Centre of a compact convex set

Abstract

We introduce the notion centre of a convex set and study the space of continuous affine functions on a compact convex set with a centre. We show that these spaces are precisely the dual of a base normed space in which the underlying base has a (unique) centre. We also characterize the corresponding base norm space. We obtain a condition on a compact, balanced, convex subset of a locally convex space, so that the corresponding space of continuous affine functions on the convex set is an absolute order unit space. Similarly, we characterize a condition on the base with a centre of a base normed space, so that the latter becomes an absolutely base normed space.

1 Introduction

Let K be a compact, convex subset of a locally convex space X and let $A_{{\mathbb{R}}}(K)$ denote the space of real valued, continuous and affine functions defined on K. Then, $A_{{\mathbb{R}}}(K)$ is a complete order unit space and the state space of $A_{{\mathbb{R}}}(K)$ is affinely homeomorphic to K. Conversely, if (V, e) is a complete order unit space with its state space S(V),  then V is unitally order isomorphic to $A_{{\mathbb{R}}}(S(V)).$ The self-adjoint part of an operator system (that is, a self-adjoint subspace containing identity) in a unital $\text{C}^*$-algebra A is an order unit space. Therefore, the above said discussion describes Kadison’s functional representation theorem [4].

Furthermore, if K and L are compact convex sets in suitable locally convex spaces, then K is affinely homeomorphic to L if and only if $A_{{\mathbb{R}}}(K)$ is unitally order isomorphic to $A_{{\mathbb{R}}}(L).$ In other words, there exists a bijective correspondence between the class of compact convex sets of locally convex spaces and the class of complete order unit spaces (see, for example, [1, Theorem II.1.8]).

In this paper, we discuss an intrinsic characterization of compact convex sets which are affine homeomorphic to balanced, convex, compact sets. We also discuss the corresponding subclass of complete order unit spaces. We introduce the notion of a centre of a convex set and prove that a compact convex set K in a locally convex space X is affine homeomorphic to a balanced, convex, compact set L in another locally convex space Y if and only if K has a (unique) centre.

Let $(V^{(\cdot )}, V^{(\cdot )+}, e)$ be the order unit space obtained by adjoining an order unit to a normed linear space V using a construction due to M. M. Day (see Theorem 2.1). Then, the corresponding cone yields a natural base $B^{(\cdot )},$ such that $(V^{(\cdot )}, V^{(\cdot )+}, B^{(\cdot)})$ is a base normed space Proposition 2.3. We prove that $B^{(\cdot )}$ is precisely a base with a centre (Theorem 4.5).

We also study the properties A(B),  the space of continuous affine functions on a compact convex set B with a centre $b_0$ in a locally convex space X. We prove that A(B) is unitally order isomorphic to $(V^{(\cdot )}, e)$ for some Banach space V (Theorem 3.1). We also discuss the order unit spaces whose state space have a centre (Theorem 5.6).

In [7], the author discussed an order theoretic generalization of spin factors obtained by adjoining an order unit to a normed linear space. Let V be a normed linear space and let $(V^{(\cdot )}, e)$ is the order unit space obtained by adjoining an order unit to V. He proved that there is a canonical absolute value defined in $V^{(\cdot )}$ and that $(V^{(\cdot )}, e)$ becomes an absolute order unit space if and only if V is strictly convex [7, Theorem 2.14]. In the present paper, we discuss an absolute value in a base normed space in which the base has a centre and find a condition under which the space becomes an absolutely base normed space (Theorem 6.7).

This paper is a part of the ongoing program to study the various aspects of absolutely ordered spaces initiated by the author [5,6,7]. In [7] we discussed a geometric aspect of an absolute order unit space. In the present paper, we have discussed a geometric aspect of an absolutely base normed space.

A summary of the paper is as follows.

In Sect. 2, we recall the adjoining of an order unit to a normed linear space and the notion of lead points of a convex set and introduce the notion of a centre of a convex set. We study some related properties.

In Sect. 3, we consider the space of continuous affine functions on a compact convex set with a centre and describe its order and norm structures. We prove that this space can be characterized as one obtained by adjoining an order unit to a Banach space.

In Sect. 4,we study the norm and order structure of a base norm space in which the base has a (unique) centre. We produce an example to show that the base of every base normed space need not have a centre.

In Sect. 5, we consider the order unit spaces having a central state space. We obtain a characterization for a state to be a centre of the state space. We study the norm and order structure an order unit space having a central state and prove that such a space is precisely obtained by adjoining an order unit to a normed linear space. Let us recall that such a space becomes an absolute order unit space if the underlying normed space is strictly convex [7, Theorem 3.14]. We introduce the Property (S) on a compact, balanced, convex subset of a real locally convex space and prove that an order unit space having a central state is an absolute order unit space if and only if the corresponding state space satisfies Property (S).

In Sect. 6, we describe an absolute value on a base normed space in which the base has a centre. This absolute value arises naturally. We characterize the condition on the underlying base under which the space becomes an absolutely ordered base normed space.

2 Centre of a convex set

2.1 Adjoining an order unit to a normed linear space

The following construction has been adopted from [3, 1.6.1] and is apparently due to M. M. Day.

Theorem 2.1

(M. M. Day) Let $(V, \Vert \cdot \Vert )$ be a real normed linear space. Consider $V^{(\cdot )} \,{:}{=}\, V \times {\mathbb{R}}$ and define

$$\begin{aligned} V^{(\cdot )+} \,{:}{=}\, \lbrace (v, \alpha ): \Vert v \Vert \le \alpha \rbrace . \end{aligned}$$

Then $(V^{(\cdot )}, V^{(\cdot )+})$ becomes a real ordered space such that $V^{(\cdot )+}$ is proper, generating and Archimedean. In addition, $e = (0, 1) \in V^{(\cdot )+}$ is an order unit for $V^{(\cdot )}$ so that $(V^{(\cdot )}, e)$ becomes an order unit space. The corresponding order unit norm is given by

$$\begin{aligned} \Vert (v, \alpha ) \Vert _e = \Vert v \Vert + \vert \alpha \vert \end{aligned}$$

for all $(v, \alpha ) \in V^{(\cdot )}.$ In particular, $(V^{(\cdot )}, e)$ is isometrically isomorphic to $V \oplus _1 {\mathbb{R}}.$ Thus, V,  identified with $\lbrace (v, 0): v \in V \rbrace ,$ can be identified as a closed subspace of $V^{(\cdot )}.$ Furthermore, $V^{(\cdot )}$ is complete if and only if so is V.

The following result can proved in a routine way.

Proposition 2.2

Let V be a real normed linear space and consider the corresponding order unit space $(V^{(\cdot )}, e).$ Then, the dual of $(V^{(\cdot )}, e)$ is a base normed space isometrically isomorphic to $(V^{*(\cdot )}, S(V^{(\cdot )})),$ where

$$\begin{aligned} S(V^{(\cdot )}) \,{:}{=}\, \lbrace (f, 1): f \in V^*\ \text{and}\ \Vert f \Vert \le 1 \rbrace \end{aligned}$$

is the state space of $(V^{(\cdot )}, e)$ as well as the base of $V^{*(\cdot )+}$ which determine the dual norm as the base norm on $V^{*(\cdot )}.$

We show that a similar construction is also available on $V^{(\cdot )}$ itself albeit with a different norm (and the same order structure).

Proposition 2.3

Let V be a real normed linear space and consider the corresponding ordered vector space $(V^{(\cdot )}, V^{(\cdot )+}).$ Put

$$\begin{aligned} B^{(\cdot )} = \lbrace (v, 1): \Vert v \Vert \le 1 \rbrace . \end{aligned}$$

Then $B^{(\cdot )}$ is a base for $V^{(\cdot )+},$ such that $(V^{(\cdot )}, V^{(\cdot )+}, B^{(\cdot )})$ is a base normed space (which is denoted by $V^{(1)}).$ The corresponding base norm is given by

$$\begin{aligned} \Vert (v, \alpha ) \Vert _B = \max \lbrace \Vert v \Vert , \vert \alpha \vert \rbrace \end{aligned}$$

for all $(v, \alpha ) \in V^{(\cdot )}.$ In particular, $(V^{(\cdot )}, B^{(\cdot )})$ is isometrically isomorphic to $V \oplus _{\infty } {\mathbb{R}}.$ Thus V,  identified with $\lbrace (v, 0): v \in V \rbrace ,$ can be identified as a closed subspace of $V^{(\cdot )}.$ Furthermore, $V^{(\cdot )}$ is complete if and only if so is V.

