# ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BAIRE INDEX FUNCTIONS

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Chaatit, Mascioni, and Rosenthal de ned nite Baire index for a bounded real-valued function f on a separable metric space, denoted by i(f), and proved that for any bounded functions f and g of nite Baire index, i(h) i(f) + i(g), where h is any of the functions f + g, fg, f ˅g, f ^ g. In this paper, we prove that the result is optimal in the following sense : for each n; k < ω, there exist functions f; g such that i(f) = n, i(g) = k, and i(h) = i(f) + i(g).

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### ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BAIRE INDEX FUNCTIONS

• 1. MATEMATIˇCKI VESNIK MATEMATIQKI VESNIK 69, 3 (2017), 207–213 September 2017 research paper originalni nauqni rad ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BAIRE INDEX FUNCTIONS Atok Zulijanto Abstract. Chaatit, Mascioni, and Rosenthal deﬁned ﬁnite Baire index for a bounded real-valued function f on a separable metric space, denoted by i(f), and proved that for any bounded functions f and g of ﬁnite Baire index, i(h) ≤ i(f) + i(g), where h is any of the functions f + g, fg, f ∨ g, f ∧ g. In this paper, we prove that the result is optimal in the following sense : for each n, k < ω, there exist functions f, g such that i(f) = n, i(g) = k, and i(h) = i(f) + i(g). 1. Introduction A real-valued function f deﬁned on a separable metric space X is called a diﬀerence of bounded semicontinuous functions if there exist bounded lower semicontinuous functions u and v on X such that f = u−v. The class of all such functions is denoted by DBSC(X). Some authors have studied this class and some of its subclasses (see, e.g. [1,3]). Chaatit, Mascioni, and Rosenthal [1] deﬁned ﬁnite Baire index for functions belonging to DBSC(X), whose deﬁnition we now recall. Let X be a separable metric space. For a given bounded function f : X → R, the upper semicontinuous envelope Uf of f is deﬁned by Uf(x) = limy→xf(y) = inf{sup y∈U f(y) : U is a neighborhood of x} for all x ∈ X. The lower oscillation oscf of f is deﬁned by oscf(x) = limy→x|f(y) − f(x)| for all x ∈ X. Finally, the oscillation osc f of f is deﬁned by osc f = Uoscf. Next, for any ε > 0, let os0(f, ε) = X. If osj(f, ε) has been deﬁned for some j ≥ 0, let osj+1(f, ε) = {x ∈ L : osc f|L(x) ≥ ε}, where L = osj(f, ε). A bounded function 2010 Mathematics Subject Classiﬁcation: 26A21, 54C30, 03E15 Keywords and phrases: Finite Baire index; oscillation index; Baire-1 functions. 207
