On the Removability Property of <span class="nowrap"><svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-4.5268pt" id="M1" height="12.3952pt" version="1.1" viewBox="-0.0657574 -7.8684 10.1561 12.3952" width="10.1561pt"><g transform="matrix(.017,0,0,-0.017,0,0)"><path id="g113-253" d="M559 271C559 395 493 448 433 448C370 448 304 394 275 297C259 245 235 121 219 37C157 54 113 110 113 201C113 310 180 386 276 429L265 457C131 410 23 326 23 186C23 83 86 5 211 -10C180 -159 167 -220 158 -252L165 -261C198 -255 236 -246 251 -218L287 -5C421 9 559 119 559 271ZM484 273C484 144 400 40 293 33L352 331C364 389 382 406 405 406C431 406 484 370 484 273Z"/></g></svg>-</span>Natural Domains in Real Banach Spaces

He, Yuehui

doi:https://doi.org/10.1155/2022/8617936

Journal of Mathematics

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Abstract Introduction Preliminaries Data Availability Conflicts of Interest Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2022 | Article ID 8617936 | https://doi.org/10.1155/2022/8617936

On the Removability Property of -Natural Domains in Real Banach Spaces

Yuehui He¹

Academic Editor: Ji Gao

Received26 Feb 2022

Accepted05 May 2022

Published20 Jul 2022

Abstract

In this paper, we investigate the removability of -natural domains in Banach spaces. Let be a real Banach space with dimension at least 2, a domain in , and let be a countable subset of which satisfies a quasi hyperbolic separation condition. Then, is -natural if and only if is -natural, quantitatively.

1. Introduction

Throughout the paper, we always assume that denotes a real Banach space with dimension of at least 2, and a domain means a nonempty open and connected set in .

The motivation of this study stems from the discussions in [1], where the authors studied the removability of uniform domains and -uniform domains in Banach spaces. The following are the main results in [1].

Theorem 1. ([see [1], Theorem 2]). A domain is a -uniform domain if and only if is a -uniform domain, where the uniformity coefficients and depend on each other.

Theorem 2. ([see [1], Theorem 1]).A domain is a -uniform domain if and only if is a -uniform domain, where the control functions and depend on each other.

Here, denotes a sequence of points of satisfying a -quasihyperbolic separation condition in . Other terminology in the above two theorems and in the rest of this section will be given in Section 2 unless otherwise stated.

We recall that uniform domains in Euclidean spaces were introduced by John in connection with his work on elasticity [2]. The term is due to Martio and Sarvas [3]. Roughly speaking, a domain is a uniform domain if it is possible to travel from one point of the domain to another along a curve which is not too close to the boundary and not too long. Up to this time, uniform domains have been studied extensively because of their own characteristics (cf. [4–6]) and their significant applications (see, e.g., [7–13]). To investigate the relationships between conformal invariants and quasi hyperbolic metrics, in [14], Vuorinen introduced the so-called -uniform domains, which are a generalization of uniform domains. See [1,15–19] for the development of -uniform domains in recent years.

In [20], to explore the relationships among relative quasimappings, Väisälä introduced a class of more general domains, i.e., the so-called -natural domains, and proved that each domain in Banach spaces with finite dimension is -natural with depending only on , see [[20], Theorem 2.7]. But this property is invalid in Banach spaces with infinite dimension. In fact, Väisälä constructed a domain, a broken tube, in Banach spaces with infinite dimension, which is not natural, see [[21], Example 4.12]. Since all uniform domains and -uniform domains are -natural, naturally, one will ask whether there is a result similar to Theorems 1 and 2 for the setting of -natural domains in . The purpose of this paper is to investigate this question. Our main result is as follows.

Theorem 3. Suppose that denotes a domain and is a countable subset of which satisfies -quasi hyperbolic separation condition. Then is -natural if and only if is -natural, where the control functions and depend on each other, together with the constants from Theorem 4 and .

This paper is organized as follows: in Section 2, necessary terminology will be introduced and a series of lemmas will be proved. In Section 3, we will prove Theorem 3.