Proof

It suffices to prove that $B^{(\cdot )}$ is a base for $V^{(\cdot )+},$ such that

$$\begin{aligned} {{\,\mathrm{co}\,}}\left( B^{(\cdot )} \cup - B^{(\cdot )}\right) = \lbrace (x, \alpha ): \Vert x \Vert \le 1, \vert \alpha \vert \le 1 \rbrace . \end{aligned}$$

Note that $B^{(\cdot )}$ is convex. If $(x, \alpha ) \in V^{(\cdot )+},$ then $\Vert x \Vert \le \alpha .$ Therefore, if $(x, \alpha ) \ne (0, 0),$ then $\alpha > 0.$ Now, letting $x_0 = \alpha ^{-1} x,$ we get that $(x_0, 1) \in B$ and $(x, \alpha ) = \alpha (x_0, 1)$ is a unique representation. We show that

$$\begin{aligned} {{\,\mathrm{co}\,}}\left( B^{(\cdot )} \cup - B^{(\cdot )}\right) = \lbrace (x, \alpha ): \Vert x \Vert \le 1, \vert \alpha \vert \le 1 \rbrace . \end{aligned}$$

Let $(x, \alpha ) \in {{\,\mathrm{co}\,}}\left( B^{(\cdot )} \cup - B^{(\cdot )}\right) ,$ say $(x, \alpha ) = \lambda (x_1, 1) - (1 - \lambda ) (x_2, 1)$ for some $x_1, x_2 \in V$ with $\Vert x_1 \Vert \le 1$ and $\Vert x_2 \Vert \le 1$ and $\lambda \in [0, 1].$ Then, $x = \lambda x_1 - (1 - \lambda ) x_2$ and $\alpha = \lambda - (1 - \lambda ) = 2 \lambda - 1.$ Thus, $\Vert x \Vert \le 1$ and $\vert \alpha \vert \le 1.$ Conversely, we assume that $x \in V$ and $\alpha \in {\mathbb{R}}$ with $\alpha \ne 0$ be, such that $\Vert x \Vert \le 1$ and $\vert \alpha \vert \le 1.$ If $\Vert x \Vert \le \vert \alpha \vert ,$ then $(\alpha ^{-1}x, 1) \in B^{(\cdot )}$ so that $(x, \alpha ) = \alpha (\alpha ^{-1} x, 1) \in {{\,\mathrm{co}\,}}\left( B^{(\cdot )} \cup - B^{(\cdot )}\right) .$ Therefore, we assume that $\vert \alpha \vert < \Vert x \Vert .$ Put $y = \Vert x \Vert ^{-1} x.$ Then, $(y, 1), (- y, 1) \in B^{(\cdot )}.$ Let $2 \lambda = \Vert x \Vert + \alpha$ and $2 \mu = \Vert x \Vert - \alpha .$ Then, $\lambda , \mu \in [0, 1]$ with $\lambda + \mu = \Vert x \Vert$ and $\lambda - \mu = \alpha .$ Thus

$$\begin{aligned} \lambda (y, 1) - \mu (- y, 1) = ((\lambda + \mu ) y, \lambda - \mu ) = (x, \alpha ) \end{aligned}$$

so that $(x, \alpha ) \in {{\,\mathrm{co}\,}}\left( B^{(\cdot )} \cup - B^{(\cdot )}\right) .$ $\square$

In this section, we shall obtain a geometric description of $B^{(\cdot )}.$ First, we recall the following notion introduced in [2].

2.2 Lead points of a convex set

Definition 2.4

[2] Let C be a convex subset of a real vector space X with $0 \in C.$ An element $x \in C$ is called a lead point of C,  if for any $y \in C$ and $\lambda \in [0, 1]$ with $x = \lambda y,$ we have $\lambda = 1$ and $y = x.$ The set of all lead points of C is denoted by $\text{Lead}(C).$

The following result was essentially proved in [2, Proposition 3.2].

Lemma 2.5

Let C be a non-empty, radially compact, convex subset of a real vector space X containing 0. Then, for each $x \in C$ with $x \ne 0,$ there exist a unique $x_1 \in \text{Lead}(C)$ and a unique $\lambda \in (0, 1],$ such that $x = \lambda x_1.$ (A non-empty subset S of X is called radially compact, if for any $x \in X$ with $x \ne 0,$ the set $\lbrace \alpha \in {\mathbb{R}}: \alpha x \in S \rbrace$ is compact in ${\mathbb{R}}.$)

We extend this result to a simple but interesting characterization of a norm on a real vector space.

Theorem 2.6

Let $C_0$ be a non-empty, radially compact, absolutely convex subset of a real vector space X. Let $X_0$ be the linear span of $C_0$ so that $X_0 = \cup _{n=1}^{\infty } n C_0.$

• 1. Then for each non-zero $x \in X_0,$ there exists a unique $c \in \text{Lead}(C_0)$ and a unique $\alpha > 0,$ such that $x = \alpha c.$

• 2. Let us write $r(x) \,{:}{=}\,\alpha$ and put $r(0) = 0.$ Then, $r: X_0 \rightarrow {\mathbb{R}}^+$ determines a norm on $X_0,$ such that $C_0$ is the closed unit ball.

Proof

(1): Fix $x \in X_0,$ $x \ne 0.$ As $C_0$ is absorbing in $X_0,$ there exists a non-zero $k \in {\mathbb{R}},$ such that $k x \in C_0.$ Since $C_0$ is absolutely convex, we may take $k > 0.$ As $C_0$ is radially compact, the set $S = \lbrace \lambda \in {\mathbb{R}}: \lambda x \in C_0 \rbrace$ is a compact set in ${\mathbb{R}}.$ In addition, $\sup S = \alpha _0 > 0$ with $\alpha _0 x \in C_0.$ Put $\alpha _0 x = c.$ We show that $c \in \text{Lead}(C_0).$ By Lemma 2.5, we have $c = \lambda c_1$ for some $c_1 \in \text{Lead}(C_0)$ and $\lambda \in (0, 1],$ so that $\alpha _0 \lambda ^{-1} \in S.$ As $\alpha _0 = \sup S,$ we get that $\lambda = 1$ and consequently, $c \in \text{Lead}(C_0).$ Next, let $x = \beta d$ for some $d \in \text{Lead}(C_0)$ and $\beta > 0.$ Then, $\beta ^{-1} \le \alpha _0.$ In addition, we have $c = \alpha _0 \beta d.$ Thus by the definition of $\text{Lead}(C_0),$ we get $c = d$ and $\beta = \alpha _0^{-1}.$

(2) We note that $r(x) \le 1$ for all $x \in C_0$ and that $r(x) = 1$ if and only if $x \in \text{Lead}(C_0).$ In addition, for $x \in X_0,$ $x \ne 0$ we have $r(x) > 0.$ Let $x \in X_0$ and $\alpha \in {\mathbb{R}}.$ Without any loss of generality, we assume that $x \ne 0$ and $\alpha \ne 0.$ Let $x = \lambda c$ be the unique representation with $c \in \text{Lead}(C_0).$ Then, $r(x) = \lambda .$ If $\alpha > 0,$ then $\alpha x = \alpha \lambda c.$ Thus, by the uniqueness of representation, we get that $r(\alpha x) = \alpha \lambda = \alpha r(x).$ For $\alpha < 0,$ it suffices to prove that $- c \in \text{Lead}(C_0)$ whenever $c \in \text{Lead}(C_0).$ Let $c \in \text{Lead}(C_0).$ Then, $- c \in C_0$ as $C_0$ is balanced. Assume that $- c = \beta d$ for some $d \in \text{Lead}(C_0).$ Then $c = \beta (- d)$ with $- d \in C_0.$ Now, by the definition of $\text{Lead}(C_0)$ we get $- d = c$ and $\beta = 1.$ Thus, $- c \in \text{Lead}(C_0).$ Finally, we show that r is sub-additive. For this, let $x, y \in X_0.$ Without any loss of generality, we assume that $x \ne 0$ and $y \ne 0.$ Let $x = \alpha c$ and $y = \beta d$ be the unique representations with $c, d \in \text{Lead}(C_0)$ and $\alpha > 0,$ $\beta > 0.$ Then, $r(x) = \alpha$ and $r(y) = \beta .$ Now, $x + y = (\alpha + \beta ) z,$ where $z = \left( \frac{\alpha }{\alpha + \beta } c + \frac{\beta }{\alpha + \beta } d \right) \in C_0.$ Thus

$$\begin{aligned} r(x + y) = r((\alpha + \beta ) z) = (\alpha + \beta ) r(z) \le \alpha + \beta = r(x) + r(y). \end{aligned}$$

This completes the proof. $\square$

Remark 2.7

The converse of Theorem 2.6 is evident. Let $(V, \Vert \cdot \Vert )$ be a normed linear space. Then, $B = \lbrace v \in V: \Vert v \Vert \le 1 \rbrace$ is a non-empty, radially compact, absolutely convex set in V with $\text{Lead}(B) = \lbrace v \in B: \Vert v \Vert = 1 \rbrace .$

2.3 Centre of a convex set

Definition 2.8

Let B be non-empty convex set in a real vector space X. An element $b_0 \in B$ is said to be a centre of B,  if for each $b \in B,$ there exists a (unique) $b' \in B,$ such that $b_0 = \frac{1}{2} b + \frac{1}{2} b'.$

Note that if C is an absolutely convex set, then 0 is a centre of C.

Proposition 2.9

Let B be a non-empty convex set in a real vector space X with a centre $b_0.$ Put $B_0 = B - b_0$ and assume that $B_0$ is radially compact.

• 1. Then $B_0$ is balanced and convex.