• 2. 208 On optimality of the index f : X → R is said to be of ﬁnite Baire index if there is an n < ω such that osn(f, ε) = ∅ for all ε > 0. Then the ﬁnite Baire index of f is deﬁned by i(f) = max ε>0 i(f, ε), where i(f, ε) = sup{n : osn(f, ε) = ∅}. Clearly, if f ∈ DBSC(X) then f is a Baire-1 function, that is, the pointwise limit of a sequence of continuous functions. Based on the Baire Characterization Theorem, Kechris and Louveau [4] deﬁned the oscillation index of real-valued Baire-1 functions. The study on oscillation index of real-valued Baire-1 functions was continued by several authors (see, e.g., [3,5,6]). We recall here the deﬁnition of oscillation index. Let C denote the collection of all closed subsets of a Polish space X. Now, let ε > 0 and a function f : X → R be given. For any H ∈ C, let D0 (f, ε, H) = H and D1 (f, ε, H) be the set of all x ∈ H such that for every open set U containing x, there are two points x1 and x2 in U ∩ H with |f(x1) − f(x2)| ≥ ε. For all α < ω1 (ω1 is the ﬁrst uncountable ordinal number), set Dα+1 (f, ε, H) = D1 (f, ε, Dα (f, ε, H)). If α is a countable limit ordinal, let Dα (f, ε, H) = α <α Dα (f, ε, H). The ε-oscillation index of f on H is deﬁned by βH(f, ε) =    the smallest ordinal α < ω1 such that Dα (f, ε, H) = ∅ if such an α exists, ω1, otherwise. The oscillation index of f on the set H is deﬁned by βH(f) = sup{βH(f, ε) : ε > 0}. We shall write β(f, ε) and β(f) for βX(f, ε) and βX(f) respectively. In fact, a function f is of ﬁnite Baire index if and only if β(f) < ∞ and then β(f) = i(f) + 1. Chaatit, Mascioni, and Rosenthal proved in [1] that if f and g are real-valued bounded functions of ﬁnite Baire index and h is any of the functions f + g, fg, f ∨ g, f ∧ g, then i(h) ≤ i(f) + i(g). In this paper, we prove that the estimate i(h) ≤ i(f) + i(g) in [1, Theorem 1.3] is optimal in the following sense : For any n, k < ω, there exist bounded real-valued functions f and g such that i(f) = n, i(g) = k, and i(h) = i(f) + i(g). We process the proof by constructing functions on ordinal spaces [1, ωn+k ] and then we extend the construction to any compact metric space K such that K(n+k) = ∅, where K(α) is the αth Cantor-Bendixson derivative of K. Note that for any function f on K, Dα (f, ε, K) ⊆ K(α) , for any α < ω1. 2. Results Before we construct functions on ordinal spaces to show that Theorem 1.3 in [1] is optimal, we prove the following fact that we will use later.
