Linear extensions of some Baire-one functions

. Let X be a Hausdorﬀ topological space, and let B 1 ( X ) denote the space of all real Baire-one functions deﬁned on X . Let A be a nonempty subset of X endowed with the topology induced from X , and let F ( A ) be the set of functions A → R with a property F making F ( A ) a linear subspace of B 1 ( A ). We give a suﬃcient condition for the existence of a linear extension operator T A : F ( A ) → F ( X ), where F means to be piecewise continuous on a sequence of closed and G δ subsets of X and is denoted by P 0 . We show that T A restricted to bounded elements of F ( A ) endowed with the supremum norm is an isometry. As a consequence of our main theorem, we formulate the conclusion about existence of a linear extension operator for the classes of Baire-one-star and piecewise continuous functions.


Introduction.
We consider real functions. Let X, A, B 1 (X), F(X), and P 0 (X) have the same meaning as in the abstract. By F b (A) we denote the subspace of F(A) consisting of all bounded functions. We shall consider F b (A) with the norm This paper deals with the following extension problem which is inspired by the classical Tietze extension theorem. Let A be a fixed (e.g., closed, G δ , etc.) nonempty subset of a topological space X, and let f 0 ∈ R A be a function with a certain property F. Can f 0 be extended to a function f ∈ R X with the same property F?
The Tietze extension theorem is of course right for a class of continuous functions. It is well known that if X is a metric space and A is a closed subset of X, the Tietze theorem can be significantly strengthened. In 1933 Borsuk [2] proved that there is a positive linear extension operator from C(A) into C(X). In 1951 Dugundji [6] generalized Borsuk's theorem for continuous mappings into locally convex linear spaces. In 1933 Kuratowski [13] obtained a result for functions of the first class defined on G δ -subsets of a metric space. Furthermore, in 2005 Kalenda-Spurný [11] and Shatery-Zafarani [19] extended Kuratowski's theorem to completely regular and perfectly normal spaces, respectively. The natural problem to consider is whether an analogy to Kuratowski's theorem can hold for a given subspace of B 1 (X). In 1990, Császár [3, Theorem 3.1] obtained a positive result in this direction: applying the method of the proof of the metric-case of the Tietze theorem, he obtained a Kuratowskitype result for the property F := C d = to be a discrete limit of a sequence of continuous functions.
In this paper we consider a larger property than the C d -the property B * Peek noticed that B * 1 is larger than C d ; indeed: with the notation of Section 1, Our main result, included in Theorem 1 in Section 4 is a general Dugundji-type theorem for property P 0 in the class of normal spaces; it reduces to property B * 1 for the class of complete metric spaces (Corollary 1): we show that if A is a zero-subset of X a normal space, then there is a linear extension operator P 0 (A) → P 0 (X).

Preliminaries. Let
A be a nonempty subset of X. The set of all real functions with a closed graph on X is denoted by U(X). It is known (cf. [8,Th. 3.6] and [20, p. 196]) that Fact 1. For every f ∈ U(X) and every compact subset F ⊂ R, the set f −1 (F ) is closed.
The symbol U + (X) stands for the set of all non-negative elements of U(X). In 1985 Doboš [5] proved that the sum of two non-negative functions with a closed graph is a function with a closed graph. Since 0 ∈ U + (X), we have thus U + (X) is a cone in R X . In this paper, we use the following characterization of closedness of the graph (see [1]).
A function f : X → R is piecewise continuous on X if there is a sequence (X n (f )) n in X, depending on f , of nonempty closed sets such that X = ∞ n=1 X n (f ) and every restriction f Xn(f ) is continuous. The set of all real piecewise continuous functions on X is denoted by P(X). 1 According to the notation of Section 1, the set of all real piecewise continuous functions on X for which every X n (f ) is additionally G δ in X, is denoted by P 0 (X). Obviously, P 0 (X) ⊂ P(X) for every X (Hausdorff) with the identity P 0 (X) = P(X) for X perfectly normal. Moreover, by the Tietze theorem, and Hence 3. The family P 0 (X). In this section, we give some significant properties of the family P 0 (X). The lemma below defines a relationship between that class and the class of functions with a closed graph.
and from the identities now, by (5), (6), and Fact 1, X\W n is F σ in X.
Proof. Let f, g : X → R be two elements of P 0 (X). Then there are sequences • the restrictions f W k and g Hj are continuous.
In 2002, Borsik [1, Theorem 2] proved that if X is perfectly normal, then P(X)(= P 0 (X)) = U + (X) − U + (X). The (key) lemma below is an extended version of this result. We use it in the proof of our Theorem 1. Since the justification for this lemma is time consuming, it is located at the end of the paper. To show equality (7) below, we use some ideas from Borsik's proof. Some gaps in Borsik's justification have been completed.

