Proof By conjugating with translations if necessary, we may assume without loss of generality that $0$ is an element of X.

Let $w\in F$ lie outside L. Since F is a compact Hausdorff space, it is Tychonoff. So, there exists a continuous function $s \colon F \to \mathbb {R}$ that separates w from L; in fact, we may choose s to take the value $0$ on all of L, the value $1$ at w, and values in $[0, 1]$ everywhere else.

Choose a continuous function $\Gamma \colon [0,1] \times [-\infty , \infty ] \to [-\infty , \infty ]$ with the following properties. We require that, for each $y \in [0,1]$ , the map $\Gamma (y, \cdot )$ is an order isomorphism of $[-\infty , \infty ]$ and agrees with the identity map outside the interval $[1, 2]$ . We furthermore require that $\Gamma (0, \cdot )$ is the identity map, and that $\Gamma (1, \cdot )$ is non-affine in the interval $[1,2]$ .

Let g be in V. The map $f \colon F \to \mathbb {R}$ defined by $f(p) := \Gamma (s(p), g(p))$ is continuous. Let $\mu $ be an affine dependency carried by F. The support of $\mu $ is a subset of L, and on this set the function s takes the value $0$ . Using then that $\Gamma (0, \cdot )$ is the identity map, we get that $\mu (f) = \mu (g) = 0$ . Applying Lemma 2.2, we see that there is an element of V agreeing with f on F. We denote this element $\phi g$ . This defines a map $\phi $ from V to itself.

For each $y\in [0, 1]$ , define the function $\bar \Gamma (y, \cdot )$ to be the inverse of $\Gamma (y, \cdot )$ . Let $y_n$ be a sequence in $[0, 1]$ converging to y in the same set, and let $x_n$ be a sequence in $[-\infty , \infty ]$ converging to $x\in [-\infty , \infty ]$ . Writing $z_n := \bar \Gamma (y_n, x_n)$ , we have $x_n = \Gamma (y_n, z_n)$ , for all $n\in \mathbb {N}$ . If any subsequence of $z_n$ converges to a limit $z\in [-\infty , \infty ]$ , then $x = \Gamma (y, z)$ , which is equivalent to $z = \bar \Gamma (y, x)$ . We conclude that $z_n$ converges to $\bar \Gamma (y, x)$ , which shows that $\bar \Gamma $ is continuous. Note that $\bar \Gamma $ also satisfies all the other assumptions we made on $\Gamma $ .

Analogously to how we defined $\phi $ , we can define a map $\bar \phi \colon V \to V$ such that

$$ \begin{align*} \bar\phi g(p) = \bar\Gamma\big(s(p), g(p)\big), \qquad\text{for all } g\in V \text{ and } p\in F. \end{align*} $$

We deduce that

$$ \begin{align*} (\bar\phi \circ \phi) g(p) = g(p), \qquad\text{for all } g\in V \text{ and } p\in F. \end{align*} $$

Since, for each $g\in V$ , both g and $(\bar \phi \circ \phi ) g$ are continuous and affine on K, the barycenter formula holds for both, and we conclude that the two functions agree on all of K, and hence represent the same element of V. Similar reasoning shows that $(\phi \circ \bar \phi ) g$ always equals g. We have shown that the maps $\bar \phi $ and $\phi $ are inverse to one another. It is clear that both $\phi $ and $\bar \phi $ are order preserving.

Let $f\in X$ . Since X is directed downward and contains $0$ , there is an element g of X such that $g \le 0$ and $g \le f$ . Note that g is a fixed point of $\phi $ . It follows that $g \le \phi f$ , and we deduce that $\phi f$ is in X, since this set is upper. This establishes that $\phi $ leaves the set X invariant. A similar argument shows that the same is true for $\phi ^{-1} = \bar \phi $ . The restriction of $\phi $ to X is the order isomorphism we require.

It remains to show that this restriction is not affine. To prove this, observe that u and $2u$ are in X, and that $\phi \big ((1-\lambda ) u + 2 \lambda u \big )(w) = \Gamma (1, 1 + \lambda )$ is not affine in $\lambda $ in the range $\lambda \in [0,1]$ .

Proof Since I is an ideal, it is directed upward, and so the map $I \to V$ defined by $\alpha \mapsto \alpha - g$ is a net in V. Observe that this net is non-decreasing. Hence, it converges pointwise on K to its supremum $\sup I - g$ . Moreover, the infimum of $\alpha - g$ over K converges to the infimum of $\sup I - g$ , which is positive, since this function is positive and lower semicontinuous and has a compact domain (see Lemma 2.1 of [Reference Walsh32]). So, we may choose $\alpha \in I$ large enough that $\alpha - g$ is positive on all of K, in other words, $\alpha> g$ . Since I is lower, we deduce that $g \in I$ .

