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Geometry of Spaces of Orthogonally Additive Polynomials on C(K)C(K)

Christopher Boyd, Raymond A. Ryan and Nina Snigireva
Abstract
00footnotetext: Keywords: Orthogonally additive; Homogeneous polynomial; Banach Lattice; Regular Polynomial; Extreme point; Exposed point; Isometry.00footnotetext: MSC(2010): 46G25; 46G20; 46E10; 46B42; 46B04; 46E27.

We study the space of orthogonally additive nn-homogeneous polynomials on C(K)C(K). There are two natural norms on this space. First, there is the usual supremum norm of uniform convergence on the closed unit ball. As every orthogonally additive nn-homogeneous polynomial is regular with respect to the Banach lattice structure, there is also the regular norm. These norms are equivalent, but have significantly different geometric properties. We characterise the extreme points of the unit ball for both norms, with different results for even and odd degrees. As an application, we prove a Banach-Stone theorem. We conclude with a classification of the exposed points.

1 Introduction

A real function ff on a Banach lattice is said to be orthogonally additive if f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) whenever xx and yy are disjoint. Non-linear orthogonally additive functions on function spaces often have useful integral representations — see, for example the papers of Chacon, Friedman and Katz [8, 12, 13], Mizel [21] and Rao [24]. In 1990, Sundaresan [26] initiated the study of orthogonally additive nn-homogeneous polynomials with particular reference to the spaces Lp[0,1]L_{p}[0,1] and p\ell_{p} for 1p<1\leq p<\infty. Building on the work of Mizel, he showed that, for every orthogonally additive nn-homogeneous polynomial PP on Lp[0,1]L_{p}[0,1] with npn\leq p, there exists a unique function ξLp~\xi\in L_{\tilde{p}}, where p~=p/(pn)\tilde{p}=p/(p-n), such that

P(x)=01ξxn𝑑μP(x)=\int_{0}^{1}\xi x^{n}\,d\mu (1)

for every xLp[0,1]x\in L_{p}[0,1]. When n>pn>p, there are no non-zero orthogonally additive nn-homogeneous polynomials on Lp[0,1]L_{p}[0,1]. He went on to show that the Banach space of orthogonally additive nn-homogeneous polynomials on Lp[0,1]L_{p}[0,1] is isometrically isomorphic to Lp~L_{\tilde{p}} where the latter space is equipped not with the usual norm, but with the equivalent norm x=max{x+p~,xp~}\|x\|=\max\{\|x^{+}\|_{\tilde{p}},\|x^{-}\|_{\tilde{p}}\}.

The next significant development was the discovery of an integral representation for orthogonally additive nn-homogeneous polynomials on C(K)C(K) spaces by Pérez and Villanueva [23] and by Benyamini, Lassalle and Llavona [3], who proved a representation of the form

P(x)=Kxn𝑑μP(x)=\int_{K}x^{n}\,d\mu (2)

where μ\mu is a regular Borel signed measure on KK. The integral representations (1) and (2) have been extended and generalized in various directions in recent years. See, for example, [22, 17, 1, 30].

Orthogonally additive nn-homogeneous polynomials are also of interest in the study of multilinear operators on Banach lattices and, more generally, on vector lattices. If E,FE,F are vector lattices, an nn-linear mapping A:EnFA\colon E^{n}\to F is orthosymmetric if A(x1,,xn)=0A(x_{1},\dots,x_{n})=0 whenever xix_{i} and xjx_{j} are disjoint for some pair of distinct indices i,ji,j. Orthosymmetric multilinear mappings are automatically symmetric [4]. In [5], Bu and Buskes prove that an nn-linear function is orthosymmetric if and only if the associate nn-homogeneous polynomial is orthogonally additive.

Let EE be a Banach lattice. For every positive element aa of EE, we may form the principal ideal

Ea={xE:|x|na for some n}E_{a}=\{x\in E:|x|\leq na\text{ for some $n\in\mathbb{N}$}\}

with lattice structure inherited from EE and the norm defined by xa=inf{C>0:|x|Ca}\|x\|_{a}=\inf\{C>0:|x|\leq Ca\}. With this norm, EaE_{a} is a Banach lattice. By virtue of the Kakutani representation theorem [16], the Banach lattice EaE_{a} is canonically Banach lattice isometrically isomorphic to C(K)C(K) for some compact Hausdorff topological space KK, with aa being identified with the unit function on KK. The Banach lattice structure of EE is uniquely determined by its principal ideals. It follows that an analysis of the orthogonally additive nn-homogeneous polynomials on C(K)C(K) is central to an understanding of the behaviour of orthogonally additive nn-homogeneous polynomials on general Banach lattices.

In this paper, we focus on the geometric properties of the spaces 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) of orthogonally additive nn-homogeneous polynomials on C(K)C(K). There are two phenomena that are of particular interest. The first is that there are two natural ways to norm the space 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}). The first is the norm of uniform convergence on the unit ball of EE, given by P=sup{|P(x)|:xE,x1}\|P\|_{\infty}=\sup\{|P(x)|:x\in E,\|x\|\leq 1\}. In this norm, 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) is a Banach space. Now 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) also has a lattice structure and so another choice of norm is the regular norm, defined by Pr=|P|\|P\|_{r}=\||P|\|_{\infty}, where |P||P| is the absolute value of PP. In this norm, 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) is a Banach lattice. The existence of these two norms was first observed by Bu and Buskes [5] and is hinted at in the paper of Sundaresan [26]. These norms are equivalent, but we shall see that they have significantly different geometric properties.

The second phenomenon is the influence of the parity of the degree nn on the structure of the space 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) for the two norms. Bu and Buskes [5] showed that, when nn is odd, the supremum and regular norms on 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) are the same and that they are equivalent when nn is even. We sharpen their results, using the strategy of working first on 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) and then extending to general Banach lattices. The integral representation (2) gives a canonical isomorphism between 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) and (K)\mathcal{M}(K), the space of regular Borel signed measures on KK. The regular norm on 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) corresponds to the usual variation norm on (K)\mathcal{M}(K), but the supremum norm is identified with a different norm on (K)\mathcal{M}(K), given by μ0=max{μ+1,μ1}\|\mu\|_{0}=\max\{\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\}. We show that ((K),0)(\mathcal{M}(K),\|\cdot\|_{0}) is isometrically isomorphic to the dual space of C(K)C(K), where C(K)C(K) is endowed with the norm xd=x++x\|{x}\|_{d}=\|x^{+}\|_{\infty}+\|x^{-}\|_{\infty} and we show that this norm is closely related to the diameter seminorm (see, for example, [6]). We use these identifications to give a complete description of the extreme points of the unit ball of 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) for both norms, extending the results in [7]. Our starting point is a characterisation of the extreme points in C(K)C(K) for the norm d\|\cdot\|_{d} and (K)\mathcal{M}(K) for the norm 0\|\cdot\|_{0}. This allows us to prove a Banach-Stone theorem for (C(K),d)(C(K),\|\cdot\|_{d}).

We finish with a study of the exposed points of the unit ball of the space 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}). The identification of this space with the space of measures (K)\mathcal{M}(K), which is a dual space for both norms, allows us to use the theory of Šmul’yan [28, 29]. Using this machinery, we characterise the weak exposed and the weak strongly exposed points of the unit ball.

Preliminaries

Let EE be a real Banach space and let nn be a natural number. A function P:EP\colon E\to\mathbb{R} is an nn-homogeneous polynomial if there exists a necessarily unique, bounded nn-linear function A:EnA\colon E^{n}\to\mathbb{R} such that P(x)=A(x,,x)P(x)=A(x,\dots,x) for all xEx\in E. We write P=A^P=\widehat{A} if PP and AA are related in this way. The space 𝒫(nE)\mathcal{P}(^{n}{E}) of nn-homogeneous polynomials is a Banach space with the supremum norm,

P=sup{|P(x)|:xE,x1}.\|P\|_{\infty}=\sup\{|P(x)|:x\in E,\|x\|\leq 1\}\,.

The Banach space (𝒫(nE),)\bigl{(}\mathcal{P}(^{n}{E}),\|\cdot\|_{\infty}\bigr{)} is a dual space. We refer to the book by Dineen [10] for this and other facts about nn-homogeneous polynomials.

Now assume that EE is a Banach lattice. A partial order is defined on 𝒫(nE)\mathcal{P}(^{n}{E}) by P=A^Q=B^P=\widehat{A}\leq Q=\widehat{B} if A(x1,,xn)B(x1,,xn)A(x_{1},\dots,x_{n})\leq B(x_{1},\dots,x_{n}) for all x1,,xn0x_{1},\dots,x_{n}\geq 0. In particular, an nn-homogeneous polymonial PP is said to be positive if P0P\geq 0 in the sense of this order and PP is regular if it is the difference of two positive nn-homogeneous polynomials. The regular polynomials are precisely those that have an absolute value, which is given by the formula

|P|(x)=sup{i1,,in|A(ui11,,uinn)|:u1,,unΠ(x)},|P|(x)=\sup\Bigl{\{}\,\sum_{i_{1},\dots,i_{n}}|A(u^{1}_{i_{1}},\dots,u^{n}_{i_{n}})|:u^{1},\dots,u^{n}\in\Pi(x)\Bigr{\}}\,, (3)

where Π(x)\Pi(x) denotes the set of partitions of xx, namely, all finite sets of positive elements of EE whose sum is xx [5].

The space 𝒫r(nE)\mathcal{P}_{r}(^{n}{E}) of regular nn-homogeneous polynomials on EE is a Banach lattice with the regular norm,

Pr=|P|.\|P\|_{r}=\|\,|P|\,\|_{\infty}\,.

We have PPr\|P\|_{\infty}\leq\|P\|_{r} and in general these norms are not equivalent on 𝒫r(nE)\mathcal{P}_{r}(^{n}{E}). Every regular nn-homogeneous polynomial PP can be decomposed canonically as the difference of two positive nn-homogeneous polynomials, so that P=P+PP=P^{+}-P^{-} and |P|=P++P|P|=P^{+}+P^{-}. We refer to the paper of Bu and Buskes [5] for further details. For example, they show that (𝒫r(nE),r)\bigl{(}\mathcal{P}_{r}(^{n}{E}),\|\cdot\|_{r}\bigr{)} is a dual Banach lattice.

Let KK be a compact, Hausdorff space. The space C(K)C(K) of continuous real functions on KK is a Banach lattice with the supremum norm, x=sup{|x(t)|:tK}\|x\|_{\infty}=\sup\{|x(t)|:t\in K\}. We denote by (K)\mathcal{M}(K) the space of regular Borel signed measures on KK. Then the Banach lattice dual of C(K)C(K) can be identified with (K)\mathcal{M}(K) under the variation norm, which we denote by 1\|\cdot\|_{1}. Thus,

μ1=|μ|(K)=μ+(K)+μ(K)=μ+1+μ1,\|\mu\|_{1}=|\mu|(K)=\mu^{+}(K)+\mu^{-}(K)=\|\mu^{+}\|_{1}+\|\mu^{-}\|_{1}\,,

where μ+\mu^{+}, μ\mu^{-} are the positive and negative parts of μ\mu.

2 Orthogonally additive nn-homogeneous polynomials

Let EE be a Banach lattice and nn a positive integer. A function P:EP\colon E\to\mathbb{R} is called an orthogonally additive nn-homogeneous polynomial if PP is a bounded nn-homogeneous polynomial with the property that P(x+y)=P(x)+P(y)P(x+y)=P(x)+P(y) whenever x,yEx,y\in E are disjoint. The space of orthogonally additive nn-homogeneous polynomials on EE is denoted by 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}). It is easy to see that 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) is a closed subspace of the space (𝒫(nE),)\bigl{(}\mathcal{P}(^{n}{E}),\|\cdot\|_{\infty}\bigr{)} of bounded nn-homogeneous polynomials with the supremum norm. Thus 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}), with this norm, is a Banach space. When n=1n=1, this space is simply the dual space EE^{\prime}, since every bounded linear functional is orthogonally additive.

We have the following integral representation for orthogonally additive nn-homogeneous polynomials on C(K)C(K) spaces, due to Pérez-García and Villanueva [23] and Benyami, Lassalle and Llavona [3] (see also [7]).

Theorem 1.

Let KK be a compact, Hausdorff topological space. For every orthogonally additive nn-homogeneous polynomial PP on C(K)C(K) there is a regular Borel signed measure μ\mu on KK such that

P(x)=Kxn𝑑μP(x)=\int_{K}x^{n}\,d\mu

for all xC(K)x\in C(K).

In general, there is no guarantee that a Banach lattice supports any non-trivial orthogonally additive polynomials of degree greater than one. Sundaresan [26] showed that there are no non-zero orthogonally additive nn-homogeneous polynomials on L1[0,1]L_{1}[0,1] for n>1n>1. In the case of 1\ell_{1}, it is easy to see that an nn-homogeneous polynomial PP is orthogonally additive if and only if there exists a bounded sequence of real numbers, (aj)(a_{j}), such that

P(x)=j=1ajxjnP(x)=\sum_{j=1}^{\infty}a_{j}x_{j}^{n}

for every x1x\in\ell_{1}, and that P=supj|aj|\|P\|_{\infty}=\sup_{j}|a_{j}|. Thus 𝒫o(n1)\mathcal{P\!}_{o}(^{n}{\ell_{1}}) is isometrically isomorphic to \ell_{\infty} for every nn.

