FUNCTIONS DEFINED ON THE INTERVAL
[0, 1]
H. R. SHATERYReceived 27 April 2004 and in revised form 26 September 2004
We prove that if the Banach algebra of bounded real Baire-1 functions (resp., small Baire classξ), defined on [0, 1], is the direct sum of two subalgebras, then one of its components contains a copy of it as a complemented subalgebra.
An important problem in topology is to determine the effects on the function space of imposing some natural topological condition on the spaceX[5].
In this way, it is practical to characterize the complemented subalgebras of Banach algebras. In our investigation, we relate the second category property of [0, 1] with certain properties of the complemented subalgebras of bounded real Baire-1 functions,β◦1([0, 1])
(bounded functions of finite Baire index,Ꮾ1([0, 1])).
We begin by recalling some definitions. Let A be a Banach algebra (resp., Banach space). Two subalgebras (resp., subspaces) M and N of A are complementary ifA=
M⊕N. A projection onAis a continuous linear operatorP:A→AsatisfyingP2=P.
IfM andN are complementary subalgebras (resp., subspaces) ofA, then there exists a projectionP onAsuch that the range ofP isM(N). The norm (sup-norm) of the pro-jectionPis always equal to or greater than 1. Norm 1 projections play a crucial role in the study of complemented subalgebras (resp., subspaces) of Banach algebras (resp., Banach spaces) (see, e.g., [4]). IfAis a finite-dimensional Banach space, then every nontrivial subspace ofAis closed and complemented inAand does not contain a copy ofA. This of course is more complicated for infinite-dimensional Banach spaces.
Pelczynski has proved that every infinite-dimensional closed linear subspace ofl1
con-tains a complemented subspace ofl1that is isomorphic tol1[4, Theorem 6, page 74]. Also
it has been proved thatC(X), the ring of continuous functions on the compact topolog-ical spaceXis the direct sum of two proper subrings if and only ifXis disconnected [5, Problem 1.B, page 20]. In this case, for every decomposition ofC(X), there is an open compact partition{A,B}ofXsuch thatC(X)=C(A)⊕C(B). We want to establish a re-sult similar to that of Pelczynski for the Banach algebra of bounded real Baire-1 functions defined on [0, 1].
Copyright©2005 Hindawi Publishing Corporation
Throughout this paper,Xis a compact subset of real numbers. The class of open (resp., closed) subsets ofXis denoted byᏳ(resp.,Ᏺ). We defineᏳδ(resp.,Ᏺσ) as the class of all of countable intersections (resp., unions) of elements ofᏳ(resp.,Ᏺ). We denote the set (Ᏻδ∩Ᏺσ) byᏴ.
LetXbe a topological space. We define the real Baire functions of class 1 as follows:
β1(X)=
f :X−→R:∃fn∞n=1⊆C(X)
such that limfn(x)=f(x), for eachx∈X. (1)
We denote the set of bounded functions inβ1(X) byβ◦1(X). The Baire-1 class,β1◦(X)
has an algebraic and isometric representation as the spaceC(ω) of all continuous func-tions on a totally disconnected compact spaceω. This representation was used to show that if the compact spaceShas an uncountable compact metrizable subset, thenβ◦1(S) is
not linearly isomorphic to any complemented subspace of the Banach spaceC(K) for σ-stonian spaceK [3]. In [1], Bade studied the linear complementation problem for the Baire classes. He proved that βα([0, 1]) is not complemented as a closed subspace ofβα+1([0, 1]) for each ordinalα < ω1.
In our investigation, we characterize the complemented topological subalgebras of the Baire-1 classes on [0, 1].
Theorem 1. If the Banach algebra of bounded real Baire-1 functions, defined on[0, 1], is the direct sum of two subalgebras, then one of its components contains a copy of it as a complemented subalgebra; that is, ifAandBare two subalgebras ofβ◦1([0, 1])such that
β◦
1
[0, 1]=A⊕B, (2)
then there exist two subalgebras,CandD, ofA(orB) such that
A=C⊕D, C∼=β◦
1
[0, 1]. (3)
Moreover, each complemented subalgebra ofβ◦1([0, 1])can be obtained by a norm1, positive
and multiplicative projection.