2. Preliminaries

In this section, necessary notions and notations in this paper will be introduced, several known results will be recalled, and a series of lemmas will be proved.

We say that a domain is called -uniform if there exists a constant with the property that each pair of points , in can be joined by a rectifiable arc in satisfying (cf. [[18], Section 6]).(1) for all ,(2),

where is the part of between and , stands for the length of , denotes the distance from to the boundary of in , and is the metric in .

The following result is from [18] due to Väisälä.

Theorem 4. ([see [18], Theorem 6.5]). For any and , all the domains , and in are -uniform with a universal , where denotes the ball in with center and radius , i.e., .

A curve is a continuous map from an interval to . If is an embedding, then it is said to be an arc. For simplicity, we also use to denote its image set . The length of is defined as usual,where the supremum is over all finite sequences . We call a curve rectifiable if its length .

Let be a rectifiable curve or arc with endpoints and . Then, there are and such that , , and for all , moreover, . The curve is the arc length parameterization of , see [[22], §2].

If is a rectifiable curve or arc in and if is continuous, the line integral of along is defined as follows:

The quasihyperbolic length of in is the number.(cf. [23]), where the quasihyperbolic distance between and is defined by the following equation:where the infimum is taken over all rectifiable arcs joining to in .

The following result is useful for the discussions in Section 3.

Theorem 5 ([see [8], Lemma 2.13]). Suppose that is -uniform domain. Then, for all , ,

Suppose that is a curve in and is a constant. It is called a -quasigeodesic (i.e., neargeodesic in [18]) if for any , ,and -quasiconvex for if for any , ,

The following result concerning the existence of quasigeodesics in is due to Väisälä.

Theorem 6. ([see [18], Theorem 3.3]).Suppose that , and . Then there is a -quasigeodesic in joining and .

Let be a homeomorphism. A domain is called -uniform if for all , ,

(cf. [14]).

For a nonempty set , we writeand

Obviously, for any domains and in , if is a nonempty set, then

Moreover, letwhere “” means “diameter,” and denotes the distance between the set and the boundary , i.e., . Then, it follows from [[20], equation (2.6)] that

Let be an increasing function. A domain is called - iffor every nonempty connected set with .

Let denote a sequence of points in . If there is a constant such that for any pair ,then we say that satisfies a -quasihyperbolic separation condition in . For such a sequence, in the following of this paper, we always use to denote it.

Lemma 1. Suppose that denotes a domain and . Then for any , with , we have

Proof. Without loss of generality, we assume thatSuppose on the contrary thatThen, it follows from [[21], Lemma 2.2(2)] and the assumption in the lemma thatwhich is the desired contradiction. Hence the lemma is true.

For simplicity, in the rest of this paper, we apply the following notations. For any , letthe boundary of is denoted by , i.e.,andwhere .

Now, we are ready to state our next lemma.

Lemma 2. For any , with , we have

Proof. Suppose on the contrary thatLetThen,which contradicts Lemma 1. Hence Lemma 2 holds true.

Also, we introduce the following notations. For a domain and a nonempty set in , letand let

Obviously, the connectedness of implies the one of the set .

We recall the reader the standard convention thatfor the empty set , where “” stands for “cardinality”.

Lemma 3. For a domain , if is a nonempty connected set in with , then .

Proof. If , then the lemma is obvious. In the following, we assume that .
Since implies , we deduce from equation (15) thatand so,For each , it follows from Lemma 2 thatwhere denotes the distance between the sets and (i.e., . Thus we getLet , . Then there are , such thatSince is connected, we getand thus,where denotes the distance from the point to the set (i.e., ). This ensures thatwhich, together with equations (31) and (33), impliesThis proves the lemma.

Lemma 4. For a domain , let . Then for any and , ,where the constant is from Theorem 4.

Proof. Let , and let . By Theorem 4, we know that is -uniform.
Since , we haveThis impliesFor any with , sincewe infer from Lemma 1 thatand thus,Sincewe deduce from equation (41) thatSimilarly, we haveBecause it follows from Lemma 2 thatwe get that (by equations (46)–(48) and Theorem 5)and hence, the lemma is proved.