Proof The set I is clearly lower and is non-empty because it contains $0$ . It is proper because it does not contain $\lambda u$ . It is also directed upward, since $L_{y, \lambda }$ is a lower semicontinuous affine function on K – see the proof of Corollary I.1.4 of Alfsen [Reference Alfsen3]. Again from the same corollary, there is a non-decreasing net of elements of I converging pointwise on K to $L_{y, \lambda }$ . Therefore, $\sup I$ is greater than or equal to $L_{y, \lambda }$ . The opposite inequality is obvious.

Proof Let $g \in I_1$ . For each $y\in K$ , we have that $g(y)$ is finite and less than or equal to $\sup I_1(y)$ , and hence strictly less than $\sup I_2(y)$ . So, by Lemma 4.1, g is in $I_2$ .

Proof Assume that $y_1 = y_2 =: y$ . Take a positive real number $\lambda $ strictly greater than both $\lambda _1$ and $\lambda _2$ . By Lemma 4.2, there is a proper ideal in $\mathcal {I}_{y, \lambda }$ , and by Lemma 4.3, this ideal contains both $I_1$ and $I_2$ .

Now assume that $y_1$ and $y_2$ differ, and that there is an ideal I containing both $I_1$ and $I_2$ . So, $\sup I$ is greater than or equal to both $\sup I_1 = L_{y_1, \lambda _1}$ and $\sup I_2 = L_{y_2, \lambda _2}$ , and hence takes the value $\infty $ everywhere on the state space K. This shows that I is the whole of C, and so not proper.

Proof Assume that I is in $\mathcal {I}_{y, \lambda }$ , with y a pure state and $\lambda \ge 0$ , and let $I_1$ and $I_2$ be proper ideals containing I. So, both $\sup I_1$ and $\sup I_2$ are greater than or equal to $\sup I$ . It follows that $I_1 \in \mathcal {I}_{y, \lambda _1}$ and $I_2 \in \mathcal {I}_{y, \lambda _2}$ , where $\lambda _1$ and $\lambda _2$ are nonnegative real numbers greater than or equal to $\lambda $ . By Lemma 4.4, there exists a proper ideal containing both $I_1$ and $I_2$ . We have shown that $I^\uparrow $ is directed upward.

Now assume that I is a proper ideal such that $I^\uparrow $ is directed upward. Let $y_1$ and $y_2$ be pure states of K where $\sup I$ takes nonnegative finite values $\lambda _1$ and $\lambda _2$ , respectively. Take $\lambda ^{\prime }_1> \lambda _1$ and $\lambda ^{\prime }_2> \lambda _2$ . By Lemma 4.2, there exist proper ideals $I_1 \in \mathcal {I}_{y_1, \lambda ^{\prime }_1}$ and $I_2 \in \mathcal {I}_{y_2, \lambda ^{\prime }_2}$ . Both of these ideals contain I, by Lemma 4.3, and so are in $I^\uparrow $ . Since $I^\uparrow $ is directed upward, it has as an element a proper ideal containing both $I_1$ and $I_2$ . By Lemma 4.4, we get that $y_1 = y_2$ . We have shown that $\sup I$ is finite at only a single pure state, and since it is lower semicontinuous and affine it follows that it takes the form $L_{y, \lambda }$ , with $y = y_1 = y_2$ and $\lambda = \lambda _1 = \lambda _2$ . We deduce that I is in $\mathcal {I}_{y, \lambda }$ .

Proof The first part follows from Lemma 4.1. The second part follows directly from the definition of $\mathcal {I}_{y, \lambda }$ .

Proof Lemma 4.5 characterizes the proper ideals having the property that they lie in $\mathcal {I}_{y, \lambda }$ , for some pure state y and $\lambda \ge 0$ . This characterization is purely in terms of the order structure on C. A similar characterization holds for the cone $C'$ . It follows that a proper ideal is in $\mathcal {I}_{y, \lambda }$ , with y a pure state of V and $\lambda \ge 0$ , if and only if its image is in $\mathcal {I}_{y', \lambda '}$ for some pure state $y'$ of $V'$ , and some $\lambda ' \ge 0$ .

Combining this with Lemma 4.4, we get that if two ideals $I_1$ and $I_2$ lie in $\mathcal {I}_{y, \lambda _1}$ and $\mathcal {I}_{y, \lambda _2}$ , respectively, then there exists a pure state $y'$ of $V'$ such that their images lie in $\mathcal {I}_{y', \lambda ^{\prime }_1}$ and $\mathcal {I}_{y', \lambda ^{\prime }_2}$ , respectively, where $\lambda ^{\prime }_1$ and $\lambda ^{\prime }_2$ are nonnegative real numbers. We define the map $\Phi $ so that $\Phi y := y'$ . We do this for each pure state y of V.