• 3. A. Zulijanto 209 Lemma 2.1. Let X, Y be Polish spaces and ε > 0 be given. If θ : X → Y is a homeomorphism and ρ : Y → R, then Dα (ρ, ε, Y ) = θ(Dα (ρ ◦ θ, ε, X)) for all α < ω1. Proof. We prove the lemma by induction on α. The statement in the lemma is true whenever α = 0 since θ is surjective. By the injectivity of θ, the lemma is also true if α is a limit ordinal. Suppose that the statement in the lemma is true for some α < ω1. Let y ∈ Dα+1 (ρ, ε, Y ). Since θ is bijective, there is a unique x ∈ X such that y = θ(x). Let U be a neighborhood of x. Since θ is a homeomorphism, θ(U) is open in Y . Therefore, there exist y1, y2 ∈ θ(U) ∩ Dα (ρ, ε, Y ) such that |ρ(y1) − ρ(y2)| ≥ ε. Let x1 = θ−1 (y1) and x2 = θ−1 (y2), then x1, x2 ∈ θ−1 (θ(U) ∩ Dα (ρ, ε, Y )). Since θ−1 (θ(U) ∩ Dα (ρ, ε, Y )) = θ−1 (θ(U)) ∩ θ−1 (θ(Dα (ρ ◦ θ, ε, X)) by the inductive hypothesis = U ∩ Dα (ρ ◦ θ, ε, X), we have x1, x2 ∈ U ∩ Dα (ρ ◦ θ, ε, X). And also, |(ρ ◦ θ)(x1) − (ρ ◦ θ)(x2)| = |ρ(y1) − ρ(y2)| ≥ ε. Therefore, x ∈ Dα+1 (ρ ◦ θ, ε, X), which implies that y = θ(x) ∈ θ(Dα+1 (ρ ◦ θ, ε, X)). Conversely, let y ∈ θ(Dα+1 (ρ ◦ θ, ε, X)). Then there exists x ∈ Dα+1 (ρ ◦ θ, ε, X) such that θ(x) = y. Let V be any neighborhood of y. Since θ−1 (V ) is open in X and x ∈ θ−1 (V ), there exist x1, x2 ∈ θ−1 (V ) ∩ Dα (ρ ◦ θ, ε, X) such that |(ρ ◦ θ)(x1) − (ρ ◦ θ)(x2)| ≥ ε. Let y1 = θ(x1) and y2 = θ(x2), then y1, y2 ∈ θ(θ−1 (V ) ∩ Dα (ρ ◦ θ, ε, X)). By the inductive hypothesis, θ(θ−1 (V ) ∩ Dα (ρ ◦ θ, ε, X)) = θ(θ−1 (V )) ∩ θ(Dα (ρ ◦ θ, ε, X)) = V ∩ Dα (ρ, ε, Y ). Therefore, y1, y2 ∈ V ∩ Dα (ρ, ε, Y ) and |ρ(y1) − ρ(y2)| = |ρ(θ(x1)) − ρ(θ(x2))| ≥ ε. Thus, y ∈ Dα+1 (ρ, ε, Y ). Besides the lemma above, we will use the following useful lemma that can be found in [5]. Lemma 2.2 ( [5], Lemma 2.1). Let U be a clopen subset of X and f : X → R is Baire-1. Then for any ε > 0 and α < ω1, we have Dα (f, ε, X) ∩ U = Dα (f, ε, U). Now we are ready to give the construction. For any m ∈ N, denote the clopen ordinal interval [1, ωm ] by Im. Note that for any nonzero countable ordinal α can be uniquely written in the Cantor normal form α = ωr1 · j1 + ωr2 · j2 + . . . + ωr · j where m ≥ r1 > . . . > r ≥ 0 and , j1, . . . , j ∈ N (see, e.g., [7] ). In the sequel we use the following function. Let a, b be any real numbers such that a = b and m ∈ N.
• 4. 210 On optimality of the index We deﬁne ϕa,b,m : Im → {a, b} by ϕa,b,m(ωr1 · j1 + ωr2 · j2 + . . . + ωr · j ) = a if j is odd b if j is even, where m ≥ r1 > . . . > r ≥ 0 and , j1, . . . , j ∈ N. The following lemma is related to the function ϕa,b,m. Lemma 2.3. For suﬃciently small ε > 0, if we let ϕ = ϕa,b,m, then Dm (ϕ, ε, Im) = {ωm } and ϕ(ωm ) = a. Proof. Take any 0 < ε < |a − b|. We prove the lemma by induction on m. Clearly, the assertion is true for m = 1 since ω is the only limit ordinal in [1, ω]. Suppose that the assertion is true for some m ∈ N. If ϕ = ϕa,b,m+1, it is clear that ϕ(ωm+1 ) = a. For each k < ω, let Lk = [ωm · k + 1, ωm · (k + 1)]. Clearly, θ : Im → Lk deﬁned by θ(ξ) = ωm · k + ξ is a