Lemma 5. Suppose that is a domain and . Then for any rectifiable curve with endpoints and ,

Proof. Since for any and ,we haveand so,This givesand thus, we getThis ensures thatand hence, the lemma is proved.

Lemma 6. Under the assumptions of Lemma 5, suppose further that is a 2-quasiconvex curve connecting two points and in , where . Then,

Proof. Since and and , we see from [[21], Lemma 2.2(3)] thatBecause for any ,we getThis completes the proof of the lemma.

Lemma 7. Suppose that is a domain and . Then, for each pair , , we have

Proof. Let , . Then it follows from Theorem 6 that there is a 2-quasigeodesic in joining and .
If , then the inequality in equation (61) easily follows from Lemma 5. In the following, we assume thatThen, . Furthermore, we prove that contains a finite number of elements. To this end, let . Then, we haveand sosince .
If contains an element with , let and . Then, we get (by Lemma 1 and equation (64))which, together with Lemma 2, implies that the length of would be infinite if contains infinite number of elements. This contradiction affirms thatNext, we construct a curve in to connect and . Since , there is an element in , denoted by , such that,(1)there are , such that does not intersect any element in and (here, recall that denotes the boundary of the ball )(2)there is a 2-quasiconvex curve joining and (cf. [[18], Lemma 2.2])If does not intersect any element in , then letOtherwise, there is an element, denoted by , in such that,(1)there are , such that does not intersect any element in and (2)there is a 2-quasiconvex curve joining and If does not intersect any element in , then letBy repeating this procedure, there are elements, (Without loss of generality, assume that ) and a curvesuch that,(1)all points and belong to the intersection for (2)all subcurves , for and of do not intersect any elements in (3)all curves for are 2-quasiconvex curves(4)By Lemma 6, we see that for each ,Since both and belong to the intersection , we deduce thatThen it follows from Lemma 5 thatSince is a 2-quasigeodesic, we obtain thatand hence, the lemma is proved.

3. Proof of Theorem 3

The purpose of this section is to prove Theorem 3. We start the proof of the theorem with two lemmas.

Lemma 8. Under the assumptions of Theorem 3, if is -natural, then is -natural, where and the constant is from Theorem 4.

Here and hereafter, the notation means that the control function depends only on , and .

Proof. Let be a nonempty connected set in with . To prove this lemma, it suffices to show thatwhere and the constant is from Theorem 4.
For the proof, let , be withIn the following, we divide the discussions into four cases according to the positions of and in .

Case 1. Suppose that there is so that , .

For this case, since , , by Lemma 4, we have

This, together with equation (75), shows that

Case 2. Suppose that , .

Under this assumption, it follows from Lemma 7 thatand thus, equation (75) gives

Finally, the assumption being -natural ensures that

Case 3. Suppose that and with .

Since , and is connected, we see that there are points and such thatand because , , and , , we infer from Lemma 4 that

Moreover, by Lemma 7, we haveand thus, the assumption being -natural guarantees that

Now, we conclude from equations (75), (82), and (84) that

Case 4. Suppose that and there is such that .

Obviously, there is a point such thatThen, equation (75) impliesand thus, the assumption being -natural, Lemmas 4 and 7 ensure that

By lettingwe conclude from the estimates equations (77), (80), (85), and (88) that equation (74) is true, and hence, the proof of this lemma is complete.

Lemma 9. Under the assumptions of Theorem 3, if is -natural, then is -natural, where and the constant is from Theorem 4.

Proof. Let be a nonempty connected set in with . In the following, we separate the discussions into three cases based on the number of the elements in the set .

Case 5. Suppose that , i.e., .

Obviously, in the case, . As the preparation, we establish a relationship between and for .

Since for any , there must be a point such thatwe know that

If , then

Now, we assume that . We consider two possibilities: and . For the first possibility, we haveand for the remaining possibility, we getHence we conclude that for any ,

Now, it follows thatand thus, the assumption being -natural gives

Case 6. Suppose that .