• 2. If $x \in \text{Lead}(B_0),$ then $- x \in \text{Lead}(B_0).$

• 3. Put $K = \lbrace x + b_0: x \in \text{Lead}(B_0) \rbrace .$ Then, for each $b \in B$ with $b \ne b_0,$ there exist a unique pair $b_1, b_2 \in K$ with $\frac{1}{2} b_1 + \frac{1}{2} b_2 = b_0$ and $\frac{1}{2} < \alpha \le 1,$ such that $b = \alpha b_1 + (1 - \alpha ) b_2.$

Proof

(1). As B is convex, so is $B_0.$ In addition, $0 \in B_0$ for $b_0 \in B.$ Let $x \in B_0.$ Then, $x = b - b_0$ for some $b \in B.$ As $b_0$ is a centre of B,  there exists a unique $b' \in B,$ such that $\frac{1}{2} b + \frac{1}{2} b' = b_0.$ Thus, $- x = b' - b_0 \in B_0.$ In addition, for $0 \le \lambda \le 1,$ we have $\lambda x = \lambda x + (1 - \lambda ) 0 \in B_0,$ so that $B_0$ is balanced as well.

(2). First observe that $- x \in B_0$ as $B_0$ is balanced. Since $x \in \text{Lead}(B_0),$ we have $x \ne 0,$ so that there exist a unique $x_1 \in \text{Lead}(B_0)$ and a unique $\lambda \in (0, 1],$ such that $- x = \lambda x_1.$ Then, $x = \lambda (- x_1).$ Now, by the definition of a lead point, we get $\lambda = 1$ and consequently, $- x = x_1 \in \text{Lead}(B_0).$

(3). Let $b = x + b_0$ for some $x \in B_0,$ such that $b \ne b_0$ (or equivalently, $x \ne 0$). Thus, there exist a unique $x_1 \in \text{Lead}(B_0)$ and a unique $\lambda \in (0, 1],$ such that $x = \lambda x_1.$ Put $\alpha = \frac{1 + \lambda }{2}.$ Then, $\frac{1}{2} < \alpha \le 1.$ For $b_1 = x_1 + b_0$ and $b_2 = - x_1 + b_0,$ we have $\frac{1}{2} b_1 + \frac{1}{2} b_2 = b_0.$ By (2), we have $b_1, b_2 \in K.$ In addition

$$\begin{aligned} b = x + b_0 = (1 - \lambda ) b_0 + \lambda b_1 = (1 - \lambda ) \left( \frac{1}{2} b_1 + \frac{1}{2} b_2 \right) + \lambda b_1 = \alpha b_1 + (1 - \alpha ) b_2. \end{aligned}$$

Next, let $b_3, b_4 \in K$ with $\frac{1}{2} b_3 + \frac{1}{2} b_4 = b_0$ and $\beta \in {\mathbb{R}}$ with $\frac{1}{2} < \beta \le 1$ be such that $b = \beta b_3 + (1 - \beta ) b_4.$ Let $b_3 = x_2 + b_0$ for some $x_2 \in \text{Lead}(B_0).$ Then, $b_4 = - x_2 + b_0$ and we have

$$\begin{aligned} x = \beta x_2 + (1 - \beta ) (- x_2) = (2 \beta - 1) x_2. \end{aligned}$$

Now, by the uniqueness of a lead representation, we get $x_2 = x_1$ and

$$\begin{aligned} 2 \beta - 1 = \lambda = 2 \alpha - 1 \end{aligned}$$

so that $\beta = \alpha .$ $\square$

Corollary 2.10

Let B be a non-empty convex set in a real vector space X and assume that $b_0 \in B$ be a centre of B. If $B_0 = B - b_0$ be radially compact, then $b_0$ is the only centre of B.

Proof

Let $b_1 \in B$ be another centre of B. Put $u_0 = b_1 - b_0.$ Then $u_0 \in B_0.$ We show that $u_0 = 0.$ As $b_1$ is a centre of B and $b_0 \in B,$ there exists $b_0' \in B,$ such that $b_1 = \frac{1}{2} b_0 + \frac{1}{2} b_0'.$ Thus, $2 u_0 = b_0' - b_0 \in B_0.$ Assume that $n u_0 \in B_0$ for some $n \in {\mathbb{N}}.$ Then, $b_2 \,{:}{=}\, - n u_0 + b_0 \in B$ as $B_0$ is balanced. Thus, there exists $c \in B,$ such that $b_1 = \frac{1}{2} b_2 + \frac{1}{2} c.$ It follows that $(n + 2) u_0 = (c - b_0) \in B_0.$ Now, by induction, $n u_0 \in B_0$ for all $n \in {\mathbb{N}}.$ Since $B_0$ is radially compact, we must have $u_0 = 0.$ Hence, $b_1 = b_0.$ $\square$

3 Affine functions of a central compact convex set

Let B be a compact and convex set in a real locally convex Hausdorff space X with a centre $b_0$ and consider the compact, balanced, convex set $B_0 = B - b_0.$ Recall that a function $f: B \rightarrow {\mathbb{R}}$ is called affine, if

$$\begin{aligned} f(\alpha b_1 + (1 - \alpha ) b_2) = \alpha f(b_1) + (1 - \alpha ) f(b_2) \end{aligned}$$

for all $b_1, b_2 \in B$ and $\alpha \in [0, 1].$ Let

$$\begin{aligned} A(B) = \lbrace f: B \rightarrow {\mathbb{R}}|\ f\ \text{is affine and continuous}\rbrace . \end{aligned}$$

As B is compact and convex, A(B) is an order unit space with the cone

$$\begin{aligned} A^+(B) = \lbrace f \in A(B): f(b) \ge 0\ \text{for all}\ b \in B \rbrace \end{aligned}$$

and the order unit $\mathbf{1}: B \rightarrow {\mathbb{R}}$ given by $\mathbf{1}(b) = 1$ for all $b \in B.$ In addition, the order unit norm is the sup-norm on A(B) as a closed subspace of C(B). (For a detailed discussion on this topic, please refer to [1, Chapter II.1].)

In this section, we describe the space of real valued, continuous, affine functions on a central, compact, convex set in a real locally convex Hausdorff space.

Theorem 3.1

Let B be central compact and convex set in a real locally convex Hausdorff space X with the centre $b_0$ and consider the closed subspace

$$\begin{aligned} A_0(B) \,{:}{=}\, \lbrace f \in A(B): f(b_0) = 0 \rbrace \end{aligned}$$

of A(B). Then, A(B) is unitally order isomorphic to the order unit space $(A_0(B)^{(\cdot )}, e).$

We shall prove this result with the help of the following lemmas.

Lemma 3.2

$A^+(B) = \lbrace f \in A(B): \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } \le f(b_0) \rbrace .$

Proof

First, we assume that $f \in A^+(B),$ so that $f(b) \ge 0$ for all $b \in B.$ Fix $b \in B$ with $b\ne b_0.$ Then, by Proposition 2.9(3), there exist a unique pair $b_1, b_2 \in K$ with $\frac{1}{2} b_1 + \frac{1}{2} b_2 = b_0$ and a unique $\alpha \in {\mathbb{R}}$ with $\frac{1}{2} < \alpha \le 1,$ such that $b = \alpha b_1 + (1 - \alpha ) b_2.$ (Here $K = \lbrace x + b_0: x \in \text{Lead}(B_0) \rbrace$ and $B_0 = B - b_0.$) Thus

$$\begin{aligned} f(b) = \alpha f(b_1) + (1 - \alpha ) f(b_2) \le \max \lbrace f(b_1), f(b_2) \rbrace . \end{aligned}$$

Again, as $\frac{1}{2} b_1 + \frac{1}{2} b_2 = b_0,$ we have

$$\begin{aligned} \frac{1}{2} f(b_1) + \frac{1}{2} f(b_2) = f(b_0) \end{aligned}$$

so that

$$\begin{aligned} 0 \le f(b) \le \max \lbrace f(b_1), f(b_2) \rbrace \le 2 f(b_0). \end{aligned}$$

Thus $\vert f(b) - f(b_0) \vert \le f(b_0)$ for all $b \in B.$ Therefore, $\Vert f - f(b_0) \mathbf{1} \Vert _{\infty } \le f(b_0).$

Conversely, we assume that $f \in A(B)$ with $\Vert f - f(b_0) \mathbf{1} \Vert _{\infty } \le f(b_0).$ Then, for each $b \in B,$ we have $\vert f(b) - f(b_0) \vert \le f(b_0)$ or equivalently,

$$\begin{aligned} 0 \le f(b) \le \max \lbrace f(b_1), f(b_2) \rbrace \le 2 f(b_0). \end{aligned}$$

Thus $f \in A(B)^+.$ $\square$

Lemma 3.3

For each $f \in A(B),$ we have

$$\begin{aligned} \Vert f \Vert _{\infty } = \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } + \vert f(b_0) \vert . \end{aligned}$$

