A Representation Theorem for Archimedean Riesz Spaces

Abstract

In previous works, Buskes and the author have made use of representations of Archimedean Riesz spaces in terms of real-valued continuous functions defined on dense open subsets of a topological space in studying tensor products. These representations may be obtained from the Ogasawara–Maeda representation by means of restriction to the set on which representing functions are real-valued, rather than infinite. In this note, we show how to obtain such a representation as a simple consequence of the Krein–Kakutani representation of an order unit space. We conclude by studying the representation of Riesz homomorphisms in this setting.

1 Introduction

The literature already contains many representation theorems which describe Archimedean Riesz spaces as spaces of continuous functions on a topological space, either real-valued or, for the more general results, extended real-valued where the values $\pm \infty$ are allowed. Several results may be found, for example, in [3]. Once the Riesz spaces fail to contain an order unit, and the representing functions can take infinite values, then difficulties appear when trying to use these representations to construct a representation of the Riesz space tensor product of two spaces. In [1], we avoided these difficulties by working in the space of functions that are real-valued but only defined on a dense open subset. Such a representation may be obtained from the Ogasawara–Maeda representation and restricting each function to the dense open set on which it takes real values. In [5], we were able to use this approach to give a short proof of the existence and basic properties of the Riesz space tensor product. In view of the potential use of such representations, and the amount of work needed to construct the Ogasawara–Maeda representation, in this note we give a simple construction of such a representation. In section 2, we summarize the properties of the space in which the representation takes values. There turns out to be little that can be said about the whole space. In section 3, we construct the representation, using the Krein–Kakutani representation of an order unit space as a starting point. In view of the importance of Riesz homomorphisms in the study of Riesz spaces, we give a (partial) representation for them that is reminiscent of (and based on) the description of Riesz homomorphisms between spaces of continuous real-valued functions on compact Hausdorff spaces.

2 The Space $S(\Sigma )$

First a topological lemma.

Lemma 2.1

If $\Sigma$ is a topological space and U, V two dense open subsets of $\Sigma$, then $U\cap V$ is a dense open subset of $\Sigma$.

Proof

It is the density of $U\cap V$ that needs proof. If not, there is an open $A\subset \Sigma$ with $A\cap (U\cap V)=\emptyset$, i.e., $A\subseteq (\Sigma \setminus U)\cup (\Sigma \setminus V)$. Let $B=U\cap A$, which is open. As $B\cap (\Sigma \setminus U)=\emptyset$, $B\subseteq (\Sigma \setminus V)$. I.e., $B\cap V=\emptyset$, which contradicts V being dense in $\Sigma$. $\square$

Let $\Sigma$ be a topological space. By $S(\Sigma )$, we mean the set of all equivalence classes of real-valued continuous functions f defined on a dense open subset, $D(f)\subseteq \Sigma$, under the equivalence relation $f\sim g$ if f and g coincide on $D(f)\cap D(g)$, which is again dense in $\Sigma$.

Lemma 2.2

If $f\in S(\Sigma )$, there is an element of its equivalence class with maximal domain.

Proof

Let $U=\bigcup \{D(g):g\sim f\}$. If $g\sim f$ and $h\sim f$, then g and h coincide on $D(g)\cap D(h)$ so we may unambiguously define u on U with $u_{|D(g)}=g_{|D(g)}$. The function u is certainly continuous and U a dense open subset of $\Sigma$. It is clear that if $g\sim u$, then $g\sim f$ so that $D(g)\subseteq U=D(u)$. $\square$

Therefore, when picking a representative of an equivalence class in $S(\Sigma )$, we may unambiguously take it to be that one with maximal domain. Nevertheless, when constructing elements of $S(\Sigma )$, it is often convenient not to have to calculate the maximal domain. $S(\Sigma )$ certainly contains $C(\Sigma )$, the continuous real-valued functions on $\Sigma$. In particular the zero function, $\varvec{0}_\Sigma$, and the constantly one function, $\varvec{1}_\Sigma$,¹ are members of $S(\Sigma )$.

It is clear that $S(\Sigma )$ becomes a real vector space when we define, for $f,g\in S(\Sigma )$ and $\lambda \in \mathbb {R}$, $(f+g)(\sigma )=f(\sigma )+g(\sigma )$ on $D(f)\cap D(g)$ and $(\lambda f)(\sigma )=\lambda f(\sigma )$ on D(f). Its zero element is $\varvec{0}$. We may introduce an order in $S(\Sigma$ by defining $f\le g$ to mean that $f(\sigma )\le g(\sigma )$ for $\sigma \in D(f)\cap D(g)$. It is routine, using continuity of elements of $S(\Sigma )$ on their domains, to verify that these are all well-defined. It is similarly routine to check that this partial order makes $S(\Sigma )$ a partially ordered vector space. It is in fact a Riesz space, with $(f\vee g)(\sigma )=f(\sigma )\vee g(\sigma )$ and $(f\wedge g)(\sigma )=f(\sigma )\wedge g(\sigma )$ on $D(f)\cap D(g)$.