To put the results of the previous paragraph in a general context, we recall that a Banach lattice EE is an AL-space if the norm is additive on the positive cone: x+y=x+y\|x+y\|=\|x\|+\|y\| for all x,y0x,y\geq 0. The Kakutani representation theorem [15, 18] states that every AL-space EE can be decomposed into a disjoint sum of copies of 1\ell_{1} and L1L_{1} spaces. Accordingly, EE is Banach lattice isometrically isomorphic to a space of the form

[1(Γ)(αAL1[0,1]mα)]1\Bigl{[}\ell_{1}(\Gamma)\oplus\bigl{(}\bigoplus_{\alpha\in A}L_{1}[0,1]^{m_{\alpha}}\bigr{)}\Bigr{]}_{1}

In this representation, the unit basis vectors eγe_{\gamma} in 1(Γ)\ell_{1}(\Gamma) are in one-to-one correspondence with the atoms in EE of unit norm. We recall that a positive element xx of EE is said to be an atom if 0yx0\leq y\leq x implies that yy is a scalar multiple of xx. We can write the second component in this representation as L1(μ)L_{1}(\mu), where μ\mu is the product of the Lebesgue measures on the sets [0,1]mα[0,1]^{m_{\alpha}}. Thus, we see that EE can be represented as the disjoint sum 1(Γ)1L1(μ)\ell_{1}(\Gamma)\oplus_{1}L_{1}(\mu), where the measure μ\mu is nonatomic.

Proposition 1.

Let EE be an AL-space and let n>1n>1. There is a non-zero orthogonally additive nn-homogeneous polynomial on EE if and only if EE contains at least one atom.

Proof.

Let 1(Γ)1L1(μ)\ell_{1}(\Gamma)\oplus_{1}L_{1}(\mu) be the Kakutani representation of EE as described above.

Suppose that EE contains an atom. Then the set Γ\Gamma in the Kakutani representation is non-empty. Choose γ0Γ\gamma_{0}\in\Gamma and define P(x)=xγ0nP(x)=x_{\gamma_{0}}^{n} for x=(xγ)1(Γ)x=(x_{\gamma})\in\ell_{1}(\Gamma) and P(x)=0P(x)=0 for xL1(μ)x\in L_{1}(\mu). Then PP is a non-zero orthogonally additive nn-homogeneous polynomial.

Conversely, suppose that PP has no atoms. Then the Kakutani representation of PP is L1(μ)L_{1}(\mu) where the measure μ\mu is nonatomic. The proof in this case can be gleaned from [26], but we can give a direct proof as follows. We treat the case n=2n=2 for simplicity. Suppose that PP is an orthogonally additive 22-homogeneous polynomial on L1(μ)L_{1}(\mu), where μ\mu is nonatomic. Let AA be the bounded, symmetric bilinear form that generates PP. Then AA is orthosymmetric: if x,yx,y are disjoint, then A(x,y)=0A(x,y)=0 [5, Lemma 4.1]. It follows from the fact that L1(μ)^πL1(μ)L_{1}(\mu)\hat{\otimes}_{\pi}L_{1}(\mu) is isometrically isomorphic to L1(μ2)L_{1}(\mu^{2}) that there exists gL(μ2)g\in L_{\infty}(\mu^{2}) such that

A(x,y)=x(s)y(t)g(s,t)𝑑μ2(s,t)A(x,y)=\int x(s)y(t)g(s,t)\,d\mu^{2}(s,t)

If we take xx, yy to be the characteristic functions of arbitrary disjoint measurable sets, this integral is zero and so we have

A(x,y)=Dx(t)y(t)g(t,t)𝑑μ2A(x,y)=\int_{D}x(t)y(t)g(t,t)\,d\mu^{2}\

for all x,yL1(μ)x,y\in L_{1}(\mu), where DD is the diagonal. However, if μ\mu has no atoms, then the product measure of the diagonal is zero. Hence P(x)=0P(x)=0 for every xx. ∎

The Banach lattices L1(μ)L_{1}(\mu), where μ\mu is nonatomic, do not support any real valued lattice homomorphisms. Our next result indicates that the existence of non-trivial nn-homogeneous orthogonally additive polynomials on a Banach lattice is closely related to the existence of lattice homomorphisms.

Proposition 2.

Let EE be a Banach lattice, let φE\varphi\in E^{\prime} and let n2n\geq 2. The nn-homogeneous polynomial defined by P(x)=φ(x)nP(x)=\varphi(x)^{n} is orthogonally additive if and only if either φ\varphi or φ-\varphi is a lattice homomorphism.

Proof.

Suppose that φ\varphi or φ-\varphi is a lattice homomorphism. Then if xx and yy are disjoint, we have either φ(x)\varphi(x) or φ(y)=0\varphi(y)=0 and so P(x+y)=P(x)+P(y)P(x+y)=P(x)+P(y).

Conversely, suppose that P=φnP=\varphi^{n} is orthogonally additive. For every xEx\in E, the vectors x+x^{+} and txtx^{-} are disjoint for all tt\in\mathbb{R}. Therefore

φ(x+)n+tkφ(x)n=P(x++tx)=j=0n(nj)φ(x+)njφ(x)jtj.\varphi(x^{+})^{n}+t^{k}\varphi(x^{-})^{n}=P(x^{+}+tx^{-})=\sum_{j=0}^{n}\binom{n}{j}\varphi(x^{+})^{n-j}\varphi(x^{-})^{j}t^{j}\,.

for every tt\in\mathbb{R}. Hence either φ(x+)=0\varphi(x^{+})=0 or φ(x)=0\varphi(x^{-})=0. If we can show that φ\varphi (or φ-\varphi) is positive, then it follows that φ\varphi (or φ-\varphi) is a lattice homomorphism.

Let aa be a positive element of EE. The principal ideal EaE_{a} generated by aa is isometrically Banach lattice isomorphic to C(K)C(K) for some compact Hausdorff topological space KK. The functional φ\varphi is represented by a regular Borel signed measure μ\mu on KK and the fact that φ(x+)\varphi(x^{+}) or φ(x)=0\varphi(x^{-})=0 for all xEax\in E_{a} implies that the support of μ\mu consists of a single point. It follows that either φ\varphi or φ-\varphi is positive on EaE_{a}. Now EE is the union of the principal ideals EaE_{a}, which are upwards directed by inclusion. Thus, if φ\varphi (or φ-\varphi) is positive on one EaE_{a}, then φ\varphi (or φ-\varphi) is positive on all of EE. ∎

A Banach lattice EE is an AM-space if the norm has the property that xy=0x\wedge y=0 implies xy=max{x,y}\|x\vee y\|=\max\{\|x\|,\|y\|\}. In contrast with AL-spaces, there is a good supply of orthogonally additive nn-homogeneous polynomials on every AM-space. The Kakutani representation theorem for AM-spaces [16] shows that the real valued lattice homomorphisms on an AM-space EE separate the points of EE. It follows that there is a rich supply of orthogonally additive nn-homogeneous polynomials of every degree on EE.


We now look at some properties of orthogonally additive polynomials on general Banach lattices. Our starting point is the fact that every orthogonally additive nn-homogeneous polynomial on a Banach lattice EE is regular. This has been shown by Toumi [27, Theorem 1]. One may also argue as follows. Let PP be an orthogonally additive nn-homogeneous polynomial on a Banach lattice EE. As EE is the upwards directed union of its principal ideals, it suffices to show that PP is regular on each of them. Since each principal ideal is Banach lattice isometrically isomorphic to a C(K)C(K), we can use the integral representation in Theorem 1. Then the Jordan decomposition of the representing measure gives a decomposition of the polynomial into the difference of two positive orthogonally additive nn-homogeneous polynomials. Therefore PP is regular.

Let P=A^P=\widehat{A} be a regular nn-homogeneous polynomial PP on EE. The absolute value of PP is given by [5, 19]

|P|(x)=sup{i1,,in|A(ui11,,uinn)|:u1,,unΠ(x)}|P|(x)=\sup\Bigl{\{}\,\sum_{i_{1},\dots,i_{n}}|A(u^{1}_{i_{1}},\dots,u^{n}_{i_{n}})|:u^{1},\dots,u^{n}\in\Pi(x)\Bigr{\}} (4)

for x0x\geq 0, where Π(x)\Pi(x) denotes the set of partitions of xx, namely, all finite sets of positive vectors whose sum is xx. In general, we have

|P(x)||P|(|x|)|P(x)|\leq|P|(|x|) (5)

for every xEx\in E and |P||P| is the smallest positive nn-homogeneous polynomial, in the sense of the lattice structure of 𝒫r(nE)\mathcal{P}_{r}(^{n}{E}), with this property. The space 𝒫r(nE)\mathcal{P}_{r}(^{n}{E}) is a Banach lattice in the regular norm,

Pr=|P|.\|P\|_{r}=\|\,|P|\,\|_{\infty}\,.

It follows from (5) that PPr\|P\|_{\infty}\leq\|P\|_{r} for every P𝒫r(nE)P\in\mathcal{P}_{r}(^{n}{E}). In general, these norms are not equivalent.

Now 𝒫o(nE)\mathcal{P\!}_{o}(^{n}{E}) is complete in the regular norm; indeed, it is even a dual Banach lattice [5, Theorem 5.4]. It follows that the supremum and regular norms are equivalent on this space. Thus, there is a sequence (Cn)(C_{n}) of positive real numbers such that PrCnP\|P\|_{r}\leq C_{n}\,\|P\|_{\infty} for every nn and every P𝒫o(nE)P\in\mathcal{P\!}_{o}(^{n}{E}). Bu and Buskes [5] show that the two norms are the same for odd values of nn. For even values of nn, they show that Cnnn/n!C_{n}\leq n^{n}/n!, the polarization constant. We shall show that, in fact, Cn=2C_{n}=2 for even values of nn and that this is sharp. This will follow from estimates we give for the value of |P||P| at positive points in EE.

If φ\varphi is a bounded linear functional on EE, then [20]

|φ|(x)=sup{|φ(y)|:|y|x}|\varphi|(x)=\sup\{|\varphi(y)|:|y|\leq x\}

for every x0x\geq 0. It would be suprising if there were such a simple formula for |P|(x)|P|(x) when PP is a regular nn-homogeneous polynomial. As a linear functional, PP acts on an nn-fold symmetric tensor power of EE and the set of vectors yy satisfying |y|x|y|\leq x is now a set of tensors, rather than elements of EE. However, if PP is orthogonally additive, it is possible to establish a relatively simple estimate for the values of |P||P|.

Theorem 2.

Let PP be an orthogonally additive nn-homogeneous polynomial on the Banach lattice EE.

  1. (a)

    If nn is odd, then

    |P|(x)=sup{|P(y)|:|y|x}.|P|(x)=\sup\bigl{\{}|P(y)|:|y|\leq x\bigr{\}}\,.

    for every x0x\geq 0 in EE.

  2. (b)

    If nn is even, then

    |P|(x)2sup{|P(y)|:|y|x}.|P|(x)\leq 2\,\sup\bigl{\{}|P(y)|:|y|\leq x\bigr{\}}\,.

    for every x0x\geq 0 in EE.

Proof.

Let x0x\geq 0. It follows from (3) that the value |P|(x)|P|(x) is unchanged if we consider PP as an nn-homogeneous polynomial on the principal ideal ExE_{x} generated by xx. Now ExE_{x} is Banach lattice isomorphic to C(K)C(K) for some compact topological space KK. Since PP is orthogonally addive there exists a regular signed Borel measure μ\mu on KK such that

P(y)=Kyn𝑑μ.P(y)=\int_{K}y^{n}\,d\mu\,.

for every yExC(K)y\in E_{x}\cong C(K). The symmetric nn-linear form on C(K)nC(K)^{n} that generates PP is given by

A(x1,,xn)=Kx1xn𝑑μ.A(x_{1},\dots,x_{n})=\int_{K}x_{1}\dots x_{n}\,d\mu\,.

Thus, for x1,,xn0x_{1},\dots,x_{n}\geq 0,

|A(x1,,xn)|Kx1xnd|μ||A(x_{1},\dots,x_{n})|\leq\int_{K}x_{1}\dots x_{n}\,d|\mu|

and it follows that

|P|(x)Kxnd|μ||P|(x)\leq\int_{K}x^{n}\,d|\mu|

for x0x\geq 0.

Now in general, for a nonnegative function wC(K)w\in C(K) we have

Kwd|μ|=sup{|Kg𝑑μ|:gC(K),|g|w},\int_{K}w\,d|\mu|=\sup\Bigl{\{}\Bigl{|}\int_{K}g\,d\mu\Bigr{|}:g\in C(K),|g|\leq w\Bigr{\}}\,,

Therefore

|P|(x)sup{|Ky𝑑μ|:yC(K),|y|xn}.|P|(x)\leq\sup\Bigl{\{}\Bigl{|}\int_{K}y\,d\mu\Bigr{|}:y\in C(K),\;|y|\leq x^{n}\Bigr{\}}\,.

where we are identifying elements of EE with continuous functions on KK. We now consider separately the cases where nn is odd and even.