Proof

Fix $f \in A(B).$ Then, $\Vert f \Vert _{\infty } \le \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } + \vert f(b_0) \vert .$ Next, as $\Vert f \Vert _{\infty } \mathbf{1} \pm f \in A^+(B),$ by Lemma 3.2, we have

$$\begin{aligned} \Vert (\Vert f \Vert _{\infty } \mathbf{1} \pm f) - (\Vert f \Vert _{\infty } \pm f(b_0)) \mathbf{1} \Vert _{\infty } \le (\Vert f \Vert _{\infty } \pm f(b_0)) \end{aligned}$$

or equivalently, $\Vert f - f(b_0) \mathbf{1} \Vert _{\infty } \le (\Vert f \Vert _{\infty } \pm f(b_0)).$ Thus

$$\begin{aligned}&\Vert f - f(b_0) \mathbf{1} \Vert _{\infty } + \vert f(b_0) \vert \\ &\quad = \max \lbrace \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } + f(b_0), \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } - f(b_0) \rbrace \\ &\quad \le \Vert f \Vert _{\infty }. \end{aligned}$$

Hence $\Vert f \Vert _{\infty } = \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } + \vert f(b_0) \vert .$ $\square$

Proof of Theorem 3.1

For $f \in A(B),$ we define $\theta (f): B \rightarrow {\mathbb{R}}$ given by $\theta (f)(b) = f(b) - f(b_0)$ for all $b \in B.$ Then, $\theta (f) = f - f(b_0) \mathbf{1} \in A_0(B),$ so that by Lemma 3.3, we have

$$\begin{aligned} \Vert \theta (f) \Vert _{\infty } = \Vert f - f(b_0) \mathbf{1} \Vert _{\infty } \le \Vert f \Vert _{\infty }. \end{aligned}$$

Thus the map $\theta : A(B) \rightarrow A_0(B)$ given by $\theta (f) = f - f(b_0) \mathbf{1}$ for all $f \in A(B)$ is a contractive, linear surjective map. We define $\chi : A(B) \rightarrow A_0(B)^{(\cdot )}$ given by $\chi (f) = (\theta (f), f(b_0))$ for all $f \in A(B).$ Then, $\chi$ is a unital linear surjection. In addition, by Lemmas 3.2 and 3.3, $\chi$ is an order isomorphism. $\square$

Example

Fix $n \in {\mathbb{N}}$ with $n \ge 2$ and consider the compact convex set

$$\begin{aligned} S_n = \left\{ (x_i) \in \ell _1^n: x_i \ge 0\ \text{and}\ \sum _{i=1}^{n} x_i = 1 \right\} . \end{aligned}$$

Then $B_n \,{:}{=}\, {{\,\mathrm{co}\,}}(S_n \bigcup - S_n)$ is the closed unit ball of the base normed space $\ell _1^n$ and is a compact convex set with the centre 0. We show that $A(B_n)$ is isometrically isomorphic to $(A(S_n)^{(\cdot )}, e).$ Note that $A(S_n)$ is isometrically isomorphic to $\ell _{\infty }^n.$

Let $f \in A_0(B_n).$ Then, $f(- \alpha ) = - f(\alpha )$ for all $\alpha \in B_n.$ In fact if $\alpha \in B_n,$ then $- \alpha \in B_n$ and we have $0 = \frac{1}{2} \alpha + \frac{1}{2} (- \alpha ).$ Thus

$$\begin{aligned} 0 = f(0) = \frac{1}{2} f(\alpha ) + \frac{1}{2} f(- \alpha ) \end{aligned}$$

so that $f(- \alpha ) = - f(\alpha ).$ Put $a_i = f(e_i),$ where $\lbrace e_i \rbrace$ is the standard unit vector basis of ${\mathbb{R}}^n.$ We show that $f((x_i)) = \sum _{i=1}^{n} x_i f(e_i)$ for every $(x_i) \in B_n.$ Fix $(x_i) \in B_n.$ Then, $\sum _{i=1}^{n} \vert x_i \vert \le 1.$ Put $\epsilon _i = \text{sign}(x_i)$ for $1 \le i \le n.$ Then, $\epsilon _i = \pm 1$ and $x_i = \epsilon \vert x_i \vert$ for each i. In addition, we have $\epsilon f(e_i) = f(\epsilon _i e_i)$ for each i. Put $\lambda _i = \vert x_i \vert$ for $1 \le i \le n$ and $\lambda _0 = 1 - \sum _{i=1}^{n} \vert x_i \vert .$ Then, $\lambda _i \ge 0$ for $0 \le i \le n$ with $\sum _{i=0}^{n} \lambda _i = 1.$ Thus as f is affine, we have

$$\begin{aligned} \sum _{i=1}^{n} x_i a_i= & {} \sum _{i=1}^{n} \vert x_i \vert \epsilon _i f(e_i) \\= & {} \sum _{i=1}^{n} \vert x_i \vert f(\epsilon _i e_i) \\= & {} \lambda _0 f(0) + \sum _{i=1}^{n} \lambda _i f(\epsilon _i e_i) \\= & {} f(\lambda _0 0 + \sum _{i=1}^{n} \lambda _i \epsilon _i e_i) \\= & {} f((x_i)). \end{aligned}$$

Next, let $f \in A(B_n).$ Define $f_0(x) = f(x) - f(0)$ for all $x \in B_n.$ Then, $f_0 \in A_0(B_n).$ Thus, for any $(x_i) \in B_n,$ we have

$$\begin{aligned} f((x_i))= & {} f_0((x_i)) + f(0) \\= & {} \sum _{i=1}^{n} x_i f_0(e_i) + f(0) \\= & {} \sum _{i=1}^{n} x_i (f(e_i) - f(0)) + f(0). \end{aligned}$$

Therefore, $f \mapsto (f(e_1),\ldots , f(e_n))$ is linear isomorphism from $A_0(B_n)$ onto ${\mathbb{R}}^n$ and $f \mapsto (f(0), f(e_1) - f(0),\ldots , f(e_n) - f(0))$ is linear isomorphism from $A(B_n)$ onto ${\mathbb{R}}^{n+1}.$

Next, let $f \in A(B_n).$ Then

$$\begin{aligned} \Vert f \Vert _{\infty }= & {} \sup \lbrace \vert f((x_i)) \vert : (x_i) \in B_n \rbrace \\= & {} \sup \left\{ \left| \sum _{i=1}^{n} x_i (f(e_i) - f(0)) + f(0) \right| : \sum _{i=1}^{n} \vert x_i \vert \le 1 \right\} \\= & {} \sup \left\{ \left| \sum _{i=1}^{n} x_i (f(e_i) - f(0)) \right| + \vert f(0) \vert : \sum _{i=1}^{n} \vert x_i \vert \le 1 \right\} \\= & {} \vert f(0) \vert + \max \lbrace \vert f(e_i) - f(0) \vert : 1 \le i \le n \rbrace \\= & {} \vert f(0) \vert + \Vert (f(e_1) - f(0),\ldots , f(e_n) - f(0)) \Vert _{\infty }. \end{aligned}$$

Hence $A_0(B_n)$ is isometrically isomorphic to $\ell _{\infty }^n$ and $A(B_n)$ is isometrically isomorphic to $\ell _{\infty }^n \oplus _1 {\mathbb{R}}.$ Since $A(S_n)$ is isometrically isomorphic to $\ell _{\infty }^n,$ we get that $A(B_n) \cong A(S_n)^{(\cdot )}$ as order unit spaces.

4 Base with a centre

Let (V, B) be a base normed space [1, Proposition II.1.12]. Then there exists a unique strictly positive $e \in V^*$ with $\Vert e \Vert = 1,$ such that $e(v) = \Vert v \Vert$ if and only if $v \in V^+.$ In particular, $B = \lbrace v \in V^+: e(v) = 1 \rbrace$ [10, Lemma 9.3 and Proposition 9.4]. In this section, we describe the base normed space with a central base.

Theorem 4.1

Let (V, B) be a base normed space. For a fixed $b_0 \in B,$ the following statements are equivalent:

• 1. $b_0$ is a centre of B;

• 2. $V^+ = \lbrace v \in V: \Vert v - e(v) b_0 \Vert \le e(v) \rbrace ;$

• 3. $B = \lbrace v \in V: e(v) = 1\ \text{and}\ \Vert v - b_0 \Vert \le 1 \rbrace .$

We use the following result to prove Theorem 4.1.

