# The Centre of the Spaces of Banach Lattice-Valued Continuous Functions on the Generalized Alexandroff Duplicate

Authors: Faruk Polat
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We characterize the centre of the Banach lattice of Banach latticeE-valued continuous functions on the Alexandroff duplicate of a compact Hausdorff space K in terms of the centre of C(K,E), the space of E-valued continuous functions on K. We also identify the centre of CD0(Q,E)=C(Q,E)+c0(Q,E) whose elements are the sums of E-valued continuous and discrete functions defined on a compact Hausdorff space Q without isolated points, which was given by Alpay and Ercan (2000).


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## Body

## 1. Preliminaries and Definitions

Throughout the paper, our terminology is mainly standard and a background on Riesz spaces and Banach lattices may be obtained from [1] or [2]. In order to avoid trivial cases, we assume that all topological spaces are nonempty and all Banach lattices are nonzero.Thecentre of a Banach lattice E, denoted by Z(E), is the lattice of the linear operators, T:E→E for which there exists a real number λ>0 such that |Tx|≤λ|x| for all x∈E. The operator norm of a central operator T is the minimum of those λ with this property. It is well known that  Z(E) equipped with the operator norm is an AM-space with order unit. The order unit is identity operator I.For a given locally compact Hausdorff spaceK and a Banach lattice E, C0(K,E) denotes the space of all continuous functions f from K into E which vanish at infinity; that is, there exists a compact set A⊂K such that ∥f(k)∥<ɛ  for each ɛ>0 and k∈K∖A. We consider this space to be normed by (1.1)‖f‖=sup⁡{‖f(k)‖:k∈K},
and ordered by (1.2)f≥g⟺f(k)≥g(k),∀k∈K.
One can show that C0(K,E) is a Banach lattice with these definitions.Ercan and Wickstead [3] showed that the centre of C0(K,E) is isometrically Riesz isomorhic to Cb(K,Z(E)s) the space of all functions f from K into Z(E) such that f is norm bounded, continuous, and f(kα)(e)→f(k)(e) in E for each e∈E whenever kα→k in K. Here, Z(E) is given the strong operator topology.IfK is a compact Hausdorff space, then C0(K,E)=C(K,E), where C(K,E) is the space of continuous functions f:K→E. Hence, the centre of C(K,E) can also be identified with Cb(K,Z(E)s). We will use this identification in the sequel.IfK is a discrete topological space, then C0(K,E) is the space of E-valued bounded functions f on K such that the set (1.3){k∈K:ɛ<‖f(k)‖}
is finite for each ɛ>0, and we will write c0(K,E) in this case.LetΣ and Γ be compact Hausdorff and locally compact Hausdorff topologies on a nonempty set K, respectively, such that Σ is coarser than Γ. These topologies on K will be denoted by KΣ and KΓ. The compact Hausdorff topology on K×{0,1} generated by the open base 𝒜=𝒜1∪𝒜2, where (1.4)A1={H×{1}:HisΓ-open},A2={G×{0,1}∖M×{1}:GisΣ-open,MisΓ-compact}
is called generalized Alexandroff duplicate of K and denoted by KΣ,Γ⊗{0,1} (see [4]). When Γ is discrete topology on K, the compact Hausdorff topological space KΣ,Γ⊗{0,1} will be denoted by A(K). The space A(K) was first considered by Engelking [5]. For K=[0,1] under the usual metric topology, A(K) was constructed by Alexandroff and Urysohn [6] as an example of a compact Hausdorff space containing a discrete dense subspace. This space is called the Alexandroff duplicate.Note thatK×{0} is a closed subspace of KΣ,Γ⊗{0,1} and the map k→(k,0) is a homeomorphism between KΣ and K×{0}.In [4], it is not proved that KΣ,Γ⊗{0,1} is a compact Hausdorff space. We give the proof here for the benefit of the reader.Theorem 1.1.
KΣ,Γ⊗{0,1} is a compact Hausdorff space.Proof.
Consider an open cover{Oi}i∈I of KΣ,Γ⊗{0,1}. By replacing each set in the cover by a union of basic open neighborhoods of all points in the set, we can assume that the cover is formed by basic open neighborhoods of the form
(1.5){Hα×{1}}α∈I∪{Gγ×{0,1}∖Mγ×{1}}γ∈Ω,
where Hα is a Γ-open set, Gγ is a Σ-open set, and Mγ is a Γ-compact set. It is easy to see that {Gγ×{0}}γ∈Ω is an open cover of K×{0}, thus there is a finite subcover Gγ1×{0},…,Gγn×{0}. Then,
(1.6)Gγ1×{0,1}∖Mγ1×{1}∪⋯∪Gγn×{0,1}∖Mγn×{1}
misses only finitely many Γ-compact sets Mγ1×{1},…,Mγn×{1}.
AsMγj (j=1,2,…,n) is compact, we have that Mγj×{1}⊂∪Hα×{1}. So, Mγj×{1}⊂∪p=1nHpj×{1}. Hence, if we add the corresponding open sets from the cover, then we obtain a finite cover of the entire space KΣ,Γ⊗{0,1}. Therefore, KΣ,Γ⊗{0,1} is compact.
To show thatKΣ,Γ⊗{0,1} is Hausdorff, it is enough to show that (k,0) and (k,1) can be separated. Let V be a Γ-open neighborhood of k such that clΓ(V) (closure of V in KΓ) is compact. Then, KΣ,Γ⊗{0,1}∖(clΓ(V)×{1}) and V×{1} are the separating open sets of (k,0) and (k,1), respectively. This completes the proof.IfKΣ is a compact Hausdorff space without isolated points and KΓ is a discrete topological space, then C(KΣ,E)∩c0(KΓ,E)={0} and CD0(KΣ,E)=C(KΣ,E)⊕c0(KΓ,E) is a Banach lattice under the pointwise ordering and supremum norm of the sums f+d, where f∈C(KΣ,E) and d∈c0(KΓ,E). We refer to [7–9] for more detailed information on these spaces. In [4], it is showed that CD0(KΣ,E) is isometrically Riesz isomorphic to C(A(K),E), where A(K) is the Alexandroff duplicate of K. We will use this identification in the sequel to characterize the centre of the space CD0(KΣ,E).