(a) We first consider the case when nn odd.
If |y|xn|y|\leq x^{n}, let v=y1/nv=y^{1/n}. Then |v|x|v|\leq x and Ky𝑑μ=Kvn𝑑μ\int_{K}y\,d\mu=\int_{K}v^{n}\,d\mu. Therefore

|P|(x)sup{|Kvn𝑑μ|:vE,|v|x}.|P|(x)\leq\sup\Bigl{\{}\Bigl{|}\int_{K}v^{n}\,d\mu\Bigr{|}:v\in E,|v|\leq x\Bigr{\}}\,.

Thus we have

|P|(x)sup{|P(y)|:|y|x}.|P|(x)\leq\sup\{|P(y)|:|y|\leq x\}\,.

and it is easy to see that the reverse inequality also holds.

(b) We now consider the case when nn even.
We have

|P|(x)sup{|Ky𝑑μ|:|y|xn}.|P|(x)\leq\sup\Bigl{\{}\Bigl{|}\int_{K}y\,d\mu\Bigr{|}:|y|\leq x^{n}\Bigr{\}}\,.

Given vExC(K)v\in E_{x}\cong C(K) satisfying |v|xn|v|\leq x^{n}, we define v1,v2C(K)v_{1},v_{2}\in C(K) by

v1(t)={v(t)1/nif v(t)00if v(t)<0v2(t)={0if v(t)0|v(t)|1/nif v(t)<0v_{1}(t)=\begin{cases}v(t)^{1/n}&\text{if }v(t)\geq 0\\ 0&\text{if }v(t)<0\end{cases}\qquad v_{2}(t)=\begin{cases}0&\text{if }v(t)\geq 0\\ |v(t)|^{1/n}&\text{if }v(t)<0\end{cases}

Then v=v1nv2nv=v_{1}^{n}-v_{2}^{n}, and so

|Kvdμ||Kv1ndμ|+|Kv2ndμ|=|P(v1)|+|P(v2)|.\Bigl{|}\int_{K}v\,d\mu\Bigr{|}\leq\Bigl{|}\int_{K}v_{1}^{n}\,d\mu\Bigr{|}+\Bigl{|}\int_{K}v_{2}^{n}\,d\mu\Bigl{|}=\bigl{|}P(v_{1})\bigr{|}+\bigl{|}P(v_{2})\bigr{|}\,.

It follows from |v|xn|v|\leq x^{n} that 0v1,v2x0\leq v_{1},v_{2}\leq x. Therefore

|P|(x)2sup{|P(y)|:0yx}=2sup{|P(y)|:|y|x},|P|(x)\leq 2\,\sup\bigl{\{}|P(y)|:0\leq y\leq x\bigr{\}}=2\,\sup\bigl{\{}|P(y)|:|y|\leq x\bigr{\}}\,, (6)

since nn is even.

To see that the bound in (6) for even values of nn is sharp, consider the example P(x)=x1nx2nP(x)=x_{1}^{n}-x_{2}^{n} on 2\mathbb{R}^{2} with any Banach lattice norm. The bound is attained for the vector x=(1,1)x=(1,1).

Corollary 1.

Let PP is an orthogonally additive nn-homogeneous polynomial on a Banach lattice EE. Then Pr=P\|P\|_{r}=\|P\|_{\infty} if nn is odd and PPr2P\|P\|_{\infty}\leq\|P\|_{r}\leq 2\,\|P\|_{\infty} if nn is even. These inequalities are sharp.

3 Orthogonally additive polynomials on C(K)C(K)

In this section, we study the supremum and regular norms on the spaces of orthogonally additive nn-homogeneous polynomials on C(K)C(K).

The integral representation for orthogonally additive nn-homogeneous polynomials on C(K)C(K) allows us to identify the vector space 𝒫r(nC(K))\mathcal{P}_{r}(^{n}{C(K)}) with (K)\mathcal{M}(K), the space of regular Borel signed measures on KK. The natural norm on (K)C(K)\mathcal{M}(K)\cong C(K)^{\prime} is the dual norm. This is the variation norm for measures: μ1=|μ|(K)\|\mu\|_{1}=|\mu|(K). We shall see that this norm corresponds to the regular norm on the spaces of orthogonally additive nn-homogeneous polynomials. However, the supremum norm on 𝒫o(nK)\mathcal{P\!}_{o}(^{n}{K}) corresponds to a different, but equivalent norm on the space of regular Borel signed measures.

The space 𝒫r(nC(K))\mathcal{P}_{r}(^{n}{C(K)}) is a Banach lattice with the regular norm, as is the dual Banach lattice (K)\mathcal{M}(K) with the variation norm. We shall see that the lattice structures of these two Banach lattices are the same. We note that the lattice structure of (K)\mathcal{M}(K) as the dual of C(K)C(K) is the same as the lattices structure of (K)\mathcal{M}(K) considered as a sublattice of the lattice of Borel signed measures on KK. In other words, a measure μ(K)\mu\in\mathcal{M}(K) is positive, in the sense that Kf𝑑μ0\int_{K}f\,d\mu\geq 0 for every nonnegative xC(K)x\in C(K), if and only if μ(E)0\mu(E)\geq 0 for every Borel subset EE of KK [25, Theorem 2.18].

Proposition 3.

Let KK be a compact Hausdorff topological space and let PP be an orthogonally additive nn-homogeneous polynomial on C(K)C(K), given by

P(x)=Kxn𝑑μ.P(x)=\int_{K}x^{n}\,d\mu\,.

Then the absolute value of PP is given by

|P|(x)=Kxnd|μ|.|P|(x)=\int_{K}x^{n}\,d|\mu|\,.
Proof.

We have seen in the proof of Theorem 2 that

|P|(x)Kxnd|μ||P|(x)\leq\int_{K}x^{n}\,d|\mu|

for every x0x\geq 0.

To prove the reverse inequality, we start with the definition of the absolute value:

|P|(x)=sup{i1,,in|A(ui11,,uinn)|:u1,,unΠ(x)}|P|(x)=\sup\Bigl{\{}\,\sum_{i_{1},\dots,i_{n}}|A(u^{1}_{i_{1}},\dots,u^{n}_{i_{n}})|:u^{1},\dots,u^{n}\in\Pi(x)\Bigr{\}}

for x0x\geq 0 Taking each of u2,,unu^{2},\dots,u^{n} to be the trivial partition {x}\{x\} gives

|P|(x)\displaystyle|P|(x) sup{i|A(ui1,x,,x)|:u1Π(x)}\displaystyle\geq\sup\Bigl{\{}\,\sum_{i}|A(u^{1}_{i},x,\dots,x)|:u^{1}\in\Pi(x)\Bigr{\}}
=sup{i|Kui1xn1𝑑μ|:u1Π(x)}=Kxnd|μ|,\displaystyle=\sup\Bigl{\{}\,\sum_{i}\Bigl{|}\int_{K}u^{1}_{i}x^{n-1}\,d\mu\Bigr{|}:u^{1}\in\Pi(x)\Bigr{\}}=\int_{K}x^{n}\,d|\mu|\,,

applying the partition form of the Riesz-Kantorovich formula for the absolute value of a linear functional [2, Theorem 1.16] to the measure dλ=xn1dμd\lambda=x^{n-1}d\mu in (K)\mathcal{M}(K) and using the fact that d|λ|=xn1d|μ|d|\lambda|=x^{n-1}d|\mu|.

Theorem 3.

Let KK be a compact, Hausdorff space. Let Jn:(K)𝒫o(nC(K))J_{n}\colon\mathcal{M}(K)\to\mathcal{P\!}_{o}(^{n}{C(K)}) be given by

(Jnμ)(x)=Kxn𝑑μ.(J_{n}\mu)(x)=\int_{K}x^{n}\,d\mu\,.
  • (a)

    For every nn, JnJ_{n} is a Banach lattice isometric isomorphism from ((K),1)\bigl{(}\mathcal{M}(K),\|\cdot\|_{1}\bigr{)} onto (𝒫o(nC(K)),r)\bigl{(}\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{r}\bigr{)}.

  • (b)

    If nn is odd, then the regular and supremum norms coincide on 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) and so JnJ_{n} is an isometric isomorphism for the supremum norm on 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}).

  • (c)

    If nn is even, then JnJ_{n} is an isometric isomorphism for the norm on (K)\mathcal{M}(K) defined by

    μ0:=max{μ+1,μ1}\|\mu\|_{0}:=\max\{\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\}

    and the supremum norm on 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}).

Proof.

Cleary, JnJ_{n} is linear and surjective. To see that it is injective, suppose that the nn-homogeneous polynomial P(x)=Kxn𝑑μP(x)=\int_{K}x^{n}\,d\mu is zero. The associated symmetric nn-linear form is

A(x1,,xn)=Kx1xn𝑑μA(x_{1},\dots,x_{n})=\int_{K}x_{1}\dots x_{n}\,d\mu

and so A(x1,,xn)=0A(x_{1},\dots,x_{n})=0 for all x1,,xnC(K)x_{1},\dots,x_{n}\in C(K). Taking x2==xn=1x_{2}=\dots=x_{n}=1, we have Kx𝑑μ=0\int_{K}x\,d\mu=0 for every xC(K)x\in C(K) and so μ=0\mu=0.

(a) Clearly, JnJ_{n} is positive. If we show that Jn1J_{n}^{-1} is also positive, then it will follow that JnJ_{n} is a lattice homomorphism [2, Theorem 7.3]. Let P=A^P=\widehat{A} be a positive element of 𝒫r(nC(K))\mathcal{P}_{r}(^{n}{C(K)}), with μ(K)\mu\in\mathcal{M}(K) satisfying Jnμ=PJ_{n}\mu=P. Then, for every nonnegative xC(K)x\in C(K), we have Kx𝑑μ=A(x,1,,1)0\int_{K}x\,d\mu=A(x,1,\dots,1)\geq 0 and so μ\mu is positive. Therefore JnJ_{n} is a lattice isomorphism for every nn.

By Proposition 3, the regular norm of P=JnμP=J_{n}\mu is Pr=|P|=|P|(1)=|μ|(K)=μ1\|P\|_{r}=\|\,|P|\,\|_{\infty}=|P|(1)=|\mu|(K)=\|\mu\|_{1}, since |P||P| is increasing on the positive cone of C(K)C(K). Therefore JnJ_{n} is both a lattice isomorphism and an isometry.

(b) This has already been proved in Corollary 1.

(c) Let μ(K)\mu\in\mathcal{M}(K) and let P=JnμP=J_{n}\mu. It follows from (a) that P+=Jnμ+P^{+}=J_{n}\mu^{+} and P=JnμP^{-}=J_{n}\mu^{-}. We have

P(x)=Kxn𝑑μ+Kxn𝑑μP(x)=\int_{K}x^{n}\,d\mu^{+}-\int_{K}x^{n}d\mu^{-}

for every xC(K)x\in C(K). As |ab|max{|a|,|b|}|a-b|\leq\max\{|a|,|b|\} for a,b+a,b\in\mathbb{R}^{+} and nn is even, it follows that Pmax{μ+1,μ1}\|P\|_{\infty}\leq\max\{\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\}.

Now let {A,B}\{A,B\} be a Hahn decomposition of μ\mu, with μ\mu positive on AA and negative on BB. If FAF\subset A is compact, then by a standard argument using Urysohn’s lemma (see, for example, [14, Theorem 12.41]) there is a decreasing sequence (xk)(x_{k}) of continuous functions on KK with values in [0,1][0,1] that converges almost everywhere with respect to |μ||\mu| to 1F1_{F}, the characteristic function of FF. Then, by the bounded convergence theorem,

Plimk|Kxkn𝑑μ|=|K1F𝑑μ|=μ+(F).\|P\|_{\infty}\geq\lim_{k\to\infty}\biggl{|}\int_{K}x_{k}^{n}\,d\mu\biggr{|}=\biggl{|}\int_{K}1_{F}\,d\mu\biggr{|}=\mu^{+}(F)\,.

It follows from the regularity of μ+\mu^{+} that Pμ+(A)=μ+1\|P\|_{\infty}\geq\mu^{+}(A)=\|\mu^{+}\|_{1}. Similarly, Pμ1\|P\|_{\infty}\geq\|\mu^{-}\|_{1}. Therefore P=μ0\|P\|_{\infty}=\|\mu\|_{0}. ∎

We summarize the identifications of the various norms, bearing in mind that the supremum and regular norms coincide for positive polynomials.

Corollary 2.

Let PP be an orthogonally additive nn-homogeneous polynomial on C(K)C(K), with corresponding measure μ(K)\mu\in\mathcal{M}(K). Then

  • (a)

    Pr=P+r+Pr=μ+1+μ1=μ1\|P\|_{r}=\|P^{+}\|_{r}+\|P^{-}\|_{r}=\|\mu^{+}\|_{1}+\|\mu^{-}\|_{1}=\|\mu\|_{1}.

