ON REAL NONISOMORPHIC BANACH SPACES WITH HOMEOMORPHIC GEOMETRIC STRUCTURE SPACES

Abstract We give an example of a pair of real Banach spaces such that they are neither linearly isomorphic nor isomorphic with respect to the structure of Birkhoff–James orthogonality, but have mutually homeomorphic geometric structure spaces.


Introduction and preliminaries
This paper is concerned with a basic problem in the theory of geometric nonlinear classification of Banach spaces.It was recently pointed out in [16,Example 4.14] that there exists a pair of complex Banach spaces (in fact, uniform algebras) (X, Y) with the following properties: X is neither linearly isomorphic to Y nor Y and X is not isomorphic to Y (nor Y) with respect to the structure of Birkhoff-James orthogonality, but X and Y have mutually homeomorphic geometric structure spaces, where Y is the complex conjugate of Y.In particular, this example solved [15,Problem 6.16] in the positive.The main objective of the present paper is to obtain a similar example in the real case.Namely, we give a pair of real Banach spaces such that they are neither linearly isomorphic nor isomorphic with respect to the structure of Birkhoff-James orthogonality, but have mutually homeomorphic geometric structure spaces.This answers [15,Problem 6.15] affirmatively.
In the rest of this paper, all Banach spaces are assumed to be real.As usual, Banach spaces X and Y are linearly isomorphic if there exists a linear homeomorphism between X and Y, in which case, we write X Y.
We recall the notion of Birkhoff-James orthogonality which was first introduced by Birkhoff [3] and was significantly developed by James [6,7].Let X be a Banach space and let x, y ∈ X.Then, x is said to be Birkhoff-James orthogonal to y, denoted by x ⊥ BJ y, if x + λy ≥ x for each λ ∈ R. Obviously, it is equivalent to the usual orthogonality in Hilbert spaces.Moreover, as the existing results indicate, the behaviour of Birkhoff-James orthogonality is closely related to the geometric structure of Banach spaces; see [1] for a comprehensive survey on generalised orthogonality types.
Very recently, the nonlinear equivalence based on the structure of Birkhoff-James orthogonality has been studied in [2,5,13,14,16].Let X, Y be Banach spaces.Then, X and Y are isomorphic with respect to the structure of Birkhoff-James orthogonality, denoted by X ∼ BJ Y, if there exists a (possibly nonlinear) bijection T : X → Y such that x ⊥ BJ y if and only if Tx ⊥ BJ Ty.In this direction, some classification results as well as descriptions of Birkhoff-James orthogonality preservers are known.
The geometric structure space S(X) of a Banach space X was first introduced in [14, Definition 3.4 and Proposition 3.5] for classifying the class of spaces of continuous functions C 0 (K), and the family of classical sequence spaces c 0 and p by their structure of Birkhoff-James orthogonality.It is defined as the set Moreover, it was shown in [14, Proposition 3.5] that: 1 ∪ S = 2 does not hold in general, that is, the closure operator S → S = does not necessarily induce a topology on S(X).The closure space S(X) is said to be topologisable if the closure operator S → S = satisfies the Kuratowski closure axioms, or equivalently, if the set {S ⊂ S(X) : S = = S} fulfils the axioms of closed sets.It is known that S(X) is not topologisable whenever X is a reflexive smooth Banach space with dim X ≥ 2 [14, Theorem 3.8], and that S(C 0 (K)) is topologisable and homeomorphic to K [14, Theorem 5.2].
The theory of geometric structure spaces was further developed in [15], where the nonlinear equivalence of Banach spaces based on geometric structure spaces was introduced and studied.Let X, Y be Banach spaces.Then, X and Y are isomorphic with respect to geometric structure spaces, denoted by X ∼ S Y, if they have homeomorphic geometric structure spaces, that is, if there exists a bijection Φ : S(X) → S(Y) satisfying Φ(S = ) = Φ(S) = for each S ⊂ S(X).It was shown in [15] that C 0 (K)-spaces are isometrically classified under '∼ S ' [15,Theorem 3.16], and that c 0 and p have mutually different geometric structure spaces [15,Theorem 6.14].These apply to classification results under '∼ BJ ' via the following fact.If X, Y are Banach spaces and if X ∼ BJ Y, then X ∼ S Y [14,Theorem 3.10].This allows us to find the difference between the structure of Birkhoff-James orthogonality of two given Banach spaces by using their geometric structure spaces.
With this notation, the main result of this paper is stated as follows.
THEOREM 1.1.There exists a pair of real Banach spaces (X, Y) such that X Y and X BJ Y, but X ∼ S Y.
Here, it should be noted that if either of X, Y is finite dimensional, or both of X, Y are reflexive and smooth, then X ∼ S Y implies that X Y [15, Theorems 4.3 and 6.5].Hence, the desired pair must be constructed outside of these classes.