homeomorphism. Therefore, by Lemma 2.1 and Lemma 2.2 we have Dm (ϕ, ε, Im+1) ∩ Lk = Dm (ϕ|Lk , ε, Lk) = θ(Dm (ϕ|Lk ◦ θ, ε, Im)) = θ(Dm (ϕa,b,m, ε, Im)) = θ({ωm }) = {ωm · (k + 1)}. Thus {ωm · k : 0 < k < ω} ⊆ Dm (ϕ, ε, Im+1). Recall that Dm (ϕ, ε, Im+1) ⊆ Dj (ϕ, ε, Im+1) for all j < m. Let j ≤ m and take any neighborhood U of ωm+1 . Then there exists an even k < ω such that ωm · k ∈ U ∩ Dj (ϕ, ε, Im+1) and |ϕ(ωm+1 ) − ϕ(ωm · k)| = |a − b| ≥ ε. Thus ωm+1 ∈ Dm+1 (ϕ, ε, Im+1), and therefore Dm (ϕ, ε, Im+1) = {ωm · k : 0 < k < ω} ∪ {ωm+1 }. Since (Dm (ϕ, ε, Im+1)) = {ωm+1 }, then it follows that Dm+1 (ϕ, ε, Im+1) = {ωm+1 }. The ordinal interval In+k = [1, ωn+k ], n, k ∈ N, can be written as a disjoint union 0≤α<ωk [ωn · α + 1, ωn · (α + 1)] ∪ {ωn · ξ : ξ ≤ ωk , ξ is a limit ordinal}. We use the function ϕa,b,m to prove the following lemma. Lemma 2.4. Let n ∈ N be ﬁxed and a, b ∈ R with a = b. If for any k ∈ N we deﬁne gk : In+k → {a, b} by gk(τ) = ϕ(α + 1) if τ = ωn · α + ξ, ξ ∈ [1, ωn ], 0 ≤ α < ωk ϕ(ξ) if τ = ωn · ξ, ξ ≤ ωk is a limit ordinal, where ϕ = ϕa,b,k : Ik → {a, b}, then Dk (gk, ε, In+k) = {ωn+k } for any suﬃciently small ε > 0. Proof. Take any 0 < ε < |b − a|. We prove the lemma by induction on k. First we prove for k = 1. Take any neighborhood U of ωn+1 , then there is an odd number < ω such that ωn · + 1 ∈ U, therefore |g1(ωn · + 1) − g1(ωn+1 )| = |ϕ( + 1) − ϕ(ω)| = |b − a| ≥ ε. Thus ωn+1 ∈ D1 (g1, ε, In+1). Furthermore, for all τ < ωn+1 can be written as τ = ωn · + ξ, where < ω and 1 ≤ ξ ≤ ωn . Therefore g1(τ) = ϕ( + 1) and since [1, ω) = ∅, it follows that D1 (g1, ε, In+1) = {ωn+1 }.
• 5. A. Zulijanto 211 Now we assume that Dk (gk, ε, In+k) = {ωn+k } and we will prove that Dk+1 (gk+1, ε, In+k+1) = {ωn+k+1 }. For each j < ω, let Lj := [ωn+k · j + 1, ωn+k · (j + 1)] and gj = gk+1|Lj . Let θ : In+k → Lj be deﬁned by θ(ξ) = ωn+k · j + ξ, clearly that θ is a homeomorphism and gk = gj ◦ θ. Therefore, by Lemma 2.1 and Lemma 2.2 we have Dk (gk+1, ε, In+k+1) ∩ Lj = Dk (gj , ε, Lj) = θ(Dk (gj ◦ θ, ε, In+k)) = θ(Dk (gk, ε, In+k)) = θ({ωn+k }) = {ωn+k · (j + 1)}. Thus,{ωn+k · j : 0 < j < ω} ⊆ Dk (gk+1, ε, In+k+1) ⊆ D (gk+1, ε, In+k+1), for all < k. Since gk+1(ωn+k+1 ) = ϕ(ωk+1 ) = a and there exists an even j < ω which implies gk+1(ωn+k · j) = ϕ(j) = b, it follows that ωn+k+1 ∈ Dk+1 (gk+1, ε, In+k+1). Since {ωn+k · j : 0 < j < ω} = ∅, we obtain Dk+1 (gk+1, ε, In+k+1) = {ωn+k+1 }. The proof is completed. Theorem 2.5 below shows that Theorem 1.3 in [1] is optimal. Theorem 2.5. For any n, k ∈ N, there exist f, g : In+k → R such that i(f) = n, i(g) = k, and i(h) = n + k, where h is any of the functions f + g,fg,f ∨ g, f ∧ g. Proof. Let a, b ∈ R with a = b, n ∈ N, and ϕ = ϕa,b,n : In → {a, b}. Deﬁne f : In+k → {a, b} by f(τ) = ϕ(ξ) if τ = ωn · α + ξ, ξ ∈ [1, ωn ], 0 ≤ α < ωk a if τ = ωn · ξ, ξ ≤ ωk is a limit ordinal. . We are to prove that i(f) = n. Take any 0 < ε < |b − a|. For any α < ωk , let Lα = [ωn · α + 1, ωn · (α + 1)] and fα = f|Lα . Let θ : In → Lα be deﬁned by θ(ξ) = ωn · α + ξ. Then it is clear that θ is a homeomorphism from In to Lα. Also, by the deﬁnition of f, clearly fα ◦ θ = ϕ. Since θ : In → Lα is a homeomorphism, then by Lemma 2.1 and Lemma 2.2 we have Dn (f, ε, In+k) ∩ Lα = Dn (fα, ε, Lα) = θ(Dn (fα ◦ θ, ε, In)) = θ(Dn (ϕ, ε, In)) = θ({ωn }) = {ωn · (α + 1)}. and fα(ωn · (α + 1)) = (ϕ ◦ θ−1 )(ωn · (α + 1)) = ϕ(ωn ) = a. It follows that Dn (f, ε, In+k) ∩ 0≤α<ωk Lα = 0≤α<ωk (Dn (fα, ε, Lα)) = 0≤α<ωk {ωn · (α + 1)}. Therefore Dn (f, ε, In+k) ⊆ 0≤α<ωk {ωn · (α + 1)} ∪ {ωn · ξ : ξ ∈ Ik, ξ is a limit ordinal} = {ωn · α : α ∈ Ik}. Since f(ωn · α) = a for all α ∈ Ik, then Dn+1 (f, ε, In+k) = ∅. This implies that β(f) = supε>0 β(f, ε) = n + 1, and therefore i(f) = n. Now, let c, d ∈ R with c = d and denote ψ = ϕc,d,k : Ik → {c, d}. Deﬁne
• 6. 212 On optimality of the index g : In+k → {c, d} by g(τ) = ψ(α + 1) if τ = ωn · α + ξ, ξ ∈ [1, ωn ], 0 ≤ α < ωk ψ(ξ) if τ = ωn · ξ, ξ ≤ ωk is a limit ordinal. Then, by Lemma 2.4, Dk (g, ε, In+k) = {ωn+k } which implies that i(g) = β(g) − 1 = supε>0 β(g, ε) − 1 = k. Let h = f + g and choose the numbers a, b, c, d such that a + c = b + d. Take any suﬃciently small ε > 0. For each 0 ≤ α < ωk , let Lα = [ωn · α + 1, ωn · (α + 1)] and hα = h|Lα . For each α < ωk and τ = ωn ·α+ξ ∈ Lα we have hα(τ) = ϕ(ξ)+ψ(α+1). Therefore, hα ◦ θ = ϕ + ψ(α + 1). Since θ : In → Lα is a homeomorphism, then by Lemma 2.1 and Lemma 2.2, for each α < ωk we have Dn (h, ε, In+k) ∩ Lα = Dn (hα, ε, Lα) = θ(Dn (hα ◦ θ, ε, In)) = θ (Dn (ϕ + ψ(α + 1), ε, In)) = θ({ωn }) = {ωn · (α + 1)} and h(ωn · (α + 1)) = (hα ◦ θ)(θ−1 (ωn · (α + 1)) = a + ψ(α + 1). Take any limit ordinal ξ ≤ ωk . Then h(ωn · ξ) = a + ψ(ξ). For any neighborhood U of ωn · ξ, there exists α < ξ such that ωn · α ∈ U ∩ D (h, ε, In+k) for all < n and h(ωn · α) = a + ψ(ξ). It follows that ωn · ξ ∈ Dn (h, ε, In+k) . Thus we obtain that Dn (h, ε, In+k) = {ωn · α : 1 ≤ α ≤ ωk } and h(ωn · α) = a + ψ(α) for each 1 ≤ α ≤ ωk . Let q : [1, ωk ] → {ωn ·α : 1 ≤ α ≤ ωk } be deﬁned by q(α) = ωn ·α. Then q is bijective and continuous (see, e.g., [7]). Since [1, ωk ] is compact and {ωn · α : 1 ≤ α ≤ ωk } is Haussdorf, then q is a homeomorphism (see, e.g., [2]). It can be observed that h ◦ q = ϕa+c,a+d,k. Therefore, by Lemma 2.1, Dn+k (h, ε, In+k) = Dk (h, ε, Dn (h, ε, In+k)) = q Dk (h ◦ q, ε, Ik) = q({ωk }) = {ωn+k }. It implies