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Journal of Function Spaces
Volume 2019, Article ID 5234507, 9 pages
https://doi.org/10.1155/2019/5234507
Research Article

An Orlicz-Besov Poincaré Inequality via John Domains

Department of Sciences, China University of Geosciences, Beijing 100083, China

Correspondence should be addressed to Hongyan Sun; nc.ude.bguc@yh_nus

Received 14 October 2018; Accepted 17 November 2018; Published 1 January 2019

Academic Editor: Stanislav Hencl

Copyright © 2019 Hongyan Sun. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Denote by the intrinsic Orlicz-Besov space, where , is a Young function, and is a domain. For and optimal , via John domains, we establish criteria for bounded domains supporting an Orlicz-Besov Poincaré inequality. This extends the known criteria for bounded domains supporting Sobolev-Poincaré inequality and its fractional analogue.

1. Introduction

Let in this paper. Recall that a bounded domain is called a John domain with respect to some and if, for each , there is a rectifiable curve parameterized by arc-length (called a John curve for ) such that , , and for all .

It is well known that John domains essentially characterize bounded domains of supporting Poincaré-Sobolev inequality and some fractional analogue in the literature. Precisely, Bojarski [1] proved that, for any , a John domain always supports the following Sobolev-Poincaré inequality: there exists a constant such that whenever the distributional derivative . In this paper, is the average of in set with . When and , as proved by Hurri-Syrjänen and Vähäkangas [2] (see also Dyda, Ihnatsyeva, and Vähäkangas [3]), a John domain always supports the following fractional Sobolev-Poincaré inequality: there exists a constant such that whenever . Moreover, as proved by Zhou [4], when and , a John domain always supports the following Hajłasz-Sobolev Poincaré inequality: there exists a constant such thatwhenever is an intrinsic -order Hajłasz upper gradient of ; that is,Conversely, suppose that is a bounded simply connected planar domain and a bounded domain that is quasiconformally equivalent to some uniform domain when . Buckley-Koskela [5] proved that if satisfies (1), then is a John domain. If either (3) or (2) holds, then is also proven to be a John domain; see [24].

The main purpose is to prove that John domains essentially characterize bounded domains of supporting an Orlicz-Besov Poincaré inequality as below. Let be a Young function in ; that is, is convex and satisfies , for and . Given any and domain , as motivated by the Orlicz-Besov spaces considered in [6] and also semi-norms that appeared in the right hand side of (2) and (3), we consider the intrinsic Orlicz-Besov space , which is defined as the collection of all measurable functions in whose (semi-)norms are finite. Modulo constant function is a Banach space. In the spirit of (1), (2), and (3), we say that a bounded domain supports an Orlicz-Besov Poincaré inequality with respect to and if there exists a constant depending on , , , and such that

Theorem 1. Let and be a Young function satisfying (i) If is a John domain, then supports the Orlicz-Besov Poincaré inequality (6).
(ii) Assume that is a bounded simply connected planar domain, or a bounded domain that is quasiconformally equivalent to some uniform domain when . If supports the Orlicz-Besov Poincaré inequality (6), then is a John domain.

The assumption (7) above guarantees and hence the nontriviality of ; see Lemma 4. For the optimality of (7), see Remark 5. The assumption (8) always fails if . Indeed, since is convex, , and for all , there exists such that for all . Then we always have For , Young functions satisfying (7) and (8) include with and or , with and also their convex combinations. For , Young functions satisfying (7) and (8) include with and or and also their convex combinations.

Moreover, for any and , let and . Then , and satisfies (7) and (8). In this case, we always have We see that (6) coincides with (2). Thus, Theorem 1 extends above criteria for bounded domains supporting fractional Poincaré inequality given in [2, 3] (see also [4]).

We prove Theorem 1 in Section 3. To prove Theorem 1 (i), we use Boman’s chain rule for John domains and an Orlicz-Besov imbedding in cubes under (7) and (8) as proved by [7] (see Lemma 8) and also borrow some ideas from [1, 4]. To prove Theorem 1 (ii), we use (7) and (8) to calculate the -norms of some cut-off functions in a precise way; see Lemmas 6 and 7. This allows us to prove the LLC(2) property of domains supporting inequality (6); see Proposition 9. By borrowing some ideas from [4, 5, 8], we then obtain Theorem 1 (ii).

Here we make some conventions on the notations or notion used in this paper. Throughout the paper, denotes positive constants, which depends only on but whose value might be changed from line to line. We write if there exists a constant such that . For any and , denotes the distance from to and denotes the diameter of the set .

2. Some Basic Properties

The following properties of Young functions are given in [7, ].