Lemma 4.2

Let (V, B) be a base normed space and assume that $b_0 \in B$ is a centre of B. Then, for each $v \in V,$ we have

$$\begin{aligned} \Vert v \Vert \ge \max \lbrace \Vert v - e(v) b_0 \Vert , \vert e(v) \vert \rbrace . \end{aligned}$$

Proof

Since $b_0$ is the centre of the base B,  for each $b \in B,$ there exists a (unique) $b' \in B,$ such that $b_0 = \frac{1}{2} b + \frac{1}{2} b'$ (Definition 2.8). Let $v = \lambda b_1 - \mu b_2$ for some $b_1, b_2 \in B$ and $\lambda , \mu \ge 0$ with $\lambda + \mu = \Vert v \Vert .$ Then, $e(v) = \lambda - \mu ,$ so that

$$\begin{aligned} v - e(v) b_0= & {} \lambda b_1 - \mu b_2 - (\lambda - \mu ) b_0 \\= & {} \lambda (b_1 - b_0) - \mu (b_2 - b_0) \\= & {} \lambda \left( \frac{b_1 - b'_1}{2} \right) - \mu \left( \frac{b_2 - b'_2}{2} \right) \\= & {} \frac{1}{2} \left\{ (\lambda b_1 - \mu b_2) - (\lambda b'_1 - \mu b'_2) \right\} . \end{aligned}$$

Thus $\Vert v - e(v) b_0 \Vert \le \lambda + \mu = \Vert v \Vert .$ Also, we have $\vert e(v) \vert = \vert \lambda - \mu \vert \le \lambda + \mu$ which completes the proof. $\square$

Proof of Theorem 4.1

1. implies 2.: Let us assume that $b_0$ is a centre of B. If $v \in V^+,$ then $e(v) = \Vert v \Vert .$ Thus, by Lemma 4.2, $\Vert v - e(v) b_0 \Vert \le e(v).$ Now, assume that $\Vert v - e(v) b_0 \Vert \le e(v)$ for some $v \in V.$ Find $b_1, b_2 \in B$ and $\lambda , \mu \ge 0,$ such that $v - e(v) b_0 = \lambda b_1 - \mu b_2$ and $\Vert v - e(v) b_0 \Vert = \lambda + \mu .$ Then

$$\begin{aligned} 0 = e(v - e(v) b_0) = e(\lambda b_1 - \mu b_2) = \lambda - \mu \end{aligned}$$

so that $2 \lambda = 2 \mu = \Vert v - e(v) b_0 \Vert \le e(v).$ Thus

$$\begin{aligned} v = \lambda (b_1 - b_2) + e(v) b_0 = \lambda b_1 + \lambda b'_2 + (e(v) - 2 \lambda ) b_0 \in V^+. \end{aligned}$$

Thus $V^+ = \lbrace v \in V: \Vert v - e(v) b_0 \Vert \le e(v) \rbrace .$

2. implies 3.: Next we assume that $V^+ = \lbrace v \in V: \Vert v - e(v) b_0 \Vert \le e(v) \rbrace .$ Let $b \in B.$ Then, $b \in V^+$ and $\Vert b \Vert = 1 = e(b),$ so that $\Vert b - b_0 \Vert \le 1.$ Put $u = b - b_0.$ Then, $e(u) = 0,$ $\Vert u \Vert \le 1$ and we have $b = u + b_0.$ Conversely, let $u \in V$ with $e(u) = 0$ and $\Vert u \Vert \le 1$ and put $b = u + b_0.$ Then, $e(b) = 1$ and $\Vert b - e(b) b_0 \Vert = \Vert u \Vert \le 1 = e(b).$ Thus, by the assumption, $b \in V^+,$ so that $\Vert b \Vert = e(b) = 1,$ that is, $b \in B.$ Therefore, $B = \lbrace v \in V: e(v) = 1 \ \text{and} \ \Vert v - b_0 \Vert \le 1 \rbrace .$

3. implies 1.: Finally, we assume that

$$\begin{aligned} B = \lbrace v \in V: e(v) = 1\ \text{and}\ \Vert v - b_0 \Vert \le 1 \rbrace . \end{aligned}$$

Let $b \in B$ and consider $b' \,{:}{=}\, 2 b_0 - b.$ Then, $e(b') = 1$ and

$$\begin{aligned} b' - e(b') b_0 = 2 b_0 - b - b_0 = b_0 - b. \end{aligned}$$

As $b \in B,$ by the assumption, we have $\Vert b' - e(b') b_0 \Vert \le 1,$ so that $b' \in B.$ Thus, $b_0$ is a centre of B. $\square$

Corollary 4.3

Let (V, B) be a base normed space with a centre $b_0$ of B. Then, for each $v \in V,$ we have $\Vert v \Vert = \max \lbrace \Vert v - e(v) b_0 \Vert , \vert e(v) \vert \rbrace .$

Proof

If $v \in V^+ \cup - V^+,$ then $\Vert v \Vert = \vert e(v) \vert .$ Thus by Lemma 4.2, we get $\Vert v \Vert = \max \lbrace \Vert v - e(v) b_0 \Vert , \vert e(v) \vert \rbrace .$ Next, let $v \notin V^+ \cup - V^+.$ Then, by Theorem 4.1, we have $\vert e(v) \vert < \Vert v - e(v) b_0 \Vert .$ Put $u = \frac{v - e(v) b_0}{\Vert v - e(v) b_0 \Vert }.$ Then, $e(u) = 0$ and $\Vert u \Vert = 1,$ so that $k = u + b_0 \in B,$ by Theorem 4.1. Also then, $k' = - u + b_0.$ Next, put $2 \lambda = \Vert v - e(v) b_0 \Vert + e(v)$ and $2 \mu = \Vert v - e(v) b_0 \Vert - e(v).$ Then, $\lambda , \mu > 0$ with $\lambda + \mu = \Vert v - e(v) b_0 \Vert$ and $\lambda - \mu = e(v).$ Thus

$$\begin{aligned} \lambda k - \mu k' = (\lambda + \mu ) u + (\lambda - \mu ) b_0 = v - e(v) b_0 + e(v) b_0 = v. \end{aligned}$$

Now, it follows that

$$\begin{aligned} \Vert v \Vert \le \lambda + \mu = \Vert v - e(v) b_0 \Vert \end{aligned}$$

so that $\Vert v \Vert = \max \lbrace \Vert v - e(v) b_0 \Vert , \vert e(v) \vert \rbrace .$ $\square$

Corollary 4.4

A centre of B,  if it exists, is unique.

Proof

Let $b_1 \in B$ be another centre of B. Then, by Theorem 4.1, $b_1 = u_1 + b_0$ for some $u_1 \in V$ with $e(u_1) = 0$ and $\Vert u_1 \Vert \le 1.$ If $b_1 \ne b_0,$ then $u_1 \ne 0.$ In which case, $b_2 = - \Vert u_1 \Vert ^{-1} u_1 + b_0 \in B.$ Since $b_1$ is a centre, by Theorem 4.1, we must have $\Vert b_2 - b_1 \Vert \le 1.$ However, this is not true as $\Vert b_2 - b_1 \Vert = 1 + \Vert u_1 \Vert > 1.$ Hence, $b_1 = b_0.$ $\square$

Combining Theorem 4.1 and Corollary 4.3, we may deduce the following

Theorem 4.5

A base normed space (V, B) with a centre $b_0 \in B$ is isometrically isomorphic to $\left( V_0^{(\cdot )}, B_0^{(\cdot )}\right) ,$ where $V_0 = \lbrace v \in V: e(v) = 0 \rbrace$ and

$$\begin{aligned} B_0^{(\cdot )} = \lbrace (v, 1): v \in V_0\ \text{with}\ \Vert v \Vert \le 1 \rbrace . \end{aligned}$$

Here e is the order unit of $V^*$ corresponding to the base B.

Example

Consider $V_n = \ell _1^n({\mathbb{R}}).$ Then, $B_n = \lbrace (\alpha _i): \alpha _i \ge 0 \ \text{and}\ \sum \alpha _i = 1 \rbrace$ is the base for $V_n,$ so that $(V_n, B_n)$ is a base normed space. For $n \ge 3,$ $B_n$ can not have a centre. To see this fix $b_0 = (\alpha _i^0) \in B_n,$ so that $\sum \alpha _i^0 = 1.$ In addition, if $\alpha _m^0 = \inf \lbrace \alpha _i^0: 1 \le i \le n \rbrace ,$ then $\alpha _m^0 \le \frac{1}{n}.$ Consider the m-th coordinate vector $e_m \in V_n.$ Then, $e_m \in B_n$ and we have

$$\begin{aligned} \Vert e_m - b_0 \Vert _1 = 1 - \alpha _m^0 + \sum _{i \ne m} \alpha _i^0 = 2 (1 - \alpha _m^0) \ge 2 \left( 1 - \frac{1}{n}\right) > 1. \end{aligned}$$

Thus $b_0$ is not a centre of $B_n.$

5 Order unit spaces with a central state space

We now describe a non-dual version of A(B),  where B is a central, compact convex subset of a real locally convex space.

Proposition 5.1

Let (V, e) be an order unit space and let $\tau \in S(V).$ Then, the following statements are equivalent:

• 1. $\tau$ is a centre of S(V);¼/Para>

• 2. $v \le 2 \tau (v) e$ for every $v \in V^+;$

• 3. $V^+ = \lbrace v \in V: \Vert v - \tau (v) e \Vert \le \tau (v) \rbrace .$

In this case, we say that $\tau$ is a central state.