Applying the same considerations to the inverse map $\phi ^{-1}$ , we see that $\Phi $ is a bijection from the pure states of V to those of $V'$ .

Proof Choose a positive number $\lambda $ such that $g_1(x) < \lambda < g_2(x)$ . By Lemma 4.2, we can find a net of proper ideals $I_\alpha $ , defined on the same directed set as $x_\alpha $ , such that $\sup I_\alpha = L_{x_\alpha , \lambda }$ for all $\alpha $ . Note that, for $\alpha $ large enough, $g_1(x_\alpha ) < \lambda < g_2(x_\alpha )$ , and so, by Lemma 4.6, $I_\alpha $ contains $g_1$ but does not contain $g_2$ . So, for $\alpha $ large enough, the ideal $\phi I_\alpha $ contains $\phi g_1$ but not $\phi g_2$ . By Lemma 5.1, for each $\alpha $ , the ideal $\phi I_\alpha $ is an element of $\mathcal {I}_{\Phi x_\alpha , \lambda ^{\prime }_\alpha }$ , with $\lambda ^{\prime }_\alpha \ge 0$ . We deduce using Lemma 4.6 that

$$ \begin{align*} \phi g_1 (\Phi x_\alpha) \le \lambda^{\prime}_\alpha \le \phi g_2 (\Phi x_\alpha), \end{align*} $$

for $\alpha $ large enough. Taking the limit as $\alpha $ tends to infinity, we get the result.

Proof Denote the limit of $x_\alpha $ by x. Since $F'$ is compact, it suffices to show that $\Phi (x_\alpha )$ has at most one limit point. Assume for the sake of contradiction that there are two distinct limit points y and z, and take subnets $y_\beta $ and $z_\gamma $ of $x_\alpha $ such that $\Phi (y_\beta )$ and $\Phi (z_\gamma )$ converge to y and z, respectively. The set $C'$ separates $K'$ , and so there exists an element h of $C'$ that takes distinct values on y and z. Switch y and z if necessary and choose $\lambda> 0$ so that $h(y) < \lambda < h(z)$ . For $\rho>0$ large enough,

$$ \begin{align*} g_1 := h + \rho u' \qquad\text{and}\qquad g_2 := (2 \lambda + \rho) u' - h \end{align*} $$

are in $C'$ , where $u'$ is the order unit of $V'$ . Observe that $g_1(y) < g_2(y)$ and $g_1(z)> g_2(z)$ . We apply Lemma 5.2 to the map $\phi ^{-1}$ twice, once for each of these inequalities. This gives that $\phi ^{-1} g_1(x) = \phi ^{-1} g_2(x)$ .

Let

$$ \begin{align*} \delta := \frac {g_2(y) - g_1(y)} {2}> 0. \end{align*} $$

The function $g_3 := g_1 + \delta u'$ is in $C'$ and satisfies the relations $g_3(y) < g_2(y)$ and $g_3(z)> g_2(z)$ . Using the same reasoning as before, we get that $\phi ^{-1} g_3$ and $\phi ^{-1} g_2$ agree at x.

Let $g_4 := g_2 + \delta u'$ . Again, $\phi ^{-1} g_3$ and $\phi ^{-1} g_4$ agree at x. Continuing in this manner, we see that the functions $\phi ^{-1} (g_1 + n\delta u')$ all take the same value at x, independently of $n\in \mathbb {N}$ . But every element of $C'$ lies below $g_1 + n\delta u'$ for some value of $n\in \mathbb {N}$ , and $\phi ^{-1}$ is an order isomorphism between $C'$ and C. It follows that no element of C takes a value strictly greater than $\phi ^{-1} g_1(x)$ at x, which is absurd.

Proof From Lemma 5.3 and the theorem from [Reference Bourbaki8] mentioned above, we get that $\Phi $ extends in a unique way to a continuous mapping between F and $F'$ . Similar reasoning shows that the inverse mapping $\Phi ^{-1}$ also extends to a continuous mapping, this time between $F'$ and F, and it is easy to see that this extension is the inverse of the extension of $\Phi $ . Thus, the extension of $\Phi $ is a homeomorphism.

The second part is proved by applying Lemma 5.2 to any net of pure states of V converging to x.