  • (b)

    If nn is odd, then P=Pr\|P\|_{\infty}=\|P\|_{r}.

  • (c)

    If nn is even, then P=max{P+r,Pr}=max{μ+1,μ1}=μ0\|P\|_{\infty}=\max\{\|P^{+}\|_{r},\|P^{-}\|_{r}\}=\max\{\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\}=\|\mu\|_{0}.

We note that the norm 0\|\cdot\|_{0} is easily seen to be equivalent to the dual (variation) norm on (K)\mathcal{M}(K). In fact, we have

μ0μ12μ0\|\mu\|_{0}\leq\|\mu\|_{1}\leq 2\,\|\mu\|_{0}

for every μ(K)\mu\in\mathcal{M}(K).

It will be useful to have an alternative expression for the norm 0\|\cdot\|_{0} on (K)\mathcal{M}(K). Using the identity max{a,b}=12(|a+b|+|ab|)\max\{a,b\}=\frac{1}{2}\bigl{(}|a+b|+|a-b|\bigr{)} for non-negative real numbers, we have

μ0\displaystyle\|\mu\|_{0} =12(|μ+1+μ1|+|μ+1μ1|)\displaystyle=\frac{1}{2}\bigl{(}\bigl{|}\|\mu^{+}\|_{1}+\|\mu^{-}\|_{1}\bigr{|}+\bigl{|}\|\mu^{+}\|_{1}-\|\mu^{-}\|_{1}\bigr{|}\bigr{)}
=12(μ1+|μ+(K)μ(K)|)=12(μ1+|μ(K)|)\displaystyle=\frac{1}{2}\bigl{(}\|\mu\|_{1}+\bigl{|}\mu^{+}(K)-\mu^{-}(K)\bigr{|}\bigr{)}=\frac{1}{2}\bigl{(}\|\mu\|_{1}+\bigl{|}\mu(K)\bigl{|}\bigr{)}

Thus, we have

μ0=max{μ+1,μ1}=12(μ1+|μ(K)|)\|\mu\|_{0}=\max\bigl{\{}\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\bigr{\}}=\frac{1}{2}\Bigl{(}\|\mu\|_{1}+\bigl{|}\mu(K)\bigl{|}\Bigr{)} (7)

These results clarify the geometric properties of the spaces 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}); for the regular norm, these spaces are all essentially the same as the dual space (K)\mathcal{M}(K) with the variation norm. The case of 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}) with the supremum norm and nn even is substantially different. To understand this, we must study the extreme point structure of the unit ball of (K)\mathcal{M}(K) for the norm 0\|\cdot\|_{0}.

4 Extreme points in 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)})

In this section, we study the extreme points of the unit ball of the space 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}). We begin with the regular norm. We have seen in Proposition 3 that there is an isometric isomorphism

(𝒫o(nC(K)),r)((K),1)\bigl{(}\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{r}\bigr{)}\cong\bigl{(}\mathcal{M}(K),\|\cdot\|_{1}\bigr{)}

where 1\|\cdot\|_{1} denotes the variation norm on (K)\mathcal{M}(K), the space of regular Borel signed measures on KK. Furthermore, when the degree nn is odd, the supremum and regular norms on 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) coincide.

It is a classical result that the extreme points of the unit ball of (K)\mathcal{M}(K) for the variation norm are the measures of the form ±δt\pm\delta_{t}, where tKt\in K (see, for example, [11, V.8.6]). The isomorphism between 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) and (K)\mathcal{M}(K) associates the polynomial P(x)=x(t)nP(x)=x(t)^{n} with the measure δt\delta_{t}. Thus, we have

Proposition 4.

Let KK be a compact Hausdorff topological space. The extreme points of the closed unit ball of the space (𝒫o(nC(K)),r)\bigl{(}\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{r}\bigr{)} are the nn-homogeneous polynomials ±δtn\pm\delta_{t}^{n}, where tKt\in K and δtn(x)=x(t)n\delta_{t}^{n}(x)=x(t)^{n}.

This result is given in [7] for the supremum norm, but the proof given there is not valid for polynomials of even degree. However, this does not affect the results that follow in [7]. In particular, their elegant proof of the integral representation still stands. Essentially, all that is required for their arguments to work is that 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}) is a dual space and that the extreme points of the unit ball are as described above.

We now turn to the geometry of 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}) for the supremum norm, where the degree nn is even. We have the isometric isomorphism

(𝒫o(nC(K)),)((K),0)\bigl{(}\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}\bigr{)}\cong\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}

where μ0=max{μ+1,μ1}\|\mu\|_{0}=\max\{\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\}. We will show that 0\|\cdot\|_{0} is the dual of a norm on C(K)C(K) that is equivalent to the supremum norm.

The norm we seek is related to the diameter seminorm on C(K)C(K), which is defined by

ρ(x)=diam(x)=sup{|x(s)x(t)|:s,tK}.\rho(x)=\operatorname{diam}(x)=\sup\{|x(s)-x(t)|:s,t\in K\}\,.

It is easy to see that we also have

ρ(x)=2inf{xα1K:α}\rho(x)=2\,\inf\bigl{\{}\|x-\alpha 1_{K}\|_{\infty}:\alpha\in\mathbb{R}\bigr{\}}

The kernel of ρ\rho is the one dimensional subspace of constant functions. As in [6], we use Cρ(K)C_{\rho}(K) to denote the quotient space C(K)/kerρC(K)/\ker\rho. It is a Banach space under the norm

π(x)ρ=ρ(x)\|\pi(x)\|_{\rho}=\rho(x)

where π:C(K)C(K)/kerρ\pi\colon C(K)\to C(K)/\ker\rho is the quotient map. Following Cabello-Sanchez [6], we note that this means that (Cρ(K),ρ)\bigl{(}C_{\rho}(K),\|\cdot\|_{\rho}\bigr{)} is isometrically isomorphic, up to a a constant factor 22, to the quotient space of (C(K),)\bigl{(}C(K),\|\cdot\|_{\infty}\bigr{)} by the subspace of constant functions. Therefore the dual space (Cρ(K),ρ)\bigl{(}C_{\rho}(K),\|\cdot\|_{\rho}\bigr{)}^{\prime} is isometrically isomorphic, up to a constant factor 1/21/2, to a subspace of (C(K),)\bigl{(}C(K),\|\cdot\|_{\infty}\bigr{)}^{\prime}, the space of regular Borel signed measures with the variation norm. This subspace is the space of measures μ\mu satisfying μ(K)=0\mu(K)=0 and on it we have [6]

μ1=2μ(Cρ(K),ρ).\|\mu\|_{1}=2\|\mu\|_{\bigl{(}C_{\rho}(K),\|\cdot\|_{\rho}\bigr{)}^{\prime}}\,.
Theorem 4 (Cabello-Sanchez [6]).

Let KK be a compact Hausdorff topological space. A regular Borel signed measure μ\mu is an extreme point of the unit ball of the dual space (Cρ(K),ρ)\bigl{(}C_{\rho}(K),\|\cdot\|_{\rho}\bigr{)}^{\prime} if and only if μ=δsδt\mu=\delta_{s}-\delta_{t}, where ss and tt are distinct points of KK.

In order to apply this result, we first need to identify the predual of the norm 1\|\cdot\|_{1} on (K)C(K)\mathcal{M}(K)\cong C(K)^{\prime}.

Theorem 5.

Let KK be a compact Hausdorff topological space. Let d\|\cdot\|_{d} be the norm on C(K)C(K) defined by

xd:=x++x=max{x,ρ(x)}\|{x}\|_{d}:=\|x^{+}\|_{\infty}+\|x^{-}\|_{\infty}=\max\bigl{\{}\|x\|_{\infty},\rho(x)\bigr{\}} (8)

where ρ\rho is the diameter seminorm. Then the dual space of (C(K),d)\bigl{(}C(K),\|\cdot\|_{d}\bigr{)} is isometrically isomorphic to the space of regular Borel signed measures on KK with the norm μ0=max{μ+1,μ1}\|\mu\|_{0}=\max\{\|\mu^{+}\|_{1},\|\mu^{-}\|_{1}\}.

Proof.

A routine calculation shows that the formula xd=x++x\|{x}\|_{d}=\|x^{+}\|_{\infty}+\|x^{-}\|_{\infty} defines a norm on C(K)C(K). To establish the second equality in (8), we consider two cases.

  1. (a)

    Suppose the function xx has constant sign. Then ρ(x)x\rho(x)\leq\|x\|_{\infty} and one of x+\|x^{+}\|_{\infty}, x\|x^{-}\|_{\infty} is zero. Therefore xd=x\|{x}\|_{d}=\|x\|_{\infty}.

  2. (b)

    If xx changes sign, then xx++x=ρ(x)\|x\|_{\infty}\leq\|x^{+}\|_{\infty}+\|x^{-}\|_{\infty}=\rho(x).

Therefore xd=max{x,ρ(x)}\|{x}\|_{d}=\max\bigl{\{}\|x\|_{\infty},\rho(x)\bigr{\}} for every xC(K)x\in C(K).

Let us denote the dual norm of 1\|\cdot\|_{1} by 1\|\cdot\|_{1}^{\prime}. If xC(K)x\in C(K) and μ(K)\mu\in\mathcal{M}(K), then

Kx𝑑μ=(Kx+𝑑μ++Kx𝑑μ)(Kx+𝑑μ+Kx𝑑μ+).\int_{K}x\,d\mu=\Bigl{(}\int_{K}x^{+}\,d\mu^{+}+\int_{K}x^{-}\,d\mu^{-}\Bigr{)}-\Bigl{(}\int_{K}x^{+}\,d\mu^{-}+\int_{K}x^{-}\,d\mu^{+}\Bigr{)}\,.

Now

0Kx+𝑑μ++Kx𝑑μx+μ+1+xμ1xdμ00\leq\int_{K}x^{+}\,d\mu^{+}+\int_{K}x^{-}\,d\mu^{-}\leq\|x^{+}\|_{\infty}\|\mu^{+}\|_{1}+\|x^{-}\|_{\infty}\|\mu^{-}\|_{1}\leq\|{x}\|_{d}\|\mu\|_{0}

and similarly

0Kx+𝑑μ+Kx𝑑μ+xdμ0.0\leq\int_{K}x^{+}\,d\mu^{-}+\int_{K}x^{-}\,d\mu^{+}\leq\|{x}\|_{d}\|\mu\|_{0}\,.

Therefore

|Kx𝑑μ|xdμ0\Bigl{|}\int_{K}x\,d\mu\Bigr{|}\leq\|{x}\|_{d}\|\mu\|_{0}

and so μ1μ0\|\mu\|_{1}^{\prime}\leq\|\mu\|_{0}.

Fix μ(K)\mu\in\mathcal{M}(K) and let ε>0\varepsilon>0. Let {A,B}\{A,B\} be a Hahn decomposition for μ\mu, where AA is a positive set and BB a negative set. Since μ\mu is regular, there exist compact sets CAC\subset A and DBD\subset B such that |μ|(AC),|μ|(BD)<ε|\mu|(A\setminus C),|\mu|(B\setminus D)<\varepsilon. By Urysohn’s lemma, there is a continuous function y:K[0,1]y\colon K\to[0,1] that takes the values 11 and 0 on the sets CC and DD respectively. Then y1=1\|y\|_{1}=1 and

Ky𝑑μ=Cy𝑑μ+ACy𝑑μ+BDy𝑑μ\int_{K}y\,d\mu=\int_{C}y\,d\mu+\int_{A\setminus C}y\,d\mu+\int_{B\setminus D}y\,d\mu

It follows that

|Kydμ|μ+(C)2εμ+(A)3ε=μ+13ε.\Bigl{|}\int_{K}y\,d\mu\Bigl{|}\geq\mu^{+}(C)-2\varepsilon\geq\mu^{+}(A)-3\varepsilon=\|\mu^{+}\|_{1}-3\varepsilon\,.

Similarly,

|Ky𝑑μ|μ13ε\Bigl{|}\int_{K}y\,d\mu\Bigr{|}\geq\|\mu^{-}\|_{1}-3\varepsilon

Thus, μ1μ03ε\|\mu\|_{1}^{\prime}\geq\|\mu\|_{0}-3\varepsilon for every ε>0\varepsilon>0

Therefore μ1=μ0\|\mu\|_{1}^{\prime}=\|\mu\|_{0} for every μ(K)\mu\in\mathcal{M}(K). ∎

4.1 The extreme points of the unit ball of (C(K),d)\bigl{(}C(K),\|\cdot\|_{d}\bigr{)}

The extreme points of the closed unit ball of C(K)C(K) with the supremum norm are the constant functions ±1\pm 1. Our next result shows that changing to the equivalent norm given in the preceding proposition leads to a different set of extreme points.

Theorem 6.

A function xx is an extreme point of the closed unit ball of (C(K),d)(C(K),\|\cdot\|_{d}) if and only if either
(i) x(t)=1x(t)=1 or 0 for every tKt\in K, or
(ii) x(t)=1x(t)=-1 or 0 for every tKt\in K
(and {t:x(t)0}\{t:x(t)\neq 0\}\neq\varnothing in each case.)

Proof.