## 2. Main Results

LetΣ and Γ be compact Hausdorff and locally compact Hausdorff topologies on K, respectively, such that Σ is coarser than Γ, and let E be a Banach lattice. Then Cb*(KΣ,Z(E)s) denotes the set of all norm bounded and continuous functions f from K into Z(E) such that rαf(kα)(e)→rf(k)(e) in E for each e∈E whenever (kα,rα)→(k,r) in KΣ,Γ⊗{0,1}.We consider the vector spaceCb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) equipped with coordinatewise algebraic operations, the order (2.1)0≤(f,d)⟺0≤f(k)(e),0≤f(k)(e)+d(k)(e)foreachk∈K,
and the norm (2.2)‖(f,d)‖=max⁡{‖f(k)+rd(k)‖:(k,r)∈K×{0,1}}.The norm defined onCb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) makes it a Banach space. This is clear, as this norm is equivalent to standard products norms (we have, e.g., (1/2)max{∥f∥,∥d∥}≤∥(f,d)∥≤(∥f∥+∥d∥)). This has no relation to Banach lattices, but it is just a property of Banach spaces. The space Cb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) is a lattice. This is proved by computing |(f,d)|=(|f|,|f+d|-|f|), where the absolute values on the right-hand side are pointwise. The norm defined on Cb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) is a Riesz norm. This is obvious from definitions. Therefore, the space Cb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) is a Banach lattice.Actually, the spaceCb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) is isometrically Riesz isomorphic to Cb(KΣ,Γ⊗{0,1},Z(E)s) the space of norm bounded, continuous functions f from K×{0,1} into Z(E) such that f(kα,rα)(e)→f(k,r)(e) in E for each e∈E whenever (kα,rα)→(k,r) in KΣ,Γ⊗{0,1} as the following shows.Theorem 2.1.
Cb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s) and Cb(KΣ,Γ⊗{0,1},Z(E)s) are isometrically Ries isomorphic spaces.Proof.
Define the map(2.3)π:Cb(KΣ,Z(E)s)×Cb*(KΣ,Z(E)s)⟶Cb(KΣ,Γ⊗{0,1},Z(E)s),
by
(2.4)π(f,d)(k,r)(e)=f(k)(e)+rd(k)(e),
for each (k,r)∈K×{0,1} and e∈E.
Let(kα,rα)→(k,r) in KΣ,Γ⊗{0,1}. Then, kα→k in KΣ so that f(kα)(e)→f(k)(e) and rαd(kα)(e)→rd(k)(e) in E for each e∈E. Hence, f(kα)(e)+rαd(kα)(e)→f(k)(e)+rd(k)(e) in E for each e∈E so that the map π is well defined. It follows immediately that π is an isometry, as π(f,d) agrees with f+d on K×{1} and with f on K×{0}. It is obvious that π(f,d)≥0⇔(f,d)≥0.
It remains to show thatπ is onto. Let h∈Cb(KΣ,Γ⊗{0,1},Z(E)s) be given. Define
(2.5)f(k)(e)=h(k,0)(e),d(k)(e)=h(k,1)(e)-h(k,0)(e),
for each k∈K and e∈E. The norm boundedness of f and d follows directly from the norm boundedness of h. If kα→k in KΣ, then (kα,0)→(k,0) in KΣ,Γ⊗{0,1} so that
(2.6)f(kα)(e)=h(kα,0)(e)⟶h(k,0)(e)=f(k)(e),
in E for each e∈E, hence f∈Cb(KΣ,Z(E)s).
