Abstract
Suppose that are -regular Banach spaces, that and are domains, and that is a homeomorphism. The aim of this study is to prove that the quasisymmetric mapping under the quasihyperbolic metric is equivalent to the freely quasiconformal mapping under the condition being -regular Banach spaces. This result is a partial solution to an open problem raised by Väisälä.
1. Introduction and Main Result
Quasiconformal mappings in the plane were introduced by Grötzsch in 1928 when he studied the most nearly conformal mappings of a square onto a rectangle, not a square, which map vertices on vertices [1]. After that, this class of mappings attracted much attention (see, e.g., [2, 3] and the references therein). The concept was later generalized to the setting of by several authors (see [4–8]). From the late 1980’s onwards, by using quasihyperbolic metrics, Väisälä established the theory of freely quasiconformal mappings in Banach spaces [9–13].
The class of quasisymmetric mappings on the real axis was first introduced by Beurling and Ahlfors [14], who found a way to obtain a quasiconformal extension of a quasisymmetric self-mapping of the real axis to a self-mapping of the upper half plane. This concept was later generalized by Tukia and Väisälä when they studied quasisymmetric mappings among metric spaces [15]. As we know, every quasisymmetric mapping is freely quasiconformal, but the converse is not true in general. As in the definition of freely quasiconformal mappings, by using quasihyperbolic metrics, in [9], Väisälä introduced a new class of mappings, that is, quasisymmetric mappings in the quasihyperbolic metric, and got the following relationship between freely quasiconformal mappings and quasisymmetric mappings in the quasihyperbolic metric.
Theorem 1. (see [9], Theorem 5.14). Suppose that and are Banach spaces and that and are domains (i.e., nonempty open and connected sets). If is freely -quasiconformal, then is a fully -quasisymmetric mapping in the quasihyperbolic metric with , where the symbol “” means that the control function depends only on the control function .
In , the converse of Theorem 1 is also true, which means that quasisymmetric mappings in the quasihyperbolic metric and freely quasiconformal mappings are quantitatively equivalent (cf. [9], Remark 3.5). (For two conditions, we say that Condition implies Condition quantitatively if Condition implies Condition , and the data of Condition depend only on the one of Condition and other given quantities. If also Condition implies Condition quantitatively, then we say that Condition is quantitatively equivalent to Condition ).
Naturally, one will ask whether the converse of Theorem 1 is true in Banach spaces. In fact, the problem was raised by Väisälä in the following form.
Problem 1. (see [13], Problem 13.2). Suppose that and are Banach spaces, that and are domains, and that is an -quasisymmetric mapping in the quasihyperbolic metric. Is freely -quasiconformal with ?
In 1998, Heinonen and Koskela [6] showed that quasiconformality and quasisymmetry are quantitatively equivalent in a large class of metric spaces, which includes Euclidean spaces, but not all Banach spaces. The spaces satisfy some controlled geometry conditions, for instance, regularity. A metric measure space is called Q-regular, , if there is a constant so thatfor all balls in with radius , where “diam” means “diameter.” Also, we say that a metric space is called -regular if there is a measure on such that (1) holds for some constants and .
With the aid of regularity, we get the following result.
Theorem 2. Suppose that and are -regular Banach spaces and that and are domains. Then, is freely -quasiconformal if and only if is an -quasisymmetric mapping in the quasihyperbolic metric with and depending only on each other and the constant .
Remark 1. Note that Banach spaces need not to be -regular; hence, the sufficiency of Theorem 2 can be regarded as a partial answer to Open Problem 1.
The rest of this paper is organized as follows. In Section 2, necessary definitions and results will be recalled, and in Section 3, Theorem 2 will be proved.
2. Preliminaries
2.1. Notation
Throughout the study, and denote real Banach spaces with dimension at least 2, and are domains, the norm of a vector in is written as , and for every pair of points , in , the distance between them is denoted by . Moreover, denotes the open ball with the center and radius , and and denote the boundary and the closure of , respectively. Let denote the cluster set of at a point over all neighborhoods of in . For , in , let denote the segment or the closed interval in with the endpoints and , respectively. And we remark and .
2.2. Quasihyperbolic Metric
Let be a domain. For a rectifiable curve in , that is, the arclength , its quasihyperbolic length is defined bywhere denotes the distance from to the boundary of .
For each pair of points , in , the quasihyperbolic distance between and is defined in the usual way:where the infimum is taken over all rectifiable curves joining to in . It is known that is a metric in , which is called quasihyperbolic metric (abbreviated QH metric).
