AN EXTENSION OF CAMBERN’S THEOREM FERNANDA BOTELHO and JAMES JAMISON The theorem of Cambern for the into isometries of

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FERNANDA BOTELHO and JAMES JAMISON

The theorem of Cambern for the into isometries of C(X, E) is generalized to complex strictly convex range spaces. This allows for an extension and also a new proof of the vector valued Banach-Stone theorem. As a consequence, complex strictly convex Banach spaces have the strong Banach-Stone property.

AMS 2000 Subject Classification: Primary 30D55; Secondary 30D05.

Key words: isometry, Banach-Stone theorem, Cambern theorem, complex strictly convex Banach space, space of vector valued continuous functions.

1. INTRODUCTION

The theorems of Holszty´nski and Cambern have been the source for our ideas and have provided a path to the generalizations which we obtain. We begin by recalling Holszty´nski’s theorem on the characterization of the into isometries between spaces of continuous functions.

Theorem 1.1 (cf. [6]). If X and Y are compact and Hausdorff spaces and T :C(X)→C(Y) is a linear isometry, then there exist a closed subset Y₀ of Y, a surjective continuous map ϕ:Y0→X, andα∈C(Y) with kαk∞= 1 and |α(y)|= 1 for everyy ∈Y0,such that

(1.1) T(f)(y) =f(ϕ(y))α(y)

for every f ∈C(X) and y∈Y₀.

The classical Banach Stone theorem gives a representation for surjective isometries between C(X) and C(Y). The usual proof of the Banach Stone theorem relies on a characterization of the extreme points of the dual unit ball, cf. [1]. We show that the Banach-Stone characterization of surjective isometries can be derived from Holszty´nski’s in a straightforward manner.

This provides a proof of the Banach Stone theorem that only depends on basic topological properties of the spaces involved and does not require information about the extreme points of the dual ball. Our motivation for following this approach is related to an extension of Holszty´nski’s theorem to spaces of vector valued continuous functions due to Cambern [2]. He proved Holszty´nski’s theorem for into isometries between C(X, E) andC(Y, E1) assuming that the

REV. ROUMAINE MATH. PURES APPL.,54(2009),3, 181–188

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Banach spaces E and E1 are strictly convex. We generalize this theorem to the case whenever E andE₁ are complex strictly convex. Moreover, following the path set forth in the scalar case, we obtain a generalization of the Banach- Stone theorem to the complex strictly convex setting. Our techniques are different from the standard approach available in the literature and, to our knowledge, the results are new.

2. REMARKS ON HOLSZTY ´NSKI’S THEOREM

Holszty´nski’s proof is given for complex valued functions, however it is mentioned in [6], as a footnote, that the same characterization is also valid for real valued functions. For completeness of exposition we provide the minor modification to Holszty´nski’s proof for the real-valued case. Before giving the modification we recall some essential notation from [6]:

Sx={f ∈C(X) :kfk= 1 and |f(x)|= 1}, x∈X, R_y ={g∈C(Y) :kgk= 1 and |g(y)|= 1}, y∈Y,

Q_x={y∈Y :T(S_x)⊂R_y}, x∈X.

If C(X) refers to real-valued continuous functions on X, then steps (i)–(vi) in Holszty´nski’s proof are valid with a minor modification necessary to show step (i). This first step asserts that if f ∈ C(X) vanishes at x ∈ X (i.e., f(x) = 0), then T(f)(y) = 0 for every y ∈Q_x. Without loss of generality we may assume that f has norm 1, and that T(f)(y) 6= 0 for some y ∈ Q_x.We set g= min{1 +f,1,1−f}. Therefore,g(x) = 1 andkgk_∞= 1.This implies that g and g−f are in S_x.Hence |T(g)(y)|= 1 and |T(g−f)(y)|= 1.This implies thatT(f)(y) = 0, contradicting our initial assumption.

We now show how Holszty´nski’s theorem leads to an easy proof of the Banach-Stone theorem. For purposes of exposition,we recall the statement of that theorem, see [5].

Theorem 2.1. If X and Y are compact and Hausdorff spaces and T : C(X) → C(Y) is a surjective linear isometry, then there exist a homeomorphism ϕ : Y → X and α ∈ C(Y) with kαk_∞ = 1 and |α(y)| = 1 for every y ∈Y, such that

(2.1) T(f)(y) =α(y)f(ϕ(y)),

for every f ∈C(X) and y∈Y.

