PhyslibAlpha.Mathematics.Order.PositiveDual.RieszKantorovich

For an ordered additive commutative group $E$, the Riesz decomposition property states that for any elements $f_1, f_2, g \in E$ such that $0 \le f_1$, $0 \le f_2$, and $0 \le g \le f_1 + f_2$, there exist $g_1, g_2 \in E$ such that $g = g_1 + g_2$ with $0 \le g_1 \le f_1$ and $0 \le g_2 \le f_2$.

Let $E$ be a lattice-ordered group (or a vector lattice). For any elements $f_1, f_2, g \in E$ such that $0 \le f_1$, $0 \le f_2$, and $0 \le g \le f_1 + f_2$, there exist $g_1, g_2 \in E$ such that $g = g_1 + g_2$ with $0 \le g_1 \le f_1$ and $0 \le g_2 \le f_2$.

Let $E$ be an ordered real vector space and let $\phi, \psi: E \to \mathbb{R}$ be positive linear maps. For any element $f \in E$, this definition calculates the Riesz–Kantorovich supremum as $$\sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$$ This value represents the best split of $f$ between the functionals $\phi$ and $\psi$.

Let $E$ be an ordered real vector space and let $\phi, \psi: E \to \mathbb{R}$ be positive linear maps. For any $f \in E$, the set $\{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$ is bounded above in $\mathbb{R}$.

Let $E$ be an ordered real vector space and let $\phi, \psi: E \to \mathbb{R}$ be positive linear maps. For any elements $f, g \in E$ such that $0 \le g \le f$, the sum $\phi(g) + \psi(f - g)$ is less than or equal to the Riesz–Kantorovich supremum $(\phi \sqcup \psi)(f)$, which is defined as $\sup \{ \phi(g') + \psi(f - g') \mid 0 \le g' \le f \}$.

Let $E$ be an ordered real vector space and let $\phi, \psi: E \to \mathbb{R}$ be positive linear maps. For any element $f \in E$ such that $0 \le f$ and any $r \in \mathbb{R}$, if $\phi(g) + \psi(f - g) \le r$ for all $g \in E$ satisfying $0 \le g \le f$, then the Riesz–Kantorovich supremum satisfies $(\phi \sqcup \psi)(f) \le r$.

Let $E$ be an ordered real vector space and let $\phi, \psi: E \to \mathbb{R}$ be positive linear maps. For any element $f \in E$ such that $f \ge 0$, the Riesz–Kantorovich supremum, defined as $(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$, is non-negative, i.e., $(\phi \sqcup \psi)(f) \ge 0$.

Let $E$ be an ordered real vector space and let $\phi, \psi: E \to \mathbb{R}$ be positive linear maps. Let the Riesz–Kantorovich supremum be defined as $$(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$$ For any non-negative elements $f_1, f_2 \in E$, it holds that $$(\phi \sqcup \psi)(f_1) + (\phi \sqcup \psi)(f_2) \le (\phi \sqcup \psi)(f_1 + f_2)$$

Let $E$ be an ordered real vector space that satisfies the Riesz decomposition property. Let $\phi$ and $\psi$ be positive linear maps from $E$ to $\mathbb{R}$. For any $f_1, f_2 \in E$ such that $0 \le f_1$ and $0 \le f_2$, the Riesz–Kantorovich supremum, defined by $(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$, is additive on the positive cone: $$(\phi \sqcup \psi)(f_1 + f_2) = (\phi \sqcup \psi)(f_1) + (\phi \sqcup \psi)(f_2)$$

Let $E$ be an ordered real vector space where scalar multiplication by positive reals is monotonic. Let $\phi, \psi : E \to \mathbb{R}$ be positive linear maps. For any positive real number $c > 0$ and any non-negative element $f \in E$ ($f \ge 0$), the Riesz–Kantorovich supremum $(\phi \sqcup \psi)$ satisfies the following homogeneity property: $$(\phi \sqcup \psi)(c \cdot f) = c \cdot (\phi \sqcup \psi)(f)$$ where the Riesz–Kantorovich supremum is defined as $(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$.

Let $E$ be an ordered real vector space satisfying the Riesz decomposition property. Given two positive linear functionals $\phi, \psi: E \to \mathbb{R}$, the Riesz–Kantorovich supremum $\phi \sqcup \psi$ is the positive linear functional on $E$ defined on the positive cone by the formula: $$(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$$ for all $f \ge 0$. The definition extends this mapping from the positive cone to the entire space $E$ to form a complete positive linear functional.

Let $E$ be an ordered real vector space satisfying the Riesz decomposition property. Let $\phi$ and $\psi$ be positive linear functionals from $E$ to $\mathbb{R}$. For any non-negative element $f \in E$ (where $0 \le f$), the value of the Riesz–Kantorovich supremum functional $(\phi \sqcup \psi)$ evaluated at $f$ is equal to the supremum of the values obtained by splitting $f$ between $\phi$ and $\psi$: $$(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \}$$

Let $E$ be an ordered real vector space that satisfies the Riesz decomposition property. For any two positive linear functionals $\phi, \psi: E \to \mathbb{R}$, the Riesz–Kantorovich supremum map $\phi \sqcup \psi$, defined for each $f \ge 0$ by the formula $$(\phi \sqcup \psi)(f) = \sup \{ \phi(g) + \psi(f - g) \mid 0 \le g \le f \},$$ is the least upper bound of the set $\{\phi, \psi\}$ with respect to the partial order on positive linear functionals.