Proof. First we note that the idempotents of the ringβ◦1([0, 1]) areχH’s forH∈Ᏼ[1]. It
is obvious thatH∈Ᏼif and only if [0, 1]−H∈Ᏼ. Suppose that the ringβ◦1([0, 1]) is the
direct sum of two subringsAandB,
β◦
1
[0, 1]=A⊕B. (4)
The constant function ˆ1 belongs to β◦1([0, 1]), therefore there existe1∈A ande2∈B
such that
ˆ1=e1+e2. (5)
Thuse1ande2are two disjoint idempotents; that is,e1e2=0, becauseA∩B= {0}, and
Suppose thate1=χH1ande2=χH2 for suitableH1andH2inᏴ. By (5),{H1,H2}is a
partition of [0, 1] byᏴsets. Therefore, we conclude thatA=β◦
1([0, 1])χH1∼=β◦1(χH1) and
similarlyB=β◦
1(χH2). Now, suppose that{H1,H2}is a partition of [0, 1] byᏴsets. Let f
andgbe inβ1◦(H1) andβ1◦(H2), respectively. We definehas follows:
h(x)=
f(x) ifx∈H1,
g(x) ifx∈H2.
(7)
SupposeFis a closed subset ofR. Then
h−1(F)= f−1(F)∪g−1(F). (8)
Hence f−1(F) andg−1(F) areᏳ
δ sets in H1 andH2, respectively, and so they are both
Ᏻδ in [0, 1]. Consequentlyh−1(F) isᏳδin [0, 1] andh∈β◦1(X) [1]. It is obvious that if
h∈β◦
1(X), thenh|H1∈β◦1(H1) andh|H2∈β◦1(H2). We now defineϕfromβ◦1([0, 1]) onto
β◦
1(H1)⊕β◦1(H2) as
ϕ(f)=f|H1,f|H2
. (9)
Thenϕis a surjective algebra isomorphism. Thus there exists a one-to-one correspon-dence between algebra decompositions ofβ1◦([0, 1]) and theᏴpartitions of [0, 1].
The interval [0, 1] is a second-category topological space, andH1andH2are countable
unions of closed sets, therefore one of them has nonempty interior. Suppose the interior ofH1is not empty. Then there exists a closed interval [a,b]⊆H1. Clearly, we have
β◦
1
[0, 1]∼=β◦
1
[a,b]. (10)
On the other hand, there exists anᏴset,H3in [0, 1] disjoint from [a,b] such that
H1=[a,b]∪H3. (11)
Therefore,
β◦
1
H1
=β◦
1
[a,b]⊕β◦
1
H3
. (12)
Thusβ1◦([0, 1]) is complemented inβ◦1(H1).
Now we prove the second assertion. LetCbe a complemented subalgebra ofβ1◦([0, 1]).
Therefore, there exists anᏴset,Hin [0, 1] such thatC=β◦
1(H). We define
P:β◦1
[0, 1]−→β◦
1
[0, 1],
P(f)= f|H. (13)
LetHc=[0, 1]−Hand f ∈β1◦([0, 1]). We have f = f|H+f|Hc. Thus,
f =maxf|H,f|Hc≥f|H. (14)
Therefore,P ≤1. It follows thatP =1 since the norm of a projection is always equal to or greater than 1. If f ∈β◦1([0, 1]) and f ≥0, then f|H≥0, and therefore,P is
The finite and small Baire classes have been studied by many people (e.g., [2,7]). We begin by recalling the definition of the indexβ. Suppose thatHis anᏴset in [0, 1], andf is a real-valued function whose domain isH. For any>0, letH0(f,)=H. IfHα(f,) is defined for some countable ordinalα, letHα+1(f,) be the set of all thosex∈Hα(f,)
such that for every openUcontainingx, there are two pointsx1andx2inU∩Hα(f,)
with|f(x1)−f(x2)| ≥. For a countable limit ordinalα, we let
Hα(f,)= ∩
α<αHα(f,). (15)
The indexβH(f,) is taken to be the leastαwithHα(f,)= ∅if suchαexists, andω1
otherwise. The oscillation index of f is
βH(f)=supβH(f,) :>0. (16)
It is known that f :H→Ris Baire-1 if and only ifβH(f)< ω1[7]. We define the set
of bounded functions of finite Baire index (resp., small Baire classξfor each countable ordinalξ) as
Ꮾ1(H)=
f ∈β◦
1(H) : βH(f)<∞
, Ꮾξ1(H)=
f ∈β◦
1(H) : βH(f)≤ωξ
.