In the rest of this section, we study the kind of Riesz space that arises in this way.

Proposition 2.3

The Riesz space $S(\Sigma )$ is Archimedean

Proof

Suppose that $f,g\in S(\Sigma )$ with $nf\le g$ for all $n\in {\mathbb {N}}$. We have $nf(\sigma )\le g(\sigma )$ for all $n\in {\mathbb {N}}$ and $\sigma \in D(f)\cap D(g)$, so that $f(\sigma )\le 0$ for all $\sigma$ in the dense open set $D(f)\cap D(g)$, meaning that $f\le \varvec{0}$. $\square$

Ideals and bands are important objects in any Riesz space. Bands behave nicely in $S\Sigma )$ but ideals are not so amenable. We will see later that matters improve in the Riesz subspaces that arise from our representations.

Proposition 2.4

If $B\subseteq S(\Sigma )$, then the following conditions are equivalent:

• (1) B is a band.

• (2) B is a projection band.

• (3) There is an open subset U of $\Sigma$ such that $B=B_U$ where $B_U=\{f\in S(\Sigma ):f_{|U\cap D(f)}\equiv 0\}$.

Proof

Assume that B is a band, so that $B=B^{dd}$. For each $b\in B$, the set $V_b=\{\sigma \in D(b): b(\sigma )\ne 0\}$ is an open subset of the open set D(b) (here it helps to assume that each b has maximal domain.) Let $V=\bigcup _{b\in B} V_b$, and $U=(\Sigma \setminus V)^\circ$. Clearly, $B\subset B_U$. If $f\in B^d$ then $f\perp b$ for all $b\in B$, so that $f\equiv 0$ on $V_b$ and, hence, $f\equiv 0$ on V, i.e., $B^d\subseteq B_V$. Conversely if $f\in B_V$, then certainly $f\perp B$ so $B^d=B_V$. The same argument yields $B^{dd}=B_U$, but $B^{dd}=B$ as B is a band. If $f\in S(\Sigma )$ define $g,h\in S(\Sigma )$, both with domain $D(f)\cap (U\cup V)$, as follows.

$$\begin{aligned} g_{|D(f)\cap V}\equiv f_{D(f)\cap V}&\quad g_{|D(f)\cap U}\equiv 0,\\ h_{|D(f)\cap U}\equiv f_{D(f)\cap U}&\quad h_{D(f))\cap V}\equiv 0.\\ \end{aligned}$$

Then $f=g+h$, with $g\in B$ and $h\in B^d$. This suffices to establish all the equivalences. $\square$

Proposition 2.5

Every disjoint family $(f_\gamma )_{\gamma \in \Gamma }$ in $S(\Sigma )_+$ has a supremum.

Proof

With the notation of the preceding proposition, let $V=\bigcup _{\gamma \in \Gamma }V_{f_\gamma }$ and $U=(\Sigma \setminus V)^\circ$, so that $U\cup V$ is a dense open subset of $\Sigma$. The function f, with domain $U\cup V$, with

$$\begin{aligned} f_{|V_{f_\gamma }}=f_{\gamma |V_{f\gamma }}\ \forall \gamma \in \Gamma \quad f_{|U}\equiv 0,\end{aligned}$$

is the desired supremum. Note that f is well-defined as the sets $V_{f_\gamma }$ are disjoint. $\square$

So, $S(\Sigma )$ has the projection property and is laterally complete. If $\Sigma$ is Stonean, then $S(\Sigma )$ is linearly order isomorphic to $C^\infty (\Sigma )$ which is universally complete. That fails to be true in general, as $S(\Sigma )$ need not be uniformly complete.

Example 2.6

Define a sequence $(f_n)_{n=1}^\infty$ in S([0, 1]) by

$$\begin{aligned}f_n=2^{-2n}\sum _{k=1}^{2^n-1}k\chi _{(k2^{-n},(k+1)2^{-n})},\end{aligned}$$

where, for an open set U, $\chi _U$ takes the value 1 on U and 0 on $([0,1]\setminus U)^\circ$. If S([0, 1]) were uniformly complete then $\sum _{n=1}^\infty f_n$ ought to converge uniformly relative to $\varvec{1}$. The limit function would have to be undefined wherever one of the $f_n$ has a (potential) jump discontinuity, so at all integer multiples of all $2^{-n}$, a dense subset of [0, 1]. That makes it impossible for there to be a limit with dense open domain.

