WorksMathematics
Plasticity of the unit sphere of the space
Abstract
In this paper, we prove that the unit sphere of the space is plastic, i.e., every non-expansive bijection from this sphere onto itself is an isometry.
Introduction
Definition. A metric space is called plastic if every bijective non-expansive map is an isometry.
Recall that a map is called non-expansive (non-contractive) if for any
A map is called non-contractive if for any the inverse inequality holds. In [2], Cascales, Kadets, Orihuela, and Wingler posed the following question:
Problem. Is it true that the unit balls of all Banach spaces are plastic?
This question has a number of partial positive results (see, for example, [2, 3, 5, 6, 13] and references therein), but is still open in full generality in infinite dimensions. In finite dimensions, the answer is positive due to the result on totally bounded metric spaces obtained by Naimpally, Piotrowski, and Wingler [10]. The problem of plasticity of the unit balls of Banach spaces may be considered as a continuation of the well-known results of Mazur–Ulam [9] extended by Mankiewicz [8].
Theorem 1.1. (Mazur–Ulam-Mankiewicz.) Every bijective isometry between the unit balls and extends to a linear bijective isometry from onto .
In 1987, Tingley [12] raised the question of whether the unit balls in the aforementioned theorem could be replaced by unit spheres:
Problem. (Tingley.) Let be a bijective isometry between the unit spheres of two Banach spaces and . Does admit an extension to a linear isometry between and ?
Despite numerous partial results for specific spaces (see, for example, [4], [11]), Tingley’s problem remains a challenging open question in dimensions three and higher. Moreover, it has been fully resolved in the two-dimensional setting by Banakh only in 2022 [1]. As its natural continuation, one may consider the plasticity problem for the unit spheres of Banach spaces, posed by V. Kadets in personal communication, 2024.
Problem. (Kadets.) Is it true that the unit spheres of all Banach spaces are plastic?
In [7] this question was answered positively for Hilbert spaces, , , and . Observe also, for the unit ball of the space only conditional plasticity was shown. Namely, a bijective non-expansive map is an isometry if is continuous [6]. It is currently unknown whether this additional condition may be omitted. As the title says, the main objective of this paper is to prove the plasticity (without any additional conditions) of .
To facilitate our analysis, we introduce the necessary notation and structural definitions used throughout the text. For a real Banach space , let and denote its unit sphere and closed unit
2020 Mathematics Subject Classification. 46B20, 46C05.
Key words and phrases. Non-expansive map, non-contractive map, unit sphere, isometry, expand-contract plastic space.
The author was supported by the National Research Foundation of Ukraine, project No. 2025.07/0369 “Qualitative methods of nonlinear analysis of heterogeneous structures”. ball, respectively. In the context of the classical sequence space , an element is represented as a sequence such that and .
We denote by the dense subset of consisting of all elements with finite support. For any vector , its support is defined as the set of indices corresponding to non-zero coordinates:
Thus, if and only if .
Furthermore, let be the standard unit vector basis of , where has 1 at the -th position and zeros elsewhere. We define as the set consisting of all standard basis vectors and their negatives:
Evidently, . For any real number , its signum function is denoted by .
The paper is organized as follows. In the second section, we collect and modify some results known for the unit spheres of some other sequence spaces or for to adapt them to the case of . These results imply that any bijective non-expansive mapping is tightly bound to a coordinate permutation and a sign-switching sequence. Roughly speaking, we “turn off” this coordinate permutation and sign-switching and obtain a function . This section also includes some previously known or slightly modified properties of this function. Section 3 is devoted to further investigation of the properties of , which finally allowed us to see that reduces to identity, which in turn means that reduces to an isometry.
Preliminaries
The analog of the first lemma of this section is Lemma 3.1 from [6], which was proved for and . The “only if” part was also proved in Lemma 4.1 in [7] for the unit spheres of and . Although this result cannot be considered as a completely new one, for the reader’s convenience, we give it with proof.
Lemma 2.1. The points satisfy the condition that for every
if and only if and .
Proof. Let us first show the “if” part. Suppose that without loss of generality , and fix arbitrary . If , then both inequalities from the lemma statement hold. If , then , otherwise .
Conversely, let us prove the “only if” part. Suppose by contradiction that . Then there exist two distinct indices and such that and and an index , which without loss of generality we may assume different from , such that . Let us construct a vector as follows: set , and . All other coordinates of are set to zero. Then we clearly obtain:
and by the choice of , we also have , which contradicts the assumption.