Proof Let x be a pure state of V, and choose $\lambda> 0$ . By Lemma 4.2, $\mathcal {I}_{x, \lambda }$ contains an ideal I. Applying Lemma 5.1 to the maps $\phi $ and $\phi '$ , we get that $\phi ' \circ \phi (I)$ is in $\mathcal {I}_{(\Phi ' \circ \Phi ) x, \lambda '}$ , for some $\lambda ' \ge 0$ . Applying the same lemma to the map $\phi ' \circ \phi $ gives that $\phi ' \circ \phi (I)$ is also in $\mathcal {I}_{\Phi " x, \lambda "}$ , for some $\lambda " \ge 0$ . This implies that $\Phi ' \circ \Phi $ and $\Phi "$ agree at x, and hence on the set of pure states of V, since x was arbitrary. Finally, we extend the conclusion to the whole of F using Proposition 5.4.

Proof Consider first the case where $c := f(p)> 0$ . Take $\epsilon \in (0, c)$ , and let $g_1 := (c - \epsilon )u$ and $g_2 := (c + \epsilon )u$ . Observe that $g_1(p) < f(p) < g_2(p)$ . So, by Proposition 5.4, $\phi g_1(q) \le \phi f(q) \le \phi g_2(q)$ , which is equivalent to $\Lambda _p(c - \epsilon ) \le \phi f(q) \le \Lambda _p(c + \epsilon )$ . Letting $\epsilon $ approach zero and using the continuity of $\Lambda _p$ gives the result.

The case where $f(p) = 0$ is similar, only now $\epsilon $ is taken in $(0, \infty )$ and the lower bounds are disregarded.

Proof Observe that the homeomorphism on F induced by $M_\lambda $ is the identity map, and by Lemma 5.5, the homeomorphisms induced by $\phi $ and $\phi ^{-1}$ , respectively, are inverses. The first part then follows using Lemma 5.5 again.

To prove the second part, fix $p \in L$ . Consider first the case where $0 < \lambda \le 1$ . Observe that in this case $M_\lambda (g) \le g$ , for all $g\in C'$ , and so $\Delta _\lambda (f) \le f$ , for all $f\in C$ . We deduce that $\Lambda ^{(\lambda )}_p(x) \le x$ , for all $x\ge 0$ . This implies that (9) holds. So, by Lemma 6.13, we get that $\Lambda ^{(\lambda )}_p$ is homogeneous of degree $1$ .

When $\lambda \ge 1$ , we apply the same argument to $\Delta _{1 / \lambda }$ , to get that $\Lambda ^{(1/\lambda )}_p$ is homogeneous of degree $1$ , and then use that $\Delta _\lambda = (\Delta _{1 / \lambda })^{-1}$ , and so $\Lambda ^{(\lambda )}_p$ is the inverse of $\Lambda ^{(1/\lambda )}_p$ , and hence homogeneous of degree $1$ .

Proof For each $\lambda>0$ , define the homogeneity defect map as in (10). By Lemma 7.1, the associated coordinate transformation maps can be written

$$ \begin{align*} \Lambda^{(\lambda)}_p(x) = \rho(\lambda) x, \qquad\text{for all } \lambda>0 \text{ and } x \ge 0, \end{align*} $$

where $\rho \colon (0, \infty ) \to \mathbb {R}_{\ge 0}$ is some function.

Note that, for all positive $\lambda $ and $\lambda '$ , we have $\Delta _{\lambda } \circ \Delta _{\lambda '} = \Delta _{\lambda \lambda '}$ . The homeomorphism induced by each of $\Delta _{\lambda }$ , $\Delta _{\lambda '}$ , and $\Delta _{\lambda \lambda '}$ is the identity map, and the coordinate transformations $\Lambda ^{(\lambda )}_p$ , $\Lambda ^{(\lambda ')}_p$ , and $\Lambda ^{(\lambda \lambda ')}_p$ are all continuous on $(0, \infty )$ . So, from Proposition 5.6, we get that $\Lambda ^{(\lambda )}_p \circ \Lambda ^{(\lambda ')}_p = \Lambda ^{(\lambda \lambda ')}_p$ . It follows that $\rho ({\lambda }) \rho ({\lambda '}) = \rho ({\lambda \lambda '})$ . The function $\rho $ is non-decreasing on its domain $(0, \infty )$ . Clearly, $\rho $ cannot be identically zero, because $\Delta _1$ is the identity map and hence $\rho (1)=1$ . According to [Reference Kannappan10, Theorem 1.49], the only other possibility is that $\rho (\lambda ) = \lambda ^c$ , for all $\lambda>0$ , where c is a nonnegative real number.

Observe that $\phi \circ \Delta _\lambda = M_\lambda \circ \phi $ , for each $\lambda> 0$ . We deduce that

$$ \begin{align*} \Lambda_p(\lambda^c x) = \lambda \Lambda_p(x), \qquad\text{for all } \lambda>0 \text{ and } x\ge 0. \end{align*} $$

Recall that $\Lambda _p$ is non-decreasing and unbounded, but not necessarily continuous. By evaluating both sides of this equation at a point x where $\Lambda _p$ is positive and continuous, taking $\lambda \neq 1$ , we see that c cannot be zero.