To show that every such function is extreme, let xd=1\|{x}\|_{d}=1, with x(t)=1x(t)=1 for tAt\in A and x(t)=0x(t)=0 for tAct\in A^{c}, where AA is a nonempty subset of KK. Suppose that

x=ay+bz,x=ay+bz\,,

where a,b(0,1)a,b\in(0,1) with a+b=1a+b=1 and yd=zd=1\|{y}\|_{d}=\|{z}\|_{d}=1. Then, for tAt\in A, ay(t)+bz(t)=1ay(t)+bz(t)=1. But |y(t)|,|z(t)|1|y(t)|,|z(t)|\leq 1 and it follows that y(t)=z(t)=1y(t)=z(t)=1 for every tAt\in A.

Now, if tAct\in A^{c}, then ay(t)+bz(t)=0ay(t)+bz(t)=0. But diam(y),diam(z)1\operatorname{diam}(y),\operatorname{diam}(z)\leq 1 and y,z=1\|y\|_{\infty},\|z\|_{\infty}=1 imply that 0y(t),z(t)10\leq y(t),z(t)\leq 1 for every tKt\in K and hence y(t)=z(t)=0y(t)=z(t)=0 for every tAct\in A^{c}. Therefore y(t)=z(t)=x(t)y(t)=z(t)=x(t) for every tKt\in K and so xx is an extreme point. The case in which xx takes values 1-1 and 0 is done in exactly the same way.

We now show that every extreme point is of this type. Let xx be an extreme point. Since xd=max{x,diam(x)}=1\|{x}\|_{d}=\max\bigl{\{}\|x\|_{\infty},\operatorname{diam}(x)\bigr{\}}=1, there are two cases to consider.

Case 1: x=1\|x\|_{\infty}=1 and diam(x)1\operatorname{diam}(x)\leq 1. Then xx takes its values either in [1,0][-1,0] or [0,1][0,1]. Suppose it is the latter. Then there is at least one point at which x(t)=1x(t)=1. Suppose there is a point sKs\in K for which 0<x(s)<10<x(s)<1. Then, by a standard argument, there is a function yC(K)y\in C(K) with values in [0,1][0,1] and supported by a neighbourhood of ss, such that x±y1\|x\pm y\|_{\infty}\leq 1. Clearly, we also have diam(x±y)1\operatorname{diam}(x\pm y)\leq 1. This implies that xx is not extreme and so we have a contradiction. Therefore xx can only have values 0 or 11.

Case 2: diam(x)=1\operatorname{diam}(x)=1 and x<1\|x\|_{\infty}<1. There exist points s,ts,t in KK such that |x(t)x(s)|=diam(x)=1|x(t)-x(s)|=\operatorname{diam}(x)=1. Without loss of generality we may assume that x(t)>x(s)x(t)>x(s). Then xx takes its values in the interval [x(s),x(t)][x(s),x(t)]. If there exists uKu\in K such that x(s)<x(u)<x(t)x(s)<x(u)<x(t), then, using the same perturbation argument as in the proof of Case 1, it follows that xx is not extreme. Therefore xx has precisely two distinct values, x(s)x(s) and x(t)x(t).

Suppose that 1<x(s)<x(t)<1-1<x(s)<x(t)<1. Then, for sufficiently small ε>0\varepsilon>0, we have x±ε1Kd=1\|{x\pm\varepsilon 1_{K}}\|_{d}=1, which implies that xx is not extreme. Therefore either x(t)=1x(t)=1 and x(s)=0x(s)=0, or x(s)=1x(s)=-1 and x(t)=0x(t)=0.

Note that if KK is connected, then the closed unit ball of C(K)C(K) for both the supremum norm and the norm d\|\cdot\|_{d} has the same extreme points — the constant functions 11 and 1-1. However, if KK has more than one connected component, then there are functions that are extreme for \|\cdot\|_{\infty} but not d\|\cdot\|_{d}, and vice versa.

4.2 The extreme points of the unit ball of ((K),0)\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}

The isometric isomorphism

(C(K),d)((K),0).\bigl{(}C(K),\|\cdot\|_{d})^{\prime}\cong\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}\,.

now enables us to identify the extreme points of the unit ball of ((K),0)\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}.

Theorem 7.

Let KK be a compact Hausdorff topological space. A regular Borel signed measure μ\mu on KK is an extreme point of the unit ball of ((K),0)\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)} if and only if it is one of the following:

  • (a)

    μ=±δt\mu=\pm\,\delta_{t}, where tKt\in K;

  • (b)

    μ=δsδt\mu=\delta_{s}-\delta_{t}, where ss, tt are distinct points in KK.

Proof.


Step 1. We show that every extreme point must be one of the types described in the statement. Let K~\tilde{K} be the space KK2K\cup K^{2}, with the sum topology, where K2K^{2} carries the product topology. For xC(K)x\in C(K), let x~\tilde{x} be the continuous function on K~\tilde{K} defined by x~(u)=x(u)\tilde{x}(u)=x(u) for uKu\in K and x~(s,t)=x(s)x(t)\tilde{x}(s,t)=x(s)-x(t) for (s,t)K2(s,t)\in K^{2}. The fact that

x~=max{sup{|x(u)|:uK},sup{|x(s,t)|:s,tK}}=xd\|\tilde{x}\|_{\infty}=\max\bigl{\{}\sup\{|x(u)|:u\in K\},\sup\{|x(s,t)|:s,t\in K\}\bigr{\}}=\|{x}\|_{d}

shows that the mapping xx~x\mapsto\tilde{x} is an isometric embedding of (C(K),d)\bigl{(}C(K),\|\cdot\|_{d}\bigr{)} into a closed subspace of (C(K~),)\bigl{(}C(\tilde{K}),\|\cdot\|_{\infty}\bigr{)}. It follows from [11, V.8.6] that every extreme point of the unit ball of (C(K),d)((K),0)\bigl{(}C(K),\|\cdot\|_{d})^{\prime}\cong\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)} is either ±δu\pm\delta_{u} for some uKu\in K, or δsδt\delta_{s}-\delta_{t} for some s,tKs,t\in K.

Step 2. ±δt\pm\,\delta_{t} are extreme points: Suppose that

δt=aμ1+bμ2,\delta_{t}=a\mu_{1}+b\mu_{2}\,,

where μ1,μ2(K)\mu_{1},\mu_{2}\in\mathcal{M}(K), μ10=μ20=1\|\mu_{1}\|_{0}=\|\mu_{2}\|_{0}=1, a,b(0,1)a,b\in(0,1) and a+b=1a+b=1. Applying δt\delta_{t} to the function 1K1_{K}, we have

aμ1(K)+bμ2(K)=1.a\mu_{1}(K)+b\mu_{2}(K)=1\,.

On the other hand, 1Kd=1\|{1_{K}}\|_{d}=1 implies that |μi(K)|1|\mu_{i}(K)|\leq 1 for i=1,2i=1,2. Therefore μ1(K)=μ2(K)=1\mu_{1}(K)=\mu_{2}(K)=1 and it follows from μi0=12(μi1+|μi(K)|)=1\|\mu_{i}\|_{0}=\frac{1}{2}\bigl{(}\|\mu_{i}\|_{1}+|\mu_{i}(K)|\bigr{)}=1 that μ11=μ21=1\|\mu_{1}\|_{1}=\|\mu_{2}\|_{1}=1. Since δt\delta_{t} is an extreme point of the unit ball of (K)\mathcal{M}(K) for the variation norm, it follows that μ1=μ2=δt\mu_{1}=\mu_{2}=\delta_{t}. Therefore δt\delta_{t} is an extreme point of the unit ball of ((K),0)\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}.

Step 3. δsδt\delta_{s}-\delta_{t} is extreme for every pair of distinct points s,tKs,t\in K: Suppose that

δsδt=aμ1+bμ2,\delta_{s}-\delta_{t}=a\mu_{1}+b\mu_{2}\,,

where μ1,μ2(K)\mu_{1},\mu_{2}\in\mathcal{M}(K), μ10=μ20=1\|\mu_{1}\|_{0}=\|\mu_{2}\|_{0}=1, a,b(0,1)a,b\in(0,1) and a+b=1a+b=1. Without loss of generality, we may assume that a=b=12a=b=\frac{1}{2}. As δsδt1=2\|\delta_{s}-\delta_{t}\|_{1}=2, we have 4μ11+μ214\leq\|\mu_{1}\|_{1}+\|\mu_{2}\|_{1}. On the other hand,

μi0=12(μi1+|μi(K)|)=1for i=1,2\|\mu_{i}\|_{0}=\frac{1}{2}\bigl{(}\|\mu_{i}\|_{1}+|\mu_{i}(K)|\bigr{)}=1\quad\text{for $i=1,2$}

and it follows that μi1=2\|\mu_{i}\|_{1}=2 and μi(K)=0\mu_{i}(K)=0 for i=1,2i=1,2. Therefore δsδt\delta_{s}-\delta_{t}, μ1\mu_{1} and μ2\mu_{2} all lie in (Cρ(K),ρ)\bigl{(}C_{\rho}(K),\|\cdot\|_{\rho}\bigr{)}^{\prime}, the space of regular Borel signed measures on KK that are zero on KK. Furthermore, these measures are all unit vectors in this space, since the variation norm is exactly twice the dual norm in (Cρ(K),ρ)\bigl{(}C_{\rho}(K),\|\cdot\|_{\rho}\bigr{)}^{\prime}. It follows from the result of Cabello-Sanchez (Theorem 4 above) that μ1=μ2=δsδt\mu_{1}=\mu_{2}=\delta_{s}-\delta_{t}. Therefore δsδt\delta_{s}-\delta_{t} is an extreme point of the unit ball of ((K),0)\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}.

We can now describe the extreme points of the unit ball of 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}) for the supremum norm. Recall that, when nn is odd, the supremum and regular norms coincide. Thus, by Propositions 3, 4 and Theorem 7 we have the following result.

Corollary 3.

Let KK be a compact, Hausdorff space.

  • (a)

    If nn is odd, then P𝒫o(nC(K))P\in\mathcal{P}_{o}(^{n}{C(K)}) is an extreme point of the closed unit ball of the space (𝒫o(nC(K)),)\bigl{(}\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) if and only if P=±δtnP=\pm\delta_{t}^{n}, for some tKt\in K, where

    δtn(x)=x(t)n.\delta_{t}^{n}(x)=x(t)^{n}\,.
  • (b)

    If nn is even, then P𝒫o(nC(K))P\in\mathcal{P}_{o}(^{n}{C(K)}) is an extreme point of the closed unit ball of the space (𝒫o(nC(K)),)\bigl{(}\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) if and only if PP is one of the following:

    • (i)

      P=±δtnP=\pm\delta_{t}^{n} for some tKt\in K, where δtn(x)=x(t)n\delta_{t}^{n}(x)=x(t)^{n};

    • (ii)

      P=δsnδtnP=\delta_{s}^{n}-\delta_{t}^{n}, where s,ts,t are distinct points in KK, and

      (δsnδtn)(x)=x(s)nx(t)n.(\delta_{s}^{n}-\delta_{t}^{n})(x)=x(s)^{n}-x(t)^{n}\,.
Example 1.

Suppose that the compact, Hausdorff space KK has just two points, α,β\alpha,\beta. Then the vector lattice C(K)C(K) can be identified with 2\mathbb{R}^{2}, where (x1,x2)2(x_{1},x_{2})\in\mathbb{R}^{2} corresponds to the function αx1\alpha\mapsto x_{1}, βx2\beta\mapsto x_{2}. The supremum norm on C(K)C(K) is identified with the supremum norm on 2\mathbb{R}^{2}. The orthogonally additive nn-homogeneous polynomials on 2\mathbb{R}^{2} have the form P(x)=a1x1n+a2x2nP(x)=a_{1}x_{1}^{n}+a_{2}x_{2}^{n}. The regular and supremum norms are

Pr\displaystyle\|P\|_{r} =|a1|+|a2|,\displaystyle=|a_{1}|+|a_{2}|\,,
P\displaystyle\|P\|_{\infty} ={|a1|+|a2|,if n is odd,max{|a1|,|a2|,|a1+a2|},if n is even.\displaystyle=\begin{cases}|a_{1}|+|a_{2}|\,,\quad\text{if $n$ is odd,}\\ \max\{|a_{1}|,|a_{2}|,|a_{1}+a_{2}|\}\,,\quad\text{if $n$ is even.}\end{cases}

The diagrams below show the unit balls for both norms.

1-1111-111x1nx_{1}^{n}x2nx_{2}^{n}x2n-x_{2}^{n}x1n-x_{1}^{n}
1-1111-111x1nx_{1}^{n}x2nx_{2}^{n}x2n-x_{2}^{n}x1n+x2n-x_{1}^{n}+x_{2}^{n}x1nx2nx_{1}^{n}-x_{2}^{n}x1n-x_{1}^{n}

Unit ball of (𝒫o(nC(K)),r)\bigl{(}\mathcal{P}_{o}(^{n}{C(K)}),\|\cdot\|_{r}\bigr{)} for any nn

and (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) for nn odd.

Unit ball of (𝒫o(nC(K)),)\bigl{(}\mathcal{P}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}\bigr{)}

for nn even.