3 A Representation of Archimedean Riesz Spaces

We start with the case that the Riesz space contains a weak order unit.

Lemma 3.1

If E is a Riesz space and $x,y,z\in E_+$ then

$$\begin{aligned}(x+y)\wedge z=(x\wedge z+y\wedge z)\wedge z.\end{aligned}$$

Proof

Certainly $(x+y)\wedge z\ge (x\wedge z+y\wedge z)\wedge z$. As $0\le (x+y)\wedge z\le x+y$, the Riesz Decomposition Property gives us $x_1,y_1$ with $0\le x_1\le x$, $0\le y_1\le y$ and $x_1+y_1=(x+y)\wedge z$. As $x_1,y_1, x_1+y_1\le z$, we actually have $(x+y)\wedge z=x_1+y_1\le (x\wedge z+y\wedge z)\wedge z$. $\square$

Lemma 3.2

Suppose that $\Omega$ is a compact Hausdorff space, E a supremum norm dense Riesz subspace of $C(\Omega )$, containing the constants. If U is a non-empty open set in $\Omega$ containing $\omega \in \Omega$ then there is $h\in E_+$ with $\varvec{0}\le h\le \varvec{1}$, $h(\omega )=1$ and $h_{|\Omega \setminus U}\equiv 0$.

Proof

By Urysohn’s lemma there is $f\in C(\Omega )$ with $-\frac{1}{2}\varvec{1}\le f\le \frac{3}{2}\varvec{1}$, $f(\omega )=\frac{3}{2}$ and f constantly $-\frac{1}{2}$ on $\Omega \setminus U$. As E is dense in $C(\Omega )$ for the supremum norm, there is $g\in E$ with $\Vert g-f\Vert _\infty \le \frac{1}{2}$. This means that $g(\sigma )\le 0$ for $\sigma \in \Omega \setminus U$, $g(\sigma )\le \frac{3}{2}$ for $\sigma \in U$ and $g(\omega )\ge 1$. Now take $h=g^+\wedge \varvec{1}$. $\square$

Theorem 3.3

If E is an Archimedean Riesz space with a weak order unit u, there is a compact Hausdorff space $\Omega$ and a lattice isomorphism $\pi :E\rightarrow S(\Omega )$ such that $\pi (u)=\varvec{1}$ and the image of the principal ideal $E_u$ is a Riesz subspace of $C(\Omega )$ that is dense for the supremum norm on $C(\Omega )$.

Proof

Take as a starting point the Krein–Kakutani representation, $x\mapsto x^\wedge$² of $E_u$ on some compact Hausdorff space $\Omega$. This maps u to $\varvec{1}$ and has supremum norm dense image in $C(\Omega )$. All we need to do is extend this representation to the whole of E. We start by extending it to $E_+$. If $x\in E_+$, then for all $n\in {\mathbb {N}}$, $x\wedge nu\in E_u$ so we consider $(x\wedge nu)^\wedge \in C(\Omega )$. As $x\wedge (n+1)u\ge x\wedge nu$, for any $\omega \in \Omega$, we have $(x\wedge (n+1)u)^\wedge (\omega )\ge (x\wedge nu)^\wedge (\omega )$. If $(x\wedge nu)^\wedge (\omega )<n$, then $(x\wedge (n+1)u)^\wedge (\omega )=(x\wedge nu)^\wedge (\omega )$, as $(x\wedge (n+1)u)\wedge nu=x\wedge nu$, so that $(x\wedge (n+1)u)^\wedge (\omega )\wedge n=(x\wedge nu)^\wedge (\omega )$, bearing in mind that $u^\wedge =\varvec{1}$. Thus, there is an open subset U of $\Omega$ on which $(x\wedge nu)^\wedge (\omega )$ is eventually constant. On that set $\omega \mapsto \bigvee _{n\in {\mathbb {N}}} (x\wedge nu)^\wedge (\omega )$ is continuous. We claim that the complement of this set, the closed set $G=\{\omega \in \Omega : (x\wedge nu)^\wedge (\omega )=n\ \forall n\in {\mathbb {N}}\}$ has an empty interior so that U is dense. We may then define $\pi (x)\in S(\Omega )$ to have the value $\bigvee _{n\in {\mathbb {N}}} (x\wedge nu)^\wedge (\omega )$ for $\omega$ in its domain U.