Substituting $\gamma := \lambda ^c$ , we get

$$ \begin{align*} \Lambda_p(\gamma x) = \gamma^{1/c} \Lambda_p(x), \qquad\text{for all } \gamma> 0 \text{ and } x \ge 0. \end{align*} $$

This shows that $\Lambda _p$ is homogeneous of degree $1/c$ .

Proof Fix $f\in V$ . For each $p\in F$ , we have that $\Lambda _p(1) = 1$ , and so $\Lambda _p(x) = x^{\alpha (p)}$ , for all $x\in \mathbb {R}_{\ge 0}$ . Since $\phi (2 u)$ is continuous on $F'$ , and $\phi (2 u)(q) = 2^{\alpha (p)}$ , for each $q\in F'$ , where $p := \Phi ^{-1}(q)$ , we see that $\alpha $ is continuous on F, and hence bounded. This implies that the second derivative of $\Lambda _p$ is bounded uniformly in p in an interval neighborhood $I \subset (0, \infty )$ of $1$ , that is, there exists $M> 0$ such that

$$ \begin{align*} \Big| \frac {d^2 \Lambda_p(x) } {dx^2} \Big| = \left|\alpha(p)\big(\alpha(p) - 1\big) x^{\alpha(p) - 2}\right| \le M, \qquad\text{for all } p \in F \text{ and } x \in I. \end{align*} $$

Taylor’s theorem gives that, for $p\in F$ and $\delta $ small enough,

$$ \begin{align*} \phi(u + \delta f)(\Phi(p)) &= \Lambda_p \big((u + \delta f) (p) \big) \qquad\qquad\text{(by Proposition 5.6)} \\ &= 1 + \delta \alpha(p) f(p) + R(\delta, p, f(p)), \end{align*} $$

with the remainder term satisfying $|R(\delta , p, f(p))| \le M \delta ^2 f(p)^2 /2$ .

This shows that the limit in the definition of the Gateaux derivative converges uniformly on $F'$ to $\alpha (p) f(p)$ . By Lemma 6.5, this means that the Gateaux derivative $d\phi (u; f)$ exists and has the required form.

The map $\mathcal {D} \colon f \mapsto d\phi (u; f)$ is obviously linear and order preserving.

Consider the inverse $\phi ^{-1}$ of $\phi $ . Each of the coordinate transformations $\Lambda ^{\prime }_q$ , with $q\in F'$ , of this map is also homogeneous, this time of degree $1/\alpha (p)$ , where $p := \Phi ^{-1}(q)$ . Using the same reasoning as for $\phi $ , we get that the Gateaux derivative of $\phi ^{-1}$ at the order unit $u'$ in the direction $f \in V'$ is given by

$$ \begin{align*} d\phi^{-1}(u'; f)(p) = \frac {1} {\alpha(p)} f(q), \qquad\text{for all } q\in F', \end{align*} $$

where $p := \Phi ^{-1}(q)$ . From this expression, we see that the map $f \mapsto d\phi ^{-1}(u'; f)$ is also linear and order preserving. Moreover, it is the inverse of the map $\mathcal {D}$ . Hence, $\mathcal {D}$ is a linear order isomorphism between V and $V'$ .

It remains to show that the homeomorphism $\Phi ^{\mathcal {D}}$ induced by $\mathcal {D}$ is the same as that induced by $\phi $ . Let p be a pure state of V, and choose $\lambda>0$ . By Lemma 4.2, the set $\mathcal {I}_{p, \lambda }$ contains an ideal I, satisfying $\sup I = L_{p, \lambda }$ . Since $\mathcal {D}$ is an order isomorphism, it maps I to an ideal $I'$ of $C'$ . We see from the explicit formula for $\mathcal {D}$ that $\sup I'$ takes the value $\alpha (p) \lambda $ at $\Phi (p)$ and the value infinity everywhere else. So, $I'$ is in $\mathcal {I}_{\Phi (p), \alpha (p) \lambda }$ . It follows using Lemma 5.1 that $\Phi ^{\mathcal {D}}(p) = \Phi (p)$ . Since this holds for an arbitrary pure state of V, the pure states are dense in F, and $\Phi ^{\mathcal {D}}$ and $\Phi $ are continuous, the conclusion follows.

