Abstract
Let be a complex Banach space and be the Banach space of all bounded continuous functions from a Hausdorff space to , equipped with sup norm. A closed subspace of is said to be an -valued function algebra if it satisfies the following three conditions: (i) is a closed subalgebra of , the Banach space of all bounded complex-valued continuous functions; (ii) for all and ; and (iii) for every and for every . It is shown that -homogeneous polynomial and analytic numerical index of certain -valued function algebras are the same as those of .
1. Introduction
In this paper, we consider only complex nontrivial Banach spaces. Given a Banach space , we denote by and its closed unit ball and unit sphere, respectively. Let be the dual space of . If and are Banach spaces, a -homogeneous polynomial from to is a mapping such that there is a -linear continuous mapping from to such that for every in . The Banach space of all -homogeneous polynomials from to is denoted by endowed with the polynomial norm . We refer to [1] for background knowledge on polynomials.
We are mainly interested in the following spaces. For two Banach spaces , and a Hausdorff topological space , where is the interior of . Then is a Banach space under the sup norm and both and are closed subspaces of . In case that is the complex scalar field , we just write , , and . Let The spatial numerical range of in is defined by and the numerical radius of is defined by
Let be a Banach space. The -homogeneous polynomial numerical index is defined in [2] by The -analytic numerical index and -analytic index are defined, respectively, by It is clear from the definitions that for all .
Choi, García, Kim, and Maestre showed [3] that and for uniform algebras . In general, it is not difficult to see that if is a (unital) function algebra on a Hausdorff space, then, by the Gelfand transform, is isometric to a (unital) uniform algebra on where is the maximal ideal space of . We present this fact in Proposition 2 for the completeness of the paper. In this paper, we introduce a -valued function algebra and the Gelfand transform does not work in this case. In the proof of [3], they used a very useful Urysohn type theorem, which was obtained by Cascales, Guiro, and Kadets [4]. Recently, Kim and the author found [5] that a Urysohn type theorem holds for some function algebras. It plays an important role in the main results of this paper. For some geometric properties on -homogeneous polynomial (analytic) numerical index, refer to [6, 7].
Let us briefly review some necessary notions. A nontrivial -closed subalgebra of of is called a function algebra on a Hausdorff space . For a Banach space , a nontrivial subspace of is said to be an -valued function algebra if it satisfies three conditions: (i) is a function algebra on ; (ii) , where for ; and (iii) for every and , where for . A subset of is said to be norming for if holds for all . By unital function algebra, we mean a function algebra containing all constant functions. A function algebra on a compact Hausdorff space is said to be a uniform algebra if separates the points of (that is, for every in , there is such that . Note that the definition of function algebra in this paper is different from the usual one in [8].
Let be an element of an -valued function algebra . The is said to be a peak function at if there exists a unique such that . A peak function is said to be a strong peak function at if and for every open subset containing we get The corresponding point is called a strong peak point for . We denote by the set of all strong peak points for . It is easy to see that if is compact, then every peak function is a strong peak function. It is worth remarking that if is a nontrivial separating separable subalgebra of on a compact Hausdorff space , then is a norming subset for [9]. There is a compact Hausdorff space such that is an empty set [10]. For more information about peak functions and points, refer to [8, 10].
For an -valued function algebra , let . Then . Indeed, if is a strong peak function at , then choose such that and it is clear that is a strong peak function in at . Therefore, . Conversely, if is a strong peak function at , then choose . Therefore, is a strong peak function at . Hence we have . In addition, if is norming for , then it is also norming for since in has the same norm as for every and .
The following lemma will be useful to get main results. In proofs of the main results, the denseness of the strong peak functions in an -valued function algebra is an important part and equivalent to the fact that the set of strong peak points is norming for . That means that the fact that every element in can be approximated by the sequence of strong peak functions is equivalent to the fact that the norm of every element in can be approximated on the set of strong peak points for . The approximation by strong peak functions will prove to be useful to deal with the geometric properties of function algebras especially those related to generalized numerical indices of Banach spaces.
Lemma 1 (see [5]). Let be a function algebra on and fix . Then, given and for every open subset containing , there exists a strong peak function such that , , and for all ,
2. Main Results
The proof of [3, Theorem 2.1] shows that if is a uniform algebra. Since a function algebra is isometric to a uniform algebra by the Gelfand transform, we have the following.
Proposition 2. Let be a function algebra on a Hausdorff space . Then it is isometric to a uniform algebra on a compact Hausdorff space and .
Proof. Let be a function algebra and be the set of all nonzero algebra homomorphisms from to . The maximal ideal space is a compact Hausdorff space with the Gelfand topology. The Gelfand transform of is defined by for . For , let be the dirac delta function by for . Fix a nonzero and let ; then for all and Since the Gelfand transform is a homomorphism, is isometrically isomorphic to the image , where is the image of the Gelfand transform. Then is a closed subalgebra of and it is separating the points of . Thus, it is a uniform algebra on the compact Hausdorff space .
For the second part, the proof used in [3, Theorem 2.1] to show can be applied to show that for uniform algebras .
Proposition 2 gives a positive answer to the third question raised by Acosta and Kim [11].
Theorem 3. Let be a Banach space and suppose that is an -valued function algebra on a Hausdorff space such that is a norming subset for . Then we have (i) for every ,(ii) and(iii).
Proof. We prove holds. The proofs for the other two cases are exactly the same. It is well-known that for all complex Banach spaces [12].
Let . Then is a function algebra. Let with and be given. Choose so that . Since is norming for , find such that . Since is continuous, there is such that for every with .