To see that G has empty interior, suppose to the contrary that $V\subset G$, V is open and $\sigma \in V$. By Lemma 3.2 there is $y\in (E_u)_+$ with $y^\wedge (\omega )=0$ for $\omega \in \Omega \setminus V$, so certainly on $\Omega \setminus G$, $y^\wedge (\omega )\le 1$ for $\omega \in G$ and $y^\wedge (\sigma )=1$. For each $n\in {\mathbb {N}}$, we have $ny^\wedge (\omega )=0\le (x\wedge nu)^\wedge (\omega )$ for $\omega \in \Omega \setminus G$, and $(ny)^\wedge (\omega )\le n=(x\wedge nu)^\wedge (\omega )$ for $\omega \in G$. I.e., $ny^\wedge \le (x\wedge nu)^\wedge$. Hence $ny\le (x\wedge nu)\le x$ for all $n\in {\mathbb {N}}$. As $y^\wedge (\sigma )=1$, $y^\wedge \not \le 0$ so that $y\not \le 0$, contradicting E being Archimedean.

If $x,y\in E_+$ we claim that $\pi (x+y)=\pi (x)+\pi (y)$. To verify this it suffices to check for equality of values at each point $\omega$ of the dense open set $D(\pi (x))\cap D(\pi (y))\cap D(\pi (x+y))$. Choose $n\in {\mathbb {N}}$ large enough that $\pi (x)(\omega )=(x\wedge mu)^\wedge (\omega )$, $\pi (y)(\omega )=(y\wedge mu)^\wedge (\omega )$ and $\pi (x+y)(\omega )=((x+y)\wedge mu)^\wedge (\omega )$ for all $m\ge n$ and $\pi (x)(\omega ), \pi (y)(\omega ),\pi (x+y)(\omega )\le n$. Using Lemma 3.1, we then have

$$\begin{aligned} \pi (x+y)(\omega )&=((x+y)\wedge 2nu)^\wedge (\omega )\\&=((x\wedge 2nu+y\wedge 2nu)\wedge 2nu)^\wedge (\omega )\\&=(x\wedge 2nu+y\wedge 2nu)^\wedge (\omega )\wedge (2nu)^\wedge (\omega )\\&=[(x\wedge 2nu)^\wedge (\omega )+(y\wedge 2nu)^\wedge (\omega )]\wedge 2n\\&=[\pi (x)(\omega )+\pi (y)(\omega )]\wedge 2n\\&=\pi (x)(\omega )+\pi (y)(\omega ). \end{aligned}$$

As it is clear that if $x\in E_+$ and $\lambda \in \mathbb {R}_+$ then $\pi (\lambda x)=\lambda \pi (x)$, we may unambiguously extend $\pi$ to a linear map of E into $S(\Omega )$ by setting $\pi (x-y)=\pi (x)-\pi (y)$ for $x,y\in E_+$.

Our next step is to show that $\pi$ is a lattice homomorphism. It suffices ([3], Definition 18.1) to prove that if $x\wedge y=0$ then $\pi (x)\wedge \pi (y)=\varvec{0}$ (the zero element of $S(\Omega )$). To prove this it suffices to check that $\pi (x)(\omega )\wedge \pi (y)(\omega )=0$ for all $\omega \in D(\pi (x))\cap D(\pi (y))$. Choose $n\in {\mathbb {N}}$ large enough that $(x\wedge nu)^\wedge (\omega )=\pi (x)(\omega )$ and $(y\wedge nu)^\wedge (\omega )=\pi (y)(\omega )$. As $(x\wedge nu)\wedge (y\wedge nu)=0$ and $z\mapsto z^\wedge$ is a lattice homomorphism on $E_u$, we have

$$\begin{aligned}\pi (x)(\omega )\wedge \pi (y)(\omega )= (x\wedge nu)^\wedge (\omega )\wedge (y\wedge nu)^\wedge (\omega ) =[(x\wedge nu)\wedge (y\wedge nu)]^\wedge (\omega )=0,\end{aligned}$$

completing this step.

The final task is to show that $\pi$ is one-to-one. If $\pi (x)=\varvec{0}$, then we now know that $\pi (x^\pm )=\pi (x)^\pm =\varvec{0}$, so it suffices to show that if $x\in E_+$ and $\pi (x)=\varvec{0}$ then $x=0$. For all $\omega \in D(\pi (x))$ we have $\pi (x)(\omega )=0$ so that $(x\wedge nu)^\wedge (\omega )=0$ for all $n\in {\mathbb {N}}$. As $z\mapsto z^\wedge$ is one-to-one on $E_u$, this means that $x\wedge nu=0$ for all $n\in {\mathbb {N}}$. But u is a weak order unit, so this forces $x=0$ as required. $\square$

It will be evident that in general our space $\Omega$ will be much smaller than the Stonean space involved in the Ogasawara–Maeda representation. It would be pleasant if $\Omega$ were minimal in some sense. Such a result seems unlikely as there is no uniqueness about $\Omega$. Even in such a simple case as C([0, 1]) we could take $u=\varvec{1}_{[0,1]}$, when $\Omega =[0,1]$, or $u(t)=t$, when $\Omega$ is the Stone–Cech compactification of (0, 1].