that Dn+k+1 (h, ε, In+k) = ∅, and therefore i(h) = β(h) − 1 = n + k. Similarly, we can prove for h = fg, h = f ∧ g, and h = f ∨ g by choosing the appropriate numbers a, b, c, and d. We may choose a, b, c, d such that ac = bd, a < b and c < d, and a > b and c > d for h = fg, h = f ∧g, and h = f ∨g, respectively. Furthermore, the result in Theorem 2.5 may be extended to any compact metric space K such that K(n+k) = ∅. For this, we use the following lemma. Lemma 2.6 ( [5], Lemma 6.8.). Let K be a compact metric space. If K(α) = ∅ for some 0 < α < ω1, then there is a subspace L ⊆ K such that L is homeomorphic to [0, ωα ]. Theorem 2.7. Let K be any compact metric space such that K(n+k) = ∅. Then there exist f, g : K → R such that i(h) = i(f) + i(g), where h is any of the functions f + g, fg, f ∧ g, f ∨ g. Proof. By Lemma 2.6, there exists L ⊆ K such that L is homeomorphic to In+k, suppose that θ : L → In+k is the homeomorphism. By Theorem 2.5, there exist ˜f, ˜g : In+k → R such that i(˜h) = i( ˜f) + i(˜g), where ˜h is any of the functions ˜f + ˜g, ˜f˜g, ˜f ∧ ˜g, ˜f ∨ ˜g. Deﬁne f, g : L → R by f = ˜f ◦ θ and g = ˜g ◦ θ. Let ψ be any of the functions ˜f, ˜g, ˜h. Then by Lemma 2.1, we have Dj (ψ ◦ θ, ε, L) = θ−1 (Dj (ψ, ε, In+k)), j ≤ n + k. It
• 7. A. Zulijanto 213 follows that i(h) = i(f) + i(g) on L, where h is any of the functions f + g, fg, f ∧ g, f ∨ g. Furthermore, by Theorem 3.6 of [5], f, g, and h can be extended onto K with preservation of the ﬁnite index i. Acknowledgement. The author would like to thank the referee(s) for valuable suggestions which lead to the improvement of this paper. References [1] F. Chaatit, T. Mascioni, H. Rosenthal, On functions of ﬁnite Baire index, J. Funct. Anal. 142 (1996), 277–295. [2] J. Dugundji, Topology, Allyn and Bacon, Boston, 1966. [3] R. Haydon, E. Odell, H. P. Rosenthal, On certain classes of Baire-1 Functions with Applica- tions to Banach space theory, in : Functional Analysis, Lecture Notes in Math., 1470, 1-35, Springer, New York, 1991. [4] A. S. Kechris, A. Louveau, A classiﬁcation of Baire Class 1 functions , Trans. Amer. Math. Soc. 318 (1990), 209–236. [5] D. H. Leung, W. K. Tang, Functions of Baire class one, Fund. Math. 179 (2003), 225–247. [6] D. H. Leung, W. K. Tang, Extension of functions with small oscillation, Fund. Math. 192 (2006), 183–193. [7] Z. Semadeni, Banach spaces of continuous functions, Vol. 1, Polish Sci. Publ., Warsawa, 1971. (received 31.10.2016; in revised form 05.05.2017; available online 13.06.2017) Department of Mathematics, Universitas Gadjah Mada, Sekip Utara, Yogyakarta 55281, Indonesia E-mail: atokzulijanto@ugm.ac.id
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