We now provide the arguments necessary to complete the proof of the Banach-Stone theorem using Holszty´nski’s theorem. The first step is to show that Y = Y₀. If there exists z ∈ Y \Y₀, then let f ∈ C(X) be a Uryshon function with values in [0,1] such that f(z) = 1 and f(Y0) = 0. Since T is

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onto, there exists h ∈C(X) such thatT(h) =f. Equation (2.1) implies that for every y∈Y₀ we have

f(y) =T(h)(y) =α(y)h(ϕ(y)) = 0.

Therefore, h(ϕ(y)) = 0.The surjectivity of ϕimplies that h is the zero function. This contradiction asserts that Y₀=Y. To show that ϕis injective, we start by assuming the existence of two points in Y,y₀6=y₁ such thatϕ(y₀) = ϕ(y1) =x(i.e.,y0, y1 ∈Qx). Letf ∈C(X) be a Urysohn function with values in the interval [0,1] such thatf(y₀) = 0 andf(y₁) = 1. Leth∈C(X) be such that T(h) = f. Thus, we have T(h)(y) = f(y) = α(y)h(ϕ(y)) for all y ∈ Y.

In particular,

α(y₀)h(ϕ(y₀)) =f(y₀) = 0 and α(y₁)h(ϕ(y₁)) =f(y₁) = 1.

These equalities are incompatible. In [6] is shown that, for every x, Qx is nonempty, then Q_x must reduce to a single point. This implies that ϕ is injective, thus a homeomorphism, which completes the proof.

3. SPACES OF CONTINUOUS VECTOR-VALUED FUNCTIONS In this section we follow the path set forward in the scalar case by first extending a vector valued version of Holszty´nski’s theorem to a new setting, complex strictly convex spaces. We then show how this result can be generalized to obtain the full Banach-Stone theorem. This demonstrates that complex strictly convex spaces have the strong Banach-Stone property [1].

The first vector valued version of Holszty´nski’s theorem is due to Cambern.

His theorem (see [2]) gives the form of an into isometry between two spaces of vector valued continuous functions, C(X, E) and C(Y, E₁), with X and Y compact topological spaces, E andE₁ Banach spaces, andE₁ strictly convex.

These spaces are equipped with the usual normk · k∞. For reference purposes, the characterizations of surjective isometries due to Jerison and Cambern’s generalization for into isometries are given in the following statement.

Theorem 3.1. 1. (Jerison [8]). If A is an isometry from C(X, E) onto C(Y, E), withE strictly convex, there exists a homeomorphismτ of Y onto X and a continuous map y→A_y from Y into the space of bounded operators on E, equipped with the strong operator topology, such that Ay is an isometry of E for all y∈Y and

A(F)(y) =A_y(F)(τ(y)) for F ∈ C(X, E), y∈Y.

2. (Cambern [2]). If E andE1 are normed linear spaces withE1 strictly convex andAan isometry fromC(X, E)intoC(Y, E₁), then there exist a subset B(A) ⊂Y and a continuous function φ:Y → B(E, E₁) such that φ(y) =Ay

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(here, B(E, E₁) denotes all bounded operators from E into E1 equipped with the strong operator topology) with kA_yk ≤1 for all y ∈ Y and kA_yk = 1 for all y∈B(A), and there exists a continuous τ from B(A) onto X such that

A(F)(y) =A_y(F)(τ(y))

for F ∈ C(X, E), y∈B(A). If E is finite dimensional, then B(A) is a closed subset of Y.

First, we show that Cambern’s characterization holds for E1 complex strictly convex. Then we derive a generalization of Jerison’s theorem from Cambern’s theorem under the hypotheses that E and E1 are strictly convex or complex strictly convex Banach spaces. It follows, as a consequence, that complex strictly convex spaces have the strong Banach-Stone property (cf. [1]).

We refer the reader to [7] for an extensive number of examples of complex strictly convex spaces. For completeness we first review relevant definitions.

Definition 3.2. A complex Banach spaceE is said to be complex strictly convex if and only if given x and y ∈ E such that kxk = ke^iθy +xk = 1 for every θ ∈ R, then y = 0. Equivalently, given x and y ∈ E such that kxk = k ±iy+xk = 1, then y = 0 (cf. [9]). A Banach space E is said to be strictly convex if and only if given x and y ∈ E of norm 1 and such that k^x+y₂ k= 1,thenx=y.