Let and be an open subset of containing . Then by Lemma 1, there is a strong peak function such that and for every , and for every .
Define by for all . It is easy to check that is well-defined and for all . Then, let , Then we have the following. Choose such that and find a complex number with and a proper satisfying . Then the function is an element of . By the maximum modulus theorem, there exists with such that takes its maximum modulus on . Hence, Let , choose with , and define the function by for . Then , and hence . Let for . Then . Then Since , there is so that Note that because . Hence we have Since is arbitrary, . This holds for all with . Therefore, we get .
A version of the Bishop-Phelps-Bollobás type theorem for holomorphic functions has been shown [5, 13]. In the following theorem, we present a similar result. However the main focus is the denseness of the set of all strong peak functions, which is different from that of the results in [5].
Theorem 4. Let be a Banach space and an -valued function algebra on a Hausdorff space . Then, given , whenever a norm-one element in and a point in satisfy , there is a norm-one strong peak function at such that .
Proof. Suppose that satisfies the prescribed conditions. Then is an open set containing . There exists . Using Lemma 1, take a strong peak function such that , , and for all . Set It is easy to check that and . Moreover, from the inequality we have that for all if we consider two cases and . Hence, we get and complete the proof since we know is a strongly norm attaining function from the fact that is a strong peak function.
From Theorem 4, we have the following consequence.
Corollary 5. Let be a Banach space and be an -valued function algebra on a Hausdorff space . Then the set is norming if and only if the set of strong peak functions in is dense.
Proof. The necessity is proved by Theorem 4. For the converse, assume that the set of strongly norm attaining functions in is dense in . Given , there is a sequence of strong peak functions in such that . For each , let be the strong peak point corresponding to . Then Thus, This means that . This shows that is a norming subset of .
Theorem 6. Let be a Banach space and be an -valued function algebra on a Hausdorff space such that is a norming subset for . Fix and define the map by for and . Suppose that is an element of for every and for every . Then we have .
Proof. By Theorem 3, we have only to show that . Consider the set Let be the natural projection. Then since is norming for , Corollary 5 shows that is dense in . Then, it is shown [14] that for every , we have Given with , we have . Indeed, is a map from to . Since is uniformly continuous on , given , there is such that if and , then . If and , then for all . Hence . This shows that is uniformly continuous on . Now it is enough to show that is -holomorphic on [15]. Fix and , and let be an open subset in the complex plane. Let for . Then is a -valued continuous function on . For each , is holomorphic. Fix and choose such that . The Cauchy integral formula shows that, for each , As a result, we have since the continuity of implies the Bochner integrability of the integral. This means that is holomorphic on and is holomorphic on [15]. We also have since there is a strong peak function at such that and is in for each . It is clear that . For every , there is such that Therefore, we get .
The same proof shows the following.
Theorem 7. Let be a Banach space and be an -valued function algebra on a Hausdorff space such that is a norming subset for . Fix and define the map by for and . Suppose that is an element of for every in and for every ). Then we have .
Proof. The main difficulty in the proof of Theorem 7 is to check that is in . Let be the corresponding continuous -linear map defining . Let by for and . Then it is easy to check that is a continuous -linear map and for . The other part of the proof is the same as the proof of Theorem 6.
Let be Banach spaces and let be either or . Notice that are -valued function algebras over . If a Banach space is finite dimensional, is the set of all complex extreme points of as observed in [16, 17]. A strongly exposed point of is a strong peak point for , so if a strongly exposed point of is dense in , then and it is norming for . It is also proved in [17] that if is locally -uniformly convex space and it is an order continuous sequence space, then is norming. The typical example of uniformly complex convex sequence space is . For the definitions related to various complex convexities and more examples, we refer to [9, 18–21].
Let be a closed convex and bounded set in a Banach space . The set has the Radon-Nikodým property if, for every probability space and every -valued countably additive measure on such that for every with , there is a Bochner measurable so that
The space is said to have the Radon-Nikodým property if its unit ball has the Radon-Nikodým property [22]. For the basic properties and useful information on the Radon-Nikodým property, see also [22–25]. It has been shown [9] that if has the Radon-Nikodým property, then is norming for .
Corollary 8. Suppose that satisfies one of the following conditions: (i) has the Radon-Nikodým property; (ii) is locally uniformly convex space; (iii) is a locally -uniformly convex order continuous sequence space. Then we have (i) for every ,(ii).
Proof. If satisfies one of the three conditions, and it is norming for . Therefore, Theorem 3 implies that , . For the case (ii), fix and define the map by for and . Then . Consequently, Theorem 6 shows that and the proof of (ii) is complete. The remaining proof (i) can be finished in the same way by Theorem 7.
By Theorem 3, we get the following.
Corollary 9. Let be a Hausdorff topological space and suppose that is norming for . If is a Banach space with , we have for all .
As we show in the next proposition, closed bounded convex sets with the Radon-Nikodým property satisfy the condition of Corollary 9.
Proposition 10. Suppose that is a nonempty closed bounded convex subset of a Banach space and has the Radon-Nikodým property. Then is norming for and the set of strong peak functions of is dense.
Proof. It is enough to show that the set of strong peak functions of is dense by Corollary 5. Given and , the Stegall perturbed optimization theorem [25] shows that there is such that the function strongly attains its norm at and . Choose a complex number such that Then it is easy to check that is a strong peak function at and . This shows the denseness of the set of strong peak functions on .
Data Availability
No data were used to support this study.
Conflicts of Interest
The author declares no conflicts of interest.
Acknowledgments
The author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2016R1D1A1B03934771).