It is worth pointing out one simple property of this representation. The proof is a trivial consequence of the definition of $\pi$.

Proposition 3.4

Let E be an Archimedean Riesz space with weak order unit u and $\pi :E\rightarrow S(\Omega )$ be the representation of Theorem 3.3. If $x,y\in E$ and $|x|\le |y|$ then $D(\pi (y))\subseteq D(\pi (x))$.

Proposition 3.5

Let E be an Archimedean vector lattice with weak order unit u and let $\pi :E\rightarrow S(\Omega )$ be the representation of Theorem 3.3. If $v\in E_+$, then v is a weak order unit if and only if $\pi (v)$ is strictly positive on a dense subset of $D(\pi (v))$.

Proof

If not, then there is a non-empty open set $U\subset D(\pi (v))$ on which $\pi (v)$ vanishes. By Lemma 3.2, there is $0\ne x\in E_+$ with $\pi (x)$ vanishing off U. Then, $\pi (x)\perp \pi (v)$ so that $x\perp v$, contradicting v being a weak order unit. The converse implication is obvious. $\square$

If E is an Archimedean Riesz space but does not have a weak order unit, then (unless it is the zero space) it has a disjoint order basis by [3], Theorem 28.5.

Corollary 3.6

If E is an Archimedean Riesz space with disjoint order basis $(u_\lambda )_{\lambda \in \Lambda }$, then there is a disjoint union $\Omega$ of compact Hausdorff spaces $(\Omega _\lambda )_{\lambda \in \Lambda }$ and a one-to-one lattice homomorphism $\pi :E\rightarrow S(\Omega )$ that maps each $u_\lambda$ to the characteristic function of $\Omega _\lambda$ and with the image of $E_{u_\lambda }$ being dense in the space of functions that are continuous on the whole of $\Omega _\lambda$ and vanish on all $\Omega _\mu$ for $\mu \ne \lambda$.

Proof

The only non-trivial point is that the map is one-to-one. It suffices to prove that if $x\ge 0$ and $\pi (x)=\varvec{0}$ then $x=0$. In this case, $\pi (x)$ vanishes on (a dense open subset of) each $\Omega _\lambda$, so that $x\wedge u_\lambda =0$ for all $\lambda \in \Lambda$ and hence $x=0$. $\square$

We turn now to identifying important subsets of an Archimedean Riesz space in its representation. We saw in Sect. 2 that bands in $S(\Sigma )$ are particularly well-behaved, so will start with them. For simplicity of statements, we restrict our results to the case that there is a weak order unit. The extension to the general case is routine.

Proposition 3.7

Let E be an Archimedean Riesz space with weak order unit u, and let $x\mapsto x^\wedge$ be its representation in $S(\Sigma )$ as in Theorem 3.3. If $B\subseteq B$ then the following conditions are equivalent:

• (1) B is a band.

• (2) There is an open subset U of $\Sigma$ such that $B=B_U$ where $B_U=\{x\in E:{x^\wedge }_{|U\cap D(x^\wedge )}\equiv 0\}$.

Proof

For (1) implies (2) repeat, essentially, the corresponding proof from Proposition 2.4. If (2) holds, by Lemma 3.2, for each $\sigma \in U$, there is $x_\sigma \in E_u$ with $(x_\sigma )^\wedge (\sigma )=1$ and $(x_\sigma )^\wedge$ vanishing on $\Sigma \setminus U$. Thus, $x_\sigma \perp B_U$ and $x_\sigma \in (B_U)^d$. If $y\perp x_\sigma$, then $y^\wedge (\sigma )=0$, so if $y\in (B_U)^{dd}$ then $y^\wedge$ vanishes on U, i.e., $(B_U)^{dd}\subseteq B_U$, showing that $B_U$ is a band. $\square$

There is, of course, no third equivalence of B being a projection band. There seems little prospect of giving a simple description of ideals in general. By analogy with the C(X) situation, we might have hoped that the functions in $S(\Sigma )$ that vanish on a fixed set form an ideal. Even when we amend this naive approach to take the actual domain into account, it fails for $S(\Sigma )$.

Example 3.8

In S([0, 1]), the set $J=\{f\in S([0,1]):\frac{1}{2}\in D(f)\text { and }f(\frac{1}{2})=0\}$ is not a lattice ideal, because $\varvec{1}_{[0,1]}\in J$, but if we define f to be 1 on $[0,\frac{1}{2})$ and 0 on $(\frac{1}{2},1]$, then $f\not \in J$ yet $0\le f\le \varvec{1}_{[0,1]}$.