4.3 The isometries of (C(K),d)(C(K),\|\cdot\|_{d})

We would like next to determine the isometries of the spaces 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)}), both for the regular and the supremum norms. Our results show that this reduces to the problem of finding the isometries between the spaces (K)\mathcal{M}(K) for the variation norm and the equivalent norm 0\|\cdot\|_{0}.

The Banach-Stone theorem [11, V.8.8] uses the classification of the extreme points of the space of regular Borel signed measures to determine the isometries of C(K)C(K) spaces with the supremum norm. We recall the statement of this theorem: if TT is an isometric isomorphism between C(K)C(K) and C(L)C(L), then there exists a homeomorhism φ:LK\varphi\colon L\to K and a function αC(L)\alpha\in C(L) with values ±1\pm 1, such that

(Tx)(s)=α(s)x(φ(s))\bigl{(}Tx\bigr{)}(s)=\alpha(s)x(\varphi(s)) (9)

for all xC(K)x\in C(K), sLs\in L. We shall say that an linear bijection, TT, from C(K)C(K) to C(L)C(L) is canonical if it has this form. In other words,

Tx=αxφ,Tx=\alpha\,x\circ\varphi\,,

where α\alpha, φ\varphi are as described above.

Consider the space ((K),0)(C(K),d)\bigl{(}\mathcal{M}(K),\|\cdot\|_{0}\bigr{)}\cong\bigl{(}C(K),\|\cdot\|_{d}\bigr{)}^{\prime}. By Theorem 7, the set of extreme points of the unit ball of ((K),0)({\cal M}(K),\|\cdot\|_{0}) is {±δu,δtδs:u,t,sK,ts}\{\pm\delta_{u},\delta_{t}-\delta_{s}:u,t,s\in K,t\neq s\}. The crucial step in showing that an isometry TT of from (C(K),d)(C(K),\|\cdot\|_{d}) to (C(L),d)(C(L),\|\cdot\|_{d}) is canonical is to establish that TtT^{t}, the transpose of TT, maps each δt\delta_{t} to ±δs\pm\delta_{s} for some ss in KK. This leads to the following proposition.

Proposition 5.

Let KK and LL be compact Hausdorff topological spaces and let T:(C(K),d)(C(L),d)T\colon\bigl{(}C(K),\|\cdot\|_{d}\bigr{)}\to\bigl{(}C(L),\|\cdot\|_{d}\bigr{)} be an isometric isomorphism. Let SL={tL:Tt(δt)=±δs, for some sK}S_{L}=\{t\in L:T^{t}(\delta_{t})=\pm\delta_{s},\mbox{ for some }s\in K\}. If SLS_{L} contains more than one point, then TT is canonical. Moreover, in addition, α\alpha will either take the constant value 11 or 1-1 on LL.

Proof.

Assume that |SL|2|S_{L}|\geq 2 and SLcS_{L}^{c} is non-empty. Choose rSLcr\in S_{L}^{c}. Then we have that Tt(δr)=δuδvT^{t}(\delta_{r})=\delta_{u}-\delta_{v} for some uu and vv in KK. Since |SL|2|S_{L}|\geq 2, there are tt and ss in LL with tst\neq s so that Tt(δt)=±δwT^{t}(\delta_{t})=\pm\delta_{w} and Tt(δs)=±δpT^{t}(\delta_{s})=\pm\delta_{p} for some ww and pp in KK with wpw\neq p. We now claim that {w,p}{u,v}\{w,p\}\not=\{u,v\} and so there tt in LL so that Tt(δt)=±δwT^{t}(\delta_{t})=\pm\delta_{w} with wu,vw\not=u,v. Without loss of generality suppose that w=uw=u and p=vp=v. Then we have Tt(δt)=±δuT^{t}(\delta_{t})=\pm\delta_{u} and Tt(δs)=±δvT^{t}(\delta_{s})=\pm\delta_{v}. Hence

Tt(δtδs)=±δuδvT^{t}(\delta_{t}-\delta_{s})=\pm\delta_{u}\mp\delta_{v}

and therefore (Tt)1(δuδv)=±(δtδs)(T^{t})^{-1}(\delta_{u}-\delta_{v})=\pm(\delta_{t}-\delta_{s}). Since (Tt)1(T^{t})^{-1} is a bijection we have δr=±(δtδs)\delta_{r}=\pm(\delta_{t}-\delta_{s}) which is impossible. Let tt in LL be such that Tt(δt)=±δwT^{t}(\delta_{t})=\pm\delta_{w} with wu,vw\not=u,v. Then δrδt\delta_{r}-\delta_{t} is an extreme point of the unit ball of ((L),0)({\cal M}(L),\|\cdot\|_{0}). However,

Tt(δrδt)=δuδv±δw.T^{t}(\delta_{r}-\delta_{t})=\delta_{u}-\delta_{v}\pm\delta_{w}\,.

Since wu,vw\not=u,v, δuδv±δs\delta_{u}-\delta_{v}\pm\delta_{s}, is not an extreme point of the unit ball of ((K),0)({\cal M}(K),\|\cdot\|_{0}). This is a contradiction. Hence, if |SL|2|S_{L}|\geq 2, then SL=LS_{L}=L.

Note that T:(C(K),)(C(L),)T\colon(C(K),\|\cdot\|_{\infty})\to(C(L),\|\cdot\|_{\infty}) is an isomorphism since the norms \|\cdot\|_{\infty} and d\|\cdot\|_{d} are equivalent. Further, since SL=LS_{L}=L, we have that for every tt in LL there is ss in KK such that Tt(δt)=±δsT^{t}(\delta_{t})=\pm\delta_{s}. Hence TtT^{t} maps extreme points of the unit ball of ((L),1)({\cal M}(L),\|\cdot\|_{1}) to the extreme points of the unit ball of ((K),1)({\cal M}(K),\|\cdot\|_{1}) in one to one manner. Hence Tt(B(L))B(K)T^{t}(B_{{\cal M}(L)})\subseteq B_{{\cal M}(K)} and (Tt)1(B(K))B(L)(T^{t})^{-1}(B_{{\cal M}(K)})\subseteq B_{{\cal M}(L)}. This gives us that T:(C(K),)(C(L),)T:(C(K),\|\cdot\|_{\infty})\to(C(L),\|\cdot\|_{\infty}) is an isometric isomorphism. Hence we can now apply Banach-Stone theorem to find a homeomorphism φ\varphi from LL to KK and a function αC(K)\alpha\in C(K) with α(t)=±1\alpha(t)=\pm 1 for all tKt\in K such that

T(x)=αxφ.T(x)=\alpha\,x\circ\varphi.

Now let us see that α\alpha is constant on LL. To see this suppose that

SL+={tL:Tt(δt)=δs for some sK}S_{L}^{+}=\{t\in L:T^{t}(\delta_{t})=\delta_{s}\hbox{ for some }s\in K\}

and

SL={tL:Tt(δt)=δs for some sK}S_{L}^{-}=\{t\in L:T^{t}(\delta_{t})=-\delta_{s}\hbox{ for some }s\in K\}

are both non empty. Choose tt in SL+S_{L}^{+} and rr in SLS_{L}^{-}. Suppose that Tt(δt)=δuT^{t}(\delta_{t})=\delta_{u} and that Tt(δr)=δvT^{t}(\delta_{r})=-\delta_{v}. Then δtδr\delta_{t}-\delta_{r} is an extreme point of the unit ball of ((L),0)({\cal M}(L),\|\cdot\|_{0}) yet Tt(δtδr)=δu+δvT^{t}(\delta_{t}-\delta_{r})=\delta_{u}+\delta_{v} is not extreme point of the unit ball of ((K),0)({\cal M}(K),\|\cdot\|_{0}). The result now follows and we get that

T(x)=±xφ.T(x)=\pm\,x\circ\varphi.

Let us now consider the case when |SL|=1|S_{L}|=1 and show that we can construct a non canonical isometry in this case. To help understand this result, we first consider the following example.

Example 2.

Let K={a,b}K=\{a,b\} and L={α,β}L=\{\alpha,\beta\}. We observe that we can identify both (C(K),d)(C(K),\|\cdot\|_{d}) and (C(L),d)(C(L),\|\cdot\|_{d}) with 2\mathbb{R}^{2}. Let xx in C(K)C(K) and set x1=x(a)x_{1}=x(a) and x2=x(b)x_{2}=x(b). Then (x1,x1)2(x_{1},x_{1})\in\mathbb{R}^{2} and the norm of (x1,x1)(x_{1},x_{1}) is given by

(x1,x2)d=max{|x1|,|x2|,|x1x2|}.\|{(x_{1},x_{2})}\|_{d}=\max\{|x_{1}|,|x_{2}|,|x_{1}-x_{2}|\}.

Now define T:(2,d)(2,d)T:(\mathbb{R}^{2},\|\cdot\|_{d})\to(\mathbb{R}^{2},\|\cdot\|_{d}) by

T(x1,x2)=(x1,x1x2)T(x_{1},x_{2})=(x_{1},x_{1}-x_{2})

Clearly, TT is a continuous linear bijection. We can also show that

Tt(δα)=\displaystyle T^{t}(\delta_{\alpha})= δa,\displaystyle\delta_{a},
Tt(δβ)=\displaystyle T^{t}(\delta_{\beta})= δaδb.\displaystyle\delta_{a}-\delta_{b}.

We have that

T(x1,x2)d=max{|x1|,|x1x2|,|x2|}=(x1,x2)d\|{T(x_{1},x_{2})}\|_{d}=\max\{|x_{1}|,|x_{1}-x_{2}|,|x_{2}|\}=\|{(x_{1},x_{2})}\|_{d}

and hence TT is an isometry. However, TT is not canonical since

(Tx)(α)\displaystyle(Tx)(\alpha) =x(a),\displaystyle=x(a)\,,
(Tx)(β)\displaystyle(Tx)(\beta) =x(a)x(b).\displaystyle=x(a)-x(b)\,.

Guided by Proposition 5 and the above example, we now have the following result.

Theorem 8.

Let KK and LL be compact Hausdorff topological spaces.

  1. (a)

    Suppose that KK and LL do not contain isolated points. Then every isometric isomorphism TT from (C(K),d)(C(K),\|\cdot\|_{d}) onto (C(L),d)(C(L),\|\cdot\|_{d}) has the form

    T(x)=±xφ.T(x)=\pm\,x\circ\varphi.

    for some homeomorphism φ:LK\varphi\colon L\to K.

  2. (b)

    Suppose that either KK or LL contains an isolated point. Let T:(C(K),d)(C(L),d)T:(C(K),\|\cdot\|_{d})\to(C(L),\|\cdot\|_{d}) be an isometric isomorphism. Then TT is one of the following types.

    1. (i)
      T(x)=±xφ.T(x)=\pm\,x\circ\varphi.

      for some homeomorphism φ:LK\varphi\colon L\to K.

    2. (ii)

      There exist pp in KK and tt in LL and a homeomorphism φ:L{t}K{p}\varphi\colon L\setminus\{t\}\to K\setminus\{p\} such that T=±T1T=\pm T_{1}, where

      (T1x)(t)\displaystyle(T_{1}x)(t) =x(p)\displaystyle=x(p)
      (T1x)(s)\displaystyle(T_{1}x)(s) =x(p)x(φ(s))for st.\displaystyle=x(p)-x(\varphi(s))\qquad\text{for $s\neq t$.}
Proof.

(a) Note that L=SLSLcL=S_{L}\cup S^{c}_{L}. We claim that, if |SL|=1|S_{L}|=1, then LL contains an isolated point. Suppose that SL={t}S_{L}=\{t\} and, without loss of generality, Ttδt=δsT^{t}\delta_{t}=\delta_{s}. Then

(T1K)(t)=δt(T1K)=(Ttδt)(1K)=δs(1K)=1.(T1_{K})(t)=\delta_{t}(T1_{K})=(T^{t}\delta_{t})(1_{K})=\delta_{s}(1_{K})=1\,.

For any rSLcr\in S_{L}^{c}, a similar calculation shows that (T1K)(r)=0(T1_{K})(r)=0. As SL=(T1K)1(1)S_{L}=(T1_{K})^{-1}(1) and SLc=(T1K)1(0)S_{L}^{c}=(T1_{K})^{-1}(0) and T1KT1_{K} is continuous, it follows that SLS_{L} and SLcS_{L}^{c} are disjoint closed sets. Therefore SL={t}S_{L}=\{t\} is an isolated point of LL. Therefore, if LL does not contain isolated points then |SL|2|S_{L}|\geq 2 and Proposition 5 gives us that TT is canonical.

(b) We only need to consider the case |SL|=1|S_{L}|=1 as otherwise Proposition 5 gives us that TT is canonical. Suppose LL contains an isolated point tt and KK an isolated point pp. For each xx in C(K)C(K) the function TxTx as defined in (b) is continuous and the mapping xTxx\to Tx is easily seen to be an isometry.