We set our notation to be used in this section as in Cambern’s paper [2].

A:C(X, E)→ C(X, E₁) denotes an isometry such that A(F)(y) =A_y(F)(τ(y)),

as stated in Cambern’s theorem. The operators A_y are defined by A_y(e) = A(E)(y) withE(x) =e, the constant function inC(X, E).

Theorem3.3. LetE andE1be Banach spaces withE1complex strictly convex and A an isometry from C(X, E) into C(Y, E₁). Then there exist a subset B(A) ⊂ Y, and a continuous function φ : Y → B(E, E₁) such that φ(y) = A_y with kA_yk ≤1 for all y ∈Y and kA_yk = 1 for all y ∈ B(A), and there exists a continuous τ fromB(A) onto X such that

A(F)(y) =A_y(F)(τ(y))

for F ∈ C(X, E), y∈B(A). If E is finite dimensional, then B(A) is a closed subset of Y.

The proof of this theorem is similar to Cambern’s proof. The difference occurs in the proof of Lemma 2 in [2]. Rather than reproduce all of Cambern’s lemmas we refer the reader to [2] and only state the crucial lemma and provide a proof of it under this new hypothesis. We replace the strict convexity of the range space E1 by the complex strict convexity. In order to make the proof

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of the lemma a little more self contained we record some of the basic notation from [2]:

F_e,x={F ∈C(X, E) :F(x) =kFk_∞·e},

B(e, x) ={y∈Y :k(A(F))(y)k=kFk_∞ for allF ∈ F_e,x}, B(x) =S

B(e, x) and B(A) =S

x∈XB(x).

We now state Lemma 2 from [2] and provide the essential changes in the proof.

Lemma3.4. Ify∈B(x)then for eachF ∈C(X, E)we have(A(F))(y) = A_y(F(x)).

Proof. If y ∈ B(x) then y ∈ B(e, x) for some e ∈ E with kek = 1.

Suppose that F ∈C(X, E) vanishes on some neighborhood U of x. We want to show first that A(F)(y) = 0. Choose a nonnegative function f ∈ C(X) such that f(x) = kfk_∞ = kFk_∞, the support of which is contained in U. Define F₁(z) =f(z)e forz ∈X. Then kF₁(z)k_E =|f(z)|.As a consequence, kF₁k∞ = max

z∈X kF(z)k_E = max

z∈X |f(z)|= f(x). This shows that F ∈ F_e,x.Let θ∈Rand setF_θ= e^iθF +F₁.Then

kF_θk= max

z∈X{ke^iθF(z) +F1(z)k_E}= max

z∈X{kF(z)k_E,kF₁(z)k_E},

and kF_θk = f(x). Since y ∈ B(e, x), we have k(AF₁)(y)k_E₁ = kF₁k = f(x) and

k(AF_θ)(y)k_E₁ =kF_θk=f(x).

Thus,

e^iθ(A(F))(y)

f(x) +(A(F₁))(y) f(x)

= 1

for all θ ∈ R. The complex strict convexity of E₁ implies that A(F)(y) = 0.

The remainder of the proof is exactly as in [2].

We now have sufficient machinery to prove that complex strictly convex Banach spaces have the “strong-Banach Stone property.”

Theorem3.5. IfEandE1are strictly convex(or complex strictly convex) Banach spaces andAis an isometry fromC(X, E)ontoC(Y, E₁),thenB(A) = Y, τ is a homeomorphism from Y onto X, and Ay is an onto isometry for every y∈Y such that

A(F)(y) =Ay(F)(τ(y)) for F ∈C(X, E) and y∈Y.

We start by proving the next result that encompasses the essential fea- tures needed for the proof of the theorem.

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Proposition 3.6. If E and E1 are strictly convex Banach spaces and A is an isometry from C(X, E) onto C(Y, E₁), thenB(A) =Y,τ is a homeomorphism from Y onto X,Ay is an onto isometry for every y∈Y.

Proof. The proof is presented in a sequence of steps (i)–(vii).

(i) For everyy∈B(A), A_y is onto.