For an ordered real vector space $E$ over $\mathbb{R}$, the property `HasLatticeDualCone E` states that the space of positive linear functionals $E \to_p \mathbb{R}$ forms a join-semilattice. Specifically, for any two positive linear functionals $\phi$ and $\psi$, there exists a positive linear functional $\chi$ that is their least upper bound (supremum), denoted $\chi = \phi \sqcup \psi$, with respect to the partial order where $\phi \leq \psi$ if $\phi(f) \leq \psi(f)$ for all $f \geq 0$ in $E$.

Let $E$ be an ordered real vector space over $\mathbb{R}$ that is a directed order. If $E$ satisfies the Riesz decomposition property—meaning that for any $f_1, f_2, g \in E$ such that $0 \le f_1, 0 \le f_2$, and $0 \le g \le f_1 + f_2$, there exist $g_1, g_2 \in E$ such that $g = g_1 + g_2$ with $0 \le g_1 \le f_1$ and $0 \le g_2 \le f_2$—then the space of positive linear functionals on $E$ forms a lattice (i.e., for any two positive linear functionals, their least upper bound exists).

Let $E$ be a real vector lattice (an ordered real vector space over $\mathbb{R}$ that is also a lattice). Then the space of positive linear functionals on $E$ forms a lattice. That is, for any two positive linear functionals $\phi, \psi: E \to \mathbb{R}$, there exists a least upper bound $\phi \sqcup \psi$ with respect to the partial order where $\phi \leq \psi$ if and only if $\phi(f) \leq \psi(f)$ for all $f \geq 0$ in $E$.

Let $E$ be an ordered real vector space such that the space of positive linear functionals on $E$ forms a lattice. For any positive linear functionals $\theta, \alpha, \beta$ satisfying $\theta \leq \alpha + \beta$, there exist positive linear functionals $\theta_1$ and $\theta_2$ such that $\theta_1 \leq \alpha$, $\theta_2 \leq \beta$, and $\theta = \theta_1 + \theta_2$.

Let $E$ be an ordered real vector space such that the positive linear functionals on $E$ form a lattice. Let $\{\psi_i\}_{i=0}^{m-1}$ be a finite family of positive linear functionals on $E$, and let $\alpha$ and $\beta$ be positive linear functionals such that their sum equals the sum of the family: $$\sum_{i=0}^{m-1} \psi_i = \alpha + \beta$$ Then there exist two families of positive linear functionals $\{\psi_{1,i}\}_{i=0}^{m-1}$ and $\{\psi_{2,i}\}_{i=0}^{m-1}$ such that for each $i \in \{0, \dots, m-1\}$, the functional splits as $\psi_i = \psi_{1,i} + \psi_{2,i}$, and the sums of these families satisfy: $$\sum_{i=0}^{m-1} \psi_{1,i} = \alpha \quad \text{and} \quad \sum_{i=0}^{m-1} \psi_{2,i} = \beta$$

Let $E$ be an ordered real vector space. The property `HasRefinement E` states that for any two finite families of positive linear functionals $\{\alpha_i\}_{i=1}^m$ and $\{\beta_j\}_{j=1}^n$ on $E$ such that their sums are equal, i.e., $\sum_{i=1}^m \alpha_i = \sum_{j=1}^n \beta_j$, there exists a family (or "table") of positive linear functionals $\{\gamma_{ij}\}_{i=1, j=1}^{m, n}$ such that: 1. The sum of each row matches the first family: $\sum_{j=1}^n \gamma_{ij} = \alpha_i$ for all $i \in \{1, \dots, m\}$. 2. The sum of each column matches the second family: $\sum_{i=1}^m \gamma_{ij} = \beta_j$ for all $j \in \{1, \dots, n\}$.

Let $E$ be an ordered real vector space with a directed order. If the space of positive linear functionals on $E$ forms a lattice (specifically, a join-semilattice, denoted by the property `HasLatticeDualCone E`), then $E$ satisfies the finite refinement property for positive linear functionals. This means that for any two finite families of positive linear functionals $\{\alpha_i\}_{i \in I}$ and $\{\beta_j\}_{j \in J}$ such that their sums are equal, i.e., $$\sum_{i \in I} \alpha_i = \sum_{j \in J} \beta_j,$$ there exists a family of positive linear functionals $\{\gamma_{i,j}\}_{i \in I, j \in J}$ such that for each $i \in I$, $\sum_{j \in J} \gamma_{i,j} = \alpha_i$, and for each $j \in J$, $\sum_{i \in I} \gamma_{i,j} = \beta_j$.

Let $E$ be an ordered real vector space that satisfies the refinement property. Suppose $\{\alpha_i\}_{i \in I}$ and $\{\beta_j\}_{j \in K}$ are two families of positive linear functionals on $E$ indexed by finite sets $I$ and $K$. If the sums of these families are equal, i.e., $$\sum_{i \in I} \alpha_i = \sum_{j \in K} \beta_j,$$ then there exists a family of positive linear functionals $\{\gamma_{i,j}\}_{i \in I, j \in K}$ such that for each $i \in I$, the sum over $j$ is $\alpha_i$ (i.e., $\sum_{j \in K} \gamma_{i,j} = \alpha_i$), and for each $j \in K$, the sum over $i$ is $\beta_j$ (i.e., $\sum_{i \in I} \gamma_{i,j} = \beta_j$).