Proof

(1) implies (2): We assume that $\tau$ is a centre of S(V). Let $v \in V^+.$ If $f \in S(V),$ then by assumption, $2 \tau - f \in S(V).$ Thus, $f(2 \tau (v) e -v) = 2 \tau (v) - f(v) \ge 0$ for all $f \in S(V),$ so that $v \le 2 \tau (v) e.$

(2) implies (1): Now, we assume that $v \le 2 \tau (v) e$ for each $v \in V^+.$ Let $f \in S(V).$ Put $f' = 2 \tau - f.$ Then, $f'$ is linear and $f'(e) = 1.$ We show that it is positive. Let $v \in V^+.$ Then, by assumption, $v \le 2 \tau (v) e.$ As $f \ge 0,$ we have $f'(v) = 2 \tau (v) - f(v) = f(2 \tau (v) e - v) \ge 0.$ Thus, $f' \in S(V).$

(2) implies (3): We again assume that $v \le 2 \tau (v) e$ for each $v \in V^+.$ Thus, if $v \in V^+,$ then $v \le 2 \tau (v) e,$ or equivalently, $\tau (v) e \pm (v - \tau (v) e) \in V^+.$ Therefore, $\Vert v - \tau (v) e \Vert \le \tau (v).$ Conversely, let $v \in V$ with $\Vert v - \tau (v) e \Vert \le \tau (v).$ Then, $\tau (v) e \pm (v - \tau (v) e) \in V^+,$ so that $2 \tau (v) e - v \in V^+.$ Thus, $V^+ = \lbrace v \in V: \Vert v - \tau (v) e \Vert \le \tau (v) \rbrace .$

(3) implies (2): Finally we assume that $V^+ = \lbrace v \in V: \Vert v - \tau (v) e \Vert \le \tau (v) \rbrace .$ Then, for $v \in V^+$ we have $\Vert v - \tau (v) e \Vert \le \tau (v).$ Thus, $\tau (v) e \pm (v - \tau (v) e) \in V^+.$ In particular, $v \le 2 \tau (v) e.$ $\square$

The norm condition on positive elements of V is somewhat minimal as we see in the next result.

Proposition 5.2

Let (V, e) be an order unit space with a central state $\tau \in S(V).$ Then, for each $v \in V,$ we have $\Vert v - \tau (v) e \Vert \le \Vert v - \alpha e \Vert$ for all $\alpha \in {\mathbb{R}}.$

Proof

Fix $\alpha \in {\mathbb{R}}.$ Then

$$\begin{aligned} \vert f(v) - \alpha \vert = \vert f(v - \alpha e) \vert \le \Vert v - \alpha e \Vert \end{aligned}$$

for every $f \in S(V).$ By Proposition 5.1, given $f \in S(V),$ there exists a unique $f' \in S(V),$ such that $f + f' = 2 \tau .$ Thus

$$\begin{aligned} \vert f(v) - f'(v) \vert \le \vert f(v) - \alpha \vert + \vert f'(v) - \alpha \vert \le 2 \Vert v - \alpha e \Vert . \end{aligned}$$

Since $f - f' = 2(f - \tau ),$ we get

$$\begin{aligned} \vert f(v - \tau (v) e) \vert = \vert f(v) - \tau (v) \vert \le \Vert v - \alpha e \Vert \end{aligned}$$

for all $f \in S(V).$ Hence, $\Vert v - \tau (v) e \Vert \le \Vert v - \alpha e \Vert$ for all $\alpha \in {\mathbb{R}}.$ $\square$

Proposition 5.3

Let (V, e) be an order unit space with a central state $\tau \in S(V).$ Then, for each $v \in V,$ we have

$$\begin{aligned} \Vert v \Vert = \Vert v - \tau (v) e \Vert + \vert \tau (v) \vert . \end{aligned}$$

Proof

Fix $v \in V.$ Then, $\Vert v \Vert \le \Vert v - \tau (v) e \Vert + \vert \tau (v) \vert .$ In addition, as $\Vert v \Vert e \pm v \in V^+,$ by (1), we have

$$\begin{aligned} \Vert (\Vert v \Vert e \pm v) - \tau (\Vert v \Vert e \pm v) e \Vert \le \tau (\Vert v \Vert e \pm v). \end{aligned}$$

In other words, $\Vert v - \tau (v) e \Vert \le \Vert v \Vert \pm \tau (v),$ so that $\Vert v - \tau (v) e \Vert + \vert \tau (v) \vert \le \Vert v \Vert .$ Hence, $\Vert v \Vert = \Vert v - \tau (v) e \Vert + \vert \tau (v) \vert .$ $\square$

Remark 5.4

Let (V, e) be an order unit space with a central state $\tau \in S(V).$ Then, $\tau$ is unique and we have

• 1. $V^{*+} = \lbrace f \in V^*: \Vert f - f(e) \tau \Vert \le f(e) \rbrace = \lbrace f + \alpha \tau : f(e) = 0\ \text{and}\ \Vert f \Vert \le \alpha \rbrace ;$

• 2. $S(V) = \lbrace f_0 + \tau : f_0 \in V^*, f_0(e) = 0 \ \text{and}\ \Vert f_0 \Vert \le 1 \rbrace ;$ and

• 3. For each $f \in V^*,$ we have $\Vert f \Vert = \max \lbrace \Vert f - f(e) \tau \Vert , \vert f(e) \vert \rbrace .$

Proposition 5.5

Let (V, e) be an order unit space with a central state $\tau \in S(V).$ Then, the Banach dual of $V_0 = \lbrace v \in V: \tau (v) = 0 \rbrace$ is isometrically isomorphic to $V_0^* \,{:}{=}\, \lbrace f \in V^*: f(e) = 0 \rbrace .$

Proof

Consider the mapping $\chi : V_0^* \rightarrow (V_0)^*$ given by $f \mapsto f|_{V_0}.$ Then, $\chi$ is linear. In addition, for $f \in V_0^*$ we have

$$\begin{aligned} \Vert f \Vert= & {} \sup \lbrace \vert f(v) \vert : v \in V \ \text{with}\ \Vert v \Vert \le 1 \rbrace \\= & {} \sup \lbrace \vert f(v - \tau (v) e) \vert : v \in V\ \text{with}\ \Vert v - \tau (v) e \Vert + \vert \tau (v) \vert \le 1 \rbrace \\= & {} \sup \lbrace \vert f(v) \vert : v \in V_0\ \text{with}\ \Vert v \Vert \le 1 \rbrace . \end{aligned}$$

Thus $\chi$ is an isometry. Furthermore, for $f_0 \in (V_0)^*,$ we define $f: V \rightarrow {\mathbb{R}}$ by $f(v) = f_0(v - \tau (v) e)$ for all $v \in V.$ Then, $f \in V_0^*$ with $\chi (f) = f_0.$ $\square$

We can summarize the findings of this section in the following.

Theorem 5.6

Let (V, e) be an order unit space with a central state $\tau \in S(V).$ Then, $V/{{\mathbb{R}} e}$ is isometrically isomorphic to $V_0 \,{:}{=}\, \lbrace v \in V: \tau (v) = 0 \rbrace$ and V is unitally and isometrically order isomorphic to $V_0 \oplus _1 {\mathbb{R}}$ via the map $v \mapsto (v - \tau (v), \tau (v)).$

5.1 Strict convexity

Definition 5.7

Let $B_0$ be a compact, balanced, convex subset of a real, locally convex, Hausdorff space X. We say that $B_0$ is said to satisfy Property (S) with respect to $X^*,$ if for each $x_0 \in \text{Lead}(B_0)$ there exists at most one $f \in X^*,$ such that

$$\begin{aligned} \sup \lbrace \vert f(x) \vert : x \in B_0 \rbrace = f(x_0) = 1. \end{aligned}$$

Remark 5.8

The notion of Property (S) is motivated by the characterization of strictly convex spaces by M. G. Krein in 1938. However, the same result was also proved independently by A. F. Ruston in 1949 [9, Theorem 1]. (Please refer to [8, 5.5.1.1] for a detailed discussion on it.)

Definition 5.9

Let (V, e) be an order unit space with a central trace $\tau \in S(V).$ We say that V is tracial, if $S(V)_0 ({:}{=}\, S(V) - \tau )$ satisfies Property (S).

Theorem 5.6 can now be restated as follows.

Corollary 5.10

Let $V_0$ be a strictly convex real normed linear space and consider the order unit space $V_0^{(\infty )}$ obtained by adjoining an order unit to $V_0$ (see, [7, Theorem 2.17]). Then, $V_0^{(\infty )}$ is precisely a tracial order unit space.

6 Central absolutely base normed space

In [7], we described the absolute value that naturally arises out of the order structure in $V^{(\cdot )}.$ We studied the corresponding absolute order unit space and showed that it is a generalization of a spin factor. In this section, we shall discuss the description of the canonical absolute value in $V^{(\cdot )}$ in the context of the corresponding base $B^{(\cdot )}.$

Let us recall the notion of absolutely ordered spaces introduced in [6]. (See also, [5].)