Let us see that if TT is not canonical then this is the form that an isometry can take. By definition and the fact that TT is invertible we have that |SK|=1|S_{K}|=1, where SKS_{K} is the set of points qq in KK for which Tt(δt)=±δqT^{t}(\delta_{t})=\pm\delta_{q} for some tLt\in L. Let SL={t}S_{L}=\{t\} and SK={p}S_{K}=\{p\}. Then Tt(δt)=±δpT^{t}(\delta_{t})=\pm\delta_{p}. Let us suppose that Tt(δt)=δpT^{t}(\delta_{t})=\delta_{p}. We claim that for each ss in SLcS_{L}^{c} we have Tt(δs)=δpδqT^{t}(\delta_{s})=\delta_{p}-\delta_{q} for some qq in K{p}K\setminus\{p\}. Otherwise we have that δtδs\delta_{t}-\delta_{s} is extreme but Tt(δtδs)=δpδu+δvT^{t}(\delta_{t}-\delta_{s})=\delta_{p}-\delta_{u}+\delta_{v} is not. The mapping Tt(δs)=δpδqT^{t}(\delta_{s})=\delta_{p}-\delta_{q} now induces a bijection φ:L{t}K{p}\varphi\colon L\setminus\{t\}\to K\setminus\{p\} so that Tt(δs)=δpδφ(s)T^{t}(\delta_{s})=\delta_{p}-\delta_{\varphi(s)}. Since the mapping L{p}((K),σ((K),𝒞(K)))L\setminus\{p\}\to({\cal M}(K),\sigma({\cal M}(K),{\cal C}(K))), sδpδφ(s)s\mapsto\delta_{p}-\delta_{\varphi(s)}, is continuous, φ\varphi will be continuous. As φ\varphi is a continuous bijection from the compact space L{t}L\setminus\{t\} to the Hausdorff space K{p}K\setminus\{p\} it is a homeomorphism. Rewriting sδpδφ(s)s\mapsto\delta_{p}-\delta_{\varphi(s)} in terms of xx, we see that (Tx)(s)=x(p)x(φ(s))(Tx)(s)=x(p)-x(\varphi(s)). When Tt(δt)=δsT^{t}(\delta_{t})=-\delta_{s}, we obtain (Tx)(s)=x(φ(s))x(p)(Tx)(s)=x(\varphi(s))-x(p). ∎

Our characterisation of the isometries of (C(L),d)(C(L),\|\cdot\|_{d}) onto (C(K),d)(C(K),\|\cdot\|_{d}) allows us to construct isometries of (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) onto (𝒫o(nC(L)),)(\mathcal{P\!}_{o}(^{n}{C(L)}),\|\cdot\|_{\infty}). Given a homeomorphism φ:KL\varphi\colon K\to L we use CφC_{\varphi} to denote the composition operator Cφ:C(L)C(K)C_{\varphi}\colon C(L)\to C(K) defined by Cφ(f)=fφC_{\varphi}(f)=f\circ\varphi for each ff in C(L)C(L). The transpose of the canonical isometry of (C(K),d)(C(K),\|\cdot\|_{d}) onto (C(L),d)(C(L),\|\cdot\|_{d}) determined by φ\varphi now gives rise to the isometry T:(𝒫o(nC(K)),)(𝒫o(nC(L),))T\colon(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty})\to(\mathcal{P\!}_{o}(^{n}{C(L),\|\cdot\|_{\infty}})) given by T(P)=PCφT(P)=P\circ C_{\varphi}.

To understand the isometries from (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) to (𝒫o(nC(L)),)(\mathcal{P\!}_{o}(^{n}{C(L)}),\|\cdot\|_{\infty}) induced by non canonical isometries of (C(L),d)(C(L),\|\cdot\|_{d}) onto (C(K),d)(C(K),\|\cdot\|_{d}) we note that if KK and LL have isolated points tt and pp respectively then we have that (C(K),)(C(K),\|\cdot\|_{\infty}) is isometrically isomorphic to (C({t}),)(C(K{t}),)(C(\{t\}),\|\cdot\|_{\infty})\oplus_{\infty}(C(K\setminus\{t\}),\|\cdot\|_{\infty}) while (C(L),)(C(L),\|\cdot\|_{\infty}) is isometrically isomorphic to (C({p}),)(C(L{p}),)(C(\{p\}),\|\cdot\|_{\infty})\oplus_{\infty}(C(L\setminus\{p\}),\|\cdot\|_{\infty}). Hence, if PP is an nn-homogeneous orthogonally additive polynomial on (C(K),)(C(K),\|\cdot\|_{\infty}\|) then we can write PP as P=λδtn+P2P=\lambda\delta_{t}^{n}+P_{2} where P2=P|C(K{t})P_{2}=P|_{C(K\setminus\{t\})}. It follows that the transpose of each non canonical isometry from (C(K),d)(C(K),\|\cdot\|_{d}) onto (C(L),d)(C(L),\|\cdot\|_{d}) gives an isometry from (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) onto (𝒫o(nC(L)),)(\mathcal{P\!}_{o}(^{n}{C(L)}),\|\cdot\|_{\infty}) of the form

T(P)=P(1)δpnP2CφT(P)=P(1)\,\delta_{p}^{n}-P_{2}\circ C_{\varphi}

where φ\varphi is a homeomorphism of K{t}K\setminus\{t\} to L{p}L\setminus\{p\}.

In a similar manner, we can construct canonical isometries from (𝒫o(nC(K)),r)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{r}) onto (𝒫o(nC(L)),r)(\mathcal{P\!}_{o}(^{n}{C(L)}),\|\cdot\|_{r}).

5 Exposed points in 𝒫o(nC(K))\mathcal{P}_{o}(^{n}{C(K)})

In this section we shall characterise the weak exposed and weak strongly exposed point of the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime}. We have an upper bound for this set. We know that it is contained in the set of extreme points of the unit ball of (C(K),d)((K),0)(C(K),\|\cdot\|_{d})^{\prime}\cong({\cal M}(K),\|\cdot\|_{0}) and that the set of extreme points of this set is equal to {±δp,δtδs:p,t,sK,ts}\{\pm\delta_{p},\delta_{t}-\delta_{s}:p,t,s\in K,t\neq s\}.

Let us begin with some definitions.

Definition 1.

Let EE be a Banach space. A point xx in the closed unit ball of EE is said to be an exposed point if there exists φE\varphi\in E^{\prime} with φ=1\|\varphi\|=1 such that

φ(x)=1 and φ(y)<1 for yB¯E\{x}.\varphi(x)=1\text{ and }\varphi(y)<1\text{ for }y\in\overline{B}_{E}\backslash\{x\}.

If this is the case then we say that φ\varphi exposes xx.

Definition 2.

We say that xx is a strongly exposed point of the closed unit ball of EE if there exists φE\varphi\in E^{\prime} such that

φ(x)=1\varphi(x)=1

and whenever (xn)n(x_{n})_{n} is a sequence in B¯E\overline{B}_{E} with limnφ(xn)=1\lim_{n\rightarrow\infty}\varphi(x_{n})=1 then (xn)n(x_{n})_{n} converges to xx in norm. We will say that φ\varphi strongly exposes xx.

If E=FE=F^{\prime} is a dual Banach space and the point xEx\in E is exposed (respectively, strongly exposed) by φ\varphi in FF we say that xx is a weak exposed (respectively, weak strongly exposed) point of EE and that φ\varphi weak exposes (respectively, weak-strongly exposes) the unit ball of EE at xx.

We also observe that if each δt\delta_{t}, tKt\in K and each δtδs\delta_{t}-\delta_{s}, t,sKt,s\in K with tst\not=s are of norm 11 in (C(K),d)(C(K),\|\cdot\|_{d})^{\prime}. Hence, if δt\delta_{t} is exposed by xx then we must have diam(x)<1{\rm diam}(x)<1. Conversely, if δtδs\delta_{t}-\delta_{s} is exposed by xx then we must have x<1\|x\|_{\infty}<1.

We note that if KK is a compact Hausdorff topological space then a net (tα)α(t_{\alpha})_{\alpha} converges to tt in KK if and only if y(tα)y(t_{\alpha}) converges to y(t)y(t) for every yy in C(K)C(K).

5.1 Gâteaux differentiability of the norm

We start with a characterisation of Gâteaux differentiability of the norm on (C(K),d)(C(K),\|\cdot\|_{d}).

Theorem 9.

Let KK be a compact Hausdorff topological space. Let tKt\in K, xC(K)x\in C(K) with xd=1\|{x}\|_{d}=1. Then the following are equivalent

  1. (a)

    The norm of (C(K),d)(C(K),\|\cdot\|_{d}) is Gâteaux differentiable at xx with derivative δt\delta_{t}.

  2. (b)
    1. (i)

      xd=x(t)=1\|{x}\|_{d}=x(t)=1 and diam(x)<1{\rm diam}(x)<1.

    2. (ii)

      If (tn)n(t_{n})_{n} is a sequence of points in KK such that limnx(tn)=1\lim_{n\to\infty}x(t_{n})=1 then (tn)n(t_{n})_{n} has a subnet, (tα)α(t_{\alpha})_{\alpha} such that (tα)α(t_{\alpha})_{\alpha} converges to tt.

  3. (c)

    tt is the unique point in KK with x(t)=1x(t)=1 and diam(x)<1{\rm diam}(x)<1.

Proof.

First observe that Šmul’yan [28, 29] (see also [9]) showed that a point xx in BC(K)B_{C(K)} weak exposes the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δt\delta_{t} if and only if the norm of C(K)C(K) is Gâteaux differentiable at xx with derivative δt\delta_{t}. Hence we have that (a) implies (c).

Let us see that (c) implies (b). Clearly we have that (c) implies (b) (i).

Suppose that (c) is true and that (b) part (ii) fails. Then there is a sequence (tn)n(t_{n})_{n} in KK with limnx(tn)=x(t)=1\lim_{n\to\infty}x(t_{n})=x(t)=1 but that for all subnets (tα)α(t_{\alpha})_{\alpha} of (tn)n(t_{n})_{n} there is yy in C(K)C(K) such that y(tα)↛y(t)y(t_{\alpha})\not\to y(t). As KK is compact, we can choose a subnet (tα)α(t_{\alpha})_{\alpha} of (tn)n(t_{n})_{n} and ss in KK so that limαtα=s\lim_{\alpha\to\infty}t_{\alpha}=s. We claim that tst\not=s. Suppose t=st=s. Then for every yy in C(K)C(K) we have that limβy(tβ)=y(t)\lim_{\beta}y(t_{\beta})=y(t) contrary to what we have assumed. As sts\not=t and xx is continuous we have x(s)=limβx(tβ)=1x(s)=\lim_{\beta}x(t_{\beta})=1 which contradicts (c). Hence, we see that (c) implies (b).

Next suppose that (b) is true and that (a) is false. Then we can find yy in C(K)C(K), ε>0\varepsilon>0 and a sequence of positive numbers (λn)n(\lambda_{n})_{n} converging to 0 so that

|x+λnydxdλny(t)|ελn\bigl{|}\|{x+\lambda_{n}y}\|_{d}-\|{x}\|_{d}-\lambda_{n}y(t)\bigr{|}\geq\varepsilon\lambda_{n}

for every positive integer nn.

Note that as x+λnyd(x+λny)(t)\|{x+\lambda_{n}y}\|_{d}\geq(x+\lambda_{n}y)(t) we actually have that x+λnydxdλny(t)\|{x+\lambda_{n}y}\|_{d}-\|{x}\|_{d}-\lambda_{n}y(t) is non-negative and therefore we have

x+λnydxdλny(t)λnε\|{x+\lambda_{n}y}\|_{d}-\|{x}\|_{d}-\lambda_{n}y(t)\geq\lambda_{n}\varepsilon

for every positive integer nn.

Each of the functions x+λnyx+\lambda_{n}y attains its norm either at a point of the form δt\delta_{t} or at a point δuδv\delta_{u}-\delta_{v}. As diam(x)<1{\rm diam}(x)<1 choosing nn sufficiently large we can assume that x+λnyx+\lambda_{n}y attains its norm at a point of the first type. Hence, for each nn in \mathbb{N}, we can find tnt_{n} in KK and βn=±1\beta_{n}=\pm 1 so that

βn(x+λny)(tn)=x+λnyd.\beta_{n}(x+\lambda_{n}y)(t_{n})=\|{x+\lambda_{n}y}\|_{d}\,.

Then we have

1\displaystyle 1 =xdβnx(tn)=βn(x+λny)(tn)βnλny(tn)\displaystyle=\|{x}\|_{d}\geq\beta_{n}x(t_{n})=\beta_{n}(x+\lambda_{n}y)(t_{n})-\beta_{n}\lambda_{n}y(t_{n})
x+λnyd|λn|yd.\displaystyle\geq\|{x+\lambda_{n}y}\|_{d}-|\lambda_{n}|\|{y}\|_{d}.

As (λn)n(\lambda_{n})_{n} is a null sequence we have that x+λnyd|λn|yd\|{x+\lambda_{n}y}\|_{d}-|\lambda_{n}|\|{y}\|_{d} converges to xd\|{x}\|_{d} as nn tends to \infty. Hence we have that βnx(tn)1\beta_{n}x(t_{n})\to 1. However, as diam(x)<1{\rm diam}(x)<1, we have x(tn)>0x(t_{n})>0 for all nn. Hence, without loss of generality, we may assume that βn=1\beta_{n}=1 for all nn and therefore we have limnx(tn)=1\lim_{n\to\infty}x(t_{n})=1.