The QH metric of a domain in was introduced by Gehring and Palka [16]. This metric was later generalized to the setting of Banach spaces and metric spaces and has become an important tool in the study of geometric function theory [9, 10, 13, 17–20].
An arc is c-quasiconvex iffor all , where is the subarc of between and , and is the arclength of .
Theorem 3. (see [9], Lemma 2.2)(1)For all , we havewhere “min” means “minimum,” and(2)For , and , ,
In addition, if , then
Theorem 4. (see [9], Lemma 2.6). For every , there is a constant such thatfor all and , where QH balls with center and radius .
Theorem 5. (see [10], Lemma 2.2). For each and , let be a sphere in . Then, each pair of points in can be joined by a 2-quasiconvex arc in .
2.3. Freely Quasiconformal Mappings and Quasisymmetric Mappings in the Metric
Definition 1. Let : be a homeomorphism, where and are domains, and let be a growth function, that is, a homeomorphism of with for . We say that is(1)-semisolid iffor all . Here and hereafter, the primes are used to denote the images, e.g., ;(2)-solid if both and are -semisolid(3)M-quasihyperbolic (abbreviated M-QH) if is -solid with (4)Fully -solid if is -solid on every subdomain of . Fully -solid mappings are also said to be freely -quasiconformal (abbreviated -FQC) mappings.
Theorem 6. (see [9], Theorem 4.7). Suppose that , , and that : is -. Then, is fully -.
Theorem 6 shows that if is -QH, then is -FQC with . To further illustrate the preciseness of Theorem 6, an example of QH mapping is given below.
Inversion 1 (see [9], Inversion 4.9). Let and let : be the inversion . Then, is 3-QH and fully 36-QH.
Definition 2. Let . For domains and , a homeomorphism : is called(1)-quasisymmetric (abbreviated -QS) if there exists a homeomorphism with such thatfor each and each triple , , of distinct points in (2)-locally -QS if the restriction is -QS whenever (3)-QHQS if is -QS in the QH metric in
Remark 2. We see from ([15], Theorem 2.2) and ([9], 3.10) that the inverse of an -QS (res. QHQS) is -QS (res. QHQS) with , and also, the inverse of an -locally -QS may not be -locally -QS. By [9] (3.4), we know that -QH is -QHQS with .
In [9], Väisälä shows that FQC is quantitatively equivalent to local QS. This result has been widely used to study the properties of freely quasiconformal mappings, see [21] and the references therein.
Theorem 7. (see [9], Theorem 5.10). For a homeomorphism : , the following conditions are quantitatively equivalent:(1) is -FQC(2) and are -locally -QS
3. Proof of Theorem 2
The purpose of this section is to prove Theorem 2. To reach this goal, we only need to prove the sufficiency of Theorem 2 since the necessity of Theorem 2 follows from Theorem 1.
Assume that both and are -regular Banach spaces with , and the homeomorphism : is -QHQS. By performing auxiliary similarities, we may assume that and are either or . By Theorem 3 and Remark 2, we know that the inversion is FQC and QHQS, and the composition of two FQC (resp. QHQS) mappings is also FQC (resp. QHQS). Hence, we can use auxiliary inversions to normalize the map so that both and are unbounded and , , .
Now, we begin to prove the sufficiency of Theorem 2, i.e., is -FQC.
To prove is -FQC, by Theorem 5, it suffices to show that both and are -locally -QS ( will be given in Lemma 3). Since is also -QHQS with , by symmetry, we only need to prove that is -locally -QS; i.e., for each , the restriction is -QS.
Before proving Theorem 2, we need some preparation. Without loss of generality, we assume that . For convenience, denote
In order to simplify the proof process, we make some simple statements.
For each , let . We take such thatwhere is a small constant satisfying
For , we denote
We denote the images of and under by and , respectively. Obviously, and . By Theorem 7, we know that and are connected. Now, the proof of Theorem 2 is mainly discussed on and .
Since , we have . We take to be the first point starting from to . Then, . Obviously, . Therefore, we get two sequences and which satisfyfor any .
Let , where means the greatest integer part of . Under these preparations, now, we are ready to establish several lemmas. The first lemma will show that does not too close to the boundary, which is the main lemma to prove Theorem 2. Denote
First, we come to compare the distances of points and to the boundary .
Lemma 1. .
Proof. We prove this lemma by a contradiction. Suppose on the contrary thatUnder the assumption “,” we first prove the following claim.
Claim 1. For each and each , we haveFor this claim, we first show that for any point ,We will prove this inequality by a contradiction. Suppose on the contrary that there exist an integer and one point such that the inequalityholds.