Definition 6.1

[6, Definition 3.4] Let $(U, U^+)$ be a real ordered vector space and let $\vert \cdot \vert : U \rightarrow U^+$ satisfy the following conditions:

• 1. $\vert v \vert = v$ if $v \in U^+;$

• 2. $\vert v \vert \pm v \in U^+$ for all $v \in U;$

• 3. $\vert k v \vert = \vert k \vert \vert v \vert$ for all $v \in U$ and $k \in {\mathbb{R}};$

• 4. If $u, v, w \in U$ with $\vert u - v \vert = u + v$ and $\vert u - w \vert = u + w,$ then $\vert u - \vert v \pm w \vert \vert = u + \vert v \pm w \vert ;$

• 5. If u, v and $w \in U$ with $\vert u - v \vert = u + v$ and $0 \le w \le v,$ then $\vert u - w \vert = u + w.$

Then $(U, U^+, \vert \cdot \vert )$ is called an absolutely ordered space.

Let (V, B) be a base normed space with a centre $b_0$ of B. Consider the set $K = \lbrace u + b_0: e(u) = 0 \ \text{and}\ \Vert u \Vert = 1 \rbrace \subset B.$ We note that if $k \in K,$ then $k' \in K.$ In fact, $k' = - u + b_0$ whenever $k = u + b_0.$ The following results show that K determines an absolute value in V.

Lemma 6.2

Let (V, B) be a base normed space with a centre $b_0$ of B and let $b \in B$ with $b \ne b_0.$ Then, there exist a unique $k \in K$ and a unique $\frac{1}{2} < \alpha \le 1,$ such that $b = \alpha k + (1 - \alpha ) k'.$

Proof

As $b \in B$ with $b \ne b_0,$ we have $0 < \Vert b - b_0 \Vert \le 1.$ Put $\alpha = \frac{1 + \Vert b - b_0 \Vert }{2}.$ Then, $\frac{1}{2} < \alpha \le 1.$ Put $u_0 = \frac{b - b_0}{\Vert b - b_0 \Vert }.$ Then, $e(u_0) = 0$ with $\Vert u_0 \Vert = 1.$ Thus, $k = u_0 + b_0 \in K$ with $k' = - u_0 + b_0$ and we have

$$\begin{aligned} \alpha k + (1 - \alpha ) k' = (2 \alpha - 1) u_0 + b_0 = \Vert b - b_0 \Vert \left( \frac{b - b_0}{\Vert b - b_0 \Vert } \right) + b_0 = b. \end{aligned}$$

Next, let $b = \beta k_1 + (1 - \beta ) k'_1$ for some $k_1 \in K$ and $\frac{1}{2} < \beta \le 1.$ Then, there exists $u_1 \in V$ with $e(u_1) = 0$ and $\Vert u_1 \Vert = 1,$ such that $k_1 = u_1 + b_0$ and $k'_1 = - u_1 + b_0.$ Then, as above $b = (2 \beta - 1) u_1 + b_0,$ so that $(2 \alpha - 1) u_0 = (2 \beta - 1) u_1.$ Since $\frac{1}{2} < \alpha \le 1$ and $\frac{1}{2} < \beta \le 1,$ we have $2 \alpha > 1$ and $2 \beta > 1.$ Now, as $\Vert u_0 \Vert = 1 = \Vert u_1 \Vert ,$ we deduce that $2 \alpha - 1 = 2 \beta - 1.$ Thus we have $\alpha = \beta ,$ so that $u_0 = u_1,$ $\square$

Theorem 6.3

Let (V, B) be a base normed space with a centre $b_0$ of B and let $v \in V {\setminus } {\mathbb{R}} b_0.$ Then, there exist $k \in K$ and $\alpha , \beta \in {\mathbb{R}},$ such that $v = \alpha k + \beta k'.$ This representation is unique in the following sense: if $v = \lambda k_1 + \mu k_1'$ for some $k_1 \in K$ and $\lambda , \mu \in {\mathbb{R}},$ then either $k_1 = k$ with $\lambda = \alpha$ and $\mu = \beta$ or $k_1 = k'$ with $\lambda = \beta$ and $\mu = \alpha .$

Proof

First consider $v \in V^+ {\setminus } {\mathbb{R}} b_0.$ Then, $v = \Vert v \Vert b$ for $b \,{:}{=}\, \Vert v \Vert ^{-1} v \in B,$ such that $b \ne b_0.$ Thus, by Lemma 6.2, there exist a unique $k \in K$ and $\frac{1}{2} < \alpha _0 \le 1,$ such that $b = \alpha _0 k + (1 - \alpha _0) k'.$ Therefore, $v = \alpha k + \beta k',$ where $\alpha = \Vert v \Vert \alpha _0$ and $\beta = \Vert v \Vert (1 - \alpha _0).$ A similar proof works for $v \in - V^+ {\setminus } {\mathbb{R}} b_0.$

Now, let $v \notin V^+ \cup - V^+.$ Then, $\vert e(v) \vert < \Vert v - e(v) b_0 \Vert .$ Put $u = \frac{v - e(v) b_0}{\Vert v - e(v) b_0 \Vert }.$ Then, $e(u) = 0$ and $\Vert u \Vert = 1,$ so that $k = u + b_0 \in K$ with $k' = - u + b_0.$ First, we assume that $e(v) \ge 0.$ Put $2 \alpha = e(v) + \Vert v - e(v) b_0 \Vert$ and $2 \beta = e(v) - \Vert v - e(v) b_0 \Vert ,$ we have $\alpha > 0,$ $\beta < 0$ with $\alpha - \beta = \Vert v - e(v) b_0 \Vert$ and $\alpha + \beta = e(v).$ Thus

$$\begin{aligned} \alpha k + \beta k' = (\alpha - \beta ) u + (\alpha + \beta ) b_0 = v - e(v) b_0 + e(v) b_0 = v. \end{aligned}$$

Next, let $v = \lambda k_1 + \mu k'_1$ for some $k_1 \in K$ and $\lambda , \mu \in {\mathbb{R}}.$ Then, $e(v) = \lambda + \mu .$ Let $k_1 = u_1 + b_0$ for some $u_1 \in V$ with $e(u_1) = 0$ and $\Vert u_1 \Vert = 1.$ Then, $k' = - u_1 + b_0$ and we have $v = (\lambda - \mu ) u_1 + e(v) b_0.$ Replacing $u_1$ by $- u_1,$ if required, we get that $\lambda \ge \mu .$ Thus, $\Vert v - e(v) b_0 \Vert = \lambda - \mu .$ Now, it follows that $\lambda = \alpha$ and $\mu = \beta$ so that $u_1 = u_0$ and consequently, $k = k_1.$

When $e(v) < 0,$ we interchange $\alpha$ and $\beta$ and replace k by $k'.$ $\square$

Remark 6.4

Let us call the unique decomposition $v = \alpha k + \beta k'$ with $k \in K$ and $\vert \alpha \vert \ge \vert \beta \vert$ as the K-decomposition of $v \in V {\setminus } {\mathbb{R}} b_0.$ Note that for $v = \lambda b_0,$ we have $v = \alpha (k + k')$ for all $k \in K,$ where $2 \alpha = \lambda .$ Thus, given $v \in V$ and a decomposition $v = \alpha k + \beta k'$ with $\vert \alpha \vert \ge \vert \beta \vert ,$ we obtain $\vert v \vert = \vert \alpha \vert k + \vert \beta \vert k' \in V^+$ which is uniquely determined by v. This defines a mapping $\vert \cdot \vert : V \rightarrow V^+.$

Proposition 6.5

Let (V, B) be a base normed space with a centre $b_0$ of B. Then, the mapping $\vert \cdot \vert : V \rightarrow V^+$ satisfies the following properties:

• 1. $\vert v \vert = v$ if $v \in V^+;$

• 2. $\vert v \vert \pm v \in V^+$ for all $v \in V;$

• 3. $\vert \lambda v \vert = \vert \lambda \vert \vert v \vert$ for all $v \in V$ and $\lambda \in {\mathbb{R}};$

• 4. If $u, v, w \in V$ with $\vert u - v \vert = u + v$ and $\vert u - w \vert = u + w,$ then $\vert u - \vert v \pm w \vert \vert = u + \vert v \pm w \vert .$

Proof

Let $v \in V$ and consider its K-decomposition $v = \alpha k + \beta k',$ where $k \in K$ and $\vert \alpha \vert \ge \vert \beta \vert .$ Then, $\vert v \vert \,{:}{=}\, \vert \alpha \vert k + \vert \beta \vert k'.$ As $k, k' \in V^+,$ verification of (1), (2) and (3) is straightforward. To prove (4), let $u, v, w \in V$ with $\vert u - v \vert = u + v$ and $\vert u - w \vert = u + w.$ Then, $u, v, w \in V^+.$ If $u - v \in V^+,$ we have $u + v = \vert u - v \vert = u - v,$ so that $v = 0.$ Thus

$$\begin{aligned} \vert u - \vert v \pm w \vert = \vert u - w \vert = u + w = u + \vert v \pm w \vert . \end{aligned}$$