To show thatd∈Cb*(KΣ,Z(E)s), let (kα,rα)→(k,r)∈KΣ,Γ⊗{0,1}. We now examine the possibilities.
Suppose first thatr=1. Then, (rα) is eventually 1. As (kα,0)→(k,0) in KΣ,Γ⊗{0,1}, we have rαd(kα)(e)→rd(k)(e) in E for each e∈E in this possibility.
Suppose now that(kα,rα)→(k,0) and assume that rαd(kα)(e) does not converge to zero in E. Then, there is a subnet (rαβ) of (rα) such that rαβ=1 and ɛ<∥d(kαβ)(e)∥ for each β and for some ɛ>0. On the other hand, since (kαβ,1)→(k,0) and (kαβ,0)→(k,0) in KΣ,Γ⊗{0,1}, we have h(kαβ,1)(e)→h(k,0)(e) and h(kαβ,0)(e)→h(k,0)(e) so that d(kαβ)(e)=h(kαβ,1)(e)-h(kαβ,0)(e)→0. This contradiction shows that d∈Cb*(KΣ,Z(E)s). It is clear that π(f,d)=h, and this completes the proof.SinceZ(C(KΣ,E)) and Z(C(KΣ,Γ⊗{0,1},E)) can be identified with Cb(KΣ,Z(E)s) and Cb(KΣ,Γ⊗{0,1},Z(E)s), respectively, we immediately have the following from the previous theorem.Corollary 2.2.
Z(C(KΣ,Γ⊗{0,1},E) and Z(C(KΣ,E))×Cb*(KΣ,Z(E)s) are isometrically Riesz isomorphic spaces.LetKΓ be a discrete topology, and let E be a Banach lattice. The set of all bounded functions f:K→Z(E) such that the set {k:ɛ<∥f(k)(e)∥foralle∈E} is finite will be denoted by c0(KΓ,Z(E)s).Lemma 2.3.
LetKΣ be a compact Hausdorff space, and let Γ be a discrete topology on K. Then, Cb*(KΣ,Z(E)s)=c0(KΓ,Z(E)s).Proof.
Letf∈c0(KΓ,Z(E)s). Suppose that f∉Cb*(KΣ,Z(E)s). Then, there exists a net (kα,1) in A(K) such that (kα,1)→(k,0)∈A(K) and ɛ<∥f(kαβ)(e)∥ for some subnet (kαβ) of (kα), ɛ>0, and for each e∈E. So, (kαβ) has finite range which is a contradiction. Conversely, assume that f∈Cb*(KΣ,Z(E)s) but f∉c0(KΓ,Z(E)s). Then, there exist some e∈E and a sequence (kn) such that ɛ<∥f(kn)(e)∥ for each n and kn≠km whenever n≠m. Then, there exists a subnet (knα) of kn such that (knα,1)→(k,0) so that f(knα)(e)→0 which is impossible and this completes the proof.By Theorem2.1 and the previous lemma, we have the following.Theorem 2.4.
LetKΣ be a compact Hausdorff space, and let Γ be a discrete topology on K. Then, Cb(A(K),Z(E)s) and Cb(KΣ,Z(E)s)×c0(KΓ,Z(E)s) are isometrically Riesz isomorphic spaces.As the centre ofCD0(KΣ,E) can be identified with Cb(A(K),Z(E)s), we immediately have Theorem  3.1 of [8] as follows.Corollary 2.5.
LetKΣ be a compact Hausdorff space without isolated points, and let Γ be a discrete topology on K. Then, the centre of CD0(KΣ,E) and Z(C(KΣ,E))×c0(KΓ,Z(E)s) are isometrically Riesz isomorphic spaces.Note that in the corollary above, if all the operatorsT∈Z(E) are norm attaining; that is, there exists some e∈E with ∥e∥=1 such that ∥T∥=∥T(e)∥, then c0(KΓ,Z(E)s) can be replaced by c0(KΓ,Z(E)).



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*Source: 604931-2011-04-06.xml*