Results
Let K be a compact Hausdorff space.Then, the symbol C(K) denotes the Banach space of all continuous functions on K, where the norm of f ∈ C(K) is defined by For each nonzero regular Borel measure μ on K, let Z(K, μ) be the hyperplane of C(K) given by The desired example will be constructed by using C(K) and Z(K, μ) for a special pair (K, μ).To be precise, we adopt the connected compact Hausdorff space K, constructed by Koszmider [9, Section 5], such that C(K) is infinite-dimensional and not isomorphic to any of its hyperplanes.In this setting, we automatically have C(K) Z(K, μ) for any nonzero regular Borel measure μ on K.
Next, we find a regular Borel (probability) measure μ on K such that C(K) BJ Z(K, μ), but C(K) ∼ S Z(K, μ).It will turn out that the required property for such a μ is nonatomicity.Let μ be a nonnegative Borel measure on K.Then, a Borel subset E of K is called an atom of μ if μ(E) > 0, and whenever F is a Borel subset of E, either μ(F) = 0 or μ(E \ F) = 0.A Borel measure μ is said to be atomic if it has an atom.As was noted by Knowles [8, page 63], if E is an atom of μ, then μ({t}) = μ(E) > 0 for some t ∈ E. Hence, μ is atomic if and only if μ({t}) > 0 for some t ∈ K.
According to [8, Theorem 1], if K is perfect, that is, if K has no isolated point, then there exists a nonatomic regular Borel probability measure on K.We remark that Koszmider's space K is perfect since it is connected and Hausdorff.Hence, we can find a nonatomic regular Borel probability measure on Koszmider's space K.
The following lemma will be needed for proving both C(K) BJ Z(μ, K) and C(K) ∼ S Z(μ, K) provided that μ is a nonatomic regular Borel probability measure on K.

LEMMA 2.1. Let K be a compact Hausdorff space and let μ be a nonatomic regular Borel probability measure on K. If t ∈ K and if U is an open neighbourhood of t, then there exists an
PROOF.First, we recall that the support for μ is given by supp(μ) = {t ∈ K : μ(U) > 0 for each open neighbourhood U of t}.
If t supp(μ), then μ(U) = 0 for some open neighbourhood U of t.In particular, each s ∈ U does not belong to supp(μ).Hence, K \ supp(μ) is an open subset of K and supp(μ) is a closed subset of K. We also note that μ(supp(μ)) = 1.Indeed, since μ is regular, for each ε > 0, there exists an open subset U of K such that supp(μ) ⊂ U and x∈F is an open covering for a compact set F and has a finite subcovering U x 1 , . . ., U x n .It follows from the monotonicity and subadditivity of μ that Therefore, μ(F) = 0, which implies that μ(U) = 1.This proves that 1 − ε < μ(supp(μ)) for arbitrary ε > 0. Hence, μ(supp(μ)) = 1 holds.Now, let t ∈ K and let U be an open neighbourhood of t.Suppose first that t supp(μ).Set F = supp(μ) ∪ (K \ U).Then, by Urysohn's lemma, there exists an Next, suppose that t ∈ supp(μ).Then, μ(U) > 0. Since V is an open neighbourhood of t} = 0 by the nonatomicity and regularity of μ, there exists an open neighbourhood V of t such that V ⊂ U and 0 < μ(V) < 2 −1 μ(U).By Urysohn's lemma, we have a g ∈ C(K) . Now, by the inner regularity of μ, we can find a compact subset Furthermore, we obtain REMARK 2.2.It also follows from the preceding lemma that Z(K, μ) separates the points of K.
Using Lemma 2.1, we can prove that C(K) BJ Z(K, μ) for nonatomic regular Borel probability measures μ on K. Recall that an element x of a Banach space X is called a right symmetric point for the Birkhoff-James orthogonality if y ∈ X and y ⊥ BJ x imply that x ⊥ BJ y.The early study on local symmetry of Birkhoff-James orthogonality can be found in [17], while the term 'symmetric point' first appeared in [11,12].It is obvious that the right symmetry of a point is stable under Birkhoff-James orthogonality preservers.Moreover, it can be shown that 1 ∈ C(K) is a right symmetric point for the Birkhoff-James orthogonality.Indeed, if f ∈ C(K) and if f ⊥ BJ 1, then f (K) contains both nonpositive and nonnegative numbers; otherwise, To summarise, it will turn out that C(K) BJ Z(K, μ) once it has been proved that Z(μ, K) contains no nonzero right symmetric points for Birkhoff-James orthogonality.Now, we are ready to prove the following theorem.THEOREM 2.3.Let K be a connected compact Hausdorf space and let μ be a nonatomic regular Borel probability measure on K.Then, Z(K, μ) contains no nonzero right symmetric points for Birkhoff-James orthogonality.Consequently, C(K) BJ Z(K, μ).