Nevertheless, we are able to identify some ideals in our representations. Recall that an ideal J in a Riesz space E is prime if $x,y\in E$ with $x\wedge y=0$ implies that either $x\in J$ or $y\in J$.

Proposition 3.9

Let E be an Archimedean Riesz space with weak order unit u, and let $x\mapsto x^\wedge$ be its representation in $S(\Sigma )$ as in Theorem 3.3. If $\sigma \in \Sigma$, then the set $J=\{x\in E:\sigma \in D(x^\wedge )\text { , }x^\wedge (\sigma )=0\}$ is a prime ideal in E.

Proof

Suppose that $x\in J$ and that $|y|\le |x|$, then $\sigma \in D(x^\wedge )\subseteq D(y^\wedge )$, by Proposition 3.4, and $|y^\wedge (\sigma )|=|y|^\wedge (\sigma )\le |x|^\wedge (\sigma )=|x^\wedge (\sigma )|=0$, so that $y\in J$ and J is an ideal. If $x\wedge y=0$ then certainly $x,y\ge 0$. If $\sigma \in D(x^\wedge )\cap D(y^\wedge )$ then $x^\wedge (\sigma )$ and $y^\wedge (\sigma )$ are real numbers with $0=(x\wedge y)^\wedge (\sigma )=x^\wedge (\sigma )\wedge y^\wedge (\sigma )$, so certainly either x or y lies in J. If, for example, $\sigma \in D(x^\wedge )\setminus D(y^\wedge )$ and $x\not \in J$ then $x^\wedge (\sigma )>0$. But $(y\wedge u)^\wedge (\sigma )=1$ so that

$$\begin{aligned}0=(x\wedge y)^\wedge (\sigma )\ge \big (x\wedge (y\wedge u)\big )^\wedge (\sigma )=x^\wedge (\sigma )\wedge 1>0,\end{aligned}$$

which is a contradiction. A similar argument deals with the case that $\sigma \not \in D(x^\wedge )\cup D(y^\wedge )$. $\square$

Corollary 3.10

Let E be an A4chimedean Riesz space with weak order unit u, and let $x\mapsto x^\wedge$ be its representation in $S(\Sigma )$ as in Theorem 3.3. If $\emptyset \ne A\subset \Sigma$ then $J=\{x\in E:A\subseteq D(x^\wedge )\text { and }{x^\wedge }_{|A}\equiv 0\}$ is an ideal in E.

Slightly unexpected is:

Proposition 3.11

Let E be an A4chimedean Riesz space with weak order unit u, and let $x\mapsto x^\wedge$ be its representation in $S(\Sigma )$ as in Theorem 3.3. If $\sigma \in \Sigma$ then $J=\{x\in E:\sigma \in D(x^\wedge )\}$ is a prime ideal in E.

Proof

That J is an ideal follows immediately from Proposition 3.4. If $x, y\ge 0$ and neither lies in J, then $(x\wedge nu)^\wedge (\sigma )=(y\wedge nu)^\wedge (\sigma )=n$ for all $n\in {\mathbb {N}}$. Hence $((x\wedge y)\wedge nu)^\wedge (\sigma )=n$ for all $n\in {\mathbb {N}}$ and $\sigma \not \in D((x\wedge y)^\wedge )$, i.e., $x\wedge y\not \in J$.

$\square$

Note that in this last result, J need not be a proper ideal, e.g., if E has a strong order unit.

4 Representation of Riesz Homomorphisms

In this section, we restrict ourselves to Archimedean Riesz spaces E and F having weak order units u and v, respectively. We fix representations of them in $S(\Sigma )$ and $S(\Omega )$, respectively, as in Theorem 3.3, both of which we will denote by $x\mapsto x^\wedge$. Our aim is to give a concrete description of a Riesz homomorphism $T:E\rightarrow F$. There is no difficulty if Tu is a weak order unit for F. Indeed if $Tu=v$ the proof is easy.

Theorem 4.1

If $Tu=v$ then there is a continuous map $\pi :\Omega \rightarrow \Sigma$ such that $(Tx)^\wedge (\omega )=x^\wedge (\pi \omega )$ for all $\omega \in D((Tx)^\wedge )$.

