Diameter preserving maps on function spaces

In this paper we describe, under certain assumptions, surjective diameter preserving mappings when defined between function spaces, not necessarily algebras, thus extending most of the previous results for these operators. We provide an example which shows that our assumptions are not redundant.

Diameter preserving mappings are indeed isometries if we consider the underlying spaces of functions endowed with the diameter (semi) norm. So, it is apparent that the description of the extreme points of the closed unit ball of the dual of the spaces endowed with the diameter norm is the key to study such maps. González and Uspenskij [15] and, independently, Cabello [8] characterized the extreme points of the closed unit ball of the dual of C(X ) (X compact) endowed with the diameter norm. However, if we consider linear subspaces of C(X ) (function spaces), the situation might be more intricate. Namely, in [14] (see also [6]), the authors describe such extreme points when the function spaces satisfy the Unique Decomposition property, although several examples confirm that a complete characterization seems to be complicated.
In this paper, we first analyse the diametral boundary of a function space, a concept which was just proposed (without any study) in [3] and which is based on the extreme points cited above. We check that it is indeed a boundary and show that it is a useful tool for the description of the diameter preserving surjections in this context. We also provide a characterization of the extreme points for several types of subspaces, which include extremely regular function spaces, Thanks to these results, we can describe diameter preserving surjections defined between such function spaces, thus extending most of the previous results for these maps.
We first need to state some preliminary definitions. Let X be a compact (Hausdorff) space and let C(X ) be the algebra of all scalar-valued continuous functions on X endowed with the supremum norm . ∞ .
A linear subspace A of C(X ) is called a function space on X if A separates the points of X and contains the set of constant functions, denoted by C. A function space A on X is called a function algebra if it is an algebra. A will denote the closure of A in (C(X ), . ∞ ).
Given a linear subspace A of C(X ), the Choquet boundary, Ch(A), of A is the set of all x ∈ X for which δ x , the evaluation functional at x, is an extreme point of the unit ball of the dual space of (A, . ∞ ). Clearly, For any function f ∈ C(X ), diam( f ) denotes the diameter of the range of f . For two function spaces A and B, a linear map T : For a function space A on a compact space X , A d stands for the quotient space A/C endowed with the diameter norm, π( f ) d := diam( f ) for all f ∈ A, where π is the quotient map π : A → A/C, and (A * d , · * d ) its dual space. Moreover, we denote the unit ball of the dual space A * d by B A * d and the set of its extreme points by ext . Let A be a function space on X and x ∈ X . A representing measure for δ x is a positive measure μ on X such that δ x ( f ) = X f dμ for all f ∈ A. We use the same notation, δ x , for the evaluation functional at x and the unit point mass at x.

Extreme points
Let us suppose, in this section, that A is a function space on a compact space X . Notice that by arguments similar to those in the proof of [14, Theorem 1], we can regard ext (B A * d ) as a nonempty subset of {αδ x,x : x, x ∈ Ch(A), x = x , α ∈ T} and that this inclusion may be proper. As a consequence, the following set, which was proposed in [3] without any study, can be defined: The diametral boundary of A is defined as follows: On the other hand, we can define the following linear subspace of C(X × X ): endowed with the supremum norm.
Proof Consider the linear operator L : Since π( f ) d = L( f ) ∞ , we have that L is a linear onto isometry. As a consequence, the adjoint map L * : (A − A) * −→ A * d sends the extreme points of the closed unit ball of (A − A) * onto the extreme points of the closed unit ball of A * d . That is, the extreme points of A * d can be identified with Ch(A − A). It is well known the Choquet boundary of each linear subspace B of continuous functions is a boundary for B in the sense that every element of B attains its maximum value on it. Hence, from Proposition 2.2, we can deduce that dch(A) is a boundary for (A d , · d ) in the sense that, given f ∈ A, then π( Let us now describe the diametral boundary of several function spaces: Indeed, this is a characterization of the Unique Decomposition property [6].
However, the description of the diametral boundary of a function space is not always so clear as the following example shows. First, we need a technical result which we shall also use in the proof of our main result: Consider the subspace E of C R (X ) spanned by Re A and u. Each element of E can be written as k + λu for a unique k ∈ Re A and λ ∈ R. Define a linear functional In particular, k(x 0 ) = kdμ for all k ∈ Re A and, consequently, f (x 0 ) = f dμ for all f ∈ A. This means that μ is a representing measure for δ x 0 and since x 0 ∈ Ch(A), then, by [17, Corollary, P. 52], μ is the point mass at x 0 . Therefore, by (*), 1 − = udμ = u(x 0 ) and so u(x 0 ) < 1, a contradiction showing that r = 1.