(17) The set of bounded functions of finite Baire index,Ꮾ1 (resp., small Baire classξ,
Ꮾξ1(H)), is a Banach algebra (with sup-norm). It is obvious that iff :H1→Ris a Baire-1
function,H2⊆H1⊆[0, 1] (H1,H2∈Ᏼ) andg= f|H2, thenβH2(g)≤βH1(f). Therefore,
if f ∈Ꮾ
1 (resp., f ∈Ꮾ
ξ
1(H1)), theng∈Ꮾ1 (resp.,g∈Ꮾ
ξ
1(H2) (but the converse is not
true). So we have the following.
Remark 2. The previous theorem is also valid for the set of bounded functions of finite Baire index (resp., small Baire classξ).
It may happen that for twoᏴsets,H1 andH2 (⊆[0, 1] such thatH2=[0, 1]−H1),
and f1∈Ꮾ1(H1) andf2∈Ꮾ1(H2), f = f1+f2does not belong toᏮ1([0, 1]). Suppose
thatH1is the Cantor setCin the interval [0, 1] andH2=[0, 1]−C. Letfi=iχHi(i=1, 2),
and f = f1+f2. It is obvious thatf /∈Ꮾ1([0, 1]). Thus,
Ꮾ
1
[0, 1]Ꮾ1
H
1⊕Ꮾ1
H
2. (18)
But ifMandNare two complementary subalgebras ofᏮξ1([0, 1]), then there exist suitable
Ᏼsets,H1andH2, such that
M=Ꮾ
1
H1
, N=Ꮾ
1
H2
. (19)
By the above argument and using the epimorphism
Θ:Ꮾ1
[0, 1]−→Ꮾ1(C),
we see that there exists a noncomplemented subalgebraᏰsuch that
Ꮾ1
[0, 1]
Ᏸ ∼=Ꮾ1(C), (21)
andᏰis not of the formᏮ1(H) for anyᏴsetH. (The algebra homomorphismΘis onto
by Tietze extension theorem [8, Theorem 3.6].)
It has been proved that for real compact spacesX andY, a linear isometry between β◦
1(X) andβ1◦(Y) induces an algebra (a ring) isometry [6]. Is this true for linear
com-plemented subspaces ofβ◦1([0, 1])? If the answer to the above question is positive, then
it should be easy to prove an analogous theorem for linear complemented subspaces of β◦
1([0, 1]).
Acknowledgment
The author is grateful to the referees for their helpful comments.
References
[1] W. G. Bade,Complementation problems for the Baire classes, Pacific J. Math.45(1973), 1–11. [2] F. Chaatit, V. Mascioni, and H. Rosenthal,On functions of finite Baire index, J. Funct. Anal.142
(1996), no. 2, 277–295.
[3] F. K. Dashiell Jr.,Isomorphism problems for the Baire classes, Pacific J. Math.52(1974), 29–43. [4] J. Diestel,Sequences and Series in Banach Spaces, Graduate Texts in Mathematics, vol. 92,
Springer-Verlag, New York, 1984.
[5] L. Gillman and M. Jerison,Rings of Continuous Functions, The University Series in Higher Mathematics, D. Van Nostrand, New Jersey, 1960.
[6] J. E. Jayne,The space of classαBaire functions, Bull. Amer. Math. Soc.80(1974), 1151–1156. [7] A. S. Kechris and A. Louveau,A classification of Baire class1functions, Trans. Amer. Math. Soc.
318(1990), no. 1, 209–236.
[8] D. H. Leung and W.-K. Tang,Functions of Baire class one, Fund. Math.179(2003), no. 3, 225– 247.
H. R. Shatery: Department of Mathematics, University of Isfahan, 81745-163 Isfahan, Iran