So we may assume that $u - v, u - w \notin V^+ \bigcup - V^+.$ Then, $u \ne 0,$ $v \ne 0$ and $w \ne 0.$ Consider the (unique) K-decomposition $u - v = \lambda _1 k_1 + \mu _1 k_1'$ for some $k_1 \in K$ and $\vert \lambda _1 \ge \vert \mu _1 \vert .$ Then, $\vert u - v \vert = \vert \lambda _1 \vert k_1 + \vert \mu _1 \vert k_1'.$ Since $\vert u - v \vert = u + v,$ we get that $u = \lambda _1 k_1$ and $v = - \mu _1 k_1'.$ Thus, $\lambda _1 > 0$ and $\mu _1 < 0.$ Again, as $\vert u - w \vert = u + w,$ we get that $u = \lambda _2 k_2$ and $w = - \mu _2 k_2'$ for some $k_2 \in K$ and $\vert \lambda _2 \ge \vert \mu _2 \vert$ with $\lambda _2 > 0$ and $\mu _2 < 0.$ Thus, $k_1 = k_2$ and $\lambda _1 = \lambda _2.$ Therefore

$$\begin{aligned} \vert u - \vert v \pm w \vert \vert = \vert \lambda _1 k_1 - \vert \mu _1 \pm \mu _2 \vert k_1' \vert = \lambda _1 k_1 + \vert \mu _1 \pm \mu _2 \vert k_1' = u + \vert v \pm w \vert . \end{aligned}$$

$\square$

Remark 6.6

It follows from the proof of Proposition 6.5 that $\vert u - v \vert = u + v$ if and only if there exist $k \in K$ and positive real numbers $\alpha$ and $\beta ,$ such that $u = \alpha k$ and $v = \beta k'.$

Theorem 6.7

Let (V, B) be a base normed space with a centre $b_0$ of B. Consider

$$\begin{aligned} V_0 = \lbrace v \in V: e(v) = 0 \rbrace = \lbrace v - e(v) b_0: v \in V \rbrace . \end{aligned}$$

Then $V_0$ is closed subspace of V and the following statements are equivalent:

• 1. $K = \text{ext}(B);$

• 2. $V_0$ is strictly convex; and

• 3. For all $u, v, w \in V^+$ with $\vert u - v \vert = u + v$ and $0 \le w \le v$ we have $\vert u - w \vert = u + w.$

Proof

(1) implies (2): First we assume that $K = \text{ext}(B).$ Let $u_0, u_1 \in V_0$ be, such that $u_0 \ne u_1$ and $\Vert u_0 \Vert = 1 = \Vert u_1 \Vert$ and assume to the contrary that $\Vert u_{\alpha } \Vert = 1,$ where $u_{\alpha } = \alpha u_1 + (1 - \alpha ) u_0$ for $0< \alpha < 1.$ Put $k_0 = u_0 + b_0,$ $k_1 = u_1 + b_0$ and $k_{\alpha } = u_{\alpha } + b_0.$ Then, by construction, $k_0, k_1, k_{\alpha } \in K$ and we have $k_{\alpha } = \alpha k_1 + (1 - \alpha ) k_0.$ Since $K = \text{Ext}(B),$ we deduce that $k_0 = k_1 = k_{\alpha }.$ This leads to a contradiction $u_0 = u_1.$ Thus, $V_0$ must be strictly convex.

(2) implies (3): Next we assume that $V_0$ is strictly convex. Consider $u, v, w \in V^+$ with $\vert u - v \vert = u + v$ and $0 \le w \le v.$ If $u - v \in V^+,$ then $v = 0,$ so that $w = 0$ and we have $\vert u - w \vert = u + w.$ If $v - u \in V^+,$ then $u = 0,$ so that $\vert u - w \vert = w = u + w.$ Thus, we may assume that $u - v \notin V^+ \bigcup - V^+.$ Then, $u \ne 0$ and $v \ne 0.$ We may also assume that $w \ne 0.$ Since $\vert u - v \vert = u + v,$ following the proof of Proposition 6.5, we can find $k \in K$ and $\alpha \ge \beta > 0,$ such that $u = \alpha k$ and $v = \beta k'.$ Since B is a base for $V^+$ and since $w \in V^+$ with $w \ne 0,$ there exist a unique $b \in B$ and $\lambda > 0,$ such that $w = \lambda b.$ By Lemma 6.2, we have $b = \gamma k_1 + (1 - \gamma ) k_1'$ for some $k_1 \in K$ and $\frac{1}{2} \le \gamma \le 1.$ Since $k, k_1 \in K,$ there exist $x, x_1 \in V_0$ with $\Vert x \Vert = 1 = \Vert x_1 \Vert ,$ such that $k = x + b_0$ and $k_1 = x_1 + b_0.$ Then, $k' = - x + b_0$ and $k_1 = - x_1 + b_0.$ As $w \le v$ we have

$$\begin{aligned} \lambda \left( \gamma (x_1 + b_0) + (1 - \gamma ) (- x_1 + b_0) \right) \le \beta (- x + b_0) \end{aligned}$$

so that $- \left( \beta x + \lambda (2 \gamma - 1) x_1 \right) + (\beta - \lambda ) b_0 \in V^+.$ Now, as $\frac{1}{2} \le \gamma \le 1$ we have $0 \le 2 \gamma - 1 \le 1.$ Thus, by Theorem 4.1, we get

$$\begin{aligned} \beta - \lambda \le \beta - (2 \gamma - 1) \lambda = \Vert \beta x \Vert - \Vert (2 \gamma - 1) \lambda x_1 \Vert \le \Vert \beta x + \lambda (2 \gamma - 1) x_1 \Vert \le \beta - \lambda . \end{aligned}$$

Therefore, we have $\gamma = 1$ and $\Vert \beta x + \lambda x_1 \Vert = \beta - \lambda .$ It follows that $b = k_1,$ so that $w = \lambda k_1.$ Since $V_0$ is strictly convex, we conclude that

$$\begin{aligned} \Vert \beta x + \lambda x_1 \Vert ^{-1} (\beta x + \lambda x_1) = \Vert \lambda x_1 \Vert ^{-1} (- \lambda x_1), \end{aligned}$$

or equivalently, $\beta x + \lambda x_1 = - (\beta - \lambda ) x_1$ as $\Vert x \Vert = \Vert x_1 \Vert .$ Thus, $x_1 = - x,$ so that $k_1 = k'$ and $w = \lambda k'.$ Now

$$\begin{aligned} \vert u - w \vert = \vert \alpha k - \lambda k' \vert \alpha k + \lambda k' = u + w. \end{aligned}$$

(3) implies (1): Finally, we assume that (3) holds. We show that $K = \text{ext}(B).$ Let $b \in B,$ such that $b \notin K.$ Then, there exists $x \in V_0$ with $\Vert x \Vert < 1,$ such that $b = x + b_0.$ As $b_0 = \frac{1}{2} k + \frac{1}{2} k' \notin \text{ext}(B),$ without any loss of generality we may assume that $x \ne 0.$ Put $x_1 = \Vert x \Vert ^{-1} x.$ Then, $b_1 \,{:}{=}\, x_1 + b_0 \in B$ and we have $x = \Vert x \Vert b_1 + (1 - \Vert x \Vert ) b_0$ is a proper convex combination in B. Thus, $b \notin \text{ext}(B).$ Therefore, $\text{ext}(B) \subset K.$

Conversely, let $k \in K$ and assume that $k \notin \text{ext}(B).$ Then, there are $b, c \in B$ and $\delta \in (0, 1),$ such that $k = \delta b + (1 - \delta ) c.$ Let $x_0, y, z \in V_0$ with $\Vert x_0 \Vert = 1,$ $\Vert y \Vert \le 1$ and $\Vert z \Vert \le 1,$ such that $k = x_0 + b_0,$ $b = y + b_0$ and $c = z + b_0.$ Then, $x_0 = \delta y + (1 - \delta ) z.$ Now

$$\begin{aligned} 1 = \Vert x_0 \Vert = \Vert \delta y + (1 - \delta ) z \Vert \le \delta \Vert y \Vert + (1 - \delta ) \Vert z \Vert \le \delta + (1 - \delta ) = 1 \end{aligned}$$

so that $\Vert y \Vert = 1 = \Vert z \Vert .$ Put $u = - x_0 + b_0,$ $v = x_0 + b_0$ and $w = \delta (y + b_0).$ Then, $u, v, w \in V^+ {\setminus } \lbrace 0 \rbrace$ with $u \in K$ and $v = u'.$ Thus, $\vert u - v \vert = u + v.$ In addition, $v - w = (1 - \delta ) (z + b_0) \ge 0$ but $\vert u - w \vert \ne u + w$ for $w \ne \gamma v'$ for any positive real number $\gamma$ which contradicts the assumption that the statement (3) holds. Thus, $K \subset \text{ext}(B).$ This proves (1). $\square$

Definition 6.8

Let (V, B) be a base normed space with a centre $b_0$ of B and assume that $K \,{:}{=}\, \lbrace k \in B: \Vert k - b_0 \Vert = 1 \rbrace = \text{ext}(B).$ Then, $(V, B, b_0)$ is said to be a central base normed space.

Now Theorem 6.7 yields in the following result.

Corollary 6.9

Let $(V, B, b_0)$ be a central base normed spaces, so that $V_0 = \lbrace v \in V: e(v) = 0 \rbrace$ is strictly convex and that V is isometrically order isomorphic to $V_0^{(1)}$ (see, Proposition 2.3).