Proof of Theorem 1.1

Without loss of generality, we may assume that the domain of $\phi $ is C and the co-domain is $C'$ – if this is not the case, then choose some element f of X and consider the map $g \mapsto \phi (f + g) - \phi (f)$ from C to $C'$ . By Lemma 2.1, we can choose order units u and $u'$ of V and $V'$ , respectively, such that $\phi (u) = u'$ .

Recall that we are using L to denote the closure of the union of the supports of the affine dependencies of V carried by F. By Lemma 7.2, for each $p\in L$ , the coordinate transformation $\Lambda _p$ is homogeneous of some positive degree. Likewise, the coordinate transformation $\Lambda ^{\prime }_q$ of the inverse map $\phi ^{-1}$ is homogeneous of positive degree, for each $q\in L'$ , where $L'$ is defined analogously to L. Observe that $\Lambda _p$ and $\Lambda ^{\prime }_q$ are inverses when $q = \Phi (p)$ . It follows that $\Lambda _p$ is homogeneous of positive degree for all $p\in \Phi ^{-1}(L')$ . Therefore, we can write

$$ \begin{align*} \Lambda_p(x) = x^{\alpha(p)}, \qquad\text{for all } p\in L \cup \Phi^{-1}(L') \text{ and } x\in\mathbb{R}_{\ge 0}, \end{align*} $$

where $\alpha $ is a positive function.

By considering the image under $\phi $ of $2u$ , where u is the unit, we see that $\alpha $ is continuous on $L \cup \Phi ^{-1}(L')$ . Since L and $L'$ are compact, $\alpha $ is bounded above and below by positive numbers on this set.

By Tietze’s extension theorem, we may extend $\alpha $ to a continuous function on the whole of F. We reuse the symbol $\alpha $ to denote this extended map. The extension can be chosen to be bounded above and below by positive numbers.

Let $h\in C$ , and consider the function $f \colon F' \to \mathbb {R}$ defined by

$$ \begin{align*} f(q) := h(p)^{\alpha(p)}, \qquad\text{for all } q\in F', \end{align*} $$

where p and q are related by $q = \Phi (p)$ . The function f is continuous. Moreover, by Proposition 5.6, it agrees with $\phi (h)$ on $L'$ , and so $\mu (f) = 0$ , for each affine dependency $\mu $ of $V'$ carried by $F'$ . Applying Lemma 2.2, we get that there is an element $g_h$ of $V'$ such that $g_h(q) = f(q)$ , for all $q\in F'$ . Since $g_h$ takes nonnegative values on $F'$ , it is an element of $C'$ .

Define the map $\kappa \colon C \to C'$ by setting $\kappa (h) := g_h$ , for all $h\in C$ . This map is clearly order preserving. Similar reasoning to that in the previous paragraph shows that one may define a map from $C'$ to C having coordinate transformations

$$ \begin{align*} \Lambda^{\prime}_q(x) = x^{1 / \alpha(p)}, \qquad\text{for all } q\in F', \end{align*} $$

where $p = \Phi ^{-1}(q)$ . This map is the inverse of $\kappa $ and is also order preserving. Hence, $\kappa $ is an order isomorphism. Observe that $\kappa (u) = u'$ . It only remains to apply Lemma 8.1 to get that there is a linear order isomorphism between V and $V'$ .

Proof By Theorem 1.1, there is a linear order isomorphism $\mathcal {D}$ from $V'$ to V. Moreover, the order isomorphism $\mathcal {D}$ exhibited in the proof of that theorem induces the same homeomorphism between $F'$ and F as $\phi ^{-1}$ . By Lemma 5.5, this homeomorphism is the inverse of the one $\Phi $ induced by $\phi $ between F and $F'$ .

Consider the map $\vartheta := \mathcal {D} \circ \phi $ . This is an order isomorphism from C to itself, and, by Lemma 5.5, the homeomorphism it induces on F is the identity map. By Lemma 7.2, each of the coordinate transformations $\Lambda _p$ ; $p\in L$ of $\phi $ is homogeneous of some positive degree. Furthermore, every coordinate transformation $\Lambda ^{\mathcal {D}}_q$ ; $q\in F'$ of $\mathcal {D}$ is homogeneous of degree $1$ . Let $\Lambda ^\vartheta _p$ ; $p\in F$ be the coordinate transformations of $\vartheta $ . Observe that, for each $p\in L$ , we have $\Lambda ^\vartheta _p = \Lambda ^{\mathcal {D}}_q \circ \Lambda _p$ , where $q := \Phi (p)$ , and hence the coordinate transformation $\Lambda ^\vartheta _p$ is homogeneous of the same degree as $\Lambda _p$ . This implies that (9) holds, and Lemma 6.13 is therefore applicable. We deduce that, for all $p\in L$ , the transformation $\Lambda ^\vartheta _p$ is homogeneous of degree $1$ , and hence so is $\Lambda _p$ .