Let ξ ∈ E1 be such that kξk_E₁ = 1. We define G : Y → E1 such that G(y) =ξ. Clearly, G∈ C(Y, E₁) and kGk_∞= 1.Since A is onto, there exists F ∈ C(X, E), kFk∞ = 1, such that A(F) = G. For every y ∈ B(A) we have A(F)(y) =G(y) =A_y(F(τ(y)).Hence ξ=A_y(F(τ(y)) orξ ∈Range(A_y).

(ii) IfG ∈ C(Y, E₁) is such thatkGk∞ = 1 and kG(y)k_E₁ = 1 for every y ∈B(A),then if F ∈ C(X, E) is such thatA(F) =G, we havekF(x)k_E = 1 for every x∈X.

Since A(F) = G for every y ∈ B(A), we have A(F)(y) = G(y) = A_y(F(τ(x))). Therefore, 1 = kG(y)k_E₁ ≤ kA_yk kF(x)k_E = kF(x)k_E. Since kFk∞= 1 andτ is onto, we havekF(x)k_E = 1 for everyx∈X.

(iii) IfE is strictly convex, thenB(A) =Y.

Let z ∈ Y \B(A) and let β :Y → [0,1] be a continuous function such that β(B(A)) = 1 and β(z) = 0. We choose a vector e₁ ∈E₁ of norm 1 and set G(y) =β(y)·e₁ for everyy∈Y.Clearly,kGk∞= 1.Since Ais onto, there exists a unique F ∈ C(X, E) of norm 1 such thatA(F) =G.By statement (ii) for everyx∈X we havekF(x)k_E = 1.DefineG∗(y) = (2β(y)−1)·e₁. There exists F∗ such that A(F∗) =G∗.LetF0 in C(X, E) be such that A(F0) =E1, with E1 the constant function everywhere equal to e₁. Set F∗ = 2F −F₀. Statement (ii) implies that kF∗(x)k_E =kF₀(x)k_E = 1 for everyx. Therefore,

(2F(x)−F0(x))+F0(x) 2

= 1 or, equivalently,

(F∗(x))+F0(x) 2

= 1. Since E is strictly convex, we have F∗(x) = F0(x) for all x. This implies that F = F0, henceAF =AF₀orG=E1.This contradicts the fact thatg(z) = 06=E1(z) = e₁ and shows thatB(A) =Y.

(iv)Ay is injective for every y∈B(A).

Suppose there existsy₀∈B(A) andξ ∈E of norm 1 such that A_y₀(ξ) = 0. Since A is an onto isometry, there exists A^∗ : C(Y, E₁) → C(X, E) such that AA^∗ = Id_C(Y,E₁₎ and A^∗A = Id_C(X,E). Define X∈ C(X, E) byX(x) = ξ.

Cambern’s theorem asserts the existence of B(A^∗)⊂X and a surjective and continuous map τ₁ :B(A^∗)→ Y such that A^∗[G](x) =A^∗_x[G(τ₁(x)] for every x ∈B(A^∗).Given y0 ∈B(A), there existsx0 ∈B(A^∗) such that τ1(x0) = y0. Therefore,

ξ=A^∗A(X)(x₀) =A^∗_x₀[A(X)(y₀)] =A^∗_x₀[A_y₀(X(τ(y₀))] =A^∗_x₀[A_y₀(ξ)] = 0.

Hence Ay0(ξ) = 0 impliesξ = 0.

(v) For everyx∈Xande∈E(of norm 1),B(x, e) =B(x) =S{B(x, e) : e∈E andkek_E = 1} has exactly one element.

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Assume thatB(x, e) has at least two elements y1 and y2. Define a continuous map β :Y →[0,1] such that β(y₁) = 1 andβ(y₂) = 0.Setf =A_y₁(e) (f 6= 0 since Ay1 is injective). Denote G(y) = β(y)·f. There exists F, kFk_∞=kfk_E₁ =kGk_∞,such thatA(F) =G.SinceA(F)(y1) =G(y1) =f = A_y₁(F(x)),we havekfk_E₁ ≤ kA_y₁k · kF(x)k_E ≤ kF(x)k_e.On the other hand, A(F)(y2) =G(y2) = 0 =Ay2(F(x)) implies that F(x) = 0 (Ay2 is injective).

