Journal of OperatorsVolume 2013 (2013), Article ID 506129, 9 pageshttp://dx.doi.org/10.1155/2013/506129

Research Article

Linear Isometries between Real Banach Algebras of Continuous Complex-Valued Functions

Davood Alimohammadi and Hadis Pazandeh

Department of Mathematics, Faculty of Science, Arak University, Arak 38156-8-8349, Iran

Received 27 February 2013; Revised 14 August 2013; Accepted 26 August 2013

Academic Editor: Jan Lang

Copyright © 2013 Davood Alimohammadi and Hadis Pazandeh. 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.

Let and be compact Hausdorff spaces, and let and be topological involutions on and , respectively. In 1991, Kulkarni and Arundhathi characterized linear isometries from a real uniform function algebra on (, ) onto a real uniform function algebra on (, ) applying their Choquet boundaries and showed that these mappings are weighted composition operators. In this paper, we characterize all onto linear isometries and certain into linear isometries between and applying the extreme points in the unit balls of and .

1. Introduction and Preliminaries

Let and denote the field of real and complex numbers, respectively. The symbol denotes a field that can be either or . The elements of are called scalars. We also denote by the set of all with .

Let be a normed space over . We denote by and the dual space of and the closed unit ball of , respectively. For a subset of , let denote the set of all extreme points of . Kulkarni and Limaye showed [1, Theorem 2] that if is a nonzero linear subspace of and , then has an extension to some .

We know that if and are normed spaces over and is a linear isometry from onto over , then is a bijection mapping between and .

Let be a compact Hausdorff space. We denote by the unital commutative Banach algebra of all continuous functions from into , with the uniform norm , . We write instead as . For , we consider the linear functional on defined by (), which is called the evaluation functional on at . Clearly, for all . It is known [2, page 441] that

Let be a real or complex linear subspace of . A nonempty subset of is called a boundary for (with respect to ), if for each the function assumes its maximum on at some . We denote by the intersection of all closed boundaries for . If is a boundary for , it is called the Shilov boundary for (with respect to ).

Let be linear subspace of containing , the constant function with value on . A representing measure for is an -valued regular Borel measure on such that for all . Let . If , then and , the point mass measure on at , is a representing measure for . We denote by the set of all for which is the only representing measure for . If is a boundary for , it is called the Choquet boundary for (with respect to ). We know that

Let be a real linear subspace of , and let be nonempty subset of . We say that is extremely regular at if for every open neighborhood of and for each there is a function with such that for all and for all .

Let be a nonempty set. A self-map is called an involution on if for all . A subset of is called -invariant if . Clearly, if is -invariant, then . A -invariant measure on is a measure on such that .

Let be a topological space. An involution on is called a topological involution on , if is continuous.

Let be a compact Hausdorff space, and let be a topological involution on . We define Then, is a unital uniformly closed self-adjoint real subalgebra of which separates the points of and does not contain , the constant function with value on . Note that for all . Moreover, and for all . In fact, the complex Banach algebra can be regarded as the complexification of the real Banach algebra . Note that if and only if is the identity self-map on . Hence, the class of real Banach algebras is, in fact, larger than the class of real Banach algebras .

Let be a compact Hausdorff space, a topological involution on , and a subset of . We say that separates the points of if for each with there exists a function in such that . It is known [3, Lemma ] that if , then separates the points of .

Let be a compact Hausdorff space. We assume that denotes , regarded as a real normed space by restricting the scalar multiplication to real scalars. Let be the topological involution on defined by () and (). Then the map defined by is an isometrical isomorphism from onto . Hence, the class of real Banach algebras is, in fact, larger than the class of real Banach algebras .

Let be a compact Hausdorff space, and let be a topological involution on . Let be a real subspace of containing . A real part representing measure for is a -invariant regular real Borel measure on such that for all . Let . If , then and the measure is a real part representing measure for . We denote by the set of all for which is the only real part representing measure for . Clearly, is a -invariant subset of . If is a boundary for , it is called the Choquet boundary for with respect to . We know that .

The real Banach algebra and its real linear subspaces were first considered by Kulkarni and Limaye in [4]. For a detailed account of several properties of , we refer to [3].

Let be a compact Hausdorff space. For each , define the map by in fact, . Clearly, .

Let be a topological involution on . For each define the map by . Kulkarni and Limaye showed [1, Proposition 3] that Applying this result, they obtained the following theorem.

Theorem 1 (see [1, Corollary 5]). Let be a compact Hausdorff space, and let be a topological involution on . Suppose that Then

The classical Banach-Stone theorem states that if is a linear isometry from onto , then there exists a homeomorphism from onto and a continuous function with such that

This well-known theorem has been generalized in several directions. In 1966, an important generalization was given by Holsztyński [5] by considering into linear isometries as the following.