Example 2.4
Let us consider the hexagon S in the plane with vertices at the points be the space of affine continuous functions on the compact convex set S, that is, the linear subspace of C(S) consisting of the functions of the form is an arbitrary intermediate point on the edge x 5 x 6 (vertices are x 5 and x 6 ). Since −1 . Analogously, the intermediate points on the edge x 2 x 3 do not belong to Ch(A(S)). Now let z 0 = (x 0 , y 0 ) be an intermediate point on the edge x 1 x 6 or x 3 x 4 (resp. x 1 x 2 or x 4 x 5 ) belonging to Ch(A(S)). Take a neighborhood U of z 0 such that (x 0 ± , y 0 ± ) ∈ S\U (resp. (x 0 ± , y 0 ∓ ) ∈ S\U ) for some > 0 and then, by a similar argument to the above, we obtain a contradiction. Therefore, Ch(A(S)) = {x 1 , Since Indeed, we claim that ext (B A(S) * d ) = {αδ x 1 ,x 4 , αδ x 2 ,x 5 , αδ x 3 ,x 6 : α ∈ T}. In order to see this, suppose that δ , and so we have such that β 1 , β 2 ∈ T, t 1 , t 2 ∈ [0, 1] and t 1 + t 2 = 1. Thus for each f (x, y) = ax + by + c, which easily gives a contradiction. Similarly,

Diameter preserving mappings
In the sequel, given a function space A on a compact space X , we set As mentioned before, we can regard ext (B A * d ) as a nonempty subset of {αδ x,x : x, x ∈ Ch(A), x = x , α ∈ T}, and consequently X is a nonempty subset of Ch(A).