Proof Let $z\in U$ . The map

$$ \begin{align*} \vartheta_z \colon C \to C', \quad f \mapsto \phi(f + z) - \phi(z) \end{align*} $$

is an order isomorphism from the cone C of V to the cone $C'$ of $V'$ . Let $\Phi $ be the homeomorphism from F to $F'$ that this induces.

Let $p\in L$ , and write $q := \Phi (p)$ . From Lemma 9.1, the coordinate transformation $\Lambda _p$ of $\vartheta _z$ is homogeneous of degree $1$ . This means that there is some $\lambda>0$ such that $\vartheta _z(f)(q) = \lambda f(p)$ , for all $f\in C$ . Note that, for all $f \ge z$ ,

$$ \begin{align*} \phi(f) = \vartheta_z(f - z) + \phi(z), \end{align*} $$

and so,

$$ \begin{align*} \phi(f)(q) &= \lambda f(p) - \lambda z(p) + \phi(z)(q) \\ &= a(f(p)), \end{align*} $$

where $a \colon \mathbb {R} \to \mathbb {R}$ is defined to be $a(x) := \lambda x + c$ , with $c := -\lambda z(p) + \phi (z)(q)$ .

Now let w be another point of U, and define $\vartheta _w$ in a similar way to how we defined $\vartheta _z$ . Again, we get a homeomorphism from F to $F'$ , taking p to some element $q'$ of $F'$ . As before, we have $\phi (f)(q') = a'(f(p))$ , for all $f \ge w$ , where $a'$ is some increasing affine function from $\mathbb {R}$ to itself.

Since V is an order-unit space, it has an element x that is greater than both z and w. Consider the set $X := \{f \in V \mid f \ge x\}$ . Every point f of this set satisfies both $f \ge z$ and $f\ge w$ . Its image is $\phi (X) = \{g \in V' \mid g \ge \phi (x)\}$ . For g in $\phi (X)$ , both $g(q)$ and $g(q')$ only depend on $\phi ^{-1}(g)(p)$ . By varying g within $\phi (X)$ , we see that q and $q'$ are equal. Comparing the expressions for $\phi (\cdot )(q)$ and $\phi (\cdot )(q')$ on X then gives that the two affine functions a and $a'$ are equal.

We have shown that the homeomorphism $\Phi $ and the affine function a do not depend on the choice of z. We conclude that the equation $\phi (f)(q) = a(f(p))$ holds for all $f\in U$ .

Proof Let $\{\lambda _i\}_i$ and $\{f_i\}_i$ be finite collections of real numbers and elements of U, respectively, such that $\sum _i \lambda _i = 1$ and $\sum _i \lambda _i f_i \in U$ .

Denote by $\Phi $ the homeomorphism from F to $F'$ given by Lemma 9.2. Let $q\in F'$ , and let $p \in F$ be such that $q = \Phi (p)$ . By assumption, p is in L. So, by Lemma 9.2 again, there exists an increasing affine function $a \colon \mathbb {R} \to \mathbb {R}$ such that $\phi (h)(q) = a(h(p))$ , for all $h\in U$ . Therefore,

$$ \begin{align*} \phi \Big(\sum_i \lambda_i f_i\Big)(q) = a\Big(\sum_i \lambda_i f_i(p)\Big) = \sum_i \lambda_i a \big(f_i(p)\big) = \Big(\sum_i \lambda_i \phi(f_i) \Big)(q). \end{align*} $$

Since this is true for all $q\in F'$ , we deduce that

$$ \begin{align*} \phi \Big(\sum_i \lambda_i f_i\Big) &= \sum_i \lambda_i \phi(f_i). \end{align*} $$

We have shown that $\phi $ is an affine map.

Proof of Theorem 1.3

If Condition 1.2 holds, then Lemma 9.3 shows that every order isomorphism between U and $U'$ is affine.

Assume now that U is directed downward, that all order isomorphisms between U and $U'$ are affine, and that there exists an order isomorphism $\vartheta $ between U and $U'$ , which is of course necessarily affine. If Condition 1.2 does not hold for the space V, then by Lemma 3.1, there exists an order isomorphism $\phi $ from U to itself that is not affine. In this case, the composition $\vartheta \circ \phi $ is an order isomorphism between U and $U'$ that is not affine. We conclude that Condition 1.2 holds for V.

Proof of Theorem 1.5

Denote by F the closure of the set of pure states, and by L the closure of the union of the supports of the affine dependencies carried by F, as in (1).