Lemma 4. Let X be a normal topological space. Then
4. The main result. Our main result presented below is a solution to the extension problem for mappings from P 0 (X).

Theorem 1. Let X be a normal topological space, and let A be a closed and
G δ subset of X. Let also f ∈ P 0 (A). Then the mapping T A : P 0 (A) → P 0 (X) given by the formula is a linear extension operator such that its restriction to P 0 b (A) is an isometry Proof. Since A is a closed and G δ -subset of X, A = [g = 0], where g is a nonnegative continuous function on X (see [7, p. 62]). Let us fix f ∈ P 0 (A). By Lemma 4, there are mappings p, for every x ∈ A. Furthermore (see [21, Theorem 1]), p and q have closed graph extensions p (A,g) , q (A,g) : X → R + given by the formulas respectively. By Lemma 4, p (A,g) − q (A,g) ∈ P 0 (X), and thus the formula Vol. 110 (2018) Linear extensions of some Baire-one functions 179 defines a linear extension mapping T A from P 0 (A) into P 0 (X). It is now obvious is an isometry.
From the above theorem and Fact 2, we immediately obtain the following Corollary 1. Let X be a complete metric space [resp. perfectly normal topological space], and let A be its nonempty closed subset. Then the mapping T A given by (8) is a linear extension operator

The proof of Lemma 4.
Proof. By equality (2), U + (X) is a cone in R X . By Lemma 2, U + (X) ⊂ P 0 (X). Thus, by Lemma 3, we have We shall show that the inverse inclusion to (9) is true. Our argumentation is a refinement the proof of Borsuk's result [1,Theorem 1]. Gaps that appeared in it we have completed in italics. Let f ∈ P 0 (X) and let (W k ) ∞ k=1 be an increasing sequence of closed and G δ subsets of X such that ∞ k=1 W k = X and the restriction f W k is continuous for every k ∈ N.
It is obvious that we only have to consider the case W k W k+1 for all k's (otherwise f would be continuous on X). Set W 0 = ∅ and E k = W k \W k−1 , k=1,2,. . . .
Let us fix k ∈ N.
Since X is normal and every W k is G δ and closed in X, W k is a zero-set. Put g1 ≡ 1. For k = 2, 3, . . ., there are continuous functions g k : X → [0, 1] such that Let h k = min{g1, g2, . . . , g k }. Every h k is a non-negative continuous function such that h k+1 h k for every k ∈ N. Furthermore, from the definition of h k , by (11) and by the fact that the sequence (W k ) ∞ k=1 is strictly increasing, it follows that Combining (12) and (13) we obtain a sharp inequality for h k : 180 W. Sieg Arch. Math.
Since all the sets E k are pairwise disjoint, the function t : X → R of the form is [by (14)] well-defined and strictly positive.
Set f1 = f + + t and f2 = f − + t, where f + = max{f, 0} and f − = max{−f, 0}. The functions f1 and f2 are obviously nonnegative. Furthermore f1 − f2 = f . The proof of our lemma will be completed once we show the function f1 has a closed graph. For this purpose we shall apply Lemma 1. Let us fix x ∈ X and m ∈ N.
From (10) it follows that there is a number k0 ∈ N such that Hence (and because h k 0 +1 is continuous on X), for every m ∈ N there is a neighborhood V

Note that for x an isolated point, we may set
Thus let x be non-isolated. Without loss of generality we can assume that From (13) and (17) Furthermore, since f + is continuous on . It is obviously an open neighborhood of x. We shall show that inclusion (3) holds true with V = V (x; m). Let us consider two cases: (a) y ∈ E k 0 ∩ V (x; m), (b) y ∈ V (x; m)\E k 0 . Case (a). Let us fix y ∈ E k 0 ∩ V (x; m). Thus y ∈ W k 0 ∩ V (x; m). Furthermore, from (18) and (19) we obtain Case (b). Note that in this case Vol. 110 (2018) Linear extensions of some Baire-one functions 181 and thus y ∈ V (x; m) and y / ∈ W k 0 \W k 0 −1 .
6. Open problems. Since we do not know if P 0 (X) = P(X) for X an arbitrary normal space, the following two problems seem to be natural.

Problem 1.
Characterize the family P 00 (X) def = U + (X) − U + (X) for X a completely regular space, at least for X a normal space.

Problem 2.
Solve the Extension Problem P 00 (A) → P 00 (X), for X a normal (or completely regular) space and A a closed or G δ subset of X.