Atomic sublattices and basic derivatives in finance

Abstract

Suppose that E is a vector lattice where the ordering and the lattice operations in E are defined pointwise by a countable family ${\mathcal {F}}=\{f_i|i\in {{\mathbf {N}}}\}$ of positive linear functional of E and Z is a sublattice of E. Based on algebraic and order properties of E we give necessary and sufficient conditions in order Z to be atomic. Especially we show the existence of a basic sequence $\{b_n\}$ of extremal points (atoms) of $Z_+$ so that for any $x\in Z_+$ a unique sequence $({\widehat{x}}(n))$ of real components of x with respect to $\{b_n\}$ exists so that $x=\sup \{{\widehat{x}}(n)b_n\;|\;n\in {{\mathbf {N}}}\}$ and also $x=sup_{n}\sum _{i=1}^n{\widehat{x}}(i)b_i$. These results give an answer to the problem of the existence of basic derivatives in financial markets.

Appendix

Proposition 4.1

If X is a separable ordered normed space with closed and normal positive cone $X_+$, or if X is the topological dual of a separable ordered normed space F, i.e. $X=F^*$ and $X_+=F^*_+$, where $F^*_+$ is the dual cone of $F_+$ in $F^*$, then there exists a countable family ${\mathcal {F}}=\{f_i\;|\;i\in {\mathbb {N}}\}$ of positive, continuous, linear functionals of X so that the ordering in X is defined pointwise by ${\mathcal {F}}$.

Proof

• (i) Suppose that X is a separable ordered normed space with closed and normal positive cone $X_+$. According to [2], Theorem 3.57, X is order isomorphic to a lattice-subspace W of C[0, 1] and suppose that $T:X\longrightarrow W$ is such an order isomorphism. Suppose that $\{r_n\}$ is the sequence of rational numbers of [0, 1] and $g_n(x)=x(r_n)$ for any $x\in C[0,1]$. Then $\{g_n\}$ is a sequence of positive, continuous linear functionals of C[0, 1] which defines pointwise the ordering in C[0, 1] and $W_+=W\cap C_+[0,1]$ is the positive cone of W. For any $x\in X$ we have:

  $$\begin{aligned}&x\in X_+\Longleftrightarrow T(x)\in W_+ \Longleftrightarrow g_n(T(x))\ge 0, \text {for any\;} n \Longleftrightarrow \\&\quad \Longleftrightarrow (T^*(g_n))(x)\ge 0, \text {for any\;} n\Longleftrightarrow f_n(x)\ge 0, \text {for any\;} n, \end{aligned}$$

  where $f_n=T^*(g_n)$ and $T^*$ is the adjoint of T. Also each $f_n$ is positive because for any $x\in X_+$ we have $f_n(x)= (T^*(g_n))(x)=g_n(T(x))\ge 0$ because $T(x)\in C_+[0,1]$. The sequence $\{f_i\;|\;i\in {\mathbb {N}}\}$ of $X^*$ defines pointwise the ordering in X because for any $x,y\in X$ we have:

  $$\begin{aligned} x\ge y\Longleftrightarrow x-y\in X_+\Longleftrightarrow f_i(x)\ge f_i(y),\;\text {for any}\;i. \end{aligned}$$

• (ii) Suppose that F is a separable ordered normed space so that $X=F^*$ and $X_+=F^*_+$. Then there exists a sequence $\{x_n\}$ of $U_+$, dense in the positive part $U_+=U\cap F_+$ of the unit ball U of F. For each n the linear functional $f_n$ of X so that $f_n(x^*)=x^*(x_n)$ for any $x^*\in F^*$, is a continuous, positive, linear functional of X and the sequence $\{f_i\;|\;i\in {\mathbb {N}}\}$ defines the positive cone $X_+$ of X i.e. $X_+=\{x^*\in X\;|\; f_i(x^*)\ge 0\;\text {for any}\;i\}$. Indeed, if $f_i(x^*)\ge 0$ for each i we have that $x^*(x_i)\ge 0$ for each i, therefore $x^*(x)\ge 0$ for each $x\in U_+$ hence $x^*(x)\ge 0$ for each $x\in F_+$ and $x^*\in X_+$. Also $f_i(x)\ge 0$ for any $x\in X_+$ and any i, therefore the ordering in X is defined pointwise by ${\mathcal {F}}$. $\square$