Proof

The restriction of T to $E_u$ gives us, via the representations, a Riesz homomorphism of a dense Riesz subspace of $C(\Sigma )$ into $C(\Omega )$. This extends by continuity to the whole of $C(\Sigma )$, so Theorem 3.2.12 of [4] gives us a continuous $\pi :\Omega \rightarrow \Sigma$ such that $(Tx)^\wedge (\omega )=x^\wedge (\pi \omega )$ for all $x\in E_u$ and all $\omega \in \Omega$. Now take $x\in E^+$ and $\omega \in D((Tx)^\wedge )$. Choose $n\in {\mathbb {N}}$ large enough that $(Tx)^\wedge (\omega )=(Tx\wedge mv)^\wedge (\omega )$ for all $m\ge n$. Then, we have

$$\begin{aligned} (Tx)^\wedge (\omega )&=\bigvee _{m\in {\mathbb {N}}}(Tx\wedge mv)^\wedge (\omega )\\&=\bigvee _{m\in {\mathbb {N}}}(Tx\wedge mTu)^\wedge (\omega )\\&=\bigvee _{m\in {\mathbb {N}}}(T(x\wedge mu))^\wedge (\omega )\\&=\bigvee _{m\in {\mathbb {N}}}(x\wedge mu)^\wedge (\pi \omega )\\&=x^\wedge (\pi \omega ), \end{aligned}$$

which also points out that we necessarily have $\pi \omega \in D(x^\wedge )$. Applying this argument to $x^+$ and $x^-$ separately and subtracting gives the general result.

$\square$

If Tu were in the ideal $F_v$ as well as being a weak order unit we could use Theorem 3.2.10 of [4] as a starting point for the following proof; however, it seems necessary to reproduce parts of its proof for the general case.

Theorem 4.2

It Tu is a weak order unit for F then there are $\alpha \in S(\Omega )_+$, for which $W=\{\omega \in D(\alpha ):\alpha (\omega )>0\}$ is a dense subset of $\Omega$, and a continuous map $\pi :W\rightarrow \Sigma$ such that

$$\begin{aligned}(Tx)^\wedge (\omega )=\alpha (\omega )x^\wedge (\pi \omega )\text { if }(\omega \in D((Tx)^\wedge )\cap W).\end{aligned}$$

If $Tu\in F_v$, then we actually have $\alpha \in C(\Omega )_+$.

Proof

Take $\alpha =(Tu)^\wedge$. The set W is dense as Tu is a weak order unit, using Proposition 3.5. If $\omega \in W$ consider the map $\phi :E_u\rightarrow \mathbb {R}$, defined by $\phi (x)=(Tx)^\wedge (\omega )$, which is defined in view of Proposition 3.4. As T is a lattice homomorphism, $\phi$ is a real-valued lattice homomorphism on $E_u$. Hence $\psi :x^\wedge \mapsto \phi (x)$ is a real-valued lattice homomorphism on $(E_u)^\wedge \subseteq C(\Sigma )$, which extends by continuity to the whole of $C(\Sigma )$, so is a multiple of evaluation at some point of $\Sigma$. Considering the action on u, we see that the multiple is precisely $\alpha (\omega )$ and we denote the corresponding point of $\Sigma$ by $\pi (\omega )$. If we had $\phi (x)=\alpha (\omega )x^\wedge (\sigma _1)=\alpha (\omega )x^\wedge (\sigma _2)$ for all $x\in E_u$ then, as $\alpha (\omega )>0$, $x^\wedge (\sigma _1)=x^\wedge (\sigma _2)$, so that $\sigma _1=\sigma _2$, and $\pi$ is unambiguously defined. To see that $\pi$ is continuous, suppose to the contrary that $(\omega _\gamma )$ is a net in W converging to $\omega \in W$, but that $\pi (\omega _\gamma )\not \rightarrow \pi (\omega )$. By considering a subnet, we may assume that there is an open set $U\subset \Sigma$ with $\pi (\omega )\in U$ and none of the $\pi (\omega _\gamma )$ lying in U. By Lemma 3.2 there is $x\in E_u$ with $x^\wedge (\pi (\omega ))=1$ and $x^\wedge$ vanishing on $\Sigma \setminus U$. Now $0=\alpha (\omega _\gamma )x^\wedge (\pi (\omega _\gamma ))=(Tx)^\wedge (\omega _\gamma )\rightarrow (Tx)^\wedge (\omega )=\alpha (\omega )x^\wedge (\pi (\omega ))=\alpha (\omega )>0$, which is a contradiction.

This gives us the desired description of T on $E_u$. It remains to extend it to the whole of E. Once that is done for $E_+$, the extension to E is routine. If $x\in E_+$, $\omega \in D((Tx)^\wedge )\cap W$, so that $\alpha (\omega )>0$, then choose n such that $n\alpha (\omega )=n (Tu)^\wedge (\omega )>(Tx)^\wedge (\omega )$. Then,

$$\begin{aligned} (Tx)^\wedge (\omega )&=(Tx)^\wedge (\omega )\wedge n(Tu)^\wedge (\omega )\\&=(Tx\wedge nTu)^\wedge (\omega )\\&=(T(x\wedge nu))^\wedge (\omega )\\&=\alpha (\omega )(x\wedge nu)^\wedge (\pi (\omega ))&\text {(as }x\wedge nu\in E_u)\\&=\alpha (\omega )[x^\wedge (\pi (\omega ))\wedge n u^\wedge (\pi (\omega ))]\\&=[\alpha (\omega )x^\wedge (\pi (\omega ))]\wedge [n\alpha (\omega ) u^\wedge (\pi (\omega ))]\\&=[\alpha (\omega )x^\wedge (\pi (\omega ))]\wedge n\alpha (\omega )&\text {(as }u^\wedge =\varvec{1}_\Sigma ). \end{aligned}$$