Theorem 2. Let and be compact Hausdorff spaces. If is a linear isometry from into , then there exists a closed subset of  , a continuous map from onto and a function with and such that

In 1975, another important generalization of Banach-Stone theorem was given by Novinger [6] for certain complex subalgebras of and considering the concept of their Choquet boundaries. In 1991, Kulkarni and Arundhathi [7] generalized the given results by Novinger for certain real subalgebras of and considering the concept of their Choquet boundaries. We can cite some other generalizations of Banach-Stone theorem, for example, the generalization obtained by Cambern [8] for space of vector-valued continuous functions, by Jeang and Wong [9] for spaces of scalar-valued continuous functions vanishing at infinity, by Araujo and Font [10] for certain subspaces of scalar-valued continuous functions considering concepts of their Choquet and Shilov boundaries, and by Jiménez-Vargas and Villegas-Vallecillos [11] for spaces of scalar-valued Lipschitz functions.

Throughout the rest of this paper, and are compact Hausdorff spaces and and are topological involutions on and , respectively.

In this paper, we characterize all linear isometries from onto and certain linear isometries from into , applying the extreme points in and . Our results in Section 2 are some of the given results by Kulkarni and Arundhathi in [7] that we obtain by applying the extreme points in and . The main result in Section 3 is a generalization of the given result by Holsztyński in [5] for certain into isometries.

2. Onto Linear Isometries

We first determine unit-preserving linear isometries from onto . For this purpose, we need the following lemmas.

Lemma 3 (see [3, Lemma ]). Let and be real linear subspaces of   and , respectively. Suppose that , , is extremely regular at and is extremely regular at . Let be a linear isometry from onto and for all . Then, for all or for all .

Lemma 4. Let be a nonempty -invariant closed subset of . Then, is extremely regular at . In particular, is extremely regular at for all in .

Proof. Assume that and is an open neighborhood of in . By Urysohn’s lemma, there exists a function with for all , for all , and for all . Define the function by . Then, , , for all and for all . Hence, is extremely regular at .

Theorem 5. Let be a linear isometry from onto with . Then, there exists a homeomorphism from onto with on such that

Proof. We claim that Let . By Theorem 1, . Since is a linear isometry from onto , we conclude that is a linear isometry from onto and so Therefore, . Hence, our claim is justified.We now show that for each there exists a unique such that Let . By (13), we have . Hence, by Theorem 1, there exists such that Since and , by (16), we have Since and , we have and so . Thus, by (16), we have From (18) we deduce that Since is a linear isometry from onto and by Lemma 4, is extremely regular at , and is extremely regular at , we conclude that for all in or for all by Lemma 3. We assume that whenever for all in and whenever for all in . Then, and we have This proves the existent of . To show uniqueness, assume that there exists such that Then, for all in . This implies that since separates the points of .Now we define the map   by   whenever for all in . In fact, we have Now we prove that on . Let . Then, . From (22), for each we have This implies that since separates the points of . Therefore, on .Continually, we prove that is bijective. Since is a linear isometry from onto with , by the above argument, we deduce that there exists a map such that To prove the injectivity of , let . For each , by (22) and (24) we have This implies that since separates the points of . Therefore, , the identity mapping on , and so is injective.To prove that is onto, let . For each , by (24) and (22) we have This implies that since separates the points of . Therefore , the identity mapping on , and so is onto.We now check that is continuous. Let , and let be a net in such that in . Then we have By (27) and definition of , we deduce that Since separates the points of , by (28), we have in . Therefore, is continuous.Since is a bijective continuous mapping, is a compact space, and is a Hausdorff space, we conclude that is continuous, and so is a homeomorphism. Hence, the proof is complete.

Corollary 6. Let be an isometry mapping from onto with and . Then is an isomorphism.

Proof. Since and are real Banach algebras and is an isometry from onto with , we conclude that is a linear isometry by Mazur-Ulam theorem (see [12]). Since , applying Theorem 5, we deduce that there exists a homeomorphism from onto such that It follows that is a homomorphism. Hence, the proof is complete.

We now study the onto case (not necessarily unit preserving).

Lemma 7. Let with . Then, for all , if for every there exists such that with implies that

Proof. Assume that there exists such that . Then, since . Let . Then, and . Take . Since , we deduce that is a -invariant open subset of and . By Urysohn’s lemma, there exists such that for all , , and for all . Let . Then, , for all , , and for all . Therefore, Hence, for each we have This completes the proof.

Lemma 8. Let be a linear isometry from onto , and let . Then, for all .

Proof. Since and , by Lemma 7, it is sufficient to show that for each there exists such that with implies that Let , and let with . The surjectivity of implies that there exists such that . Since is an isometry, we have . This implies that there exists such that . Therefore, It follows that Hence, Since and , we have It is enough that we choose .

In the following result, we show that every linear isometry from onto is a weighted composition operator.

Theorem 9. Let be a linear isometry from onto . Then, there exist a function with for all and a homeomorphism from onto with on such that

Proof. Assume that . Then . By Lemma 8, we have for all . Clearly, for all and . We define the map by . Then, is a linear isometry from onto , and . By Theorem 5, there exists a homeomorphism from onto with on such that This implies that Hence, the proof is complete.

3. Into Linear Isometries

We formulate our main result in this section which is a version for into linear isometries of -spaces of a known Holsztyńskis’s theorem (Theorem 2) for into linear isometries of -spaces. We first study unit-preserving linear isometries from into .