The notion of s-property has been defined in [11] and has been generalized in the ws-property in [6] as follows:

Definition 4.2

Suppose that Y is an ordered vector space, the ordering in Y is defined pointwise by a countable family ${\mathcal {G}}=\{g_i\;|\;i\in {{\mathbf {N}}}\}$ of positive linear functionals of Y and X is an ordered subspace of Y. If for any $x\in X_+$, $x\ne 0$ and for each $i\in {{\mathbf {N}}}$ the set $D= \{y\in [0,x]_X \mid g_i(y)=0\}$ has at least one maximal element³, we say that X has the ws-property (with respect to ${\mathcal {G}}$).

If in the above definition the set D has a maximum element then, according to [11], X has the s-property. The next two results are in fact from [6] which, for the sake of completeness, we present with proof.

Theorem 4.3

Suppose that Y is an ordered vector space, the ordering in Y is defined pointwise by a countable family ${\mathcal {G}}=\{g_i\;|\;i\in {{\mathbf {N}}}\}$ of positive linear functionals of Y and suppose that X is an ordered subspace of Y. If a linear topology $\tau$ of X exists so that

• (i) $X_+$ is $\tau$-closed,

• (ii) each increasing net of $X_+$, order bounded in X, has a $\tau$-convergent subnet,

• (iii) for each i the set $N_i^+ =\{y\in X_+\;|\;g_i(y)=0\}$ is $\tau$-closed, then X has the ws-property.

Proof

Suppose that $x\in X_+$ and $i\in {{\mathbf {N}}}$. We shall show that the set $D=[0,x]_X\cap N_i^+$ has maximal elements. The order interval $[0,x]_X$ is $\tau$-closed because $X_+$ is $\tau$-closed, therefore D is $\tau$-closed. Suppose that $\Gamma$ is a totaly ordered subset of D. For each finite subset A of $\Gamma$, denote by $x_A$ the maximum of A. Then $(x_A)$ as an increasing, order bounded net of D has a subnet $(x_{A'})$ convergent to a vector y of D. We have $x_{A'}\le y$ because for any, temporarily constant, $A'_0$ we have $x_{A'}-x_{A'_0}\ge 0$, for each $A'\supseteq A'_0$ therefore $y\ge x_{A'_0}$, because $X_+$ is $\tau$-closed. This implies that y is an upper bound of $\Gamma$ and by Zorn’s lemma the set D has maximal elements. $\square$

Recall that a Banach lattice E has order continuous norm if and only if each decreasing sequence $\{x_n\}$ of $E_+$ with infimum 0, is convergent to 0, or equivalently the order intervals of E are weakly compact, Theorem 4.9, page 186, of [1].

Proposition 4.4

Suppose that Y is an ordered vector space, the ordering in Y is defined pointwise by a countable family ${\mathcal {G}}=\{g_i\;|\;i\in {{\mathbf {N}}}\}$ of positive linear functionals of Y.

• (i) If Y is a Banach lattice with order continuous norm, then any ordered subspace X of Y with closed positive cone $X_+$, has the ws-property.

• (ii) If Y is the topological dual of a separable ordered normed space F, i.e. if $Y=F^*$ and $Y_+=F^*_+$, where $F^*_+$ is the dual cone of $F_+$ in $F^*$ and the functionals $g_i$ are weak-star continuous, then any weak-star closed ordered subspace X of Y with normal positive cone $X_+$ has the ws-property (therefore any weak-star closed ordered subspace of $\ell _\infty$ has the ws-property with respect to the family of the coefficient functionals $f_i$ of $\ell _\infty$).

Proof

Statement (i) follows by Theorem 4.3 because $X_+$ is closed, the order intervals of X are weakly compact and the functionals $g_i$, as positive linear functional of a Banach lattice are continuous. For statement (ii), note first that $X_+$ is weak-star closed. Therefore for any $x\in X_+$, the order interval $[0,x]_X$ is weak-star closed and ||.||-bounded because the cone $X_+$ is normal, hence $[0,x]_X$ is weak-star compact, by Alaoglou’s theorem. Therefore each increasing and order bounded net of $X_+$ has a weak-star convergent subnet, therefore X has the ws-property, by Theorem 4.3. $\square$