We denote the unique element in by . Since for any ,andwe have

Furthermore, Theorems 4 and 5 guarantee that

Then, we deduce from equation (100) that

If , then equation (102) gives

In the following, we assume thatThen, .

Next, we are going to establish a relation between the quantities and as indicated in equation (133). To this end, we consider two possibilities: and .

For the former possibility, it follows thatand thus,

Furthermore, since is connected, the similar reasoning as in the discussions in Case 5 demonstrates that

Still, we need a relationship between and as stated in equation (114). For this, let . Then it follows thatand

These implyand thus, we infer from that

Therefore,which, together with equation (106), implies that

Hence equation (15) ensures that

By combining equations (107) and (114), we obtain that

For the latter possibility, that is, , to reach the goal, let , and letthe closure of the ball . Then it follows from the assumption thatand so, for any , we haveand

Then for any in , from Lemma 1, we know

This and equation (118) imply

In the sequel, we are going to establish a relation between the quantities and as stated in equation (126). For this, let and . Then it follows from equation (121) thatand

Since , we see thatwhich implies

Then, the fact and the assumption being -natural guarantee that

Next, let us estimate the quantity in terms of as indicated in equation (131). Because for any ,andwe obtain that

This, together with the fact , shows that

Then, it follows from equation (15) thatand thus, equation (126) implies

Now, we conclude from equations (115) and (132) that

Thus, equation (102) gives

Case 7. Suppose that .

This assumption implies that . As the preparation, we are going to get an estimate on the quantity with the aid of as shown in equation (154).

To this end, let , be such that

If , then

In the following, we assume that at least one of and belongs to some element of . Obviously, it suffices to consider the two cases: both and belong to some element(s) of , or and belongs to an element of .

For the first case, there must be two elements , in so that

If . Since is connected, we see that . Let

Since , by [[21], Lemma 2.2(2)], we get

On the other hand, by Lemma 1, we haveand

Therefore [[21], Lemma 2.2(1)] leads towhich, together with equation (139), giveswhere .

Similarly, we have

These estimates, together with equation (135), ensure that

If , i.e., , then by [[21], Lemma 2.2(2)], we obtain

Since , we see that contains two elements of and such that . LetThen, the similar reasoning as in the arguments of equation (142) giveswhich, together with equation (146), implies thatand thus, equation (135) leads to

For the second case, that is, and for , since , there is an element such that . Because is connected, we take

Then, the similar reasoning as in the arguments of equation (143) giveswhich implies

Now, we deduce from equations (136), (145), (150), and (153) that

Still, we need a relationship between the quantity and the one as indicated in equation (176). Recall that the set is connected.

First, we establish a relationship between the quantities and as indicated in equation (165).

By Lemma 3, we see that . Let , be such that

Then, there are and such that

We shall show that . Suppose on the contrary that . Then,This, together with equation (155), implies

Since , there must an element in such that . Then,where the last inequality follows from Lemma 1. This contradicts equation (158). Hence .

It follows that (by Lemma 1)

Let

Then (by Lemma 1),and thus,This, together with equation (155), givesand so,

Also, we need a relationship between the quantities and as indicated in equation (174). To reach this goal, let be such thatand then, there is such thatand let

Then, we haveand

These implyand

Therefore,and thus, we obtain

Now, it follows from equations (165) and (174) thatand so, equation (15) gives

Now, we are ready to finish the proof of the lemma. Since the set is connected, we infer from the similar arguments as in Case 5 thatand thus, equation (176) gives

Then, we know from equation (154) that

By lettingwe conclude from the estimates equations (97), (103), (134), (150), and (179) that the lemma is true.

Proof. of Theorem 3. The necessity part of the theorem follows from Lemma 8, and its sufficiency part follows from Lemma 9. Hence the theorem is proved.

Data Availability

The data used to support the study are included in the paper.

Conflicts of Interest

The author declares that there are no conflicts of interest.

Acknowledgments

The author was partly supported by NNSFs of China under Grant nos. 12071121, 11571216, and 11720101003.

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Copyright © 2022 Yuehui He. 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.

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