Theorem 3.1 Let A i , i = 1, 2, be (complex or real-valued) function spaces on compact spaces X i satisfying the following conditions:
Assume that T : A 1 → A 2 is a surjective diameter preserving map. Then there are a homeomorphism ψ : X 2 → X 1 , a scalar λ ∈ T and a linear functional L : A 1 → C such that T f (y) = λ f (ψ(y)) + L( f ) for all f ∈ A 1 and y ∈ X 2 .
Proof Since T is a diameter preserving map then it induces a linear isometry T d : A 1d → A 2d under the diameter norm defined by T d (π( f )) = π(T f ) for all f ∈ A 1 . Hence the adjoint of T , T * d : A * 2d → A * 1d , is a linear bijective isometry and, so,
We must now prove that ϕ(x) ∈ {x, x } for each x ∈ X 1 . Contrary to what we claim, assume that there exists x 3 ∈ X 1 \{x, x 1 , x 2 } such that ϕ(x) does not belong to {x, x 3 }. Reasoning as in the above paragraph and by condition (i), there exist distinct points y 1 , for some β ∈ T. Then {x 1 , x 2 } = {y 1 , y 2 }, which contradicts the injectivity of . Therefore, we may conclude that there exist β ∈ T and an injective map ϕ : . It is not difficult to see ϕ( X 1 ) = X 2 since A 2 satisfies condition (i). Now let ψ := ϕ −1 : X 2 → X 1 . Then, ψ is a bijective map such that for all y, y ∈ X 2 , and so for each f ∈ A 1 and y, y ∈ X 2 , T f (y) − T f (y ) = λ( f (ψ(y)) − f (ψ(y ))). Now, by fixing y ∈ X 2 , we can define a linear functional L : Finally we shall show that ψ is a homeomorphism. To see the continuity of ψ, let (y i ) i be a convergent net in X 2 to y 0 such that (ψ(y i )) i converges to x 0 in X 1 . It is enough to show that x 0 = ψ(y 0 ). Contrary to what we claim, let us suppose that there is a neighborhood U of ψ(y 0 ) such that x 0 ∈ X 1 \U . Take u ∈ C(X 1 ) with 0 ≤ u ≤ 1 on X 1 , u(ψ(y 0 )) = 1 and u = 0 on X 1 \U . Since ψ(y 0 ) ∈ Ch(A 1 ), we know, by Lemma 2.3, that sup{Reh(ψ(y 0 )) : h ∈ A 1 , Reh ≤ u} = 1. Hence, it is not difficult to see that sup{Reh(ψ(y 0 )) : h ∈ A 1 , Reh ≤ u} = 1. Then there exists h ∈ A 1 with Reh ≤ u and Reh(ψ(y 0 )) > 3 4 . We can consider i 0 such that, for all , then, for a sufficiently large index i, |T h(y i ) − T h(y 0 )| > 1 4 , which is a contradiction. Thus x 0 = ψ(y 0 ) and, consequently, ψ is continuous.
Similarly, we prove that ϕ (ψ −1 ) is continuous. Suppose, on the contrary, that (x i ) i is a convergent net in X 1 to x 0 such that (ϕ(x i )) i converges to y 0 in X 2 and y 0 = ϕ(x 0 ). As above, we can choose k ∈ A 2 with Rek(y 0 ) ≤ 0 and Rek(ϕ(x 0 )) > 3 4 . Then taking h ∈ A 1 such that k = T h , from we get a contradiction. Therefore, ψ is a homeomorphism. shows. Indeed, it is a routine matter to verify that δ x 1 = δ x 2 − δ x 3 , which proves that A(S) does not satisfy condition (i). Condition (ii) in Theorem 3.1 is true if, for instance, ext (B A * 1d ) = {αδ x,x : x, x ∈ Ch(A 1 ), x = x , α ∈ T}, although it is not true for general function spaces as can be deduced from Example 2.4 too.
2. The following example, adapted from [7, p.134], shows that assumptions (i) and (ii) in our main result (Theorem 3.1) cannot be relaxed: Let Then, according to [7, p.134], there is a linear diameter preserving bijection between A(K ) and A(S), while it is clear that there is no homeomorphism between between K and S. As a consequence of the above paragraph, it is apparent, taking A 1 = A(K ) and A 2 = A(S), that condition (i) in Theorem 3.1 is not redundant. Furthermore, condition (ii) cannot be removed either. Indeed, if we take A 1 = A(S) and A 2 = A(K ), we note that condition (i) for A 1 in such theorem is used only to obtain the injectivity of in the proof, but in this particular case we can check directly that is injective (because card(dch(A(S))) = card(dch(A(K ))) = 3 and T * d is a bijective isometry). Hence, since the result is not valid in this case, we conclude the necessity of condition (ii).
We are now showing that Theorem 3.1 can be applied to several types of function spaces.
It is said (see, e.g., [10]) that a function space A on X is extremely regular if for each x ∈ X , each neighborhood V of x and each scalar with 0 < < 1, there exists a function f ∈ A with f ∞ = f (x) = 1 and | f | < on X \V .

Corollary 3.3 Let A and B be (real-valued) extremely regular function spaces on
compact spaces X and Y , respectively, and T : A → B be a surjective linear diameter preserving map, then there are a homeomorphism ψ : Y → X , a scalar λ ∈ {±1} and also a linear functional L : A → R such that for all f ∈ A, y ∈ Y , T f (y) = λ f (ψ(y)) + L( f ).
Proof Let α 1 , . . . , α n ∈ R\{0} and different points x 1 , . . . , x n ∈ X such that α 1 δ x 1 + · · · + α n δ x n = 0 on A. Let V be a neighborhood of x 1 such that x j / ∈ V for all j ∈ {2, . . . , n}. Choose 0 < β < min(|α 1 |, 1). Then for = β n i=1 |α i | , there exists a function f ∈ A with f (x 1 ) = 1 = f ∞ and | f | < on X \V . Consequently, and, then, β > |α 1 |, which is a contradiction. This implies that A (resp. B) satisfies condition (i) in Theorem 3.1. Furthermore, by [13], ext (B A * d ) = {±δ x,x : x, x ∈ X, x = x } and, thus, A and, similarly, B satisfy the conditions in Theorem 3.1. n for all n ∈ N and f (∞) = 0. Assume that A is the R-linear span generated by the set {1, f, δ n : n ∈ N}. Then, clearly, A is an extremely regular function space on N ∞ which is not an algebra. Indeed, if A is an extremely regular function algebra on a compact space X and f is a function in C(X ) such that f = f 2 and f, f 2 / ∈ A, then the R-linear span generated by the set { f, A} is an extremely regular function space which is not an algebra.