Theorem 10. Let be a linear isometry from into satisfying the following conditions: (i),(ii)if and there is a point such that for all , then is extremely regular at . Then, there exist a -invariant closed boundary for and a continuous map from onto with on such that

Proof. Let . Since is a linear isometry, we deduce that is a uniformly closed linear subspace of . Moreover, implies that . We define the map by . Then, is a linear isometry from the real Banach space onto the real Banach space . Therefore, is a linear isometry from onto and so Now we define We first show that is nonempty. To prove this fact, we show that the following statement, namely (I), holds.(I) For each there exists such that .Let . Then by Theorem 1. From (42), there exists such that Since is a linear subspace of , there exists such that By Theorem 1, there exists such that From (44), (45), and (46), we have Since and , by (47) we have This implies that since . Hence, by (47) we have and so statement (I) holds.Since for all in , statement (I) implies that is nonempty.We next show that the following statement, namely, (II), holds.(II) For each there exists a unique point such that for all in .Let . Then, . Hence, by Theorem 1, there exists such that Since and , by (50) we have This implies that since . Thus, by (50) we have From (52) we give It follows that According to (54) and condition (ii), is extremely regular at . On the other hand, is extremely regular at by Lemma 4. Hence, by Lemma 3, we deduce that for all in or for all in . We assume that whenever for all in and whenever for all in . Then, and we have This proves the existent of . To show uniqueness, assume that there exists such that From (55) and (56) we have for all in . This implies that since separates the points of . Thus, statement (II) holds.Now we define the map by whenever for all in . The statement (II) implies that is well-defined. By definition of , we have We prove that in . Let . For each , by (57) we have This implies that since separates the points of . Hence, on .Continually, we show that is a -invariant. Let . For each we have This implies that Hence, by Theorem 1. Therefore, and so is -invariant.Now we show that is surjective. Let . According to statement (I), we deduce that there exists a point such that It follows that and we have Statement (II) implies that there exists a point such that From (62) and (63), we have for all in . This implies that since separates the points of . If , then by (63) and definition of we have . If , then by (63) and definition of we have for all in , and so . Therefore, is surjective.Continually, we show that is a boundary for . Let . Then, there exists a function such that . Since is a continuous complex-valued function on the compact space , there exists a point such that . The surjectivity of implies that there exists a point such that . Hence, , and so Therefore, is a boundary for .We now check that is continuous. Let , and let be a net in such that in . Then, for each we have in . This implies that for each in . Since separates the points of , we have in . Therefore, is continuous.Finally, we show that is closed in . Let , the closure of in . Then, there exists a net in such that in . Then, for each we have , and so in . Since the net converges for all in , we conclude that the net converges in to a point . Hence, for each we have . Therefore, for all in . This implies that for each we have Hence, This implies that , by Theorem 1. Therefore, . So is closed in .

Note that if is a linear isometry from onto , then and so is extremely regular at for all in . Therefore, if is a linear isometry from on with , then conditions (i) and (ii) in Theorem 10 hold.

We now study the into case (not necessarily unit-preserving).

Theorem 11. Let be a linear isometry from into satisfying the following conditions: (i) for all , (ii)if and there is a point such that for all in , then is extremely regular at . Then, there exist a -invariant closed boundary for , a function with for all , and a continuous map from onto with on such that

Proof. Assume that . Then, , and by (i) we have for all . Clearly, for all , and . We define the map by . It is easy to see that is a linear isometry from into such that .Applying (ii), we can easily show that if and there exists a point such that for all in , then is extremely regular at . Let . Then, . We define by and by . Clearly, is a linear isometry from onto , and is a linear isometry from onto . Now, we define It is easy to see that By given arguments in the proof of Theorem 10, is a -invariant closed boundary for , and there exists a continuous function from onto with such that We now show that is a boundary for . Let . Then, . Hence, there exists a point such that Since for all , we conclude that From (72) and (73), we have Therefore, is a boundary for .Finally, we show that Let and . From (73), we have Applying (76), we deduce that Hence, the proof is complete.

Note that if is a linear isometry from onto , then for all in by Lemma 8 and is extremely regular at for all in by Lemma 4, and so the conditions (i) and (ii) in Theorem 11 hold.

The following result shows that the triple associated to isometry in Theorem 11 possesses a universal property.

Corollary 12. Let be a linear isometry from into satisfying the following conditions: (i) for all , (ii)if and there is a point such that for all in , then is extremely regular at . Let and be as in Theorem 11. If is a nonempty -invariant subset (not necessarily closed) of   and with for all and is a continuous map (not necessarily onto) with on such that then , and .

Proof. Let . Since , taking in the above expression of , we have . Thus, .Let and defined by . Then, Let . Then, the above expression of reads as follows: From (80), for each we have This implies that From (82), we deduce that is an extreme point of by Theorem 1. Therefore, . So .We now show that . Let . By (80), we have Since , we deduce that , and so by Theorem 11 we have Since , by (83) and (84) we have This implies that since separates the points of . Thus, . Hence, the proof is complete.

Acknowledgment

The authors would like to thank the referee for some valuable and useful comments.

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