Assume that the state space is not a Bauer simplex, and let $\phi $ be an order isomorphism between $\operatorname {\mathrm {int}} C$ and a non-empty upper subset U of V. We must show that U does not equal V. Since the state space is not a Bauer simplex, there exists an element p of L. Let $\Phi $ be the homeomorphism on F given by Lemma 9.2, and write $q := \Phi (p)$ . According to the same lemma, there is an increasing affine function $a \colon \mathbb {R} \to \mathbb {R}$ such that $\phi (f)(q) = a(f(p))$ , for all $f\in \operatorname {\mathrm {int}} C$ . Observe that $f(p)$ is positive for all $f\in \operatorname {\mathrm {int}} C$ . It follows that $\phi (f)(q)$ is bounded below by $a(0)$ as f varies in $\operatorname {\mathrm {int}} C$ , which means that the map $g \mapsto g(q)$ is bounded below on U. Since this map is not bounded below on V, we deduce that U and V differ.

Now assume that the state space is a Bauer simplex, that is, that the set of affine dependencies carried by F is empty. Define the map $\phi \colon \operatorname {\mathrm {int}} C \to V$ , $x\mapsto \phi (x)$ by

$$ \begin{align*} \phi(x)(p) := \log(x(p)), \qquad\text{for all } p \in F. \end{align*} $$

Since, for each $x\in \operatorname {\mathrm {int}} C$ , the function $\phi (x)(\cdot )$ is continuous on F, we have by Lemma 2.2 that $\phi (x)$ is a well-defined element of V. It is clear that $\phi $ is order preserving. Similar reasoning shows that the coordinate-wise exponential function defined on V is also well-defined and order preserving. Since this function is the inverse of $\phi $ , the proof is complete.

Proof We first consider the two-dimensional problem, that is, we study the order isomorphisms of the set $H_2 := \{(x,y) \mid x + y \ge 0\}$ in the space $\mathbb {R}^2$ . Let $\phi $ be such a map. For each $a\in H_2$ , denote by $U_a$ the set of elements of $H_2$ that are greater than or equal to a. If a is a point of $H_2$ , then the restriction of $\phi $ to $U_a$ is an order isomorphism between $U_a$ and $U_{\phi (a)}$ . Such maps are known to be of the form $(x, y) \mapsto P(\phi _1(x), \phi _2(y))$ , where P is a permutation of the coordinates, and $\phi _1$ and $\phi _2$ are order isomorphisms between upper sets of $\mathbb {R}$ . Note that, for a and $a'$ in $H_2$ , the intersection of the sets $U_a$ and $U_{a'}$ has non-empty interior. So, we see that when a is varied over $H_2$ , all the permutations obtained are the same, and we can write $\phi (x, y) = P(\phi _1(x), \phi _2(y))$ , with $\phi _1$ and $\phi _2$ being order isomorphisms of $\mathbb {R}$ . Since $\phi $ preserves the set $\big \{(x, -x) \mid x\in \mathbb {R} \big \}$ , we have that $\phi _1(x) = -\phi _2(-x)$ , for all $x\in \mathbb {R}$ .

Returning to the three-dimensional case, we use similar reasoning to show that every order isomorphism $\phi $ from H to itself can be written in the form $\phi (x, y, z) = P(\phi _1(x), \phi _2(y), \phi _3(z))$ , where P is a permutation of the three coordinates and $\phi _i$ ; $1\le i\le 3$ are order isomorphisms of $\mathbb {R}$ such that $\phi _1(x) + \phi _2(y) + \phi _3(z) = 0$ whenever $x + y + z = 0$ . Without loss of generality, we may assume that P is the identity permutation, and that $\phi (o) = o$ , where o is the origin.

The map $\phi $ then acts as an order isomorphism on the set of points $(x, y, z) \in H$ such that $z = 0$ . Applying the two-dimensional result of the first paragraph, we get that $\phi _1(x) = -\phi _2(-x)$ , for all $x\in \mathbb {R}$ . Similar considerations lead us to conclude that $\phi _1(x) = -\phi _3(-x)$ , for all $x\in \mathbb {R}$ , and we deduce that $\phi _2 = \phi _3$ on $\mathbb {R}$ . By reasoning in a similar way about the other coordinates, we see that $\phi _1 = \phi _2 = \phi _3$ .

We now evaluate $\phi $ at the point $(\alpha + \beta , -\alpha , -\beta )$ , with $\alpha $ and $\beta $ in $\mathbb {R}$ . From our results so far, $\phi _1(\alpha + \beta ) = \phi _1(\alpha ) + \phi _1(\beta )$ . So, $\phi $ is additive. Since it is also monotonically increasing, it follows that it is linear (see [Reference Kannappan10, Theorem 1.2]). Likewise, the same is true for $\phi _2$ and $\phi _3$ . We have proved that $\phi $ is affine.