But $(Tx)^\wedge (\omega )<n\alpha (\omega )$, so that $(Tx)^\wedge (\omega )=\alpha (\omega )x^\wedge (\pi (\omega ))$ as claimed. $\square$

If Tu is not a weak order unit for F, then we would still have the same description of T as in Theorem 4.2, but now only on a set W that is not dense. If we had a complete analog of Theorem 3.2.10 of [4], then we would have all $(Tx)^\wedge$ vanishing on (the interior of) the complement of W. That is not the case in our setting.

Example 4.3

We take as the domain, E, of our operator the linear span of C(0, 1]) and the extended real-valued function $\phi (t)=1/|t-\frac{1}{2}|$. It is not difficult to see that this is a linear space under the usual operations. What is not quite obvious is that it is a Riesz space under the pointwise operations. By 2.2.1 of [2], it suffices to prove that if $f+\alpha \phi \in E$, where $f\in C([0,1])$ and $\alpha \in \mathbb {R}$, then the pointwise positive part function $t\mapsto \big (f(t)+\alpha \phi (t)\big )^+$ lies in E. If $\alpha \le 0$ then the (extended real-valued) continuous function $f+\alpha \phi$ is bounded above by a real number, so its positive part lies in $C([0,1])\subset E$. If $\alpha >0$ then let $M=\Vert f\Vert _\infty$ and choose $\delta >0$ such that $\alpha \phi (t)\ge M$ if $|t-\frac{1}{2}|<\delta$. Outside the interval $(\frac{1}{2}-\delta ,\frac{1}{2}+\delta )$, the continuous function $(f+\alpha \phi )^+-\alpha \phi$ is continuous and bounded by 3M. In the intervals $(\frac{1}{2}-\delta ,\frac{1}{2})\cup (\frac{1}{2},\frac{1}{2}+\delta )$, $(f+\alpha \phi )^+-\alpha \phi$ coincides with f, so certainly extends continuously to $(\frac{1}{2}-\delta ,\frac{1}{2}+\delta )$. This shows that $(f+\alpha \phi )^+\in E$.

Note that for all $f+\alpha \phi \in E$, we have

$$\begin{aligned}\lim _{t\rightarrow \frac{1}{2}} |t-1/2|(f(t)+\alpha \phi (t))=\lim _{t\rightarrow \frac{1}{2}}|t-1/2|f(t)+\lim _{t\rightarrow \frac{1}{2}}|t-1/2|\alpha /|t-1/2|=\alpha .\end{aligned}$$

Define $T:E\rightarrow C([0,2])$ by

$$\begin{aligned}T(h)(t)={\left\{ \begin{array}{ll}|t-\frac{1}{2}|h(t)&{}(t\in [0,\frac{1}{2}))\\ \lim _{t\rightarrow \frac{1}{2}}|t-\frac{1}{2}|h(t)&{}(t\in [\frac{1}{2},\frac{3}{2}])\\ (|t-\frac{3}{2}|h(t-1)|&{}(t\in (\frac{3}{2},2]). \end{array}\right. } \end{aligned}$$

If we carry out the construction of Theorem 4.2, using the weak order units $\varvec{1}_{0,1]}$ and $\varvec{1}_{[0,2]}$, then

$$\begin{aligned}\alpha (t)={\left\{ \begin{array}{ll}\frac{1}{2}-t&{}(t\in [0,\frac{1}{2}))\\ 0&{}(t\in [\frac{1}{2},\frac{3}{2}])\\ t-\frac{3}{2}&{}(t\in (\frac{3}{2},2]), \end{array}\right. }\end{aligned}$$

and

$$\begin{aligned}\pi (t)={\left\{ \begin{array}{ll}t&{}(t\in [0,\frac{1}{2}))\\ t-1&{}(t\in [\frac{3}{2},2]). \end{array}\right. }\end{aligned}$$

Note that in this example, we can even extend $\pi$ continuously to the whole of [0, 2] by giving it the value $\frac{1}{2}$ on $[\frac{1}{2},\frac{3}{2}]$. In spite of $\alpha$ vanishing on $[\frac{1}{2},\frac{3}{2}]$, $T(\phi )$ is constantly one there.