Open Journal of Applied Sciences
Vol.04 No.09(2014), Article ID:48591,3 pages
10.4236/ojapps.2014.49042
Some Rearrangement Inequalities on Space of Homogeneous Type
Tiejun Chen
Yiyang Medical College Hunan Pro of China, Yiyang, China
Email: cwwlove@sina.com
Copyright © 2014 by author and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/



Received 9 June 2014; revised 22 July 2014; accepted 2 August 2014
ABSTRACT
Let
be a
Muckenhoupt weight. In this paper we get the estimate of rearrangement
in homogeneous space that is
. The similar estimate is obtained only on space of
.
Keywords:
Rearrangement, Homogeneous Space,
Weight

1. Introduction
We first recall some basic notions about the homogeneous space and the weights we are going to use.
Definition 1 [1] . (Homogeneous space X). Let X be a set. A function d:
is called a quasi- distance on X if the following conditions are satisfied:
1) for every x and y in X,
, and
if and only if x = y,
2) for every x and y in X,
,
3) there exists a constant K such that
for every x, y and z in X.
Let μ be a positive measure on the
-algebra of subsets of X generated by the d-balls
, with
and r > 0. Then a structure (X, d, μ), with d and μ as above, is called a space of homogeneous type.
We say that (X, d, μ) is a space of homogeneous type regular in measure if μ is regular, that is for every measurable set E, given
, there exists an open set G such that
and
A non-negative locally integrable on homogeneous space X function 
weight function we call the measure
X, define its non-increasing rearrangement 


Definition 2 (



holds for every ball B and every measurable set

2. Basic Lemmas
Denote doubling condition D, a weight 



Lemma 1 [3] . Let (X, d, μ) be a space of homogeneous type. Let 





some
Lemma 2. (C-Z decomposition) [4] [5] . Let (X, d, μ) be a space of homogeneous type such that the open balls are open sets. Let f be a nonnegative integrable function defined on X, then for every 




1)
2) 


Lemma 3. 





Bi cover E and
Proof: If
Letting
then
For every ball B centered at
i.e.

If 





this is a contradiction.
Then 

3. Inequalities Conclusion
Theorem 1. 


Proof: The proof is similar to Lerner [5] - [7] ,
From [6] , We get two collections of balls
Fix X, with



joint balls
Which contains
Then
Select from 






Note that
Since
Then

i.e.

We have
Taking supremum over all 

Fund
A project supported by scientific research fund of Hunan provincial education department in China (NO:
References
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