Then we have that

ελn\displaystyle\varepsilon\lambda_{n}\leq x+λnydxdλny(t)\displaystyle\|{x+\lambda_{n}y}\|_{d}-\|{x}\|_{d}-\lambda_{n}y(t)
=\displaystyle= (x+λny)(tn)x(t)λny(t)\displaystyle(x+\lambda_{n}y)(t_{n})-x(t)-\lambda_{n}y(t)
=\displaystyle= x(tn)x(t)+λn(y(tn)y(t))\displaystyle x(t_{n})-x(t)+\lambda_{n}\left(y(t_{n})-y(t)\right)
\displaystyle\leq λn(y(tn)y(t))\displaystyle\lambda_{n}\left(y(t_{n})-y(t)\right)

However this means that there is no subnet (tα)(t_{\alpha}) of (tn)n(t_{n})_{n} so that y(tα)y(t_{\alpha}) converges to y(t)y(t) and so (b) (ii) is false. ∎

We recall that a function xC(K)x\in C(K) is said to peak at a point tKt\in K if tt is the unique point at which xx attains its maximum.

Lemma 1.

Let KK be a compact Hausdorff topological space and tKt\in K. Then there is xx in C(K)C(K) which peaks at tt if and only if {t}\{t\} is a GδG_{\delta} subset of KK.

Proof.

We first suppose that {t}\{t\} is a GδG_{\delta} subset of KK. Then we can find a sequence of open sets (Un)n(U_{n})_{n} so that {t}=n=1Un\{t\}=\bigcap_{n=1}^{\infty}U_{n}. As KK is compact and Hausdorff it is completely regular. Hence, for each nn\in\mathbb{N} we can find a continuous function xn:K[0,1]x_{n}\colon K\to[0,1] such that xn(t)=1x_{n}(t)=1 and xn(Unc)=0x_{n}(U_{n}^{c})=0. Now let x:K[0,1]x\colon K\to[0,1] be defined by

x(t)=6π2n=11n2xn(t).x(t)=\frac{6}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}x_{n}(t).

Then we have x(t)=1x(t)=1 and x(s)<1x(s)<1 for sKs\in K, sts\not=t. So xx peaks at tt.

Conversely, if there is xx in C(K)C(K) which peaks at tt, for each nn\in\mathbb{N} let Un={sK:x(s)>11nU_{n}=\{s\in K:x(s)>1-\frac{1}{n}. Then {t}=n=1Un\{t\}=\bigcap_{n=1}^{\infty}U_{n}. As each UnU_{n} is open, {t}\{t\} is a GδG_{\delta} set.∎

The weak exposed points of the ball of the form δt\delta_{t} are characterised by the following proposition.

Proposition 6.

Let KK be a compact Hausdorff topological space. Then {±δt:tK}\{\pm\delta_{t}:t\in K\} is contained in the set of weak exposed points of the unit ball of ((K),0)({\cal M}(K),\|\cdot\|_{0}) if and only if KK is first countable.

Just as we have characterised the weak exposed points of the ball of the form δt\delta_{t} we now characterise weak exposed points of the form δtδs\delta_{t}-\delta_{s}. Replacing δt\delta_{t} with δtδs\delta_{t}-\delta_{s} in Theorem 9 we obtain the following result.

Theorem 10.

Let KK be a compact Hausdorff topological space. Let t,sKt,s\in K, xC(K)x\in C(K) with xd=1\|{x}\|_{d}=1. Then the following are equivalent

  1. (a)

    The norm of (C(K),d)(C(K),\|\cdot\|_{d}) is Gâteaux differentiable at xx with derivative δtδs\delta_{t}-\delta_{s}.

  2. (b)
    1. (i)

      xd=x(t)x(s)=1\|{x}\|_{d}=x(t)-x(s)=1 and x<1\|x\|_{\infty}<1.

    2. (ii)

      If (tn)n(t_{n})_{n} and (sn)n(s_{n})_{n} are sequences of points in KK such that limnx(tn)x(sn)=1\lim_{n\to\infty}x(t_{n})-x(s_{n})=1 then (tn)n(t_{n})_{n} and (sn)(s_{n}) have subnets (tα)α(t_{\alpha})_{\alpha} and (sα)α(s_{\alpha})_{\alpha} which converge to tt and ss respectively.

  3. (c)

    t,st,s is the unique pair of points in KK with x(t)x(s)=1x(t)-x(s)=1 and x<1\|x\|_{\infty}<1.

As the proof of the following lemma is similar to that of Lemma 1 we omit it.

Lemma 2.

Let KK be a compact Hausdorff topological space and t,sKt,s\in K with tst\not=s. Then there is xx in C(K)C(K) such that x(t)=12x(t)=\frac{1}{2}, x(s)=12x(s)=-\frac{1}{2} and 12<x(u)<12-\frac{1}{2}<x(u)<\frac{1}{2} for uK{t,s}u\in K\setminus\{t,s\} if and only if {t}\{t\} and {s}\{s\} are GδG_{\delta} sets.

The weak exposed points of the ball of the form δtδs\delta_{t}-\delta_{s} are now characterised by the following proposition.

Proposition 7.

Let KK be a compact Hausdorff topological space and nn be an even integer. Then {δtδs:t,sK,ts}\{\delta_{t}-\delta_{s}:t,s\in K,t\neq s\} is contained in the set of weak exposed points of the unit ball of ((K),0)({\cal M}(K),\|\cdot\|_{0}) if and only if KK is first countable.

Propositions 6 and 7 can be rephrased in terms of spaces of orthogonally additive polynomials. Since we have canonically identified the space 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) with the space (K)\mathcal{M}(K), we may transfer the weak topology on (K)=C(K)\mathcal{M}(K)=C(K)^{\prime} to the space 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}). References to the weak topology on 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) should be understood in this sense. It is easy to see that this is the topology of pointwise convergence on 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}).

Proposition 8.

Let KK be a compact Hausdorff topological space and nn be an even integer. Then {±δpn,δtnδsn:p,t,sK,ts}\{\pm\delta_{p}^{n},\delta_{t}^{n}-\delta_{s}^{n}:p,t,s\in K,t\neq s\} is equal to the set of weak exposed points of the unit ball of (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) if and only if KK is first countable.

5.2 Fréchet differentibility of the norm

We now characterise Fréchet differentiability of the norm on (C(K),d)(C(K),\|\cdot\|_{d}).

Theorem 11.

Let KK be a compact Hausdorff topological space. Let tKt\in K, xC(K)x\in C(K) with xd=1\|{x}\|_{d}=1. Then the following are equivalent.

  1. (a)

    The norm of (C(K),d)(C(K),\|\cdot\|_{d}) is Fréchet differentiable at xx with derivative δt\delta_{t}.

  2. (b)
    1. (i)

      xd=x(t)=1\|{x}\|_{d}=x(t)=1 and diam(x)<1{\rm diam}(x)<1.

    2. (ii)

      If (tn)n(t_{n})_{n} is a sequence of points in KK such that limnx(tn)=1\lim_{n\to\infty}x(t_{n})=1 then (tn)n(t_{n})_{n} is eventually equal to tt.

  3. (c)

    xx weak strongly exposes the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δt\delta_{t}.

Proof.

First observe that Šmul’yan [28, 29] (see also [9]) showed that a point xx in BC(K)B_{C(K)} weak strongly exposes the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δt\delta_{t} if and only if the norm of C(K)C(K) is Fréchet differentiable at xx with derivative δt\delta_{t}. Thus (a) and (c) are equivalent.

If the norm of (C(K),d)(C(K),\|\cdot\|_{d}) is Fréchet differentiable at xx with derivative δt\delta_{t} then it is Gâteaux differentiable at xx with derivative δt\delta_{t}. Theorem 9 now implies that (b) (i) holds.

Suppose that (c) is true. Then xx in BC(K)B_{C(K)} weak strongly exposes the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δt\delta_{t}. If limnx(tn)=1\lim_{n\to\infty}x(t_{n})=1 then limnδtn(x)=δt(x)=1\lim_{n\to\infty}\delta_{t_{n}}(x)=\delta_{t}(x)=1. As ff weak-strongly exposes the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δt\delta_{t} we have that limnδtn=δt\lim_{n\to\infty}\delta_{t_{n}}=\delta_{t} in norm. However, as δuδv0=1\|\delta_{u}-\delta_{v}\|_{0}=1 whenever uvu\not=v we see that only way we can have (δtn)n(\delta_{t_{n}})_{n} converge to δt\delta_{t} is that the sequence (tn)n(t_{n})_{n} is eventually equal to tt.

The implication (b) implies (a) is similar to the corresponding part of the proof of Theorem 9 where instead of using the fact that (tn)n(t_{n})_{n} has a subsequence that converges to tt we use the fact that (tn)n(t_{n})_{n} has a subsequence so that it is eventually equal to tt.∎

Corollary 4.

Let KK be a compact Hausdorff topological space and tKt\in K. Then δt\delta_{t} is a weak strongly exposed point of the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} if and only if tt is an isolated point of KK.

Proof.

If tt is an isolated point of KK then the function given by

x(s)={1,s=t1/2otherwisex(s)=\begin{cases}1,&s=t\\ 1/2&\hbox{otherwise}\\ \end{cases}

is continuous on KK. Moreover, if x(tn)1x(t_{n})\to 1 then (tn)n(t_{n})_{n} is eventually equal to tt.

Conversely, if tt is not an isolated point of KK. Choose a sequence of points (tn)n(t_{n})_{n} with tntt_{n}\not=t, all nn, so that tnt_{n} converges to tt. Let xx be any function in C(K)C(K) with xd=1\|{x}\|_{d}=1 and x(t)=1x(t)=1. Then we have that x(tn)x(t)=1x(t_{n})\to x(t)=1. However, as (tn)n(t_{n})_{n} is not eventually equal to tt we see that condition (b) (ii) of Theorem 11 is not satisfied and therefore no xx in C(K)C(K) with xd=1\|{x}\|_{d}=1 can expose the unit ball (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δt\delta_{t}.∎

Theorem 12.

Let KK be a compact Hausdorff topological space. Let t,sKt,s\in K, xC(K)x\in C(K) with xd=1\|{x}\|_{d}=1. Then the following are equivalent

  1. (a)

    The norm of (C(K),d)(C(K),\|\cdot\|_{d}) is Fréchet differentiable to xx with differential δtδs\delta_{t}-\delta_{s}.

  2. (b)
    1. (i)

      xd=x(t)x(s)=1\|{x}\|_{d}=x(t)-x(s)=1 and x<1\|x\|_{\infty}<1.

    2. (ii)

      If (tn)n(t_{n})_{n} and (sn)n(s_{n})_{n} are sequences of points in KK such that limnx(tn)x(sn)=1\lim_{n\to\infty}x(t_{n})-x(s_{n})=1 then (tn)n(t_{n})_{n} and (sn)n(s_{n})_{n} are eventually the constant sequences tt and ss respectively.

  3. (c)

    xx weak strongly exposes the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} at δtδs\delta_{t}-\delta_{s}.

Proof.

The proof is similar to Theorem 11 and therefore omitted.∎

Corollary 5.

Let KK be a compact Hausdorff topological space and t,sKt,s\in K with tst\not=s. Then δtδs\delta_{t}-\delta_{s} is a weak strongly exposed point of the unit ball of (C(K),d)(C(K),\|\cdot\|_{d})^{\prime} if and only if tt and ss are isolated points of KK.

We can rephrase these results in terms of spaces of orthogonally additive polynomials as follows.

Proposition 9.

Let KK be a compact Hausdorff topological space, let nn be an even integer and let s,ts,t be distinct points in KK.

  • (a)

    δtn\delta_{t}^{n} is a weak strongly exposed point of the unit ball of (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) if and only if tt is an isolated point of KK.

  • (b)

    δtnδsn\delta_{t}^{n}-\delta_{s}^{n} (sts\neq t) is a weak strongly exposed point of the unit ball of (𝒫o(nC(K)),)(\mathcal{P\!}_{o}(^{n}{C(K)}),\|\cdot\|_{\infty}) if and only if tt and ss are isolated points of KK.

In particular, we see that if KK has no isolated points, then the unit ball of 𝒫o(nC(K))\mathcal{P\!}_{o}(^{n}{C(K)}) does not contain any weak strongly exposed points.


Acknowledgements

We thank Dirk Werner and Tony Wickstead for helpful discussions.

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Christopher Boyd, School of Mathematics & Statistics, University College Dublin, Belfield, Dublin 4, Ireland.
e-mail: Christopher.Boyd@ucd.ie

Raymond A. Ryan, School of Mathematics, Statistics and Applied Mathematics, National University of Ireland Galway, Ireland.
e-mail: ray.ryan@nuigalway.ie

Nina Snigireva, School of Mathematics, Statistics and Applied Mathematics, National University of Ireland Galway, Ireland.
e-mail: nina.snigireva@nuigalway.ie