Since and , we haveThis, together with the assumption “,” impliesand sowhich contradicts with (13). Hence, (20) is true.
Again, since (20) and , we haveThen, Claim 1 holds.
Next, we come to prove the following inequality:Because and , we see from Theorem 3 thatFrom the above two inequalities, we getSince , we see from Theorem 3 thatFor any point , by Claim 1, we getMeanwhile, we also get from Claim 1 thatThe above two inequalities show thatIt follows from the condition “ being -” thatThe inequality (34) is obtained by (28), (29) and (32)where . Therefore, (26) holds.
By (26), for any , we haveObviously,And soThis impliesfor any . By this fact, we have the following claim.
Claim 2. Suppose that , and points and . Then,We shall prove this claim by a contradiction. Suppose on the contrary that there exist two integers , such that there are two points and which satisfyBecause points , and by Theorem 3, we getThese show thatThe choice of point , (20) and (26) and Theorem 3, impliesThus,It follows from the condition “ being -” thatThen, (42) and (44) implyOn the other hand, we come to prove the following inequality:To this end, without loss of generality, we may assume that . The choice of points and yields thatTherefore, by Theorem 3, we getThese implyAccording to the assumption , we can find one point such thatAgain, we see from Theorem 3 thatThen, we haveIt follows from (50) and (53) and the condition “ being -” thatThus, (47) holds. But it contradicts with (46). Hence, we complete the proof of Claim 2.
Claim 3. There must exist an integer such thatfor each point , where .
Also, we will prove this claim by a contradiction. Suppose on the contrary that for any integer , there are some points such thatDenote . For any , by Claim 2, we haveNow, we will prove that all of are contained in .
For any point , we haveThen,Since is a -regular Banach space, we haveThis is the desired contradiction, and hence, we complete the proof of Claim 3.
Claim 3 shows the relationship between the distance of point to the boundary and the distance of point on to the boundary . Now, we come to compare the distance of point to the boundary with the diameter of .
Let and . Then, we will show thatTo this end, we illustrate the proof of (61) in two cases.
If , then . Since and are separated by , and is a homeomorphism and , we know that and are separated by . Let be a line which contains . Then, divides into two parts and . No loss of generality, we assume that contains . Since and are separated by , we know . And so there exists one point . Therefore, the condition impliesIf , then there exists some point with . Since and is connected, we get from (35) that there exists some point. . Again by (35), we haveHence, (61) holds.
For each , it follows from (61) that there exists some point such thatTaking withNext, let be the first intersection point of with starting from to .
From the above construction, we get two sequences and in . Furthermore, we also get two corresponding sequences and in satisfying and for all . Based on the above facts, we have the following claims.
Claim 4. For any , we haveTo prove this claim, without loss of generality, we may assume that .
Based on the construction of two sequences and , we getIt follows from Claim 3 thatThe above two inequalities show thatBy Theorem 2, we haveThese, together with the condition “ being -,” implyHence, Claim 4 holds.
Claim 5. For any , we havewhere .
According to the condition points and , by Theorem 2, we getThese, together with Claim 4, implywhere .
Therefore,So Claim 5 holds.
Combined with the above claims, now we continue to complete the proof of Lemma 1.
Suppose that , and points and , respectively. Then, we haveThe last inequality holds since , and soSince is a -regular Banach space, and by Claim 5, we getThis is the desired contradiction, and hence, we complete the proof of Lemma 1.
Next, we come to estimate the quasihyperbolic distances between the point and any point in .
Lemma 2. For any point , we havewhere .
Proof. We see from Theorem 2 and thatThe above two inequalities show thatSince , and by Lemma 1, we getThis shows thatIt follows from (81) and (83) and the condition “ being -” thatFinally, we will discuss that is -locally -QS.
Lemma 3. The restriction is - with , where and are the constants of Theorem 4.
Proof. We take three points , , and in . By Theorem 2, we havefor each . Then, .
It follows from Theorem 4 that there is a constant such thatThese show thatMoreover, we infer from and Lemma 2 thatfor . Then, .
It follows from Theorem 4 and the condition “ being -” that there is another constant such thatThus, (87) and (89) implyTherefore, is - with . Lemma 3 holds.
Hence, we complete the proof of Theorem 2.
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that there are no conflicts of interest.
Acknowledgments
The authors were partially supported by the NNSF of China (no. 12071121), SRF of Hunan Provincial Education Department (No. 22B0034) and NSF of Hunan Province (no. 2022JJ30365).