PROOF.Let f be a nonzero element of Z(K, μ).Since Birkhoff-James orthogonality is homogeneous, we may assume that f ∞ = 1.Moreover, from f dμ = 0 and the intermediate value theorem, f (t) = 0 for some t ∈ K. Let U − = {s ∈ K : f (s) < 0} and U + = {s ∈ K : f (s) > 0}, and let for each n ∈ N. We divide the argument into the two cases.
To this end, we begin with the following well-known fact which can be proved by a combination of the Mazur separation theorem and Milman's partial converse to the Krein-Milman theorem.LEMMA 2.4.Let K be a compact Hausdorff space and let M be a closed subspace of C(K).Then, ext(B M * ) ⊂ {±δ t |M : t ∈ K}, where B M * is the unit ball of the dual space M * of M and δ t is the evaluation functional at t ∈ K, that is, We make use of the preceding lemma for identifying the support functionals for maximal faces of B Z(K,μ) .For this purpose, we need the notion of weak peak points for subspaces of C(K).Recall that a point t ∈ K is called a weak peak point for a subspace The Bishop-de Leeuw theorem states that the set of weak peak points for a uniform algebra coincides with its Choquet boundary (see, for example, [10,Section 8]).
In analogy with [16, Lemma 3.2], we have the following lemma (see also [4,Lemma 3.2]).LEMMA 2.5.Let K be a compact Hausdorff space, let M be a closed subspace of C(K) and let t ∈ K be a weak peak point for M.Then, F t = δ −1 t (1) ∩ B M is a maximal face of B M and Φ * (F t ) = {δ t |M}.
PROOF.Let F be a proper face of B M containing F t and let f ∈ F. Further let However, this is impossible by the choice of g.Hence, t ∈ A f .Now, we note that f (t We need another auxiliary lemma.LEMMA 2.6.Let X be a Banach space and let B be a weakly * closed subset of B X * such that x = max{|ρ(x)| : ρ ∈ B} for each x ∈ X.Then, for each convex subset C of S X , there exists a ρ ∈ B such that C ⊂ ρ −1 (1) ∩ B X or C ⊂ ρ −1 (−1) ∩ B X .In particular, each maximal face of B X has the form ρ −1 (1) ∩ B X or (−ρ) −1 (1) ∩ B X for some ρ ∈ B.
PROOF.We first note that B 0 = {±ρ : ρ ∈ B} is also weakly * closed.Take arbitrary finitely many elements x 1 , . . ., x n of C. Since B x is a nonempty weakly * closed subset of B 0 .Moreover, the family (B x ) x∈C has the finite intersection property.Therefore, x∈C B x ∅ by the weak * compactness of B 0 .Now, we obtain C ⊂ ρ −1 (1) ∩ B X for an arbitrary ρ ∈ x∈C B x ⊂ B 0 , as desired.
We conclude this paper with the following theorem which completes the construction of an example of a pair of Banach spaces (X, Y) such that X Y and X BJ Y, but X ∼ S Y. THEOREM 2.7.Let K be a compact Hausdorff space and let μ be a nonatomic regular Borel probability measure on K.Then, S(Z(K, μ)) = {I t : t ∈ K}, where I t = ker(δ t |Z(K, μ)).Moreover, C(K) ∼ S Z(K, μ).PROOF.By Lemma 2.1, each t ∈ K is a weak peak point for Z(K, μ).From this and Theorem 2.5, it turns out that F t = δ −1 t (1) ∩ B Z(K,μ) is a maximal face of B Z(K,μ) with Φ * (F t ) = {δ | Z(K, μ)} for each t.Hence, ker ρ ∈ S(Z(K, μ)) for each t ∈ K, that is, S(Z(K, μ)) ⊃ {I t : t ∈ K}.
For the converse, we note that B = {δ t : t ∈ K} is a weakly * closed subset of B *
Next, let ι(t) = I t for each t ∈ K. Since Z(K, μ) separates the points of K, the mapping ι is a bijection from K onto S(Z(K, μ)).Suppose that A ⊂ K.If t 0 ∈ A, then there exists a net (t a ) a ⊂ A that converges to t 0 .It follows that f (t 0 ) = 0 whenever f ∈ t∈A I t , which implies that I t 0 ∈ {I t : t ∈ A} = .Hence, {I t : t ∈ A} ⊂ {I t : t ∈ A} = .Conversely, if t 0 A, then setting U = K \ A yields an open neighbourhood of t.By Lemma 2.1, there exists an f ∈ Z(K, μ) such that f (t 0 ) = f ∞ = 1 and f (K \ U) = {0}.Since K \ U = A, we see that f ∈ t∈A I t \ I t 0 .Therefore, I t 0 {I t : t ∈ A} = .This shows that {I t : t ∈ A} ⊃ {I t : t ∈ A} = , that is, ι(A) = ι(A) = .From this, ι is a (closure space) homeomorphism from K onto S(Z(K, μ)).Moreover, S(C(K)) and K are homeomorphic (as closure spaces) by [14,Theorem 5.2].Thus, S(C(K)) and S(Z(K, μ)) are also homeomorphic.This completes the proof.