer satisfying and let us define the functional as follows:
Then, for an arbitrary , there exists an element which belongs to the following subspace of :
such that
Remark 1.4.
The proof of this lemma (see Section 3) does not apply to the case , since the relation
may fail to hold for (see (3.4)–(3.6)). Hence the problem to obtain for (essentially) still remains.
The existence of the maximizer of can be seen in the proof of Theorem 1.1, where we construct it concretely, but here we would like to see this briefly though the proof of the following lemma.
Lemma 1.5.
Assume that the assertion of Lemma 1.3 holds, then the maximizer of exists.
Proof.
Let be sufficiently large, and let and be as
From Lemma 1.3, we see that if the maximizer exists, it is the element of . So, we can restrict the definition domain of to . Since is convex and strongly closed (by Sobolev inequality) in , it is weakly closed. In addition, is weakly compact, so is also weakly compact. Moreover, is weakly lowersemicontinuous in , and hence attains its minimum in . This proves the lemma.
Finally, we introduce some studies related to the present paper. When (Hilbertian Sobolev space case), the best constants for the embeddings of into for various conditions were treated in Richardson [2], Kalyabin [3], and [4–8]; see also references of these literatures. On the other hand, for the case , few literature seems to be available. In [9], Kametaka, Oshime, Watanabe, Yamagishi, Nagai, and Takemura obtained the best constant of (1.2) when belongs to a subspace of which consists of periodic functions
as
where is a Bernoulli polynomial, and is an unique solution of the equation
in the interval . Moreover, in [1], Oshime obtained the best constant and . Other topics on this subject, especially the best constant of Sobolev inequalities on Riemannian manifolds, are seen in Hebey [10].
2. Proof of Theorem 1.1
First, we prepare the following lemma.
Lemma 2.1.
Let and be a function satisfying
then, it holds that
where is an arbitrary constant (later, in Lemma 2.3, one fixes the value of to satisfy (1.6)).
Proof.
By integration by parts, we obtain the result.
From Lemma 1.3, to obtain the best constant of (1.2), we can restrict the definition domain of the functional to the nonzero element of . Now, let , then from Lemma 2.1 and Hölder's inequality, we have for ,
So, we have
and the equality holds in (2.4) if and only if there exists satisfying
To confirm the existence of such , we use the following lemmas.
Lemma 2.2.
Let satisfy
then the solution of
exists in .
Proof.
Let us define as
Clearly is and
Moreover, from the assumption, it holds that .
Lemma 2.3.
The solution of (1.6) uniquely exists.
Proof.
Let
Since
the assertion is proved.
Using Lemma 2.2 and 5, we obtain the following lemma.
Lemma 2.4.
Let be a solution of (1.6) (when ), then the solution of (2.5) belongs to for .
Proof.
First, we prove the case and . For simplicity, let us put . Note that in these cases is a continuous function on .
In the case , is an even function, so integration of over the interval vanishes. In addition,
holds, so the integration of over the interval also vanishes. Hence, by Lemma 2.2, the solution of (2.5) belongs to . Properties and follow from the fact that is an even function and (2.12). So, we have proven .
In the case, is an odd function, so integrations of and over the interval vanish. Moreover, from (1.6), the integration of over the interval also vanishes. Hence, again by Lemma 2.2, we have the solution of (2.5) which belongs to. The remaining part is the same as case (i).
In the case , let us define as follows:
Clearly it holds that satisfies (2.5) (a.e.), , and . To see , let be an arbitrary element of . Since
we have, in a distributional sense,
Therefore, . This proves the case .
Proof of Theorem 1.1.
From Lemma 1.3, and the argument of this section, especially Lemma 2.4, . This proves Theorem 1.1.
Proof.
In this case, (1.6) becomes
So, we can explicitly solve this equation with respect to . Substituting to (1.5), we obtain the result.
3. Proof of Lemma 1.3
Now, all we have to do is to prove Lemma 1.3.
Proof of Lemma 1.3.
To avoid the complexity of notation, in the follwings, we fix . Let be an arbitrary element of (), and let
Here, we can assume , since if it does not, can be replaced by .

(i)
There is the case
Let us define as
We have
when , (3.4) and (3.5) when , and (3.4) when . Further, let us define
Then , since , . Moreover, from (3.2), we have . In addition, clearly . So, in the case (i), we have proven the lemma.

(ii)
There is the case
Let be an element satisfying , and let
Further, let be
So,
and hence we obtain
By putting the righthand side of (3.12) becomes
Similarly, let us put
and define as
So,
and hence we obtain
By putting the righthand side of (3.17) becomes
Let us put
and define
Note that
The derivative of is

(a)
The case
In this case, we have
Since
from the assumption (3.23), is monotone increasing. So, we have
But, from (3.23), it holds that
So, if we put
as case (i), we have and . In addition, = . So, we have proven the case (ii)(a).

(b)
The case
In this case, we have
Moreover
since we have (3.29) and the assumption (3.8) (), respectively. Therefore, there exists such that . Let us define the constant as
then . Now we have
since
Let us define as
Then, again as case (i), we have and by (3.33), . In addition, . So, we have proven the case (ii)(b). This completes the proof.
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Watanabe, K., Kametaka, Y., Nagai, A. et al. Symmetrization of Functions and the Best Constant of 1DIM Sobolev Inequality. J Inequal Appl 2009, 874631 (2009). https://doi.org/10.1155/2009/874631
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Keywords
 Unique Solution
 Riemannian Manifold
 Arbitrary Element
 Sobolev Inequality
 Distributional Sense