Thus, . Since , this implies for some , meaning . By symmetry, we also get , so . If , we can easily find to violate the condition. Finally, if , then choosing yields , which is another contradiction. Therefore, we have . □
The next lemma was also proved in [7] (Lemma 4.2) for and , however, it also works for .
Lemma 2.2. Let be a non-expansive bijection, and let . Then and .
Proof. For every either or , and the non-expansiveness of implies that for every . is bijective, so for every . Now all that remains is to apply the previous lemma. □
The next lemma was extracted from Lemma 3.2 in [6], where the same was proved for .
Lemma 2.3. Let be a non-expansive bijection. Then there is a bijection and a sequence such that for every and the following holds:
if , then ;
if and (), then ();
if and (), then ().
Proof. Let us fix an arbitrary . By Lemma 2.2, for the elements and , we have or , and correspondingly, or for some . Define and . The mapping defined in this way is injective due to the injectivity of .
In what follows, we show that conditions 1)–3) hold for all and . Since is dense in and is continuous, it suffices to verify these conditions only for , that is, for with .
For every define . Observe that for every , either or there exists such that . To demonstrate this, let us fix and suppose for every . By the definition of , this means for every and thus, . Since is non-expansive and bijective, for every , either or there is such that . Further, we will construct a special sequence and show . By the construction of , the latter inequality will automatically imply items (2) and (3). Let
Observe by its definition, and for every , so , and thus items (2) and (3) are proved.
It remains to prove item (1). Let us fix and change the definition of such that . We still have and for every . Thus, , which implies . On the other hand, setting , we have the opposite inequality . Thus, item (1) is proved.
The last step is to prove surjectivity of . Let us argue by contradiction. Suppose there is . From the above we have for every . Using the continuity of and density of in once again, we obtain for every . This contradiction to the surjectivity of completes the proof.
:::
The preceding lemma yields a bijection and a sequence . Let us define the following isometric automorphism of by
for all . Denote the restriction of this automorphism to by . Observe that is a bijective isometry of onto itself. To prove the plasticity of , it suffices to show , or equivalently, to show that is the identity map on . We shall establish this fact in what follows. By its definition, is a non-expansive bijection from onto itself. Furthermore, Lemma 2.3 implies that acts as the identity on and satisfies the following properties (also extracted from analogous results for proved in [6]).
Lemma 2.4. For each and ,
if , then ;
if , then ;
if , then .
Lemma 2.5. For each and , if , then , and if , then .
The next lemma is extracted from Lemma 4.7 from [7]. This property was established for and . However, almost the same argument works for .
Lemma 2.6. Let , . Then for every the following inequalities hold:
and
Proof. Let be such that , i.e. . The non-expansiveness of then implies that , i.e. , and
This proves (1) for all such that , and (1) is obviously true for all such that . The claim (2) follows directly from Lemma 2.4. □
Main results
Let us consider the set of points with finite support, such that for any and any , .
Lemma 3.1. The map is the identity map on .
Proof. We will proceed by induction on the cardinality of the support. The base case of induction holds since is the identity on . Now, suppose for any with , . Further, let us argue by contradiction. Without loss of generality we may consider
where for all , and suppose . By the previous results, one has
where, at least for some , and all have the same signs as the corresponding . Consider the following element
with some . By Lemma 2.6, , for all . Moreover,
Thus, in particular,
and since can be chosen arbitrarily small, we can achieve . We obtain the contradiction , which completes the proof. □
Lemma 3.2. Let and differ in one single coordinate. Moreover, and . Then .
Proof. By Lemma 2.6, we have
and they have the same signs. On the other hand
Together, these conditions give us the needed conclusion. □
Theorem 3.1. is the identity map on .
Proof. Again, since is continuous and is dense in , it suffices to consider only elements , that is, . Let us show gradually increasing the number of nonzero coordinates . If has only one nonzero coordinate, then for some and there is nothing to prove. If has two nonzero coordinates, the case with moduli of these coordinates equal to 1 is covered by Lemma 3.1. Now consider of the following form
with and . Then is obviously equal to 1. Consider with and use Lemma 3.2. We have .
Analogously, by changing only one coordinate to its sign at each step, we may show for any with arbitrarily many coordinates of modulus 1 and any finite number of coordinates whose moduli are less than 1.
Thus, holds for all . Finally, by the density of and the continuity of , we conclude that is the identity map on the entire sphere .
Department of Mathematics and Informatics, V. N. Karazin Kharkiv National University, 61022 Kharkiv, Ukraine.
Email address: [email protected]
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