Lemma 2. Let and be a Young function.
(i) If satisfies (7), then (ii) If satisfies (8), then as for any and

Next we show that intrinsic Orlicz-Besov functions are locally integrable.

Lemma 3. Let and be a Young function. For any bounded domain , we have as sets.

Proof. Let . For , we have By Fubini’s theorem, for almost all we have For any ball with , one may choose such satisfying above inequality so that Thus this implies that By Jensen’s inequality, we have Notice that as . Then which implies that That is, as desired. The proof of Lemma 3 is complete.

The assumption (7) guarantees the nontriviality of intrinsic Orlicz-Besov spaces in bounded domains.

Lemma 4. Let and be a Young function satisfying (7). For any bounded domain , we have as sets.

Proof. Let and . By (7) we have Since the right hand side is less than when is sufficiently large enough, we know that ; that is, as desired. The proof of Lemma 4 is complete.

Remark 5. We remark that the assumption (7) is optimal to get or the nontrivial of in the following sense. For and , let . By a direct calculation, the Young function satisfies (7) if and only if . On the other hand, let . If , then , and observe that (10) also holds for all measurable functions in . For any ball , write as the space of all functions satifying By [9, &4.2] and , we know that only contains constant functions. If , then for any ball with . Thus is a constant on every ball with , and hence by the connectedness of , is a constant function in .
To end this section, we calculate -norms of some special functions, which will be used in Sections 2 and 3.
Let and . If is connected, we write ; if is disconnected, we let be any component of . For with , define a function in by

Lemma 6. Let and be a Young function satisfying (7) and (8). For any bounded domain and with , if is a component , with , we have Here is a constant depending only on , , and .

Proof. Write for simplicity. For any and , implies that at least one of lies in . ThusSince for all with , and for all with , we further write By a change of variables and applying (7), we obtain Similarly, by a change of variables and applying (8), we obtain Therefore,If by the convexity of , we have Thus as desired. The proof of Lemma 6 is complete.

Let , , and such that . Define where the infimum is taken over all rectifiable curves joining and .

Lemma 7. Let and be a Young function satisfying (7) and (8). There exists a constant such that, for all bounded domains , , and satisfying , we have

Proof. Assume that is disconnected. One has . Note that if and , we have . Indeed, for such , we have and hence Therefore, . Letting be the line segment joining contained in , we have . For any , we know that gives a curve joining and and Thus we have . Similarly, .
For any , we have Indeed, letting be the line segment joining , we have . Since connects and for any , we have Similarly, by changing the roles of , we also have Therefore, we obtain Note that, for , one has By (7), we obtain for some . Thus for any , by Jensen’s inequality, we have , which implies that as desired.

3. Proof of Theorem 1

With the aid of the following result given in [7, ], we are able to prove Theorem 1 (i) as below. For any cube and any measurable functions in , we write

Lemma 8. Let and be a Young function satisfying (7) and (8). Then there exists a constant depending on such that

Proof of Theorem 1. (i) Assume that is a bounded John domain. By Boman [10] and Buckley et al. [11], enjoys the following chain property: for every integer , there exist a positive constant and a collection of cubes such that
(a) for all , and (b) is fixed cube; for any other , one can find a sequence from satisfying that, for all , we have Set . By given in (a), we write By Lemma 8, we have and hence Let , for with . For any , by as in (a), we have . Thus by (12) and convexity of , we get We then obtain and hence .
To estimate , for each write By (b), Hölder inequality, and Lemma 8, we have Since , given in (b), we have Therefore, by (a), Recall that by Lemma 4.2 in [1] (see also [12]), for any and , we have Thus, Since (a) implies the bounded overlaps of cubes in , by , we have for all . From this, it follows thatBy an argument same as , we have
Combining the estimates for and together, we arrive at Since this yields and we obtain (6) as desired. This completes the proof of Theorem 1 (i).

To prove Theorem 1 (ii), we need the following result.

Proposition 9. If is a bounded domain supporting the -imbedding, then has the LLC(2) property; that is, there exists a positive constant depending only on , and such that, for all and , any pair of points in can be joined in .

Proof. Fix a point so that We claim that there exists such that are contained in the same component of whenever for some and .
Note that this claim gives LLC(2) properties of . Indeed, assume that this claim holds for the moment. First, letting , we show that are contained in the same component of whenever for some and . Indeed, if , then, for any , we have and hence, by , we have ; that is, are contained in the connected set . If and , are contained in the connected set . If and , letting , by , we see that . By the above claim, are contained in the same component of . Noting that we see that are contained in the same component of as desired.
Let for some and . If , then . Since , by the above argument, we know that for some component of . Similarly, since , . Thus . If , then , which implies , and hence