This proves that B(x, e) has at most one element. It was shown in [2] that B(x, e) is nonempty. Observe that a similar proof allows us to conclude that B(x) =S

{B(x, e) :e∈E and kek_E = 1} has exactly one element.

(vi)τ :B(A)→X is a homeomorphism.

First, observe that τ and τ1 :B(A^∗) → Y are continuous and bijective.

We have shown in (v) that B(x) ={y}and B(y) ={x₀}.We now prove that x0 = x. Given F ∈ C(X, E) such that F(x) = kFk_∞·e for some norm 1 vector e∈ E, we have kA(F)(y)k = kFk_∞ = kAFk_∞. Therefore, kF(x)k = kF(x₀)k= kA^∗(A(F))(x0)k =kA(F)k∞, which implies that x =x0, B(y) = {x},τ ◦τ₁ = Id_B(A^∗₎, andτ₁◦τ = Id_B(A). Henceτ is a homeomorphism and B(A) is a compact subset of X. Statement (iii) implies that B(A) =X.

(vii) For everyy∈Y,Ay is an isometry.

We need to show that we havekA_y(e)k=kekfor every e∈E. Without loss of generality, we assumeeto be of norm 1. Suppose that there existy₀and esuch thatkA_y₀(e)k<1.Denote byF the constant function everywhere equal to e(kFk_∞ =kek_E = 1). Since A is an isometry, we havekA(F)k_∞= 1 and A(F)(y) = Ay(e).Therefore, sup_ykA_y(e)k_E₁ = 1. Assuming thatkA_y₀(e)k<

1,there exists an open neighborhoodW_y₀ such that kA_z(e)k< 1 +kA_y₀(e)k

2 <1

for every z ∈ W_y₀. There also exists an open neighborhood O_y₀ such that y₀ ∈ O_y₀ ⊂ O_y₀ ⊂ W_y₀.Select β continuous, with values in the interval [0,1], such that β(x) = 1 for every x∈τ(O_y₀) and β(x) = 0 for everyx /∈ τ(W_y₀).

Set G(x) =β(x)·e. Therefore, we have kGk_∞= 1 =kA(G)k_∞ and A(G)(y) =A_y(G(τ(y)) =

( β(τ(y))·A_y(e) ify∈ W_y₀, 0 ify /∈ W_y₀. This implies that kA(G)k_∞<1 since

kA(G)(y)k=







|β(τ(y))| kA_y(e)k ≤ 1 +kA_y₀(e)k

2 <1 ify∈ W_y₀,

0 ify /∈ W_y₀.

This contradiction proves the statement.

We have shown that given an isometry A from C(X, E) onto C(Y, E₁), with E a strictly convex Banach space,X and Y compact topological spaces,

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we have B(A) = Y, τ is a homeomorphism from Y onto X, and A(F)(y) = A_Y(F(τ(y)) for every F ∈ C(X, E) and y ∈ Y. Moreover, A_y is an onto isometry in B(E, E1) for everyy∈Y. This completes the proof.

We recall that strictly convex Banach spaces are also complex strictly convex. As a consequence, to complete the proof of Theorem 3.5, we only need to modify the argument in statement (iii) in the proof of Proposition 3.6.

Proof. If kF(x)k = 1 for every x, then kiF(x)k = 1. Therefore, we have k(−i)(F(x)−F∗(x)) + iF(x)k = kiF_∗(x)k = 1 and ki(F(x)−F∗(x)) + iF(x)k =k2F(x)−F∗(x)k = 1. Since E is complex strictly convex, we have F =F∗. The remainder of the proof of Theorem 3.5 follows from the previous proposition.

Remark 3.7. It would be interesting to know whether Cambern’s theorem holds for range spaces E which have trivial multipliers, cf. [1].

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[3] M.M. Day,Normed Linear Spaces. Springer-Verlag, New York–Heidelberg, 1973.

[4] J. Dugundji,Topology. Allyn and Bacon, Inc. Boston, 1970.

[5] R.J. Fleming and J.E. Jamison, Isometries on Banach Spaces: Function Spaces.Chap- man & Hall, 2003.

[6] W. Holszty`nski,Continuous mappings induced by isometries of spaces of continuous functions. Studia Math.26(1966), 133–136.

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Received 26 November 2008 The University of Memphis Department of Mathematical Sciences

Memphis, TN 38152 mbotelho@memphis.edu jjamison@memphis.edu